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ABCDA and the arrowheads on CD and DA must point as shown.

CD (1.25 inches = 7,500 pounds) represents the force exerted on joint (2) by 24; since it acts toward the joint the force is a push and member 2 t is in compression. DA (0.43 inches = 2,580 pounds) represents the force exerted on the joint by member 23; since the force acts toward the joint it is a push and the member is in compression. Member 23 therefore exerts a push on joint (3) as show n in Fig. 19 (a).

At joint (3) there are four forces, 7,800 pounds, 2,580 pounds, and the forces exerted on the joint by members 34 and 36. To determine these, construct the polygon for the four forces. Thus, AB (1.3 inches long with arrowhead pointing to the left) represents the 7,600-pound force and BC (0.43 inches long with arrowheads pointing down) represents the 2,580-pound force. Next draw from A and C two lines parallel to the unknown forces and mark their intersection D; then the force polygon is ABCDA and the arrowhead on CD and DA must point upward and to the right respectively. CD (0.43 inches = 2,580 pounds) represents the force exerted on the joint by member 34; since the force acts away from the joint it is a pull and the member is in tension. DA (0.87 inches = 4,920' pounds) represents the force exerted upon the joint by the member 36; since the force acts away from the joint, it is a pull and the member is in tension.

We have now determined the amount and kind of stress in members 12, 13, 23, 24, 34 and 36. It is evident that the stress in each of the members on the right-hand side is the same as the stress in the corresponding one on the left-hand side; hence further analysis is unnecessary.

2. It is required to analyze the truss represented in Fig. 20 (a), the truss being supported at the ends and sustaining two loads, 1,800 and 600 pounds, as shown. (For simplicity we as sumed values of the load; the lower one might be a load due to a suspended body. We shall solve algebraically.) The right and left reactions equal 900 and 1,500 pounds as is shown in Example 1, Page 56. At joint (1) there are three

forces, namely, the reaction 1,500 pounds and the forces exerted by members 13 and 14, which we will denote by and F, respect ively.• The three forces are represented in Fig. 20 (b) as far as they are known. These three forces being in equilibrium, their horizontal and their vertical components balance. Since there are but two horizontal components and two vertical components it follows that (for balance of the components) must act downward and F2 toward the right. Hence member 13 pushes on the joint and is under compression while member 14 pulls on the joint and is under tension. From the figure it is plain that the horizontal component of F, = F, cos 53° 8' = 0.6 F,*, the horizontal component of F2 = the vertical component of F, = F, cos 36° 52' = 0.8 Fi, and the vertical component of the reaction = 1,500.

Hence 0.6 F, = and 0.8 F, = 1,500; 1,500 or, F, 0.8 = 1,875 pounds, and F2 = 0.6 X 1,875 = 1,125 pounds.

Since members 14 and 13 are in tension and compression respect ively, 14 pulls on joint (4) as shown in Fig. 20 (e) and 13 pushes on joint (3) as shown in Fig. 20 (cl).

The forces acting at joint (4) are the load 600 pounds, the pull 1,125 pounds, and the forces exerted by members 34 and 24; the last two we will call F3 and F4 respectively. The four forces being horizontal or vertical, it is plain without computation that for balance must be a pull of 1,125 pounds and F3 one of 600 pounds. Since members 11 and 43 pull on the joint they are both in tension.

Member 43, being in tension, pulls down on joint (3) as shown in Fig. 20 (d). The other forces acting on that joint are the load 1,800 pounds, the push 1,875 pounds, the pull 600 pounds, and the force exerted by member 32 which we will call F5. The only one of these forces having horizontal components are 1,875 and F5; hence in order that these two components may balance, F5 must act toward the left. is therefore a push and the member 32 is under compression.