Question

501 50 1 150b 3 in. 12 in. I)

Calculate and plot the shear force V and bending moment M everywhere on the beam; Calculate the maximum normal stress and maximum shear stress experienced by the beam anywhere along its length; What would the tensile yield strength o^y need to be and what would the shear yield strength t^y need to be in order for the beam to have a Safety Factor of 3 against failure in both tension and shear

Answer 1

Now sigma F_y = 0 R_A + R_B = 750 + 750+(150*12) R_A + R_B = 3300 1b (1) => sigma M_A = 0 R_B * 12 = (750*4) + (750*8)+(150*12*6) R_B = 16501b R_A = 1650 1b To calculate bending moment every where (BM)_A = 0 (BM)_B = 5400 1b ft (BM)_D = 5400 1b ft (BM)_B = 0 (BM)_MAX = 5700 1b ft (SF)_A = 1650 1b (SF)_C = 1050 1b (SF)_C = 300 1b (SF)_D = -300 1b (SF)_D = -1050 1b (SF)_B = -1650 1b (SF)_A = 0 # Shear force diagram: # Bending moment diagram => maximum shear stress Tow _max = 3/2 * (SF)_max/b*h = 3/2 * 1650/ 12*3 Tow_max = 68.75 Psi \r\n Maximum bending stress (sigma_b)_max = BM_ max * Y/I = 5700*12*(12/2)/12*3^3/12 (sigma_ b__max = 15200 Psi