this, together with W which rests directly on 1, makes up the load of W, already mentioned. (3.) Smaller Secondary Trusses 34 5, 9 10 2. Each of the points 4 and 10 sustains a load of W, from which the stresses on the bars of those smaller trusses can be determined. One-half of the load on 4, that is ' W, hangs by the suspensionrod 65; and this, together with W, which rests directly on 6, makes up the load of W on that point, formerly mentioned. The same remarks apply to the suspension-rod 8 9. (4.) Resultant Stresses. The pull between 5 and 9 is the sum of those due to the primary and larger secondary trusses; that between 5 and 3, and between 9 and 2, is the sum of the pulls due to the primary, larger secondary, and smaller secondary trusses. The thrust on 1 6 is due to the primary truss alone; that on 64 to the primary and larger secondary truss; that on 43 to the primary, larger secondary, and smaller secondary trusses; and similarly for the divisions of the other rafter. Example III. Suppose that instead of only three divisions, there are n divisions in each of the rafters 1 3, 1 2, of fig. 78; so that besides the middle suspension-rod 1 7, there are n-2 suspension-rods under each rafter, or 2 n 4 in all; and n 1 sloping struts under each rafter, or 2 n 2 in all. There will thus be 2n - 1 centres of resistance; that is, the ridge-joint 1, and n each rafter; and the load directly supported on each of these W 1 on The total load on the ridge-joint, 1, will be as before,; that The total load on the upper joint of any secondary truss, distant from the ridge-joint by m divisions of the rafter, will be, W; 4 n The stresses on the struts and tie of each truss, primary and secondary, being determined as in Article 149, are to be combined as in the preceding examples. 160. Compound Trusses-Several frames, without being distinguishable into primary and secondary, may be combined and con nected in such a manner, that certain pieces are common to two or more of them, and require to have their stresses determined by the Theorem of Article 158. Example I. In fig. 79, 89 represents part of the horizontal platform of a suspension bridge, supported and balanced by being hung from the top of a central pier, 1, by pairs of equally inclined rods or ropes, viz :—1 8 and 1 9; 1 6 and 1 7; 14 and 1 5; 1 2 and 1 3. 2 Fig. 79. Here 8 19 is to be considered as a distinct triangular frame, consisting of a strut 89, and two ties 1 8 and 1 9, loaded with equal weights at 8 and 9, and supported at 1. Let x denote the height of the point of suspension I above the level of the loaded points, ys ya, the distance of those points on either side of the middle of the pier, P the load at each point, Rg R, the pull on each of the ties, 1 8, 1 9, T., the thrust between 8 and 9 along the platform. Then we have = and similar equations for each of the other distinct frames 617, 4 1 5, 2 1 3. Then using a similar notation in each case, the thrust along the platform and so on for as many pairs of divisions as the platform consists of. Example II. Fig. 80 represents the framework for supporting one side of a timber bridge, resting on two piers at 1 and 4. consists of four distinct trusses, viz., It but all those trusses have the same tie-beam, 14; and the pull along that tie-beam is the sum of the pulls due to the four trusses. 161. Resistance of Frame at a Section.—THEOREM. If a frame be acted upon by any system of external forces, and if that frame be conceived to be completely divided into two parts by an ideal surface, the stresses along the bars which are intersected by that surface, balance the external forces which act on each of the two parts of the frame. This theorem, which requires no demonstration, furnishes in some cases the most convenient method of determining the stresses along the pieces of a frame. The following consideration shows to what extent its use is limited. CASE 1. When the lines of resistance of the bars, and the lines of action of the external forces, are all in one plane, let the frame be supposed to be intersected anywhere by a plane at right angles to its own plane. Take the line of intersection of these two planes for an axis of co-ordinates; say for the axis of y, and any convenient point in it for the origin O; let the axis of a be perpendicular to this, and in the plane of the frame, and the axis of 2 perpendicular to both, and in the plane of section. the The external forces applied to the part of the frame at one side of the plane of section (either may be chosen) being treated as in Article 59, give three data, viz., the total force along x = -F.; total force along y = F, and the moment of the couple acting round z = M; and the bars which are cut by the plane of section must exert resistances capable of balancing those two forces and that couple. If not more than three bars are cut by the plane of section, there are not more than three unknown quantities, and three relations between them and given quantities, so that the problem is determinate; if more than three bars are cut by the plane of section, the problem is or may be indeterminate. The formula to which this