Ait eas CORNELL UNIVERSITY LIBRARY BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND GIVEN IN I89I BY HENRY WILLIAMS SAGE ‘ornell University Library eflections and statistically indetermin WORKS OF CLARENCE W. HUDSON PUBLISHED BY JOHN WILEY & SONS Notes on Plate-Girder Design. 8vo, vii+75 pages, figures throughout the text and 2 folding plates. Cloth, $1.50 net. Deflections and Statically Indeterminate Stresses. Small 4to, xiii +258 pages, profusely illustrated with figures in the text, and full page plates Cloth, $3.50 net. DEFLECTIONS AND —~—J MINATE STRESSES STATICALLY INDET BY CLARENCE W. HUDSON M. Am. Soc. C. E. | Professor of Civil Engineering, Polytechnic Institute of Brooklyn, New York, Consulting Engineer FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS Lonpon: CHAPMAN & HALL, Limitep 1911 Coprricut, 1911, BY CLARENCE W. HUDSON THE SCIENTIFIC PRESS ROBERT DRUMMOND AND COMPANY BROOKLYN, N. Y. PREFACE Previous to the past year very few text-books in English had been published on the subject of statically indeterminate stresses, the books of Martin and Hiroi being the only ones with which the author is familiar. For many years the author has had in mind and has been preparing matter for the publication of a text-book which would give the very few underlying prin- ciples of the subject and show their application to a number of the problems to which they apply. While the principle of the work of deformation has been freely used by the author in many of the general demonstrations and in the solution of some of the special problems, the determination of the statically indeterminate stresses and reactions has generally been made directly from the proportionality between load and deformation as the basis. The deformations which an elastic body undergoes when subjected to load may be determined in several ways if the deformations are so small that the loaded structure remains practically similar to the unstressed structure. For a structure built of an elastic material having a unit elongation, for the unit stress used, of 0.0006, the methods of this book would have a high degree of precision, but for a material just as perfectly elastic having a unit elongation of 0.06 they would have no great value. Several methods of finding the distortions of elastic structures under loading are developed and the importance of the subject, particularly in bridge erection, is shown. The method of finding deflections by means of the work due to an auxiliary load of unity is given much prominence both for solid and open webbed structures, as it appears to be the simplest of all the methods of computing deflections, when the steps in the operation of computing deflections are con- sidered, as they properly should be, a part of computations which must be made in actually building a structure. The method of writing the equations for unknown stresses, forces, and reactions in terms of certain elastic deformations due to unit loads has led to new formulas for finding these quantities for certain structures. Such formulas iii iv PREFACE for unknown crown forces for both the solid webbed and braced arch have been published for the first time by the author. The space devoted to a comparison of the exact and approximate methods of integration in finding the deflections of certain structures it is believed is warranted and should be of value to persons reading this subject for the first time. The author wishes to acknowledge that Art. 68 has been prepared by Henry W. Hopes, Art. 69 has been prepared from material furnished by Paun L. Wo.FeL, part of Art. 70 has been prepared by the Fort Pirr Bripcz Works, and Art. 71 has been prepared by Francis P. WITMER. In addition to the books previously alluded to the author desires to acknowledge his indebtedness to the works of Cain, CHurcu, M. A. Hows, Merriman, Morey, and Swain. Much of the work of computation in connection with the numerical solution of the problems of this book has been done by my assistants, H. P. Kirxuam, Wo. F. Carson, and E. J. Squire. Throughout the body of the book where methods have been used which are known by the author to be the result of the labors of others, credit for such work has there been given. The scope of such a text-book as this may be indefinitely extended by many useful examples, but it is believed that enough examples have been given to show how the solution of problems in deflections and statically indeterminate struc tures may be made in terms of the elastic properties of the materials. CLARENCE W. Hupson. Brooxiyn, N. Y., June 1, 1911. CONTENTS CHAPTER I ELEMENTARY INDETERMINATE FORMS ‘ PAQGH ARTICER? le INTRODUCTORY oan by ears anche eG Den De aa eli nda Rede teas 1 Conditions of static equilibrium. Character of supports necessary to render a structure statically determinate. Arrangement of bars to render a trussed structure statically determinate. ArvicLeE 2, ELmMeNTARY CONCEPTIONS... .....0.0.0000 cc0ee censeeceseeeece 4 Elastic properties of materials. Table of moduli of elasticity for various materials. ARTICLE 3. ComposiITeE MEMBERS..............0-:0 eee eee ee teen tee ee tees 5 A column composed of three materials. A tension member consisting of a riveted and an eye-bar portion. ARTICLE 4. JOINTED INDETERMINATE FORMS.............000000 0c cece eee eeuues 7 Separation of a simple indeterminate form into two determinate forms. Com- putation of the deflections by geometry. Computation of the indeterminate ‘ stresses. ArticLE 5, JOINTED INDETERMINATE FORMS.........0 0.0... 0000 0 ceeee ene eeee 10 Separation of a simple indeterminate form into two determinate forms. Com- putation of deflections by geometry.. Computation of the indeterminate stresses. ARTICLE 6. BEAMS AND TRUSSES......... 0.000000 cece eee ccc e teen eee eens 16 Reactions for certain straight beams. The number and character of supports of a structure, as affecting the distortion of the structure. CHAPTER II DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES WITH SOLID WEBS Articte 7. DEFLECTIONS oF 4 STRaicHT Bram By MEANS oF THE DIFFERENTIAL : Equation oF Irs AXIS.........00.00 0... c cece cece ces nnneeeeees 19 Derivation of the most general equation of the elastic line. vi ARTICLE 8. ARTICLE 9. ARTICLE 10. ARTICLE 11. ARTICLE 12. ARTICLE 13. ARTICLE 14. ARTICLE 15. ARTICLE 16. ARTICLE 17. CONTENTS DEFLECTIONS OF STRAIGHT BEaMs BY MEANS OF THE WorK DUE TO AN AuxtLiaRy LoapD oF UNITY. ........6. fees Derivation of the differential equation for finding the deflection of any point of the axes of a beam due to the longitudinal flexural stresses. DEFLECTION OF STRAIGHT BEaMs DUE TO SHEAR.................. General equation for deflection due to shear. Application of general equation to special cases. DEFLECTION OF A STRAIGHT CANTILEVER BEAM FOR A LOAD AT ANY Point, THE Moment oF INERTIA BEING CONSTANT... .........-. By means of the second differential equation. DEFLECTION OF A STRAIGHT SIMPLE BEAM FOR A LoaD AT ANY POINT, THE Moment oF INERTIA BEING CONSTANT... ..........2-00 000 By means of the second differential equation. Tue THEOREM OF THREE MOMENTS.... 1.0.0.0 ccc cece e eee eee eee Derivation of the three-moment equation. Reactions FOR Two EquaL SPANS BY THE THEOREM OF THREE NOMEN IT Scio an deren nities” meee aginst ein eee eh AT Three-moment equation applied to a beam on three supports to find the reac- tions. REACTIONS FOR THREE SPANS BY THE THEOREM OF THREE MOMENTS Three-moment equation applied to a beam on four supports to find the reac- tions. REACTIONS OF BEAMS ON THREE SuPPORTS BY MEANS OF THE WorK Done sy AN AuxiniaRY Loap oF UNITY.... . ....... Application of same to finding swing bridge reactions. DEFLECTION OF A CANTILEVER BEAM BY MEANS OF THE DIFFERENTIAL Equation oF Its Evastic LINE, WHEN THE MomENT oF INERTIA OF THE BEAM IS NOT CONSTANT............. ....0-eee bake oink are oh Make up of the plate girder used for illustration. The values of the ordinates and tangents at points where the moment of inertia changes in amount. Tue HorizontaL DEFLECTION OF THE Upper Lerr CORNER OF A SIMPUB ABRAM ct esl cogs ea aosite » aculg inn alsin isien shy es Sagkein, p eal eoaw italien Computation of the amount of such deflection for a special case. PAGE 20 22 25 26 29 33 34 36 37 ARTICLE 18. ARTICLE 19. ARTICLE 20. ARTICLE 21. ARTICLE 22. ARTICLE 23. ARTICLE 24, ARTICLE 25. ARTICLE 26. ARTICLE 27. CONTENTS vii PAGE Tue VERTICAL DEFLECTION OF THE CENTER OF A GIRDER WITH VARYING MoMENT oF INERTIA UNDER UNIFORM LoaD........... 42 THE VERTICAL DEFLECTION OF THE CENTER OF A GIRDER WITH Varying Moment or INERTIA UNDER A CONCENTRATED LOAD AT Any POINT IN THE SPAN. ois ccna cea a4 aE Teese Nee ee aes 43 Tse VERTICAL DEFLECTION AT THE LOADED PoINT oF A GIRDER witH VARYING MoMENT oF INERTIA...... steal Scgh it teeth ala Te eet 45 Tue VERTICAL DEFLECTION AT THE END OF A CANTILEVER GIRDER OF VaRYING MomENT oF INERTIA UNDER UNIFORM LoaD......... 46 THE VERTICAL DEFLECTION AT THE END OF A CANTILEVER GIRDER OF VARYING MoMENT oF INERTIA FOR A LOAD AT THE END....... 47 TRUE REACTIONS FOR PLATE-GIRDER SWING BRIDGES.............-. 48 With the effect of shearing stresses neglected. CoMPARISON OF METHODS FOR CoMPUTING SWING-BRIDGE Reactions 49 Reactions for plate girders with variable moment of inertia. Reactions for plate girders with constant moment of inertia. Comparison of results. Moments for the two cases and comparison of results. SHEARING STRESSES IN PRopUCING DEFLECTIONS FOR PLATE GirRDERS 51 Shear on the web. Shear on the flanges. Coefficient for shear distribution. The effect of shear in producing reaction and moment. SprcraL PRoBLEM BY MEANS OF THE FRAENKEL FORMULA........... 56 DETERMINATION OF DEFLECTION BY MEANS oF APPROXIMATE MeruHops OF INTEGRATION... 0... (ccc cece eee eee eee 56 Comparison of the accuracy of exact and approximate method of integration. CHAPTER Il DEFLECTIONS AND STRESSES FOR CURVED STRUCTURES WITH SOLID WEBS ARTICLE 28. DEFLECTIONS OF CURVED BEAMS...........0ccccsecececccuveucuce 59 General formulas for: The horizontal deflection of any point on the axis, the vertical deflection of any point on the axis, and the angular rotation of any plane cross-section. vill ARTICLE 29. ARTICLE 30. ARTICLE 31. ARTICLE 32. ARTICLE 33. ARTICLE 34. ARTICLE 35. ARTICLE 36. CONTENTS Tur Dertections or A Crrcutar Ring uNDER Two EQuaL AND Opposttm FORCES. s..4c¢0 064 cous ensseds ater ieecvans meas neaee The horizontal deflection. The vertical deflection. The angular rotation of a plane cross-section. Tur GeneraL Equations oF CONDITION FOR FINDING THE STRESSES IN A CrrcuLAR Ring UNDER Two Equal AND OpposiITE Forcss. Writing the equations of condition for the unknown forces and moments in terms of unit deflections. Solution of the equations of condition for values of the unknown forces and moments. Tue Drsian oF A PrpE CULVERT. 