What are the steps in steel beam design per AISC 360?

The standard steel beam design workflow per AISC 360-22: Step 1 — Determine factored loads (1.2D + 1.6L per ASCE 7 for LRFD). Step 2 — Calculate required strength: Mu = wu x L2/8 for uniform load on simple span, Vu = wu x L/2. Step 3 — Select trial section from AISC Steel Manual tables (Table 3-2 for W shapes). Step 4 — Check flexural strength Mr = phi x Zx x Fy (compact section) or phi x Sx x Fy (noncompact). Step 5 — Check shear strength Vr = phi x 0.60 x Fy x d x tw. Step 6 — Check deflection at service loads. Step 7 — Verify bracing and LTB if applicable. Step 8 — Check vibration for floor beams per AISC DG 11.

What load combinations are used for steel beam design?

Per ASCE 7-22 for LRFD: combination 1.2D + 1.6L governs for typical floor beams. For roofs: 1.2D + 1.6Lr (or 1.2D + 1.0Lr + 0.5S for snow). For beams supporting snow: 1.2D + 1.6S + 0.5L. For beams in seismic systems: combinations from ASCE 7 Section 12.4 including overstrength factor Omega_o when required. The 1.2D + 1.6L combination typically produces the highest factored load for floor beams. For serviceability (deflection) checks, use unfactored loads: D + L for floor beams, D + Lr for roof beams. Wind uplift can produce net negative moment that must be checked.

How do I check if lateral-torsional buckling governs?

Per AISC 360-22 F2: if unbraced length Lb Lr, elastic LTB governs: Mr = phi x Fcr x Sx where Fcr = Cb x pi2 x E / (Lb/rt)2. Lp and Lr are tabulated in AISC Manual Table 3-2 for each W shape. For a W21x44, Lp = 6.5 ft and Lr = 18.8 ft. Continuous bracing from a concrete slab typically eliminates LTB concerns for gravity beams.

How is beam vibration checked per AISC?

AISC Design Guide 11 provides the methodology for floor vibration. The key metric is peak acceleration under walking excitation, which must be below 0.5%g for offices and 0.2%g for sensitive equipment. Parameters: floor natural frequency fn = (pi/2L2)sqrt(EIt/g x w), effective panel weight, and damping ratio (typically 3-5% for steel-framed floors). For a 30 ft W21x44 at 10 ft spacing, fn is typically 3.5-5 Hz. Frequencies below 3 Hz risk resonance with walking harmonics. Increasing beam depth, reducing spacing, or adding a thicker slab raises fn and reduces vibration.

How do I document a beam design calculation?

A complete beam design calculation document should include: (1) design criteria — governing code (AISC 360-22), design method (LRFD), steel grade, deflection limits; (2) loading summary — dead load breakdown, live load, load combinations; (3) structural analysis — span, bracing points, moment and shear diagrams; (4) section properties — d, bf, tf, tw, Zx, Sx, Ix, rts from AISC Manual Table 1-1; (5) limit state checks — flexure (Mr vs Mu with utilization ratio), shear (Vr vs Vu), deflection (delta vs limit), LTB if applicable; (6) conclusion — section adequacy statement with utilization ratios. Each check references the specific AISC equation number.

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. This example uses a different beam and loading scenario from other worked examples on this site to ensure unique educational value.

Problem Statement

Design a typical interior floor beam for a 6-story office building with composite steel deck flooring. The beam spans 32 ft simply supported between girders. The floor framing plan places beams at 10 ft on center. The floor slab is 3-1/4 in. normal weight concrete (145 pcf) on 3 in. deep, 20 gage composite metal deck (Verco W3 FormLok or similar). The beam must support the tributary floor area between the beam centerlines.

Loads:

Load Type	Value	Source
Concrete slab + deck	55 psf	3-1/4 in. NWC + 3 in. deck
Steel framing (self-weight)	5 psf	Estimated, will verify
MEP + ceiling	10 psf	Mechanical, electrical, plumbing, suspended ceiling
Partitions	15 psf	Allowance for movable partitions
Total dead load	85 psf	Excluding beam self-weight
Live load (office)	50 psf	ASCE 7-22 Table 4.3-1, office buildings
Live load reduction	Per ASCE 7-22 Eq. 4.7-1	KLL = 2 (interior beam), AT = 32 x 10 = 320 ft^2

Live load reduction per ASCE 7-22 Section 4.7.2:

Lo = 50 psf. KLL x AT = 2 x 320 = 640 ft^2.

