STRAINS IN TRUSSES. LONDON: PRINTED BY qPOTTISWOODE AND CO., NEW-STREET SQUARE AND PARLIAMENT STREET THE STRAINS IN TRUSSES COMPUTED BY MEANS OF DIAGRAMS: WITH TWENTY EXAMPLES DRAWN TO SCALE. BY FRANCIS A. RANKEN, M.A. C.E. LECTURER AT THE HARTLEY INSTITUTION, SOUTHAMPTON; FORMERLY ASSISTANT ENGINEER ON THE CAMBRIAN RAILWAYS, AND SCHOLAR OF GONVILLE AND CAIUS COLLEGE, CAMBRIDGE. D. APPLETON AND CO. NEW YORK. 1872. PREFACE. IN THE FOLLOWING PAGES I have endeavoured to set in a clear light the theory and method of computing by diagrams the Strains in Trusses bearing a constant load, without taking into consideration the subject of transverse strain. While treating only Balancing Systems of forces in one plane, each system acting on a point, I have purposely avoided introducing the Principle of Moments, wishing to confine myself to the results to be deduced from the Parallelogram of Forces. I have tried to avoid everything that might give the appearance of difficulty, and I hope that what I have written may be found intelligible even by those who have no previous knowledge of Statics. In doing so I have made use of the assistance of others when necessary, and I have to thank Professor FULLER, of University College, for several suggestions. CONTENTS. PAGE I. INTRODUCTION......... 1 II. BALANCING SYSTEMS OF THREE FORCES..... 8 III. ROOF TRUSSES........ 13 IV. BALANCING SYSTEMS OF MORE THAN THREE FORCES... 28 V. ROOF TRUSSES (continued)....... 33 VI. GIRDERS, OR BRIDGE TRUSSES...... 52 VII. APPENDIX......... 60 THE STRAINS iN TRUSSES. I. INTRODUCTION. 1. I PROPOSE to investigate certain cases in which the external forces acting on structures can be reduced to separate systems of forces, each system acting on a point; and by the term structure I mean to be understood a combination of distinct pieces of one or more materials which is designed to remain in equilibrium under the forces that may act upon it. 2. A structure is in equilibrium, so far as we are now concerned, if it retains as a whole a fixed position with respect to the earth, and bodies once placed at rest have the property of remaining without motion until some cause acts upon them so as to change their position. This tendency to remain at rest is called inertia, and causes tending to produce change of position in bodies, or parts of bodies, are called forces. Matter, or the substance of which bodies are composed, is 2 THE STRAINS IN TRUSSES. that on which forces act, and bodies containing different amounts of matter require different amounts of force to produce in them the same velocity in the same time, the amount of force being proportional to the amount of matter acted upon. Thus we say that the inertia of a body is proportional to its mass, or quantity of matter, and a body composed of wood, whose mass is less than an equal bulk of iron, possesses also less inertia in the same proportion. The forces with which we shall have to do are the attraction of the earth for all bodies, which is called weight, and the resistance offered by substances on the application to them of forces which are in general caused by the weights of other bodies. 3. A force always acts in a straight line, called its direction, which passes through its point of application, where it acts on a body. These, together with the magnitude, are called the elements of a force, and may be represented geometrically by a finite straight line. When two'forces act simultaneously on the same point their tendency will be to move the point in a direction lying between the directions of the two forces, if these directions are different, and this tendency will have a magnitude bearing some relation to the magnitude of the forces. There will therefore be some single force whose tendency would be identical with that of the two forces, and this force is called the resultant of the two forces, which with reference to their resultant are called components.. INTRODUCTION. 3 It is obvious that if a force equal and opposite to the resultant acted on the point in combination with the two original forces, the point would remain at rest under the influence of the three forces. Such a system of forces is called a balancing system. 4. It is necessary also to consider systems of parallel forces, for the attraction of the earth on bodies is an example of forces which are sensibly parallel. The resultant of a system of parallel forces is a single force in a direction parallel to them. The direction of gravitation, or the attraction of the earth at any point, is called the vertical, and any line perpendicular to it is said to be horizontal. The magnitude of this attraction on any body is called the weight of the body. Weight, or the force of gravity, which if unresisted would produce in all bodies equal velocities in equal times, is proportional to the mass or quantity of matter a body contains (Art. 2). Hence, we have the means of measuring both matter and force: for if two bodies weigh the same, they will contain the same quantity of matter, however different they may look, and they will also require the same amount of force to counteract their weight and keep them from falling to the earth. We arrive, then, at the common mode of measuring forces by pounds, tons, kilogrammes, &c., a force of one pound meaning a force of magnitude sufficient to overcome the weight of a quantity called a pound of matter, and keep it suspended above the earth's surface, in a certain locality. 4 THE STRAINS IN TRUSSES. 5. A balancing system may consist of only two forces, as in the case of the suspended weight just referred to, and in this case the two forces must be equal and opposite, and must of course act through the same point, but a much more general case is that of three or more forces acting on a point and keeping it at rest. The fundamental proposition of the science of forces treats of this case, and the proof of this proposition, which is called the parallelogram offorces, will be found in mathematical works. The enunciation is as follows:-'If two forces acting on a point be represented in direction and magnitude by two straight lines, and a parallelogramn be constructed having these two straight lines as adjacent sides, the diagonal of the parallelogram which passes through the point will represent the resultant in direction and magnitude.' Exanmple: B /A 8 Draw two straight lines A B =7, A C 8, including an angle of 60~ at the point A. Then the diagonal A D will be 13. Hence, by the above proposition, a force of 13 acting INTRODUCTION. 5 (like R) in the direction D A, will, with the original forces 7 and 8, form a balancing system keeping the point A at rest. A machine with pulleys may be made to illustrate this practically. Readers acquainted with trigonometry will be able to calculate the magnitude of A D thus: A D2=A B2 + A C2 + 2 A B. A C. COs 60 =49 + 64 + 2 x 7 x8 x 2 169 - 132, AD = 13 The angle D A c is found thus: SinD A C — 7 sin 60~. 6. The Triangle of Forces.