Solution Manual Steel Structures Design And Behavior Site

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 = 0.75 ) → ( 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} )

This manual provides detailed step-by-step solutions for end-of-chapter problems. Solutions follow the AISC Specification for Structural Steel Buildings (ANSI/AISC 360) – current edition. References to provisions (e.g., Section D2, Table D3.1) refer to the AISC Specification. Chapter 2: Tension Members Problem 2.17 (Sample Problem) solution manual steel structures design and behavior

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).

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). Assume failure path: tension on net area across

[ A_n = A_g - \sum (d_h \cdot t) + \sum \left( \frac{s^2}{4g} \cdot t \right) ]

[ P_{n, yielding} = F_y \cdot A_g = 36 \cdot 3.75 = 135 \text{ kips} ] References to provisions (e

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).

Yielding: LRFD 121.5 kips, ASD 80.8 kips Fracture: LRFD 97.1 kips, ASD 64.8 kips →

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.

So ( R_n = 191 \text{ kips} ) (lower governs). This is much higher than tensile fracture or yielding – thus block shear does not control.