-- | ----- | ----- | | d | 23.6 | in. | | bf | 7.01 | in. | | tf | 0.505 | in. | | tw | 0.395 | in. | | Sx | 114 | in.^3 | | Zx | 134 | in.^3 | | Ix | 1,350 | in.^4 | | Iy | 29.1 | in.^4 | | J | 1.08 | in.^4 | | Cw | 3,870 | in.^6 | | rts | 1.73 | in. | | ho | 23.1 | in. | | ry | 1.34 | in. | | A | 16.2 | in.^2 |

Loads: Dead load = 65 psf x 8 ft trib = 520 plf (slab + deck + MEP + ceiling); Live load = 50 psf x 8 ft trib = 400 plf (office per ASCE 7-22 Table 4.3-1, except lobbies and corridors)

Factored Load (LRFD): w_u = 1.2 x 520 + 1.6 x 400 = 1,264 plf = 1.264 klf

Step 1: Required Strength

Maximum factored moment for simply supported uniformly loaded beam:

M_u = w_u x L^2 / 8 = 1.264 x (36)^2 / 8 = 204.8 kip-ft

Maximum factored shear:

V_u = w_u x L / 2 = 1.264 x 36 / 2 = 22.8 kips

Step 2: Flexural Capacity — Compactness Check (AISC 360-22 Table B4.1b)

Flange compactness (lambda_f = bf / 2tf):

lambda_f = 7.01 / (2 x 0.505) = 6.94
lambda_pf = 0.38 x sqrt(E/Fy) = 0.38 x sqrt(29,000/50) = 9.15
lambda_f = 6.94 < lambda_pf = 9.15 → Flange is COMPACT

Web compactness (lambda_w = h/tw):

h = d - 2 x k_des = 23.6 - 2 x 0.875 = 21.85 in. (approximate)
lambda_w = 21.85 / 0.395 = 55.3
lambda_pw = 3.76 x sqrt(E/Fy) = 3.76 x sqrt(29,000/50) = 90.6
lambda_w = 55.3 < lambda_pw = 90.6 → Web is COMPACT

Section is compact — plastic moment Mn can be developed.

Step 3: Nominal Flexural Strength (AISC 360-22 Chapter F2)

Since the compression flange is continuously braced, LTB does not govern. For compact W-shapes:

M_n = M_p = Fy x Zx = 50 x 134 = 6,700 kip-in. = 558.3 kip-ft

Design flexural strength:

phi_b x M_n = 0.90 x 558.3 = 502.5 kip-ft

Check: M_u = 204.8 kip-ft << phi_b Mn = 502.5 kip-ft. Flexure OK. Utilization = 204.8/502.5 = 0.408.

Step 4: Shear Capacity (AISC 360-22 Chapter G2)

Web shear buckling coefficient (unstiffened web, h/tw = 55.3):

k_v = 5.34 (for unstiffened webs, a/h > 3.0)

Check shear buckling threshold:

1.10 x sqrt(k_v x E / Fy) = 1.10 x sqrt(5.34 x 29,000 / 50) = 1.10 x 55.7 = 61.2
h/tw = 55.3 < 61.2 → C_v = 1.0 (shear yielding, no buckling)

Nominal shear strength (AISC G2-1):

V_n = 0.6 x Fy x Aw x C_v = 0.6 x 50 x (d x tw) x 1.0 = 0.6 x 50 x (23.6 x 0.395) x 1.0
V_n = 0.6 x 50 x 9.32 x 1.0 = 279.7 kips

Design shear strength:

For rolled I-shapes with h/tw = 55.3 > 2.24 x sqrt(E/Fy) = 53.9, AISC 360-22 Section G1 specifies phi_v = 0.90 (not 1.00; the phi_v = 1.00 limit applies only to stockier webs with h/tw <= 53.9 for 50 ksi steel).

phi_v x V_n = 0.90 x 279.7 = 251.7 kips (LRFD)
V_n / Omega_v = 279.7 / 1.67 = 167.5 kips (ASD)

Check: V_u = 22.8 kips << 251.7 kips. Shear OK. Utilization = 22.8/251.7 = 0.091.

