Solved Problems

Example 1: Basic Nominal Capacity

Determine the design flexural strength (ϕbMn\phi_b M_n) of a continuously braced W16x31 beam of A992 steel (Fy=50F_y = 50 ksi). The plastic section modulus ZxZ_x is 54.0 in354.0 \text{ in}^3.

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Case Studies

Case Study 1: Temporary Bracing during Construction

A 40-foot span steel beam is designed to support a composite concrete floor slab.

Scenario: During the concrete pouring phase, before the concrete cures and bonds to the shear studs, the steel beam must support the entire wet weight of the concrete by itself.

Solution: Once the concrete cures, it provides continuous lateral bracing to the top flange of the beam (Lb=0L_b = 0). However, during construction, the unbraced length (LbL_b) is the full 40-foot span.

Because 40 ft>Lr40 \text{ ft} > L_r, the bare steel beam would fail via elastic lateral-torsional buckling (LTB) under the wet concrete load. To solve this, the contractor must install temporary lateral bracing (such as bridging or strut members) at 10-foot intervals. This reduces LbL_b to 10 feet, keeping it within Zone 2 (inelastic LTB) and ensuring the beam has sufficient temporary strength until the concrete hardens.

Case Study 2: Retrofitting a Non-Compact Section

An existing older building is being evaluated for a change in use that will significantly increase floor loads.

Scenario: The existing floor beams are older rolled shapes where the flanges are classified as "non-compact" according to modern AISC specifications.

Solution: Because the flanges are non-compact, they are susceptible to local buckling before the entire cross-section can reach its full plastic moment capacity (MpM_p). The nominal moment capacity (MnM_n) is governed by Flange Local Buckling (FLB) rather than yielding.

To increase the flexural capacity without replacing the beams, the engineers design a retrofit. They specify welding continuous steel cover plates to the top and bottom flanges. This not only increases the section modulus but also effectively thickens the flanges, changing their classification from non-compact to compact and allowing the modified section to reach its full plastic capacity.


Additional Solved Problems

Example 2: Interpolating Capacity with $C_b$

Determine the nominal flexural strength (MnM_n) of a W18x50 beam (Fy=50 ksiF_y = 50 \text{ ksi}) if it is braced at its ends and at midspan. The total span is 30 ft. Given properties: Lp=5.83 ftL_p = 5.83 \text{ ft}, Lr=16.9 ftL_r = 16.9 \text{ ft}, Mp=379 kip-ftM_p = 379 \text{ kip-ft}, and for Cb=1.0C_b = 1.0, the inelastic buckling capacity at Lb=15 ftL_b = 15 \text{ ft} is calculated as MLTB=285 kip-ftM_{LTB} = 285 \text{ kip-ft}. The actual moment gradient modifier CbC_b for this segment is 1.301.30.

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Example 3: Flange Local Buckling Interpolation

A built-up I-shaped beam has a non-compact flange but a compact web. The yield stress is Fy=50 ksiF_y = 50 \text{ ksi}. The plastic moment capacity is Mp=600 kip-ftM_p = 600 \text{ kip-ft}. The flange width-to-thickness ratio is λ=10.5\lambda = 10.5. The limits are λp=9.15\lambda_p = 9.15 and λr=24.1\lambda_r = 24.1. The yield moment accounting for residual stresses (0.7FySx0.7 F_y S_x) is 410 kip-ft410 \text{ kip-ft}. Determine the nominal moment capacity MnM_n assuming continuous bracing (Lb=0L_b = 0).

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Example 4: Elastic LTB (Zone 3)

A W14x30 beam is subjected to bending with an unbraced length Lb=20 ftL_b = 20 \text{ ft}. The steel is A992 (Fy=50 ksiF_y = 50 \text{ ksi}). Assume Cb=1.0C_b = 1.0. The limiting lengths are Lp=5.26 ftL_p = 5.26 \text{ ft} and Lr=15.0 ftL_r = 15.0 \text{ ft}. The critical buckling stress FcrF_{cr} has been calculated as 38.5 ksi38.5 \text{ ksi} for this unbraced length. The section modulus Sx=42.0 in3S_x = 42.0 \text{ in}^3 and Zx=47.3 in3Z_x = 47.3 \text{ in}^3. Determine the nominal moment capacity MnM_n.