L = Lo x (0.25 + 15 / sqrt(KLL x AT)) = 50 x (0.25 + 15 / sqrt(640)) L = 50 x (0.25 + 15 / 25.30) = 50 x (0.25 + 0.593) = 50 x 0.843 = 42.1 psf

Minimum reduced live load for members supporting one floor: L_min = 0.50 Lo = 25 psf. Our reduced value (42.1 psf) exceeds the minimum. Use L = 42.1 psf.

Tributary width = 10 ft (beam spacing).

Step 1: Determine Factored Loads

LRFD Load Combination per ASCE 7-22 Section 2.4.1:

Controlling combination for gravity: 1.2D + 1.6L

Uniform dead load on beam: w_D = (85 psf) x 10 ft = 850 plf = 0.850 klf

Uniform live load on beam: w_L = (42.1 psf) x 10 ft = 421 plf = 0.421 klf

Factored uniform load: w_u = 1.2 x 0.850 + 1.6 x 0.421 = 1.020 + 0.674 = 1.694 klf

Maximum factored moment (simply supported): M_u = w_u x L^2 / 8 = 1.694 x 32^2 / 8 = 1.694 x 128 = 216.8 kip-ft = 2,602 kip-in

Maximum factored shear: V_u = w_u x L / 2 = 1.694 x 16 = 27.1 kip

Step 2: Select Trial Section

From AISC Manual Table 3-2 (W-shapes selection by Zx):

Zx_req >= M_u / (phi_b x Fy) = 2,602 / (0.90 x 50) = 2,602 / 45 = 57.8 in^3

Check W21 sections. A W21x44 has Zx = 95.4 in^3, well above the minimum. The section also satisfies the compactness criteria for flexure (flange and web slenderness both within lambda_p limits for A992 steel).

Trial section: W21x44 (ASTM A992, Fy = 50 ksi, Fu = 65 ksi)

Section properties:

d = 20.7 in., bf = 6.50 in., tw = 0.350 in., tf = 0.450 in.
Ix = 843 in^4, Sx = 81.6 in^3, Zx = 95.4 in^3
r_ts = 1.69 in., h_o = 20.2 in., J = 0.44 in^4, Cw = 8,040 in^6
A = 13.0 in^2

Self-weight = 44 plf, close to the 5 psf allowance (5 psf x 10 ft = 50 plf). The difference is negligible. Recalculate factored load with the actual self-weight:

w_D_actual = (55 + 10 + 15) x 10 + 0.044 = 800 + 0.044 = 0.844 klf (the 5 psf allowance for steel was within 6 plf of actual) w_u_actual = 1.2 x 0.844 + 1.6 x 0.421 = 1.013 + 0.674 = 1.687 klf

Redo M_u = 1.687 x 128 = 216.0 kip-ft = 2,592 kip-in (negligible change -- proceed with original).

Step 3: Flexural Capacity (AISC 360 Section F2)

The W21x44 is a compact I-shaped member. Check flange and web slenderness.

Flange slenderness (Table B4.1, Case 1): lambda_f = bf / (2 x tf) = 6.50 / (2 x 0.450) = 6.50 / 0.900 = 7.22 lambda_pf = 0.38 x sqrt(E / Fy) = 0.38 x sqrt(29,000 / 50) = 0.38 x 24.08 = 9.15

lambda_f = 7.22 < lambda_pf = 9.15. Flange is compact.

Web slenderness (Table B4.1, Case 9): lambda_w = h / tw = (20.7 - 2 x 0.450) / 0.350 = 19.8 / 0.350 = 56.6 lambda_pw = 3.76 x sqrt(E / Fy) = 3.76 x 24.08 = 90.5

lambda_w = 56.6 < lambda_pw = 90.5. Web is compact.

