Steel Beam Capacity

Check steel beam flexural strength (phiMn), shear capacity (phiVn), and lateral-torsional buckling (LTB) per AISC 360, AS 4100, EN 1993, and CSA S16. Enter a section, steel grade, and unbraced length to get instant capacity results with utilization ratios.

Example result: W16x40 (Fy = 50 ksi, A992), Lb = 10 ft, Cb = 1.14 -- phiMn = 261 kip-ft (inelastic LTB governs), phiVn = 146 kips. If Mu = 200 kip-ft, utilization = 0.77 (passes).

Quick Reference -- Common W-Shape Beam Capacities

Capacities below are for A992 steel (Fy = 50 ksi), Cb = 1.0, and fully braced conditions (Lb <= Lp). phi = 0.90 for flexure, phi = 1.00 for shear.

When Lb > Lp, capacity drops due to LTB. Use the calculator above for the LTB-reduced capacity at your specific unbraced length.

How the Beam Capacity Calculator Works

The calculator determines the design flexural strength (phiMn) and design shear strength (phiVn) of a steel beam, then compares these capacities to the factored demands. Flexural capacity is governed by the interaction of three phenomena: section yielding, local flange or web buckling, and lateral-torsional buckling (LTB). The tool classifies the section as compact, non-compact, or slender based on width-to-thickness ratios, determines the unbraced length regime (plastic, inelastic LTB, or elastic LTB), and selects the appropriate nominal moment equation.

For compact sections with adequate lateral bracing (Lb less than or equal to Lp), the full plastic moment Mp = Fy * Zx is achieved. Between Lp and Lr, the capacity is linearly interpolated between Mp and 0.7*Fy*Sx (the onset of elastic buckling). Beyond Lr, elastic LTB governs and capacity drops rapidly. The Cb modification factor accounts for non-uniform moment gradient, allowing higher capacity when the moment diagram is not constant along the unbraced length.

Shear capacity for most rolled W-shapes with stocky webs is phi*Vn = phi * 0.6 _ Fy _ Aw, where phi = 1.00 and Aw = d * tw. For slender-web plate girders, web shear buckling reduces capacity and tension field action may be invoked.

Key Equations

Plastic moment (AISC 360-22 Eq. F2-1):

phi*Mn = phi * Mp = phi * Fy * Zx

Where phi = 0.90, Fy = yield strength, Zx = plastic section modulus.

Inelastic LTB (AISC 360-22 Eq. F2-2, Lp < Lb ≤ Lr):

phi*Mn = phi * Cb * [Mp - (Mp - 0.7*Fy*Sx) * (Lb - Lp)/(Lr - Lp)]  ≤  phi * Mp

Elastic LTB (AISC 360-22 Eq. F2-3, Lb > Lr):

Mn = Fcr * Sx
Fcr = (Cb * pi^2 * E) / (Lb/rts)^2 * sqrt(1 + 0.078 * (J*c)/(Sx*ho) * (Lb/rts)^2)

Limiting unbraced lengths:

Lp = 1.76 * ry * sqrt(E/Fy)
Lr = 1.95 * rts * (E/(0.7*Fy)) * sqrt(J*c/(Sx*ho) + sqrt((J*c/(Sx*ho))^2 + 6.76*(0.7*Fy/E)^2))

Cb factor (AISC 360-22 Eq. F1-1):

Cb = 12.5*Mmax / (2.5*Mmax + 3*MA + 4*MB + 3*MC)

Where Mmax = maximum moment, MA/MB/MC = moments at quarter points within the unbraced segment.

Shear strength (AISC 360-22 Eq. G2-1):

phi*Vn = phi * 0.6 * Fy * Aw * Cv1

Where phi = 1.00 for most rolled shapes (Cv1 = 1.0 when h/tw ≤ 2.24*sqrt(E/Fy)), Aw = d * tw.

