i J. Jtanfe I pan, (ttmntll Itotemig ptatg THE GIFT OF ^yyirsjz., k. ^.......^..v^A^wv^.. s\,£.b.TQ.n.(3 z.a-\B\u 6896-2 Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924022800746 THE AMERICAN HOUSE-CARPENTER A TEEATISE ON THE ART OF BUILDING, THE STRENGTH OF MATERIALS. by E. G. HATFIELD, Architect, MEM. AM. INST. OF ARCHITECTS. BBVENTH EDITION, REVISED AND ENLARGED WITH ADDITIONAL ILLUSTRATIONS. NEW YORK: JOHN WILEY & SON, 15 ASTOB PLACE. J872. m\ j) i "Entered according to Act of Congress, in the year 185T, by E. G. HATFIELD, In the Clerk's Office of the District Court of the United States, for the Southern District of >Tew York. PREFACE. This book is intended for carpenters — for masters, journeymen and apprentices. It has long been the com- plaint of this class that architectural books, intended for their instruction, are of a price so high as to be placed beyond their reach. This is owing, in a great measure, to the costliness of the plates with which they are illus- trated: an unnecessary expense, as illustrations upon wood, printed on good paper, answer every useful pur- pose. Wood engravings, too, can be distributed among the letter-press ; an advantage which plates but partially possess, and one of great importance to the reader. Considerations of this kind induced the author to undertake the preparation of this volume. The subject matter has been gleaned from works of the first autho- rity, and subjected to the most careful examination. The explanations have all been written out from the figures themselves, and not taken from any other work ; and the figures have all been drawn expressly for this book. In doing this, the utmost care has been taken to make everything as plain as the nature of the case would admit. The attention of the reader is particularly directed to the following new inventions, viz; an easy method of describing the curves of mouldings through three given I V PKEFACE. points ; a rule to determine the projection of eave cor> nices; a new method of proportioning a cornice to a larger given one ; a way to determine the lengths and bevils of rafters for hip-roofs ; a way to proportion the rise to the tread in stairs ; to determine the true position of butt-joints in hand-rails ; to find the bevils for splayed- work ; a general rule for scrolls, &c. Many problems in geometry, also, have been simplified, and new ones intro- duced. Much labour has been bestowed upon the sec- tion on stairs, in which the subject of Land-railing is presented, in many respects, in a new, and it is hoped, more practical form than in previous treatises on that subject. The author has endeavoured to present a fund of use- ful information to the American house-carpenter that would enable him to excel in his vocation ; how far he has been successful in that object, the book itself must determine. New York, Oct. 15, 1844. FIFTH EDITION". SmcE the first edition of this work was published, I have received numerous testimonials of its excellent practical value, from the very best sources, viz. from the workmen themselves who have used it, and who have profited by it. As a convenient manual for reference in respect to every question relating either to the simpler operations of Carpentry or the more intricate and PREFACE. abstruse problems of Geometry, those who have tiied it assure me that they have been greatly assisted in using it. And, indeed, to the true workman, there is, in the study of the subjects of which this volume treats, a con- tinual source of profitable and pleasurable interest. Gentlemen, in numerous instances, have placed it in the hands of their sons, who have manifested a taste for practical studies ; and have also procured it for the use of the workmen upon their estates, as a guide in their mechanical operations. I was not, then, mistaken in my impressions, that a work of this kind was wanted ; and this evidence of its usefulness rewards me in a measure for the pains taken in its preparation. Nm York, Oct. 1, 1852. SEVENTH EDITION. It is now thirteen years since the first edition of the, American House Carpenter was published. The attempt to furnish the recipients of this book with a fund of useful information in a compact and accessible form, has been so far successful that the sixth edition was exhausted nearly a year ago. At that time it was determined, before issuing another edition, to make a thorough revision of the work. The time occupied in this labour has been unexpectedly prolonged by at least six months, and this has resulted from various causes, but more especially from the absorbing nature of my professional duties. A large portion of the work has been rewritten, VI PEEFACE. about 130 pages of new matter introduced and many new cuts inserted. The most important additions to the work will be found in the section on Framing or Construction. Here will be found, now first published, the results of experi- ments on such building materials as are in common use in this country, and an extended series of rules for the application of this experimental knowledge to the prac- tical purposes of building. Some of the rules are new, while others heretofore in use have been simplified. This section has been much improved, and it is hoped that it will be of service, not only to the house carpenter but also to the architect and civil engineer. In preparing the original work, a desire to state the subjects treated of in terms suited to the comprehension of all classes of workmen, precluded the use of algebra- ical symbols and formulae. In this edition, however, it has been deemed best to introduce them wherever they would contribute to the clearer elucidation of the sub- ject ; but care has been taken to state them in a simple form at first, and so to explain the symbols as they are introduced that those heretofore uninstructed in regard to them, may comprehend what little is here exhibited, and at the same time be induced to pursue the study more fully in works more strictly mathematical. But for those who may not succeed in comprehending the algebraical formulae, it maybe stated that all the practical deductions derived from them are written out in words at length, so as to be fully understood without their assistance. E. G. H. New York, Sept. 1, 185T. TABLE OF CONTENTS. iMTBODUcnoN. — Directions for Drawing 1-14 SECTION I.— PBACTICAL GEOMETEY Definitions .... Problems on Lines and Angles Problems on the Circle . Problems on Polygons Problems on Proportions . Problems on the Conic Sections Demonstrations. — Definitions, Axioms, &c. Demonstrations. — Propositions and Corollaries 15-70 71-80 81-92 93-106 107-110 111-128 130-139 140-167 SECTION IL— AECniTECTUEE. History Styles. — Origin, Definitions, Proportions Grecian Orders. — Doric, Ionic and Corinthian Roman Orders. — Doric, Ionic, Corinthian and Composite Egyptian Style Buildings generally Plans and Elevation for a City Dwelling Principles of Architecture. — Requisites in a Building Principles of Construction. — The Foundations, Column Principles of Construction. — The Wall, Lintel, Arch Principles of Construction. — The "Vault, Dome, Roof 168-181 182-196 197-211 212-215 216, 317 218-233 223, 224 225-239 230-232 233-235 23C-238 Vlll TABLE OF CONTENTS. SECTION III.— MOULDINGS, COBNICES, &0. ABIS Mouldings.— Elements, Examples 239-250 Cornices. — Designs 251 Cornices. — Problems 252-256 SECTION IV.— FEAMING, OE CONSTEUCTION. First Principles. — Laws of Pressure Resistance of Materials. — Strength, Stiffness Resistance to Compression. — "Various kinds . Results of Experiments on American Materials, Tables I., Practical Rules and Examples Resistance to Tension .... Results of Experiments on American Materials, Table TTT. Practical Rules and Examples Resistance to Cross Strains. — Strength, Stiffness Resistance to Deflection. — Stiffness, Formulae Practical Rules and Examples Table IV. — "Weight on Beams, Formulas Praetieal Rules and Examples Table V.— Dimensions of Beams, Formula? Resistance to Rupture. — Strength Results of Experiments on imerican Materials, Table VI Table VII. — Safe "Weight on Beams, Formulae Practical Rules and Examples Table VIH. — Dimensions of Beams, Formulae Practical Rules and Fxamples Systems of Framing, Simplicity of Designs . Floors. — Various, Cross-furring, Reduction of Formulae Practical Rules and Examples Bridging-strips, Girders, Precautions Partitions. — Examples, load on Partitions, 4c. Hoofs. — Stability, Inclination Load. — Roofing, Truss, Ceiling, "Wind, Snow Btrains. — Vertical, Oblique, Horizontal Resistance of the Material in Rafter and Tie-beam Dimensions. — Rafter, Braces, Tie-beam, Iron Rods Practical Rules and Examples .... Table 12. — Weight of Roofs, per Foot IL 257-282 283-286 287-290 291-293 294-305 306 307, 30S 309-316 317-319 320-322 323-326 326 327-329 329 331 331 333 333 834 334 335 336, 337 338-344 345-349 350-353 354, 355 356-358 359-368 363 370-374 375-383 376 TABLE OF CONTENTS. IX Examples of ftoofs AKT& 384-3£fi Problems for Hip-rafter 387-38a Domes. — Examples, Area of Ribs 389-391 Problems in Domes 392-398 Bridges. — Examples , 399-401 Rules for Dimensions . . 401-405 Abutments and Piers . , . 406, 407 Stone Bridges, Centreing Joints in Timberwork . ■ 408-417 418-427 ■*ron Work. — Pins, Nails, Bolts, Straps Iron-Girders. — Cast Girder, Bow-string, Brick Arch Practical Rules and Examples .... 428 429-435 431-435 SECTION V.— DOOES, WINDOWS, to. Doors. — Dimensions, Proportions, Examples Windows. — Form, Size, Arrangement, Problems 436-441 442-448 SECTION VI.— STAIES. Principles, Pitch Board Platform Stairs, Cylinders, Rail, Face Mould Winding Stairs, Falling Mould, Face Mould, Joints Elucidation of Butt Joint , . Quarter-circle Stairs. — Falling Mould, Face Mould Face Mould. — Elucidation Face Moulds. — Applied to Plank, Bevils, kc Face Moulds.^-Another method .... Scrolls, Rule^Falling and Face Moulds, Newel Cap 449-45G 457-463 469-476 477 478-480 481 483-481 485-488 489-498 SECTION VII.— SHADOWS. Shadows on Mouldings, Curves, Inclinations, Ac. Shadows.— Reflected Light .... 499-fiSI 521 APPENDIX. Algebraical Signs Trigonometrical Terms PiOl a 6 X TUiLE OF CONTENTS. PAQl Glossary of Architectural Terms • 7 Tables of Squares, Cubes and Roots .18 Rules for Reduction of Decimals .27 Table of Areas and Circumferences of Circle* 29 Table of Capacity of "Wells, Cisterns, 4c. 83 Table of Areas of Polygons, &c ...... . 34 Table of Weights of Materials ....... . M INTRODUCTION. Art. 1. — A knowledge of the properties and principles of lines can best be acquired by practice. Although the various problems throughout this work may be understood by inspection, yet they will be impressed upon the mind with much greater force, if they are actually performed with pencil and paper by the student. Science is acquired by study — art by practice : he. therefore, who would have any thing more than a theoretical, (which must of necessity be a superficial,) knowledge of Carpentry, will attend to the following directions, provide himself with the articles here specified, and perform all the operations described in the follow- ing pages. Many of the problems may appear, at the first read- ing, somewhat confused and intricate ; but by making one line at a time, according to the explanations, the student will not only succeed in copying the figures correctly, but by ordinary attention will learn the principles upon which they are based, and thus be able to make them available in any unexpected case to which they may apply. 2. — The following articles are necessary for drawing, viz : a drawing-board, paper, drawing-pins or mouth-glue, a sponge, a T-square, a set-square, two straight-edges, or flat rulers, a lead pencil, a piece of india-rubber, a cake of india-ink, a set of draw- ing-instruments, and a scale of equal parts. 3. The size of the drawing-board must be regulated accord- ing to the size of the drawings which are to be made upon it. Yet for ordinary practice, in learning to draw, a board about 15 1 £ AMERICAN HOUSE CARPENTER. by 20 inches, and one inch thick, will be found large enough, and more convenient than a larger one. This board should be well-seasoned, perfectly square at the corners, and without clamps on the ends. A board is better without clamps, because the little service they are supposed to render by preventing the board from warping, is overbalanced by the consideration that the shrinking of the panel leaves the ends of the clamps project- ing beyond the edge of the board, and thus interfering with the proper working of the stock of the T-square. When the stufl is well-seasoned, the warping of the board will be but trifling ; and by exposing the rounding side to the fire, or to the sun, it may be brought back to its proper shape. 4. — For mere line drawings, it is unnecessary to use the test drawing-paper ; and since, where much is used the expense will be considerable, it is desirable for economy to procure paper of as low a price as will be suitable for the purpose. The best paper is made in England and marked " Whatman." This is a hand-made paper. There is also a machine-made paper at about half-price, and the Manilla paper, of various tints of rus- set color, is still less in price. These papers are of the various sizes needed, and are quite sufficient for ordinary drawings. 5. — A drawing-pin is a small brass button, having a steel pin projecting from the under side. By having one of these at each corner, the paper can be fixed to the board ; but this can be done in a much better manner with mouth-glue. The pins will pre- vent the paper from changing its position on the board ; but, more than this, the glue keeps the paper perfectly tight and smooth, thus making it so much the more pleasant to work on. To attach the paper with mouth-glue, lay it with the bottom side up, on the board ; and with a straight-edge and penknife, cut off the rough and uneven edge. With a sponge moderately wet, rub all the surface of the paper, except a strip around the •edge about half an inch wide. As soon as the glistening of the •water disappears, turn the sheet over, and place it upon the INTRODUCTION. board just where you wish it glued. Commence upon one of the longest sides, and proceed thus : lay a flat ruler upon the paper, parallel to the edge, and within a quarter of an inch of it With a knife, or any thing similar, turn up the edge of the papei against the edge of the ruler, and put one end of the cake of mouth-glue between your lips to dampen it. Then holding it upright, rub it against and along the entire edge of the paper that is turned up against the ruler, bearing moderately against the edge of the ruler, which must be held firmly with the left hand. Moisten the glue as often as it becomes dry, until a sufficiency of it is rubbed on the edge of the paper. Take away the ruler, restore the turned-up edge to the level of the board, and lay upon it a strip of pretty stiff paper. By rubbing upon this, not very hard but pretty rapidly, with the thumb nail of the right hand, so as to cause a gentle friction, and heat to be imparted to the glue that is on the edge of the paper, you will make it adhere to the board. The other edges in succession must be treated in the same manner. Some short distances along one or more of the edges, may afterwards be found loose : if so, the glue must again be applied, and the paper rubbed until it adheres. The board must then be laid away in a warm or dry place ; and in a short time, the sur- face of the paper will be drawn out, perfectly tight and smooth, and ready for use. The paper dries best when the board is laid level. When the drawing is finished, lay a straight-edge upon the paper, and cut it from the board, leaving the glued strip still attached. This may afterwards be taken off by wetting it freely with the sponge ; which will soak the glue, and loosen the paper. Do this as soon as the drawing is taken off, in order that the board may be dry when it is wanted for use again. Care must be taken that, in applying the glue, the edge of the paper does not become damper than the rest : if it should, the paper must be laid aside to dry, (to use at another time,) and another sheet be used in its place. 4 AMERICAN HOUSE CARPENTER. Sometimes, especially when the drawing board is new, the paper will not stick very readily ; but by persevering, this diffi- culty may be overcome. In the place of the mouth-glue, a strong solution of gum-arabic may be used, and on some accounts is to be preferred ; for the edges of the paper need not be kept dry, and it adheres more readily. Dissolve the gum in a sufficiency of warm water to make it of the consistency of linseed oil. It must be applied to the paper with a brush, when the edge is turned up against the ruler, as was described for the mouth-glue. If two drawing-boards are used, one may be in use while the other is laid away to dry ; and as they may be cheaply made, it is advisable to have two. The drawing-board having a frame around it, commonly called a panel-board, may afford rather more facility in attaching the paper when this is of the size to suit ; yet it has objections which overbalance that con sideration. 6 — A T-square of mahogany, at once simple in its construc- tion, and affording all necessary service, may be thus made, Let the stock or handle be seven inches long, two and a quarter inches wide, and three-eighths of an inch thick: the blade, twenty inches long, (exclusive of the stock,) two inches wide, and one-eighth of an inch thick. In joining the blade to the stock, a very firm and simple joint may be made by dovetailing it — as shown at Fig. 1. Pig. i. INTRODUCTION. 7. — The set-square is in the form of a right-angled triangle ; and is commonly made of mahogany, one-eighth of an inch in thickness. The size that is most convenient for general use, is six inches and three inches respectively for the sides which con • tain the right angle ; although a particular length for the sides is by no means necessary. Care should be taken to have the square corner exactly true. This, as also the T-square and rulers, should have a hole bored through them, by which to hang them upon a nail when not in use. 8. — One of the rulers may be about twenty inches long, and the other six inches. The pencil ought to be hard enough to retain a fine point, and yet not so hard as to leave ineffaceable marks. It should be used lightly, so that the extra marks that are not needed when the drawing is inked, may be easily rubbed off with the rubber. The best kind of india-ink is that which will easily rub off upon the plate ; and, when the cake is rub- bed against the teeth, will be free from grit. 9. — The drawing-instruments may be purchased of mathe- matical instrument makers at various prices : from one to one hundred dollars a set. In choosing a set, remember that the lowest price articles are not always the cheapest. A set, com- prising a sufficient number of instruments for ordinary use, well made and fitted in a mahogany box, may be purchased of the mathematical instrument-makers in New York for four or five dollars. But for permanent use those which come at ten or twelve dollars will be found to be the best. 10.— The best scale of equal parts for carpenters' use, is one that has one-eighth, three-sixteenths, one-fourth, three-eighths, one-half, five-eighths, three-fourths, and seven-eighths of an inch, and one inch, severally divided into twelfths, instead ot being divided, as they usually are, into tenths. By this, if it be required to proportion a drawing so that every foot of the object represented will upon the paper measure one-fourth of an inch, use that part of the scale which is divided into one-fourths of an 6 AMERICAN I10USE-CARPENTEU. inch taking for every foot one of those divisions, and for every inch one of the subdivisions into twelfths ; and proceed in like manner in proportioning a drawing to any of the other divisions of the scale. An instrument in the form of a semi-circle, called a protractor, and used for laying down and measuring angles, is of much service to surveyors, but not much to carpenters. 11. — In drawing parallel lines, when they are to be paraliel to either side of the board, use the T-square; but when it is required to draw lines parallel to a line which is drawn in a direction oblique to either side of the board, the set-square must be used. Let a b, (Fig. 2,) be a line, parallel to which it is Fig a desired to draw one or more lines. Place any edge, as c d, ot the set-square even with said line ; then place the ruler, g h, against one of the other sides, as c e, and hold it firmly ; slide the set-square along the edge of the ruler as far as it is desired, as at/; and a line drawn by the edge, if, will be parallel to a b. 12.— To draw a line, as k I, {Fig. 3,) perpendicular to another, as a b, set the shortest edge of the set-square at the line, a b ; place the ruler against the longest side, (the hypothenuse of the right-angled triangle ;) hold the ruler firmly, and slide the set- square along until the side, e d touches the point, k ; then the line, I k, drawn by it, will be perpendicular to a b. In like INTRODUCTION. manner, the drawing of other problems may be facilitated, as will be discovered in using the instruments. Fig. a. 13. — In drawing a problem, proceed, with the pencil sharpened to a point, to lay down the several lines until the whole figure is completed ; observing to let the lines cross each other at the several angles, instead of merely meeting. By this, the length of every line will be clearly "denned. With a drop or two of vater, rub one end of the cake of ink upon a plate or saucer, until a sufficiency adheres to it. Be careful to dry the cake of ink ; because if it is left wet, it will crack and crumble in pieces. With an inferior camel's-hair pencil, add a little water to the ink that was rubbed on the plate, and mix it well. It should be diluted sufficiently to flow freely from the pen, and yet be thick enough to make a black line. With the hair pencil, place a little of the ink between the nibs of the drawing-pen, and screw the nibs together until the pen makes a fine line. Beginning with the curved lines, proceed to ink all the lines of the figure ; being careful now to make every line of its requisite length. II they are a trifle too short or too long, the drawing will have a ragged appearance ; and this is opposed to that neatness and accuracy which is indispensable to a good drawing. When the ink is dry, efface the pencil-marks with the india-rubber. If 8 AMERICAN HOUSE-CARPENTER. the pencil is used lightly, they will all rub off, leaving those lines only that were inked. 14. — In problems, all auxiliary lines are drawn light ; while the lines given and those sought, in order to be distinguished at a glance, are made much heavier. The heavy lines are made so, by passing over them a second time, having the nibs of the pen separated far enough to make the lines as heavy as desired. If the heavy lines are made before the drawing is cleaned with the rubber, they will not appear so black and neat; because the india-rubber takes away part of the ink. If the drawing is a ground-plan or elevation of a house, the shade-lines, as they are termed, should not be put in until the drawing is shaded ; as there is danger of the heavy lines spreading, Avhen the brush, in shading or coloring, passes over them. If the lines are inked with common writing-ink, they will, however fine they may be made, be subject to the same evil ; for which reason, india-ink is the only kind to be used. THE AMERICAN HOUSE-CARPENTER. SECTION I.— PRACTICAL GEOMETRY. DEFINITIONS. 15. — Geometry treats of the properties of magnitudes, 16. — A point has neither length, breadth, nor thickness. 17. —A line has length only. 18. — Superficies has length and breadth only. 19. — A plane is a surface, perfectly straight and even in every direction ; as the face of a panel when not warped nor winding. 20. — A solid has length, breadth and thickness. 21. — -A right, or straight, line is the shortest that can be drawn between two points. 22. — Parallel lines are equi-distant throughout their length. 23. — An angle is the inclination of two lines towards one another. {Fig. 4.) Fig. 4. Fig. 5. Fig. & 2 10 AMERICA1F HOUSE-CARPENTER. 24. — A right angle has one line perpendicular to the othei. (Fig. 5.) 25. — An oblique angle is either greater or less than a right angle. {Fig. 4 and 6.) 26. — An acute angle is less than a right angle. (Fig- 4.) 27. — An obtuse angle is greater than a right angle. (Fig. 6.) When an angle is denoted by three letters, the middle one, in the order they stand, denotes the angular point, and the other two the sides containing the angle ; thus, let a b c, (Fig. 4,) be the angle, then b will be the angular point, and a b and b c will be the two sides containing that angle. 28. — A triangle is a superficies having three sides and angles. (Fig. 7, 8, 9 and 10.) Fig. 7. Fig. e. 29. — An equi-lateral triangle has its three sides equal. (Fig. 7.) 30. — An isosceles triangle has only two sides equal. (Fig. 8.) 31. — A scalene triangle has all its sides unequal. (Fig. 9) Fig. ». Fib. in. 32. — A right-angled triangle has one right angle. (Fig. 10.) 33. — An acute-angled triangle has all its angles acute. (Fig. 7 and 8.) 34. — An obtuse-angled triangle has one obtuse angle. (Fig. 9.) 35. — Aquadrangle has four sides and four angles. (Fig. 11 to 16.) PRACTICAL GEOMETRY. 11 Fig. 11. Fig. 12. 36.— A parallelogram is a quadrangle having its opposite * sides parallel. (Fig. 11 to 14.) 37"- — A rectangle is a parallelogram, its angles being right angles. (Fig. 11 and 12.) 38. — A square is a rectangle having equal sides. (Fig. 11 .) 39. — A rhombus is an equi-lateral parallelogram having ob- lique angles. (Fig. 13.) Fig. 13. 7 Fig. 14. 40. — A rhomboid is a parallelogram having oblique angles. (Fig.U.) 41. — A trapezoid is a quadrangle having only two of its sides parallel. (Fig. 15.) Fig. 15. Fig. 16. 42. — A trapezium is a quadrangle which has no two of its sides parallel. (Fig. 16.) 43. — A polygon is a figure bounded by right lines. 44. — A regular polygon has its sides and angles equal. 45. — An irregular polygon has its sides and angles unequal. 46. — A trigon is a polygon of three sides, (Fig. 7 to 10 ,) a tetragon has four sides, (Fig. 11 to 16 ;) a pentagon has 12 AMERICAN HOUSE-CARPENTER. five, (Fig. 17 ;) a hexagon six, (Fig. 18 ;) a heptagon seven, (Fig. 19 ;) an octagon eight, (Fig. 20 ;) a nonagon nine ; a decagon ten ; an undecagon eleven ; and a dodecagon twelve sides. Fig. 17. Fig. 16. Fig. 19. Fig. 20. 47. — A circle is a figure bounded by a curved line, called the circumference ; which is every where equi-distant from a cer- tain point within, called its centre. The circumference is also called the periphery, and sometimes the circle. 48. — The radius of a circle is a right line drawn from the centre to any point in the circumference, (a b, Fig. 21.) All the radii of a circle are equal. Fig. 21. 49. — The diameter is a right line passing through the centre, and terminating at two opposite points in the circumference. Hence it is twice the length of the radius, (c d, Fig. 21.) 50. — An arc of a circle is a part of the circumference, (c b or bed, Fig. 21.) 51. — A chord is a right line joining the extremities of an arc. (hd,Fig.2\.) PRACTICAL GEOMETRY. 13 52. — A segment is any part of a circle bounded by an arc and its chord. (A, Fig. 21.) 53. — A sector is any part of a circle bounded by an arc and two radii, drawn to its extremities. (B, Fig. 21.) 54. — A quadrant, or quarter of a circle, is a sector having a quarter of the circumference for its arc. (C, Fig. 21.) 55. — A tangent is a right line, which in passing a curve, touches, without cutting it. (f g, Fig. 21.) * 56. — A cone is a solid figure standing upon a circular base diminishing in straight lines to a point at the top, called its vertex. (Fig. 22.) Fig. 22. 57._The axis of a cone is a right line passing through it, from the vertex to the centre of the circle at the base. 58.— An ellipsis is described if a cone be cut by a plane, hot parallel to its base, passing quite through the curved surface. (a b, Fig. 23.) 59.— A parabola is described if a cone be cut by a plane, parallel to a plane touching the curved surface, (c d, Fig. 23— c d being parallel to/ g.) 60.— An hyperbola is described if a cone be cut by a plane, parallel to any plane within the cone that passes through its vertex, (e h, Fig. 23.) 61.— Foci are the points at which the pins are placed in de- scribing an ellipse. (See Art. 115, and/, /, Fig. 24.) 14 AMERICAN HOUSE-CARPENTER. 62. — The transverse axis is the longest diameter of the ellipsis, (a b, Fig. 24.) 63. — The conjugate axis is the shortest diameter of the ellipsis ; and is, therefore, at right angles to the transverse axis. (c 4, Fig. 24.) 64. — The parameter is a right line passing through the focus of an ellipsis, at right angles to the transverse axis, and termina- ted by the curve, (g h and g t, Fig. 24.) 65. — A diameter of an ellipsis is any right line passing through the centre, and terminated by the curve, (k I, or m n, Fig. 24.) 66. — A diameter is conjugate to another when it is parallel to a tangent drawn at the extremity of that other — thus, the diame- ter, m n, {Fig. 24,) being parallel to the tangent, o p, is therefore conjugate to the diameter, k I. 67. — A double ordinate is any right line, crossing a diameter of an ellipsis, and drawn parallel to a tangent at the extremity of that diameter, [i t, Fig. 24.) 68. — A cylinder is a solid generated by the revolution of a right-angled parallelogram, or rectangle, about one of its sides ; and consequently the ends of the cylinder are equal circles. [Fig. 25.) PRACTICAL GEOMETRY. 15 Fig. 25. Fig. 26. 69. — The axis of a cylinder is a right line passing through it, from the centres of the two circles which form the ends. 70. — A segment of a cylinder is comprehended under three planes, and the curved surface of the cylinder. Two of these are segments of circles : the other plane is a parallelogram, called by way of distinction, the plane of the segment. The circular segments are called, the ends of the cylinder. [Fig. 26.) i\ r . B. — For Algebraical Signs, Trigonometrical Terms, &c, see Appendix. PROBLEMS. KCGHT LINES AND ANGLES. 71. — To bisect a line. Upon the o a -4 J £ / C t 1 s ->■ ^s Tig. 47 a. 88 a. — Points in the circumference of a circle may be ob- tained arithmetically, and positively accurate, by the calcula- tion of ordinates, or the parallel lines, 1, 2, 3, 4. (Fig. PRACTICAL GEOMETRY. 29 47 a.) These ordinates are drawn at right angles to the chord line, a I, and they may be drawn at any distance apart, either equally distant or unequally, and there may be as many of them as is desirable ; the more there are the more points iD the curve will be obtained. If they are located in pairs, equally distant from the versed sine, c d, calculation need be made only for those on one side of c d, as those on the opposite side will be of equal lengths, respectively ; for example, 1, on •he left-hand side of c d, is equal to 1 on the right-hand side, 2 on the right equals 2 on the left, and in like manner for the others. The length of any ordinate is equal to the square root of the difference of the squares of the radius and abscissa, less the difference between the radius and versed sine. (Art. 166.) The abscissa being the distance from the foot of the versed sine to the foot of the ordinate. Algebraically, y — vV— x 2 — (r _ .y) } where y is put to represent the ordinate ; x, the ab- scissa ; v, the versed sine ; and r, the radius. Example. — An arc of a circle has its chord, a h, (Fig. 47 a,) 100 feet long, and its- versed sine, c d, 5 feet. It is required to ascertain the length of ordinates for a sufficient number of points through which to describe the curve. To this end it is requisite, first, to ascertain the radius. This is readily done in (e-Y 8 + v* 50 2 -f 5 2 accordance with Art. 87 a. For, ^- , becomes — — — = 2v 2x5 252'5 = radius. Having the radius, the curve might at once be described without the ordinate points, but for the impracti- cability that usually occurs, in large, fiat segments of the circle, of wetting a location for the centre ; the centre usually being inaccessible. The ordinates are, therefore, to be calculated. In Fig. 47 a the ordinates are located equidistant, and are 10 feet apart. It will only be requisite, therefore, to calculate those on one side of the versed sine, c d. For the first ordinate, 1, the form ula, y - Vt^ x* - (r — v) becomes y = V252-5' - 10' - (2 52'5 - 5). = V63756-25 - 100 - 247'5. = 252-3019 - 247-5. = 4-8019 = the first ordinate, 1. 30 AMMtlCAJS HOITSB -CAEPENTEE. For tne second — y = -i/252-5 2 - 20 2 -(252-5 - 5). = 251-7066 - 24:7-5. = 4-2066 = the second ordinate, 02. For the third— y = V252-5 2 - 30 s - 247-5. = 250-7115 - 247-5. = 3-2115 = the third ordinate, 03. For the fourth — y = V252-5 2 - 40 2 - 247*5. = 249-3115 - 247-5. = 1-8115 = the fourth ordinate, 04. The results here obtained are in feet and decimals of a foot. To reduce these to feet, inches, and eighths of an inch, proceed as at Reduction of Decimals in the Appendix. If the two-feet rule, used by carpenters and others, were decimally diyided, there would be no necessity of this reduction, and it is to be hoped that the rule will yet be thus divided, as such a reform would much lessen the labor of computations, and insure more accurate measurements. 5-0 = ft. 4-8019 = 4-2066 = 3-2115 = 1-8115 = Ordinates, Versed sine, c d, = ft. 01,= " 2,= " 3,= " 4,= 5 - inches. 4 - 9f inches nearly. 4-2§ inches nearly. 3 -2 J inches nearly. 1-91 inches nearly. Fig. 48. 89. — In a given angle, to describe a tanged cwoe. Let a b c, {Fig. 48,) be the given angle, and 1 in the line, a b, and 5 in the line, b c, the termination of the curve. Divide 1 b and b 5 into a like number of equal parts, as at 1, 2, 3, 4 and 5 ; join 1 and 1, 2 and 2, 3 and 3, &c. ; and a regular curve will be formed that will be tangical to the line, a b, at the. point, 1, and to b c at 5. This is of much use ir.. stair-building, in easing the angles formed between the wall-string and the base of the hall, also PRACTICAL GEOMETKY. 31 between the front string and level facia, and in many other instances. The curve is not circular, hut of the form of the parabola, {Fig. 93 ;) yet in large angles the difference is not perceptible. This problem can be applied to describing the Fig. 49. curve for door heads, window-heads, &c, to rather better ad- vantage than Art. 87. For instance, let a b, {Fig. 49,) be the width of the opening, and c d the height of the arc. Extend c d, and make d e equal to c d ; join a and e, also e and b ; and proceed as directed above. 90. — To describe a circle within amy gwen triangle, so that the sides of the triangle shall be tangical. Let a b c, {Fig. 50,) be the given triangle. Bisect the angles a and b, according to Art. Yf ; upon d, the point of intersection of the bisecting lines, with the radius, d e, describe the required circle. Fig .11. 32 AKEEICAN HOTTSE-CAKPENTER. 91. — About a given circle, to describe an equi-lateral tri- angle. Let a dbc, {Fig. 51,) be the given circle. Draw the diameter, c d; upon d, with the radius of the given circle, describe the arc, a e b; join a and b ; draw/?, at right angles to d c ; make/ c and c g, each equal to a b ; from /, through a, draw/ h, also from gr, through b, draw ^ A; then fgh will be the triangle required. 92. — To /?«? a right line nearly equal to the circumference of a circle. Let abed, {Fig. 52,) be the given circle. Draw the diameter, a c ; on this erect an equi-lateral triangle, a e c, according to Art. 93 ; draw g / parallel to a c ; extend e c to /, also e a to g ; then g f will be nearly the length of the semi-circle, a d c ; and twice gf will nearly equal the circum- ference of the circle, a b c d, as was required. Lines drawn from e, through any points in the circle, as o, o and o, to p, p and p, will divide g f in the same way as the semi-circle, a d c, is divided. So, any portion of a circle may be transferred to a straight line. This is a very useful pro- blem, and should be well studied ; as it is frequently used to solve problems on stairs, domes, &c. 92, a. — Another method. Let a bf c, {Fig. 53,) be the given circle. Draw the diameter, a c ; from d, the centre, and at right angles to a c, draw db ; join b and c; bisect b c at e j from d, through e, draw df ; then ef added to three times the diameter, will equal the circumference of the circle sufficiently near for PRACTICAL GEOMETRY. 33 Fig. 53. many uses. The result is a trifle too large, If the circumfer- erence found by this rule, he divided by 648-22, the quotient will be the excess. Deduct this excess, and the remainder will be the true circumference. This problem is rather more curious than useful, as it is less labor to perform the operation arithmetically: simply multiplying the given diameter by 3-1416, or where a great degree of accuracy is needed by 3-1415926. POLYGONS, AC. 93, — Upon a given line to construct cm egui-laieral. triangle. Let a I, {Fig. 54,) be the given line. Upon a and b, with a b for radius, describe arcs, intersecting at c ; join a and c, also c and o ; then a cl will be the triangle required. 94. — To describe cm egui-lateral rectangle, or square. Let a b, {Fig. 55,) be the length of a side of the proposed square. Upon a and o, with a b for radius, describe the arcs a d and he; bisect the arc, a e, in /; upon e, with e f for radius, de- 5 34 AMERICAN HOtrSE-CARFENTEB. r. a Fig. 55. scribe the arc, of d ; join a and o, c and d, d and o ; then a o d I will be the square required. h r. ^y/ £ }\ e f 5f. 95. — Within a given circle, to inscribe an equi-lateral tri- angle, hexagon or dodecagon. Let a i c d, {Fig. 56,) be the given circle. Draw the diameter, b d ; upon b, with the radius of the given circle, describe the arc, a e c ; join a and c, also a and d, and c and d — and the triangle is completed. For the hexagon: from a, also from c, through e, draw the lines, a f and c g ; join a and b, b and c, c and f, &c, and the hexagon is completed. The dodecagon may be formed by bisecting the sides of the hexagon. Each side of a regular hexagon is exactly equal to the radius of the circle that circumscribes the figure. For the radius is equal to a chord of an arc of 60 degrees ; and, as every circle is supposed to be divided into 360 degrees, there is just 6 times 60, or 6 arcs of 60 degrees, in the whole circum- ference. A line drawn from each angle of the hexagon to the centre, (as in the figure,) divides it into six equal, equi-lateral triangles. 96. — Within a square to inscribe an octagon. Let abed, PRACTICAL GEOMETEY. 35 Fig. 57. {Fig. 57,) be the given square. Draw the diagonals, a d and be; upon a, b, o and d, with a e for radius, describe arcs cut- ting the sides of the square at 1, 2, 3, 4, 5, 6, 7 and 8 ; join 1 and 2, 3 and 4, 5 and 6, &c, and the figure is completed. In order to eight-square a hand-rail, or any piece that is to be afterwards rounded, draw the diagonals, a d and b e, upon the end of it, after it has been squared-up. Set a gauge to the distance, a e, and run it upon the whole length of the stuff, from each corner both ways. This will show how much is to be chamfered off, in order to make the piece octagonal. , {Art. 159.) Fig. 58. Fig. 59. Fig. 60. 97 _ Within a given circle to inscribe any regular polygon. Let abc 2, {Fig. 58, 59 and 60,) be given circles. Draw the diameter, ac; upon this, erect an equi-lateral triangle, a e c, according to Art. 93 ; divide a c into as many equal parts as the polygon is to have sides, as at 1, 2, 3, 4, &c. ; from e, 36 AMEEICAU- H0USE-CAEPENTEE. through each even number, as 2, 4, 6, &c, draw lines cutting the circle in the points, 2, 4, &c. ; from these points and at right angles to a c, draw lines to the opposite part of the circle ) this will give the remaining points for the polygon, as b,f, &c. In forming a hexagon, the sides of the triangle erected upon a o, (as at Fig. 59,) mark the points b and f. This method of locating the angles of a polygon is an approximation suffi- ciently near for many purposes ; it is based upon the like prin- ciple with the method of obtaining a right line nearly equal to a circle. {Art. 92.) The method shown at Art. 98 is accurate. Fig- 61. Fig. 62. Tig. 63. 98. — ■ Upon a given line to describe any regular polygon. Let a o, (.Fig. 61, 62 and 63,) be given lines, equal to a side of the required figure. From b, draw b c, at right angles to a I) ; upon a and b, with a b for radius, describe the arcs, a o d and f eb ; divide a o into as many equal parts as the polygon is to have sides, and extend those divisions from c towards d; from the second point of division counting from c towards a, as 3, {Fig. 61,) 4, (Fig. 62,) and 5, (Fig. 63,) draw a line to b ; take the distance from said point of division to a, and set it from b to e; join e and a ; upon the intersection, o, with the radius, o a, describe the circle a f d b j then radiating lines, drawn from 5 through the even numbers on the arc, a d, will cut the circle at the several angles of the required figure. In the hexagon, (Fig. 62,) the divisions on the arc, a d, are not necessary ; for the point, o, is at the intersection of the arcs, a d an&fb, the points, /and d, are determined by the intersection of those arcs with the circle, and the points above, g and h, can be found by drawing lines from a and 5, through the centre, o. In polygons of a greater number of sides than the hexagon, the in- tersection, o, comes above the arcs ; in such case, therefore, the PRACTICAL GEOMETRY. 37 lines, a e and b 5, {Fig. 63,) have to be extended before they will intersect. This method of describing polygons is founded on correct principles, and is therefore accurate. In the circle equal arcs subtend equal angles, (Arts. 86 and 162.) Although this method is accurate, yet polygons may be described as accu- rately and more simply in the following manner. It will be observed that much of the process in this method is for the pur - pose of ascertaining the centre of a circle that will circumscribe the proposed polygon. By reference to the Table of Polygons in the Appendix it will be seen how this centre may be obtained arithmetically. This is the Rule. — Multiply the given side by the tabular radius for polygons of a like number of sides with the proposed figure, and the product will be the radius of the required circumscribing circle. Divide this circle into as many equal parts as the polygon is to have sides, connect the points of division by straight lines, and the figure is complete. For exam- ple : It is desired to describe a polygon of T sides, and 20 inches a side. The tabular radius is 1*1 523 824. This multiplied by 20, the product, 23-04:7648 is the required radius in inches. The Rules for the Reduction of Decimals, also in the Appendix, show how to change decimals to the fractions of a foot or an inch. From this, 23 - 047648 is equal to 23 T V inches nearly. It is not needed to take all the decimals in the, table, three or four of them will give a result sufficiently near for all ordinary practice. Fiff. 64. 99. — To construct a triangle whose sides shall be severalty equal to three given Unes. Let a, b and c, (Fig. 64,) be the given lines. Draw the line, d e, and make it equal to c ; upon e, with b for radius, describe an arc at /; upon d, with a for radius, describe an arc intersecting the other at/; join d and/, also / and e ; then df e will be the triangle required. Fig. 65. 38 PRACTICAL GEOMETRY. 100. — To construct a figure equal to a g .ven, right-lined figure. Let abed, [Fig. 65,) be the given figure. Make ef, {Fig. 66,) equal to c d ; upon /, with d a for radius, describe an arc at g ; upon e, with c a for radius, describe an arc intersecting the other at g ; join g and e ; upon / and g, with d b and a b for radius, describe arcs intersecting at h ; join g and h, also h and / ; then Fig. 66 will every way equal Fig. 65. So, right-lined figures of any number of sides may be copied, oy first dividing them into triangles, and then proceeding as above. The shape of the floor of any room, or of any piece of land, &c, maybe accurately laid out by this problem, at a scale upon paper ; and the contents in square feet be ascertained by the next. 101. — To make a parallelogram equal to a given triangle. Let a b c, (Fig. 67,) be the given triangle. From a, draw a d, at right angles to be; bisect a d in e ; through e, draw / g, parallel to b c ; from 6 and c, draw b /and c g, parallel to d e; then bfgc will be a parallelogram containing a surface exactly equal to that of the triangle, a b c. Unless the parallelogram is required to be a rectangle, the lines, 6 /and c g, need not be drawn parallel to d e. If a rhomboid is desired, they may be drawn at an oblique angle, provided they be parallel to one another. To ascertain the area of a triangle, multiply the base, b c, by half the perpendicular height, d a. In doing this, it matters not which side is taken for bp%e. ^^ e /^ C / Fig. 68. AMERICAN HOUSE-CARPENTEH. 39 102.— A parallelogram being given, to construct anothet equal to it, and having a side equal to a given line. Let A {Fig. 68,) be the given parallelogram, and B the given line Produce the sides of the parallelogram, as at a, b, c and d ; make e d equal to B ; through d, draw c /, parallel to g b ; through e, draw the diagonal, c a ; from a, draw a f, parallel to e d then C will be equal to A. (See Art. 144.) A a ^ v » s > , B 103. — To make a square equal to two or more given squares . Let A and B, (Fig. 69,) be two given squares. Place them so as to form a right angle, as at a ; join b and c ; then the square, C, formed upon the line, b c, will be equal in extent to the squares, A and B, added together. Again : if a b, {Fig. 70,) be equal to the side of a given square, c a, placed at right angles to a b, be the side of another given square, and c d, placed at right angles to 40 PRACTICAL GEOMETRY. c b, be the side of a third given square ; then the square, A, formed upon the line, d b, will be equal to the three given squares. (See Art. 157.) The usefulness and importance of this problem are proverbial. To ascertain the length of braces and of rafters in framing, the length of stair- strings, &c, are some of the purposes to which it may be applied in carpentry. (See note to Art. 74, 6.) If the length of any two sides of a right-angled triangle is known, that of the third can be ascertained. Because the square of the hypothenuse is equal to the united squares of the two sides that contain the right angle. (1.) — The two sides containing the right angle being known, to find the hypothenuse. Rule. — Square each given side, add the squares together, and from the product extract the square- root : this will be the answer. For instance, suppose it were required to find the length of a rafter for a house, 34 feet wide, — • the ridge of the roof to be 9 feet high, above the level of the wall-plates. Then 17 feet, half of the span, is one, and 9 feet, the height, is the other of the sides that contain the right angle Proceed as directed by the rule : 17 9 17 9 119 81 = square of 9. 17 289 = square of 17. 289 - square of 17. 370 Product. 1 ) 370 ( 19-235 + = square-root of 370 ; equal 19 feet, 2} in. 1 1 nearly : which would be the required — length of the rafter. 29 ) 270 9 261 382)- -900 2 764 3843 ) 13600 3 11529 38465)- 207100 (By reference to the table of square-roots 192325 in the Appendix, the root of almost any number may be found ready calculated ; also, to change the decimals of a foot to inches and parts, see Rules for the Reduction of Decimals in the Appendix.) AMERICAN HOUSE-CARPENTER. 