So ( R_n = 191 \text{ kips} ) (lower governs). This is much higher than tensile fracture or yielding – thus block shear does not control.
Given edge distance = assume 1.5 in (standard), spacing = 3 in, hole diameter = 1 in, thickness = 0.5 in. solution manual steel structures design and behavior
LRFD: ( \phi_t = 0.75 ) → ( P_d = 0.75 \times 129.5 = 97.1 \text{ kips} ) ASD: ( \Omega_t = 2.00 ) → ( P_a = 129.5 / 2.00 = 64.8 \text{ kips} ) So ( R_n = 191 \text{ kips} ) (lower governs)
Assume failure path: tension on net area across the end row, shear on two net areas along both sides of bolt group. LRFD: ( \phi_t = 0
Tension net area across last bolt row = (gage distance – one hole) * t = ( (2.0 - 1.0)*0.5 = 0.5 \text{ in}^2 ) per plane? Two planes? For single angle, block shear occurs in the connected leg only.