Module 7: Steel Compression Members - Examples & Applications

Local vs. Global Buckling

Basic: Checking Section Slenderness (Flange)

A W14x90 column made of A992 steel (Fy=345 MPaF_y = 345 \text{ MPa}, E=200,000 MPaE = 200,000 \text{ MPa}) is proposed for a design. Determine if the flange is considered slender or non-slender under pure compression.

Given Section Properties for W14x90:

  • Flange width (bfb_f): 368 mm368 \text{ mm}
  • Flange thickness (tft_f): 18 mm18 \text{ mm}

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Conceptual: Residual Stresses and Column Strength

An engineer tests two identical steel columns. Column A was perfectly annealed (stress-free) after manufacturing. Column B is a standard hot-rolled shape containing typical residual cooling stresses. They are both tested in axial compression. Will their load-deflection curves be identical up to the point of failure? Explain why.

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Axially Loaded Column Design

Intermediate: Design Compressive Strength (Inelastic Buckling)

Determine the LRFD design compressive strength (ϕPn\phi P_n) of a pinned-pinned W-shape column.

Given: Ag=10,000 mm2A_g = 10,000 \text{ mm}^2, Fy=345 MPaF_y = 345 \text{ MPa}, E=200,000 MPaE = 200,000 \text{ MPa}. The critical slenderness ratio is KL/r=50KL/r = 50. Assume the section is non-slender.

LRFD Factor: ϕc=0.90\phi_c = 0.90.

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Advanced: Design Compressive Strength (Elastic Buckling)

Determine the LRFD design compressive strength (ϕPn\phi P_n) of a slender, fixed-free (cantilever) pipe column supporting an awning.

Given: Ag=3,500 mm2A_g = 3,500 \text{ mm}^2, Fy=250 MPaF_y = 250 \text{ MPa}, E=200,000 MPaE = 200,000 \text{ MPa}. The actual unbraced length is L=4.5 mL = 4.5 \text{ m}. The radius of gyration is r=50 mmr = 50 \text{ mm}. Assume the pipe wall is non-slender.

LRFD Factor: ϕc=0.90\phi_c = 0.90.

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