Solution Manual Steel Structures Design And Behavior -

[ P_{n, yielding} = F_y \cdot A_g = 36 \cdot 3.75 = 135 \text{ kips} ]

[ R_n = 0.6 F_u A_{nv} + U_{bs} F_u A_{nt} \quad \text{with } U_{bs}=1.0 \text{ (uniform tension)} ] [ R_n = 0.6 \times 58 \times 5.0 + 1.0 \times 58 \times 0.5 = 174 + 29 = 203 \text{ kips} ] But limited by ( 0.6 F_y A_{gv} + U_{bs} F_u A_{nt} = 0.6 \times 36 \times 7.5 + 29 = 162 + 29 = 191 \text{ kips} ) solution manual steel structures design and behavior

Yielding: LRFD 121.5 kips, ASD 80.8 kips Fracture: LRFD 97.1 kips, ASD 64.8 kips → [ P_{n, yielding} = F_y \cdot A_g = 36 \cdot 3

Given edge distance = assume 1.5 in (standard), spacing = 3 in, hole diameter = 1 in, thickness = 0.5 in. spacing = 3 in

Block shear rupture strength (AISC Eq J4-5):