-This important proposition is derived immediately from the preceding, being in fact only another way of stating it. Consider the triangle A C ID; A C represents 8 in direction and magnitude, D C represents 7 in magnitude and is parallel to its direction, and A D represents 13 in direction and magnitude, and 8, 7, and 13 acting in these directions upon A, keep it at rest. Hence the proposition:' If three forces keep a point at rest, three straight lines parallel to their directions, taken iln order, will form a triangle whose sides will be proportional to the magnitudes of the forces.' If, then, we know the directions of three forces, forming a balancing system, and the magnitude of one of them, the magnitudes of the others are obtained by a simple geometrical construction. The importance of this in designing structures will shortly appear. *BJ 3 6 THE STRAINS IN TRUSSES. The proposition can easily be extended to the case of more than three forces by substituting for any two of the forces their resultant and combining it with another of the forces; and so on. Thus: replace P and Q, in the figure, by B D S h their resultant A D, and combine A D with another force R. The resultant of these two will be a force s, represented in direction and magnitude by A F; P, Q, R, and s will therefore form a balancing system keeping A in equilibrium, and the sides of the figure with thick lines, A B, B D, D F, F A, forming a closed polygon, are parallel and proportional to P, Q, R, and s; and so on. Hence, The Polygon of Forces. — If any number of forces in one plane keep a point at rest, straight lines parallel to their directions, taken in order, will form a polygon whose sides will be proportional to the magnitudes of the forces.' 7. One more preliminary proposition alone need be noticed.' If two parallel forces act at two points, A and B, in the INTRODUCTION. 7 same direction, their resultant is equal to their sum and acts at a point c, so that c divides A B inversely as the forces.' A C B 8 7 Thus, if 3 act at A, and 4 at B, 7 acts at c, and A C -4 ofA B, 78~~~~~~~~~7 BC = L f A B. When the forces are equal, c is half-way between A and B. Now, if a line A B is uniformly loaded throughout, we may take equal pairs of forces equally distant from the middle and replace them by their sum acting at the middle, till finally we get the sum of all the load acting in the middle of the line, and this load we may divide into two halves, one half at each end of the line. The proof of this proposition may be studied in mathematical works, and it may be illustrated practically by easy experiments. 8 TIE STRAINS IN TRUSSES II. BALANCING SYSTEMS OF THREE FORCES. 8. WE may now proceed to the solution, by means of diagrams, of problems illustrating the principles which have been enunciated in the Introduction. (i.) Forces of 8 and 15 lbs. act on a point A in directions including a right angle. Find a third force which in combination with the two given forces shall form a balancing system at the point A. Draw B C equal and parallel to the force 8, and B D equal and parallel to the force 15. Join D C; then D C will C 8 15 B D 17 DIAGRAM. be equal and parallel to the required force. This follows from Art. 6, and if we draw through A a line parallel to c D, we shall have the direction of the required force. The SYSTEMS OF THREE FORCES. 9 diagram is drawn to a scale representing 10 lbs. by one inch, and to this scale c D measures 17. Therefore the magnitude of the required force is 17 lbs. Hence, 8, 15, and 17 lbs., acting as represented in the figure, form a balancing system at the point A. (ii.) Two forces of 144 and 145 lbs. act at a point A, and a third force forming a balancing system makes a right angle with the force 144. Find the magnitude of this force. Draw B C equal and parallel to 144, and cD at right angles to it. From the centre B describe a small arc of a circle of radius 145 cutting c D in D. Then c D will be equal and parallel to the required force. c D measures 17, therefore 144, 145, and 17 lbs., acting as in the figure, will form a balancing system at A. 17 141 141 D B =C DIAGRAM. (iii.) Two forces of 31 lbs. act on a point A including an angle of 60~. Find a third force which will form a balancing system at A. Draw B C, B D equal and parallel to the given forces. Then 10 THE STRAINS IN TRUSSES. c D, which is approximately 6 lbs., will be the required force. D A 8 C 3~ B DIAGRAM. (iv.) Find two forces making angles of 120~ and 150~ with a force of 13'8 lbs., and forming with it a balancing system at A. Draw B C equal and parallel to 13'8, and through B and c draw straight lines in the given directions. They will include a right angle at D, and be equal to the required forces. Their magnitudes are 12 and 6'9 lbs. e1 9D 13.8 "' DIAGM a. (v.) Three forces making angles of 120~ with one another will form a balancing system if they are equal. SYSTEMS OF THREE FORCES. 11 D A B C DIAGRAM. (vi.) Two forces of 50 lbs. each Include an angle of 150~. Find the third force of the balancing system. Result: 25'88 lbs. 05 VO Bq, A 2588 D DIAGRAM. (vii.) A weight of 20 lbs. is hung by two cords, each 10 20 lbs. DIAGRAIM. 12 THE STRAINS IN TRUSSES. feet long, to two points, on the same level 19 ft. 6 in. apart. Find the force with which each cord is pulled. Draw the directions of the cords and weight, as in the figure, and draw c D equal and parallel to the weight of 20 lbs. Then B c, and B D, drawn parallel to the cords, will be found to represent 37'5 lbs. each, the required force of tension. Note.-When lines in the diagrams are not exactly of the proper length, the error must be attributed to unequal contraction of the paper after printing. ROOF TRUSSES. 1 III. ROOF TRUSSES. 9. THE preceding examples are probably sufficient to illustrate the method of finding by diagrams the relations between three forces forming a balancing system at a point. We may now proceed to the consideration of the forces acting on the particular kind of structures which are known as trusses, used in the construction of roofs, &c. Definition. —A truss is the term applied to a triangular or polygonal frame to which rigidity is given by staying and bracing, so that its figure shall be incapable of alteration by turning of the bars about their joints.* In the cases about to be considered it will always be assumed that the load acts only upon the joints. Where this is not the case, the several pieces bearing a distributed load must possess sufficient stiffness to sustain the load without danger, and the calculations for determining their dimensions for this purpose must be made according to the * Rankine's Applied Mechanics. 