Module 8: Steel Beams (Flexural Members) - Examples & Applications

Worked-example data provenance

Unless an example explicitly cites a code table, manufacturer report, or material specification, numerical material properties and adjustment factors are problem-supplied inputs. They demonstrate the calculation procedure and must not be reused as universal NSCP design values for another species, grade, steel grade, section, or product.

Basic: Calculating the Plastic Moment Benchmark

Determine the plastic moment benchmark (MpM_p) of a W16x36 beam of A992 steel (Fy=345 MPaF_y = 345 \text{ MPa}). Then interpret what additional conditions must be satisfied before MpM_p can govern nominal flexural strength.

Given Section Properties:

  • Plastic Section Modulus (ZxZ_x): 1.05×106 mm31.05 \times 10^6 \text{ mm}^3

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Intermediate: Moment Capacity of a Laterally Supported Beam

Determine the LRFD design flexural strength (ϕbMn\phi_b M_n) of the same W16x36 beam (Mp=362.25 kN⋅mM_p = 362.25 \text{ kN}\cdot\text{m}). Assume the compact section is connected to a concrete floor system that provides the effective continuous lateral restraint required by the design model.

LRFD Factor: ϕb=0.90\phi_b = 0.90 for flexure.

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Advanced: Calculating Inelastic LTB Capacity (Zone 2)

Determine the nominal moment capacity (MnM_n) of a W14x90 beam of A992 steel (Fy=345 MPaF_y = 345 \text{ MPa}) with an unbraced length Lb=6.0 mL_b = 6.0 \text{ m}. Assume the moment gradient yields a bending coefficient Cb=1.14C_b = 1.14.

Given Section Properties and Limits:

  • Plastic Moment (MpM_p): 600 kN⋅m600 \text{ kN}\cdot\text{m}
  • Yield Moment modified for residual stress (0.7FySx0.7 F_y S_x): 420 kN⋅m420 \text{ kN}\cdot\text{m}
  • Limiting unbraced length for plastic yielding (LpL_p): 4.0 m4.0 \text{ m}
  • Limiting unbraced length for inelastic buckling (LrL_r): 12.0 m12.0 \text{ m}

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Conceptual: The Effect of Unbraced Length on Moment Capacity

An engineer designs a roof using standard W-shape steel beams spaced 3 meters apart. Initially, the metal roof deck was planned to be directly fastened to the top flanges of the beams, providing continuous lateral support. However, to save money, the contractor proposes removing the direct fastening and only bracing the beams at their ends (12 meters apart). Explain the structural consequences of this change regarding the beam's moment capacity.

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Basic: Checking Shear Strength

A W24x68 steel beam (Fy=345 MPaF_y=345\text{ MPa}) has a maximum factored shear Vu=450 kNV_u=450\text{ kN}. For the stated worked case, take Cv=1.0C_v=1.0 and ϕv=1.00\phi_v=1.00.

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Intermediate: Checking Deflection Limits

A simply supported floor beam spans 8.0 meters8.0 \text{ meters}. Under service (unfactored) loads, the immediate elastic deflections are calculated as: ΔDead=12 mm\Delta_{\text{Dead}} = 12 \text{ mm} and ΔLive=18 mm\Delta_{\text{Live}} = 18 \text{ mm}. For this worked example, the project serviceability criteria specify a live-load limit of L/360L/360 and a total-load limit of L/240L/240.

Verify if the beam satisfies serviceability requirements.

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Advanced: NSCP Combined Axial Compression and Flexure

A W14x90 column is subjected to factored axial compression Pu=1200 kNP_u=1200\text{ kN} and factored major-axis moment Mux=250 kN⋅mM_{ux}=250\text{ kN}\cdot\text{m}. The supplied available strengths are ϕcPn=3000 kN\phi_cP_n=3000\text{ kN} and ϕbMnx=550 kN⋅m\phi_bM_{nx}=550\text{ kN}\cdot\text{m}, with no y-axis moment. Evaluate the stated NSCP Section 508 interaction check.

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Example 8 — Architectural benefit of reducing unbraced length

A compact steel beam is controlled by lateral-torsional buckling. The architecture permits a credible brace at midspan. What structural change should be evaluated before selecting a heavier beam?

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Example 9 — Calculate Lp and Cb before selecting an LTB equation

For a compact doubly symmetric I-shape take Fy=345 MPaF_y=345\text{ MPa}, E=200,000 MPaE=200{,}000\text{ MPa}, and ry=40 mmr_y=40\text{ mm}. The quarter-point absolute moments in one unbraced segment are MA=80M_A=80, MB=60M_B=60, MC=40 kN⋅mM_C=40\text{ kN}\cdot\text{m} and Mmax=100 kN⋅mM_{max}=100\text{ kN}\cdot\text{m}.

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Example 10 — Slender-web shear coefficient and available strength

For a stated unstiffened web case, take h/tw=150h/t_w=150, kv=5.0k_v=5.0, E=200,000 MPaE=200{,}000\text{ MPa}, Fy=345 MPaF_y=345\text{ MPa}, overall depth d=600 mmd=600\text{ mm}, and tw=4 mmt_w=4\text{ mm}. Determine CvC_v, VnV_n, and the LRFD available shear strength using ϕv=0.90\phi_v=0.90 for this non-special G2 case.

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