2.1.0. cece een eee The general equations of condition. Computation of the necessary deflections. Solution of the equations. Special examples. STRESSES IN THE STEEL FRAMING FOR A TUNNEL LINING............ The general equations of condition. Computation of the necessary deflections. Solution of the equations. Special example. CHAPTER IV ARCHES WITH SOLID WEBBED RIBS Tur DETERMINATION OF LIvE- AND D&aAD-LOAD STRESSES IN THE Exastic ARCH WITH A SOLID WEB, AND WITHOUT HINGES Definition of the necessary nomenclature. Equations of condition. Solution of the equations of condition. General formulas for thrust, shear, and moment at the crown. TEMPERATURE STRESSES ............0.000. General formulas for thrust and moment. Values of the thrust and moment for a special case. Location of the line of action of the thrust for the special case. Stresses Dur To RiB SHORTENING. General formulas for thrust and moment. Values of the thrust and moment for a special case. Location of the line of action of the thrust for the special case. Liuitine PosiTions ror THE THRUST, FOR STRESSES OF THE SAME CHARACTER AS THE THRUST ON Any SECTION oF AN ArcH Rib... For the plain concrete or masonry rib of rectangular cross-section. For the steel or iron rib of I-beam cross-section. For the reinforced concrete rib. PAGB 61 65 69 75 83 85 87 CONTENTS 1x PAGB ARTICLE 37. APPLICATION OF THE Previous Mersuop To DETERMINING THE STRESSES IN AN ARCH RING............. 00. cece eee eens 90 The data for the problem. Computation of the necessary deflections. Solution of the general equations for crown thrust and moment. Influence lines and tables for maximum thrust and moment at all points on the arch riig. Computation of the maximum unit stresses. ARTICLE 38. Erection or Masonry oR PLAIN CoNCRETE ARCHES. ........... 104 Erection of the arch rib. Proper order of erecting the various portions of the spandrel walls and floors. ARTICLE 39. Tue Two-Hincep Arco Rip witH Sotip WEB'...... ...........-. 119 General equation for horizontal thrust. Computation of necessary deflections for determining the horizontal thrust. Influence lines and tables for loadings giving maximum stresses. ARTICLE 40. SHEARING STRESSES AT ANY CROSS-SECTION. ........-.0-000.0 0c eee 127 Method of finding the shear at any cross-section of an arch rib. Computa- tion of the shear at a definite section. CHAPTER V DEFLECTIONS OF STRUCTURES WITH EITHER SOLID OR OPEN WEBS ARTICLE 41. Tue First THEOREM OF CASTIGLIANO... ..........0005. tee... 129 Statement of the theorem. Demonstration of the theorem. Method of applying the theorem. ARTICLE 42. Tue SECOND THEOREM OF CASTIGLIANO...........0 000 cee cee eeeee 133 Statement of the theorem. Demonstration of the theorem. Use of the theorem. ARTICLE 43. MaxweELu’s RECIPROCAL THEOREM....... 0.0.0... eee s eee cence 135 Statement of the theorem. Demonstration of the theorem. CHAPTER VI DEFLECTIONS AND STRESSES IN STRUCTURES WITH OPEN WEBS Articte 44, DEFLECTIONS OF OPEN FRAMEWORKS BY THE METHOD oF WoRK..... 137 Derivation of the.general formula for finding the deflection of any point in any desired direction. ARTICLE 45. ARTICLE 46. ARTICLE 47. ARTICLE 48, ARTICLE 49. ARTICLE 50. ARTICLE 51. ARTICLE 52. CONTENTS PAGE Tue Errect or Puay or Pin Hotes AND ERRORS IN PRODUCING DEFLECTIONS. ...0 0. cece eee ee cee nee RAG 2. phon s Ata cnea ie hea ans 141 Arbitrary changes in the length of any member or members in producing deflection. Solution of a special example. GrapuicaL Meruop or FINDING THE DISPLACEMENTS OF THE PANEL Points OF A SYMMETRICAL STRUCTURE, SYMMETRICALLY DEFORMED 143 Selection of a point and member of reference. Development of the method. GrapuicaL Meruop or FINDING THE DISPLACEMENTS OF THE PANEL Points oF STRUCTURES WHICH ARE UNSYMMETRICALLY DEFORMED 146 Correction of graphically determined deflections for rotation of the reference member. CoMPARISON OF THE ANALYTICAL AND GRAPHICAL METHODS OF DETERMINING THE DISPLACEMENTS OF THE PANEL POINTS OF A SmAci CANTILEVERS. “ _ 41x60 2 80 3 X25,000,000 75,000,000’ Ps os nm 1 x60 = 60 3X 13,000,000 39,000,000’ nde i 1x60 60 ~ 3 X8,000.000 24,000,000" 6 ELEMENTARY INDETERMINATE FORMS The shortening of the different materials under their actual stresses are For P, =. 6081 -__1,56081 oF £1 99,000,000 2,340,000,000’ P, = 6082 _ __1,87282 275 000,000 2,340,000,000’ p, -608:__ ___3,6008s 339,000,000 — 2,340,000,000’ 60S. 5,850S4 P, ~ 24,000,000 2,340,000,000° These from the nature of the problem are equal, and also the sum of the stresses equals the load, Si +S2+Ss +84 = 100,000, or 156081 , 156051 , 156051 _ 199 ogo, Sit+Fg75 +3600 +5850 or Si +.83381 + .433S8, + .267S1 = 100,000, _ 100,000 S1= 9.533 = 39,500, and solving similarly rer S2= 3 040 = 32,900, 100,000 _ Ss =F ogg = 17,100, 100,000 _ 8s=—F 59 = 10,500. A member in a bridge which during the passage of the live load will be subject to alternate tensile and compressive stress is designed for the tensile stress of 2,000,000 at a unit stress of 20,000 lbs. net section. If the member as designed consists of a riveted part of 60 sq.ins. gross, giving 50 sq.ins. net section and several eye-bars of 50 sq.ins. gross and net, what is the actual tension on the net section of the riveted member if one-sixth of the length of the riveted member is taken out for rivet holes? The pins at the end may be taken as rigid. Let Si be the total stress carried by the built part of the member, and S2 be the total stress carried by the eye-bars. A stress of unity will elongate the built part an amount Bs Pie N= a OU Ex6x50° EX6x60 18002’ JOINTED INDETERMINATE FORMS 7 and will elongate the eye-bar part an amount _ 4 =a The actual changes in length of the built and eye-bar parts of the member es 318.1 one Sol 18002 50H be equal. Also respectively, and these from the nature of the problem must : +Sz2 =2,000,000. Substituting for S2 its value= ahg 1, we have: 36 Si +2558; = 2,000,000, and Si =1,074,600. 1,074,600 = 21,500 per sq.in. 50 The maximum unit stress on this section is probably considerably greater than 21,500 lbs., as it is believed that built up sections of a given net area do not have as great strength as a bar of uniform section of the given area, and that built tension members do not have the same strength per unit of net area The tension in the net section = as eye-bars. PROBLEMS No. 8a. A column of concentric annular portions of concrete of 15 sq.ins., aluminum 10 sq.ins., phosphor bronze 6 sq.ins., wrought iron 3 sq.ins., and high carbon steel 3 sq.ins., is stressed by a total load of 200,000 Ibs. What part of the load is carried by each material? No. 3b. A bridge member consists of six eye-bars 8 ins, X1 in.=48 sq.ins. gross and net, and a built-up portion of 36 sq.ins. gross, giving 28 sq.ins. net section. The gross section is 7/9 and the net section 2/9 of the length for the built-up part. Under a stress in the member of 1,500,000 Ibs. what is the unit stress at the net section of the built-up part? ART. 4. JOINTED INDETERMINATE FORMS Let the members 71, T2, T3, and 7's of Fig. 4a, be assumed to be in a vertical plane and to be connected to points, in a horizontal plane, about which they may freely rotate, and that they connect to a similar point and support a load P at their bottom ends. This problem is very similar to those of Articles 2 and 3, though at first glance it may not seem so. Let 71, T2, T3, and 74 be plain rods of structural steel. It is perfectly clear that no matter what the actual stresses are, owing to the symmetry of the figure, the stress in 7; =stress in 7, and in T2=T3. 8 ELEMENTARY INDETERMINATE FORMS Experience in connection with tests of steel bars and observation of steel structures under passage of loads teaches beyond all question that the deforma- tion of the structure of Fig. 4a will be very slight, provided P be such as to produce stresses less than the elastic limit of the material. The elongation of a bar 30 ft. long when stressed up to the elastic limit will not exceed 3 in. The drop of the load P for any rationally designed structure similar to Fig. 4a will not exceed the maximum stretch of a bar 30 ft. long. Let the members of Fig. 4a be so divided as to form two structures as shown in b and ¢ of Fig. 4, each of which is statically determined. Let the portion of P carried by b be denoted by Pi and the portion carried by c be denoted by Po. 49's} —10—sfe —410 9} | Gull oe =) a Sec. 6 = 1.01379 yt Lo nm oO ~ = Sy Sec. &, =1.11804-¥2 L Fia. 4. Let di be the vertical deflection of O under a vertical load of 10,000 Ibs. at O for b of Fig. 4 and let dz be the same deflection under the same load for O of c. Then the total deflection of O of b is Pid; and of O of cis Peds. If we consider 6 and c brought together again, O of both figures must be the same point, P, Pi and P2 are tobe taken in units of 10,000 lbs., as d1 « Pid,=Pod2 and dz are computed for 10,000 lIbs., in order that and as P:+P2=P | _ these deflections may be stated in a small number of decimal places. All the data to solve the problem when the numerical values of di and de are known is given. These values will be found. The stress in 7; and 7's under a vertical load of 10,000 at O of b of Fig. 4 is 5590, and their elongation _ 5590 X33.5410 = 30,000,0001- = 0.00625 {t. JOINTED INDETERMINATE FORMS 9 The lengths of the two sides are now 33.5410 +.00625 =33.54725 each. As the sides are equal the new position of O is vertically below its former position as is shown in Fig. 4d, and the desired deflection d, =H —30, and H =‘ (83.54725)?— (15)? =30.00696, ae d:=0.0070 ft. The deflection dz of O of c is found in the same manner to be .0026 ft., whence P, =.271P, P2=.729P, T, and Tz=+.154P and Te. and T3=+.368P. For P =50,000 T, and T4=+7700 and T2 and T3=+18400. Fie. 4d. The deflections di and dz may always be computed, as has just been done, but the method is far too laborious. The increment or decrement to y due to a corresponding increment or decrement to x may be readily obtained by the use of the calculus. The general relation between x and y of Fig. 4e is y? =x? —H? =x? —225, dy =2dz. ? y That is, to obtain the deflection, multiply the change in length of x by the sec. $1. dy =d, =.00625 x1.118 =.0070, as was obtained by the method of geometry. When the deflected figure does not differ greatly from the original figure the ratio of ; is almost exactly the same in both cases. PROBLEM No. 4a. If a vertical member T’;, consisting of a 2in. X2in. square bar, be added to Fig. 4a, and the load P=100,000 lbs., compute the stresses in T,, T,, T;, T,, and T;. 