Question

Q3) Calculate the ultimate moment capacity of the irregular shape RC sections shown in Fig. 3, use f'-4000 psi and f,-60000 psi. 4 in 4 in 4n 7 in 10 in A.-4 #7 Fig. 3

Answer 1

Given, F_c' = 4000 Psi f_y = 60000 psi b = 4 in + 4 in + 4 in = 12 inch depth d_t = 17 inch Area of steel Ast = 4 times pi/4 (0.275)^2 Ast = pi times (0.275)^2 Ast = 0.237 in^2 \r\n Neutral axis of the section is given by ultimate compression (or) ultimate, 0.36F_c times b times x_u = 0.87 f_y Ast 0.36 times 4 times 12 times x_u = 0.87 times 60 times 0.237 x_u = 0.87 times 60 times 0.237/0.36 times 4 times 12 x_u = 0.715 inches Hence ultimate moment capacity of beam is given by M = (0.36F_c times b times x_u) times (d - 0.42 x_u) = (0.36 times 4 times 12 times 0.715) times (17 - 0.42 (0.715)) = 12.3552 times (17 - 0.3003) = = 12.3552 times 16.6999 = 206.04 therefore ultimate moment capacity of beam is = 206.04 kips.inch