Given edge distance = assume 1.5 in (standard), spacing = 3 in, hole diameter = 1 in, thickness = 0.5 in.
Tension member connected to gusset plate – check block shear along bolt group.
[ 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
Better to follow AISC manual example: For L4×4×½ connected with 3 bolts, block shear strength:
Assume failure path: tension on net area across the end row, shear on two net areas along both sides of bolt group. Given edge distance = assume 1
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.
Path 1: straight line through both holes (no stagger effect since in same leg, but stagger formula still applies if line zigzags – here, holes are in same leg, so stagger not applied unless crossing to other leg? For angles, net section often through holes in same leg, stagger effect negligible for two holes on same line. However, typical solution uses two holes: ( A_n = A_g - 2 \cdot (d_h \cdot t) ) = ( 3.75 - 2 \cdot (1.0 \cdot 0.5) = 3.75 - 1.0 = 2.75 \text{ in}^2 ). Tension net area across last bolt row =
Block shear rupture strength (AISC Eq J4-5):
For L4×4×½: ( \bar{x} = 1.13 \text{ in} ) (from AISC Manual). Length of connection ( L ) = distance between first and last bolt = 2 pitches = 6 in.
So ( R_n = 191 \text{ kips} ) (lower governs). This is much higher than tensile fracture or yielding – thus block shear does not control.