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Example 5: Calculating the Moment Gradient Factor ($C_b$)

An unbraced beam segment of length Lb=20 ftL_b = 20 \text{ ft} has the following moment distribution under factored loads: maximum moment Mmax=150 kip-ftM_{max} = 150 \text{ kip-ft} at midspan. The moment at the quarter point (MAM_A) is 75 kip-ft75 \text{ kip-ft}, the moment at the midpoint (MBM_B) is 150 kip-ft150 \text{ kip-ft}, and the moment at the three-quarter point (MCM_C) is 75 kip-ft75 \text{ kip-ft}. Determine the moment gradient factor CbC_b.

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Example 6: Biaxial Bending Interaction

A W10x49 column in a frame is subjected to bending about both principal axes. The factored required moments are Mux=120 kip-ftM_{ux} = 120 \text{ kip-ft} and Muy=30 kip-ftM_{uy} = 30 \text{ kip-ft}. The design flexural capacities have been determined as ϕMnx=227 kip-ft\phi M_{nx} = 227 \text{ kip-ft} and ϕMny=105 kip-ft\phi M_{ny} = 105 \text{ kip-ft}. Verify if the section is adequate using the simplified AISC linear interaction equation.

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Example 7: Hole Reduction in Tension Flange

A W18x50 beam has a tension flange spliced using 1-inch diameter bolts. There are two bolts in the cross-section. The gross area of the flange is Afg=bf×tf=7.50 in×0.570 in=4.275 in2A_{fg} = b_f \times t_f = 7.50 \text{ in} \times 0.570 \text{ in} = 4.275 \text{ in}^2. The net area AfnA_{fn} is Afg2×(1.0 in+0.125 in)×0.570 in=2.99 in2A_{fg} - 2 \times (1.0 \text{ in} + 0.125 \text{ in}) \times 0.570 \text{ in} = 2.99 \text{ in}^2. The steel is A992 (Fy=50 ksiF_y = 50 \text{ ksi}, Fu=65 ksiF_u = 65 \text{ ksi}). Determine if a reduction in nominal moment capacity is required.

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Example 8: Plastic Hinge Moment Redistribution Concept

A propped cantilever beam (fixed at one end, simply supported at the other) supports a uniform load ww. Elastic analysis shows the maximum moment occurs at the fixed support, M=wL2/8M = wL^2 / 8, and the positive moment is 9wL2/1289wL^2 / 128. When the moment at the fixed support reaches the plastic moment MpM_p, what happens if additional load is applied?

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Example 9: Beam Selection Using $Z_x$ Tables

A simply supported beam with a span of 24 ft24 \text{ ft} must carry a factored uniform load of wu=3.0 kip/ftw_u = 3.0 \text{ kip/ft} (including beam self-weight). The beam is continuously braced (Lb=0L_b = 0). Select the lightest W-shape of A992 steel (Fy=50 ksiF_y = 50 \text{ ksi}) from the AISC manual that can safely support the load.

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Example 10: Inelastic LTB Capacity Check

A W21x44 beam (Fy=50 ksiF_y = 50 \text{ ksi}) has an unbraced length Lb=10 ftL_b = 10 \text{ ft}. The plastic moment is Mp=399 kip-ftM_p = 399 \text{ kip-ft}. The properties are: Lp=5.25 ftL_p = 5.25 \text{ ft}, Lr=14.8 ftL_r = 14.8 \text{ ft}, and 0.7FySx=241 kip-ft0.7 F_y S_x = 241 \text{ kip-ft}. Assume Cb=1.0C_b = 1.0. Determine the nominal moment capacity MnM_n.

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