CORNELL UNIVERSITY LIBRARY GIVEN TO THE COLLEGE OF ENGINEERING by the hiachine-design dept. “Ta ass M36 Date Due 8D scenes aw SIBLEY scp CORNELL uN; VERSITy | KECEIVED J-425-h bz, fa +1 eewemee names, cae! ELEMENTARY MACHINE DRAWING AND DESIGN ‘He Graw Ill Book Qs he PUBLISHERS OF BOOKS FORU Hectrical World v Engineering News-Record Power v Engineering and Mining Journal-Press Chemical and Metallurgical Engineering Electric Railway Journal v Coal Age American Machinist + Ingenieria Internacional Electrical Merchandising v BusTransportation Journal of Electricity and Western Industry Industrial Engineer TMT TT TT Ti TT PO ELEMENTARY MACHINE DRAWING AND DESIGN BY WILLIAM C. MARSHALL, M.E., C.E. Engineer Remington Arms Union Metallic Cartridge Co., Bridgeport Works; Formerly Professor of Machine Design and Descriptive Geometry in the Sheffield Scientific School of Yale University Member A. S.M.E.. Member S. A. E., Member S. P. E. E. First EpItion SrxTeentH IMPRESSION McGRAW-HILL BOOK COMPANY, Ine. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 &8 BOUVERIE ST., E. C. 4 1912 Copyricut, 1912, BY THE McGraw-Hitut Book Company, Inc. PRINTED IN THE UNITED STATES OF AMERICA THE MAPLE PRESS COMPANY, YORK, PA. CONTENTS PAGE List OF “Toons REQGIRED Ai sw asc nea eaves eal Saeed neal mee es 8 Directions ror InpivipuaL PLates at ENp of Each CHAPTER..... TESTS AND EXAMINATION PAPERS......... 84, 117, 118, 171, 172, 242, 267 Metric CoNVERSION TABLE. ... 0.0.00... cece cece aes 2 LocGaritHMic TABLES................. Pr hana aint ne ha et omen ee 3 LEST (QUESTIONS hy 5 .cstyy'd ata dog aie eee ae BE HS vA nd a SS 291-313 TINDER dria o § do encased Saahcea Vin saausls amscgar Wereaiets ammiuenk Bien Nae 315 CHAPTER I. WorkKING DRAWINGS... 00.00.0000 cece eee ee 21 T.. PASPENINGS). RIVETS: 2. so scaid ca ie bo Achaemenid dae dome aed ae 45 III. Pipes anp Pire FITTINGS.....0 2.2.0.2 eee 54 IV. Screw THREADS AND SPRINGS.............. 000 cee eee eee 67 V. FasTentnes (SCREWS AND BOLTS)............000000eeeeue 79 VI. Keys AND: COTTersss.¢2.¢ s34.0004908 Beat RS CANS REEL I Seee TES 98 VII. Suarrine AND SHAFT COUPLINGS............. 0000000 eee eee 104 VIII. Srurring Boxes... 2.0... ee eee Raseysna fete 112 IX. Saarr Bearinas, JOURNALS, HANGERS... ........ BienGink patans 119 X. Pistons anp Piston Rops...............--. Dien hands ethece 150 XI. CROSSHEADS. . 0.00002 eee eee dees *.. 161 XII. Connectine Rops sews 173 XIII. Enainge Cranxs anp EcCENTRICS...............-5- Elias 189 XIV. Purteys anp Bevrine. ........0.0ss0¢esd ss eeereeweeteevanye 199 DEV SBUR (GRARIN Gis ies sng a Gohan dro nen nesses Ameena SANS Seis Mees 215 XCVI2 BEVEIN GEARING s « 6 acgdd hoagie navanel aeiee anes uarnuen Mase nas: ocbue guuian EE XVII. Worm GBaRinG 2 ccc cacc aan wed ead ata deh gnome eave eee 258 XVI. Vauvis;, Cocks; BPG. vwwisiew sore Sar geamae wna s awn ve wee y 268 TABLES PAGE Userut INrorMATION—DEcIMAL EQUIVALENTS..........02+---0058 1 1. Bott anp Cap Screw Heaps anp Bout STRENGTH............. 9 2. INTERNATIONAL AND Metric SYSTEM............00000 0020s 10 3. Comparative AREAS AT Root or Vaxious Systems or THREAD.. 11 4. Dimensions oF ROLLED BEAMS.............0 0.00: e eee eee eee 12 he SEESAINGES UN EONS ia fe ice scgadsrcptcatnn bs ob tanto a BASOaSIR dois began cha edysusng gh bs awe 61 6. FricTION OF WATER IN PIPES.......... 000000 c cece eee nee 63 7. FRICTION OF WATER IN ELBOWS...........0000 000 2c eee eee 64 8. Sizes or NippLes, CoupLINGS, CAPS, ETC...........0 00 cece 13 9. StanparD W. I. Pipz DIMENSIONS.............0.0 0.0 c eee eens 14 LOM OIG AC UPS's yz gets anata geste aie SAGA eased ue eRe e eaten ee mene eames ene 16 11. StanNDARD STEAM FLANGES........0... 0000 ccc cue eee eee eenees 15 12, Spire’ PINS s<5 seven cue eh pede ee ancy awk hard Pet gunn amiedion a nine 16 PSS CADIR: RUNS 5 iiss: gpet ese ears dana Basa ceet acide eaaIAe Aegan emdapetndwene 16 14. A. L. A. M. Botts anp NOTs..... 0.0.0.0... 0.0 eee 17 Ld, WOODRUEH IRB YSsi <5 sive naw di done d Sais RG OREN Gedy EE SEAS 18 P56: “WINGHINUTS a xk ise ochent ws manta ee eatmiin asain Sa paate a rpecetealen atin mm aad A 88 16. BearInc PRESSURES FOR BEARINGS.............0.00. 000 ee ce eee 19 17. COEFFICIENTS OF FRICTION. ........... 00.000 c cee cee ee eeeueeues 19 18. WEIGHT OF SUBSTANCES. 2.0.0.0... 0c cece cece cece e eens eeaes 20 19. STRENGTH OF MATERIALS... 2.2... c eee cee ccc ee neces uvace 20 20), HACTORS OF ISABITY 2 ccjce 5 ccaspaisiadeaeeote wc anaeoa dye e awe ceed ee peas 20 vili USEFUL INFORMATION Area of surface between a circular are and its chord is approximately 2 the circumscribed rectangle. Vol. of frustum of cone or pyr. when axis is perp. to bases is V =(sum of base areas+ V product of base areas) <} height of frustum. Vol. of sphere = D? x .5236. A gallon =231 cu.in. A cubic foot =7.48 gals. A gallon of water at 39.2° F. weighs 8.3389 Ibs. A cubic foot of water at 62° F. weighs 62.355 Ibs. A cubic foot of water at 212° F. weighs 59.833 lbs. A column of water 1 ft. high at 39.2° F. exerts a pressure of .4335 lh. per square inch. A column of water 2.307 ft. high at 39.2° F. exerts a pressure of 1 Ib. per square inch. 1 atmosphere =14.7 Ibs. pressure per square inch =29.922 ins. of mercury = 33.9 ft. of water. DECIMAL EQUIVALENTS Decimals. 32ds. | 16ths. Decimals. | 32ds. | 16ths. | 03125 | 53125 | ¥ | 0625 dy 5625 $s | .09375 hs 59375 i i 125 k 625 | | 4 15625 & 65625 a | 1875 tr 6875 i i 21875 | de T1875 a | 25 i 75 4 23125 | 78125 a | 8125 fs 8125 # | 34375 % 84375 aq 375 : 875 : i 40625 B . 90625 8 4375 & 9375 rr 46875 | 96875 ¥ 5 4 1.00 2 ELEMENTARY MACHINE DRAWING AND DESIGN METRIC CONVERSION TABLE Millimeters X .03937 = inches. Millimeters + 25.4 =inches. Centimeters X .3937 = inches. Centimeters + 2.54 =inches. Meters X39.37 =inches. (Act Congress.) Meters X3.281 = feet. Meters X 1.094 = yards. Kilometers X .621 = miles. Kilometers + 1.6093 = miles. Kilometers X 3280.8693 = feet. Square millimeters x .00155 =square inches. Square millimeters + 645.1 =square inches. Square centimeters X.155 =square inches. Square centimeters + 6.451 =square inches. Square meters X 10.764 =square feet. Square kilometers x 247.1 =acres. Hectare X2.471 =acres. Cubic centimeters + 16.383 = cubic inches. Cubic centimeters +3.69 =fluid drams (U.S.P.). Cubic centimeters + 29.57 = fluid ounces (U.S.P.). Cubic meters X 35.315 =cubic feet. Cubic meters X 1.308 = cubic yards. Cubic meters X 264.2 =gallons (231 cu.in.). Liters X61.022 =cubie inches. (Act Congress.). Liters X 33.84 = fluid ounces (U.S.P.). Liters X .2642 = gallons (231 cu.in.). Liters +3.78 = gallons (231 cu.in.). Liters + 28.316 =cubic feet. Hectoliters X3.531 = cubic feet. Hectoliters X 2.84 =bushels (2150.42 cu.in.). Hectoliters X .131 =cubic yards. Hectoliters + 26.42 =gallons (231 cu.in.). Grammes X15.432=grains. (Act Congress.) Grammes +981 = dynes. Grammes (water) +29.57 =fluid ounces. Grammes + 28.35 = ounces avoirdupois. Grammes per cubic centimeter + 27.7 =pounds per cubic inch. Joule X.7373 =foot-pounds. Kilo-grammes X 2.2046 = pounds. Kilo-grammes X 35.3 = ounces avoirdupois. Kilo-grammes per square centimeter X 14.223 = pounds per square inch. Kilo-gram-meters X 7.233 =foot-pounds. Kilo-grammes per meter X.672 =pounds per foot. Kilo-grammes per cubic meter x .062 =pounds per cubic foot. Tonneau X 1.1023 =tons (2000 lbs.). Kilo-watts < 1.34 =horse-power. Watts +746 =horse-power. Watts X .7373 =foot-pounds per second. Calorie X3.968 =B.T.U. Cheval vapeur + .9863 = horse-power. (Centigrade X 1.8) +32 =degree Fahrenheit. Frane <.193 =dollars. Gravity Paris =980.94 centimeters per second. USEFUL INFORMATION 3 TABLE OF LOGARITHMS WAG) | Bt we la | Bo) ee Pe ee 0 0000 | 3010 | 4771 | 6021 | 6990 | 7782 | 8451| 9031/9542; 4. gt 1| 2.2) 2.1 1} 0000 | 0414 | 0792 | 1139 | 1461 | 1761 | 2041 | 2304 | 2553 |2788| 9| 4/4] 4/2 213010 | 3222 | 3424 | 3617 | 3802 | 3979 | 4150 | 4314 | 4472 | 4624 | 3] 6.6] 6.3 3] 4771 | 4914 | 5051 | 5185 | 5315 | 5441 | 5563 | 5682 | 5798 | 5911 | 3] 85) 16:3 4} 6021 | 6128 | 6232 | 6335 | 6435 | 6532 | 6628 | 6721 | 6812 | 902 | $ | 13-2 | 17-8 5] 6990 | 7076 | 7160 | 7243 | 7324 | 7404 | 7482 | 7559 | 7634 | 7709 | 8| 17.6 | 16.8 6].7782 | 7853 | 7924 | 7993 | 8062 | 8129 | 8195 | 8261 | 8325 | 838e | 9 19.8 | 18.9 7] 8451 | 8513 | 8573 | 8633 | 8692 | 8751 | 8808 | 8865 | 8921|8976| | 29 1 19 8] 9031 | 9085 | 9138 | 9191 | 9243 | 9294 | 9345 | 9395 | 9445 | 9494/1] 2.0] 1.9 9] 9542 | 9590 | 9638 | 9685 | 9731 | 9777 | 9823 | 9868 | 9912 | 9956 | 2] 4.0] 3.8 4| 8.0| 7.6 10 J 0000 | 0043 |.0086 | 0128 | 0170 | 0212 | 0253 | 0294 | 0384 /0374 | 5| 10.0 9:5 11} 0414 | 0453 | 0492 | 0531 | 0569 | 0607 | 0645 | 0682 | 0719 | 0755 | 3] 169] 123 12] 0792 | 0828 | 0864 | 0899 | 0934 | 0969 | 1004 | 1038 | 1072 | 1106 | 9] 18.0] 17:1 13] 1139 | 1173 | 1206 | 1239 | 1271 | 1303 | 1335 | 1367 | 1399 | 1430 14] 1461 | 1492 | 1523 | 1553 | 1584 | 1614 | 1644 | 1673 | 1703 |1732] | 4% | 4% 151761 | 1790 | 1818 | 1847 | 1875 | 1903 | 1931 | 1959 | 1987 | 2014| 2] 3.6] 3.4 16] 2041 | 2068 | 2095 | 2122 | 2148 | 2175 | 2201 | 2227 | 2253| 2279 | 3| 5.4) 5.1 17] 2304 | 2330 | 2355 | 2380 | 2405 | 2430 | 2455 | 2480 | 2504 | 2520 | 5| 9.0 8.5 18 2553 | 2577 | 2601 | 2625 | 2648 | 2672 | 2695 | 2718 | 2742 | 2765 | 3) 198/179 19] 2788 | 2810 | 2833 | 2856 | 2878 | 2900 | 2923 | 2945 | 2967 | 2980 | &| 14-4) 13.6 20 } 3010 | 3032 | 3054 | 3075 | 3096 | 3118 | 3139 | 3160] 8181] 3201) 21 | 3222 | 3243 | 3263 | 3284 | 3304 | 3324 | 3345 | 3365 | 3385 |3404| 3] 3S] 32 22 | 3424 | 3444 | 3464 | 3483 | 3502 | 3522 | 3541 | 3560 | 3579 | 3598] 3| 4:8] 4:5 23 | 3617 | 3636 | 3655 | 3674 | 3692 | 3711 | 3729 | 3747 | 3766 | 3784| 4] 6-4) 6.0 24 | 3802 | 3820 | 3838 | 3856 | 3874 | 3892 | 3909 | 3927 | 3945 | 3962 | 8 | 9-6] 9.0 25 | 3979 | 3997 | 4014 | 4031 | 4048 | 4065 | 4082 | 4099 | 4116 | 4133 | §| 15's | 12.0 26 | 4150 | 4166 | 4183 | 4200 | 4216 | 4232 | 4249 | 4265 | 4281 | 4298 | 9] 14.4 | 13.5 27 | 4314 | 4320 | 4346 | 4362 | 4378 | 4393 | 4409 | 4425 | 44404456] | 14) 4g 28 | 4472 | 4487 | 4502 | 4518 | 4533 | 4548 | 4564 | 4579 | 4594/4609 1] 14] 43 29 | 4624 | 4639 | 4654 | 4669 / 4683 | 4698 | 4713 | 4728 | 4742 | 4757 | 2| 2'8| 26 30] 4771 | 4786 | 4800 | 4814 | 4829 | 4843 | 4857 | 4871 | 4886 |4900| 4] 3:6] 32 6} 8.4] 7.8 31] 4914 | 4928 | 4942 | 4955 | 4969 | 4983 | 4997 | 5011 | 5024] 5038 | 7) 9.8] 9-1 32 15051 | 5065 | 5079 | 5092 | 5105 | 5119 | 5132 | 5145 | 5159 | 5172 | 9 | 19'6 | 11:7 33 [5185 | 5198 | 5211 | 5224 | 5237 | 5250 | 5263 | 5276 | 5289 | 5302 34 [5315 | 5328 | 5340 | 5353 | 5366 | 5378 | 5391 | 5403 | 5416 5428) || 12, | 11, 35 | 5441 | 5453 | 5465 | 5478 | 5490 | 5502 | 5514 | 5527 | 5539 | 5551) 3] 9°7| 9'2 36 | 5563 | 5575 | 5587 | 5599 | 5611 | 5623 | 5635 | 5647 | 5658 5670| 3] 3.6) 3.3 37 [5682 | 5694 | 5705 | 5717 | 5729 | 5740 | 5752 | 5763 | 5775 | 5786 | 5| 6.0} 5.5 38 | 5798 | 5809 | 5821 | 5832 | 5843 | 5855 | 5866 | 5877 | 5888 | 5900 | 6] 7.2| 8.6 39 | 5911 | 5922 | 5933 | 5944 | 5955 | 5966 | 5977 | 5988 | 5699 | 6010 | | o%6| sis 9110.8| 9.9 40 | 6021 | 6031 | 6042 | 6053 | 6064 | 6075 | 6085 | 6096 | 6107 | 6117 Wine) il} e@)eiha|)se}e;y | se] se Log = =0.4971 n?= 9.8696 Log x?=0.9943 Qn = 6.2831 Log 2x =0.7981 Via =1.7724 Loz V x =0.2485 4 ELEMENTARY MACHINE DRAWING AND DESIGN N.JL.o/ 1 | 2/3 / 4/5 | 6/)]7 4/8 4] 9 P. P. 40 | 6021 | 6031 | 6042 | 6053 | 6064 | 6075 | 6085 | 6096 | 6107 | 6117 9 8 4116128 | 6138 | 6149 | 6160 | 6170 | 6180 | 6191 | 6201 | 6212 | 6222} 1! 0.9] 08 42 1 6232 | 6243 | 6253 | 6263 | 6274 | 6284 | 6294 | 6304 | 6314 | 6325 z zs a 43 | 6335 | 6345 | 6355 | 6365 | 6375 | 6385 | 6395 | 6405 | 6415 | 6425 | 3] 3%] 3° 44] 6435 | 6444 | 6454 | 6464 | 6474 | 6484 | 6493 | 6503 | 6513 | 6522 | 2| $3| 72 45 | 6532 | 6542 | 6551 | 6561 | 6571 | 6580 | 6590 | 6599 | 6609 | 6618] 7| 63) 5/6 46 | 6628 | 6637 | 6646 | 6656 | 6665 | 6675 | 6684 | 6693 | 6702 | 6712] 8| 72) 6-4 47 | 6721 | 6730 | 6739 | 6749 | 6758 | 6767 | 6776 | 6785 | 6794 | 6803 48 ] 6812 | 6821 | 6830 | 6839 | 6848 | 6857 | 6866 | 6875 | 6884 | 6893 49 | 6902 | 6911 | 6920 | 6928 | 6937 | 6946 | 6955 | 6964 | 6972 | 6981 : 50 | 6990 | 6998 | 7007 | 7016 | 7024 | 7033 | 7042 | 7050 | 7059 | 7067 plicwe 2) QF 5147076 | 7084 | 7093 | 7101 | 7110 | 7118 | 7126 | 7135 | 7143 | 7152 4} 3.6 5297160 | 7168 | 7177 | 7185 | 7193 | 7202 | 7210 | 7218 | 7226 | 7235 | 48 53 | 7243 | 7251 | 7259 | 7267 | 7275 | 7284 | 7292 | 7300 | 7308 | 7316 7] 63 8] 7.2 54] 7324 | 7332 | 7340 | 7348 | 7356 | 7364 | 7372 | 7380 | 7388 | 7396 9} 81 55] 7404 | 7412 | 7419 | 7427 | 7435 | 7443 | 7451 | 7459 | 7466 | 7474 56 | 7482 | 7490 | 7497 | 7505 | 7513 | 7520 | 7528 | 7536 | 7543 | 7551 57 7559 | 7566 | 7574 | 7582 | 7589 | 7597 | 7604 | 7612 | 7619 | 7627 % 58 | 7634 | 7642 | 7649 | 7657 | 7664 | 7672 | 7679 | 7686 | 7694 | 7701 1| 08 59] 7709 | 7716 | 7723 | 7731 | 7738 | 7745 | 7752 | 7760 | 7767 | 7774 a) Ae 4] 3. 60}. 7782 | 7789 | 7796 | 7803 | 7810 | 7818 | 7825 | 7832 | 7839 | 7846 5| 20 6| 4.8 61] 7853 | 7860 | 7868 | 7875 | 7882 | 7889 | 7896 | 7903 | 7910 | 7917 ae 62] 7924 | 7931 | 7938 | 7945 | 7952 | 7959 | 7966 | 7973 | 7980 | 7987 9| 7.2 63 | 7993 | 8000 | 8007 | 8014 | 8021 | 8028 | 8035 | 8041 | 8048 | 8055 64] 8062 | 8069 | 8075 | 8082 | 8089 | 8096 | 8102 | 8109 | 8116 | 8122 65] 8129 | 8136 | 8142 | 8149 | 8156 | 8162 | 8169 | 8176 | 8182 | 8189 > 6618195 | 8202 | 8209 | 8215 | 8222 | 8228 | 8235 | 8241 | 8248 | 8254 1| 0.7 67 | 8261 | 8267 | 8274 | 8280 | 8287 | 8293 | 8299 | 8306 | 8312 | 8319 2 2 68 | 8325 | 8331 | 8338 | 8344 | 8351 | 8357 | 8363 | 8370 | 8376 | 8382 4] 28 69 J 8388 | 8395 | 8401 | 8407 | $414 | 8420 | 8426 | 8432 | 8439 | 8445 S| oe 7| 49 70 8451 | 8457 | 8463 | 8470 | 8476 | 8482 | 8488 | 8494 | 8500 | 8506 a| oe 7148513 | 8519 | 8525 | 8531 | 8537 | 8543 | 8549 | 8555 | 8561 | 8567 72] 8573 | 8579 | 8585 | 8591 | 8597 | 8603 | 8609 | 8615 | 8621 | 8627 73 ] 8633 | 8639 | 8645 | 8651 | 8657 | 8663 | 8669 | 8675 | 8681/8686] - | 74 | 8692 | 8698 | 8704 | 8710 | 8716 | 8722 | 8727 | 8733 | 8739 |8745| 3 | 9-8 75 | 8751 | 8756 | 8762 | 8768 | 8774 | 8779 | 8785 | 8791 | 8797 |8802] 3 | 1% 76 | 8808 | 8814 | 8820 | 8825 | 8831 | 8837 | 8842 | 8848 | 8854 | 8859 4 | 2-4 7 | 8865 | 8871 | 8876 | 8882 | 8887 | 8893 | 8899 | 8904} 8910/8915} 6! 3.6 78 8921 | 8927 | 8932 | 8938 | 8943 | 8949 | 8954 | 8960|8965|8971| 7 ve 79] 8976 | 8982 | 8987 | 8993 | 8998 | 9004 | 9009 | 9015] 9020) 9025/ 9 | 37% 80] 9031 | 9036 | 9042 | 9047 | 9053 | 9058 | 9063 | 9069 | 9074 | 9079 N.JLo} 1 / 2]3 ]41/5!]6)7]s8]o0| | USEFUL INFORMATION L.0 80 81 82 83 85 86 87 88 89 90 91 92 93 94 95 96 97 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 9085 9138 9191 9243 9294 9345 9395 9445 9494 9090 9143 9196 9248 9299 9350 9400 9450 9499 9096 9149 9201 9253 9304 9355 9405 9455 9504 9101 9154 9206 9258 9309 9360 9410 9460 9509 9106 9159 9212 9263 9315 9365 9415 9465 9513 9112 9165 9217 9269 9320 9370 9420 9469 | 9518 9117 9170 9222 9274 9325 9375 9425 9474 9523 9122 9175 9227 9279 9330 9380 9430 9479 9528 9128 9180 9232 9284 9335 9385 9435 9484 9533 9133 9186 8238 9289 9340 9390 9440 9489 9538 9547 9552 9557 9562 9566 9571 9576 9581 9586 9590 9638 9685 9731 9777 9823 9868 9912 9956 9595 9643 9689 9736 9782 9827 9872 9917 9961 9600 9647 9694 9741 9786 9832 9877 9921 9965 9605 9652 9699 9745 9791 9836 9881 9926 9969 9609 9657 9703 9750 9795 9841 9886 9930 9974. 9614 9661 9708 9754 9800 9845 9890 9934 9978 9619 9666 9713 9759 9805 9850 9894 9939 9983 9624 9671 9717 9763 9809 9854 9899 9943 9987 9628 9675 9722 9768 9814 9859 9903 9948 9991 9633 9680 9727 9773 9818 9863 9908 9952 9996 0000 0004 0009 0013 0017 0022 0026 0030 0035 0039 0043 0086 0128 0170 0212 0253 0294 0334 0374 0048 0090 0133 0175 0216 0257 0298 0338 0378 0052 0095 0137 0179 0220 0261 0302 0342 0382 0056 0099 0141 0183 0224 0265 0306 0346 0386 0060 0103 0145 0187 0228 0269 0310 0350 0390 0065 0107 0149 0191 0233 0273 0314 0354 0394 0069 0111 0154 0195 0237 0278 0318 0358 0398 0073 0116 0158 0199 0241 0282 0322 0362 0402 0077 0120 0162 0204 0245 0286 0326 0366 0406 0082 0124 0166 0208 0249 0290 0330 0370 0410 0414 0418 0422 0426 0430 0434 0438 0441 0445 0449 0453 0492 0531 0569 0607 0645 0682 0719 0755 0457 0496 0535 0573 0611 0648 0686 0722 0759 0461 0500 0538 0577 0615 0652 0689 0726 0763 0465 0504 0542 0580 0618 0656 0693 0730 0766 0469 0508 0546 0584 0622 0660 0697 0734 0770 0473 0512 0550 0588 0626 0663 0700 0737 0774 0477 0515 0554 0592 0630 0667 0704 0741 0777 0481 0519 0558 0596 0633 0671 0708 0745 0781 0484 0523 0561 0599 0637 0674 0711 0748 0785 0488 0527 0565 0603 0641 0678 0715 0752 0788 0792 0795 0799 0803 0806 0810 0813 0817 0821 0824 CONGR WHe OWNAMIR WHE CONGR wWhe WWNNHNEREOS PRO WNLE SO RRO WNN HE O SNoRoawan™ nonononont nononuoanon™ L.0 6 ELEMENTARY MACHINE DRAWING AND DESIGN L. 0 PP. 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 0792 0795 0799 0803 0806 0810 0813 0817 0821 0824 0828 0864 0899 0934 0969 1004 1038 1072 1106 0831 0867 0903 0938 0973 1007 1041 1075 1109 0835 0871 0906 0941 0976 101] 1045 1079 1113 0839 0874 0910 0945 0980 1014 1048 1082 1116 0842 0878 0913 0948 0983 1017 1052 1086 1119 0846 0881 0917 0952 0986 1021 1055 1089 1123 0849 0885 0920 0955 0990 1024 1059 1093 1126 0853 0888 0924 0959 0993 1028 1062 1096 1129 0856 0892 0927 0962 0997 1031 1065 1099 1133 0860 0896 0931 0966 1000 1035 1069 1103 1136 1139 1143 1146 1149 1153 1156 1159 1163 1166 1169 1173 1206 1239 1271 1303 1335 1367 1399 1430 1176 1209 1242 1274 1307 1339 1370 1402 1433 1179 1212 1245 1278 1310 1342 1374 1405 1436 1183 1216 1248 1281 1313 1345 1377 1408 1440 1186 1219 1252 1284 1316 1348 1380 1411 1443 1189 1222 1255 1287 1319 1351 1383 1414 1446 1193 1225 1258 1290 1323 1355 1386 1418 1449 1196 1229 1261 1294 1326 1358 1389 1421 1452 1199 1232 1265 1297 1329 1361 1392 1424 1455 1202 1235 1268 1300 1332 1364 1396 1427 1458 1461 1464 1467 1471 1474 1477 1480 1483 1486 1489 1492 1523 1553 1584 1614 1644 1673 1703 1732 1495 1526 1556 1587 1617 1647 1676 1706 1735 1498 1529 1559 1590 1620 1649 1679 1708 1738 1501 1532 1562 1593 1623 1652 1682 1711 1741 1504 1535 1565 1596 1626 1655 1685 1714 1744 1508 1538 1569 1599 1629 1658 1688 1717 1746 1511 1541 1572 1602 1632 1661 1691 1720 1749 1514 1544 1575 1605 1635 1664 1694 1723 1752 1517 1547 1578 1608 1638 1667 1697 1726 1755 1520 1550 1581 1611 1641 1670 1700 1729 1758 1761 1764 1767 1770 1772 1775 1778 1781 1784 1787 1790 1818 1847 1875 1903 1931 1959 1987 2014 1793 1821 1850 1878 1906 1934 1962 1989 2017 1796 1824 1853 1881 1909 1937 1965 1992 2019 1798 1827 1855 1884 1912 1940 1967 1995 2022 1801 1830 1858 1886 1915 1942 1970 1998 2025 1804 1833 1861 1889 1917 1945 1973 2000 2028 1807 1836 1864 1892 1920 1948 1976 2003 2030 1810 1838 1867 1895 1923 1951 1978 2006 2033 1813 1841 1870 1898 1926 1953 1981 2009 2036 1816 1844 1872 1901 1928 1956 1984. 2011 2038 2041 2044 2047 2049 2052 2055 2057 2060 2063 2066 WOONOCR Whe COON Whe OwWNOTRWNE NNNERROOO WOWNNNHEROS Pee EOOCOO DORNODARNWY SwarRoawan™ NaH OUNmoaw” L.0 USEFUL INFORMATION L. 0 owe 160 2041 2044 2047 2049 2052 2055 2057 2066 161 162 163 164 165 166 167 168 169 2068 2095 2122 2148 2175 2201 2227 2253 2279 2071 2098 2125 2151 2177 2204 2230 2256 2281 2074 2101 2127 2154 2180 2206 2232 2258 2284 2076 2103 2130 2156 2183 2209 2235 2261 2287 2079 2106 2133 2159 2185 2212 2238 2263 2289 2082 2109 2135 2162 2188 2214 2240 2266 2292 2084 2111 2138 2164 2191 2217 2243 2269 2294 2092 2119 2146 2172 2198 2225 2251 2276 2302 170 2304 2307 2310 2312 2315 2317 2327 171 172 173 174 175 176 177 178 179 2330 2355 2380 2405 2430 2455 2480 2504 2529 2333 2358 2383 2408 2433 2458 2482 2507 2531 2335 2360 2385 2410 2435 2460 2485 2509 2533 2338 2363 2388 2413 2438 2463 2487 2512 2536 2340 2365 2390 2415 2440 2465 2490 2514 2538 2343 2368 2393 2418 2443 2467 2492 2516 2541 2345 2370 2395 2420 2445 2470 2494 2519 2543 2353 2378 2403 2428 2453 2477 2502 2526 2550 180 2553 2555 2558 2560 2562 2565 2567 2574 181 182 183 184 185 186 187 188 189 190 2577 2601 2625 2648 2672 2695 2718 2742 2765 2579 2603 2627 2651 2674 2697 2721 2744 2767 2582 2605 2629 2653 2676 2700 2723 2746 2769 2584 2608 2632 2655 2679 2702 2725 2749 2772 2586 2610 2634 2658 2681 2704 2728 2751 2774 2589 2623 2636 2660 2683 2707 2730 2753 2776 2591 2615 2639 2662 2686 2709 2732 2755 2778 2598 2622 2646 2669 2693 2716 2739 2762 2785 2788 2790 2792 2794 2797 2799 2801 2808 191 192 193 194 195 196 197 198 199 2810 2833 2856 2878 2900 2923 2945 2967 2989 2813 2835 2858 2880 2903 2925 2947 2969 2991 2815 2838 2860 2883 2905 2927 2949 2971 2993 2817 2840 2862 2885 2907 2929 2951 2973 2965 2819 2842 2865 2887 2909 2931 2953 2975 2997 2822 2844 2867 2889 2911 2934 2956 2978 2999 2824 2847 2869 2891 2914 2936 2958 2980 3002 2831 2853 2876 2898 2920 2942 2964 2986 8008 200 3010 3012 3015 3017 3019 3021 3023 3030 DOONOAAWN eS SwoOWNHMbwoOnNe NNNEREROOSO HReEHHOOSO NAH DUNE” DOPNoOwWo mt” L. 0 2 8 ELEMENTARY MACHINE DRAWING AND DESIGN LIST OF TOOLS REQUIRED T-square, 24” blade (no more or no less). 60° triangle, 11” side, celluloid preferred. 45° triangle, 9” side, celluloid preferred. Kelsey triangle 30°-60°-45° with knob and erasing slot or Zange triangle. Set of drawing instruments, comprising 6’’ compasses, with pencil, pen, lengthening bar, dividers (hair spring or bow spacing), bow pencil, bow pen, ruling pen (Alteneder or Riefler preferred). Curved rulers (1 large and 1 small). Pencil eraser (Emerald, or any hard make, not a soft one). Ink eraser (Faber’s typewriter). 12” triangular architects’ scale or flat scale divided into 16ths. Art gum or sponge rubber (for cleaning only). 1-4H, 1-6H drawing pencil, or a holder with leads. 4H leads for compasses. 6 thumb tacks. Cake of magnesia, or chalk. 1 bottle of black (waterproof) drawing ink with filler on cork. 1 pen holder of good size around. 1 piece of cotton cloth about the size of a handkerchief, for wiping off drawings. 1 pencil sharpener (sandpaper, or flat file). 3 writing pens (Gillott’s 404, Lady Falcon, ball pointed). 1 protractor (Penfield’s). 1 pen wiper. 1 blotter. 1 soapstone metalworkers’ crayon (flat). 1 loose leaf note book for calculations 9110 I-P. 1 sheet, 6’’6’’, of celluloid .005 thick, dull on both sides (used for an erasing shield). 1 slide rule (K and E, or Faber), 10’ long. 22 sheets $ Imperial size drawing paper (22’’15”’), (size No. 3). buff or K and E Normal, and 15 sheets (size No. 1), (8X 103"), with holes punched for above note book. The student’s name or initials ought to be placed on each article. The following instruments, while not required, will be found very useful: Horn center, needle prick point. Triangular scale guard. Thumb tack lifter. Arkansas oil stone 2’ 4’’X7,"" (for sharpening pens). Section liner (G and J type), the simpler the better. TaBLE I Strength of Bolts. USEFUL INFORMATION Oo Sizes of Bolt and Cap Screw Heads and Nuts. GAR SSSSSSSAASSr yews . 0 Ae AHR BOO SPAOANwWDOLr a ty gSSSSAAAARYSRSAGRSS 4 mm RB ANN mpeg sed aL ae SES EO Oe Oe a oh ay 1 ‘poet Jo A lescsedsuasg “oe an hehe! [2 ieeegegegs sei Se eS ‘urbg sod gL Lat aonnna on IS wc 19 x “sary 0001 "300% Bi igwrseanaoads 48 YQAU9I4g s[ISUa,T, aT A AVSSRASSSAS2QSSSaARR SAA HOOD eS A ary a Yo) a> 1. ft N10 Nw ww uw 1p uw . NOOO nt Ee MOCO ww it Apog jo loyoureiq] TS AAAKRSAERBSSREOAABSSE wees Se Se Ss eS * 3) Hu tereonrreowoedton « - (1 Sw SSSs No Oo) HO O Q vou ye siouma] $73 |= DIA ARR SAR SR RSS SASS esos et est oe qd: *SOOrnrMHnnnnorotdnnrywHo * ca NH OANOCS mid D> mo JOO" 7B Barly oe: (SSSSASaRSTBESRSAGE S eae ia 2 eoonmeo me wouy sod epvoras) 13 | PFZLRVEHMOANANSCPHMNOONH ‘ureId a mal? [2 of? pe eie Wee 3 Bg ow Sree PS ‘peoy uoing ep = Beek Fo one he Sn Boe tee ds ear “ured S dea ce nis ye eis aN ey ES Rie “peoy 381d {hy = loo re wi ha ALE ao roe CEN ar ELLE Re 3 “q3deq Eh rte TE pe IE cto LE rice FE sa cat eo coy By Se ONES q a 2 ole of of8 se Big “wed rail} Set etre coo let Met alco cat et elo lo ie hs PO8 ole aoe ae uo RM 5 ais “‘qydeq tootootaw “ff onfoo 8 nin olf wnleo cast r=j0 riko rit eo Tt & 8 aes ¥ 5 “"UIPLAA|: = 5 of eho fs rales og alco as calst [00 rileo iit colo via wnlo SST a 2 aes * 3| maeg st tte IE co IS ais UF ao ct rio wont Ft tt q mere es re] : > 2 S Ae eilea off wnloo Blatt Sho oo et aa. 2 es & 8 wpa) CS 5: Sse sass vas | ane Fils © ALAS OS AS aw AS ASO AS aw SAS se aoc ; akas aig o ee ~ Ji wales ali jee eos wo lo ese ee AS20) wpm] Q)|s 2 ine Ra aw AB ne ene Fo: “‘qqydeq = Sofas ase as a as = an|oo ae fe ale oft ol ol ed oofLoleaw 18 ero AS ein ofS eisai cepa ish0, ikon AW. soko een,cal wD 2a rss es eS et ot ae : P14 “UIP Q = oot ote. as Ee as re 86 wie ie EE oo A wie I . He EH NNANANAN 4 wl ef? alo “a-a Jo rojoIEI = ie AS 15 capo PE eee P15 snjo cole m0 ie rt ke ries se be 15 16 14 9 10 11 12 138 8 10 ELEMENTARY MACHINE DRAWING AND DESIGN TABLE 2 INTERNATIONAL SYSTEM 1 2 3 4 5 6 7 8 Diem of) pink, | emt) pich, || Beet) pitch, || DEES) Piton, m/m. m/m. m/m m/m m/m. m/m. m/m. m/m. 3 0.55 il 1.5 30 3.5 60 5.5 33 0.55 12 1.75 33 3.5 64 6 4 0.7 14 2 36 4 68 6 5 0.85 16 2 39 4 72 6.5 6 1.0 18 2.5 42 4.5 76 6.5 7 1.0 20 2.5 45 4.5 80 7.0 8 1.25 22 2.5 48 5 9 1.25 24 3 52 5 10 1.5 27 3 56 5.5 METRIC SYSTEM Diam. . Diam. Diam. noky,| Pited | Poot." | nh, Paton, | Piamst | of] Pisce. | Pigm, et m/m. m/m m/m m m. m/m m/m m/m m/m m/m 3 0.5 2.35 16 2.0 13.40 38 4.0 32.80 4 0.75 3.03 18 2.5 14.75 39 4.0 33.80 5 0.75 4.03 20 2.5 16.75 40 4.0 34.80 6 1.0 4.70 22 2.5 18.75 42 4.5 36.15 7 1.0 5.70 22 3.0 18.10 44 4.5 88.15 8 1.0 6.70 24 3.0 20.10 45 4.5 39.15 8 1.25 6.38 26 3.0 22.10 46 4.5 40.15 9 1.0 7.70 27 3.0 23.10 48 5.0 41.51 9 1.25 7.38 28 3.0 24.10 50 5.0 43.51 10 1.5 8.05 30 3.5 25.45 52 5.0 45.51 11 1.5 9.05 32 3.5 27.45 56 5.5 48 .86 12 1.5 10.05 33 3.5 28.45 60 5.5 52.86 12 1.75 9.73 34 3.5 29.45 64 6.0 56.21 14 2.0 11.40 36 4.0 30.80 68 6.0 60.21 72 6.5 63.56 USEFUL INFORMATION 11 TABLE 3 COMPARATIVE AREAS AT ROOT OF VARIOUS SCREW THREAD SYSTEMS 1 2 3 4 5 6 #7 8 9 10 11 12 #138 «14 V Thread. | U. S. Stand./A. L. A. M.| Whitworth. aoe piped. " ge ser: |e iat 2 he tele |e le 8 |s , ee eee eee hi tome. ceca | tees ae p Gael cages gel 3 |28| 2 |EE| 2/23) 2/22] 3 [EEl4| |: 2) 2 las] & jae] 2 | ea} 2 | BA 2 |s4/ 5] 8 3 A <6 <6 off fSe Bef aienitelisin & 7 se oures os j led wo |20un [00 62] =H 09] Hoa| ea] Heol Ss ee Ce eT res Sy ee se H}O0 a0 4] et 09]00 09/00 rales cof or Si, oe oe we te AN! od AeA RHA 0 | seoriwenponior ats | my | aeasswertaataclaels Mees HNANN RANANANS Tepe rNn Seances MOR WDAONAN al ~ | £282228s)a BAS we efettola “ote fe, iter cole mie CUINN N69 08 HD sodue speuueya TABLE 5 FLANGE UNIONS ON PAGE 61 TABLE 6 FRICTION OF WATER IN PIPES ON PAGE 63 TABLE 7 FRICTION OF WATER IN ELBOWS ON PAGE 64 USEFUL INFORMATION 13 TABLE 8 SIZE OF PIPE FITTINGS . Standard Wrought Iron Nom. Nipple Lengths. Pipe Couplings, Black or Caps. Pipe Galvan. Size. Close. Short. Outside Dia.| Length. Length. Dia. 3 i 13 8 3 4 3 13 53 5 a1. li i ; 2 13 i lize 13 3 1y¥6 3 i 2 lis 1g 3 1} 1 13 2 18% z 1h 1§ 24 2 2h 14 2 Wl ow 24 t i 13 24 2 2 23 235 23 13 22 23 | 23 3 3% i 1§ 3t 3 $ 3 38 3% 3 4 33 24 4 4395 33 2 43 4 3 4 bv E 2 54 Plugs. Bushings (Fig. 29). Locknuts, Pie | Zeneth. | Kengtiy | Side of | “ites.” | Across | HEREC: | auiek, | eeha Size. | Thread.| Head. | Head, | Head. | Flats. (B). Flats. pe (A). (C). 3 3 i z i ws + a t & 3 2 ts é 7 i 3 is 1 4 | ee Pe a) ee | et | i) # | @ |) wile } 4b | € 1 i ve ie 3 ly 2 6 3 a | ¢ | 3 2 | 4# | 1 | 3 s | 2 ee ee ee ee 2 1 ¥6 lve 3 23 3 z 3 2h 1 § 13 16 3 te 3 45 3 z 3 1x 2 33 1 z 3 aie ae ne ee ee a ee 4 13 t i 3 53 1 : ELEMENTARY MACHINE DRAWING AND DESIGN 14 *gT eded uo OT 91qRL 66882 490 OF 8 089 °T 089°T OOT 0 SPO OT SPP OT 99¢8°0 OSZ’ OT 610 OT Or G6L°c9 | TOL’ ES 8 OLg T OLg'T O0T'O L096 LEE 6 Fre 0 889 '6 000 '6 6 Tc0'0G | LL1°83 8 O9F 'T O9F 'T O0T'0 PES '8 PEE 8 6ce 0 GZ9'8 6862 8 SPL 88 | 1L42°8s 8 09€ TL 098 'T OOL'O Org’ 2 OFE 2 TOO GeO L €60'2 4 988° 8% 692° 8T 8 096 'T 092 °T OOT "0 9r¢'9 9FE'9 082 0 S299 ¢90°9 9 $8661 POS FT 8 O9T T OOT'T OOT'O 06F 'S 06z°¢ 6g< 0 99°¢ SFOS g 096 °ST OPE CT 8 OOT'T OOT'T Oot 0 [86°F TEL P 9F6 0 000°S 809 °F ip 0&2 GT G99 OT 8 0g0'°T 0S0°T OoT'0 bev Pv VEG Ft 2&3 0 009 7 90°F v 988 °6 T00 6 8 000 °T 000°T OOT'O 8e6°¢ 8EL'¢ 9260 000 °F 8S € #6 6882 gsc’ 2 8 0¢6°0 0960 OOT'O TPP € T¥G'€ L120 00g € 290° € § 082 °F 6eL°¢ 8 068 °0 0680 OOT'O 618°S 619% $020 G18 °G 89F'G 4% goes 609 '°§ $11 08s 0 0890 690 °0 6E°S 102% FST 0 SLES L90°S G $€0°S 829 °% 11 oss 0 0s¢'0 690°0 998 'T 8éL'T SPIO 006 °T O19 T 41 G6F'T PPS S FI Ors 0 Orc '0 690°0 “969 'T 88h T OFT 0 099 T Ose T 1 098°0 899'T age org 0 OTs 0 690°0 €86'T VI 'T VET 0 STe'T 8F0'T IT gé¢'0 SILT tI OOF 0 OOF 0 290°0 gc0'T 116'0 sIT'0 0S0'T ¥68'0 t s08 0 LE8'0 tI 068 '0 06¢'0 2490'0 S180 10L°0 601 0 080 8690 t O6T 0 6g¢°0 8T 008 °0 008 °0 $00 9990 899 °0 160°0 ¢$19°0 b6P 0 e FOTO OcFr 0 81 06¢ 0 062 0 $0 '0 6c 0 e&h'0 880°0 Org '0 PIE O - t 990°0 T¥%'0 L1G 0619 O6T 0 620°0 €6¢°0 Fee 0 890°0 SOF 0 02420 T ‘yaularenbg| ‘spunog *‘raquInNN Jeouour “soqouy ‘soyouy “sayouy *saqouy ‘sayouy *soyouy *sayouy ‘saqouy Bu MBS! |, “spBeIqL | “Spverq1 F ‘apisuy ‘pied 3 ; s194u2 qoaytag =| SPB2ITL JO puq u10330q * - SpIsInG ‘ pisuy. SaDLatet) aranoan eck sqouy sod | “9SIC 1890], | YySueT= Wed= 7 ye -viq=sqiy “eiq=t seoUIOGL a i | eae eee apwuy prepunyg | SP8IUL é 7 0 “spug pawalog q-a ‘aqny, jo JayaUIBIq, éL It or 6 8 L 9 g v € & T ddId GHCTHM NOYILHDNOUM AO SNOISNHWIGC GCUVANVLS 6 FIavy 15 USEFUL INFORMATION I él +P 9¢ Eng j z £69 $1g t LVS SF I 1 98 $6r $8 j BE fo ECS _1g : 28°T CF 1 1 ze EGP oP j eL of Eo fpr t TL'T 9¢ I 1 8Z 9€ $e z Ghee | eee ¥8e 8€ t SPT 0g I I 8% ¥E see EG | MeL ese 498 9¢ ¥ 8e'T 8% ¥1 T ¥ #1 ¥I€ z min jie THE Eee ; 0g 'T 9 1 1 06 62 16% ET tL. [eee oe #18 t GZ'T ¥ ¥I I 02 reg. eres reed Wi hey | eee ee FOB | t SIT G 1 1 02 ne Ele e hee ile eons HS fo a 60°T 02 § I I 9L 3 ZS ee eee ¢ Sty wg eee ae CZ StF a ae st ZO" I SI I i 91 HIG] eee SEL eS 1 BEG! [res er G60 91 I : ral Het | eet ey. il eee z ig..| ##ere st 98°0 #1 ; z ae | eee Be) | sekens ; BE | eeseee st 620 aT ; g 7 ey. | eaenas gt, | seuss 2 Gy | aeeers ee se a g a Bop fe ey | se eeed fy er) |nreeen st 89°0 8 § 8 topo $y. | eae ey gep of occ 8 $9°0 g : : 8 OL | EE pee il for | eee : 09°0 4 ; é 8 ‘6 6} pees 41 II _ ¢ 9¢°0 9 BU ES 86 ee ee Be feces | ee PU Veeget) 2 YP ede a a : : 8 EL oahey ae | Woe deed gt ee Nea kacay bee 7 02'0 Y : v 7 2 ie a ere gl z 6 =e me e e 1 6 a 0 m4 g ¥ Dm | eas a i 48 oa t Lb'0 ¥e E : ne re | Weta a H ip i : eo : i : goof goof ee i Gf ean i oF'0 &z ; : Bf ee Be Western ee : S| teens : TED z “841 006 “Sql SL ‘salog 70 “BqI 00% “sq[ GL “8qT 006 “sql SL iota SI 002 SqI cL ‘gen a rea tora: igen e ean “ag * 7 ssau Hor ean apuUTyA: arom I On : gor yo wiser [ae cae ecg, [| 8BueiL tovemmra | PP | sePaHAD | sepunAg (SHHGNITAD NOUI-LSVD YO4) GAD IL TIav NVTI WVALS GCUVAGNVLS 16 ELEMENTARY MACHINE DRAWING AND DESIGN TABLE 10 DIMENSIONS OF PLAIN OIL CUPS (BRASS) Nom. pee | a | | ee | DP we | ae |e Thread. z - a” ” a” a i ” ae Wy" 16” te de a |e ae Gee LE eS be lee |G dea |e gabe |e |e ee da BY (28 122 |25 | [le ]le | a [se [4a fe TaBLE 12 SPLIT PIN OR SPRING COTTER ” B L Wire Gage No. B L Wire Gage No. a a 13 i 1-4 4 a 4-2 12 ts 14 1 4 4-22 11 3 134 & +2 10 ts £5 u 24 9 i 2'-6 h 42} 8 i 3 -6 ie Bh 7 Lengths vary by } up to 4” and by He ae 6 1” from 4’ to 6”. a 1-3 5 TaBLE 13 TAPER PIN A Taper 4’ per Ft. ¥ Number... ccs ccasus 0 1 S 3 4 5 6 7 8 9 10 D actual........ . 156] . 172} . 193] .219] .250] .289] .341].409] .492 | .591 |.706 D approx........ a3 see | az | 4 | a 3 eR TR os che en neg a $-1 |$-14/2-14| 4-12) 2-2 |3-23}2-34)1-33]/12-42)12-52/13-6 Lengths vary by quarters between limits given. USEFUL INFORMATION TABLE 14 SCREW THREADS AND BOLTS 17 New Sranparps RECOMMENDED BY THE MECHANICAL BRANCH OF THE AssoclaTION oF LiceNsED AUTOMOBILE MANUFACTURERS ae: LDS B refers to all nuts and screwheads. P=pitch of thread. d=diameter of cotter pins. P . — =flat top. DX1.5=length of threaded portion. 8 TABLE OF DIMENSIONS D P A A, B C E H I K d 3 | 28) az is as e vs a5 is ve ve | 24 | 3 a 2 35 ee St wa ve Te a | 94 5 L 9 2 s a lal Hlelela lal elel ele 4/20!) & ve i vs s 3 3 32 33 ws | 18 | #& ar t vs v2 a s ve t $ | 18 | # a té z vz a 3 vs t % | 16 | && 3 1 i v2 cra s ve $ | 16) 8 ra lie | ¢ oa 6 s vz 5 z| 14] 3 #7; 3 ao | Oa $ a2 3 1 | 14} 1 3 lie | 2 35 i 3 v2 3 13) 12 | 1 a 13 vs aE a aE 32 oa Wy) 12] lys 18 | a3 Te aa 32 a 12 | 12 | 148 | 1 2 3 4 ly | 3 te | i 13 | 12 | 13 1s 23 | 3 t 13 2 ie u 2 100,000 pounds per square inch tensile strength. High-grade steel 60,000 pounds per square inch elastic limit. 18 ELEMENTARY MACHINE DRAWING AND DESIGN TABLE 15 WOODRUFF KEYS SMALL Sizes No. of Key D C B No. of Key D Cc B 1 » | a | @ Bo} 1 f& | 2) ¢ | & | & 6 | oo} ow | & 3 | 4 t | & | ow los | & 4 ; & | a 18 } i & 5 | § bol ox c tl & | & 6 | @ | & | & | 1 low | & 7 : t | x || 2% ti) se | & 8s) # | & | w& | 21 Be oe Nh) ae 9 | ? | w | & || D tlw | & 0 | & | & | & | 2 i bs M1 i & | & || 2 3 i es 12 ] a ts 23 3 ts s AG t | & | F Bl) 2) | oe 3] 1 | gw | we || 24 } 1 | & 14 1 3a ys 25 $ is ea 15 1 i te G } i wa Larce SIzes No. of Key F E D C B A 26 133 133 25 ts aE 33 27 133 133 = 7 aE ve 28 133 133 23 te a2 a3 29 133 aE 25 3 ae a2 30 3 SE 33 3 ie ve 31 3 38am 33 vs 3 3 32 j 3a 3h 3 B + 33 25 Séz | 88 ts Ht is 34 3 8a 33 3 13 3 Woodruff keys are shown on page 100. USEFUL INFORMATION Tasie 16 BEARING PRESSURES FOR BEARINGS 19 Pressure in Pounfds per Square Inch. Crank pins of shearing machines.......:........ 3000 Crosshead neck bearings (intermittent ee Mie clsaoe 800-2100 Wrist pins (gas engines)..................... 800-1000 Crank pins (small engines), land (slow speed) . 150- 200 (marine engines).............. 400— 500 . (land engines (fast)).......... 500- 800 cs torpedo-boats........ re aatat 850-1000 ee DO ON cise dow ao ena mieten meee ere el naw Oe 7 1200-1800 ss GASONE CN IINES = ses cse- sissies seeay sv oargpaianyay sien atcemsa faves aiy a Aamir 350-1206 Main bearings (gasoline engines). .3..00..cncaeeus noun end ea ebeb ead whe 350— 400 slow speed stationary... 0.0.00... eee c eee cee eee 280- 540 ua high speed stationary. . 210— 370 marine ,engine (naval) . 275— 400 me (merchant) . oi 200-— 300 Flywheel shaft bearings. ........ 00. ccc cece ee teen eee een nneee 150-— 250 HGOOWONG SNOB VOR 65 co 9 ooo wd omega ak wee dd avaine RE RA EM AEd SE BWR oe 60—- 140 Heavy line shaft (brass or Babbitt) 100- 150 Light line shaft (cast-iron bearings) . 15- 25 Line shaft on gun metal bearings.......... 200 Generator and dynamo bearings isi hia onrase aif 30- 80 Pivots, wrought-iron shaft on gun metal step....... 200- 700 Collar thrust bearings for propeller shafts............ Aga 50- 80 Slides, cast-iron on. Babbitt. cs. 0ccamecscgdigade havent FERaweE oS ane 200-— 300 Oo cast iron on cast iron... ee tte teas 25- 100 TABLE 17 COEFFICIENTS OF FRICTION Cast Iron on Bronze, 720 Feet per Minute. (THurston.) Oils Pressure per Square Inch 16 Ibs. 48 lbs. Sperm, lard..... She eas Veo E 0.138-0.192 | 0.077-0.144 Olive, cotton-seed, rape. ............0 eee 0.107-0.245 | 0.079-0.131 Cod and menhaden...............000eeeeee 0.124-0.167 | 0.081-0.122 MUG al ss case ctnesenre acest sorte sce e HARON ld Anca 0.145-0.233 | 0.094-0.222 COEFFICIENT OF FRICTION OF VARIOUS SURFACES Surfaces Coefficient of Friction Wood OD WOORLET Vi ce gun creas td gen nea ped emciaw ade ih 0.25-0.5 ne BOAPED spa haves dle how tea seathnr threes Pee Rene 0.2 -0.04 Metals on dry wood (oak). ... 2.0.2... .0 0 eee ee eee 0.5 -0.6 a wet wood (oak)... 1.0.0.0... eee eee eee 0.24-0.26 es SOAPY ons doctsaeauhiienis or neatne daly ieee ohare 0.2 Leather on metals ice tae lating: aig con oath onee ae 0.56 ne SW Obi siea circadian, Ral dotalas vetlicanians ne 0.36 ad ae ASV oj.acia ce dee SAE Ey eee GAS 0.23 ao ales ee ee ee 0.15 Metals on metals, dry. ........000 060 e ee eee ee 0.15-0.2 os ne WE biscuits tguea weSnrases 0.3 Smooth surfaces continuously greased...........--- 0.05 S a best results... .........00..000000- 0.03-0.036 0.27-0.38 Leather O90 Ole, p00 neva bers SRee eae ee eR aE NE 20 ELEMENTARY MACHINE DRAWING AND DESIGN TABLE 18 WEIGHT OF SUBSTANCES IN POUNDS Cubic Inch Cubic Foot. Brass) cc ae. de G4 RAE Od Hela ea ees 0.3 520 Bronze ge. 5:h'g gcin og Bly ae vate so aematleee Sect 0.319 552 Brick (ecommon)..............06.00 06: 0.064 112 OES (AINE) oes 's Gavan Oc yeblawignnd kaaswalg 0.08-0.09 140-150 COPPEP sss eae e eee tee aes ena 0.318 550 Cast inn cscs cars acae OMS KORG PAG EE ew RS 0.26 450 PAT UH27) each chal aac cua teenes babeaa Saran ean 0.046 80-100 GIBBS ai ures casos nian Mand ae atanehactue Sa ee 0.095 165 GrAniGG Sodas Maysa Bie’ Sew A am gla Rages 0.092 160-170 TCH iis ahve deen eenwin Wa ene a EARS 0.411 710 Sarid i340 Jae ted be Shs ee ee Be ed 0.052 90-106 Sandstoneiss <4 :e.acoe cain goo anaes C.08-0.09 140-150 Sheet has, sceasand wea eauaren ae eannkes as 0.283 490 Trap rock: .c.ceeweaeddh new eters sskeas 0.1 170-200 Water ccxdatnarolinde Gand Lewin uae ele e 0.036 62.35 Wrought iron. ........ 0.0.00. cece eee 0.282 490 DAG 3 3 Sil oan Ge be MAY Ganesh Mobos oS 0.248 428 TaBLe 19 MEAN STRENGTH OF MATERIALS IN POUNDS PER SQUARE INCH Tension Compression. Shear. Wrought iron............ 55,000 50,000 45,000 Cast iron................ 18,000 90,000 11,000 Cast brass.............. 18,000 10,000 * DteG loin. ig eset Pala weds a sud ened 65,000 |... eee eee 48,000 StONGs sadee's sa eaey bea wae 9000-12,000 3000-18,000 BHC cacccecanca chess 280-300 800- 4,000 TABLE 20 FACTORS OF SAFETY (AveraGe) Factor For. Live, oF Verving Material. puaPoak eee Subjected ead Load. a o Varying Equal ait. Load: d Suiting | Sarees of | Shocks: Kinds Cast iron..............006 4 6 10 15 Wrought iron and steel...... 3 5 8 12 Timber.............. Bale we 7 10 15 20 Brickwork and masonry... .. 20 30 ELEMENTARY MACHINE DRAWING AND DESIGN CHAPTER I WORKING DRAWINGS 1. Working Drawings are divided into two general classes, viz.: Assembly Drawings and Detail Drawings. The assembly drawing shows the machine with all its parts in their proper position in the machine. A few of the principal dimensions are given and the parts may be indicated by dis- tinguishing marks as A, B, C, etc., to serve as a guide to the erector of the machine. The overall dimensions are serviceable in determining the space required for setting up the machine. Detail drawings give minute particulars regarding the form and construction of each part of a machine. They are usually drawn to a larger scale than assembly drawings and are sent into the shop for the use of the workmen. Consequently, they are often called shop drawings. An example of an assembly drawing is the globe valve in Fig. 180, facing page 283. Detail drawings will be found on nearly all the plates in these notes. 2. A Detail Drawing is one which contains all the informa- tion the workmen may need who are concerned in the making of the object delineated. It may be a sketch, or an isometric projection or an orthographic projection made on paper or cloth in pencil or ink, or it may be in the form of a photographic print. Orthographic projections only are referred to in these notes. 21 22 ELEMENTARY MACHINE DRAWING AND DESIGN Above all things a working drawing should be clearly made and self-explanatory so that no questions need be asked regarding the object represented. All the dimensions should be clearly and accurately shown, the material of which each part is made should be stated and the finish required on the different surfaces, also the number of pieces required of the drawing shown, the sizes and kinds of holes, etc. If the drawing is to be used by several workmen in the various stages of manufacture of the object drawn, the information which one workman may need is often of no value to another. The pattern maker may need some instruction and dimensions which the machinist does not need as he has the object itself to work on. However, the drawing must give him the proper information as to what work he shall do on the object. A drawing should be neatly made and the notes printed on it should be in a clear legible style of printing. Many a drawing has been spoiled in appearance by the printing on it. 3. The steps to be taken in the making of a working draw- ing may be presented as follows: (1) PENCILLING ON PAPER. Layout of views, blocking in, completion of outline, dimen- sioning, lettering. (2) TRAcING ON CLOTH. (3) DupLicaTION OF ORIGINAL, by photo printing. Under (1), more in detail, we have the following: (a) Choice of views; (b) Choice of scale and size of paper; (c) Location of views on sheet; (d) Center lines of views; (e) Drawing of principal lines in all views; (f) Drawing of other or secondary lines in all views; (g) Location of bolts, keys, shafts, etc.; (h) Dimension lines, extension lines, arrows, dimensions; (¢) Section lining, notes, title, border, trimming line. Under (2), tracing is comprised: (a) Stretching cloth over pencil work, powdering cloth; (b) Inking in circles, ares, irregular curves; (c) Inking in horizontal lines with T square and vertical lines with triangle; (d) Inking in inclined lines, shade lines (if desired); WORKING DRAWINGS 23 (e) Dimension lines, extension lines, center lines; (f) Arrows, figures, notes, title, section lines, border, trimming, cleaning. Do not ink in more views on tracing cloth than can be- finished in a day. All the drawings comprised in these notes can be drawn on five sizes of paper. No. 1, 8’X102”, No. 2, 11”X15; No. 3, 22X15"; No. 4, 20X27”; and No. 5, 4027”. The dimension first given is the side of the sheet to be placed horizontal. Nearly all the objects can be drawn on the first three sizes of paper. Before commencing to make a drawing find out how many views are required of the objects to be drawn, the scale of the drawing, and how much room each view will need. Draw in the center lines of each view as fine, light, continuous lines. SCALE OF DRAWING 4. Scales. In laying off the dimensions on a drawing which is not full size, it is customary to use scales which are made to correspond to the size of drawing which is wanted These scales, although reduced in length, are always divided like a foot rule. Thus a quarter size scale is represented by 3 inches divided into twelve parts to represent inches, and these inches divided again into sixteen parts to represent sixteenths of an inch. When the dimension is laid off by this scale it is spoken of always as the full size dimension. In reducing to half size it is customary to use the full size scale and consider each half inch as if it were an inch, each eighth as a quarter, each sixteenth as an eighth, and each thirty-second as a sixteenth. In order to lay off 243” half size, count off two 34” spaces, six 1,’ spaces, and one #;’’ space just as if each half inch was marked 1-2-3, etc., instead of the inches. See Fig. 1 (B). Never divide the full size dimension by two to lay it off half size, but make the scale itself do this and always think and talk of the full size dimensions. The triangular scales, which are called architects’ scales, have two reduced scales at each end of each edge except the edge which has sixteenths on it. Each of these scales is marked at 24 ELEMENTARY MACHINE DRAWING AND DESIGN each end by a number, as 3, 13, 2, 2, etc., meaning that the scale so marked has 3”, or 14”, or 3”, divided in the same manner that a foot rule is divided, viz.: into twelve parts and each of those parts subdivided into halves, quarters, and eighths. Only one space of 3’’-13”, etc., on each scale is divided off into sub- divisions at each end of the ruler, but each division of 3’’-1}’-2”, etc., length is marked with a number to correspond to the number of that division counting from the subdivided division. Fig. 1 (A) shows a scale of 13’’=1 foot, which is an eighth size scale. One foot, 62’’, is shown laid off on this scale. When taking dimensions from a scale, lay its edge along the line to be measured and mark the distance with a pencil or needle point. Never use the scale as a ruler for drawing lines. The scales commonly used for machine drawing are: Full size, half size, quarter size, neal (B oe |! [e ee fu fe 4 cs £4 "2 Ltt ibbbbnitibbbbth 1163" | A Fic. 1 (A). and eighth size. The scale used should always be noted on the drawing, as indicated above or as follows: 6” =1’, 3’ =1', 13” =1’, 1"=1’, ete. Use full size scale if possible or advisable, if not, use half size, quarter size, eighth size, twelfth size, etc. All of these reduced scales will be found on the triangular architect’s scales commonly used in draughting offices. All views of the same object must be drawn to the same scale. Different objects may be drawn to different scales on the same sheet, but the scale should be noted in each case which differs from full size. 5. Blocking in. After locating the principal center lines of the views the outline is built up around them by putting in the principal lines in each view, then secondary lines. Draw straight lines first then the curves which join them. Circles can be put in first where they are principal lines. Work all views of one object at the same time, and divide the work into the WORKING DRAWINGS 25 constructive and finishing stages, the former consisting of light lines, the latter of heavier lines over the light ones. The lines drawn during the second stage, the arrows, the dimensions, and printing will show more clearly and can be traced with less strain on the eyes if a softer drawing pencil (HHHH) is used than the one used (6H) during the first stage. Use a conical pencil point for the finishing stage and do not sharpen it to a prick point when making arrows and figures. The shaft coupling of Figs. 1 and 2 shows how the lines are drawn in the two stages and how arrows are made before putting in dimension numerals. Pulley Coupling Pencil Layout Section of Pulley Coupling Fic. 1. Fia, 2. Dimension lines are fine continuous lines broken to allow the dimension to be inserted. The arrows touch the lines to which the dimension is measured. These may be lines of the drawing or extension lines drawn from them. See Fig. 3. 6. A section of an object is often shown instead of an outside view. It is obtained by a plane which cuts the object in such a direction that the section made shows the interior more clearly than the outside view taken from the same position. The material ‘cut by the section plane is cross hatched or section lined by drawing parallel lines across it. These lines are drawn at 45° to the horizontal and from vy” to }”’ apart, depending on the area 26 ELEMENTARY MACHINE DRAWING AND DESIGN and scale of the section. Various combinations of light, heavy, and broken lines are often used to designate different materials, but there is no universal standard. The standard sections of the U.S. Navy Dept. are shown in Fig. 4, page 27. The safest way to indicate the material is by a note on the drawing giving the material of which each part is made. Sections of parts of a machine cut by the same plane, whether of the same or of different materials, which touch, should have the section lines slant in opposite directions. (See section of pulley coupling, Fig. 2.) When the scale of the drawing is much smaller than full size and the material cut by the section plane is thin, as the L ‘ Niu ~S ee _ , 4 2 at eo A 5 i n 3" f y 3 a f Bre 477 : 4 ‘Dimension lines very FineL~ \eetenain lines fine) Fic. 3.—Specimen Drawing. plates of a boiler or tank, the section is often blacked in as shown in Fig. 5. Breaks are often made in a long bar to reduce its length on the drawing as well as to indicate its general shape. Such breaks as are conventional and standard are shown below in Fig. 6. Cylinders are generally represented as in (A) and (B) and pipes at (A), (B), and (C). 7. It is not customary to section every part lying in and cut by the plane of a section. The exceptions which are not section lined are bolts, bolt heads, and nuts, screws of all kinds, shafts, keys, rivets, oil cups, pipes and all cylinders whose axes lie in the section plane. If the section plane cuts the axes of the parts mentioned above, at right angles, then the part cut is section lined. Arms of pulleys and gears are not sectioned when the section plane contains the axis of the arm. Keys and only when cut by planes perpendicular to their greatest length. 8. Arrows on draw- ings should be put in as soon as the dimension and extension lines have been drawn. arrows by two strokes towards the point. Make them sharp pointed and black, with the barbs close to the shank and not spread out, and use a cone pointed pencil. The arrows point away dimension in WN crowded spaces when they point towards it. Arrows always touch. QS the extension lines on s the outside as in (D), SSS Fig. 7, and in Fig, 2) RX s of the pulley . ling, or on the inside asin (A), (B), (C), of Make coup- 9. Dimension nu- merals are put in after the arrows are made. They should be clearly made and rather blacker than other lines of the drawing. See Fig. 8. The style of numeral should be open and made as shown in Plate 2 last line, or in Fig. 9 WORKING DRAWINGS gibs are section lined Copper Babbitt Cast Steel ~ SN WM Q_°Vv ¥ RAQN § SN) ae Ui ' Leather Stone. Standard Cross Sections of US Navy Department AAT NAP RO S (tH eeaeeen x ThE Ba: Sarda, RN ” PSN wl S Hr herd OQ PeeN age Vulcanite Wood Brick Earth 27 Fic. 4. 28 ELEMENTARY MACHINE DRAWING AND DESIGN Gol Plates Plate Fig.1 and Angles Fig.2 I Beam Fig.3 Fia. 5. I Beam Channel Angle Rolled Sections pa ee) a ra Rectangular Bar Cylinwer LZ Sa ee ~ ey ES 7) CLL LLL ESS =o Cylindrical Pipe Cylinder Fig. 6.—Conventional Breaks. (A) K 3" Dimension Line Continuous (B) e--------- 3" ---------- >| Dimension Line Broken (C) ip Arron only P 3 Like this orthis--+3 (E) yf <— This or bettie 3 Northis--r- + These Fic. 7.—Arrows and Dimension Lines. Fie. 8.—Sample Drawing. WORKING DRAWINGS 29 line (B). On a full sized drawing the numerals should be about 3” high and fractions about 33;”".. All numerals for horizontal or vertical dimensions must be placed to read from the bottom or right hand side of the drawing when the person reading is facing the bottom or right hand edge of the sheet respectively. abedetghijkinnopgrstuvwxyz. AP abe het o hij t hmenmoe gra run el (2324567890 ~ ©) abc defghijklmnoparstuvwxyz ©) aocdet Se (0) / abcde Imnopgrsiuvy, WxeGZz = abiderghklmnapgrstuv 5 WXYZ. Fie. 9. Oblique lines should be dimensioned to read from one or the other of these two edges. Fig. 10 shows how to place the figures for the lines in various positions which occur on drawings. 10. The dimensions placed on a drawing should be expressed in inches and common or vulgar fractions up to 24’’ and in feet and. inches above that. When fractions of an inch are used the denominators are usually multiples of 2, 4, and 8, as 3, 4, $ 6) 3m) eT: The numerator is always a smaller number than the denominator. The division line between numerator and denominator must always have the same direction as the dimension line belonging to the fraction, as 2, am, se, never 3/4, XX. The Fia. 10. fraction should always be reduced to its lowest terms, that is, 14 should be } and ,%; be §. The dimensions on a drawing bear no relation whatever to the scale of the drawing. They are always fuil size and nothing but full size dimensions are ever spoken of or ever written. When dimensions are expressed in feet and inches a dash ismade between the feet number and the inches number, as 4’-63”’ and if 3) ELEMENTARY MACHINE DRAWING AND DESIGN there are no inches to be expressed a cipher is made in the inch place as 6’-0’.. Avoid fractional numbers in long distance measurements. If the distance between two principal centers should happen to be 6’-734” very likely 6’-72” would answer as well, or even 6’-8” might be used, but if it is the sum of several dimensions it must not be changed. Dimensions are always placed on dimension lines, never on a line of the drawing outline. 11. Special forms of dimensions are shown in Fig. 11. Numerals which are located on surfaces crossed by section lines, are placed in spaces and are not crossed by the section lines. See the coupling of Fig. 2. The longest dimensions of an object should be placed farthest from the object, the shorter ones nearest to it, as in Fig. 8. Fic. 11.—Special Dimensions. Diameters of circles are usually given instead of radii. If the radius only can be shown when a diameter is wanted, the diameter numeral is given with a small (d) written after it and one arrow head is shown touching the part of the circle which is visible, as in (M) Fig. 11. In a view half in section the dimen- sion is given as shown at (NV) #’d, and 142’d. Ares of circles have their radii given as at (Z) or at (K). On pencil drawings a small free-hand circle is made around the center of the are to assist the tracer in finding it. When a tube is shown in section as at (J) a question might arise as to whether the arrows touched the inside or outside of the tube. This is obviated by placing the arrows for the outside dimension on the outside of the tube those for the inside on the inside. 12. Holes in circular flanges are located on a circle called the bolt circle whose diameter is always given. The distarce WORKING DRAWINGS 31 from this circle to the outside of the flange should also be given, especially if the flange is a turned one. (P) of Fig. 12 shows such a flange. (Q) of the same figure shows how holes are located in a plate by giving the distances from adjacent edges to the e (Q) Hales in Plate State whether Holesare Tapped or Drilled Fig. 12. centers of the holes. Whenever it is necessary to express dimen- sions for extreme accuracy as in thousandths of an inch, they are expressed in decimals, as 1.3278. For cut gears the outside diameters are often expressed in this way, also on work which is turned or ground either for force, shrink, or running fits. In the following Fig. 13, is shown a working drawing of a crank shaft for a triplex power pump. ett [> J (3 B ere yi eer y Y Tt ni y %& me 7 Lf “| 1 har lf 3 23 be ‘2 ate F ‘ a i P 45 ss > \ lt meen | eer 4 -f-allover . | 2 Detail ofCrank Shaft R27 et 2° e274 9" 12" Triplex Power Pump Scale:6%!' Mar: 10-1911 Fic. 13.—Sample Drawing. 18. Explanatory notes are placed on a drawing to supple- ment the information given by the dimensioned outline of the 382 ELEMENTARY MACHINE DRAWING AND DESIGN object. They comprise such information as the name of the object, the number wanted, the material of which it is made, the kind of finish desired, whether holes are to be drilled, bored, tapped, or reamed, or if cylinders are turned, ground, polished, etc. The weight of an object or the area of an irregular opening or a note indicating certain assumptions, is often very useful information to one who is called upon to pass on a drawing. 14. Abbreviations in common use on drawings are given below. MarTERIALS: C. I. =cast iron, Med. S. =medium steel, W. I. =wrought iron, Mal. I.=malleable iron, St. =mild steel, Bz. =bronze, T. S.=tool steel, Br. =brass, M. S.=machinery steel, Bab. =babbitt. S. C.=steel castings, OTHER ABBREVIATIONS: Pes. = pieces, Sq. =square, Pi. = pitch or threads per inch, Hex. = hexagonal, ft. =feet, Oct. =octagonal, in. =inch, D, d, or diam. =diameter, ins. = inches, lg. =long, R=Radius, cyl. =cylinder, r=radius, T =teeth, Thds. = threads, 0” =square inch, D. P. =diametral pitch, XO’ =square feet, P. D.=pitch diameter, ° = degrees, f=finish, H. P. =horse-power, '=feet, K. W. =kilowatts, "’=inches, cm. = centimeters, R.P.M. =revolutions per minute, B. & S.=Brown & Sharpe, m/m =millimeters, Se. =screw, m = meters, Std. =standard, U.S.St.=United States Standard. A.S. & W.=American Steel & Wire Hd. =head, Gauge. B.W. G. = Birmingham Wire Gauge, C. to C.=center to center, C. to E. =center to end. Ssop OPERATIONS are marked as follows: z+’ pipe tap = thread this hole to fit a 3” pipe thread, wl ml ” tap = thread this hole to fit a 3” screw thread, 3 tap-10 =thread this hole for a 3’ bolt having 10 threads per inch, 2” drill =drill this hole 3” diameter, 3” bore =bore this hole to 3” diameter, 4” turn =turn this cyl. to 4” diameter, 1” core =use a core 1” diam. for this hole. WORKING DRAWINGS 33 15. A finish mark is used on a drawing to indicate a surface which is to be machined. It consists of a lower case f placed on the line which shows the edge view of the surface to be finished. The cross bar of the f makes a slight angle with this line as shown in Fig. 12 preceding. The degree of finish desired is expressed by the word scrape, buff, polish, grind, etc., placed parallel to the line denoting the surface. If the object is entirely machined the f marks are omitted and a note is placed below the object, viz., —f— all over, as in Fig. 13. The allowance required on the pattern or forging for material which is to be cut away in finishing is left to the judgment of the pattern maker or blacksmith. The dimensions given on the drawing are those of the finished object. 16. Titles. Every drawing sheet should have a title on it containing more or less information regarding the object or objects thereon delineated. This title is nearly always placed in the lower right hand corner of the sheet to facilitate finding the sheet when placed in a drawer with other drawings. It should contain the following matter: The subject matter on the sheet, The person or company for whom the drawing was made, The scale of the drawing, The date of completion of the drawing, The name of the draughtsman, The name of the person who checks or approves the work, The filing number of the drawing, The number of the sheet and the total number of sheet in the set. As the drawings of an elementary character in a school or college are made for the school, it is advisable to substitute the school name for No. 2, of the title. The remainder of the title is approximately the same as that above. In the layout of a sheet the space necessary for a title should be determined and the views placed so as not to encroach on it. A space of three inches by five inches ought to be sufficient for a title on any one of the plates in this machine drawing work. This space will be in the lower right hand corner of all sheets just inside the border line. A border line will be drawn one-half inch inside the edges of the drawing paper. It is not necessary to make this line elaborate on machine drawings, therefore a 08 SS or HE ee 34. ELEMENTARY MACHINE DRAWING AND DESIGN single continuous line will be sufficient. A few titles are shown on Pl. 1, as well as the form of title recommended for students in the Sheffield Scientific School. 17. A shop bill or bill of material is often placed on the drawing instead of on a separate sheet. A bill of this sort gives the name of each part of the machine which is represented on the sheet, the material it is made of, the number of parts of each kind required to help make a complete machine, the mark of each part, and any data conveniently expressed, as the length of a bolt, its grip, location in the machine, etc. Such a bill for a cast iron shaft coupling might read as follows: Britt oF MATERIAL Pes. |Name of Part] Material. Mark. Dimensions. Remarks. 