Question

2) The beam shown is supported by steel member BC as shown. Due to the forked ends on the member, consider the supports at B and C to act as pins for x-x axis buckling and as fixed supports for y-y axis buckling. Member BC is made of A992 steel with a modulus of elasticity of 29,000 ksi and a yield strength of 50 ksi. a) Calculate the maximum allowable load, P, on the system (in kips) considering buckling and yielding. [Ans. To check: Pallow bucking - 23.85 kip] b) Suppose instead the load in part (a) was P 180 kips. Design the cross-section by calculating the required moment of inertia of member BC to prevent yielding and buckling. Be sure to check weak and strong axes of bending. A (in2) 5.33 8.33 12 b (in) h (in) 4 1-2/3 2 3 ft 3 in. 1 in

Answer 1

Sigma m_A = 0 F_BC (3/5) (4) - p (8) = 0 F_BC = 3.33 p PCr = pi^2 E I_4/L^2 = pi^2 (2900) (3 times 1^3/12)/1 times 5 times 12^2 Pcr = pi^2 EI_4/(K.L)^2 3.33 p = pi^2 (29000) (3 times 1^3/12)/(0.5 (5 times 12))^2 3.33p = 79.51 p = 23.87 kip \r\n \F_BC = 3.33 (180) = 600 kip about 4 - 4 axis 600 = pi^2 E I_y/(K.L)^2 600 = pi (29000) I_y/(0.5 (5 times 12))^2 I_y = 1.8867 in^4 about x - x axis 600 pi^2 (29000) I x/(1.(5 times 12))^2 I_x = 7.546 in y therefore weak axis is I_y therefore I_y is critical use h = sin, b = 1 - 2/3 in I_4 = (5 (1.667)^/12 = 1.93 in > 1.8867 in therefore use h = sin, b = 1 - 2/3 in A = 8.33 in^2