Question

3%in 314 in 3,4 in Calculate the LRFD Design Strength of the section below for a 10' span length. Hint-check the general requirement and calculate the design strengths for both Yielding on Gross Section and Rupture on Net Section and select the controlling. 8. A WT6 x 22.5, A992 steel, attached through the flange with six 7s-in bolts wT6 × 22.5 5%in l J78 in bolts 2h in 3h in 3'% in

Answer 1

Given that wT = 6 times 22.5 d = 6.03 in b = 8.05 in t_u = .326 in b_f = 0.575 A_g = 6.56/in^2 r_y = 1.95 in LRFD Gross yielding design strength = phi in = phi A_g f_y = 0.90 times 6.56 times 50 gross yielding strength = 295.2n kips Net section strength = phi P_n = phi A_c f_u = 0.75 times A_c times 65 ksi for allotted connection A_e = u. A_n A_n = A_s - (no of holes) times (diameter of hole) = 6.56 - 6 times (7/6 + 1/8] times 0.575 = 3.11 in^2 \r\n u = 1 - x/L for W T 6 times 22.5, x = 1.13 in span length, L = 10� u = 1 - 1.13/10 times 12 = 0.99 therefore A_e = u.A_n = 0.99 (3.11) A_e = 3.08 in^2