Steel Beam Design Example — AISC 360-22 LRFD Worked Problem

Complete step-by-step steel beam design following AISC 360-22 LRFD provisions. This worked example covers load determination, member selection, flexural capacity (Chapter F), shear capacity (Chapter G), deflection verification, and a vibration check per AISC Design Guide 11. All calculations are shown explicitly so you can follow every step.

Problem statement

Design a simply supported floor beam for a 5-story office building.

Given:

• Beam span: L = 30 ft (9.14 m)
• Beam spacing: s = 10 ft (3.05 m) center-to-center
• Slab: 3.25 in. lightweight concrete on 2 in. composite steel deck (total slab depth = 5.25 in.)
• Steel grade: ASTM A992 (Fy = 50 ksi, Fu = 65 ksi)
• Lateral bracing: Continuous by steel deck (top flange fully braced)
• Building code: IBC 2021 with ASCE 7-22 load combinations
• Deflection limits: L/360 (live load), L/240 (total load)
• Occupancy: Office (50 psf live load per ASCE 7-22 Table 4.3-1)

Step 1 — Determine loads

Dead load

Live load

Live load reduction (ASCE 7-22 Section 4.7)

Tributary area A_T = 30 ft x 10 ft = 300 SF. K_LL = 2 (interior beam). Influence area = K_LL x A_T = 2 x 300 = 600 SF > 400 SF. Reduction applies.

L_reduced = L_0 x (0.25 + 15 / sqrt(K_LL x A_T))
L_reduced = 50 x (0.25 + 15 / sqrt(600))
L_reduced = 50 x (0.25 + 0.612)
L_reduced = 50 x 0.862 = 43.1 psf

Minimum: 0.50 x L_0 = 25 psf. Use L_reduced = 43.1 psf.

Line loads on beam

Assume estimated beam self-weight = 50 lb/ft (verify after selection).

w_D = (70 psf x 10 ft) + 50 lb/ft = 750 lb/ft = 0.750 kip/ft
w_L = (43.1 psf x 10 ft) = 431 lb/ft = 0.431 kip/ft

Factored load (LRFD, ASCE 7-22 Combination 2)

w_u = 1.2 x w_D + 1.6 x w_L
w_u = 1.2 x 0.750 + 1.6 x 0.431
w_u = 0.900 + 0.690 = 1.590 kip/ft

Step 2 — Determine required moment and shear

Maximum moment (simply supported, UDL)

M_u = w_u x L^2 / 8
M_u = 1.590 x 30^2 / 8
M_u = 1.590 x 900 / 8
M_u = 178.9 kip-ft = 2,147 kip-in

Maximum shear

V_u = w_u x L / 2
V_u = 1.590 x 30 / 2 = 23.9 kips

Step 3 — Select trial section

Required plastic section modulus

For a compact section with full lateral bracing (Chapter F, Section F2):

Z_req = M_u / (phi x Fy)
Z_req = 2,147 / (0.90 x 50) = 47.7 in^3

Trial selection from AISC Manual Table 3-2

The W18x35 provides phi*Mn = 0.90 x 50 x 66.5 / 12 = 249 kip-ft, which exceeds 178.9 kip-ft. Utilization = 178.9/249 = 72%.

Select W18x35. Verify self-weight: 35 lb/ft vs. estimated 50 lb/ft. The lighter beam reduces demand.

Revised factored load

w_D = (70 x 10) + 35 = 735 lb/ft = 0.735 kip/ft
w_u = 1.2 x 0.735 + 1.6 x 0.431 = 0.882 + 0.690 = 1.572 kip/ft
M_u = 1.572 x 900 / 8 = 176.8 kip-ft (negligible change)

Step 4 — Flexural capacity check (AISC Chapter F, Section F2)

Section properties — W18x35 (AISC Table 1-1)

Compact section check (Table B4.1b)

Flange: lambda = bf/(2tf) = 7.06. Compact limit lambda_p = 10.79 (for Fy = 50 ksi). 7.06 < 10.76 — compact.