2 | Flanges C.I. (1) or F 1 | Sunk key | Steel (2) or K | 3” wide, }’thk., 4” lg. 1 |Sunk key | Steel (3) or A, | 3” wide, }” thk., 32” lg. 4 | Bolts W.I. | Hex. hd. | 3” diam., 2” grip, 22” lg (U.S.St.) Hex. nut For a screw cap stuffing box: Britt oF MATERIAL nee Name of Part. Material. Mark. Remarks. Finish. 1 Screw cap Brass Cr Casting 1 Gland Brass F Casting 1 Casing Brass A Casting 18. Lettering and Printing. A drawing will not attract favorable attention unless it has good printing or lettering on it. It may be inaccurate or poorly drawn, but if it is well lettered the untrained eye unconsciously judges it by this. On the other hand a drawing may be accurate and the lines well made yet be condemned by careless or unsightly lettering. Drawings for machine shops, pattern shops, forge shops, etc., do not require such elaborate titles as one finds on maps, nor does the lettering need to be made with instruments of pre- cision. Free-hand printing, if well made, is good enough for WORKING DRAWINGS 35 Title suiti ble for a school drawing. a DETAIL Sflanged ShattC. oupling of ME Oepl Sheftield Scientific School ARMA pra SHAFT Yale Unive rsily SOKW OC Generator Westinghouse Electric Mfg. Co. Sedle 6=/' Yan.119/2. Scale hal/fsize ‘Yev 20, 1/902 Drawn by D. Swift — Class of 1912 ME. Approved by. 77. —P/ of (20) PIs. Orawn by... — Checked by _ Approved by________ 10742 ppB = ee bridge 520 £19 Kun, Evansville, Arizona. IO. K and T. Ry. o0Leé Div. J Spans, Single Track, Through 17820'c.¢.End Fins. 7OP_CHORDS #0 END POSTS. Scole V4'= tt? AMERICAN BRIDCE Co., EDGEMOOR PLANT. “MYMOM-3DaGINE NOS FTL A. B. Co. Contr. No...292___...In Charge or Wilson... Checked py. 4:4: silt eel pate 9/27/09 Rev.—_...--.-- Orver No. 42:30 “Sneet NO. /2__. 1 Made by. Lo wale. Date. 2/25/00 Rev....920... Cut blualrint onthishie standard lille of American Bridge Co. SIDE, ELEVATION CARRIAGE. rox SIX INCH B-L: RIFLE DRAWN UNDER. THE. DIRECTION OF GOL:CWLARNED CADET QRIMTHER_ Scale, 16 Title used at U.S Military Academy Dec. 1903. PLATE 1.—TItT1ss. 36 ELEMENTARY MACHINE DRAWING AND DESIGN notes and titles on shop drawings, besides taking much less time and skill to make. The following system is recommended for beginners and will give good results with a little practice. It applies to lower case letters. The curved letters are based on the letter (0) which is made with one stroke of the pen or pencil as in making a cipher. The other lines of this system are either horizontal, vertical or slightly inclined from the vertical. The body of certain letters as a—b-d-g—p-q is a complete cipher with a straight line added for a—-b-d—p and straight line ending in a curve for g and g. The c-e and 2 are incomplete ciphers the (e) having a horizontal line added. The x is com- posed of two incomplete ciphers touching at their sides with openings opposite their common tangent point. The letters n—m-r-s—u and y are modified ciphers with straight lines added. The other letters f-—j-k-I-t-v-w-z are straight lines to which we add curves inf and 7. No other “ tails’ or lines are added at top or bottom. abcde fghijklarn opgrstuvwxyz. AP abcdet Gg as pcanh gatas Fuse pee (23456278970 ~ (B) eee hijklmnopagrstuvwxyz ©) aocdet Mpa ay Sasi (0) abe Kl nopgrstu vz, vs WA Se Tabet detighy yee a WLYZ. Fic. 14, Fig. 14 (A) and (B) show these letters in the finished state as well as in the process of making. This alphabet may easily be varied by changing the slope of the letter as in (C) which is vertical and (D) which is back hand, or by changing the normal spacing of the letters to the extended spacing as in (Z) or the compressed as in (F). 19. The capital letters shown at (£) and (F) on Plate 2, give an idea of the Gothic type suitable for use with the above system of lower case letters. The inclination with the vertical is about 3 to 8. The other styles of lettering on Plate 2 are more fancy and more frequently found on architectural drawings than on 37 WORKING DRAWINGS 6 ALVId wo POC BLOGLEE! ZH | MAMINIDL SRAOCHONMM 1 Eh hy ACO 2LE Wh (2 ELOGEEAZAXMASIILSY OSONNIMIIHIATOIGY (q) ZAXMANLSHYODDONWIYNPSIHOISIAIOGYV (2) ZAXMANLSYODdGONW IAS IHOAIIOSV C6QLOSVEZI i _7h&xmanyabdouwpy{iyByepoqe ZAXMANLSAIOMONWNTY PIHOIIAOAV 3B 06QLOSLLZI y — akxmangeubdouunylighjapoqv ZAXMMIUSYCUONW ISTH AFTIOIV 38 ELEMENTARY MACHINE DRAWING AND DESIGN machine drawings. (A) and (B) are especially used on work of an architectural type. (C) and (D) are used in titles and notes where lower case letters are not desired, all the letters being small cayntals. Attention is called to the figures in alphabet (A) and (Ff) and the method indicated for making the lines which compose them. These are numbered in the order in which they are made. 20. The beginner is recommended always to draw limiting lines or guide lines with the T-square for the tops and bottoms of all letters and especially for lower case letters. These lines should be drawn so fine, with the pencil, as scarcely to be visible, otherwise they will mar the appearance of the printing if left on the drawing. Lower case letters should be from 3" to 4” high and the capitals which accompany them about ;” high to #3;”", depending on the scale of the drawing and the character or importance of the note. Use a conically pointed pencil for all lettering and not of too hard a grade, 2H to 4H being hard enough. Too sharp a point is not advisable. Beginners will find it much easier to make good looking printing if they do not attempt the vertical style, for it very difficult to make all straight lines vertical and any line out of the vertical is at once detected. 21. In the making of a title certain words are more important than others and require larger letters to give them more prom- inence. Usually the name of the object delineated on the sheet of drawing is the most important portion of a title. The scale and date being of least importance are in the smallest type. Intermediate sized letters would be used for other information such as name of machine of which the object drawn was a com- ponent part, name of concern building the machine, name of draughtsman, chief engineer, checker, etc. As a means of giving prominence to a word, wide spacing is often employed, also large or black faced letters. As an illustration take a title composed of the following: Detail of armature shaft for 30 K.W. generator for the Westinghouse Electric and Mfg. Co. Seale full size. Jan. 1, 1911, drawn by ......... traced by.......... approved by... sheet...of....sheets, No. 10,430. This will be found on Plate 1. The student who wishes to study the subject of lettering still more is advised to consult such books on lettering WORKING DRAWINGS 39 as Reinhardt’s “Freehand Lettering,’ Daniell’s ‘ Freehand Lettering,” and Jacoby’s Textbook on Plain Lettering, in which the spacing of letters, arrangement of subject, width of letters, styles of alphabets, etc., are treated at length. In many large manufacturing plants the titles are printed or stamped on the drawings and the draughtsman fills out the blank spaces as needed for the special sheets. 22. Shade Lines are used to make the object drawn appear in relief or stand out from the paper. Shade lines are heavier than the ordinary lines of an unshaded drawing while the lines not shaded are lighter to produce more of a contrast without requiring excessively heavy shade lines. Shade lines are the dividing lines between the lighted surfaces and shaded surfaces of the object when the parallel rays of light come from above the upper left hand corner of the drawing board at an angle of 45° with the plane of the paper. Each view is shaded inde- pendently of the others but the light rays come from the same direction always. All surfaces in the plane of the paper or parallel to this plane are considered as lighted. Vertical surfaces are either lighted or dark depending on whether the rays strike them or not. An inclined surface may or may not be lighted, depending on either the angle of its slope with the plane of the paper or whether it is cut off from the light by an intervening part of the object. One rule for shading commonly used is to shade the lower and right hand edges of a surface. The following Fig. 15 will illus- trate the use of shade lines to Top View represent depressions (B and D) and projections (A and). A circle “ [1] l is shaded after drawing a light | T | | | | a Loeresad circle, by moving the compass point along a 45° line zy away from the Front View center of the circle a distance equal His: 18 Shade Tines: to the thickness of shaded circle desired. Without changing the radius draw a semi-circle to join the original circle at z and z, Fig. 15. The shade lines of a hole are made heavy outside the hole 40 ELEMENTARY MACHINE DRAWING AND DESIGN in order to preserve the original diameter and enable the dimension to be given inside. The opposite is true if the circle represents a cylinder. 23. Bolt heads and nuts are shaded in various ways, but Fig. 16 represents average practice in this respect. Other plates such as the straight arm pulley, eccentric and strap, cross-head end of connecting rod, and bearing box, are examples of shading. Detail drawings are not shaded as a rule but assembled machines or “picture” drawings are often treated in this manner. It takes more time and often is not worth the trouble, although it greatly enhances the clearness of a drawing in the eyes of persons who cannot readily read drawings made by orthographic projection. C i T : Le a NK Shade Lines on Hexagon Nuts Fig. 16. TT he | 24. Tracings. Whenever it is necessary to duplicate a drawing which is drawn in pencil it is the custom to use some form of photo reproduction, which requires an original drawing to consist of opaque lines on a transparent background or the reverse. The background material employed is cloth or paper, which has been rendered transparent by a preparation of wax or paraffin. A drawing on such paper or cloth is made with India ink and is called a tracing, because it is generally made by placing the cloth over another drawing and tracing on it the lines which show through from the drawing beneath. As it is sometimes difficult to see through this transparent paper or cloth it is sometimes advantageous to make a pencil drawing directly on WORKING DRAWINGS 4] the tracing cloth and then ink over the pencil lines. It makes it easier on the eyes and saves paper. To obtain copies from a tracing, it is placed in a frame, in contact with sensitized paper and exposed to the sunlight or strong electric light for several minutes. The sensitized paper is then washed in water or some chemical solution which dissolves the chemical in the paper which was protected from the light rays by the lines on the tracing and we get white lines on blue ground or black ground. These are known as blue prints or black prints, and serve for shop or field use in engineering work. The prime requisites for a tracing are clearness and opaqueness of outline. Light thin lines on a tracing produce faint lines on a blue print and may even disappear if the exposure to the light rays is too long. A poor tracing will never give a blue print of as good quality as the tracing, always worse. 25. In the making of a tracing, always observe the following points and rules: Tracing cloth is always used in preference to tracing paper because it is tougher and endures rough handling. It comes in rolls and always has a tendency to curl up when unrolled and cut in sheets. This is due to the rolling in a roll and to the fact that the transparent substance makes it curl up on that side on which it has been applied, called the smooth side. If the inking is done on the smooth side, the tendency to curl is still further increased, but if the dull side is used the tendency to curl is diminished and the tracing lies flat. Pencil lines are easily made on the dull side, while on the smooth side it is almost impossible to make them. This accounts for the use of the dull side in all shops where the tracings have to be “ checked,” or corrected. The only advantage the smooth side has over the dull side is the doubtful one of lending itself better to erasures. Inking is more difficult as the ink is more apt to run on the smooth side. At any rate the dull side is recommended to all beginners, especially in technical schools. 26. The tracing cloth is stretched on the drawing board over the drawing to be traced, dull side up. Place a thumb tack in the middle of the top edge on large tracings and run the hand down over the cloth to the middle of the bottom edge where another tack is placed. Then place a tack in each corner after first sliding the hand towards the corner from one of the center 42 ELEMENTARY MACHINE DRAWING AND DESIGN tacks. The cloth should be large enough for a margin outside the border line of the pencil drawing. The cloth is now treated to a sprinkling of chalk or magnesia, which is rubbed in with the fingers or a cloth to remove all oily spots which might prevent the even distribution of the ink. This powder is carefully wiped off with a cloth, leaving the surface free from powder and ready to receive the ink. Do not use too much powder. As tracing cloth does not take ink in the same manner as drawing paper it is recommended to try the drawing pen on a scrap of tracing cloth before beginning on the work of tracing. Do not aim to save ink on a tracing but make heavy lines without attempting to shade them. Draw large circles first, small circles, irregular curves, all in some systematic manner, as from the top of the sheet downwards or from left to right. Do not ink in circles or curves on more views than you expect to complete on the same day, for tracing cloth has a way of absorbing moisture which makes it expand in a disconcerting manner, throwing centers and lines out of true with the drawing beneath. Horizontal lines are then drawn with the T-square, working down from the top of the sheet, then vertical lines are drawn with the triangle by moving it along the T-square blade from left to right. Draw straight lines towards the curves they are to join. Ink in inclined lines in all directions and then finish up any lines relating to the outline of the object such as breaks or lines to be drawn with a writing pen. 27. We now come to the lines on a drawing other than the lines bounding the object. These are center lines, dimension lines, and extension lines, all drawn with a straight line drawing pen. They can be drawn either with red ink or with black ink. Drawn with red ink they are continuous, that is, not broken. When drawn with black ink they are made as follows: Center lines are very fine and are continuous or dot and dash lines. The latter lines are composed of dots and dashes from three- quarters to one inch long. Dimension lines are fine light lines continuous or broken. Extenston lines are light and continuous, as fine as a pen will draw. After these lines are all drawn the arrows are put in with a writing pen and black India ink. Then put in the dimensions also in black and with a writing pen. Notes and printed directions and titles are then made, WORKING DRAWINGS 43 followed by the section lining and the border line. Section lines are fine, black and continuous. Sometimes in place of section lining the area to be shown in section is darkened by rubbing black pencil or colored crayon on it which gives a light blue effect when this surface is reproduced on a blue print. A tracing can be cleaned of dirt and pencil lines by using a sponge rubber, art gum, or a cloth dampened with gasoline. The ink lines are not affected by these processes. Draw all ink lines from left to right and do not press too hard against the ruling edge, as this closes the nibs of the pen and changes the width of the line. 28. Erasing from a tracing may be done most satisfactorily by using an ink eraser and sliding a celluloid triangle underneath the tracing cloth at the point where the eraser is used. If a continuous line is to be made broken a sharp knife can be used instead of an eraser for making the spaces in the continuous line. After using a knife, however, the glazed surface must be restored as nearly as possible by rubbing the roughened spots with pumice powder, pumice stone, soapstone crayon, or some smooth hard substance like the end of a pocket knife. This prevents dirt from collecting on the erased surface, and enables a new ink line to be drawn over it without spreading. Before printing on tracing cloth always draw light pencil guide lines for the tops and bottoms of all letters, even if there are guide lines on the pencil drawing underneath. Do not try to trace the freehand printing of a pencil drawing. The writing pens used should have rather stiff nibs and the stroke of the pen when making the letters should be made with a constant light pressure on the cloth, in order to avoid changing the width of the line during the stroke. Do not use a fine pointed pen. Gillott’s 404 or a Lady Falcon are good pens for ordinary notes and figures. A Falcon or ball pointed pen will be found more suitable for freehand title printing. An old pen works better than a new one because the nibs become slightly separated by long use and the ink flows better. ‘Too much ink on a pen will cause a blot on the tracing, therefore it is always better to shake a pen after dipping it in the ink bottle, to shake off the surplus ink. Wipe a pen often with a good pen wiper of cloth, for ink will not flow freely unless a pen is clean. 44 ELEMENTARY MACHINE DRAWING AND DESIGN 29. A résumé of the order of operations in tracing is as follows: (a) Preparing surface of cloth. (Art. 26.) (b) Inking circles, ares, irregular curves. (Arts. 26-27.) (c) Inking horizontal lines with T square and vertical lines with triangle (work from top to bottom and left to right). (d) Inking of inclined lines, shade lines (if desired). (Art. 27.) (e) Center lines, dimension lines, extension lines. (Art. 27.) (f) Arrows, figures, notes, freehand pen work, title. (Art. 27.) (g) Section lining border lines, trimming, cleaning. (Art. 27.) Ink in only as many views as can be completed on the same day. 30. Checking. When a drawing is completed, it should be checked, either by the draughtsman himself, or preferably by some one especially appointed for this work. If it is necessary for the draughtsman to check his own drawing a few points especially should be borne in mind. They are as follows: 1. Views. Be sure that the views completely represent the object to be made and that they are properly arranged with regard to each other. 2. Notes. See that these give information not supplied by the drawing of the object such as: nature of material, pieces wanted, finish of surfaces, kinds of holes, etc. 3. Dimensions. In checking, dimensions may be scaled to check accuracy of drawing, but should be verified by compu- tation. In checking over-all dimensions, add together the subdivisions which compose them. Dimensions of parts which go together, such as a shaft and pulley, should be compared. Be sure that all dimensions which each department of the shop may need are on the drawing. A workman is never allowed to scale or calculate a missing dimension. See that the same dimension is not repeated on two or more views of the same object. 4. Finisu. See that proper finish marks are clearly indicated, and the character of finish properly and fully stated. 5. StanparRps. Be sure that the sizes and forms of bolts, screws, and other standardized parts, conform to the accepted standards of the same. 6. TituE. See that the title is complete, yet concise, and contains all the necessary details that should be recorded under this head. 7. After corrections have been made the initials .of the checker should be placed in the title space. CHAPTER II FASTENINGS—RIVETS 31. FasTENINGS are necessary: for holding in place the compo- nent parts of a machine. It may be that none of these parts will ever require moving so that fastenings of a permanent character can be used. The most common permanent fastenings are rivets. They are used to fasten together the plates of a boiler, or a tank, or the iron beams of a bridge, or of a building, or of a roof truss. They are of wrought iron or steel. A rivet is made by upsetting the end of a cylindrical iron or steel bar to form a head and then cutting off the bar at some distance from the head. The rivet is then heated, pushed through the holes in the plates to be connected and another head or point formed on the opposite end. During the process of forming the new head the rivet is squeezed into the hole in the plates and completely fills it. The new head alse presses the plates tightly against the first formed head. The new head is formed by a riveting machine or by hand hammers. The four common forms of heads are shown below in Fig. 17. / 3 13 D—+ D ie at AT y ke re yt || ee [RAS =F S196 = | 4 Button Steeple countersunk, fons Head Head Head Fic. 17.—Rivet Heads. 32. The length of a rivet is measured from the under side of the head to the end of the shank, before riveting. This length 45 46 ELEMENTARY MACHINE DRAWING AND DESIGN is equal to the combined thickness of all the plates to be con- nected (called the grip) plus enough more to provide the material necessary for the new head and for filling the rivet hole, which is made #,”’ larger than the rivet. In boiler and tank riveting all four kinds of heads are used, but in riveting structural ironwork the button and counter- sunk heads are the only ones used. The conventional represen- a 4S = | T0> Conv. Button Countersunk Pan Head Q 8 ~N Fig. 18.—Conventional Rivet Heads. tation of these two is shown above in Fig. 18. A pan head is also shown. Fig. 19 is a scale for obtaining the dimensions of rivet heads. _ mm 15 re 1813 13 16 LL 1 4 Mu Ts 16 eS HJ] 1 16 12 Z Ts 16 _—— 18. ——— Rivet Diam. 3 7 ; 5 i 5 73 715 Als ; Multiples of d 5 FiRd 14146 1.6 1g iz 2d Fic. 19.—Scale for Rivet Head Dimensions. 33. The plates which are joined by rivets may overlap or touch at the edges. In the first case the joint is called a lap joint, in the second case a butt joint. Lap joints have one or two rows of rivets on lines parallel to the edges of the plates. The lines are called gauge lines. FASTENINGS—RIVETS 47 The distance apart of the rivets on one of the lines is called the pitch, denoted by P in the following figures. A joint with one row of rivets is called a single riveted joint, with two rows a double riveted joint, and with three rows a triple riveted joint. If two of the rivets in parallel rows in a double or triple riveted joint are on a line perpendicular to the line of the joint, the joint is a chain riveted joint. If the rivets in one row are placed opposite the spaces in another row they are staggered. Butt joints need one or two cover plates to connect the main plates. In boiler work two cover plates are used, the inner one sometimes being wider than the outer one. Riveted joints may be classed as follows: Lap { Single D ae Riveting. Arrangement of rivets {Chain \ Staggered. Single Arrange- One cover pl. | er Chain Butt | Double } Riveting. ment Two cover pls. Triple of saveta { Staggered. 34. In the examples, shown in Figs. 21, 22, 23, for drawing purposes, the following letters and formulae will be used for boiler joints only. T =thickness of plates to be joined. T, =thickness of cover plate when one is used =13T. T>2=thickness of cover plate when two are used =3T. d=diam. of rivet hle=KVT=VK°T. D=diam. of rivet (usually 75” less than d). K=variable depending on kind of joint=1.5 for single and double lap joints=1.3 for butt joint with two cover plates. P=pitch of rivets measured along the gauge line. P,=the diagonal distance between rivets on different gauge d lines=0.6(P —d)+d or 3P +3 (Kennedy). G=grip of rivet. E= P-—d= Fig. 20 is a diagram for obtaining the diameter of rivet hole when the thickness of plate (¢) is known and the value of (K) is either 1.5 or 1.3, depending on kind of joint. 48 ELEMENTARY MACHINE DRAWING AND DESIGN --Use this circle when K=L5 -- Use this ci when Ke I. 7 \ bee Gia UL y Wy Nn RQ x Q ¥ \ SN }<—— / op ——>} Lap Fic. 22.—Staggered Double Riveted Lap Joint. Fia. 21.—Single Riveted Lap Joint. FASTENINGS—RIVETS 49 VaLuEs or (P—d)"” ror Dirrerent THICKNESSES OF PLATE aND VARIOUS Pressures (b) T, ins. b=120 b=150 b =225 4 2.53 | 2.39 | 2.16 ts 2.99 | 2.83 | 2.55 3 3.43 | 3.24 | 2.93 a 3.85 | 3.63 | 3.29 2 4.25 | 4.02 | 3.63 WW | ANSs SE NN NAY Double Riveted ButtJoint Fig. 23." 35. In riveted structures it becomes necessary to connect plates in other ways differing from those shown above. The plates may be at right angles or parallel or oblique to each other. In these cases rolled bars of steel are used of various forms, those most commonly used being shown in Fig. 24, and 50 ELEMENTARY MACHINE DRAWING AND DESIGN called angles, Z bars, channels, and I beams. Columns, beams, and girders are formed by riveting these bars together with plates in different combinations. These are known as built sections, the simplest type being shown below, composed of a plate and two angles. “ Angles” are made with equal or unequal legs and of varying thicknesses. An angle is designated by giving the distance F 1S 7-225 CLLR ae Oy ANA © ) cane N LU | RIE WY River Angle Iron Aare Built Section CZ f \ yt y i t EE?) JAt Ww [ Web” * “Ye Flange Cc» +10 Dips : ke Web Ws me F ; Channel Fie. 24.—Rolled Structural Shapes. and F, and the thickness t as 3’ 21""*3,". In Z bars the depth D is given, also F and t in the following order, 6” X3” x3”. I beams and channels require D to be given and the weight per lineal foot as 10”, 833%, I beam. The other dimensions of these rolled bars are usually found in tables furnished by the mills which roll them. An angle may used to connect two plates, placed at right angles, FASTENINGS—RIVETS 51 by means of rivets passing through each leg of the angle and the adjacent plate. The lines along which the rivets are placed on a rolled beam are called gauge lines. The distance of the gauge line is given from the back of the angle, or channel or Z bar and is a standard for each length of leg or width of flange. On I beams the distance between the two gauge lines is given. Fig. 25 shows two plates con- nected by an angle iron which is riveted to both of them. The pitch P is a vari- able quantity depending on the diameter of rivet and the length (F) of the angle leg. The minimum C— j Y} me Ce -"rivets,- "holes F he ISX Fic. 25. value of L is given below for various values of C. The minimum value of P is about three times the diameter of rivet. G. 1h iB ib | a ie |e 1 Se | | 1 | 3” rivet | 14 | 135 | 13 | lds 15 z 2 7 3 0 L privet} 1g} as} iuplim fi }1 fe} el Rl a | Fig. 26 shows an I beam fastened to two parallel plates. The rivets are staggered to avoid cutting the flange by two holes in the same vertical plane perpendicular to the web of the beam. A channel or Z bar is often used in place of an I beam for con- necting two parallel plates. The drawing exercise may be varied by making either sub- stitution in Fig. 26. 52 ELEMENTARY MACHINE DRAWING AND DESIGN Assume values of P in Figs. 25, 26. Dimensions of structural shapes will be found in Table 4. Page 12. { —tL WD FX J WS ———— Ute i Sc 54 c -"~* [T—>-, ~"rwet's,-"holes ’ rH >» 7 a Fic, 26.—I-Beam, INSTRUCTIONS Plate 1, No. 2 paper, 2 hours allowed in class. Av. time necessary 32 hours. Fig. 1. Two plates whose thickness is T=( )’ are to be joined by means of a double riveted lap joint. The pressure they are to stand isb=( ) Ibs. Calculate the size of rivet to use the pitch of the rivets. and make two views (section and top view) of the joint, as in Fig. 22, FASTENINGS—RIVETS 53 Use the first formula for P;. Draw full size. Use a (a) head on one end of rivet and (c) head on the other. Fig. 2. Make a section and top or side view of two plates 7” thick meeting at right angles and connected by a F’ XF’’ xt” angle iron as shown in Fig. 25. Button head rivets. Fig. 3. Make two views (one a cross-section) of a ( ”) ( ). with a plate 3” thick riveted on each flange by button head rivets. Draw half size if maximum dimension of section exceeds 5’. Put all dimensions on the three figures above. Name of plate to be Riveted Joints and Rolled Sections. Bring to class fully dimensioned sketches of the above figures and draw from them. Text books not to be used in class. Reduce all decimals to nearest common fraction. (See Art. 10.) TaBLE OF ASSIGNMENTS FOR RIVETING 1 2 3 4 5 6 7 8 9 10 TT) ¢ ts g is a 7 te | 3 vs 3 Fic. 1 b | 120 | 120 | 120} 120 | 120 | 150 | 150 | 150 | 150 | 150 B. a| Pan | Pan | Pan | Cone |Cone |Cone |Cone |Cone| Csk. | Csk. c |Steep.| St. | But. | Cone | Csk.| But.| But.) St. | But. | St. T\) 3 3 3 $ $ 3 3 | & | we | ve Fig. 2) F 3 22 24 2 24 3 3 3 24 4 t t te | ve a |] 3 | fw | 4 4 3 Fig.3 D| 5 6 6 5 3 4 5 7 i 8 Kind I |Chan.| I |Chan.| Z Z Z I |Chan.} I 11 12 13 14 15 16 17 18 19 rl) 4 vs a ve 3 vs 3 2 ve Fie. 1 b | 225 225 225 225 225 225 225 150 225 B- |e But. | Cone | Pan | But. | Cone | Csk. | Csk. | Csk. | Cone c | Cone | Pan | Cone | Csk. | But. | Cone] But. | Pan | St. Tl) vw | t+ | 2] 4 |e iw | w | we | & Fig.2) F | 3% 3 2 23 22 24 3 5 6 bla lel Ee | a] Bae ae Fig.3 D/| 6 9 9 10 8 6 5 5 3 Kind Z I |Chan. I |Chan.| Z I |Chan.| Z Give on the drawing the name of each joint or structural shape, dimensions of each kind of rivet head, size of each angle iron, Z bar, I beam or channel and its weight, as 3” <3” x?” L 10% or 3’-I-10#. CHAPTER III PIPING AND PIPE FITTINGS 36. THE pipes in common use to convey water, steam, gas, etc., are made of wrought iron and are known on the market by their nominal diameter. This nominal diameter is nearer the actual internal than the actual external, and in the larger sizes is the same as the actual internal. In small sizes the nominal inside diameter is much smaller than the actual inside, due to the fact that the early manufacturers of pipe made the small pipes too thick. Later, when they desired to change the thick- ness, it was considered better to increase the inside than to change the outside, for the latter would have necessitated changing all pipe fittings then on the market. The inside was increased, but. the size of the original inside was still used as the name of the pipe. Table 9 gives the nominal inside, actual inside, actual outside, thickness of metal, threads per inch, etc. The nominal diameter of a pipe is always used in speaking of the pipe and is printed on a drawing where necessary to give the size of pipe. It should never be used as a dimension for the inside or outside of a pipe. The representation of a pipe on a drawing always necessitates the drawing of two parallel lines a distance apart equal to the actual outside diameter of the pipe. This outside diameter is found in Table 9, but the dimension found there is never given on the drawing. If it is necessary to show the inside of the pipe the thickness may also be taken from the table, but must not be given as a dimen- sion on the drawing. In all calculations involving the carrying capacity of pipes the actual inside diameter is always used, or the area of the circle whose diameter is the inside diameter of the pipe. 37. Since straight sections of pipe are only made from 16 to 20 feet long it is necessary to couple them together by short 54 PIPING AND PIPE FITTINGS 55 tubes threaded on the inside and called couplings. The dimensions of these couplings will be found in Table 8. They screw on the pipe ends according to the distances given in Table 9, Column 9. If a coupling has right and left hand threads in it so that it will draw together the pipes which it couples, it is distinguished from a plain coupling by having ribs on the outside running lengthwise. See Fig. 27. These ribs are four in number up to and including 1” if pipe and six in number above that. They Pipe are 3" wide measured on the circumference and extend 7%” above the outside of the - coupling. : | Pipes are also joined by screwing them R&L. into cast iron flanges and then bolting the sn NENG flanges together. Valves are often placedin 4 R&L.Coupling a pipe line by bolting to cast iron flanges Fic. 27. which are screwed on the ends of the pipes. Pipes or valves so connected can be disconnected readily by taking out the bolts which pass through the flanges. Fig. 28 shows a cast iron flange and below the table of dimensions adopted by the American Society of Mechanical Engineers, the Master Steam Fitters us - Association and the manufacturers of ee iG, fittings. The diameter of bolts given are CastironPipeFlange “ for pressures under 80 lbs. Above 80 lbs. Fig. 28. use bolt diameters 4" greater. The width of flange between the bolt centers and hub may be taken equal to the width of flange outside the bolt centers, or just enough to allow the nut to clear the fillet by #5”. TaBLe For Cast Iron Pipe FLANGES For use with Fig. 28 ips, | Bae 8 | ae |e ae | Se Le |e | 8 L 6 |7 | 73/8! |9 | 94 | 10 | 41 | 123 | 13% M 42 |53 | 6 |7 | 72 | 72 | 83 4) 102 | 112 8 3] #] | te) re] re} re | 10 | lve] ls N ie | | lel aee #) fl €) 8 No.bolts} 4 |4 | 4 |4 |4 /|8 | 8 | 8] 8 | 8 F 1 {1s | 1¢ ]13 | 18 | aa | ue | we] uw | a 56 ELEMENTARY MACHINE DRAWING AND DESIGN When pipes or cylinders are made of cast iron the flanges are cast on the cylinders themselves. The diameter, thickness, bolt circle diameter, number and diameter of bolts, etc., are given in Table 11. 38. The thread used on pipe and in pipe fittings is shown and described in the chapter on screw threads. Short pieces of pipe threaded on both ends are called nipples. If one of the threaded ends is cut with a left hand thread the nipple is called an R and L nipple. See Fig. 27. When a nipple is so short that the threaded portions meet at the middle, it is called a close nipple. If there is a short unthreaded portion in the middle the nipple is a short nipple. Lengths of nipples will be found in Table 8. The amount of thread on a nipple which enters a fitting will be found in column 9, Table 9. 39. By pipe fittings we mean such fittings as are necessary to make connections between two or more pipes of the same size or of different sizes which may meet end to end or at various angles. They are made of wrought iron, malleable iron, cast iron, or brass. Couplings only are made of wrought iron. Malleable and cast iron fittings are used on steam, gas and water pipe for medium and heavy pressures, up to 250 lbs. pressure. Among the fittings in use may be mentioned caps, plugs, locknuts, bushings, reducers, elbows, crosses, tees, Y-branches, unions, bends. If two pipes of different size enter a fitting the dimensions of the fitting are usually determined from the larger pipe. A cap is a fitting used to close the end of a pipe which has had a connection removed. It is drawn from the outside dimen- sions given in Table 8. It is usually of wrought iron or malleable iron. A plug is used to close an opening in a fitting from which a pipe has been removed or into which a pipe will be screwed at some future time. It has a square head and a threaded body whose outside diameter equals the outside diameter of a wrought iron pipe of the same nominal size. It is tapered like a pipe end. The dimensions of the square head and length of that portion beyond the head (all of which is threaded) will be found in Table 8. A plug is shown in Fig. 31 as it appears when screwed into a fitting. PIPING AND PIPE FITTINGS 57 A locknut is used to make a tight joint between a pipe and a fitting Whenever it has been necessary to cut perfect threads on the pipe ends for more than the standard distance. These threads are cut in order to let it enter the fitting far enough to permit the other end of the pipe to be introduced into a fixed fitting. The locknut simply presses against the face of the fitting and draws the threads on the pipe against those in the fitting, which prevents the liquid or. gas within from escaping between the threaded surfaces. The dimensions of locknuts for pipes are given in Table 8. Locknuts are hexagonal in shape and the distance across flats is the distance between the parallel sides of the hexagon. A groove is made in one of the sides of the locknut and filled with wicking so that when this side is pressed against the face of the fitting, all leaking at the joint will be stopped. A locknut in position is shown in Fig. 31. A bushing is used when a pipe of any size must be screwed into a fitting which fits a larger size pipe. The threaded outside of the bushing fits the threaded hole in the fitting. The hole if the bushing corresponds to the size on pipe which is to be screwed into it. The bushing has a hexagonal head whose distance across flats and thickness will be found in Table 8. The size of a bushing depends on the diameter of two pipes but the principal dimensions ae . depend on the size of the larger pipe. z™ x*y" Bushing A 1” by 3” bushing for instance will Fia. 29. be proportioned in external dimensions from the 1” size but the hole in it will take a 3” pipe. Fig. 29 shows a bushing. The dimensions A, B, and C can be taken from Table 8 for the x dimension. 40. Cast Iron Fittings are used on steam and water pipe connections to enable them to be removed in case of rusting on, by breaking them with a heavy hammer. They can be replaced by new ones at small expense. The standard pattern is used for pressures up to 150 lbs. per square inch, the extra heavy from 150 to 250 lbs. The most common of these fittings are the elbows (90° and 45°), called ells, tees, crosses, Y branches or bends. As it is often necessary for a draughtsman to represent pipe 58 ELEMENTARY MACHINE DRAWING AND DESIGN lines with their fittings and yet not dimension them, nor indeed know their dimensions, the following scale was devised by Prof. Charles B. Richards and the writer. See Fig. 30. Scale for Cast Iron Fittings for 7" La ee Eee Wrought Iron Pipe : Lb 4 = bx | 2 = ax Designed by Prof.C.B.Richards& 6 ta B=-be |B=be Prof. we Marshall 5" EA eae C=be |C=ae eS, ewe [A é Z-o7 | 2749 Cee E-gx | £=9x ee” F =bf | F=aF ea 6 =hb| Gha oat z ee z a4 is # -be | # -ac 452 Outside diam. of pipe_ | beams. eae ie me r vs ae x FATE eel stat ara os" gts 7? at The commercial Nomina! Diameter of Pipe dimensions of fit- tings of the Wal- 4 to ap worth Mfg. Co., AgCk = oC Z 4 which areshown be- uy = S | +Bre OS | : | 1 | low the scale, were rr a a is used as a basis. The re a Pra = scale can be laid | mie lB-cl SS] Sif out full size from ene rey “teed | the table on the Standard Ell & Tee WE G, next page, the di- a Tw © Reducing ee-Releay- mensions given : being laid off above Me the x line on the pipe diameter or- o aD e daeled dinate indicated at Standard Y Branch the top of each Fic. 30. column. After con- structing and let- tering the scale, the table or “key” at the right is then put in as well as the printing below it, which enables the outside diameter of any pipe to be found from the scale directly without atable. The fittings shown at the bottom of Fig. 30 can then be drawn from the scale and key. As the b, e, g lines are used PIPING AND PIPE FITTINGS 59 most often for standard fittings they are made blacker than the others to attract attention. TaBLe For LayinG out SCALE ee au at 41” 13” AR 5” 8” Pipe ax B # 1.0 | 13 143 33 3k 58 ins. bx z 33 e | li 1g 335 33 bi . ex 3 i6 te) lve | ls | 238 | 38% | 43 7 dz tz defee | aeeie econ wee) seeneae |} eee 455 ce ex ao $ ve te i 2Ps 233 Avs . fx we i ve ie a 2 233 3i8 . qa ee s 3 oy oe 5 s ve “ ha ve de ve ve v6 te 1.0 13h os A fitting is generally known by the pipe which fits it but it is often necessary to consider more than one size of pipe. An ell may have one size opening in one side and another size in the other, in which case two ordinates must be used for A, B,C, D, and FE. A tee may have all openings of one size or only two alike or all different. In describing tees, the “1un” is named first then the “ outlet.” If there are two sizes on the Ber Wye “ren” the larger is given first, as in the one shown here | which would be called a 1’ 3X3” tee. If the two opposite open- ings are alike and larger than the outlet the tee is called a reducing tee, the run being given first. If the outlet is larger than the run the tee is called a “‘ bull head ”’ tee. In the reducing tee shown in Fig. 30 the letters with the subscript r are laid off from one ordinate of the scale and those with the subscript o from another ordinate. The tapped holes in the fittings have the same diameter as the outside of the pipes which fit them. The 45° ell and the Y branch use the same letters for dimen- sions not shown as those given for the tees and 90° ell. A cross has four openings whose center lines are 90° apart and the pipes which fit them may be of one, two or three sizes. The outlets are always alike while the ‘‘1uns” may be alike or different. Fig. 31 shows a cross with all openings the same and called a 2” cross. If the openings are as indicated here the outlets alike and runs different, it will be called a 13’ X1X* 2” 60 ELEMENTARY MACHINE DRAWING AND DESIGN cross. The lettered dimensions in Fig. 30 may be used for the cross dimensions. The cross in Fig. 31 has a pipe and locknut at one outlet and a plug at the other. Both runs are open. 41. A union is used for connecting the sections of a pipe line when the line has to be taken down occasionally or to make the line easier to assemble as well as to dispense with locknuts. It is made of malleable iron or brass. A union is composed of three pieces. Two of these, (D) and (C), are screwed firmly on the ends of the pipes to be con- nected and the third piece called the “nut” draws them tightly together by means of the threads on (D) and the flange X-thrds per in. (D) (C) ‘Hex —t ee ' K | | ait, 3 au" 4 Plugs’ > sgsate Fia. 31. Fig. 32. Locknut t on (C). Fig. 32 shows a union holding together two pipes (A) and (B). The view shows a half section above the center line. The black substance between (C’) and (D) is a rubber gasket which keeps the joint from leaking. Sometimes this is omitted and the joint ground to a spherical form. The dimensions of four sizes of unions will be found in the table following. It is recommended to omit the pipes from the drawing which can be either assembled or detailed. Flange Unions are used to join pipes by screwing them on the pipe ends and bolting together. The Table for dimensions of such unions is given on page 61, and the drawing of the union on page 62. (See Fig. 33.) PIPING AND PIPE FITTINGS TaBLE FoR DIMENSIONS oF UNIONS rile ue a oe coal teen ay & oy et cal ane o> lt enon onl o é ec N xs ° a _ a o rilo0 1} ce wl is (s fale eo mln cols a wD oe Sl N oO mo of | 0 colt oa Pt colt rao reo oO H o mo nN 0 N coal ma min 28 loo po anjeo ~ N ha ralea elo | ay = oO NANNY cols walso oOo ©: ho colt ewes oO eo a a ae rile |O cao aie ae ae ries fo alt sn|00 wd ao awd rin colt [0 cols mile anloo ug olf +g t}o0 HH oO a ~ ad Yen) mo 8 ai x onl a let iB wn) foo os rile ae a ade a mln mit colt loo Anjo et ve) aa a 100 eo] =H rl reo 6 2/00 oo| <4 riled Paleo lt anjoo mit enlao 1n|o0 re 5 Co re HH ei a wiles cole olf ol ils nao 8 lea Hs ofS z mt bo H cals ait ah Boo Im nina N sH roo HAS te iS a niles snlao uno lt creo on N sH QL ee N A 3 loo joo of oS = = [| 5 12 tm 1 SSIS g 8 S5gk, 8 °. o GN a ay. ae aOR, * Boi Sl/phe ha sol un Pie Oo) sgakepegaegee sar cals alt ralea o Rasa Peas koo gs Sen S| Seat Gy ee sO N ga ZIlIA FF WA A ZA BZ 62 ELEMENTARY MACHINE DRAWING AND DESIGN 42. Pipes are used for conveying water, steam, gas, or air. The volume delivered depends on the pressure, the size of the pipe, the length of pipe as well as the number of elbows, tees, and valves in the line. Le The frictional resistance of the pipe and fittings y amounts to a reduction in pressure at the outlet of the pipe which may reduce the flow greatly. The velocity of flow may be calculated ap- - proximately for water Tk pipe discharge by the following formule. Flanged Union Fig. 33. hd SN T+ 54d v=mean velocity in feet per second; m= coefficient from table below; d=diameter of inside of pipe in feet; ° h=total head in feet; L=total length of line in feet. If the pressure is given in pounds per square inch instead of the head in feet, divide d by 0.434, which is the pressure per square inch exerted by a column of water one foot high, and the value obtained will be A required. VALUES OF COEFFICIENT m a Diam. of'Pipe in Feet. L+54d .05 .O1 .05 1 1.5 2 3 4 .005 29 31 33 Bt) 37 40 44 47 .O1 34 35 37 39 42 45 49 53 .02 39 40 42 45 49 52 56 59 .03 41 43 47 50 54 57 60 63 .05 44 47 52 54 56 60 64 67 .10 47 50 54 56 58 62 65 70 .20 48 51 55 58 60 64 67 70 PIPING AND PIPE FITTINGS 63 The friction of water in long pipes requires an increase in pressure at the entrance of the pipe to give the same discharge at the outlet which would be obtained in a short pipe with the given pressure. The increased pressure required to overcome this friction is given in Table 6. The friction loss in elbows is given in Table 7. TABLE 6 Friction oF WATER IN PIPES * Pressure in Pounds per Square Inch to be added for each 100 feet of Clean Iron Pipe Gallons Nominal Pipe Sizes. Minute Bee | ga a | i | ot | oe |-oe | om | oe | a el ee 5 | 3.3} .84) .31) .12) .04) .02 10 )13.0) 3.16! 1.05) .47/ .12) .04| .02 15 128.7) 6.98) 2.38} .97| .25) .08} .04) .02 20 |50.4)12.3 | 4.07) 1.66} .42) .14) .06) .03 25 |78.0/19.0 | 6.40} 2.62) .62) .21/ .10) .04) .02 30 .. /27.5 | 9.15) 3.75} .91) .380) .13) .06) .03 35 .. 187.0 112.4 | 5.05) 1.22 40; .17| .09) .05) .02; 40 .. (48.0 {16.1 | 6.52) 1.60 53} .23) 11) .06) .02) 45 .. |... [20.2 | 8.15} 1.99 66} .28) .14) .07) .03 50 .. |... 124.9 |10.0 | 2.44) (81) .35) .17) .09] .04 60 .. |... {86.0 |14.0 | 3.50) 1.17} .50) .24] .13) .05) .02 70 .. |... [48.0 |20.0 |} 4.80} 1.50} .60) .38) .19} .07) .03 75 .. | ... [56.1 ]22.4 | 5.32) 1.80} .74 80 fess ... 164.0 {25.0 | 6.30) 2.00) .90) .41] .23) .08) .03 90 ie ... |80.0 |32.0 | 7.80) 2.58) 1.10) .54) .26) .09) .04 100 .. | we. |... (89.0 | 9.46) 3.20) 1.31) .64) .33) .12) .05} .02 125 .. | wee Jou | ee. [14.9 | 4.89) 1.99) 96} .49] .17) .07) .03 150 .. | wee |oeee | ee. [21.2 | 7.00) 2.85) 1.35} .69} .25) .10) .04 175 are Nh Sia ... |... [28.1 | 9.46) 3.85] 1.82) .93]) .384) .13) .05 200 doer fe tees ... |... [87.5 |12.47| 5.02) 2.38/1.22) .42) .17) .07 250 es week ... |... |... [19.66) 7.76) 3.70)1.89] .65) .26) .12 300 oo dee | wee [owes | oe. [28.06)11.2 | 5.04/2.66) .93] .37) .17 350 Cod eee fee Joaee [wee fee. [15.2 | 7.10/3.65/1.26} .50) .23 400 “a Pree wo. | wee | wee | ee [19.5 | 9.25)/4.73)1.61) .65) .30 450 of eee dee Poaee [wee fee. [25.0 J11.70/6.01/2.00) .81) .37 500 co | cee | eee | eee Joeee | ee [80.8 [14.5 |7.43]2.40) .96) .45 750 og Weems ecb sepa | see | ae Bees) wee f weer awe es. a 2H E038 PAO Ue eke. lage Tee I Awe- ll oe tane akcer | eee Ice RL HORM Tcl cee Vacca cee Luv pee | -ooce | -whe 4 ces | ce (OOB8 WROD co Ncnce Neva bee | mae ll|'ane |) ag: ll ine He Aone tenon aoe Table is based on Ellis and Howland’s experiments. To find “Friction Head” in feet multiply figures by 2.3. 64 ELEMENTARY MACHINE DRAWING AND DESIGN TABLE 7 FRIcTION OF WATER IN ELBows Pressure in Pounds per Square Inch to be added for each Elbow Gallons Nominal Pipe Sizes, per Minute ‘red, | # | 2 | 1t | 14] 2 | 22 | 3 | se] 4 | 8 | 6 5 .07] .027| .008] .005} .002 10 .28] .094} .031] .018} .006! .003 15 .63 | .212) .069} .04 | .014] .005 20 |1.12} .376) .123] .069; .025} .012) .005 25 |1.74]| .585) .194) .108} .038) .02 | .008 30 gare .845) .278) .157) .055| .028) .011 35 1.15 | .380) .215) .076) .037; .015) .009 40 1.50 | .495) .278) .098} .049) .02 | .011] .007 45 -90 |. 626) .352) .125) .062) .026) .015) .009 50 med .77 | .43 | .153) .08 | .032) .017| .01 60 3.38 |1.11 | .62 | .22 | .112} .044) .026) .015} .006) .003 70 4.60 |1.52 | .86 | .304) .148} .06 | .035) .021} .009) .004 75 5.30 |1.74 | .98 | .85 | .172) .072} .04 | .024) .01 | .005 80 6.00 |1.98 |1.11 | .3892) .196} .08 | .044) .027} .012) .005 90 7.60 |2.50 |1.41 | .50 | .248) .104) .06 | .035} .014| .007 100 3.08 |1.72 | .612} .382 | .128) .068] .043) .017| .008 125 ... (2.72 | .97 | .48 | .20 | .112) .067) .027| .013 150 3.92 |1.39 | .685} .286} .16 | .096] .039] .019 175 5.32 |1.90 | .935] .390) .218) .132] .053]) .026 200 6.88 |2.44 |1.28 | .512| .272) .172] .068) .032 250 ... |3.86 |1.91 | .80 | .446} .268) .109} .052 300 5.56 |2.74 |1.14 | .64 | .384) .156) .076 350 3.77 |1.58 | .88 | .530) .215) .103 400 5.12 |2.05 |1.09 | .688| .272) .128 450 6.20 |2.58 |1.45 | .870| .352) .170 500 7.64 |3.20 |1.78 |1.07 | .436} .208 750 oes ... |2.42 | .970| .470 1000 4.28 |1.74 | .832 1250 6.70 |2.71 |1.31 1500 9.68 |3.88 |1.88 Table is based on Weisbach’s formula for very short bends, or with a radius equal to the radius of the pipe. multiply figures by 2.3. To find “Friction Head” in feet 43. The flow of steam in pipes can be computed from the following formula: 7 | P,— Pod) = 56.68 (Pi— Pod?) Q Lw ’ PIPING AND PIPE FITTINGS 65 where the symbols are as given below: Q=quantity in cubic feet per minute; d=pipe diameter in inches; L=length of pipe in feet; P\=pressure of steam per square inch at entrance of pipe; P2=pressure of steam per square inch at exit of pipe; w= weight per cubic foot of steam at pressure Pi. 20h d=0.199 5f Qo Pi—P2 The resistance of a globe valve is about the same as that _ 144Xdiam. of pipe ~ 1+(3.6+diameter)’ The resistance of a right angled elbow or a tee is ? that of a globe valve. The resistance to entrance of a pipe is the same as for a globe valve. in an additional length of pipe in feet 44. Oil cups are usually made with pipe threads of standard dimensions to fit pipe tapped holes, but the inside diameters of these threaded ends do not agree with the inside dimensions of pipes having the same outside diameter. The inside is smaller than that of the pipe and nearly agrees with the nominal inside of the pipe. That is, a 3’ pipe thread on an oil cup will have the actual outside diameter of a quarter inch pipe but the inside will be 1’’ diameter instead of .364’’. Oil cup dimensions will be found in Table 10, page 16. INSTRUCTIONS Plate 1. No. 1 paper, 2 hours allowed. Average total hours required, 2.9. Draw two views full size of each of the following pipe fittings. An (h)” locknut, an (f)’’ R and L coupling, an (7) X(7)” bushing, a cast iron pipe flange for an (m)"’ pipe to carry 100 lbs. pressure, a (g)’’ cap, an (e)” plug, and a 1” R and L short nipple. Put on all the allowable dimensions and name the plate Pipe Fittings. Plate 2. Construct outside the class the scale for pipe fittings from the table in Art. 40. Use a sheet of drawing paper size No. 2, placing it on the board with the 11’ side parallel to the T square blade. Draw the scale for fittings at the top of it with the “ key” above and to the ® 66 ELEMENTARY MACHINE DRAWING AND DESIGN left of the scale. Omit the table used in laying out the scale. Below the scale draw in class the following fittings, putting the letters on the drawings as they are given in Fig. 30. Astandard Elland Tee (combined in one drawing) for an (a”) pipe. A (b’’) reducing Tee (extra heavy). A (c’’) standard 45° Ell and a (d”) Y branch. Name under each fitting. The scale is to be drawn outside of class and the fittings in class during one period of two hours. Average total hours required, 2.8. Name the plate “Scale for C. I. Pipe Fittings.” Plate 3. No.1 paper, 2 hours allowed. Average total hours required 3.5. Draw in detail each part of a (7) Union. Two views of each will be required, one half of the side view in section. Give all the dimensions of each part and name the plate, “Details of a —’’ Union.” Make a bill of material on the drawing. TABLE OF Sizes For Pipe Firrinc Puates 1) Be We a ee) ee ae et Pitele] 2 4.1 ie | ae Poe) ala lm | te ae 1 1 fj 1 t/ 2/4] 1 yj} | 3 cane i |l ge) eta a Pee ae) oe) ae) ah) a |g hl up | eja | ¢] ub] a]t ja | 4] e | we] 4 | & 14 | ae) ge a le) Se ae ye i) b| 4] 3] #1 el a] ala al] a] ata ml 2 13 |3h| 4 | 4k 16 |e |e |e | Bt ba tay Platela| (4a 18! So 1) ee ea 214. i) ae. ae 2 |b [lax 31x Hex 3/1 x SLE XSL X Bax HX SLE SEX TLE xa] Xd e|] #/ tl) 1 1m [ay eli a) et] a ae d| we | 44) 4 ee | Se hae aed ae) gee | Pee Sey Al) a ee ee ae a ae ae Prob. 1. Calculate the velocity of flow of water at the outlet of a line of 2” pipe 200 feet long, which has a head of 150 ft., at the entrance. There are 4 elbows in the pipe line. Also calculate the volume in gallons delivered at the outlet. Prob. 2. An engine uses 1000 cu.ft. of steam per hour. The length of pipe between boiler and engine is 100 feet with two elbows and two globe valves. The boiler pressure is 100 lbs. and pressure in engine cylinder 90 lbs. w=0.225. What diam. of pipe will be necessary and at what velocity will the steam enter the steam chest of the engine? CHAPTER IV SCREW THREADS AND HELICAL SPRINGS 45. Screw threads are composed of helicoidal surfaces formed by a tool, which cuts into a rotating cylinder and is made to move in a line parallel to the axis of the cylinder while cutting. Each point on the tool traces a helix whose construction is shown in Fig. 34. &e | Leen, (B) | ] Sey, | i 4, E, CG & < 7D Tan.a@ = Fic, 34. If the surface of the cylinder is laid out flat, the helix appears on this development as a straight line. The angle « is the angle which the helix makes with a plane perpendicular to the axis of the cylinder. The distance H is the pitch of the helix and D is the diameter of the cylinder. The shape of thread produced depends on the form of tool used. The forms of tools are such as to produce threads known by the following trade names, viz.: Vee, Sellers or U. S. Standard, Modified Square or Acme, Buttress, Whitworth, Pipe, Square, International or Metric, etc. 67 68 ELEMENTARY MACHINE DRAWING AND DESIGN These threads may be divided into two classes, viz.: those used for fastenings and those used for transmitting power. Under the first head are grouped the V thread, U. S. Standard, Whitworth, International, pipe, and others whose form is of a V shape. Under the second, or power transmission head, are found the square, buttress, modified square or Acme. x —¥ 60 zs V Thread §% 8 2 88 Fia. 35. Fic. 36.—Helices on V Thread. 46. In order to correctly represent a thread, we must know its pitch, the shape of the tooth or thread, as well as the outside diameter of the cylinder on which it is formed. The helices must be constructed as shown in Fig. 34. A V thread has a profile as shown in Fig. 35. The construction of a V thread oer Ste Fig. 37. tit <> 10 pi. Fic. 38.—Conventional Thread. Fic. 39.—U. S. St. Thread. is shown in Fig. 36, and the completed representation in Fig. 51 (A). In order to lessen the labor of drawing the helices they can usually be represented by straight lines as in Fig. 37. A further simplification omits the V profile of the thread and SCREW THREADS AND HELICAL SPRINGS 69 shows merely the lines joining the roots and the lines joining the points or crests. The root lines are quite heavy and the crest lines light as shown in Fig. 38, which is the conventional repre- sentation of V and U. S. St. screw threads. The pitch of a V thread is =P=x, where N=No. of threads per inch. The depth of thread=PX.866. Diam. at root=D—1.73P. The V thread has the disadvantage of being easily damaged, which led to the adoption of the U. S. Standard thread, which has the tops and bottoms of the threads flattened to 4 the pitch. This thread profile is shown in Fig. 39 together with the outside view of a threaded cylinder, the helices being replaced by straight lines. The conventional representation is shown in Fig. 38 the same as a V thread. When the helices are actually constructed the thread appears as shown in Fig. 51 (B), but this is rarely done. The pitch (P) of these threads is determined from the formula P=0.24V/D+0.625—0.175= 5. 0.65 The depth of thread = A =0.65P =. N=the number of threads per inch and will be found in column 16, Table 1. The diameter at root=D—24 =D—+ which is given in column B, Table 1. The diam. at root and depth of thread are used on bolt calculations involving tensile strength and not in drawing ; P the profile of thread. For drawing purposes P, D, and "g are the only dimensions needed. 47. The Whitworth thread is the standard thread of Great Britain, its profile being similar to the U. S. Standard ‘-P «Fitch >| “ie The angle between the sides is , 55° instead of 60°, the tops and Lp bottoms being rounded off § the SE depth instead of being flattened Whitworth | off 3 the depth. The proportions Fra. 40. are shown in Fig. 40. The pitch P=5, where N=number of threads per inch=0.08D+0.04 70 ELEMENTARY MACHINE DRAWING AND DESIGN 9D +67 8D+2 thread. The depth of tooth is & where S=depth of a V thread having the same pitch. The diameter at root= D.=D-*, where D=the outside diameter of thread. The area of the circle D; diameter is the area used for calculating the tensile strength of the cylinder whose outside diameter is D. 48. The International Standard thread is the system recom- mended by the congress held in Zurich on October 24, 1898. The thread is a modification of the metric system in use in France, the bottom being rounded instead of flat. See Fig. 41. The ao nearly, or N= nearly. D=the outside diameter of / f B YOY OR pee = P \ ‘= Px International ce, wo Standard Thread 7 (B.A) Thread. Fig. 41. Fie. 42. metric svstem thread is the same as the U. 8. Standard in Fig. 39. P=pitch, d=PX.6495. All dimensions for this system are given in millimeters. The pitch of the threads for different diameters of metric and international systems will be found in Table 3, page 10. 49. The British Association Thread (B. A.) Standard was adopted in Great Britain for use on such work as small precision machinery and electrical fittings for small screws below 1” diameter. The size is denoted by a number. The angle between the sides of the thread is 47.5° and the depth of thread is .6 the pitch. The tops and bottoms are rounded off. Fig. 42 shows the profile of this thread and the proportions used in making it. 50. British makers through their standards committee have recommended a thread of finer pitch than the Whitworth standard but of the same form. These threads are used for bolts subjected SCREW THREADS AND HELICAL SPRINGS 71 to excessive shock and vibration, as on crossheads, connecting rod ends, main bearings, piston rods, valve rods, etc. The diameter at root D,;=0.95D—0.07. When D=the outside diameter of thread; pitch=P=7,/D? for sizes up to 1”. From 1” to 6” P =v. D*. The number of threads per inch N is given in Table 3. These are called the British Standard Fine Threads. In the United States the Association of Licensed Automobile Man- ufacturers have adopted a finer pitch for automobile use than the U.S. Standard, although the thread has the same proportions as the U.S. St. The number of threads per inch will be found in Table 3. 51. The influence of the pitch of thread on the strength of a cylinder is quite marked. In some cases by doubling the number of threads per inch the strength can be increased 20% or even more. Fine threads are more apt to be injured in handling and erecting. Table 3 shows the relative strength of cylinders of given diameters having threads of varying pitch. Whenever tightness is desired rather than any other feature it is advisable to use finer threads than the number given in the U. S. Standard table, that is, approximately the number given for the British Standard fine thread. Pier THREAD 52. Pipe thread is used on the conical ends of wrought iron pipe and in the holes and fittings into which these pipes enter. The pitch is finer than the standard pitch for a cylinder of the same diameter as the outside of the pipe. This is for the purpose of insuring tightness of joints rather than for strength. The shape of thread as designed by Mr. Briggs is shown in Fig. 43. The outside diameter of the pipe is Do which is not the same as the nominal diameter of pipe. The taper of the end of the pipe is 1 in 32 to the axis. The angle of the thread is 60°. The pitch P is found from Table 9 which gives the number of threads per inch N. Instead of having a depth equal to .866P, as in a V thread, the threads are rounded off at top and bottom 1 so that the height H is reduced to 0.857 where N=threads per inch. The length of perfect thread extends from the end of 0.8D0+4.8 : the pipe back a distance T oa Do=external diameter 72 ELEMENTARY MACHINE DRAWING AND DESIGN of pipe. Then come two threads with perfect bottoms and flat tops followed by several imperfect threads with flat tops and flat bottoms, which are produced by the chamfered mouth of the threading die. The thickness S of the material below the root of the thread at the end of the pipe will be found in Table 9 or it can be calculated from the formula S=0.0175Do+.025”. Fie. 43.—Section of Thread for Pipe. The diameter at the end of the pipe is also given in Table 9, as well as the depth of thread. The bottom line of perfect thread is parallel to taper of end of pipe, then changes to slope up to the outside of pipe at a point 5P away, measured parallel to the axis of the pipe. 53. The threads used for transmitting power are either square or modifications of the square. The object is to reduce the tendency to burst the nut, by taking the pressure on a surface

