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.

Example 1: W14x90 Column (Case 1)

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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Example 2: 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.

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Example 3: 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.

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Example 4: W12x65 Column

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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Example 5: W10x49 Bi-axial Bending

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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Example 6: Axial Tension and Bending Interaction

A structural member is subjected to a factored axial tensile load Pu=120 kipsP_u = 120 \text{ kips} and a factored bending moment Mux=80 kip-ftM_{ux} = 80 \text{ kip-ft}. The member is a W10x30.

The available axial tensile strength is ϕtPn=260 kips\phi_t P_n = 260 \text{ kips} and the available flexural strength is ϕbMnx=137 kip-ft\phi_b M_{nx} = 137 \text{ kip-ft}. There is no weak-axis bending (Muy=0M_{uy} = 0). Evaluate the adequacy of the member using the AISC interaction equations.

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Example 7: Determining Governing Interaction Case

An engineer is checking a W12x50 column in a moment frame. The column carries a high bending moment due to lateral wind loads, but relatively low axial gravity load because it is on the top story.

The factored axial compression is Pu=45 kipsP_u = 45 \text{ kips}, and the strong-axis bending moment is Mux=150 kip-ftM_{ux} = 150 \text{ kip-ft}. The section capacities are ϕcPn=500 kips\phi_c P_n = 500 \text{ kips} and ϕbMnx=265 kip-ft\phi_b M_{nx} = 265 \text{ kip-ft}.

Determine which interaction equation governs and calculate the interaction value.

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Example 8: $B_1$ Moment Amplifier Calculation

A W14x99 column in a non-sway (braced) frame is subjected to a factored axial compressive load of Pu=600 kipsP_u = 600 \text{ kips}. A first-order elastic analysis indicates the column is bent in single curvature with equal end moments of M1=M2=180 kip-ftM_1 = M_2 = 180 \text{ kip-ft} due to gravity loads. There are no transverse loads between supports.

The column length is L=14 ftL = 14 \text{ ft}, with effective length factor K1=1.0K_1 = 1.0. The strong axis moment of inertia is Ix=1110 in4I_x = 1110 \text{ in}^4 and the modulus of elasticity is E=29000 ksiE = 29000 \text{ ksi}.

Determine the amplified moment MuxM_{ux} accounting for PδP-\delta effects.

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