4} Agair, : suppose it be required, in a frame building, to rind the length of a brace, having a run of three feet each way from the point of the right angle. The length of the sides containing the right angle will be each 3 feet : then, as before — 3 3 9 = square of one side. 3 times 3 = 9 = square of the other side. 1 8 Product : the square-root of which is 4 - 2426 -f- ft., or 4 feet, 2 inches and |ths. full. (2.) — The hypothenuse and one side being known, to find the other cide. Rule. — Subtract the square of the given side from the square of the hypothenuse, and the square-root of the product will be the answer. Suppose it were required to ascertain the greatest perpendicular height a roof of a given span may have, when pieces of timber of a given length are to be used as rafters. Let the span be 20 feet, and the rafters of 3 X 4 hemlock joist. These come about 13 feet long. The known hypothenuse, then, is 13 feet, and the known side, 10 feet — that being half the span of the building. 13 13 39 13 169 = square of hypothenuse. tO times 10 '= 100 = square of the given side. 69 Product : the square-root of which is 8 •3066 + feet, or 8 feet, 3 inches and *-ths. full. This will be the greatest perpendicular height, as required. Again : suppose that in a story of 8 feet, from floor to floor, a step-ladder is re- quired, the strings of which are to be of plank, 12 feet long ; and it is desirable to know the greatest run such a length of string will afford. In this case, the two given sides are— hypothenuse 12, perpendicular 8 feet. 12 times 12 = 144 = square of hypothenuse. 8 times 8 = 64 = square of perpendicular. 80 Product : the square-root of which is 8-9442 •+- feet, or 8 feet, "*1 inches and ftths. — the answer, as required. 6 42 PRACTICAL GEOMETRY. Many other cases might be adduced to show the utility of this problem. A practical and ready method of ascertaining thd length of braces, rafters, &c, when not of a great length, is to apply a rule across the carpenters'-square. Suppose, for the length of a rafter, the base be 12 feet and the height 7. Apply the rule diagonally on the square, so that it touches 12 inches from the corner on one side, and 7 inches from the corner on the other. The number of inches on the rule, which are intercepted by the sides of the square, 13f nearly, will be the length of the rafter in feet ; viz, 13 feet and fths of a foot. If the dimensions are large, as 30 feet and 20, take the half of each on the sides of the square, viz, 15 and 10 inches ; then the length in inches across, will be one-half the number of feet the rafter is long. This method is just as accurate as the preceding ; but when the length of a very long rafter is sought, it requires great care and precision to ascertain the fractions. For the least variation on the square, or in the length taken on the rule, would make perhaps several inches difference in the length of the rafter. For shorter dimensions, however, the result will be true enough. rig. 7i. 104. — To make a circle equal to two given circles. Let A and B, {Fig. 71,) be the given circles. In the right-angled tri- angle, a b c, make a b equal to the diameter of the circle. B, and c 6 equal to the diameter of the circle, A ; then the hypothenuse ; Fig. 72. AMERICAN HOUSE-CARPENTER. 43 a c, will be the diameter of a circle, C, which will be equal in area to the two circles, A and B, added together. Any polygonal figure, as A, {Fig. 72,) formed on the hypo- thenuse of a right-angled triangle, will be equal to two similar figures,* as B and C, formed on the two legs of the triangle. Fig. 73, 105. — To construct a square equal to a given rectangle. Let A, (Fig. 73,) be the given rectangle. Extend the side, a b, and make b c equal to b e; bisect a c in/, and upon/, with the radius, / a, describe the semi-circle, age; extend e b, till it cuts the curve in g ; then a square, b g h d, formed on the line, b s, will be equal in area to the rectangle, A. e I A c g Tig. 74. 105, a.— Another method. Let A, {Fig. 74,) be the given rectangle. Extend the side, a b, and make a d equal to a c, * Similar figures are such as have their several angles respectively equal, and theii suljs respectively proportionate. 44 PRACTICAL GEOMETRY. bisect a d in e ; upon e, with the radius, e a, describe the semi- circle, afd; extend g b till it cuts the curve in/; join a and f ; then the square, B, formed on the line, af will be equal in area to the rectangle, A. (See Art. 156 and 157.) 106.— To form a square equal to a given triangle. Let a b, (Fig. 73,) equal the base of the given triangle, and b e equal half its perpendicular height, (see Fig. 67 ;) then proceed a? directed at Art. 105. L07. — Two right lines being given, to find a third propor- tional thereto. Let A and B, (Fig. 75,) be the given lines. Make a b equal to A ; from a, draw a c, at any angle with a b ; make a c and a d each equal to B ; join c and b ; from d, draw d e, parallel to c b ; then a e will be the third proportional re- quired. That is, a e bears the same proportion to B, as B does to A. Fig. 76. 108. — Three right lines being given, to find a fourth pro portional thereto. Let A, B and C, (Fig. 76,) be the given lines. Make a b equal to A ; from a, draw a c, at any angle with a b; make a c equal to B, and a e equal to C ; join c and 6 ; from e, draw e/, parallel to c 6; then af will be the fourth proportional required. That is, a f bears the same proportion to C, as B does to A. AMERICAN HOUSE-CARPENTER. 45 To apply this problem, suppose the two axes of a given ellipsis and the longer axis of a proposed ellipsis are given. Then, by this problem, the length of the shorter axis' to the proposed ellip- sis, can be found : so that it will bear the same proportion to the longer axis, as the shorter of the given ellipsis does to its longer. (See also, Art. 126.) A- B- 109. — A line with certain divisions being given, to divide another, longer or shorter, given line in the same proportion. Let A, {Fig. 77,} be the line to be divided, and B the line with its divisions. Make a b equal to B, with all its divisions, as at 1, 2, 3, &c. ; from a, draw a c, at any angle with a b ; make a c equal to A ; join c and b ; from the points, 1, 2, 3, &c, draw lines, parallel to c b ; then these will divide the line, a c, in the same proportion as B is divided — as was required. This problem will be found useful in proportioning the mem- bers of a proposed cornice, in the same proportion as those of a given cornice of another size. (See Art. 253 and 254.) So of a pilaster, architrave, &c. Fig. 78. 110. — Between two given right lines^ to find a meajipio- portional. Let A and B, (Fig. 78,) be the given lines. On the line, a c, make a b equal to A, and b c equal to B ; bisect a cine ; upon e, with e a for radius, describe the semi circle, a d 46 AMERICAN HOUSE-CARPENTER. c; at b, erect b d, at right angles to a c; then b d will be the mean proportional between A and B. That is, a b is to b d as b d is to b g. This is usually stated thus — a b '• b b ::b d :b c, and since the product of the means equals the product of the extremes, therefore, ab xb o = b d . This is shown geometri- cally at Art. 105. CONIC SECTIONS. 111. — If a cone, standing upon a base that is at right angles with its axis, be cut by a plane, perpendicular to its base and passing through its axis, the section will be an isosceles triangle ; 'as a b c, Fig. 79 ;) and the base will be a semi-circle. If a cone be cut by a plane in the direction, e f, the section will be an ellipsis; if in the direction, m I, the section will be a, parabola; and if in the direction, r o, an hyperbola. (See Art. 56 to 60.) If the cutting planes be at right angles with the plane, a b c, then— 112.— To find the axes of the ellipsis, bisect ef, {Fig. 79,) in g; through g, draw h i, parallel to ab; bisect h i inj; upon j, with j h for radius, describe the semi-circle, hlci; from g, draw g h, at right angles to h i; then twice g 7c will be the conjugate -axis, and e/the transverse. AMERICAN HOUSE-CARPENTER. 47 113.— To Jind the axis and base of the parabola. Let m I, {Fig. 79,) parallel to a c, be the direction of the cutting plane! From m, draw m d, at right angles to a b ; then I m will be the axis and height, and m d an ordinate and half the base; as at Fig. 92, 93. 114— To find the height, base and transverse axis of an hyperbola. Let o r, {Fig. 79,) be the direction of the cutting plane. Extend o r and a c till they meet at n ; from o, draw o p, at right angles to a b; then r o will be the height, nr the transverse axis, and o p half the base ; as at Fig. 94. Fig. 80. 115. — The axes being given, to find the foci, and to describe an ellipsis with a string. Let a b, {Fig. 80,) and c d, be the given axes. Upon c, with a e or 6 e for radius, describe the arc, f f ; then / and /, the points at which the arc cuts the transverse axis, will be the foci. At/ and /place two pins, and another at c ; tie a string about the three pins, so as to form the triangle, ffc ; remove the pin from c, and place a pencil in its stead ; keeping the string taut, move the pencil in the direction, eg a.; it will then describe the required ellipsis. The lines, fg and g f show the position of the string when the pencil arrives at g. This method, when performed correctly, is perfectly accurate; but the string is liable to stretch, and is, therefore, not so good to use as the trammel. In making an ellipse by a string or twine. that kind should be used which has the least tendency to elasticity, For this reason, a cotton cord, such as chalk-lines are commonly made of, :s not proper for the purpose : a linen, or flaxen cord is mu:h better. 4S PRACTICAL GEOMETRY. Fig. 81 116. — The axes being given, to describe an ellipsis with a trammel. Let a b and c d, {Fig. 81,) be the given axes. Place the trammel so that a line passing through the centre of the grooves, would coincide with the axes ; make the distance from the pencil, e, to the nut,/, equal to half c d ; also, from the pen- cil, e, to the nut, g, equal to half a b ; letting the pins under the nuts slide in the grooves, move the trammel, e g, in the direction, c b d ; then the pencil at e will describe the required ellipse. A trammel maybe constructed thus : take two straight strips ot board, and make a groove on their face, in the centre of their width ; join them together, in the middle of their length, at right angles to one another ; as is seen at Fig. 81. A rod is then to be prepared, having two moveable nuts made of wood, with a mor- tice through them of the size of the rod, and pins under them large enough to fill the grooves. Make a hole at one end of the rod, in which to place a pencil. In the absence of a regular tram- mel, a temporary one may be made, which, for any short job, will answer every purpose. Fasten two straight-edges at right angles to one another. Lay them so as to coincide with the axes of the proposed ellipse, having the angular point at the centre. Then, in a rod having a hole for the pencil at one. end, place two brad-awls at the distances described at Art. 116. While the pencil is moved in the direction of the curve, keep the brad-awls hard against the straight-edges, as directed for using the tram- mel-rod, and one-quarter of the ellipse will be drawn. Then, by shifting the straight-edges, the other three quarters in succes- sion may be drawn. If the required ellipse be not too large, a carpenters'-square may be made use of, in place of the straight- edges. An improved method of constructing the trammel, is as fol lows : make the sides of the grooves bevilling from the face oi the stuff, or dove-tailing instead of square. Prepare two slips of wood, each about two inches long, which shall be of a shape to 'ust fill the groove when slipped in at the end. These, instead ot AMERICAN HOUSE-CARPF>-TER. 49 pins, are to be attached, one to each of the moveable nuts with a screw, loose enough for the nut to move freely about the screw as an axis. The advantage of this contrivance is, in preventing the nuts from slipping out of their places, during the operation of describing the curve. Fig. 82. 117. — To describe an ellipsis by ordinates. Let a b and c a, (Fig. 82,) be given axes. With c e or e d for radius, de- scribe the quadrant, f g h ; divide f h, a e and e b, each into a like number of equal parts, as at ] , 2 and 3 ; through these points, draw ordinates, parallel to c d tmifg ; take the distance, 1 i, and place it at 1 1, transfer 2 j to 2 m, and 3 k to 3 n ; through the points, a, n, m, I and c, trace a curve, and the ellipsis will be completed. The greater the number of divisions on a e, &c, in this and the following problem, the more points in the curve can be found, and the more accurate the curve can be traced. If pins are placed in the points, n, m, I, B 1 / 'K , are equal. Thus, it is proved for the angle at d; it is also true for any other angle ; for, since all other inscribed angles on that side of the chord line, cf, equal the angle A, (Art. 161,) therefore the angle formed by a chord and tangent equals any angle in the opposite segment of the circle. This being proved for the acute angle, D, it is also true for the obtuse angle, a cf ; for, from any point, n, (Fig. 114,) in the arc c nf draw lines to d,f and cy now, if it can be proved tMt the angle a cf equals the angle f n c, the entire proposition is proved, for the angle f n o equals any of all the inscribed angles that can be drawn on that side of the chord. (Art. 161.) To prove, then, that a cf equals c nf: the angle a cf equals the sum of the angles A and B ; also the angle c nf equals the sum of the angles C and D. The angles B and D, being inscribed angles on the same chord, df are equal. The angles C and A being right angles, (Art. 156,) are likewise equal. Now, since A equals C, and B equals J), therefore the sum of A and B equals the sum of O and D — or the angle a cf equals the angle c nf 164. — Proposition. Two chords, a o and c d, (Fig. 115,) intersecting, the parallelogram or rectangle formed by the two parts of one is equal to the rectangle formed by the two pails of the other. That is, c e mi Itiplied by e d, the product is 72 PRACTICAL GEOMETEY. Fig. 115. equal to the product of a e multiplied by e b. The triangle A is similar to the triangle B, because it has corresponding an- gles. The angle i equals the angle e, {Art. 148 ;) the angle at c equals the angle at a because they stand upon the same chord, d b, (Art. 161 ;) for the same reason the angle b equals the angle d, for each stands upon the same chord, a o. There- fore, the triangle A having the same angles as the triangle B, the length of the sides of one are in like proportion as the length of the sides in the other. So, e d : a e :: e i : o e. Hence, a e multiplied by e i is equal to e d mutiplied by c e — or the product of the means equals the product of the ex- tremes. 165. — Proposition. In any circle, when a segment is given, the radius is equal to the sum of the squares of half the chord and of the versed sine, divided by twice the versed sine. Let a b, {Fig. 116,) be the chord line, and v the versed sine of the segment. By the preceding article the triangle A is shown to be like the triangle B, having equal angles and proportionate AMEKICAU HOTJSE-CAHPENTEE. 73 Fig 116. length of sides. Therefore, v : n::m : i, or — = i ; that is, i is v equal to the square of n (or n X ri) divided by v. This result being added to v equals the diameter o x, which may be indi n* cated by the letter d ; thus, \-v = i + v = d ; and the half BF Fig. 127. 102 AMERICAN HOUSE-CARPENTER. ovei the centre of the column; and introducing horizontal instead of inclined mutules in the cornice, and in some instan ces dispensing with them altogether. The Ionic was modified by diminishing the size of the volutes, and, in some specimens, introducing a new capital in which the volutes were diago- nally arranged, {Fig. 127.) This new capital has been termed modern Ionic. The favorite order at Rome and her colonies was the Corinthian, {Fig. 128.) Eut this order, the Roman artists in their search for novelty, subjected to many altera- tions — especially in the foliage of its capital. Into the upper part of this, they introduced the modified Ionic capital; thus combining the two in one. This change was dignified with the importance of an order, and received the appellation, Composite, or Roman: the best specimen of which is found in the Arch of Titus, {Fig. 129.) This style was not much used among the Romans themselves, and is but slightly appreciated now. 214. — The Tuscan Order is said to have been introduced to the Romans by the Etruscan architects, and to have been the only style used in Italy before the introduction of the Grecian orders. However this may be, its similarity to the Doric order gives strong indications of its having been a rude imita- tion of that style : this is very probable, since history informs us that the Etruscans held intercourse with the Greeks at a remote period. The rudeness of this order prevented its ex- tensive use in Italy. All that is known concerning it is from Vitruvius — no remains of buildings in this style being found among ancient ruins. 215. — For mills, factories, markets, barns, stables, &c, where utility and strength are of more importance than beauty, the improved modification of this order, called the modern Tuscan, {Fig. 130,) will be useful ; and its simplicity recommends it where economy is desirable. 216. — Egyptian Style. The architecture of the ancient ARCHITECTURE. 103 fill. 1: 104 . eg AMERICAN HOUSE-CAKPENTEE. 3&. is** BO* »IJ4 ms 39 - end!* K*;M^e»'««K*KiK*ss^ S|g ?ft? w '" "^ " Fl£. 129, TUSCAN ORDER. 105 106 ARCHITECTURE. Egyptians -to which that of the ancient Hindoos bears some re- semblance — is characterized by boldness of outline, solidity and grandeur. The amazing labyrinths and extensive artificial lakes, the splendid palaces and gloomy cemeteries, the gigantic pyramids and towering obelisks, of the Egyptians, were works of immen- sity and durability ; and their extensive remains are enduring proofs of the enlightened skill of this once-powerful, but long since extinct nation. The principal features of the Egyptian Style of architecture are — uniformity of plan, never deviating from right lines and angles ; thick walls, having the outer surface slightly deviating inwardly from the perpendicular ; the whole building low ; roof flat, composed of stones reaching in one piece from pier to pier, these being supported by enormous columns, very stout in proportion to their height ; the shaft sometimes polygonal, having no base but with a great variety of handsome capitals, the foliage of these being of the palm, lotus and other leaves ; entablatures having simply an architrave, crowned with a huge cavetto orna- mented with sculpture ; and the intercolumniation very narrow, usually 1 J diameters and seldom exceeding 2|. In the remains of a temple, the walls were found to be 24 feet thick ; and at the gates of Thebes, the walls at the foundation were 50 feet thick and perfectly solid. The immense stones of which these, as well as Egyptian walls generally, were built, had both their inside and outside surfaces faced, and the joints throughout the body of the wall as perfectly close as upon the outer surface. For this reason, as well as that the buildings generally partake of the pyramidal form, arise their great solidity and durability. The dimensions and extent of the buildings may be judged from the temple ot Jupiter at Thebes, which was 1400 feet long and 300 feet wide- exclusive of the porticos, of which there was a great number. It is estimated by Mr. Gliddon, U. S. consul in Egypt, that not less than 25,000,000 tons of hewn stone wert employed in the erection of the Pyramids of Memphis alone, — or enough to con- struct 3,000 Bunker-Hill monuments. Some of the blocks are 40 EGYPTIAN STYLE. H. P. Fig. 181. 1.«fc*j^* . fHfr. $$M Fi Cnder-CelU r. AECHITECTUEE. Ill 223. — Dwelling houses are built of various dimensions and Btyles, according to their destination ; and to give designs and directions for their erection, it is necessary to know their situa- tion and object. A dwelling intended for a gardener, would require very different dimensions and arrangements from one intended for a retired gentleman — with his servants, horses, &c. ; nor would a house designed for the city be appropriate for the country. For city houses, arrangements that would be convenient for one family might be very inconvenient for two or more. Fig. 132, 133, 134, 135, 136, and 137, represent the ichnographical projection, or ground-plan, of the floors of an ordinary city house, designed to be occupied by one family only. Fig. 139 is an elevation, or front-view, of the same house : all these plans are drawn at the same scale — which ia that at the bottom of Fig. 139. Fig. 132 is a Plan of the Under-Cellar. a, is the coal-vault, 6 by 10 feet. o, is the furnace for heating the house. c, d, are front and rear areas. Fig. 133 is a Plan of the Basement. a, is the' library, or ordinai-y dining-room, 15 by 20 feet. 0, is the kitchen, 15 by 22 feet. c, is the store-room, 6 by 9 feet. d, is the pantry, 4 by 7 feet. e, is the china closet, 4 by 7 feet. f, is the servants' water-closet. g, is a closet. h, is a closet with a dumb-waiter to the first story abovfi. i, is an ash closet under the front stoop. j, is the kitchen-range. Te, is the sink for washing and drawing water 1, are wash trays. 112 AMERICAN H0U6E-CAKPENTEK. AKCHTTECTTJItE. 113 Fig. 134 is a Plan of the First Story. a, is the parlor, 15 by 34 feet. b, is the dining-room, 16 by 23 feet. c, is the vestibule. e, is the closet containing the dumb-waiter from the basement, f, is the closet containing butler's sink. g, g, are closets. A, is a closet for hats and cloaks. i, j, are front and rear balconies. Fig. 135 is the Second Story. a, a, are chambers, 15 by 19 feet. b, is a bed-room, 1h by 13 feet. c, is the bath-room, 74 by 13 feet. d, d, are dressing-rooms, 6 by 7 J feet. e, e, are closets. f,f, are wardrobes. g, g, are cupboards. Fig. 136 is the Third Story. a, a, are chambers, 15 by 19 feet. b, b, are bed-rooms, TJ by 13 feet. c, c, are closets. d, is a linen closet, 5 by 7 feet. e, e, are dressing-closets. f,f, are wardrobes. g, g, are cupboards. Fig. 137 is the Fourth Story. a, a, are chambers, 14 by 17 feet. b, b, are bed-rooms, 8£ by 17 feet. c, o, c, are closets. d, is the step-ladder to the root. 15 114 AMERICAN HOUSE-OAEPENTEE. AKCHITECTUKE. U5 Fig. 138 is the Section of the House showing the heights of the several stories. Fig. 139 is the Front Elevation. The size of the house is 25 feet front by 55 feet deep ; this is about the average depth, although some are extended to 60 and 65 feet in depth. These are introduced to give some general ideas of the prin- ciples to be followed in designing city houses. In placing the chimneys in the parlours, set the chimney-breasts equi-distant from the ends of the room. The basement chimney-breasts may be placed nearly in the middle of the side of the room, as there is but one flue to pass through the chimney-breast above ; but in the second story, as there are two flues, one from the basement and one from the parlour, the breast will have to be placed nearly perpendicular over the parlour breast, so as to receive the flues within the jambs of the fire-place. As it is desirable to have the cbimney-breast as near the middle of the room as possible, it may be placed a few inches towards that point from over the breast below. So in arranging those of the stories above, always make provision for the flues from below. 224. — In placing the stairs, there should be at least as much room in the passage at the side of the stairs, as upon them; and in regard to the length of the passage in the second stoiy, there must be room for the doors which open from each of the principal rooms into the hall, and more if the stairs require it. Having assigned a position for the stairs of the second story, now generally placed in the centre of the depth of the house, let the winders of the other stories be placed perpendicularly over and under them; and be careful to provide for head- room. To ascertain this, when it is doubtful, it is well to draw a vertical section of the whole stairs ; but in ordinary cases, this is not necessary. To dispose the windows properly, the 116 AMERICAS' HOTJSECABPENTEK. Fie. 139. •Aevaticn. ARCHITECTURE. 117 middle window of each story should be exactly in the middle of the front ; hut the pier between the two windows which light the parlour, should be in the centre of that room ; be- cause when chandeliers or any similar ornaments, hang from the centre-pieces of the parlour ceilings, it is important, in order to give the better effect, that the pier-glasses at the front and rear, be in a range with them. If both these ob- jects cannot be attained, an approximation to each must be attempted. The piers should in no case be less in width than the window openings, else the blinds or shutters when thrown open will interfere with one another ; in general practice, it is well to make the outside piers § of the width of one of the middle piers. When this is desirable, deduct the amount of the three openings from the width of the front, and the re- mainder will be the amount of the width of all the piers ; divide this by 10, and the product will be $ of a middle pier ; and then, if the parlour arrangements do not interfere, give twice this amount to each corner pier, and three times the same amount to each of the middle piers. PRINCIPLES OF ARCHITECTURE. 225. — In the construction of the first habitations of men, frail and rude as they must have been, the first and principal object was, doubtless, utility — a mere shelter from sun and rain. But as successive storms shattered the poor tenement, man was taught by experience the necessity of building with an idea to durability. And when in his walks abroad, the symmetry, proportion and beauty of nature met his admiring gaze, contrasting so strangely with the misshapen and dispro- portioned work of his own hands, he was led to make gradual changes ; till his abode was rendered not only commodious and durable, but pleasant in its appearance ; and building became a fine art, having utility for its basis. 118 AMERICAN HOT/SE-CABPENTEK. 226. — In all designs for buildings of importance, utility, durability and beauty, the first great principles of architec- ture, should be pre-eminent. In order that the edifice be useful, commodious and comfortable, the arrangement of the apartments should be such as to fit them for their several des- tinations ; for public assemblies, oratory, state, visitors, retir- ing, eating, reading, sleeping, bathing, dressing, &c. — these should each have its own peculiar form and situation. To accomplish this, and at the same time to mate their relative situation agreeable and pleasant, producing regularity and harmony, require in some instances much skill and sound judgment. Convenience and regularity are very important, and each should have due attention ; yet when both cannot be obtained, the latter should in most cases give place to the former. A building that is neither convenient nor regular, whatever other good qualities it may possess, will be sure of disapprobation. 227. — The utmost importance should be attached to such arrangements as are calculated to promote health: among these, ventilation is by no means the least. For this purpose, the ceilings of the apartments should have a respectable height ; and the sky-light, or any part of the roof that can be made moveable, should be arranged with cord and pullies, so as to be easily raised and lowered. Small openings near the ceiling, that may be closed at pleasure, should be made in the partitions that separate the rooms from the passages — espe- cially for those rooms which are used for sleeping apartments. All the apartments should be so arranged as to secure their being easily kept dry and clean. In dwellings, suitable apart- ments should be fitted up for oathing with all the necessary apparatus for conveying the water. 228. — To ii sure stability in an edifice, it should be designed upon well-known geometrical principles : such as science has de- monstrated to be necessary and sufficient fc r firmness and dura AMERICAN HOUSE-CARPENTER. 119 bility. It is well, also, that it have the appearance of stability as well as the reality ; for should it seem tottering and unsafe, the sensation of fear, rather than those of admiration and pleasure, will he excited in the beholder. To secure certainty and accu- racy in the application of those principles, a knowledge of the strength and other properties of the materials used, is indispensa- ble ; and in order that the whole design be so made as to be capable of execution, a practical knowledge of the requisite mechanical operations is quite important. 229. — The elegance of an architectural design, although chiefly depending upon a just proportion and harmony of the parts, will be promoted by the introduction of ornaments — provided this be judiciously performed. For enrichments should not only be of a proper character to suit the style of the building, but should also have their true position, and be bestowed in proper quantity. The most common fault, and one which is prominent in Roman archi- tecture, is an excess of enrichment : an error which is carefully to be guarded against. But those who take the Grecian models for their standard, will not be liable to go to that extreme. In ornamenting a cornice, or any other assemblage of mouldings, at least every alternate member should be left plain ; and those that are near the eye should be more finished than those which are dis- tant. Although the characteristics of good architecture are utili- ty and elegance, in connection with durability, yet some buildings are designed expressly for use, and others again for ornament : in the former, utility, and in the latter, beauty, should be the gov- erning principle. 230. — The builder should be intimately acquainted with the principles upon which the essential, elementary parts of a build- ing are founded. A scientific knowledge of these will insure certainty and security, and enable the mechanic to erect the most extensive and lofty edifices with confidence. The more important parts are the foundation, the column, the wall, the lintel, the arch, the vault, the dome and the roof. A separate description of the 120 ARCHITECTURE. peculiarities of each, would seem to be necessary ; and cannol perhaps be better expressed than in the following language of a modern writer on this subject. 231. — "In laying the Foundation of any building, it is ne- cessary to dig to a certain depth in the earth, to secure a solid basis, below the reach of frost and common accidents. The most solid basis is rock, or gravel which has not been moved. Next to these are clay and sand, provided no other excavations have been made in the immediate neighbourhood. From this basis a stone wall is carried up to the surface of the ground, and constitutes the foundation. Where it is intended that the super- structure shall press unequally, as at its piers, chimneys, or columns, it is sometimes of use to occupy the space between the points of pressure by an inverted arch. This distributes the pressure equally, and prevents the foundation from springing be- tween the different points. In loose or muddy situations, it is always unsafe to build, unless we can reach the solid bottom below. In salt marshes and flats, this is done by depositing tim- bers, or driving wooden piles into the earth, and raising walls upon them. The preservative quality of the salt will keep these timbers unimpaired for a great length of time, and makes the foundation equally secure with one of brick or stone. 232. — The simplest member in any building, though by no means an essential one to all, is the Column, or pillar. This is a perpendicular part, commonly of equal breadth and thickness, not intended for the purpose of enclosure, but simply for the sup- port of some part of the superstructure. The principal force which a column has to resist, is that of perpendicular pressure. In its shape, the shaft of a column should not be exactly cylin- drical, but, since the lower part must support the weight of the superior part, in addition to the weight which presses equally on the vvhole column, the thickness should gradually decrease from bottom to top. The outline of columns should be a little curved, so as to represent a portion of a very long spheroid, or paraboloid, AMERICAN HOUSE-CARPENTER. 121 rather than of a cone. This figure is the joint result of two cal- culations, independent of beauty of appearance. One of these is, that the form best adapted for stability of base is that of a cone ; the other is, that the figure, which would be of equal strength throughout for supporting a superincumbent weight, would be generated by the revolution of two parabolas round the axis of the column, the vertices of the curves being at its ex- tremities. The swell of the shafts of columns was called the en- tasis by the ancients. It has been lately found, that the columns of the Parthenon, at Athens, which have been commonly sup. posed straight, deviate about an inch from a straight line, and that their greatest swell is at about one third of their height. Columns in the antique orders are usually made to diminish one sixth or one seventh of their diameter, and sometimes even one fourth. The Gothic pillar is commonly of equal thickness throughout. 233. — The Wall, another elementary part of a building, may be considered as the lateral continuation of the column, answer- ing the purpose both of enclosure and support. A wall must diminish as it rises, for the same reasons, and in the same propor- tion, as the column. It must diminish still more rapidly if it ex- tends through several stories, supporting weights at different heights. A wall, to possess the greatest strength, must also con- sist of pieces, the upper and lower surfaces of which are horizon- tal and regular, not rounded nor oblique. The walls of most of the ancient structures which have stood to the present time, are constructed in this manner, and frequently have their stones bound together with bolts and cramps of iron. The same method is adopted in such modern structures as are intended to possess great strength and durability, and, in some cases, the stones are even dove-tailed together, as in the light-houses at Eddystone and Bell Rock. But many of our modern stone walls, for the sake ol cheapness, have only one face of the stones squared, the inner half of the wall being completed with brick ; so that they can, 16 122 ARCHITECTURE. in reality, be considered only as brick walls faced with stone Such walls are said to be liable to become convex outwardly, from the difference in the shrinking of the cement. Rubble walls are made of rough, irregular stones, laid in mortar. The stones should be broken, if possible, so as to produce horizontal surfaces The coffer walls of the ancient Romans were made by enclosing successive portions of the intended wall in a box, and filling it with stones, sand, and mortar, promiscuously. This kind of structure must have been extremely insecure. The Pantheon, and various other Roman buildings, are surrounded with a double brick wall, having its vacancy filled up with loose bricks and cement. The whole has gradually consolidated into a mass of great firmness. The reticulated walls of the Romans, having bricks with oblique surfaces, would, at the present day, be thought highly unphilosophical. Indeed, they could not long have stood, had it not been for the great strength of their cement. Modern brick walls are laid with great precision, and depend for firmness more upon their position than upon the strength of their cement. The bricks being laid in horizontal courses, and continually overlaying each other, or breaking joints, the whole mass is strongly inter- woven, and bound together. Wooden walls, composed of timbers covered with boards, are a common, but more perishable kind. They require to be constantly covered with a coating of a foreign substance, as paint or plaster, to preserve them from spontaneous decomposition. In some parts of Prance, and elsewhere, a kind of wall is made of earth, rendered compact by ramming it in moulds or cases. This method is called building in pise, and is much more durable than the nature of the material would lead us to suppose. Walls of all kinds are greatly strengthened by angles and curves, also by projections, such as pilasters, chimneys and buttresses. These projections serve to increase the breadth of the foundation, and are always to be made use of in large buildings, and in walls of considerable length. AMERICAN HOUSE-CARPENTER. 123 234. — The Lintel, ox beam, extends in a right line over a vacant space, from one column or wall to another. The strength of the lintel will be greater in proportion as its transverse vertical diameter exceeds the horizontal, the strength being always as the square of the depth. The floor is the lateral continuation or connection of beams by means of a covering of boards. 235. — The Arch is a transverse member of a building, an- swering the same purpose as the lintel, but vastly exceeding it in strength. The arch, unlike the lintel, may consist of any num- ber of constituent pieces, without impairing its strength. It is, however, necessary that all the pieces should possess a uniform shape, — the shape of a portion of a wedge, — and that the joints, formed by the contact of their surfaces, should point towards a common centre. In this case, no one portion of the arch can be displaced or forced inward ; and the arch cannot be broken by any force which is not sufficient to crush the materials of which it is made. In arches made of common bricks, the sides of which are parallel, any one of the bricks might be forced inward, were it not for the adhesion of the cement. Any two of the bricks, however, by the disposition of their mortar, cannot collective- ly be forced inward. An arch of the proper form, when com- plete, is rendered stronger, instead of weaker, by the pressure of a considerable weight, provided this pressure be uniform. While building, however, it requires to be supported by a centring of the shape of its internal surface, until it is complete. The upper stone of an arch is called the key-stone, but is not more essential than any other. In regard to the shape of the arch, its most simple form is that of the semi-circle. It is, however, veiy fre- quently a smaller arc of a circle, and, still more frequently, a por- tion of an ellipse. The simplest theory of an arch supporting itself only, is that of Dr. Hooke. The arch, when it has only its own weight to bear, may be considered as the inversion of a chain, suspended at each end. The chain hangs in such a form, that the weight of each link or portion is held in equilibrium by 12i ARCHITECTURE. the result of two forces acting at its extremities ; and these forces, or tensions, are produced, the one by the weight of the portion of the chain below the link, the other by the same weight increased by that of the link itself, both of them acting originally in a ver- tical direction. Now, supposing the chain inverted, so as to con- stitute an arch of the same form and weight, the relative situa- tions of the forces will be the same, only they will act in contrary directions, so that they are compounded in a similar manner, and balance each other on the same conditions. The arch thus formed is denominated a catenary arch. In common cases, it differs but little from a circular arch of the extent of about one third of a whole circle, and rising from the abut- ments with an obliquity of about 30 degrees from a perpendicu- lar. But though the catenary arch is the best form for support- ing its own weight, and also all additional weight which presses in a vertical direction, it is not the best form to resist lateral pressure, or pressure like that of fluids, acting equally in all direc- tions. Thus the arches of bridges and similar structures, when covered with loose stones and earth, are pressed sideways, as well as vertically, in the same manner as if they supported a weight of fluid. In this case, it is necessary that the arch should arise more perpendicularly from the abutment, and that its general figure should be that of the longitudinal segment of an ellipse. In small arches, in common buildings, where the disturbing force is not great, it is of little consequence what is the shape ot the curve. The outlines may even be perfectly straight, as in the tier of bricks which we frequently see over a window. This is, strictly speaking, a real arch, provided the surfaces of the bricks tend towards a common centre. It is the weakest kind of arch, and a part of it is necessarily superfluous, since no greater portion can act in supporting a weight above it, than can be included be- tween two curved or arched lines. Besides the arches already mentioned, various others are in use The acute or lancet arch, much used in Gothic architecture, is AMERICAN HOUSE-CARPENTER. 125 described usually from two centres outside the arch. It is a strong arch for supporting vertical pressure. The rampant arch is one in which the two ends spring from unequal heights. The horse-shoe or Moorish arch is described from one or more centres placed above the base line. In this arch, the lower parts are in danger of being forced inward. The ogee arch is concavo-con- vex, and therefore fit only for ornament. In describing arches, the upper surface is called the extrados, and the inner, the in- trados. The springing lines are those where the intrados meets the abutments, or supporting walls. The span is the distance from one springing line to the other. The wedge-shaped stones, which form an arch, are sometimes called voussoirs, the upper- most being the key-stone. The part of a pier from which an arch springs is called the impost, and the curve formed by the upper side of the voussoirs, the archivolt. It is necessary that the walls, abutments and piers, on which arches are supported, should be so firm as to resist the lateral thrust, as well as vertical pressure, of the arch. It will at once be seen, that the lateral or sideway pressure of an arch is very considerable, when we recol- lect that every stone, or portion of the arch, is a wedge, a part of whose force acts to separate the abutments. For want of atten- tion to this circumstance, important mistakes have been committed, the strength of buildings materially impaired, and their ruin ac- celerated. In some cases, the want of lateral firmness in the walls is compensated by a bar of iron stretched across the span ot the arch, and connecting the abutments, like the tie-beam of a roof. This is the case in the cathedral of Milan and some other Gothic buildings. In an arcade, or continuation of arches, it is only necessary that the outer supports of the terminal arches should be strong enough to resist horizontal pressure. In the intermediate arches, the lat- eral force of each arch is counteracted by the opposing lateral t force of the one contiguous to it. In bridges, however, where individual arches are liable to be destroyed by accident, it is desi 126 ARCHITECTURE. rable that each of the piers should possess sufficient horizontal strength to resist the lateral pressure of the adjoining arches. 236. — The Vault is the lateral continuation of an arch, serving to cover an area or passage, and bearing the same relation to the arch that the wall does to the column. A simple vault is con- structed on the principles of the arch, and distributes its pressure equally along the walls or abutments. A complex or groined vault is made by two vaults intersecting each other, in which case the pressure is thrown upon springing points, and is greatly increased at those points. The groined vault is common in Gothic architecture. 237. — The Dome, sometimes called cupola, is a concave cover- ing to a building, or part of it, and may be either a segment of a sphere, of a spheroid, or of any similar figure. When built of stone, it is a very strong kind of structure, even more so than the arch, since the tendency of each part to fall is counteracted, not only by those above and below it, but also by those on each side. It is only necessary that the constituent pieces should have a common form, and that this form should be somewhat like the frustum of a pyramid, so that, when placed in its situation, its four angles may point toward the centre, or axis, of the dome. During the erection of a dome, it is not necessary that it should be supported by a centring, until complete, as is done in the arch. Each circle of stones, when laid, is capable of supporting itself without aid from those above it. It follows that the dome may be left open at top, without a key-stone, and yet be perfectly secure in this respect, being the reverse of the arch. The dome of the Pantheon, at Rome, has been always open at top, and yet has stood unimpaired for nearly 2000 years. The upper circle of stones, though apparently the weakest, is nevertheless often made to support the additional weight of a lantern or tower above it. In several of the largest cathedrals, there are two domes, one within the other, which contribute their joint support to the lan- tern, which rests upon the top. In these buildings, the dome AMERICAN HOUSE-CARPENTER. 127 rests upon a circular vail, which is supported, in its turn, by arches upon massive pillars or piers. This construction is called building upon pendentives, and gives open space and room for passage beneath the dome. The remarks which have been made in regard to the abutments of the arch, apply equally to the walls immediately supporting a dome. They must be of sufficient thickness and solidity to resist the lateral pressure of the dome, which is very great. The walls of the Roman Pantheon are of great depth and solidity. In order that a dome in itself should be perfectly secure, its lower parts must not be too nearly vertical, since, in this case, they partake of the nature of perpendicular walls, and are acted upon by the spreading force of the parts above them. The dome of St. Paul's church, in London, and some others of similar construction, are bound with chains or hoops of iron, to prevent them from spreading at bottom. Domes which are made of wood depend, in part, for their strength, on their in- ternal carpentry. The Halle du Bled, in Paris, had originally a wooden dome more than 200 feet in diameter, and only one foot in thickness. This has since been replaced by a dome of iion (See Art. 389.) 238. — The Roof is the most common and cheap method of covering buildings, to protect them from rain and other effects of the weather. It is sometimes fiat, but more frequently oblique, in its shape. The flat or platform-roof is the least advantageous for shedding rain, and is seldom used in northern countries. The pent roof, consisting of two oblique sides meeting at top, is the most common form. These roofs are made steepest in cold cli- mates, where they are liable to be loaded with snow. Where the four sides of the roof are all oblique, it is denominated a hipped roof, and where there are two portions to the roof, of different ob- liquity, it is a curb, or mansard roof. In modern times, roofs are made almost exclusively of wood, though frequently covered with incombustible materials. The internal structure or carpen- try of roofs is a subject of considerable mechanical contrivance 128 architecture:. The roof is supported by rafters, which abut on the walls on each side, like the extremities of an arch. If no other timbers existed, except the rafters, they would exert a strong lateral pres- sure on the walls, tending to separate and overthrow them. To counteract this lateral force, a tie-beam, as it is called, extends across, receiving the ends of the rafters, and protecting the wall from their horizontal thrust. To prevent the tie-beam from sagging, or bending downward with its own weight, a king- post is erected from this beam, to the upper angle of the rafters, serving to connect the whole, and to suspend the weight of the beam. This is called trussing. Queen-posts are sometimes added, parallel to the king-post, in large roofs ; also various other connecting timbers. In Gothic buildings, where the vaults do not admit of the use of a tie-beam, the rafters are prevented from spreading, as in an arch, by the strength of the buttresses. In comparing the lateral pressure of a high roof with that of a low one, the length of the tie-beam being the same, it will be seen that a high roof, from its containing most materials, may produce the greatest pressure, as far as weight is concerned. On the other hand, if the weight of both be equal, then the low roof will exert the greater pressure ; and this will increase in propor- tion to the distance of the point at which perpendiculars, drawn from the end of each rafter, would meet. In roofs, as well as in wooden domes and bridges, the materials are subjected to an in- ternal strain, to resist which, the cohesive strength of the material is relied on. On this account, beams should, when possible, be of one piece. Where this cannot be effected, two or more beams are connected together by splicing. Spliced beams are never so strong as whole ones, yet they may be made to approach the same strength, by affixing lateral pieces, or by making the ends overlay each other, and connecting them with bolts and straps of iron. The tendency to separate is also resisted, by letting the two pieces Into each c>ther by the process called scarfing. Mortices, in- AMERICAN HOUSE-CARPENTER. 