Design And Behavior: Solution Manual Steel Structures

A single-angle tension member, L4×4×½ (A36 steel), is connected to a gusset plate with 7/8-inch diameter bolts as shown in Figure P2.17 (three bolts in one leg, staggered: 3" on center along length, 2" gage). Compute the design tensile strength (LRFD) and allowable tensile strength (ASD).

Block shear rupture strength (AISC Eq J4-5): solution manual steel structures design and behavior

Gross shear length = ( 1.5 + 3 + 3 = 7.5 \text{ in} ) (from edge to last bolt). Net shear length = ( 7.5 - 2.5 \times d_h = 7.5 - 2.5 = 5.0 \text{ in} ) (since 2.5 holes along shear path? Actually 2.5 holes for two lines? Need precise – typical simplified: net shear area = ( (7.5 - 2.5*(1.0))*0.5 = 2.5 \text{ in}^2 ) per plane, two planes = 5.0 in²). A single-angle tension member, L4×4×½ (A36 steel), is

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 ). Net shear length = ( 7

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

Assume failure path: tension on net area across the end row, shear on two net areas along both sides of bolt group.

Thickness ( t = 0.5 \text{ in} ). Two hole diameters in the failure path (assuming worst path goes through both holes in the same leg – check path 1-2-3).