“jp £1 TTT SquareThread Fie. 44. Fie. 45. Fie. 46. Sq. Thread. Conventional Sq. Thread. which is at right angles to the axis of the screw. A square thread is shown in Fig. 44. It has half as many threads per inch as a V or U. §. Standard of the same diameter and is consequently only half asstrong. Fig. 51 (C) shows the square thread accurately drawn, while Fig. 45 shows the helices replaced by straight lines. The conventional method of representing the square thread is shown in Fig. 46. The depth of space equals : and the width between the inclined lines is also - SCREW THREADS AND HELICAL SPRINGS 73 54. The perpendicular sides of the square thread make it difficult to cut and also render difficult the engaging and dis- engaging of a split nut which is often used in power transmission, as on the lead screw of a lathe. For these reasons it is desirable to slope the sides of the square thread at a small angle. The thread so formed is called the Modified Square or Acme Standard jefe Pitch ——» Rig at Fitch 3 P v \ 29° SIN kw + ¢ —sl ae | Acme Standard Thread Modified Square S Fig. 47. Fig. 48. thread. The angle of its sides is 29° and the width of flat at the top isf=.3707P. The depth is d=5 +01" Fig. 47 shows the shape of the thread profile as it actually is, while Fig. 48 shows the thread as it is usually represented on a drawing. The repre- sentation with helices drawn is shown at (D) in Fig. 51. When these are replaced by straight lines we have the representation as shown in Fig. 49, which is the conventional representation. Re ean tl F Fie. 49.—Conventional Acme or Buttress Thread Modified Sq. Thread. Fra. 50. After taking the outside diameter D the mean diameter D, is then s laid out equal to (>), and the teeth constructed on the lines drawn through the extremities of the D; diameter parallel to the axis. The slope of the sides can be 15° but the real angle 143° must be given on the drawing. The number of threads per inch is given in the following list. 1/14/2/33]4/5] 6 | 7 | 8 | 9 | 10] (Thds. per in.) 1| .75] .5| .666] .25] .2| .166| .142] .125] .1111] .10] Pitch of single thread. 74 ELEMENTARY MACHINE DRAWING AND DESIGN 55. The Buttress thread is used where pressure is exerted in one direction only as in screw presses or cannon breech blocks. The profile is shown in Fig. 50 and the dimensions are S=P. eS ~§ 8? side being perpendicular to the axis. a the angle of one side with the vertical is 45°, the other st £8 ve. (A) V Thread & 2— Given dram. * Ye . (B) U.S.Standard te & = & ee : he £ Giver NAM. _—— oe = + (C) Square Thread ¥ aN - Saye Modified Square Screw Threads Fig. 51. 56. So far the threads considered have been single threads, and the pitch has been the distance from one crest to the next. If we wish to give the nut in which the screw rotates, an axial movement twice as great, we must either increase the pitch by cutting a coarser thread, which means a deeper cut in the SCREW THREADS AND HELICAL ‘SPRINGS 75 cylinder, or we must cut two parallel threads on the cylinder. The cutting of a coarse thread will reduce the strength of the cylinder. If we cut two parallel threads the depth of thread and distance between the points or crests will remain the same as for a single thread, but the pitch will be the distance between every other crest. In Pie +P a single thread a crest on one side of the cylinder is opposite a root on the opposite side. In a double thread the crest of one thread is opposite the crest of the other. Fig. 52 shows the difference in appear- Fic. 52. ance of a single thread and double threads cut on the same cylinder. P in the second figure is double the P in the first. It is customary to indicate on a drawing the pitch wanted by giving the number of threads per inch as 20 p.t. The word double, triple, quadruple, etc., is used Pie also in case of multiple threads, as: 10 p.t. double. Threads are either right hand or left hand and are represented by right or left hand slopes as shown in Fig. 53; (a) is (a) (b) aright hand and (6) is left hand. The Fia. 53. conventional representation for V_ or Right and Left Threads. U.S. St. threads is like Fig. 38, instead of that shown in Fig. 53. 57. Calculations which concern the tensile strength of threaded cylinders are based on the area of the circle whose diameter is that the root of the thread, not the outside diameter. Table 3 gives the areas at the root of cylinders threaded with various pitched threads according to. different systems. The strength of U.S. St. threads for a strain of 1000 lbs. per square inch of area at the root will be found in column F, Table 1. For any other strain desired, multiply the tabular value by the required strain divided by 1000. As an example suppose it is required to find the load a bolt 1” diameter will sustain if the material is to be strained to 5000 lbs. per square inch. The table gives 551 lbs. for a strain of 1000 Ibs. per square inch. This will be multiplied by 2988, which gives 551X5=2755 lbs. 76 ELEMENTARY MACHINE DRAWING AND DESIGN HELIcAL SPRINGS 58. The representation of helical springs properly comes up in connection with screw threads only because they are both Mean Diam. —————>| Serpentine Fig. 54. based on the helix. The spring made of wire whose cross-section is a circle is delineated by constructing the helix whose pitch is the pitch of one coil and diameter equal to the mean diameter Helical Spring Fie. 55. of the coil of wire. A sphere with a diameter equal to the diameter of the wire is then drawn in a number of positions which center on the hclix. The lines tangent to these circles representing the spheres, will be the lines forming the apparent SCREW THREADS AND HELICAL SPRINGS 77 contour of the spring. The construction of a coil of wire by this method is shown in Fig. 54. This is too laborious for showing springs as usually required on drawings. An easier, but fully as representative method, will be found in Fig. 55. The helices of Fig. 54 have been replaced by straight lines drawn tangent to circles whose centers are on the cylinder with diameter Dm=the mean diameter. P is the pitch, D the outside diameter and D, the inside diameter of the coil. The diameter of wire used in making the spring is W. The view shows both section and outside. If the wire has a square cross-section the circles in Fig. 55 will be replaced by squares. 59. Besides the screw threads mentioned above, which are used for screwing metal into metal, there are threads used for screwing metal into wood. The wood é Length screw threads differ from the threads ies / aunt already shown in the shape and dis- MAA REA tance apart of the threads. They are Lag Screw thin, sharp and far apart, which Fig. 56. enables them to cut their way into the wood and give plenty of wood between for holding. The body of the screw is tapering and gimlet pointed to enable it to enter the wood easily. The form of thread is shown approxi- mately by the lag screw of Fig. 56. INSTRUCTIONS One Exercise of 2 hours allowed in class. Av. total hours required, 2.6. Scale full size. Paper No. 1. Fig. 1 to be a profile of a (a) thread, (2 crests and 1 space) whose pitch is (0). Dimension this figure and name beneath. Fig. 2 to be a profile of a (c) thread (2 crests etc.), pitch (d”). Other directions as above. Fig. 3 to be a section of thread for a (e’’) pipe (scale double size). Dimension completely with decimals, name and scale below it. Fig. 4 to be a profile of an (f) thread full size, (g) threads per inch, dimension and name. Fig. 5. Make an outside view of a cylinder (h”) diam. threaded with an (i) thread, (2 crests), (x) threads per inch, full size, dimension and name. 78 ELEMENTARY MACHINE DRAWING AND DESIGN Fig. 6. Draw a conventional V thread as cut on a (J’’) diam cyl. (m) hand. Fig. 7. Draw a helical spring of (n”) pitch, (3 coils) Dm=( "), W=( ”) half in section and half elevation, full size dimensioned. “Screw Threads and Spring” is the name of the plate. TaBLE oF Data FoR PLATE ON ScREW THREADS AND SPRING Fig. 1. Fig. 2, |Fi8) Fig. 4. Fig. 5. Fig. 6, Fg. 7. : 2s a 6” Cc dq’ el t g”’ het t k” l’ m” nl” Dag" Ww” 1/vli lu.s.s} 2] 2]8q./ 4 | 1 |Acmel 1] 2] Ret 1] 2 | 3 via & |i lo |e | al! lata tote al ak] 4 SB ¥) i) ae ee te aa a ae NE Ob] eg ayy ep | giggle la bel ee faa ela. |e) ele Biv) a] @ | Bla le pie) ol ba ala. PE] ela eo) Vig) * [tala «| iele) @ | el Time |e] 24 @| Vi 4g) © [aelal ia fag] * | ola] b. |] 2} 3 8) Viia| | ag ease fa 1 ag) | 2) ial Rm. |e] 81 9|/Whit. 1, Met. % | 3 {B.A} 1” | 1 |Inter} —}| 1 |Doub;, 1 Sq. R 10} «© 13] Met. | 1 | 2a] «| agra] “© |-| 1g{Doubl 1} 2 | 2 L CHAPTER V SCREWS AND BOLTS 60. Ir the parts of a machine have to be removed or adjusted it becomes necessary to fasten them in place by fastenings easily disconnected. The simplest and most common form of fastening of this kind is a cylinder threaded at one end and with a head of some kind formed on the other. These fastenings are called SCREWS or BoLTS. Under the first heading will be found Cap Screws, Machine Screws, Set Screws, Lag Screws, and Wood Screws; under Lent the second, Tap Bolts, Coupling Bolts (with nuts), U. S. St. Bolts and nuts, Myfg.’s Standard Bolts and Stud Bolts and nuts. There Filister Head Machine Screw are many shapes of heads used on id both screws and bolts. Those screws whose heads are cylindrical are provided with slots in order to turn them with a screw driver. These are shown in Fig. 57 and are called Cap Screws and Machine Screws. If the curve on the head of the filister head screw is omitted, the head becomes a round head. The button head is a hemisphere Flat Head capping a short cylinder of the Fra. 57.—Machine Screws. same diameter. The flat head screw becomes a French head when the top is curved as shown by the dotted curve. 61. If the diameter of these screws is less than half an inch they are called machine screws and the diameter is given by a screw gauge number. The dimensions of these heads and slots are as follows: Button Head 79 80 ELEMENTARY MACHINE DRAWING AND DESIGN Fiuister Heap: d as given in column 12, Table 1, h=D. slot V=38, S=3D. Burron Heap: Radius of hemisphere=%, C=D. B is given in Table 1, column 15, slot same as for filister heads. * Frat Heap: d is given in Table 1, column 14, slope of sides as shown in Fig. 57, or angle between sloping sides=82°. Slot same as for other screws. The radius of the curved top of the filister and French heads is=3D. This can be easily obtained by measuring D three times with the compasses. This is quicker ) than multiplying D by 3 and laying off the a tS. dimension with a scale. The threaded ends Finishof Bolt Ends of cylinders used for screws and bolts are Fic. 58. finished off on drawings by rounding with a radius equal to 2D as shown at (A), Fig. 58, or by beveling them at 45° as shown at (B). The threads are not drawn beyond the end of the cylindrical portion. In general, conventional threads only are represented on drawings. 62. If the heads of screws and bolts are not made cylindrical they are either square or hexagonal and a wrench or spanner is used to turn the screws into the threaded hole which holds eR ne ao # ot Po le oF (B) == (A) po bag he ig0 Bolt Enters Section & Side View End View ig 0-~| Hole Tapped Fra. 59.—Tapped Hole. Fia. 60. them. This threaded hole is called a tapped hole because the thread is cut in it with a tap. The hole is drilled before tapping, the conical form of the bottom of the hole being due to the drill point. A tapped hole is shown partly in section in Fig. 59. The depth is 13 times the diameter. If there is no bolt end in the hole the threads of the hole (when in section) appear left handed. SCREWS AND BOLTS 81 The distance the screwed end enters the hole is shown in Fig. 60. Compare also the end views of a tapped hole without the bolt in it, as in Fig. 59, with that containing the bolt end in Fig. 60. 63. Cap Screws having other heads than round ones are the Sq. Hd. and Hex. Hd. Screws shown in Fig. 61. The dimen- H 74s / - Hex.HeadCap Screw Sq. Head Cap Screw Fie. 61. sions of the sq. hd. are given in Table 1, columns 10 and 11, or may be computed from the following formule: E=D+}" for D=1” to 2” (inc.). E=D-+2" for D=?" and upwards. H=D for all sizes. The dimensions of the hex. hd. cap screws will be found in columns 8 and 9, Table 1, or may be calculated from the formule, G=D+ 3," for D=}” to x4” (ine.). G=D+%" for D=4” to 13” (inc.). H=D for all sizes. | The square heads are rounded with i Ps a radius equal to 3D as shown in Fig. : 62. This view shows one face of the ‘ at square while the two faces are visible |! \ 7 in Fig. 61. The dotted lines in Fig. | 62 show where the corners will appear t i when two faces are visible. The hexag- onal head cap screw is chamfered (bev- . 1 eled) by a cone. This protects the Son ined corners from injury. The angle of the Sq. Head cone surface with the edges of the head Fig. 62. is taken as 60° or 45°. The curves on the faces called chamfer curves are hyperbolas since they are 82 ELEMENTARY MACHINE DRAWING AND DESIGN sections of a conical surface cut by planes parallel to the axis. These curves are drawn with circular arcs as shown in Fig. 69, which illustrates the construction and delineation of the U. S. St. hex. hd. for a bolt. The cap screw head is smaller across flats than the U. 8. St. bolt head and requires less room for turning. 64. Whenever a hexagon is drawn it is circumscribed about a circle whose diameter is the distance across flats. To draw a hexagon proceed as follows, referring to Fig. 63 for numbers. (F) is for a hex. whose long diameter is parallel to the T square blade, while (G) is for the position of the long diameter per- pendicular to the blade. In (F) draw 1-2-3 lightly with T sq., then draw 4 and 5 heavy with triangle, reverse triangle and draw 6 and 7 heavy. Then go over 1 and 8 with T square, making them heavy between 4 Method of drawingHex. and 6 and 5 and7. In (@) all Fic. 63. the lines are drawn with triangle instead of with Tsq. and triangle. The long diameter or diameter across corners of a hexagon equals 1.155 times the distance across flats. That of a square is 1.414 times the distance across flats. This shows that a square head requires 23% more space for clearance in turning than a hex. hd. having the same distance across flats. The governing dimensions of the screws previously mentioned are, 1, the diameter of body (D). 2, style of head. 3, length used in ordering as indicated by L or word length in preceding figures. 4, length of threaded portion (usually=2 the length given by 3). 5, number of threads per inch if different from standard number. The length 3 is determined from the thickness of the loose part which the screw holds in position, called the grip, plus the amount which enters the tapped hole equal to 14D. 65. Lag Screws are used for fastening machines to wooden floors, wooden joists, etc., and have sq. heads chamfered at 45°. The body is threaded with a wood screw thread to give it more holding power in wood. The proportions of head are shown in Fig. 64 as well as the shape and general method of repre- sentation. M=%D. The chamfer circle in the end view is usually SCREWS AND BOLTS 83 drawn tangent to the sides of the square and the chamfered corners omitted in the front view. 66. Wood Screws usually have flat heads like that shown in Fig. 57, with the angle between the sloping sides of the cone equal BS / Length to 82°. There are other styles of q in LASS head. The diameter of body is *— Lag Screw given by a screw gauge number, Fic. 64. the sizes running from 0 to 30. The lengths run from 3” to 6” with threaded portion equal to zy the total length of screw. 67. In all the screws thus far mentioned the mode of fastening is to drill a hole through the smaller of the parts to be joined and tap a hole in the larger part. The screw is then put through the drilled hole and screwed into the tapped hole until its head pinches the two parts together. 68. If a part of a machine must be often removed, the screw must be taken: out of the tapped hole at each removal, thus wearing the threads in the hole. In order to obviate the wear in cases of this kind a stud bolt is used. A stud bolt is a cylinder threaded at both ends. One end is firmly screwed into the tapped hole and remains there permanently. The other end projects through the loose part of the machine far enough to take a nut which screws on the end and presses the loose piece against the stationary part. By removing the nut the parts can easily be separated and the wear of the unscrewing and screwing will come on the threads of the stud and nut. A stud bolt with hex. nut is shown in Fig. 65. The threaded end C screws into a Tn the tapped hole and has a length iy of thread equal to 13D. B is Bak equal to the thickness of loose Stud Ae and Nut part (grip) minus 7’. A equals Fra. 65. the depth of nut (H) plus 1”) H=D,F=3D+%". Ahex. nut can be drawn by the conventional mnethoa shown in Fig. 69 instead of the method shown in Fig. 67. When the nut is tightly screwed against the part which it holds in place, its top should be flush with the cylindrical end of the bolt, not as represented in Fig. 65. 84 ELEMENTARY MACHINE DRAWING AND DESIGN Stud bolts are used in steam engine and pump cylinder heads, steam chest covers, and in all places where frequent removals would tend to wear the threads in a tapped hole. Test No. 1. Arts. 31-68. (2 hours allowed) 1. Show to scale the cross-section of a 5’ channel. The flange width is 12”, thickness of web 33”, gauge line 13’ from back of web. Rivet 3”. 2. Sketch the outline of 3 styles of rivet heads and name each one. 3. Sketch a single riveted lap joint and indicate the ‘‘pitch” of the rivets. : 4. Using your scale for C. I. fittings, construct full size a 2’ x13” reducing tee (standard pattern). Dimension the drawing completely so that the tee can be made from your drawing. 5. Show a short length of a cyl. 2” diam. threaded with a double V thread, right hand, 3” pitch. (Do not construct helices.) 6. Make a working drawing of a stud bolt without a nut. Diam. 2”, grip =12”, give total length, distance threaded and not threaded. Test No. 2. Arts. 31-69. (2 hours allowed.) 1. Show to scale the cross-section of a 4” x22” Z bar. Web thick- ness = }’’, gauge line 13” from back of web. Rivet =2”. 2. Sketch a double riveted lap joint when P —d =3”, P: =.6(P—d) +d, The plates are 8” thick. Use button head rivets. 3. Using your scale for C. I. fittings for W. I. pipe, construct full size a 3’’X1}” reducing tee (Standard pattern). Show half the front view in section. Dimension your drawing so that the tee can be made from it. 4, Show a short length of cyl. 22” diam. threaded with a mod. sq. thread of 3” pitch. Thread is left hand double (use straight lines for helices). 5. A tap bolt is to be made fora 14” grip. 3” diam. 3 1” oct F h= 5. Make a drawing of this bolt showing its length, length of thread, head, ete. 69. Tap Bolts are like cap screws in having a hex. head on one end of a threaded cylinder. The head is thinner and wider SCREWS AND BOLTS 85 across flats than the cap screw head. Fig. 66 shows a tap bolt head. The formula for F is the U. S. Standard F =$D+}4". The body is threaded 3 its length. The dimensions of these heads will be found in Table 1, columns 4 and 5. The manufacturers of bolts make a tap bolt with a head so proportioned as to permit of its being made at one upset thus reducing the cost. This is Fic. 66. called the Manufacturers’ Standard tap bolt and F=13}D. The dimensions of these heads are given in Table 1, columns 6 and 7. 70. A Through Bolt is used to fasten two pieces together when there is room for the bolt to pass entirely through both pieces and take a nut on the projecting end. The head is either hexagonal or square and the nut is the same. Square heads and nuts are used on rough work principally. The dimensions of through bolt heads and nuts are based on formule adopted by manufacturers throughout the United States and called U. S. Standard formule. Fia. 67.—U. 8. Standard Bolt and Nut. Fig. 67 shows a complete through bolt with hex. hd. and nut. The formule for obtaining the values of the letters when D is known are F T=D, F=1.5D+}", h=z. These formule hold for square or hexagonal forms. The long diameter of the hex.=1.155F, that of the square 1.414 F. In Table 1, columns 2, 3, 4, 5 will be found the dimensions of heads and nuts for values of D given in column 1. Fig. 68 86 ELEMENTARY MACHINE DRAWING AND DESIGN is a scale for obtaining the dimensions of a U. 8. 8. bolt head and nut, by measuring with the compass. For conventional ty % | yo GSl [x5 12 a cols ry elo - 6 fou alu ae. z An | cs -IN + Mw cB KX ‘ | a ow oo 0 ie ay yn 2 a a -jt Y CO es gs DOCE at G no w SS So 25 ne @ oY oO ols —r a | Fig. 70.—Wing Nut. Fia. 71.—Sq. Nut. flat which must be allowed for in drawing the curve of the top. =48D+2” for U. S. 8. nuts and bolt heads. H=D for nuts, => for heads. 88 ELEMENTARY MACHINE DRAWING AND DESIGN 72. Wing nuts are often used on bolts from } to 3” diameter. They are designed to be turned by the thumb and finger and are sometimes called thumb nuts. Their dimensions as shown by letters in Fig. 70 will be found in Table 15) below. TABLE 156 Wine Nots D A B Cc E F G R 19 11 1 5 7 9 $ 33 a2 a 32 yd 32 3a 5 13 1 s 1 1 1 2 32 16 32 16 4 8 4 a2 x 5 3 3 2. t 13 3 3 3 s ve 16 5 1 3 1 7 5 J a Te 13 a z i6 a 2 16 3 13 te is 3 2 3 16 15 7 5 9 5 5 1 is lie 3 3 is aa a 2 3 235 lis $ i v3 < i 73. The length of a bolt is the distance measured from the under side of the head to the end of the threaded shank and does not include the rounded or beveled end. The grip is the thickness of the material to be clamped between the head and nut. In listing or ordering bolts the following dimensions and data are necessary: 1, number wanted; 2, diameter; 3, kind of head; 4, length; 5, grip; 6, location. 74. Besides the bolts already mentioned, there are Coupling 5 Bolts, whose heads have the same dimensions as U. 8. St. nuts. They x ie are provided with U. 8. St. nuts and Lipdak | are used on flanged shaft couplings. BO “Woe ele Tee Head Bolts used to fasten 1 down work on planer carriages, (A) le D + testing floors, ete., where slots are made with wider spaces beneath. 5g. The heads are made as at (A) : Q in Fig. 72 so as to pass through D LY the slots and then turn through (B) 90° in the space, or the bolts Fre, 72.—Tee Head Bolts, 22¢ pushed into the slot from the end and have a square on the shank to prevent turning in the slot as (B). D SCREWS AND BOLTS 89 Swing Eye Bolts have a head cylindrical in shape through which a pin can be passed perpendicular to the axis of the bolt. The dimensions may be taken as A= 13D, B=D, C=11D. This isshown in Fig. 73. a as 75. Foundation Bolts are used for hold- a ing machine frames, engine beds, roof Ds cy a trusses, etc., to concrete or masonry foun- dations. They are divided into two Fic. 73. classes, viz., those imbedded in the Swing Eye Bolt. masonry and those passing through it. The Lewis bolt and rag bolt belong to the first class and the cottered bolt to the second. The rag bolt shown in Fig. 74 has the shank tapered of pyramidal form with jagged edges. A hole wider at the bottom than at the top is cut in the stone. The bolt is then placed in position and molten lead or sulphur poured in the space between the stone and head. Of course this bolt is difficult to remove and is used for permanent fastenings. F = x ZA | Masory. = > Y SN GDL q § N Rag Bolt S$ “130-4 N é NS Fia. 74. Fic. 75.—Lewis Bolt. The Lewis Bolt has a shank of rectangular section. One side of the head is parallel to the center line of the bolt and the other tapering as shown in Fig. 75. To fix the bolt in position it is dropped in the hole and moved over against the tapering side of the hole, after which the key is dropped in. Tightening the nut wedges the head and key firmly in the hole. There should be some clearance below the head to facilitate the removal of 90 ELEMENTARY MACHINE DRAWING AND DESIGN the bolt. The taper is about 14” to 12”, Key thickness T D equal to z" The Cottered Bolt passes clear through the masonry, and has a washer and cotter at its lower end. Dy, is the diameter at the root of thread when the outside diameter at the threaded end is D. Do isthe diameter at the cottered end, which is greater than D, to compensate for the material cut out for the cotter hole. Fia. 76.—Cottered Bolt. Fic. 77.—Lifting Eye Bolt. If Di is the diameter of the rod at root then the thickness T of cotter may be 4Di1, D2=14D,. The other dimensions are given on the sketch shown in Fig. 76. Lifting Eye Bolts are used for the purpose of attaching a hoisting rope or hook. Fig. 77 shows the general outline. The dimensions are as given in the table. Eye Bott DIMENSIONS A B Cc D E F G $ oo ; eA 18 - a 2 8 8 16 4 : 2 i 23 Ss lt it Z 8 2 25 te lye 1} i ; 4 | 2H i i | ou 1 : i 3t 16 iy 1 SCREWS AND BOLTS 91 76. When bolts and nuts are used on machines subject to excessive vibration the nuts tend to work loose and then work off the bolt. To prevent this, various methods of locking them have been devised, the best known and most common one being the locknut. This is an extra nut screwed tightly against the regular nut to jam or lock it on the bolt. The nut used for this locking is usually half as thick as a U.S. St. nut and is chamfered on both ends. Since the top nut takes all the load it should be the larger one; but owing to the difficulty of turning a thin nut with a standard wrench, when placed beneath a standard nut, in practice the thin nut is placed on top. Quite often when two nuts are used as above, the standard nut and the lock- nut are each made one-half the total thickness of the two together, that is, each one is ? as thick as a standard nut. Sometimes the end of the bolt which projects beyond the nut is turned down to the diameter at the root of the threads and a hole is drilled through it close to the nut. A split cotter pin is put through this hole and the split ends spread apart. See Fig. 78. This prevents the nut from working loose. The dimensions of split pins will be found in Table 12, page 16. 77. The Association of Licensed Automobile Manufacturers has recommended for automobile work a bolt with threads of finer pitch than the U.S. St. number, and also a castle nut or capstan nut. The material of the bolt and nut is high grade steel having a tensile strength of 100,000 lbs. per sq.in. and an elastic limit of 60,000 lbs. per sq-in. The forms of nut and head and their dimensions are shown in Table 14. The cotter pins are those shown in Table 12 and called split pins. A hole is drilled through the bolt so that when the nut is screwed on tight, this hole lines up with one of the slotted holes in the nut. The split pin is then pushed through nut and bolt and opened at the split end to prevent it from backing out of the hole. This prevents the nut from loosening. On the bolts for connecting rod ends and on piston rod ends a Penn 92 ELEMENTARY MACHINE DRAWING AND DESIGN or Ring nut is often used. See Fig. 79. The part (A) of the rod is counterbored to take the lower part of the nut which is turned to fit and also grooved. A set screw passes through A and bears against the bottom of the groove, thus locking the nut. The prin- cipal dimensions are also given in Fig. 79. The diameter of the turned part of the nut is a trifle less than the distance across flats. 78. A spring washer is also used to prevent a nut from working off. It consists of a piece of spring steel forming a portion of a helicoid. This is slipped over the bolt and the nut is screwed on pressing the washer down between the nut and fixed surface of (C). The sharp edge of the washer cuts into the under | | s Im | "IS le | % | | Is V4 ta LE t { ~Iy (Cc) Att | | | Nut | | T | 6 ly fo om Fic. 79.—Penn or Ring Nut. Fia. 80.—Spring Washer. side of the nut, preventing any reversal of motion with consequent working off. A and B of Fig. 80 show the-washer before and after screwing up the nut. 79. Washers. When the surface against which a nut presses is rough or uneven, a washer is used to provide a smooth surface for the nut to turn on. It is also used to distribute the pressure of the nut over a larger area, when the material, against which it bears, is not strong enough to resist the pressure. Washers are also used under heads or nuts which bear against wood. The diameters of U. 8. St. washers are as follows: The thickness is given by the wire gauge number or by the nearest fraction following. From d=} to 3” (incl.) D=2d+}” | d=diam. of bolt, From d= 4," to %” (incl.) D=2d+3” } D=outside diameter From d=” to 2}” (inel.) D=2d+3" of washer. SCREWS AND BOLTS . 93 Thickness of washer= 7. For d= 4" and ¥5" T=No. 16=7;” 6c d= 3” 66 5” T=No. 14= 8,” 66 duc a” 66 3” T=No. 12= 3,” ee, oo aes T=No. 10= }” ‘ d= # to 14” T=No. 9=%;” SS Gale S" ae T=No. 8=14" ‘1 d=2y" T=No. 6=34" 80. Set Screws. Besides the bolts and screws already men- tioned, there is a class of screws used to prevent relative motion between two machine parts in contact. This relative motion may be prevented by tapping a hole through one part and screwing a set serew through this hole until its point presses hard against the surface of the other part. It is evident that the pressure thus produced will not be sufficient to overcome any very great tendency to move, therefore set screws are used principally in cases where adjustment is needed, or some part is to be held in its proper position. They are made of iron or steel, case hardened. The point of the set screw may have to bear against a flat or curved surface; accordingly there are a variety of points. In Fig. 81 an oval point and cup point are shown. The radius < Length. H > > F . : y= Oval Point Cup Point Loth SquareHead Set Screw Fia. 81. of the oval point is equal to D. The radius of the cup may be D : : z 2? the slope of the beveled part being 60° with the sides of the threaded part. Instead of the rounded cup, a conical cup is often used which is made by a drill point. A cone point is also used, the angle at the point being 60°. A flat point is occasionally found in adjusting screws, the end D being beveled at 45* and the length of bevel being 3" 94 ELEMENTARY MACHINE DRAWING AND DESIGN Heads of set screws are square, with a rounded top whese De radius is 21D. The length of head H is equal to D, or 9° Siving the names high head and low head respectively. The head is separated from the shank by a neck whose approximate diameter is equal to the diameter of the threads at the root. The length of this neck is about twice the depth of thread. The under side of the head is generally rounded with a curve whose radius is the same as the top, viz., 24D. The diameters of set screws begin with 3’’, vary by jg to 3”, then by 3” to 1”. The shank is threaded its whole length by U.S. St. threads, the number of threads per inch being given in column 16, Table 1. The length of neck may be taken equal to the difference between columns 18 and 19, in Table 1. In cases where it is objectionable or contrary to law to have the head of a set screw project above the surface of the piece into which it is screwed, it is made without a head, as, shown in Fig. 81 on the right. A slot is made for a screw driver in one end to enable it to be turned into place, or it 13 made as shown in Fig. 82 which shows a hex. hole in the screw in which a key can be inserted. The Fic. 82. bolts and screws with their fittings which have Set Screw, been treated thus far are the principal ones used in ordinary, machine design. There are many others for special work, many of them being shown and described in the American Machinist’s Handbook, by Colvin and Stanley. 81. As previously stated, in calculations which involve the tensile strength of bolts or screws it is customary to consider the area of the bolt or screw at the root of the thread. In Table 1, Column A, this area is given for each bolt when the thread is U. 8. St. If we consider this area to resist tension at the rate of 1000 lbs. per sq.in. of area then column D shows the load in pounds which the bolt will carry. If we allow 2000 Ibs. per sq.in. then the bolt will carry a load twice as great as the load of column D. By the same reasoning a strain of n thousand lbs. per sq.in. will permit loading the bolt with n times the load given in column D. Small bolts are more liable to be overstrained by screwing up the nuts too tightly, especially in SCREWS AND BOLTS 95 cases where joints are kept tight by means of the bolt tension. When calculations of this kind are made, the strain at the root section should be proportioned somewhat as follows: Large bolts and studs; iron, 6000; steel, 8000 to 10,000. Bolts }” and under, 2500 to 5000; steel, 6000 to 8000. Bolts on cylinders under 10” diam., 3000 for iron, 5000 for steel. Bolts under 3” diameter are to be avoided in places where they are subjected to much pressure in screwing up. A workman can easily twist off a 4” bolt by applying his whole force at the end of an ordinary wrench whose handle has been lengthened by a convenient piece of gas pipe. The theoretical length of the wrench handle is about 15 times the bolt diameter (d) and the average force exerted by a workman will be 40 lbs. With this leverage and pull, a half inch bolt will be strained to 15,000 lbs. per sq.in. at the root. The strength of a bolt to resist shear will be found in column £, Table 1, for a strain of 1000 lbs. per sq.in. For (n) thousand lbs. per sq.in. multiply the tabular value by (n) to find the strength of the bolt INSTRUCTIONS Screws and Bolts Plate 1, No. 1 paper. 2 hours allowed. Av. total time required 2.9 hours. Fig. 1. Draw side (partly in section) and end view of a tapped hole fora bolt D=() with a bolt end in the hole for the max. distance. The material in which the hole is tapped may be sectioned for C. I. Fig. 2, make side and end views of a (a) head machine screw, D =(’’). Grip =("). Fig. 3, make side and end views of a (b) head machine screw. D=("), Grip =("). Fig. 4, make side and end views of a (c) head cap screw D=(”) Grip =(). Show the max. number. of faces in the side view. Fig. 5, make side and end views of a stud bolt with a hex. nut (3 faces showing). D=(’'). Grip =(’’). Fig. 6, make side (2 faces) view and end view of a tap bolt. Grip = ("). D=("). Name of plate to be “Screws, Tap and Stud Bolts,”’ 96 ELEMENTARY MACHINE DRAWING AND DESIGN Taste Givinc Data For PuiatTE 1 Fig. 1. Fig. 2. Fig. 3. Fig. 4. Fig. 5. Fig. 6. 3) ee we pee s lpr |&| ¢ |p| &| pr | 2] py | 4 * 5 5 5 5 o 1} 1 |Fillister2 | 1 |Button] 1 | 3 | Hex. 3 z/2 |1/%/ & ag] « wll “ |elilsa la | aye jt]a | § set *@ ala) * lalg ies! a |) ] 1) | a oe | alae] @ | bla [ea le | ale {ale | 5| 14 So | “ ) Se | Hex) 2 [1 12/8 t ue | “1 | ab | Flat] 4]e) 8a. /$ ite]? |e] e | 3 7] %| “ wl] el “| al ed] Beeld |] 1 [a] a ft gia | gel de] fe] 4| Sa |e] ele [42 ft aa] « afl f [ale | Bele | w]e fa] a ft wp 2 | Lab] Tela] 8a a | ala a] ee] Plate 2. No. 1 paper. 2 hours allowed. Av. total time required 2.4 hrs. Fig. 1, draw a side view, front view and two end views of a U. S. St. hex. hd. bolt with hex. nut. D=(’’). Grip=(’’). In the front view show two faces of head and three faces of nut. End views each side of the front view. Fig. 2, draw a front view and two end views of a U. S. St. sq. hd. bolt with sq. nut. D=("). Grip =(). Show one face of the head and two faces of the nut in the front view. Use spherical finish for both head and nut. Name of plate to be “U. 8. St. Bolts and Nuts.’’ Give all dimensions of heads and nuts, length of bolts, ete. TaBLE For Data For Pu. or U. 8. St. Bouts No 1 2 3 4 5 6 7 8 9 10 p=i| +) ¢] a] %]1fuju | w| xs Fig. 1 oii [aloe ae la [ef [o a t = = . Day) atk | [lv] wel le | wll | a Fig. 2° ai al oi ‘ p | Cope | ae | a] | | a jlis ea : j a i Prob. 1. (a) Calculate ‘the load é ae” steal bolt will support in tension when the material i striined ‘to 600 Ibs! per sq.in. (b) Same for a 1” steel bolt. , SCREWS AND BOLTS 97 Prob. 2. How many bolts and of what diam. will you use to sustain a load of 30,000 Ibs. in tension? Plate 3. No. 1 paper. 2 hours allowed. Av. total hours required 2.94, Fig. 1, draw side and end views of a sq. hd. (a) point set screw. D=("). Length=(’’). Show two faces of head. Fig. 2, draw two views, front and either top or side of a lifting eye bolt whose shank diameter A =(’). Fig. 3, draw the top and side view of a Penn nut for a bolt whose diam. d=("). Show the nut as it appears on a bolt passing through a machine part with the set screw in position. Put a split pin through the bolt end. Fig. 4, draw the front and top view of a cottered bolt with a hex. nut. The material through which the bolt passes as well as washer and cotter are to be shown in this view. D=('’). Grip to be assumed. Make a side view of the cottered end. ASSIGNMENT TABLE FOR PuatE 3 Fig. 1. Fig. 2. Fig. 3. Fig. 4. No. a D Length. A d D 1 oval & 1 - i 3 2 cup 6 3 3 1} 1 3 oval 8 1 3 t 1x 4 cup 3 i 2 ly 3 5 oval a 1 t 13 V3 6 cup t 1 1 1 a PROFZATY CEFARTMENT MACHINE DESIGN _ SIBLEY SCHOOL - CORNELL UNIVERSITY oe CHAPTER VI KEYS, COTTERS, ETC. 82. WueEN two machine parts are to be fixed in such a manner that one part cannot rotate around the other, they are generally keyed together. Examples of this are found in wheels placed on shafts and keyed to them, rocker arms or levers used to turn shafts, cranks of engines, hand wheels on valve stems, etc. Saddle Key (F) Flat Key @) Pin Key(H) Fia. 83 There are many ways of keying and the draughtsman must . choose the style best adapted to the case in hand. A Saddle key is used for light work and acts by friction on the shaft. A Flat key rests on a flattened surface on the shaft and is made with a slight taper on the top in the direction of its length and driven into the groove in the hub from the end of the groove. hub. These are shown in Fig. 83 by sections perpendicular to Feather Key Side two gib heads fitted on both sides Fig. 84. of the hub, the object being to Fig. 84. The sides of the key are parallel, also the top and bottom. 98 The Pin key is a tapered pin driven into a tapered hole drilled the axes of the shaft. allow the wheel to slide along a half in the shaft and half in the Sic auai dane cena a ee The Feather key is made with shaft and still turn with it. The side view of this key is shown in KEYS, COTTERS, ETC. 99 83. Sunk keys are let half into the shaft and half into the hub which is to be keyed. The sides of such keys are always parallel while the top is on a slope with the bottom. The sides of these keys fit tightly against the sides of the grooves in the shaft and hub, but the tops may or may not fit. The bottoms fit if the key is placed in the shaft and the hub driven on. A sunk key is shown in Fig. 85 (end and side views) together with if | Key andSeat Fia. 85. the shaft and hub which it keys. The key has a length L, width W and thickness 7. Good proportions for W and T are given by the following formule: W=3D4+}", T=3;D+4", D=the diameter of shaft. When a sunk key is tapered it is usually provided with a gib head to facilitate its withdrawal from the shaft. The dimensions of this head are shown in Fig. 86. WY” Tver dfiy" WK and T are the same as above. ’ oe 4 84. The grooves in the shaft ; ; and hub = called key slots or Drive Key (Gib head) key ways, and the two methods Fic. 86. of cutting them in the shaft are shown in Fig. 85 and Fig. 87. In Fig. 85 two holes are drilled in the shaft W’” diameter and L’” apart. The metal between these holes is then cut out, leaving a slot or keyway W ee 6 hes o-W of width (W) and depth 5 with semicircular ends. The key may 100. ELEMENTARY MACHINE DRAWING AND DESIGN have semicircular ends to fit the keyway or be cut square with a length equal to LZ only. If the key seat is cut with a milling cutter it appears as shown in Fig. 87 (D) side view. The cutter wn Keyway cut by Cutter Fi, 87. 3 Values of D (diam. of shaft) Fria. 88.—Scale for Values of W and T. 2 Keys projectabove shaft & Large Sizes 26-36 Fic. 89.—Woodruff Keys. Ue uw has a thickness equal to (W) and cuts a keyway > deep whose Jength is (L). This keyway has a curved bottom surface at the ends, the radius of the curve being equal to the radius of the cutter, say from 13” to 3’. In Fig. 87 (C) is a top view KEYS, COTTERS, ETC. 101 of the shaft showing the key in the keyway. Fig. 88 is a scale for finding values for 7 and W for different diameters of shafts. The letters correspond to those in Figs. 85, 86, 87. 85. Woodruff Keys are used for light work and are very easy to fit. The key consists of part of a disk as shown in Fig. 89. The key seat in the shaft is made by sinking a milling cutter into the shaft. The diameter of the cutter is D. The key projects above the shaft one-half its thickness. If the hub to be keyed is long, then two or more keys are used. The keyway in the hub is made in the ordinary way of width C and depth . The dimensions for different sizes of keys are given in Table 15, page 18. For light work it is often convenient to use taper pins instead of keys. The holes for such pins, after drilling through shaft and nut, are reamed to the taper of the pin and the pin driven in until the small end comes through the hub. The pin can be easily removed by striking on this end and driving it out. 86. When two rods are to be connected rigidly in such a way as to transmit force in the direction of their length only, the most convenient joint for the purpose is the cottered joint. The cotter is a flat bar, wedge shaped, and driven into a slot in such a way as to firmly hold the parts against either tension or compression. The end of one rod is enlarged to form a socket into which the end of the other rod fits. Both of these rod ends are provided with slots which are in line when one rod end is fitted into the socket of the other. The cotter is then driven through these slots, making the complete joint as shown in Fig. 90. It can be shown that when the joint is in tension its parts will be of the same strength when the proportions are as follows: De D,=1.21D, D=.82D;, De=1.75D, B=1.31D, ao m=N=from 3Dto D, Dg=2.42D, Ds=1.4D, te=.42D. There must be clearance in a cottered joint if the cotter is to draw the rod end of £ tightly into the socket F. This clearance is provided at (cl) in the rod end and at / in the socket. The cotter is driven home in this kind of a joint and the clearance 102 ELEMENTARY MACHINE DRAWING AND DESIGN varies from 7” to 2’... The total taper of the cotter must not exceed 9°, to prevent the cotter from slipping back when the surfaces in contact are greasy. This corresponds to a taper of 1 in 7. In order to be on the safe side, the taper is made less than this, viz., from 1/’ per foot to 3” per foot. If there a eae X Fe { | fs Ni WN ~ FE ci, NAN Y 1S] SVE 2 LBV bez ok Y/)4 f as to Fia. 90.—Cottered Joint. is some arrangement for keeping the cotter from loosening, this taper may be increased to 1 in 7. Cotters are used to fasten uprights to bed plates, pistons to piston rods, piston rods to cross- heads, connecting rod stubs to straps, for foundation bolts with C. I. washers, etc. 87. If a cotter AD was driven so as to draw the straps CB onto the connecting rod F, Fig, 91, the friction of the cotter y A GB ew oar cotter E F Gib acl. = rfl lH D Wo 27 SS Z Fig. 91. Fia. 92. against the strap at H would cause it to take the position shown by dotted lines at B. To prevent this a gib is used as in Fig. 92. In this case the taper is on cotter and gib, and the holes KEYS, COTTERS, ETC. 103 in the parts held together, have parallel sides. Two gibs are sometimes used, the taper being equally divided between them or on one only. The distance B is the same as for cotter alone as given in Fig. 90. INSTRUCTIONS Keys and Cotters Plate 1. No. 1 paper. 2 hours allowed. Av. total hours required 3.13. Fig. 1, draw the end, top and front views of a shaft. D=(_ )’’ con- taining a sunk key. All three views must show the key in position in the keyway. Keyway to be placed as shown in Fig. 87 but made as shown in Fig. 85. Use letters for dimensions of key except for L which = ae Fig. 2, draw 2 views of a section of shaft 14’ diam. and show a Woodruff key No. (__) as it would appear in this shaft. Fig. 3, make 3 views of a cottered joint as shown in Fig. 90 when the value of D=( )”. Name of plate is ‘‘Keys and Cotters.” TABLE FOR PLATE ON Keys anp CoTrers 2 Fig. 2. . Fig. 2. . No. -™ te Woodrutt Pig, 3. No. pe A Woodruff Fig. 3. 1 1k 2 10 13 7 j|1%] 21 1 2 13 2 11 13 8 1 13 E 13 3 2 2 12 15 9 |182 25 14 4 Qt 12 13 13 10 | 13812] 26 14 5 23 2 15 13 11 275 12 31 1is5 6 3 2 19 13 12 | 23,2 32 45 MACHINE Design SIBLE Y SCH OOL CORNELL UNI VERSITy RECEIVED. _ teeta eee ee ame, CHAPTER VII SHAFTING AND SHAFT COUPLINGS 88. A BAR arranged to rotate about its axis and transmit power is called a shaft. The cross-section of this bar perpen- dicular to the axis of rotation is in practice either a square or a circle, although at the points of support it is always a circle. When a shaft is too long to be made in one length two or more lengths are used, joined together by couplings. The diameter of a shaft depends on the power it transmits, the load on it, the distance apart of the bearings which support it, as well as the kind of material of which it is made. Shafts are accordingly divided into two classes: (a) Those which transmit a uniform torque and are not subjected to bend- ing, and (b) those which transmit torque and are subjectéd at the same time to bending due the weight of the rotating parts to which it transmits or from which it receives power. Case (a) is the only one which we will consider here, as it is much more common than case (b) besides being less difficult of calculation without a knowledge of mechanics. 89. If a force F acts at the end of and perpendicular to a lever (as in Fig. 93, (a)), whose length is R, or at the rim of a t Pulley e (c) ys 5 7 (h-T3)R gear whose radius is R, the product FR is called the twisting moment or torque acting to twist the shaft d, that is T=FR. Let T =the torque, d=diameter of shaft in inches, f,=maximum 104 SHAFTING AND SHAFT COUPLINGS 105 shear stress in the shaft. N=R.P.M. of shaft. H=horse-power transmitted by the shaft. Z=—=modulus of the section of shaft. xd? : ; Z=-—~ for a circular section. 16 Then d?. df, T= 7f,= f= Fh, ee ee or Pte ce oe also 5.17 aa . ° . . . + . . . . (3) The value of f, varies from 6000 lbs. for wrought iron shafts, to 18,500 for steel shafts. The horse-power transmitted is equal to H - but V=2xRN, where R is the radius in feet to the point of application of F. But T=FR where RF is in inches, _ FX2xRN .. 2. nn : _FRN and H= 1233000 (if R is taken in inches) from which H= 63000" Substituting for FR its value T we have, EXS800 7, and substituting again in (2) this value of T gives us a= 5.1H X 63000 NXfs If fs= 9000, H =2 93/42 : then d=3.3N/y, + + 2 ee ee (4) If fs= 13500, then d=2.87.8/7, age Shee Bede Gl (CB) N 106 ELEMENTARY MACHINE DRAWING AND DESIGN We thus have the means of determining the diameter of a shaft to transmit a given horse-power at a given number of R.P.M. or the converse. Shaft diameters vary by sixteenths of an inch and are 7," less than the nominal size; that is, a 2” shaft is 148” in diameter. 90. There are several kinds of shaft couplings in use, divided into two classes, (a) fast or permanent, comprising muff and flange couplings, and, (b) disengaging couplings of the claw coupling type. Flange Couplings consists of two cast iron flanges or disks keyed to the ends of the shafts and held together by bolts. | _WUS|Standardor 4 Coupling Bolt Pas oe) -f-all over Cast Iron Flange Coupling Fic. 94. The whole power of the shaft is transmitted through these bolts which are in shear. Fig. 94 shows a flange coupling whose bolt heads and nuts are not protected. Each flange is turned true with its own shaft after being keyed on. The faces of the flanges are then brought together, the alignment of the shafts being insured by allowing one shaft to pass through its own flange and enter the shaft hole of the other flange about 2”. The two shafts in this case must be of the same diameter. The ends of the shafts which enter the flanges are often turned down to a smaller SHAFTING AND SHAFT COUPLINGS 107 diameter to provide a shoulder for the flange to fit against. The keys are driven in from the face of the flange, before the flanges are trued and brought together. The bolts are carefully fitted to the holes in the flanges as they are in shear.. The dotted lines near the joint of the flanges represent recesses in each flange face which do not have to be machined as the flanges only come in contact beyond them. Otherwise the surfaces of the flanges are finished in the lathe. The fillet joining the flange to its hub is made with as large a radius as possible but not so large as to bring the curve under the hex. nut and prevent it from turning freely on a flat surface. The outer edges of flange and hub are rounded off to prevent the sharp edges from injuring the workmen. The radius of fillets varies from 4’ to $’. As a rule rounded external edges are for safety and looks. while internal angles are filleted for strength. The number of bolts used depends on the diameter of shaft but not in a direct ratio, as the number is usually not less than three and increases by even numbers from four upwards. There is no real reason why odd numbers of bolts cannot be used, but they are not as a rule. If N equals the number of bolts, then N may vary from 2D+2, to D+2, where D is the shaft diameter (nominal). 91. Since the bolt diameter influences the diameter of the outside of the flange it is better to have more bolts of smaller diameter than fewer of large diameter. 0.423D Vi +.3”, using the nearest U.S.St. diameter above the value of d obtained. As the bolts take the strain of transmission through the coupling we may determine the shear in them by the following: A _D*x + XPXNXG = 6 If d=bolt diameter then we may take d= po Dishixdr2 pa Die ~ 16xXxd?xAXN’ 2d2AN’ Where f=shear in the bolts and 4 =radius of bolt circle given below. jf.=shearing strain allowed in the shaft. Low and Bevis give the following values for the lettered dimensions of Fig. 94. 108 ELEMENTARY MACHINE DRAWING AND DESIGN C~A _0.423D gi=2= 5 d= Vi +.3”, D=diameter of shaft, e=varies from 3’ to 3”, B=1.8D+0.8", F=0.35D+.35", A=B+43.2d, L=1.2D+0.8", C=B+6d, g=2" to 3”. t and w are standard key dimensions given by the scale in Chapter VI. All calculations to be reduced to nearest sixteenth of an inch. 92. The principal objection to flange couplings like the one shown in Fig. 94 is the liability of an accident from the pro- jecting bolt heads or nuts. The objection is overcome by mak- ing an external flange deep enough to guard the heads and nuts. A coupling of this type is called a pulley coupling and is shown in Fig. 95. The flanges are keyed to their respective shaft ends and bolted together. The principal dimensions taken from D. A. Low’s Mechanical Engineer’s Pocket book are given below: 0.425D B=1.8D+0.8” d= —— +03” + VN +0.3", L=1.2D+0.8", D=diameter of shaft. H=0.5D+1” but not less than 0.3D+1.3d+0.3”, h=0.3D+0.3", 6=0.1D+0.2", n=1.5d, a e=¥", to", N=from 3D+2, to D+2; but not less than 3, and usually the nearest even number, W =key width and t=key thickness from chapter on keys. (Use diagram.) SHAFTING AND SHAFT COUPLINGS 109 Sizes must be taken to the nearest sixteenth except in taking the value of d, which must be the nearest standard diameter given in Table 1. 93. Flexible Couplings are used to connect shafts which are slightly out of line or when one shaft is rigidly held from end play and the other is free to move a short distance. A Universal Joint Coupling is used to connect two shafts when their axes are intersecting and inclined to each other. It is used on milling machines, automobile propeller shafts, etc. The coupling is called a Hooke’s joint or Cardan joint and is of the form shown in Fig. 96. The forked ends are often (8) Section of Pulley Coupling Fig. 95. Fic. 96.—Hooke’s Joint. forged on the shafts to be connected. The dimensions are deter- mined from the following formule, taken from Spooner’s Machine Construction and Drawing: D == Y =0.05d, d 9? L=14D, T=1.2D, 1-2, D=diam. of shaft. > ll Q - 110 ELEMENTARY MACHINE DRAWING AND DESIGN A double joint of this type will preserve a constant velocity ratio between the shafts connected, but a single joint will not. 94. Claw couplings are used for connecting or disconnecting shafts when the change is to be made often. The usual form of this coupling consists of two flanges; one flange keyed to one shaft and the other flange free to slide on the other shaft but made to rotate with it by a feather key. The faces of these flanges are provided with teeth which engage when driving. Fig. 97 shows the component parts of such a Fig. 97.—Claw Coupling. coupling when transmitting power. To disengage the coupling the left hand flange A is moved to the left a distance greater than g. The claws of one flange fit loosely in the recesses of the other to allow their engagement more easily. The groove in A is made to receive a forked lever for sliding A along the shaft. The dimensions given by Unwin are as follows: D=2.6d+ .5” e=0.4d +0.85” b=1.2d4+1.45” f=0.16d+0.375” c=14d+1.6” g=0.4d +0.5” INSTRUCTIONS Couplings Plate 1, No. 3 paper, 2 hours allowed. Av. total hours required, 3.55. Make a front view (lower half in section) and end view of a flange coupling (a) type for a shaft whose diam. D=( ”). If D is greater than 32” use half size scale. The front view is to be half in section below the center line, the section plane passing through the axis of the shaft. The only material which this plane cuts, and necessary to be shown in section, is that of the cast iron flanges. The bolts, nuts, keys and shaft are not to be section lined. Begin spacing the bolts on a vertical diameter in the end view. Place the key in one of the flanges 90° from that in the other and have the side view of one key appear in the half section view. Use (b) bolts and make a bill of material on the drawing. SHAFTING AND SHAFT COUPLINGS 111 Use common fractions in dimensioning this plate, not decimals. Be sure and give the following dimensions: Length of shaft of D, diam., outside diam. of pulley coupling, diam. at edge of face, inside of rim, diam. of bolt circle, length of each key, length of bolt, radii of fillets, depth of recess for bolt heads and nuts. List each key in bill of material, also give diam., length, grip and kind of bolts used. : Mark the parts on the drawing with the mark given in the bill of material. Take D, =D — +," for 14” shaft. Take D, =D — #2” for 2” to 3” shaft. Take D, =D —}" for 3” to 4” shaft. Take D, =D —#;" for 4” to 5” shaft. Use 3 bolts on 13” coupling. Use 4 bolts up to 32” coupling. Use 6 bolts above 32” coupling. Use a standard diam. of bolt. Set the keys on a shaft D, diam. end view first. TABLE FoR CoUuPLING SIzEs No.| 1 | 2 ajai[s|e|7/s| 9 10 | 1 | 12 | 13 23 | 2 3 4 4 D | 13 woo 3 | 8 wim oO co ra 43 al 1g 2/2 > lH a |non-protected for all except those with an ¢ after the assignment No. who will take Fig. 95 instead of Fig. 94. b | U.S. St. bolts for those without an (f) after the assignment No. Those with an (f) after the No. will use coupling bolts. Prob. 1. Calculate the value of f for the bolts of your coupling when the value of fs for the shaft is 7500 lbs. per sq.in. Prob. 2. Make 3 views of a Hookes joint for a 13” diam. shaft. Dimension carefully. Av. time required 3.25 hours. Prob. 3. Make 2 views of a claw coupling for a 2” shaft. CHAPTER VIII STUFFING BOXES 95. WHENEVER a reciprocating or rotating rod or spindle passes through the wall of a vessel containing a fluid or gas it becomes necessary to use a stuffing bor to prevent leakage along the rod. _ Examples of this may be seen on steam engines and steam pumps where the piston rods pass through the ends of cylinders containing steam or water. The valve stem of a globe valve or angle valve used for controlling the flow of steam or water is another case where a stuffing box is necessary to prevent leakage. The propeller shaft of a vessel passes through a stuffing box as it issues from the vessel. 