[ 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} )
Check alternative staggered path through first hole in one leg then to hole in opposite leg? For L4×4, gage between legs (distance from back of one leg to center of holes in other leg) ≈ 2.5 in (AISC gage for angles). But given gage = 2.0 in, stagger term: ( s^2/(4g) = 3^2/(4 2) = 9/8 = 1.125 ). For one diagonal path: ( A_n = A_g - 2 (d_h t) + (1.125 t) ) = ( 3.75 - 1.0 + 0.5625 = 3.3125 \text{ in}^2 ) → larger than 2.75, so critical net area = 2.75 in².
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
[ U = 1 - \frac{1.13}{6} = 0.812 ]
[ A_n = A_g - \sum (d_h \cdot t) + \sum \left( \frac{s^2}{4g} \cdot t \right) ] [ R_n = 0
Block shear rupture strength (AISC Eq J4-5):
[ A_{gv} = 2 \times ( \text{shear length along bolt line}) \times t = 2 \times 7.5 \times 0.5 = 7.5 \text{ in}^2 ] [ A_{nv} = A_{gv} - 2 \times (2.5 \times d_h \times t) \quad \text{(2.5 holes per shear plane)} = 7.5 - 2 \times (2.5 \times 1.0 \times 0.5) = 7.5 - 2.5 = 5.0 \text{ in}^2 ] [ A_{nt} = ( \text{gage} - d_h) \times t = (2.0 - 1.0) \times 0.5 = 0.5 \text{ in}^2 ] For one diagonal path: ( A_n = A_g - 2 (d_h t) + (1
[ P_{n, yielding} = F_y \cdot A_g = 36 \cdot 3.75 = 135 \text{ kips} ]