129 tended to truss or suspend one piece by another, should be formed upon similar principles. Roofs in the United States, after being boarded, receive a se- condary covering of shingles. When intended to be incombustible, they are covered with slates orearthern tiles, or with sheets of lead, copper or tinned iron. Slates are preferable to tiles, being lighter, and absorbing less moisture. Metallic sheets are chiefly used for flat roofs, wooden domes, and curved and angular surfaces, which require a flexible material to cover them, or have not a sufficient pitch to shed the rain from slates or shingles. Various artificial compositions are occasionally used to cover roofs, the most com- mon of which are mixtures of tar with lime, and sometimes with sand and gravel." — Ency. Am. (See Art. 354.) 17 SECTION III.— MOULDINGS, CORNICES, &c. MOULDINGS. 239. — A moulding is so called, because of its being ot the same determinate shape ulong its whole length, as though the whole of it had been cast in the same mould or form. The regular mouldings, as found in remains of ancient architecture, are eight in number ; and are known by the following names : Fig. 140. ^] Annulet, band, cincture, fillet, listel or square. Fig. 141. J Astragal or bead. Fig. 142. L Fig. 143. Torus oi tore. Scotia, trochilus or mouth. Fig. 144. Ovolo, quarter-round or echinus. AMERICAN HOUSE-CARPENTER. 131 Cavetto, cove or hollow. Cymathrm, or cyma-recta. Ogee. Fig. 14T. Inverted cymatium, or cyma-reversa Some of the terms are derived thus : fillet, from the French word 7^, thread. Astragal, from astragalos, a bone of the heel — or the curvature of the heel. Bead, because this moulding, when properly carved, resembles a string of beads. Torus, or tore, the Greek for rope, which it resembles, when on the base of a column. Scotia, from shotia, darkness, because of the strong shadow which its depth produces, and which is increased by the projection of the torus above it. Ovolo, from ovum, an egg, which this member resembles, when carved, as in the Ionic capi- tal. Cavetto, from cavus, hollow. Cymatium, from kumaton a wave. 240. — Neither of these mouldings is peculiar to any one of the orders of architecture, but each one is common to all ; and al- though each has its appropriate use, yet it is by no means con- fined to any certain position in an assemblage of mouldings The use of the fillet is to bind the parts, as also that of the astra- gal and torus, which resemble ropes. The ovolo and cyma-re- versa are strong at their upper extremities, and are therefore used to support projecting parts above them. The cyma-recta and cavetto, being weak at their upper extremities, are not used as supporters, but are placed uppermost to cover and shelter the other parts. The scotia is introduced in the base of a column, to 132 MOULDINGS. CORNICES, &C. separate the upper and lower torus, and to produce a pleasing variety and relief. The form of the bead, and that of the torus, is the same ; the reasons for giving distinct names to them are, that the torus, in every order, is always considerably larger than the bead, and is placed among the base mouldings, whereas the bead is never placed there, but on the capital or entablature ; the torus, also, is seldom carved, whereas the bead is ; and while the torus among the Greeks is frequently elliptical in its form, the bead retains its circular shape. While the scotia is the reverse of the torus, the cavetto is the reverse of the ovolo, and the cyma- recta and cyma-reversa are combinations of the ovolo and cavetto. 241. — The curves of mouldings, in Roman architecture, were most generally composed of parts of circles ; while those of the Greeks were almost always elliptical, or of some one of the conic sections, but rarely circular, except in the case of the bead, which was always, among both Greeks and Romans, of the form of a semi-circle. Sections of the cone afford a greater variety ol forms than those of the sphere ; and perhaps this is one reason why the Grecian architecture so much excels the Roman. The quick turnings of the ovolo and cyma-reversa, in particular, when exposed to a bright sun, cause those narrow, well-defined streaks of light, which give life and splendour to the whole. 242. — A profile is an assemblage of essential parts and mould- ings. That profile produces the happiest effect which is com- posed of but few members, varied in form and size, and arranged so that the plane and the curved surfaces succeed each other al- ternately. 243. — To describe the Grecian torus and scotia. Join the extremities, a and b, {Fig. 148;) and from/, the given projection of the moulding, draw/ o, at right angles to the fillets ; from b, draw b h, at right angles to a b; bisect a bine; join / u^d c, and upon c, with the radius, c/ describe the arc, /A, cutting b h in h ; through c, draw d e, parallel with the fillets ; make dc and c e, each equal to b h ; then d e and a b will be conjugate diame- AMERICAN HOUSE-CARPENTER. 133 Fig. 148. ters of the required ellipse. To describe the curve by intersec- tion of lines, proceed as directed at Art. 118 and note ; by a trammel, see Art . 116 ; and to find the foci, in order to describe it with a string, see Art. 115. Fig. 149. d \ a Fig. 150. 244. — Fig. 149 to 156 exhibit various modifications of the Grecian ovolo, sometimes called echinus. Fig. 149 to 153 are IBi MOULDINGS, CORNICES, &C. [^ d w I Fig. 151. Fig. 152. ,•••"■•■■■ ^4 V > Fig. 158. Fig. 154 c J Fig. 155. Fig. 156. elliptical, a b and 6 c being given tangents to the curve ; parallel to which, the semi-conjugate diameters, a d and d c, are drawn. In Fig. 149 and 150, the lines, a d and d c, are semi-axes, the tangents, a b and 6 c, being at right angles to each other. To draw the curve, see Art. 118. In Fig. 153, the curve is para- bolical, and is drawn according to Art. 127. In Fig. 155 and 156, the curve is hyperbolical, being described according to Art. 128. The length of the transverse axis, a b, being taken at pleasure in order to flatten the curve, a b should be made short in propor« tion to a c. AMERICAN HOUSE-CARPENTER. m Fig. 58 Fig. 16T. 245. — To describe the Grecian cavetto, {Fig. 157 and 158,) having the height and projection given, see Art. 118. a Fig. 159. Fig. 160. 246. — To describe the Grecian cyma-recta. When the pro- jection is more than the height, as at Fig. 159, make a b equal to the height, and divide abed into 4 equal parallelograms ; then proceed as directed in note to Art. 118. When the projec- tion is less than the height, draw d a, {Fig. 160,) at right angles to a b ; complete the rectangle, abed; divide this into 4 equal rectangles, and proceed according to Art. 118. 1 ____ ^5 &t Y>^r ^ ^ p^* Fig. 161. d Fig. 162. 247. To describe the Grecian cyma-reversa. When the 136 MOULDINGS, CORNICES, &C. projection is more than the height, as at Fig. 161, proceed as di rected for the last figure ; the curve being the same as that, the position only being changed. When the projection is less than the height, draw a d, (Fig. 162,) at right angles to the fillet : make a d equal to the projection of the moulding : then proceed as directed for Fig. 159. 248. — Roman mouldings are composed of parts of circles, and have, therefore, less beauty of form than the Grecian. The bead and torus are of the form of the semi-circle, and the scotia, also, in some instances ; but the latter is often composed of two quad- rants, having different radii, as at Fig. 163 and 164, which re- semble the elliptical curve. The ovolo and cavetto are generally a quadrant, but often less. When they are less, as at Fig. 167, the centre is found thus : join the extremities, a and b, and bisect a b in c ; from c, and at right angles to a b, draw c d, cutting a level line drawn from a in d ; then d will be the centre. This moulding projects less than its height. When the projection is more than the height, as at Fig. 169, extend the line from c until Fig. 163. Fig. 164. Fig. 165. Fig. 166. AMERICAN HOTJSE-CARPENTER. 137 Fig. 167. a Fig. 169. Fig. 1T0. Fig. 17L Fig. 172. Fig. 178. Fig. 174 18 133 MOULDINGS, CORNICES, &C, \ 1 \ Fig 175. Fig. 176. Fig. 177. it cuts a perpendicular drawn from a, as at d / and that will be the centre of the curve. In a similar manner, the centres are found for the mouldings at Fig. 164, 168, 170, 173, 174, 175, and 176. The centres for the curves at Fig. I'll and 178, are found thus : bisect the line, a b, at c / upon a, c and 5, suc- cessively, with a o or c b for radius, describe arcs intersecting at d and d / then those intersections will be the centres. 249. — Fig. 179 to 186 represent mouldings of modern inven- tion. They have been quite extensively and successfully used in inside finishing. Fig. 179 is appropriate for a bed-moulding under a low projecting shelf, and is frequently used under man- tle-shelves. The tangent, i h, is found thus : bisect the line, a b, at o, and b c at d; from d, draw d e, at right angles to e 5 / from b, draw bf, parallel to e d; upon b, with bd for radius, describe the arc, df ; divide this arc into 7 equal parts, and set one of the parts from s, the limit of the projection, to o / make o h equal to o e ; from h, through c, draw the tangent, hi ; divide bh,ho,ci and i a, each into a like number of equal parts ; and draw the in- AMERICAN HOUSE-CARPENTER. 139 Fig. 178. Kg. 181- 140 MOULDINGS, CORNICES, &C. Fig. 182. Fig. 1S3. Fig 184. Fig. 185. Fig. 186. tersecting lines as directed at Art. 89. If a bolder form is desired, draw the tangent, * h, nearer horizontal, and describe an elliptic curve as shown in Fig. 148 and 181. Fig. 180 is much used on base, or skirting of rooms, and in deep panelling. The curve is found in the same manner as that of Fig. 179. In this case, however, where the moulding has so little projection AMERICAN HOUSE-CARPENTER. 14? in comparison with its height, the point, e, being found as in the last figure, h s may be made equal to s e, instead of o s as in toe last figure. Fig. 181 is appropriate for a crown moulding of a cornice. In this figure the height and projection are given ; the direction of the diameter, a b, drawn through the middle cf the diagonal, e /, is taken at pleasure ; and d c is parallel to a e. To find the length of d c, draw b h, at right angles to a b ; upon o, with o / for radius, describe the arc,/, h, cutting b h in h ; then make o c and o d, each equal to b h* To draw the curve, see note to Art. 118. Fig. 182 to 186 are peculiarly distinct from ancient mouldings, being composed principally of straight lines ; the few curves they possess are quite short and quick H. P. H. P. ! 5 15 4 m ~J S 11 1 9 10* 10 Fig. 187. 15 14} Hi 17 Fig. 188. 250. -Fig. 187 and 188 are designs for antas caps. The * The manner of ascertaining the length of the conjugate diameter, d c, in this figure, and also in Itg. 148, 198 and 199 is new, and is important in this application. It !■ founded upon well-known mathematical principles, viz = All the parallelograms that may be circumscribed about an ellipsis are equal to one another, and consequently any on. is equal to the rectangle of the two axes. And again : the sum of the sq-iares of every pair of conjugate diameters is equal to the sum of the squares of the two azes. 142 AMEEICAK HOTJSE-CAItPENTEK. diameter of the ante is divided into 20 equal parts, and the height and projection of the members, are regulated in accord- ance with those parts, as denoted under if and P, height and projection. The projection is measured from the middle of the antse. These will be found appropriate for porticos, door- ways, mantel-pieces, door and window trimmings, &c. The height of the ante for mantel-pieces, should be from 5 to 6 diameters, having an entablature of from 2 to 2J diameters. This is a good proportion, it being similar to the Doric order. But for a portico these proportions are much too heavy ; an antse, 15 diameters high, and an entablature of 3 diameters, will have a better appearance. CORNICES. 251. — Fig. 189 to 197 are designs for eave cornices, and Fig. 198 and 199 are for stucco cornices for the inside finish of rooms. In some of these the projection of the uppermost member from the facia, is divided into twenty equal parts, 7 ^3 MOULDINGS. CORNICES. &0. 143 and the various members are proportioned according to thoao parts, as figured under iTand P. Fig. 190. Fig. Ml. 344 AMERICAN HOUSE-CARPENTER. a 7 ■y r ^^ w Fig. 193. Kg. 193. MOTJLDINGS, COENIOES, &0. 145 Fig. 194. H. P. I 16; S* Fig. 195. 19 U6 AMERICAN HOUSE-CARPENTER. H. P. H20 a i* 2i n Fig. 196. H.P . 33 20 11' li 1\. J^OvvCvvO Fig. 197. \\ MOULDINGS, CORNICES, &C. 147 lOOOODOOOOO Fig. 198. Fig. 199. 143' AMERICAN HOUSE-CARPENTER. d b 12 3 4c Fig. 200. 252. — To proportion an eave cornice in accordance with the height of the building. Draw the line, a c, (Fig. 200,) and make b c and b a, each equal to 36 inches ; from b, draw b d, at right angles to a c, and equal in length to f of a c ; bisect b d in e, and from a, through e, draw a f; upon a, with a c for radius, describe the arc, cf, and upon e, with e/for radius, describe the arc,/rf; divide the curve, df c, into 7 equal parts, as at 10, 20, 30, &c, and from these points of division, draw lines to b c, pa- rallel to d b ; then the distance, b 1, is the projection of a cornice for a building 10 feet high ; b 2, the projection at 20 feet high ; b 3, the projection at 30 feet, &c. If the projection of a cornice for a building 34 feet high, is required, divide the arc between 30 and 40 into 10 equal parts, and from the fourth point from 30, draw a line to the base, b c, parallel with b d ; then the distance of the point, at which that line cuts the base, from b, will be the projec- tion required. So proceed for a cornice of any height within 70 feet. The above is based on the supposition that 36 inches is the proper projection for a cornice 70 feet high. This, for general purposes, will be found correct ; still, the length of the line, b c, maybe varied to suit the judgment of those who think differ- ently. Having obtained the projection of a cornice, divide it into 20 equal parts, and apportion the several members according to its destination — as is shown at Fig. 195, 196, and 197. MOULDINGS, CORNICES, &C. b 149 Fig. 201. 253. — To proportion a cornice according to a smaller given fine. Let the cornice at Fig. 201 be the given one. ' Upon any point in the lowest line of the lowest member, as at a, with the height of the required cornice for radius, describe an intersecting arc across the uppermost line, as at b ; join a and b : then b 1 will oethe perpendicular height of the upper fillet for the proposed cor- nice, 1 2 the height of the crown moulding — and so of all the members requiring to be enlarged to the sizes indicated on this line. For the projection of the proposed cornice, draw a d, at right angles to a b, and c d, at right angles to be; parallel with c d, draw lines from each projection of the given cornice to the line, ad; then e d will be the required projection for the proposed cornice, and the perpendicular lines falling upon e d will indicate the proper projection for the members. 254. — To proportion a cornice according to a larger given one. Let A, {Fig. 202,) be the given cornice. Extend a o to b, and draw c d, at right angles to a b ; extend the horizontal lines of the cornice, A, until they touch o d ; place the height of the proposed cornice from o to e, and join f and e ; upon o, with the projection of the given cornice, o a, for radius, describe the quad- rant, a d ; from d, draw d b, parallel tofe; upon o, with o b for radius, describe the quadrant, be; then o c will be the proper pro- jection for the proposed cornice. Join a and c ; draw lines from the 150 AMERICAN HOUSE-CARPENTER. Fig. 802. projection of the different members of the given cornice to a o ; parallel to o d ; from these divisions on the line, a o, draw lines to the line, o c, parallel to a c ; from the divisions on the line, of, draw lines to the line, o e, parallel to the line, f e ; then the di- visions on the lines, o e and o c, will indicate the proper height and projection for the different members of the proposed cornice. In this process, we have assumed the height, o e, of the proposed cornice to be given ; but if the projection, o c, alone be given, we can obtain the same result by a different process. Thus : upon o. with o c for radius, describe the quadrant, c b ; upon o, with o a for radius, describe the quadrant, ad ; join d and b ; from/, draw f e, parallel to d b ; then o e will be the proper height for the pro- posed cornice, and the height and projection of the different mem- bers can be obtained by the above directions. By this problem, a cornice can be proportioned according to a smaller given one as well as to a larger ; but the method described in the previous article is much more simple for that purpose. 255. — To find the angle-bracket for a cornice. Let A, {Fig. 203,) be the wall of the building, and B the given bracket, which, for the present purpose, is turned down horizontally. The angle- bracket, C, is obtained thus : through the extremity, a, and paral- MOULDINGS, CORNICES, &C. 151 A e I- h \|i7 o ./• B o o\ 1 i\ \o 2 2\\g /^ a 3\ Fig. 203. Fig. 204. lei with the wall,/tf, draw the line, a b ; make e c equal a J, and through c, draw c 6, parallel with e d; join d and b, and from the several angular points in B, draw ordinates to cut d b in 1, 2 and 3 ; at those points erect lines perpendicular to d b ; from h, draw A g, parallel to/ a ; take the ordinates, 1 o, 2 o, &c, at B, and transfer them to C, and the angle-bracket, C, will be defined. In the same manner, the angle-bracket for an internal cornice, or the angle-rib of a coved ceiling, or of groins, as at Fig. 204, can be found. 256. — A level crown moulding being given, to find the raking moulding and a level- return at the top. Let A, {Fig. 205,) be the given moulding, and A b the rake of the roof. Divide the curve of the given moulding into any number of parts, equal or unequal, as at 1, 2, and 3 ; from these points, draw horizontal lines to a perpendicular erected from c; at any convenient place on the rake, as at B, draw a c, at right angles to A b ; also, from b, draw the horizontal line, b a; place the thickness, d a, of the moulding at A, from b to a, and from a, draw the perpendicular line, a e ; from the points, 1, 2, 3, at A, draw lines to C, parallel to A b ; make a 1, a 2 and a 3, at B and at C, equal to a 1, &c, at A ; through the points, 1, 2 and 3, at B, trace the curve — this will be the proper form for the raking moulding. From 1, 2 and ioi AMERICAN HOUSE-CARPENTER. Fig 205. 3, at C, drop perpendiculars to the corresponding ordinates from 1,2 and 3, at ^./through the poirts of intersection, trace the curve — this will be the proper, form ior the return at the top. SECTION IV.— FRAMING. 257. — This subject is, to the carpenter, of the highest impor- tance ; and deserves more attention and a larger place in a volume oi this kind, than is generally allotted to it. Something, indeed, has been said upon the geometrical principles, by which the seve- ral lines for the joints and the lengths of timber, may be ascer- tained ; yet, besides this, there is much to be learned. For how- ever precise or workmanlike the joints may be made, what will it avail, should the system of framing, from an erroneous position of its timbers, &c, change its form, or become incapable of sus- taining even its own weight? Hence the necessity for a know- ledge of the laws of pressure and the strength of timber. These being once understood, we can with confidence determine the best position and dimensions for the several timbers which compose a floor or a roof, a partition or a bridge. As systems of framing are more or less exposed to heavy weights and strains, and; in case of failure, cause not only a loss of labour and material, but frequently that of life itself, it is very important that the materials employed be of the proper quantity and quality to serve their des tination. And, on the other hand, any superfluous material is not only useless, but a positive injury, it being an unnecessaiy load upon the points of support. It is necessary, therefore, to know 20 154 AMERICAN HOUSE-CARPENTER. the least quantity of timber that will suffice for strength. Tho greatest fault in framing is that of using an excess of material. Economy, at least, would seem to require that this evil be abated. Before proceeding to considei the principles upon which a sys- tem of framing should be constructed, let us attend to a few of the elementary laws in Mechanics, which will be found to be of great value in determining those principles. 258. — Laws of Pressure. (1.) A heavy body always exerts a pressure equal to its own weight in a vertical direction. Example: Suppose an iron ball, weighing 100 lbs., be supported upon the top of a perpendicular post, {Fig. 220 ;) then the pressure exerted upon that post will be equal to the weight of the uall; viz., 100 lbs. (2.) But if two inclined posts, {Fig. 206,) be substituted for the perpendicular support, the united pressures upon these posts will be more than equal to the weight, and will be in proportion to their position. The farther apart their feet are spread the greater will be the pressure, and vice versa. Hence tremendous strains may be exerted by a comparatively small weight. And it follows, therefore, that a piece of timber intend- ed for a strut or post, should be so placed that its axis may coin- cide, as near as possible, with the direction of the pressure. The direction of the pressure of the weight, W, {Fig. 206,) is in the vertical line, b d ; and the weight, W, would fall in that line, if the two posts were removed, hence the best position for a support w Fig. 206. FRAMING. 156 for the weight would be in that line. But, as it rarely occurs in systems of framing that weights can be supported by any single resistance, they requiring generally two or more sup- ports, (as in the case of a roof supported by its rafters,) it be- comes important, therefore, to know the exact amount of pres- sure any certain weight is capable of exerting upon oblique supports. Now it has been ascertained that the three lines of a triangle, drawn parallel with the direction of three concur- ring forces in equilibrium, are in proportion respectively to these forces. For example, in Fig. 206, we have a represen- tation of three forces concurring in a point, which forces are in equilibrium and at rest ; thus, the weight, W, is one force, and the resistance exerted by the two pieces of timber are the other two forces. The direction in which the first force acts is vertical — downwards ; the direction of the two other forces is in the axis of each piece of timber respectively. These three forces all tend towards the point, b. Draw the axes, a b and b o, of the two supports ; make b d vertical, and from d draw d e and d f parallel with the axes, b g and b a, respectively. Then the triangle, b d e, has its lines parallel respectively with the direction of the three forces ; thus, b d is in the direction of the weight, W, d e parallel with the axis of the timber b c, and e b is in the direction of the timber a b. In accordance with the principle above stated, the lengths of the sides of the triangle, b d e, are in proportion respectively to the three forces aforesaid ; thus — As the length of the line, b d, Is to the number of pounds in the weight, W, So is the length of the line, b e, To the number of pounds' pressure resisted by the timber, a b. Again — As the length of the line, b d, Is to the number of pounds in the weight, W, 156 AMERICAN HOUSE-CARPENTEB. So is the length of the line, d e, To the number of pounds' pressure resisted by the timber, be. And again — As the length of the line, b e, Is to the pounds' pressure resisted by a b, So is the length of the line, d e, To the pounds' pressure resisted by b c. These proportions are more briefly stated thus — 1st. bd : W::be:P, P being used as a symbol to represent the number of r ounds' pressure resisted by the timber, a b. 2nd. bd : W :: de: Q, Q representing the number of pounds' pressure resisted by the timber, b c. 3d b e: P :: de: Q. 259. — This relation between lines and pressures is important, and is of extensive application in ascertaining the pressures induced by known weights throughout any system of framing. The parallelogram, b e df, is called the Parallelogram of Forces ; the two lines, b e and b f, being called the compo- nents, and the line b d the resultant. "Where it is required to find the components from a given resultant, {Fig. 206,) it is not needed to draw the fourth line, df, for the triangle, b d e, gives the desired result. But when the resultant is to be ascertained from given components, {Fig. 212,) it is more con- venient to draw the fourth line. 260r — The Resolution of Forces is the finding of two or more forces, which, acting in different directions, shall exactly balance the pressure of any given single force. To make a practical application of this, let it be required to ascertain the oblique pressure in Fig. 206. In this Fig. the line b d measures half an inch, (0-5 inch,) and the line b e three- tenths of an inch, (0'3 inch.) Now if the weight, W, be sup- FRAMING. 157 posed to be 1200 pounds, then the first stated proportion above, bd : Wv.be'. P, becomes 0-5 : 1200 :: 0-3 : P: And since the product of the means divided by one of the extremes gives the other extreme, this proportion may be put in the form of an equation, thus — 1200 X 0-3 0-5 = P. Performing the arithmetical operation here indicated, that is, multiplying together the two quantities above the line, and dividing, the product by the quantity under the line, the quo- tient will be equal to the quantity represented by P, viz., the pressure resisted by the timber, a b. Thus — 1200 0-3 0-5')360-0 720 = P. The strain upon the timber, a b, is, therefore, equal to 720 pounds ; and the strain upon the other timber, b o , is also 720 pounds ; for in this case, the two timbers being inclined equally from the vertical, the line e d is therefore equal to the line b 9. Fig. 207. 158 AMERICAN HOUSE-CAEPENTER. 261.— In Fig. 207, the two supports are inclined at different angles, and the pressures are proportionately unequal. The supports are also unequal in length. The length of the sup- ports does not alter the amount of pressure from the concen- trated load supported ; but generally long timbers are not so capable of resistance as shorter ones. They yield more readily laterally, as they are not so stiff, and shorten more, as the com- pression is in proportion to the length. To ascertain the pres- sures in Fig. 207, let the weight suspended from b d be equal to two and three-quarter tons, (2-75 tons.) The line b d mea- sures five and a half tenths of an inch, (0 - 55 inch,) and the line i e half an inch, (0 - 5 inch.) Therefore, the proportion id : W::be:P, becomes 0-55 : 2-75 :: 0-5 : F, , 2-75 X 0-5 ry and — -— — = P. - 5o 2-75 0-5 0-55)1 -375(2-5 = P. 110 275 275 The strain upon the timber, 5 e, is, therefore, equal to two and a half tons. Again, the line e d measures four-tenths of an inch, (0-4 inch ;) therefore, the proportion Id : W:: ed: Q, becomes 0-55 : 2-75 :: 0-4 : Q, , 2-75 x 04 _ aDd — (R5— = Q - 2-75 0-4 0-55)1-100(2 = Q. 110 FRAMING. 155 The strain upon the timber, if, is, therefore, equal to two ions. 262. — Thus it is seen that the united pressures exerted by a weight upon two inclined supports always exceed the weight. In the last case 21 tons exerts a pressure of 2£ and two tons, equal together to 4| tons; and in the former case, 1200 pounds exerts a pressure of twice 720 pounds, equal to 1440 pounds. The smaller the angle of inclination to the horizon- tal, the greater will be the pressure upon the supports. So, in the frame of a roof, the strain upon the rafters decreases gra- dually with the increase of the angle of inclination to the horizon, the length of the rafter remaining the same. 263. — This is true in comparing systems of framing with each other; but in a system where the concentrated weight to be supported is not in the middle, (see Fig. 207,) and, in consequence, the supports are not inclined equally, the strain will be greatest upon the support that has the greatest inclina- i'nn to the horizon. 364. — In ordinary cases, in roofs for example, the load ia not concentrated but is that of the framing itself. Here the amount of the load will be in proportion to the length of the rafter, and the rafter increases in length with the increase of the angle of inclination, the span remaining the same. So it is seen that in enlarging the angle of inclination to the horizon in order to lessen the oblique thrust, the load is increased in consequence of the elongation of the rafter, thus increasing the oblique thrust. Hence there is a limit to the angle of inclina- tion. A rafter will have the least oblique thrust when its angle of inclination to the horizon is 35° 16' nearly. This angle is attained very nearly when the rafter rises 8£ inches per foot ; or, when the height, B C, {Fig. 216,) is to the base, A C, as 8j is to 12, or as 0-7071 is to 1-0. 265. — Correct ideas of the comparative pressures exerted upon timbers, according to their position, will be readily 160 AMERICAN HOUSE-CAEPENTEE. formed by drawing various designs of framing, and estimating the several strains in accordance with the parallelogram of forces, always drawing the triangle, b d e, so that the three lines shall be parallel with the three forces, or pressures; re- spectively. The length of the lines forming this triangle is unimportant, but it will be found more convenient if the line drawn parallel with the known force is made to contain as many inches as the known force contains pounds, or as many tenths of an inch as pounds, or as many inches as tons, or tenths of an inch as tons : or, in general, as many divisions of any convenient scale as there are units of weight or pressure in the known force. If drawn in this manner, then the num- ber of divisions of the same scale found in the other two lines of the triangle will equal the units of pressure or weight of the other two forces respectively, and the pressures sought will be ascertained simply by applying the scale to the lines of the triangle. For example, in Fig. 207, the vertical line, b d, of the tri- angle, measures fifty-five hundredths of an inch, (0 - 55 inch ;) the line, b e, fifty-hundred ths, (0"50 inch ;) and the line, e d, forty, (0 - 40 inch.) Now, if it be supposed that the vertical pres- sure, or the weight suspended below b d, is equal to 55 pounds, then the pressure on b e will equal 50 pounds, and that on e d will equal 40 pounds ; for, by the proportion above stated, bd : W::be:P, 55 : 55 :: 50: 50; and so of the other pressure. 266. — If a scale cannot be had of equal proportions with the forces, the arithmetical process will be shortened somewhat by making the line of the triangle that represents the "known weight equal to unity of a decimally divided scale, then the other lines will be measured in tenths or hundredths ; and in the numerical statement of the proportions between the lines and forces, the first term being unity, the fourth term will bo FEAMING. 101 ascertained simply by multiplying the second and third terms together. For example, if the three lines are 1, 0-7 and 1-3, and the known weight is 6 tons, then ' id : W::le:P, becomes 1 : 6 :: 0-7 : P = 4-2, equals four and two-tenths tons. Again id : W:: ed: Q, becomes 1 : 6 ;: 1-3 : Q = 7-8, equals seven and eight-tenths tons. Fig. 208. 267. — In Fig. 208 the weight, W, exerts a pressure on the struts in the direction of their length ; their feet, n n, have, therefore, a tendency to move in the direction n o, and would so move, were they not opposed by a sufficient resistance from: the blocks, A and A. If a piece of each block be cut off at the horizontal line, an, the feet of the struts would slide away from each other along that line, in the direction, n a ; but if,, instead of these, two pieces were cut off at the vertical line, n i, then the struts would descend vertically. To estimate the' horizontal and the vertical pressures exerted by the struts, let n ohe made equal (upon any scale of equal parts) to the num- 21 162 AMERICAN HOUSE-CARPENTER. ber of tons with which the strut is pressed ; construct the parallelogram of forces by drawing o e parallel to a n, and of parallel to h n ', then nf, (by the same scale,) shows the num- ber of tons pressure that is exerted by the strut in the direc- tion n a, and n e shows the amount exerted in the direction n i. By constructing designs similar to this, giving various and dissimilar positions to the struts, and then estimating the pressures, it will be found in every case that the horizontal pressure of one strut is exactly equal to that of the other, how- ever much one strut may be inclined more than the other ; and also, that the united vertical pressure of the two struts is exactly equal to the weight, W. (In this calculation the weight of the timbers has not been taken into consideration, simply to avoid complication to the learner. In practice it is requisite to include the weight of the framing with the load upon the framing.) Fig. 209. 268.— Suppose that the two struts, B and B, {Fig. 208.) were rafters of a roof, and that instead of the blocks, A and A, the walls of a building were the supports: then, to prevent the walls from being thrown over by the "thrust of B and B, it would be desirable to remove the horizontal pressure. This FRAMING, 163 may be done by uniting the feet of the rafters with a rope, iron rod, or piece of timber, as in Fig. 209. This figure is similar to the truss of a roof. The horizontal strains on the tie-beam, tending to pull it asunder in the direction of its length, may be measured at the foot of the rafter, as was shown at Fig. 208 ; but it can be more readily and as accu- rately measured, by drawing from f and e horizontal lines to the vertical line, i d, meeting it in o and o / then f o will be the horizontal thrust at B, and e o at A ; these will be found to equal one another. When the rafters of a roof are thus connected, all tendency to thrust the walls horizontally is removed, the only pressure on them is in a vertical direction, being equal to the weight of the roof and whatever it has to support. This pressure is beneficial rather than otherwise, as a roof having trusses thus formed, and the trusses well braced to each other, tends to steady the walls. Fig. 211. 269. — Fig. 210 and 211 exhibit methods of framing for sup- porting the equal weights, IF and W. Suppose it be required 164 AMERICAN HOUSE-CAEPENTER. to measure and compare the strains produced on the pieces, A B and A 0. Construct the parallelogram of forces, 6 bfd, according to Art. 258. Then If will show the strain on A B, and h e the strain on A 0. By comparing the figures, b d be- ing equal in each, it will be seen that the strains in Fig. 210 are about three times as great as those in Fig. 211 : the posi- tion of the pieces, A B and A 0, in Fig. 211, is therefore far preferable. C Fig. 212. 270. — The Composition of Forces consists in ascertaining the direction and amount of one force, which shall be just capable of balancing two or more given forces, acting in different directions. This is only the reverse of the resolution of forces, and the two are founded on one and the same principle, and may be solved in the same manner. For example, let A and B, {Fig. 212,) be two pieces of timber, pressed in the direction of their length towards b — A by a force equal to 6 tons weight, and B equal to 9. To find the direction and amount of pres- sure they would unitedly exert, draw the lines, o e and of in a line with the axes of the timbers, and make b e equal to the pressure exerted by B, viz., 9 ; also make b f equal to the pressure on A, viz., 6, and complete the parallelogram of forces, e hfd; then b d, the diagonal of the parallelogram, will be the direction, and its length, 9 - 25, will be the amount^ FRAMING. 165 of the united pressures of A and of B. The line, b d, is termed the resultant of the two forces, bf and I s. If J. and B are to be supported by one post, G, the best position for that post will be in the direction of the diagonal, b d ; and it will require to be sufficiently strong to support the united pressures of A and of B, which are equal to 9 '25 or 9J tons. Fig. 21?. 271. — Another example : let Fig. 213 represent a piece of framing commonly called a crane, which is used for hoisting heavy weights by means of the rope, B bf which passes over a pulley at b. This is similar to Fig. 210 and 211, yet it is materially different. In those figures, the strain is in one direction only, viz., from b to d ; but in this there are two strains, from A to B and from A to W. The strain in the direction A B is evidently equal to that in the direction A W. To ascertain the best position for the strut, A 0, make b e equal to bf, and complete the parallelogram of forces, e bf d ; then draw the diagonal, b d, and it will be the position re- quired. Should the foot, 0, of the strut be placed either higher or lower, the strain on A would be increased. In constructing cranes, it is advisable, in order that the piece, B A, may be under a gentle pressure, to place the foot of the 166 AMERICAN HOUSE-CARPENTER. strut a trifle lower than where the diagonal, i d, would indi- cate, but never higher. 7\ *"a/UM, Fig. 214. L vwV 272.— Ties and Struts. Timbers in a state of tension are called ties, while such as are in a state of compression are termed struts. This subject can be illustrated in the following manner : Let A and B, {Fig. 214,) represent beams of timber support- ing the weights, W, W and W; A having but one support, which is in the middle of its length, and B two, one at each end. To show the nature of the strains, let each beam be sawed in the middle from a to 5. The effects are obvious : the cut in the beam, A, will open, whereas that in B will close. If the weights are heavy enough, the beam, A, will break at 5y while the cut in B will be closed perfectly tight at a, and the beam be very little injured by it. But if, on the other hand, the cuts be made in the bottom edge of the tim- bers, from o to b, B will be seriously injured, while A will scarcely be affected. By this it appears evident that, in a piece of timber subject to a pressure across the direction of its length, the fibres are exposed to contrary strains. If the tim- ber is supported at both ends, as at B, those from the top edge down to the middle are" compressed in the direction of their length, while those from the middle to the bottom edge are in a state of tension ; but if the beam is supported as at A, the contrary effect is produced ; while the fibres at the middle of either beam are not at all strained. The strains in a framed FRAMING. 167 truss are of the same nature as those in a single beam. The truss for a roof, being supported at each end. has its tie-beam in a state of tension, while its rafters are compressed in the direction of their length. By this, it appears highly important that pieces in a state of tension should be distinguished from such as are compressed, in order that the former may be pre- served continuous. A strut may be constructed of two or more pieces ; yet, where there are many joints, it will not resist compression so well. 273. — To distinguish ties from struts. This may be done by the following rule. In Fig. 206, the timbers, a b and b c, are the sustaining forces, and the weight, W, is the straining force; and, if the support be removed, the straining force would move from the point of support, 5, towards d. Let it be required to ascertain whether the sustaining forces are si/retched or pressed by the straining force. Rule : upon the direction of the straining force, b d, as a diagonal, construct a parallelo- gram, e bf d, whose sides shall be parallel with the direction of the sustaining forces, a b and o d; -through the point, b, draw a line, parallel to the diagonal, efj this may then be called the dividing line between ties and struts. Because all those supports which are on that side of the dividing line, which the straining force would occupy if unresisted, are com- pressed, while those on the other side of the dividing line are stretched. In Fig. 206, the supports are both compressed, being on that side of the dividing line which the straining force would occupy if unresisted. In Fig. 210 and 211, in which A B and A O are the sustaining forces, A G is compressed, whereas A B is in a state of tension ; A being on that side of the line, h i, which the straining force would occupy if unresisted, and A B on the opposite side. The place of the latter might be supplied by a chain or rope. In Fig. 209, the foot of the rafter at A is sustained by two forces, the wall and the tier 168 AMERICAN HOUSE-CAEPENTEE. beam, one perpendicular and the other horizontal : the direc- tion of the straining force is indicated by the line, b a. The dividing line, h i, ascertained by the rule, shows that the wall is pressed and the tie-beam stretched. Fig. 215. 274. — Another example : let E A B F, {Fig. 215,) represent a gate, supported by hinges at A and F. In this case, the straining force is the weight of the materials, and the direction of course vertical. Ascertain the dividing line at the several points, G, B, I, J, H and F. It will then appear that the force at G is sustained by A G and G F, and the dividing line shows that the former is stretched and the latter com- pressed. The force at iZis supported by A BTaxxd H F — the former stretched and the latter compressed. The force at B is opposed by H B and A B, one pressed, the other stretched. The force at F is sustained by G F and F F, G F being stretched and F F pressed. By this it appears that A B is in a state of tension, and F F, of compression ; also, that A B: and G F are stretched, while B H and G E are compressed : which shows the necessity of having A H and G F, each in one whole length, while B H and G E may be, as they are shown, each in two pieces. The force at J is sustained by G J and J II, the former stretched and the latter compressed. FRAMING. 169 The piece, G D, is neither stretched nor pressed, and coul 1 be dispensed with if the joinings at J and I could be made as effectually without it. In case A B should fail, then C D would be in a state of tension. 275. — The centre of gravity. The centre of gravity of a uniform prism or cylinder, is in its axis, at the middle of its length ; that of a triangle, is in a line drawn from one angle to the middle of the opposite side and at one-third of the length of the line from that side ; that of a right-angled triangle, at a ooint distant from the perpendicular equal to one-third of the base, and distant from the base equal to one-third of the per- pendicular ; that of a pyramid or cone, in the axis and at one- quarter of the height from the base. 276. — The centre of gravity of a trapezoid, (a four-sided figure having only two of its sides parallel,) is in a line joining the centres of the two parallel sides, and at a distance from the longest of the parallel sides equal to the product of the length into the sum of twice the shorter added to the longer of the parallel sides, divided by three times the sum of the two parallel sides. Algebraically thus — _ I (2 a + V) a ~ 3{a+,i) where d equals the distance from the longest of the parallel aides, I the length of the line joining the two parallel sides, and a the shorter and o the longer of the parallel sides. Example. — A rafter, 25 feet long, has the larger end 14 inches wide, and the smaller end 10 inches wide, how far from the larger end is the centre of gravity located ? Here, 1 = 26, a = {%, and b = \ f , , l(2a + h) 25 (2 x H + if) 25 x j \ hence d = -^j-— ^ - — (|| + H) »- S x H - 2±*Z± = 85° = = n feet . ncheg ne 3 x 24 72 In irregular bodies with plain sides, the centre of gravity 22 170 AMERICAN HOUSE-C&JRPENTER. may be found by balancing them upon the edge of a prism— upon the edge of a table — in two positions, making a line each time upon the body in a line with the edge of the prism, and the intersection of those lines will indicate the point required. Or suspend the article by a cord or thread attached to one corner or edge ; also, from the same point of suspension, hang a plumb-line, and mark its position on the face of the article; again, suspend the article from another corner or side, (nearly at right angles to its former position,) and mark the position of the plumb-line upon its face ; then the intersection of the two lines will be the centre of gravity. Fig. 216. 277. — The effect of the weight of inclined beams. An in- clined post or strut, supporting some heavy pressure applied at its upper end, as at Fig. 209, exerts a pressure at its foot in the direction of its length, or nearly so. But when such a beam is loaded uniformly over its whole length, as the rafter of a roof, the pressure at its foot varies considerably from the direction of its length. For example, let A B, {Fig. 216,) be a beam leaning against the wall, B c, and supported at its foot by the abutment, A, in the beam, A c, and let o be the centre of gravity of the beam. Through o, draw the vertical line, b d, and from B, draw the horizontal line, B b, cutting b d in b ; join b and A, and b A will be the direction of the thrust. To prevent the beam from loosing its footing, the joint at A should be made at right angles to b A. The amount of pressure will be found thus : let b d, (by any scale of equal FRAMING. 171 jaits,) equal the number of tons upon the beam, A B y draw d e, parallel to B i / then b e, (by the same scale,) equals the pressure in the direction, b A / and e d, the pressure against the wall at B — and also the horizontal thrust at J., as these are always equal in a construction of this kind. 278. — The horizontal thrust of an inclined beam, {Fig. 216,) — the effect of its own weight — may be calculated thus : Bule. — Multiply the weight of the beam in pounds by its base, A C, in feet, and by the distance in feet of its centre of gravity, o, (see Art. 275 and 276,) from the lower end, at A / and divide this product by the product of the length, A B, into the height, B C, and the quotient will be the horizontal thrust in pounds. This may be stated thus : H — —rj-, where d equals the distance of the centre of gravity, o, from the lower end ; b equals the base, A O ; w equals the weight of the beam ; h equals the height, B C; I equals the length of the beam ; and i? equals the horizontal thrust. Example. — A beam, 20 feet long, weighs 300 pounds; its centre of gravity is at 9 feet from its lower end; it is so inclined that its base is 16 feet and its height 12 feet ; what is the horizontal thrust ? rr dbw , 9x16 x 300 9x4x25 . . K Here — -— becomes — — = = 9x4x5 hi 12 X 20 5 = 180 = H = the horizontal thrust. This rule is for cases where the centre of gravity does not occur at the middle of the length of the beam, although it is applicable when it does occur at the middle; yet a shorter rule will suffice in this case, — and it is thus : — Bule. — Multiply the weight of the rafter in pounds by the base, A C, {Fig. 216,) in feet, and divide the product by twice the height, B O, in feet ; and the quotient will be the horizon tal thrust, when the cer tre of gravity occurs at the middle of the beam. 172 AMERICAN HOUSE-OABPENTEB. If the inclined beam is loaded with an equally distributed load, add this load to the weight of the beam, and use this total weight in the rule instead of the weight of the beam. And generally, if the centre of gravity of the combined weights of the beam and load does not occur at the centre of the length of the beam then the former rule is to be used. Fig. 217. 