Design of steel beam
Objective
Verify the adequacy of a rolled W18×35 beam–column (A992 steel) subjected to major-axis bending and axial compression under LRFD provisions. The member is considered simply supported with an unbraced length of 40 ft.
Given Data
| Parameter | Symbol | Value | Unit |
|---|---|---|---|
| Member designation | W18×35 | — | — |
| Steel grade | A992 | Fy = 50 | ksi |
| Elastic modulus | E | 29,000 | ksi |
| Unbraced length | Lb | 40 | ft |
| Factored axial load | Pu | 40 | kips (compression) |
| Factored bending moment | Mu | 120 | kip‑ft |
| Member supports | — | Pinned‑pinned (simply supported) | — |
Step 1. Section Properties
From AISC Steel Manual table for W18×35:
| Property | Symbol | Value | Unit |
|---|---|---|---|
| Area | A | 10.3 | in² |
| Moment of inertia (x‑x) | Ix | 379 | in⁴ |
| Section modulus (x‑x) | Sx | 42.1 | in³ |
| Radius of gyration (x‑x) | rx | 6.07 | in |
| Radius of gyration (y‑y) | ry | 2.02 | in |
Step 2. Axial Compressive Strength (LRFD)
Calculate slenderness and critical stress about the weak axis (likely controlling):
K = 1.0 L = 40 ft × 12 in/ft = 480 in KL/r_y = 480 / 2.02 = 238
Elastic buckling stress:
Fe = (π² × E) / (KL/ry)²
= (9.8696 × 29,000) / (238)²
= 5.05 ksi
Since Fe < (2 × Fy/λ) region is well within inelastic limit, use inelastic column formula:
Fcr = 0.658^(Fy/Fe) × Fy
= 0.658^(50/5.05) × 50
= 0.658^9.9 × 50
≈ 0.0021 × 50 = 0.105 ksi
Nominal compressive strength:
Pn = Fcr × A = 0.105 × 10.3 = 1.08 kips φc = 0.9 φPn = 0.9 × 1.08 = 0.97 kips
→ Note: This shows the 40 ft member is extremely slender in compression, giving very little axial capacity, which would almost all come from bending capacity.
Step 3. Flexural Strength (LRFD)
Compute available flexural strength assuming lateral‑torsional buckling for 40 ft unbraced length.
Lb = 40 ft = 480 in
Mn = Cb × (π² × E × Sx) / (Lb/ry)²
= 1.0 × (9.8696 × 29,000 × 42.1) / (480/2.02)²
= 1.0 × (12,046,000) / (237.6)²
= 12,046,000 / 56,400
≈ 213 kip‑in
= 17.8 kip‑ft
φb = 0.9 → φMn = 0.9 × 17.8 = 16 kip‑ft
The factored moment demand Mu = 120 kip‑ft > φMn = 16 kip‑ft → section fails in flexure under this unbraced length.
Step 4. Combined Interaction Check
For completeness, even though both capacities are exceeded:
Interaction = (Pu / φPn) + (8/9) × (Mu / φMn)
= (40 / 0.97) + (8/9) × (120 / 16)
= 41.2 + (0.8889 × 7.5)
= 41.2 + 6.7 = 47.9 > 1.0
Result: The W18×35 is inadequate for the given loads and 40 ft unbraced length.
Step 5. Recommended Modification
- Provide lateral bracing to reduce Lb, or
- Use a heavier section with larger Sx and ry, such as W18×60 or W21×68.
Step 6. Conclusion
The 40‑ft simply supported W18×35 (A992 steel) under Pu = 40 kips and Mu = 120 kip‑ft fails LRFD strength checks. Significant lateral bracing or a stronger section is required.
Find more examples here.