Since both flange and web are compact, the nominal flexural strength is Mn = Mp = Fy x Zx:

Mn = 50 ksi x 95.4 in^3 = 4,770 kip-in = 397.5 kip-ft phi_b x Mn = 0.90 x 397.5 = 357.8 kip-ft

Check: M_u = 216.8 kip-ft < phi_b Mn = 357.8 kip-ft. OK.

Flexural utilization ratio = 216.8 / 357.8 = 0.606.

Check lateral-torsional buckling (LTB) limit state (Section F2.2):

Unbraced length Lb = 32 ft (full span, the deck provides continuous lateral bracing to the top flange in compression). Wait -- composite deck provides lateral bracing at the top flange only. For the unshored construction condition, before the concrete has cured, the top flange of the bare steel beam is in compression (positive moment) and is laterally unbraced by the uncured concrete. The unbraced length for this condition depends on the erection sequence.

For this example, assume the deck is attached with puddle welds or shear studs at a maximum spacing of 12 in., providing continuous bracing per AISC 360 Appendix 6 Section 6.3. Lb = 0 (continuously braced). LTB does not govern.

If the beam were unbraced: For Lb = 32 ft, the limiting lengths:

Lp = 1.76 x r_y x sqrt(E / Fy) = 1.76 x 1.27 x 24.08 = 53.8 in. = 4.48 ft (where r_y = 1.27 in. for W21x44) Lr = from AISC Manual Table 3-2: 11.5 ft for W21x44

Since Lb = 32 ft >> Lr = 11.5 ft, elastic LTB would govern. The nominal moment for Lb > Lr:

Mn = Fcr x Sx = (Cb x pi^2 x E / (Lb / r_ts)^2) x sqrt(1 + 0.078 x J/(Sx x h_o) x (Lb/r_ts)^2) x Sx

This produces Mn significantly less than Mp. For unbraced conditions, a deeper or heavier section would be required. The lesson: lateral bracing is essential for efficient beam design.

Step 4: Shear Capacity (AISC 360 Section G2)

The web shear capacity for an unstiffened I-shaped member:

h / tw = 56.6 (from above)

For webs without transverse stiffeners, the shear buckling coefficient kv = 5.34 (AISC 360 Section G2.1(b)).

Check C_v1: 1.10 x sqrt(kv x E / Fy) = 1.10 x sqrt(5.34 x 29,000 / 50) = 1.10 x sqrt(3,097) = 1.10 x 55.65 = 61.2

h/tw = 56.6 <= 61.2, so Cv = 1.0. The web yields in shear before buckling.

Nominal shear strength: Vn = 0.6 x Fy x Aw x Cv = 0.6 x 50 x (d x tw) x 1.0 = 0.6 x 50 x (20.7 x 0.350) = 0.6 x 50 x 7.245 = 217.4 kip

phi_v x Vn = 0.90 x 217.4 = 195.6 kip

Check: Vu = 27.1 kip << phi_v Vn = 195.6 kip. OK.

Shear utilization ratio = 27.1 / 195.6 = 0.139. Shear is not controlling for this long, lightly loaded beam. Flexure governs.

Step 5: Deflection Check

Deflection limits per AISC Design Guide 3 and IBC Table 1604.3:

Live load deflection: L/360 = (32 x 12) / 360 = 1.07 in.
Total load deflection: L/240 = (32 x 12) / 240 = 1.60 in.

Live load deflection (service level, not factored): w_L_service = 0.421 klf (unfactored live load)

delta_LL = 5 x w_L x L^4 / (384 x E x I) delta_LL = 5 x (0.421/12) x (32 x 12)^4 / (384 x 29,000 x 843) delta_LL = 5 x 0.03508 x (384)^4 / (384 x 29,000 x 843)

Let me compute this step by step:

L = 384 in.
L^4 = 384^4 = 2.174 x 10^10 in^4
5 x w x L^4 = 5 x (0.421/12) x 2.174 x 10^10 = 5 x 0.03508 x 2.174 x 10^10 = 3.814 x 10^9
384 x E x I = 384 x 29,000 x 843 = 384 x 24,447,000 = 9.388 x 10^9

delta_LL = 3.814 x 10^9 / 9.388 x 10^9 = 0.406 in.

Check: delta_LL = 0.406 in. < L/360 = 1.07 in. OK.