Web: lambda = h/tw = 53.5. Compact limit lambda_p = 137.27. 53.5 < 137.27 — compact.

W18x35 is compact for Fy = 50 ksi.

Lateral-torsional buckling check (F2)

The top flange is continuously braced by the composite deck, so Lb = 0.

Since Lb = 0 < Lp, the beam reaches full plastic moment:

phi x M_n = phi x M_p = phi x Fy x Zx
phi x M_n = 0.90 x 50 x 66.5 = 2,993 kip-in = 249 kip-ft

Flexural check: phi x M_n = 249 kip-ft > M_u = 176.8 kip-ft (OK). Utilization = 71%.

Step 5 — Shear capacity check (AISC Chapter G, Section G2.1)

Shear without tension field action

phi x V_n = phi x 0.60 x Fy x Aw x Cv

where Aw = d x tw = 17.70 x 0.300 = 5.31 in^2.

Check web slenderness: h/tw = 53.5 < 260 (for unstiffened webs per G2.1(b)).

So Cv = 1.0, and:

phi x V_n = 1.0 x 0.60 x 50 x 5.31 x 1.0 = 159 kips

Shear check: phi x V_n = 159 kips > V_u = 23.9 kips (OK). Utilization = 15%.

Shear is far from governing. This is typical for uniformly loaded floor beams.

Step 6 — Deflection check (serviceability)

Live load deflection

Using the unfactored (service) live load: w_L = 0.431 kip/ft = 0.03592 kip/in.

delta_L = 5 x w_L x L^4 / (384 x E x I)
delta_L = 5 x 0.03592 x (360)^4 / (384 x 29,000 x 510)
delta_L = 5 x 0.03592 x 16,796,160,000 / 5,672,640,000
delta_L = 0.531 in

Live load limit: L/360 = 360/360 = 1.00 in. Since 0.531 < 1.00 in — OK.

Total load deflection

w_total = w_D + w_L = 0.735 + 0.431 = 1.166 kip/ft
delta_total = delta_L x (w_total / w_L)
delta_total = 0.531 x (1.166 / 0.431)
delta_total = 0.531 x 2.705 = 1.44 in

Total load limit: L/240 = 360/240 = 1.50 in. Since 1.44 < 1.50 in — OK, but tight.

Camber recommendation

Dead load deflection: delta_DL = delta_L x (w_D / w_L) = 0.531 x (0.735/0.431) = 0.906 in.

Recommended camber = 75-80% of dead load deflection = 0.75 x 0.906 = 0.68 in. Use 3/4 in. camber.

Step 7 — Vibration check (AISC Design Guide 11)

Natural frequency estimation

Effective weight for vibration:

w_eff = w_DL + 11 psf (per AISC DG11 for office floors)
w_eff = (70 + 11) x 10 + 35 = 845 lb/ft = 0.845 kip/ft

Beam deflection under effective weight:

delta_eff = 5 x 0.07042 x 360^4 / (384 x 29000 x 510)
delta_eff = 1.04 in

Natural frequency:

f_n = 0.18 x sqrt(g / delta_eff) = 0.18 x sqrt(386 / 1.04) = 0.18 x 19.3 = 3.5 Hz

f_n = 3.5 Hz is marginal. Target is 4.0 Hz for office floors.

Remediation options

Recommendation: Use composite action with shear studs. A W18x35 with 19 studs per half-span achieves I_eff = 1,200+ in^4, giving f_n > 5 Hz. This is the most economical solution.

Step 8 — Summary of results

W18x35 passes all checks. Add 3/4 in. camber and composite studs for vibration performance.

Design summary comparison — alternative sections

Code comparison for beam design

The design procedure is essentially the same across all four codes. The main differences are in resistance factors and deflection limits.

Common mistakes in beam design

1. Forgetting to check deflection when strength governs. Many engineers select a beam that passes the strength check and stop. For spans over 25 ft, deflection often governs and requires a larger section.