10 ELEMENTARY INDETERMINATE FORMS ART. 5. JOINTED INDETERMINATE FORMS The determination of the stresses of the structure of Fig. 4a was very much simplified by making 7;=74 and T2=T3 in cross-section. Let 71, T2, 73, and T,sbe 1X}4in., 1X1in., 1}X1 in, and 2X1 in. in cross-section, respectively. Let it be required to find the stresses in the various members. The portions of P carried by each of the two statically determined figures, a and b of Fig. 5, into which the main figure is divided, is now very different from that of Fig, 4a. It is clear that any vertical load P acting on a will elongate 7, four times as much as it will 74 and that therefore the point O will take some new position such as is shown by the intersection of the dotted lines. Fie. 5. In the same manner P will cause the point O of b to take some position shown in the same manner. It can readily be shown by geometrical compu-- tation for different loadings that for loads producing stresses less than the elastic limit the horizontal and vertical motions of the points O are proportional to the loads that produce them. Let P1 (see Fig. 5) be the portion of P carried by a when a and } act together; Pz (see Fig. 5) be the portion of P carried by b when a and 6 act together. ; 4, be the vertical deflection of O produced by P when P is carried entirely by a; 43 be the horizontal deflection of O produced by P when P is carried entirely by a; d,; be the vertical deflection of O of a due to a vertical load of unity at O; JOINTED INDETERMINATE FORMS 11 dz be the horizontal deflection of O of a due to a vertical load of unity at O; dio be the vertical deflection of O of a due to a horizontal load of unity at O; dso be the horizontal deflection of O of a due to a horizontal load of unity at O; dz be the vertical deflection of O of b due to a vertical load of unity at O; ds be the horizontal deflection of O of b due to a vertical load of unity at O; ‘ dzo be the vertical deflection of O of b due to a horizontal load of unity at O; dao be the horizontal deflection of O of b due to a horizontal load of unity at O; now d3 =dio0 and d4=dz20, which does not as yet appear. Since the ratio of the horizontal to the vertical motion of O of a due toa vertical load is not in general equal to the same ratio for O of }, there will be a horizontal component of the force produced by connecting O of a to O of 6b, when under ver- tical loading. It is clear that the final stresses in a when O of a and 6 are connected are those due to P when a acts alone modified by P2 and H, the forces generated at O of a by joining O of a to es O of b, and the stresses in b are those due to ; P, and H. . ne YEH That is, to join Oa to Ob of Fig. 5c, when i Oa is supporting P, will develop a force the hori- yp zontal component of which may be represented Fre. Be. by H and the vertical component by P2. This problem may be solved in many ways and simpler equations than the following may be written, but it is desirable to make as general a solution as possible. Three equations will now be written from which Pi, Pz, and H will be found. Deflections of point O when a is supposed to carry P. The downward deflection of Oa = 4, —P2di —Hdio, The downward deflection of 0b = P2d2+Hdaz0. 12 ELEMENTARY INDETERMINATE FORMS These must be equal when the structure acts as a whole. Po(ditde)+H(diotds)=41 . - - +--+ + QQ) The deflection to the right of Oa = 43 —P2d3 —Hd30, The deflection to the right of Ob =P2ds+Hds0, i: P2(ds3 +ds) +H (dso +da4o) =49 2. «© «© © % ww % (2) and Peter. ck eA Oe eee S er From these three equations all the unknown quantities may be found in terms of easily determined deflections. i 4,(d3 +d) —43(di +d2) ee. . (4) (dio +d20)(d3 +d) — (dso +d40) (di +dz) Po= A1(d30+d4o) — 43(dio+da0) ne ee (5) (di +dz)(d30+d40) — (dz +d4)d10+d20) , Pea Pa Pes ie te ee oe. Sm Owe Bee SD These may be written in simpler form by making di+d2 =a, d3+d4=b, diotd20=4a1, dso t+dao =bi, Then _ 416 — Asa = ~ a,b —ab,’ P _ 41b, — 4301 eS hea In order to save labor it will be well to tabulate the results of the various steps of the computations as they are made. TasLe No. 5a COMPUTATIONS OF STRESSES IN THE MEMBERS OF Fic. 4a WHEN THE CROSS-SECTION OF THE MEMBERS ARE AS SHOWN IN Fia. 5 Stresses Stresses Stresses Stresses Stresses 2% Members. | “Due to Due to Jy X 1000. 1000. | Duet D Final De: Un. Y a P=50000. H.” | Px=asaaag. | Stresses. T, +0.559 | +1.118 |+0.001293|+0.002586) +27,950 0 — 18,630 + 9,320 T, +0.507 | +3.041 |+0.000532]+0.003190] ...... 0 +16,900 | +16,900 T; +0.507 —3.041 |+0.000354| 0.002124; ...... 0 +16,900 +16,900 T, +0.559 —1.118 |+0.000323]—0.000646) +27,950 0 — 18,630 + 9,320 JOINTED INDETERMINATE FORMS 13 U. =vertical load of unity; U,,=horizontal load of unity acting to the right; 4, =elongations for stresses due to Uy; 4, =elongations for stresses due to U;,,. E in computing 4, and 4, is taken at 29,000,000. The first six columns of Table No. 5a may be computed at once. Having thus determined the changes in length of the various members due to horizontal and vertical forces of unity at O, the deflections d: to ds and dio to dao will be determined by geometry. For a sufficient degree of accuracy, to make the determination of the amount of the deflections valuable by this method, the length of the members should be taken to at least five figures for the decimal part, the computation of d2o only will be recorded for illustration. 0. =80.41168 Sipceaiae = 1 30.4138 — Qo OF ote 2 Fie. 5. The sides and angles used in computing dzo are those of the dotted figure of g of Fig. 5: d2o =sin A x AO —30.00000 ft., 2/8(S —AC)(S —CO)(S — at) ACXAO sin A = in which — 35.41435. 4 2V/8(S—AC) (S— CO)(S—A0) 2V/8(S — AOS CO)(S—AO) Sin A.AO= ACXAO AO= log S =1.549179, log S—AC =1.405079, gs 14 ELEMENTARY INDETERMINATE FORMS log S—CO = .698739, 699201 log S—AO =F 359198" This +2 =2.176099. 2 1 Toe (Fe -5) = .698970 log = 1.477129 number =30.00055 ft. =sin A.AO, whence d2o =.00055 ft. a more exact calculation gives d2o =.00054 ft. With the deflections found by geometry and of the amounts stated on d, e, f, and g of Fig. 5, the unknowns P2 and H may be found: 50,000 _ 4, =.00090 XF000- =.045 ft., 50,000 _ 43 =.00108 000 = .054 ft., (d; +dz) = (.00090 +00045) x am = .00000185 ft. =a, (ds +ds) =.00000162 =6, (dio +d20) =.00000162 =ay, (dso +440) =.00001981 =b,. Whence He= .045 X.00000162 — .054 x .00000135 = 729 —729 -0 00000162 x .00000162 —.00001981 x.00000135 .026244 —.267435 and P,= .045 x .00001981 —.054 x .00000162 _ 8914.5 —874.8 200000135 x .00001981 — .00000162 x.00000162 .267435 —.026244 _ 8089.7 _ ‘ ~ 241191 88 8855, and P, =16666.7. The correctness of these values for H, P:, and Pz may be seen by noting that di, =2d2 from (d) and (e) of Fig. 5, d3 =2d4 from (d) and (e) of Fig. 5. The true stresses are now given in the final column of Table No. 5a. The fact that H =0 is a somewhat remarkable coincidence, as the data for the problem were taken at random. fe JOINTED INDETERMINATE FORMS 15 Let us suppose 72 and T3 of Fig. 5 to be interchanged in position; the complete tabulated computation of the stresses is as follows: Tasie No. 5b Stresses Stresses Stress Stresses Stress Members. Due to Due to Ay X 1000. AnX 1000. Due to Due to Due to Final Stress. v. Un. 50000. H=1840. | P2=32600. T, +0.559 +1.118 |+0.001293/+0.002586) +27,950 — 2050 —18,200 + 7,700 T, +0.507 +3.041 |+0.000532/+0.002124) ...... +5580 +16,500 +22,080 T, +0.507 —3.041 |+0.000354/—0.003190} ...... ~ —65580 +16,500 +10,920 Ly +0.559 —1.118 |+0.000323/—0.000646| +27,950 +2050 —18,200 +11,800 The expressions for H and Ps, using values of d2o and d4 with signs opposite to that heretofore, are: 4, (dg —dz) —43(di +de) —486 H= i(d3 —d4 3 = ej ; (Gini) —d) ideo Pa Gs) ees ani 41(d30+d 40) — 43(di0 —d20) 8622.9 P= 1(d30 _ =i89 eee destda) ada iny oe PROBLEMS No. 5a. Find the stresses in the members of Fig. 52 by dividing the main figure into the two static figures 7 and 7 by deriving new general expressions for P, and H, when T,, T,, T;, and T, are Lin. X4in., Lin. X1in., lin. X1}in., and 1 in. X2 in. respectively in cross-section. <—10 104 104 Li UZ Z \ | ! | Te Ts 1 a Ty [ | | | i z I a | | | | 3 ! | 9 | | —— ra 50000 Fig. 5. No. 5b. The same as No. 5a except that 7,, Tz, T;, and T, are 1 in. X}in., 14 in. x1 in., 1 in. X1 in. and 1 in. x2 in. respectively. No. 5c. What are the final horizontal and vertical motions of point O of Problem No. 5a? No. 5d. What are the final horizontal and vertical motions of point O of Problem No. 5b? No. 5e. Compute by geometry 4, and 4; of Fig. 5a when P=25,000 lbs. No. 6f. Compute by geometry 4, and 4, of Fig. 5a when P=50,000 lbs. 16 ELEMENTARY INDETERMINATE FORMS ART. 6. BEAMS AND TRUSSES In a structure which rests on three supports of the nature indicated in Fig. 6a, the conditions of static equilibrium do not furnish enough conditions to enable the reactions Ri, Re, and R3 to be determined. If the interior support be considered slowly lowered until it no longer touches the structure, the structure deflects until the internal stresses in its component parts are sufficient to establish equilibrium and there is given a structure on two supports, or a structure statically determinate with reference to the exterior forces. Now with the center support considered removed, if we can find the deflec- tion of b, with reference to its former position under the action of any loads, P, which will be called 4 and if at the same time we can find the deflection of b under a load of unity at b, which will be called d, then Rez =5 for an elastic structure. le y F A Th TH Th | T T Ry A, Re a Rs Datum Plane Fic. 62. Fig. 6b. In a structure fixed at one end and supported at the other such as is shown in Fig. 6b, to determine the reaction R, and thereby Re, consider the support at R, removed and let 4 be the deflection under the given loading of this point of support, then if d be the deflection of the point of support under a load of unity when R: is removed, Ri =5 for the given loading and the support in place. As has been previously stated, any support to a structure other than those required by the condition of static equilibrium, renders the outer forces indeter- minate. This indeterminateness may properly be considered as due to the inability of the structure to take the natural distortions it would undergo if subject to statically determinate reactions. Suppose the structure of Fig. 6r to be held in equilibrium by two supports capable of exerting only vertical reactions. The distorted structure assumes somewhat the form of the dotted line figure of s. If the structure is supported in the manner indicated, in either t, u, or », the distorted figure will be somewhat like that of the respective dotted line figures of each case. BEAMS AND TRUSSES 17 Each support fixed in position is capable of exerting a reaction which must pass through the point of support, and may always be represented by horizontal and vertical component forces. These component forces for each reaction are a direct function of the horizontal and vertical displacements which would take place under