reasoning leads are as follows:-Let a be positive in a direction from the plane of section towards the part of the structure which is considered in determining F, F,, and M; lety lie to the right of + when looking from z; let angles measured from Ox towards+y, that is, towards the right, be positive; and let the lines of resistance of the three bars cut by the plane of section make the angles i, ią, is, with x. Let m, n, ng, be the perpendicular distances of those three lines of resistance from O, distances towards the positive right of O a being considered as { negative} left Let R, R, R, be the resistances, or total stresses, along the three bars, pulls being positive, and thrusts negative. Then we have the following three equations : F= R1 cos i + R2 cos i + R3 cos i;; F, R, sin + R2 sin i + R, sin is; = - M = R1 n1 + Rq ng + Rg Ng; .(1.) from which the three quantities sought, R1, R2, R3, can be found. Speaking with reference to the given plane of section, F. may be called the normal stress, F, the shearing stress, and M the moment of flexure or bending stress; for it tends to bend the frame at the section under consideration. y CASE 2. When the bars of the frame, and the forces applied to them, act in any direction, the forces applied to one of the two divisions of the frame are to be reduced to rectangular components; and the three resultant forces along these rectangular axes, F., F,, F, and the three resultant couples round these three axes, M., M, M,, are to be found as in Article 60. Those forces and couples must be equal and opposite to the corresponding forces and couples arising from the stresses along the bars cut by the section; and thus are obtained six equations between those stresses and known quantities; so that if the section cuts not more than six bars, the problem is determinate; if more, it is or may be indeterminate. The equations are obtained as follows:-Let R denote the stress along any one of the bars, pull being positive and thrust negative. Let 2, 3, 7, be the inclinations of the line of resistance of that bar to the axes of x, y, z. Let n be its perpendicular distance from O. Conceive a plane to pass through O and through the line of resistance of the bar, and a normal to be drawn to that plane in such a direction, that looking from the end of that normal towards O, the bar is seen to lie to the right of O, and let a,,, be the angles of inclination of that normal to the three axes. Let & denote the summation of six corresponding quantities for the six bars. Then the six equations are, F,= · R cosa; F, · R cos ß; F, = − M2 = 2 · R n cos λ ; My Σ 2. R cos; = R n cos μ; = R n cos"; (2). from which the six stresses sought can be computed by elimination. The plane of y z being as before, that of the section, F, is the total direct stress on it; F, and F, are the total shearing stresses; M, and M, are bending couples, and M, a twisting couple. REMARKS.-Every problem respecting the equilibrium of frames can be solved by the method of sections explained in this which Article, can also be solved by the method of polygons explained in the previous Articles; and the choice between the two methods is a question of convenience and simplicity in each particular case. The following is one of the simplest examples of the solution of a problem in both ways. Fig. 81 represents a truss of a form very common in carpentry (already referred to in Article 156), and consisting of three struts, A C, CC, C A, a tiebeam A A, and two suspension-rods, CB, C B, which serve to suspend part of the weight of the tie-beam from the joints CC, and also to stiffen the truss in the manner mentioned in Article 156. Let i denote the equal and opposite inclinations of the rafters A C, CA, to the horizontal tie-beam A A; and leaving out of consideration the portions of the load directly supported at A A, let P, P, denote equal vertical loads applied at CC, and P, -P, equal upward vertical supporting forces applied at A A, by the resistance of the props. Let H denote the pull on the tiebeam, R the thrust on each of the sloping rafters, and T the thrust on the horizontal strut C C. Proceeding by the method of polygons, as in Article 153, we find at once, (Thrusts being considered as negative.) To solve the same question by the method of sections, suppose a vertical section to be made by a plane traversing the centre of the right hand joint C; take that centre for the origin of co-ordinates; let a be positive towards the right, and y positive downwards; let x, y, be the co-ordinates of the centre of resistance at the right hand point of support A. When the plane of section traverses the centre of resistance of a joint, we are at liberty to suppose either of the two bars which meet at that joint on opposite sides of the plane of section to be cut by it at an insensible distance from the joint. First, consider the plane of section as cutting CA. The forces and couple acting on the part of the frame to the right of the section are F1 = 0 ; F, — — P F. = Then, observing that for the strut A C, n = 0, and that for the tie A A, n = y1, we have, by the equations 1 of this Article |