14 THE STRAINS IN TRUSSES. theory of beams, because the dimensions calculated for them as component parts of a truss will not be sufficient. The theory of beams, however, does not come within the scope of this book. 10. The load on a roof, now to be considered, may be made to bear on certain points of the principal frames or trusses. It consists of the slates, or other covering material, fastened to small, or secondary rafters, which, placed at short intervals apart, rest on horizontal purlins, on the ridgepiece of the roof, and on the pole-plates. These horizontal pieces which support the rafters bear upon the principal trusses, which therefore have to bear the weight of all the superincumbent materials, their own weight, and the pressure of the wind, making altogether the total load. The principal trusses are placed at intervals along the roof, and the amount of load each of them bears will depend on their distance apart. Referring to Art. 7, we see that a uniformly distributed load, like the weight of roofing-materials between two principal trusses, is borne equally by two supports at its ends, so that each principal truss will bear half the load on that portion of the roof between it and the next truss; and if the trusses are equally distant from one another, each will bear a load equal to one division of the roof, except the end trusses, which will only bear half as much. To this must be added the weight of the truss, and the pressure of the wind on one division of the roof, the three forces forming the load on one truss. ROOF TRUSSES. 15 The values of these quantities are obtained by measuring the dimensions of the various'portions of the roof, and ascertaining what their weight will be by reference to tables of weights of materials, the allowance made for the wind being about 40 lbs. on every square foot of the surface of the roof. To understand the method of computing strains by diagrams, so far as the principal trusses are concerned, we shall not need to know the actual load on any roof, for the principle will be sufficiently explained by taking any number or even a symbol to represent it. I shall therefore take the number 12, for the sake of convenience, and treat it as a symbol representing the load on one roof truss. 11. The simplest kind of truss consists of three essential pieces, viz.: two principal rafters, or principals, connected with one another at the top, and fastened to the extremities A B C of a horizontal piece called the tie-beam, whose use is to keep the lower ends of the principals from separating. These three form a triangle ABC; AB and AC are the principals, and B c the tie-beam, the ends of which rest on the walls of the building. It is necessary, however, to keep the tie-beam from bending down out of a horizontal or level line, and this is done by suspending its middle 16 THE STRAINS IN TRUSSES. point by a piece called the king-post from the point A. With this addition the truss is called a king-post truss. Very little consideration is necessary to see that the load resting by means of the purlins, &c., upon the principals will tend to push them downwards. They are therefore said to be in compression, and any piece of a structure which is compressed is called a strut. The tie-beam, on the other hand, which has to keep the ends of the principals from separating, is pulled, and is said to be in tension, and pieces in tension are called ties. F' B D C The accompanying figure will help to explain the construction of the king-post truss. IK and H L are two of ROOF TRUSSES. iT the small rafters, resting on the ridge-piece H, the purlins Xw and N, and the pole-plates K and L. By means of these supports the superincumbent load is transmitted to the principal truss, which consists of two principals A B and A c, the tie-beam B c (resting on wall-plates B and c), the kingpost A D, and two struts E F and E o to keep the principals from bending. The king-post may be fastened to the tie-beam by means of an iron strap, or an iron king-bolt may be used instead of it. (viii.) The diagram is commenced by drawing the centre lines of all the pieces along which the forces of resistance will act. Now, taking 12 as a symbol to represent the load, one quarter of it will act on each of the four equal divisions K M, MH, H N, N L, SO that each of these divisions bears a load represented by 3. Again, 3 acting on x M may be replaced by 1'5 acting at x, and 1'5 acting at M; and 3 acting on M H may be replaced by 1P5 acting at M, and 1P5 at H (Art. 7). Hence we have two forces of 1P5 acting at M, that is to say, 3 acts at M. Similarly, 3 acts at H and N, and 1[5 at L. Draw a b in the direction of the load at F, and make a b -=3. From a and b draw a c and b c parallel to the supporting forces at F, that is to say, a c parallel to the principal and b c to the strut F E. Then these supporting forces will be represented by the lengths of a c and b c, that 18 THE STRAINS IN TRUSSES. is, by 3'6 and 3'5 respectively. Now draw c d parallel to the tie-beam, and consider the two triangles a c d and b c d. The former is the diagram for the point B, having its sides parallel to the push down the principal a c, the pull along the tie-beam c d, and the supporting pressure of the wall a d. Now, since a b c is the diagram for the balancing system at ~ F - G.,, B 8'4 D b'4 C, 3'2 DIAGRAM. r, and a c is the thrust down the principal, c d will be the pull along the tie-beam, and a d the pressure of the wall in the balancing system at B, and these are 3'2 and 1'75 respectively. Again, at 1 we have a balancing system caused by the thrust down each of the struts F E and G E, and the pull of the king-post. The diagram for this point ROOF TRUSSES. 19 will be double of the triangle b c d, that is, it will be the side b c parallel to F E, another side parallel to G E, and a third side parallel to A E. Hence, the forces along F E and G E will each be equal to b c, or 3 5, and that along the king-post will be twice b d, or 2'5. We have now done with the effects of the load at F in producing strains or forces along the adjacent pieces of the truss, and we have partly considered the load at G which causes the thrust along G E. Further consideration of the load at G is unnecessary, because its treatment will be the same as that of the load at F. Now, the strain on the king-post 2 5, above mentioned, is transmitted to A, and must be added to the load 3 already existing there to form the total load at that point, which is balanced by the thrusts of the two principals. We thus get the load at A equal to 5'5, nearly half the whole load. Draw e f parallel and equal to 5 5, and from e andf draw e g and f g parallel and therefore equal to the thrusts down the principals caused by the load at A. These lines both measure 5'9. Now draw g h parallel to the tie-beam, and we get f g h, the diagram for B, and e g h that for c, so far as the load at A is concerned in producing balancing systems at those points. We may now add up the various strains on the pieces produced by the load distributed at the various points of the roof, noticing that when two strains are transmitted through the same piece, the total strain will be the effect of both. c2 20 THE STRAINS IN TRUSSES. Strain on principalsbetween A and F eg, orf g 5.9 or A and G between F and B)f or oand c) Strain on struts E F, or E G, b c = 35. These are the compressive strains. Strain on tie-beam, h g + c d = 8'4 Strain on king-post, 2 b d = 2'5. N.B.