the given loading if the support in question were removed, provided the structure is perfectly elastic and the displacements very small. Further, as the displacements at any point in a structure are in general a function of all the loads, the displacements at each support, when enough redundant conditions have been considered removed to render the structure deter- minate, may be written in terms of the loads and the component forces at the other supports. By i - writing equations for the displacements at each b € point of support in succession as stated, a sufficient F number of equations, taken with the three static h i conditions, may be written to enable all the unknown forces to be found. For structures reteset. which contain more members than are necessary for stability, the stresses in the redundant bars may be found the same as unknown outer forces, —— Zi subject to the limits of the problem. ee eae The method of thus determining the inde- u terminate forces and stresses will be illustrated 7 SSS SS y by many examples in a later part of this book. Sa tn The exactness of the solution of statically 7 indeterminate structures depends on the strict- ness of the relation between loads and their deformations and nothing else. It may be stated that the design of a structure statically indeterminate either with reference to the outer or inner forces may be made, provided expres- sions for the redundant forces may be written in terms of the elastic properties of the materials. As the solution: of problems in connection with statically indeterminate structures depends on the determination of the elastic deformations of the struc- ture, taken under such conditions as are imposed by the nature of the problem, a study of the best means of determining elastic deformations is absolutely essen- tial. For this and the further reason that a knowledge of the elastic deflections of structures is essential for the proper handling of much engineering construction, considerable space will be devoted to this subject in the following pages. For- Fia. 6. 18 ELEMENTARY INDETERMINATE FORMS tunately many works on Applied Mechanics and Structures contain enough on the subject to make additional study easy. PROBLEM No. 6a. Consult any text-book on Applied Mechanics or a later part of this book, and from the equation of the axis of a beam determine the center deflection 4 of the beam due to a load P at a distance kl from the left end (k being a fraction of J). Then find the center deflection d due to a load of unity at the center. From the values of 4 and d thus found, derive a general formula for the center reaction for a continuous beam on level supports with two equal spans and loaded at any point. CHAPTER II DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES WITH SOLID WEBS ART. 7. DEFLECTIONS OF A STRAIGHT BEAM BY MEANS OF THE DIFFERENTIAL EQUATION OF ITS AXIS THE curve formed by the axis of a bar when subject to flexure is called the elastic line. There are several general differential equations of this axis. The equation of the elastic line, which perhaps has the most general applicability to the problems of flexure, is EJ a = M, in which M is the moment of the stress couple on any cross-section expressed as a function of 2, the effect of the shearing stresses in pro- ducing distortion being neglected. Let the dotted portion of Fig. 7a repre- sent a part of any beam subject to flexure of which the full line MN is the neutral axis, and let AB and CD be any two plane sec- tions distant ds before bending. Let the relative motion of these two planes, which takes place under flexure, be considered with reference to the section AB. Then ds =r.d¢, Fig. 7a. and from experience ds =dz with a high degree of accuracy, for all the materials used in flexure in engineering practice. Whence at d d and further as-a(%), for such values of ¢ as will be met in engineering practice. pa at __ de _ de? a(34) dy dy dx dx 19 20 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES From Fig. 7a, the angles mon and DnD’ being equal, we have om:mn::nD: DD’, or r:dv::c:e, in which c¢ is the distance from the neutral axis of the piece to the fiber whose elongation is e. e for this case, where S is the unit stress of the fiber, SL dx —E SE whence pe gil Se — e@. Sdx oS’ S M 2 aa 2 BIeY —M. “wv? dy mM’ dx? For any special beam, M in the preceding differential equation is stated in terms of x measured from a selected origin and the expression integrated twice, the constants of integration being determined for the limits of the special case. ART. 8. DEFLECTIONS OF STRAIGHT BEAMS BY MEANS OF THE WORK DUE TO AN AUXILIARY LOAD OF UNITY! The differential equation of the elastic line given in Art. 7, while the most general in its applicability to the problem of flexure for straight beams, is on account of its form not the one best suited to Oe da Guslereatany aes . certain of these problems. In many cases ooo Sse only the deflection at a certain point of a si 1 structure is desired without the necessity iia: Ge of knowing the deflection at all points. The general expression for the deflection at any definite point of a straight beam of length 1 subject to flexure will now be derived. Let 4=the deflection of any point in the desired direction for a given loading. Let kl=distance from some known point to the point whose deflection is required, & being a fraction and less than unity. Let unity be an auxiliary force acting at the point and in the direction for which 4 is desired. Let M be the bending moment at any point on the neutral axis produced by the given loading. ‘The formula of this article was first derived by Professor Fraenkel. BY MEANS OF WORK DUE TO AN AUXILIARY LOAD OF UNITY 21 Let m be the bending moment due to the auxiliary load of unity at the point where M is taken. Then the stress in any fiber of an area da and distant z from the neutral axis due to M aie z da Pegs and the same due to m MC 2 =F 5 a The elongation of any fiber for a length dz due to M is Me 2 g, a2 _M , ds Ic ‘Eda I" E° The work in this fiber of length dz due to moving the fiber stress for the auxiliary load of unity when gradually applied through the elongation of the fiber due to M M _dxm Mmez?da =—.2.—.7..2.da= and for the entire cross-section of the beam the work _Mmdx - QET”’ and the internal work throughout the beam = ’Mmdz lo 2EI* The external work of the auxiliary force in moving through the deflection due to the given loading 4 =ly4=-, The external and internal work are equal. 4a (Mindz _ "Mmde , ’Mmdx - 0 EI 0 EI kl EI : In the derivation of the previous general formula if M be taken as the bending moment due to a load of unity acting in the same manner as the load ha? of unity which develops the moment m then the value of 4= f ae 0 It may seem clearer to the student to include the deformation due to the auxiliary load of unity in writing the expressions for both the internal and external work due to this auxiliary force. To thus include the effect of this load, 22 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES Let 41: =the deflection of the point of application of the force of unity due to its own action; medz EI’ 4, =the change in length of any fiber due to the force of unity; = as was shown for the given loading. The total change of length of any fiber now _ Mzdzx 4 mada - ET EI and the internal work throughout the beam, ‘Mmde Umeda We (one or ee ee The total deflection desired, now =4+ 41, aa the external work done by the force of unity zdx = (M +) ar Sie a a Ok ew ee ae Making Eq. (a) = Eq. (6), we have ’ Mmdx ‘mda | Beh ae +f EI” aa med ‘Md mdx mdx 4,= |, ET? therefore a= ET PROBLEM No. 8a. Derive the general expression for the deflection in the direction of the line of action of the load, of the loaded point in a beam under one load. ART. 9. DEFLECTION OF STRAIGHT BEAMS DUE TO SHEAR The value of 4 just determined is that due to the longitudinal flexural stresses; this deflection will be in- creased by the effect of the shearing stresses. The deflection 4, due to shear for a rectangular prismatic section will now be determined. Let a and b of Fig. 9 represent a portion of the length, aoe and the cross-section of such a beam. Let S be the shear at any section due to the given loading. Let s be the shear at the same section for the auxiliary unit load. Let Q be the intensity of shear at a distance z from the neutral axis due to S. Let q be the intensity of shear at a distance z from the neutral axis due to s. Let Qs be the amount of shear on an elementary area dist. z from the neutral axis due to S. —— 1—__>| | DEFLECTION OF STRAIGHT BEAMS DUE TO SHEAR 23 Let gs be the amount of shear on an elementary area dist. z from the neutral axis due tos. Then s(t) -5(@-2", Ib\2 2 21 \4 and fe 12 pi geet, Qr (4 zZ Joe The transverse deformation for an element of length dz @ Qrdx s(q-#)_ S (? _» Pode oF Ree ate. ral . ) in which it should be noted that EF is the modulus of elasticity for shear. Also gs= 5 ag - #) xb-dz and if g; is gradually developed, the work done by unity on an element of the cross-section and dz in length =dz. — (¢ — 2) bed For the section chosen, this is readily integrable. For an J section, the cross- section should be divided up into several parts and the average values for S and s for each part used in determining the work on the entire section and dz in length. For the rectangular beam taken, the work on the entire cross-section for a length dx ; ; "2 8s /@ . 188s," 27dtz d2z3_, 28 =a f san (x ~*) t= gas alee +51- or Sonn ( | Now let 5z2 be the vertical deflection of the ‘init load caused by the shearing stress on the elementary length, then if unity be gradually applied the external 18Ss work =" = and this = 2.30 ee or — dy 18 S8 _ a, 6 St 15 Ebd "5 EA’ and for the whole length oe dz © S88 ‘5 EA In the immediately following is given the maximum deflection due to the shearing stresses for four cases of loading for a beam of rectangular cross-section. Case I. Cantilever with concentrated load at end. i. af, mat 6 ('Pldx 5 EA 5Jo EA Px_ 6Pl “5 5 ea|~ bEA’ 24 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES Case II. Cantilever with uniform load of w per unit of length. ql worm, 2) te Bie Led & glLOH A) EA Case III. Simple beam with concentrated load at the center. 1 oP 1. ¢ (Pf , 6 Fee 5 ( =)( 5 ae 6P'T x 6Pl 4,= 0 + j SSS ea a eee = | san a EA 5 E,A °5 EA 20 wl Fig.9c S 1 P ano 1 % «<———¢ >} x | : 7 per lin. tts x x ts x 9 oO foe ! L y | Fig.9e Y Fig.of Case IV. Simple beam with uniform load of w per unit of length. i ; ‘Gonpeep G=)CD ——we )-=-dxz ——wer).( —=)dx oof a ee 2 3° L.A 5) PROBLEM No. 9a. Find the deflection at the center, due to shear, of a 24-in.-80-Ib.