2. Using LRFD factored loads for deflection. Deflection is a serviceability check. Use unfactored (working) loads, not the 1.2D + 1.6L combination.

3. Not checking vibration on spans over 25 ft. AISC DG11 requires a vibration check. Beams with f_n < 4 Hz in offices will have occupant complaints.

4. Ignoring the construction (unbraced) condition. Before the deck is placed, the top flange is unbraced. Check Lb = full span for the construction load case (beam weight + wet concrete).

5. Specifying 100% camber. Over-camber causes the floor to be high at midspan. Use 75-80% of dead load deflection for camber.

6. Using the bare steel I for composite deflection. If the beam has shear studs and composite deck, use the effective composite I (typically 1.5-2.5x the steel I) for post-composite deflections only.

Frequently asked questions

How do I select the right W-shape for a floor beam? Calculate the required plastic section modulus Z_req = M_u / (phi x Fy). Then look up the lightest section from AISC Manual Table 3-2 that meets or exceeds Z_req. Always verify deflection and vibration separately.

What is the minimum beam size for a 30 ft office floor span? W18x35 non-composite or W16x26 composite. The W16x26 at 30 ft with 10 ft spacing requires about 18 studs per half-span for composite action.

When does deflection govern over strength? For typical office loads (50 psf LL, 70 psf DL), deflection governs when the span exceeds approximately 25 ft for non-composite beams. Composite beams can span 30-35 ft before deflection governs.

Do I need to check shear for floor beams? Yes, but it almost never governs for uniformly loaded beams. Shear typically governs only at heavily loaded transfer beams, coped ends, or short-span beams with point loads. This example shows shear utilization of only 15%.

What is the AISC phi factor for flexure? phi = 0.90 for flexure (Chapter F). For shear, phi = 1.00 (Chapter G, unstiffened webs). For bearing-type bolts, phi = 0.75.

Should I design for composite or non-composite action? For spans over 25 ft with typical office loads, composite design is more economical. A W21x44 composite beam at 30 ft has 82% more capacity than non-composite. For short spans (< 20 ft) with light loads, non-composite may save more in stud costs than beam weight.

Complete AISC F2 beam design — W18x50 under UDL

This worked example demonstrates a full AISC 360-22 Section F2 beam design for a W18x50 carrying a uniformly distributed load.

Given

• Beam: W18x50, ASTM A992 (Fy = 50 ksi, Fu = 65 ksi)
• Span: 30 ft, simply supported
• Unbraced length for LTB: Lb = 30 ft (no lateral bracing except at supports)
• Dead load: 0.80 kip/ft (includes beam self-weight)
• Live load: 1.20 kip/ft

Section properties (W18x50)

Step 1 — Factored load

wu = 1.2 x 0.80 + 1.6 x 1.20 = 0.96 + 1.92 = 2.88 kip/ft
Mu = wu x L^2 / 8 = 2.88 x 30^2 / 8 = 324 kip-ft = 3,888 kip-in
Vu = wu x L / 2 = 2.88 x 30 / 2 = 43.2 kips

Step 2 — Compact section check (Table B4.1b)

Flange: lambda_p = 0.38 x sqrt(E/Fy) = 0.38 x 24.08 = 9.15
        bf/(2tf) = 6.58 < 9.15 -- compact

Web:    lambda_p = 3.76 x sqrt(E/Fy) = 3.76 x 24.08 = 90.5
        h/tw = 46.2 < 90.5 -- compact

W18x50 is compact. Use Section F2.

Step 3 — Moment capacity (AISC F2.1)

Plastic moment:

phiMpx = phi x Fy x Zx = 0.90 x 50 x 101 = 4,545 kip-in = 378.8 kip-ft

Step 4 — Lateral-torsional buckling check (F2.2)

Lp and Lr determination:

Lp = 1.76 x ry x sqrt(E/Fy) = 1.76 x 1.65 x 24.08 = 69.8 in = 5.82 ft

For Lr, calculate with bf/(2tf) = 6.58 and h/tw = 46.2:

Lr = 1.95 x rts x (E/0.7Fy) x sqrt(1 + (0.7Fy/(0.7Fy)) x (bf/(2tf)) / sqrt(Iy/Cb x J/Sx))

Using simplified approach (AISC Manual Table 3-2): Lr for W18x50 = 15.9 ft.