96. Fig. 98 represents a stuffing box of the simplest kind. (A) is the case formed on the wall through which the rod D passes, and contains the packing (B). This packing is generally some soft material such as hemp rope, canvas covered wicking or specially prepared material well greased to allow the rod to move freely with as little friction as possible. The packing is forced inwards by means of a loose metal sleeve called a gland, which is pressed down by turning the nuts on the stud bolts. These studs screw tightly into the lugs on the casing and project through lugs on the gland. The bottom of the packing space as well as the bottom of the gland is beveled in such a way as to press the packing against the rod when the gland is forced down on it. The width of the annular packing space S depends on the diameter of the rod D but does not vary directly as ‘D. Professor Charles B. Richards originated the following formula which gives suitable values of S for rods varying in diameter from 3” to 24”. S=(0.8/D+i")—3". 112 STUFFING BOXES 113 Between certain limits however this formula can be replaced by straight line formule which are easier of solution, as S=3;D+ ;" for rods from 1” to 1” diameter, S= D+?” oe 6c 12 to OM ec S= 4D+2” oe ce ou to 4” ce S=3;D+3" be oe 4” to 8” 6c a) ie nh | 2 ! iY | i ek © BA ales fis G a 4 Fe ,™ 8 30° Amt ge B 4 Ss] e Sl E ask ss ¥ y = Stuffing Box Fie. 98. Use the nearest thirty-second of an inch to the calculated value. The scale shown in Fig. 99 is much easier to use, however, and saves all calculation. The depth of packing space (H) varies from 48 to 68, according to conditions of speed of rod, pressure tending to cause leakage, and convenience in repacking. The largest. value is used when the leakage is apt to be the worst and the wear on packing the greatest as well as the length 114 ELEMENTARY MACHINE DRAWING AND DESIGN of time the box must last without repacking. The length of gland exclusive of flanges should always be 28 less than the length of packing space (H). This prevents the packing from being compressed in the box to more than 3 its original volume. There is no fixed rule for the diameter (a) of the stud bolts but since the pressure thev must stand depends on the size of packing and length of packing space it seems reasonable to use some such ratio as the following: Use the largest constant for the smallest box. When D varies from 3’ to 1”, a=1.2S to 1.18, When D varies from 1” to 2’, a=1.18 to S. When D varies from 2” to 4”, a=S to .98, When D varies from 4” to 8”, a=.9S to .8S. Ordinates Values of (5) 2 Values of (D) Fig. 99. The thickness (¢) of the casing wall may be made equal to S in nearly all cases, without danger of rupture, especially on small boxes. On large boxes this gives too small values of (t) and it can then be increased to t=.04D+#”. In Fig. 98 only two stud bolts are used for forcing in the gland, but on large shafts the gland is so large that three or more are used. In that case the bolt lugs on both gland and casing become cylindrical flanges instead of two lugs on opposite sides. The thickness of the lug on the gland is 138, while the lug on the casing varies from 1ja to 1ja. The distance between the centers of the studs may be taken as D+4S-+a, which will bring the inside of the bolt tangent to the outside of the casing surrounding the packing space. The top views of the lugs of the gland and casing is the same, the outside of the lug coming a sixteenth of an inch beyond the corner of the nut. The studs should be long enough to project two threads above the upper surface of the gland STUFFING BOXES 115 lug when the gland is just entering the packing space. The minimum thickness of material at the bottom of the packing space should not be less than S but can be made greater if necessary. 97. A stuffing box of the type shown in Fig. 98 occupies considerable space on account of the flanges on the gland. In many cases it is desirable to reduce this space, and for this reason the stuffing box shown in Fig. 100 is often used. (D) f a Lat © iL ss — 2thrds, Rly a. B ESR Baie o % (B) ms 1 5 ortarger | WY YW +4 oNoae ED) I ft Syn I Lf ee ie Ea ") ed "Caps Fie. 100.—Screw Cap Stuffing Box. Fig. 101. It consists of three parts: the cap, the gland, and the casing. The casing may or may not be part of the wall of the vessel. The gland has a slight enlargement at the top to prevent it from entering the packing space too far. This enlargement is beveled on the under side to facilitate removal. The gland is pressed against the packing by a cap which screws over the end of the casing. This cap is formed on its exterior to fit the 116 ELEMENTARY MACHINE DRAWING AND DESIGN jaws of a wrench or the pin of a spanner. In the first case part of the cap is made hexagonal, as at (C). In the second case the cap has holes in the circumference as shown in Fig. 101 (D) or slots as shown at (£). The hexagon of Fig. 100 may be placed at the outer end of the cap or near the center and varies in length from 4 the cap length on short caps to 3 the length on long ones. The holes in Fig. 101 (D) vary from 3% to 3%” in diameter and about the same in depth. The slots of Fig. 101 (Z) are about 2” wide by 4” deep. The hexagon on the cap is circumscribed about the cylindrical part. The length and width of packing space and gland is the same as for boxes with flanged glands. The threads on casing and cap are of fine pitch, varying from 16 per inch on small boxes to 6 per inch on large ones. Even numbers of threads per inch are used as: 14, 12, 10, 8. 98. A casing which screws into a cylinder wall may or may not be provided with a hexagonal portion, for convenience in removing it or putting it in place. In the first case it is formed as in Fig. 100 (B) with a flange below a hexagon which has the same distance across flats as the hexagon on the cap. In the second case shown as (A) Fig. 100 there is a flange only, the placing in position being accomplished by a pipe wrench used on the flange. Sometimes there are holes in the inside end of the casing below the packing space to take a spanner. The dimensions of these screw cap stuffing boxes are given in the drawings of Fig. 100. Aside from the holes and slots the caps of Fig. 101 have the same dimensions as the cap of Fig. 100. INSTRUCTIONS Stuffing Boxes Plate 1,No.2paper. 2hoursallowed. Av. total hours required, 2.88. Draw a stuffing box for a rod whose diam. D=( ”). Box to be of the type requiring bolts to press the gland against the packing. Make 3 views, front and side views half in section, top view complete outside. Use letters for dimensions where they are given but numerals elsewhere. Give formule for letters used arranged in tabular form. Name of plate is ‘Bolted Gland Stuffing Box.” STUFFING BOXES 117 Plate 2, No.1 paper. 2 hours allowed. Av. total hours required, 3.0. Draw in detail the parts of a screw cap stuffing box forarod D=(_ ") diam. Use the style of cap marked ( ). Completely dimension each part, using numerals entirely. Use the style of casing shown at (_ ). Name of plate, “ Details of Screw Cap Stuffing Box for ( ’’) Rod.” TABLE FOR STUFFING-BOX PLATES No. tf; Se) ae ee) eR | we ee PL t).D% suse. 2 | 24 | oi | ef | ot la | aa | | oe) D 1 t i $ | %& | li g i | Pl. 1,| Cap.. E\¢\|oije¢loulelei\bpie Casing] A | 4 | B| A | B B|B|B|A No. u | 12] 13 | 14/1 15 | 16 | 17 | 18 | 19 | 20 BL Se elie 1g | 12 | 13 | 2 | 2e5| 2a | Qe | 1 | 1 | 1s D ol] de | iv] 1 | ise | ike] i | le jis] 3 Pl2| Cap...) ¢ | D| ¢ . c|p|eB C Cain! Ate | £2) e) al el ee | Ble On Pl. 2 make two views of each part. Do not section the gland. Indicate number of threads per inch on cap and casing. Use conventional system of threading wherever possible. In giving dimensions pay special attention to diameters. Give all the dimensions on each part, place the name and mark of each part below it. Make a bill of material. (see Art. 17.) Test No. 3. Arts. 70-98 1. Draw 3 views of a U.S. St. Sq. nut fora 1” diam. bolt. (8 threads.) Give all dimensions. 2. What is a cottered joint? A screw cap stuffing box? A split cotter pin? A drive key (gib head)? 3. Draw end, top, and side views of a 2” shaft in which a keyway has been cut to make a sunk key 3” x 3%’ X2” long. Give the dimen- sions of the keyway. 4, Make a sketch drawing of the gland for a “bolted gland” stuffing box indicating by dimension lines and arrows all the dimensions nec- essary for making the gland. Choose your own views. 118 ELEMENTARY MACHINE DRAWING AND DESIGN Test No. 4. Arts. 70-98 1. Draw 3 views of a U. S. St. Hex. Nut for a 1” diam. bolt (con- ventional method) and give dimensions. (8 threads.) 2. What is a Penn nut? A cottered bolt? A Woodruff key? A set screw. 3: Draw end, top and side views of a shaft 2” diam. showing the keyway for a sunk key whose dimensions are 3” x 3’’ X2” long. Give dimensions for the keyway only. 4. Make a sketch drawing of one of the flanges of a shaft coupling indicating what dimensions you would give and where, for making the flange. Choose your own views. CHAPTER IX BEARINGS, JOURNALS, ETC. 99. Tux parts of a rotating shaft which touch the supports of the shaft are called journals. The supports are called bear- ings. The simplest form of a bearing is a cylindrical hole in the frame of a machine as shown at (A) Fig. 162. This cannot be renewed without renewing the whole bearing. At (B) Fig. 102, Pp LAL ISSA SALTS AAS Journal _ } ees : Fic. 102. is shown a bushing which can be renewed when worn. The load on the shaft is acting in the direction of the arrows P. If P acts in a line parallel to the axis of rotation, as at (A) or (B), Fig. 103, the bearing is called a collar or thrust bearing. The pressure is then resisted by one or more collars. If the pressure acts as shown in Fig. 103 (C) the pressure is taken on the end of the shaft and the bearing is called a footstep or pivot bearing. 100. The effective area of a bearing is the area of the bearing when projected on a plane perpendicular to the direction of the load. In Fig. 102 itis LXD. In Fig. 103 (A and B) it is aD? _=D°\ V_® (p2_ pe where N equals the number of collars. 119 120 ELEMENTARY MACHINE DRAWING AND DESIGN In Fig. 103 (C) the projection is a circle of diameter D whose area is mD? “4° The load supported by a bearing is the product of the working pressure per square inch and the projected area of the bearing in square inches. If we call p the pressure per square inch, A the projected area of bearing and W the total load on the bearing, then W=pA. A is given above for each of the three cases preceding. \y Ss & (A) (B) Fig. 103. 101. Bearings for horizontal shafts supporting vertical loads are made with two or more parts to facilitate the introduction of the shaft as well as to allow for adjustment in case of wear. One of the simplest of these divided bearings is that found on the frame of a machine and called a frame bearing or bearing bor. Fig. 104. The cap is removable and is held in place by stud bolts and nuts or by cap screws which pass through its sides and are screwed into the frame beneath. This bearing may be lined either with a brass box or babbitt metal. The brass box is made in halves, one of them so formed as to prevent the box from rotating in the frame. The babbitt is cast in a recess in both frame and cap. A bearing box with the babbitt lining is shown in Fig. 104. _ The babbitt does not extend quite the whole length of the box and is beveled at the ends to enable it to be chiseled out when renewal is necessary. The cap is fastened to the frame by stud bolts and nuts. The cap is made thicker where the bolts pass through it by making “bosses ” which extend upwards far enough to give a flat surface BEARINGS, JOURNALS, ETC. 121 for the nuts to rest against. Two bolts are sufficient except in cases where L=4S when four are used. The cylindrical part of the bolt boss is made a trifle larger than the long diameter of the hex. nut. If the nut strikes the curved surface of the *< 2S ToAS=L ! ' { t ' \ ' ' ' ' 1 ‘ ! ' | ! ' | | ' 1 ! ! | ' Fig. |. Bearing Box Fig. 104. cap before resting on the boss, the cap is counterbored down to the level of the boss, the diameter of the bore being an eighth of an inch or so larger than the nut across corners. The oil for lubrication is contained in the cavity cast in the cap and flows to the bearing through holes drilled in the cap and babbitt. 122 ELEMENTARY MACHINE DRAWING AND DESIGN This cavity has its bottom made in the form of a trough and is filled with waste to prevent the oil from flowing too freely. A cover is placed over the cavity to keep out dirt. Instead of a cavity we may use an oil cup tapped in the top of the cap. This tapped pipe hole and boss for it are shown in the sketch marked Fig. 3 in Fig. 104. Fig. 1 of Fig. 104 is a half end view of the box and a half section by a vertical plane containing the axis of the stud bolt. Fig. 2 of Fig. 104 is a half top view, on the left, of the box assembled with part of the oil box cover removed. The right hand half is a top view of the frame when the cap of the bearing has been removed. 102. The important dimensions are lettered, their values being found by substituting in the straight line formula the letter wanted and values given for « and @ under that letter in the following table. S is the diameter of shaft for which the box is to be drawn. The formula is: required dimen. = «S+8. a b c e g h v n m a} .93 | .73 875 .02 | .1875 | .28 .453 | .125 125 8} .844, .3828 | .25 .07 | .3125 | .093 |—.265 | .0625 | .375 d=, fae+ze” The formula above is used for proportioning the parts of a series of machines when two sizes have been constructed and it is desired to introduce intermediate sizes. That is, suppose some part of a machine whose nominal size was say 30’, had a dimension of 6’’, while the same part on a 70’ machine had a dimension of 83’... The values of a and @ can be determined from the above straight line formula so that the dimension of this part for any sizes between 30 and 70 could be calculated as above. This determination is made as follows: 6” =30a+6, 84" =70a-+6. Solving these two equations gives the value of «=.062 and @=4.14. The required dimension of the part for a 40” machine would then be A=40X.062+4.14=6.62". When the sizes of BEARINGS, JOURNALS, ETC. 123 S run by fractional numbers it is somewhat tedious to make the multiplications of « and S. 103. To lessen the labor of calculation as well as to enable the value to be easily measured on a drawing it is often advan- tageous to construct a scale in which the base line is divided into equal spaces, the points of division being numbered according to equal values of (S). On ordinates erected at the ends of this line are measured (from the base line) values found from the formula when the value of (S) used corresponds to the values of (S) which mark the points at the ends of the base line. Joining like points on these ordinates gives an inclined line. The distance from the base line to this line, measured on any ordinate, gives the value of the dimension required for the value of (S) taken on the base line where the ordinate was erected. 104. Bearings are often made to be bolted entire to some sort of foundation or support; as a stone base, a vertical wall of a building or the ceiling of a room. In the first case they are called pillow blocks, in the second, wall or bracket bearings, in the third, hangers. An ordinary pillow block is shown in Fig. 105 below and consists of the following parts: Block, cap, brasses or steps, cap bolts, and foundation bolts. The dimensions for the letters given may be taken from the following formule: D is the unit for most of the formule, L for the rest. Take L=2D, e=1.6D+1.5" A=3.6D+5", W2=.7L, H=1.05D+.5”" T=.3D+.4", Wi=.8L, d=.25d+.25" C=2.7D+4.2 T,=.383D+.3", t=.07D+2". 105. The “ brasses’’ or “steps” used in these pillow blocks are made in halves whose plane of contact is usually horizontal. The load is perpendicular to this horizontal plane except in the crank shaft bearings of engines and similar machines. In these cases the dividing plane is perpendicular to the load. The steps are made with flanges at each end to prevent endwise motion 124 ELEMENTARY MACHINE DRAWING AND DESIGN along the shaft. Motion of rotation is prevented by various devices such as stop pins, stop lugs, by making the backs of the steps octagonal or square or by curving the backs with a X Tl) pi ee Tee ed Ws "ig = 71 @ ~ Ny = Ds Ig Caeey aer T*;, 2 4 ty i so | Y i fi, 1 ‘x Ye | Y K y/ ff \ LAN = 2) . mars x 9 ISS FN is T | pufi N i “19 \ le rH r CZ, | S R | is | His L i y NS = | WN Ca AEP op Fd” Flor * Pillow Block for D"Shaft Fia. 105. circular arc whose center is eccentric to the shaft. Fig. 106 (A) shows a lower step made with an octagonal back, and Fig. 106 (B) with a rectangular back. In both cases the upper step D (not shown) may have a cylindrical back whose radius is g tit. The flange will be like those on the steps shown. BEARINGS, JOURNALS, ETC. 125 “ Brasses”’ or “steps” are made of gun metal, phosphor or manganese bronze, white metal or “ anti-friction”’ metal, quite as often as of brass. In many cases the steps are lined with babbitt metal which is run into spiral or longitudinal grooves, or holes drilled in the inner surface of the step. 106. The materials used for bearings vary and depend on the pressure per square inch of bearing surface. Cast iron makes a good bearing for a steel shaft when the pressure does not exceed 300 Ibs. per sq.in. and the linear velocity of rubbing does not exceed 150 ft. per minute. The harder bronzes are used for high pressures and velocities. The amount and quality B Fic. 106.—Brasses. of lubrication greatly affects the life of a bearing as well as the amount of heating which occurs. The pressure per sq. inch allowed in bearing surfaces depends on the constancy of the load it carries, the method of lubrication and the quality of lubricant. 107. Table 16 gives the pressures used for bearings of different kinds. The efficiency of a bearing depends on the resistance it offers to the rotation of the shaft running on it. This frictional resistance to motion between two surfaces in contact is called the coefficient of friction (u) and depends on the pressure, kind of surface, and velocity of rubbing. The coefficients of friction are given for different surfaces in Table 17. The foot-pounds of work lost in friction of a bearing are equal to the load on the bearing in pounds times the coeffi- 126 ELEMENTARY MACHINE DRAWING AND DESIGN cient of friction times the distance passed through by the rubbing surface in feet per minute. PyuV =ft.-lbs. of work, u.=coeff. of friction. For shafts, QarN V= 2? where r=radius of shaft in inches, N=R.P.M. of shaft. . arN The foot-lbs. of work per minute = 7-25" If we divide this by 33,000 ft.-lbs. (the equivalent of one horse- power) the horse-power (H) lost in friction will be expressed by the formula: 2PunrN 4 =75%33000" u=coeff. of friction, P=load on bearing in lbs., r=rad. of shaft in inches, N=R.P.M. of shaft. Test No. 5. Arts. 1-107 (3 hours allowed for this paper) 1. Make neat freehand sketches of the following objects: (a) a cone head rivet. (b) an angle iron with equal legs (cross section). (c) a pipe fitting bushing. (d) a R and L coupling. (e) a standard T. (f) make a title only for a drawing sheet in the Sheffield Scientific School on which has been drawn a stuffing box gland (half size) for a 1” rod. 2. With the aid of sketches describe how the following four kinds of fastenings are used in connecting together parts of machines and quote an example of each one where it is preferable to use that kind of fastening in preference to the others, giving your reasons. (a) a bolt and anut. (b) arivet. (c) acotter. (d) a key. 3. Draw a spring whose inside diameter is 23’’, the diameter of the wire being 3” and the pitch 1’... The length of the spring is 6”. 4. How are short lengths of shafting coupled together to form a long straight shaft? Make a neat sketch of the parts used to couple these shafts and enumerate them. 5. What is a bearing? Into what classes are bearings divided? What is the effective area of a bearing? How would you calculate the BEARINGS, JOURNALS, ETC. 127 bearing area necessary to support a load acting at right angles to the axis of rotation of a shaft? 6. Calculate the foot-pounds of work expended in turning a shaft 2” diam. 200 R.P.M. in a bearing 5” long when the load supported is (V0) 1000 lbs. and the value of w= The load acts as in question 5. c=.43, v=rubbing veloc. in ft. per. sec. 7. What use is made in machine design of the straight line formula in proportioning the parts in a series of machines? Illustrate by a concrete example. 8. Make a tracing of the drawing furnished you, inserting all the omitted dimensions and carefully lettering any directions which may be necessary for a working drawing. 9. Make a drawing (dimensioned) of a stud bolt 1’ diam. whose grip is 23”, 108. Hangers are used to support shaft bearings from ceiling joists or beams. They are made of cast iron in two distinct parts, viz.: the frame and the bearing. The frame has an inverted base which spreads in a plane perpendicular to the shaft and is provided with bolt holes through which pass the hanger bolts for supporting it from the overhead beams. The frame is of a U or J form extending down from the beams as far as may be necessary to provide a support for the shaft bearing. This bearing is free to swivel around a vertical axis and turn in a vertical plane at the same time. This permits it to adjust itself to a true alignment with the shaft. In the Seller’s System the bearing is supported by two spherical cups which hold the spherical central portion of the bearing so that it can adjust itself to any direction of the shaft. The cups are formed in the ends of two tubes whose exterior surfaces are threaded for part of their length in order to engage with spaces on the inside of two cylinders formed in the frame. By turning these adjusting screws the bearing can be raised, lowered or clamped tightly together. These adjusting screws are locked in position by set screws. The bearing itself is made in halves, the upper one containing two receptacles for oiled waste. Beneath the hanger is placed an oil drip dish to catch the excess oil which works out of the bearing. The drip dish may be made with hooks which hang from two lugs formed on the lowest part of the frame. The split pins pass through the hooks 128 ELEMENTARY MACHINE DRAWING AND DESIGN into these lugs and act as a safety device to prevent the dish from being knocked off the lugs. The shaft is introduced into the hanger or the hanger is placed in position after the shaft is up. In either case a U hanger must have some opening for placing the shaft in the central space from the side or from below. This is accomplished by having a removable block in one side of the U or by removing the lower half of the hanger. In the first case the hanger has one parting bolt and is called an open side hanger. In the second case there are two bolts and the hanger is an open end hanger. The construction of these hangers is shown in Fig. 107 and Fig. 108 and was evolved by Prof. Charles B. Richards. A hanger with the single or J type of frame is shown in Fig. 109, A, B, C. This hanger was originally designed by Mr. William Mason, of New Haven, Conn.,of the Winchester Repeating Arms Company. It was first modified for use in the Sheffield Scientific School by Prof. Charles B. Richards and is shown in a still further modification in Fig. 109. The dimensions can be worked out for a series of “drops” and shaft diameters by means of the table of constants and the formula accompanying them. The bearing in the J hanger is supported by horizontal cylindrical trunnions which rest in a yoke free to turn around a horizontal axis. Fig. 109, D, E. 109. The two dimensions on a hanger which control all the others are the “drop” and shaft diameter. The “drop”’ is the perpendicular distance from the foot pads to the center of the shaft and is indicated by D on all the hangers in sketches 107-109. The shaft diameter is indicated by d on the open side and open end hangers and by A on the J hanger. If the “drop” and diameter of shaft are known the other dimensions of the hanger can be calculated by using the following tables on pages 130, and 131 with the formule accompanying them. As an example suppose d is 2’”’ and D is 20”. If LZ is desired for an open end hanger the formula L=ad+@D+y gives by sub- stitution of the values of «, 6 and y found in table opposite L, the expression. : L=(4X2”)+(.7X20”) +.8” =22.8", 128 ELEMENTARY MACHINE DRAWING AND DESIGN into these lugs and act as a safety device to prevent the dish from being knocked off the lugs. The shaft is introduced into the hanger or the hanger is placed in position after the shaft is up. In either case a U hanger must have some opening for placing the shaft in the central space from the side or from below. This is accomplished by having a removable block in one side of the U or by removing the lower half of the hanger. In the first case the hanger has one parting bolt and is called an open side hanger. In the second case there are two bolts and the hanger is an open end hanger. The construction of these hangers is shown in Fig. 107 and Fig. 108 and was evolved by Prof. Charles B. Richards. A hanger with the single or J type of frame is shown in Fig. 109, A, B, C. This hanger was originally designed by Mr. William Mason, of New Haven, Conn.,of the Winchester Repeating Arms Company. It was first modified for use in the Sheffield Scientific School by Prof. Charles B. Richards and is shown in a still further modification in Fig. 109. The dimensions can be worked out for a series of “drops” and shaft diameters by means of the table of constants and the formula accompanying them. The bearing in the J hanger is supported by horizontal cylindrical trunnions which rest in a yoke free to turn around a horizontal axis. Fig. 109, D, E. 109. The two dimensions on a hanger which control all the others are the “drop” and shaft diameter. The ‘drop”’ is the perpendicular distance from the foot pads to the center of the shaft and is indicated by D on all the hangers in sketches 107-109. The shaft diameter is indicated by d on the open side and open end hangers and by A on the J hanger. If the “drop” and diameter of shaft are known the other dimensions of the hanger can be calculated by using the following tables on pages 130: and 131 with the formulz accompanying them. As an example suppose d is 2” and D is 20”. If L is desired for an open end hanger the formula L=ad+ @D+y¥ gives by sub- stitution of the values of a, 8 and y found in table opposite L, the expression. : L=(4X2")+(.7X20") +.8” = 22.8”, ly Section at-zz- t toed oe ' Hook Detail of Hook Detail of Drip Dish for 0.E.Type Drip Dish for 0.S.Type ; (H) (K) u, Section at-z,z,- Ady. 2E: x Hub (3) of frame SI S.Std. ‘ 8 DripPan v Pee for Open End (B) ‘ordvenside (C) Open End (D) OpenSide (E) = wwe] KSI, C iO ly a » er aw y>ly! fen | (1) y NI Foot Arranged for two Frame for Open End Frame for Open Side Bolts,when Drop 2 25" Open Side and Open End Hangers - Sellers System Fia. 107, (To face page 128) BEARINGS, JOURNALS, ETC. 129 which is the length of foot required. Nominal diameter of shafts increase by fourths of an inch and the actual diameters are a, Pitch of threads 3 forallsizes (A) Fia. 108. Detail of Bearing and Adjusting Screw. usually one-sixteenth less than the nominal. The nominal is used for all calculations. 130 ELEMENTARY MACHINE DRAWING AND DESIGN Open Enp AND OPEN Sipe Hancers (Sellers System). Fig. 107 and Fig. 108. Table for determining the lettered dimensions. sion=ad+@D+y in which d=the diameter of the shaft and D=the drop of the hanger; both in inches. Formula: Required dimen- Dimen-| 4 B ae Remarks. Dimen-| g | ¥ Remarks. sion. ins. slon. ins. A /1.6 0 |0.4 | Width opening inframe.....| a@ {1.5 |0.25] Diam. shaft box at middle B {1.6 0 |1.0. | Depth opening in frame.....| 6b 1.25]0.25} ditto at end Cc f1.0 0 |1.0 | Diam. adj. screw boss...| ¢ [1.0 |0.75 Cc’ 10.5 0 |8.0 | Depth of adj. screw boss... D |Givendrop}|.... | ............ d Given diam. of shaft E |0.19| 0 j0.5 | Thickness of frame web...| ¢ |0.95/0.5 F 1.0 0 (0.5 | Width of frame at hub...... t 0.9 |0.25} Diam. oil cellars G |0.19 | O |0.375} Diam. parting bolt......... f’ |=about + of (f) H (0.375) O 1.5 A’ 10.25] 0 |0.5 | ............ g =about 3 of (f) H” |d++4", Side opening in frame I 13J, End of hub to set screw 0.5 |1.25) Diam. of adj. screw (outside) J th, Diam. of set screw h' \=h—3” | Diam. of adj. screw at root of L (|4.0 |0.7 |0.8 | Leth. of foot thread M |1.0 [0.18 |2.0 | Width of foot zt |0.330.6 | Diam. of adj. screw (inside) N {1.25 | O |2.0 | Lgth. of foot pad......... O- {2.75 |0.7 |—0.2] C. to ec. lag se. Ll |=4d Lgth. of box P 0.4 0 (0.5 | Thick. of foot at lagse.....} m | =1+1" Leth. of oil pan Q same as H Thick. of foot} n |1.25|1.75| Width of oil pan FR 10.1 OF 1058 als aad sea A ave 0 = in Depth of oil pan S (0.6 O- Leds it yuateee tewese p | is optional thick. of oil pan S’ 10.3 0 (0.75 T |0.75 0 (2.25 U_ (0.19 | O {0.375] Lag se. diam. The pitch of the threads on U' |U+}%”, width of lag sc. hole the adjusting screws of Vv U+4", length of lag sc. hole the above Table is 4”’ for W | 3 width of hole=4U’ all sizes of hangers. X {0.4 |0.025/0.8 | Thick. of frame at foot Y 1.0 (0.05 [0.5 | Width of frame at foot Center Hole x 4 (B) Flange“ \ Foot \ Pad\|s— _G" |——| ES) 5 y e SS Gt | footlad t (at a pie ell, | pafBolt Boss, Front Rib Frame Web thes 4 Ai I ot Ee ‘< Ze, a3 yn \ x a7 / J" Frame Hanger Designed by William Mason (A) Winchester Repeating Arms Co. New Haven, Conn. Yoke Spin Bead: Fig. 109, 4 ed Drain Plug Clamping Screw (C) Boltlug Bearing Box (E) T J 2 ee ke A | cap dolty section atxx Zolt i Pieter ug" oe Yoke End- Yj Song zl UL [hatter ne ENT L ae oe : MH : UY AS > b “J 7 Adj Bol v Ady.Screw J a s aunmibn < Drip \Drs rugnion v Sl bac arid YokeArch Yoke Rib k Swivel Yoke Yoke Spindle i — mR] a4 Detail for Swivel Yoke and Bearing Box for J" Hanger (To face page 181) BEARINGS, JOURNALS, ETC. 131 TaBLE FoR DETERMINING THE LETTERED Dimensions or (J) HaNcErs, Fig. 109 Formula, The required dimension equals «4+, in which A is the diam. of the shaft in inches for which the hanger is adapted. Values of Values of Values of Dimen- Dimen- Dimen- sion. slon, sion. a . a < B a . 1 ins. ins. 1ns, B 1.1 | 5.4 || a | 0.06 | 0.4 n |0 3.38 B 1.0 3.5 a’ =a—-z% 0 0.05 0.35 Cc 1.0 4.0 a’ =a+3” Dp 0.125 | 0.31 H 0.5 | 3.0 al” =a+}" q 0.2 | 08 I 1.75 5.25 b =a r 0.25 1.13 J 1.25 0.62 c =a+}”" Ss 0.9 1.07 M 1.12 2.6 d 0.25 0.875 s! 0.9 2.2 P 1.5 5.75 a’ 0.5 0.75 t 0.06 1.04 Q 0.25 2.73 e 0.25 2.6 wu 0.12 1.0 E=H'+H+23f+1.65” || f 0.19 0.6 u’ 0.2 0.5 E’ =H +13f+0.82” f’ =f+8" ul” =17-—(u+u’) F(see Table 2 below) g =k+i" v 1.25 2.4 G=E'+3g+3" h =iF w =2J +3" G'=E' t =l w! =4J+q5A K=U+4+3" j 0.38 0.94 x 0.44 |] 0.71 K’=k k 0.5 1.25 y 0 1.75 L=4A I 0.5 1.75 |] y’ 0.2 | 0.3 N =0.82A +0.083D+4” i 0.7 1.8 z = 45 of width of frame O=1D Us =K+a H' (see Table 2 below) Ye = at = 35 of width of frame m | 0.125 | 0.3 TaBLe (2). Formula: The required dimension=eA+$+D(yA+8), in which A= diam. of shaft and D= the drop of the hanger, both in inches. Dimension. a 6B Y 8 F 0.75 2.87 0.031 0.078 H’ 5.1 0.6 —0.15 0.72 182 ELEMENTARY MACHINE DRAWING AND DESIGN The Table below gives the drops of hangers for various shaft diameters. 1 2 3 4 5 6 a Actual shaft diameter.....| x | 1x5 | 148 | 235 ; 228 | 3x5 | 338 y | 126 | 2% | 2H | 345 |..... 435 | 4h Drop of hanger, D........ a 8 b 10 ; 10 | 10 c 13 | 13 | 13 | 13 ] 138 e 16 | 16 | 16 | 16 | 16 |] 16 | 16 f 20 20 20 20 20 20 20 g 25 25 25 25 25 25 h 30 | 30 | 30 | 30 | 30 | 30 a ee 36 | 386 | 36 | 36 In assigning a size to be drawn if the column number is given as 3, the line x or y determines the diameter of shaft to use in column 3, and the letter a . . . 7, denotes the line to use for the drop. As, assignment 4ye would indicate a diameter of shaft of 333;’’ and a drop of 16”. Drop hangers are also used in an inverted position and are then called floor stands. They answer the same purpose as a pillow block but the distance from the foot support to the center of the shaft is greater. 110. Post hangers or post boxes are used to support a shaft from a vertical post or frame work. The framework of a post hanger may be designed in the same manner and have similar shapes as the frames of drop hangers. The adjusting screws, bearing boxes, etc., can be made like those of drop hangers. Fig. 110 shows the frame for a post hanger of the open side type. The open end type can be made to open at the end like the open end drop hanger. The U frame can also be adapted for post use, the swivel and bearing box being the same as the J hanger. A post hanger containing a ball bearing is shown in Fig. 1103. The extension from the face of the post support to tke center of the shaft varies as follows, reference being made to Table 2, Art. 113. Bearings 1 and 2 have extensions of 7” and 8”. Bearings 3, 4 and 5 have extensions of 8” and 9”. Bearings 6 and 7 have extensions of 9” and 103’. Bearings BEARINGS, JOURNALS, ETC. 133 8, 9 and 10 have extensions of 103” and 12’. Bearings 11, 12, 13 have extensions of 12’. The main dimensions of post hangers are given in a Table of Dimensions on page 136 referring to Fig. 1103. 111. A Wall Bracket is used when the distance from a vertical support to the center of the shaft is too great to allow a post hanger to be used. It consists of a casting in the form of + Past Hanger Sellers Type Fia. 110 a right angled triangle with a short leg vertical against the support and another leg horizontal supporting a pillow block. The bracket in Fig. 111 is taken from Spooner’s Mach. Const. and Design. It is a good illustration of the ordinary type of wall bracket. The values of A and Wi are regulated by the length and breadth of the pillow block supported. The unit for the dimensions is D+3/’.. D=the diameter of shaft supported by the pillow block. The pillow block dimensions can be found in Art. 104. 184 ELEMENTARY MACHINE DRAWING AND DESIGN | “Fia. 1104. BEARINGS, JOURNALS, ETC. 1385 Bolt Diameters =. 25 unit 42 La PN =F Ss N <— W, —> Yo Section on d-b apie it Hf 1 i} | |! \ WT ae din HI Wall Bracket Heir To Support Pillow Block..."Shaft 1) [i St || Design from H.J.Spooner | a x He Scale 641’ Mar:26-1912 ee Hyde I TT | i TLL é 4 PI J i ie 2a) y y Wipe a So] oI K T x A oo pg 20 Unit for proportions = D+ a Fia. 111. 136 ELEMENTARY MACHINE DRAWING AND DESIGN Main Dimensions oF Batt Bearinc Post HancEers Extension. A B Cc D E | F G H I No. Diam Inches. Bolts. | Bolts. 7 or8 8 6 3 2% | 5 7 5 | 3 1} 3 ? 8 or9 9+ 7i | 33:13 = !16 8 6 | 33 | 14 4 3 9 or 103 | 11 | 83 | 44 | 33/63] 93] 7]4 | 18] 4 3 103 or 12 12 93 | 53) 4 4 | 103 | 8 | 43 | 13 4 3 12 14 103 | 7 42 | 74 | 124) 9 | 43] 12 4 1 112. The bearings shown thus far have been of the plain type without special devices for automatic lubrication. In such bearings the oil is supplied from an oil cup in the top through an oil hole in the cap. Fig. 112. Self oiling devices include Chain Oilers, Ring Oilers, and Collar Oilers. The first two are similar in their principle of operation. A ring of metal or a flexible chain rests on and is rotated by the shaft. The chain or ring dips into a reservoir of oil below the shaft, lifts the oil to the top of the shaft, and allows it to flow over the bearing. See Fig. 112. Types from H. L. Caldwell. & Son Co., Chicago. Any surplus oil is removed by scrapers or wipers at each end of the bearing and returned to the oil reservior. These wipers are sometimes made of lead and pressed against the shaft by springs. The oil is thus used over again and there is not the need of constantly renewing the oil as in the case of plain oiled bearings. Collar oiled bearings are similar in principle to the ring oiled bearings, the ring being replaced by a small collar BEARINGS, JOURNALS, ETC. 137 fastened to the shaft. This dips into the oil reservoir and raises the oil to the top of the shaft. 113. There are other types of bearings for rotating shafts in which the shaft is supported on rollers or balls. The object of these bearings is to reduce the bearing friction and lessen the cost of maintenance, both tending towards increased efficiency in power transmission. The coefficient of friction for loads ranging from the maximum to half the allowable load, varies from 0.0018 to 0.0017. Fic. 1133. For smaller loads it is somewhat greater. Fig. 113 shows a hanger, made by the Hess-Bright Mfg. Co., which contains an annular ball bearing. Fig. 1134 is a view of a dissembled hanger and a post hanger. These hangers are designed for loads from 550 Ibs. to 10,000 Ibs. The ball races are curved with a radius of $ to An times that of the ball, according to the system of Prof. Stribeck. The safe working load on a ball for a two point bearing recom- mended by Prof. Stribeck is P=2100d? (d=diam. of ball). The PN . total load on a bearing of this kind is ee where W is the number of balls in the race and P the heaviest load on any one ball. 138 ELEMENTARY MAGHINE DRAWING AND DESIGN BEARINGS, JOURNALS, ETC. 139 TABLE X BALL-BEARING LINE SHAFT CEILING HANGERS Main Dimensions of Hangers, Standard Hanger ‘Feet’? Dimensions Dia.” wr aw wm te ae ae tt wa wt Ni " f Pattern No. |Drop’’| A B C D’ | £E Pr’ | @’ | Bolts. at 5465 | 8]15 | 4 | 12 13/1 | 92/33| 2 5 5466 | 10 | 153 | 42 | 122 14/1 | 92/33) 2 8 5467 | 12 | 164 | 43 | 13 13/1 | 93/33] 2 5 cue 5468 | 15 | 173 | 42 | 133 2 )1 | 9$/33] 2 | § 5469 | 18|18 |5 | 143 2 |1 | 92/33] 2 8 5470 | 21/19 | 54 | 154} 22/2 | 12 | 92/33] 4 $ 5471 | 24/20 | 52) 16 | 22]2 | 12] 92/38] 4 5 (5480 } 10] 18 | 53] 15 13. | 13 ]114/5 | 2 § 5481 | 12 | 193 | 53 | 16 2] 12 /113/5 | 2 5 5482 | 15 | 21 | 53) 173 | .. | 13] 12 }1d]5 | 2 8 Hanger | 5483 | 18] 222]6 | 192]... | 14] 12 ]11d/5 | 2 5 “B” 15484 | 21 | 244 | 62) 21 | 3 | 2 | 14 /11a]5 | 4 $ 5485 | 24 | 263 | 64 | 222 | 3 | 2 | 12 }11d]5 | 4 8 5486 | 27 | 282 | 63) 24 |3 [2 | 12 }11e]5 | 4 5 5487] 30/30 | 7 | 2523/3 |2 | 12 Jia]5 | 4 5 5495 | 12 | 22 2/19 | .. | 13] 18 }13815 | 2 a 24/6 | 193]... | 13 | 14 /133/5 | 2 3 Se? (ae os Not |. a ulisls | 2 | 4 _ 5498 | 21 | 274 | 7 | 222 | 32 | 22 | 12 ]132]5 | 4 a sanger.! 5499 | 24 | 29 | 74 | 24 | 34] 24 | 19 | 1383/5 | 4 3 c” | 5500 | 27 | 303 | 72 | 25 | 34 | 23 | 13 |13a/5 | 4 | 3 5501 | 30 | 324 | 84 | 27 | 33] 24) 12 |1384)5 | 4 8 5502 | 33 | 342 | 83 | 282 | 32 | 22 | 12 |132/5 | 4 8 5503 | 36136 |9 | 30 | 32) 22] 12/133/5 | 4 3 5419 | 15 | 25 | 73 | 21 | 341) 2 | 13 |168|7 | 4 x 5420 | 18 | 263 | 8 | 223 | 32) 2 | 13 |16817 | 4 z 5421 | 21 | 28 | 84) 233 | 32) 2 | 12 |168/7 | 4 z Hanger | 5422 | 24 | 293 | 84 | 243 | 33/2 | 12|168/7 | 4 1 “Dp” 15423 | 27 | 314 | 83] 26 |4 | 3 | 12 ]168/7 | 4 1 5424 | 30 | 323|9 | 2721/4 | 3 | 132 |168)7 | 4 z 5425 | 33 | 343 | 94 | 2823 | 4 | 3 | 12]168;/7 | 4 | F 5426 | 36/36 | 93) 30 |4 |3 | 12)168)7 ] 4 | F (5429 | 15 | 262 | 73 | 22 | 34/13 )2 J188]}7) 4 | 1 5430 | 18 | 272 | 8 | 234 | 32} 12] 2 |188)7 | 4 | 1 5431 | 21] 29 | 84) 244] 32])2 | 2 |188]7 ] 4 ] 1 Hanger | 5432 | 24 | 303 | 82 | 253 | 43 | 2 | 2 |183/7 ]) 4 1 «Re 45433 | 27/312 |9 | 262] 43]2 | 2 |188/7] 4 | 1 5434 | 30 | 322 | 93 | 272 | 43/3 | 2 |188]7 | 4 | 1 5435 | 33 | 342/93] 29 | 43/3 | 2 |188}7 ] 4 | 1 [5436 | 36) 36 [10 | 30 | 43/3 |2 |182)/7 | 4 | 1 140 ELEMENTARY MACHINE DRAWING AND DESIGN TaBLE Y Drop of Hanger (inches). No. | Brg. |itTbs| gen Sizela} b | ec | d|]e|fig|hi|tyk il 1} 48M| 850} A /8} 10 | 12] 15 | 18) 21 | 24 2 /13;M| 1100) A |8/ 10] 12 | 15} 18 } 21 |} 24 3 /14;M| 1750) B 10 | 12 | 15 | 18 | 21 | 24 | 27 | 30 4 144M); 2100} B }..}| 10 | 12 | 15 | 18 | 21 | 24 | 27 | 30 5 |138M| 2400) B }..| 10 | 12 | 15 | 18 | 21 | 24 | 27 | 30 6 |23;M| 3300] C |..| .. | 12 | 15 | 18 | 21 } 24 | 27 | 30 | 33 | 36 7 |27;M)| 4400) C }..| .. | 12 | 15 | 18 | 21 | 24 | 27 | 80 | 33 | 36 8 |23:M| 5000} D .. | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 9 |238M| 5700) D 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 10 /333;M| 6400) D 15 | 18 | 21 | 24 | 27 | 30 | 83 | 36 11 |33;M} 7700} E 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 12 38H4M| 8400) EF 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 13 )3481|10000| £ 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 8 2 Che D +e E # Outer tC Zz face Light Type Van. ® Nin i F ap pias ef} Se sy OF /nner ‘Race Medium Type Fic. 114. Hess-Bright Ball-Bearing Adapter. The ball races are usually made to hold from 8 to’20 balls. The inner race is fastened to a split bushing firmly clamped to the shaft. The outer race is free to move in an endwise direction. A detail of the bearing as used on line and jack shafts is shown BEARINGS, JOURNALS, ETC. 141 in Fig. 114. The dimensions of these bearings will be found in the Table Z below. The diameter of balls in these bearings will be found in this Table also. The number of balls in each one may be calculated from the formule in Art. 113. The principal dimensions of the hangers are given in the Tables X, Y, following Fig. 113. TaBLe Z DIMENSIONS OF ADAPTER BEARINGS FOR BOTH CEILING AND POST HANGERS M Bearings for General Work aes, | oA |] Bo | | Se ae |e) ae) |e | eae ins. ins. ins. ins. ins. ins. ins. | ins. ins. ins. ins. ing. 48 12.83] .81].375) .75| .5381)1.18] 33 33 t zs | 306 13; |3.14] .93] .375] .83]| .420|1.37 i 4 + 3 307 14, |3.93/1.31].375] .98| .379/1.77| 38 | 42] 52] & | 309 134 | 4.33 ]1.50| .3875 | 1.06 | .534|1.96 3 4 a # | 310 148 |4.72/1.62| .3875)1.14] .447 | 2.16 t a 6 4 311 28 |5.51|1.87| .375|1.30] .528/2.56| 43 £) 6h 2 | aia 275 |6.29|)2.12) .875)1.46| .509 | 2.95 3 3 3 1 315 2 16.69] 2.31| .3875)1.53) .595 | 3.15 - 8 z | lve | 316 218 |7.08|2.43] .875|1.61! .640 | 3.34 3 3 9 14 317 33 |7.48|2.43] .375)1.69| .654)3.54] 55 9 zs 13; | 318 375 | 8.64 | 2.62 | .375/ 1.85) .681 | 3.93 3 10 104 1% | 320 84: |8.85|2.81] .375)1.93 | .726 | 4.13 8 108 | 113 13 321 348 |9.44|2.87| .375)1.97 .718 | 4.33 4 1124} 112] 13 322 114. Roller bearings are used in place of plain bearings whenever the conditions demand less friction and the loads carried are very heavy. The frictional resistance is less than in plain bearings due to rolling contact. The rollers are separated at the ends by rings into which the ends of the rollers are fitted by turning them down. The shaft is fitted with a hardened steel sleeve carefully ground. A similar liner fits the casing of the bearing, the two forming a hard and smooth path for the rollers. The rollers themselves are made of tool steel hardened and ground. The dimensions of these rollers may be determined from the following formule taken from Spooner’s Machine Construction and Drawing. d=diam. of roller, D=diam. of 142 ELEMENTARY MACHINE DRAWING AND DESIGN shaft, d=0.08D+ 3” (nearest #” being taken) used for shaft diameters up to about 6”. The safe load per inch of the total length of rollers may=2000d? Ibs. assuming that one-third of the rollers support the load. Section atC-D Roller -8 RollerBearing for 3"Shaft D SectionatA yu 5z— a?NL ‘ W =33400 =e where W=the total safe load on a journal bearing in lbs. (with not less than six rollers) ; L=length of each roller in inches; N=number of rollers; Fra. 1158. BEARINGS, JOURNALS, ETC, 143 d=diam. of rollers in inches; S=linear velocity of convex bearing surface (sleeve) in feet per minute for values above 50 ft. per minute; D=diam. of shaft (or bore of sleeve). In Fig. 115 is shown a roller bearing pillow block. 115. The distance apart of bearings designed to support shafting, depends on the number and position of the pulleys and gears which take power from the shaft. The bearings must be close enough to prevent undue sagging or deflection of the shaft between them. This deflection is produced by the weight of the pulleys or gears, the pull on the belts, weight of shaft itself, ete. Pulleys are placed as close to bearings as possible. The amount of deflection between bearings depends on the rotative speed of the shaft. The critical speed in R.P.M. for a given deflection 8 is R.P.M.=200+/1+%, where 8 is in fraction of an inch. For shafts carrying a fair proportion of pulleys the span (S) in feet between hangers may be taken as S=6+/d, d=diameter of shaft in inches. If the shaft is used for transmission only the above distance may be increased 50%. For high speed unloaded shafting qd, s=1754/4 N=R.P.M.; d=shaft diam. in inches; S=span in feet (not to be exceeded). INSTRUCTIONS Bearing Boxes Before coming to class construct a scale to use for obtaining the lettered dimensions on bearing boxes from 1” to 5’. Use No. 1 paper and hard pencil. Lay off the values of a—-b-c—d-e-g-h—w above the base line, m and n below it. Mark all the inclined lines in the scale by their respect- ive letters. Scale ordinates are to be full size and the dimensions for the box you draw are to be taken from the scale with dividers. The 144 ELEMENTARY MACHINE DRAWING AND DESIGN dimension numerals finally placed on the completed bearing box drawing can be obtained by measurement on the drawing itself. To find the diameter of shaft to use in drawing the bearing box, take the following data: The vertical load carried by the box is W=(__) lbs. The length of box L=2S. The pressure per sq.in. of projected area is p=( ) lbs. The diam. S calculated from above datais ”. Plate 1, No. 3 paper. 2 hours allowed for drawing the views shown in Fig. 104. Exercise 2 on Plate 1. 2 hours allowed for drawing a side view of the box with the left hand half representing a vertical longitudinal section through the axis of the shaft. Use the method of oiling by (a) oil cup, (6) cast receptacle in cap. Give all dimensions which are indicated by letters or assumed by the draughtsman in the making of the drawing. Those who have a prime No. for their assignment No. may use oiling method (a), all others use method (b). Total av. time for this plate is 5.75 hours. Also make either: (c) a working drawing of the cap alone or (d) make a projection of the box when viewed at an angle of 45° with the horizontal. Place this view above and to the right of the front view by projecting from the front view with a 45° triangle. Name of this plate to be chosen by the student. TaBLE oF Data For BEariInGc Box PLATE No.| 1] 2 | 3] 41/5 | 6 | 7 | 8 | 9 |10}11/12] 13] 14/ 15) 16 W | 400 | 600 | 400 | 600 | 400 600 400 | 700 |500|/700/500)700}500)/700/500)700 Prob. 1. Calculate the strain per sq.in. in the threaded portion of the cap bolts if the load W is doubled and acts upward. Prob. 2. Calculate the horse-power lost in friction in your bearing box, if the shaft makes 200 R.P.M. and the value of » is taken as a eT c=constant obtained from experiment =.43. » =rubbing veloc. in ft. per sec. Prob. 3. Draw the cap of a pillow block for a 2” shaft. Give all dimensions. Prob. 4. Draw the base of a pillow block for a 23” shaft. Complete dimensions. Prob. 5. Draw the steps of a 3” pillow block. Fully dimension. BEARINGS, JOURNALS, ETC. 146 Hangers Open side hanger (Sellers System) (A). Open end hanger (Sellers System) (B). J hanger (C). The hanger to be drawn will be a (_ ) hanger (the line to use above being indicated in the table below). Drop=(_ ), shaft diam.=(_ ). It will be drawn either assembled or in detail, the views required being indicated under the respective heads of assembly and detail. Time allowed in class 10 hours. Av. time for O.E. or O.S8. hanger is 13 hours, for J hanger 11 hours, for ball bearing hanger 11 hours. ASSIGNMENT TABLE FOR HANGERS Assign. No. 1 2 3 / 4 5 6 7 8 9 |} 10) 11 , 12 Kind of Hanger..... 4 | B|A|B ;C|A|BICIlAIlB CIA Detailed.....] .. | .. | yes}|yes}] .. | .. | .. | yes] .. | yes Assembly....]| yes} yes] .. | .. | yes} yes |yes| .. | yes] .. | yes | yes Column No. in Text (Art. 109)..] 1 1 2 | 2 2 3 3 2 | 4 3 2 5 Line for diam. shaft.| 2 | y xily x x y y x x | y | 2 Line for D...| c¢ e e tf e f e e g | e f h ASSIGNMENT TABLE For Hancers (Cont.) | Assign No. | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 Kind of Hanger.....| B| A | B}]|C|]A|B)|C|A!]B/ClAIB Detailed.....] yes] .. | -. | -- ]yes}|yes}yes| .. | .- | -. | yes Assembly .. yes |yes}yes| .. | .. | -. | yes}|yes | yes] .. | yes Column No. in Text (Art. 109). .| 4 1 4 3 5 4 3 2 3 3 3 3 Line for diam. shaft y y x x x y y ¥ y x y zx LineforD...) e | f | f | f |g g | @ e|ftle gig 146 ELEMENTARY MACHINE DRAWING AND DESIGN Views of Hangers Hanger Assembled (1) Front elevation (looking at the end of the shaft) showing the right hand half by an outside view and the left hand half by a section of the box and part of adj. screws. (2) Side elevation, showing the left hand half in section. (3) Bottom view, half of drip dish removed. (4) Frame Sections. Cut the frame by one or more horizontal planes, also in the (J) hanger show a section of the yoke. A bill of material must be placed on the sheet of drawings (Art. 17). Hanger Detailed (1) Frame (including closing piece and its bolts). (a) Front view one-half (partly) in section. (b) Side view one-half (partly) in section. (c) Bottom views. (d) Horizontal cross-section of leg. (2) Apsustine Screw. (a) Half longitudinal section (half outside). (b) End view. (3) Bearinc Box. Upper and lower halves together. (a) Half lengthwise section of upper half and outside of lower half. (b) Top view. (c) End view, half in section (right or left hand side). (4) Drip Disa. Side and top and end views, sectioned where necessary. Following details in addition to above for (J) hanger only: (5) Yoxr. Front, side and bottom views and section of yoke. (6) Apsustine Screws (for yoke and for journal box). Detail drawings must be completely dimensioned, finished surfaces marked (f), material noted for each part, the number wanted of each and the name of each part printed beneath it, as well as its number. A bill of material must be placed on the drawing (Art. 17). The scale of the drawing must be such as to allow the assembled views to be drawn on a No. 3 sheet. The details can be drawn to various scales depending on the size of the part in question. Use No. 3 paper for details. In case the assignment number as posted on the bulletin board has a number in parenthesis above it the student will draw a detail of that part of the hanger which is opposite the number in parenthesis in the list above under hanger detailed, corresponding to the number above the assign- BEARINGS, JOURNALS, ETC. 147 ment No., viz.: ® means draw a detail of the bearing box for an open end hanger whose dimensions will be found by reference to the table in Art. 109 on page 132. Prob. 1. Instead of plain bearing, design ring oiled bearing. Post Hangers or Post Boxes Prob. 1. Design a post hanger for a (d)’’ shaft when the distance from the face of the post to center of shaft is (A). Use the open side type of U frame, as in Fig. 110, No. 3 paper. Prob. 2. Same as above, but using “J” type of frame and swivel yoke, No. 3 paper. ASSIGNMENT TABLE No. d A B Cc | D E F H mee ae i | ig 5 | 143 | 33 | é | | ag 2/18] 6/18 | 4 | oF | 15 | 3 a | z 3 | 1 | 8 | 233 | 53 | 7 | rR | 4 2 a a) 3s | 9 |) oe (se | |e la 2/2 B | 9 | 3 1 oe | @ | 8 | om | ae: 6 | sa | 1 | 29% | ef } of | at | ae | ae] 4 | Prob. 3. Draw the two views of a ball bearing post hanger shown in Fig. 1103, adding a top view. Put on all dimensions and a bill of material. Use No. 3 paper: ASSIGNMENT TABLE FoR B. B. Post HancEr Assign. No. 1 2/3 }4 {5 )6 1! 74 8 | 9 | 10 Extension....| 8 9 10% | 104 | 103 | 103 | 12 12 12 12 Line No. Table Y, Art. 113...) 2 3 5. 7 8 9 10 11 12 13 148 ELEMENTARY MACHINE DRAWING AND DESIGN Wall Bracket Design a wall bracket to hold a pillow block for a (A)” shaft when the distance from bracket support to center of shaft is (b)’”. The dimen- sions A and W may be obtained from Art. 104. Make side view, top view, and section on (db) also view from left side. Completely dimension. Use No. 3 paper. ASSIGNMENT TABLE FoR WALL BRACKET 10} 11 | 12 | 13 | 14 ag on o nr oo © No. | 1 2 3 27| 2 hb nw nw Nie Ww w ren no wo rr ow w lH rN ie nw leo ow > ray Ball Bearing Hanger Design a ball bearing hanger for the diameter of shaft ( )’’ and drop ( ”) as given in Table Y, Art. 113, in line ( ) and column (_ ). The dimensions of adapter bearings for this hanger will be found in Table Z, Art. 113, for the proper size of shaft. The number of balls must be calculated from the diameter given and load on bearing. Make a front view (A), left hand in section, a bottom view and a side view (B) left hand in section. Put on all the dimensions and make a bill of material on the drawing. Use No. 3 paper. ASSIGNMENT TABLE For B. B. Hancers (Table Y, Art. 113) Assign. No. | 1] 2/3]4/!5;/6/ 7) 8] 9 |10)11)12)13]14] 15 | 16 Line No...) 2] 2]2]4;4/4]6/6]6/8]8}|8]10]10] 10 | 12 Column...Jb}e/gilejel/gle|flhid\|flhlielg | t g . If the assignment number is followed by a (+) take the odd numbered line in the hanger table next in order down the page below the line number given in assignment table, but keep the same column letter as given in the assignment table. BEARINGS, JOURNALS, ETC. 149 Roller Bearing Draw the roller bearing pillow block shown in Fig. 115, and calculate the load it will carry safely at 150 R.P.M. of shaft. No. 3 paper. Prob. 1. Draw a detail of the... of the roller bearing pillow block of Fig. 115. No. 2 paper. Prob. 2. (a) Design a roller bearing pillow block to support a 23” shaft which carries a load of 4400 lbs. No. 3 paper. (b) Same when shaft diam. =3” and load =5700 lbs. (c) Same when shaft diam, =22’’ and load = 2650 lbs. Prob. 3. Calculate the distance apart of the hangers for a (a) 23” shaft. (b) 3” shaft. (c) 33” shaft. (d) 2}” shaft. CHAPTER X PISTON AND PISTON RODS 116. A Piston is a cylindrical body fitted within a hollow cylinder so that it can slide in it under pressure and still prevent the escape of the fluid which presses against it. Examples may be found in steam engines, steam pumps, and gas engines. The piston is ordinarily fastened to a piston rod which passes through a stuffing box in the end of the cylinder in which it slides. If the piston is provided with valves which allow the fluid to pass through it during one of its strokes, it is called a bucket. When the diameter of the piston is reduced to the diameter of the piston rod or when the piston rod diameter is increased to that of the piston we have what is called a plunger. A piston which receives pressure on one side only and which carries a pin passing through one end of the connecting rod is called a trunk piston. They are commonly found on internal combustion engines. 117. A plain piston, however well fitted at first, soon would become leaky. To prevent this, it is customary to fit packings to pistons in such a manner that the piston is kept from rubbing against the walls of the cylinder, the wear coming on the packing * alone. , A piston should be designed and constructed in such a way that it is sufficiently strong, keeps tight, has few parts, its bolts and nuts are securely fastened, and it works with as little friction as possible. The simplest piston packing is that which consists of metal rings which are sprung into grooves turned in the circumference of the piston and called Ramsbottom rings. These rings, before springing into place, are slightly larger in diameter than the inside of the cylinder in which the piston is to fit, so that 150 PISTON AND PISTON RODS 151 their tendency to spring back when in place, presses them against the inside surface of the cylinder and makes a tight joint. They are slotted across their circumference to spring them into place. Fig. 117 shows several cross-sections of pistons in contact with a cylinder wall showing the rings in place. The width of Fia. 117.—Piston Packing Rings. the rings parallel to the axis of the cylinder may be taken as T=0.03D to 0.06D and their thickness as W=0.025D to 0.03D where D=diameter of piston. The disadvantage of the spring rings is that it is necessary to remove the piston from the cylinder in order to get at or replace the rings and they cannot be used on large pistons. oT: KARA Ve a a 4 ON“ WY (8) aS 4 NY: 4 2 Fig. 118.—Piston Rings. To overcome these objections the piston is made in two parts as shown in Fig. 118, (2) is the piston body and (1) the follower plate or junk ring. The rings (3) are backed up by other rings 152 ELEMENTARY MACHINE DRAWING AND DESIGN as (4) which press them against the cylinder wall. In (A) the follower plate is held against the piston body by the nut on the end of the piston rod. At (B) the follower plate is held on by several tap bolts (5). The follower plate is often called the junk ring. & | oe pry = y [nN sa a eo fe] Fie. 119.—Piston Ring Joints. 118. When a single spring ring is used, the joint in the ring, which is necessary to allow it to be sprung into place, must be designed to prevent leakage of steam. A common form of joint Gr WY Uy > EET) €) Fic. 120.—Connections of Piston Rods to Pistons. is shown at (A), Fig. 119, where (1) is a tongue fastened to one end of the ring and sliding in a slot in the other end. FISTON AND PISTON RODS 153 At (B) is shown the method employed when there are two rings in the packing space. (C) is still another method commonly used. (D) shows the joint for a small ring. 119. Pistons are fastened to piston rods in a number of ways, the most important being shown in Fig. 120. (A) is used on small pistons when it is not necessary to remove the piston from the rod. The taper of the conical part in (B) varies from 1 in 4 to 1 in 7. (D) is used when the follower plate is held on by the nut of the piston rod. (£) is used on conical steel pistons. 120. Pistons are made either of cast iron or of cast steel, pressed steel, or forged steel. Where lightness is not of special 7) D =o] : IN WEZ NSS NIES } At Z = Me {| > x Z BY! | yp | , YA 4 HE: ES LE, t L HAL Fig. 121.—C. I. Piston. importance they are made of cast iron and are cylindrical in form. When lightness is important they are made of cast steel and are conical in shape, as this gives greater strength and rigidity, with 30 to 35% less weight. Piston rods are made of steel and their strength is calculated at the point of least sectional area, usually at the bottom of the thread. Finer threads than the U. S. Standard are often used in order to keep the diameter of the rod down. 121. The proportions of pistons vary, depending on the class of work and kind of piston. The dimensions of a piston of cast steel, conical in shape, will differ very much from the dimensions of a cast iron piston for a steam pump. In Fig. 121 is shown a form of C. I. piston used on cylinders from 6” to 16” diameter and having the dimensions given in the accompanying Table from Haeder and Powell’s Hand Book on the Steam Engine. 154 ELEMENTARY MACHINE DRAWING AND DESIGN In this type of piston the junk ring is a sort of cover held in place by the nut on the piston rod. TaBLF For C. I. Pistons up To 16” Dram. Diers. b c d e f g h t k l Cyl. 6 | 3 3 13 3 rc te gw | 13 | We fl 8 33 se 13 3 3 3 ve | 13 | 2 li 10 38 8 13 Ped 7 s | 1g | 28 | 1a 12 | 4 | 2 3 a 2 e [la a | 1g 14 | 43 i 24 é te 3 g | 23 z | 2 16 | 43 a 23 is t 3 & | 2% | 33 | 23 Cast steel pistons conical in form are used in marine engines largely. The piston shown in Fig. 122 and the proportions KF ecco gL WT _ K —>=./9010.22D aLead 4 f be dz —> i oH —1 9. 