279. — In Fig. 217, two equal beams are supported at their feet by the abutments in the tie-beam. This case is similar to the last ; for it is obvious that each beam is in precisely the position of the beam in Fig. 216. The horizontal pressures at JB, being equal and opposite, balance one another ; and their horizontal thrusts at the tie-beam are also equal. (See Art. 268— Fig. 209.) When the height of a roof, (Fig. 217,) is one-fourth of the span, or of a shed, (Fig. 216,). is one-half the span, the horizontal thrust of a rafter, whose centre of gravity is at the middle of its length, is exactly equal to the weight distributed uniformly over its surface. FBAMING. 173 280. — In shed, or lean-to roofs, as Fig. 216, tie horizontal pressure will be entirely removed, if the bearings of the raft- ers, as A B, (Fig. 218,) are made horizontal — provided, how- ever, that the rafters and other framing do not bend between the points of support. If a beam or rafter have a natural curve, the convex or rounding edge should be laid uppermost. 281. — A beam laid horizontally, supported at each end and uniformly loaded, is subject to the greatest strain at the mid- dle of its length. Hence mortices, large knots and other de- lects, should be kept as far as possible from that point ; and, in resting a load upon a beam, as a partition upon a floor beam, the weight should be so adjusted, if possible, that it will bear at or near the ends. Twice the weight that will break a beam, acting at the centre of its length, is required to break it when equally dis- tributed over its length ; and precisely the same deflection oi sag will be produced on a beam by a load equally distributed, that five-eighths of the load will produce if acting at the centre of its length. 282. — "When a beam, supported at each end on horizontal bearings, (the beam itself being either horizontal or inclined,) has its load equally distributed, the amount of pressure caused by the load on each point of support is equal to one half the load ; and this is also the case, when the load is concentrated at the middle of the beam, or has its centre of gravity at the middle of the beam ; but, when the load is unequally distri- buted or concentrated, so that its centre of gravity occurs at some other point than the middle of the beam, then the amount of pressure caused by the load on one of the points of support is unequal to that on the other. The precise amount on each may be ascertained by the following rule. Rule. — Multiply the' weight w, (Fig. 219,) by its distance, CB, from its nearest point of support, JB, and divide the product by the length, A B, of the beam, and the quotient wiJl be the 174 AMERICAN HOUSE-CARPENTER. Pig. 210. amount of pressure on the remote point of support, A. Again, deduct this amount from the weight, w, and the remainder will be the amount of pressure on the near point of support, B ; or, multiply the weight, w, by its distance, A 0, from the remote point of support, A, and divide the product by the length, A B, and the quotient will be the amount of pressure on the near point of support, B. When I equals the length, A B ; a = A C; b — G B, and w = the load, then vj i j = A = the amount of pressure at A, and w a —j— = B — the amount of pressure at B. Example. — A beam, 20 feet long between the bearings, has a load of 100 pounds concentrated at 3 feet from one of the bearings, what is the portion of this weight sustained by each bearing ? Here w = 100 ; a, 17 ; I, 3 ; and I, 20. w I 100 x 3 Hence A =- I w a And^=-r-r= 20 100 x 17 = 15. •=85. I ~ 20 Load on A = 15 pounds. Load on B = 85 pounds. Total weight = 1( pounds. FRAMING. 1T5 RESISTANCE OF MATERIALS. 283. — Before a roof truss, or other piece c f framing, can be properly designed, two things are required to be known. The one is, the effect of gravity acting upon the various parts of the intended structure ; the other, the power of resistance possessed by the materials of which the framing is to be con- structed. In the preceding pages, the former subject having been treated of, it remains now to call attention to the latter. 284. — Materials used in construction are constituted in their structure either of fibres (threads) or of grains, and are termed, the former fibrous, the latter granular. A.11 woods and wrought metals are fibrous, while cast iron, stone, glass, &c, are gra- nular. The strength of a granular material lies in the power of attraction, acting among the grains of matter of which the material is composed, by which it resists any attempt to sepa- rate its grains or particles of matter. A fibre of wood or of wrought metal has a strength by which it resists being com- pressed or shortened, and finally crushed ; also a strength by which it resists being extended or made longer, and finally sundered. There is another kind of strength in a fibrous mate- rial ; it is the adhesion of one fibre to another along their sides, or the lateral adhesion of the fibres. 285. — In the strain applied to a piece of timber, as a post supporting a weight imposed upon it, {Fig. 220,) we have an instance of an attempt to shorten the fibres of which the tim- ber is composed. The strength of the timber in this case is termed the resistance to compression. In the strain on a piece of timber like a king-post or suspending piece, (J., Fig. 221,) we have an instance of an attempt to extend or lengthen the fibres of the material. The strength here exhibited is termed the resistance to tension. "When a piece of timber is strained like a floor beam, or any horizontal piece carrying a load, {Fig. 222,) we have an instance in which the two strains of 176 AMEKIOAN HOUSE-CAEPENTEK. Fig. 221. Fig. 222. compression and tension are brought into action ; the fibres of the upper portion of the beam being compressed, and those of the under part being stretched. This kind of strength of tim ber is termed resistance to cross strains. In each of these three kinds of strain to which timber is subjected, the power of resistance is in a measure due to the lateral adhesion of the fibres, not so much perhaps in the simple tensile strain, yet to a considerable degree in the compressive and cross strains. But the power of timber, by which it resists a pressure acting compressively in the direction of the length of the fibres, tend- ing to separate the timber by splitting off a part, as in the case of the end of a tie beam, against which the foot of the rafter presses — is wholly due to the lateral adhesion of the fibres. 286. — The strength of materials is that power by which they resist fracture, while the stiffness of materials is that quality which enables them to resist deflection or sagging. A know- ledge of their strength is useful, in order to determine theii FRAMING. 177 jiinits of size to sustain given weights safely ; but a knowledge of their stiffness is more important, as in almost all construc- tions it is desirable not only that the load be safely sustained, hut that no appearance of weakness be manifested by any sen- sible deflection or sagging. I. RESISTANCE TO COMPRESSION. 287. — The resistance of materials to the force of compression may be considered in four several ways, viz. : 1st. When the pressure is applied to the fibres longitudi nally, and on short pieces. 2d. When the pressure is applied to the fibres longitudi- nally, and on long pieces. 3d. When the pressure is applied to the fibres longitudi- nally, and so as to split off the part pressed against, causing the fibres to separate by sliding. 4th. When the pressure is applied to the fibres trans- versely. Posts having their height less than ten times their least side will crush before bending ; these belong to the first case : while posts, whose height is ten times their least side, or more than ten times, will bend before crushing ; these belong to the second case. 288. — In the above first and fourth cases of compression, experiment has shown that the resistance is in proportion to the number of fibres pressed, that is, in proportion to the area. For example, if 5,000 pounds is required to crush a prism with a base 1 inch square, it will require 20,000 pounds to crush a prism having a base of 2 by 2 inches, equal to 4 inches area ; because 4 times 5,000 equals 20,000. Experiment has also shown that, in the third case, the resistance is in proportion to the area of the surface separated without regard to the form of the surface. 289. — In the second case of compression, the resistance is in 23 178 AMERICAN HOTJSE-CABPENTER, proportion to the area of the cross section of the piece, multi plied by the square of its thickness, and inversely in propor- tion to the square of the length, multiplied by the weight. When the piece is square, it will bend and break in the direc- tion of its diagonal ; here, the resistance is in proportion to the square of the diagonal multiplied by the square of the dia- gonal, and inversely proportional to the square of the length multiplied by the weight. If the piece is round or cylindrical, its resistance will be in accordance with the square of the dia- meter multiplied by the square of the diameter, and inversely proportional to the square of the length, multiplied by the weight. 290. — These relations between the dimensions of the piece strained and its resistance, have resulted from the discussion of the subject by various authors, and rules based upon these relations are in general use, yet their accuracy is not fully established. Some experiments, especially those by Prof. Hodgkinson, have shown that the resistance is in proportion to a less power of the diameter, and inversely to a less power of the height ; yet the variance is not great, and inasmuch as the material is restricted in the rules to a strain decidedly within its limits of resistance, no serious error can be made in the use of rules based on the aforesaid relations. 291.- — Experiments. In the investigation of the laws appli- cable to the resistance of materials, only such of the relations of the parts have been considered as apply alike to wood and metal, stone and glass, or other material, leaving to experi- ment the task of ascertaining the compactness and cohesion of particles, and the tenacity and adhesion of fibres; those quali- ties upon which depend the superiority of one kind of material over another, and which is represented in the rules by a constant number, each specific kind of material having its own special constant, obtained by experimenting on specimens of that peculiar material. FRAMING. 179 292. — The following table exhibits the results of experiments on such woods as are in most common use in this country for the purpose of construction. The resistance of timber of the TABLE I. COMPRESSION. Kind of Material. White wood, Mahogany (Baywood), Ash, . Spruce, Chestnut, . White pine, Ohio pine, Oak, . Hemlock, . Black walnut, Maple, Cherry, White oak, Georgia pine, Locust, Live oak, . Mahogany (St. Domingo), Lignum vitas, Hickory, . •397 ■439 •517 •369 •491 •388 •586 •612 •423 •421 •574 •494 ■774 •613 •762 •916 •837 1-282 •877 ■9-3 Pounds per in. 2432 3527 4175 4199 4791 4806 4809 5316 5400 5594 6061 6477 6660 6767 7652 7936 8280 8650 9817 600 880 1040 1050 1200 1200 1200 1330 1350 1400 1515 1620 1665 1700 1910 1980 2070 2160 2450 Q^s fa Sli 5 3 © "Sis g-ag Pounds per in. 470 690 490 3S8 780 540 510 1180 & 160 230 160 130 260 180 170 400 Pounds per in. 600 1300 2300 600 950 600 1250 1900 600 1600 2050 1900 2000 1700 2100 6100 4300 5800 3100 8S •a .2 '*> o ffl 9 °« £ 300 650 1150 250 475 300 625 950 300 800 1025 950 1000 850 1050 2550 2150 2900 1550 same name varies much ; depending as it obviously must on the soil in which it grew, on its age before and after cutting, on the time of year when cut, and on the manner in which it has been kept since it was cut. And of wood from the same tree, much depends upon its location, whether at the butt or towards the limbs, and whether at the heart or at the sap, or at a 'point midway from the centre to the circumference of the tree. The pieces submitted to experiment were of ordinary good quality, such as would be deemed proper to be used in framing. The prisms crushed were 2 inches long, and from 1 inch to 1J inches square ; some were wider one way than the 180 AMERICAN HO0SE-CARPENTEK. other, but all containing in area of cross section from 1 to 2 inches. There were generally three specimens of each kind. The weight given in the table is the average crushing weight per superficial inch. In the preceding table the first column contains the specific gravity of the several kinds of wood, showing their compara- tive density. The weight in pounds of a cubic foot of any kind of wood or other material, is equal to its specific gravity multiplied by 62-5 ; this number being the weight in pounds of a cubic foot of water. The second column contains the weight in pounds required to crush a prism having a base of one . inch square ; the pressure applied to the fibres longitudi- nally. The third column contains the value of C in the rules ; C being equal to one-fourth of the crushing weight in the preceding column. The fourth column contains the weight in pounds, which, applied to the fibres longitudinally, is required to force off a part of the piece, causing the fibres to separate by sliding, the surface separated being one inch square. The fifth column contains the value of S in the rules, H being equal to one third of the weight in the preced- ing column. The sixth column contains the weight in pounds required to crush the piece when the pressure is applied to the fibres transversely, the piece being one inch thick, and the surface crushed being one inch square, and depressed one twentieth of an inch deep. The seventh column contains the value of P in the rules ; P being the weight in pounds applied to the fibres transversely, which is required to make a sensible impression one inch square on the side of the piece, this being the greatest weight that would be proper for a post to be loaded with per inch surface of bearing, resting on the side of the kind of wood set opposite in the table. A greater weight would, in proportion to the excess, crush the side of the wood under the post, and proportionably derange the framing, if not cause a total failure. It will be observed that the measure of FRAMING. 181 tnis resistance is useful in limiting the load on a post accord- ing to the kind of material contained, not in the jpost, but in the timber upon which the post presses. 293. — In Table II. are the results of experiments made tc test the resistance of materials to flexure: first, the flexure produced by compression, the force acting on the ends of the fibres longitudinally; secondly, the flexure arising from the effects of a cross strain, the force acting on the side of the fibves transversely, the beams lseing laid on chairs or rests. Of white oak, No. 1, there were eight specimens, of 2 by 4 inches, and 3£ feet long, seasoned more than a year after they were prepared for experiment. Of the other kinds of wood there were from three to five specimens of each, of I4 by 24 inches, and from H to 2£ feet long. Of the cast iron there were six specimens, of 1 inch square and 1 foot long; and of the wrought iron there were five specimens of American, three of f by 2 inches, and two of 1£ inches square, and three specimens of common English, 3 by 2 inches ; the eight speci- mens being each 19 inches long, clear bearing. In each case the result is the average of the stiffness of the several specimens. The numbers contained in the second column are the weights producing the first degree of flexure in a post or strut, where the post or strut is one foot long and one inch square; so, likewise, the numbers in the fifth column, and which are represented in the rules by 2?, are the weights required to deflect a beam one inch, where the beam is one foot long, clear bearing, and one inch square. — (See remarks upon this, Art. (321.) The numbers in the third column are equal to one-half of those in the second. The numbers con- tained in the fourth column, and represented by n in the rules, show the greatest rate of deflection that the material may be subjected to without injury. This rate multiplied by . the length in feet, equals the total deflection within the limits of elasticity. 382 AMERICAN HOUSE-CARPENTER. TABLE H. FLEXURE. Kind of Material. Specific Gravity. Under Compression. Pounds pro- ducing the first degree of flexure. Talue of B in the Rules. Under Cross Btiain. Valne of ■n in the Rules. Talue af JE ir the li idea Hemlock, Spruce, . White pine, Ohio yellow, pine, . Chestnut, . White oat, No. 1, . White oak, No. 2, . Georgia pine, . Locust, Cast iron, Wrought iron, common English Wrought iron, American, . 0-402 •432 •407 •586 •52 •82 •805 •755 •863 7-042 7-576 7-576 2640 4190 2350 6000 7720 6950 9660 10920 1320 2095 1175 3000 3860 3475 4830 5460 08794 0-09197 0-1022 0-049 007541 009152 0-0567 0-07723 0-06615 00148 0-03717 0-04038 1240 1550 1750 1970 2330 2520 2590 2970 3280 S0500 45500 51400 PRACTICAL KULES FOR COMPRESSION. First Case. 294. — To find the weight that can be safely sustained by a post, when the height of the post is less than ten times the diameter if round, or ten times the thickness if rectangular, and the direction of the pressure coinciding with the axis. Rule I. — Multiply the area of the cross-section of the post, in inches, by the value of C in Table I., the product will be the required weight in pounds. A G=w. (1.) Example. — A Georgia, pine post is 6 feet high, and in cross- section, 8 x 12 inches, what weight will it safely sustain ? The area = 8 x 12 = 96 inches ; this multiplied by 1700, the value of C, in the table, set opposite Georgia pine, the result, 163,200, is the weight in pounds required. It will be observed that the weight would be the same for a Georgia pine post of any height less than 10 times 8 inches = 80 inches = 6 feet 8 FRAMING. 183 inehes, provided its breadth and thickness remain the same, 12 and 8 inches. 295. — To find the area of the cross-section of a post to sus- tain a given weight safely, the height of the post being less than ten times the diameter if round, or ten times the least side if rectangular ; the pressure coinciding with the axis. Rule II. — Divide the given weight in pounds by the value of C, in Table I., and the product will be the required area in inches jf = A. (2.) Example. — A weight of 38,400 pounds is to be sustained by a white pine post 4 feet high, what must be its area of section in order to sustain the weight safely ? Here, 38,400 divided by 1200, the value of O, in Table I., set opposite white pine, gives a quotient of 32 ; this, therefore, is the required area, and such a post may be 5 x 6"4 inches. To find the least side, so that it shall not be less than one-tenth of the height, divide the height, reduced to inches, by 10, and make the least side to exceed this quotient. The area, divided by the least side so determined, will give the wide side. If, however, by this process, the first side found should prove to be the greatest, then the size of the post is to be found by Rule VII., VIII., or IX. 296. — If the post is to be round, by reference to the Table of Circles in the Appendix, the diameter will be found in the column of diameters, set opposite to the area of the post found in the column of areas, or opposite to the next nearest area. For example, suppose the required area, as just found by the example under Rule II., is 32 ; by reference to the column of areas, 33\L83 is the nearest to 32, and the diameter set opposite is 6'5. The post may, therefore, be 6£ inches diameter. Second Case. 297.- -To ascertain the weight that can be sustained safely 184 AMERICAN HOUSE-CARPENTER. by a post whose height is, at least, ten times its least side if rectangular, or ten times its diameter if round, the direction of the pressure coinciding with the axis. Rule III. — When l he post is round the weight may he found- by this rule : Multiply the square of the diameter in inches by the square of the diameter in inches, and multiply the product by 0*589 times the value of B, in Table II., divide this product by the square of the height in feet, and the quo- tient will be the required weight in pounds. 0-589 BD 2 D' 0589 B B* w = jf = — ^ (3.) Example. — What weight will a Georgia pine post sustain safely, whose diameter is 10 inches and height 10 feet ? The square of the diameter is 100 ; 100 x 100 = 10,000. And 10,000 by 0-589 times 4830, the value of B, Table II., set opposite Georgia pine, = 28,448,700, and this divided by 100, the square of the height, equals 284,487, the weight required, in pounds. Rule TV. — If the post be rectangular the. weight is found by this rule : Multiply the area of the cross-section of the post by the square of the thickness, both in inches, and by the value of B, Table II. Divide the product by the square of the height in feet, and the quotient will be the required weight in pounds. At B b t 3 B w = — h —=-Ji- (*■) Example. — What weight will a white pine post sustain safely, whose height is 12 feet, and sides 8 and 12 inches re- spectively ? The area = 8 x 12 = 96 inches ; the square of the thickness, 8, = 64. The area by the square of the thick- ness, 96 x 64, = 6144 ; and this by 1175, the value of B, for white pine, equals 7,219,200. This, divided by 144, the square of the he : ght = 50,133|, the required weight in pounds. FRAMING. IS. J Hule V. — If the post be square, the weight is found by this rule : Multiply the value of B, Table II., by the square of the area of the post in inches, and divide the product by the square of the height in feet, and the quotient will be the required weight in pounds. A 2 B D*B w =—r=-ir- < 5 -> Example. — What weight will a white oak post sustain safely, whose height is 9 feet, and sides each 6 inches ? The value of B, set opposite white oak, is 3475 ; this, by (36 x 36 =) 1296, the square of the area, equals 4,503,600. This pro- duct, divided by 81, the square of the height, gives for quo- tient, 55,600, the required weight in pounds. 298. — To ascertain the size of a post to sustain safely a given weight when the height of the post is at least ten times the least side or diameter. Hule VI. — When, the post is to be round or cylindrical, the size may be obtained by this rule : Divide the weight in pounds by 0-589 times the value of B, Table II., and extract the square root of the product ; multiply the square root by the height in feet, and the square root of this product will be the diameter of the post in inches. D=J ^\/-~ = (/^- w - (6.) V V -589.S V-5S9B K ' Example. — What must be the diameter of a locust post, 10 feet high, to sustain safely 40,000 pounds? Here 0'589 times 5,460, the value of B for locust, Table EL, equals 3215-9. The weight, 40,000, divided by 3215-9, equals 12-438. The square root of this, 3 - 5268, multiplied by 10, the height, equals 35-268, and the square root of this is 5*9386 or 5|f inches, the required diameter of the post. Hule VII. — If the post is to be rectangular, the size may be obtained by this rule : Multiply the square of the height in 24 186 AMERICAN HOt'SE-CAEPENTEE. feet by the weight in pounds, and divide the product by the value of B, Table II. Now, if the* breadth is known, divide the quotient by the breadth in inches, and the cube root of this quotient will be the thickness in inches. But if the thick- ness is known, and the breadth desired, divide, instead, by the cube of the thickness in inches, and the quotient will be the breadth in inches. - = A (T-i D-434/Z) (23.) and (21) becomes A = ™ (24.) T- 0-434/ 1, K ' and (22) becomes R = A, (25.) 313. — These rules stated in words at length are as fol- lows : — To ascertain the weight that may be suspended safely from & vertical beam, when the weight of the beam itself is to be taken into account, and when a portion of the fibres are cut in framing. Rule XVI. — From the sectional area of the beam, dednct FEAMING. 195 the sectional area of uncut fibres, and divide the remainder by the sectional area of the uncut fibres, and to the quotient add unity; multiply this sum by 0-434 times the specific gravity of the beam, and by its length in feet ; snbstract this product from the value of T, Table III., and the remainder, multiplied by the sectional area of the uncut fibres, will be the required weight in pounds. w = A (T- 0-434 (n + 1)/ I) (20.) Example. — A white pine beam, set vertically, 5x9 inches and 30 feet long, is so cut by mortices as to have remaining only 5x6 inches sectional area of uncut fibres : what weight will such a beam sustain safely, as a tensile strain ? The uncut fibres, 5x6 = 30, deducted from the area of the beam, 5x9 = 45, there remains 15. This remainder, divided by 30, the area of the uncut fibres, the quotient is - 5. This added to unity, the sum is 1-5. This, by - 434 times 0*458, the specific gravity set opposite white pine in Table III., and by 30, the length of the beam in feet, the product is 8-95. This product, deducted from 2,367, the value of T set opposite white pine in Table III., the remainder is 2,358 - 05. This remainder multi- plied by 30, the sectional area of the uncut fibres, the product, 70,741-5, is the required weight in pounds. 314. — When the beam is uncut for mortices or other pur- poses, the former part of the rule is not needed ; the weight will then be found by the following rule. Rule XVII. — Deduct 0*434 times the specific gravity of the beam, multiplied by its length in feet, from the value of T, Table III. ; the remainder, multiplied by the sectional area of the beam in inches, will be the required weight in pounds. w = A (T- 0-434/ 1). (23.) Example. — A Georgia pine beam, set vertically, is 25 feet long ami 7x9 inches in sectional area: what weight will it sustain safely, as a tensile strain ? By the rule, 0-434 times 196 AMEEICAN HOUSE-CAEPENTER. - 65, the specific gravity of Georgia pme, as in Table III., mul tiplied, by 25, the length in feet, the product is 7*05. This product, deducted from 2,833, the value of T, Table III., set opposite Georgia pine, and the remainder, 2,825 - 95, multiplied by 63, the sectional area, the product, 178,034-85, is the : . required weight in pounds. 315. — To ascertain the sectional area of a vertical beam that will safely sustain a given tensile strain, where the weight of the beam itself is to be considered. Rule XYIII. — "Where the beam is cut for mortices or other purposes, let the relative proportion of the uncut fibres to those, that are cut, be as 1 is to n, (n being a decimal to be fixed on at pleasure.) Then to the value of n add unity, and multiply ing the sum by 0-434 times the specific gravity in Table III., and by the length in feet. Deduct this product from the value of T, Table III., divide the given weight in pounds by. this remainder, and the quotient will be the area of the uncut fibres in inches. Add unity to the value of n, as above, and multiply the sum by the area of the uncut fibres ; the product will be the required area of the beam in inches. r-0-434(» + l)/J, K > R = A(n + l), (22.) ExaTwple. — A vertical beam of white oak, 30 feet long, is required to resist effectually a tensile strain of 80,000 pounds : what must be its sectional area ? The relative proportion of the uncut fibres is to be to those that are cut as 1 is to 04. To 0-4, the value of n, add 1- ; the sum is 1*4. This, by 0-434 times -774, the specific gravity of white oak in Table Hi., and by 30, the length, the product is 14-109. This, deducted from 2,833, the value of J 1 for white oak in Table III, the remainder is 2,818-891. The given weight, 80,000, divided by 2,818-891, the remainder, as above, the quotient, 28-38, is the area of the uncut fibres. This multiplied by the sum of 0-4 and 1-, (or FEAMTNG. 197 the value of n and unity = 14,) the pre duct, 39732, is the required area of the beam in inches. 316. — "When the fibres are uncut, then their sectional area equals the area of the beam, and may be found by the follow- ing rule. Rule XIX, — Deduct 0-434 times the specific gravity in Table III., multiplied by the length in feet, from the value of T, Table ITT., and divide the weight in pounds by the remain- der. The quotient will be the required area in inches. A = T -0434/7- m Maample. — A vertical beam of locust, 15 feet long, fibres all uncut, is required to sustain a tensile strain equal to 25,000 pounds : what must be its area ? Here 0-434 times -794, the specific gravity for locust in Table III., multiplied by 15, the length in feet, is 5-17. This, from 2,650, the value of T for locust, Table III., the remainder is 2,644-83. The given weight, 25,000, divided by 2,644-83, the remainder, as above, the quotient, 9-45, will be the required area in inches. ITI. RESISTANCE TO CKOSS-STBAINB. 317. — A load placed upon a beam, laid horizontal or in- clined, tends to bend it, and if the weight be proportionally large, to break it. The power in the material that resists this bending or breaking, is termed the resistance to cross-strains, or transverse strains. While in posts or struts the material is compressed or shortened, and in ties and suspending-pieces it is extended or lengthened ; in beams subjected to cross-strains the material is both compressed and extended. (See Art. 254.) When the beam is bent, the fibres on the concave side are compressed, while those on the convex side are extended. The line where these two portions of the beam meet — that is, the portion compressed and the portion extended — the hori- zontal line of juncture, is termed the neutral line or plane. It 198 AMEKICAN HOUSE-CAKPENTEK. is so called because at this line the fibres are neither com pressed nor extended, and hence are under no strain -whatever. The location of this line or plane is not far from the middle of the depth of the beam, when the strain is not sufficient to injure the elasticity of the material ; but it removes towards the concave or convex side of the beam as the strain is increased, until, at the period of rupture, its distance from the top of the beam is in proportion to its distance from the bot- tom of the beam as the tensile strength of the material is to its compressive strength. 318. — In order that the strength of a beam be injured as little as possible by the cutting required in framing, all mor- tices should be located at or near the middle of the depth. There is a prevalent idea among some, who are aware that the upper fibres of a beam are compressed when subject to cross-strains, that it is not injurious to cut these top fibres, provided that the cutting be for the insertion of another piece of timber — as in the case of gaining the ends of beams into the side of a girder. They suppose that the piece filled in will as effectually resist the compression as the part removed would have done, had it* not been taken out. Now, besides the effect of shrinkage, which of itself is quite sufficient to prevent the proper resistance to the strain, there is the mecha- nical difficulty of fitting the joints perfectly throughout ; and, also, a great loss in the power of resistance, as the material is so much less capable of resistance when pressed at right angles to the direction of the fibres, than when directly with them, as the results of the experiments in the tables show. 319. — In treating upon the resistance to cross-strains, the subject is divided naturally into two parts, viz. stiffness and strength : the former being the power to resist deflection or bending, and the latter the resistance to rupture. 320. — Resistance to Deflection. "When a load is placed upon a beam supported at each end, the beam bends more or FEAMING. 191) less ; the distance that the beam descends uuder the operation of the load, measured at the middle of its length, is termed its deflection. In an investigation of the laws of deflection it has been demonstrated, and experiments have confirmed it, that while the elasticity of the material remains .uninjured by the pressure, or is injured in but a small degree, the amount of deflection is directly in proportion to the weight producing it, and is as the cube of the length ; and, in pieces of rectangular sections, it is inversely proportional to the breadth and the cube of the depth : or, inversely proportional to the fourth power of the side of a square beam or of the diameter of a cylindrical one. Or, when I equals the length between the supports, w the weight or pressure, o the breadth, d the depth, and j? the deflection ; then — equals a constant quantity for beams of all dimensions made from a like material. Also, 7J=* <»> where s equals a side of a square beam ; and 0-589 &p = % (28 -) where D equals the diameter of a cylindrical beam. The constant here is less than in the case of the square and of the rectangular beams. It is as much less as the circular beam is less stiff than a square beam whose side is equal to the diame- ter of the cylindrical one. The constant, E, is therefore mul- tiplied by the decimal - 589. 321. — It may be observed that _E*in (26) and (27) would be equal to w, in case the dimensions of the beam and the amount of deflection were each made equal to unity ; and in (28) equal to w divided by 0-589. That is, when in (26) the length is 1, the breadth 1, and the depth 1, then E wou'.d be 200 AMERICAN HOUSE-CARPENTER. equal to the weight that would depress the beam from its ori ginal line equal to 1. Thus — „_ V w ^ V x w the dimensions all laken in inches except the length, and this taken in feet. This is an extreme state of the case, for in most kinds of material this amount of depression would exceed the limits of elasticity ; and hence the rule would here fail to give the correct relation among the dimensions and pressure. For the law of deflection as above stated, (the deflection being eqxial for equal weights,) is true only while the depressions are small in comparison with the length. Nothing useful is, therefore, derived from this position of the question, except to give an idea of the nature of the quantity represented by the constant, E '/ it being in reality a measure of the stiffness of the kind of material used in comparing one material with another. Whatever may be the dimensions of the beam, E, calculated by (26,) will always be the same quantity for the same material; but when various materials are used, E will vary according to the flexibility or stiffness of each particular material. For example, i^will be much greater for iron than for wood ; and again, among the various kinds of wood, it will be larger for the stiff woods than for those that are flexible. 322. — If the amount of deflection that would be proper in beams used in framing generally, (such as floor beams, girders and rafters,) were agreed upon, the rules would be shortened, and the labor of calculation abridged. Tredgold proposed to make the deflection in proportion to the length of the beam, and the amount at the rate of one-fortieth of an inch (= 0-025 inch) for every foot of length. He was undoubtedly right in the manner and probably so in the rate ; yet, as this is a mat- ter of opinion, it were better perhaps to leave the rate of de- flection open for the decision of those who use the rules, and then it may be varied to suit the peculiarities of each case FRAMING. 20 1 that may arise. Any deflection within the limits of the elas- ticity of the material, may be given to beams used for some purposes, while others require to be restricted to that amount of deflection that shall not be perceptible to a casual observer. Let n represent, in the decimal of an inch, the rate of deflec- tion per foot of the length of the beam ; then the product of n, multiplied by the number of feet contained in the length of the beam, will equal the total deflection, =nl. Now, if n I be substituted for p in the formulas, (26,) (27) and (28,) they will be rendered more available for general use. For example, let this substitution be made in (26,) and there results — V w T w ^ = bd'nl = YWH? ^ 9,) where I is in feet, and i, d and n in inches ; and for (27) — 8 nl s n' ^ ' also for (28)— W- Vw - Vw tti \ ^ - 0-589 D l nl~ 0-589 Z> 4 n> ^ 6l -> where the notation is as before, with also s and D in inches. In these formulas, w represents the weight in pounds concen- trated at the middle of the length of the beam. If the weight, instead thereof, is equally distributed over the length of the beam, then, since f of it concentrated at the middle will de- flect a beam to the same depth that the whole does when equally distributed, (Art. 281,) therefore — ^ = 0-589 J?' n> v**-) where w equals the whole of the equally distributed load. Again, if the load is borne by more beams than one, laid parallel to each other — as, for example, a series or tier of floor 26 202 AMERICAN HOUSE-CAKPENTEK. beams — and the load is equally distributed over the supported surface or floor ; then, if/ represents the number of pounds of the load contained on each square foot of the floor, or the pounds' weight per foot superficial, and o represents the dis- tance in feet between each two beams, or rather the distance from their centres, and I the length of the beam in feet, in the clear, between the supports at the ends ; then c I will equal the area of surface supported by one of the beams, and/c I will represent the load borne by it, equally distributed over its length. Now, if this representation of the load be substituted for w in (32,) (33) and (34) there results— *=%*? = %£-*, m h rP m. h ft? ■». ' x ' n' E JJ4ll = llll\ (36 .) 8 n s n ' jfolF $ foP ^ ~ 0-589 D* n ~ : 589 D l n KOi '> Practical Rules and Examples. 323. — To ascertain the weight, placed upon the middle of a beam, that will cause a given deflection. Mule XX. — Multiply the area of the cross-section of the beam by the square of the depth and by the rate of the deflec- tion, all in inches ; multiply the product by the value of E, Table II., and divide this product by the square of the length in feet, and the quotient will be the weight in pounds required. Example. — What weight can be supported upon the middle of a Georgia pine girder, ten feet long, eight inches broad, and ten inches deep, the deflection limited to three-tenths of an inch, or at the rate of 0'03 of an inch per foot of the length ? Here the area equals 8 X 10 = 80 ; the square of the depth equals 10 X 10 = 100 : 80 x 100 = 8,000 ; this by 0-03, the rate of deflection, the product is 240 ; and this by 2970, the value of i?for Georgia pine, Table II., equals T12,800. This TEAMING. 203 product, divided by 100, the square of the length, the quotient, 7,128, is the weight required in pounds. Mule XXI. — Where the beam is square the weight may bo found by the preceding rule or by this : — Multiply the square of the area of the cross-section by the rate of deflection, both in inches, and the product by the value of E, Table II., and divide this product by the square of the length in feet, and the quotient will be the weight required in pounds. Example. — What weight placed on the middle of a spruce beam will deflect it seven-tenths of an inch, the beam being 20 feet long, 6 • inches broad, and 6 inches deep ? Here the area is 6 X 6 = 36, and its square is 36 X 36 = 1296 ; the rate of deflection is equal to the total deflection divided by the length, =^k = 0-035 ; therefore, 1296 x 0-035 = 45-36, and ° ' 20 this by 1550, the value of E for spruce, Table II., equals 70,308. This, divided by 400, the square of the length, equals 175-77, the required weight in pounds. Rule XXII. — When the beam is round find the weight by this rule : — Multiply the square of the diameter of the cross- section by the square of the diameter, and the product by the rate of deflection, all in inches, and this product by 0-589 times the value of E, Table II. This last product, divided by the square of the length in feet, will give the required weight in pounds. Example. — What weight on the middle of a round white pine beam will cause a deflection of 0-028 of an inch per foot, the beam being 10 inches diameter and 20 feet long? The square of the diameter equals 10 X 10 = 100 ; 100 x 100 = 10,000 ; this by the rate, 0-028, = 280, and this by 0-589 x 1750, the value of E, Table II., for white pine, equals 288,610. This last product, divided by 400, the square of the length, equals 721'5, the required weight in pounds. 324. — To ascertain the weight that will produce a given de 204: AMERICAS' HOUSE-CARPENTER. flection, when the weight is equally distributed over the length of the beam. Rule XXIII.— The rules for this are the same as the threa preceding rules, with this modification, viz., instead of the square of the length, divide by five-eighths of the square of the length. 325. — In a series or tier of beams, to ascertain the weight per foot, equally distributed over the supported surface, that will cause a given deflection in the beam. Rule XXIY. — The rules for this are the same as Eules XX., XXL, and XXIL, with this modification, vizr., instead of the square of the length, divide by the product of the distance apart in feet between each two beams, (measured from the centres of their breadths,) multiplied by five- eighths of the cube of the length, and the quotient will be the required weight in pounds that may be placed upon each superficial foot of the floor or other surface supported by the beams. In this and all the other rules, the weight of the material com- posing the beams, floor, and other parts of the constructions is understood to be a part of the load. Therefore from the ascer- tained weight deduct the weight of the framing, floor, plaster- ing, or other parts of the construction, and the remainder will be the neat load required. Example. — In a tier of white pine beams, 4 x 12 inches, 20 feet long, placed 16 inches or \\ feet from centres, what weight per foot superficial may be equally distributed over the floor covering said beams — the rate of deflection to be not more than 0*025 of an inch per foot of the length of the beams. Proceeding by Eule XX. as above modified, the area of the cross-section, 4 x 12, equals 48 ; this by 144, the square of the depth, equals 6912, and this by 0-025, the rate of deflection, equals 172-8. Then this product, multiplied by 1750, the value of E, Table II., for white pine, equals 302,400. The distance between the centres of the beams is 1J feet, the cube FRAMING. 205 of the length is 8,000, and/i by I of 8,000 equals 6,6663. The above 302,400, divided by 6,666g, the quotient, 45-36, equals the required weight in. pounds per foot superficial. The weight of beams, floor plank, cross-furring, and plastering oc- curring under every square foot of the surface of the floor, is now to be ascertained. Of the timber in every 16 inches by 12 inches, there occurs 4x12 inches,, one foot long; this equals one-third of a cubic foot. Now, by proportion, if 16 inches in width contains £ of a cubic foot, what will 12 inches 1 v 19 12 in width contain ? 3 1 — = fy — i of a cubic foot The floor plank (Georgia pine) is 12 x 12 inches, and 1J inches thick, equal to y| of a cubic foot, equals ^, equals ^ ¥ . Of the furring strips, 1x2 inches, placed 12 inches from centres, there will occur one of a foot long in every superficial foot. Now, since in a cubic foot there is 144 rods, one inch square and one foot long, therefore, this furring strip, 1 x 2 x 12 inches, equals T f T = T ' ¥ of a cubic foot. The weight of the timber and furring strips, being of white pine, may be esti- mated together : 1 + T V = || + 7 V = -if of a cubic foot. "White pine varies from 23 to 30 pounds. If it be taken at 30 pounds, the beam and furring together will weigh 30 x || pounds, equals Y'92 pounds. Georgia pine may be taken at 50 pounds per cubic foot ;* the weight of the floor plank, then, is 50 X T S T = 5*21 pounds. A superficial foot of lath and plastering will weigh about 10 lbs. Thus, the white pine, 7'92, Georgia pine, 5 "21, and the plastering, 10, together equal 23-13 pounds ; this from 45-36, as before ascertained, leaves 22-23, say 22i pounds, the neat weight per foot superficial that may be equally distributed over the floor as its load. * To get the Weight of wood or any other material, multiply its specific gravity by 625. For Sp« cine Uraviiies see Tables L, H., and III. and the Appendix for "Weight of Materials. 206 AMERICAN HOUSE-CAEPENTEE. 326. — To ascertain the weight when the beam is laid not horizontal, but inclined. Rule XX V . — In each of the foregoing rules, multiply the result there obtained by the length in feet, and divide the pro- duct by the horizontal distance between the supports in feet ; and the quotient will be the required weight in pounds. The foregoing Rules, stated algebraically, are placed in the following table : — When the When the weight is When the beam Is beam is laid Eect- angular. Square. Round. Horizontal Concentrated at middle, 10, in pounds, eqnals Equally distributed, «?, in pounds, equals By the foot superficial,/ in pounds, equals Concentrated at middle, to, in pounds, equals Equally distributed, w, in pounds, equals By the foot superficial,/ in pounds, equals (3S.) Enb d' I' (39.) En, a' (40.) •589 EnD* I' Horizontal (41.) Enbdi (43.) En s' HI' (48.) ■MIA EnD* V Horizontal (44.) Enid' (45.) Ens' He I' (46.) •9424 En D* Inclining (47.) Enid' Ih (48.) Ens' Ih (49.) •589 EnD' Ih Inolining (50.) Enbd< y a Ih (51.) Ens' %lh (52.) •9424 EnD* Ih Inclining (58.) Enbdi % cl'h (54) Ens' %cl>h (55.) ■942 4 EnD' cl*h In the above table, h equals the breadth, and d equals the depth of cross-section of beam ; s equals the breadth of a side of a square beam, and D equals the diameter of a round beam ; n equals the rate of deflection per foot of the length ; KBAMTNG. 207 D, s, b, d and n 3 all in inches ; I equals the length, c equals the distance between two parallel beams measured from the centres' of their breadth ; h equals the horizontal distance between the supports of an inclined beam ; I, c and h in feet ; w equals the weight in pounds on the beam ; f equals the weight upon each superficial foot of a floor or roof supported by two or more beams laid parallel and at equal distances apart ; E is a constant, the value of which is found in Table II. ; r is any decimal, chosen at pleasure, in proportion to unity, as b is to d, from which proportion b equals d r. 327. — To ascertain the dimensions of the cross-section of a beam to support the required weight with a given deflection. Mule XXVI. — Preliminary. When the weight is concen- trated at the middle of the length. Multiply the weight in pounds by the square of the length in feet, and divide the pro- duct by the product of the rate of deflection multiplied by the value of E, Table II., and the quotient equals a quantity which may be represented by M- — referred to in succeeding rules. w F nr Inn •> -rr- = M. (56.) En K ' Rule XXVII. — Preliminary. When the weight is equally distributed over the length. Multiply five-eighths of the weight in pounds by the square of the length in feet, and divide the product by the rate of deflection multiplied by the value of E, Table II., and the quotient equals a quantity which may be represented by iV^-referred to in succeeding rules. *£^ = N- (57.) En ' Mule XXVIII.— Preliminary. When the weight is given per foot superficial and supported by two or more beams. Multiply the distance apart between two of the beams, (mea- sured from the centres of their breadth,) by the cube of the length, both in feet, and multiply the product by five-eighths . of the weight per foot superficial ; divide this product by the 208 AMERICAN HOUSE-CAEPENTEE. product of the rate of deflection, multiplied by the value of E Table II., and the quotient equals a quantity which may be represented by TJ- — referred to in succeeding rules. $&? = V. (58.) Hula XXIX.