-To this latter must be added the weight of the tie-beam, which will also increase the load at A in practice, but for simplicity I have omitted this quantity. These are the tensile strains. Now, the pressure on each wall is evidently half the whole load, that is to say, 6 (Art. 7). This is obtained by adding together for the point B, 1'5 the load resting on the pole-plate K, a d, or 1'75, andf h, or 2'75. 1'5 + 1'75 -+ 2'75 = 6. The same process gives 6 as the pressure on,the wall at c. I have investigated this case of the king-post truss somewhat at length to make it as plain as possible. Succeeding cases will be treated more briefly, for the principle of drawing diagrams for the balancing systems at the various points being once understood, its application may always be made in the same manner. In the examples, for the sake of convenience, the diagrams are made small, but in practice, to insure the proper accuracy, the scale for the outline ROOF TRUSSES. 21 of the truss should not be less than 5 feet to the inch, and for the diagrams 10 tons to the inch, the actual load on the roof being treated exactly as the symbol 12 is in this book. (ix.) The figure represents the outline of a queen-post roof-truss. A C and B D are principals, c D the tie-beam, A E and B F the queen-posts, A B the straining-piece, G E and F H struts. Taking 12 to represent the load, we consider.ih distributed over c A, A B, and B D in the proportion of their lengths, so A 7 B 8' 9 7 DIAGRAM. that A B bears 5, while c A and B D bear 3-5 each. c G and G A bear half of this each, that is, 1P75. Half the load on c G may be considered to be supported by c, i.e. *875, and half by G, which point also supports half the load on G A, making the total load on G equal to 1. 75. 22 THE STRAINS IN TRUSSES. Draw a b equal and parallel to 1'75, the load at G, and a c, b c parallel to the principal G c and the strut G E respectively. Then a b c is the diagram for the point G. Draw c d parallel to the tie-beam c E; then a c d'is the diagram for the point c, and b c d for the point E. From these last two diagrams we find a pressure of'875 on the wall at c, and an equal pressure on E transmitted by the queen-post to A. The same construction gives the strains caused by the load at H, which is equal, and acts in every respect like to the load at G. The load at A is made up of half the load on A B-that is, half of 5 —half the load 1'75 on A o, and the pressure transmitted by the queen-post from the point E. 2'5 +'875 +'875 = 4'25, the whole load at A. Draw e f equal and parallel to 4'25, the load at A, and draw e g and f g parallel to the principal and tie-beam respectively; e f g is then the diagram for the point A, and it is also the diagram for the point c, and gives the strains at that point caused by the load at A. We now have the strains on all the pieces of the truss, as follows: Strain on principalsbetween A and G, or B and i, e g = 8'2 between G and c, or H and D, eg + a c = 9'95 Straining beam A B, f g = 7 Struts G E, or F H, b c = 1'75 ROOF TRUSSES. 23 Tie beambetween E and F, fg = 7 between c and E, or F and D, f g + c d = 8'5 Queen-posts A E, or B F, b d = -875. The pressure on the walls is obtained by adding a d and e f to the pressure at c caused by half the load on c G, that is, to.875. Thus'875 +'875 + 4'25 = 6, or half the total load on the truss, which it evidently must be (Art. 7). (x.) The next figure represents a roof truss containing a king-post A F, and two queen-posts D G and E H. The load 12 is distributed over six equal portions of the two principals; we shall therefore have a weight represented by 2 resting on each of the five points K, D, A, E, and L, while the points B and c bear 1 each. Make a b equal and parallel to 2, the load at K, and a c, b c parallel to K B, K G. Then a b c is the diagram for K. Draw c d parallel to B G, make e f equal and parallel to the load at D, which consists of 2 and b d equal to 1, therefore ef = 3. Draw e g, f g parallel to D F, D B, and g h parallel to B F. Then ef g is the diagram for D. Make k I equal and parallel to the load at A, which consists of 2 and twice e h equal to 4, therefore k I = 6, twice e h being taken because of the two struts D F, E F, as in the king-post truss. Draw k m, I m parallel to the principals A C, B c, and m n parallel to the tie-beam B c. Then k m I is the diagram for A. 24 THE STRAINS IN TRUSSES. 12:6 (G 10.1 F 10'1 H 126 a e \ 25 7,6.DLGRAM. Compressive Strains: Principals, A D, or A E, 8'2 D K, or E L, 8'2 + 2'8 = 11 K B, or L c, 8'2 + 2.8 + 2.8 = 13.8 Struts, K G, or H L, 2'8 D F, or E F, 3'2 Tensile Strains: Tie-beam, F G, or F H, 7'6 + 2'5 = 10'1 G B, or H C, 7'6 + 2'5 + 2'5 = —12'6 Ties, D G, or E H, 1 AF 4 Pressure on each wall: 1 + 1 + 1 + 3 = 6. ROOF TRUSSES. 25 (xi.) The next roof is a king-post truss with two sloping tie-rods instead of a horizontal tie-beam. Iron is a suitable material, the struts being of T or angle iron, and the ties iron bars. Castings are used for the junctions of ties with struts. It will be observed that the sloping tie-rods exert a downward pull independent of their weight on the kingbolt, which is not the case when there is a horizontal tiebeam. 19 6 C 4,7 14 8.7 DOGSS~8' Compressive Strains: Principalsbetween A and E, or A and F, F g = 9-2 between E and B, or F and c, eg + ac -- 92 + 4'9 =14:1 Struts E D, or F D b c = 4'8 26 THE STRAINS IN TRUSSES. Tensile Strains: Tie-rods B D, or C D, h g + c d = 8'7 + 4'7 = 13'4 King-bolt A D, 2 b d = 32 Pressure on one wall: 1'5 + e h + hl + a d = 6. The diagram for E or F is a b c; for D, twice b c d; for A, efg; for B, gf k, g k I, and a c d. The only difference between this case and the former king-post truss is that here we have sloping tie-rods which introduce the new triangle g h k, the sides of which are parallel to B D, D C, D A. It will therefore be seen that the force 3'2, with which D A is pulled, is due partly to the struts E D, F D, and partly to the ties B D, C D, the former causing a strain on A D of 2 6, and the latter a strain of'6. - The pressure on the walls is derived from i of the load together with the vertical strains due to the three component trusses A B C, D B C, and EB D. (xii.) The next truss is constructed on the same principles as the last one, which are capable theoretically of indefinite extension, and for this reason the diagrams will be drawn without any explanation. If 12 represents the whole load, the load at E is _ = 1 5, that at G is 1'5 + b d = 2'3, that at K is 1'5 + fh = 3'05, that at A is 1'5 + 2 1 n = 6. The pressure on the wall is ~75 + ad + eh + nk + or + rt= 6. ~~~ cc*4 ~b~~~~ 0 -J4~~~~~~~~~p p-4 155 cI~~~~~~~ & ~ ~ M X3 a, fi q t Ace;L. ~Q.T A5 o 4~~~5 R~~ A~~~~~~~p O ~ 77 28 THE STRAINS IN TRUSSES. IV. BALANCING SYSTEMS OF MORE THAN THREE FORCES. 