7010.80; +=} ry De Diam. HP Cylinder és K WP. "me 7 Unit for other hii wi ove © S& =0T02+5 (See yeu?) S'S 100 Edie Fig. 122.—Cast Steel Piston. given are taken from D. A. Low’s pocket book and are based on examples from triple expansion marine engines. The unit for proportions is pve where D is the diam. of high pressure cylinder in inches and P the boiler pressure in Ibs. per sq.in. above atmosphere. K=.19D to .22D; d, = .75d2. PISTON AND PISTON RODS 155 Calculate dg for the H.P. cyl. and use on the others D,=.7D2 to 8D; L=.039VD+ 35", G=.017D+3". unit = PVP for the following dimensions: A=.48 unit for H.P. piston; C=.33 unit for H.P.; A=.54 unit for IP. piston; C =.34 unit for LP.; A=.64 unit for L.P. piston; C=.38 unit for L.P.; B=1.8 to3.1;\ av.=2.2; F=.74 unit. E=3.8 to 5.4 unit av. =4.6, or such as will make the sloping part of the L.P. piston inclined at 20°. H=1.5 to 2.7, av.=1.7 unit. The boss may be threaded for a short distance or made with a flange to facilitate the removal of the piston from the cylinder as shown by the dotted lines at x in Fig. 122. In a compound or triple expansion engine B, FE, F, and H are the same for all pistons. The diameter of the junk ring bolts may be am a(2xvoti) +1", p in this case is taken as the effective steam pressure, that is, half the boiler pressure for high pressure pistons, quarter boiler pressure for I.P. pistons and the boiler pressure divided by the ratio of low pressure to high pressure piston diameters for L.P. pistons. D=the diameter of the piston. The pitch of these bolts is from 5d to 10d. They may be tap bolts with heads set in a recess in the junk ring, or pro- jecting from the upper surface of the ring. Bolts of this kind are usually locked in some manner to prevent working loose and dropping off into the cylinder. The nuts which are screwed into the piston body to receive the junk ring bolts are kept from turning by taper pin keys. (See Art. 82, Fig. 83 (H).) 122. Piston Rods are made of wrought iron, steel and bronze, their length depending on the stroke of the piston, thickness 156 ELEMENTARY MACHINE DRAWING AND DESIGN of cylinder end, stuffing box depth and kind of crosshead used to guide the end of the rod which works outside the cylinder. The calculation of the diameter (d) of the rod depends on the unbalanced pressure on the piston, the diameter of the piston and the stress per sq.in. allowed.in the metal of the rod. Some writers treat the rod as a long column, but as most rods have a length less than 10d, the general method appears to be to use a moderate working stress and calculate the area of cross- section from that. The total pressure on the piston in a cylinder whose diameter is D” will be the area of the end of the piston times the pressure pD?x 4 of a piston rod to the pressure on it, due to the piston pressure, per sq.in. (p) onit. This can be expressed as The resistance 2, will be eh where d=the rod diameter and f=the strain per sq.in. which the metal is seagil to receive. mane the load : : pD?r _ to the resistance gives 4 = ae from which d=D - Since the stress in the rod alternates from a pull to a push or from tension to compression, the value of (f) lies between 2000 and 4500, the smaller values being used for long stroke engines. For marine engines using Siemens-Martin steel rods the value of (f) varies from 2500 in freight steamers to 7000 in small cruisers and torpedo boat destroyers. The maximum stress in the threaded end of the rod must not exceed 4500 in the first case to 13,000 in the last. 123. The length of a piston rod cannot be given arbitrarily on account of the many varied conditions which govern it. If the distance from the under side of the piston to the center of the crosshead pin is given, then the manner of connecting it to both crosshead and piston will determine the shape at each end as well as its length over all. Some of the ways of connecting the piston rod to the piston are shown in Fig. 120. The method of connecting the rod to a crosshead will be found in the chapter following this. 124. Pistons used for internal combustion engines are designed to receive pressure on one end only, and as they take the place of piston rod and crosshead they are open at the end opposite PISTON AND PISTON RODS 157 the pressure end, and contain a wrist pin on which the con- necting rod oscillates. They are made of cast iron or steel, turned slightly smaller at the closed end to allow for expansion when heated by the burning gases and fitted with several Ramsbottom rings to insure tightness during the working stroke. The wrist pin is placed near the center of length of the piston and is supported by bosses cast on the inside of the piston. The ~—— M ——-> Inoik be F +E te ‘ F aS r N Bae IZIOLT = Side N87 : OF V¥e. i aN { G ar Web, por >| | 8 ssi ts s 53L—> - dD DR L a 8 4-by4ty ead a = Fic. 123.—Trunk Piston for Gasolene Engine. length of the piston varies from L=D in small engines to L=1.5D. Fig. 123 shows a piston for an engine of the internal combustion type using gasoline as fuel. The dimensions can be obtained from the following formule from Louis Lacoin’s book on Auto- mobile Motors where L=D: A=.05D; B=.05D; h=0.035D; E=0.05D; K=0.05D; F=0.04D; M=.A4D; d=.2D ; G=.3D; J=i1D; N=.1D; C=.023D. 158 ELEMENTARY MACHINE DRAWING AND DESIGN For gas engine pistons the length L is greater than D, being equal to 1.5D to 2D. Other dimensions differ from those given above for gasolene engine pistons. The thickness of the flat end of the piston may be taken as .06D+2’’, thickness of sides=.06D+ 4”, the length of wrist pin between supporting bosses=.5D. The diameter of supporting bosses=1.5d+4’.. The diameter of wrist pin for small engines using a max. piston pressure of 350 lbs. per sq.in.=0.16D light to 0.28D ee 450 lbs. per sq.in.=0.21D 8 to 0.3D a To give strength to the piston it should be ribbed on the inside by ribs at right angles extending across the flat end and down the sides nearly to the wrist pin; 5 to 7 Ramsbottom rings are used for packing, the thickness being S and width 2” to 3” for small engines, 3’”’ to #” for large ones. 125. The weight of a piston in fast running engines has such a great influence on the crank pin pressure that pistons of this kind are made as light as possible. Pressed steel pistons for aeroplane motors are made weighing 1.87 lbs. for a bore of 3.93” (including the packing rings and wrist pin). As an exercise in estimating weights it is recommended to calculate the weight of one of the foregoing examples of pistons having in view the calculation of weight of all the reciprocating parts of some particular engine for which this piston is suitable. INSTRUCTIONS Pistons and Piston Rods No. 3 paper. Cast iron pistons, 2 hours allowed in class. Av. time required, 24 hours. Draw a side view (half in section) and an end view from opposite end from piston rod of an (a)” diam. piston suitable for a steam engine cylinder. Use the kind of piston packing rings shown in (c) =Fig. 117 (A), (d) =Fig. 118 (A), (e) =Fig. 118 (B), using the dimensions given in the table for Fig. 121 as far as possible. When the packing of Fig. 117 is used the junk ring will be omitted and the piston will be hollow with a hole in the top for removing the sand core. The hole is to be threaded and fitted with a plug after- PISTON AND PISTON RODS 159 wards. Dot in piston rod. Give all dimensions. If Fig. 118 (B) is used, the nut of the piston rod will press against the top of the hollow piston. The junk ring bolt diameter can be figured from the formula in Art. 121, assuming the boiler pressure as (P) lbs. Calculate the pressure per sq.in. on the piston rod. Dot in invisible lines in end view. Cast Steel Pistons No. 3 paper. 2 hours allowed in class. Av. time required is 32 hours. Draw a side view (half in section by plane through axis of piston rod) and an end view of a cast steel piston. Use Fig. 122 as a basis for drawing this piston. The piston is to be (a)” diam. for a (b) H.P., (c) LP., (d) L.P., cylinder for a (e) compound, (f) triple exp. steam engine using (g) lbs. per sq.in. in the H.P. cylinder. The diameter of the cylinders of this engine areH.P.=( )”,1LP.=( )”,L.P.=( )”. (In a compound the I.P. cylinder is omitted.) Show the piston rod and nut by dotted lines and give all dimensions on the drawing. The boss around the piston rod is to be provided with a flange on top for lifting the piston from the cylinder. The packing ring is to be arranged as shown in Fig. 119A to prevent leakage past the joint. The steam pressure for triple expansion engine work varies from 160 to 180 lbs. per sq.in. above atmosphere and for compounds from 120 to 150 lbs. Calculate the pressure per unit area of piston rod at root of thread and in the body of rod. Calculate all dimensions before class. Dot invisible lines in end view and give name of piston, for what cyl., and kind of engine, Internal Combustion Engine Piston No. 2 paper. (A). 2 hours allowed in class. Av. time required is 3 hours. Draw a side view (half in section by a longitudinal cutting plane) and a half end view of the open end of a (a)” piston for an internal combustion engine. Show the Ramsbottom rings in place and the wrist pin with the set screw. Use the proportions given in Fig. 123. Give all the dimensions on the drawing. Calculate all dimensions before class. (B) Same views as above in (A), but diam. of piston is (b)” and designed for gas engine work when the max. piston pressure is (c) lbs. Omit the wrist pin in the drawing. Make the length of piston equal to 1.5 its diameter. Put in all invisible lines in end view and any necessary dimensions. Give size of engine in title. 160 ELEMENTARY MACHINE DRAWING AND DESIGN ASSIGNMENT TABLE FoR PISTONS Cast Iron Pistons (Steam Engines) “ 1 2 3 4 5 6 7 8 9 | 10] 11 | 12 | 13; 14 a 6 8 | 10; 12 | 14} 16] 6 8 | 10] 12] 14 | 16 | 10 | 12 ec |d|e|d|e}ejd|ejle}te}d|c}ldje P ; 90 | 110 | 100 | 90 | 110 | 100 | 115 | 120 | 130 | 110 | 100 | 120 | 120 | 100 Cast Steel Pistons (Marine Engines) a@110)16/ 9 |143]223] 10 | 16 | 25 | 12 | 224/183} ]153] 22 | 15 b | d blel}d]ob e|dj;bj|]di/ib ble b eee er Pa Pe pce Pee tees bee op ae H.P.| 10/10) 9 9 9 | 10 10 | 10 | 12 | 12 |184}153] 13 | 15 1p. | 143 | 143 }142; 16 | 16 | 16 22 | 244 LP| 16 | 16 | 223 | 223 | 223 | 25 | 25 | 25 | 221] 223 | 332 | 283 | 333 | 33 a 120 | 120 | 180 | 180 | 180 | 180 | 180 | 180 | 125 | 125 | 150 | 150 | 200 | 180 Internal Combustion Engine (A) (Gasoline), (B) Gas a| 4 | 43) 49) 5 | 5) ) 53 | 53) 6 | 3¢ | 43) 48 | 42 | 5g | 48 b 8 |; 10/12] 14) 7 9 |; 11] 138) 15 | 8 |} 10 | 12 |] 15 | 16 c | 350 | 350 | 350 | 350 | 350 | 350 | 350 ; 350 350) 450 | 450 | 450 | 450 | 450 Prob. 1. Calculate the stress per sq.in. in the piston rod which fits your piston for the section at root of thread and for the main part of the rod. Prob. 2. Calculate the pressure per inch of projected area on the wrist pin on (a) the part within the bosses and (6) the part between the bosses when the pressure on the piston is 350 lbs. per sq.in. Prob. 3. Calculate the weight of your piston without the piston rod (steam engine). Prob. 4. Calculate the weight of an internal combustion engine piston with rings and wrist pin (gas and gasoline engine), CHAPTER XI CROSSHEADS 126. A cRossHEAD is that part of a machine which moves in a straight line, with a reciprocating motion, and carries the cylindrical pin on which oscillates one end of a connecting rod. Steam engines, power pumps, and air compressors are some of the most common machines using crossheads. Fig. 124 shows the principal parts of a horizontal steam engine: (1) piston, (2) piston rod, (3) crosshead, (4) upper guide 3 10 5 RF \2 Fic. 124.—Horizontal Steam Engine. bar, (5) wrist pin, (6) connecting rod, (7) crank pin, (8) crank, (9) erank shaft, (10) lower guide, (11) cylinder, (12) bed. The crosshead is provided with one or more surfaces which slide in contact with constraining surfaces called guides. There may be one, two or four guides, the number serving as an indication of the class to which the crosshead belongs. Stationary engines have two or four guides, while marine engines are usually fitted with slipper guides which comprise two guiding surfaces. The method of fixing the wrist pin in the crosshead and the kind of wrist pin used also help to classify crossheads. In engines of the type which run “ over” (in the direction indicated in Fig. 124), the lower guide bar takes all the down- ward pressure due to the steam acting on the piston. There is so little pressure ever coming on the upper guide that it is 161 162 ELEMENTARY MACHINE DRAWING AND DESIGN negligible. The area of the surface of the crosshead in contact with the guides during the forward stroke (away from the cylinder) can be calculated if the following are known; the pressure (R) and the pressure per sq.in (p1) which the surface is per- mitted to carry. This latter pressure varies (according to the speed of the crosshead), from 25-100 on stationary engines to 55-120 on marine engines. 127. The calculation of the maximum force (f) acting on the guides depends on the force Q, the length of the connecting rod, the radius of the crank, the diameter of piston, and the steam pressure on the piston. The radius of the crank is, of course, half the stroke. Let P=steam pressure per sq.in.; D=diameter of piston in ins.; L=length of connecting rod in terms of the stroke as nl(l=stroke) ; Q= px The position of the connecting rod which gives the maximum value of (R) is that which occurs when the crank is perpendicular to the line of motion of crosshead, ost as shown in Fig. 125. The forces ep yiz-r? acting at the wrist pin are marked Q, R, S, and their values are al oe determined by the relation of the Fic. 125. sides of the triangle which is called the force diagram. That is Q: R= VLI?—? : 1, or Qr=RV L272, but L=nl and rad. Therefore Q_p fine? 3 =R,| (ni) —— Since PrD? Q= 4” CROSSHEADS 163 we have PrD*l _ x aaa 8 = Ry |r? -y therefore PrD? ono _ Prd? 4 =RV4ni —1 and ie age This value of (R)+by the value of (p1) used, depending on the speed of the crosshead, will give the area of the surface of the crosshead in contact with the guide bar below. If the ratio of the length to the breadth of this surface is assumed, the breadth can be calculated. ase Oil Grooves, | We ALLS ZAIN oS)’ C WS ¢ ) F 3 iY Tat vere ” oy pi ay | yf K | AWE | Ae A Ee RT = odd ab : aes ( i pet) | 4 | Heaton Sp | 7 tar & [el MW Li \\ yi | eh | i) \\ 1 eS LI ieee =] STP nae 4 ‘ \ | “4 I | Section on-xy- f ! oles Piston kod a Crosshead le Fic, 126, Top View In Fig. 126 is shown a crosshead for a small horizontal high speed engine. There are four guides, two above and two below. The bearing surfaces are planes whose total area is 8m?. The ratio of length to breadth in this case is 4:1. If we calculate 164 ELEMENTARY MACHINE DRAWING AND DESIGN (R) and assume a value of (p1) we shall have 8m? > from which (m) is easily obtained. 128. The piston rod connection to the crosshead is made either by threading the rod and screwing it into the end of a cylindrical boss on one end of the crosshead (Fig. 126) or by means of a cotter passing through rod and boss. In some cases the piston rod and part of the crosshead are made of one piece, but this is found on small engines as a rule or on engines with slipper crossheads. See Fig. (129). The proportions of these connections will be found in the various figures. 129. The wrist pin or crosshead pin, or gudgeon pin as the pin is called, which holds the connecting rod to the crosshead, may be made in either of the ways shown in Fig, 127, The 1 vb os Lal | "3 PAs (A) (B Fic. 127.—Wrist Pins. he SS — pin at (A) is designed to receive the thrust of the connecting rod between the supports (C) and (D) while the pin at (B) projects from a central support (#) and takes the thrust on both sides. In the latter case the connecting rod end must be forked. The diameter of the pin (d) depends on four things, the length (1), the greatest load on the pin, the allowable stress of the material, and the pressure per sq.in. allowed on the bearing surface. If we call (T) the load (f) the stress allowed in the material, (2) the length of the pin and (d) the diam., the following formula will express the diameter of the pin on the basis of strength: 1.273T1 oa d= ae: 27371 CROSSHEADS 165 or if the ratio of I to d is fixed as 5, then Lae 12er 4 fd f d’ L23F fi d=, ete ie aman If the bearing surface is to be considered, the following requirements must be met: pXlXd=T, where p=pressure per sq.in. allowed on the bearing. (See Art. 100 and Table 16.) d? If l=ad where x is assumed, then ane which is the expression generally used for pin diam. If the strength is taken into account and we substitute =z in the equation for (d) above we get, d=+/ 7 a This gives /=about 2d to 2.5d according to the values assumed for f and p. This, however, is greater than the values convenient to use for this purpose. Values of (x) used in practice vary from 1 to 2. If the pin is wrought iron the value of f is 4500-5000, while for steel it may be taken from 4500-6500. The wrist pin dimensions used in trunk pistons for gas engine work will be found in Art. 124. The value of T may be obtained from the value of Q given in Art. 127 by multiplying it by the secant of the maximum angle made by the connecting rod with the line of the piston rod called 6, or Q sec. 6=7. 6 can be determined if the lengths of connecting rod and crank are known. (See Fig. 125.) 130. A type of crosshead used on an engine with two guide bars is shown in Fig. 128. The guides are bored out to an inside diameter of D’’, two shoes are fitted to the crosshead and turned to the same diameter as the guides. The piston rod is fastened to the crosshead by screwing it in. In Fig. 129 is shown a marine type of slipper crosshead taken from Marine Engineering, by Hermann Wilda, in which the piston rod and head are in one forging. The brasses are round in the upper half and square in the lower and are lined with white metal. 166 ELEMENTARY MACHINE DRAWING AND DESIGN 0% X 02-61 al Tj) 9 | St | 8@ | et] $2] ¢ |) FY] G] ee) fr) & | er; & | & | & | o2xoz-er SIX 8I-L1 T| = | €¢ |#er ) e@/ ett] ge | fe} fe | ep] felt] Ss] ty) SPS] ee] ostxet-zt 9IX 9I-ST i ¢ | ip | ot | ' | e0r] © | | fe] pl tr} er] ¢ | et % | ¢ | € |] 9LX9I-st FLX FI-SIT t | € | |%or | | 6 | & | ge | &) $e] St [Mr] 2] Ee | SE] ) Ee | PTX PFI-er ZIX €I-ZI-11 s | € |e | 6 | #]#2 ) 42} te] 22) e@] tr] tr] ) 8 | & | & |S) erxet-er-ait OTXOI-6 fe | ee € '%2 | #1 !8o | fr! $ oo} dr} | SE | St | a] | (OLXOT-6 8X 8-2 £.8 |¢¢ | a lat |: arlazi{ar| c/2 |e] ei |] te] ea] exes LS | YO} dO|NW| TM ier HO|d¥|\|doO\qv| Z| A) X\|M) Al] al oS “srapulpAg *srapulfAd jo 8aZ1g jo BIZIS *g10}49'T 04} JO san[eA 0% X 02-61 t | 6 |S | | 2 eet} $9] f9 | ot] fn] 9) $¢ | 48] 82] St] Zor] st] ¢| ozxXoz-6r SIX8I-LT g St | #% | $9 | ézt| $9 | $9 | Zor] Fr | go | Fr | 82 | So] For] $6 | fOr] #7 | SIXS8I-ZT 91 X 9T-ST | #3 | #9 | P11] % ¢|36 | #1 | #h | P| #9] ES] 1/88 | eet] F] OLX9I-CT FLX#FI-EST g } | 20 | ip |te | tp] dp) es |] ti] tps 8 9| SlEcL} tL | Est} #8 | FIX PFI-€L GIXEI-ZI-11_ | & } | 81) #7188 | fy | $e [32 | & | $e] fe) $9 | 8th] TTF Il] €| &tXI-eI-11t OLX 01-6 & |eb | t | et | te | tz | $e] $e | $9] & | $e) 8a | $6 | Mel te | ee | %6 | $z | OLXOT-6 8X8-L Sf | #1) 8% | to | 2 | #2 | ¢ | 8 | 80 | 82] fe) ta} $2 lap |e z 8X8-L S)/Y|/O;/o|N|wiT]Mlerli)/a#lpalalalaloalaly ‘srapulyAD J | ‘sapuryAD JO 89ZIg jo 89ZIg *6109}9'T 94} JO san[eA ‘urbs sad ‘sqy Sz jo alnsseIg Wag SUIYIOM & IO} porns (821 ‘BZ) T ‘ON ‘dVaHSSOUD TAINO qduOg YO SNOISNAWIQ] JO ATAV, "I ‘ON USIsaq peezy ssol.y—"QzT ‘OI \ bubu04 jays - Utd 4Siuy DUIpudd xg fe l dae veGer= pla a ee ot Epp) pe fae tlae) ee --— 2 bre rea e aps oe vPro 1 -- 7 & | s ' ' | a it as ei: q= === x | | | A g : _| ! f Il ee ies or - SAD | ; | ' LSP t ee ea eel a Vy | | : = Ve Z! hod V1 Es, = oy ! ye ape Reet chard ef LAT NY aga de I eee ee reek + UOH {$B. sa ae St Oe Wages SES 4522 Fay - re ~-f---\,---1----}----b--L,4.--T-4 el ‘ ua K g i NW pokoor------ -SSBIG+ — ---- 5-H (ep ie -Frt---- pos SSS aoa a= ro ++ iy ! 1 YO 20-TH. i Buiyso)' (94S L * Ty / wT, / = 4 jf | ‘ Tt {of we Se Rec at 0 perc x é wt 1 f wi AN ' youl sedspusgy | 155 - : -4 eae é | : See \ 4 & 1 x \ Qs 7 ey av a ~ \ “A et 1 fhe ee Le ree ee rhe XBR WB CLE. Ze Zz LE DEL Kz LL “Vn ZZ SELLE Lae “Zz AE 168 ELEMENTARY MACHINE DRAWING AND DESIGN The proportions given are in terms of (d) the diameter of the crosshead pin. This is for engines of large power. m=not less than 1.6d, w=.4 to .5d, T =.5d to .55d. Tig. 129.—Slipper Crosshead. diam. of bolts determined from f =4300 to 5000 Ibs. per sq.in. for W.I. = 6400 to 7100 lbs. per sq.in. for steel. Distance between bolt centers = 2d. Make the body of the bolts the same diameter as the root of thread of the bolts, but use a finer pitch thread than the standard. CROSSHEADS 169 Thickness of brasses 8 =.03d+3,”. H=3.5d; to 4.5dz; C=.98” to 32”; g=22" to 53”; e=.6" to 12”; So = .4" to 3”; l=d to 2d. B=2.5d: to 2.9de; Calculate piston rod diameter for 5000 lbs. per sq.in. when load on rod is half that given as per table of assignments. = P. d,y=1.07D 7, p=unbalanced working steam pressure in lbs. per sq.in.; D=diam. of cyl. of engine; k,=4300 to 7000 Ibs. per sq.in. INSTRUCTIONS Crossheads Example 1. No. 3 paper; 4 hours allowed in class. Av. time required is 4.35 hours. Draw a crosshead like the one shown in Fig. 126, adding to the views there shown a third view looking at the top view from the end containing the piston rod. Calculate the value of (m) when the steam pressure in the cyl. is 100 lbs. per sqin. Engine cyl. is 93’ diam. Stroke of engine=12”. Conn. rod length=3xstroke. Max. Pressure on bearing surfaces of crosshead is 25 lbs. per sq.in. Calculate (A) and (B) when B=1.24A and the max. bearing pressure on the pin is to be 1180 lbs. per inch of projected area. C' is to be made large enough to allow the strap of the connecting rod to clear by 3”. (See Fig. 137.) The oil grooves may be taken not larger than 35” wide witha semi- circular cross-section. EE may be calculated from the diam. of the engine cyl. and steam press. by allowing a tensile strain of 4700 lbs. per sq.in. at the root of thread and using 8 threads per in. on the threaded portion. The locknut is a standard nut across flats and the depth of the hex. part is standard. The sleeve of the nut protects the threads on the 170 ELEMENTARY MACHINE DRAWING AND DESIGN E piston rod and is made about 2 high above the hex. The other end of the piston rod is to be shown, the connection to the piston being of the kind shown in Fig. 120 (A). Thickness of piston=4}”. It is advisable to break the rod between the crosshead and piston and omit the long cylindrical portion on the drawing. The total length can be indicated by a dimension line only. Calculate the strain on the cross- head pin, using the formula in Art. 129. Pl. can be traced in 3.25 hours. Av. time required is 6 hours for complete plate. Prob. 1. No. 3 paper. 4 hours allowed in class. Av. time required is 6 hours. Draw to scale and dimension carefully the crosshead shown in Fig. 128, 3 views; bottom view to be a half sec. above center line. This will be calculated for an engine (Bore= ), Stroke= . Length of connecting rod=3 Xstroke. Steam at 125 lbs. pressure. Find the pressure per inch of bearing surface of shoe. Calculate the diam. at root of threads on piston rod for a unit stress =4500 lbs. per sq.in. Use 10 threads per inch for threading piston rod. Find the press. per sq.in. of projected area of wrist pin. Calculate the unit stress in the wrist pin from the formule in Art. 129. Make a note on the drawing giving all the above information, as Steam pressure in cyl. = 125 lbs. per sq.in. Max. press. on shoe = 000 lbs. per sq.in. Unit stress in piston rod =000 lbs. per sq.in. Pressure per sq.in. of projected area of wrist pin = > Marine Engine.” Assignment Table follows. ASSIGNMENTS FOR Pros. 2 No. 1 2 3 | 4 5 | 6 7 8 9 10 Bore....... 123 | 11] 143) 14] 15] 16] 17); 17); 19) 21 Stroke..... 17} | 22) 24 24} 27) 30} 24} 27) 39 | 36 Steam...... 160 | 160 | 160 | 160 | 150 | 160 | 150 | 120 | 160 | 160 Use half the given steam pressure in calculating d;, d and bolt diameter. Use !=2td and 1000 lbs. per inch of projected area of wrist pin. For bolts use f=5500 lbs. and make the thread 10 or 12 pitch. Use maximum values of constants in working out (B) and (H). Use minimum values for (g) and (c). Put 3 to 5 sections of babbitt in the shoe-bearing surface. No. 6 Test. Arts. 116-130. (2 hours allowed.) 1. (2) What is a piston? (b) How is leakage past it prevented? (c) Sketch two ways of fastening a piston to a piston rod. (d) What is a trunk piston and where is it used? 2. A gasoline engine has 4’ bore, 6” stroke, length of connecting rod=2} times stroke. The piston length equals its diameter. The wrist pin diameter =0.2D, and its length between supports is 2D. If the explosion pressure in the cyl. is 300 lbs. per sq.in. calculate (a) the max. thrust of the piston against the cyl. walls (in lbs. per inch of pro- jected piston bearing area). Calculate (6) the pressure per inch of projected area on the wrist pin. 3. A piston rod 13” diam. is screwed into a crosshead by being threaded with 10 thds. p.i. (depth of thd.=.065”). What steam pressure can be used if the piston diameter is 10” and the max. unit stress in the rod is not to exceed 4500 lbs. per sq.in.? What will be the unit stress in the unthreaded portion of the rod? 4, What area of crosshead bearing surface will you allow in Question 3 if the pressure on the guides is 30 lbs. per sq.in.? (Length of conn. rod =4 times the stroke, which is 12”.) 172 ELEMENTARY MACHINE DRAWING AND DESIGN No. 7 Test. Arts. 116-130. (2 hours allowed.) 1. Describe in detail (a) the various forms of pistons, (b) how they are fastened to the piston rods, and the (c) methods of keeping them tight. (d) What is the function of a wrist pin? 2. An engine has a cyl. bore of 10” and a stroke of 12’... The con- necting rod length is 23 times the stroke. The piston rod is 2” diam. threaded 10 thds. p. i. (depth of thread =.065’’). The crosshead bearing surfaces have an area of 60 sq.in. When these surfaces are carrying a max. pressure of 30 lbs. per sq.in. calculate (a) the steam pressure in the cyl., (6) the unit stress in the piston rod at least section, (c) the diameter and length of the wrist pin (J=2d) when the max. pressure per in. of projected area is 1000 lbs. 3. A gasoline engine is 44’ bore X6” stroke and the connecting rod is 15” long. The length of piston =its diam. Find (a) the max. pressure of the piston against the cyl. wall when the explosion pres- sure =350 lbs. per sq.in. (b) Find the pressure per sq.in. of projected area of wrist pin when the pin diam. =.2D and its length is 0.5D. CHAPTER XII CONNECTING RODS 131. A conneEcTING rod is a link used to connect two machine parts whose motions are different in character. A part moving with reciprocating motion may thus be connected to a part having circular motion; an example of this is found in the con- nection between the crosshead pin and the crank pin of a steam engine. (See Fig. 124.) The length of a connecting rod is the distance between the centers of the pins which it connects. In engine or pump connecting rods this length is usually a multiple of the length of movement of the piston or plunger in one direction. This movement is called the stroke of the engine or pump. The length of connecting rod varies from 2 to 4 times the stroke on land engines to 13 to 23 times the stroke on marine engines. In gas engines the length of rod is from 2 to 4 times the stroke. 132. A connecting rod has each end shaped to receive a pin in much the same manner that a bearing receives a rotating shaft except that in the latter case the shaft rotates while the bearing is fixed. The pressure on the bearing surfaces of the pins in a connecting rod is used to determine the size of these pins and consequently the size of the ends of the rod. The size of the pin at the end of the rod having reciprocating motion and called the wrist or crosshead pin can be found by reference to Art. 124, Fig. 123, for gas engines, and Art. 129 for steam engines. The end of the connecting rod turning on the wrist pin is called the little end or crosshead end while the other end is the crank end or big end. The crank pin dimensions may be determined by reference to Art. 142 and Table 16. When the diameters and lengths of the pins have been determined for each end and the length of the rod from c. to c. of pins, it will then be necessary to find the area of the least transverse section of the rod. The point of least sectional area is near the crosshead end of the rod. The 173 174. ELEMENTARY MACHINE DRAWING AND DESIGN shape of the section at this point may be a circle, a rectangle with the longer side parallel to the plane of motion of the rod, or an I section with the web parallel to this plane. See Fig. 130. 133. The calculation of this section as well as other sections nearer the crank pin requires an extended knowledge of mechanics, therefore it has been considered advisable to give empirical formule which will give results close enough for drawing purposes. The size of the least section depends on the pressure on the rod due to the working fluid in the cylinder as well as the strain on the rod due to the speed at which it runs. The maximum push or pull on the rod may be found by reference to Arts. 127, 129 and Fig. 125. In gas and gasoline engine work the dimensions of the rod at the wrist pin end may be calculated from the following formule. For a circular section from Prof. Lucke the diameter d=.011Dv/p to 0.014D-/p, D=diam. of cyl. and (p) =initial press. per sq.in. (say 350). For plain rectangular rods the thickness (4) is 0.008D~/p and the width b at the piston end is 1.6. At the crank end the width b is 2.3¢ If the section is of the I shape the width of flanges may be 1.3¢ as found above, and the web thickness (S) =0.6¢. For steam engines the diam. (d) at a point .4 the length of the rod from the crank end may be d=0.0164-~/mPP, where P=total unbalanced pressure on piston, /=length of rod in inches and (m) has the following values: Piston speed ft. per min., 200 400 600 800 m 30 20 15 10 The diameter at the small end=.9d; at the large end, 1.1d. For rectangular rods of breadth (6) and thickness (f) see Fig. 130 (B) and (C). Haeder and Powles give the following: (When b= -X¥Y-U0'9aS 1 178 ELEMENTARY MACHINE DRAWING AND DESIGN being screwed on. The rod and the rod end as well as the cap are designed to be turned in a lathe. The curves at (A) and (B) of Fig. 1 are plotted as the intersections, (A) of a surface of revolution with a cylindrical surface and (B) with a plane similar to the curve on the strap end. The other curves on the cap in Fig. 1 and Fig. 3 are circles or arcs or circles, one view giving the radii for the curves of the other. In Fig. 1 the width of the end of the rod may be found from the known length c. to c., the taper of each side and the diameter at the strap end where this taper begins. The curves which join the sloping: sides of the rod to the forked end through which the bolts pass, may be drawn with a curved ruler. The horizontal dimension indicated but not given in Fig. 1, just above the horizontal Shape of Cross-section J J 16 12" Crosshead \Bin) c.toc.= 3*Stroke a= Sees Connecting Rod Fic. 133. center line, may be assumed to bring the vertical straight line, to the right of zz, to the left of the bolt heads. In Fig. 2 the radius of the curve joining the stub to the head is also to be assumed. This rod end was designed for a connecting rod to be used on a 93’ X12” horizontal engine to run at 275 R.P.M. The entire rod is shown in Fig. 133. The stroke of the engine being 12” the length of rod c. to c. is easily found. 137. Another type of marine end is shown in Fig. 134. The dimensions for several sizes will be found in the table on page 179. The length c. to c. may be taken as 3Xstroke of engine given, which is the last number in the first or last columns. 138. Solid Ends are used on crank pins of the overhung type, that is, a pin supported at one end only. These ends are lighter than other forms and are also less liable to breakdowns. The end is forged solid and the rectangular hole is then cut out having rounded corners. Fig. 135 shows the general form of these ends. The adjustment is obtained by a wedge actuated 179 CONNECTING RODS $2 X 02-61 T|/H# |] Hs) 2 g 9| #1]|/#e] 9] 9} #8) 8] t) 8} for | = | 8 | 9 | OXOS-6T ZZ X 8I-LT TT] | 4¢) Ho] ¢ #¢ | HI | ge | &¢ | é¢ 8 | #8 | #8) #2] = SE) #2] $9 | GX8I-LT 02 X 9I-ST t Sip] #9 |] 4¢|) te] $1] fe | fp | ? #21) EL] 8 i & | 8.) fp | OZX9I-ST SIX PFI-81 t a ae ao Ht i StL | a | 3p | te | Yo] #9 | 2t2 z SE | $9 | fp | SIXFI-8T SIX E-ZI-11 | % Se | tt ae ee ott | $8 | $ otf ig 9 t é EG | $€ atXel-CItl ZI XOI-6 : t g 7 | te ¢| Z| ee] ¢| Hp] ip} ¢ ; SE | ¢| € | ZLXOI-6 OLX 8-2 S| | §o | Sel] Ha] Fol] & | st] te] é fe | pl tp : ¥ | fe | §% | OLX8-L Piseiaes m\qvi Zz} aA | xX] aM] AlaAlSS|O|WI|TI/MV)I\| Ola | aja ; : "s19puy jo sazig coe “(pus pyos) gE] o1NSIg Oo sierze'T JO son[eA 0z X 02-61 % | 3a | $20) $0} Fz ¥ 8 | #9 | t91 F fe | 491 | tH | ter 6 | 02X0Z-6I SIX 8I-L1 oj} at jer} Rot] 2] & tL | ¢ 79 | #¢| € @| it] ¢ Il | 8 | 8EX8I-LT 9TX 9T-ST #1 | $1 | for} #1] 79 | F #9 | ¢ #¢ G| tp | Ha] tet} S | or] 2 | 9txgt-st FLXFI-81 fr | 41] 6 | Sr] é¢ | 2 t $ Gg | | # | $11} ¢ | 2 ¥9 | FIXPI-€T GIXeI-ZI-11| #1 | iT | F 41) ty | ? G |] é yp) fe} F @| for; = | € gg | ZLXSI-ZI-Il OLX 0I-6 Ts} ft | é Ht | ¢ t 3 fe | Sel] is | Ht é sr | 3 fp | OLXOI-6 OLX 8-2 T| ? 4% e | 7] & ie t @] | ? a1 |? go |e $e | OLX8-L fl | HD|do| 4 | 0] 2 | a | N|Wi|T/N J) er |ol|las|aila “srapulpsAg “siapulfAD jo 8aZIS jo 8OZIS, *(pue oueUI) FET INFIY WO s19}}987T Jo son[eA ‘sq'T GZ = 9mNsseig WreAIg SNOISAq dOY wOd SNOISNAWIGC] 4O SsaTavy, 180 ELEMENTARY MACHINE DRAWING AND DESIGN by two bolts. Dimensions are given in the table on page 179 for these ends. I THU ——rh — a= \ Babbitt is secured to Dia, GH 'DOx elther by tinning Dia l/ i by "T'slots. ‘ ra See ee LU : 4 H— — — 7 «| © Q S Vt? > 7mm cj oe es Hee ANY \SteelCasting | “L_) ¥y i<——- ff —___»>| es ja MarineType Design No.3 Fia. 184.—Connecting Rod End. 139. The little end of a connecting rod is usually made of a type called a strap end. CONNECTING RODS 181 Reference to Fig. 133 will make clear the parts which com- pose it. The end of the rod itself is made rectangular and with a hole to take a gib and cotter. The brasses are made Cy x T a Te | = T / Th ET me MIS Tt AT oe \ Y ly i ‘a cS + — (i t 7T J &%} xX Babbitt is secured to aKL boxes either by tinning pats or by’T's/ots Wy “W ‘ou nas AS4 aa Uap for Set Screw \e +0;/Pipelap *" > Pp > yy + Z N31 BAN ~ Ls. a 7) 2 iy \ S> NS $ LM 7 ! Ah< Fic. 136.—Connecting Rod End. wo b=D44" t02"; to= 9 or ; 2bt =area of circle D’’ diam.; nomen ot? | tg=1.17t to 1.5t1; tz = 1.2t, to 1.5t, or such that area through cotter hole 2 bt. Area of cross-section of cotter and gib2 bt:; 2 area D area of D ts = 1.331; “3-¢ 73 ee tg=3d4+}3"; t,=.5t to t. 1 2 area of D : S t= .08d +79 S = 4t, or 3 34. Se diam. = Zz CONNECTING RODS 183 Often the thickness fg is used in place of t; and the strap is uniform in thickness except where ¢2 is shown, where it is made 1.25t3. 140. In Fig. 137 is shown, more in detail, the strap end of the rod already shown in Fig. 133. The crosshead pin is of that type shown in Fig. 126, which requires a strap end. The brasses extend the whole length of the pin, which is 21” diameter. The strap encloses the brasses and is fastened to the stub end of the rod by means of a gib and cotter or key held in place by an oval point set screw bearing in the bottom of a shallow groove in the side of the cotter. The oil cup shown is designed to wipe off a drop of oil at each forward stroke. The wiper is a thin piece of brass No. 16 gauge. There are certain dimensions which are given in order to calculate certain others denoted by the letters a—m-s-n. By reference to the sub Figs. 1-2-3—4 of Fig. 137 it will be clear how the various parts would fail if there was not enough material in them at these points. Suppose the rod in tension and the resistance to the tensile stress was 20,000 lbs. per sq.in. of area in the circular cross section of the smallest part of the rod (at A). The total resistance would be 20,000 xr?.. The strap would fail in tension as shown in Fig. 1, unless the transverse sectional area through the slots for the gib and cotter was equal to that at (A). Since the area is in four sections, each of the four areas must equal eS but each area isa rectangle whose sides are (a) and the half width of strap at 2 the side of the slot. quate the rectangle area and ae and we can solve for (a). The strap can now fail by shearing as in Fig. 2. The strength of the material for shear may be taken as 15,000 Ibs. per sq.in. This means the area sheared at the end of the strap must be 3 more than the area in tension or 4amX15000=xr?X 20000. The end of the rod end may also fail by shearing out under the pressure from the cotter. Since 2 areas fail, their sum must equal the area just used in finding (m). The cotter and gib also will fail by shearing off in two places. Their sheared area must equal the preceding sheared area of the end of the rod. The thickness of gib and cotter is given in Fig. 6, from which (S) can be calculated. The con- 184 ELEMENTARY MACHINE DRAWING AND DESIGN “LET COLT ybig POY bul,oeuu0) Jo pug. Pred aS gb 4 MOA Juos4 10) io ns 4 | ug 4 sf = 4 1 . bunsvoys hg buyn| | | = | = 419 PuOKaY | Bolle J ) al < osey ar) if? So “weuNen Me lot Lussways kq Bui/2y poy, 7 ‘ sl Paar ‘ ae b tt : NE HRA ayy ady 2 Z oe “Aly ul SS ; ee J df dad fr peo . ¥ 4 g /b14 MalA dol ny 0005005 FT| SE CONNECTING RODS 185 struction of the curve of intersection on the stub end is shown in Fig. 138, The dimensions of a~m-n-s worked out above should be changed from decimals to the nearest thirty-second of an inch. End Fic. 138.—Intersection on Rod. INSTRUCTIONS Connecting Rods Example 1. No. 2 paper. 2 hours allowed in class. Av. time required is 3.6 hours. Prob. 1. Draw three views of a connecting rod for a gasoline engine, bore, , stroke , length c. to c. =2} Xstroke. Max. pressure in cyl. =300 Ibs. per sq.in. Style of rod shown in Fig. 131. Calculate the size of wrist pin from Art. 124. The cross-section of rod will be like Fig. 180 (D), the depth of small end, width of flanges and web being calculated from Art. 133. The diameter (d:) and length (i) of crank pin may be calculated by assuming a pressure per sq.in. of projected area = lbs. and a length ,=Xd2, wherex= . Calculate the weight of the rod if made of steel weighing 0.20 lb. per cu.in. W. I. at 0.27 lb., babbitt at 0.3 Ib. and C. I. at 0.28 1b. Title to be ‘“Connecting Rod for X Gasoline Engine.”’ Put a note on drawing giving data used and results of calculations. Make calculations before class. No sketches needed. ASSIGNMENT TABLE FOR GASOLINE ENGINE Rops No. 1/2/)3 }4)]5 |6!]7);) 8/9 ! 10; 11! 12] 13 Bore...... 32 4| 44) 44) 42 5] 53] 53) 53) 6] 48) 42) 52 Stroke....| 5 6| 6 53] 6 ON CF 4} 73 8] 6 7 7 Trish. te hed toh 1 oJL.2]2.1 41.2 | 1) 41.2 41.25)1.3 |1.3 [1.4/1.1 1.2 }1.2 186 ELEMENTARY MACHINE DRAWING AND DESIGN Prob. 2. 2 exercises at 2 hours each. Av. time required is 7 hours. No. 3 paper. Draw three views of the crank end of a connecting rod fora X12” engine (length c. to c.=3 times stroke), shown in Figs. 132 and 133. Section the upper half of the front view, the upper half of the end view and the lower half of the bottom view. The boxes are brass, the rod and cap are steel, the bolts W. I. and the shims of C. I. Supply all omitted dimensions, and calculate the stress per sq.in. at the root of thread on bolts when the steam pressure is 100 Ibs. per sq.in. Also calculate the pressure per sq.in. of projected area of crank pin, ASSIGNMENT TABLE FOR PROBLEM 2 Bore... . 9 ee 10 103 11 113 12 13 Prob. 3. Marine end connecting rod. No. 3 paper. 4 hours allowed in class. Av. time required is 7 hours. Make three views of the end shown in Fig. 134. The third view will be a view looking towards the cap. The size of engine is X . Steam pressure in cyl. =125 lbs. Use the dimensions given in the table and calculate the max. stress per sq.in. on the bolts. Also calculate the pressure per sq.in. of projected area of crank pin. Make the upper half of the front view in section, and the lower half of the end view. Give dimensions for working drawing. Title to be “Crank End of Connecting Rod for xX Engine.” “Marine Type.” In a note give steam pressure, thrust on rod and pressure per inch of projected area of crank pin, and stress on bolts. ASSIGNMENT TABLE FOR PROBLEMS 3 AND 4 no. [1 f2|afels[ol7]s| 9 w]a|rs}rs| sts 11)12/13) 14 11/12)14; 14 15 16 | 17 | 18 | 19 | 20 | 13 Bore....| 7 | 8 | 9 0 16 16] 18! 18 | 20 | 20 | 12 Stroke ..| 8 | 8 ]1 Prob. 4. Solid end connecting rod. No.3 paper. 4 hours allowed in class. Draw three views, front, top and end. Make front view half in section above center line and top view half in section below center line. The size of engine for which the rod is suitable will be a x. Steam pressure 125 Ibs. per sq.in. Use the dimensions given in the table for Fig. 135. Calculate the pressure per sq.in. of projected CONNECTING RODS 187 area of crank pin. Calculate the stress per sq.in. at least section of the end (where the bolts go through). Calculate the shearing stress per sq.in. at the extreme end of the rod. A note is to be placed on the drawing giving the results of these calculations. Give all dimensions on the drawing. Title to be “Connecting Rod End (Solid Type) for a X _ Engine.” Standard title for remainder. Work out calcula- tions before class. No sketches needed. Assignments in preceding Table. Prob. 5. Strap end of connecting rod. No. 2 paper. Make two views of a strap end of a connecting rod like that shown in Fig. 136. Make the front view a half outside below the center line and make the top view a half section below the center line. Calculate the dimensions when the diameter D= ",d= ” and l=2xd, wherex= . Art. 86, Fig. 92, shows cotter and gib to use. Make a sketch of the rod end, with dimensions inserted, before class. ‘‘Brasses’’ are described in Art. 105 (Fig. 106). Set screws to be sq. hd. oval point. Calculate the stress per sq.in. at the weakest part of the strap, at (A), through gib and cotter and at (B) when the pull on (D) is 7000 lbs. per sq.in. Also calculate the pressure on the projected area of wrist pin. Put a note on the drawing giving the results of these calculations and the data used. Title to be “Strap End of Connecting Rod.” 4 hours allowed in class, ASSIGNMENT TABLE FOR PROBLEM 5 Tell as op | a 5 6 7 8 9 10 p | i | 18 | ae | ae | ag | el ew i z | 14g Mg PS eh a | ae | sae tae & || 2h | 24 e | 4} dati 1.8] 2 Ted | 4201 12254 ae) ace | 18 Prob. 6. Strap end of connecting rod. No. 3 paper. 6 hours allowed in class. Av. time required for first two ex. requirements is 5.85 hours. Fig. 137. Three views of this end. Front, top, and end (from the right hand side). Make the front view a half section above the center line, the top view a half section below the center line, and the end view a half vertical section on right of center line by a plane containing the axis of the wrist pin. Do not section the oil cup in the front view and end view, nor the gib and cotter. Pipe length will be found in oil cup Table 10. Plot intersection curve from Fig. 138. Calculate the values of a-m-n and s, using the nearest thirty-second of an inch on the drawing. Art. 140 explains these calculations. Calculate the pressure per inch of pro- 188 ELEMENTARY MACHINE DRAWING AND DESIGN jected area of wrist pin when the length of connecting rod is 36’ and the steam pressure is 100 Ibs. per sq.in. and the diam. of steam cyl. is D’= . Calculate the stress per sq.in. at smallest sec. of rod end, at least sec. of strap, at (n), at cross-section of key and gib and at end of strap. Put a note on the drawing, giving your results and the data used. Omit Figs. 1-2-3-4 from the drawing. Give all dimensions and make a standard title, calling the object drawn a ‘Strap End of Connect- ing Rod for a D” X12’"Engine.” No sketches are required for the first exercise, but the calculations for a-m-s and n are to be made before class. For the second exercise the other calculations are to be made. For the third exercise a dimensioned detail drawing of either the (a) strap or a (b) brass or (c) key and gib are to be made on No. 1 paper to scale. ASSIGNMENT TABLE FOR PROBLEM 6 No 1 2 3 4 5 6 7 8 9 10 | 11 12 Dread sect ies 93 | 10 | 103 | 11 | 114 | 12 | 94 | 10 | 103 | 11 } 113 | 12 Detail a b c a b c c a b c a | b Prob. 7. Draw to scale, quarter size, on No. 1 paper, a detail of the body or stub only of the connecting rod whose ends are shown in Figs. 182 and 137. The complete rod is shown in Fig. 133. CHAPTER XIII ENGINE CRANKS AND ECCENTRICS 141. Tue crank of an engine is used for transmitting the recip- rocating pressure on the piston of an engine into a pressure tending to rotate the crank shaft. The connecting rod is attached at one end to the crank pin at the other end to the crosshead. The crank is a single lever keyed at one end to the crank shaft. At the other end the crank is provided with a projecting pin called the crank pin. The distance between the centers of the crank pin and crank shaft is the “throw” of the crank and is equal to one-half the stroke of the piston. Reference to Fig. 124 will make clear the relation of the parts of an engine. The crank pin moves at a constant velocity _2nRN 42 N=R.P.M., but the piston moves at a velocity which varies from 0 at the end of the stroke to a velocity near the center of the stroke approximately equal to the velocity of the crank pin. The mean velocity of the crank pin is 1.57 times the mean velocity of the piston. The connecting rod exerts a pressure on the crank pin in direct line with the rod. This pressure on the pin is resolved into a tangential component which produces a turning effort on the crank and into a component acting along a radial line through the center of the crank pin circle. The first component multiplied by the throw of the crank gives the turning moment on the crank shaft. 142. Overhung Cranks. Fig. 139 (A) shows a cast-iron crank with a steel crank pin fastened in by a nut and taper pin through the bolt end. =ft. per min., where R= throw or radius in inches and R=radius of crank or throw; d=diam. of crank pin; l=length of crank pin =2d. Where z lies between 1 and 2, 189 190 ELEMENTARY MACHINE DRAWING AND DESIGN Unwin gives the following values for e: e=0.08d+0.2” to 0.12d+0.2”; v=1.5e. Spooner gives t=.6D for C. I. cranks. The diameter and length of the crank pin is determined by the load it must carry and the pressure per inch of bearing area. This latter pressure depends on the kind of machine it Mel + be Ld K/Z 0 eo i—-t | fl Se rol Spey et s | let a x4 | jf t " t Ww Pen ! \ y 2 TATE, xt Sec.on XX ce on 135D (8) Cast Iron C Ra Forged Steel or W. 1. Crank Fig. 139. is on. The pressures per sq.in. allowed for engines will be found in Table 16. If we have for example a load (F) on a crank pin and the allowable pressure per sq.in. (p) is 300 lbs., the area of pin bearing surface, (A) (see Art. 100), would be expressed as Ane If we say the length of crank pin (J) is to be 13d, the expression for its diameter (d) can be written as F 1.5) Fig. 139 (B) shows a steel or W.I. crank with a crank pin fastened in by a cotter. This method of fastening in the pin may be used on a C.I. crank but is not so good as the method in Fig. (A), eed af or d= Pp ENGINE CRANKS AND ECCENTRICS 191 A third method of securing the pin to the crank is shown in Fig. 140. The pin is forced into a hole in the disc and the end riveted over. Sometimes a key is used instead of forcing the pin into the hole. Fig. 140 shows a C. I. disc crank. The portion of the disc opposite the crank pin is made heavier to balance the weight of the crank pin and part of the weight of the connecting rod. 143. Besides the overhung crank pins set in cranks keyed to crank shafts, we find crank pins made integral with the crank Fic. 140.—Cast Iron Dise Crank. shafts to which they are joined by a web on each side. Fig. 13 is a crank shaft for a triplex power pump. On each end of this shaft are keyed disc cranks similar to that shown in Fig. 140. Crank shafts and pins are often forged under the steam hammer as one forging, comprising several crank pins, webs and bearings. Small shafts of this kind are largely used in automobile and marine motors of the internal combustion type with two, four, six or eight cylinders. 144. The calculation of the thickness (t) of the web for a crank of forged steel or wrought iron of the type shown in Fig. 139 (B) involves the force (F), length (R), distance (x), and width W. 192 ELEMENTARY MACHINE DRAWING AND DESIGN The twisting moment on the shaft is FR=T, from which the diameter D of the shaft may be obtained (see Art. 89). 35-17. fs” proximately by equating the twisting moment of the shaft to (1.5D)?tf _ Df, 6 Bet D= If W=1.5D then we may find the value of (é) ap- the bending moment on the web, that is, which __ 6D fs___—.523 Df. () ~ 5.10.5D)2f f~o oo aoe where (f) and (f.) may be 4500 to 5500 and 5000 to 13,500 respectively. FR,=B=bending moment through the web at (W). Fxr=T=twisting moment. Then the equivalent moment B.=3(B+~V B?+T?), and equating this to the resisting moment t 2 of the section at (W) ut we have a 2 a/ R242 (B+V ETT) =t or Ere? . @ where (f) is the same factor used above for the shaft. The value of (t) in equation (1) enables us to find an approximate value of (x) which is necessary in obtaining (7), used in equation (2) for the final value of (¢). 145. Eccentric and Strap. An eccentric is a modified crank used when the crank radius is so small that the crank pin and shaft, about which its center turns, interfere with each other; in fact, the crank pin surrounds the shaft. The eccentric is used to con- vert circular motion into reciprocating motion and is chiefly employed to drive the slide valve of a steam engine. The eccen- tric is keyed to the shaft which drives it and imparts motion to the valve by means of the eccentric strap and eccentric rod. The strap is made in two parts to enable it to be placed on the eccentric and the eccentric rod is fastened to the strap by bolts or screws. Fig. 143 shows an eccentric with its strap and part of the rod. To prevent the strap from slipping off the eccentric its inner surface is made either hollowed or projecting to fit the outside surface of the eccentric. Fig. 141 shows several sections of eccen- tric and strap bearing surfaces. The most common form is the one shown at (1). In Fig. 143 this style is used. In order io ENGINE CRANKS AND ECCENTRICS 193 make the eccentric lighter, two irregular openings are made in it separated by a rib which is usually placed opposite the key at the widest part between the shaft hole and outside of strap. The diameter of the eccentric may be made equal to (d) obtained from d=1.2D+2r+3”, where D=the diameter of shaft, and (r)=the eccentricity of the eccentric center. (A) is made equal to B , where (B) is the breadth of the eccentric, measured along the shaft. The value of (B) depends on the resist- ance which the slide valve offers to movement by the eccentric. Fig. 141, As this depends on the steam pressure which produces friction of the valve, on the area of the bearing surface and the coefficient of friction, its calculation is somewhat difficult; therefore the value of (B) will be given. 146. The strap is made in two parts which are held together by two bolts passing through lugs on the outside. The thickness of the strap is taken as (-4 The center line of the strap bolt is drawn tangent to the outside circumference of the strap. The diameter of these bolts may be taken equal to .4B, using the nearest standard diameter above the dimension found by calculation. In order to stiffen the strap on the side opposite 194 ELEMENTARY MACHINE DRAWING AND DESIGN the eccentric rod a rib is cast on the outer surface 3 the width of the strap and projecting a maximum distance equal to F=C. This rib narrows as it approaches the bolt lugs until it is 7’ or less in height. The surface of the bolt lug against which the nut bears must be at such a distance from the parting line of the strap halves (xy) that there will be a clearance of 7’ between the nut corners and the stiffening rib. Use a U.S. St. hex. head bolt with a standard nut and a locknut or with two nuts of the same thickness each equal to ? the bolt diameter. 147. The eccentric rod is fastened to the strap by means of a key, or by threading the end of it or by making a T head and pass- ing screws through it. These three styles are shown in Fig. 142 and in Fig. 1 of Fig. 143. In all these cases the hole or holes which are drilled in the strap should stop at the circle which defines the outside of the strap. In the case where the eccentric rod Ecce. Strap Ecc. Strap Ece. Rod Case (I) © Case(2) Fig. 142. end has a T head the screws used are cap screws whose diameter is the same as that of the strap bolts. The center lines of these screws are placed as near as possible to the center line of the rod and yet allow the heads of the screws to clear the fillet between the rod and its T head. This fillet may have a radius equal to 4” or more. The thickness of the T measured parallel to the center line of the rod may be equal to 0.45B. The diameter (F) of the eccentric rod is taken equal to 0.9B. By locating the point at the bottom of the tapped hole for the cap screws the surface of the strap can be found against which the T is fastened. 148. The rubbing surfaces of the eccentric and strap are oiled 195 ENGINE CRANKS AND ECCENTRICS e614 ee ee eae [= | oa ee ant te 45 = ~N S4U¥4I07- ‘EFT ‘DIY Oia x Gmaingdo9 COT aK | db adliy ge te x 2‘b14 LOIS Ul f/BU JAMO/ AYO T ——V——_c—S—o0MS0—0 Tt ‘7 LH >| : | : ia $Y AD14{S PUD BADBUS I4,UsI905 AX pT poesia ky. pets py bis U0 UOIIASJ/OH | aceasta | i MAlA PUT HOH 196 ELEMENTARY MACHINE DRAWING AND DESIGN by means of an oil cup which is screwed into a boss on top of a strap bolt lug. The cup shown in Fig. 143 is one designed to receive oil dropping from a fixed cup. The length of the cup ina line parallel to the eccentric rod must be such that the oil will drop into the cup in any position of the eccentric. 2r-+4’’ will be long enough. The inside width of the cup at right angles to the long dimension above will be +” and the thickness of its sides=7 5”. The hex. portion between the cup and threaded pipe end will have a diameter across flats as given in Table 10 (oil cups). The other dimensions below the hex. may be taken from the same Table 10. In order to permit the oil to flow around the strap bolt, the latter is grooved to a depth equal to the depth of thread on it. See Table 1, Col. 18, for diameter at root of threads of U.S. St. bolts. INSTRUCTIONS Cranks Arts. 141-143. Example 1. No. 2 or 3 paper. 2 hours allowed in class. Av. time required is 4 hours. Draw two views of a crank and crank pin. The pin is to be fastened to the crank in the manner shown in Fig. The crank is to be like that shown in Fig. . Calculate the crank pin dimensions from the data given in the Table below. The load (F) = lbs. The value of p= lbs., the length 1=zd where z= . The radiusR = ”. The finished plate should be a working drawing with all dimensions given, standard title, etc. Calculate the diam. D, using f=9000. (Art. 89.) ASSIGNMENT TaBLE (CRANKS) No. 1 2 3 4 5 6 7 8 9 10 Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. Pin |139A |139A |139A | 189B | 139A | 140 | 140 | 139B | 139B | 139B Crank.| Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. | Fig. 189A |139A |139B | 139B | 139 B| 139B | 139A | 139B | 139B | 1394 6300 | 9800 |14200) 15400 | 25000 | 25000 | 39500 | 6300 | 9800 | 14200 1000 | 1090 } 1100; 1200} 1200} 1000} 1000} 900 | 900 | 1000 1 1 1 1.4.) 2.1 1.2) 1.2 | 1.1 1.1 1.1 6 8 10 12 14 16 18 7 9 11 Areva ENGINE CRANKS AND ECCENTRICS 197 It may be necessary in some cases to increase the value given for (R), if the hubs interfere. Prob. 1. Draw the crank shaft shown in Fig. 13, and show on each end of it a disc crank having the same throw as the forged crank. The cranks are to be placed at 120° from each other. Make the crank pin length for the disc pins equal their diam., the pressure on them being the same as the center pin. Prob. 2. Calculate the thickness (¢) of the web for a W. I. crank, for assignment No. in the preceding table, using f;=5500 and f =5000. Prob. 3. Draw aC. I. crank of the following size. R=12’", D=53, d =3",1=34’’, using a nut on the inside end of crank pin. t=2’, Length of hub=D. No. 2 paper. Prob. 4. Draw a W. I. crank with the following dimensions: R =18”, D=8", d=5”", 1=54”", t=4’, length of hub =63”, D,=143". Use a cotter to fasten the pin to crank whose width is 23” at large end and thickness =1”.. Use No. 2 paper. Prob. 5. Draw a C. I. disc crank with the following dimensions: R=12", D=63", D, =2'-8", F=5", d=4", 1=43". No. 2 paper. Prob. 6. Draw a C. I. disc crank with the following dimensions: R=10", D=6'’, D, =2’-6”, F =43", d=3}3'",1=4"". Use No. 3 paper. Eccentric and Strap Size paper No. 3. Drawing to be in pencil only; time allowed 2 exer- cises in class =4 hours. Av. total hours required, 5.42. Views to be drawn are: Front view (Fig. 1) of Fig. 143, upper half in section. Top view (Fig. 2), lower half in section. End view (Fig. 4) from the left hand end showing the lower half in section by a plane zy (Fig. 1). Omit Fig. 3 on the drawing but use it for reference in making the views called for. Strap is cast iron, ecc. and rod of steel, oil cup brass, bolts and screws W. I. Place the oil cup on the strap in the top and end views and also in the front view, if there is room on it. If not, then place it to the right of and near the end view, from which it is projected. The following dimensions must be placed on the finished drawing in addition to those shown on Fig. 143. c. to c. of strap bolts; c. to ce. of cap screws; length of bolt lug from vertical center line, dis- tance from vertical center line to c. line of oil cup, and to end of ecc. rod T end. Thickness of T head on ecc. rod and width of same. Depth of tapped cap screw hole. Plot intersection curve /. Make a bill of material for eccentric and strap without the ecc. rod, (on the drawing). Put on a standard title instead of the one shown. The data for drawing the Plate are as follows: D=( "), r=( ”), B=( ”) found from assigned column below. Place ecc. center (z) 198 ELEMENTARY MACHINE DRAWING AND DESIGN the center of shaft to the (y) of a vertical and on a line making (z)° with the horizontal. TABLE OF DIMENSIONS OF ECcENTRICS 1 2/3/4/5/6/7/8]9 10 |12/13/14)15|16/17/18 D\ 3 3 13 |35 1/33/33 | 33/3 [3 3 18 (8% [8% |8% | 2% | 23 /82 ei) A El1R/ 1 | FMM] D | oS) Th aR Lae) Beda) 18) 12) 2 By) 14 |13|13 u 15 }15) 18) 1S) 1 | 1s ls [Late]L elas] 15] 15 flat x {Above |ab.