— Preliminary. When the learn is laid not horizontal, but inclining. In Rules XXVI. and XXYII., instead of the square of the length multiply by the length, and by the horizontal distance between the supports, in feet. And in Eule XXVIII., instead of the cube of the length, multiply by the square of the length, and by the horizontal distance between the supports, in feet. From (56) From (57) From (58) if^ = ir,. (60.) liule XXX. — When the beam is rectangular to find the dimensions of the cross-section. Divide the quantity repre- sented by M, W or TJ, (in preceding preliminary rules,) by the breadth in inches, and the cube root of the quotient will equal the required depth in inches. Or, divide the quantity represented by M, iVor TJ, by the cube of the depth in inches, and the quotient will equal the required breadth in inches. Or, again, if it be desired to have the breadth and depth in proportion, as r is to unity, (where r equals any required deci- mal,) divide the quantity represented by M, N or TJ, by the value of r, and extract the square root of the • quotient : and the square root extracted the second time, will equal the depth in inches. Multiply the depth thus found by the value of r, and the product will equal the breadth in inches. TEAMING. 209 Example. — To find the depth. A beam of spruce, laid on supports with a clear bearing of 20 feet, is required to support a load of 1674 pounds at the middle, and the deflection not to exceed - 05 of an inch per foot ; what must be the depth when the breadth is 5 inches. By Rule XXVI. for load at middle : the product of 1674, the weight, by 400, the square of the length, equals 669,600. The product of - 05, the rate of de- flection, multiplied by 1550, the value of E, from Table II., set opposite spruce, is 77 - 5. The aforesaid product, 669,600, divided by 77-5, equals 8640, the value of M. Then by Eule XXX., 8640, the value of M, divided by 5, the breadth, the quotient is 1728, and 12, the cube root of this, found in the table of the Appendix, equals the required depth in inches. Example. — To find the breadth. Suppose that in the last example it were required to have the depth 13 inches ; in that case what must be the breadth ? The value of M, 8640, as just found, divided by 2197, the cube of the depth, equals 3-9326, the required breadth — nearly 4 inches. Example. — To find both breadth and depth, and in a certain proportion. Suppose, in the above example, that neither the breadth nor the depth are given, but that they are desired to be in proportion as 0-5 is to 1-0. ISTow, having ascertained the value of If, by Rule XXVI., to be 8640, as above, then, by Rule XXX, 8640, divided by 0'5, the ratio, gives for quotient 17,280. The square root of this (by the table in the Appen- dix,) is 131-45, and the square root of this square root is 11-465, the required depth. The breadth equals 11-465 X 0-5, which equals 5-7325. The depth and breadth may be 11 \ by 5| inches. In cases where the load is equally distributed over the length of the beam, the process is precisely the same as set forth in the three preceding examples, except that five-eighth* of the weight is to be used in place of the whole weight ; and hence it would be a useless repetition to give examples to illustrate such cases. 27 210 AMERICAN HOUSE-CARPENTER. Example. — When the weight is per foot superficial to find the depth. A floor is to be constructed to support 500 pounds on every superficial foot of its surface. The beams to be ot white pine, 16 feet long in the clear of the supports or walls, placed 16 inches apart, from centres, to be 4 inched thick, and the amount of deflection not objectionable provided it be within the limits of elasticity. Proceeding by Rule XXYIIL, the product of \\ feet, (equal to 16 inches,) multiplied by 4096, the cube of the length, equals 5461£. This, multiplied by 312-5, (equal to £ of the weight,) equals 1,706,666. The largest rate of deflection within the limits of the elasticity of white pine is 0-1022, as per Table II. This, multiplied by 1750, the value of E for white pine, Table II., equals 178-85. The former product, 1,706,666, divided by the latter, 178-85, equals 9,542-5, the value of V. Now, by Rule XXX., this value of TT, 9,542-5, divided by 4, the breadth, equals 2385-6, the cube root of which, 13-362, is the required depth — nearly 13f inches. Example. — To find the breadth. Suppose, in the last exam- ple, that the depth is known but not the breadth, and that the depth is to be 13 inches. Having found the value of V, as before, to be 9542-5, then by Rule XXX, dividing 9542-5, the value of U, by 2197, the cube of the depth, gives a quotient of 4-3434 and this equals the breadth — nearly 4f inches. Example. — To find the depth and breadth in a given propor- tion. Suppose, in the above example, that the breadth and depth are both unknown, and that it is desired to have them in proportion as 0-7 is to TO. Having found the value of Zf, as before, to be 9542-5, then by Rule XXX., dividing 9542-5, the value of V, by - 7, the quotient is 13,632, th,e square root of which is 116 - 75, and the square root of this is 10-805, the depth in inches. Then 10-805, multiplied by 0-7, the product, 7-5635, is the breadth in inches. The size may be 7 r V by 10J# inches. 328.— Example. — In the case of inclined be ims to find the FRAMING. 2ii hpth. A beam of white pine, 10 feet long in the clear of the bearing, and laid at such an inclination that the horizontal distance between the supports is 9 feet, is required to support 12,000 pounds at the centre of its length, with the greatest allowable deflection within the limits of elasticity ; what must be its depth when its breadth is fixed at 6 inches ? By refer- ence to Table II. it is seen that the greatest value of n, within the limits of elasticity, is 0-1022. By Eule XXVI, for con- centrated load, and Eule XXIX., for inclined beams, 12,000, the weight, multiplied by 10, the length, and by 9, the hori- zontal distance, equals 1,080,000. The product of 0-1022, the greatest rate of deflection, by 1750, the value of E, Table II., for white pine, equals 178-85. Dividing 1,080,000 by 178-85, the quotient is 6038-58, the value of M. Now, by Eule XXX., for rectangular beams, 6038-58, the value of M, divid- ed by 6, the breadth, the quotient is 1006-43. The cube root of this, 10-02, a trifle over 10 inches, is the depth required. Example. — In case of inclined beams to find the breadth. In the last example suppose the depth fixed at 12 inches; then by Eule XXX., 6038-58, the value of M, as above found, divided by 1728, the cube of the depth, equals 3-4945, or nearly 3£ inches — the breadth required. Example. — Again, in case the breadth and depth are to be in a certain proportion j as, for example, as 0-4 is to unity. Then by Eule XXX., 6038-58, the value of M, found as above, divided by 0-4, equals 15,096-45, the square root of which is 122-87, and the square root of this square root is 11-0843, a trifle over 11 inches — the depth required. Again, 11 multi- plied by the decimal 0-4, (as above,) equals 4-4, a little over 4f inches — the breadth required. In the three preceding examples, the weight is understood to be concentrated at the middle. If, however, the weight had been equally distributed, the same process would have been used to obtain the dimensions of the cross-section, with 212 AMERICAN HOUSE-CAKPENTEE. only one exception ; viz. f of the weight instead of the whole weight would have been used. (See Rule XXVII.) Example. — In case of inclined beams ; the weight per foot superficial, and borne by two or more beams. A tier of spruce beams, laid with a clear bearing of 10 feet, and at '20 inches apart from centres, and laid so inclining that the horizontal distance between bearings is 8 feet, are required to sustain 40 pounds per superficial foot, with a deflection not to exceed 002 inch per foot of the length ; what must be the depth when the breadth is 3 inches ? Proceeding by Rule XXIX for inclined beams, and bj- Rule XXVIIL, If, (= 20 inches,) the distance from centres, multiplied by 100, the square of the length, and by 8, the horizontal distance between bearings, equals 1,333-J ; this, by f x 40, five-eighths of the weight, equals 33,333£. This, divided by 0-02 x 1550, the rate of deflection, by the value of E, Table II., for spruce, equal to 31, equals 1075-27, the value of IT. Now by Rule XXX. for rectangular beams, 1075-27, divided by 3, the breadth, equals 358-42, the cube root of which, 7 - l, is the required depth in inches. Example. — The same as the preceding ; but to find the breadth, when the depth is fixed at 8 inches. By Rule XXX., 1075-27, the value of JJ, divided by 512, the cube of the depth, equals 2-1 — the breadth required in inches. Example. — The same as the next but one preceding ; but to find the breadth and depth in the proportion of OS to 1*0, or 3 to 10. By Rule XXX, 1075-27, the value of U, divided by 0-3, the value of r, equals 3584-23. The square root of this is 59-869, and the square root of this square root is 7 - 737 — the depth required in inches. This 7'737, multiplied by 0-3, the value of r, equals 2-3211 — the required breadth in inches. The dimensions may,- therefore, be 2fV by 7J inches. Rule XXXI. — When the beam is square to find the side. Extract the square root of the quantity represented by M, JV FRAMING. 213 or U, in preliminary Eules XXVI., XXVII. and XXVIII., and the square root of this square root will equal the side required. Example. — A heam of chestnut, having a clear bearing of 8 feet, is required to sustain at the middle a load of 1500 pounds ; what must be the size of its sides in order that the deflection shall not exceed 0-03 inch per foot of its length ? By Kule XXVI., 1500, the load, multiplied by 64, the Square of the length, equals 96,000. This product divided by 0-03 times 2330, the value of E, Table II., for chestnut, gives a quotient of 13734, the quantity represented by M. Now by Rule XXXI., the square root of 1373-4 is 37*05, and the square root of this is 6-087. The beam must, therefore, be 6 inches square. In this example, had the load, instead of being con- centrated at the middle, been equally distributed over its length, the side would have been equal to the side just found, multiplied by the fourth root of £ or of 0-625, equal to 6-087 X 0-889 = 5-4 inches. (See Eules XXVII. and XXXI.) Example. — In the case where (he weight is per foot superfi- cial and borne by two or more beams. A floor, the beams of which are of oak, and placed 20 inches or 1^ feet apart from centres, and which have a clear bearing of 20 feet, is required to sustain 200 pounds per superficial foot, the deflection not to exceed 0-025 inch per foot of the length, and the beam to be square. By Rule XXV ILL, If, the distance from centres, multiplied by 8000, the cube of the length, equals 13,333§ ; and this by 125, (being f of 200 pounds,) equals l,666,666f. Dividing this by 0-025 times 2520, the value of E, Table II, for oak, the quotient is 26,455 — a number represented by IT. Now by Rule XXXL, the square root of this number is 162-65. and the square root of this square root is 12-753— the required side. The beam may be 12f inches square. Example. — Inclined square beams, load at middle. A. bar of cast-iron, 6 feet long in the clear of bearings, and laid 214 AMERICAN HOUSE-CARPENTER. inclining so that the horizontal distance between the bearings is 5 feet, is required to sustain at the middle 3000 pounds, and the deflection not to exceed 001 inch per foot of its length ; what must be the size of its sides ? By Eule XXVI. for load at middle, modified by Eule XXIX. for inclined beams ; 3000, the weight, multiplied by 6, the length, and by 5, the horizontal distance between bear- ings, equals 90,000. The rate of deflection, 0-01, by 30,500, the value of E, Table II., for cast-iron, equals 305 ; and 9000 divided by 305, equals 295-082, the value of M. Now by Eule XXXI. for square beams, the square root of 295-082 is 17-18, the square root of which is 4-145 — the size of the side required ; a trifle over 41, the bar may, therefore, be 4J- inches square. Example. — Same as preceding, hut the weight equally distm- Jjuted. By Eule XXVII. f of the weight is to be used instead of the weight ; therefore 295-082, the value of M, as above, multiplied by I, will equal 184-426, the value of iV. By Eule XXXI. the square root of 184-426 is 13-58, the square root of which is 3'685 — the size of the side required ; equal to nearly 3}^ inches square. Example. — Same as preceding case, hut the weight per foot superficial, and sustained by 2 or more bars, placed 2 feet from centres, the load being 250 pounds per foot superficial. By Eule XXVIII., modified by Eule XXIX., the distance from centres, 2, multiplied by 36, the square of the length, and by 5, the horizontal distance, equals 360. This by 156-25, five-eighths of the weight, equals 56,250. The rate of deflec- tion, 0-01, by 30,500, the value of E, Table II., for cast-iron, equals 305. The above 56,250, divided by 305, equals 184-426, the value of U. Now by Eule XXXI. the square root of 184-426. the value of If, is 13-58, the square root of which is 3-685 — the size of the side required. It will be observed that this result is precisely like that in the last example. This is as it should be, for each beam has to sustain the weight on 2 x 6 FRAMING. 215 ^ 12 superficial feet, equal to 12 X 250, equal 3100 pounds; and all the other conditions are parallel. Rule XXXII. — When the learn is round to find the diame- ter. Divide the value of M, JST or U, found by Eules XXVI., XXVII. or XXVIII., by the decimal 0-589, and extract the square root: and the square root of this square root will be the diameter required. Example. — In the case of a concentrated load at middle A round bar of American iron, of 5 feet clear bearing, is required to sustain 800 pounds at the middle, with a deflection not to exceed 0-02 inch per foot ; what must be its diameter ? By Eule XXVI. for load at middle, 800, the weight, multiplied by 25, the square of the length, equals 20,000. The rate of deflection, 0-02, by 51,400, the value of E, Table II, for Attn-- rican wrought iron, equals 1028. The above 20,000, divided by 1028, equals 19-4552, the value of M. Now, by Eule XXXIL, 19-4552, the value of M, divided by 0-589 equals 33-03, the square root of which is 5*747, and the square root of this is 2-397, nearly 2-4, the diameter required in inches, equal to 2f large. Example. — Same case as the preceding, hut the load equally distributed. By Eule XXVII., five-eighths of the weight is to be used instead of the whole weight ; therefore the above 33 - 03, multiplied by I, equals 20-64375, the square root of which is 4-544, and the square root of this square root is 2-132, the diameter required in inches, 2-J- inches large. Example. — When the weight is per foot superficial, and sus- tained by two or more oars or beams. The conditions being the same as in the preceding examples, but the weight, 100 pounds per foot, is to be sustained on a series of round rods,, placed 18 inches apart from centres, equal 1*5 feet. By Eule XXVIII., for weight per foot superficial, 1*5, the distance from centres, multiplied by 125, the cube of the length, and , by 62 - 5, five-eighths of the weight, equals 11,718-75. Thia 216 AMERICAN HOUSE-CABPENTEE. divided by 1028, the product of the rate of deflection by the value of E, as found in the preceding example, equals ll - 4, the value of TJ. Now by Eule XXVII., 114, the value of U, divided by 0-589, equals 19-42, the square root of which is 4407, and the square root of this square root is 2-099, the diameter required — very nearly 2 T \ inches. Example. — When the beam is round and laid inclining, the weight concentrated at the middle. A round beam of white pine, 20 feet long between bearings, and laid inclining so that the horizontal distance between bearings is 18 feet, is required to support 1250 pounds at the middle, with a deflection not to exceed 0'05 inch per foot; what must be its diameter? By Kule XXVI. for load at middle, modified by Eule XXIX. for inclined beams, 1250, the weight, multiplied by 20, the length, and by 18, the horizontal distance, equals 450,000. The rale of deflection, 0-05, multiplied by 1750, the value of E, Table II., for white pine, equals 87'5. The above 450,000 divided by 87-5, equals 5142-86, the value of M. Now by Eule XXXII. for round beams, 5142-86, the value of M, divided by 0-589, equals 8731-5, the square root of which is 93-44, and the square root of this square root is 9 - 667, the diameter re- quired — equal to 9§ inches. Example. — Same as in preceding example, hut the weight equally distributed. By Eule XXVII., five-eighths of the weight is to be used instead of the whole weight, therefore 8731-5, the result in the last example just previous to taking the square root, multiplied by ■§, equals 5457'2, the square root of which is 73-87, and the square root of this square root is 8-59, the diameter required — nearly 8f inches. Example. — Same as in the next but one preceding example, tut the weight per foot superficial, and supported by two or more beams. A series of round hemlock poles or beams, 10 feet long clear bearing, laid inclining so as that the horizontal distance between the supports equals 7 feet, and laid 2 feet FRAMING. 217 and 6 inches apart from centres, are required to support 20 pounds per superficial foot without regard to the amount of deflection, provided that the elasticity of the material be not injured ; what must be their diameter ? By Eule XXVUT. for weight per foot superficial, modified by Eule XXIX. for inclined beams, 2-5, the distance from centres, multiplied by 100, the square of the length, and by 7, the horizontal distance between bearings, and by five-eighths of the weight, 12-5, equals 21,875. The greatest value of n, Table II., for hem- lock, 0-08794, multiplied by 1240, the value of E, Table II., for hemlock, equals 109-0456. The above 21,875, divided by 109-0456, equals 200-6, the value of U. Now by Eule XXXE., the above 200-6, divided by 0-589, equals 340-6, the square root of which is 18-46, and the square root of this square root is 4-296, the diameter required — equal to 4 t V inches nearly. 329. — The greater the depth of a beam in proportion to the thickness, the greater the strength. But when the difference between the depth and the breadth is great, the beam must be stayed, (as at Fig. 228,) to prevent its falling over and break- ing sideways. Their shrinking is another objection to deep beams ; but where these evils can be remedied, the advantage of increasing the depth is considerable. The following rule is, to find the strongest form for a heam out of a given quantity of timber. Rule. — Multiply the length in feet by the decimal, 0'6, and divide the given area in inches by the product ; and the square of the quotient will give the depth in inches. Example. — What is the strongest form for a beam whose given area of section is 48 inches, and length of bearing 20 feet ? The length in feet, 20, multiplied by the decimal, 0*6, gives 12 ; the given area in inches, 48, divided by 12, gives a quotient of 4, the square of which is 16 — this is the depth in inches ; and the breadth must be 3 inches. A beam 16 inches 28 218 AMERICAN HOTJSE-CARPENTEK. by 3 would bear twice as much as a square beam of the same area of section ; whicb shows how important it is to make beams deep and thin. In many old buildings, and even in new ones, in country places, the very reverse of this has been practised ; the principal beams being oftener laid on the broad side than on the narrower one. The foregoing rules, stated algebraically, are placed in the following table. TABLE V. STIFFNESS OF BEAMS : DIMENSIONS. When the beam is laid When the weight is Bectangular. Square. Bound. Value of depth. Value of breadth. When b=dr, value of d. "Value of a side. Value of th« diameter. Concentrated at middle Equally distributed By the foot superficial (62.) Enb (63.) ID V- End* (64.) Enr (65.) %Jwl<- En (66.) «/ wl> ■SmEn a o ■§ o w (67.) y X « l> Enb (68.) },'«!■ End* (69.) Enr (70.) En (71.) ■9424 En (72.) Enb (78.) End' (74.) X/HfcP Enr (75.) En (76.) ■em En Concentrated at middle Equally distributed By the foot superficial (77.) l/wlh Enb (78.) w I K Ends (79.) \twlh Enr (80.) i/w IK En (81.) \r wiK •589 En bo g '3 3 (82.) %/% w I A Enb (88.) ¥s w IK End* (84.) i/% tolh ~Enr C85.) En (86.) «/ wlh ■942i En (87.) \/%fePh Enb (88.) Hfal'h Entp (89.) Enr (90.) X/fcl'h En (91.) V/cZ'A ■9424 £■» In the above table, 5 equals the breadth, and d the depth of cross-section of beam ; n equals the rate of deflection per foot of the length ; 5, d and n, all in inches. Also, I equals the length, c the distance between two parallel beams measured from the FRAMING. 219 centres of their breadth, and h equals the horizontal distance between the supports of an inclined beam ; Z, c and h, all in feet. Again, w equals the weight on the beam,/ equals the weight upon each superficial foot of a floor or roof, supported by two or more beams laid parallel and at equal distances apart ; w and / in pounds. And r is any decimal, chosen at pleasure, in proportion to unity, as b is to d — from which pro- portion b — dr. E is a constant the value of which is found in Table II. 330. — To ascertain the scantUng of the stiffest beam, that can be cut from a cylinder. Let d a cb, {Fig. 223,) be the section, and e the centre, of a given cylinder. Draw the diameter, a b / upon a and b, with the radius of the section, describe the arcs, d e and «c; join d and a, a and c, c and b, and b and d y then the rectangle, d a c b, will be a section of the beam required. Tig. 223. 331. — Besistance to Rupture. — The resistance to deflection having been treated of in the preceding articles, it now re- mains to speak of the other branch of resistance to cross strains, namely, the resistance to rupture. "When a beam is laid horizontally and supported at each end, its strength to resist a cross strain, caused by a weight or vertical pressure at the middle of its length, is directly as the breadth and square of the depth and inversely as the length. If the beam is square, or the depth equal to the breadth, then the strength is directlj 220 AMERICAN KOUSE-CAKPENTEK. as the cube of a side of the beam and inversely as the length, and if the beam is round the strength is directly as the cube of the diameter and inversely as the length. "When the weight is concentrated at any point in the length, the strength of the beam is directly as the length, breadth, and square of the depth, and inversely as the product of the two parts into which the length is divided by the point at which the weight is located. When the beam is laid not horizontal but inclining, the strength is the same as in each case above stated, and also in proportion, inversely as the cosine of the angle of inclination with the horizon, or, which is the same thing, direetly as the length and inversely as the horizontal distance between the points of support. When the weight is equally diffused over the length of a beam, it will sustain just twice the weight that could be sus- tained at the middle of its length. A beam secured at one end only, will sustain at ■ the other end just one-quarter of the weight that could be sustained at its middle were the beam supported at each end. These relations between the strain and the strength exist in all materials. For any particular kind of material, — = S- (92) 8, representing a constant quantity for all materials of like strength. The superior strength of one kind of material over another is ascertained by experiment ; the value of S being ascertained by a substitution of the dimensions of the piece tried for the symbols in the above formula. Having thus obtained the value of S, the formula, by proper inversion, becomes use- ful in ascertaining the dimensions of a beam that will require a certain weight to break it ; or to ascertain the weight that will be required to break a certain beam. It will be observed in the preceding formula, that if each of the dimensions of the FRAMING. 221 beam equal unity, then w — S. Hence, S is equal to the weight required to break a beam one inch square and one foot long. The values of S, for various materials, have been ascertained from experiment, and are here recorded : — TABLE VI. STRENGTH. Materials. Green plate-glass Spruce ... Hemlock .... Soft white pine Hard white pine Ohio yellow pine Chestnut .... Georgia pine Oak Locust .... Oast-iron (from 1550 to 2280) Number of Experiments. The specimens broken were of various dimensions, from one foot long to three feet, and from one inch square to one by three inches. The cast-iron specimens were of the various kinds of iron used in this country in the mechanic arts. S may be taken at 2,000 for a good quality of cast-iron. It is usual in determining the dimensions of a beam to suppose it capable of sustaining safely one-third of the breaking weight, and yet Tredgold asserts that one-fifth of the breaking weight will in time injure the beam so as to give it a permanent set or bend, and Hodgkinson says that 'cast-iron is injured by any weight however small, or, in other words, that it has no elastic power. However this may be, experience has proved cast- iron quite reliable in sustaining safely immense weights for a long period. Practice has shown that beams will sustain safely from one-third to one-sixth of their breaking weight. If the load is bid on quietly, and is to remain where laid, at rest, beams may be trusted with one-third of their breaking weight, but if the load is moveable, or subject to vibration, 222 AMERICAN HOT/SE-CARPENTEK one-quarter, one-fifth, or even, in some cases, one-sixth is quite a sufficient proportion of the breaking load. 332. — The dimensions of beams should be ascertained only by means of the rules for the stiffness of materials, {Arts. 320 ; 323, et seq.,) as these rules show more accurately the amount of pressure the material is capable of sustaining without injury. Yet owing to the fact that the rules for the strength of mate- rials are somewhat shorter, they are more frequently used than those for the stiffness of materials. In order that the proportion of the breaking weight may be adjusted to suit cir- cumstances it is well to introduce into the formula a symbol to represent it. The proportion represented by the symbol may then be varied at discretion. Let this symbol be a, a decimal in proportion to unity as the safe load is to the break- ing load, then S a will equal the safe load. Hence, Said 2 ,. . w — - — j (93.) for a safe load at middle on a horizontal beam supported at both ends ; and 2 Said' ,-, . w = j (94.) for a safe load equally diffused over the length of the beam : and „ 2 S a b d' , f= cV (95.) for the load, per superficial foot, that can be sustained safely upon a floor supported by two or more beams, c being the dis- tance in feet from centres between each two beams, and f the load in pounds per superficial foot of the floor. Generally, in (93,) (94,) and (95,) w equals the load in pounds ; S, a constant, the value of which is found in Table VI. ; a a decimal, in pro- portion to unity as the safe load is to the breaking load ; I the length in feet between the bearings ; and o and d the breadth and depth in inches. FRAMING. 223 TO FIND THE WEIGHT. 333. — The formulas for ascertaining the weight in tho seve- ral cases are arranged in the following table, where c, f, w, S, a, I, b and d represent as above ; and also s equals a side of a square beam ; D equals the diameter of a cylindrical beam ; m and n equal respectively the two parts into which the length is divided by the point at which the weight is located ; and h equals the horizontal distance between the supports of an inclined beam. TABLE VTI. STRENGTH OF BEAMS; SAFE WEIGHT. When the When the weight Is ."When the beam is beam 1b laid Ecctangular. Square. Bound. Concentrated at middle, w, in pounds, equals Equally distributed, w, in pounds, equals By the foot superficial, / in pounds, equals Concentrated at any point in the length, 10, in pounds, equals (96.) I (9T.) San* I (98.) ■BS9I^8a I 3 a o (99.) 28ab

4ni7i (107.) ■MDa Sal m n Concentrated at middle, w, in pounds, equals Equally distributed, w, in pounds, equals By the foot superficial, f, in pounds, equals. Concentrated at any point in the length, w, in pounds, equals (108.) Sabd? A (109.) 8 a & h (110.) •589 D*8a h a i (111.) 2Sab

g k + <£c.) ■147 Sal (14T.) Sar (148.) So" (149.) •589 5 a (152.) (158.) 2£o" (154.) 1-178 £ a (157.) y/cA? 2(Sror (158.) \/fehl 2Sa (159.) •/ fr.hl 1-178 £ a (162.) */4 hwmn Sari' (163.) */4 hwmn Sa U~ (164.) \J hwmn •147 3 oY' (167.) */4 K(Rmn + Pgfc + Sxs.) Sari' (168.) »/4 h(Smn + Pgk + Ac.) Sal (169.) >/h(Rma + P gin-Ac.) ■lit Sal' 230 AMEKICAN HOUSE-CAKPENTEB. Practical Rules and Examples. Rule XXXVHI. — Preliminary. When the weight is con- centrated at the middle. Multiply the weight, in pounds, by the length, in feet, and divide the product by the value of S, Table VI., multiplied by a decimal that is in proportion to unity as the safe weight is to the breaking weight, and the quotient is a quantity which may be represented by J, referred to in succeeding rules. Rule XXXIX. — Preliminary. When the weight is equally distributed. One-half of the quotient obtained by the preced- ing rule is a quantity which may be represented by K, referred to in succeeding rules. w I K (171.) 2Sa~ Rule XL. — Preliminary. When the weight is per foot su- perficial. Multiply the weight per foot superficial, in pounds, by the square of the length, in feet, and by the distance apart from centres between two parallel beams, and divide the pro- duct by twice the value of S, Table VI., multiplied by a deci- mal in proportion to unity as the safe weight is to the break- ing weight, and the quotient is a quantity which may be re- presented by Z, referred to in succeeding rules. Rule XLI. — Preliminary. When the weight is concentrated at any point in the length. Multiply the distance, in feet, from the loaded point to one support, by the distance, in feet, from the same point to the other support, and by four times the weight in pounds, and divide the product by the value of 8, Table VI., multiplied by a decimal in proportion to unity as the safe weight is to the breaking weight, and by the length, FRAMING. 231 in feet ; and the quotient is a quantity which may be repre- sented by Q, referred to in the rules. 4: w m n „ "^T = , {Fig. 227,) called a header. The beams, a a, into which the header is framed, are called trimmers or carriage-teams. These framed beams require to be made thicker than the common beams. The header must be strong enough to sustain one-half of the weight that is sustained upon the tail beams, c c, (the wall at the opposite end or another header there sustaining the other half,) and the trimmers must each sustain one-half of the weight sustained by the header in addition to the weight it supports as a common beam. It is usual in practice to make these framed beams one inch thicker than the common beams for dwellings, and two inches thicker for heavy stores. This practice in ordinary cases answers very well, but in extreme cases these dimensions are not proper. Rules applicable gene- rally must be deduced from the conditions of the case — the load to be sustained and the strength of the material. 342. — For the header, formula (68,) Table V., is applicable. The weight, represented by w, is equal to the superficial area of the floor supported by the header, multiplied by the load on every superficial foot of the floor. This is equal to the length of the header multiplied by half the length of the tail beams, and by 86 pounds for dwellings and ordinary stores, or FRAMING. 243 by 500 pounds for warehouses. Calling the length of the tail beams, in feet, g, formula (68,) becomes " Ed* n Then if/ equals 86, and n equals 0-03, there results x 900 gV „_ This in words, is, as follows : Rule L. — Multiply 900 times the length of the tail beams by the cube of the length of the header, both in feet. The product, divided by the cube of the depth, multiplied by the value of E, Table II., will equal the breadth, in inches, for dwellings or ordinary stores. Example. — A header of white pine, for a dwelling, is 10 feet long, and sustains tail beams 20 feet long, its depth is 12 inches, what must be its breadth ? By the rule, 900 X 20 x 10' = 18,000,000. This, divided by (12 3 X 1750 =) 3,024,000, equals 5 - 95, say 6 inches, the required breadth. For heavy wareliouses the rule is the same as the above, only using 1550 in the place of the 900. This constant may be varied, at discretion, to anything between 900 and 5000, in accordance with the use to which the floor is to be put. 343. — In regard to the trimmer or carriage beam, formula (136,) Table VITI., is applicable. The load thrown upon the trimmer, in addition to its load as a common beam, is equal to one-half of the load on the header, and therefore, as has been seen in last article, is equal to one-half of the superficial area of the floor, supported by the tail beams, multiplied by the weight per superficial foot of the load upon the floor; therefore, when the length of the header, in feet, is represented by/, and the length of the tail beam by n, w equals ^- x „- x/i equals \fj n, and therefore (136,) of Table VIII., becomes x fJ m "' °~ SadH 2 14 AMERICAN II0USE-CAEPENTEE. equals the additional thickness to be given to a common beam when used as a trimmer, and for dwellings when f equals 86 and a equals 0-3, this part of the formula reduces to 286 1, or, for simplicity, call it 300, which would be the same as fixing/ at 90 instead of S6. Then we have b = 300 J?, n ' ( 18 °0 ode This, in words, is as follows : Hide LI. — For dwellings. Multiply 300 times the length of the header by the square of the length of the tail beams, and by the difference in length of the trimmer and tail beams, all in feet. Divide this product by the square of the depth in inches, multiplied by the length of the trimmer in feet, and by the value of /S, Table YL, and the quotient added to the thick- ness of a common beam of the floor, will equal the required thickness of the trimmer beam. Example. — A tier of 3 x 12 inch beams of white pine, hav- ing a clear bearing of 20 feet, has a framed well-hole at one side, of 5 by 12 feet, the header being 12 feet long, what must be the thickness of the trimmer beams ? By the rule, 300 x 12 x 15 2 x 5, divided by the product of 12 2 x 20 x 390, equals 3 - 6, and this added to 3, the thickness of one of the common beams, equals 6"6, the breadth required, 6J inches. For stores and warehouses the rule is the same as the above, only the constant, 300, must be enlarged in proportion to the load intended for the floor, making it as high as 1600 for heavy warehouses. 344. — When a framed opening occurs at any point removed from the wall, requiring two headers, then the load from the headers rest at two points on the carriage beam, and here for- mula (141.) Table VIII., is applicable. In this special case this formula reduces to 300 j (1^91+ 7c- g) b swj < 181 -> FEAMING. 24-5 where o equals the additional thickness, in inches, to be given to the carriage beam over the thickness of the common beams ; j, the length of the header, in feet ; m and Tc the length, in feet, respectively, of the two sets of tail beams, and m + n = k + g = l. The constant in the above, (181,) is for dwellings ; if the floor is to be loaded more than dwelling floors, then it must be increased in proportion to the increase of load up to as high as 1600 for warehouses. Rule LII. — Trimmer learns for framed openings occurring so as to require two headers. Multiply the square of the length of each tail beam by the difference of length of the tail beam and trimmer, all in feet, and add the products ; multiply their sum by 300 times the length, in feet, of the header, and divide this product by the product of the square of the depth, in inches, by the length, in feet, and by the value of S, Table VI. ; and the quotient, added to the thickness of a common beam of the tier, will equal the thickness of the trimmer beams. Example. — A tier of white pine beams, 4 X 14 inches, 20 feet long, is to have an opening of 5 x 10 feet, framed so that the length of one series of tail beams is 7 feet, the other 8 feet, ■what must be the breadth of the trimmers ? Here, (7 s X 13) + (8 a X 12) equals 1405. This by 300 x 10 equals 4,215,000. This divided by 1,528,800 (= 14 2 + 20 X 390) equals 2-75, and this added to 4, the breadth, equals 6'75, or 6|, the breadth required, in inches. 345. — Additional stiffness is given to a floor by the insertion of bridging strips, or struts, as at a a, (Fig. 228.) These pre- vent the turning or twisting of the beams, and when a weight is placed upon the floor, concentrated over one beam, they prevent this beam from descending below the adjoining beams to the injury of the plastering upon the underside. It is usual to insert a course of bridging at every 5 to 8 feet of the length 246 AMEEIOAN HOUSE-CAKPENTEB. Fig. 228. of the beam. Strips of board or plank nailed to the underside of the floor beams to receive the lathing, are termed cross- furring, and should not be over 2 inches wide, and placed 12 inches from centres. It is desirable that all furring be narrow, in order that the clinch of the mortar be interrupted but little. "When it is desirable to prevent the passage of sound, the openings between the beams, at about 3 inches from the upper edge, are closed by short pieces of boards, which rest on cleets^ nailed to the beam along its whole length. This forms a floor, on which mortar is laid from 1 to 2 inches deep. This is called deafening. 346. — "When the distance between the walls of a building is great, it becomes requisite to introduce girders, as an addi- tional support, beneath the beams. The dimensions of girders may be ascertained by the general rules for stiffness. For- mulas (72,) (73,) and (74,) Table V., are applicable, taking f, at 86, for dwellings and ordinary stores, and increased in pro- portion to the load, up to 500, for heavy warehouses. When but one girder occurs, in the length of the beam, the distance from centres, c, is equal to one-half the length of the beam. 347. — When the breadth of a girder is more than about 12 inches, it is recommended to divide it by sawing from end to end, vertically through the middle, and then to bolt it to FKAMING. 217 gether with the sawn sides outwards. This is not to strength- en the girder, as some have supposed, but to reduce the size of the timber, in order that it may dry sooner. The opera- tion affords also an opportunity to examine the heart of the stick — a necessary precaution; as large trees are frequently in a state of decay at the heart, although outwardly they aro seemingly sound. "When the halves are bolted together, thin slips of wood should be inserted between them at the several points at which they are bolted, in order to leave sufficient space for the air to circulate between. This tends to prevent decay ; which will be found first at such parts as are not exactly tight, nor yet far enough apart to permit the escape of moisture. 348. — When girders are required for a long bearing, it is usual to truss them ; that is, to insert between the halves two pieces of oak which are inclined towards each other, and which meet at the centre of the length of the girder, like the rafters of a roof-truss, though nearly if not quite concealed within the girder. This, and many similar methods, though extensively practised, are generally worse than useless ; since it has been ascertained that, in nearly all such cases, the ope- ration has positively weakened the girder. A girder may be strengthened by mechanical contrivance, when its depth is required to be greater than any one piece oi Fig. 229. 248 AMERICAN HOUSE-CARPENTER. timber will allow. Fig. 229 shows a very simple yet mvalu able method of doing this. The two pieces of which the gir- der is composed are bolted, or pinned together, having keys inserted between to prevent the pieces from sliding. The keys should be of hard wood, well seasoned. The two pieces should be about equal in depth, in order that the joint be- tween them may be in the neutral line. (See Art. 317.) The thickness of the keys should he about half their breadth, and the amount of their united thicknesses should be equal to a trifle over the depth and one-third of the depth of the girder. Instead of bolts or pins, iron hoops are sometimes used ; and when they can be procured, they are far preferable. In this case, the girder is diminished at the ends, and the hoops driven from each end towards the middle. 349. — Beams may be spliced, if none of a sufficient length can be obtained, though not at or near the middle, if it can be avoided. (See Art. 281.) Girders should rest from 9 to 12 inches on the wall, and a space should be left for the air to circulate around the ends, that the dampness may evapo- rate. Floor-timbers are supported at their ends by walls of considerable height. They should not be permitted to rest upon intervening partitions, which are not likely to settle as much as the walls ; otherwise the unequal settlements will derange the level of the floor. As all floors, however well- constructed, settle in some degree, it is advisable to frame the beams a little higher at the middle of the room than at its sides,' — as also the ceiling-joists and cross-furring, when either are used. In single floors, for the same reason, the rounded edge of the stick, if it have one, should be placed uppermost. If the floor-plank are laid down temporarily at first, and left to season a few months before they are finally driven together and secured, the joints will remain much closer. But if the edges of the plank are planed after the first laying, they will FRAMING. 249 shrink again ; as it is the nature of wood to shrink after every planing however dry it may have been before. PARTITIONS. 350. — Too little attention has been given to the construction of this part of the frame-work of a house. The settling of floors and the cracking of ceilings and walls, which disfigure to so great an extent the apartments of even our most costly houses, may be attributed almost solely to this negligence. A square of partitioning weighs nearly a ton, a greater weight, when added to its customary load, such as furniture, storage, &c, than any ordinary floor is calculated to sustain. Hence the timbers bend, the ceilings and cornices crack, and the whole interior part of the house settles ; showing the necessity for providing adequate supports independent of the floor- timbers. A partition should, if practicable, be supported by the walls with which it is connected, in order, if the walls set- tle, that it may settle with them. This would prevent the separation of the plastering at the angles of rooms. For the same reason, a firm connection with the ceiling is an im- portant object in the construction of a partition. 351. — The joists in a partition should be so placed as to dis- charge the weight upon the points of support. All oblique pieces in a partition, that tend not to this object, are much better omitted. Fig. 230 represents a partition having a door in the middle. Its construction is simple but effective. Fig. 231 shows the manner of constructing a partition having doors near the ends. The truss is formed above the door-heads, and the lower parts are suspended from it. The posts, a and b, are halved, and nailed to the tie, c d, and the sill, e f. The braces in a trussed partition should be placed so as to form, as near as possible, an angle of 40 degrees with the horizon. In partitions that are intended to support only their own weight, 32 250 AMEEICAN HOtTSE-CAEPENTEE. n -* i II N \\A 1 T ^ I IHTi Y L) u Fig. 230. ■«* ».■' U^ m — ■ TOiU r ~~ i ill WW a & L Fig. 281. the principal timbers may be 3 X 4 inches for a 20 feet span, 3| X 5 for 30 feet, and 4 x 6 for 40. The thickness of the filling-in stuff may be regulated according to what is said at Art. 345, in regard to the width of furring for plastering. The filling-in pieces should be stiffened at about every three feet by short struts between. All superfluous timber, besides being an unnecessary load upon the points of support, tends to injure the stability of the plastering ; for, as the strength of the plastering depends, in a great measure, upon its clinch, formed by pressing the mortar FEAMING. 251 through the space between the laths, the narrower the surface, therefore, upon which the laths are nailed, the less will be the quantity of plastering unclinched, and hence its greater secu- rity from fractures. For this reason, the principal timbers of the partition should have their edges reduced, by chamfering off the corners. Fig. 2S2. 352. — "When the principal timbers of a partition. require to be large for the purpose of greater strength, it is a good plan to omit the upright filling-in pieces, and in their stead, to place a few horizontal pieces ; in order, upon these and the principal timbers, to nail upright battens at the proper dis- tances for lathing, as in Fig. 232. A "partition thus con- structed requires a little more space than others ; but it has the advantage of insuring greater stability to the plastering, and also of preventing, to a good degree the conversation of one room from being heard in the other. "When a partition is required to support, in addition to its own weight, that of a floor or some other burden resting upon it, the dftnensions of the timbers may be ascertained, by applying the principles which regulate the laws of pressure and those of the resistance of timber, as explained at the first part of this section. The following data, however, may assist in calculating the amount of pressure upon partitions : 252 AMERICAN HOUSE-CARPENTER. "WTr.te pine timber weighs from 22 to 32 pounds per cubic foot, varying in accordance with the amount of seasoning it has had. Assuming it to weigh 30 pounds, the weight of the beams and floor plank in every superficial foot of the flooring will be, when the beams are 3x8 inches, and placed 20 inches from centres, 6 pounds. 3 X 10 " " " 18 " " " 1% " 3 x 12 " " " 16 " " " 9 " 3 x 12 " " " 12 " " " 11 " 4 x 12 " " " 12 " " " 13 4 x 14 " " " 14 " " " 13 " In addition to the beams and plank, there is generally the plastering of the .ceiling of the apartments beneath, and some- times the deafening. Plastering may be assumed to weigh 9 pounds per superficial foot, and deafening 11 pounds. Hemlock weighs about the same as white pine. A parti- tion of 3 x 4: joists of hemlock, set 12 inches from centres, therefore, will weigh about 2-J- pounds per foot superficial, and when plastered on both sides, 20-J pounds. 353. — "When floor beams are supported at the extremities, and by a partition or girder at any point between the extre- mities, one-half of the weight of the whole floor will then be supported by the partition or girder. As the settling of parti- tions and floors, which is so disastrous to plastering, is fre- quently owing to the shrinking of the timber and to ill-made joints, it is very important that the timber be seasoned and the work well executed. Where practicable, the joists of a partition ought to extend down between the floor beams to the plate of 'the partition beneath, to avoid the settlement con- sequent upon the shrinkage of the floor beams. ROOFS.* 354. — In ancient b.iildings, the -Norman and the Gothic, the • See also Art. 238. FRAMING. 