Since Lb = 30 ft > Lr = 15.9 ft, the beam is in the elastic LTB range.

With Cb = 1.0 (uniform moment):

Fe = Cb x pi^2 x E / (Lb/ry)^2 x sqrt(1 + 0.078 x J/(Sx x ho) x (Lb/ry)^2)

From AISC Manual Table 3-2, phiMn for W18x50 at Lb = 30 ft with Cb = 1.0:

phiMn = 161 kip-ft  (elastic LTB governs)

Check: phiMn = 161 kip-ft < Mu = 324 kip-ft -- FAILS with Cb = 1.0

With Cb = 1.67 (UDL on simply supported beam, per ASCE 7 moment gradient):

For a simply supported beam with UDL, Cb = 1.14 (AISC F1 commentary). However, with lateral bracing at supports only:

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

For UDL: Mmax = wL^2/8, MA = 7wL^2/128, MB = wL^2/8 (at 0.5L), MC = 7wL^2/128.

Cb = 12.5 x Mmax / (2.5Mmax + 3(7/128)Mmax(128/8) + 4Mmax + 3(7/128)Mmax(128/8))
Cb = 12.5 / (2.5 + 2.0625 + 4 + 2.0625) = 12.5 / 10.625 = 1.176

phiMn = 1.176 x 161 = 189.3 kip-ft

Check: phiMn = 189.3 kip-ft < Mu = 324 kip-ft -- STILL FAILS

W18x50 is inadequate for 30 ft unbraced length with this load. The beam needs either lateral bracing or a larger section.

Step 5 — Shear capacity (AISC G2.1)

h/tw = 46.2
Cv = 1.0 (since h/tw <= 2.24*sqrt(E/Fy) = 107)
phiVn = phi x 0.6 x Fy x Aw = 0.90 x 0.6 x 50 x (18.00 x 0.355) = 172.5 kips

Shear check: phiVn = 172.5 kips > Vu = 43.2 kips (OK). Utilization = 25%.

Step 6 — Deflection check

Service live load: wLL = 1.20 kip/ft
delta_LL = 5 x w x L^4 / (384 x E x I)
         = 5 x (1.20/12) x (360)^4 / (384 x 29,000 x 800)
         = 5 x 0.10 x 1.68e10 / 8.95e9 = 0.938 in.

LL deflection limit (L/360): 30 x 12 / 360 = 1.00 in.

delta_LL = 0.938 in. < 1.00 in.  (OK)

Step 7 — Revised design with mid-span brace

Adding a lateral brace at mid-span: Lb = 15 ft.

From AISC Manual Table 3-2, phiMn at Lb = 15 ft with Cb = 1.14:

phiMn = 290 kip-ft (inelastic LTB range, Cb = 1.14 applied)

Check: phiMn = 290 kip-ft < Mu = 324 kip-ft -- CLOSE but still insufficient

Adding two braces (at third points, Lb = 10 ft):

phiMn = phiMpx = 378.8 kip-ft (fully braced, Lb < Lp)

Check: phiMn = 378.8 kip-ft > Mu = 324 kip-ft (OK). Utilization = 86%.

D/C summary table

The W18x50 works with third-point bracing. Moment capacity at 86% and live load deflection at 94% are close to limits. Consider W18x55 for additional margin.

Run this calculation

• Beam Calculator
• Section Properties Database

Related references

• Steel Floor Beam Design
• Beam Formulas
• Beam Sizes
• Deflection Limits
• Floor Vibration
• Composite Beam Design
• Beam Design Guide
• Steel Grades
• How to Verify Calculations

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