|ab.|ab.|ab.|ab.|ab./ab.|...| Below| B| B| B)| B/ B|B|B y |Right} R| AR...) b) 0) 0) 0) DL \Left L\|L IL...) R|RIR z | 30 |60/ 75 | 90} 75 | 60 | 45 | 30 380 | 45 | 60 | 75 | 90 | 75 | 60 | 45 1 Prob. 1. Compare the strength of the two strap bolts at root of thread with that of the eccentric rod. Use the dimensions you have on the drawing. Prob. 2. Make a detail working drawing of (a) the half of an eccentric strap which is not to be connected to the ece. rod. .@ working drawing of the half of an eccentric strap which is to receive the ecc. rod cap screws. eccentric sheave. FR -_ se =, Wd bea ox DEF ARTMENT MACHINE DESIGN SIBLEY SCHOOL CORNELL UNI VERSITY RECEIVED (6) Make (c) Make a detail working drawing of the CHAPTER XIV , PULLEYS AND BELTING 149. There are two principal systems used for transmitting power from a prime mover to machinery, namely, (1) toothed gearing and (2) wrapping connectors. Under the latter head are found belt gearing and rope gearing. Belts are carried by pulleys which they either cause to rotate or receive motion from. The frictional resistance to slipping of the belt on the surface of the pulley face determines the power transmitted. Belts are made of leather, cotton, cotton leather, rubber, rawhide and leather links. The material most commonly used is leather from the hides of steers, taken from along the back. Belts are called “ single ” when the thickness is that of a single hide and double when two or three hides are cemented together. The thickness of single belts varies from %;"" to ~;’’ but the average is 35’’ or 3’, while the thickness of double belts ranges from 3” to .33’. Rubber belts vary in thickness from .18” for 3 ply to .45” for 8 ply. The weight of leather belts is about 55 Ibs. per cu.ft. The smallest diameter of pulley on which a belt will run fully loaded without excessive wear, is about 35 times the belt thickness and preferably greater. The width of belts does not go below 2” as arule. Up to 4” the width increased by quar- ters of an inch, above 4” up to 7” by half inches and by inches between 7” and 20”. The ends of belts are joined to make a continuous connector. The joint may be laced, sewed, cemented or riveted, the actual strength of the joint being from 60% to 75% that of the belt. The ultimate strength of good leather belting varies from 3500 to 6000 lbs. per sq.in. but the working strength is taken as 200 to 300 lbs. per sq.in. 150. When a belt is: wrapped around the rims of the pulleys which it connects, there will be a certain tension in it which is con- 199 200 ELEMENTARY MACHINE DRAWING AND DESIGN stant when the belt is at rest. If one pulley is rotated, the initial tension in the belt will be increased in the belt between the pulleys on one side and decreased in the belt on the other side. The difference between the tensions in the tight and loose sides is called the driving force. The belt does not slip around the pulleys because of the frictional resistance between the belt and the pulley faces. If the driving force is less than the frictional resistance the pulley will turn when the belt is pulled away from one pulley toward another. As the driving force depends on the difference in tension between the two sides of the belt, it is necessary to know the relation between 7; and 7;. T,=tension on tight side. T;=tension on slack side. 1;=initial tension (belt at rest). Then DEBT, or Te+T,=20. The ratio of 7; to T; depends on the speed of the belt (which involves the centrifugal force at the pulley rim), the coefficient of friction between belt and pulley and the length of the arc of contact between belt and pulley. By the use of calculus it can be proved that T a= E+, 8 where E=2.718, w=coeff. of friction, 6=the whole angle of con- tact in circular measure. This expression may be reduced to the following expressions which are sometimes more convenient to use. Common log Ls .434.6, if 6=circular measure i a ‘““ =0.007578u. if 6 is in degrees mS ‘“ =2.729uN if N =fraction of circum. embraced by the belt. The value of yu, varies from .29 to .46, depending on the condition of the belt and pulley. It may be as low as .15 on oily pulleys PULLEYS AND BELTING 201 or even as high as .76 for special paper pulleys. For iron pulleys and leather belts the average value of u is .3. The value of 0 for open belts depends on the distance apart of the pulley axes, and the radii of the pulleys connected by the belt. For an open belt 6=180°—26, where 6 is found from sin = 8 R=rad. of larger pulley, r=rad. of smaller pulley, and 1=distance between pulley centers, all in inches. For crossed belts 6=180°+2¢ and ¢ is obtained from Bin ont, The arc of contact in a majority of cases is 180° and this is taken often in the computation of tables for belting, a multiplier being used to increase or decrease the width of belt obtained from the tables. When a belt is transmitting power it tends to stretch more on the tight than on the slack side and the driving pulley continually receives more length of belt than it delivers, therefore the velocity of the circumference is faster than that of the belt. The driven pulley on the other hand receives a less length than it delivers and its surface moves slower than the belt. This phenomenon is called the creep of a belt. This creep together with the slip of belts on the pulleys is from 1% to 3%, the former value being permissible, the latter not to be exceeded. As the slip increases with the belt speed larger values of u may be taken for higher speeds. Unwin gives the following rule for max. values of wu: u=0.2+.004./V V=vel. ft. per min. From the above equations the value of x can be calculated for an assumed value of (uw) when the values of (R), (r), and (I) are given. Unwin gives in the last edition of his Machine Design, Vol. I, the following data regarding the initial tension which has proved, in a number of experiments, not to agree with the 202 ELEMENTARY MACHINE DRAWING AND DESIGN accepted theory of Ti+ + T;=2T, but does agree with the expres- sion W7,+-~/T, =2/7;, (tensions being in Ibs. per sq.in.). As P= BM =K, we can substitute it in the above formula, $ which gives VIi+4|¢=2VT or 4K a 7 (VK+1)? ” and 4 = (VK+1)? * If we take the values of u=.3 and 6 for an arc of contact of 165° we have K=2.37, and oe =1.18 instead of 2.5 by the tty ordinary theory. 151. The driving force P=T,—Ts; and a rule which gives good results is to allow for each inch of belt width for a single belt a value of 40 lbs. The horse-power which a belt transmits depends on the force P and the velocity of the belt, that is PV A= 33000" where H =horse-power, V =vel. of belt in ft. per min., P=effective pull on the belt. The economical speed of belts is from 4000 to 5000 ft. per min. If a belt runs around a pulley of diameter D’’ which makes (N) R.P.M. the expression for H will be PxDN A= 33000312? PULLEYS AND BELTING 203 or if W=width of belt in inches, 40WxDN WDN A= 3300012 3152.8 or 3152.8 W=Hx DN” The effective pull on a belt may be reduced by centrifugal force at high speeds to such a point as to prevent the belt from driving. The weight of leather belting is about 0.43 lb. per sq.in. per foot of length. If w=this weight, then the centrifugal force of one foot length of belt one inch square running on a pulley of 0 feet diam. at a speed of v feet per sec. is P= wer 12? = eo 7D baie This force on one-half of the pulley circumference produces in each of the straight belt segments a tension T,=0.014v? lbs. per sq.in. The following table gives the tension per square inch for different velocities. v= 10 15 25 50 75 100 125 ft. per sec. V = 600 900 1500 3000 4500 6000 7500 ft. per min. T = 1.4 3.1 8.8 35 79 140 218 lbs. per sq.in. As the centrifugal tension balances part of the tension pro- ducing friction on the pulleys, then T,—7; and T;—T, are the effective tensions preventing slip. If we take K ile we now have go as or if P=driving T; Ts = T. effort=7,—T; in Ibs. per sq.in. of section, T= GP + T., which shows how the value of 7; is increased by centrifugal tension. 152. In order to calculate the speed (R.P.M.) of pulleys connected by a belt it is necessary to know their diameters and the 204 ELEMENTARY MACHINE DRAWING AND DESIGN R.P.M. of one of them. If there was no slip of the belt the velocity of the circumference of one pulley would equal that of the other pulley, that-is if D; and De are the diameters of the two pulleys and Nj and N¢ their respective R.P.M. we would have TDN, = TDo2Ne. Cancel x and we have DiNi =D2N2. If there are several pulleys in a train of belting, as in Fig. 144, let Ni=R.P.M. of the first driver (K), and Di=its diameter. Neand De2=R.P.M. and diameter of the first follower (A). Then N1D,=Ne2D2. The second driver (B) is on the same shaft as (A) therefore has the same R.P.M. as (A), viz.. N2=Ns3, but its diam. being D3, we have NeDo=N3D3= Fia. 144.—Belting Diagram. N2D3. The second follower (C) has a diam. Da and makes Na R.P.M. and N3D3=N4 Ds. Substituting for N3D3 and N2De we have NiXDiXDs3 _ N D2XDe from which we see that the continued product of the diameters of the drivers multiplied by the R.P.M. of first driver and divided by the continued product of the follower diameters gives the R.P.M. of the last follower. This relation holds for any number whatever of drivers and followers. If all but one of the diameters and R.P.M. are given, the missing number may be determined by solving the above equation. 153. Pulleys. Pulley is the name given to a wheel which transmits or receives power by means of belting. It has a smooth face, wider than the belt (from } to 1”), which is tapered from the center towards each edge to keep the belt from running off if no other prevention exists such as a belt shifter or flanges on each side of the face. This taper is called the crown of the pulley and is indicated in Fig. 146 by the letter (c), the distance being exaggerated for PULLEYS AND BELTING 205 clearness in the drawing. The rim has a thickness at the edge denoted by (f) and tapers towards the center 3” per ft. to facilitate drawing the pattern from the mould in the foundae This same taper is found on the outside of the hub and for the same purpose. _B+6 , D iy, Values for (t) and (c) may be taken as C= 09> t=ao9 +3 for pulleys made of cast iron. The rims of C.I. pulleys should not be run at a greater speed than 100 ft. per sec. to be safe against bursting. Sek ee 154. The arms of a pulley are made of elliptical or segmental plane of revolution of the center lines YD 2 to 3h, whichever looks better on : Y/, in diam. if the width of face is 6” or cross section as shown in Fig. 145. The dimension (h) is taken in the ARS of the arms and at the circumference of the hub. At the rim (h’) is from the pulley considered. The number Sedheental Avari Sectfon of arms may be taken as 4 up to 20” less, and 6 arms for sizes above this. Unwin gives the rule for number DB of arms=3+ 150 the nearest whole Elliptical Arm Section number being taken, not less than Fia. 145. four. 155. The hub of a pulley may have a diameter at center of twice the shaft diameter and decrease in diam. towards the ends 1” per ft. The length of the hub varies from 3B to {B depending on whether it is a loose pulley or one keyed to the shaft. The key is placed under an arm to avoid weakening the hub. A set screw is often used to keep the key from being displaced relative to the hub, or vice versa. The set screw in Fig. 146 (D) is shown at right angles to the key, but it is often placed so its point will bear on the key instead of the shaft. If the pulley is small a hole may have to be drilled in the rim over the set screw to allow the hole to be drilled and tapped in the hub to receive the screw. Some small pulleys are occasionally fastened to shafts by two set screws without a key, but this practice is not to be recommended. The diameter of set screw lies between 2” for small pulleys to 2” for large ones. The center of the 206 ELEMENTARY MACHINE DRAWING AND DESIGN screw may be placed half way between the arm and the end of hub. 156. The size of pulley arms depends on the force exerted on the rim of the pulley tending to make it rotate. h' Fig.D Straight, Arm Pulley t “ie Fab" Zz DI | ey | Key~| pez | ie 2/2 ERS 7 | ! Fig. A Fig. 146. Steppep Conge PULLEY. This force tends to break the arms at the hub so it is only necessary to determine their dimensions at this point. The turning effort (P) at the rim may be considered as 45 B Ibs. for single belts up to 90 B lbs. for double belts. B=width of face. PULLEYS AND BELTING 207 The moment of this force about the center of the pulley is ERD. It is somewhat less at the circumference of the hub but this may be neglected. The resistance to breaking at the hub is divided between the arms. The resistance offered to the breaking strain depends on the shape of the arm cross-section, its distribution about an axis parallel to the shaft passing through the center of area of section, and the kind of material. Since the section of pulley arms is elliptical or segmental and the material cast iron, the resistance of the section may be expressed as Si where S=the stress per sq.in. allowed for C. I. Z=moment of inertia of the section and c=distance from the neutral axis to the extreme fibre of the section. The expression for the load on each arm is 2x? where n=number of arms. The resistance of an arm of pele section whose minor axis=0.4 major axis (h), is 8X35 5 xh? x0. 4h, which equals .0393h3 XS. This is practically the same for a segmental section. We then have for the expression of bending and resistance to bending a = .0393Sh3. The value of (S) should be from 1800 to 2250 lbs. per sq.in. for C. I. arms. For S=1800, we have h= 0.192,4°2. n h= oareg PP, z In case any other value of S is desired the general formula ie ee 0786Sn For S= 2250, may be used, 208 ELEMENTARY MACHINE DRAWING AND DESIGN The value of (P) is often not definitely known, so the arms must be designed to resist the maximum driving force which is rarely in excess of 4 the greatest tension in the belt. This reduces to the expressions for (P) given previously as 45B for single belts, 90B for double belts. B=breadth of pulley face. 157. The arms of pulleys are either straight as shown in Fig. 146 (D), or curved as in Fig. 147. The object of curving the arms is to prevent fracture when the casting cools. With modern methods of casting and properly proportioned straight arms, this danger is greatly lessened. At the present time there are many other kinds of pulleys besides C. I. ones. Wrought iron and pressed steel pulleys are Fie. 147.—Curved Arm Pulley. used for high speeds and large diameters owing to their lightness and freedom from cooling strains. Wood pulleys are used where lightness is important as they are 30-70% lighter than C. I. and 30-40% lighter than steel and W. I. pulleys. Some C. I. pulley rims are made with cork inserts which give a higher efficiency than plain rims. Paper pulleys with and without cork inserts are also used to obtain an increase of efficiency over plain C. I., W. I. or steel pulleys. Split Pulleys are used when there would be difficulty in placing a pulley on a shaft on account of bearings or other pulleys already in place, or because a pulley would be too large to either handle PULLEYS AND BELTING 209 or ship if made in one piece. Such pulleys are subjected to the same forces as solid pulleys when once bolted in place on the shaft. 158. The diameter of shaft hole in a pulley does not depend on the diameter nor on the width of the face of a pulley, therefore the hub of a pulley is usually made of such a size as to allow the pulley to be used on several shafts up to a maximum above which it will be necessary to use another size hub. Bushings are used for smaller shafts. The diameter of hub may be assumed for drawing purposes in some such proportion as follows: From 9” to 12” diam. pulley make hub 33’ diam. when face =2” og tp 19" eR a ies ee ge te oft te 12” te a ayy eg gt Bie i " a sy” Eg ne 1B he 16” i . ae ah RD ee ie a tn 16" e tt ayy egg 18” 4-16” ee fi ayy gM eT te Oa” He - ayy eg gt eg to oa ad at ah epMarR! i Deg 80” fi tt a” Og (39 to 83” e a aay eg ger 159. Cone Pulleys or Stepped Cone Pulleys are used when varying speeds are desired of a shaft which is run by a belt from a shaft turning at constant speed. Fig. 146 (4) shows a stepped cone pulley having 5 steps. The large end of the cone is filled by a plate which is keyed to the shaft and fastened to the shell of the cone by flat head screws. The screws should not be less than ?”’ diameter and 4 or 6 in number. The rim thickness and crowning are calculated as for a straight arm pulley. The thickness of the sides of the steps may equal the sum of (¢) and (c), or t+75”. The plate may either be set flush with the large end of the cone as at (B) or placed against the last step as at (A). On large pulleys this plate is replaced by a spider made with arms to save weight. 210 ELEMENTARY MACHINE DRAWING AND DESIGN 160. The weight of C. I. pulleys often needs to be known approximately and the following formule are given for making the calculation. D. K. Clarke gives the weight in lbs. per inch of face width as W=7.6D—1.5 up to 12D—9.5 for rough castings where D=diam. of pulley in feet. An American rule is wgt. of pulley in lbs. = W = (.0175D! 8" +3) B+ .0362D?—2. D=diam. in inches. B=face width (inches). 161. The following books and articles are recommended to those students who wish to study the subject of belting still further. Machine Design, Vol. 1, W. C. Unwin. Machine Design Const. & Drawing, H. J. Spooner. Machine Design Const. & Drawing, D. A. Low and A. W. Bevis. Machine Design, Kimball and Barr. Pulley and. Belt Transmission, Rockwood Mfg. Co. catalogue. Articles on Belt Transmission, Proc. A.S.M.E. Lathe Design, Nicholson. Machine Design, Chas. L. Griffin. Machine Design, A. W. Smith. INSTRUCTIONS Pulleys Example 1. Paper No. 2 size. 2 hrs. in class for drawing a pulley in pencil. Scale, full, half, or quarter size. A pulley larger than 11” diam. up to 21” diam. to be drawn half size. Above 21” to be drawn quarter size. Place the front view at the bottom of sheet and section at top of sheet. If it is necessary break away some of the front view to avoid interference with the section. Draw a straight arm pulley whose - diam. D=( )”, B=( )", d=( )". Take the diam. of hub as given in Art. 158 in Chapter XIV on Pulleys. Use the type of arm section shown in Fig. 145 ( ) making the calculation for its dimensions from Art. 156, using a ( ) belt and a value of S=( ) lbs. The number of PULLEYS AND BELTING 211 arms may be taken as in Art. 154, t, c and h, will be found in Art. 153 and 156. Draw in dotted lines a curved arm meeting the rim at the same place as one of the straight arms. Use method shown in Fig. 147. Make a note on the sheet giving the data assumed in calculating the size of arm, as “ Arm Calculation for Belt.” ‘Strainin Arm S= lbs. per sq.in.” On the pulley itself give all the dimensions in numerals. Work out all dimensions before coming to class, and make a sketch with dimensions on it, TasBLe ror Givinc Sizes Usep 1n Drawine STRAIGHT AND CURVED ARM PULLEYS No 1 2 3 4 5 6 7 8 9 D 16 16 16 16 16 16 17 17 17 B 4 5 6 7 8 9 5 6 7 d ee ee eee es a Arm | Seg. EL. Seg. Ell. | Seg. | Ell. | Seg. | Ell. | Seg. Belt | Sing. | Doub. | Sing. | Doub.| Sing. | Doub.} Sing. | Doub.| Sing. S 1800 1900 2000 | 2250 | 1800 ! 2000 | 1900 | 2200 | 1800 No. | 10 11 12 13> jae Os I TB ae | D | 17 18 18 18 | 19 | 19 19 | 19 | 20 B 8 6 7 8 6 7 8 | 9 6 d 2 22 22 oi | 22 | 2] 2 | at | at 8 Arm | Ell. Seg. Ell. Seg. | Ell. | Seg. | Ell. | Seg. | Ell Belt | Doub. | Sing. | Doub. | Sing. | Doub.| Sing. | Doub.| Sing. | Doub. S 2000} 1900 2000 | 1900 | 2200 | 18C0 ! 2000 | 2200 | 2250 No. | 19 20 21 22 23 24 25 26 27 28 D 20 20 20 21 21 21 22 22 24 30 B 7 8 9 5 6 7 6 7 8 8 d 2h 24 2} 23 23 24 23 23 28 22 Arm | Seg. | Ell. | Seg.| Ell. | Seg.} Ell. | Seg. | Ell. | Seg. | Ell. Belt. | Sing. | Doub.| Sing. |} Doub.|Sing.| Sing. | Doub.! Doub.] Doub.} Doub. S |1900! 2250 | 2000] 2000 | 1900; 1800 | 2000 | 2250 | 2000 | 2250 An assignment No. with an e after means take all data from the column given except the last two lines which will be taken from the next column to the right. Example 2. Cone Pulley. No. 2 Paper. 2 hrs. for pencil in class. Scale half size. Place the cone pulley with its axis vertical, small end 212 ELEMENTARY MACHINE DRAWING AND DESIGN near bottom of paper. Make a view of the large end near the top of the sheet. Omit the shaft in both views. Use the same diam. steps as given: in Fig. 146. Make B=( )”, d=shaft=( )”. Use the style of end plate shown in Fig. (__) of Fig. 146. Make the screws 2” diam. 4 in number. Other dimensions as shown in Fig. 146 to be placed on drawing. TaBLeE FoR Cone PuLuey AssIGNMENT No. ta |S 14) el ete > el & | told | il we B 2/2 /2 }2 |2 | 2/2 |2 | 23] 22] at | 22 | 22 d iz} iz] 1g}13]} 13] 2] a3) 12} 2 | 2] 12] 13] 2 Style |A/B/A/B/]A|B|A|B/]B]B|B|Al|B B 92 | 2: | 2 | OF | ob | 2a] 22 [at] 3 | oe | 22 | on] 2 d Ste Tio |S |e To ot) ori os) on fat] Style |B|A\A/|A!B]A\B]B|B]B|Al]A|A Pulley Calculations Prob. 1. In Fig. 144 is shown a train of belting connecting an engine flywheel (K) with a generator pulley (H). The flywheel belts to a jack shaft on which is a pulley (A) and a pulley (B). From (B) the belt runs to a line shaft carrying two pulleys (C) and (£). From pulley (£) a belt runs to a counter shaft on which are two pulleys (F) and (G), the latter belting direct to the generator pulley (H). Suppose the R.P.M. of (K) and (H) are known, the horse-power of the motor and the diameters of all the other pulleys except one. It is desired to find the proper diameter and width of face of the pulley to be used in the place of the missing pulley. The R.P.M. of (K) and (H) as well as a series of diameters of the intermediate pulleys are given below. Omit the diameter of some pulley (the one indicated in the assignment table following) and calculate the diameter and width of face of the pulley which should be used in place of the one omitted. Use driving force =40 lbs. per in. of belt width. DIAMETERS cF PuLusys Pulley. K A B Cc E F G A Diameter........ 60 18 20 22 24 20 19 16 PULLEYS AND BELTING 213 Use column ( ) below for obtaining the necessary values of X, Y, horse-power, etc. albile|dlelfig{|hlijk|limi|n|o|pl|laqir x 50) 53} 58] 60) 65) 70} 79)100/120/150/175)190/200} 90)110/130)160 Y 215)228/246|258/280/301/340/430|516/645|751|817/860|387|475|560|/687 mit |A)/H|B\|C\EH|F/\G\|A\|B\C\|E\F\|G|H|A|BIC H.LP. 6} 6} 8} 9} 10) 10) 10) 14) 15) 16} 18] 19) 10) 14) 15) 16} 17 Prob. 2. A pulley (A”) diam. making (B) R.P.M. drives a second pulley which makes (C’) R.P.M. and the H.P. transmitted is (D). Find the diam. and face of second pulley when the values of A-B-C-D are given as below in column (__s). TABLE FOR PROBLEM 2 a b c d e f g|hj})tijlk{| li myn 10 | 12) 14] 16] 18] 20] 22] 24] 26) 28) 30} 32] 34 200 | 190 | 180 | 160 | 150 | 140 | 140/120]110| 100) 90| 80] 70 70} 90 | 100 | 120 | 140 | 120 | 150] 140} 130 | 120) 110/100] 90 6 7 8 9 8 7 6}; 5] 6] 7} 8; 9] 10 VaAwe Prob. 3. A (A”’) diam. pulley makes (B) R.P.M. and its face is (C”) wide. What H.P. will it transmit when using a single belt. Use data in column (_) below. Prob. 4. Take the pulley given in Problem 3, suppose it transmits (D) horse-power. How many R.P.M. ought it to make if the face is the same width (C’’) as above. Use column (_) below. TaBLE FOR PROBLEMS 3 AND 4 a b c dje|{f\g|h o4| 24| 26 | 28| 30; 32| 34| 36| use this line for 3 and 4. 200 | 190 | 180 | 190 | 170 | 160 | 140] 150| use this line for 3 only. 6 7 8 9 8 9} 10]| 11] use this line for 3 and 4. 5 6 7 8 9) 10] 11] 12) use this line for 4 only. VaAwar Prob. 5. Two pulleys are (A). ft. apart and their diameters are (B”’) and (C”). If the driving force is (D) lbs. find the tension on the tight and loose sides of the belt and the initial tension when the belt is at 214 ELEMENTARY MACHINE DRAWING AND DESIGN rest. Coeff. of friction =0.3. What H.P. is transmitted when the larger pulley makes (EZ) R.P.M.? Use Table below for values of A~B-C-D-E. TABLE FOR PROBLEM 5 25} 28) 29] 30] 32] 34] 24] 26) 31] 27] 33} 35] 37] 39 12} 14} 16] 16] 17] 20} 20) 22] 22; 22} 18; 18] 18] 18 30] 32] 31] 36] 30] 30; 34] 36] 24] 30] 26; 28 300 | 400 | 500 | 600 | 300 | 400 | 500 | 600 | 300 | 200 | 400 | 500 | 600 | 700 100 | 120 | 130 | 140 | 100 | 110 | 120 | 130 | 140 | 150 | 160 | 170 | 100 ; 110 Soanh we 6 oo w Prob. 6. Taking the dimensions A-B-C-D from the above Table, find the values of T;7,T; using the theory taken from Unwin in Art. 150. Prob. 7. Find the tensions, 7,7; and T; when the centrifugal force is taken into account (Art. 151). Use the data from the above Table. In order to obtain a velocity of belt take the R.P.M. of the larger pulley as three times the value of (£) given in the Table above. Prob. 8. Calculate the weight of your C. I. straight arm pulley as closely as possible and then compare this weight with the weight as figured by D. K. Clarke’s rule and by the American rule. The weight of C. I. =.26 lb. per cu.inch. Prob. 9. Calculate the tension per inch of width in a belt 3,” thick due to centrifugal force when the belt is running 5000 ft. per min. Prob. 10. Calculate the unit stress in the shaft of your pulley as given in assignment table for Ex. 1 supposing it to be of wrought iron. (Art, 89). CHAPTER XV SPUR GEARING 161. TooruEp wheels are used to communicate motion between shafts which are (1) parallel, (2) non-parallel, but in the same plane, (3) non-parallel and non-intersecting, (4) at right angles. To the 1st class helong “ spur gears,” to the second “ bevel gears,” to the third “spiral gears’ and to the fourth (which is a special case of the third) “ screw gears.’’ In all these cases the working surfaces of the teeth transmit motion by sliding contact. The demonstration of this as well as the theory of the form of these surfaces will be found in the text books on mechanism. Spur gears, Bevel gears and Worm gears will be taken up in this order in this course. 162. Spur Gears. These are cylindrical wheels having teeth on the periphery which roll in contact with the same velocity ratio as two cylinders in contact or as two pulleys connected by a belt. The circle of the gear which is used for calculating this velocity ratio is called the pitch circle, the depth of a tooth being taken partly outside and partly inside this circle. The circle which passes through the tops of the teeth and which has the greatest diameter of any part of the wheel is called the addendum circle. It is the outside diam. of the cylinder into which the cutter enters to form the teeth on cut gears. The circle which passes through the bottom of the spaces between the teeth is called the dedendum or root circle. See Fig. 151. The curves which form the faces and the flanks of a tooth are either cycloidal (as in Fig. 151) or involutes of a base circle which is inside the pitch circle, as shown in Fig. 150. 163. The width of a tooth on the pitch circle is governed by the pressure which the gear transmits and the length of the tooth. The distance measured along the pitch circle from a point on the face of one tooth to the corresponding point on the next tooth is called the circular pitch. It includes a tooth and a space. In 215 216 ELEMENTARY MACHINE DRAWING AND DESIGN cut gears the tooth and space are equal in width but in gears’ having cast teeth the space is wider than the tooth, the difference in width being called backlash. The addendum (in cast gears) is found by multiplying the circular pitch (P) by some fraction which will give a practical result as (.8P). The dedendum equals the addendum plus a small amount called the clearance, which allows the tops of the teeth of one gear to clear the bottoms of the spaces on the other. This clearance is taken as (0.08P) on cast gears. On gears with cut teeth the depth of the tooth is determined from a term called diametral pitch. This term is the number obtained by dividing the number of teeth in the gear by its pitch diameter. The commercial diametral pitches begin with 1 and increase by fourths up to 3, then 33, 4 and by whole numbers up to 12, then by even numbers to 32, 36, 40, 48. The product of the circular and diametral pitch equals 3.1415=z, so if one pitch is known, the other one can be easily found. 164. In speaking of diametral pitch the terms three pitch, eight pitch, etc., are used, whereas circular pitch is expressed in inches and decimals of an inch. Diametral pitch is a ratio; circular pitch is a linear distance. The smaller the diametral pitch the larger the tooth. The addendum (A) of a cut tooth equals the reciprocal of the diametral pitch. See Fig. 151. The clearance (cl) equals ;5 the addendum and the dedendum (E) equals the sum of addendum and clearance. Since the depth of a tooth is the sum of addendum and dedendum, we can write an expression for the depth in terms of the diametral pitch (p), viz. 2. de 9 depth= +5 tig" pip 10p’ The outside diameter of the gear, which is the same as the diam- eter of the circle passing through the tops of the teeth (called the addendum circle) is found by adding twice the addendum to the pitch circle diameter. If a pitch diameter ae the outside diameter Do=D+2A; but A a a nen Do=D+=. Since D= = where N=number of teeth, we on by sink: N+2 N ,2 Do=—+— or Do= OD |p 0 SPUR GEARING 217 The following table gives the relation between diametral and circular pitches. Diametral Circular Diametral Circular Diametral Circular Pitch. Pitch. Pitch. Pitch. Piteh. Pitch. 1 3.141 5 0.628 20 0.157 1} 2.513 6 0.524 22 0.143 1} 2.094 7 0.449 24 0.131 2 1.795 8 0.393 26 0.121 2 1.571 9 0.349 28 0.112 23 1.396 10 0.314 30 0.105 23 1.257 11 0.286 32 0.098 2 1.142 12 0.262 36 0.087 3 1.047 14 0.224 40 0.079 35 0.898 16 0.196 48 0.065 4 0.785 18 0.175 165. In order to train the eye to identify teeth of different pitches a number of teeth are shown full size in Figs. 148, 149 and marked with their proper pitch number. 166. In machine drawings the outlines of wheel teeth are not shown as a rule, but in case they may be required, a method is shown for both involute and cycloidal systems. In the involute system the tooth outline is a single circular arc between the adden- dum and base circles (Fig. 150). The angle of action is taken as 15° and (y) is the point on the line of centers of the gears in mesh. The base circle is tangent to the line drawn through (y) making 15° with the common tangent to the two pitch circles through (y). The point (x) on this 15° line of action is found by drawing a line from the center of the pitch circle perpendicular to the line of action. (x) is the center of the arc forming the face of the tooth whose radius R=ay. This arc terminates at the base circle which is a circle tangent to the line of action and has its center at the center of the wheel. From the base circle to the root circle the line (mo) is radial. This distance is usually so short that the fillet completes the part of the tooth inside the base circle. The width of a tooth equals the space, each one being half the circular pitch. 167. If the teeth are cycloidal in shape the outline is composed of two curves, one for the face, the other for the flank, as shown in Fig. 151. The radii for the curves are obtained from the table from Grant’s Odontics by multiplying the tabular value by the circular pitch or dividing the values given for diametral pitch 218 ELEMENTARY MACHINE DRAWING AND DESIGN by the diametral pitch of the wheel. The centers of the arcs for the faces are taken inside the pitch circle on a circle whose a. 5 és e oO. oo a = QO: a oO: oO o oO Q: s\s a © a: a: S ~~ radius is less than the pitch circle by an amount called (dis.) whose tabular value is changed according to the pitch of the wheel. Fig. 148, SPUR GEARING 219 The flank are centers are on a circle of greater radius than the pitch circle. The increase of radius is the tabular (dis.) given aw 6 2 Ss Ca ON ® oO- > on b To cs LY of & OZ S& aS £ “ Y a: } ~ SPUR GEARING 227 174. The strength of an elliptical arm is given in Art. 156, the pull (P) on the belt corresponding in gears to the pressure (F) on the teeth. If (f) is taken at 3000 Ibs. unit stress we have the following expression for the size of the elliptical arms. FD _ % 5» On 73904 O00: oie we Ge SY In this case y is taken as equal to - giving the ellipse axes a and 6 at gs center of wheel. 3 If (6) =e «this reduces to *? - 7 3000, from which we have FD FD — 3 = 3; a=W = 18. nendhs Hee “a eh 175. Applying this method to the calculation of an arm of + section gives us the width of the arm called (B) in Fig. 152. The strength of the arm is nearly all due to the part which is in the plane of bending, viz.: the plane of rotation of the wheel, the ribs being of very little value to resist the bending action. If we disregard the ribs we have left to calculate a section, rectangular in shape, whose width (¢) corresponds to (b) in the table of Art. 173, and (B) corresponds to (h), from which ED = f=) for’, es be pa a BD F If the letters of Fig. 152 are substituted in (1) it will appear as pa. As = and (f) is taken equal to 3000 lbs. per : dich cao de ee : sq.in., substitution gives a B =i000n’ from which RE B=y|00 = 04 Bo eee es 2) 228 ELEMENTARY MACHINE DRAWING AND DESIGN This value of (B) is laid off at the center of the wheel as shown in Fig. 153 (C) and the arms are tapered towards the rim. The widths of the arm at the points where it joins the hub and the rim are to be given on the drawing in preference to (B) alone. The value of (F) must be calculated from the teeth dimensions by formula (3) in Art. 180. The calculation for an arm of 7 section is made in a similar manner and dimensions are laid off on the drawing similarly. 176. Henry Hess gives the following formule for arm dimen- sions for spur gear wheels, in which the letters used are as follows: Z=modulus of resistance of arm section, P=circular pitch, p=diametral pitch, R=ratio of face width to circular. pitch =%, f= face width, N=number of teeth, n=number of arms, 3 = Z ee for circular pitch, . . . . (1) and mR(N—7), : ‘ Z eae for diametral pitch, . . . (2) Applying these formule to arms whose cross-section is an ellipse having a minor axis (Z) and major axis (2) gives (N —7)P3P a3 2 E= ae Oek =, wer’, for circular pitch . (3) (N—7)R — 3{(N—7)r2b : ‘ a|( “ier = {aoe for diametral pitch. (4) These are taken from The American Machinist Gear Book, by C. H. Logue, in which graphical charts are also given for their solution. SPUR GEARING 229 177. The weights of gear wheels are often estimated by some empirical rule which depends on the circ. pitch, width of face, etc. Reuleaux gives the following: W =0.0357 bP?(6.25 N-+0.04 N?) where b=face; P=circular pitch; N=number of teeth; W =weight of gear blank. (Deduct 30% for finished wet.) As (P?) is not found to be a correct factor and the values of the constants for the No. of teeth run parallel with the No. of teeth, etc., Mr. Logue finds the wet. W=NXbXK where (K) is a constant given for the pitch = 2 2 Am oo WaigXNxh 2... This formula cannot be used for low numbers of teeth nor above 33” circular pitch. Another formula from Machinery reduces to much the same as the preceding, being W=KXP?bN. (K) is a coefficient which may be taken as 0.35 for pinions and 0.45 for gears, or if the diameter (D) is known and (P), the weight of pinion=3.1 DP?. Weight of gear=4.2 DP?. The price of gears varies so much that a formula for calculating it cannot be given. The type of formula can be expressed as: Price = (coeff.x PN-+coeff.xP) the coeff. to be determined for each manufacturer. Cast iron cut gears cost about 20% more than cast teeth, and cast steel gears from 50% to 75% more than cast iron gears of the same size. 178. The efficiency of spur gears varies from 90% to 98% depending on the quality of workmanship, lubrication and material used. The teeth should be as near the perfect theo- retical shape as possible, and mounted accurately and firmly in position. The lubrication should be continuous, of sufficient quantity and the gears should work in a closed casing. The material should be hard and close grained and different for the two gears. Hardened steel will give the best results with phos- phor bronze. The pitch should be as fine as possible, without too wide a face, as this tends to increase the efficiency. Any- thing which tends to shorten the line of contact and confine it to the pitch point increases the efficiency. 230 ELEMENTARY MACHINE DRAWING AND DESIGN 179. Strength of Gear Teeth. The load which a gear trans- mits is divided among several teeth depending on the angle of approach and recess and on the quality of workmanship. The load might be brought on the corner of the top of one tooth and would tend to break that tooth along a surface running from the root of the tooth to some point on the top. This is shown in Fig. 154 by the portion abed. As this would only occur in very badly fitted gears, the case of the load being uniformly distributed along the top of the tooth is the condition most used in calculations for strength. The tendency is then to break off the tooth along the surface (abcd) shown in Fig. 155. Py Fig. 154. Fig. 155. This surface is a rectangle whose sides are (b) the width of face and a side equal to the width of a tooth at the root which is nearly - This value x is too large for wheels having less than 24 teeth and too small for more than 24 teeth. The bending moment of the force (F) will be (Fh) and the resisting moment at the root (fz). z is the modulus of the section (a rectangle) which is given in the table in Art. 173, as tbh?. As (h) equals = in Fig. 155, we have z=,bP?. In Art. 168 the involute system gives the depth of a tooth equal to .6866P, therefore Fh=FX.6866P and equating the bending moment and resisting moment gives -6866FP = x¢bP?f, from which F=0.607bPf, or for any other system of gearing F=kbPf where (k) is a variable which depends on the shape of the teeth, or on the number of teeth in the gear. SPUR GEARING 231 Mr. Wilfred Lewis, an American engineer, made a large number of experiments to determine the value of this variable (k) for the cycloidal system as well as the 15° and 20° involute systems, by drawing a large number of teeth and measuring their thickness at the root. From these measurements he evolved in 1893 what he calls a tooth factor. His formula was F=SPby, in which F=the load taken at the pitch point, S=the working stress in the material, b=the width of face, and y=the tooth factor. The tooth factor for 15° involute and cycloidal teeth = (0.124 — | , for 20° involute teeth y= (0.154 - oe) . For 0.276 radial flanks y = 0.075 -——y— . Nz=number of teeth. Below will be found a table of values of (y) calculated from these formule for various numbers of teeth. VALUES OF Factor (y) y y No. me Teeth.| Invotute, | Myles | Radiat |lreeth.| Iavotute, Inyolute, | Radial 20°. Cyeloidal. | Flanks. 20°. Ceyloidal, | Flanks. 12 0.078 0.067 0.052 27] O.111 0.100 0.064 13 0.083 0.070 0.053 30 |} 0.114 0.102 0.065 14 0.088 0.072 0.054 34 {| 0.118 0.104 0.066 15 0.092 0.075 0.055 38 | 0.122 0.107 0.067 16 0.094 0.077 0.056 43 | 0.126 0.110 0.068 17 0.096 0.080 0.057 50 | 0.130 0.112 0.069 18 0.098 0.083 0.058 60 | 0.134 0.114 0.070 19 0.100 0.087 0.059 75 | 0.138 0.116 0.071 20 0.102 0.090 0.060 100 | 0.142 0.118 0.072 21 0.104 0.092 0.061 150 | 0.146 0.120 0.073 23 0.106 0.094 0.062 300 | 0.150 0.122 0.074 25 0.108 0.097 0.063 jj rack} 0.154 0.124 0.075 The value of (S) in the formula depends on the material of the teeth and on the velocity of the teeth. The load on a tooth must be reduced as the speed increases on account of the impact. If the working stress allowed when the wheel is at rest, called static stress, is multiplied by ae the result will be the allow- able working stress (S) when the velocity of a point on the pitch circle is (V) ft. per min. This multiplier was introduced by 232 ELEMENTARY MACHINE DRAWING AND DESIGN Carl G. Barth and gives nearly the same result as the table originally given by Mr. Lewis for the working stresses for different speeds of teeth. If the static stress S, at (0) velocity is taken as unity, the following table gives the multipliers for finding the working stress a use for different speeds when modified by the expression —— io ae The value of the static stress’ S; may be taken as follows: Cast iron, ordinary workmanship = 6,000 - high grade workmanship = 8,000 Steel, ordinary workmanship = 15,000 g “high grade workmanship =20,000 |" Phosphor bronze, ordinary workmanship = 9,000 “cc “ce high grade workmanship = 12,000 TaBLE oF STRENGTH Factors Vel. in ft. per min. =V! 0 | 100) 200] 300) 450} 600} 900} 1200) 1800 | 2400 Strength factor.......]1.0].857| .75].666].571} 0.5] 0.4/0.333] 0.25 0.2 180. The formule for calculating teeth for various speeds will be found below in their proper order for consecutive calculations. (1) To Find Velocity per Min. at Pitch Circle, Multiply together the pitch diam. in inches (D), the R.P.M. (R), and 0.262 or V=0.262DR. (2) To Find Allowable Unit Stress at Given Velocity, Multiply the allowable static stress by 600 and divide by 600 the velocity in ft. per min.+600 or S=S,X =~ V-+600" (3) To Find Safe Tangential Load at Pitch Circle, Multiply together the allowable unit stress for the given’ velocity, the width of face, the circular pitch and tooth factor or F=SPby. (4) To Find Mar. Safe Horse Power, Multiply the safe load at the pitch line by the velocity of pitch circle in feet per minute and divide by 33,000 or _ FV H.P. ~ 33000’ SPUR GEARING 233 181. A rule for the width of face (6) of a gear suggested by an English engineer and giving good results in practice is as follows: b= o1sv y+ that is, multiply the sq. root of the velocity of the pitch line in feet per min. by 0.15, add 9 to the product and divide by the diametral pitch: The width of face in practice varies from 2 to 8 times the circular pitch with 3 as a fair average for gears moving rapidly and having cut teeth. The speed of gear teeth is limited by the noise they make. This noise is objectionable when ordinary cut gears are run at a higher velocity than 1200 ft. per min. at the pitch line. 2000 ft. per min. is about the practical pitch line speed limit, although gears of special design may run even higher than this. The maximum speed to avoid fracture would be for different gears somewhat as follows: 4,000 ft. per min. for C. I. of 8,000 Ibs. per sq.in. sta. Strength 8,000 “ “Cast St. of 16,000 “ = “ 10,000 . “Mach. St. of 25,000 ‘ . + The gears to attain such speeds must be exceptionally accurate and well balanced. A formula giving the safe speed depending on the number of teeth in contact (n), circular pitch (P) and a factor (k) is given below. For general use the estimated factor is also given. Safe speed = Pnk. Style of Gear. Value of (*). Spur Gear, pattern molded........ 0 to 300 ‘« machine “........ 110 to 450 ue «« | commercial cut........ 600 cc «* with exact cutters...... 700 cc" cut stepped teeth...... 820 ee EE AbOhs Sine yeeceeae es 900 mo SE @ FAWN O si se cic segy oe sanes tee 1000 The number of teeth in contact may be found graphically by dropping perpendiculars from the centers of the gears to the line of pressure. If the points of intersection of these perpen- diculars with the line of pressure fall outside the points where the addendum circles cut the line of pressure, the No. of teeth between the latter points is used. If they fall inside, the No. 234 ELEMENTARY MACHINE DRAWING AND DESIGN of teeth is taken between the feet of the perpendiculars. The distance between the points divided by the circular pitch gives the No. of teeth in contact approximately. 182. In making calculations for gear tooth strength the smallest gear in the train must be calculated if all the gears are of the same material. The teeth of small gears are weaker than wheels having the same pitch but a greater number of teeth, as will be evident by inspecting the tooth factor in the Lewis formula. If the pinion can be made of a stronger material than the gear which meshes with it, then the calculation for the tooth strength will be based on the material of the larger wheel. 183. Gear calculations can be simplified by the use of diagrams, especially in cases where the first calculation to find the pitch of a gear is only an approximation to the final result. Fig. 156 is a diagram for finding the value of (S) in the Lewis formula when the static strength of the material is known and the velocity at the pitch circle in ft. per min. It is often required to find the pitch when the pitch diameter and speed of pitch circle are known but the number of teeth is unknown. The face of the gear is known or assumed by Art. 181. If the Lewis formula is used for 15° involute or cycloidal systems then F=SPby=SPo(.124-), but TT TT P=p 08 a, F=bx2xs(.124-7), Pp Dp or 389 2.15 = S —_-———. - us( P De) If we call (w) =F = the load per inch width of face of gear and substitute in the last expression, we have 389 71) as le. oe ek Ok ee Gee (5 Dp? (D _S 2.15 from which W and y to this lineor Project the intersection For various values of y Ww Circular Pitch P ® Diametral Pitch p TOD eao oa. 89 5 4 2 4 . u 4 '00,—1,200 ~O ae lads 000 rr z wwe 900 weer 7 ene 00; a ee 800 eseeeet tr Sa 9,000 700 2.64 10,000 —F- 909 J F400 2 300 M24 250 uw 4 ——F 200 8 4 3 1. 150 Ey x ae o I 2 1.04 a oe -30,000 io 0.84 a 0.7, Q L-40,000 z [50,000 - n 28 a 8 2 o od 5 > a DIAGRAM FOR THE LEWIS FORMULA FOR STRENGTH OF GEAR TEETH By H. L. Szwarp The Lewis formula for the strength of gear teeth is We=SPby; or, if b=KP; W=SP2Ky. A line from S to W crosses the center line at the same point as a line from X to P. EXAMPLE. Given: Cast Iron gear, velocity of teeth =150 ft. per min. W =3500 lbs, =load carried by teeth. Y= K=4. Required: P As shown by dotted lines, P=o.o6s, To face page 234. Fig. 158. a Values of y aac oof ine or vice versa, $1 2A 1 Values of Factor (4) y ¥y Rp" [Lavetute} Eavolute| Razin |] No Tavolute) Involute| padtal eetn| 20° [cyctutdai| Flanks || posen| 20° oy Flanks 12 | 0.078 | 0.067 | 0.052 27 | 0.111 | 0.100 | 0.064 13 | 0.083 | 0.070 | 0.058 30 | 0.114 | 0.102 | 0.065 14 | 0,088 | 0.072 | 0.054 34 | 0.118 | 0.104 | 0,066 15 | 0.092 | 0.075 | 0.055 38 | 0.122 | 0.107 | 0.067 16 | 0.094 | 0.077 | 0.056 43 | 0.126 | 0.110 | 0.068 17 | 0.096 | 0.030 | 0.057 50 | 0.130 | 0.112 [ 0.069 18 | 0.098 | 0.083 | 0058 60 | 0.134 } 0.114 | 0.070 19 | 0.100 | 0.087 | 0.059 75 | 0.138 | 0.116 | 0.071 20 | 0.102 | 0.090 | 0.060 |} 100 | 0.142 | 0.118 | 0.072 21 $ 0.104 | 0.092 | 0.061 |] 150 | 0.146 | 0.120 | 0.073 a 0.106 | 0.094 | 0.062 || 300 | 0.150 | 0.122 | 0.074 2 0.108 | 0.097 | 0.063 || rack | 0.154 | 0.124 | 0.075 ae, AY M5 e ie 28 a ¢ 1,000 3 ao 1,500 fe " £2 2,000 4 2000 3 1,200 F2500F Y t00 1,000 -3,0004-900 E 600 10 “OST “OL SPUR GEARING 235 Sale Pressure in Pounds for Cast tron = nm o wn a =! o Oo 3° °o oO oO ° °o So 236 ELEMENTARY MACHINE DRAWING AND DESIGN Fig. 157 from Kimball and Barr’s Machine Design is based on formula (1) above and enables the diametral or circular pitch to be easily found when the width of face (6) is known, the diameter of the pitch circle (D), the stress allowed in the teeth (S) (found from Fig. 156) and the load (F) on the teeth. Enter the diagram at load (w) per inch of face on scale (C) come down vertically to the inclined line denoting the stress allowed (S), called P in diagram). Run along horizontally from this point to meet the vertical line erected from the gear diameter at the bottom of the diagram at scale (D). The curved line nearest to this point of intersection indicates the nearest diametral pitch. Diagram Fig. 158 was designed by Mr. H. L. Seward to facilitate calculations involving strength of gears based on the Lewis formula. The directions for its use appear on the diagram. ; 600 The values of (S) have not been modified by the term 600+" but remain as originally given by Mr. Lewis. In this diagram the width of face can be determined for any given pitch in terms of the pitch as b=KP either (P) or (K) being changed to suit the design of gear under consideration. The rule for width of face given in Art. 181 may be used as a trial width. 184. The horse-power transmitted by a train of gears depends on the load on the teeth, the velocity of the pitch circle in ft. per min. and the efficiency of the gears. A single pair of gears will transmit H.P. = (4) X90 to 98% depending on the grade of workmanship and fitting together. The velocity (V) is taken in ft. per min. as ya D=pitch diam. in inches, N=R.P.M. (F) =force on teeth at pitch line obtained from the Lewis formula. 185. The problem of finding the velocity ratio of a pair of gears is the same as for two pulleys (Art. 152). To compute the speed ratio of the first and last gears in a train use the continued product of all the drivers as a single driver and the continued product of all the followers as a single follower and proceed as above. if an idler is placed in the train it does not need to be considered, for it figures both as a follower and a driver. In designing gear trains it is best to divide the reduction as evenly as possible among the different pairs of gears forming the » where 237 SPUR GEARING YH SYMON 602 ye “LST “ON i = SAYIL Ul 109 JO AAaUIOIG - Yf 2DIS 09 0s Ov 0g 02, | Ss 0 Q S y . 00s 2 2¢ Ne j g ooo & 3 ® & 2 00s! 0002 'o 00s 000! o0si 0002 o0s2 oon 2 GID{ JO YIU, Jad POOT -Da/o2g 238 ELEMENTARY MACHINE DRAWING AND DESIGN train. For a reduction of say 64 if two sets of gears were used the reduction in each set would be 1/64=8 whereas in a triple reduction, the reduction in each set would be 2/64 =4. That is, three pairs of gears would be used, each pair having a velocity ratio of 4 to 1. 186. The relative powers of a train of gears are inversely porportional to the velocity of their pitch circles. In adrum and gear hoist as shown in Fig. 159 we have a pulley (G) on a counter- shaft which belts to pulley (F) on the shaft of gear (A). Gear (A) meshes with an idler (B) which turns a gear (C) and the drum (D) to which (C) is rigidly connected. If the drum is rotated it takes up or lets out the rope which is fastened to the weight (W), thus raising or lowering it. The load (W) on the rope produces a pressure (F) on the teeth of gear (C) inversely proportional to the lever arm of the gear, or WXrad. of drum=(F) Xrad. of gear (C). The ‘pressure on the = teeth of (B) to turn (C) will be greater than (F) on account of the Fic. 159.—Hoisting Train. friction between the teeth. If the frictional loss of each pair is 5% then we will need 10% more power in gear (A) than is required to raise the weight (W). That is, the teeth of the gears must be proportioned to stand 10% more than the load (F) requires. The horse-power (H1) supplied at the pulley (F) to the gear shaft is decreased 10% by frictional resistance before the drum is reached, therefore,the power available to raise the weight is 10% less than the initial horse-power, or H=.9H, that is, WY noe BV 33000.9 ~~ * 33000’ Vi=the velocity of pitch circle of gear (A), and F=load on teeth, of (A) V =velocity of wgt. (W). From this equation =F. If (V) is known the R.P.M. of SPUR GEARING 239 the gear (A) are known and the velocity ratio of (D) and (A) can be determined. From this the diameter of gear (A) can be found, if (C) is assumed, which will give at once the load (F) on the gear teeth, enabling the pitch to be determined for all gears which run together. A more extended study of gear transmission will be found in the following books, which have been consulted in preparing this brief treatise. Spur Gearing, Machinery Reference Series No. 15. Elements of Machine Design—Part 1, by W. C. Unwin. American Machinist Gear Book, Charles H. Logue. Elements of Machine Design, Kimball and Barr. Mechanical Engineers’ Pocket Book, Kent. INSTRUCTIONS Spur Gearing Three two hour exercises, No. 3 paper, allowed in class, scale half size. Total time req’d on pl. is 7 hrs. The gears shown in Fig. 153 are to be drawn in the reverse position to that shown, that is, (A) and (C) change places. If any change is made in vertical arrangement it must be by moving the center of the pinion towards the top of the paper. Its pitch circle must be kept in contact with the pitch circle of (B). Fig. 1 may be omitted but the remaining figures are to be drawn, also a section of oval or + arms. The pitch diameters of (A), (B) and (C) will be respectively ( )’, ( )”, ( ). A is to be a pinion in all cases. If the diam. of (B) is 13” or more it is to have oval arms, 4 or 6 in number. (C) is to have + shaped arms, the number being (7) in the assignment table. The diametral pitch (p) is ( ). The width of face b=( )”. The diameters of the shafts in wheels (A) and (B) are to be calculated from Art. 89 using a value of (f) =13,500. That in wheel (C) from formula in Art. 172 increased by 3”. The gears are of cast iron. Keyways to be shown and dimensioned in all gears. Use the empirical formule for oval arms given in Art. 172 or 176 for drawing the arms, then cal- culate the strain (f) in the arms by using formule in Art. 174. For + arms use formule in Art. 172 for drawing, then calculate the stress (f) for the arms, using formule in Art. 175. The hub diam- eters may be taken as twice the shaft diameters except in the wheel 240 ELEMENTARY MACHINE DRAWING AND DESIGN (C) where + arms are drawn. Then take the hub diameter as indicated in. Art. 172, since the shaft diameter will be larger than that necessary to transmit the power from the gear alone. Take the set screw diam. =3” for all cases. Give the dimensions indicated on the drawing in Fig. 153, also the distance c. to c. of shafts of each pair of gears, the depth from top of teeth to inside of rim, diam. and length of set screw, diam. of hubs, width of arm at rim and at hub. For each gear make a note giving the following— T, — DP, — PD, — Face,— OD. Also make a table giving the data for cutting all three gears as illustrated in Art. 169. Calculate the H.P. which the gears will transmit based on the pinion teeth when the pinion makes ( ) R.P.M. (given in assignment table) and when the friction loss is 5 per cent between pinion and gear (C). See Art. 180. Calculate the weight of each gear, giving the name of formula used. ASSIGNMENT TABLE FOR SPuR GEAR PLATE 1/}/2/)3)4!15;,6)7 {819 ) 10/11) 12) 13] 14] 15 8} 8] 8}12}12} 8] 8| 8] 9] 9} 9} 9] 8] 8] 8 12) 12] 12) 16) 16} 12] 12] 12) 14) 12] 13] 14) 12) 12] 12 32 | 36 | 38 | 32 | 24 | 24) 34| 32 | 33 | 32 | 32 | 32 | 32 2 [12/15] 15] 13] 23] 23/22] 3| 2] 2) 2/23} 13] 13 Pitch diam. ( of | Ses BQDE w bo w bo No. of arms. . 4/6/616,)6)4)4)4/6/4)4/4)414]4 Face....... 2/52/6|73|] 6 | 4] 4 | 33) 3 | 43) 33] 33) 33] 53) 5% R.P.M...... 200)215}242|200|190 210/220) 200)190'200)190}180)170)190/180 Example. A student who has column (1) draws an 8” pinion, 12” gear (B), 32” gear (C). Diam. Pitch =2. Gear (C) has 4 arms. Width of face of all gears equals 43 inches. R.P.M. of pinion to be 200. Gear Train Design Prob. 1. In Fig. 159 of the text is shown a train of hoisting mechanism. The table below gives the dimensions of the pulleys and gears, the R.P.M. of (@) and the weight of (W) in pounds. It is required to determine the circular and diametral pitches necessary to use in these gears when they are made of cast iron. The efficiency of the gears is taken as 95%. Determine the number of teeth in each gear, its outside diam. and width of face. The width of face is to be assumed as given below the following table. SPUR GEARING 241 TABLE FoR Spur Gear TRAIN CALCULATIONS 1/2/3/4ls 6|7 8 | 9 | 10] 11] 12] 13} 14) 15 G, Diameter... .| 40) 39) 40} 39] 40) 38] 38] 38) 39] 37] 36) 35) 34] 32) 30 F, ....| 20} 20} 19} 19} 18} 20; 19) 18) 18} 20} 18) 17} 16] 15} 17 A, e weeefees[eee}...] 8’7| for} alll casles B, a ceea{eee[e..]...[12”] for] all! casles C, “ cese{eee[e..[..-|82’7 for} all] casles D 7 : 48’’| for| all! casjes R.P.M. of @. . . .100/110]115)120'125|130}150/160 170] 180] 190 200 210|220|200 Pounds lifted. . . |470'600|700|560/650/490'5001520'540'620|7001800 900/700'800 The width of face of gears to be (3 times P) for those men who have a small (a) after the number of column given them. Those with small (6) use face width =24P “cc 6c (c) ce 6c =2P ce cc (ad) be 6c“ =33P oe ce (e) ce ee =21P 6c 6c (f) 6c “co =23P Prob. 2. Calculate the proper diametral pitch to use on two gears (A) and (B) of the above table if (A) makes the R.P.M. given for (C) and the load on the teeth is twice the number of lbs. given in the last line. Use the Lewis formula and width of face as proportioned by one of the small letters a-b-c above. Work from the smaller gear always. Prob. 3. What should be the pitch and width of face of a steel pinion 4” pitch diam. 15° involute system making 750 R.P.M. and transmitting 10 H.P. The workmanship is high grade. The width of face is to be proportioned according to Art. 181. Prob. 4. A 16 tooth pinion of 1}” pitch and 34” face drives a 75 tooth wheel at a speed of 200 ft. per min. The tooth is 15° involute. (a) Find their working strength if both gears are cast iron. (b) Find the strength if pinion is made of steel and wheel of C. I. (c) Find the horse-power of both combinations. Prob. 5. A pair of 20° involute gears transmit 8 horse-power. The distance between centers is 12” and velocity ratio is 3 to 5. If pinion shaft makes 200 R.P.M. and face width is 12” find the pitch and number of teeth on the gears. Prob. 6. A C.I. gear makes 120 R.P.M. and transmits 20 horse- power. The pitch diam.=30’. What will be (a) its diametral pitch and (b) its number of teeth? 242 ELEMENTARY MACHINE DRAWING AND DESIGN Prob. 7. A C. I. gear 44” diam. is to transmit 21 H.P. at 50 R.P.M. If the diametral pitch is 2.5 find the least width of face. Prob. 8. What gears will be required to lift a load of 2400 Ibs. at a uniform rate of speed employing a 10 horse-power motor, running at 1120 R.P.M. driving with a rawhide pinion 4” pitch diameter. (Safe static stress =5000). Hoisting drum diameter = 10’. Prob. 9. A load of 2400 lbs. is to be raised at a uniform speed of 115 ft. per min, What size motor and gears will be required? Test No. 8. Arts. 116-186. 33 Hours allowed. 1. An engine with 12” bore, 14” stroke, carries steam at 100 lbs. pressure. The connecting rod is 40’ long. Calculate the following dimensions: (a) Diam. of piston rod (outside). (f;=4500). 