253 walls and buttresses were erected so massive and firm, that it was customary to construct their roofs without a tie-beam; the walls being abundantly capable of resisting the lateral pressure exerted by the rafters. Bat in modern buildings, the walls are so slightly built as to be incapable of resisting scarcely any oblique pressure; and hence the necessity of constructing the roof so that all oblique and lateral strains may be removed; as, also, that instead of having a tendency to separate the walls, the roof may contribute to bind and steady them. 355. — In estimating the pressures upon any certain roof, for the purpose of ascertaining the proper sizes for the timbers, calculation must be made for the pressure exerted by the wind, and, if in a cold climate, for the weight of snow, in addition to the weight of the materials of which the roof is composed. The weight of snow will be of course according to the depth it acquires. Snow weighs 8 lbs. per cubic foot, and more when saturated with water. In a severe climate, roofs ought to be constructed steeper than in a milder one, in order that the snow may have a tendency to slide off before it becomes of sufficient weight to endanger the safety of the roof. The inclination should be regulated in accordance with the qualities of the material with which the roof is to be covered. The following table may be useful in determining the smallest in- clination, and in estimating the weight of the various kinds of covering : Material. Tin Copper Lead . Zinc Short pine shingles Long cypress shingles Slate . Inclination. Rise 1 inch to a foot 1 " 2 inches " O it It 5 " " Weight upon a square foot. | to 1 J lbs. 1 to \\ " 4 to 7 " 1£ to 2 " 1 1 to 2 " 2 to 3 " 5 to 9 " U54 AMERICAN HO'fSE-CARPENTEK. The weight of the covering, as above estimated, is that of the material only, with the weight of whatever is used to fix it to the roof, such as nails, &c. What the material is laid on, such as plank, boards or lath, is not included. The weight of plank is about 3 pounds per foot superficial ; of boards, 2 pounds ; and lath, about a half pound. 356. — The weights and pressures on a roof arise from the roofing, the truss, the ceiling, wind and snow, and may be stated as follows : First, the Hoofing. — On each foot superficial of the inclined surface, Slating will weigh about *l lbs. Eoof plant, 1£ inches thick . . " " " 1*1 " Eoof beams or jack rafters . . " " " 2'3 " Total, 12 lbs. 'This is the weight per foot on the inclined surface ; but it is desirable to know how much per foot, measured horizontally, this is" equal to. The horizontal measure of one foot of the inclined surface is equal to the cosine of the angle of inclina- tion. Therefore, cos. : 1 :: p : w = -±— ; cos. where p represents the pressure on a foot of the inclined sur- face, and w the weight of the roof per foot, measured horizon- tally. The cosine of an angle is equal to the base of the right- angled triangle divided by the hypothenuse, which in this case would be half the span divided by the length of the rafter, or — , where s is the span, and I the length -f the rafter. Hence, p _ p _%lp cos. " ± "" ~T" 5 or, twice the pressure per foot of inclined surface, multiplied FRAMING. 255 T>y the length of the rafter, and divided by the span, will give the weight per foot measured horizontally ; or, 24 - = w (182.) equals the weight per foot, measured horizontally, of the roof beams, plank, and covering for a slate roof. Second, the Truss. — The weight of the framed truss is nearly in proportion to the length of the truss, and to the distance apart at which the trusses are placed. w = 5-2 cs (183.) equals the weight, in pounds, of a white pine truss with iron suspension rods and a horizontal tie beam, near enough for the requirements of our present purpose ; where s equals the length or span of the truss, and c the distance apart at which the trusses are placed, both in feet. It is desirable to know how much this is equal to per foot of the area over which the truss is to sustain a covering. This is found by dividing the weight of the truss by the span, and by the distance aDart from centres at which the trusses are placed ; or, ^f = 5-2 = w (184) cs equals the weight in pounds per foot to be allowed for the truss. Third, the Ceiling. — The weight supported by the tie beams, viz. : that of the ceiling beams, furring and plastering, is about 9 pounds per superficial foot. Fourth, the Wind. — The force of wind has been known as high as 50 pounds per superficial foot against a vertical sur- face. The effect of a horizontal force on an inclined surface is in proportion to the sine of the angle of inclination, the ef- fect produced being in the direction at right angles to the in- clined surface. The force thus acting may be resolved into forces acting in two directions — the one horizontal, the other vertical ; the former tending, in the case of a roof, to thrust 256 AMEEICAK H0TTSE-CAKPENTEK. aside the walls on which, the roof rests, and the latter acting directly on the materials of which the roof is constructed — this latter force heing in proportion to the sine of the angle of inclination multiplied by the cosine. This is the vertical effect of the wind upon a roof, without regard to the surface it acts upon. The wind, acting horizontally through one foot superficial of vertical section, acts on an area of inclined sur- face equal to the reciprocal of the sine of inclination, and the horizontal measurement of this inclined surface is equal to the cosine of the angle of inclination divided by the sine. This is the horizontal measurement of the inclined surface, and the vertical force acting on this surface is, as above stated, in pro- portion to the sine multiplied by the cosine. Combining these, it is found that the vertical power of the wind is in proportion to the square of the sine of the angle of inclination. There- fore, if the power of wind against a vertical surface be taken at 50 pounds per superficial foot, then the vertical effect on a roof is equal to w = 50 sin. 2 = 50 ~ (185.) for each piece of the inclined surface, the horizontal measure- ment of which equals one foot ; where I equals the length of the rafter, and A the height of the roof. Fifth, Snow.— The weight of snow will be in proportion to the depth it acquires, and this will be in proportion to the rigour of the climate of the place at which the building is to be erected. Upon roofs of most of the usual inclinations, snow, if deposited in the absence of wind, will not slide off. When it lias acquired some depth, and not till then, it will have a tendency, in proportion to the angle of inclination, to slide off in a body. The weight of snow may be taken, there- fore, at its weight per cubic foot, 8 pounds, multiplied by the depth it is usual for it to acquire. This, in the latitude of New York, may be stated at about 2^ feet. Its weight would, FEAMtNQ. 257 therefore, be 20 pounds per foot superficial, measured horizon- tally. 357. — There is one other cause of strain upon a roof; namely, the load that may be deposited in the roof when used as a room for storage, or for dormitories. But this seldom occurs. When a case of this kind does occur, allowance is to be made for it as shown in the article on floors. But in the general rule, now under consideration, it may be omitted. 358. — The following, therefore, comprehends all the pressures or weights that occur on roofs generally, per foot superficial ; For roof beams, plank, and slate (182) " the truss (184) " ceiling " wind (185) " snow, latitude of New York 24 - lbs. s 52 " 9 " 60 j," 20 Having found the weight per foot, the total weight for any part of the roof is found by multiplying the weight per foot by the area of that part. This process will give the weight sup- ported by braces and suspension rods, and also that supported by the rafters and tie beam. But in these last two, only half of the pressure of the vjind is to be taken, for the wind will act only on one side of the roof at the same time. The vertical pressure on the head of a brace, then, equals W = 4 en(6 l - + SSS + 12-5 1) (186.) And W = cp n, where p equals 4u>- + 8-55 + 12-5 j 2 j, equals the weight per foot. And the aggregate load of the roof on each truss equals W = 4 o s (6 1 + 8-55 + 6-25 1) (187.) 7 7i2 And W = o q s, where q = 4(6 - + 8-55 + 6-25 -^), equals the 33 25S AMERICAN II0U8E-CABPENTEK. weight per foot ; where c equals the distance apart from centres at which the trusses are placed ; n the distance horizontally between the heads of the braces, or, if these are not located at equal distances, then n is the distance horizontally from a point half-way to the next brace on one side to a point half-way to the next brace on the other side ; I the length of the rafter ; a the span, and h the height — all in feet. 359. — By the parallelogram of forces, the weight of the roof is in proportion to the oblique thrust or pressure in the axis of the rafter, as twice the height of the roof is to the length of the rafter; or, W : R :: 2 h : I, or W1 2A : I :: W : R = ~-j-, (188.) where R equals the pressure in the axis of the rafter. And the weight of the roof is in proportion to the horizontal thrust in the tie beam, as twice the height of the roof is to half the span; or, W : Hv.Zh :i-, or 2A:|:: W: H=¥j., (189.) where H equals the horizontal thrust in the tie beam ; the value of W in (188) and (189) being shown at (187), and (18?) being compounded as explained in Art. 356. The weight is that for a slate roof. If other material is used for covering, or should there be other conditions modifying the weight in any particular case, an examination of Art. 356 will show how ,to modify the formula accordingly. 360. — The pressures may be obtained geometrically, as shown in Fig. 233, where A B represents the axis of the tie beam, A C the axis of the rafter, D E and F B the axes of the braces, and D G, FF, and OB, the axes of the suspension rods. In this design for a truss, the distance A B is divided into three FRAMING. 259 equal parts, and the rods located at the two points of division, G and E. By this arrangement the rafter A G is supported at .equi-distant points, D and F. , The point D supports the rafter for a distance extending half-way to A and half-way to F, and the point F sustains half-way to D and half-way to 0. Also, the point G sustains half-way to F and, on the other rafter, half-way to the corresponding point to F. And because these points of support are located at equal distances apart, there- fore the load on each is the same in amount. On D Gf make Da equal to 100 of any decimally divided scale, and let Da represent the load on D, and draw the parallelogram aiDc. Then, by the same scale, D b represents {Aft. 258) the pressure in the axis of the rafter by the load at D / also, D o the pressure in the brace D F. Draw o d horizontal ; then D d is the ver- tical pressure exerted by the brace D E at E. The point F sustains, besides the common load represented by 100 of the scale, also the vertical pressure exerted by the brace DE/ therefore, since D a represents the common load on D, F, or O, make Fe equal to the sum of Da and D d, and draw the parallelogram Fg ef. Then Fg, measured by the scale, is the pressure in the axis of the rafter caused by the load at F, and Ff is the load in the axis of the brace F B. Draw fh horizontal ; then Fh is the vertical pressure exerted by the brace F B at B. The point G, besides the common load re- presented by D a, sustains the vertical pressure Fh caused by the brace FB. and a like amount from the corresponding brace on the opposite side. Therefore, make Gj equal to the sum of D a and twice Fh, and draw j h parallel to the opposite rafter. Then Ch is the pressure in the axis of the rafter at G. This is not the only pressure in the rafter, although it is the total pressure at its head G. At the point F, besides the pressure Ck, there is Fg. At the point D, besides these two pressures, there is the pressure D b. At the foot, at A, there is still an additional pressure : while the point D sustains the load half- 260 AMERICAN HOUSE-CARPENTER. way to F and half-way to A, the point A sustains the load half-way to D. This load is, in this case, just half the load at D. Therefore draw A m vertical, and equal to 50 of the scale, or half of D a. Extend A to I ; draw m I horizontal. Then A. I is the pressure in the rafter at A caused by the weight of the roof from A half-way to D. Now the total of the pressures in the rafter is equal to the sum of A I + D i f Eg added to CJc. Therefore make Tc n equal to the sum of A I + D o + Fg, and draw n o parallel with the opposite rafter, and nj hori- zontal. Then Co, measured by the same scale, will be found equal to the total weight of the roof on both sides of B. If Da — 100 represent the portion of the weight borne by the point D, then Co, representing the whole weight of the roof, should equal 600, (as it does by the scale,) for D supports just one-sixth of the whole load. As Cn is the total oblique thrust in the axis of the rafter at its foot, therefore nj is the horizon- tal thrust in the tie beam. 361. — In stating the amount of pressures in the above as being equal to certain lines, it was so stated with the under- standing that the lines were simply in proportion to the weights. To obtain the weight represented by a line, multiply its length (measured by the scale used) by the load resting at D, (or at i^or O, as these are all equal in this example,) and divide the product by 100, and the quotient will be the weight required. For, as 100 of the scale is to the load it represents, so is any other dimension on the same scale to the load it represents. 362.— Example. Let A B {Fig. 233) equal 26 feet, OB 13 feet, and A O 29 feet, and AG-, G E, and EB, each 8f feet. Tet the trusses be placed 10 feet apart. Then the weight on D, for the use of the braces and rods, is, per (186), equal to 4:cnn>l + 8-55 + 12£^) = 4 x 10 X 8=(6 x g + 8-55 + 12* X g-J FEAMING. 2G; = 346$ x 14-398 = 4991-3. This is the common load at the points D, F, and O, and each of the lines denoting pressures multiplied by it and divided by 100, or multiplied by the quotient of t!j£lL? = 49-913, the pro- duct will be the weight required. 49-913 may be called 50, for simplicity ; therefore the pressure in the brace DE equals 112 x 50 = 5600 pounds, and in the brace FB, 140 x50^- 7000 pounds, and in like manner for any other strain. For the rafters and tie beam the total weight, as per (187), equals 4cs(6^+8-55 + 6i^) = 4 x 10 x 52(6 x g + 8-55 + 6* X g) = 2080 x 13-148 = 27343-68 pounds. This is the total weight of the roof supported by one truss. The oblique thrust in the rafter A G is, per (188), equal to IW_29 x 27343-68 2 A 2x13 = 30498-72 pounds. To obtain this oblique thrust geometrically : Go (Fig. 233) represents the weight of the roof, and measures 600 by the scale ; and the line Gn, representing the oblique thrust, measures 670. By the proportion, 600 : 670 :: 27343-68 : 30533-8, = the oblique thrust. The result here found is a few pounds more than the other. This is owing to the fact that the line Gn is not exactly 670, nor is the length of the rafter precisely 29 feet. Were the exact dimensions used in each case the results would be identical ; but the result in either case is near enough for the purpose. The horizontal strain is, per (189), equal to 262 AMERICAN HOUSE-CARPENTER. Ws _ 27343- 68 X 52 4A ~ 4x 13 = 27343-68 pounds. The result gives the horizontal thrust precisely cquai iu w« weight. This is as it should be in all cases where the height of the roof is equal to one-fourth of the span, hut not other- wise ; for the result depends (189) upon this relation of the height to the span. Geometrically, the result is the same, for Co and nj {Fig. 233,) representing the weight and horizontal thrust, are precisely equal by measurement. 363. — The weight at the head of a brace is sustained partly at the foot of the brace and partly at the foot of the rafter. The sum of the vertical effects at these two points is just equal to the weight at the head of the brace. The portion of the weight sustained at either point is in proportion, inversely, to the horizontal distance of that point from the weight ; there- fore, V=W$-, (190.) where V equals the vertical effect at the foot of the brace ; W, the weight at the head of the brace ; g, the horizontal distance from the foot of the rafter to the head of the brace ; and a, the distance from the same point to the foot of the brace. 364. — For the oblique thrust in the brace : from the triangle Ffh {Fig. 233,) Fh : Ff :: sin. : rad. sin. : rad. :: V : T; therefore, T = Z- = rL (191.) sin. A ' where T equals the oblique thrust in the brace ; V, the verti- cal pressure caused by T at the foot of the brace (190) ; a I and h the length and height respectively of the brace. 365. — Example. Brace 1) E, Fig. 233. In this case, equals the product of the weight per superficial foot, m^^ ; FEAMING. 263 plied by the area supported at the point D, equals 5000 pounds, {Art. 362.) The length g equals 8f feet, and a equals 17* feet. Therefore (190), V= W$- = 5000 X ^| = 2500 pounds equals the vertical pressure at E caused by the brace D E. Ther for the oblique thrust, I equals 9-6 feet, and h equals 4-3 feet. Therefore, from (191), T = v\ = 2500 x ?4 = 65814 pounds equals the oblique thrust in the brace D E. In Art. 362 it was found to be 5600. The discrepancy is owing to like causes of want of accuracy in the case of the rafter, as explained in Art. 362. Another Example. — "Brace FB, Fig. 233. In' this case, W equals the product of the weight per superficial foot, multiplied by the area supported by the point F, added to the vertical strain caused by the brace J) E. From Art. 362 the weight of roof on F equals 5000 pounds, and the vertical strain from brace D E is, as just ascertained, = 2500, total 7500, equals W. The length, g, equals two-thirds of 26 feet, equals 17*, and a equals 26 feet. Therefore, from (190), V= W$- = 7500 x ~ = £000 a 26 equals the vertical effect at B caused by the brace FB. Then, for the oblique thrust in the brace, I equals 32 - 2, and h equals 8f . Therefore, from (191) 7 19*9 T= V~ = 5000 x ==f- = 7038-5 h 8f equals the oblique thrust or strain in the axis of the brace. It was 7000 by the geometrical process, (Art. 362.) 366. — The strain upon the first rod, D G, equals simply the weight of the ceiling supported by it, added to the part of the tie beam it sustains. The weight of the tie beam will equal 264 AMERICAN HOUSE-CAKPENTEB. fie. 288. FRAMING. 265 about one pound per superficial foot of the ceiling. The weight per foot for the ceiling is stated (see Art. 356 third, and 358,) at 9 pounds. To this add 1 pound for the tie beam, and tho sum is 10. Then JV= 10 en. (192.) The strain on the second tie rod equals the weight of ceiling supported, = ]¥, added to the vertical effect of the strain in the brace it sustains, [see (190)] or equal to = 10 en + V. (193.) The strain on the third rod is equal to i\T, added to the ver- tical effect of the strain in the brace it sustains, and this is the strain on any rod. The first rod has no brace to sustain, and the middle rod sustains two braces. In this case the strain equals U = 10 c n + 2 V. (194.) It may be observed that V represents the vertical strain caused by that brace that is sustained by the rod under consi- deration ; and, as the vertical strain caused by any one brace is more than that caused by any other brace nearer the foot of the rafter, therefore the V of (193) is not equal to the V of (194). Hence a necessity for care lest the two be confounded and thus cause error. 367.— Examples. The rod D G [Fig. 233) has a strain which equals (192) N= 10 e n = 10 x 10 x 8$ = 867 pounds. The strain on rod FE equals (193) O = 10 on + V- 867 + 2500 = 3367 pounds. The strain on rod OB, the middle rod, equals (194) U= 10 c n + 2 V= 867 + 2 x 5000 = 10867 pounds. 368. — The load, and the strains caused thereby, having been discussed, it remains to speak of the resistance of the ma- terials. First, of the Rafter.— Generally this piece of timber is so pinioned by the roof beams or purlins as to prevent any late- 34 266 AMERICAN HOUSE-CARPENTER. ral movement, and the braces keep it from deflection ; there- fore it is not liable to yield by flexure. Hence the manner of its yielding, when overloaded, will be by crushing at the ends, or it will crush the tie beam against which it presses. The fibres of timber yield much more readily when pressed toge- ther by a force acting at right angles to the direction of their length, than when it acts in a line with their length. The value of timber subjected to pressure in these two ways is shown in Art. 292, Table i., the value per square inch of the first stated resistance being expressed by P, and that of the other by 0. Timber pressed in an oblique direction yields with a force exceeding that expressed by P, and less than that by 0. When the angle of inclination at which the force acts is just 45°, then the force will be an average between P and G. And for any angle of inclination, the force will vary in- versely as the angle ; approaching P as the angle is enlarged, and approaching G as the, angle is diminished. It will be equal to G when the angle becomes zero, and equal P when the angle becomes 90°. The resistance of timber per square inch to an oblique force is therefore expressed by M = P + ^(G~P), (195.) where A° equals the complement of the angle of inclination. In a roof, A° is the acute angle formed by the rafter with a vertical line. If no convenient instrument be at hand to measure the angle, describe an arc upon the plan of the truss — thus: with GB {Fig. 233) for radius, describe the arc B g, and get the length of this arc by stepping it off with a pair of dividers. Then where a equals the length of the arc, and h equals B G, the height of the roof. Therefore, FRAMING. 267 M=P + 0-63^{O-P) (196.) equals the value of timber per square inch in a tie beam, C and P being obtained from Table L, Art. 292. When C for the kind of wood in the tie beam exceeds set opposite the kind of wood in the rafter, then the latter is to be used in the rules instead of the former. 369. — Having obtained the strain to which the material is subjected in a roof, and the capability of the material to resist that strain, it only remains now to state the rules for determin- ing the dimensions of the material. 370. — To obtain the dimensions of the rafter : — It has been shown that the strain in the axis of the rafter equals (188), 2 A This is the strain in pounds. Timber is capable of resisting effectually, in every square inch of the surface pressed (196), P + 0-63| | (O - P) pounds. And when the strain and resistance are equal, P = bd[P + 0-63| | (C - P)l where b and d are respectively the breadth and depth of the rafter. Hence ld = - a -. (197.) P + 0-63f f(<7-P) Example. — {Fig. 233.) The strain in the axis of the rafter m this example, ascertained in Art. 362, is 30498-72 pounds. If the timber used be white pine, then P = 300 and G— 1200, The length of the arc Bg is 14J feet, and h = 13. Therefore bd = 30498 - 7 f 4 ,. = 32-8. 300 + (0-63f x tt x 900) This is the area of the abutting surface at the tie beam — say 6 by 5^- inches. At least half this amount should be added 268 AMERICAN HOUSE-CARPENTER. to allow for the shoulder, and for cutting at the joints for braces, &c. The rafter may therefore be 6 by 9 inches. The above method is based upon the supposition that the rafter is effectually secured from flexure by the braces and roof beams. Should this not be the case, then the dimensions of the rafter are to be obtained by rules in Art. 298, for posts. ^Nevertheless, the abutting surface in the joint is to be deter- mined by the above formula (197). 371. — To obtain the dimensions of the braces: — Usually, braces are so slender as to require their dimensions to be ob- tained by rules in Art. 298 ; the strain in the axis of the brace having been obtained by formula (191), or geometrically as in Art. 360. The abutting surface of the joint of the brace is to be ob- tained, as in the case of the rafter, by formula (195) ; A° be- ing the number of degrees contained in the acute angle formed by the brace and a vertical line, for the joint at the tie beam ; but for the joint at the rafter, A° is the number of degrees contained in the acute angle formed by the brace and a line perpendicular to the rafter, or it is 90, diminished by the num- ber of degrees contained in the acute angle formed by the rafter and brace. Example. — Fig. 233, Brace D E, of white pine. In this brace the strain was found {Art. 362) to be 5600 pounds, the length of the brace is 9 - 6 feet. By Art. 298, the brace is therefore required to be 4 - 18 X 6 inches. For the abutting surface at the joints, for white pine, P equals 300 and, C1200. The angle D EF equals 63° 26'. By (197) and (195), ld= T 5600 ~ P + f {JO - P) ~ 300 + [-;f 6 ' x (1200 - 300)] 5600 . . . = 9-3TB = 6 mcheS - r This is the area of the abuttirg surface of the joint at the tie FRAMING. 269 beam. To obtain the joint at the rafter, the angle FD F. equals 53° 8', and hence T 5600 7>d = P + ^(C-P) 300 + [—Jr 1 - x (1200 - 300)] 5600 D „_ . . ■ = 8-375 inches. 300 + (^- x 900) " This is the area of the abutting surface of the joint at the rafter. Anotlier Example. — Brace FB, Fig. 233, of white pine, 12-2 feet long. The strain in its axis is (AH. 362) 7000 pounds. By Art. 298, the brace is required to be 5J x 6 inches. For the abutting surface of the joints, P equals 300, C equals 1200, and the angle I B G equals 45° ; there/ore, i .7 woo nl . , od = = = 9$ inches. 300+ [| x(1200-300)] This is the area of the abutting surface at the tie beam. For the surface at the rafter, the angle G FB equals 71°, and 90 — 71 = 19, equals the angle to be used in the formula ; therefore, hd = — ^ = 14'3 inches, nearly. 300 + J X (1200-300)] This is the area of the abutting surface of the joint at the rafter. 372. — To obtain the dimensions of the tie beam : — A tie beam must be of such dimensions as will enable it to resist effectually the tensile strain caused by the horizontal thrust of the rafter and the cross strains arising from the weight of the ceiling, and from any load that may be placed upon it in the roof. From (17), Art. 310, . w H A ~T~Y where H equals the horizontal thrust, and from (189), 270 AMERICAN HOUSE-CARPENTER. therefore, A _B__ Ws T ~ ±hT where W equals the weight of the roof in pounds, as shown at (187); s, the span ; h, the height, both in feet; and T, a constant set opposite the kind of material, in Table III. ; and A equals the area of uncut fibres in the tie beam. About one-half of this should be added to allow for the requisite cut- ting at the joints ; or, the area of the cross section of the tie beam should be equal to at least -| of the area of uncut fibres ; or, when h d equals the area of the tie beam, then ^=tS- ■ (198) - Example. — The weight on the truss at Fig. 233 is shown to be (Art. 362) 27343-68 pounds, say 27500 pounds ; the span is 52 feet, the height 13, and the value of T for white pine is (Table III.) 2367, therefore , , . Ws . 27500 x 52 .,„.,., 1 d = f TT = f x IslTW = m mches equals the area of cross section of the tie beam requisite to resist the tensile strain. This is smaller, as will be shown, than what is required to resist the cross strains, and this will be found to be the case generally. The weight of the ceiling is 9 pounds per superficial foot; the length of the longest unsup- ported part of the tie beam is 8f feet ; then, if the deflection per lineal foot be allowed at - 015 inch, the depth of the tie beam will be required ((72), Table Y.) to be 6'M inches. But in order effectually to resist the strains tie beams are subjected to at the hands of the workmen, in the process of framing and elevating, the area of cross section in inches should be at least equal to the length in feet. Were it possible to guard against this cause of strain, the size ascertained by the rule, 6 TEAMING. ift 1 by 6-14, would be sufficient ; but to resist this strain, the size should be 6 by 9. There is yet one other dimension for the tie beam required, and that is, the distance at which the joint for the rafter must be located from the end of the tie beam, in order that the thrust of the rafter may not split off the part against which it presses. This may be ascertained by Rule XI., Art. 302, for all cases where no iron strap or bolt is used to secure the joint ; but where these fastenings are used the abutment may be of any convenient length. And in using irons here, care should be exercised to have the surface of pressure against the iron of sufficient area to prevent indentation. 373. — To obtain the dimensions of the iron suspension rods. By Art. 310, (IT), A - w and T varies (Table III.) from 5000 to 17000, according to the diameter inversely ; for the smaller rods are stronger in pro- portion than the larger ones. Example.— Taking T equal 5000, then the area of the rod D O {Fig. 233) requires {Art. 367) to be equal to corresponding to 0469 inch diameter. This rod may be half inch diameter. Another Example.— The rod FE {Fig. 233) is loaded with {Art. 367) 3367 pounds, therefore equals the area of the rod, the corresponding diameter of which is 0-925. This rod may be one inch diameter. Again, a third example; the rod C B. This rod is loaded with {Art. 367) 10867 pounds, therefore 272 AMERICAN HOUSE-CARPENTER. . 10867 01fr0 . , equals the required area of the rod, the diameter correspond- ing to which is 1*66. This rod may therefore be If inches diameter. 374. — While discussing the principles of strains in roofs and deducing rules therefrom, the truss indicated in Fig. 233 has been examined throughout. The result is as follows : rafter, 6x9; tie beam, (6 X 6, or) 6x9; the first brace from the wall, 4J x 6 inches, with an abutting surface at the lower end of 6 inches, and at the upper end of 8f inches; the other brace, 5J x 6 inches, with an abutting surface at the lower end of 9£ inches, and at the upper end of 14 T 3 „ inches ; the shortest rod, % inch diameter ; the next, 1 inch diameter ; and the middle rod, If inches diameter. PRACTICAL RULES AND EXAMPLES. For Roofs Loaded as per Art. 356. 375. — Rule Lin. To obtain the dimensions of the rafter Multiply the value of R (Table IX., Art. 376) by the span of the roof, by the length of the rafter, and by the distance apart from centres at which the roof trusses are placed, all in feet, and divide the product by the sum of twice the height of the roof multiplied by the value of P, Table I., set opposite the kind of wood used in the tie beam, added to the difference of the values of C and P in said table multiplied by 14 times the length of the arc that measures the acute angle formed between the rafter and a vertical line, the arc having the height of the roof for radius (see arc B G, Fig. 233), and the quo- tient will be the area of the abutting surface of the joint at the foot of the rafter. To the abutting surface add its half, and the sum will be the area of the cross section of the rafter FRAMING. 273 This rule is upon the presumption that the rafter is secured from flexure by the roof beams and by braces and ties at short intervals, as in Fig. 233. In roofs where the rafter does not extend up to the ridge of the roof but abuts against a horizontal straining beam (c, Fig. 237), in the rule for rafters, take for the length of the rafter the distance from the foot of the rafter to the ridge of the roof; or, a distance equal to what the rafter would be in the absence of a straining beam. The area of cross section of the straining beam should be made equal to that of the rafter, as found by the rule so modified. Example. — Find the dimensions of a rafter for a roof truss whose span is 52 feet, and height 13 ; the length of the rafter being 29 feet, the trusses placed 10 feet apart from centres, and the arc measuring the angle at the head of the rafter (having the height of the roof for radius) being 144 feet, white pine being used in the tie beam. The height of this roof being in proportion to the span as 1 to 4, the value of S, in Table IX. is 52 - 6 ; multiplying this, in accordance with the rule, by 52, the span of the roof, and by 29, the length of the rafter, and by 10, the distance between the roof trusses, the product is 793208. The value of P for white pine in Table I. is 300 ; multiplying this by 2 x 13 = 26, twice the height of the roof, the product is 7800. The. value of C for white pine, (Table I.) is 1200, hence the difference of the values of O and P is 1200-300 = 900; this multiplied by li, and by 144, the length of the arc, the product is 16031 ; this added to the 7800 aforesaid, the sum is 23831. The aforesaid product of 793208, divided by this 23831, the quotient, 33"3, equals the area in inches of the abutting surface of the joint at the tie beam. To this add 16'7, its half, and the sum, 50, equals the area of cross section of the rafter. This divided by the thick- ness of the rafter, say 6 inches, the quotient, 8&, is the breadth. The rafter is therefore to be 6 X Si inches. It may be made 6x9, avoiding the fractions. 35 su AMERICAN HOJSE-CAKrENTER. 376. — The following table, calculated upon data in Art. 358, presents the weight per foot for roofs of various inclinations, and covered with slate. TABLE IX When height of roof The yertical strain per foot of surface supported, measured horizontally, is to 6pan as on rafters •= JR = on braces = Q =* 1 to 8 1 " 7 48 pounds 48-6 " 49 o pounds. 60-5 I " 6 49-4 " 51-9 ( 1 " 5 50-6 " 54- < 1 " 4 52-6 " 57-6 ( 1 « 3 66-3 " 64' ' 1 " 2 63'7 " 76-2 c 1 " 1 81- " 101- To get the proportion that the height bears to the span, di- vide the span by the height ; then unity will be to the quotient as the height is to the span. In case the quotient is not a whole number, the required value of It or Q will not be found in the above table, but may be obtained thus : multiply the decimal part of the quotient by the difference of the values of It set opposite the two proportions, between which the given proportion occurs as an intermediate, and subtract the product from the larger of the two said values of It ; the remainder will be the value of It required. The process is the same for the values of Q. Example. — A roof whose span is 60 feet, has a height of 25 feet. Then 60 divided by 25 equals 24. The proportion, therefore, between the height and span is 1 to 2-4. This pro- portion is an intermediate between 1 to 2 and 1 to 3. The values of It, opposite these two, are 63*7 and 56 - 3. The dif- ference between these values is 7'4 ; this multiplied by 04, the decimal portion of the quotient, equals 2-96 ; this subtracted from 63-7, the larger value of It, the remainder, 60*74, is the required "Mine of It. FRAMING. 275 The values of R and Q are those for a roof covered with slate weighing 7 pounds per superficial foot of the roof sur- face. When the roof covering is either lighter or heavier, sub- tract from or add to the table values, the difference of weight between 7 pounds and the weight of the covering used, and the remainder, or sum, will be the value of R or Q required. 377.— Mule LIY. To obtain the dimensions of braces. Multiply the value of Q (Table IX., Art. 376) by the distance apart in feet at which the roof trusses are placed, and by the horizontal distance in feet from a point half-way to the next point of support of the rafter on one side of the brace, to a corresponding point on the other side. The product will be the weight in pounds sustained at the head of the brace. To this add the vertical strain (Art. 360) on the suspension rod located at the head of the brace, and make a vertical line dropped from the head of the brace, as Fe, Fig. 233, equal, by any convenient scale, to this sum, and draw the parallelo- gram Ffe g. Then Ff, measured by the same scale, equals the pressure in the axis of the brace F B. Multiply this pres- sure in pounds by the square of the length of the brace in feet, and divide the product by the breadth of the brace in inches multiplied by the value of B (Table II., AH. 293). The cube root of the quotient will be the thickness of the brace in inches. If this cube root should exceed the breadth of the brace, the result is not correct, and the calculation will have to be made anew, taking a larger dimension for the breadth. Example.— The brace FB (Fig. 233) is of white pine, and is required to sustain a pressure in its axis of 7000 pounds (Art. 362). The length of the brace is 12 feet and its breadth 6 inches, what must be its thickness ? Here 7000, the pressure, multiplied by 144, the square of the length, equals 1008000. The value of B is' 1175 ; this by 6, the breadth of the brace, equals 7050. The product 1008000 divided by the product 7050 equals 143, the cube root of which, 5-23, is the required 276 AMERICAN HOTJSE-CAKPENTEE. thickness of the brace in inches. The brace will therefore be 5 - 23 by 6 inches, or 5$ by 6. 378. — Rule LY. To obtain the area of the abutting sur- face of the ends of braces. Divide the number of degrees contained in the complement of the angle of inclination by 90, and multiply the quotient by the difference of the values of/7 and P, set opposite the kind of wood in the tie beam or rafter, in Table I., Art. 292 ; and to the product add the said value of P, and by the sum divide the pressure in the axis of the brace, and the quotient will be the area of the abutting sur- face. The complement of the angle of inclination referred to is, for the foot of the brace, the acute angle contained between the brace and a vertical line ; and for the head of the brace, the acute angle contained between the brace and a line per- pendicular to the rafter. Example. — To find the abutting surface of the ends of the brace FB {Fig. 233). The complement of the angle of incli- nation, for the foot of the brace, is that contained between the lines F B and F E, and measures by the protractor, 45°. The tie beam is of white pine, and the values of P and C for this wood are 300 and 1200 respectively, and the pressure in the axis of the brace is 7000 pounds. Now by the rule, 45 di- vided by 90 equals - 5, this by the 900, the difference of the values of O and P, equals 450 ; to this add 300, the value of P, and the sum is 750. The pressure in the axis of the brace, 7000, divided by this 750, equals 91, the required area of the abutting surface at the foot of the rafter. The complement of the angle of inclination for the head of the brace is that con- tained between the lines B F and Fp, and measures by the protractor 19°. The rafter being of white pine, the values of P and O are as before. By the rule, 19 divided by 90 equals 0'2£, and this multiplied by 900, the difference of the values of P and C, equals 190 ; to this add 300, tbe value v^P, and FRAMING. 277 the sum is 490. The pressure, 7000, divided by this 490, equals 14-3 inches, the required area of the abutting surface at the head of the brace. 379. — To obtain the dimensions of the tie beam. Tie beams are subjected to two kinds of strain — tensile and transverse. Rule LYI. — To guard against the tensile strain, multiply the value of R (Table IX., Art. 376) by three times the' distance apart at which the trusses are placed, and by the square of the span of the truss, both in feet. Divide this product by the value of T, (Table III., Art. 308) set opposite the kind of wood in the tie beam, multiplied by 8 times the height of the roof in feet, and by the breadth of the tie beam in inches. The quotient will be the required depth in inches. The result thus obtained is usually smaller than that re- quired to resist the cross strain to which the tie beam is sub- jected. The dimensions required to resist this strain, where there is simply the weight of the ceiling to support, may be obtained by this rule : Rule LVII. — Multiply the cube of the longest unsupported part of the tie beam by 400 times the distance apart at which the trusses are placed, both in feet ; and divide the product by the breadth of the tie beam in inches, multiplied by the value of E, (Table II., Art. 293) set opposite the kind of wood in the tie beam, and the cube root of the quotient will be the re- quired depth of the tie beam in inches. The result thus obtained may not be sufficient, in some cases, to resist the strains to which the tie beam is subjected in the hands of the workmen during the process of framing. Rule LVin. — To resist these strains the area of cross sec- tion in inches should be at least equal to the length in feet. Example. — The tie beam in Fig. 233. For this case we have the value of R 52*6, the trusses placed 10 feet from centres, the span 52 feet, the height 13 feet, the breadth 6 inches, and the value of T 2367. Then by the rule, 52-6 X 3 x 10 x 52* 278 AMERICAN HOUSE-CARPENTER. = 4266912, and 2367 X 8 x 13 x 6 = 1477008 ; the formei product divided by the latter, the quotient equals 2 - 9, equals the required depth of the tie beam in inches. The other strains will require the depth to be more. To resist the cross strains, we have the longest unsupported part of the tie beam 8f feet, (this dimension is frequently greater than this,) distance from centres 10 feet, and breadth 6 inches. Then, by the rule, 8| 3 x 400 x 10 = 2603852, and 6 x 1750 = 10500 ; the former product divided by the latter, the quotient is 248, the cube root of which, 6 - 28, equals the required depth in inches. The tie beam therefore is to be 6 by 6'28 inches, or 6 X 7 inches. But if not guarded against severe accidental strains from careless handling this size would be too small. It would, in this case, require to be 52 inches area of cross section, say 6x9 inches. 380. — To obtain the diameter of the suspension rods, when made of good wrought iron. Rule LIX. — Divide the weight or vertical strain, in pounds, by 4000. The square root of the quotient will be the required diameter of the rod in inches. Example. — A suspension rod is required to sustain 16000 pounds, what must be its diameter ? Dividing by 4000, the quotient is 4 ; the square root of which, 2, is the required diameter. The vertical strain on any rod is equal to the weight of so much of the ceiling as is supported by the rod, added to the vertical strain caused by each brace that is footed in the tie beam at the rod. The weight of the ceiling supported by a rod, is equal to ten times the distance apart in feet at which the trusses are placed, multiplied by half the distance in feet between the two next points of support, one on either side of the rod. The vertical strain caused by the braces can be as- certained geometrically, as in Art. 360. 381.— When the suspension rods are located as in Fig. 233, dividing the span into equal farts, the diameter of the rods FRAMING. 279 may be obtained without the preliminary calculation of the strain, as follows : Rule LX. — For the first rod from the wall. Multiply the distance apart at which the trusses are placed by the distance apart between the suspension rods, and divide the product by 400. The square root of the quotient will be the required diameter of the rod. Example. — Eod D G, Fig. 233. In this figure the rods are located at 8$ feet apart, and the distance between the trusses is 10 feet. Therefore, 10 x 8f = 86g ; this divided by 400, the quotient is 0-2167, the square root of which, 0*465, is the required diameter. The diameter may be half an inch. Rule LXI. — For the second rod from the wall. To the value of Q (Table IX., Art. 376) add 20, and multiply the sum by the distance apart at which the trusses are placed and by the distance between the rods, both in feet, and divide the pro- duct by 8000. The square root of the quotient will be the re- quired diameter. Example. — Eod EE, Fig. 233. The distances apart in this case are as stated in last example. The value of Q is 57'6, and when added to 20 equals 77"6. Therefore, 77-6 X 10 x 8f — 6673£ ; this divided by 8000, the quotient is 0-8341, the square root of which, 0'91, is the required diameter. This rod may be one inch diameter. Rule LXII. — For the centre rod. To the value of Q (Table IX., Art. 376) add 5, and multiply the sum by the distance apart at which the trusses are placed and by the distance apart between the rods, both in feet, and divide the product by 2000. The square root of the quotient will equal the required diameter. Example. — Eod OB, Fig. 233. The distances apart as be- fore, and the value of Q the same. To Q add 5, and the sum is 62-6. Then 62*6 x 10 x 8| = 5425$ ; this divided by 2000, the quotient is 2-7126, the square root of which, 1-647, equal? the required diameter. This rod may be If inches diameter. 280 ameeican house-caepentee. 382. — For all wrought iron straps and bolts the dimensions may be found by this rule. Rule LXIII. — Divide the tensile strain on the piece, in pounds, by 5000, and the quotient will be the area of cross section of the required bar or bolt, in inches. 383. — Roof-beams, jack-rafters, and purlins. All pieces of timber subject to cross strains will sustain safely much greater strains when extended in one piece over two, three, or more distances between bearings ; therefore roof-beams, jack-rafters, and purlins should, if possible, be made in as long lengths as practicable ; the roof-beams and purlins laid on, not framed into, the principal rafters, and extended over at least two spaces, the joints alternating on the trusses ; and likewise the jack-rafters laid on the purlins in long lengths. The dimen- sions of these several pieces may be obtained by the following rule : Rule LXIY.— From the value of Q (Table IX., Art. 376) deduct 10, and multiply the remainder by 33 times the distance from centres in feet at which the pieces are placed, and by the cube of the distance between bearings in feet ; divide the pro- duct by the value of E (Table II., Art. 293) for the kind of wood used and extract the square root of the quotient. The square root of this square root will be the required depth in inches. Multiply the depth thus obtained by the decimal 0-6, and the product will be the required breadth in inches. Example.- — Roof-beams of white pine placed 2 feet from cen- tres, resting on trusses placed 10 feet from centres, the height and the span of the roof being in proportion as 1 to 4. In this case the value of Q is 57"6. By the rule, 57-6 — 10 = 47-6, and 47-6 x 33 x 2 x 10 s = 3141600. This divided by 1750, the quotient is 1795-2, the square root of which is 42-37, and the square root of 42-37 is 6'5, the required depth. This multi- plied by 0-6 equals 3-9, the required breadth. These roof- beams may therefore be 4 by 6£ inches. FRAMING. 281 384. — Five examples of roofs are shown at Figs. 234, 235, 236, 237, and 238. In Fig. 234, a is an iron suspension rod, & 5 are braces. In Fig. 235, a, a, and i are iron rods, and dd,oc, are braces. In Fig. 236, a i are iron rods, d d braces, and c the straining beam. In Fig. 237, aa,ih, are iron rods, ee, dd, are braces, and c is a straining beam. In Fig. 238, purlins are located &tPP, &c. ; the inclined beam that lies upon them is the jack-rafter; the post at the ridge is the king post, the others are queen posts. In this design the tie beam is increased in height along the middle by a strengthen- ing piece [Art. 348), for the purpose of sustaining additional weight placed in the room formed in the truss. 385. — Fig. 239 shows a method of constructing a truss having a built-rib in the place of principal rafters. The proper form for the curve is that of a parabola, {Art. 127.) This curve, 36 282 AMERICAN HOUSE-CARPENTER. when as flat as is described in the figure, approximates so near to that of the circle, that the latter may be used in its stead. The height, a I, is just half of a c, the curve to pass through the middle of the rib. The rib is composed of two series of abutting pieces, bolted toge- ther. These pieces should be as long as the dimensions of the tim- ber will admit, in order that there may be but few joints. The sus- pending pieces are in halves, notched and bolted to the tie- beam and rib, and a purlin is framed upon the npper end of each. A truss of this construc- tion needs, for ordinary roofs, no diagonal braces between the sus- pending pieces, but if extra strength is required the braces may be added. The best place for the suspending pieces is at the joints of the rib. A rib of this kind will be sufficiently strong, if the area of its section contain about one-fourth more timber, than is required for that of a raf- ter for a roof of the same size. The proportion of the depth to the thickness should be about as 10 is to 7. 