Design Code Requirements

Key difference: AISC uses a linear interpolation between Lp and Lr for inelastic LTB, while AS 4100 uses the alpha_s member slenderness reduction factor applied to the section capacity (alpha_s * Ms). EN 1993 uses buckling curves (chi_LT) from Table 6.3 and 6.4.

Step-by-Step Example

Problem: Check the flexural capacity of a W16x40 beam (Fy = 50 ksi, Grade 50) with an unbraced length Lb = 10 ft under uniform load. Cb = 1.14 (simply supported, uniform load).

Step 1 -- Section properties: W16x40: Zx = 72.9 in^3, Sx = 64.7 in^3, ry = 1.57 in, rts = 1.73 in, J = 0.794 in^4, ho = 15.5 in, d = 16.0 in, tw = 0.305 in.

Step 2 -- Limiting unbraced lengths: Lp = 1.76 _ 1.57 _ sqrt(29000/50) = 1.76 _ 1.57 _ 24.08 = 66.5 in = 5.54 ft. Lr: complex formula, but for W16x40 Lr approximately equals 16.0 ft (from AISC tables).

Step 3 -- Determine LTB regime: Lb = 10 ft. Lp = 5.54 ft < Lb = 10 ft < Lr = 16.0 ft. Inelastic LTB governs.

Step 4 -- Compute phi*Mn: Mp = 50 * 72.9 = 3,645 kip-in = 303.8 kip-ft. phiMn = 0.90 _ 1.14 _ [3645 - (3645 - 0.75064.7)(10-5.54)/(16.0-5.54)] = 0.90 _ 1.14 _ [3645 - (3645 - 2264.5)(4.46/10.46)] = 0.90 _ 1.14 _ [3645 - 1380.5 * 0.4264] = 0.90 _ 1.14 _ [3645 - 588.7] = 0.90 _ 1.14 _ 3056.3 = 3,135.8 kip-in. Check cap: phiMp = 0.90 * 3645 = 3,280.5 kip-in. 3,135.8 < 3,280.5, so no cap. phi*Mn = 3,136 kip-in = 261.3 kip-ft.

Step 5 -- Check shear: phi*Vn = 1.0 * 0.6 _ 50 _ (16.0 _ 0.305) = 1.0 _ 0.6 _ 50 _ 4.88 = 146.4 kips.

Result: phiMn = 261 kip-ft, phiVn = 146 kips. If factored moment demand Mu = 200 kip-ft, utilization = 200/261 = 0.77. OK.

Common Design Mistakes

• Using Zx when Sx should be used for elastic LTB: For Lb > Lr, the nominal moment is Fcr _ Sx, not Fcr _ Zx. Using the plastic section modulus for the elastic buckling regime overestimates capacity by approximately 12% for typical W-shapes.
• Assuming Cb = 1.0 for all cases: Cb = 1.0 is the most conservative assumption (constant moment). For typical uniform loads on simply supported beams, Cb = 1.14, and for midspan point loads Cb = 1.32. Using Cb = 1.0 wastes 10-30% of available capacity.
• Ignoring that shear phi = 1.00 for rolled shapes: Many engineers reflexively use phi = 0.90 for shear, but AISC 360-22 Section G2.1 specifies phi = 1.00 for compact-web rolled I-shapes. The phi = 0.90 value applies only to built-up girders and certain plate girder conditions.
• Not checking web crippling at concentrated loads: At bearing points (supports and concentrated load locations), local web yielding and web crippling must be checked per AISC J10. These are often the controlling limit state for short-span heavily loaded beams.
• Forgetting to check the compression flange slenderness: While most standard W-shapes are compact, built-up sections, welded plate girders, and some HSS shapes can be non-compact or slender, reducing flexural capacity below Mp.
• Double-counting camber as deflection reduction: Camber offsets dead-load sag only. It does not reduce live-load deflection, which is the serviceability criterion that typically governs beam selection.