10 threads per in. (6) Area of crosshead bearing surface in inches (press. per inch allowed =30 lbs.). (c) Diam. and length of crosshead pin (bearing pressure =1200 Ibs. per inch). Pin to have a length =1} xdiam. (d) Diam. and length of crank pin (bearing pressure =1000 lbs. per inch). Length of pin =1.5 xdiam. (e) Calculate diam. of crank shaft when f, =9000; Dif ere ques./d Fic. A. (f) If the crank has the dimensions of (Fig. A) find the value of (f) when 3(B4VBe4T? WT 2. Two pulleys are 34” and 20” diam. and their centers are 36’-0” apart. What H.P. will be transmitted if the larger pulley makes SPUR GEARING 243 110 R.P.M. and the initial tension is 460 Ibs.u=.3. How wide a belt will you use? T Log = =0.007578u8 0=180°—26. sing=———. T,+T,=2T 8 3. A hoist is composed of two C.I. gears and drum placed as shown in the accompanying sketch (Fig.B). The maximum load to be Fic. B. hoisted is 5000 Ibs. at 200 ft. per min. The max. R.P.M. of motor pinion to be 400. Find the diametral pitch of the gear teeth, the diam. of pinion and number of teeth in each gear, also the H.P. of motor. CHAPTER XVI BEVEL GEARS 187. In the preceding chapter, gears have been considered whose axes of rotation were parallel. The teeth were constructed on cylinders and the line of contact was parallel to the axes of rotation. The pitch had a constant value for the entire length of the tooth. If the axes of two shafts which carry a pair of gears are inter- secting instead of parallel, the pitch surfaces of the gears change from cylinders to cones. These pitch cones have a common vertex at the point of intersection of the shafts. The teeth formed on these cones must then be changed from those found on spur gears and will be found to change in width as the vertex of the cone is approached. The shape of the teeth will be clearly seen by reference to Fig. 161. The portion of the slant side of the pitch cone used for teeth is not more than 4 and is measured from the base of the pitch cone towards its vertex. The shafts (A) and (B) in Fig. 160 meet at (O) and the bevel gears connecting them are represented by (F) and (Ff). (C) and (Ci) are the respective pitch cones whose vertices lie at (0). The large ends of the teeth lie on the surface of the sphere shown by the dotted circle, but it is difficult in practice to lay them out as spherical. A method which gives a sufficiently close approximation substitutes for the sphere the surface of two cones, called normal cones, whose sides are perpendicular to the sides of the pitch cones. Their vertices are at (O01) and (O2) but their bases are coincident with those of the pitch cones. This method is called Tredgold’s Approximate Method. By developing the surface of a normal cone we show the ends of the ‘bevel teeth lying in it so that they look like spur gear teeth on a cylinder whose radius is the same as the slant height of the normal cone. (1) and (Rz) in Fig. 160 show the radii of the developed 244 BEVEL GEARS 245 normal cones containing the big ends of the teeth on gears (F) and (F 1). In this position the teeth are constructed precisely as though they were spur gear teeth on a wheel of radius (fi) or (Re). The circular pitch is the same as that lying in a circle whose diameter (D) is that of the base of the pitch cone. The diametral pitch is the same also but the number of teeth governing the selection of a cutter is obtained from the circle whose diameter is (2R2). So also if the cycloidal system of teeth is used and Grant’s Odont- ograph Table employed to draw the tooth outline, the number of teeth in the circle (2R2) must be used in that table. Se —, orm att go , pe / oa BS / F) + Base ft 7] \y / XN C! : vo) \y ow ! NS , 8 or g & I ME a QLSIS\ | ' Pitch Cones=< | Gy Sis cy 4 \ Ol ce = ! $ 7 2 \ NJ ro fl & ‘ \ 798 \ ; A \ N\ > 7 ~ » 7 ee ie? 2 Fic. 160. 188. Shafts intersect at so many angles that a great variety of combinations of pinion and gear occurs in practice. The velocity ratio may vary with each angle of shaft intersection so that a proper knowledge of how to lay out bevel gears must comprise a great variety of cases. They are fortunately so much alike that a knowledge of the general method involved in laying out one will be sufficient for the others. The velocity ratio and angle between the shafts are usually given so that the first determination will be the pitch cone angle (9). In Fig. 161 (Fig. 1) isa diagram for finding (6) when the angle (oc) between two shafts (A) and (B) is known as well 246 ELEMENTARY MACHINE DRAWING AND DESIGN Se Bevel Gears Fie. 161. BEVEL GEARS 247 as the velocity ratio (x) and (y) between the two shafts. That is (A) makes (y) R.P.M. and (B) makes (xz) R.P.M. If a parallelogram is formed as shown, the diagonal will be the common element of the pitch cones which will roll in contact at the required velocity ratio. This diagram might better be laid off on the lines of the two shafts themselves at the point where the gears are to be drawn to save the trouble of trans- ferring 6 from Fig. 1 to Fig. 3 (see Fig. 161). In Fig. 162 are shown a number of styles of gear pairs as they often occur with various angles (x) and various velocity ratios. 0 Po A USGNay YT Vy Y /] \ po y Uj ad y | 3 SS fp pay zl N TY Say By | Uj Say ZA (D) Z Z ox ° (E) A Z 5 (F) 4 Acute 462° ~~ CrownGear |e? InternalGea Angle and Pinion , ® andPinion Fig. 162, Fig. 161 represents the type of intersecting shafts shown at (B) Fig. 162. 189. After finding (6) for the pitch cone the diameter of one gear may be assumed and laid off perpendicularly to its own shaft axis at a point which will permit the diameter to form the base line of its pitch cone. From the end of this pitch diameter lying on the common element of the two gears a perpendicular can be dropped to the axis of the other gear which will give the pitch radius of that gear. The normal cones of each gear can then be drawn and their vertices determined. From these vertices as centers, circles are drawn with the slant heights of the normal .248 ELEMENTARY MACHINE DRAWING AND DESIGN cones as radii (See Fig. 160). On these circles as pitch circles teeth are laid out in the same manner as for spur gears. By swinging these teeth back to their respective normal cones the depth of teeth at the big end is shown and lines are drawn to the vertices of the pitch cones. The length of face of the teeth is laid off on the pitch cone, thus fixing the depth of tooth at the small end. If it is desired to show the teeth in end view (Fig. 4 of Fig. 161) the circles passing through the tops, bottoms and pitch points can be drawn for both big and little ends. On the circles for the big end lay off first distances on the pitch circle equal to half the circular pitch, then on addendum and dedendum circles respectively the width of the teeth at top (a) and bottom (6), working each side of the center lines of the teeth. From these points lines are drawn towards the center of the shaft stopping at their intersection with the corresponding addendum, dedendum and pitch circles of the little end. Curves are then put in through the points thus laid out, to give the outlines of the ends of the teeth. The lower half (outside view) of Fig. 3 is then determined by projecting the points of Fig. 4 back to Fig. 3. The lines representing the top edges as well as the bottoms of the teeth are conical elements and pass through the point of intersection of the shafts. 190. The thickness of rim is the same as for spur gears and is taken at the big end of the teeth. The inside surface of the rim is a conical surface with vertex at the pitch cone vertex. If a web is used to connect rim and hub, its thickness may be taken as +8". If arms are used they are made like those shown for spur gears of the 1 cross section. (See Fig. 152, B.) The dimension (B) is taken in the plane of rotation of the wheel. The feathers of these arms are placed on the side of the arm towards the big end of the gear and support that part of the rim between the arm and end of tooth. The hub and shaft diameters are obtained from the proportions of the gear by the same method used for spur gears. The load (F) acting at the rim to produce bending of the arms at the circumference of the hub is obtained from the “ Lewis ”’ formula modified as follows: For the circular pitch use the circular pitch of the teeth at the center of their length, that is, a mean circular pitch. For the tooth factor (y) use the number of teeth in the circle which has the slant BEVEL GEARS 249 height of the normal cone as a radius. In finding a value for (S) use the velocity of the pitch point half way between big and little ends. The width of face of a bevel gear according to Brown and Sharpe should not exceed 5P or = where P=circular pitch, p= diametral pitch. Neither should the width of face exceed 4 the pitch cone slant height. Having found (F) the load on the teeth (Art. 193) and its lever arm, which is Di+D, 4 where (D,=pitch diam. at large end, D,=pitch diam. at small end, the bending moment on an arm becomes ea n 4 , where (n=number of arms=.55\+/PXNo. of teeth). The re- Sitb? 6 rib). (S:=allowable unit stress in the material of the arm.) Since the expression for bending moment equals the moment of resistance, we have sisting moment of an arm of 1 section= (neglecting the F Bis Bs) 2S n 4 ~ 6 from which B can be obtained. The diameter of shaft and hub are calculated like those on a spur gear. The length of hub may be from 1 to 13 the width of face. The hub on the side of the gear towards the small end should not project far enough to interfere with cutting the tooth. If the pitch cone angle is so great that the feather would be narrow at the rim it may be made as shown in Fig. 163. 191. It is customary to draw bevel gears in section and to give on the drawing the dimensions for turning the blank. The 250 ELEMENTARY MACHINE DRAWING AND DESIGN “SOT “OIL UOIUld PUD 4099 |aAeg Vis JIAO //D-f- $9 YODW-UOlUld rh \ > = 7 Ze ZZ uooa8 way) Vi aT | rca Uj PHB PLU SO-F- y, "|"9-spag | -z. xy y Le Y = AQMABY BEVEL GEARS 251 directions for cutting the teeth are given in tabular form on the drawing as follows: Data for Cutting Teeth. Gear. Pinion. Number of teeth............ 60 15 Cutting angle............... — —° No. of cutter................ Diametral pitch............. 3 Shape of teeth.............. 15° involute Addendum................. 0.333 Whole length of tooth........ 0.719 Tooth thickness............. 0.524 Addendum at small end...... 0.204 Tooth thickness small end... . 0.320 The dimensions required on the drawing for the pattern maker and machinist are given in Fig. 163. The pattern for the gear “blank ” must be made before the machinist can receive the casting of the wheel on which he is to form teeth, therefore many dimensions are used which are of no value to the machinist. 192. There are a number of angles needed on a bevel gear drawing which can be calculated from the principal dimensions more accurately than they can be measured with a protractor. Bevel gear angles and dimensions are named as follows (see Fig. 164): 6=center angle; F=face angle= 0+ J; C=cutting angle=@—K; J =angle increment; K=angle decrement; V =diameter increment; Y =backing; D,;=pitch diameter large end; D,=pitch diameter at small end; Do=outside diam. =D;+2V (at large end); P=circular pitch; p =diametral pitch; a=apex distance; 252 ELEMENTARY MACHINE DRAWING AND DESIGN H=distance from pitch line to apex; H,=distance from point of tooth to axis of mating gear; R=face measured parallel with axis; M =depth of rim at front end; b=face; S=addendum = my n=number of teeth; aye, Ss t > INE Ws Lp 0 RS = | Dimensions for Bevel Gear —Fore > Fie. 164. For MITRE GEARS 6=45°; 1414 tan. j=" or : a n tan. K mY n V=0.7078; Y=0.7078; a=D;X0.707; R=bcosF; D;=D,—2(b0.7071). When the axes are not at right angles there are four ether eam- binations, viz.: (1) angle axes less than 90°, (2) axes greater than BEVEL GEARS 253 90°, (8) crown bevel gears, and (4) internal bevel gears. The center angles for these combinations are found as follows: a=angle between shaft axes; 6=center angle of gear; 61=center angle of pinion; N2=No. of teeth in gear; Ni=No. of teeth in pinion; Case (1) sin a Tan 6= 6:=a¢—A. ’ 1 Net 8 a Case (2) sin (180° —a) Tax t= hp eh 1 — Wy cos (180—«) Case (3) 6=90° 6; =a—90°. Case (4) Tan te ae 0:=8-—«. 193. The strength of bevel gear teeth is based on the Lewis formula modified to suit the changing pitch found in them. The Lewis formula is 600 F=8S,Pby and modified by Barth to F=S,Pby 600-7" In using this formula (P) is taken as the average pitch (P,) of the tooth; that is at the middle point of the face. To determine 254 ELEMENTARY MACHINE DRAWING AND DESIGN this we must first find the apex distance (a) which is obtained from ees 2 sin 60° The average pitch is then found from the following formula: P(a-3) P,=——_—.. 7 a The value of (V) is determined from the average pitch diameter when the R.P.M. of the gear are known. Da=NP,0.3183 (N =No. of teeth), Va=0.2618D, X (R.P.M.). The tooth factor (y) is determined by the number of teeth in the circle whose radius is the slant height of the normal cone (See Fig. 160) (Art. 187). This number will be greater than the actual number of teeth cut on the gear itself. The expression for the number of teeth (V1) to use is, NV SG where (N) = actual No. of teeth in gear. S, is the static stress to be allowed in the teeth. This is 8000 lbs. for cast iron and 20,000 lbs. for steel. The modified Lewis formula for bevel gears will finally read as 600 F=S,P, by 600-+-V.° The H.P. which a pair of bevel gears will transmit will depend . FV, i on the load (F) and the velocity (Va) or H.P.= 33559" (Va is the velocity in ft. per min.) The load (Ff) must be taken for the wheel having the weaker teeth, usually those of the pinion if it is made of the same material as the gear. If a drawing of a gear is at hand when making the calculations for strength it is usually quite accurate enough to scale the dimen- sions (a), (Da) and slant height of normal cone, and use them in the calculations. V. can be easily calculated from (D,) and the R.P.M. of the gear in question. BEVEL GEARS 255 194. In ordering bevel gears the following information should always be given either for a new pair or to replace worn gears: GEAR PINION Number of teeth =Ne2| Number of teeth =N, Circular pitch =P | Pitch iP Face =b | Face =b Bore of shaft Bore Pitch diameter =D, | Pitch diameter =D, Outside diameter =D, | Outside diameter =D, Backing =Y | Backing =Y Pt. of tooth to center of Pt. of tooth to center of pinion shaft =H,| gear shaft =H, Length of hub =L | Length of hub =L Diam. of hub Diam. of hub Key seat Key seat Material Material Shaft angles assumed at 90° unless otherwise stated. State whether keyway is straight or tapered, and if tapered, from which side it drives. It is advisable to send a paper impression of the teeth if replacing old gears. The following books treat the subject of bevel gears more in detail: Elements of Machine Design, by W. C. Unwin. American Machinist Gear Book, by Charles H. Logue. Machinery Reference Book No. 37, by Ralph E. Flanders. I. C. 8. Pamphlets. The Constructeur, by Reuleaux. INSTRUCTIONS Bevel Gears Three exercises of 2 hrs. each. Paper No. 3. Make a drawing of a bevel gear and pinion both in section in view containing the shaft axes. Draw an outline of half the teeth in the pinion by the Tredgold method, and show two views of half the pinion teeth as represented in Figs. 3 and 4 of Fig. 161. Use 15° involute system. The drawing of the gear and pinion is to be like Fig. 163 with the Table added as given in Art. 191. The angles are to be given to degrees and minutes and dimen- 256 ELEMENTARY MACHINE DRAWING AND DESIGN sions to hundredths, except in cases where they can be given in sixteenths, cighths, quarters, etc. Teeth dimensions, pitch diameters, backing, outside diameter of blank, addendum, are some of the dimensions to be given in decimals. Caleulate the diameter of shaft for each gear (same as for spur gears) using (2) and force (F) and wrought iron as material. Draw the large gear with arms of 1 section if the pitch diam. is 18” or more. If the pinion has 15 teeth or less make it of steel. All the gears will be of cast iron except as above noted. The pitch diameter given is for the larger of the two gears. Average time required is 6.2 hours. TABLE TO BE USED FOR Data REQUIRED IN DRAWING THE BEVEL GEARS 1/2)3)4{]5 /]6) 7/8 | 9 | 10] 11] 12 a shaft angle®..} 90 | 85 | 80 | 75 | 70 | 65 | 60 | 70 | 75 | 80! 85 | 90 Velocity. . saw 1.5/1.5 |1.71)2.07/1.66)1.94) 2.2 |2.36) 2 /1.92/2.25 Ratio. .... (x)} 1 1 1 1 1 1 1 1 1 1 1 1 D (Pitch diam.)...... 7.6] 12 | 12 | 12 | 12 | 10] 12] 11} 11 | 12 | 12 | 18 p (diametral pitch)....... 21 | 22 | 23) 2 | 23 | 22 | 22] 3 | 3 | 23 | 22] 2 b (Width face)..| 23 | 24 | 3 | 34] 32] 3 | 22 | 23 | 22 | 23 | 32] 8 R.P.M. pinion .| 145 | 150 | 160 | 170 | 200 | 170 | 200 | 220 | 240 | 210 | 220 | 240 13 | 14) 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 a shaft angle...} 85 | 80 | 75 | 70 | 65 | 60 | 70 | 80 | 90 | 110 | 120] 100 Veloc. .... § (y)|1.94/1.66] 3 | 3 | 4 |3.33) 2 | 5 | 4 |] 2 | 38) 5 Ratio..... ()i Te) Ped | | Pk) bap kh | ot | to D (Pitch diam.)...... 12 | 10 | 24 | 24 | 24 | 20 | 26 | 25 | 24 | 26 | 20 | 25 p (Diametral pitch)....... 22 (2 1 2b 1+ 2) 2) 2-02) 2 | BRL a | oe b (Width face)..) 23 | 3 | 33] 4 | 4 | 3 ry » ale e othe w Alo »~ ale w een ee ole R.P.M. pinion .| 210 | 200 | 310 | 320 | 400 | 350 | 310 | 250 | 260 | 250 | 350 | 250 BEVEL GEARS 257 When a (’) mark is placed after the number of column given on the bulletin board assignments, it means take the angle alpha in the column under the number, but take all the other data from the next column to the right. Prob. 1. Calculate the H.P. transmitted by the gears you are draw- ing when the pinion makes the number of R.P.M. given in the column you are using. First find the value of (F) in the Lewis formula applied to bevel gears. Velocity calculated at center of face and Pa taken at the same point. Prob. 2. Given the angle between two shafts, the velocity ratio between them, R.P.M. of one of them and pressure on the teeth to deter- mine the proper pitch to use and the diameters of the gears for both shafts. Use angle a, velocity ratio and R.P.M. given in the above table. For pressure (F) in lbs. use twice the number given for R.P.M. Prob. 3. A pair of C.I. bevel gears axes at 90° are to transmit 65 H.P. The ratio of their diameters is 5 to 6. The smaller gear makes 190 R.P.M. Width of face limited to 53’. Its pitch diameter is 30”. Calculate the pitch and No. of teeth in each gear. Prob. 4. Find the working strength of a pair of C. I. ‘mitre gears making 200 R.P.M. Each gear has 55 teeth, 2” pitch, 6” face and 15° involute system. Prob. 5. The angle between shafts of two bevel gears is 60° and the diameters of the gears are 30” and 36”. Calculate the center angle of the smaller gear. Prob. 6. Angle between shafts =90°, p=3, No. of teeth in pinion =15, in gear 60, width of face=4’’. Make the necessary calculations for dimensions of pinion and gear. Prob. 7. What power can be transmitted by a pair of mitre gears having the following dimensions? 30 teeth—2 inch pitch—5 inch face— 19.107” pitch diameter, at 50 R.P.M. Made of cast iron. CHAPTER XVII WORM GEARING 195. Worm gearing is a modification of screw gearing in which the shafts are at right angles, axes non-intersecting. The velocity ratio of screw gearing is independent of the radii of the pitch surfaces, the great advantage of screw gearing being that a high velocity ratio can be obtained with comparatively small wheels. The disadvantage is that friction and wear are greater than with toothed gearing. The efficiency of a worm and worm wheel depends on the coefficient of friction between the rubbing surfaces on the thread angle and the velocity of the rubbing surfaces. An efficiency of 98% has been obtained under ideal conditions, but ordinary conditions of use will reduce this to from 60% to 80%. 196. The worm consists of a threaded cylinder several inches long having a thread with a profile of the Acme form. The linear pitch of the thread is the axial distance from a point on a tooth to the corresponding point on the next one. The lead is the distance a point on a thread advances axially in one revolution of the worm. A single-threaded worm will have its pitch and lead equal, while a triple-threaded worm will have a lead three times the pitch. To find the lead of a worm multiply the linear pitch by the number of threads. 197. The velocity ratio of worm and worm wheel is the same as the number of teeth in the wheel if the worm is single-threaded. If the worm is double or triple-threaded the velocity ratio is respectively 4 and 3 the number of teeth in the wheel. The teeth of the worm wheel when shown in section as in Fig. 165 are like the teeth of a spur gear, involute system, line of action 144°. The worm teeth, when cut by a section plane containing the axis, are like the teeth of a rack. Circular pitch is used for worm gearing because it is more convenient to cut the worm teeth to this dimension. The strength of the wheel teeth deter- 258 WORM GEARING 259 mines the value of the pitch to use. The number of threads in the worm does not affect the circular pitch of the wheel. 198. The length of the worm varies from 3P to 6P although, as only two teeth are in contact at one time, 3P is enough. To find the minimum length of worm advisable for complete action with wheel, subtract four times the addendum of the worm thread from the throat diameter of the wheel, square the remainder and subtract the result from the square of the throat diameter of the wheel. The square root of the result is the minimum length of worm advisable. The pitch diameter of the worm varies from 2P to 6P but is generally as small as possible, except when the angle of the worm thread approaches 45°, when the conditions can be reversed. The addendum may be taken as 0.3183P and the whole depth of tooth as 0.6866P. The helix angle of the worm determines the efficiency to a large extent. Angles below 9° have proved to be unsuccessful while all above 12° 30’ have proved successful according to Mr. Wilfred Lewis, Mr. Christie and other engineers. This angle is the angle between the helix and a perpendicular to the axis of the worm. According to recent experiments a maximum efficiency is obtained with an angle of 45°, but the efficiency changes very little if this angle is decreased to 30° or increased to 60°. Prof. Unwin gives for the efficiency (e) of a worm and wheel = 1l—y cot « I+ tan a’ Prof. Barr gives the efficiency (e) (neglecting the friction due to end thrust) as _ tan «(1—yu tan a) 7 tan atu ’ which is a maximum when tan «=~+/ 1+y2 —yv. w=coeff. of fric- tion. Substituting u=.05 in above gives the angle «=43° 34’. If the friction of the worm step is considered _ tan a(1—y tan a) ~ tanat2n ”’ which is a maximum, when tan «=+/2+4y2—2u. If (u) =0.05, a=52° 49’, the angle for max. efficiency. A change in the value of (uw) in the preceding formule will change the value of the effi- ciency, reducing it considerably when the angle («) is small. 260 ELEMENTARY MACHINE DRAWING AND DESIGN The lower the efficiency the greater the wear. The angle may be increased in two ways, increasing the pitch with a constant diameter or reducing the diameter with a constant pitch. To find the helix angle (a) of the worm, multiply the pitch diameter of the worm by (x) and divide the lead by this product. The quotient is the tangent of the thread angle (a). 199. The speed of the rubbing surfaces of worm and wheel when the gear is worked up to the limit of its strength should not greater than 200 to 300 ft. per min. according to Mr. Lewis. If this limit is exceeded abrasion occurs. Higher velocities can be used if the load is decreased. Before the limit is reached there is a pressure at which the friction suddenly increases, due to the squeezing out of the lubricant and consequent rise in tem- perature. The sliding velocity in ft. per min. at the pitch line is expressed by 3 adn V= 72 = 0.262dn, where d=pitch diam. of worm in inches and n=R.P.M. of worm. This is true only for small values of («) and should be multiplied by the secant («) for thread angles of 20° or more. In order to make a worm and gear self locking the tangent of the angle («) must be less than the coefficient of friction (u), that is _P d=pitch diam. of worm Tan a= a< Se . If w is assumed to have an average value of 0.05 then the limiting value of (P) will be P=0.05rd=0.157d at which the system will be interlocking at 300 ft. per min. velocity. 200. The resistance to turning the worm is due to the force at the pitch line of the worm wheel called (F), the coefficient of friction (wu) and the tan of « If we call this resistance (W) we have W=F tan (a+) where ® is the angle whose tan is the coeff. of friction. The reaction between the rubbing surfaces is the tooth pres- sure expressed as Pig ge a sin (a+) cos («+6)" The end thrust on the worm is the pressure on the rubbing surfaces multiplied by the cos (a). WORM GEARING 261 The side thrust on the worm=F; sin a. The load (W) can be used to determine the diameter of the worm shaft, for aT HLP. transmitted. V=velocity of worm pitch line in ft. per min. or di =k & ae, N=R.P.M. of worm. k=constant= 3.3 for wrought iron at 9000 Ibs. per sq. in. stress, and k =2.87 for steel at 13,500 lbs. stress. If the load (W) is taken as acting at vt c from the axis of the shaft, We is the twisting moment which is equaled by the resisting eo to torsion of a circular shaft of diameter di, or We =0.196d13f, where (f) =max. stress allowed in the shaft material =9000 for_ W. I., 13,500 for steel. This equation reduces to di=ki 3 we ki =.083 for W. I., ky =.072 for steel, d=pitch diameter of worm. 201. The worm wheel is much like a spur gear wheel excepting the teeth and rim. The rim is curved to follow the circumference of the worm and the teeth are usually made as shown in Fig. 3 of Fig. 165. The angle (6) may vary from 30° to 45°, but 30° is commonly used. The worm wheel may have thirty or more teeth, less than 30 being considered undesirable. If the velocity ratio is less than thirty the worm must have more than one thread. The throat radius of a worm gear is the same as the outside diam- eter of a spur gear of the same number of teeth and pitch, or= AW T)P.0818 The width of rim of the worm gear at the do+(0.34P) 2 widest part when 0=30°, is, b= » where do =outside diam. of worm. The rim and hub may be joined by a solid web, a web with holes in it to lighten the wheel or by arms of elliptical section. In the last case their cross section may be determined as for a spur gear having the same pitch and diameter except the width of face used in the Lewis formula must be the width of teeth at the root. This will be found to be 2n (2+.05 P1) sec a ——— 262 ELEMENTARY MACHINE DRAWING AND DESIGN PE tbe iP << —— 5 Ll. —— “GOL “OL JOQUM WOM PUD WUO/A }k——— SEJPYS IO JOD poy apissng —_» [| “OUOYAHIOI9A “""UOH!d << oS, | el Wi Ly 7 fs 2 yt we gt 2 “ a“ WORM GEARING 263 (when §=30°), do=outside diam. of worm, Pi1=pitch of worm wheel perpendicular to rubbing surface, equal to P cos « Fj must be used in place of F in the Lewis formula and P in place of P. The shaft diameter of the wheel is calculated as for a spur gear. The length of hub and its diameter may be taken according to the proportions given for spur gears. 202. The teeth of worm wheels are generally cut by means of a cutter called a hob, which is a duplicate of the worm made of steel with flutes cut in it parallel to its axis. The hob is mounted in a frame with the worm wheel occupying the same position relatively to it which the worm will occupy but with centers farther apart. The hob and wheel are then rotated with the same velocity ratio for which the worm and wheel were designed and the hob is fed against the wheel. The teeth are first partially formed by a milling cutter previous to cutting with the hob so the final cutting is really a finishing cut. There are certain fine points regarding worm wheel tooth cutting which are out of place in a work of this kind, such as relieving of hobs, number of flutes, length of hob, etc. Books on gear cutting treat of these points in detail and should be read by the designer and student of worm gearing. 203. It is customary to make the worm of different material from the wheel as the wear on the former is greater. Worms are made of hardened steel while the worm wheels are made of cast iron, bronze and steel. Sometimes the rim is made of bronze and fastened to a C. I. central portion to lessen expense. Worm bearings must be arranged to take end as well as side thrust with provision for copious lubrication if efficiency is desired. Ball bearing thrust and annular bearings require less attention than plain bearings besides reducing the coeff. of friction. Such bear- ings should be used whenever possible, especially in cases where the worm wheel sometimes drives the worm, as in automobile rear axles. 204. In ordering a worm and worm wheel it is always necessary to give the following data for worm: Distance c. to ¢. of shafts— Length of worm—Length of thread—Outside diam. of worm— Pitch of thread—Pitch diam. of worm—Diam. of worm shaft— Key seat of worm shaft—Lead of worm—Right or left hand— Hub projection of worm—Material of worm. For Gear: Pitch diam. of gear—No. of teeth—Pitch—Width of rim—Length 264 #LEMENTARY MACHINE DRAWING AND DESIGN through hub—Hub protection—Bore of hub—Key seat of shaft— Hub diam.—Material of wheel. 205. Fig. 166 shows the drawing for a worm, indicating the various dimensions as referred to in the preceding articles. Fig. Hap }«—____ length Frosect Lengthof Thr rasect, Pitch 7g nate Pitch Diarr. Hub dn Outside Diam. Helix a Angle Worm Double - Right Hand Lead "Pitch ” Mater Fie. 166. 167 is a worm gear with similar dimensions on it. It is not necessary on worm and gear drawings to show more than a section of the teeth by a plane through the worm axis. If it is desired to represent the teeth of the wheel as they appear behind this section, it may be done approximately by the method shown in Fig. 168. (D) is the center of the end view of the worm, Db is the pitch radius of the worm and Ri Rez Rz are the radii of the addendum, pitch and dedendum circles of the part of the tooth lying behind the section and on the line DCi1._ Radial lines are drawn through the three points 1-2-3 until they intersect the circles drawn with Ri Re Rz respectively as radii. From these points of intersection (7) is laid off as shown, locating three WORM GEARING 265 points on the outline of the rear part of the tooth. A radius is then found by trial which will draw an arc through these three rp —lead of worm oints. points 3 , when 0=30°. Outside Diam. (to sharp corners) ———> Diameter (trimmed) Throat Diam. Pitch Diarn. a J r [4 4 aie A | | S “ AS | ie Yi ‘ Angle le t ' Center Distance———> Worm Wheel No. of Teeth __._ Circ. Pitch____- Angle of Cut_____- Material_____. Worm- Double Right Hand Outside Diam..____ Fic. 167. D 7} Worm Wheel ° Tooth ay ©» i a b Cn R; PR. c iar (A) Fig. 168.—3 Worm Wheel Tooth. The curves on the end of the worm shown in Fig 165 are the intersections of the helicoids forming the thread surfaces with the plane of the end of the worm. 266 ELEMENTARY MACHINE DRAWING AND DESIGN INSTRUCTIONS Worm Gearing To be drawn on No. 3 paper in two class exercises of 2 hours each. The views to be drawn will be the same as shown in Fig. 165 (Figs. 2 and 8), omitting Fig. 1. The dimensions and data in the form of notes must include all that which is found in Figs. 166 and 167, as well as the data used in calculating the various dimensions, viz., speed of rubbing surfaces, load on teeth, coeff. of friction, end thrust on worm, calculated efficiency, R.P.M. of worm, velocity ratio, helix angle of worm, H.P. transmitted at R.P.M. of worm. Av. time reqd. for this plate is 6.2 hours. The involute system, 144° worm tooth profile is to be used. The velocity of rubbing surface of worm to be 250 ft. per min. TaBLE FoR Worm aNnpD WorM WHEEL PLATE No. 1}/2)/3/4/5)]6]7/] 8] 9 |10)11)12]13) 14/15) 16 Cire. pitch... 1) 1] 1] 1} 1 )18)18])13)18)18]14)12]12)14) 14511 R.orL.Hand RI! L)|R!LIR/|LIR| LI RIL RI|LIRI|LIRIL Lead........ W/7}2)1 73) 2 )1$)19] 22) 23 )12)15)12!14)14)14] 2 No. Teeth. . .| 62 | 49 | 50 | 51 | 62 | 48 | 44 | 48 | 52 | 38 | 38 | 40 | 41 | 45 | 35 | 50 No. 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |26} 27 | 28 | 29) 30] 81 | 32 Cire. pitch.| 1] % | | & | 38] 28] 48 [Lve]l fe] ve}. 187 1.5, 1/1 RorLH’d R|R/|R|R| RR|R|R|R|R) RR RI RIRIL Lead...... 1) 13 | 28 | 34 | 1% [248] 32 | 25 )3835| 44 | 4.75 (2.7.2.4, 3} 313 bn or oO a> Oo I o fi oO No. Teeth .| 40 | 50 | 60 | 7 42/40] 32 | 40] 50 | 50/30 | 33 Prob. 1. Design a worm and gear to have a velocity ratio of 3 to 1 with a linear pitch of .75’’ (about). Center distance between shafts =5’’. What is the efficiency? Prob. 2. A double threaded steel worm drives a phosphor bronze worm wheel with a velocity ratio of 25 to 1. Center distance of shafts =13.5”. If the worm makes 350 R.P.M. find the circular pitch of the worm wheel, the pitch diameters, and the helix angle (not to be less than 15°). Also determine the max. tangential pressure allowable on the worm wheel, the H.P. transmitted and the efficiency of the worm and wheel when vu. =0.05. WORM GEARING 267 Test No. 9. Pulleys, Belting and Gearing Arts. 149-205. (2 hours allowed.) 1.* A hoisting train is shown in Fig. A. The motor is 15 H.P. at 775 R.P.M. The weight is to be raised at a velocity of 500 ft. per rit ot min, The gears are C.I., the \_j_J’*%" el > teeth to run at 800 ft. per min. L/ 1 4e* | | The velocity ratio of the gears is 4to1l. The width of face =5P. The pulley on the motor is 11” diam. The drum is 30” diam. Find the following (a): Width of belt, (0) Diam. of pulley at B, (c) Diam. of gears, (d) Wgt. of load raised (efficiency 95%), (e) Diametral and circular pitches of gear teeth, width of face and Fig. A. teeth in each gear. 2.* A worm and wheel are used to drive the rear axle of a truck. If the resistance to slipping the rear wheels is 5200 lbs., and the dimensions are taken for rear wheel, worm wheel and worm, as shown in Fig. B, find the pitch to use on the worm wheel, the number of threads on worm, number of teeth in worm wheel, helix angle, and efficiency of the worm and wheel. The velocity ratio is 8:1. The worm wheel is phosphor bronze and the load on the wheel is taken by two teeth. The coeff. of friction may be taken as 0.02. Fig. C. 3. How much H.P. will the C. I. bevel gears shown in Fig. C transmit when the smaller one makes 250 R.P.M. Velocity ratio 3:1 p=2, 15° involute system. The pinion has a 13” diam. shaft of W.I. Is this shaft large enough to transmit the strength of the gear teeth? Prove your conclusion. * Take 1 or 2 and 3. Use books in this test. CHAPTER XVIII VALVES 206. Vatves are placed in pipe lines in order to control the movement of gases or fluids through the pipes. They can be divided into two general classes, viz.: (A) Automatic, (B) Con- trolled. The (A) class may be further divided into (1) valves opened and closed by the machine of which they are parts, and (2) valves which are operated by the action of the fluid or gas itself. Class (B) comprises valves opened or closed by hand or by inde- pendent mechanisms. Steam valves of engines or pumps and poppet valves of gas ergines belong to (A, 1). Water valves of pumps, valves in buckets, check valves and safety valves belong in (A, 2). Globe, angle, and gate valves, stop cocks, and pet cocks, belong to class (B). As it is not the purpose of this treatise to treat of automatic machines as a whole, class (A, 1) will be omitted and class (A, 2) examined. 207. Check valves, as previously stated, belong to class (A, 2) and are used to stop the flow in pipes in one direction. They are used on pump delivery and suction pipes in either a horizontal or a vertical position. The moving part in these valves may be a ball, a swinging flap or a flat plate lifting and falling. The ends of the valve bodies are tapped to fit the pipes they connect or flanged to fit the flanges on the pipes. A check valve for horizontal pipes is shown in Fig. 169. It is called a “‘ swing check ”’ from the kind of motion of the clapper in opening and closing. The body is of bronze with tapped ends. There are three other openings in the body, the largest one for introducing the moving parts into the body. It is closed by a plug with hexagonal end. A second opening, at right angles to 268 VALVES 269 the valve seat, enables the disc on the swinging part to be ground on the seat before assembling the valve. This opening is closed by a threaded plug which acts as a stop for the clapper. The third opening is at the side of the body and is designed to admit the pin on which the clapper swings. It is closed by a screw plug after the valve is assembled. 9 Globe Swing Check Valve Fic. 169. The clapper consists of four parts—the rotating disc, the swing- ing part or hinge, the stud and the nut which fasten the disc to the hinge. The two last parts are loosely joined together so that the action of the fluid will cause the disc to rotate and thus prevent any particle from lodging on the seat. It also permits the disc to seat properly in case the hinge is out of true with the seat. 208. A direct lift check valve which forms an angle in a pipe line is shown in Fig. 171. This was first designed by Prof. Charles 270 ELEMENTARY MACHINE DRAWING AND DESIGN B. Richards and does away with air pockets, which are detri- mental to the efficient working of a check valve. The threaded ends of the body are like threaded pipe ends and have standard dimensions for perfect thread, imperfect thread, taper and out- side diameter. The lifting part of the valve is guided by three projections cast on the inside of the body and by the stem working in a hole in the screw cap. Ail Sec.on-Xx- Pipe thread, xm i ty 3 Me 4 f L—— | ih i cy Bronze Direct Lift Check Valve Fig. 171. The stem is cut away on three sides to allow the fluid in the cavity of the cap to flow out when the valve opens. The theoretical lift of the valve must be enough to give an area of cylindrical surface between valve and seat equal to the area of the opening in the seat. 2 If D=diam. of seat opening, ah =area of opening. If h= lift of valve hXxD=area of cylindrical surface between valve and seat. VALVES 271 2 As hrD a we have h =? =theoretical lift of valve. 4 This check valve may be made with the lower end vertical, thus allowing it to be used as an elbow as well as a valve. 209. Safety Valves belong in class (A, 2) as they are designed to be opened by the pressure of steam or gas within them when- ever that pressure exceeds the pressure for which they are set. There are two kinds of safety valve, one controlled by a spring, the other by a dead weight hung from a lever. One of the latter TABLE oF DIMENSIONS FOR GLOBE AND ANGLE VALVES (From 3” to 8’ nominal size.) Flanged cast iron bodies for pressures up to 150 Ibs.per sq.in. valve. Required dimension =«D+6. D=nominal size of De || ae ae PR cies || BRR ee. | op A same as D Cale. for 6000 lbs. O”” q" 0.25 | 0.8 c | 0.9 | 0.5 d” | 0.26 | 0.9 D=given diam. a | at 150 lbs. 0” q in valve * 3 thickness of a, =a-+pitch of thread u E= diaphragm t= ONT “~~ ) meas. in vert. plane a b ae See wu | 0.12 | 0.55 c ; F v= G 0.7 0.5 c= | 0.2E H 0.6 0.5 d 0.02 | 0.1 Pitch of thread on stem : ee ae ne is 7’ for valves up to F : = 4g 10 in., and 3” above L=T g = Sin Art. 96 stuff. that size. M |06 | 0.1 boxes. P=2Xstud diam.-+ j=a +29+2t+diam. Zs” of gland stud. Q [1.0 | 1.75 m | 0.66 | 2.52 R=Stand. flange for C. I. pipes, Art. 37 o | 0.03 | 0.15 S=Lor 13 Xstud _—™ 5” diam. Ps +s T =thickness of C. I. stand. pipe flange Art. 37 W=2Q m =2a, +3" ee NB oo ND 272 ELEMENTARY MACHINE DRAWING AND DESIGN TABLE FoR SAFETY VALVES Safety Valve dimensions not given below can be taken from the Table for globe valves. P| ae |) Se ea) cae |||) a | 0.055 | 0.14 k 0.38 0.24 || Flange bolts (see a, =3 of (a) h 0.22 0.18 C. I. standard d =diam. bonnet stud m 0.09 0.08 flanges) Art. 37. +H" Bonnet studs A n=f—Z5" same as flange eo 0 0.1 0.57 bolts in number f 0.25 0.625 Dp 0.65 1.3 and diam. Do h 0.16 | 0.8 q 0.27 0.57 not place any g 0.17 0.35 r 0.09 0.2 flange bolts on h 0.28 0.32 s’ 0.22 0.3 the vertical a 0.5 0.5 u 0.6 1.0 diam. of flange. j 0.09 0.08 v 0.25 0.75 is shown assembled in Fig. 172. It is designed to stand a maxi- mum pressure of 100 lbs. per sq.in. A shell of cast iron is provided with three openings for pipe connections through two of which (1) and (2) steam may pass freely from boiler to engine. The third opening (3) is provided for the escape of steam from the valve when the pressure within causes the valve to open. An opening is provided at the top of the shell through which the valve disc is introduced as well as the seat on which it rests. A bonnet closes this opening and is held in place by stud bolts and nuts. The stem passes through the bonnet and transmits pressure from the valve disc to a lever whose fulcrum is a pin at its left hand end. The upward pressure of the stem against this lever tends to move it in a vertical direction about the fulcrum. To counteract this movement, a ball is hung on the lever to the right of the stem and at a distance from the fulcrum corresponding to the pressure to be balanced within the valve. The lever is supported and guided by a yoke which slips over the central part of the bonnet and is locked in any position by a locknut. The valve stem is fastened to the disc by a nut of special form made to fit loosely around the stem and screw into the disc. A flange on the stem enables the nut to draw the disc and stem tightly together. A cylindrical rod formed on the bottom of the dise passes through a guide supported by two arms cast with the seat. This VALVES ais y ‘© Soper Area= (C) U/ Half Sec.at-yy- Wat> Ibs. : v & ) : X-Inch 8 Safety Valve te Fig. 172. 274 ELEMENTARY MACHINE DRAWING AND DESIGN guide insures exact seating of the disc on the seat. The seat is screwed into place by fine threads, viz.: 14 to 8 per inch depend- ing on the size of valve; the finest on the smallest valve. The flanges of these valves are made of standard size to fit C. I. pipe flanges as given in Table 11 or in Chap. III. The drilling of these flanges for bolt holes is determined from the same table. The spacing is symmetrical with respect to a vertical diameter without placing any bolt hole on this diameter. Use the same number of bolts to hold the bonnet on. They may have the same diameter as the flange bolts but are studs with hex. nuts. The flange bolts will have sq. heads and hex. nuts. 210. In order to obtain the proper area through the narrowest portion of the valve it is necessary to find the true shape of the section taken by a plane —yy-. The area of this section should not be less than the internal cross sectional area of the pipe con- nected to the valve. This latter area will be that of a circle whose diameter is (A). The construction of the outline of the section yy is shown in Fig. 178, being obtained as follows: Several planes are passed 8 y (0) not on «aff? a af cect, esi ‘Ns x re of are Ew A ; (A) E C @ p ve Fi See w ° ae = ae eS x — —— x b a“ i Ss % ae % Construction Diagram forobtaining x | Area of Least Opening in aS ae: & Globe Valve S 5 (c) g zZ 3| z; 6 zy Fia. 173. perpendicular to the axis of the pipe (View C) to cut circles from the surface of revolution. The circles are then shown in VALVES 275 the end view (B) as (tt) whose radius (ot) equals (cd) the dis- tance from the axis to the point where the auxiliary plane cuts the surface of revolution (in View C). The point (as f) where these auxiliary planes cut the section plane yy is then projected to the same end view (B) to obtain the distance (a) from the ver- tical center plane of the valve (oe) to the inside of the shell. By revolving the plane yy about its intersection with the center plane until it lies in the plane of the paper, we obtain the true outline of the section cut by it from the valve. The revolution is shown at (A) where BB is the axis of symmetry of the section. Points to the right of BB on the curve (BC) are points on the inside of the shell behind the center plane of the valve. Distances (a) are found from the end view (B) and laid off perpendicular to BB on lines drawn from the various points (f) on yy. The curve DE is found by passing planes (xx), (1171) in view (C) perpendicular to the vertical axis (FG) of the conical surface. These planes cut circles zz, 2121 from the surface shown in plan with radii (vz) (v121)._ The distances (c), (6), etc., are then found by dropping perpendiculars from (g), (h), to meet the correspond- ing ares (zz), (2121). These distances are then laid off from BB in view (A) perpendicular to BB. 211. The area of this section is then found by the trapezoidal rule as follows: In Fig. 174 divide the base (AC) into any number Fic. 174. of equal parts, and number the ordinates erected at these points. Take 4 the sum of the lengths of the end ordinates—(1) and (9) in the figure—add the sum of the length of all the intermediate 276 ELEMENTARY MACHINE DRAWING AND DESIGN ordinates to the half sum of the end ordinates and multiply the total by the space (e) between two adjacent ordinates. The result. will be the area within the curves (BC), (DE) the base (EC) and the ordinate (BD). It is more accurate to first find the area bounded by the outer curve (BC), the base line (AC) and the ordinate AB and then subtract the shaded area ADE. The accuracy of the result is increased by increasing the number of ordinates. 212. The weight of the ball is calculated by the principle of moments about the fulcrum pin (O) as a center. P=the total pressure exerted upwards by the steam at 100 lbs. per sq.in. on the valve disc. p=the weight of the moving parts which rise when the valve opens. P—p=the upward force acting at (A), Fig. 175. The weights acting downwards are the ball i Pe re. 9 fuleru Fig. 175. (W) and the lever (w) acting at its center of mass. These upward and downward forces are in equilibrium, that is, the algebraic sum of their moments equals zero. (P—p)i—wl—WL=0, from which waSopinul The lever is wrought iron, the ball cast iron and the moving parts bronze, their weights per cubic inch being W. I.=.28, C. I.=.26, bz.=0.30. JD and i are given either in the table of dimensions or on the plate. The handle of the ball can be designed first, its weight determined and subtracted from the weight (W) thus giving the weight of the ball alone. Its diameter can then easily be calculated. 213. The fulcrum pin diameter can be assumed 3f! and its strength checked by calculating it for double shear, which should not exceed 7000 lbs. per sq.in. as the pin is of W. I. The sectional area of the supports for the pin taken by a horizontal plane VALVES 277 through the axis of the pin should be checked for stress in tension. As they are C. I. this stress should not exceed 2500 lbs. per sq. in. of section. The stress in the bonnet bolts (at root of thread) should be determined for a pressure of 150 lbs. per sq.in. inside the valve. The shell is C. I. and should be checked for stress not to exceed 2000 Ibs. per sq.in. in tension for a steam pressure of 150 lbs. per sq.in. The spherical part is treated like a cylinder of equal diameter, viz., 150 X 2rC' = 2K X 2000 X 2rC. 214. A list of the parts of the valve with a note of their material, etc., should be placed on the drawing in a table somewhat as shown below. List or Parts ror —” Sarety VALVE No. . | ed Required. Mark. | Material. Name. Description. 1 1070 C.I. Shell Flanges faced and turned 1 1071 Bz. Seat 1 1072 Bz. Disc 1 1073 Bz. Disc nut 1 1074 Bz. Stem -f- all over 1 1075 C. I. Stem cap 1 1076 C.I. Bonnet 1 1077 C.I. Yoke 1 1078 W.I. Lock nut 1 1079 W.I. Fule. pin —f- all over 1 1080 C. I. Ball —Rough- 1 1081 W. I. Lever ase) If sant eeens St. Bonnet studs | U.S. St. nuts —” diam. — grip. harnites St. Flange bolts | Sq. hd. hex. nut, —’” diam., —_N lg., —"grip. Directions as to views, etc., will be found in the instructions. The dimensions for the parts can be worked out from the tables in Art. 209 for any size of valve wanted from 3” to 8” diameter. They should be changed from decimals to common fractions before using them on the drawing. 215. In class (B) taken in the order of their construction from the standpoint of complication, we find first pet cocks or air cocks as they are called. They are used on pipe lines to allow air or water to escape, either for the purpose of increasing the efficient flow or as an indication that the line is under pressure. 278 ELEMENTARY MACHINE DRAWING AND DESIGN In Fig. 176 is shown a pet cock threaded with a 2” pipe thread. They can also be obtained in a larger size with a 3” pipe thread. The other dimensions are given on the drawing. og st. 7 /6\ 16 Ss Mul PetCock Fia. 176. 216. Stop cocks are used to regulate the flow of fluids in a pipe line. The regulation is obtained by turning a taper plug, fitting in the body, the axis of rotation intersecting and at right angles to the axis of the pipe line. The taper plug is pierced by a slot which is either brought in line with the openings in the body or turned at right angles to them, thus permitting flow of the fluid or cutting it off entirely. One end of the plug is squared to allow the use of a wrench for turning it. A mark on the end of this square indicates whether the slots are turned to allow the passage of the fluid or not. In Fig. 177 is shown an assembled stop cock. The dimensions for several sizes are given in the following table. VALVES 279 TaBLe oF Stop Cock DIMENSIONS Nom) 4) Bilc|D\E\F\|@|lalr\|s|/Kim|nio|Pl/e@ rs BR illgs] ¢ ] $e] 1 | te) ells] afl |e] 213) bps] Bl g |i | & [de | 1a 18 [tae fae [eae ug] ele] ela] eda 1 fase] # [die f 1b |te [19 fre |aalagefia | ale] ef a | a] 4 12/13 | 3 [13 | 12 [15 |1f[2t | 18 ]1E |e | o | della] el] 2 | 14 | 148] 4 [18 | 25 2s [12 25 | 18/22 [LE | F] tells | ae] vel 2 2 (22 | 4 [1s | 23 [233/28 |3e 2 jazglos|z (1 [tga] asl ae 3 I «Side of sq A P 1 a Stop Cock Fic. 177. 217. Angle and globe valves are used to provide a ready means of controlling by hand the flow of fluids by means of a disc which is raised and lowered by a hand wheel and screw thread on the spindle attached to the disc. The disc rests on a seat within the valve when the flow is arrested. When the flow is to continue the disc is raised from the seat, leaving a free passage between them. An angle valve is used in place of a 90° elbow, the fluid turning through a right angle within the valve itself. The shell for an angle valve is shown in Fig. 178 (B). The other parts of the 280 ELEMENTARY MACHINE DRAWING AND DESIGN valve are the same as those in the globe valve shown in Fig. 179. Valves of this type are provided with tapped ends for small sizes and with flanged ends for larger ones. The general outside of the globe and angle valves with tapped ends is shown in Fig. 178A. The inside construction and detail is shown by the sec- tion in Fig. 178 (B) of the shell of an angle valve, and by Fig. 179 (B), the section of a globe valve. ote s © Thds. Bi. “The = Pipe Ta a S 2 ; ‘pe Tap Ee Tae 3 4 Section of Angle Valve Shell @ a Tapped Ends Fic. 178 (A). Fie. 178 (B). The various dimensions of these valves can be obtained from the straight line formula, the constants for which are given in the table on page 281 for each lettered dimension. These were all determined by Prof. Charles B. Richards. The required dimension = «D+8D=nominal size of valve. 218. Globe valves with flanged bodies are in general con- structed as shown in Fig. 180. The shell is spherical in shape having flanges of standard size on opposite sides connected to the spherical portion by cylindrical pipes. In the central portion of the shell is a circular opening into which is screwed a seat. A circular disc closes the opening in this seat. The disc is attached VALVES 281 TABLE FoR VALVES WITH TapPeD ENDS; ALL BRONZE. Sizes 3 To 3” Values of Dimension. |————— > Remarks. 8 .0 Diam. opening in valve seat 4 Inside radius of shell thle gawine: | etext Nominal diam. (given) 12 Depth of valve seat 2 Radius of V shell diaphragm 21 Radius of V shell diaphragm 3 Radius of V shell diaphragm ee ee eee Radius of fillet, globe to pipe 1 Thickness of shell 5 C. of globe to bonnet flange 24 Diam. of opening bonnet seat oo oo ke Oo one lea) > C. of globe to flanged ends Dist. across flats, flanged ends Thickness of tapped ends Diam. of screw on valve stem Thickness of nut on valve disc Depth of nut on valve disc Thickness of valve disc io saeke ce! loners Chamfer valve seat, Thickness of flange on bonnet seat : Width of stuffing box =39 Depth of stuffing box =2.59 Depth of packing gland Diam. stuffing box screw Height of bonnet Depth of stem screw in bonnet Dist. across flats on bonnet hex. Thickness of arm of hand wheel Diam. of rim of hand wheel =2Q Outs. diam. of hand wheel e2e0ceCenrr @ © COO EO OOO OD > oo = oS na oo Oo ort 48 ol, 382 9 a 15 rn Oo ooooss WWWwan SSETVRISSeTHAWAt aac ce yQn|WDOVOSRMMOWBWAIAw eooor oe Pitch of thread on stem =0.04D+0.11”. tq the flanged end of a stem by a nut, with hexagonal head, which slips over the flange and screws into the disc. The stem passes out of the valve through a stuffing box in the bonnet. The bonnet covers the opening in the shell above the seat and also provides a support for the stationary nut through which passes the threaded portion of the stem. A hand wheel is keyed to the stem at its upper end and provides a means of turning the stem in order to raise or lower the disc and thus open or close the valve. 282 ELEMENTARY MACHINE DRAWING AND DESIGN Fie. 180 (A), Frye. 180 (B). thdsgtin a isc NuttHex. ee thas = < ™ SC. | é YG Shell Globe Valve Complete Fig| Aa, 169 arms Hand Whee rd ea 15. per irt. i Stem Nut. Hex. Hd) ep Gland Serre ta | Stud J x >| pp he Rea} ty el _/ w \ kg I J Side View of Bonnet : Flange Bolts and Bonnet Studs Standard No.andDiam. Pp & nds Flange —L ValveShell Fig.3 ig. | Angle Valve . Fia. 180. (To face page 283) VALVES 283 TaBLE OF DIMENSIONS FOR GLOBE AND ANGLE VALVES (From 3” to 8” nominal size.) Flanged cast iron bodies for pressures up to 150 Ibs. per sq.in. Required dimension =«D+8. D=nominal size of valve. Mei “8 | 6 | ion |e 8 | a | 8 A same as D Cale. for 6000 lbs. 0” g’ | 0.25 | 0.8 c | 69 | 0.5 qd” | 0.26 | 0.9 D=given diam. a at 150 Ibs. O” q in valve e 3 thickness of a,=a+pitch of thread u Ee diaphragm t=5 —0.07 meas. in vert. plane << 2 ll te ier. |e, ee G 0.7 0.5 c=| 0.2F : ae ra fe 02 | 0.1 ae lee on nar K | 0.04 | 0.25 | s=| 29 ae ee : 10 in., and 3” above L=T g = Sin Art. 96 stuff.|| that size. M |06 | 0.1 boxes. P=2Xstud diam. j=a+t2g+2v+diam. 7” of gland stud. Q | 1.0 | 1.75 m | 0.66 | 2.52 R=Stand. flange for C. I. pipes, Art. 37 o | 0.03 | 0.15 S=L or 14 Xstud _™ 1 5” diam. a mM =2a,+%3” T =thickness of C. I. stand. pipe flange Art. 37 q 0.14 | 0.6 WwW =29Q q 0.17 | 0.75 The section of the passage by a plane (yy) is constructed as described in Art. 210 and illustrated in Figs. 173 and 174. The method of finding the area is explained in Art. 211. The threads on the stem nut and yoke are finer than standard for that diam- eter, say 12 p.i. The yoke nut has a hexagonal head whose long diameter is less than (m1). This enables the short diameter to be found. The diameter of the threaded portion is less than the short diameter. The gland of the stuffing box has two stud bolts and nuts for forcing it against the packing in the box. 284 ELEMENTARY MACHINE DRAWING AND DESIGN 219. The forces acting on the stem are: the compression due to the internal pressure in the valve and the torsion produced by turning the hand wheel hard when the disc is pressing against the seat. The former produces a stress in the stem f- equal to 2 2 the area of stem ae divided into the total pressure . D, against the disc, where S- radius of seat opening and p= pressure 2 per sq.in. in valve fae . Torsion is produced by the force (F) ‘ j : F on the hand wheel rim. F times the radius (5) equal to it =moment of the twisting force. This twisting force must over- come the friction of the threads of the stem in the stem nut caused by the pressure against the disc. P ! Xx] TOm Fre. 181. Let the pitch of the thread be (P), the mean diameter (Dn) of the thread be taken as equal to (2 $ a 2 = P_R. The tangent of the helix angle of the thread is found by dividing the pitch by the circumference of the circle whose diameter is (Dm), that is, tan an. The force (T) at the middle of the screw thread necessary to turn the stem in the nut is the same as that necessary to slide the load (R) up a plane inclined at the angle («) with the horizontal (Fig. 181). The load (R) is resolved into two forces, one (NV) normal to the incline, the other (F) along the incline. The latter is the frictional ), and the upward resistance VALVES 285 resistance and equals the normal force (N) times the coefficient of friction or F=Nuy, but nal oes so Ene ? obtained trom T cosa—y~N=Rsine as the algebraic sum of the components of the forces T and R and the resistance which (N) offers to the movement along the plane, see Fig. 181. The forces resolved perpendicular to the plane give components Tsin« and R cosa, which equal (VN) or N=R cosa+T sin a. Substituting the value of N obtained above gives T cos «—R sin « i ——————— =F cos «+T sin a, ye or _ p(sina+u cos a r=R(Se a—u sin *). (1) Since tDm ; cosa=T and sin w=F, we have ef P 4 ee r= R(p eee boy tis le. eZ) This force (7) acts with a lever arm Pn and its moment 5 equals the moment of the force required at the hand wheel i Lali or rim (——}; m TDn FW Stee ee B) from which, by substituting the value of T obtained in (2), we have DnR(P+uxDm) _ FW (4) “teDeaph) Be In order to prevent the valve from opening under a load the tana must be less than the value of yp. If w=.15 then the angle « must not be greater than 8°-30’. 286 ELEMENTARY MACHINE DRAWING AND DESIGN 220. The area of the stem cross section at the root of thread is obtained by equating the total pressure on the disc to the compressive resistance of the metal times the area strained. That is, oe pao fe, where (A) is the diameter of the seat opening, (p) the pressure per sq.in. acting on the disc, (a) is the diameter of stem at root and (f,) the compressive stress allowed per unit area of root section. An additional strain is brought on the stem when the disc comes in contact with the seat and the valve operator increases his pull on the hand wheel to make the disc press very hard against the seat. The actual torsion pro- duced is the torque produced by the operator minus the friction of the threads. From experiments made in screwing up nuts on bolts, the resultant stress on the bolt due to the load and to the stress of screwing up produces a resultant stress from 15% to 20% more than the stress due to direct load. Therefore, in calculating the area of stem at root of thread it is necessary to either decrease the value of f, or add 15% to the area at root, after calculating it to resist only the pressure on the valve disc. If the working strength of good bronze is taken as 7000 lbs. per sq.in. for shear and we increase the area 15% this will be equivalent to using a value of 6000 lbs. per sq.in. and calculating the area of (a) for compression alone. 