12. IN all the examples as yet given the strains have been treated as forming at each joint a balancing system of only three forces, and this has been possible because in all instances where the strains were to be ascertained, if more than three forces acted at a point, the additional number were in the same straight lines as one or more of the others, and could be added to, or subtracted from them. This method of treatment makes it possible to apply the principle of the triangle of forces to every point supposed to be kept at rest by three forces only, and though errors in drawing may easily arise in the diagrams unless great care is taken, a check can be applied in the cases given by ascertaining if the supporting forces of the walls obtained from the diagrams correspond with what we know them to be from Art. 7. When more than three forces keep a point at rest, the principle of the polygon of forces (Art. 6) gives the relation between them, and if we know the directions of the forces ard-th: mragnitudes of all except two, the magnitudes of the SYSTEMS OF MORE THAN THREE FORCES. 29 two which are unknown can be found by drawing a diagram in accordance with this principle. This method can also be applied to the case of forces in one plane which do not pass through a point, for the resultant of two of them can be combined with a third, and so on till we have left only three forces, two of them being those whose directions only are known. The triangle of forces will then give the magnitudes of these two forces, so that the whole process is equivalent to drawing a polygon with sides equal and parallel to the forces which are known, taken in order, and the two remaining sides being drawn parallel to those whose directions only are known will give their magnitudes also. The following simple example will show the application of these principles in a general case. Let a frame, as in the figure, be loaded with given forces, P1, P2, P., at the joints, acting in given directions, and be supported by two forces R1, R2, whose directions also are given. Construct the polygon A B C E D, having its sides parallel to the forces and so that A D, D E, E C represent P1, P,, P3 respectively in magnitude. Then A B will represent R1, and -B c will represent R2 in magnitude. Draw A F, F B parallel to the pieces lettered al, b, respectively, so that A F B is the triangle of forces for the joint numbered 3. In the same way draw B G c, the triangle of forces for the joint 4. Now, at the point 1 four forces act, but as the magnitudes of two of them are known we can construct the diagram A H X D having its sides equal and 30 TIIHE STRAINS IN TRUSSES. I C1 a, 22 R 1 P1 4 a 3 DIAGRAM. C/ b/ \,C /~L M RR,/ I 13~~~~~~~~~~~~~~~~~~C~ Ii / b~2f'/ j/ SYSTEMS OF MORE THAN THREE FORCES. 31 parallel to these four forces taken in order, so that A H will represent the strain on cl, and D K that on a2. In the same way we may draw the diagram c L M E for the joint 2. It now only remains to draw the diagram for the joint 5, where P2 acts. This is done by drawing E o equal and, parallel to B F, or bl, and D Q equal and parallel to B G, or b2,; o s and Q s being then drawn equal and parallel to A H and c L respectively must meet in a point if the drawing is correctly done, and this forms a sufficient check on the process. The following is a list of the polygons of forces for the several points: Joint Diagram 1........ AHKD 2........ CLME 3........ AFB 4........ CGB 5........ EOSQD Whether any piece is a strut or a tie will depend on the direction in which the corresponding line in the diagram is drawn. For example, in the figure A H K D, which is the diagram for the joint 1, beginning with D A, which represents P1, A H represents cl, which therefore exerts a pull on the joint 1, and is a tie; H K represents a,, which is therefore a strut; K D represents a2, also a strut. Hence P, is balanced by two struts and one tie. By this method, we see that a1, a2, a. are struts, and the other pieces are ties. 32 THE STRAINS IN TRUSSES. I have taken a very simple case in order that the diagram may not be confused by a great number of lines, but it is evident that the same method may be extended to many cases of frames consisting of triangles. loaded at the joints, and that the accuracy of the drawing is checked by observing that the polygons whose sides are parallel to the forces must all be closed figures. ROOF TRUSSES. 33 V ROOF TRUSSES (continued). 13. I WILL now, before giving other examples, revert to example (x.) to show how the method which has just been explained may be applied to that case, though the former 3 a, J+ g I" i a, c a 6 b, 8 b~ b i0, a, II II V'. DIAGRAM. DL~cr~,aM. D 34 THE STRAINS IN TRUSSES. method will probably'be thought easier, and when carefully checked will prove accurate enough. It must first be noticed that, taking 12 for the load, a load 1 rests on each wall, and there remains only 10 to be supported due to the load on the other joints of the frame. We must therefore, in considering the strains caused by these loads on the pieces, take R1 and R2 each to be equal to 5. I have numbered all the joints, and put letters to the pieces for convenience in referring to them. Now, since both the supporting forces and all the loads are vertical forces, the polygon becomes a straight line A F formed by the segments A B, B C, C D, D E, E F, in one direction, which represent PI, P4, Ps, P2, P1, each equal to 2, and in the other direction the forces R1, R2 are represented by the segments F G, G A, each equal to 5. The method of constructing the figure is exactly the same as in the preceding general case. I shall therefore only enumerate the diagrams for the several joints in the form of a table. Joints Diagrams 1........ EFHK 2........ DEKLG D 3........ GLN 4........ CBMNGC 5........ BAHM 6........ FGH 7........ AGH 8..... CGHK 10........ DG H M ROOF TRUSSES. 35 The diagram for point 9 exists in two halves, which would form a closed figure if c and D coincided, and also K and M. By measuring the sides which give the strains on the several pieces, these will be found to correspond very closely with the strains formerly found in this example. (xiii.) In this example, we may consider three forces P1, P,, Pr, each equal to three, acting at the joints 4, 1, 5, and upward pressures R,, R2, each equal to 4'5, acting at 2 and 3. 2 b, 6 b2 7 b, 3 A L coNC - f - Draw the polygon of forces A E c F B, a straight line, as D 2 36 THE STRAINS IN TRUSSES. in case (x.), because all the forces are parallel, and draw A D C, the triangle of forces for the point 2, giving the magnitudes of the strains on a, and bl. Draw D G parallel to cl, and G E parallel to a2, then A D G E is the polygon of forces for the point 4.. Draw G I parallel to b2, and I c parallel to dl, then c D G I is the polygon of forces for the point 6. The rest of the figure is constructed in the same way, beginning at the point 3. Now, it will be observed that no polygon of forces has been drawn for the point 1, though the strains on the pieces a2, a3, d, d2 are determined; but if F L, E M are drawn parallel to d1, d2, and L N, M N parallel to a., a2, a closed figure is obtained, which is evidently the polygon of forces for the point 1, having its sides equal and parallel to the forces acting at that point. The various strains can now be arranged in a table, as follows: — Compression Principals, a1, or a,, 15'7 a2, or a3, 14'8 Struts, cl, or c2, 2'9 Tension: Tie-rods, bi, or b3, 15'1 b2, 9'9 d,, or d2, 5'2 ROOF TRUSSES. 