Solved Problems

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

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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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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