221. The efficiency of the screw may be obtained by using _ tan a(1—u tan a) + tan ety (a) =helix angle, u=coeff. of friction. The thread on the stems of valves of this type is usually single and }” pitch for valves up to 10”. Above that it is #’. From the dimensions given in the table on page 283 the efficiency of the screw can be easily deter- mined, provided the value of (u) is assumed. The preceding calculations are based on a square thread. If the thread is of any other form, the pressure (N) will become (N1)=N sec. where ($) is half the angle between adjacent thread faces. In the case of an Acme thread this angle 6=7}° and the secant = 1.007, which does not increase the normal pressure enough to consider. 222. The pressure at the root of the threads of the stud bolts used to fasten the bonnet to the shell can be calculated first by dividing the total pressure on the bonnet (due to the pressure the formula proposed by Prof. Barr, viz.: e ? VALVES 287 in the valve) by the number of bolts. This will give the load on each bolt which divided by the area of the bolt at root of thread gives the stress per sq.in. of section, due to direct load. 20% increase will give the stress due to both load and screwing up of the nut. The stress in the yoke uprights can be calculated by finding the maximum load that the stem will carry and dividing it by twice the area of the section (vr). See Fig. 180 (Fig. 4). 223. The valves may be drawn either assembled or detailed or both. In the latter case it is recommended to have one student draw an assembled view and have the detail work done by some other student. Further directions for drawing assembled and detail views will be found in the instructions. A list of parts arranged as shown for a safety valve in Art. 214, should be placed on the drawing. Dimensions of parts of valves from 3” to 8” can be obtained from the table on page 283. They should be changed from decimals to common fractions before using them for drawing purposes. The parts are made of different materials as follows, and require machining as described below: Shell, bonnet, wheel and gland are’ of cast iron. Bushing, nuts, disc, stem and seat of bronze. Shell and bonnet flanges to be faced and turned, disc, yoke and nuts to be turned and threaded; spindle turned; gland turned and faced on top, bush- ing turned. Stuffing box bored out for bushing; bonnet flange turned to fit bored hole in shell and faced; valve disc threaded and faced for stem nut and turned for seat bearing; seat bushing turned and threaded on outside only; seat in shell to be bored, threaded and faced on upper surface; hand wheel bored for spindle and keyed on. Bolts are to be studs with U.S. St. hex. nuts for bonnet and gland and U.S. St. sq. head bolts with hex. nuts for the shell flanges. Detail drawings are to be fully dimensioned and finished surfaces indicated. The material of which each part is made must be designated below the part, also its name and assembly number. 288 ELEMENTARY MACHINE DRAWING AND DESIGN INSTRUCTIONS VALVES Prob. 1. Draw full size on No. 2 paper three views of an assembled (A) check valve for a (B)” pipe. The views are to be a longitudinal section, a top view and a view look- ing at the end of the valve. Give all the lettered dimensions as well as others which were assumed in the making of the drawing. Make a list of parts as indicated in Art. 17. Four hours allowed in class for this. Prob. 2. Draw full size on No. 2 paper the details of a (C)’’ diam. (D) check valve. Each part to be fully dimensioned with finish marks and material indicated. Three views of the shell will be necessary but only two of each of the other parts. A bill of material must be placed on the drawing (see Art. 17). 6 hours allowed in class for this. ASSIGNMENTS FOR Pros. 1 anp 2 (CHECK VALVES) 1 | 2 3 4 5 6 7 8 9 10 | il 12 A |Sw.] .. |D. LJ] .. .[Sw.} ... [D. Ly}... | Swe}... |D. L. B By ce DE | ae | 2 | ees 2 ee e | ees | 2a Cc : Fl vewes fg as 2 ew (pe te ee |e Bel wie Pee D Sw.| ... |D. L. Sw.]... /D. Ly ... | Swe]... [D. L. Sw. =swing check. D. L. =direct lift. Prob. 8. Stop Cock. Draw an assembled stop cock for an (A)’” pipe making the views full size and as follows. 1. Front view (right hand half in section). 2. Top view (lower half a horizontal section through the axis of pipe). 3. End view (right hand half in section). Give all dimensions and make a bill of material on the drawing. Use No. 2 paper. 6 hours allowed in class. Prob. 4. Stop Cock. Make detail drawings of a stop cock for a (5)” pipe, full size. Show the body half in section in three views. Make outside views of the other parts, dotting in the interior. Give all dimensions, finish marks and make a bill of material on the drawing. VALVES 289 Use No. 2 or No. 3 paper depending on the size of the cock. 6 hours allowed for this work in class. Prob. 5. Trace a check valve or stop cock. 4 hours allowed. ASSIGNMENT TABLE FoR Srop Cocks. (Pross. 3 anp 4) 1 2 3 4 5 6 ml 9 | 10 A (Prob. 3)... 1 140) 14 2 B (Prob. 4)iu.| ec2 | wee | eas et sdehs 3 1 wl 13] 13 2 seated detail drawings of a (a)’’, (b) valve with (c) ends. No separate part to be drawn to a smaller scale than half size. No assembled view to be smaller than half size. Details must be completely finished ready for tracing and for shop drawings. Assembled valves are to be dimensioned as shown in Figs. 178, 179 and 180. A bill of material similar to that given in Art. 214 is to be placed on one of the sheets. The views required of each part or of the assembled valve are enumer- ated on the following page. The time reqd. for detailing the S.V. is 17 hours, for the G.V. 15.2 hours. All the calculations which apply to the kind of valve drawn are to be made and handed in with the finished drawings. These calculations are described in Arts. 210-213 and 218-222. Abbreviations used in Table below are As=assembly, D =detail, G=Globe, S=Safety, A = angle, F =flanged, Sc =screwed. The table for assignment is as follows: Prob. 6. Make on one or more sheets of No. 3 paper the ( No.}1/2/3/),41]5]61/71]8 | 9 |10/11/12/13 )14/15 /|16/17/18 a |3/38}4/441516) 7) 8 | 8 |383)23|] 2 113] 3 133] 4 }43) 5 bIGI|G\|G\|G)\/E\/G\|4|E4\/AlAIGIEG/E|S|S|S)Sj)S c |\F\|FIF\F\F\F\F\F FF IScj|Sce|\Sc|F|F |) F\F | PF As As As|D|As|D|D|D |D|D|D|As| D|As|} D|As No. | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 a |23} 2 |14] 4 |43] 5] 6 8|3/33}/4/5|6)7/8]/6483 b |A|AIALA|AIlAIA AIS|S;iS/|S|S|{|S|\|GIG |G c |Sc\Sce|\Sc|F\|F\|F\|F\F |F|Sce|Sc| F\|F | F\|F | F|F Sc D|D|D\|D|As|D|D|As|D|D|As| D| D |\As| D |As|As| D 290 ELEMENTARY MACHINE DRAWING AND DESIGN Globe, Angle and Safety Valves Valve Assembled. Views to be drawn. 1. 4 vertical longitudinal section of valve through axis of pipe. 4 outside view. 2. Top view with } wheel removed (G or A) 4 top view for safety valve. 3. End view with left half in section by vertical plane containing axis of stem. 4. Section by plane (yy). Valve Detailed. Views to be drawn. Globe or Angle Valve (A) Bopy or Seti. (Cast Iron.) (1) Side elevation half (right or left) in section. Dot in interior on the half not sectioned. (2) End elevation, right or left half in section. (3) Top view. (4) Section of passageway by plane (yy). Area to be found and placed on drawing. (B) Bonnet. (Cast Iron.) (1) Front view (right or left hand in section). (2) Side view (right or left hand in section). (3) Top view (upper or lower half showing section on xz). (C) Yoxzr Nut. (2 views) (bronze.) (D) Guanp. (Cast Iron.) (1) Top view. (2) Side view showing greatest length. (EZ) Hanp Wueet. (Cast Iron.) (1) Top view. (2) Side view (half in section). (F) Vatve Disc. (Bronze.) (1) Top view. (2) Side view (half in section). (G) Vatve Seat. (Bronze.) (1) Top view. (2) Side view (half in section). (H) Vatve Stem (Bronze) 2 views. (I) Disc Nut (Bronze) 2 views. (J) Botr List, giving diam. length, number reqd., description and location. VALVES 291 Safety Valve Views to be drawn. (A) Same as globe valve. (B) o Hf (views 1 and 3 only). (C) Yoxs (Cast Iron), 3 views. : (D) Lock Nur. (W. 1.), 2 views. (EZ) Lever. (W. I.) a (F) Same as globe valve. (G) ee ce (H) 6c 6c (J) ce 6c (J) ee 6c (K) Fulerum pin, (Steel) 2 views. (L) Stem top (C. 1.) 2 views. (M) Ball (C, L.) 2 views, with wet, of ball given, Test Questions (Working Drawings) 1. (a) What constitutes a working drawing ? (b) Dimension fully the sketch (Fig. 1), scaling all dimensions. 2. Give method of procedure fol- lowed in tracing a drawing. 3. What is a working drawing? Give an outline of the opera- C) cS) tions performed in making one. 4. In tracing a pencil drawing what kind of line will you use for: center lines, section Fic. 1. lines, dimension lines, invisi- ble lines, outlines, extension lines. How do you place numerals for different directions of dimension lines? Illustrate and use fractions. 5. (a) Describe the kind of line to use for each of the following cases, center line, extension line, dimension line, section line, border line. (b) Give a list, in order of procedure, of the operations in making a tracing of a working drawing. 6. What is a working drawing? Explain the method of making a tracing and order of procedure in inking in. 7. What is a working or shop drawing? Illustrate by a sketch drawn to scale. 292 ELEMENTARY MACHINE DRAWING AND DESIGN Test Questions (Rivets) 1. (a) Sketch in good proportion these rivet heads—cone, pan, counter- sunk, button. (b) Draw two views, to scale, of a single riveted lap joint, plates 3” thick, calculating first the size of rivet and pitch. Rivets are button headed. P—d=3.43”. (c) Sketch a Z bar, channel and angle iron and give the terms used in speaking of each of them. 2. What heads are used for boiler riveting? for structural iron work? 3. What is a lap joint? a butt joint? single riveting? double riveting? Which ones are used in boiler joints? 4. What is a gauge line? How is it located on a channel? on a Z bar? on an I beam? 5. Sketch the connection between two plates placed at right angles. 6. What is the difference in diameter between a rivet and the hole in the plates which it fastens together? Which diam. is the one calculated in the formula d=KV7? If K=1.5 show how to find the diameter d from a diagram when the value of (7) is given. Test Questions (Pipes and Fittings) 1. (a) What is the difference between the actual and nominal inside diameter of a pipe, and what caused this difference. (b) In the following system of piping (Fig. 1) specify the sizes and give the Torte jonk: \. names of fittings needed to properly connect up the same. (ce) Optional (what fittings are needed, and where would you place them—so that the tanks can be filled wi ye 2 2 7 AVI'FB bi CG I S A separately or both at This ena) the same time?). =e” z stopped (d) Sketch a plug, a cap, a Z 4 close nipple. What is the purpose of the union? 2. (a) What use is made of cast, Fra. 1. ‘iron fittings for W. I. pipe and what are the names of the common ones? Sketch a reduc. tee. VALVES 293 (b) How is the size of a pipe expressed and what is the meaning of the name 1” x}” reducing tee? 3. Explain the cause of the difference between the nominal size of a pipe and the inside and outside diameters and give the name of each fitting that you have drawn and explain the use of each one. 4. Sketch and describe (a) a union; (6) a C.I. pipe flange. 5. The system of wrought iron pipes of Fig. 2 is to be connected by cast iron fittings. Where would you place them and what names would you use for each one if you had to order them? 3" — 3! 7 ye La; i Mir Bra ” |? C é, . 2 G Fic. 2. 6. (a) What are pipe fittings and of what use is a scale of fittings? (b) What is the meaning of the term “ nominal size of a pipe” and what use is made of it? 7. Indicate the pipe fittings (names) to be used on the accompanying sketch of piping (Fig. 3). All pipe 2’”’ except where noted. Fia. 3. Fria. 4. 8. Give the names of the fittings required on the pipe line shown in Fig. 4 at A, B, C, D, E, F, G, in order to connect pipes of the sizes indicated. Test Questions (Screw Threads) 1. (a) Into what two general classes are screw threads divided? Give two examples of each class. Define and explain in general terms how the pitch of a screw thread is determined. (b) Draw to scale and dimension a U.S.St. screw thread, pitch 13”. 294 ELEMENTARY MACHINE DRAWING AND DESIGN 10. 11. 12. (c) Show the conventional method of representing L.H. screw threads. How cana screw of given diameter be made stronger? Give reasons why, and explain how, a pipe thread differs from a screw thread. . (a) What are the principal forms of threads used in machine con- struction? Draw the outlines of the threads, using any pitch. (b) What is the difference between a single and double thread? . How does a pipe thread differ from the thread on a bolt and why? . Draw the outline of each of the various kinds of thread you are familiar with and give the name of each one. Pitch=1” in all cases. . Draw to scale a U.S.St. thread, V thread and mod. sq. thread, pitch =1”. . Draw 2 threads of a V threaded screw, 3?” pitch, 13’ diam. Draw 2 threads of a square threaded screw, 3” pitch, 13’ diam. Draw the conventional V thread on a ?” diam. cyl. . (a) Explain ‘‘ pitch.” (b) Explain what is meant by “ triple thread,” and show how the pitch is measured. . (a) Make a neat free-hand sketch of a modified square thread, giving correct proportions of all dimensions. (b) Make a sketch of a left handed screw end, conventional method. . (a) What is the “ pitch ” of a thread. (b) What is a “ double ” thread? (c) What is a left hand thread? (d) What “lead ” has a double thread when there are 8 threads per inch? (a) Draw, free hand the profile of a U.S. St. thread of 1” pitch, giving the relative proportions. (b) Show how to represent a conventional sq. thread on a bolt. (a) Draw, freehand, the profile of a modified sq. thread, 1’ pitch, giving relative proportions. (6) How is a sq. thread drawn conventionally on a cylindrical rod. What is “ pitch,” triple thread. How do you indicate V or U.S.St. threads on a drawing to show left or right hand and number of threads per inch. Test Questions (Bolts and Screws) . Draw a 2” hex. hd. U.S. St. bolt without nut when the grip is 14’. Give all dimensions and make the three views commonly used on drawings. . Draw a 2” stud bolt (without nut) having a grip of 13”. Thread VALVES 295 it by conventional method, give length of thread and indicate where nut is placed. 3. Show two views (end and side) of a §” set screw, oval pt. (dimen- sioned). 4. Make a neat sketch of the following: (a) Fillister head screw. (6) French head screw. (c) Rivet with button head and pan point. 5. Make a sketch (freehand) of a locknut for a bolt and give the relative dimensions (3 views). Same for set screw (cup point) two views. 6. (a) Give the formula for F (distance across flats) of a U.S. St. bolt head and nut, and thickness T for the nut. (b) How far does a bolt enter a tapped hole. 7. Draw the three views usually used to represent a U.S. St. hex. nut for a 1” bolt (chamfer conventionally) and give dimensions. 8. What is a cap screw? a stud bolt? a tap bolt? and how do they differ? 9. (a) Describe and name the different machine fastenings on which screw threads are used. (b) What types of threads are used commonly. (c) What is the difference between a tap bolt and a stud bolt and in what cases is one preferable to the other. 10. A 2” stud bolt without nut, is needed to replace a broken one which had a U.S.St. nut. The grip is 1”. How would you order the new one by sketch, knowing only what is given above? 1. Draw 3 views of a U.S. St. nut for a 2” bolt (10 thrds.), using the conventional method and giving all the dimensions necessary for making it. 12. (a) What is a U.S. St. bolt and what is the thickness of nut? (b) What is a set screw used for? Make a sketch of one. (c) What is a stud bolt and a tap bolt and how do they differ? (d) What is “ chamfer ” and why are bolt and screw heads ‘“‘ cham- fered ”’? 13. The distance across flats of a bolt head on a 1” bolt is 18”. That on a 4’ bolt head is 63’... Find the constants to use in a straight line formula, for finding this distance on other bolt heads. 14, What is the difference between a hex. hd. cap screw and a tap bolt, and why is a cap screw preferable? 15. Show 2 views of a sq. hd. bolt with hex. head. U.S. St. Diam. =1”. d=3D+2". -*. 16. What three types of machine screws have you drawn, and how far is it customary to tap them into metal when used for fastening? 296 ELEMENTARY MACHINE DRAWING AND DESIGN 17, 18. 19. 20. 21. 22. 23. 24, 25. (a) What is a Penn nut and why is it used? (b) Sketch two forms of set screw, one cof them used to safeguard workmen from injury. (c) Sketch a fillister hd. machine screw. (d) What determines the length of a stud bolt. (e) How does a hex. hd. cap screw differ from a tap bolt? (f) How are bolt ends finished (sketch). (g) When is a stud bolt used in preference to a U.S. St. bolt and nut? (a) Construct three views of a hexagonal head bolt and nut D=1”. (b) What is the difference between a tap bolt and hexagonal head cap screw? (a) Show an inch of V thread (left hand) on a 1}” bolt. (b) Show an inch of modified square thread on a 2” bolt. (c) Show three views of a hexagonal nut for a 1’ bolt, using the conventional method. (d) State the differences between a stud bolt, tap bolt, set screw and hexagonal head cap screw. Draw a tap bolt 3?” in diam. threaded 1” on the end, 23 long, sq. thread. Show top and side view of a 3” stud bolt and nut. The stud to be of the proper length to fasten a plate 13” thick to a flat surface. Describe a hexagonal head bolt and nut and make a drawing of a 13” bolt with a grip of 1”. . How does a tap bolt differ from a stud bolt and how far is it customary to tap into metal to receive the end of the bolt? Draw top and side views of a U. 8. St. nut fora 1” bolt. How do the dimensions differ on a U.S. St. bolt head and a nut? Name types of fastenings used in machine drawing. Give dimen- sions of a tapped hole, also give formule for dimensions of standard keyways. Test Questions (Keys) . (a) Sketch a sunk key in a 2” shaft (end view) and give its dimensions. (b) What is a Woodruff key? Sketch snape of it. . What is a sunk key? When used? What proportions? Illustrate how set in a shaft (end view). . Draw to scale a sunk key for 2” shaft when W = 4D+1, T=4D+2” and key is 3’”’ long. (Dimension this.) . Construct three views of a shaft 2’ in diameter, having a key 3” long set in it. Give the dimensions of key. . Show the depth of keyway for a 2” shaft and where this depth would be measured. . Show 2 views of a keyway, 2” long, in a 2” shaft VALVES 297 . The width of a key for a 2” shaft is 3” and its thickness is #,”. For a 4’’ shaft the corresponding dimensions are 3” and 3’. Deduce the formule to be used to determine the width and thickness of keys for any diameter of shaft from 1” to 4’. | W=3D+}" t=%D+4"|° . Draw a key for a 13” shaft set in the keyway of the shaft. Length of key 2”. W=%D+3", T=%D43". . Deduce the formule W= D+ 3” and T=3D-+4” knowing values of W and T as follows: For 2” shaft W =4", T = 5", for 4” shaft W=2", T=21". . (a) Make sketch of plain key (sunk), with parallel sides L” long, W" wide, T’’ thick. (b) Give the formulas for W and T, Test Questions (Shafts and Couplings) . The diam. of the hub of a flange coupling for a 2” shaft is 4.4”. The diam. of the hub for a 5” shaft is 9.8’. Deduce the constants for the straight line formula B=eD+¢8, and find the diam. of ‘hub for a 3” shaft. . What is a flanged coupling and how many pieces are there in one where four bolts are used? . If a shaft coupling for a 2” shaft has a distance 62” from center to center of bolt holes and for a 43” shaft coupling a distance of 123”, what distance must a shaft coupling of 3” have? Solve this by the deduction and use of the straight line formula. . Calculate the diameter of a W. I. shaft to transmit 10 H.P. at 300 R.P.M. . Calculate the diam. of shaft to transmit a force of 1000 lbs. applied at the end of a lever 20’ long when the lever is keyed to the shaft. Shaft of W. I. . Calculate the shearing strain on the bolts of a coupling for a 23” shaft. . Calculate the shearing strain on the bolts in a Hookes joint for a 13” W. I. shaft. . Sketch a claw coupling and explain its function. Test Questions (Stuffing Boxes) . How is it possible to prevent leakage of steam around the piston rod of an engine without excessive friction? Make a sketch of one of the component parts of the device for reducing this friction when 298 ELEMENTARY MACHINE DRAWING AND DESIGN 2. 10. 11. the piston rod is 2” diam. S=1D+3", H =48,a=S. (No section required, outside views only.) Explain the function of a stuffing box and illustrate either one of the two types of box. Draw to scale and dimension the gland only of a screw cap stuffing box fora l’ rod. H=48,S=3D+}?". (Make outside views only.) . Describe a stuffing box and illustrate by sketches what the function of the gland is. . State the conditions which require the use of a stuffing box and give two cases where boxes are used. What is the gland and what relation does its length bear to the length of the packing space? . A stuffing box is designed for a 1” shaft. S=%D+%”. Show a gland for this box when H =5S. . Describe a stuffing box and draw the gland of a screw stuffing box fora 2’ rod. Dimension this completely. . Draw and dimension the gland of a screw cap stuffing box for a 1” shaft. Length of packing space =H =48. . What is a stuffing box? and what styles do you know about? . Enumerate the parts of a screw cap stuffing box and sketch each one. What is the unit for the principal dimensions? When is the maximum length of gland and packing space used? When the minimum? Sketch approximately to scale the gland for a bolt type of stuffing box. How is the length of stud bolt determined in a bolted gland stuffing box? Test Questions (Bearing Box) . The distance from the center line of a 2’ bearing box to the center of stud bolt is 1.788”. The corresponding distance for a 5’’ box is 3.978”. Deduce the constants for the straight line formula b=aD +8, and find this distance for a 3” box. . A bearing box cap for a 2” diam. shaft is 4” thick. A bearing box cap for a 5” diam. shaft is 13” thick. Find the constants a and 6 for the straight line formula to use in obtaining intermediate sized caps. . The stem of a 3” valve is 3’ diam., while the stem of a 6” valve is 1” diam. Find the constants 2 and @ to use in a straight line formula. Find the size of stem for a 4” valve and check the value by a scale. . Explain how the formula (required dimension =aS+) is obtained and its use in machine design. on 10. 11. 12. 13. VALVES 299 . Construct a scale for bolt heads and nuts (U. 8. St.). . Using the straight line formula d=«D+6, show how the values of « and @ may be found in the formula d=$D+41, taking two values of d from the tables. . What use is made of the straight line formula in designing machine parts and how would you determine the constants to use in it for a particular dimension? . Explain a bearing box, also construction and use of the scale of dimensions. . Construct a scale for width and thickness of keys for shafts from 1” to 5”, w=%D+34". t=3D+4+2". Calculate the foot-lbs. of work lost in a bearing when the shaft diam. is 3”, length of bearing =9”, pressure per inch of projected area =75 lbs. R.P.M.=200. Shaft horizontal, load vertical. c¢=.43. (V2) v=veloc. in ft. per sec. w= : Make a working sketch of the cap of a bearing box assuming all dimensions, in good proportion. Show three views and put on dimension lines. (a) How does a bearing box differ from a pillow block? (b) What metal is used as a bearing surface in the former? What is a pillow block and what arrangement is made for adjust- ment across the line of the shaft? Test Questions (Hangers, Ball and Roller Bearings) . Determine the distance apart for hangers to support a 24” shaft, making 200 R.P.M. What will be the deflection? . What is a hanger? How is it adjustable and for what purpose? What is a ‘‘Sellers’” hanger? How does the bearing of a hanger adjust itself to the line of the shaft? What is the “drop” of a hanger? . Enumerate and describe the component parts of a hanger. . Describe the arrangement of parts in a ball bearing hanger. What are the great advantages of ball bearing hangers over plain bearing hangers? . What provision is made in the Sellers type of hanger for placing the shaft in the hanger or vice versa? . How are hanger bearings automatically lubricated? Sketch a system of automatic lubrication. . What is a post hanger? A wall bracket? . Describe a roller bearing? How is it applied to a pillow block? Enumerate the parts of a ball bearing pillow block. 300 ELEMENTARY MACHINE DRAWING AND DESIGN 9. 10. Draw a section of the inner and outer races of a ball bearing and show how the inner race is tightly fastened to the shaft. Explain the forms of contact of the revolving shaft with the three kinds of bearings, plain, ball, and roller. Test Questions (Pistons and Piston Rods) . Describe in detail (a) the various forms of pistons, (b) how they are fastened to the piston rods and the (c) methods of keeping them tight. (d) What is the function of a wrist pin? . A piston rod 13” diam. is screwed into a crosshead by being threaded with 10 thds. p. 7. (depth of thread =.065”). What steam pressure can be used if the piston diameter is 10’”’ and the max. unit stress in the rod is not to exceed 4500 lbs. per sq.in.? What will be the unit stress in the unthreaded portion of the rod? . (a) What is a piston? (b) How is leakage past it prevented? (c) Sketch two ways of fastening a piston to a piston rod. (d) What is a trunk piston and where is it used? . An engine has a cyl. bore of 10” and a stroke of 12”. The con- necting rod length is 2} times the stroke. The piston rod is 2’ diam. threaded 10 thds. p. i. (depth of thread =.065). The crosshead bearing surfaces have an area of 60 sq.in. When these surfaces are carrying a max. pressure of 30 lbs. per sq.in. cal- culate (a) the steam pressure in the cyl., (6) the unit stress in the piston rod at least section. . An engine with 12” bore, 14” stroke, carries steam at 100 lbs. pressure. The connecting rod is 40” long. Calculate the diam. of piston rod (outside) (f;=4500) 10 threads per in. . What are Ramsbottom rings? How are they prevented from leaking? Of what material are they made? . Calculate the weight of a C. I. piston 10” diam. made as in Fig. 121. . Calculate the weight of a C. I. trunk piston 6’ diam. made according to the proportions in Fig. 123. Test Questions (Crossheads) . Calculate the area of bearing surfaces for a crosshead for a 11x14” horizontal engine whose connecting rod length is 3 times the stroke. The steam pressure is 100 lbs. per sq.in., the maximum pressure on the bearing surfaces of the crosshead is 30 lbs. per sq.in., and the ratio of length to width of these surfaces is 4 to 1. The type of crosshead to be like the one you have drawn which requires two guide bars on each side. 7. 8 VALVES 301 . The sliding surfaces of an engine crosshead have an area of 60 sq.in. The engine has a cyl. 10” diam. and the stroke is 14”. The steam pressure is 100 lbs. per sq.in. The length of conn. rod is 3 Xstroke. Find the max. pressure per sq.in. on the sliding surfaces, . Calculate the area of sliding surface for the crosshead for an engine whose cyl. is 10” diam., stroke=14’’., Steam pressure 100 lbs. per sq.in. Maximum pressure on sliding surfaces to be 30 lbs. per sq.in. Length of connecting rod =3 times stroke. . An engine is 10’ X14”. Length of connecting rod is 5Xcrank. Pressure on piston =90 lbs. per sq.in. Allowed maximum: pressure on bearing surface of crosshead is 30 lbs. per sq.in. Find the area of the bearing surface of the crosshead and the maximum pressure on the crank pin. . (a) How long c. to c. would you make a connecting rod for a 12” X14” engine if ratio of conn. rod to crank is 7 to 1? (6) On same engine what area of bearing surface would you have, allowing a maximum pressure on guides of 30 Ibs. per sq.in., where the pressure on the piston is 120 lbs. per sq.in.? . What would be the area required for the rubbing surface of a cross- head in a 12” X12’ engine, the length of connecting rod being 4 times the stroke, steam pressure being 100 lbs. and allowable pressure 40 lbs. per sq.in.: An engine 8’.x10” has a conn. rod 6xXthe stroke. Pressure on the piston is 100 lbs. per sq.in. Find max. pressure on guides. . A gasoline engine is 43” bore x6” stroke and the connecting rod is 15” long. The length of piston =its diam. Find (a) the max. pressure of the piston against the cyl. wall when the explosion pressure =350 lbs. per sq.in. (6) Find the pressure per sq.in. of projected area of wrist pin when the pin diam.=.2D and its length is 0.5D. . A gasoline engine has 4” bore, 6’ stroke, length of connecting rod =22 times stroke. The piston length equals its diameter. The wrist pin diameter =0.2D. Its length between supports is 43D. If the explosion pressure in the cyl. is 300 lbs. per sq.in., calculate (a) the max. thrust of the piston against the cyl. walls (in lbs. per inch of projected piston bearing area). Calculate (b) the pressure per inch of projected area on the wrist pin. Test Questions (Connecting Rods) 1, The diameter of a connecting rod at the smallest place is 2”. The strap is 23’’ wide and the key which passes through the strap is 3” thick. 302 ELEMENTARY MACHINE DRAWING AND DESIGN Calculate the thickness (a) of the strap and the length (m) of strap between the gib and end of strap. The area for shear is $ the area for tension. 2. The strap of a connecting rod is 23” wide in a direction parallel to the axis of the crosshead pin. The key and gib are 32” thick in the same direction. Compute the thickness of strap and the combined width of key and gib when the least diameter of the connecting rod is 23”. The area to resist shear=# the area to resist tension. 3. The strap for the crosshead end of a connecting rod is 23” wide, measured parallel to the axis of the crosshead pin. The hole for the key and gib is 2?” wide measured in the same direction. The smallest diameter of the connecting rod is 2’. Calculate the thickness of strap and length of hole for key and gib, mak- ing a sketch to show what you have calculated. Area for shear is } area for tension. 4, Calculate the dimensions a-b-c-d of the end of the conn. rod shown in Fig, 1, Shearing strength =75% of tensile strength. wv Lol) 3 A ey ( |e dP) 8 eS Fig. 1. be 3 a 5. Calculate the dimensions (a) and (b) for the conn. rod end shown in Fig. 2. The shearing strength is $ the tensile strength. OQ 1} iy, -e Y S 8 A “I! ~ Y) >The 4, 3t y2 > 4 Fst. enh Ly =f 4 Sy NN Fig, 2. VALVES 303 6. Calculate the values of m-n-o-p on the end of the connecting rod shown in Fig. 3. The area in shear is to be 1.4 the area in tension. 7. Calculate the dimensions a-b-c-d for the rod end shown in Fig, 1. Shearing strength to be 60% of the tensile strength. A Siva ces "| 4 = — y's (oe, he 3" > n -F- ~~ — ¥ I xv y ° “ys LY vi_ m se 3 3 ar ay | WY EF L rt Vio (2) Fig. 1. 2, A crank has the proportions given in Fig. 2: (a) Calculate the load the crank pin will carry if the bearing pressure on it is 900 lbs. per inch of projected area. (b) Calculate the stress in the web with the above load. (c) Calculate the value of f, in the shaft when the above load is applied to the crank pin. pBtVBT) ap Dy. Wt 5.1° . Determine the proportions of web of a C. I. crank for a 12” radius The shaft is 43’ diam. The crank pin is 3” diam., 33” long and carries 1000 lbs. per inch of projected area. The width of web where it joins the hub is 83”. Hm OO a VALVES 305 Test Questions (Eccentrics) . An eccentric rod is 2” in diam.; what size bolts would be necessary to fasten it to the eccentric strap if 2 bolts were used and allowance was made for threads cut on bolt. 5 threads per inch. . What is an eccentric sheave? Draw an eccentric sheave for a 3” shaft, the eccentricity being 1’ and width of eccentric =2”, B d=1.2D+2r+3", A =F Make a view looking at the end of shaft hole, also a side view. . Compare the strength of the two eccentric strap bolts and the eccen- tric rod. The bolts have a diameter of 2” and the rod a diameter of 13”... . Explain the method of oiling an eccentric, in motion, from a stationary oil cup. . Draw to scale approximately, the half of an eccentric strap which fastens to the ecc. rod. Test Questions (Belting and Pulleys) . A belt 5” wide transmits 6 H.P. while running over a 24’ pulley. Assuming the effective pull per inch of width =40 lbs., how many R.P.M. does the pulley make? . Given, the diameter (16’) and rev. per min. (100) of a pulley and the rev. per min. of a pulley (90) driven from the first, also the horse-power transmitted (4). Find the diameter and width of face of the second pulley. . What cross sections are used for the arms of pulleys and gears? . An 8 H.P. motor runs at 600 R.P.M. From it a belt runs to a 48” pulley on a shaft which makes 150 R.P.M. What diameter pulley would you use on the motor and what width of face would it have. Driving pull on belt to be 40 lbs. per inch of width. . Sketch a section of the rim of a 24” pulley to carry a belt to transmit 10 H.P. when it makes 200 R.P.M. D #1” ‘ B+6 t=—-+—. Crowning =—— 200 8 200 . A pulley on a countershaft making 90 R.P.M. is belted to a 30” pulley which makes 120 R.P.M. (ona pump). The pump requires 2 H.P. to run it. What diam. of pulley is required for the counter- shaft and what width of belt is required? : 306 ELEMENTARY MACHINE DRAWING AND DESIGN 7. A belt 10” wide runs on a pulley 24” diam. which makes 100 R.P.M. What horse-power is it transmitting when the driving pull on the belt is 40 lbs. per inch of width? 8. A 3 H.P. pump is to be driven by a belt from a countershaft pulley to a pulley 30” diam. on the pump. The countershaft makes 90 R.P.M. and the pump pulley 120 R.P.M. Find the diam. of countershaft pulley and width of belt to use. 9. A pulley 24” diam. makes 100 R.P.M. The width of face is 63”. What H.P. is transmitted by the belt running over this pulley when the effective driving pull is 40 Ibs. per inch of width? 10. Calculate width of belt to run on a 20” pulley, making 200 R.P.M. and transmitting 5 H.P. 11. (a) Why is the face of a pulley sometimes crowned? (b) Why is a set screw used in addition to a key for securing a pulley to a shaft? 12. Find the R.P.M. of 39” and 60” pulleys when the belt between them is 6” wide and transmits 4 H.P. 13. In the sketch shown a circular saw is to run at 1000 R.P.M. and deliver 6 H.P. Bis a pulley on a (8) (© GHP gasoline engine making 450 R.P.M. C aa wae and D are on a countershaft. D is 24’ Selt ‘m8 diam. B is 22” diam. FE is 12” diam. ai What must be the diameter of C? What (2) (é) will be the width of single belt to use be- tween B and C and between D and E? 14. A belt 5” wide transmits 6 H.P. while running over a 24” pulley. Effective pull per inch of width=40 lbs. How many R.P.M. does the pulley make? 15. A pulley on a countershaft makes 90 R.P.M. and the belt from it runs a 2 H.P. pump pulley 30” diam. 120 R.P.M. What diam. pulley is needed on the countershaft and what width belt will you use? 16. A pulley 16” diam. making 100 R.P.M. transmits 8 horse-power. What is the width of face? 17. Two pulleys are 34” and 20” diam. and their centers are 36’-0” apart. What H.P. will be transmitted if the larger pulley makes 110 R.P.M. and the initial tension is 460 lbs. w=.3. How wide a belt will you use? T, R-r Log coe 6=180°—2. sin p= na Tet T, = 27: 8 VALVES 307 Test Questions (Spur Gears) 1. In the hoist shown in Fig. 1 the C. I. pinion makes 150 R.P.M. The face of the gears is to be 3P. Determine the circular and diametral pitches and H.P. the gears will transmit. Formule for Gearing: 600 600+V’ = 194 0884 N F =SPby ? 2. The drum of a hoisting engine is 36” in diameter. A gear fastened to the end of the drum shaft is 42” pitch diameter. The driving gear which meshes with this is 14’’ pitch diameter and makes 200 R.P.M. The maximum load to be lifted at this speed is 1200 lbs. Width of face of gears is 2 XP. Determine the circular pitch, diametral pitch, No. of teeth in each wheel, addendum and dedendum, and draw the blank for the small gear. Diametral pitches are 1, 13, 19, 2, 24, 23, 22, 3, 4, 5, ete. | 2000 Ibs. Fic. 1. Fia. 2. 3 A hoisting engine has the arrangement of drum and gears in Fig. 2. Max. load to be lifted =6000 Ibs. Pinion makes 200 R.P.M. Find P, p, 6, number of teeth, and outside diameter of both gears. b =width of face =3P. WV, 600 =SPb geet =33000° 600g 308 ELEMENTARY MACHINE DRAWING AND DESIGN 4, The following train is used for hoisting stone blocks of 2 tons weight at a velocity of 100 ft. per min. On the engine crank shaft is a pinion which gears with a gear on the shaft carrying the hoisting drum. This drum is 36” in diameter. The gear ratio of the pinion and gear is 1:8. What diameters of pinion and gear would you use if 12 was the smallest number of teeth allowed in the pinion and 6 feet the greatest diameter allowed for the gear. Find P, p, number of teeth in each gear, addéndum, deden- dum, width of face, and outside diameter of the larger gear. 5. A C. I. gear 36” diam., 3” face, 3 pitchis keyed to a hoisting drum shaft which is used to hoist loads of 2000 Ibs. 600 ft. per min. The drum is 42” diam. Are the gear teeth strong enough for this work? Prove your statement. 6. Construct a cycloidal tooth. Diameter of pitch circle 12”. Describing circle 2” diam. Pitch =1”. 7. (a) Find the diam. of pulley on shaft (A) and width of belt to use to run between the pulleys. Fig. 3. Belt (@) Gear. 6 a) y : oO » Pulley 4200 RPM. O-7) Fic. 3. Wot 1500 Ibs. (b) What will be the circ. pitch and diametral pitch of the gears? Width of face =3P. (c) What will be the vel. of the wgt. and the H.P. transmitted? 8. A weight of 2100 lbs. is raised by a chain passing around a drum 36” diam. making 40 R.P.M. The spur gear on the drum shaft is 30” diam. and is driven by a gear 24’ diam. Find for the latter its diametral and circ. pitches, number of teeth, width of face, addendum, dedendum, and outside diam. Width of face =3P. 9. Two gears 20” and 32’ diam. mesh. The smaller makes 100 R.P.M. The pressure on the teeth is 1000 Ibs. What H.P. is transmitted, and what is P, p, number of teeth, and outside diam. of each wheel. Width of face =3 XP. 10. Design a spur gear rim for a gear to transmit 20 H.P. when it is 36” diam. and makes 100 R.P.M. Width of face=b=2P. Give circ. pitch, diametral pitch, number of teeth, outside diam., depth of teeth. VALVES 309 11. Find the circ. pitch, diam. pitch, outside diam., number of teeth, depth of teeth, width of face of gear C, when W =2000 lbs., and A makes 200 R.P.M. Width of face =2P. Fig. 4. 12. Calculate the allowable load on aspur gear tooth of 4 diam- etral pitch 30” in diameter. using a static stress of 8000 and a tooth length of 3P. What H.P. will this gear transmit, when rotating at 100 R.P.M.? 13. A spur gear 20” diameter transmits 10 H.P. when it makes 90 R.P.M. Find the circular pitch, diametral pitch, addendum, dedendum, and outside diameter of gear blank. 14, A spur gear transmits 10 H.P., pitch diameter 20 inches, width of face 63 inches, circular pitch 3.14 inches. How many R.P.M. does it make? What is its diametral pitch, addendum, depth of tooth and outside diameter? 15. Explain the involute system of gear teeth profile construction with sketch. Fia. 4. Test Questions (Bevel Gears) 1. A train of hoisting mechanism consisting of two spur gears, 2 bevel gears and a drum, is to raise weights at a velocity of 470 ft. per min. Find the max. load which can be raised at this speed with the sizes of cast iron gears, shown in sketch, neglecting friction. 300 REM. a | : | Spur }-4P-607, / (ay? oe 32-457 /8T- t Bevel drum | 8 Gears Y We (" Fig. 1. Find diam. of spur pinion (A) and calculate the diam. of shaft to use in it. 0.684 EP. =SPby. y=0.124—-*, d= 2.87, [——. Weshay.y N N 310 ELEMENTARY MACHINE DRAWING AND DESIGN 2. A pair of bevel gears with shafts making an angle of 75° has 36 10. 11. 12. teeth in large gear, 20 teeth in small gear, 2 pitch. Make a layout of above, + scale, indicating by note, pitch cone, normal cone. Calculate pitch diameters of both gears, and velocity ratio, . The shafts of two bevel gears intersect at an angle of 60°. The velocity ratio of the shafts is 3:1. The larger gear is 24” diameter. 2 pitch. The shaft of the smaller gear is 1’ diameter. Thick- ness of rim-at base of teeth at the large end is 0.56 P. Draw the blank for the small gear, giving dimensions. Width of face =2P. . Two bevel gears 12” and 18” diam., 3” face, 4P, are transmitting power between two shafts at 90° and the larger gear is making 90 R.P.M. What H.P. will they safely transmit? Prove all statements. . Two shafts are connected by bevel gears. The vel. ratio is 3:1. The shafts meet at an angle of 75°. The larger gear is 18” diam. and has 54 teeth. Width of face is 3”. Draw the blank for the smaller gear. Calculate the H.P. of these gears. . Two bevel gears connect two shafts whose velocity ratio is 3 : 2. The angle between the shafts is 75°. The larger gear is 12’ diam., 3P. Find the diam. of smaller gear and No. of teeth in each gear, also the angle the pitch cone of the larger gear makes with the axis of the shaft. Two bevel gears have a vel. ratio of 2:1, and their shafts meet at an angle of 75°. Small gear shaft is 13” diam. Width of face = 3P. Depth of rim below teeth at large end is 0.56 P. Larger gear is 12” diam. 3 pitch. Draw small gear blank. . Give the outside diam. of a bevel gear, 3 pitch, 45 teeth, when the angle § between the axis and pitch cone is 30°. What is the tan- gent of the angle the face makes with the axis? . Two shafts meet at right angles and their velocity is 2:1. The larger of the bevel gears connecting the shafts is to have 30 teeth, 2P. How many teeth will the small gear have and what pitch diam.? Draw the blank for the small gear when width of face =2XxP, depth of metal under teeth at large end=.5P, and the shaft is 2” diam. A bevel gear has 30 teeth, 3 pitch, and the angle between the side of the pitch cone and the axis is 30°. Length of face is 2P. Shaft diam.=13”. Depth of metal under teeth=.57P. Draw and dimension the blank. Show the blanks for two bevel gears. Shafts meet at an angle of 75°. Larger gear has 34 teeth, 2 pitch and veloc. ratio is 1 to 2. Construct the outline of two bevel gears in position for trans- mitting motion. One has 36 teeth, the other 12 teeth; 2 diametral pitch. b=twice the circular pitch. Angle of shafts =60°. VALVES 311 13. Make a drawing of the gear blank, without hub, for a bevel gear whose pitch diameter is 16 inches and which has 48 teeth. Length of face of teeth equals twice the circular pitch. The angle of the pitch cone at the vertex with the axis is 30 degrees. 14, A bevel gear has 30 teeth, 3 diametral pitch, and the angle between the side of the pitch cone and the axis is 30°. The length of face of the gear is 2xXcircular pitch. The diameter of shaft is 13’".. Make a drawing of the gear blank and give the angle the face of the blank makes with the axis of the shaft. 15. Draw the gear blank for a bevel gear having 12 teeth; 2 DP. Angle at vertex of pitch cone =60°. Diam. of shaft =13”, width of face =2P. Depth of metal under teeth =,57 P, Test Questions (Worm and Gear) 1. Two shafts A and B meet at 75° and their velocity is as 3:1. The larger bevel gear has 54 teeth, 18’ PD. The width of face is 3P. Find the H.P. the bevel gears will transmit, the diam. of shaft A. Draw the blank for the bevel gear for that shaft. The pitch of the worm is 1.1” and its lead is 1.1”. Find the H.P. the worm and wheel will transmit; how large a weight can be raised by a drum 24” diam., and how large a shaft will be necessary in the worm wheel. (See Fig. 1.) 2. Vel. ratio of A to B=3 to 1. Circ. pitch of worm=1.1". Bevel gears and shafts shown in Fig. 2. (a) How many R.P.M. must (B) make? (b) Which combination (bevel gears or worm and wheel) will trans- mit the most H.P.? How much is this? Give all the cal- culations necessary to reach your conclusion. 312 ELEMENTARY MACHINE DRAWING AND DESIGN 3. Construct the outline of a tooth on a worm wheel. Pitch =1}”. Velocity ratio of worm and worm wheel=40:1. Use involute system, approximate method. 4. Worm is 3’ PD and the worm wheel 16” PD. P of worm is 1”, What is their vel. ratio and No. of teeth in wheel? 5. A worm wheel is 14” pitch diameter with teeth of 1” pitch. What is the velocity ratio between the wheel and the worm and what is the outside diameter of the worm if its pitch diameter is 3P? 6. A worm and wheel gear together, their velocity ratio being 48 : 1. If the worm is 1” pitch what will be the pitch diameter of the worm wheel? 7. The pitch diam. of a worm is 3XCP=3”. Pitch diam. of worm wheel is 15.3”. What is the velocity ratio? 8. Velocity ratio of bev. gear shafts is 3:1. Worm wheel has 64 T and the worm has a double thread. Find efficiency of worm and wheel. (See Fig, 3.) The diam. of worm is 23P, +120, 36T (a) What H.P. will the bevel gears transmit? Explain carefully how you obtain your answer. Efficiency of bevel gears 96%. (b) Is the worm wheel shaft large enough to transmit the power which the worm wheel will deliver? (Explain by calculation.) (c) Is the belt large enough to run both the sets of gearing? Explain how you reached your conclusion. Test Questions (Cocks and Valves) 1, Name and describe the parts of an angle or a globe valve, and state the material of each part. 2. Explain a globe, angle, or safety valve and its working. 3. What should be the “ lift” of the valve disc in an angle or globe valve to allow the fluid to pass with the same velocity which it has in passing through the circular opening of the seat? 4. Explain the characteristics of the various classes of valves and give an example of each kind of valve in the different classes. 10. 11. 12, 13. 14, 15. 16. 17. 18. 19, 20. 21. VALVES 313 . What is a check valve? What are the names of each type? Explain the action by means of a sketch. . Make a proportional sketch of each of the parts of a check valve. . Show by means of a diagram the action of a safety valve. Sketch the moving parts. . Explain how to obtain the area of the least opening in a globe or safety valve. What information does this give? . Calculate the weight of ball for a 6” safety valve. Weight of stem, etc., is 8 lbs. Pressure in valve 100 Ibs. per sq.in. Length of lever to 100 Ibs. mark is 50” from fulcrum, Depth of lever at fulcrum is 2”. At 100 lbs. mark it is 1.5”. The lever is §” thick and made of W. I. The steam pressure acts on the lever 33” from the fulcrum. Calculate the diam. of fulcrum pin in question 9. The bonnet opening in a globe valve is 7.4” diameter. The bonnet is fastened to the shell by 8 3” bolts. Find the max. stress in a bolt when the pressure in the valve is 150 lbs. per sq.in. What is a pet cock? A stop cock? Sketch them both. Explain the difference between a valve with flanged ends and one with tapped ends. What is an angle valve? A globe valve? A safety valve? Sketch the general outline of an angle valve with tapped ends. Explain the forces acting on the stem of a globe valve. How do you arrive at the proper diameter of the stem? Find the force necessary to use on the hand wheel of a globe valve necessary to force the disc against the seat with a total pressure 75 lbs. greater than a given pressure of steam in the valve. If the stress due to screwing up is 15% more than the direct stress due to direct load on a valve stem, calculate the outside diameter of a stem for a 6” valve when the thread is }” pitch and the stress per sq.in. on the stem is not to exceed 6000 lbs. (Acme thread.) Calculate the efficiency of the screw on a valve stem 1}” diam., _ tan « (1—w tan a) : tan «+. ; Sketch either a bonnet or a shell for a globe valve and indicate the finished surfaces by finish marks, Calculate the weight of shell or bonnet of the valve sketch accom- panying this. ;” pitch when p =.2. A Abbreviations, 32 Acme thread, 73 Addendum, 215 sik circles, 215 Air cocks, 271 A. L. A. M thread, 71 bolts and nuts, 71 se ‘« Table 14, 17 Alphabets, 35, 36, 37 Angle irons, 50 «¢ valves, 279 Angles on bevel gears, 251 Architects, scale, 23, 24 Ares, radius of, 30 Arms, bevel gear, 248 ‘« pulleys, 205 ‘¢ spur gears, 223 ‘« worm wheel, 261 Arrows, pencil, 25 «« ink, 42 Assembled drawings, 21 B Backlash, 216 Ball bearings, 137 “« races, 140 Beams, rolled, 50 Bearing box, 120 us ‘¢ formulae, 120 of ‘scale, 123 ‘pressure, 125 e ‘“ Table, 16, 19 Bearings, 119 Belts, leather, 199 Bevel gears, 244 ae gear angle, 253 = arms, 248 «plank, 251 a an dimensiond, 254 es ‘drawing, 245 INDEX Bevel gear H. P., 254 hub, 248 oe cs layout, 245 ce pinion, 247 yim, 248 a ‘« ‘shop drawings, 251 as ‘* teeth strength, 253 Bill of material, 34 Blank, bevel gear, 251 Blocking in, 24 Blue printing, 41 Bolts, 74 and nuts, shading, 40 ‘* coupling, 88 ‘€ hex. hd., sq. hd., 81 ‘« lists, 79 ‘“* mfs. stand., 79 ‘« miscellaneous, 79 ‘* rounding ends, 81 ‘* seale U.S. St., 79 ‘¢ strength, 75 “« stud, 79 «« tap, 79 ‘« Tee hd., 88 “U.S. St., 79 Borders, 33 Brasses, 123 Breaks, conventional, 26 British Asso. thread, 70 Bushing, pipe, 57 Button hd. cap screw, 80 Buttress thread, 74 Cc Caps, pipe, 56 Cap screw, 79 ce &* slots, 79 Cast-iron flanges, 55 se fittings, 57 Center lines ink, 42 " “« pencil, 23 315 316 Center lines principal, 23 “supplementary, 24 Chamfer, 87 ne curves, 87 Channels, 50 Checking, 44 Check valves, 268 Circle drawings, 30 ‘« shading, 39 Circular pitch, 216 ba ‘¢ cale., 236 Claw coupling, 110 Cleaning tracings, 43 Clearance, gear teeth, 216 Cocks, 279 Coefficients of friction, 125, 137, 260 os Table 17, 19 Cone pulleys, 209 Conn.-rods, 173 PE crank end, 176 eS gasolene eng., 176 es intersec., 185 OF lengths, 173 a sections, 173 bs strap end, 183 Conventional thread, 75 Conversion table, metric, 2 Cotters, 102 ‘« Spring, Table 12, 16 Cottered bolt, 90 ae joint, 101 Coupling bolts, 88 fe pipe, 56 ee pipe, table 8, 13 oe shaft, 106 re shaft formulae, 107 Crank pin, 189, 190 Cranks, 189 «« web, 191 Crosshead, engine, 161 es pins, 164 ee ‘forces, 162 ee slipper, 165, 168 tf turned, 165 Crowning of pulleys, 204 Curve centers, 30 Curves, 24 Cutting gear teeth, 222 Cycloidal teeth, 219 Cylinders, shading of, 39 D Decimal equivalents, 1 Dedendum, 215 INDEX Depth of tapped hole, 80 Detail drawings, 21 Diagrams, gear, 237, 238 ae gear speed, 235 Diameters, to give, 30 a of pipe, 54 Diametral pitch, 216 Dimensioning, 29, 42 Dimension, how to, 27, 29 at full size, 26 ee lines, 25 ee numerals, 27 Dimensions, how expressed, 29 te ‘« placed, 29 ae of, holes, 30 ay overall, 29, 44 a reduced, 26 oe special, 29 Directions, end of each chapter Draughtsman, name of, 33 Drawings, 21 ee ink lines, 44 ee instruments, 8 oe layout, 24 ad numbering, 33 ad paper, 23 Driving force, belts, 202 Drop hangers, 128 E Eccentric oiling, 194 «rod, 194 ee strap, 193 Eccentrics, 192 Efficiency, spur gears, 229 Equivalents, decimal, 1 Erase, how to, 44 Erasing, 43 ‘« ink, 43 ‘« from tracing, 44 ‘« shield, 8 Extension lines, ink, 44 Eye bolts, 89, 90 F Face of spur gear, 231 Factors of safety, 20 «Table 20, 20 Fastenings, 45 Feathers on gear arms, 224 Figures, 27 Finish marks, 33 | Fillister hd. cap screw, 80 Flanged unions, 62 Flanges, pipe, Table 11, 15 Flat hd. cap screws, 80 Flexible coupling, 109 Flow in pipes, 62 Follower plate, 151 Footstep bearing, 119 Formulae, couplings, 109 ue bevel gears, 251 Foundation bolts, 89 Fractions, how made, 29 ne height of, 29 French hd. cap screw, 79 Friction of bearings, 125 G Gas engine pistons, 156 Gasoline eng. pistons, 157 ‘« to clean trac., 43 Gearing, 215 Gears, arms of, 226 «¢ arm calcul., 226 ‘bevel, 244 ‘¢ “spur, 215, 239 “« teeth formulae, 221 «¢ worm, 258 Gib and key, 102 Gib head key, 99 Globe valves, 283 Grip of bolts, 88 Guide bars, pressure on, 163 Grant’s Odontograph, 220 H Half size scale, 23 Hanger, 127 Height of figures, 29 «« fractions, 29 «letters, 36 Helix, 67 Hexagon construction, 82 Hex. hd. bolt, 83 sc «e ¢* Gap screw, 81 “nut, 82 Hob for worm wheel, 263 Holes, centers of, 31 ‘« dimensions of, 31 ‘¢ tapped, 30 Horse-power belts, 202 fe gear, 236 Hub, bevel gear, 249 “« pulley, 205 ‘« ‘spur gear, 249 INDEX 317 I I-Beams, 50 Information, useful, 1 Ink, 42 Inking arcs, 42 ‘« order of, 42 Instruments, drawing, iv International St. Thd., 70 Involute gear teeth, 217 J Journals, 119 ae area, 119 ‘« pressures, 19 K Keys, 97 ‘* “sunk, 99 ‘* Woodruff, 101 Keyway, 99 ee cutting, 100 ‘« gib head, 99 L Lag screws, 82 Layout of drawings, 24 Lead of worm, 258 Length, bolts and screws, 88 Lettering, 34, 36 ae pens, 36 ee titles, 33 Lewis bolt, 89 ae formulae, 231 Lifting eye bolts, 89, 90 Lines, 42 ‘« angle of sec., 25 ‘« border, 33 ‘* dimension, 25 ‘« dimen. ink, 42 ‘« extension, 42 guide, 38 ‘« how to draw ink, 43 ‘« section, 26, 40 Locknut, 91 e pipe, 56 Logarithm Tables, 3 “ec M Machine screws, 79 Making drawings, 21 «¢ letters, 36 Marks, finish, 33 318 Material bill, 84 of strength, Table 19, 20 Metric Table, 2 Mfgs. Stand. Bolt hd., 79 Mod Sq. thread, 72 Multiple threads, 75 N Name of draughtsman, 33 Nipples, pipe, 56 Notes, abbreviations, 32 ‘“« on drawings, 31 Numerals, 29 es height of, 29 Nut, locking, 91 Nuts hex. and sq., 87 ‘¢ shading of, 40 U.S. St., 85 oO Odontograph Table, 220 Oil-cups, 44 Order of making drawing, 21 Overall dimensions, 29, 44 P Paper, drawing, 23 Pencils, 25 ‘¢ points, 25 Penn nut, 91 Pens, 43 Pillow block, 123 Pin key, 101 Pinions, 223 Pins, crosshead, 164 Pipe caps, 56 “* couplings, 55 ‘« fittings, C. I., 56 es “« scale, 59 a ‘« Table 8, 13 ‘« Table 9, 14 “* flanges, 55 a ‘« Table, 11, 15 ‘« for water and steam, 54 ‘¢ friction, 63, 64 ‘« nipples, 56 “plugs, 56 ‘* tables, 55, 61 ‘thread, 56 ‘* unions, 60 ‘« W. I. for water or gas, 14 INDEX Pistons, C. I., 153 ‘¢ east steel, 153 ‘c rings, 150 “rods, 153 we ‘« calculations, 154 Pitch, calculations, 221 «« circle, 222 “¢ eircular, 215 ‘« diametral, 217 ‘of gears, 217 ‘of rivets, 51 ‘« of thread, 68 ‘¢ of worm, 258 Plates, see end of Index Plugs pipe fittings, 56 Plunger, 150 Post hangers, 130 Powder, talcum, 42 Preparing tracing, 41 Printing on drawings, 34 ae pencils, 38 ae pens, 43 “titles, 33 Pull on belts, 203 Pulleys, 199 «¢ arms, 205 «« ealculations, 199 “« cone, 209 ‘« diameters, 204 ‘« face, 204 ‘« hub, 205 “rim, 205 ‘« speeds, 202 R Rag bolt, 81 Radii of ares, 30 Ramsbottom rings, 150 Reading figures, 29 Reduced dimensions, 23 ss scales, 23 Reproducing drawings, 40 Revolutions of worm, 260 Rim, bevel gear, 248 ‘pulley, 205 ‘ ‘spur gear, 224 worm wheel, 261 Ring oiled bearing, 136 Rivet scale, 46 Riveted joints, 46 Rivets, 45 Rolled sections, 50 vt “« Table 4, 12 4c INDEX 319 Roller bearings, 141 Round hd. cap sc., 79 Ss Safety valves, 271 Sandpaper, 8 Scale, changed, 23 ‘* of drawings, 23 Scales, 23 ‘« architects, 23 ‘« fittings, 58 “‘ for bearing boxes, 123 for bearings, 123 ‘* for bolts, 89 ‘* for keys, 100, 101 ‘* for pipes, 58 “* reduced, 23 ‘¢ rivets, 45 Screw threads, 67 Screws, cap, 79 ‘¢ dimensions, 81 «« lag, 82 ‘* machine, 79 ‘eset, 93 ‘« slotted hd., 79 «« sq. hd. cap, 81 «« wood, 83 Section, half, 23 ‘« lining, 36, 40 oy «« lines U. 8. Navy, 26 views, 25 Set screws, 92 Seward’s gear chart, 237 (folder) Shade lines, 39 Shading drawings, 39 ‘« bolts and nuts, 40 «¢ circles, 39 Shaft calculations, 104 ‘« couplings, 106 supports, 143 worm wheel, 261 Slipper crosshead, 165 Slots in screw hds., 79 Soapstone, use of, 43 Speed of gear teeth, 233 Special dimensions, 30 Split cotter pin, Table 12, 16 Spoken dimens., 29 Sponge eraser, 43 Springs, 76 Spring washer, 92 Spur gears, 215 oe 6c 6c Spur gear arms, 222, 229 ce “bub, 224 6c “cc rim, 224 “e *‘teeth, 216 «train, 238 Square heads, 81 ‘« hd. cap screw, 81 “* nuts, 87 ‘* thread, 72 Stages of drawing, 221 «« ‘construction, 25 “finishing, 25 Stop cock, 278 Strap end, conn. rod, 183 “* eccentric, 192 Strength of bolts, 94 i “ gear teeth, 230 de “ material, 20 ns ‘« thread, 71 Stretched tracings, 41 Stud bolt, 83 Stuffing boxes, 112 o ‘« scale, 113 Sunk keys, 99 Swing eye bolts, 89 Synopsis of making drawings, 21, 22 T Tables: ‘« cap screws, 9 cast-iron pipe flanges, 15 comparative areas, 11 a «ce «in elbows, 64 friction of water, 63 ‘« Grant’s Odontograph, 220 ‘« Mfg. Stand. bolts, 9 ‘« ~ oil cups, 16 pipe fittings, 13 rolled beam sections, 12 ‘« Table 13, 16 «¢ taper pins, 101 ‘threads Internat. Syst., 10 as ‘« metric syst., 10 per inch, 9 unions, 61 “U.S. St. bolts, 9 “washers, 92, 93 «« W. I. pipes, 54 Tap bolt, 84 Tapped holes, 79 Tee-head bolts, 88 Teeth outlines, 217 Test questions, end of book 6c “ce “ 4c “ce ce oe 320 Threads, A. L. A. M.,, 71 B. A., 70 oe buttress, 74 ce 11 forms of, 68 a internat. syst., Table 2, 11 fe Internat St., 70 a metric syst.; 10 multiple, ‘75 per inch, 71 as «« «Table 1, 9 ce “ce ce 4c pipe, 71 a “per inch, 13 uf right and left, 75 He single, double, 75 ue sq. and mod. sq., 72 ae U.S. standard, 69 os V, 68 os Whitworth, 69 Through bolts, 85 Titles, composition of, 33, 38 ‘¢ where place, 33 Tools for drawing, list of, 8 Tooth factors, 231 Tracing, 40 ‘* cloth, 41 e ‘« “how placed, 42 ‘« order of making lines, 42 Triangular scale, 23 Trunk piston, 150, 157 U Union, pipe, 62 Universal joint coupling, 109 U. S. Standard bolts, 85 Ae Navy sections, 26 ws ae nuts, 85 a a thread, 70 Vv Valves, 268 «« angle, 283 ‘¢ check, 270 comparative areas at root, INDEX Valves, globe, 283 «« safety, 271 Views on drawing, 21 «« number of, 22 «¢ placing of, 22 V thread, 68 Ww Wall bracket, 143 Washer, spring, 92 Washers, cut, 92 Weight of pulley, 210 «« “* spur gear, 229 ** substances, 20 Whitworth thread, 69 Wing nuts, 89 Woodruff keys, 101 of ‘« Table 15, 18 Wood screws, 83 Working drawing, 21 fe making of, 21 Worm and worm wheel, 258 ‘« and wheel H. P., 261 ‘¢ helix angle, 260 “* length and diam., 259 ‘« velocity, 260 ‘¢ wheel diam., 261 ne ce «* drawing, 262, 263 ee wheel rim, 261 .s strength, 261 a ‘« shaft, 261 teeth, 261 threads, 258 and wheel vel. ratio, 258 ‘« “efficiency, 259 Wrist pin, 164 Writing pens, 43 Wrought i iron pipes, 54 ye Table, 14 “é ce sc ce 4c ce ce PLATES Alphabets, 37 Titles, 35 “« ‘sizes of in U.S. M. 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