Total load deflection: w_total = w_D_service + w_L_service = 0.850 + 0.421 = 1.271 klf

delta_total = 0.406 x (1.271 / 0.421) = 1.23 in. < L/240 = 1.60 in. OK.

The beam is stiffer than required. Deflection does not control.

Step 6: Vibration Check (AISC Design Guide 11)

For office floor beams, walking-induced vibration can be perceptible and annoying. Check per DG11 Chapter 4 for walking excitation.

Estimated natural frequency: f_n = (pi / (2 x L^2)) x sqrt(E x I_t / m)

where I_t is the transformed moment of inertia (composite action increases stiffness), m is the effective mass per unit length.

For a preliminary check (non-composite, conservative): m = w_total / g = 1.271 / (32.2 x 12) = 0.00329 kip-sec^2/in/in

f_n = (pi / (2 x 384^2)) x sqrt(29,000 x 843 / 0.00329) f_n = (3.1416 / 294,912) x sqrt(24,447,000 / 0.00329) f_n = 1.065 x 10^-5 x sqrt(7.431 x 10^9) f_n = 1.065 x 10^-5 x 86,200 f_n = 0.918 Hz

The calculated frequency of 0.92 Hz is well below the recommended minimum of 3 Hz for walking vibration (DG11 Section 4.3). This suggests the beam may experience perceptible vibration. However, this analysis is for a bare steel beam. In the actual composite condition, the concrete slab and continuity with adjacent beams significantly increase the effective stiffness and mass, raising the natural frequency substantially above 3 Hz. A full composite vibration analysis per DG11 Chapter 5 would account for these effects and would likely show acceptable performance.

Step 7: Bearing at Supports

Check web local yielding and web crippling at the supports. Assume the beam bears on a 6 in. wide bearing plate at each girder.

Web local yielding (AISC 360 Section J10.2): lb = 6.0 in. (bearing length)

Rn = Fy x tw x (5k + lb) for interior conditions k = 0.875 in. for W21x44 (from AISC Manual dimensions)

Rn = 50 x 0.350 x (5 x 0.875 + 6.0) = 50 x 0.350 x (4.375 + 6.0) = 50 x 0.350 x 10.375 = 181.6 kip

phi = 1.00. phi Rn = 181.6 kip >> Ru = 27.1 kip. OK.

Web crippling (J10.3):

Rn = 0.80 x tw^2 x [1 + 3 x (lb/d) x (tw/tf)^1.5] x sqrt(E x Fy x tf / tw)

Rn = 0.80 x 0.350^2 x [1 + 3 x (6.0/20.7) x (0.350/0.450)^1.5] x sqrt(29,000 x 50 x 0.450 / 0.350) Rn = 0.80 x 0.1225 x [1 + 3 x 0.290 x 0.683] x sqrt(29,000 x 50 x 1.286) Rn = 0.0980 x [1 + 0.594] x sqrt(1,864,700) Rn = 0.0980 x 1.594 x 1,365 = 213.3 kip

phi = 0.75. phi Rn = 160.0 kip >> Ru. OK.

The bearing plate at 6 in. is adequate. A standard 6 in. wide bearing plate with 1/2 in. thickness is sufficient.

Summary of Design

Item	Design Value
Beam section	W21x44, ASTM A992
Span	32 ft, simply supported
Tributary width	10 ft
Factored moment, Mu	216.8 kip-ft
Design moment capacity, phi Mn	357.8 kip-ft
Moment utilization	61%
Factored shear, Vu	27.1 kip
Design shear capacity, phi Vn	195.6 kip
Live load deflection	0.41 in. (L/940 -- well within L/360)
Total load deflection	1.23 in. (L/312 -- within L/240)
Bearing plate	PL 1/2 x 6 x 8 (A36)
Composite shear studs	3/4 in. dia. x 4-7/8 in. long, 1 per rib (by others, composite design not covered)

Related References

Disclaimer

This page is for educational and reference use only. This worked example illustrates a typical design approach but does not replace project-specific engineering analysis. All designs must be verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) for the specific project. The site operator disclaims liability for any loss arising from the use of this information. Results are PRELIMINARY -- NOT FOR CONSTRUCTION.