Step 5: Deflection Check (Serviceability)

Live load deflection for simply supported uniform load:

delta_LL = 5 x w_LL x L^4 / (384 x E x Ix)
         = 5 x (400/12) x (36 x 12)^4 / (384 x 29,000 x 1,350)
         = 5 x 33.33 x (432)^4 / (384 x 29,000 x 1,350)

Compute stepwise:

L^4 = 432^4 = 3.484 x 10^10 in.^4
w_LL = 400/12 = 33.33 lb/in.
5 x 33.33 x 3.484e10 = 5.806 x 10^12
384 x 29,000 x 1,350 = 1.504 x 10^10
delta_LL = 5.806e12 / 1.504e10 = 0.386 in.

Allowable live load deflection (per IBC Table 1604.3 for floor members):

L/360 = 36 x 12 / 360 = 432/360 = 1.20 in. (live load)
delta_LL = 0.386 in. < L/360 = 1.20 in. → OK

Total load deflection:

delta_TL = 5 x [(520+400)/12] x 432^4 / (384 x 29,000 x 1,350)
         = 5 x 76.67 x 3.484e10 / 1.504e10 = 0.888 in.
L/240 = 432/240 = 1.80 in. → OK

Deflection OK.

Step 6: Lateral-Torsional Buckling Verification (If Unbraced)

For illustrative purposes, if the beam were unbraced over the full span (Lb = 36 ft = 432 in.), check LTB:

L_p = 1.76 x ry x sqrt(E/Fy) = 1.76 x 1.34 x sqrt(29,000/50) = 1.76 x 1.34 x 24.08 = 56.8 in. = 4.73 ft

c = 1.0 (doubly symmetric I-shape)
L_r = 1.95 x rts x E/(0.7Fy) x sqrt(J x c / (Sx x ho)) x sqrt(1 + sqrt(1 + 6.76 x (0.7Fy x Sx x ho / (E x J x c))^2))

Compute rts factor:

rts^2 = sqrt(Iy x Cw) / Sx = sqrt(29.1 x 3,870) / 114 = sqrt(112,617) / 114 = 335.6 / 114 = 2.94
rts = 1.715 in. (matches tabulated value closely)

Since Lb = 432 in. >> Lr (typically ~200 in. for W24x55 with typical Lr around 15-18 ft), elastic LTB would govern if unbraced. With continuous bracing, Lb is effectively zero (top flange braced at every point), so Mn = Mp.

Step 7: Bearing at Supports

Required bearing length for web local yielding (AISC J10.2):

R_n = Fy x tw x (2.5k + lb) for lb <= d (interior)

For end support with lb = 6 in. bearing length:

k = 1.01 in. (tabulated for W24x55)
R_n = 50 x 0.395 x (2.5 x 1.01 + 6.0) = 19.75 x (2.525 + 6.0) = 19.75 x 8.525 = 168.4 kips

Design bearing strength:

phi x R_n = 0.90 x 168.4 = 151.6 kips (LRFD)

Check: R_u = 22.8 kips << 151.6 kips. Bearing OK.

Step 8: Summary of Design Checks

All checks pass. The W24x55 is adequate for the design loading with significant reserve capacity. A W21x44 could also be investigated for potential weight savings.

AISC 360-22 Beam Design Overview

AISC 360-22 Chapter F governs the flexural design of structural steel members. The chapter is organized by cross-section type:

• F2: Doubly symmetric compact I-shaped members (the most common case — this example)
• F3: Doubly symmetric I-shaped members with compact webs and noncompact or slender flanges
• F4: Other I-shaped members with compact or noncompact webs
• F5: I-shaped members with slender webs
• F6: HSS and box-shaped members
• F7: Square and rectangular HSS and box members
• F8: Round HSS
• F9: Tees and double angles
• F10: Single angles
• F11: Rectangular bars and rounds
• F12: Unsymmetrical shapes

The three flexural limit states are: yielding (Mn = Mp for compact), lateral-torsional buckling (LTB, reduces Mn when Lb > Lp), and flange local buckling (FLB, for noncompact/slender flanges). For compact beams with continuous lateral bracing, yielding governs and Mn = Mp = Fy x Zx.