386 — Some writers have given designs for roofs similar to Fig. 240, having the tie-beam omitted for the accommodation of an arch in the ceiling. This and all similar designs are se- FRAMING. 283 riously objectionable, and should always be avoided ; as the small height gained by the omission of the tie-beam can never Fig. 240. compensate for the powerful lateral strains, which are exerted by the oblique position of the supports, tending to separate the 284 AMERICAN H0BT3E-CARPENTEE. walls. "Where an arch is required in the ceiling, the best plan is to carry up the walls as high as the top of the arch. Then, by using a horizontal tie-beam, the oblique strains will be en- tirely removed. Many a public building, by my own obser- vation, has been all but ruined by the settling of the roof, consequent upon a defective plan in the formation of the truss in this respect. It is very necessary, therefore, that the hori- zontal tie-beam be used, except where the walls are made so strong and firm by buttresses, or other support, as to prevent a possibility of their separating. 387. — Fig. 241 is a method of obtaining the proper lengths and bevils for rafters in a hip-roof: a b and b c are walls at the angle of the building ; b e is the seat of the hip-rafter and gf of a jack or cripple rafter. Draw e h, at right angles to b e, and make it equal to the rise of the roof; join b and h, and h b will be the length of the hip-rafter. Through e, draw d i, at right angles to b o / upon b, with the radius, b h, describe the arc, h i, cutting di in i / join b and i, and extend gf to meet b i mj • then gj will FRAMING. 285 be the length of the jack-rafter. The length of each jack-rafter is found in the same manner — by extending its seat to cut the line, b i. From/, draw fk, at right angles to fg, also f I, at right angles to be; make/ k equal to fl by the arc, I k, or make g k equal to gj by the arc, j k ; then the angle at-will be the top- bevil of the jack-rafters, and the one at k will be the down-bevil* 388. — To find the backing of the hip-rafter. At any con- venient place in b e, {Fig. 241,) as o, draw m n, at right angles to be; from o, tangical to b h, describe a semi-circle, cutting b e in s ; join m and s and n and s ; then these lines will form at s the proper angle for beviling the top of the hip-rafter. DOMES.f Fig. 242. Fig. 243. • The lengths and bevils of rafters for roof-valleys can also be found by .he above process t See also Art. 23V. 286 AMERICAN HOUSE-CARPENTER. 389. — The most usual form for domes is that of the sphere, the base being circular. When the interior dome does r ot rise too high, a horizontal tie may be thrown across, by which any de- gree of strength required may be obtained. Fig. 242 shows a section, and Fig. 243 the plan, of a dome of this kind, a b being the tie-beam in both. Two trusses of this kind, (Fig. 242,) pa- rallel to each other, are to be placed one on each side of the open- ing in the top of the dome. Upon these the whole framework is to depend for support, and their strength must be calculated accord- ingly. (See the first part of this section, and Art. 356.) If the dome is large and of importance, two other trusses may be intro- duced at right angles to the foregoing, the tie-beams being pre- served in one continuous length by framing them high enough to pass over the others. Fig. 244. Fi*. 245. 390. — When the interior dome rises too high to admit of a leve.. FRAMING. 287 tie-beam, the framing may be composed of a succession of ribs standing upon a continuous circular curb of timber, as seen at Fig. 244 and 245, — the latter being a plan and the former a sec- tion. This curb must be well secured, as it serves in the place of a tie-beam to resist the lateral thrust of the ribs. In small domes, th°se ribs may be easily cut from wide plank ; but, where an extensive structure is required, they must be built in two thicknesses so as to break joints, in the same manner as is descri- bed for a roof at Art. 385. They should be placed at about two feet apart at the base, and strutted as at a in Fig. 244. 391. — The scantling of each thickness of the rib may be as follows : For domes of 24 feet diameter, 1x8 inches. " " 36 " lJxlO " " ' 60 " 2x13 " " " 90 " 2^x13 " " " 108 " 3x13 " 392. — Although the outer and the inner surfaces of a dome may be finished to any curve that may be desired, yet the framing should be constructed of such a form, as to insure that the curve of equilibrium will pass through the middle of the depth of the framing. The nature of this curve is such that, if an arch or dome be constructed in accordance with it, no one part of the structure will be less capable than another of resisting the strains and pressures to which the whole fabric may be exposed. The curve of equilibrium for an arched vault or a roof, where the load is equally diffused over the whole surface, is that of a parabola, (Art. 127 ;) for a dome, having no lantern, tower or cupola above it, a cubic parabola, {Fig. 246 ;) and for one having a tower, &c, above it, a curve approaching that of an hyperbola must be adopted, as the greatest strength is required at its upper parts. If the curve of a dome be circular, (as in the vertical section, Fig. 244,) the pressure will have a tendency to burst the dome outwards at about one-third of its height. Therefore, when this form is used 288 AMERICAN HOUSE-CARPENTER. in the construction of an extensive dome, an iron band should be placed around the framework at that height ; and whatever ma3 be the form of the curve, a bond or tie of some kind is necessary around or across the base. If the framing be of a form less convex than the curve of equilibrium, the weight will have a tendency to crush the ribs in- wards, but this pressure may be effectually overcome by strutting between the ribs ; and hence it is important that the struts be so placed as to form c ntinuous horizontal circles. a j i kg i 393. — To describe a cubic parabola. Let a b, {Fig. 24:6,) be the base and b c the height. Bisect a b at d, and divide a d into 100 equal parts ; of these give d e 26, ef 18i, / g 14 \, g h 12J, h i 10 J, ij 9J, and the balance, 8|, to j a; divide b c into 8 equal parts, and, from the points of division, draw lines parallel to a b, to meet perpendiculars from the several points of division in a b, at the points, o, o, o, &c. Then a curve traced through these points will be the one required. 394. — Small domes to light stairways, Practical Rule and Fsmample. 434. — To obtain the .diameter of wrought iron tie-rods foi heel plates, as in Fig. 279, proceed by this rule. Rule LXX. — Multiply the weight in pounds equally distri- buted over the arch by the length of the tie-rod in feet, cleai of the heel plates, and divide the product by the height of the arc in inches, (that is, the height at the middle, from the axis of the tie-rod to the centre of the depth of the brick arch,) theD, if there is to be but one tie- rod, divide the quotient by 3000 ; 316 AMERICAN HOUSE-CARPENTER. but if two, then divide by 6000, and the square root of the quotient, in either case, will be the required diameter. Example. — The weight to be supported on a brick arch, equally distributed, is 24000 pounds ; the length of the tie-rod, clear of the heel plates, is 10 feet ; and the height, or rise, of the arc is 10 inches. Now by the rule, 24000 X 10 = 240000. This divided by 10, equals 24000. Upon the presumption that one tie-rod only will be needed, divide by 3000, and the quo- tient is 8, the square root of which is 2 - 82 inches. This is rather large, therefore there had better be two rods. In this case the quotient, 24000, divided by 6000, equals 4, the square root of which is 2, the diameter required. The arch should, therefore, have two rods of 2 inches diameter. Two rods are preferable to one. The iron is stronger per inch in small rods than in large ones, and the rules require no more metal in the two rods than in the one. Fig. 280. 435. — The Bow-string Girder, as per Fig. 280, has little to recommend it, (see Art. 433,) yet because it has by some been much used, it is well to show the rules that govern its strength, if only for the benefit of those who are willing to be governed by reason rather than precedent. To resist the horizontal thrust of the cast arch, the diameter of the rod must equal (204) IDS 3000 h ! But the cast iron arch has a certain amount of strength to re FRAMING. 317 6ist cross strains : this strength must be considered. Upon the presumption that the cross section of the cast arch at the mid- dle is of the most favorable form, as in Fig. 277, or at least that it have a bottom flange, (although the most of those cast are without it), the strength of the cast arch to resist cross strains is shown by formula (199), when I, its length, is changed to s, its span. The weight in pounds equally diffused over the arch will then equal 9000 t a d w — .. s This is the weight borne by the cast arch acting simply as a beam. Deducting this weight from the whole weight, the re- mainder is the weight to be sustained by the rod. Calling the whole weight w, then 9000 tad ws — 9000 tad = W w — = — Therefore, fiom (204), the diameter equals = V 3000 A )s Ws ooo ; w » - 9000 tad\ 3000 A where D equals the diameter of the rod in inches ; w, the weight in pounds equally diffused over the arch ; s, the span of the arch in feet ; A, the rise or height of the arc at middle, in inches ; d, the height or depth of the cross section of the cast arch in inches ; a, the area of the bottom flange of the cross section of the cast arch in inches ; and t, a decimal in proportion to unity as the safe weight is to be to the breaking weight. The rule in words at length, is Rule LXXI. — Multiply the decimal in proportion to unity 318 AMERICAN HOUSE-CARPENTER. as the safe weight is to be to the breaking weight, by 9000 times the depth of the cross section of the cast arch at middle, and by the area of the bottom flange of said section, both in inches, and deduct the product from the weight in pounds equally diffused over the arch multiplied by the span in feet, and divide the remainder by 3000 times the height of the arc in inches, measured from the axis of the tie-rod to the centre of the depth of the cast arch at middle, and the square root of the quotient will be the diameter of the rod in inches. Example. — The rear wall of a building is of brick, and is 40 feet high, and 21 feet wide in the clear between the piers of the story below. Allowing for the voids for windows, this wall will weigh about 63000 pounds ; and it is proposed to support it by a bow-string girder, of which the cross section at middle of the cast arch is 8 inches deep, and has a bottom flange containing 12 inches area. The rise of the curve or arc is 24 inches. What must be the diameter of the rod, the por- tion of the breaking weight of the cast arch, considered safe to trust, being three-tenths or 0-3 ? By the rule, 0-3 x 9000 X 8 X 12 = 259200 ; then 63000 x 21 - 259200 = 1063800. This remainder divided by (3000 x 24 =) 72000, the quotient equals 14'Y75 ; the square root of which, 3 - 84, or nearly 2>\ inches, is the required diameter. This size, though impracticably large, is as small as a due regard for safety will permit ; yet it is not unusual to find the rods in girders intended for as heavy a load as in this exam- ple, only 2i and 2J inches ! Were it possible to attach the rod so as not to injure its strength in the process of shrinking it in — putting it to its place hot, and depending on the con- traction of the metal in cooling to bring it to a proper bearing — and were it possible to have the bearings so true as to induce the strain through the axis of the rod, and not along its side, (Art. 308,) then a less diameter than that given by the rule would suffice. But while these contingencies remain, the rule FRAMING. 319 cannot safely be reduced, for, in the rule, the value of T, for wrought iron, (Table III., Art. 308,) is taken at nearly 6000 pounds, a point rather high in consideration of the size of the rod and the injuries, before stated, to which it is subjected. Tn cases where a girder wholly of cast iron {Fig. 276) is not preferred, it were better to build a brick arch resting on heel plates, {Fig. 279,) in which the metal required to resist the thrust may be divided into two rods instead of one, thus render- ing the size more practical, and at the same time avoiding the injuries to which rods in arch girders are subjected. The heel- plate arch is also to be preferred to the cast arch on the score of economy ; inasmuch as the brick which is substituted for the east arch will cost less than iron. For example, suppose the cross section of the iron arch to be thus : the blade or up- right part 8 by 1£ inches, the top flange 12 by 1J inches, and the bottom flange 6 by If- inches. At these dimensions, the area of the cross section will equal 12 + 15 + 10£ = 37-J inches. A bar of cast iron, one foot long and one inch square, will weigh 3 - 2 pounds; therefore, 37J X 3-2 = 120 pounds, equals the weight of the cast arch per lineal foot. The price of castings per pound, as also the price of brickwork per cubic foot, of course will depend upon the locality and the state of the market at the time, but for a comparison they may be stated, the one at three and a half cents per pound, and the other at thirty cents per cubic foot. At these prices the cast arch will cost 120 x 3£ = $4 20 per lineal foot; while the brick arch — 12 inches high and 12 inches thick — will cost 30 cents per lineal foot. The difference is $3 90. This amount is not all to be credited to the account of the brick arch Proper allowance is to be made for the cost of the heel plates, and of the wooden centre ; also for the cost of a small addi- tion to the size of the tie rods, which is required to sustain the strain otherwise borne by the cast arch in its resistance to a cross strain {Art. 435). Deducting the cost of these items, 320 AMEBIC AN HOUSE-CAEPENTEE. the difference in favor of the brick arch will be about $3 pei foot. This, on a girder 25 feet long, amounts to $75. The difference in all cases will not equal this, but will be sufficiently great to be worth saving. SECTION V— DOORS, WINDOWS, &c. DOORS. 436. — Among the several architectural arrangements of an edi- fice, the door is hy no means the least in importance ; and, if pro- perly constructed, it is not only an article of use, but also of or- nament, adding materially to the regularity and elegance of the apartments. The dimensions and style of finish of a door, should be in accordance with the size and style of the building, or the apartment for which it is designed. As regards the utility of doors, the principal door to a public building should be of suffi- cient width to admit of a free passage for a crowd of people ; while that of a private apartment will be wide enough, if it per- mit one person to pass without being incommoded. Experience has determined that the least width allowable for this is 2 feet 8 inches ; although doors leading to inferior and unimportant rooms may, if circumstances require it, be as narrow as 2 feet 6 inches ; and doors for closets, where an entrance is seldom required, may be but 2 feet wide. The width of the principal door to a public building may be from 6 to 12 feet, according to the size of the building ; and the width of doors for a dwelling may be from 2 feet 8 inches, to 3 feet 6 inches. If the importance of an apart- ment in a dwelling be such as to require a door of greater width 41 322 AMERICAN HOUSE-CAIIPENTER. than 3 feet 6 inches, the opening should be closed with two doors, or a door in two folds ; generally, in such cases, where the opening is from 5 to 8 feet, folding or sliding doors are adopted. As to the height of a door, it should in no case be less than about 6 feet 3 inches ; and generally not less than 6 feet 8 inches. 437. — The proportion between the width and height of single doors, for a dwelling, should be as 2 is to 5 ; and, for entrance- doors to public buildings, as 1 is to 2. If the width is given and the height required of a door for a dwelling, multiply the width by 5, and divide the product by 2 ; but, if the height is given and the width required, divide by 5, and multiply by 2. Where two or more doors of different widths show in the same room, it is well to proportion the dimensions of the more important by the above rule, and make the narrower doors of the same height as the wider ones ; as all the doors in a suit of apartments, except the folding or sliding doors, have the best appearance when of one height. The proportions for folding or sliding doors should be such that the width may be equal to f of the height ; yet this rule needs some qualification : for, if the width of the opening be greater than one-half the width of the room, there will not be a sufficient space left for opening the doors ; also, the height should be about one-tenth greater than that of the adjacent single doors. 438. — Where doors have but two panels in width, let the stiles and muntins be each | of the width ; or, whatever number of panels there may be, let the united widths of the stiles and the muntins, or the whole width of the solid, be equal to j of the width of the door. Thus : in a door, 35 inches wide, containing two panels in width, the stiles should be 5 inches wide ; and in a door, 3 feet 6 inches wide, the stiles should be 6 inches. If a door, 3 feet 6 inches wide, is to have 3' panels in width, the stiles and muntins should be each 4£ inches wide, each panel being 8 inches. The bottom rail and the lock rail ought to be each equal in width to T V of the height of the door ; and the top rail, and all DOORS, WINDOWS, &C. 32,: others, of the same width as the stiles. The moulding on the panel should be equal in width to | of the width of the stile. 1_& Fig. 281. 4:39. — Fig. 281 shows an approved method of trimming doors : a is the door stud ; b, the lath and plaster ; c, the ground ; d, the jamb ; e, the stop ; /and g, architrave casings ; and h, the door stile. It is customary in ordinary work to form the stop for the door hy rebating the jamb. But, when the door is thick and heavy, a better plan is to nail on a piece as at e in the figure. This piece can be fitted to the door, and put on after the door is hung ; so, should the door be a trifle winding, this will correct the evil, and the door be made to shut solid. MO. — Fig. 282 is an elevation of a door and trimmings suita- ble for the best rooms of a dwelling. (For trimmings generally, see Sect. III.) The number of panels into which a door should be divided, is adjusted at pleasure ; yet the present style of finish- ing requires, that the number be as small as a proper regard for strength will admit. In some of our best dwellings, doors have been made having only two upright panels. A few years expe- dience, however, has proved that the omission of the lock rail is at the expense of the strength and durability of the door ; four-panel door, therefore, is the best that can be made. 441. — The doors of a dwelling should all be hung so as to open into the principal rooms ; and, in general, no door should be hung to open into the hall, or passage. As to the proper edge of the door on which to aflix the hinges, no general rule can be assigned. 324 AMEBICAN HOUSE-CARPENTEB. Fig. 283. WINDOWS. 442. — A window should be of such dimensions, and in such a position, as to admit a sufficiency of light to that part of the apartment for which it is designed. No definite rule for the size DOORS, WINDOWS, &c. 325 can well be given, that will answer in all cases ; yet, as an ap- proximation, the following has been usecLfor general purposes. Multiply together the length and the breadth in feet of the apart- ment to be lighted, and the product by the height in feet ; then the square-root of this product will show the required number of square feet of glass. 443. — To ascertain the dimensions of window frames, add 4J inches to the width of the glass for their width, and 6f inches to the height of the glass for their height. These give the dimen- sions, in the clear, of ordinary frames for 12-light windows ; the height being taken at the inside edge of the sill. In a brick wall, the width of the opening is 8 inches more than the width of the glass— 4J for the stiles of the sash, and 3J for hanging stiles — and the height between the stone sill and lintel is about 10 £ inches more than the height of the glass, it being varied according to the thickness of the sill of the frame. 444. — In hanging inside shutters to fold into boxes, it is ne- cessary to have the box shutter about one inch wider than the flap, in order that the flap may not interfere when both are folded into the box. The usual margin shown between the face of the shutter when folded into the box and the quirk of the stop bead, or edge of the casing, is half an inch ; and, in the usual method of letting the whole of the thickness of the butt hinge into the todge of the box shutter, it is necessary to make allowance for the throw of the hinge. This may, in general, be estimated at i of an inch at each hinging ; which being added to the margin, the entire width of the shutters will be 1^ inches more than the width of the frame in the clear. Then, to ascertain the width of the box shutter, add 1£ inches to the width of the frame in the clear, between the pulley stiles ; divide this product by 4, and add half an inch to the quotient ; and the last product will be the re- quired width. For example, suppose the window to have 3 lights in width, 11 inches each. Then, 3 times 11 is 33, and 4£ added for the wood of the sash, gives 37£— —- 37i and U is 39 326 AMERICAN HOUSE CARPENTER. and 39. divided by 4, gives 9| ; to which add half an inch, and the result will be 10£ inches, the width required for the box shutter. 445. — In disposing and proportioning windows for the walls of a building, the rules of' architectural taste require that they be of different heights in different stories, but of the same width. The windows of the upper stories should all range perpendicularly over those of the first, or principal, story ; and they should be disposed so as to exhibit a balance of parts throughout the front of the building. To aid in this, it is always proper to place the front door in the middle of the front of the building ; and, where the size of the house will admit of it, this plan should be adopted. (See the latter part of Art. 224.) The proportion that the height should bear to the width, may be, in accordance with general usage, as follows : The height of basement windows, 1J of the width. " " principal-story " 2\ " " " second-story " \\ " " " third-story " 1| ' " " fourth-story " U " " " attic-story " the same as the width. But, in determining the height of the windows for the several stories, it is necessary to take into consideration the height of the story in which the window is to be placed. For, in addition to the height from the floor, which is generally required to be from 28 to 30 inches, room is wanted above the head of the window for the window-trimming and the cornice of the room, besides some respectable space which there ought to be between these. 446. — Doors and windows are usually square-Jieaded, or termi- nate in a horizontal line at top. These require no special direc- tions for their trimmings. But circular-headed doors and win- dows are more difficult of execution, and require some attention. If the jambs of a door or window be placed at right angles to the face of the wall, the edges of the soffit, or surface of the head, would be straight, and its length be found by getting the DOOES, WINDOWS, &C. 321 stretch-out of the circle, (ArL 92;) but, when the jaubs are placed obliquely to the face of the wall, occasioned by the de- mand for light in an oblique direction, the form of the soffit will be obtained by the following article : and, when the face of the wall is circular, as in the succeeding one. Fig. 283. 447. — To find the form of the soffit for circular window heads, when the light is received in an oblique direction. Let abed, {Fig. 283,) be the ground-plan of a given window, and ef a, a vertical section taken at right angles to the face of the jambs. From a, through e, draw a g, at right angles to a b ; obtain the stretch-out of ef a, and make e g equal to it ; divide e g and e f a, each into a like number of equal parts, and drop perpen- diculars from the points of division in each ; from the points of intersection, 1, 2, 3, &c, in the line, a d, draw horizontal lines to meet corresponding perpendiculars from eg; then those points of intersection will give the curve line, d g, which will be the one required for the edge of the soffit. The other edge, c h, is found in the same manner. 448. — To find the form of the soffit for circular window- heads, when the face of the wall is curved. Let abed, {Fig. 284,) be the ground-plan of a given window, and ef a, a vertical section of the head taken at right angles to the face of the jambs. 323 AMERICAN HOUSE-CARPENTER. * [3 2 7 7"; |2 3 Fig. 284. (. Proceed as in the foregoing article to obtain the line, d g ; then that will be the curve required for the edge of the soffit; the other edge being found in the same manner. If the given vertical section be taken in a line with the face of the wall, instead of at right angles to the face of the jambs, place it upon the line, c b, (Fig. 283 ;) and, having drawn ordinates at right angles to c b, transfer them to ef a ; in this way, a section at right angles to the jambs can be obtained. SECTION VI.— STAIRS. 449. — The stairs is that mechanical arrangement in a build- ing by which access is obtained from one story to another. Their position, form and finish, when determined with discriminating taste, add greatly to the comfort and elegance of a structure. As regards their position, the first object should be to have them near the middle of the building, in order that an equally easy access may be obtained from all the rooms and passages. Next in im- portance is light; to obtain which they would seem to be best situated near an outer wall, in which windows might be construc- ted for the purpose ; yet a sky-light, or opening in the roof, would not only provide light, and so secure a central position for the stairs, but may be made, also, to assist materially as an ornament to the building, and, what is of more importance, afford an op- portunity for better ventilation. 450. — It would seem that the length of the raking side of the pitch-board, or the distance from the top of one riser to the top ot the next, should be about the same in all cases ; for, whether stairs be intended for large buildings or for small, for public or for pri- vate, the accommodation of men of the same stature is to be con- sulted in every instance. But it is evident that, with the same effort, a longer step can be taken on level than on rising ground 42 330 AMERICAN HOUSE-CARPENTER. and that, although the tread and rise cannot be proportioned merely in accordance with the style and importance of the build- ing, yet this may be done according to the angle at which the flight rises. If it is required to ascend gradually and easy, the length from the top of one rise to that of another, or the hypothe nuse of the pitch-board, may be long ; but, if the flight is steep the length must be shorter. Upon this data the following pioblerr is constructed. 451. — To proportion the rise and tread to one another. Make the line, a b, {Fig. 285,) equal to 24 inches ; from b, erect b c, at right angles to a b, and make b c equal to 12 inches ; join a and c, and the triangle, a b c, will form a scale upon which tc graduate the sides of the pitch-board. For example, suppose a very easy stairs is required, and the tread is fixed at 14 inches. Place it from b to/, and from/; dmwfg, at right angles to a b ; then the length of f g will be found to be 5 inches, which is a proper rise for 14 inches tread, and the angle, f b g, will show the degree of inclination at which the flight will ascend. But, in a majority of instances, the height of a story is fixed, while the length of tread, or the space that the stairs occupy on the lower floor, is optional. The height of a story being determined, the height of each rise will of course depend upon the number into which the whole height is divided ; the angle of ascent being more easy if the number be great, than if it be smaller. By dividing STAIRS. 331 the whole height of a story into a certain number of rises, sup- pose the length of each is found to he 6 inches. Place this length from b to h, and draw h i, parallel to a b ; then h i, or b j will be the proper tread for that rise, and j b i will show the angle of as- cent. On the other hand, if the angle of ascent be given, as a b l,(b I being 10J inches, the proper length of run for a step- ladder,) drop the perpendicular, I k, from I to lc ; then I k b will be the proper proportion for the sides of a pitch-board for that run. 452. — The angle of ascent will vary according to circum- stances. The following treads will determine about the right in- clination for the different classes of buildings specified. In public edifices, tread about 14 inches. In first-class dwellings " 12£ " In second-class " "11 " In third-class " and cottages " 9 " Step-ladders to ascend to scuttles, &c, should have from 10 to 11 inches run on the rake of the string. (See notes at Art. 103.' 453.— The length of the steps is regulated according to the ex- tent and importance of the building in which they are placed, varying from 3 to 12 feet, and sometimes longer. Where two per- sons are expected to pass each other conveniently, the shortest length that will admit of it is 3 feet ; still, in crowded cities where land is so valuable, the space allowed for passages being very small, they are frequently executed at 2 J feet. 454. — To find the dimensions of the pitch-board. The first thing in commencing to build a stairs, is to make the pitch-board ; this is done in the following manner. Obtain very accurately, in feet and inches, the perpendicular height of the story in which the stairs are to be placed. This must be taken from the top ol the floor in the lower story to the top of the floor in the upper story. Then, to obtain the number of rises, the height in inches thus obtained must be divided by 5, 6, 7, 8, or 9, according to the quality and style of the building in which the stairs are to be 332 AMERICAN HOUSE-CARPENTER. built. For instance, suppose the building to be a fiist-class dwelling, and the height ascertained is 13 feet 4 inches, or 160 inches. The proper rise for a stairs in a house of this class is about 6 inches. Then, 160 divided by 6, gives 26§ inches. This being nearer 27 than 26, the number of risers, should be 27. Then divide the height, 160 inches, by 27, and the quotient will give the height of one rise. On performing this operation, the quotient will be found to be 5 inches, | and T V of an inch. Then, if the space for the extension of the stairs is not limited, the tread can be found as at Art. 451. But, if the contrary is the case, the whole distance given for the treads must be divided by the number of treads required. On account of the upper floor forming a step for the last riser, the number of treads is always one less than the number of risers. Having obtained this rise and tread, the pitch-board may be made in the follow- ing manner. Upon a piece of well-seasoned board about f of an inch thick, having one edge jointed straight and square, lay the corner of a carpenters'-square, as shown at Fig. 286. Make a b Fig 286. equal to the rise, and b c equal to the tread ; mark along those edges with a knife, and cut it out by the marks, making the edges perfectly square. The grain of the wood must run in the direction indicated in the figure, because, if it shrinks a trifle, the rise and the tread will be equally affected by it. When a pitch-board is first made, the dimensions of the rise and tread should be pre- served in figures, in order that, should the first shrink, a second could be made. 455. — To lay out the string. The space required for timbei STAIRS. 333 Fig. 287. and plastering under the steps, is about 5 inches for ordinary stairs ; set a gauge, therefore, at 5 inches, and run it on the lowet edge of the plank, as a b, {Fig. 287.) Commencing at one end, lay the longest side of the pitch-board against the gauge-mark, a b, as at c, and draw by the edges the lines for the first rise and tread: then place it successively as at d, e and/, until the re- quired number of risers shall be laid down. EJ Q Fig. 268. 456. — Fig. 238 represents a section of a step and riser, joined after the most approved method. In this, a represents the end of a block about 2 inches long, two of which are glued in the corner in the length of the step. The cove at b is planed up square, glued in, and stuck after the glue is set. PLATFORM STAIRS. 457. — A platform stairs ascends from one story to another in two or more flights, having platforms between for resting and to cnange their direction. This kind of stairs is the most easily constructed, and is therefore the most common. The cylin- 334 AMERICAN HOUSE-CARPENTEE, Fig. 289. der is generally of small diameter, in most cases about 6 irjches. It may be worked out of one solid piece, but a better way is to glue together three pieces, as in Fig. 289; in which the pieces, a, b and c, compose the cylinder, and d and e represent parts of the strings. The strings, after being glued to the cylinder, are secured with screws. The joining at o and o is the most proper for that kind of joint. 458. — To obtain the form of the lower edge of the cylinder. Find the stretch-out, d e, {Fig. 290,) of the face of the cylinder a b c, according to Art. 92 ; from d and e, draw d f and e g, at right angles to d e ; draw h g, parallel to d e, and make hf and g i, each equal to one rise; from i and/, draw ij a.ndfk, paral- lel to h g ; place the tread of the pitch-board at these last lines, and draw by the lower edge the lines, k h and i I ; parallel to these, draw m n and o p, at the requisite distance for the dimen- sions of the string ; from s, the centre of the plan, draw s q, parallel to df; divide h q and q g, each into 2 equal parts, as at v and w ; from v and w, draw v n and w o, parallel tofd; join n and o, cutting q s in r ; then the angles, u n r and rot. being eased off according to Art. 89, will give the proper curve for the bottom edge of the cylinder. A centre may be found upon Avhich to describe these curves thus : from u, draw u x, at right angles to m n ; lrom r, draw r x, at right angles to no ; then x will bo the centre for the curve, u r. The centre for the p.urve, r t, is found in the same manner. STAIRS. An'' Fig. 290. 459. — To find the position for the balusters. Place the centre of the first baluster, (b. Fig. 291,) i its diameter from the face of the riser, c d, and $ its diameter from the end of the step, e d ; and place the centre of the other baluster, a, half the tread from the centre of the first. The centre of the rail must be placed over the centre of the balusters. Their usual length is 2 feet 5 inches, and 2 feet 9 inches, for the short and the long balusters respectively. S=Gk Fig. 291. 336 AMERICAN HOUSE-CARPENTEK. Fig. 292. 460. — To find the face-mould for a round hand-rail to plat' form stairs. Case 1. — When the cylinder is small. In Fig. 292, j and e represent a vertical section of the last two steps of the first flight, and d and i the first two steps of the second flight, of a platform stairs, the line, e /, being the platform ; and a b c is the plan of a line passing through the centre of the rail around the cylinder. Through i and d, draw i k, and through 7 and e, draw j k ; from k, draw k I, parallel to / e ; from b, draw b m, parallel tog d; from I, draw I r, parallel to k j ; from n, draw n t, at right angles toj k ; on the line, o b. make o t equal to n t ; join c and t : on the line, j c, {Fig. 293,) make e c equal to e n at Fig. 292 ; from c, draw c t, at right angles toj c, and make c t STAIRS. 337 i I equal to c t at Fig. 292 ; through t, draw p I, parallel toj c, and make / I equal to 1 1 at Fig. 292 ; join I and c, and complete the parallelogram, eels; find the points, o, o, o, according to Art. 118 ; upon e, o, o, o, and £, successively, with a radius equal to half the width of the rail, describe the circles shown in the figure ; then a curve traced on both sides of these circles and just touch- ing them, will give the proper form for the mould. The joint at I is drawn at right angles to c I. 461. — Elucidation of the foregoing method. This excellent plan for obtaining the face-moulds for the hand-rail of a platform stairs, has never before been published. It was communicated to me by an eminent stair-builder of this city : and having seen rails put up from it, I am enabled to give it my unqualified re- commendation. In order to have it fully understood, I have in- troduced Fig. 294 ; in which the cylinder, for this purpose, is made rectangular instead of circular. The figure gives a per- spective view of a part of the upper and of the lower flights, and a part of the platform about the cylinder. The heavy lines, i m, m c and cj, show the direction of the rail, and are supposed to pass through the centre of it. When the rake of the second flight is the same as that of the first, which is here and is gene- rally the case, the face-mould for the lower twist will, when re- versed, do for the upper flight : that part of the rail, therefore, which passes from e to c and from c to I, is all that will need ex- planation. Suppose, then, that the parallelogram, e ao c, represent a plane lying perpendicularly over e abf, being inclined in the direction, e c, and level in the direction, c o ; suppose this plane, e a o e, 43 838 AMERICAN HOUSE-CARPENTER. Fig. 294, be revolved on e c as an axis, in the manner indicated by the arcs, o n and a x, until it coincides with the plane, e r t c ; the line, a o, will then be represented by the line, x n ; then add the paral- lelogram, xrt n, and the triangle, ctl, deducting the triangle, e r s ; and the edges of the plane, e s I c, inclined in the direction, ec, and also in the direction, c I, will lie perpendicularly over the plane, e abf. From this we gather that theline, co, beingat right angles to STAIRS. 339 c c, must, in order to reach the point, I, be lengthened the distance, n t, and the right angle, e c t, be made obtuse by the addition to it of the angle, t c I. By reference to Fig. 292, it will be seen that this lengthening is performed by forming the right-angled triangle, cot, corresponding to the triangle, c o t, in Fig. 294. The line, c t, is then transferred to Fig. 293, and placed at right angles to e c; this angle, e c t, being increased by adding the an- gle, t c I, corresponding to t c I, Fig. 294, the point, I, is reached, and the proper position and length of the lines, e c and c I ob- tained. To obtain the face-mould for a rail over a cylindrica 1 well-hole, the same process is necessary to be followed until the the length and position of these lines are found ; then, by forming the parallelogram, eels, and describing a quarter of an ellipse therein, the proper form will be given. Fig 295. 462. — Case 2 — When the cylinder is large. Fig. 295 re- 340 AMERICAN HOUSE-CARPENTER, presents a plan and a vertical section of a line passing through tha centre of the rail as before. From b, draw b k, parallel toed ; ex- tend the lines, i d an&j e, until they meet kb'mk and/; from n, draw n I, parallel to o b ; through I, draw 1 1, parallel tojk, from k, draw k t, at right angles to j k ; on the line, o b, make o t equal to k t. Make e c, {Fig. 296.) equal to e k at Fig. 295 ; from c, Fig. 296. draw c t, at right angles to e c, and equal to c t at Fig. 295 • from t, draw if p, parallel to c e, and make 1 1 equal to 1 1 at jFig - . 295 ; complete the parallelogram, eels, and find the points, o, o, o, as before ; then describe the circles and complete the mould as in Fig. 293 The difference between this and Case 1 is, that the line, c t, instead of being raised and thrown out, is lowered and drawn in. (See note at page 381.) Fig. Wl. c 463. — Case 3. — Where the rake meets the level. In Fig STAIRS. 341 297, a b c is the plan of a line passing through the centre it tne rail around the cylinder as before, andj and e is a vertical section of two steps starting from the floor, h g. Bisect e h in d, and through d, draw df, parallel to h g ; bisect/ n in I, and trom I, draw 1 t, parallel to nj; from n, draw n t, at right angles tojn_ on the line, o b, make o t equal to n t. Then, to obtain a mould for the twist going up the flight, proceed as at Fig. 293 ; making c c in that figure equal to e n in Fig. 297, and the other lines of a length and position such as is indicated by the letters of reference in each figure. To obtain the mould for the level rail , extend h o, (Fig. 297,) to i ; make o i equal to/ I, and join i and c ; ma^n c i, (Fig. 298,) equal to c i at Fig. 291 ; through c, draw c d r at d Fig. 298. right angles to c i ; make rf c equal to d/ at .FJg - . 297, and com plete the parallelogram, o dc i; then proceed as in the previous cases to find the mould. 464. — All the moulds obtained by the preceding examples have been for round rails. For these, the mould may be applied to a plank of the same thickness as the rail is intended to be, and the plank sawed square through, the joints being cut square from the face of the plank. A twist thus cut and truly rounded will hang in a proper position over the plan, and present a perfect and graceful wreath. 465. — To bore for the balusters of a round rail before round- ing it. Make the angle, o c t, (Fig. 299,) equal to the angle, o c t, at Fig. 292 ; upon c, describe a circle with a radius equal to half the thickness of the rail ; draw the tangent, b d, parallel to t c, and complete the rectangle, e b df having sides tangica! to the circle ; from c, draw c a, at right angles to oc ; then, b d being the bottom of the rail, set a gauge from b to a, and run it the whole length of the stuff; in boring, place the centre of tha M2 AM E R I C IN EI OUS E- CARPEKTEB. h bit in the gauge-mark at a, and bore in the direction, a c. To do this easily, make chucks as represented in the figure, the bottom edge, g h, being parallel to o c, and having a place sawed out, as ef, to receive the rail. These being nailed to the bench, the rail win be held steadily in its proper place for boring vertically. The distance apart that the balusters require to be, on the under side of the rail, is one-half the length of the rake-side of tha pitch-board. Fig. 800. STAIRS. 343 466. — To obtain, by the foregoing principles, the face-mould for the twists of a moulded rail upon platform stairs In Fig. 300, a b c is the plan of a line passing through the centre of the rail around the cylinder as before, and the lines above it are a vertical section of steps, risers and platform, with the lines for the rail obtained as in Fig. 292. Set half the width of the rail from b to f and from b to r, and from f and r, draw/ e and r d, parallel to c a. At Fig. 301, the centre lines of the s d rai", k c and c n, are obtained as in the previous examples. Make c i and c j, each equal to c i at Fig. 300, and draw the lines, i m andy g, parallel to c k ; make n e and n d equal tone and n d at Fig. 300, and draw d o and e I, parallel to n c ; also, through k, draw s g, parallel to n c ; then, in the parallelograms, m s d o and g s el, find the elliptic curves, d m and e g, according to Art. 118, and they will define the curves. The line, d e, being drawn through n parallel to k c, defines the joint, which is to be cut through the plank vertically. If the rail crosses the platform rather steep, a butt joint will be preferable, to obtain wlich see Art. 498. 344 AMfc-RICAN HOUSE-CARPENTER. 467. — To apply the mould to the plank. The mould obtained according to the last article must be applied to both sides of the plank, as shown at Fig. 302. Before applying the mould, the edge, ef must be bevilled according to the angle, c t x, at Fig 300 ; if the rail is to be canted up, the edge must be bevilled at an obtuse angle with the upper face ; but if it is to be canted down, the angle that the edge makes with the upper face must be acute. From the spring of the curve, a, and the end, c, draw vertical lines across the edge of the plank by applying the pitch- board, a b c ; then, in applying the mould to the other side, place the points, a and c, at b and/ ; and, after marking around it, saw the rail out vertically. After the rail is sawed out, the bottom and the top surfaces must be squared from the sides. 468. — To ascertain the thickness of stuff required for the twists. The thickness of stuff required for the twists of a round rail, as before observed, is the same as that for the straight ; but for a moulded rail, the stuff for the twists must be thicker than that for the straight. In Fig. 300, draw a section of the rail be- tween the lines, d r and e f and as close to the line, d e, as possi- ble ; at the lower corner of the section, draw g h, parallel to d e ; then the distance that these lines are apart, will be the thickness required for the twists of a moulded rail. The foregoing method of finding moulds for rails is applicable to all stairs which have continued rails around cylinders, and are without winders. WINDING STAIRS. 469. — Winding stairs have steps tapering narrower at one end than at the other. In some stairs, there are steps of parallel width incorporated with tapering steps ; the former are then called flyers and the latter winders. 470. — To describe a regular geometrical winding stairs. In Fig. 303, abed represents the inner surface of the wall en- closing the space allotted to the stairs, a e the length of the steps, and efgh the cylinder, or face of the front string. The line, STAIRS. 345 Fig. 803. U^ a e, is given as the face of the first riser, and the point, j, for the limi; of the last. Make e i equal to 18 inches, and upon o, with oi for radius, describe the arc, ij; obtain the number of risers and of treads required to ascend to the floor at j, according to Art. 454, and divide the arc, ij, into the same number of equal parts as there are to be treads ; through the points of division, 1, 2, 3, &c, and from the wall-string to the front-string, draw lines tend- ing to the centre, o ; then these lines will represent the face ot each riser, and determine the form and width of the steps. Allow the necessary projection for the nosing beyond a e, which should be equal to the thickness of the step, and then a elk will be the dimensions for each step. Make a pitch-board for the wall-string having a k for the tread, and the rise as previously ascertained ; with this, lay out on a thicknessed plank the several risers and treads, as at Fig. 287, gauging from the upper edge of the stricg for the line at which to set the pitch-board. Upon the back of the string, with a 1£ inch dado plane, make 44 346 AMERICAN HOUSE-CARPENTER. a succession of grooves 1J inches apart, and parallel with the lines for the risers on the face. These grooves must be jut along the whole length of the plank, and deep enough to admit of the plank's bending around the curve, abed. Then construct a drum, or cylinder, of any common kind of stuff, and made to fit a curve having a radius the thickness of the string less than o a ; upon this the string must be bent, and the grooves filled with strips of wood, called keys, which must be very nicely fitted and glued in. After it has dried, a board thin enough to bend around on the outside of the string, must be glued on from one end to the other and nailed with clout nails. In doing this, be careful not to nail into any place where a riser or step is to enter on the face. After the string has been on the drum a sufficient time for the glue to set, take it off, and cut the mortices for the steps and risers on the face at the lines previously made ; which may be done by boring with a centre-bit half through the string, and nicely chiseling to the line. The drum need not be made so large as the whole space occupied by the stairs, but merely large enough to receive one piece of the wall-string at once — for it is evident that more than one will be required. The front string may be constructed in the same manner ; taking e I instead of a k for the tread of the pitch-board, dadoing it with a smaller dado plane, and bending it on a drum of the proper size. 471. — To find the shape and position of the timbers neces- sary to support a winding stairs. The dotted lines in Fig: 303 show the proper position of the timbers as regards the plan : the shape of each is obtained as follows. In Fig. 304, the line, 1 a, is equal to a riser, less the thickness of the floor, and the unes, 2 m, 3 n, 4 o, 5 p and 6 q, are each equal to one riser. The STAIRS. 