37 (xiv.) The next example is very similar. 1 I~ ap aJ R P, P3 R. 22 ~~~~~~~~~~8 F E d, 12 P. C P, a4 i~~~~~~~~~~~ H b, ~~L DIPIRA~M. The figure is constructed in the same way as in the previous example, but the different inclination of the tie' 38 THE STRAINS IN TRUSSES. rods makes it seem at first sight to bear but a small resemblance to the former figure. The scale of loads A B is small, owing to the limited size of the paper, so that it will be well in this and in other cases to draw the diagram to a much larger scale. Joints Diagrams 2........ AD C 3..... BKC 4.. ADEG 5.. BKLH 6.. CDEF 7...... CKLM (xv.) This example differs from the preceding one in having two vertical struts, from 6 to 8 and from 7 to 9. They cause a different distribution of the load by introducing two more joints 6 and 7, but the whole load resting on these joints is transmitted along the struts, and the strain on the principal is therefore the same on each side of the joints 6 and 7. If we suppose the load distributed along the pieces of the principals in the proportion of their lengths, we may take r~ and P5 equal to 2'25, P2 and P, equal to 1'7, and P3 equal to 1'5, the remaining load 1 3 resting directly on each of the two walls. The resultant supporting force is therefore 6 - 1'3 = 4 7. Construct the polygon of forces, which is the straight line A B as before, and the polygons for the several joints. Now, since the load on 6 is transmitted by the strut c2 to 8, D R ROOF TRUSSES. 39 1, P ^ A B P, 5 - O7 Hto b, F xp, = 15 R b, G:px = 2'25 v v. DUGRAPM. must be drawn equal to P Q which represents this load, and R L equal and parallel to c D or b, must also be drawn in order to get the complete figure D R L H F which is the polygon of forces for the point 8. The same remarks apply to the other side of the truss. There is no diagram for either of the points 6 and 7, for the reason before stated; and it is not necessary, except as a check, to draw the diagram for the point 1, because there are in the figure sufficient lines to determine the strains at that joint. 40 THE STRAINS IN TRUSSES. Joints Diagrams 2.. - -.. AD 3 BE C 4..ADFN 5....... BEGO 8........ D LHF 9.... ERMKG (xvi.) This example is like a roof designed for the works of the drainage of London. a, d, %K/" "~\~;~: n6 R a, P. b, bE D P/ DIIAGIUAM. The diagram is constructed as in the former cases, and is like the diagram of example (xiii.), as might be expected. ROOF TRUSSES 41 It will be noticed that, in order to form complete polygons for the several points, two lines have been drawn to represent the strain on each of the pieces b., d1, and d,. The following is a list of the polygons: Joints Diagrams 2...... ACG 4...... ACDE 5..... EHK L 6... GCDF 7....... FDHM No diagram is drawn for point 1, as in the last case, and the other diagrams are similarly situated to those given. (xvii.) This is an example of what is called secondary trussing, 1, 2, 3 being considered the primary truss, 1, 2, 8 a secondary truss, 2, 7, 5 and 1, 9, 5 smaller secondary trusses, and the same on the other side of the frame. Regarding it in this way, each truss may be treated separately after calculating the additional load at the joints produced by the trusses one upon the other. In Rankine's'Civil Engineering' the load is first distributed so that onehalf rests at the point 1, one-quarter at 5, and one-eighth at 4 and at 6, and the trusses are then treated separately; but I believe the accuracy of the drawing to be better checked by the method I have used. In a small work on roofs, by Campin, the strains are expressed by formula involving the angles of inclination of the pieces, but this requires measurements which cannot be more accurate than the measure ments of a diagram like that which I have drawn. 42 THE STRAINS IN TRUSSES. azk P1 a X This diagram is drawn by the same rules as before, but at the point 5 we are met by a difficulty which has not before arisen. Having deduced from the diagrams for the joints 4 and 7 the strains on a2 and di, and knowing the load r2, we have to draw a polygon whose sides shall be equal and parallel to these forces, and also to the three unknown strains on as, c2, and d2. Now, it has been said that a polygon of forces can be constructed for a point when the directions of the forces keeping it at rest are known, and directions of the forces keepinor it at rest are known, and ROOF TRUSSES. 43 the magnitudes of all but two, and here we have apparently three unknown magnitudes. It appears, however, that the position of d2 with respect to P3 is similar to that of b1 with respect to P1; the strain on d2 will therefore be the difference between c G and D F, which is the strain on b1 caused by the load P,. This will also be found to be equal to F G, the strain on d,. The diagram for point 5 will therefore be thus drawn. Since E D represents the strain on a2, draw E M to represent the pull on d2, L M to represent P2, L K the pull on dc. Then K H drawn parallel to as, and H D to c2, will give the strains on those pieces. K H is equal and parallel to o Q, which also represents a3, and s H represents a4 in the diagram for point 6. For the point 9 we must draw o T, H V parallel to d3 and d4, and v T will be found equal and parallel to d2. I ought again to urge how necessary it is that all these diagrams should be most carefully drawn to a large scale, and how little dependence can be placed on small diagrams for the proper length of the lines, especially if they have been reproduced in printing. The diagrams for half the frame are as follows: Joints Diagrams 2.... ACG 4...A C D FE.5.....EDHKL \M 6....... OHSQ 7.........GCDF f3.... FDHN 9....... -OT V H 44 THE STRAINS IN TRUSSES. In practice it will be better to draw the diagrams for both halves of the frame, and their symmetrical position will check the accuracy of the drawing. (xviii.) Distributing the load in this case in proportion to the lengths between the joints in the principals, and constructing the diagram in the same way as before, it only needs to be remarked that in the diagram for the point 7, di is represented by L D, which is a part of the line E D, the whole of which represents a2. 1 aa 2 b6 6 b, 3 B Q P. = 2'5 dd P =1G b, K, P, = 25 DIAGRAM. ROOF TRUSSES. 