Beam Optimization Principles

When designing steel beams for economy, consider these principles:

• Span-to-depth ratio: Floor beams typically range from L/18 to L/24 for composite floor systems. The W24x55 at 36 ft gives L/d = 18.3, which is within the normal range. Deeper beams use less steel per foot but increase floor-to-floor height.
• Composite action: Composite beams (steel beam + concrete slab connected by shear studs) are 20-40% more efficient than non-composite beams. The concrete slab acts as the compression flange, increasing the effective flexural strength and stiffness.
• Cambering: For long-span beams with heavy dead loads, cambering (pre-bending the beam upward by 75-80% of the dead load deflection) eliminates the visual sag and improves ceiling flatness. W24x55 camber is typically specified for spans over 25-30 ft.
• Vibration: For office floors, the AISC Design Guide 11 walking vibration criterion often governs over strength for longer spans. The fundamental frequency should exceed 3-4 Hz to avoid perceptible vibration under foot traffic.
• Bolt hole effect on net section: In regions of high moment, bolt holes in tension flanges reduce the gross area. For compact sections, the plastic moment Mp is typically based on the gross section, and net section checks are only required for tension members (Chapter D) unless the tension flange is coped or significantly reduced.

LTB Intermediate Bracing Requirements

When intermediate bracing is provided at intervals rather than continuously:

These are from AISC 360-22 Appendix 6. For the W24x55 example with top flange continuously braced by metal deck, the deck attachment must resist 0.005 x Mr/ho = 0.005 x 6,700/(23.1) = 1.45 lb/in. perpendicular to the beam — easily met by puddle welds or powder-actuated fasteners at 12-18 in. on center.

Frequently Asked Questions

How do I account for unbraced length in continuous beams?

For continuous beams with the top flange braced by the slab, the unbraced length for negative moment regions (bottom flange in compression near supports) is the distance between points where the bottom flange is braced — typically at support stiffeners, cross-frames, or moment connections. Bottom flange bracing at supports is critical: a 10 ft unbraced length near a support can reduce the negative moment capacity by 40-60% compared to a fully braced condition.

When should I use AISC Chapter F4 versus F2?

Use Chapter F2 for doubly symmetric I-shapes with compact webs (h/tw <= 90.6 for 50 ksi steel). Chapter F4 applies when the web is noncompact or slender (h/tw > 90.6), or when the section is singly symmetric (different top and bottom flanges). Most rolled W-shapes in A992 steel are compact through W30 and above through moderate weights — check the AISC Manual Table 1-1 for compactness flags.

How much reserve capacity should a beam have?

A utilization ratio (demand/capacity) of 0.75-0.90 is typical for efficient designs. Utilization above 0.95 leaves no margin for field modifications, future loads, or construction tolerances. Utilization below 0.50 suggests the beam is oversized — investigate a lighter section. For vibration-sensitive occupancies (hospitals, labs), keeping utilization under 0.70 also provides additional stiffness that improves vibration performance.

Does the W24x55 work for a 40 ft span?

At 40 ft: Mu = 1.264 x 40^2/8 = 252.8 kip-ft, still within phi_b Mn = 502.5 kip-ft (utilization 0.503). However, deflection becomes: delta_LL = 0.386 x (40/36)^4 = 0.386 x 1.524 = 0.588 in. Allowable L/360 = 480/360 = 1.33 in. — still OK. The W24x55 works at 40 ft but utilization rises to 0.503 for flexure. Consider W24x62 or W21 series for better efficiency at longer spans.

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Related References

• Beam Formulas Reference
• Steel Beam Span Guide
• Lateral Torsional Buckling Guide
• Beam Calculator Landing
• Steel Allowable Stress Design
• AISC Steel Construction Tables
• How to Verify Calculations
• Snow Load Calculation
• Live Load Reference

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This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against the applicable standard and project specification before use. The site operator disclaims liability for any loss arising from the use of this information.