-I beam 24 ft. long, loaded at the center with 40,000 Ibs., and compare this to deflection due to the longitudinal flexural stresses. STRAIGHT CANTILEVER BEAM FOR A LOAD AT ANY POINT 25 ART. 10. DEFLECTION OF A STRAIGHT CANTILEVER BEAM FOR A LOAD AT ANY POINT, THE MOMENT OF INERTIA BEING CONSTANT The equation of the elastic line of a cantilever beam under a concentrated load at any point will now be determined. Fie. 10a. Let Fig. 10a represent such a beam. The equation will consist of two parts. The elastic line for the portion kl is straight. For the curved portion to the left of P we have EI*Y _M = —P(|—2—kl = —Pl+Px-+PH, dx 2 and integrating EI = —Plr +2 + Phe +0 but for x = =o, # =0, and therefore C =0, and hence BI = —Ple +52 + Phin. Integrating, again Ely = - PPE pyle +0, but for x =0, y =0, and therefore C =0, and hence y= le 31+3kl), which is the equation of the line for all points to the left of P, and calling 4, the deflection at the load where x =(1—kl) =1(1—k), PP(1—k)? PR(1—k)? GET 6ET 3EI The deflection at any ne = the right of P and dist. kl from P and J from the wall = +(—4;)+(—42) =4=— 2s —k}? —kl tan 0. then A= (1 —k) —31+3k1] = ([—21+2Ky = — 2” 1 as, sa Now tan 022! seesaw. dx 26 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES It has been seen that zr toe +e +Phla, and this gives for when xz =1—kl=l1(1—k) Wy PT qm 4POS®? ap Fr (1405942) dz ET Pd —k)+ +kl2(1—k) | ===(1—k) 5 +k PR, ,,(-1+k)__ PP : =a opr pis -Frl- kp -7a- k)kl = -7- ~k)?[2(1 —k) +3] = -7aR- ~2h-+3k\(1—k)? = -coa- —k)2(2+k) = al = Qk +k?) (24h) = -ZaR- Ale + 2k? +h —2h? +13] = ~ EE (2-3 +48). ART. 11. DEFLECTION OF A STRAIGHT SIMPLE BEAM FOR A LOAD AT ANY POINT, THE MOMENT OF INERTIA BEING CONSTANT Let Fig. 1la show such a beam, the notation being sufficiently explained by the figure. Fig. lla. On the left of the load a BE PU ee ee ee es ee OD dy BI =P(1- iS Baas eh GA ye He Bs TS 3 Ely =P E+Cin + Cs, Soe ay Sete sees 2 C2 =0, for « =0 when y =0. STRAIGHT SIMPLE BEAM FOR A LOAD AT ANY POINT On the right of the load M =P(1—k)x —P(x—kl) =Px—Pkx —Px+Pkl=Pk(l—z), dy _ _ pe) Pkt EIT. =Pk(l—2x) =Pkl—Pkz, . BI = Phin 4.0,, . dx 2 Ely =" ae — Pha AC pebe ys _ Pk , Pk Cya=- 5 + — Cal, from (6) for y=0 when x =1. 3 Ci=—- os, PUREE +O, = Pk? PEE +Cs, from (2) and (5), . POTDRE + Cikt= PRE _ PRE Cok — FE — sl, from (3),(6) and (7). . Dividing (9) by kl, P(1—-k) RE? PRP PR _PP C3 6 +C, = 9 6 + 3 3 a Subtracting (10) from (8), P(L-K)P _ PHP | PP PR | Cs 3 2 3 3 ke? 2P(1 —k)k82 =3PH2 +2PKl? —2PK42 +6Cs, 6C3 =2PK2 —2PHP +2Pk*? —2Ph? —3PK2 = — Pk? —2PKE, and or Cg = —P RP _ PRP a rs PUKE, _ ppop PHP Phel? Ph from (8) by substituting the value of C3 in (11), or _ _ PRP PRP op PP PhP Pk Ci= 3 + 3 +P ge _PP PRP PkP_ PP, ; Bogs og Ok 8h? +18), 27 (4) (5) (6) (7) (8) (9) (10) (11) 28 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES Substituting Ci, in (3), remembering that C2 =0, we have Ely ™ —K)a8 —F (2k — 3h? +h8)Bx (1—h)at —_(2k 32 448)P2, 2... (12) a= RI 6EI Let & be supposed greater than 4, then the maximum deflection will be some- where to the left of the load. To find where the deflection is a maximum for any special value of k, substitute in (12) the value of & and form the first derivative of y with respect to x and place equal to 0 and solve for x. From (12) dy _ eg 2_ PP 24143 ia 0= 2 7(1— k)a (2k —3k? +k3), 6EI and _k(-BQ-B)P _ (2k _ ke Ok he ea oa 3) pam Ws (3 ay Substituting this value of x in (12) and calling the maximum value of y, y1, we have: m= -Zha- (F- =) Substituting in Eq. (6) the values determined for C3 and C's there is given _Pklz? Pha? Phx Phx , PkeP BIg 5 6 6 3 6” ~ as the equation of the elastic line on the right of the load. When xz =< bol yee OBB. ee ee es When Roe aud =o, PR Y= ~A8ET’ . . . ° . . . . . . . . . (15) THE THEOREM OF THREE MOMENTS 29 ART. 12. THE THEOREM OF THREE MOMENTS One of the most useful applications of the equation of the elastic line is in finding the reactions for continuous beams. To show one of the most general applications of this let Fig. 12a show any two adjacent spans of a continuous beam having a constant moment of inertia. The beam is considered to be without longitudinal restraint, that is, that the reactions for any system of vertical loading are vertical, and also that the position of the neutral axis at the supports are maintained always at the same elevation. pean (ga H Seas Sees sSe SSeS ae = ! I 1 I jetty , ! emake . ‘ \ hy y ale at etn thcatstea tae bea cS + ea a ph eee pant a 2-—-——-— > Fie. 12a. Let the heavy line of Fig. 12a, represent the neutral axis of any two adjacent spans of a continuous beam. Let M denote the bending moment at any point, and Me, M3 and Mz be the bending moments at supports 2, 3 and 4 respectively. Let V2,V3 and V4 be the shears just to the right of supports 2, 3 and 4 respect- ively. The remainder of the notation is sufficiently defined by the figure. dy M dx2 ro Er . . . . . . . * ° . . . (1) M =M3+V32 —P3(x% —kls), i Soe x: Ge ED in general for any point to right of P3. When x=l3, M becomes M4. Msz=M3+4+Vsls —Psls3 +Pokls, Va Me™ + Pal 2, be eR ee and also M=M; +Vs3z, in general for any point to the left of Ps. Substituting this valve of M in (1), d?y _M3+V3e dy aa ae or EIT 3 =Ms+Vae. 30 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES Integrating, we have BIE = Myo +3Vo0®+Cs, ee ee eee Ely =4M32?+$V32+Ci¢+Co. Po eC RE (5) Substituting in (1) the value of M in (2) we have d?y ETT =M34+V3x—P3(«—kls), and integrating, BIS! = Mex +4V a0? —4Pa0? +Psklge +Cs, o doe. gf leh. ae. 28 (4’) Ely =3M32? +4V323 - 4P 323 +4P3kl3x2 + C3x +C4, Oy tes d (5’) when x =0 in (5), y =hs, C2 =EThs. When x =/s in (5’) y =ha, C4=EIh4—43Mals? —§V 3l33 +4P 338 —4-P3kls3 — Cols. Substituting in (5) and (5’) the values of Cz and C's respectively, we have, Ely =4M32? +4V323 +Ci¢+Elhsz, oy. Sis Ms? ee, Sev ete a. Ge (51) Ely =4M3(a? —s?) +$V3(23 —1s8) —§P3(a? — 1s?) + Ca(a —Is) ABT AAP chlal?—1e), ss sw a x oC when x =kls, in (4) and (4’), wu is the same in both equations. Moklz+3V 3k7ls? +C1 =M3kl3 +4V 3h7l32 —$P3k2l32 +P3k2ls2+C3z,. . (a) when x =kls in (51) and (51’) y is the same in both equations. 4M 3k?l3? +4V sk8ls? +Cikls +E Ihs =4M3l32(k? —1) +$Vala3(k8 —1) —§Pals3(k3 —1) +Csls(k —1) +4Pskls3(k2 -1)+EIhs. . (a1) From (a) Cy — Cg =P hls? —4Pah?le? =A maa | C1 =C3 +43P3k?ls2, Ri aie WE EA Oe ae (b) C3 =Ci—4P3k?lg2.. 2. a ae ee a (b1) From (ai), Cikls —C3kl3 +C3l3 = —4M3l32— $V 3133 — $Pals3(k3 — 1) +4P3kl33(k? —1) +HIh4,—ETIh3. THE THEOREM OF THREE MOMENTS 31 Substituting in (a1) for C; its value in (6), we have Calls +F54s198 ~Cokls +Cols = —}M la? —4V ala? —4$Pols3(k? —1) +4P3kl33 (k? —1) + EIhs—Elhsa, or Cs ls aaa —3M3 132 —4V 3133 —4$P3k3133 —4P3k3138 +2P3ls38 +4P3kl33(k? = 1) +ET (ha —hs), and El(ha—h hs) Cy = —4Msls —4V als? — $Psls?(4i8 —1) + 4Pakle?(k? —1) +8 Substituting the value of C3 in (61) for C3 in (a1), we have Cykl3 — Cikl3 +4 P3kls3 + Cils —$P3k2l33 = —4M ols? —§V ala? —$Pals2(k8 —1) , +4P3kle3(k2—1) + EIha—Elha, or Cils =—4.Msl3?—4V 3133 —4P3ls3(k3 — 1) +4P3kl33(k?2 — 1) —4P3k?l33(k -—1) + EI (ha—hea), and C= —4Msls— $V als? — tPsls?(k3 —1) +4Pskls?(k? —1) —3P3k*ls?(k —1) 5 ots, es) B Substituting the value of C1 in (4), Ere = Mgx-+4V 922 —4Mals —4V ala? — Pols2(k® —1) + 4Pshls?(k2 —1) ~4P3k?ls2(t —1) jet), 3 Substituting in the above for V3 its value in (8), UE Ig — tPols2(I —1) + SPokle?(? 1) — bPok*ls?(—1) + ET), 3 Et ~My += | Mee a +Po(1—#) | —3Msls— | Ms—Ms + py 19] : 3 The value of dy at support No. 3 is obtained by making x =0. dx dy mat _ Mals P3kl3? _ Pals? _ Pskils? Pls? _ P2kls? _ Pskils? fo ee eg ge aS 2 4 Pakils? i Ey htsaha) 4 Pakils? 2 ls 2" whence dy _ha—hs_2Msls+Mals + Pals*(k? — 3k? +2k) (6) dx ls 6EI rhe which is the value of the tangent to the elastic curve at support No. 3. 32 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES Substituting the value of C3 in (4’) we have ! Ere = M3x +4V 32? —4P32? +Pkl3x = 4Msls = LV 3132 = £P 3132 (4k3 = 1) + }Pokls?(h? 1) +E) 3 and substituting in the above for V3 its value in (3), BI” = Mee +h0? [a Mts 3 dx —4z 2| Ma Ms and when x =13, +P3(1 -1) | Pee Pie aM Ty pts Ds -1)|- pPls?(Ak® —1) +4P hls? (2 —1) + BT"), 3 RY = = Nth + Mae Msls3 4 Pals? _ Pskls? at =e 4+P3 kes? a 4\Mals _ Mals 4Mals dx 2 2 2 6 6 2 3 ].2 — ae — ‘Pst e +$Pols? (Poi a +3P3ks3ls? — 3P3kls? — EI Cate) E is) _Msls , 2Mals | Pakls? _Pskls? (ha —hs) ~ 6 - 6 * 6 6 ace lg? and the value of the tangent to the elastic curve at support No. 4 is when the subscript (3) is added to & to indicate which span it applies to. dy _ha—hs , Mala +2M4l3 + Pals?(k3 —ks®) (6’) a i SET Sos ee ke Diminishing all the subscripts in equation (6’) by unity, we have the tangent at support No. 3 in terms of load P2 on span le. dy _hs—he 4 Mole +2Male + Pals? (ke — ks?) 7 dx le 6EI ie He eae The values of wy in (6) and (7) are equal when the subscript three (3) is added to k in (6) as they are for the same point. _ ha—he 4 Male +2M le + Pole?(ke — ke?) "ds 6ET _ha—hs _2Mels+Mals + Pals? (ks? —3ks? +2ks) 8 Is GET A Tf all the supports are at the same elevation (8) reduces to Mel2+2Mal2+Pol2?(ke —k2’) = —2Mal3 — Mals — P3l3?(ks3 —3k3? +2ks), or Mol2+2Ma(l2 +s) +Mals = — Polo?(k2 —ko?) —Pls?(ks? —3k32+2k). . (9) REACTIONS FOR TWO EQUAL SPANS 33 Eq. (8) is the most general form of the relation between the moments at the three supports for any two adjacent spans of a straight or nearly straight pris- matic beam, all the quantities of which are known but the moments. One. such equation can be written for each support of a continuous beam, hence as many equations as there are unknown moments can be written, and the unknown moments found. With the moments known at the supports, the reactions may readily be found and the beam investigated by the laws of static equilibrium. ART. 13. REACTIONS FOR TWO EQUAL SPANS BY THE THEOREM OF THREE MOMENTS. For two equal spans with a load in the first span only, as indicated in Fig 13a, we have from (9) of Art. 12, when J; =l2 =1. M, Pp ike Me Mg 1 i i Fig. 13a. | 4M2= —P(k—k?)l, also M2=R,l—P(l-kl), from statics 3 Ril—PL+PH+ PH PRI _o, Ry Ry=7 4+ bbb) = PLB) PB), soe oe + + (10) also M.2=Resl. Pkl = Pkl ag Re=2(8-b)=-Tk-B). 6 ee Now Rhi+R2+R3 =P or R2=P—(Ri+Rs), Rs =P F(a +h i iW fi, Ra = Eh —k) =P+E(b-H). aa ss CD PROBLEM No. 13a. Compute the amount of R,, R,, and R; for P=10,000 lps., «=4, and 1, =1,=10 ft. 34 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES ART. 14. REACTIONS FOR A CONTINUOUS BEAM OF THREE SPANS, BY THE THEOREM OF THREE MOMENTS For three spans, the end spans being equal and the center span nl with a load in the first span as indicated in Fig. 14a, n being the ratio of the centre to the end span. eg { *| i Ms Mg | fr, TRe fre i \ I. SSSSSH Ss i--s--—- pe ip ate Sao i--------— >| Fia. 14a. M,=0 and M.=0. From (9) 2M2(l+nl)+Manl=—PP(kK-k), . . . . . . (a) remembering that M,=0, and Monl+2Ma(nl+)=0,. . . 2... 