45 The following table contains the diagrams for half the frame: Joints Diagrams 1....... KNOQS 2. GAC 4.... ACDE 5.... EDHK 6..... DF 7..... DHML (xix.) This truss and the following one are designed for the roofs of engine-houses for pumping engines. The number of pieces and their varying inclinations make the diagram appear complicated. I will therefore show the order in which the lines must be drawn. Take A B to represent the external forces, and divide it into segments representing them in magnitude. Draw A D parallel to al, c D parallel to bl. Also draw D E parallel to c, and G E to a2. From E draw E F parallel to b6, and draw C F parallel to d. For the point 5 draw M L equal and parallel to c F, draw L K parallel to as, and G H parallel to d2, and between these two lines place H K equal and parallel to G E. Then L K represents the strain on a,. Draw M N equal and parallel to L K, N 0 parallel to c2, and Q o to a4. Then Q o will represent a4. Draw o T and c T parallel to b3 and d3 respectively, and make N s and c s equal and parallel to E F and G H respectively. These, if the figure is properly drawn, will meet in the point s, and complete the polygon for the point 8, forming a check on the drawing. 46 THE STRAINS IN TRUSSES. B V a,=1 A further check is obtained by drawing u v equal and 2........ ADO P8 =' P, 2/ / 0 A further check is obtained by drawing u v equal and parallel to c T, q w parallel to d,, and equal to c?, w x equal and parallel to q o, and v x parallel to a, and equal to Q o. These last two lines must meet in the point x, and they complete the polygon for the point 1. x must also lie on a horizontal line passing through c. Joints Diagrams 1........UVXWQ 2 ADC ROOF TRUSSES. 47 Joints Diagrams 4..... ADEG 5....... GHKL M 6........ MNOQ 7........ CDEF 8....... TONS (xx.) This diagram is drawn in nearly the same order as the last. The diagram for the point 1 has been omitted to 5 P5 6 a a4 a0123~ c B F a ID1~ ~ ~K P, = 164 dC PP, N4o[1 N, P., 1'5 48 THE STRAINS IN TRUSSES. save space, but it should be drawn as a check in the follow ing manner. Diraw from x and Q lines parallel to d3 and d6 respectively, and make them equal to s T. From their extremities draw lines parallel to a5 and a, respectively, and equal to o Q. The extremities of these can then, if the figure is properly drawn, be joined by a line equal and parallel to u v, and a horizontal line through c will bisect it. Joints Diagrams 2........ ADC 4......... ADEG 5.... GHKLM 6........ MNOQ 7........ CDEF 8........ TSON 9..... OSUVW (xxi.) This truss, and modifications of it, may frequently be seen in railway stations, and similar buildings. Though the diagram is constructed on exactly the same principles as those which precede, the number of pieces, making different angles with the horizon, causes an appearance of confusion which will be considerably lessened when it is drawn out to a large scale. The diagrams for half the frame are as follows:Joints Diagrams 1.'.......ADC 2........ AEFG 3.... GKLMN 4........ NSTUV 8...... CDFH vs......(: CHO Q ROOF TRUSSES. 49 1 7 ~Pi~~~, d@W Ps - P.C DIAGRAM. A check is obtained by drawing the diagram for the point 4, and observing that the point T must lie on a horizontal line passing through c. (xxii.) Modifications of the next truss are also used for the roofs of stations, &c. It may of course consist of a great number of divisions, but I have taken the load as resting only on four joints, exclusive of the points of support, in order that the diagram may look as clear as possible. The diagrams for all the points of half the frame are constructed as before, having their sides parallel to the forces acting at each point taken in order. 1 I~~~~~ 50 THE STRAINS IN TRUSSES. 3 a5 4 7 ~~~~~~~~~~~~1 6 ~X~~~~~~~~d a2 ad K 3 b3 Y P, d, P, DIAGRaM. Joints Diagrams 1........ AD C 2.... AEFGL 3...... LKNC 7....... CDH 8........CHKM The truss is theoretically in equilibrium for a uniform steady load, but the quadrilateral figure 3, 4, 8, 9 would be subject to distortion unless the load were perfectly uniform, as is also the case with the quadrilateral in the queen-post truss. Diagonal braces 3, 9, and 4, 8, are therefore intro ROOF TRUSSES. 51 duced in this, and frequently in the queen-post truss, though in the latter case, if the load is fairly uniform, the stiffness of the tie-beam may be relied on to prevent change of form. Diagonals are also introduced in the other divisions to guard against unequal loading, in which case they will bear some of the strains. A frame thus constructed is said to be counter-braced, and the term is also applied to pieces of a structure which are constructed so as to be able to bear either tension or compression, according to the position of the load. E2 52 THE STRAINS IN TRUSSES. VI. GIRDERS, OR BRIDGE TRUSSES. 14. I HAVE hitherto treated examples of trusses which are loaded uniformly with a steady load, as is usually the case with roofs. Bridge trusses are girders constructed so far like roof trusses in that the load is applied at the joints of the frame, and the pieces are therefore strained chiefly in the direction of their lengths. When the load on any truss is steady but not uniform, the diagrams will of course be different from those I have given, though constructed on the same principles; but when the load is moving, different diagrams will be necessary for all the possible arrangements of the load. This will be the case with trussed girders for bridges, and it is not my intention to enter at length upon the discussion of the effects of moving loads on bridges. I shall, however, give two examples of ordinary trussed girders for a uniform steady load, to show that the method of diagrams can be applied to them; but as there are generally in these girders a number of parallel pieces, the strains on them can usually be found more conveniently by formula involving trigonometrical functions, and the reader can consult other BRIDGE TRUSSES. 53 works on the subject. For large spans the form of girder called the bow-string is well suited, and I shall give an example of one kind of bow-string girder, to show how the strains upon it, caused by a steady load, are found by diagrams. When the span is large, the moving load bears a smaller proportion to the fixed load due to the weight of the girder, and when this is the case the different pieces of the girder will not need to be counter-braced to resist opposite kinds of strain due to different positions of the load. As the subject of counter-bracing has been alluded to, I shall here give a simple case of a counter-braced queenpost bridge truss, to show how in this case the strains on the braces are determined, but it is not my object to go further into the subject. (xxiii.) The figure represents a queen-post truss of the same shape as the roof truss of the same name, but reversed in position, so that the struts of the former case become ties, and vice versd. It is counter-braced by the diagonals 2, 6, and 3, 5. Let us suppose a load P = 3 resting on the joint 2, and no load resting on 3. Then we know, from Art. 7, that if the three divisions a1, a2, a3 are all equal, the supporting forces R1 and R2 will be 2 and 1 respectively. Construct the diagram on the same principles as before. Then at point 5 we have two ties b1, d1, which may bear all the pressure of the strut c1, and the piece b, suffer no strain. 