The method of "equable distribution of the load,” (B) section 89, gives the following results: W 8 W is sustained at A, K, E, F, L and B, and is sustained at C. 4 W W Hence the value of R in (3) is or according as we 6 8' adopt the method (A), or the method (B). The strength and dimensions of the strut DE or DF, are found exactly in the same way as in the preceding section. TO FIND THE TENSION OF THE KING-POST CD. 100. The king-post CD supports the vertical components of the compressions on the struts ED and FD, a part of the weight of the tie-beam and its load, together with its own weight, and the weights of the two struts ED and FD. Let us first find what part of the weight of the tie-beam the king-post CD sustains. Let W' be the weight of the tie-beam and that of the floor which it supports; then the pressure at G arising from the weight W' may be supposed to consist of one-half the load between A and G, and one-half the load between G and D, that is, one-half the 1 W' W' For the same reason load between A and D, or of = 2 2 4 the pressure at Harising from the load W" is : Again the pressure at D consists of one-half the load between G and D, and one-half the load between H and D, that 1 W' W is, of of = 2 3 6 And as the pressure at A or B is evidently equal to one-half W' W' 3 6 the weight W' of the tie-beam, and that of the floor ing which it supports, are distributed in the following manner : W W 4 at G and H; and at D, A and B. > Hence, if W" be the weight of the king-post and the struts DE, DF, W' the weight of the tie-beam and the load which it supports, and R the vertical pressure at E or F, then the tension of the king-post CD will be found from the expression The strength and dimensions of the king-post CD are found as in Art. 100. Obs. It seems unnecessary to enter into further details with respect to this truss. The methods of finding the straining forces on the different parts of the truss, and the strength and dimensions of these parts, are exactly similar to the methods employed in the preceding section. We must keep in mind that the transverse strain on the tie-beam, in this case, is between the points A, G; G, D; D, H; and H, B Examples. 1. In Fig. 42, the weight of the roof supported by the truss ACB is 24 cwt., and each of the angles at A and B is 30°; find the compression on the strut KG or LH, on the hypothesis of an "equable distribution of the load." Ans. 3 cwt. strut ED of last Ans. √7 cwt. or 296 lbs. 2. Determine the compression on the example. 3. Find the tension of the king-post in Example 1, when the suspension rods EG, FH, and the struts GK, HL, are removed, the weight of the tie-beam being cwt., and that of the kingAns. 779 lbs. post and struts ED, DF, 140 lbs. 4. In the last example find the compression on the upper portion EC and FC of the principal rafters. Ans. 1,675 lbs. 5. In Example 3, find the compression on the lower portion EA and FB of the principal rafters. Ans. 1,899 lbs. 6. Find the horizontal thrust on the tie-beam in Example 3. Ans. 1,644.5 lbs. 7. In Example 1, find the compression on the strut KG, on the hypothesis (4) Art. 88, of the distribution of the load. Ans. 4 cwt. 8. Determine the compression on the strut ED in Example 2, on the hypothesis (4) of the distribution of the load. Section 5. Ans. √7 cwt. THE BROKEN ROOF. 101. When a roof consists of a number of beams AB, BC, &c., as in Fig. 43, the first or highest beam AB, resting upon a second beam BC, the second upon D Wa W2 B a third CD, and so on; this kind of roof is called a Broken Roof. The equilibrium of a roof of this sort depends upon the position of the several beams with respect to one another. Let Fig. 43 represent the half of a broken roof, the other half of which we may suppose to be represented by a horizontal force P. If the roof consists of one half only, then P will represent the reaction of the wall. Denote the angles which the beams AB, BC, &c., reckoned from the highest one AB, make with the horizon by a1, a2, as, &c., their weights by W1, W2, Wa, &c., and their lengths by la, la, la, &c. Fig. 43. Let B (Fig. 44) be the lower extremity of the highest beam. R?A P Denote by R, and R, the horizontal and vertical components of the pressure produced at B by the beam AB. Then resolving horizontally and vertically, and taking moments about B, we have for the equilibrium of the beam AB the equations, 1 B Fig. 44. If we suppose the forces W1 and W1 cot a1 = P, to act vertically and horizontally at B, then we may neglect the first beam, and proceed in like manner with the extremity C of the second beam BC (Fig. 45). Hence, for the conditions of equilibrium of the second beam, we have the equations P = R12 R2 B Pl2 sin a2 = W1 l2 cos a2 + W2 1⁄2 l1⁄2 cos a2. Wherefore 1 2 1 2 If again we suppose the forces W1 + W2 and P to act vertically and horizontally at C, we get for the equilibrium of the third beam the equations Hence, proceeding in this way from beam to beam, we readily see that, for the equilibrium of the roof in question, the horizontal thrusts at the extremities of the different beams are all equal to one another, while the difference of the vertical pressures at any two points is equal to the sum of the weights of the beams intercepted between these points. Wherefore the constant horizontal thrust is 102. When the several beams, as in the preceding article, are of equal length, and consequently of equal weight, then Similarly, W1 cot a1 = (W1+ W1) cot a2, 1 or cot a1 = 3 cot a2. cot a 5 cot. a,; and so on. Hence the cotangents of the angles of inclination of the beams of a broken roof to the horizon are as the odd numbers 3, 5, 7, &c., the beams being of equal length and of equal weight. GEOMETRICAL CONSTRUCTION. 103. When the beams of a broken roof are all equal, and the position and magnitude of the highest beam are given, the roof can be determined by geometrical construction. Let AB be the highest beam of the half of a a broken roof, and a1 the inclination of AB to the horizon. From B draw BM perpendicular to the vertical line MAab, and produce MA till Ma A = 3 MA, Mb = 5 MA; and so on. |