2s. Of) as M,=0. From (f) Mo= sar Substituting this in (a), we have M onl? = 2(nl +1) nl 2(nl +1) 4nl +-41+4n?l —n?l | Me | as Sa nl qe _2(nl+) k-K)P _ _ 2P(nl+1)(k—k*) 8n+3n?2+4 4(n+1)2—n2 ’ Mz = 2mPl+D(k—W8) __ nlP(k-28) ° 2(nl-+) (n+)? —n3} (n+? =n? 2M2(1+nl) — —PP(k —k2), Mp [2-+2n— == Pip), = —Pl(k—k?), REACTIONS FOR A CONTINUOUS BEAM OF THREE SPANS 35 Now having the values of M1, Mz, Ms, and Mz, we can find the values of Ri, Ro, Rs, and R4. _—R1—P(—f)) = —2P wt -) M2=R,l—P(l-kl) = Wale on letting 4(n +1)? —n? =m, Ri, =PA—K— SEP (hb = es Me =Ma+Senl aan —k8)__ 2P(nl + (hI) gn = ee +(R2—P+Ry)nl. nlP (k —k3) _ _ 2P(nl +1) (k —k?) m - m Ran) =P + DEE) _ Pnl(1—k) +Pnl+2t*" py Uk —K2) ae +Ranl—Prl-+Prl(1—h) —2+*" py l(k—13), Re _2P(n +1) (k—K) — Pnm(1 —k) +Pnm +21 +n)Pntk — k?) + Prk —k*) nm _P(2nk—2nk? +2k— 2k + knm—nm+nm + 2nk— 2nk? + 2n?k—- on2ks +nk + nk) nm = P(5nk —5nk3 +2k —2k? +knm +2n?k —2n?k?) nm : Ba = Ph +2 te 42" po a M, -Rg-2Pe-) m ape. 275 a ee ee) R3=P—R,—R2—R.=P—P(1— —h) +242" Pc 78) 2 P(E —#8) =< PR La k8)P, 7.220 eis), nm =o TE ee), nm _ _ 2482+ 51, 73 R3= aA P(k —k?) - (16) 36 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES ART. 15. REACTIONS OF BEAMS ON THREE SUPPORTS BY MEANS OF THE WORK DONE BY AN AUXILIARY LOAD OF UNITY One of the simplest and most useful of the applications of the method of Art. 8 is that of finding the reactions of the drawbridge of two equal spans. The reactions of such a structure are three and are one more than can be found by the laws of statics. If one of the reactions be found by some other method, then the remaining two may be found from the laws of statics. { | Ry | tas i kK—kIl—> ! Fia. 15a. Let Fig. 15a represent a simple beam with the support at the left end con- sidered removed. Let 4=the deflection of the left end due to a load P at a distance kl from the left end; 8 =the deflection of the left end due to a load of unity at that end; R, =the reaction at the left end due to P when the support is in place. oy f Mnaes +HI _f Mae ae mmdx + EI ie mdz [0.0.d2 + [ Pee—kade+ "PU -Hz.2.de 7 af ie. adx Pf (¢-% klx? +P (7- +) = af" 2 3 [F cP OPP ee oP | Then ls 2. 3 33 3. =T14—5k +k'], which is the same value as obtained in Art. 13 by special application of the theorem of three moments. DEFLECTION OF A CANTILEVER BEAM 37 ART. 16. DEFLECTION OF A CANTILEVER BEAM BY MEANS OF THE DIFFER- ENTIAL EQUATION OF ITS ELASTIC LINE, WHEN THE MOMENT OF INERTIA OF THE BEAM IS NOT CONSTANT The application of the differential equation of the elastic line to finding the deflection of a beam subject to flexure, when the beam has a varying cross- section will now be shown. For this purpose let us take a center-bearing plate-girder draw-bridge of two equal arms. This very light, single track, deck structure is shown in Fig. 16a, in sufficient detail for the purpose of illustration. Let it be required to find the deflection at the center of the end bearings. 7246. too,Ls 68" 7 68- 138'0’over all Fie. 16a. Make up of Girder. Overall Length. Effective Length. 1 web pl. 72 xX 138 ft. 136 ft. 2 top angles 5X5 X18.33 138“ 136 “ 1 top pl. 12x 98 * 96 “ 1 “s 12x ay 20 “ ai cc 12 xs 10 “ 8g « 2 bottom angles 5X5 X18.33 138“ 136 “ 1 bottom plate 12x32 98“ 96 “ 1 “ 12 xs 90 O« 20 “ 1 fs 12 x3 10 “* g I, 38,284 ins# 1.85 ft.4 In 58,335“ 2.81 “ Is 79,081“ 3.81 “ I, 100,526“ 4.85 “ 38 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES As the structure is symmetrical about the center line, the deflection of both arms will be the same, and the tangent to the elastic line at the center will be horizontal. We will assume that the girder has a uniform load of w=500 lbs. per ft. of length. Let Fig. 16b show one-half the deflected girder with notation so shown that further definition is unnecessary. L=68’ Fia. 16d. The general equation of the elastic curve is Py _ M dx? EI’ Whence, if L =length of arm and the origin be taken at the center, Py _w(Lb—-x) _ w(L? —2Lax +2?) dx? Q2EI 2EI ; ty. wl Ba Fa =| nom ae when x =0, dy _ dx” p=, and a | ee be v=snrl 2 F +5 +02, when x =0, y =0,, C2=0. DEFLECTION OF A CANTILEVER BEAM 39 By aid of the values for a and y, we will now proceed to find the deflection at the end of the plate-girder draw span. The deflection ys at A, the end of the first cover plate, will be the same as if the girder was of constant moment of inertia, Z4, throughout its entire length L. The deflection of a point on the axis at B, the end of the second cover plate, will equal that due toits position on a beam of length (£—a) with uniform load w and constant moment of inertia Iz, plus the deflection caused by the angle 0 which the beam axis makes with the horizontal at A, and which is equal to way —a), plus the distance ys which the beam has deflected at A. Continuing this operation at the end of each cover plate or at each change in the moment of inertia we may finally obtain the deflection at the end of the girder. The coordinates of a point on the axis for the portion having a moment of inertia I4 are ys and x4, and similarly for the coordinates of other points. w [Sure —4L23 fee) U4 OAR ij 613223? —4L32x33 tu] Fal Ys =u+5¥e| Ts + aes / 6 L222 22 — ales 3) [sus es, ya=Yat sal 7% Tl gies 614221? pst ter | dyes ql, nawetsia| Ti +| on on. Tn 2 oe dix 2k I4 ; Be 2 03° ays_w EE] aye t3 2E Ts dza’ Bip = 2 v2 dys _w ae dys dix2 2H Ts dix3 Substituting in the above equations the values of L, x, and I, we have, 682 x4 —68 es) dys _ | _ 3593.70 dia OB 4.85 2h’ 40 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES 3593.7w _ 9458.3 2° OF dy3_ w dr3 2H ae es 3 ee er 381 + 9458.3w 31,654.30 2E 22 ”’ dy2_ w dx2 2E 58? X38 — a 2.81 T a x68? x4? —4 x68 x48 if] _ 87,990.1w 4 4k 4.85 4b” _87990.1w , _w ae) Ys" ~“S4e '24B 3.81 3593.7w X12 x6 _ 564,777.4w 24H 24H’ 564,777.4w ° x58? x38? —4 x58 X38 +38" a 2.81 + y2= 4 9458.3w X12 X38 _ 11,461,580.0w 246 ~ oan? _ 11,461,580.0w se lace ] Y1 ~~ 948 248 1.85 31,654.3w X12 X20 _ 19,318,071.4w + 24E O4E When w =500 lbs. per ft. as it does in this case, 19,318,071.4 x500 ~ 34x14 X29,000,000 ~ 2264 ft. ART. 17. THE HORIZONTAL DEFLECTION OF THE UPPER LEFT CORNER OF A SIMPLE BEAM The previous article shows how the second differential equation of the elastic line may be used to find the deflection of a beam with a varying moment 'M mdx o EI in deflection is much simpler than that by using the second differential equation E oy 5 = M, as for the former the integration is only once performed and between given limits, thus obviating the long and tedious operations necessary to deter- of inertia. The application of the formula 4 = to the solution of problems HORIZONTAL DEFLECTION OF THE UPPER LEFT CORNER 41 mine the constants of integration when the latter method is used. In this and several of the following articles the application of the formula of Art. 8 to finding deflections of straight beams will be quite fully illustrated. As a first illustration of the general application of this formula, suppose it be required to find the horizontal deflection of the upper corner of a simple I-beam span, loaded at the center with the weight P. Fig. 17a will make the problem more clear. qo a 1 fn Onan SEER S eee ATT > #h | Unity > i b, 2 i <—Zero fea ee ie ‘2 Fig. 17a. Taking each half of the beam separately, and taking the ends of the neutra axis of the beam as the origin of x, we have: 1 2 Mmdz l \Mmdx _ (? Mmdx ‘ 4 -{ er be for the left half + | for the right half, in which, M = +4 xa, for both right and left halves of the beam, m= +2 Xz, for the left half of the beam, and m= —h xe +5, for the right half of the beam. Substituting these values for M and m in the general equation, we have: a ie L a ga (0 Pha'da _ (? Phatde if 2 Phada _ f > Phadx _ PhP Jo «| 4LET o ALET o 4HI 0 4EI 32kT If we make P =29,000, 1=360 ins., and J =1140 (ins.4**), this being the value for the moment of inertia of a 20-in. I-beam (192 lbs. per yd.), then: d 29,000 x 20 x (360)? 1 “=39 29,000,000 X1140 14°” nearly. 42 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES PROBLEM No. 17a. Find the angular rotation of the plane of the left end of the beam of this article due to moment of 200,000 inch-pounds acting in an anticlockwise direction on the left end, when held in equilibrium by an equal and opposite moment at the right end. ART. 18. THE VERTICAL DEFLECTION AT THE CENTER OF A GIRDER WITH VARYING MOMENT OF INERTIA UNDER UNIFORM LOAD In the problems of practice it is usually the vertical deflection of some point in a horizontal beam under vertical loading that is desired. The five following articles are deemed sufficient to show the application of the general formula to finding such deflections. The girders in the following problems have four values for the moment of inertia; the resulting expressions for the deflec- a pounds per linear unit \ tions, however, are of such form that expressions for the value of the deflection for a girder with a greater or less number of values of the moment of inertia can be written by inspection. For girders with inclined flanges, values of the moment of inertia can be determined at stated intervals, and these values considered constant for the intervals taken. Let it be required to find the vertical deflection at the center of a plate- girder span with three cover plates, the girder being loaded uniformly throughout its length, 1, with a vertical load of w pounds per unit of length. Fig. 18a will make the problem more clear. As the moment of inertia of the girder has four values, the second part of at Hees ‘Mmdsx _» 2 Mmdz 0 EI 0 EI the general equation, VERTICAL DEFLECTION AT THE CENTER OF A GIRDER 43 will have four parts: hw m Cee? — a ji ade mh ee 4 © (In? — a8) de 4 F (x? —a8)de 3B, Ty TOR 2E ), ~~ de oF) 8 =o wv (Sa ae ae tS _¢—bt IB-8e) L—16et | Ho 4, 2. 