Frequently Asked Questions

What is the difference between compact, non-compact, and slender sections in AISC 360? Section compactness classifies a beam’s cross-section based on the width-to-thickness ratios of its flanges and web. A compact section can reach the plastic moment Mp before local buckling occurs. A non-compact section can develop some inelastic capacity but buckles locally before reaching Mp. A slender section buckles elastically before yielding and requires an effective section reduction. The vast majority of standard W-shapes in Grade 50 steel are compact for flexure, so local buckling typically only becomes a concern for built-up girders, plate girders, or light HSS sections with high width-to-thickness ratios.

What are Lp and Lr, and when does lateral-torsional buckling (LTB) control? Lp is the limiting unbraced length below which the full plastic moment Mp can be achieved — the beam is fully braced and LTB does not reduce capacity. Lr is the limiting unbraced length above which the beam buckles elastically. Between Lp and Lr, capacity is reduced linearly (inelastic LTB). Above Lr, capacity follows an elastic buckling curve and can drop well below Mp. For practical W-shapes, Lp ranges from about 6–15 feet depending on the section — bracing a beam more closely than Lp eliminates all LTB reduction and is the most direct way to recover capacity.

How does the Cb factor affect lateral-torsional buckling capacity? The uniform moment case (constant moment along the unbraced length) is the most critical condition for LTB, and Cb = 1.0 covers this case. For non-uniform moment diagrams, Cb > 1.0, allowing higher capacity. The Cb factor is computed from the moment diagram shape using the maximum moment and quarter-point moments within the unbraced segment. A simply supported beam with uniform load has Cb ≈ 1.14, while a beam with a concentrated midspan load has Cb ≈ 1.32. For cantilevers with tip loads, Cb is taken conservatively as 1.0 unless a more refined analysis is performed.

What is the difference between φMn (flexural capacity) and φVn (shear capacity)? φMn is the design flexural strength — the product of the resistance factor φb = 0.90 and the nominal moment capacity Mn, which accounts for yielding, LTB, and local buckling. φVn is the design shear strength — for most W-shapes with stocky webs (h/tw ≤ 2.24√(E/Fy)), φv = 1.00 and Vn = 0.6FyAw, giving an unusually high resistance factor. Shear rarely controls beam design for typical W-shapes at normal spans, but it becomes critical for short heavy-load beams, coped beams, and built-up plate girders with slender webs.

What is shear lag, and when does it reduce beam or connection capacity? Shear lag occurs when not all elements of a cross-section are directly connected at a joint — for example, when only the web of a W-shape is bolted, the flanges are not fully engaged at the connection. This reduces the effective net area in tension. AISC 360 accounts for shear lag through the shear lag factor U, which multiplies the net area to give an effective net area. For standard beam end connections, shear lag primarily affects the connection element check rather than the beam midspan capacity, but it can govern at coped sections with reduced depth.

When does lateral-torsional buckling control over yielding for typical floor beams? For standard floor framing with the compression flange continuously braced by a concrete deck or closely spaced bridging, LTB rarely governs — the effective unbraced length is very short. LTB becomes critical for roof beams with no deck, crane runway girders, transfer beams in parking structures, and any beam where the compression flange is unrestrained over a long segment. As a rough guide, if your unbraced length exceeds about d/2 in feet for typical W-shapes (where d is depth in inches), check whether LTB is reducing your capacity significantly.

What is the maximum unbraced length for a W16×40 (Fy = 50 ksi) to achieve its full plastic moment? The limiting unbraced length Lp = 1.76 × ry × √(E/Fy). For a W16×40: ry = 1.57 in, E = 29,000 ksi, Fy = 50 ksi. Lp = 1.76 × 1.57 × √(29,000/50) = 1.76 × 1.57 × 24.08 = 66.5 in = 5.5 ft. Any unbraced length at or below 5.5 ft allows the W16×40 to reach its full φMp = φ × Fy × Zx = 0.90 × 50 × 72.9 = 3,281 kip-in = 273 kip-ft without LTB reduction. Beyond Lp, capacity reduces linearly until Lr (approximately 16 ft for this section), where elastic LTB begins.

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