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000 X 7 3 3

pounds, stress, fig, truss, loads, determine, left and joint

(3,000 X 7) + (3,000 X 17.5) + (1,500X 28)—(6,000 X 28) or F,"— 7 = — 7,500 The minus sign means that is a push, hence the member 24 is under compression of 7,500 pounds, a result agreeing with that previously found.

2. It is required to find the stress in the member gh of the truss represented in Fig. 25, due to the loads shown.

If we pass a section cutting bg, gh and he, and consider the left part, we get Fig. 40, the forces on that part being the 2,000 pound load, the left reaction, and the forces F2 and F3 exerted by the right part on the left. To compute it is simplest to use the condition that the algebraic sum of all the vertical components equals zero. Thus, assuming that is a pull, and since its angle with the vertical is 30', F, cos 30°— 2,000 + 3,000 = 0; or, , = 2,000 — 3,000 — 1,154.

cos 30° — 0.866 The minus sign means that F2 is a push, .hence the member is under a compression of 1,154 pounds.

3. It is required to determine the stress in the member bg of the truss represented in Fig. 25, due to the loads shown.

If we pass a section cutting bg as in the preceding illustra tion, and consider the left part, we get Fig. 40. To compute is simplest to write the moment equa tion for all the forces using joint 5 as origin. From a large scale drawing, we measure the arm of to be 13.86 feet hence, assuming to be a pull, F, x 13.86 — 2,000 X 8 + 3,000 X 16=0; or, 2,000 X 8 — 3,000 X 16 — 32,000 13.86 — 13.86 = — 2,427.

The minus sign means that is a push; hence the member is under a compression of 2,308 pounds.

The section might have been passed so as to cut members bg, fg, and fe, giving Fig. 41 as the left part, and the desired force might be obtained from the system of forces acting on that part (3,000, 2,000, F2, and To compute we take moments about the intersection of F2 and Fa, thus This is the same equation as was obtained in the first solution, and hence leads to the same result.

4. It is required to determine the stress in the member 12 of the truss represented in Fig. 26, due to the loads shown.

Passing a section cutting members n and 14, and consider ing the left part, we get Fig. 42. To determine we may write a moment equation preferably with origin at joint 4, thus: x 4.47 + 4,000 X 10 = 0*; – or, =4,000 X 10 4.47 —8,948 pounds,the minus sign meaning that the stress is compressive.

might be determined also by writing the algebraic sum of the vertical components of all the forces on the left part equal to zero, thus: 1. Determine by the method of sections the stresses in mem bers 23, 25, and 45 of the truss represented in Fig. 26, due to the

loads shown.

( Stress in-23 = – 5,600 pounds, Ans. Stress in 25 = – 3,350 pounds, ( Stress in 45 = + 8,000 pounds.

2. Determine the stresses in the members 12, 15, 34, and 56 of the truss represented in Fig. 27, due to the loads shown.

in 12 11,170 pounds, Ans. Stress in 15 = ± 10,000 pounds, Stressin 34 = — 8,940 pounds, Stress in 56 = + 6,000 pounds.

Complete Analysis of a Fink Truss. As a final illustration of analysis, we shall determine the stresses in the members of the truss represented in Fig. 43, due to permanent, snow, and wind loads. This is a very coin mon type of truss, and is usually called a " Fink " or "French" truss. The trusses are assumed to be 15 feet apart; and the roof covering, including purlins, such that it weighs 12 pounds per square foot.

The length from one end to the peak of the truss equals hence the area of the roofing sustained by one truss equals (33.54 X 15) 2 = 1,006.2 square feet, and the weight of that portion of the roof equals 1,006.2 X 12 = 12,074 pounds.

the end loads equal of the total, or 950 pounds, and the other apex loads equal 4 of the total, or 1,900 pounds.

Dead Load Stress. To determine the dead load stresses, construct a stress diagram. Evidently each reaction equals one-half the total load, that is 7,600 pounds; therefore ABCDEFGHIJKA (Fig. 44b) is a polygon for all the loads and reactions. First, we draw the polygon for joint 1; it is KABLK, BL and LK repre senting the stress in bl and lk (see record Page 72 for values). Next draw the polygon for joint 2; it is LBCML, CM and ML representing the stresses in cm and ml. Next draw the polygon for joint 3; it is KLMNK, MN and NK representing the stresses in mn, and nk.

At each of the next joints (4 and 5), there are three unknown forces, and the polygon for neither joint can be drawn. We might draw the polygons for the joints on the right side corresponding to 1, 2, and 3, but no more until the stress in one of certain mem bers is first determined otherwise. If, for instance, we determine by other methods the stress in rk, then we may construct the poly gon for joint 4; then for 5, etc., without further difficulty.

To determine the stress in rk, we pass a section cutting rk, qr, and eq, and consider the left part (see Fig. 44c). The arms of the loads with respect to joint 8 are 7.5, 15, 22.5, and 30 feet; and hence, assuming to be a pull, x 15 –1,900 X 7.5 –1,900 X 15 –1,900 X 22.5 – 950 X30 + 7,600 X 30 = 0; or,