22 ts 2 2 wie | eee 2 eee "38401, GE, foi Ie In is Ef SER fo ih In i ef’ which is the expression for the value of the desired deflection. If I; =I2=I3 =I, that is, if the moment of inertia of the beam be constant, the expression becomes: _ dult ~ 884EL which is the well-known formula for the deflection at the center of a simple beam of constant moment of inertia, under uniform loading. ART. 19. THE VERTICAL DEFLECTION AT THE CENTER OF A GIRDER WITH VARYING MOMENT OF INERTIA UNDER A CONCENTRATED LOAD AT ANY POINT IN THE SPAN Let it be required to find the vertical deflection at the center of a plate- girder span with three cover-plates, due to a vertical load, P, at any point in the span. Fig. 19a. The load P will be placed on the first cover-plate, and the expression found for the value of the deflection. The expression for the deflection for any desired position of P can then be written by inspection. DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES 44 _ ('Mmdx ~ J, EET _(7Pa ak )aPde # P(1—k)a?dx Ls = 2?) da ° Pk(la —x)dx = 2EI, i 2ET> kl 2ET, b 2ET3 0 l fe 2 Pk(lx —#?) de ° Pkatdx < { Pka?da ¢ Pka?dx +f P Pkz?dz, é 2KI4 ) 2H, 2ET, + » 2ET3 2EI, ~PU=D (ot_o° HP) PLY Ie Ut, P_ Bie - 6H (F rte) gel 7. Ei i aa Pk/b? 68 3 k323\ Pk/az Bb Bb & 8 B eon acre Te) tear tr, onan a s *-¢) — Ber +tp-nth-r PB ~6E\I, Is] 12EIg ° 4E\Ig Iz I3 Is] 16EI4’ which is the expression for the value of the desired deflection For a load P on the second cover-plate, by inspection, can be written Iz Is) ' 16EI4° )- PkI8 ~~ Pkl/ce2 c2\. Pk 12E1; | 4E \T3 T) 16EI, Pink of. w : zt Ele tale If we make J; =I2=IJ3 =I4, in either of the above values for 4, we have Pk : 4 = F740"), and if k =4, that is, for P in the center of the girder, _ PB ~ 48ET PROBLEM 19a. Find the vertical deflection at the center for the girder of this article when loaded No. at the center with a single concentrated load P. VERTICAL DEFLECTION AT THE LOADED POINT OF A GIRDER 45 ART. 20. THE VERTICAL DEFLECTION AT THE LOADED POINT OF A GIRDER WITH VARYING MOMENT OF INERTIA Let it be required to find the vertical deflection at the loaded point of a plate-girder span with three cover-plates, due to a vertical load, P, at any point in the span. P will be placed on the first cover-plate. oe ee ease 3 | P Ic i --------- I-kl —-------~- = \ Unity SS ! bee he ee ae: — SS tk Pee ea eS ea ee a Se | —— a Fic. 20a, 1- (oe o EI _(°Pa ak)Parda PO ak)PaPde ee eee 0 Ely la : El. Tel El, Jo ETs 1 7 : 2 PH2(] —7)2 a PIp2y2 b Pp2+2 ¢ PI22 ‘2 Php2p2 af PRL x)Pdx a Pk?2x?dx { Pk?x?dx J Els | ET, El, '), Els Eli Or (ta -F) Pee 6 kl a _& 3E I, Ie To Ee\le2 In I3 I3 214 I ee ae ee Ig Ig Ig Ig 404 In) 3H \Ig Ie Ig Ig SIa Ia E a (F, Bb &@ & &. B ) 3H \T, iis i, te de Sl Ts -PBE(1 2B) PUD (0 _o) PHP’ Do 3H (; Ta +5 )+ 3E (F l)* E (7. ie i) Pui Ba) SP (da Boe 2) iG Gtk ie lh Gon a hoy which is the expression for the value of the desired deflection. 46 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES If we make Tr, =I, =I3 =I4, then 7 PPh)? 3EIT ” and, in addition, if we make k =}, then PRB me 48ET" ART. 21. THE VERTICAL DEFLECTION OF THE END OF A CANTILEVER GIRDER OF VARYING MOMENT OF INERTIA UNDER UNIFORM LOAD Let it be required to find the vertical deflections of the ends of a plate-girder span with three cover-plates, supported at the center only, and loaded with a uniform vertical load, w, per unit of length. w pounds per linear unit Unity ewedz , (°° waidx °wardx 'wardx =, a | sme | sort | omre pe SE Tr, Is Io Tz I3 Is T4 - sors tan i --5) SEI, 8E\I, Io Ie Ig'I3 Ia)’ which is the expression for the value of the desired deflection. Wd y=lg=t =I 4, then, _ wit - SET VERTICAL DEFLECTION AT THE END OF A CANTILEVER GIRDER 47 ART. 22. THE VERTICAL DEFLECTION AT THE END OF A CANTILEVER GIRDER OF VARYING MOMENT OF INERTIA FOR A LOAD AT THE END Let it be required to find the vertical deflection of the ends of a plate-girder span with three cover-plates, supported at the center only, and loaded at each end with a load, P. P . Pp ot ee eee Of Unity o EI _ f* Pxtde 7 Pade , -(¢° Pxedx i t Px2dx , Ele b FIs « Hl, b ET, . af (ef Fe PF -5(F Is eo rou r) eee es 8EI4 38H qy Ts Is T3 T3 Is . which is the expression for the value of the desired deflection. If Ty =I», =I. =I4, then, Pls A=, 3EI By substituting for 4, in the above expression, the distance it is desired to raise the end of a plate-girder draw, and solving for P, the uplift necessary to raise the end the desired amount is obtained. In applying the results of the preceding problems, a deduction of 3 or 4 ft. from the over-all lengths of the cover-plates should be made, for the reason that 48 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES enough rivets to fully develop the value of a cover-plate to resist stress are not generally contained in a less distance than the first 14 or 2 ft. at the end of the cover plate. ART. 23. TRUE REACTIONS FOR PLATE-GIRDER SWING BRIDGES By the aid of the results of Arts. 21 and 22 the reactions for center-bearing plate-girder draw spans, with equal arms under uniform loading, can be obtained. Let R; =R3 be the reactions at the ends of a two-span plate-girder with three cover-plates. From the result of Art. 22, we have, for the vertical motion, 4, due to a force, Ri =R3, at the ends, the following: _ Rib 4-35T, safe a 8 Boe 2 oG deh oe ' If, at the same time, we make the value for 4 given by the result of Art. 21 equal to the value for 4 above, and solve the equation for Ri, we have: wit w/at at bt bt ct et i“ at at bt bt ct ot nth nth) Petey elas Su i ie ie i i Ue fee yo ¢ _¢) SFG a BB eo 8l4 38\T, Ie Ie Is Is I I, Iy Ing Ie I3 Ig Ia which is the expression for the value of the end reactions for a plate-girder of two equal spans having four values for the moment of inertia, and loaded uniformly throughout its length by w pounds per unit of length. If fy =I2 =I3 =I4, then R, =R3 = wil. It would not be a difficult matter to derive values for the reactions for a more general case of loading and span lengths; but the ordinary plate-girder draw has equal arms, and the only place, under the general practice in designing, where it is necessary to consider the effect of continuous loading is near and over the center support. The maximum values for shear and moment at the center support are given under conditions closely approximating continuous uniform load, and this is the problem which has been solved. METHODS FOR COMPUTING SWING BRIDGE REACTIONS ART. 24. COMPARISON OF METHODS FOR COMPUTING SWING-BRIDGE REACTIONS In order to see how much the ordinary method (using a constant moment of inertia) of computing the reactions and the moments over the center supports of plate-girder swing bridges differs from this more exact method, the comparison will be made from two widely different cases. Case I—A very light, single-track, center-bearing, plate-girder draw, 72h inches deep, from out to out of flange-angles. 1 web plate, 72 X# ins. 1388 ft. and 4 flange-angles, 5x5 “ 1388 - (55 Ibs. per yard.) 2 cover-plates, 2xt © OB # “ 2 ce cc 12 a 6c 22 cc ce 2 bc 6é 12% “ce 10 ee “ce I, = 38,248 ins.4 =1.85 ft.* and a =20 ft. Ig= 58,385 “ =2.81 “ b=58 “ Is= 79,081 “ =3.81 “ c= 64 “ I,=100,526 “ =4.85 “-. 1=68 “ a. a3 at K = 4,324.3 a 2.847.0, i 8,648.6, %_ 94349, “= 51,2105, = 4,027,2003, Te : ig 7 Ip c — = 4.2, — = 64,050.83, == 4,403,468.8, r 68,80 ; Ts B It Se = = 4,408,531.1, le 64,831.3, i + =207,394.7 — =108,107.8 + =12,847,854.8 Length over All. Effective Length. i at aot bf bt ct ot 2 syle i bh fats Ts 8 Big @& Bh ¢ a a Lk Gh he o 136 ft. | 360) 96 be 20 cb 3 8 cc | te Fig. 24a 4 i, = 5,694.0, 2 pa b4 7, 72:970,209.0, . =3,459,219.8, — =6,435,122.8 12,847,854.8 —6,435,132.8 -3 w4.59 =24.22w, 207,394.7 —108,107.8 49 50 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES The moment over the center support for this case is: M =68w(24.22 —34.00) = —665.04w ft.-lbs. The moment over the center support for constant moment of inertia is: M = —twlh? = —578w ft.-lbs, which shows that this more exact method gives a moment over the center support for this case 15 per cent greater than the usual method. Case II.—A very heavy, double-track, center-bearing, plate-girder draw, 101: inches deep, from out to out of flange angles. Length over All. Effective Length. _ === 1 web plate, 101 x# ins., 159 ft. and 156 ft. | i 4 side flange-plates, 18x? “ 159 “ “ 156 “ 4 flange angles, 8x6 “ 159 * “ 156 “ (125 lbs. per yard.) 2 cover-plates, a Cee 156 “ 2 ce ce 20 x} cc 84 6c cc 82 ce 2 cc ce 20 x} ce 24 ce ce 22 6c 2 6 6c 20 x2 cc 16 ce ce 14 ce I, =358,750 ins.4 =17.06 ft.4 and a =37 ft. I,=484,500 “ =20.96° “ 6=67 “ Fig. 24b. I3=517,600 “ =24.96 ‘“ ec=71 “ I,=6038,100 “ =29.08 ‘“ J=78 “ * _oooi, “seee7, “= woss7s, {= saainx I, a ) ot) Ts —_ ) ene qT, ) oy Is a ? ey 5 ea % _i4g494, "=120499, = gera0ss, = 807,336.06, Ts T3 IT Is 3 3 ct e —= j — =12,307.8, — =1,018,096.2, —= ; Ts 14,339.3, lk Te Ll 873,854.2, B 4 — =16,318.8, = =1,272,869.9, Pe — Is ees a + =47,976.6 — =26,774.4 + =3,362,231.9 — =1,770,606.9 R, _3, 3,362,231.9 —1,770,606.9 —3 975.07 =28 15. 8” 47,976.6 26,7744 8 SHEARING STRESSES IN PRODUCING DEFLECTION 51 The moment over the center support, for this case, is: M =78w(28.15 —39.00) = —846.30w ft.-lbs. The moment over the center support, for constant moment of inertia, is: M = —1{wl? =760.5w ft.-lbs. Which shows that this more exact method gives a moment over the center support for this case 11 per cent greater than the usual method. The two plate-girder drawbridges selected for investigation may be said to represent fairly the results obtained by the method commonly used in designing such structures. ART. 25. SHEARING STRESSES IN PRODUCING DEFLECTION FOR PLATE GIRDERS In order to determine the effect of the shearing stresses in producing deflection, the following investigation is made: Let there be taken any portion of a beam in equilibrium under the action of transverse loading, as shown in Fig. 25a. If P be the resultant in position, direction and amount of all the forces to the right of the section mn, then it is clear that the portion of the beam under consideration would be kept in equilib- rium by the couple, Fa, and the shear, S. = In general, no matter what may be the actual distribution of the stresses on any section of a beam under a given loading, these stresses may be resolved Fic. 25a. into an equivalent couple and shear. The effect of the couple in doing work has been determined, in Art. 8, in accordance with the common theory of flexure. Going back to Art. 8, and rewriting the expression for the work of the internal stresses, we have for the total internal work tie 'Mmdx he DEEL for the work of the flexural stresses) Ssdx . +(f, EA for the work of the shearing stresses). 52 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES To derive this expression for the work done by the shearing stresses: Let S =shear at any point, due to the given loading; s =shear at the same point, due to a load of unity acting at the point where deflection is desired; A =total area of the cross-section; EL, =coefficient of elasticity for shear; and let dx be an infinitesimal portion of the length of the beam; that is, let it be considered as the distance between two consecutive sections of the beam. Then, under the action of the shearing stress, S, the two consecutive sections will have a relative motion of ——~ Sde if we assume that the shearing stress is distrib- AE,’ uted uniformly over the cross-section. Now, as for the flexural stresses, we will suppose the force of unity to be applied gradually; then the work done by the shearing stress, on an elementary portion of the length of the beam, due to the force of unity, is 5 x5) and, over the entire ; ’ Ssdx length of the beam, the work is { 2EA’ Making the expressions for the external and internal work equal, we have: A !Mmdx di 1 Ssdx a” i ger |}. oR a’ and ae oe +f 'Ssdx 0 B.A’ The shearing stress is not distributed uniformly over the cross- 1. section, and hence the term for the deflection due to the shearing 7. stress should be modified by a coefficient. The value of this coeffi- ae cient for a rectangular section has been shown in Art. 9 to be £. TT ' For a section, as shown in Fig. 25b, which is a close enough des approximation to plate-girder cross-sections, for all practical pur- at poses, the deflection caused by the shearing stresses on two flanges Fe, _U8(h+d)5—15(h+.d)4h +10(h +d)2h? —3h5] i Ssdx 301? oe’ and on the web eovea?