Question

Problem 4 Member A is a steel rod with circular cross-section. The modulus of elasticity is 29000 ksi. Member B is very stiff compared to Member A, so ignore its deformations. a. Set up the virtual force system for vertical displacements DEAD 20K LIVE 10 k at Node 2. Determine the minimum diameter of rod that can be used if the deflection due to live load cannot exceed 0.5 in Assume that rods are available in 1/8 in. increments, such as 1/4 in., 3/8 in., 1/2 in. and so on. The truss is to be constructed so that Member B is level after the dead load has been applied. This is to be accomplished by fabricating Member A slightly shorter than shown in the figure. How long should Member A be fabricated? For this calculation, use 0.5 in. diameter for Member A b. 3 c.

Answer 1

Given data The member is a steel rod with circular cross-section The would of elasticity is = 29000 ksi Member B is very stiff compared member A Given the sum below: 2V_0 = 0 30 = PH times sin theta P_A = 30/sin 36 degree 32' = 50 K(T) Lh = squareroot 2^2 + 6^2 = 10 Given the diagram tan theta = 6/3 theta = tan^-1(6/3) = 36 degree 32' \r\n S_2 = 0.5 inches = P_A K L_A/P_A times L_A H_A = (50 times 1.67 times (10 times 12)/29 times 1000 times 0.5) = (83.5 times(10 times 12)/29 times 1000 times 0.5) = (83.5 times 120/29 times 1000 times 0.5) = 10,020/29 times 100 times 0.5 = (10020/29000 times 0.5) = (10020/14.500) = 0.6896 in^2 pi/4 (dn)^2 = 8.6896 \r\n d_a = 0.937 inch TildeEqual 1 inch dA = 1 inch