54 THE STRAINS IN TRUSSES. R,=2 R,=1 a, 2 a 3 3 a 1 4 P-3 5 b 6 CA B d2 c a 6 A. D E a, F DIAGRAM. The diagrams are as follows:Joints Diagrams 1....... AD C 2..A E...F G rB 3........ MNG 4........ B C K 5.....CDH 6........ CKL In the two examples which follow, a load represented by the symbol I will be considered to act at each loaded joint. (xxiv.) The figure represents what is called a Warren girder, supporting a uniform load on each joint or apex of the upper chord, and resting at the extremities of the lower chord or flange on two abutments. I have introduced this and the following example merely to show that the same method of drawing diagrams applies to trussed girders, but BRIDGE TRUSSES. 55 1 a, 2 a, 2 a, 4 a, 5 a. 6 a, 7 P P2 P23 P d. 8 f1 9 -b2 -10 b3' 11 b, 12 b 13 b6 14 b7 16 PB P, PS ptz a4 y,.ID e Ck 1......... AEG 11........ LO \I a, it is both easier and more accurate to find the strains from formulae when there are a number of parallel pieces whose inclinations are known. The polygons are as follows: Joints Diagrams I.A.......EFG 2 G KL LM N 3..QSTU 4.. T TXYZ 8.A..ACD 9........CDHF 10..HOL 11........LO W 56 THE STRAINS IN TRUSSES. When more than one system of triangular bracing is introduced, the truss is called a lattice-girder, and each system of bracing must be treated separately. (xxv.) This system of bracing may be called right-angled, to distinguish it from the isosceles system of the last example. In each case the load, and also the supporting forces, may be applied to either the upper or lower boom. In each case also, if the load is variable, separate diagrams will be necessary for all the different positions of the load, and the present example will require counter-bracing with additional pieces, -R R., a, 2 a, 3 a, 4 a4 5 a, 6 a, 7 a, S a, Pi P, P P P d, C d, c, d, c d4 cl d 4 Cd 4 7d, C7 d, 10 b, 1i b, 12 b, 13 b, 14 b, 16 b, 16 B P7 P5Za, Y b, E4 P4 a, T' a, __N oad x( ~, dls dw IF a, G DLAGRAM. BRIDGE TRUSSES. 57 whilst the former method of construction will require that some or all of the existing diagonals should be counterbraced to bear both tension and compression. Joints Diagrams 1........ ADC 2....... AFGHK 3........ LMNO 4...... OQST B' 5........ B' TYZ 10....... CDE 11........ CKHE 12........ KUMH 13........ UVWXM (xxvi.) Bowstring Girder.-The figure of this example, without the diagram, will be found in Stoney on' The Theory of Strains in Girders,' where a table of the strains is given, both for a uniform and a moving load, said to be obtained by using the principle of the triangle of forces for all the joints in succession. As I have only considered the case of a uniform load, leaving the reader to construct for himself the diagrams for the moving loads, I here subjoin Mr. Stoney's results for that case, to be compared with the measurements in my diagram. The load is taken as 10 tons at every joint of the string, or lower flange. 58 THE STRAINS IN TRUSSES. aa a, 6 a, 1'' 11 12 18 14 1 16 17d 51 b 4 b5 ba b4 b, b b, b., 10 P, P2 P, P. P. P. P, B P2 P. B' b4 x dO _~ a2, / bs,/d A b, E DIAGRAM.,Strains of Compression: a, = 78'9, a2- 86' 3, a -83'2 4= 82, a5= 81'6 Strains of Tension: b-70'5, b2 762, bG = 781, dl=1l, d2 6'6, d3 7, Fd 53, d-= 114, d2=-5, d3= 53, for a uniform load. Ill the accompanying diagram, which is drawn to a scale BRIDGE TRUSSES. 59 of thirty tons to the inch, the following are the polygons of forces: Joints Diagrams 1........ ADC 2........ ADK 3........ KDQO 4........ QDRP 5........ RDZY 11........ AEFGH 12........ H LMND' 13..... D' S T U C' 14....... C'VWXB' Having now by a number of cases exemplified the method of drawing diagrams to compute the strains in trusses, it only remains for me to recommend the reader to apply the same method to other examples which he may meet with, and to extend those I have given to the circumstances of a load which is not uniform. In every instance, when all the strains are obtained, the practical engineer will endeavour so to proportion the materials used that the sectional area of the pieces strained may have a proper ratio to the amount of strain. This ratio will depend conjointly on the substance of which the piece is composed, and on the nature of the strain, whether tension or compression, which it is to bear, and many experiments have been made to determine the proper ratio in every case of common occurrence. This belongs, however, to the subject of the'strength of materials,' on which there is no need that I should enter at present. 60 THE STRAINS IN TRUSSES. VII. APPENDIX 15. REFERRING to the enunciation of the polygon of forces in Art. 6, the reader will find it there stated that the sides of the diagrams must be drawn parallel to the directions of the forces, taken in order. By following this rule we avoid all doubt whether any piece is a strut or a tie, for the directions in which the sides of the polygons are successively drawn are the same as those in which the forces they represent act upon the point under consideration, so that it is always evident whether any force is a push or a pull upon the point. I have therefore, in all the preceding examples, drawn the diagrams so as to show all the forces acting at each point taken in order, but by disregarding the order it is often possible to reduce the number of lines in the diagrams, and so make them look neater. The instructions for drawing a diagram in a general case, like that before given, will be as follows, the order of the forces at a point being disregarded, and the word polygon, or closed figure, being taken in a modified sense. APPENDIX. 6.P, a2X P DIAGRAM. Let a frame, as in the figure, consisting of triangles, be loaded with given forces P,, P2, P3 at the joints, acting in any given directions, and be supported by two known forces R1, R2. Draw a polygon, or closed figure A E D C B, having its sides equal and parallel to the forces in the order in which they act upon the external pieces of the frame. From the extremities of A B, representing R1, draw A H, B H, parallel to a,, b,, and from the extremities of D c, representing R2, draw 62 THE STRAINS IN TRUSSES. D G, C G, parallel to as, b,. Draw H F parallel to c1, and G F to c2, and join F E, which must be parallel to a2. There are now closed figures to represent the forces acting at each point of the frame. The reader may apply this method to the preceding examples, and compare the results with what he has obtained by drawing the former diagrams to a large scale. I shall conclude by giving the diagram for example (xiii.) drawn in this way. B. ~ at, xxd, d~ em ~ 2 b, 7- b2 7'b.~: B P-3 P, = G DIAGRAM. APPENDIX. 63 Joints Diagrams 1...... FEGKH 2........ ADC 3...... BDC 4...... ADGE 5........ BDHF 6......D C KG 7........ D CKH. LONDON: PRINTED BY BPOmTTBWOODE AND CO., NEW-STLREET SQUARE AND PARLIAMENT STREET