Design Procedure

Procedure

  1. Calculate factored loads (Pu,Mux,MuyP_u, M_{ux}, M_{uy}). Include second-order effects via direct analysis or moment amplification (B1,B2B_1, B_2).
  2. Select a trial section from the beam-column tables.
  3. Determine available strengths:
    • Axial: ϕPn\phi P_n (considering the largest KL/rKL/r).
    • Flexural: ϕMn\phi M_n (considering LbL_b and CbC_b).
  4. Calculate the ratio Pu/PcP_u/P_c.
  5. Select and evaluate the appropriate AISC interaction equation (Case 1 or Case 2).
  6. Verify that the interaction value is 1.0\le 1.0. If it is greater, select a larger section and repeat.

Solved Problems

A W14x90 column in a braced frame supports a factored axial load Pu=400 kipsP_u = 400 \text{ kips} and a factored strong-axis moment Mux=250 kip-ftM_{ux} = 250 \text{ kip-ft} (including second-order effects). The available axial strength ϕPn=1000 kips\phi P_n = 1000 \text{ kips} and available flexural strength ϕMn=574 kip-ft\phi M_n = 574 \text{ kip-ft}. There is no weak-axis bending (Muy=0M_{uy} = 0).
Determine if the member is adequate based on the AISC interaction equations.

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Case Studies

Case Study 1: Wind Loading on Exterior Columns

A structural engineer is designing the exterior columns of a single-story commercial building. The roof structure applies a significant gravity load.
Scenario: During a hurricane, high wind pressures act directly on the exterior walls, which transfer the load horizontally to the columns.
Solution: The columns can no longer be designed purely as compression members. They must be designed as beam-columns because they are subjected to a high axial load (PuP_u) from the roof and a high lateral flexural load (MuxM_{ux}) from the wind pressure acting along their length.
The interaction equation ensures that the combination of these stresses does not exceed the capacity of the steel. In many cases, an exterior column that passes a gravity-only check will fail the interaction check under wind loading, requiring a section with a much larger moment of inertia (IxI_x) or a thicker web to resist the combined stresses.

Case Study 2: The P-Delta Effect in Soft-Story Buildings

A multi-story steel frame building with a tall, open first floor (a "soft story") undergoes lateral displacement (Δ\Delta) during a seismic event.
Scenario: The building sways laterally. The massive weight of the upper floors is now displaced horizontally from the foundation.
Solution: This displacement creates a Second-Order Effect (PP-Δ\Delta). The gravity load (PP) acting on the laterally displaced column (Δ\Delta) creates an additional secondary bending moment (Msecondary=P×ΔM_{secondary} = P \times \Delta).
This secondary moment adds to the primary moments caused by the earthquake, further increasing the total moment on the beam-column. To account for this, the AISC Direct Analysis Method requires engineers to apply an amplification factor (B1B_1 and B2B_2) to the primary moments when evaluating the interaction equations, ensuring the columns are designed for the true, amplified forces they will experience.

Additional Solved Problems

A W12x65 column (Fy=50 ksiF_y = 50 \text{ ksi}) in a braced frame supports a factored axial load Pu=150 kipsP_u = 150 \text{ kips} and a factored strong-axis moment Mux=120 kip-ftM_{ux} = 120 \text{ kip-ft}. The available axial strength ϕcPn=600 kips\phi_c P_n = 600 \text{ kips} and available flexural strength ϕbMnx=356 kip-ft\phi_b M_{nx} = 356 \text{ kip-ft}. There is no weak-axis bending (Muy=0M_{uy} = 0).
Determine if the member is adequate based on the AISC interaction equations.

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A W10x49 beam-column is subjected to bi-axial bending along with an axial load. The factored loads are Pu=80 kipsP_u = 80 \text{ kips}, Mux=100 kip-ftM_{ux} = 100 \text{ kip-ft}, and Muy=40 kip-ftM_{uy} = 40 \text{ kip-ft}. The capacities are ϕcPn=450 kips\phi_c P_n = 450 \text{ kips}, ϕbMnx=227 kip-ft\phi_b M_{nx} = 227 \text{ kip-ft}, and ϕbMny=108 kip-ft\phi_b M_{ny} = 108 \text{ kip-ft}. Evaluate the adequacy of the member.

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