Next, conceive the section to cut CC at an insensible distance to the left of C. Then the equal and opposite applied forces + P at C, and P at A, have to be taken into account; so that F= 0; F, = 0; M = — Pж1; from the first of which equations we obtain In the example just given, the method of sections is tedious and complex as compared with the method of polygons, and is introduced for the sake of illustration only; but in the problems which are to follow, the reverse is the case, the solution by the method of sections being by far the more simple. 162. A Half-Lattice Girder, sometimes called a "Warren Girder," is represented in fig. 82. It consists essentially of a horizontal upper bar, a horizontal lower bar, and a series of diagonal bars sloping alternately in opposite direc tions, and dividing the space between the upper and lower bars into a series of triangles. In the example to be considered, the girder is supposed DE Fig. 82. to be supported by the vertical resistance of piers at its ends A and B, and loaded with weights acting at or through the joints at the angles of the several triangles. This girder might be treated as a case of secondary trussing, by considering the upper and lower and endmost diagonal bars as forming a polygonal truss like fig. 81, but inverted, supporting a smaller erect truss of the same kind, which supports a still smaller inverted truss, which supports a still smaller erect truss, and so on to the smallest truss, which is the middle triangle. But it is more simple to proceed by the method of sections, which must be applied successively to each division of the girder. The load at each joint being known, the two supporting forces at A and B, are to be determined by the principles of the equilibrium of parallel forces in one plane (Articles 43, 44). Let PA, PB, denote those supporting forces, upward forces being treated as positive, and downward as negative; and let - P denote the load at any joint, which may be a constant or a varying quantity for different joints. Suppose now that it is required to find the stress along any one of the diagonals, such as C E, along the top bar immediately to the right of C, and along the bottom bar immediately to the left of E. Conceive the girder to be divided by a vertical plane of section CD, at an insensibly small distance to the right of C; take the intersection of this plane with the line of resistance of the top bar for the origin of co-ordinates, which sensibly coincides with C. Let x denote the distance of any one of the joints to the left of the plane of section, from that plane. Let x, be the distance of the point of support A to the left of the same plane. Let y be positive upwards; so that for the joints of the upper bar, y = = 0, and for those of the lower bar, y h, h denoting the vertical depth between the lines of resistance of the upper and lower bars. Let i be the inclination of the diagonal CE to the horizontal axis of x. In the present instance this is positive; but had CE sloped the other way, it would have been negative. Let the symbol ZAP denote the sum of the loads acting at the joints between the plane of section and the point of support A, the load at the joint C being included. Then for the total forces and couple acting on the division of the girder to the left of the plane of section, we have,-direct force, F, 0, because the applied forces are all vertical ;-shearing force, F, 2. P; a force positive or = = PA which is {negative or downward } according as the plane of section lies { { nearer to } farther from the point of support A, than a plane which divides the load into two portions equal respectively to the supporting pressures;-bending couple M Px; which is upward, and right-handed with respect to the axis of z. Now let R, denote the stress along the upper bar at C, R2 that along the lower bar at D, and R, that along the diagonal CE; then the equations 1 of Article 161 become the following: R1 R2+ R, cos i 0; or R1 + R, cos i = R2...(a.) that is, the stress along the upper bar, and the horizontal component HALF-LATTICE GIRDER. 155 of the stress along the diagonal, are equal and opposite to the stress ang the lower bar; R, sin i = F, P-P;................ (b.) = that is, the vertical component of the stress along the diagonal, balances the shearing force; - = — R2 y = R2 h = M P1 x P x ;.........(c.) that is, the couple formed by the equal and opposite horizontal stresses of equation (a), acting at the ends of the arm h, balances the bending couple. Finally, from the equations (a), (b), (c), are deduced the following values of the stresses: = — — (P. x, — ô · P x) — cotan i (P ̧ — z^ · P). A Another, and sometimes a more convenient form, can be found for the second and third of those expressions. Let s denote the length of the diagonal CE, and x, the horizontal distance of its lower end E from the point of support A; then 39 which substitutions having been made, give كم — — — { P ̧ o, — xê · P x + (x,' — 2) (P▲ — 24 · P) } } (3.) in which a' is taken to denote the horizontal distance of any joint to the left of a vertical plane traversing E. The last expression for R is exactly what would have been obtained by supposing the plane of section to traverse E instead of C. 1 with Any given diagonal is {atie according as it slopes against} strut the direction of the shearing force F, acting on a plane of section traversing it. 163. Half-Lattice Girder—Uniform Load.~CASE 1. Every joint loaded. When the joints of a half-lattice girder are at equal distances apart horizontally, and loaded with equal weights, the equations take the following form: Let N denote the even number of divisions into which vertical lines drawn through the joints divide the total length or span between the points of support. Let be the length of one of these divisions, so that N is the total span. The total number of loaded joints is N-1; this must be an odd number, and there must be a middle joint dividing the girder into two halves, symmetrical to each other in every respect, figure, load, support, and stress, so that it is sufficient to consider one half only; let the left hand half be chosen. Let the middle joint be denoted by O, and the other joints by numbers in the order of their distances from the middle joint, so that the joint numbered n shall be at the distance nl from O. The even numbers denote joints on the same horizontal bar with O; the odd numbers those on the other. The total load on the girder is -(N-1) P, of which one-half is supported on each pier; that is to say, The stress on the upper bar is everywhere a thrust;—that on the lower bar a pull. Diagonals which { towards the ends are { ties rise fall } from the middle struts. By these principles the kind of stress on each piece is determined; it remains only to compute the amount. Let n be the number of any joint; it is required to find the stress along the diagonal which runs from that joint towards the middle of the girder, and the stress along that part of either of the horizontal bars which is opposite the joint. Suppose a vertical section to be made at an insensible distance from the joint, intersecting the diagonal in question and the hori zontal bars. Between O and either pier there are N 1 loaded joints; be tween 0 and the plane of section in question, there are n-1 joints; hence between the plane of section and the pier there are So that it increases at an uniform rate from the middle towards the ends. The distance of the nth joint from the pier is x, = Hence the upward moment of the supporting force is The downward moment of the load at the joints between the plane of section and the pier is found from the consideration, that the leverage of the nearest portion of that load is nothing, and -n), so that the mean leverage is that of the farthest (1-7), so that the 1/N 2 1—n)1; which being multiplied by the load that is to say, it is proportional to the product of the segments into which the plane of section divides the length of the girder, and is N2 greatest at the middle, where it is Pl |