(n +5) + 40bdh? (r +5) 4 Shit = t 2 2 ' Ssdx 3022 a SHEARING STRESSES IN PRODUCING DEFLECTION 53 The total deflection for the shearing stresses on the entire cross-section is the sum of these, and = [sc 4-d)5—15bh(h +d) +10bh8(h +d)? —3bh5 +60bea! (r +) d L Ssdx 3 5 = +40bdh (: +5) +8h lf, 302, This expression is of such an involved nature that a simple inspection of the quantity before the integral sign gives very little information as to the relative amount of the deflection due to the shear on the flanges or web to that of the deflec- tion due to the shear on the entire cross-section. For the purpose of getting a definite idea of this relation, and, at the same time, for reinvestigating the plate girders of Case I and Case II of Art. 24, approximate sections for ‘the end and center of each of these girders will be assumed, and the actual value of the quantity before the integral sign divided by 30J?, which will be called C, computed. Case I. Approximate end-section: Approximate center-section: b=12 ins. b =12 ins. d= 1 * f= 2" h=36 “ h=36 “ j= 8 i= -™ Actual coefficient C =.086+..... .........24 037 + SEER Ty TORT tee etc OBTH Case II. Approximate end-section: Approximate center-section: b =20 ins. b =20 ins. qa 2" d= 4 “ h=50 “ h=50 “ i= 2? “ pe Actual coefficient C =.0138+......... 0.00000. 013+ : Mesto dca Ooi ue ha gees ie area of web 75 54 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES From this it is seen that the amount of deflection due to the shear on the flanges is a very small part of the deflection due to shear on the entire cross-section. A similar investigation on a section more closely approximating the U0 actual cross-section of a plate girder, as shown in Fig. 25c, would show a considerably larger amount of deflection due to the flanges; but it would still be very small. Therefore, in investigating the effect of shear in producing distortion on plate girders it will be very nearly correct to fil assume that the web alone resists all the shearing stresses and that the . Fic. 25¢, Shear is uniformly distributed. This assumption will give deflections a little too great, as the flanges of a plate girder do take a small-part of the shear—for a plate girder such as that in Case II the vertical flange plates and the vertical legs of the flange angles take considerable shear. The general equation for the deflection of a plate girder of varying cross-section j= (ee ’ Mmdx dat ' Ssdx 0 op HA’ in which Aw =area of web. Making FE, =3£, the deflection 'Mmdz 5 ('Ssdzx fa | Saee f EI '2), BA,’ now becomes From this the expression for the value of the deflection for the Problem of Art. 21, becomes = sel, CPP -F) sw 8H I, T, To To Tz T3 I4 4 FA,’ and for the Problem of Art. 22 it becomes ogee es Se OH\la ty de’ de da da Inf 2A The expression for the value of the end reactions for a plate girder of two equal spans, the girder having four values of I, and loaded uniformly throughout its length by w pounds per unit of length, is: iz at_at bt bt ot ais pe ae 3w I, Tr, To To T3 Ts I, Aw 8 (7 a a? bf BB =) 151° + aie 7 AG. ih te de i Using this expression in investigating Case I and Case II: For Case I: Re ww”? 412,722 +246,613 3 3 a” 9098749,700 ge ORB 2d 4810; SHEARING STRESSES IN PRODUCING DEFLECTION 55 and the moment over the center support, M =68w(24.48 —34.00) = —647.36w ft.-lbs. This moment is only 2.6 per cent less than that found by the first investigation for Case I. For Case IT: 3 1,591,625+116,813 3 7 and the moment over the center support, M =78w(28.70 —39.00) = —803.4w ft.-lbs. This moment is 5 per cent less than that found by the first investigation for Case IT. The result of this reinvestigation for Case I and Case II shows: For the shorter and lighter draw-span, that the deflection of the end of the span, due to the shearing stresses caused by uniform loading, is about 4 per cent of that due to the flexural stresses from the same cause; that the deflection of the end of the span due to the shear caused by a load at the end is about 3 per cent of that due to the flexure stresses from the same cause; and that the calculated moment over the center sup- port, obtained by a consideration of the flexural stresses alone, is about 2.6 per cent greater than that obtained by taking into account the effect of the shearing stresses. For the longer and heavier draw-span, the consideration of the effect of the shearing stresses shows differences about twice as great, in each instance, as was shown for the shorter and lighter structure. That the effect of the shearing stresses should be greatest in the long girder, does not seem to be in accord with the fact that, for a beam of constant cross-sec- tion, the relative effect of the shearing stresses is less the longer the span; an inspec- tion of all the elements of the problem shows that the much larger moment of inertia and the relatively less area in the web of the long girder are more than sufficient to overcome the effect of its greater length. In view of the results of this investiga- tion it would seem to be necessary to consider the effect of the shearing stresses, in modifying the effect of the flexural stresses in designing, only for very deep girders subject to very heavy loads. The amount of labor required for a proper considera- tion of the shearing stresses is very small, so that whenever great accuracy is required they may be considered. 56 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES ART. 26. SPECIAL PROBLEM BY MEANS OF THE FRAENKEL FORMULA The example of Art. 16 will now be solved by the general equation Mmdsz 4= EI ’ awit we ee a al "SEI SHU; Ga'Je Is is dul 280, Bo ie fe da te i a ee o_o) wile ejb bee! 1a, i ee oe) Sel dy Gls GY 1,444,688.8w | 123,905.8w _ 697,676.5w , 1,511, 354.4w _ 1,576,876.5w eae ye E | & z _ 805,396 X500 _ — {44 x 29,000,000 ~ 29° * which gives almost exactly the same result as the former method with not more than one-tenth the labor. | ART. 27. DETERMINATION OF DEFLECTIONS BY MEANS OF APPROXIMATE METHODS OF INTEGRATION For many structures subject to flexure the elastic line is a curve or series of curves which may not be represented by any known equation. In such cases l i Mee cannot be determined except by an approximation. For such cases it is 0 10 000 lbs. | -g-+-2- 4-3} 4-1 sufficiently exact to divide the structure in a certain number of small parts and take M and m as constant for these parts, thus finding the deflection due to the parts, which may be called partial deflections, and by adding these partial detlegrions thus L obtain {4 ae. MEANS OF APPROXIMATE METHODS OF INTEGRATION 57 l To show the degree of accuracy of the method take a case where f “nee 0 can readily be integrated by the calculus and compare the result with the approxi- mate method. To illustrate, take a 12-inch steel I—31.5 lbs. projecting 10 ft. from a wall and loaded with 10,000 lbs. at the end as illustrated in Fig. 27a. Suppose the beam to be divided into 10 equal parts each 1 ft. long. Then ae for each part is as follows: Mmdzx___60,000x 6x12, for No. 1, er FT ; se 2 Mmdz __ 180,000 x18 x12, * EI EI : a 3 Mmdzx __ 300,000 x30 x12, * EI EI a & 4 Mmdzx __ 420,000 x42 x12, * EI EI a i 5 Mmdz __ 540,000 x54 x12, 7 EI EI 3 a 6 Mmdz__ 660,000 x66 x12, EF EI : ne 7 Mmdz __780,000 x78 x12, > EI EI , # 8 Mmdzx__ 900,000 x90 x12, t E EI : i 9 Mmdz _ 1,020,000 x102 x12, . ? “ET EI ; tt 10 Mmdz _ 1,140,000 x114 x12 - “ET EI : For a case where I varies it should be used as a constant over the length of each portion. For this case J is constant, so we may write: Mmdx _12 EI £1 + 180,000x 18 = + 3,240,000 + 300,000x 30 = + 9,000,000 + 420,000x 42 = + 17,640,000 (60,000 6 =(+ 360,000 58 DEFLECTIONS AND REACTIONS OF STRAIGHT STRUCTURES + 540,000 54 = + 29,160,000 + 660,000 66 = + 43,560,000 + 780,000 78 = + 60,840,000 + 900,000 90 = + 81,000,000 +1,020,000 102 = +104,040,000 41,140,000 X 114) = +129,960,000) = 12 = 478,800,000 x72 ___ 5,745,600,000 ger ' Mmdx The value of —_—" for this well-known case is 0 EI - PB _ 10,000 X120 x 120X120 _ 5,760,000,000 3EI 3 EI ' which shows that the approximate method has a high degree of accuracy where the divisions are taken small. PROBLEM No, 27a. Find the deflection of the beam of Fig. 27a by the approximate method and taking each division 24 ins. long. CHAPTER III DEFLECTIONS AND STRESSES FOR CURVED STRUCTURES WITH SOLID WEBS ART 28. DEFLECTIONS OF CURVED BEAMS TuHE determination of the deflections of a curved bar subject to flexure cannot be made with the same degree of precision as for straight beams. The lengths of the different fibers for any small portion of the beam are not of the same length as the corresponding portion of the axis. Experience and experiment on the deflection of such bars subject to flexure shows, however, that very accurate theoretical determinations of the deflections may be made by considering all the fibers of the beam to be of the same length =" \ OT a Fig. 28a. and equal to a corresponding length of the axis. Let Fig. 28a represent any curved beam whose deflection is required, subject to a bending moment M at any cross-section. And let m be the bending moment at the same cross-section for which M is taken and due to a load of unity acting in the direction in which the deflection is desired. Then 4= ane just as was shown in Art. 8 for the straight beam. . { The horizontal deflection of e due to any given loading will be t= Mmids EI ’ 59 60 DEFLECTIONS AND STRESSES FOR CURVED STRUCTURES in which M=bending moment due to the given loading on an element of the length of the beam; and m,=bending moment due to a horizontal force of unity applied at e on the same element for which M is taken. The other quantities do not need definition. The vertical deflection of e due to the given loading will be Moats EI ’ in which M is as before defined and m,=the bending moment due to a vertical force of unity at e for the same section for which M is taken. The angular motion of the plane of the end of the beam _ (Mas oe ee 4,= A .in which M is as before. _ For let M=bending moment due to the given loading on any element of the length of the beam ds, which causes a change in length of the upper fibers of this element of the beam of nn’, and of the lower fibers of 00’. Then the angular motion 0, of the plane no due to the moment M over the length ds _ 00’ _nn’ 0 Ten but 00"