347 line, a 2, is equal to a m in Fig. 303, the line, m 3 to m n in that figure, &c. In drawing this figure, commence at a, and make the lines, a 1 and a 2, of the length above specified, and draw them at right angles to each other ; draw 2 m, at right angles to a 2, and m 3, at right angles to m 2, and make 2 m and m 3 of the lengths as above specified ; and so proceed to the end. Then, through the points, 1, 2, 3, 4, 5 and 6, trace the line, lb; upon the points, 1, 2, 3, 4, &c, with the size of the timber for radius, describe arcs as shown in the figure, and by these the lower line may be traced parallel to the upper. This will give the proper shape for the timber, a b, in Fig. 303 ; and that of the others may be found in the same manner. In ordinary cases, the shape of one face of the timber will be sufficient, for a good workman can easily hew it to its proper level by that ; but where great accuracy is desirable, a pattern for the other side may be found in the same manner as for the first. 472. — To find the falling-mould for the rail of a winding stairs. In Fig. 305, a cb represents the plan of a rail around half the cylinder, A the cap of the newel, and 1, 2, 3, &c, the face of the risers in the order they ascend. Find the stretch-out, ef, of a c b, according to Art. 92; from o, through the point of the mitre at the newel-cap, draw o s ; obtain on the tangent, e d, the position of the points, s and h\* as at t and/ 2 ; from e tf' 2 and f, draw e x, t u,f* g 2 and f h, all at right angles to e d ; make e g equal to one rise and/ 2 g 1 equal to 12, as this line is drawn from the 12th riser ; from g, through g*, drawg- i; make g x equal to about three-fourths of a rise, (the top of the newel, x, should be 3^ feet from the floor ;) draw x u, at right angles to e x, and ease off the angle at u ; at a distance equal to the thickness of * In the above, the references, a\ b% &c, are introduced for the first time. During the lime taken to refer to the figure, the memory of the/orm of these may pass from the mind, while that of the sound alone remains ; they m£ y then be mistaken for a 2, b 2, &c. This •on be avoided in reading by giving them a sotnd corresponding to their meaning, which la tecond a second b, &c. or a second, b second. 34-8 AMERICAN HOUSE-CARPENTER. Fig. S05. the rail, draw v w y, parallel to x u i ; from the centre of the plan, o, draw o I, at right angles to e d ; bisect h n in p, and through p, at right angles tog i, draw a line for the joint ; in the same manner, draw the joint at k ; then x y will be the falling-mould for that part of the rail which extends from s to b on the plan. 473. — To find the face-mould for the rail of aioinding-stairs. From the extremities of the joints in the falling-mould, as k, z and y, {Fig. 305,) draw k a 2 , z 6 2 and y d, at right angles to e d ; make b e 1 equal to / d. Then, to obtain the direction of the joint, a 5 c s , or 6 s d\ proceed as at Fig 306, at which the parts are STAIRS. 349 Fig. 806. shown at half their full size. A is the plan of the rail, and B is the falling-mould ; in which k z is the direction of the butt-joint. From k, draw k b, parallel to I o, and k e, at right angles to k b ; from b, draw b f, tending to the centre of the plan, and from/, draw / e, parallel to b k ; from I, through e draw I i, and from i, draw i d, parallel to ef; join d and b, and d b will be the proper direction 150 AMERICAN HOUSE-CARPENTER. for the joint on the plan. The direction of the joint on the other side, a c, can be found by transferring the distances, x b and o d to x a and o c. (See Art. 477.) Fig. 80T. Having obtained the direction of the joint, make s r db, {Fig. 307,) equal to s r d l 6 2 in Fig. 305 ; through r and d, draw t a , through a- and from d, draw t u and d e, at right angles to t a ; make t u and d e equal to t u and & 2 m, respectively, in Fig. 305 ; from u, through e, draw u o ; through b, from r, and from as many other points in the line, t a, as is thought necessary, as/, h and j draw the ordinates, r c,f g, hi,j k and a o ; from u, c, g, i, k, e and o, draw the ordinates, u 1, c 2, g 3, i i, k 5, e 6 and o 7, at right angles to u o ; make u 1 equal to t s, c 2 equal to r 2, g- 3 equal to/ 3, &c, and trace the curve, 1 7, through the points thus found ; find the curve, c e, in the same manner, by transfer- ring the distances between the line, t a, and the arc, r d ; join 1 and c, also e and 7 ; then, 1 c e 7 will be the face-mould required for that part of the rail which is denoted by the letters, s r d 1 V, on the plan at Fig. 305. To ascertain the mould for the next quarter, make a cje, (Fig BTAIRR. Fig. 808. 308,) equal to a* c'j e 1 at Fig. 305 ; at any convenient height on the line, d i, in that figure, draw q i 2 , parallel to e d ; through c and J, {Fig. 308,) draw b d ; through a, and from 7, draw b k and / o, at right angles to b d ; make b k and j equal to i 2 k and q i, respectively, in Fig. 305 ; from k, through 0, draw kf; and proceed as in the last figure to obtain the face-mould, A. 474. — To ascertain the requisite thickness of stuff. Case 1. — When the falling-mould is straight. Make h and k m, (Fig. 308,) equal to i y at Fig. 305 ; draw h i and m n, parallel to b d j through the corner farthest from kf, as n or i, draw n i, parallel to kf; then the distance between kf and n i will give the thickness required. 475. — Case 2. — When the falling-mould is curved. In Fig. 309, s r d b is equal to s r d 2 b 1 in Fig. 305. Make a c equal to the stretch-out of the arc, s b, according to Art. 92, and divide a c and s b. each into a like number of equal parts ; from a and c, and from each point of division in the line, ac, draw ak, e I, &c, at right an- gles to ac , make a Arequal to^win i^. 305, and c /equal to b 2 m 352 AMERICAN HOUSE-CARPENTER. '/ » in that figure, and complete the tailing-mould, k j, every way equal to u m in Fig. 305 ; from the points of division in the arc, sb, draw lines radiating towards the centre of the circle, dividing the arc. r d, in the same proportion as s b is divided ; from d and b, draw d t and b u, at right angles to a d, and from j and v, draw j u and v w, at right angles to j c ; then x t uw will be a vertical projection of the joint, d b. Supposing every radiating line across s r d b — corresponding to the vertical lines across k j — to represent a joint, find their vertical projection, as at 1, 2, 3, 4, 5 and 6 ; through the corners of those parallelograms, trace the curve lines shown in the figure ; then 6 u will be a helinet, or vertical projection, of sr db. To find the thickness of plank necessary to get out this part of the rail, draw the line, z t, touching the upper side of the helinet in two places : through the corner farthest projecting from that line, as w, draw y w, parallel to z t ; then the distance between those lines will be the proper thickness of stuff for this part of the rail. The same process is necessary to find the thickness of stuff in all cases in which the falling-mould is in any way curved. 476. — To apply the face-mould to the plank. In Fig. 310, A represents the plank with its best side and edge in view, and B the same plank turned up so as to bring in view the other side STAIRS. 353 Fig. 810. and the same edge, this being square from the face. Apply the tips of the mould at the edge of the plank, as at a and o, (A,) and mark out the shape of the twist ; from a and o, draw the lines, a b and o c, across the edge of the plank, the angles, e a b and e o c, corresponding with kfd at Fig. 308 ; turning the plank up as at B, apply the tips of the mould at b and c, and mark it out as shown in the figure. In sawing out the twist, the saw must be be moved in the direction, a b ; which direction will be perpen- dicular when the twist is held up in its proper position. In sawing by the face-mould, the sides of the rail are obtained ; the top and bottom, or the upper and the lower surfaces, are ob- tained by squaring from the sides, after having bent the falling- mould around the outer, or convex side, and marked by its edges. Marking across by the ends of the falling-mould will give the position of the butt-joint. 477. — Elucidation of the process by which the direction of the butt-joint is obtained in Art. 473. Mr. Nicholson, in his Carpenter's Guide, has given the joint a different direction to that here shown ; he radiates it towards the centre of the cylin- der. This is erroneous — as can be shown by the following operation : In Fig. 311, a r j i is the plan of a part of the rail about the joint, s u is the stretch-out of a i, and g p is the helinet, or ver- tical projection of the plan, a r j i, obtained according to Art 45 354 AMERICAN HOUSE-CARPENTER. Fig. 811. ±75. Bisect r t, part of an ordinate from the centre of the plan, and through the middle, draw c b, at right angles to g v ; from b and c, draw c d and 6 e, at right angles to * u ; from d and e, draw lines radiating towards the centre of the plan : then d o and e m wiil be the direction of the joint oh the plan, according to Nicholson, and c b its direction on the falling-mould. It will be admitted that all the lines on the upper or the lower side of the rail which radiate towards the centre of the cylinder, as d o, e m or ij, are level : for instance, the level line, w v, on the top of the STAIRS. 355 rail in the helinet, is a true representation of the radiating line, j i on the plan. The line, b h, therefore, on the top of the rail in the helinet, is a true representation of e m on the plan, and k c on the bottom of the rail truly represents d o. From k, draw k I, parallel to c b, and from h, draw hf, parallel to b c ; join I and b, also c and// then c k lb will be a true representation of the end of the lower piece, B, and c fh b of the end of the upper piece, A ; and/ k or h I will show how much the joint is open on the inner, or concave side of the rail. 356 AMERICAN HOUSE-CARPENTER. To show that the process followed in Art. 473 is correct, let do and e m, {Fig. 312,) be the direction of the butt-joint found as at Fig. 306. Now, to project, on the top of the rail in the helinet, a line that does not radiate towards the centre of the cylinder, as^' k, draw vertical lines fromj and k to w and h, and join w and h ■ then it will be evident that wh is a true representation in the helinet of j k on the plan, it being in the same plane as ; k, and also in the same winding surface asw v. The lii.e, I n, also, is a true repre- sentation on the bottom of the helinet of the line,^' k, in the plan. The line of the joint, e m, therefore, is projected in the same way and truly by i b on the top of the helinet ; and the line, d o, by c a on the bottom. Join a and i, and then it will be seen that the lines, c a, a i and i b, exactly coincide with c b, the line of he joint on the convex side of the rail ; thus proving the lower tnd of the upper piece, A, and the upper end of the lower piece, B, to be in one and the same plane, and that the direction of the joint on the plan is the true one. By reference to Fig. 306 it will be seen that the line, I i, corresponds to x i in Fig. 312 ; and that e k in that figure is a representation of/ b, and i k of d b. F: S . 818. In getting out the twists, the joints, before the falling-mould is stairs 357 applied, are cut perpendicularly, the face. mouii. being longen&jgh to include the oveiplus necessary for a butt-joint. The face-mould for A, therefore, would have to extend to the line, i b ; and that for B, to the line, y z. Being sawed vertically at first, a section of the joint at the end of the face-mould for A, would be represented in the helinet by bifg. To obtain the position of the line, b i, on the end of the twist, draw i s, (Fig. 313,) at right angles to */, and make i s equal to m e at Fig. 312 ; through s. draw s g, pa- rallel to i f, and make s b equal to * b at Fig. 312 ; join b and i ; make if equal to i fat Fig. 312, and from f, draw fg, parallel to -i b ; then i b gf will be a perpendicular section of the rail over the line, e m, on the plan at Fig. 312, corresponding to i b gfin the helinet at that figure ; and when the rail is squared, the top, or back, must be trimmed off to the line, i b, and the bottom to the line,/ g. 478. — To grade the front string of a stairs, having winders in a quarter-circle at the top of the flight connected with flyers at the bottom. In Fig. 314, a b represents the line of the facia along the floor of the upper story, bee the face of the cylinder, and c d the face of the front string. Makeg- b equal to J of the diameter of the baluster, and draw the centre-line of fherail,/g-, g k i and ij, parallel to a b, b e c and c d ; make g k and g I each equal to half the width of the rail, and through k and I, draw lines for the convex and the concave sides of the rail, parallel to the centre-line ; tangical to the convex side of the rail, and parallel to k m, draw no; obtain the stretch-out, q r, of the semi-circle, k p m, according to Art. 92 ; extend a b to t, and k m to s ; make c s equal to the length of the steps, and i u equal to 18 inches, and de- scribe the arcs, s t and u 6, parallel to mp ; from t, draw t w, tend- ing to the centre of the cylinder ; from 6, and on the line, 6 u x, run off the regular tread, as at 5, 4, 3, 2, 1 and v ; make u % equal to half the arc, u 6, and make the point of division nearest to x, as r, the limit of the parallel steps, or flyers ; make r o equal to m z ; from o, draw o a*, at right angles to n o, and equal to one rise ; AMERICAN HOUSE-CARPENTER. Fig. 814. from a°", draw a? s, parallel to n o, and equal to one tread ; from * through o, draw s f. Then from w, draw w c 2 , at right angles to n o, and set up, on the line, w c 1 , the same number of risers that the floor, A, is above the first winder, B, as at 1, 2, 3, 4, 5 and 6 ; through 5, (on the arc, 6 «,) draw d? e 2 , tending to the centre of the cylinder ; from e\ draw e 2 / 2 , at right angles to n o, and through 5, (on the line, stairs. 359 w *,) draw g 2 f, parallel to n o ; through 6, (on tho line, w c 2 ,) and/ 2 , draw the line, h 2 b 2 ; make 6 c 2 equal to half a rise, and from c 2 and 6, draw c 2 i 2 and 6/, parallel to n o ; make h~ v equal to A 2 / 2 ; from i\ draw i 2 k 2 , at right angles to v A 2 , and from/ 2 , draw/ 2 k 2 , at right angles to f A 2 ; upon If, with k 2 / 2 for radius, describe the arc,/ 2 i 2 ; make 6 2 f equal to b 2 f 2 , and ease off the angle at b 2 by the curve,/ 2 P. In the figure, the curve is de- scribed from a centre, but in a full-size plan, this would be imprac- ticable ; the best way to ease the angle, therefore, would be with a tanged curve, according to Art. 89. Then from 1, 2, 3 and 4, (on the line, w c 2 ,) draw lines parallel to n o, meeting the curve in m 2 , n 2 , o 2 and p 2 ; from these points, draw lines at right angles to n o, and meeting it in x"; r 2 , s 2 and f ; from x 2 and r 2 , draw lines tending to u 2 , and meeting the convex side of the rail in gf and/ h, were taken off; then a line drawn upon the bevilled surface from d, at right angles to k d, would show the true position of the joint, because it would be in the direction of the board for the other side ; but a line so drawn would pass through the point, o, — thus proving the principle correct. So, if a line were drawn upon the bevilled surface from d, at an angle of 45 degrees to k d, it would pass through the point, n. 485. — Another method for face-moulds. It will be seen by reference to Art. 481, that the principal object had in view in the preparatory process of finding a face-mould, is to ascertain upon it the direction of a horizontal line. This can be found by a method different from any previously proposed ; and as it requires fewer lines, and admits of less complication, it is probably to be preferred. It can be best introduced, perhaps, by the following explanation . In Fig. 324, j d represents a prism standing upon a level base, b d. its upper surface forming an acute angle with the face, b I, as at Fig. 318. Extend the base line, b c, and the raking line,j I, to meet at/; also, extend e d and g a, to meet at k; from / through k, draw / m. If we suppose the prism to stand upon a level floor, ofm, and the plane, j g a I, to be extended to meet that floor, then it will be obvious that the intersection oetween that plane and the plane of the floor would be in the line, f k; and the line, f k, being in the plane of the floor, and also in the inclined plane, j g kf, any line made in the plane, j g kf, parallel to fie, must be a level line. By finding the position of a perpendicular plane, at right angles to the raking plane, jf k g, we shall greatly shorten the process for obtaining ordinates. 36S AMERICAN HOUSE-CARPENTER. Fig. 824. This may be done thus : from/ draw/o, at right angles tofm; extend e b to o, and gj, to t ; from o, draw o t, at right angles to of, and join t and/; then t of will be a perpendicular plane, at right angles to the inclined plane, t g kf; because the base of the former, of is at right angles to the base of the latter,/ k, both these lines being in the same plane. From b, draw b p, at right angles to of, or parallel tofm ; fiovnp, draw p q, at right angles to of, and from q, draw a line on the upper plane, parallel tofm, or at right angles to tf; then this line will obviously be drawn to the point, j, and the line, qj, be equal top b. Proceed, in the same way, from the points, s and c, to find x and I. Now, to apply the principle here explained, let the curve, b s c, (Fig. 325,) be the base of a cylindric segment, and let it be re- quired to find the shape of a section of this segment, cut by a plane passing through three given points in its curved surface : one perpendicularly over 6, at the height, bj; one perpendicu- larly over s, at the height, s x ; and the other over c, at the height, c / — these lines being drawn at right angles to the chord of the base, b c. Fromj, through I, draw a line to meet the chord line extended to/; from s, draw s k, parallel to b / and from x, draw x k, parallel to jf; from/ through k, draw/wi; then/m will be the intersecting line of the plane of the section with the STAIRS. 369 Fig. 825. plane of the base. This line can be proved to be the intersection of these planes in another way ; from b, through s, and from j, through x, draw lines meeting at m ; then the point, m, will be in the intersecting line, as is shown in the figure, and also at Fig. 324. From/", draw/;?, at right angles tofm; from b and c, and from as many other points as is thought necessary, draw ordinates, parallel tofm; make p q equal to bj, and join q and/; from the points at which the ordinates meet the line, qf, draw others at right angles to q f; make each ordinate at A equal to its cor- responding ordinate at C, and trace the curve, g n i, through the points thus found. Now it may be observed that A is the plane of the section, B the plane of the segment, corresponding to the plane, q pf,oi Fig. 324, and C is the plane of the base. To give these planes their projer position, let A be turned on q f as an axis until it £70 AMERICAN HOCSE-CARPENTER. stands perpendicularly over the line, qf and at right angles to the plane, B ; then, while A and B are fixed at right angles, let B be turned on the line, p f as an axis until it stands perpendicu- larly over pf and at right angles to the plane, C; then the plane, A, will lie over the plane, C, with the several lines on one corres- ponding to those on the other ; the point, i, resting at I, the point, n, at x, and g atj ; and the curve, g n i, lying perpendicularly over b s c — as was required. If we suppose the cylinder to be cut by a level plane passing through the point, I, (as is done in finding a face-mould,) it will be obvious that lines corresponding to qfa.n6.pf would meet in I ; and the plane of the section, A, the plane of the segment, B, and the plane of the base, C, would all meet in that point. 486. — To find the face-mould for a hand-rail according to the principles explained in the previous article. In Fig. 326, a e cf is the plan of a hand-rail over a quarter of a cylinder ; and in Fig. 327, a b c d is the falling-mould ; / e being equal to the stretch-out of a df in Fig. 326. From c, draw c h, parallel to ef; bisect h c in i, and find a point, as b, in the arc, df {Fig. 326,) corresponding to i in the line, he; from i, {Fig. 327,) to the top of the falling-mould, draw ij, a.t right angles to Ac; at Fig. 326, from c, through b, draw c g, and from b and c, draw bj and c k, at right angles to g c ; make c k equal to h g at Fig. 327, and bj equal to ij at that figure ; from k, through j, draw k g, and from g, through a, draw g p ; then gp will be the intersecting line, corresponding to fm in Fig. 324 and 325 ; through e, draw p 6, at right angles to gp, and from c, draw c q, parallel to gp ; make r q equal to h g at Fig. 327 ; joinp and q, and proceed as in the previous examples to find the face-mould, A. The joint of the face-mould, u v, will be more accurately determined by finding the projection of the centre of the plan, o, as at w ; joining s and w, and drawing u v, parallel to 5 w. It may be noticed that c k and b j are not of a length corres- ponding to the above directions : they are but J the length given. BTAlItS,. £t: Fig. 826. 372 AMERICAN HOUSE CARPENTEH. Fig.82T. The object of drawing these lines is to find the point, g, and that can be done by taking any proportional parts of the lines given, as well as by taking the whole lines. For instance, supposing c k and b j to be the full length of the given lines, bisect one in i and the other in m; then a line drawn from m, through i, will give the point, g, as was required. The point, g, may also ba STAIRS. 373 obtained thus : at Fig. 327, make h I equal to c b in Fig. 326 Irom I, draw I Je, at right angles to A c ; from j", drawj k, parallel to he; from g, through k, draw g n ; at i^g\ 326, make 6 j» equal to I n in jFig\ 327 ; then g will be the point required. The reason why the points, a, b and c, in the plan of the rail ai Fig. 326, are taken for resting points instead of e, i and/, is this : the top of the rail being level, it is evident that the points, a and e, in the section a e, are of the same height ; also that the point, i, is ot the same height as b, and c as/. Now, if a is taken for a point in the inclined plane rising from the line g p, e must be below that plane ; if b is taken for a point in that plane, i must be below it ; and if c is in the plane,/ must be below it. The rule, then, for taking these points, is to take in each section the one that is nearest to the line, g p. Sometimes the line of intersection, g p, happens to come almost in the direction of the line, e r : in such case, after finding the line, see if the points from which the heights were taken agree with the above rule ; if the heights were taken at the wrong points, take them according to the rule above, and then find the true line of intersection, which will not vary much from the one already found. h. a i ^ J J / i f ' e Fig. : 495. — To describe the face-mould. At Fig. 330, from/fc, draw k r 2 , at right angles to j- 5 d ; at Fig. 331, make A r equal to A 2 r' in Fig. 330, and from r, draw r s, at right angles to r A ; from the intersection of r s with the level line, m q, through i, draw s t ; at Fig. 330, make A 2 6 2 equal to q t in .Fig-. 331, and join 6 J and r 2 ; from a 2 , and from as many other points in the arcs, a" I and k d, as is thought necessary, draw ordinates to r 2 d, at right angles to the latter ; make r b, (Fig. 332,) equal in its length and in its divisions to the line, r 2 b\ in Fig. 330 ; from r, n, o, p, q STAIRS. 379 and I, draw the lines, r k, n d, oa,pe, qf and I c, at right an- gles to r b, and equal to r* A, «P s a , /* a a , «fcc, in Fig. 330 ; through the points thus found, trace the curves, k I and a c, and complete the face-mould, as shown in the figure. This mould is to be applied to a square-edged plank, with the edge, I b, parallel to the edge of the plank. The rake lines upon the edge of the plank are to be made to correspond to the angle, s t h t in Fig. 331. The thickness of stuff required for this mould is shown at Fig. 331, between the lines s t and u v — u v being drawn pa- rallel to * t. 496. — All the previous examples given for finding face-moulds over winders, are intended for moulded rails. For round rails, the same process is to be followed with this difference : instead of working from the sides of the rail, work from a centre-line. After finding the projection of that line upon the upper plane, describe circles upon it, as at Fig. 293, and trace the sides of the moulds by the points so found. The thickness of stuff for the twists of a round rail, is the same as for the straight ; and the twists are to be sawed square through. h f v h 380 AMERICAN HOUSE-CARPENTER 497. — To ascertain the form of the newel-cap from a section of the rail. Draw a b, (Fig. 333,) through the widest part of the given section, and parallel to c d ; bisect a b in e, and through a, e and b, draw h i,fg, and k j, at right angles to a b ; at a con- venient place on the line, fg, as o, with a radius equal to half the width of the cap, describe the circle, i j g ; make r I equal to e b or e a ; join I and _;', also I and i ; from the curve, f b, to the line, I j, draw as many ordinates as is thought necessary, parallel to f g ; from the points at which these ordinates meet the line, I j, and upon the centre, o, describe arcs in continuation to meet op; from n, t, x, &c, draw n s, t u, &c, parallel to / g ; make n s, t u, &c, equal to e f, iv v, &c. ; make x y, &c, equal to z d, &c. ; make o 2, o 3, &c, equal to on, o t, &c. ; make 2 4 equal to n s, and in this way find the length of the lines crossing o m ; through the points thus found, describe the section of the newel-cap, as shown in the figure. 498. — To find the true position of a butt joint for the twists of a. moulded rail over platform stairs. Obtain the shape of the mould according to Art. 466, and make the line a b, Fig. 334, equal to a c, Fig. 300 ; from b, draw b c, at right angles to a b, and equal in length to n m, Fig. 300 ; join a and c, and bisect a c in o ; through o draw e /, at right angles to a c, and d k, parallel to cb; make o d and o k each equal to half e h at Fig. 300 ; through e and /, draw h i and g j, parallel to a c. At Fig. 301, make n a equal to e d, Fig. 334, and through a, draw r p, at right angles tone; then rp will be the true position on the face-mould for a butt joint, as was required. The sides must be sawn verti STAIRS. 381 cally as described at Art. 467, but the joint is to be sawn square through the plank. The moulds obtained for round rails, {Art. 464,) give the line for the joint, when applied to either side of the plank ; but here, for moulded rails, th r ; line for the joint can be obtained from only one side. Whe j the rail is canted up, the joint is taken from the mould laid on the upper side of the lowei twist, and on the under side of the upper twist ; but when it is canted down, a course just the reverse of this is to be pursued. When the rail is not canted, either up or down, the vertical joint, obtained as at Art. 466, will be a butt joint, and therefore, in such a case, the process described in this article will be unnecessary NOTE TO ARTICLE 462. Platform stairs with a large cylinder. Instead of placing the platform-risers at the spring of the cyl- inder, a more easy and graceful appearance may be given to the rail, and the necessity of canting either of the twists entirely obviated, by fixing the place of the above risers at a certain distance within the cyl- inder, as shown in the annexed cut — the lines indi- cating the face of the risers cutting the cylinder at k and I, instead of at p and q, the spring of the cylin- der. To ascertain the position of the risers, let a b c be the pitch-board of the lower flight, and c de that of the upper flight, these being placed so that 6 c and c d shall form a right line. Extend a c to cut d e in /; draw / g parallel to d 6, and of indefinite length : draw g o at right angles to / g, and equal in length to the radius of the circle formed by the centre of the rail in passing around the cylinder ; on o as centre describe the semicircle j g i; make o h equal to the radius of the cylinder, and describe on o the face of the cylinder phq ; then extend d b across the cylinder, cutting it in I and k — giving the position of the face of the risers, as required. To find the face-mould for the twists is simple and ob- vious : it being merely a quarter of an ellipse, hav- ing o j for semi-minor axis, and the distance on the rake corresponding to og, on the plan, for the semi-major axis, found thus, — extend t ;' to meet a f, then from this point of meeting to / is the semi-major axis. SECTION VII.— SHADOWS. 499. — The art of drawing consists in representing solids upon a plane surface : so that a curious and nice adjustment of lines is rrade to present the same appearance to the eye, as does the human figure, a tree, or a house. It is by the effects of light, in its reflection, shade, and shadow, that the presence of an object is made known to us ; so, upon paper, it is necessary, in order that the delineation may appear real, to represent fully all the shades and shadows that would be seen upon the object itself. In this section I propose to illustrate, by a few plain examples, the simple elementary principles upon which shading, in architectural sub- jects, is based. The necessary knowledge of drawing, prelim- inary to this subject, is treated of in the Introduction, from Art. 1 to 14. 500. — The inclination of the line of shadow. This is always, in architectural drawing, 45 degrees, both on the elevation and the plan ; and the sun is supposed to be behind the spectator, and over his left shoulder. This can be illustrated by reference to Fig. 335, in which A represents a horizontal plane, and B and C two vertical planes placed at right angles to each other. A rep- resents the plan, C the elevation, and B a vertical projection from the elevation. In finding the shadow of th i plane, B, the SHADOWS. 383 Fig. 835. line, a b, is drawn at an angle of 45 degrees with the horizon, and the line, c b, at the same angle with the vertical plane, B. The plane, B, being a rectangle, this makes the true direction of the sun's rays to be in a course parallel to d b ; which direction has been proved to be at an angle of 35 degrees and 16 minutes with the horizon. It is convenient, in shading, to have a set-square with the two sides that contain the right angle of equal length ; this will make the two acute angles each 45 degrees ; and will give the requisite bevil when worked upon the edge of the T- square. One reason why this angle is chosen in preference to another, is, that when shadows are properly made upon the draw- ing by it, the depth of every recess is more readily known, since the breadth of shadow and the depth of the recess will be equal. To distinguish between the terms shade and shadow, it will be understood that all such parts of a body as art not exposed to the direct action of the sun's rays, are in shade ; while those parts which are deprived of light by the interposition of other bodies, are in shadow. 384 AMERICAN HOUSE-CARPENTER. iJillK 1 i I 1 III Fig. Fig. 387. Tig. 888. 501. — To find the line of shadow on mouldings and other ho- rizontally straight projections. Fig. 336, 337, 338, and 339, represent various mouldings in elevation, returned at the left, in the usual manner of mitreing around a projection. A mere in- spection of the figures is sufficient to see how the line of shadow is obtained ; bearing in mind that the ray, a b, is drawn from the projections at an angle of 45 degrees. Where there is no return at the end, it is necessary to draw a section, at any place in the length of the mouldings, and find the line of shadow from that. 502. — To find the line of shadow cast by a shelf. In Fig. 340, A is the plan, and B is the elevation of a shelf attached to a wall. From a and c, draw a b and c d, according to the angle previously directed ; from b, erect a perpendicular intersecting c d at d ; from d, draw d e, parallel to the shelf ; then the lines, c d and d e, will define the shadow cast by the shelf. There is another method of finding the shadow, without the plan, A. Extend the lower line ct tne shelf to/, and make cf equal to the projection of the shelf SHADOWS. 385 Fig. 840. from the wall ; from/, draw fg, at the customary angle, and from c, drop the vertical line, c g, intersecting / g at g ; from g, draw g e, parallel to the shelf, and from c, draw c d, at the usual angle ; then the lines, c d and d e, will determine the extent of the shadow as before. 1 e ! I'M' "' b c Fig. 841 ; p~» B 503. — To find the shadow cast by a shelf, which is wider at one end than at the other. In Fig. 341, A is the plan, and B the elevation. Find the point, d, as in the previous example, and from any other point in the front of the shelf, as a, erect the perpen- dicular, a e ; from a and e, draw a b and e c, at the proper angle, and from b, erect the perpendicular, b c, intersecting e c in c ; 49 386 AMERICAN HOUSE-CARPENTER. from d, through c, draw do; then the lines, i d and d o, will give the limit of the shadow cast by the shelf. Fig. 343. 504. — To find the shadow of a shelf having one end acute or obtuse angled. Fig. 342 shows the plan and elevation of an acute-angled shelf. Find the line, e g, as before ; from a, erect the perpendicular, a b ; join b and e ; then b e and e g will define the boundary of shadow. '^ iiii Hi 11 c niiiiiiiiiMiiiiiiinii nnmimimii liiiiiiini III! /i Fig. 848. 505. — To find the shadow cast by an inclined shelf. In Fi<*. 343, the plan and elevation of such a shelf is shown, having also one end wider than the other. Proceed as directed for finding the shadows of Fig. 341, and find the points, d and c ; then a d an I d c will be the shadow required. If the shelf had been SHADOWS. 387 parallel in width on the plan, then the line, d c, would have been parallel with the shelf, a b. 1 ^111 III,, ' e c Fig. 844 Fig. 845. 506. — To find the shadow cast by a shelf inclined in its ver- tical section either upward or downward. From a, {Fig. 344 and 345,) draw a b, at the usual angle, and from b, draw b c, parallel with the shelf; obtain the point, e, by drawing a line from d, at the usual angle. In Fig. 344, join e and i ; then i e and e c will define the shadow. In Fig. 345, from o, draw o i, parallel with the shelf ; join i and e ; then i e and e c will be the shadow required. The projections in these several examples are bounded by straight lines ; but the shadows of curved lines may be found in the same manner, by projecting shadows from several points in the curved line, and tracing the curve of shadow through these points. Thus — ^ 1 o o ■ 1 11 Bi finiinniti mini lllllllllllll 11 iiiiiiiiiiiiiiiiiiiiiinii a a / a Fig. 847. Fig. 846. 388 AMERICAN HOTjSE-CARPENTER. 507- — To find the shadow of a shelf having its front edge, on end, curved on the 'plan. In Fig. 346 and 347, A and A show an example of each kind. From several points, as a, a, in the plan, and from the corresponding points, o, o, in the elevation, draw rays and perpendiculars intersecting at e, e, &c. ; through these points of intersection trace the curve, and it will define the shadow. e ej ill! I e lipiilliiimiini iiiiiiiiiiiim.il iiiiiiiiiiiniii Fig. 848. 508. — To find the shadow of a shelf curved in the elevation. In Fig. 348, find the points of intersection, e, e and e, as in the last examples, and a curve traced through them will define the shadow. The preceding examples show how to find shadows when cast upon a vertical plane ; shadows thrown upon curved surfaces are ascertained in a similar manner. Thus — SHADOWS. 389 509. — To find the shadow cast upon a cylindrical wall by a rejection of any kind. By an inspection of Fig. 349, it will be seen that the only difference between this and the last examples, is, that the rays in the plan die against the circle, a b, instead of a straight line. Fig. 350. 510. — To find the shadow cast by a shelf upon an inclined wall. Cast the ray, a b, (Fig. 350,) from the end of the shelf to the face of the wall, and from b, draw b c, parallel to the shelf; cast the ray, d e, from the end of the shelf ; then the lines, d e and e c, will define the shadow. These examples might be multiplied, but enough has been given to illustrate the general principle, by which shadows in all instances are found. Let us attend now to the application of this principle to such familiar objects as are likely to occur in practice. 390 AMERICAN HOUSE-CARPENTER. 511. — To find the shadow of a projecting horizontal beam From the points, a, a, &c, (Fig. 351,) cast rays upon the wall , the intersections, e, e, e, of those rays with the perpendiculars drawn from the plan, will define the shadow. If the beam be in- clined, either on the plan or elevation, at any angle other than a fight angle, the difference in the manner of proceeding can be seen by reference to the preceding examples of inclined shelves &c. Fig. 85a. 512. — To find the shadow in a recess. From the point, a, (Fig. 352,) in the plan, and b in the elevation, draw th • rays, a c and b e ; from c, erect the perpendicular, c e, and frt.m e, draw the horizontal line, e d; then the lines, c e and e d, will show the extent of the shadow. This applies only where the back of the recess is parallel with the face of the wall. Fig. 853. 513. — To find the shadow in a recess, when the face of the wall is inclined, and the back of the recess is vertical. In Fi°-. 353, A shows the section and B the ele ration of a recess of this SHADOWS. 391 kind. From b, and from any other point in the line, bo. as a draw the rays, b c and a e ; from c, a, and e, draw the horizonta' lines, c g, a f, and eh; from d and /, cast the rays, d i and/ h ; from i, through h, draw i s ; then s i and z g- will define the shadow. d Fig. 854. 514. — To find the shadow in a fireplace. From a and b, {Fig. 354,) cast the rays, a c and b e, and from c, erect the per- pendicular, c e ; from e, draw the horizontal line, e o, and join r and d ; then c e, e o, and o d, will give the extent of the shadow. Fig. S55. 515. — To find the shadow of a moulded window-lintel. Cast rays from the projections, a, o, & ;., in the plan, {Fig. 355,) and d, e, &c, in the elevation, anc' draw the usual perpendiculars in- tersecting the rays at i, i, and i ; these intersections connected 392 AMERICAN HOUSE-CARPENTER. and horizontal lines drawn fiorr them, will define the shadow The shadow on the face of the lintel is found by casting a rai back frcm i to 5, and drawing the horizontal line, s n. Fig. 356. 516. — To find the shadow cast by the nosing of a step. From a, {Fig. 356,) and its corresponding point, c, cast the rays, a b and c d, and from b, erect the perpendicular, b d ; tangical to the curve at e, cast the ray, e /, and from e, drop the perpendicular, e o, meeting the mitre-line, a g, in o ; cast a ray from o to i, and from i, erect, the perpendicular, i f ; from h, draw the ray, h k; from f to d and from d to k, trace the curve as shown in the figure ; from k and h, draw the horizontal lines, k n and h s ; then the limit of the shadow will be completed. 517. — To find the shadow thrown by a pedestal upon steps. From a, {Fig. 357,) in the plan, and from c in the elevation, draw the rays, a b and c e ; then a o will show the extent of the shadow on the first riser, as at A ; f g will determine the shadow on the second riser, as at B ; c d gives the amount of shadow on the first tread, as at C, and h i that on the second tread, as at D ; which completes the shadow of the left-hand pedestal, both on the plar and elevation. A mere inspection of the figure will be suf- SHADOWS. 393 Fig. 857. ficient to show how the shadow of the right-hand pedestal is obtained. Pig. i»8. Fig. 859. 518. — To find the sliadow thrown on a column by a square abacus. From a and b, (Fig. 358,) draw the rays, a c and b e, and from c, erect the perpendicular, c e ; tangical to the curve at d, draw the ray, d f, and from h, corresponding to / in the plan, draw the ray, ho; take any point between a and/, as i, and from this, as also from a corresponding point, n, draw the rays, i r and n s ; from r, and from d, erect the perpendiculars, r s and do; through the points, e, s, and o, trace the curve as shown in the figure ; then the extent of the shadow will be defined. 519. — To find the shadow thrown on a column by a circular abacus. This is so near like the iast example, that no explanation will be necessary farther than a reference to the preceding article 50 394 AMERICAN HOUSE-CARPENTER. Fig. 860. 520. — To find the shadows on the capital of a column. This may be done according to the principles explained in the examples already given ; a quicker way if doing it, however, is as follows If we take into consideration one ray of light in connection with all those perpendicularly under and over it, it is evident that these several rays would form a vertical plane, standing at an angle of 45 degrees with the face of the elevation. Now, we mav sup pose the column to be sliced, so to speak, with planes of this SHADOWS. 393 Fig. 861. nature — cutting it in the lines, a b, c d, &c, (Fig. 360,) and, in the elevation, find, by squaring up from the plan, the lines of sec- tion which these planes would make thereupon. For instance : in finding upon the elevation the line of section, a b, the plane cuts the ovolo at e, and therefore /will be the corresponding point upon the elevation ; h corresponds with g, i with j, o with s, and I with b. Now, to find the shadows upon this line of section, cast from m, the ray, m n, from h, the ray, h o, &c. ; then that part of the section indicated by the letters, m f i n, and that part also be- tween h and o, will be under shadow. By an inspection of the figure, it will be seen that the same process is applied to each line of section, and in that way the points, p, r, t, u, v, w, x, as alsc 1, 2, 3, &c, are successively found, and the lines of shadow traced through them. Fig. 361 is an example of the same capital with all the shadows finished in accordance with the lines obtained on Fig. 360. 521. — To find the shadow thrown on a vertical wall by a column and entablature standing 'n advance of said wall. Cusl 396 AMERICAN HOUSE-CARPENTER Fig. 862. rays from a and 6, {Fig. 362,) and find the point, c, as in the previous examples ; from d, draw the ray, d e, and from e, the horizontal line, e /; tangical to the curve at g and k t draw the rays, g j and h i, and from i and j, erect the perpendiculars, * / and; k; from m and n, draw the rays, wz/and n k, and trace the curve between k and /; cast a ray from o to p, a vertical line from p to s, and through 5, draw the horizontal li le, st; the shadow as required will then be completed. SHADOWS. 39V (■■■■I NNlllllII ■MM i I: it PI in 1 ■llllllllllllllllllllllllllllllllllllllllll ii Fig. ! Fig. 363 is an example of the same kind as the last, with all the shadows filled in, according to the lines obtained in the pre- ceding figure. Fig. 864 522 — Fig. 364 and 365 are examples of the Tuscan cornice. The manner of obtainir g the shadows is evident. 398 AMERICAN HOUSE-CARPENTER. Fig. 866. 523. — Reflected light. In shading, the finish and life of he object depend much on reflected light This is seen to advantage in Fig. 361 and On the column in Fig. 363. Reflected rays are thrown in a direction exactly the reverse of direct rays ; therefore. on that part of an object which is subject to reflected light, the shadows are reversed. The fillet of the ovolo in Fig. 361 is an example of this. On the right-hand side of the column, the face of the fillet is much darker than the cove directly under it. The reason of this is, the face of the fillet is deprived both of direct and reflected light, whereas the cove is subject to the latter. •Other instances of the effect of reflected light will te seen in the other examples. APPENDIX ALGEBKAICAL SIGNS. +, plus, signifies addition, and that the two quantities between which it stands are to be added together ; aso + i, read a added to b. — , minus, signifies subtraction, or that of the two quantities between which it occurs, the latter is to be subtracted from the former ; as a — 6, read a minus b. X, multiplied by, or the sign of multiplication. It denotes that the two quantities between which it occurs are to be multiplied together ; as a x b, read a multiplied by 6, or a times b. This sign is usually omitted between symbols or letters, and is then understood, as ab. This has the same meaning as a x b. It is never omitted between arithmetical numbers ; as 9 x 5, read nine times five. — , divided by, or the sign of division, and denotes that of the two quanti- ties between which it occurs, the former is to be divided by the latter ; as a -T- b, read a divided by b. Division is also represented thus : % in the form of a fraction. This signifies that a is to be divided by * b. When more than one symbol occurs above or below the line, or both, as , it denotes that the product of the symbols above cm the line is to be divided by the product of those below the line. =, is equal to, or sign of equality, and denotes that the quantity or f quantities on its left are equal to those on its right ; as a — b — c, read a minus b is equal to c, or equals c ; or, 9 — 5 = 4, read nine minus five equals four. This sign, together with the symbols on each side of it, when spoken of as a whole, is called an equation. a 2 denotes a squared, or a multiplied by a, or the second power of o, and a 3 denotes a cubed, or a multiplied by a and again multiplied by a, or the third power of a. The small figure, 2, 3, or 4, &c, is termed the index or exponent of the power. It indicates how many times the symbol is to be taken. Thus, a" = a a, a' =aaa, a* = aaaa, 7 is the radical sign, and denotes that the square root of the quantity following it is to be extracted, and 51 € APPENDIX. ■^ denotes that the cube root of the quantity following it is to be ex- tracted. Thus, V9 — 3, and \/2l = 3. The extraction of roots is also denoted by a fractional index or exponent, thus a* denotes the square root of a, a 5 * denotes the cube root of a, a* denotes the cube root of the square of a, &c TRIGONOMETRICAL TERMS. Fig. 864 In Fig. 366, where A B is the radius of the circle BOH, draw a line A F, from A, through any point, C, of the arc B G. From draw D perpendicular to A B ; from B draw B E perpendicular to A B ; and from G draw G F perpendicular to A G. Then, for the angle FA B, when the radius A O equals unity, CD is the sine ; AD the cosine ; D B the versed sine ; BE the tangent , G F the cotangent ; AE the secant ; and A F the cosecant. 6 APPENDIX. But if the angle be larger than one right angle, yet less than two right angles, as B A H, extend HA to K and EB to E, and from S draw H J perpendicular to A J. Then, for the angle BAH, when the radius A H equals unity, HJ is the sine ;AJ the cosine ; BJ the versed sine ; B K the tangent ; and A K the secant. When the number of degrees contained in a given angle is known, then the value of the sine, cosine, <&c, corresponding to that angle, may be found in a table of Natural Sines, Cosines,