Worked Example: Steel Tension Members Design

This worked example demonstrates how to determine the design tensile strength of a structural steel member according to AISC LRFD principles, checking both gross section yielding and net section rupture.

1. Problem Statement

A structural steel plate is subjected to a factored tensile load $P_u = 800,000 \text{ N}$ ( $800 \text{ kN}$ ). The plate has a gross area $A_g = 4000 \text{ mm}^2$ and an effective net area $A_e = 2800 \text{ mm}^2$ (after accounting for bolt holes and shear lag). The steel is Grade A36, with yield stress $F_y = 250 \text{ MPa}$ ( $\text{N/mm}^2$ ) and ultimate stress $F_u = 400 \text{ MPa}$ ( $\text{N/mm}^2$ ). Determine the design tensile strength and if the member is safe according to LRFD principles.

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2. Given Data

Yield Stress (Fy)

N/mm²

Ultimate Stress (Fu)

N/mm²

Gross Area (Ag)

mm²

Effective Net Area (Ae)

mm²

Yielding Factor (phi_y)

Rupture Factor (phi_t)

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3. Formula and Calculation

Check Tensile Yielding (Gross Section): $\phi_y P_{ny} = \phi_y \times F_y \times A_g$ For initial values: $0.90 \times 250 \times 4000 = 900,000 \text{ N}$

Check Tensile Rupture (Net Section): $\phi_t P_{nt} = \phi_t \times F_u \times A_e$ For initial values: $0.75 \times 400 \times 2800 = 840,000 \text{ N}$

Determine controlling design strength ( $\phi P_n$ ):

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Conclusion: The controlling design strength is the smaller value: $840,000 \text{ N}$ ( $840 \text{ kN}$ ), governed by net section rupture. Since the design strength ( $840,000 \text{ N}$ ) is greater than the required strength $P_u$ ( $800,000 \text{ N}$ ), the member is safe.

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