Cold-Formed Steel Wall Stud Design: AISI S100-16 Worked Example (600S162-54)

Cold-formed steel (CFS) wall studs carry axial loads from floors and roofs above, plus out-of-plane wind pressure on the wall face. Unlike hot-rolled steel shapes where local buckling is often precluded by generous width-to-thickness limits, CFS members are thin-walled and therefore governed by local buckling, distortional buckling, and global buckling interaction under the AISI S100 North American Specification. The effective width method, codified in AISI S100 Appendix 1, is the primary design approach for CFS studs under combined axial compression and bending.

This worked example walks through a complete design check for a 600S162-54 CFS wall stud — a 6-inch-deep stud with 1-5/8-inch flanges and 54-mil base steel thickness — subjected to combined axial dead-plus-live load and out-of-plane wind pressure. You will see the full effective section property derivation, global buckling check, local buckling effective width reduction, distortional buckling check, and the AISI S100 combined loading interaction equation.

PRELIMINARY — NOT FOR CONSTRUCTION. This example is for educational use. All results must be independently verified by a licensed Professional Engineer before use in any design.

Design parameters

Section properties — 600S162-54 (gross)

CFS studs are designated by depth, flange width, lip length, and base steel thickness. The 600S162-54 designation means: 6.00-inch web depth (outside-to-outside), 1.625-inch flange, 0.500-inch lip return, and 54 mil base steel (0.0566 in design thickness after coating deduction per AISI S240).

All gross properties are computed per AISI S100-16 Chapter B using the centerline method with inside bend radius R = 1.5t for a stud formed from 180-degree cold-bent corners.

Step 1: Factored loads (LRFD)

Per ASCE 7-22 Section 2.3.1 and AISI S100-16 Chapter C, the governing LRFD load combinations for combined axial compression and bending on a wall stud are:

LC1 (axial only): Pu1 = 1.2 PD + 1.6 PL = 1.2 × 1,200 + 1.6 × 1,800 = 1,440 + 2,880 = 4,320 lb

LC2 (axial + wind): Pu2 = 1.2 PD + 1.0 PL + 1.0 W = 1.2 × 1,200 + 1.0 × 1,800 = 3,240 lb

The out-of-plane wind moment on a stud spanning between floor and roof diaphragms (simply supported strip) is:

w_wind = 25 psf × (16 in / 12 in/ft) = 33.3 lb/ft
Mu_wind = w_wind × L² / 8 = 33.3 × 10² / 8 = 416 lb-ft = 4,990 lb-in

Step 2: Global buckling — strong-axis compression

Per AISI S100-16 Section E2, the nominal axial strength Pn is the minimum of flexural buckling, torsional buckling, and flexural-torsional buckling strength. For a doubly symmetric C-section, torsional and flexural-torsional buckling do not govern when the shear center and centroid coincide.

The elastic critical buckling stress for the strong axis is:

Kx = 1.0 (pinned-pinned)
Lx = 10 ft = 120 in
Fe_x = π² E / (Kx Lx / rx)² = π² × 29,500 / (120 / 2.16)²
Fe_x = 291,300 / (55.6)² = 291,300 / 3,086 = 94.4 ksi

Check if inelastic buckling applies (Fe ≥ 2.78 Fy?):

Fy = 50 ksi → 2.78 Fy = 139 ksi
Fe_x = 94.4 ksi < 139 ksi → elastic buckling governs

The critical buckling stress for global flexural buckling:

λc_x = √(Fy / Fe_x) = √(50 / 94.4) = √0.530 = 0.728
Since λc_x ≤ 1.5:
Fn_x = (0.658^(λc_x²)) × Fy = (0.658^0.530) × 50 = 0.801 × 50 = 40.1 ksi

Step 3: Effective width — local buckling reduction

Per AISI S100-16 Appendix 1 (Effective Width Method), the flat width of the web and flange elements must be reduced when the width-to-thickness ratio exceeds the slenderness limit λ.

Web element (stiffened, uniform compression):

The flat width of the web (centerline dimensions):

h = D - 2(t + R) = 6.00 - 2 × (0.0566 + 0.0849) = 6.00 - 0.283 = 5.717 in
w/t = 5.717 / 0.0566 = 101.0

The plate buckling coefficient for a stiffened element under uniform compression is k = 4.0.

λ_web = (1.052 / √k) × (w/t) × √(f / E)
λ_web = (1.052 / √4.0) × 101.0 × √(40.1 / 29,500)
λ_web = 0.526 × 101.0 × 0.0369 = 1.96 > 0.673

Since λ > 0.673, the effective width is:

ρ_web = (1 - 0.22/λ) / λ = (1 - 0.22/1.96) / 1.96 = (1 - 0.112) / 1.96 = 0.888 / 1.96 = 0.453
be_web = ρ_web × h = 0.453 × 5.717 = 2.59 in

Flange element (unstiffened, uniform compression):

The flat width of the flange:

w_flange = B - 2(t + R) - d = 1.625 - 2 × (0.0566 + 0.0849) - 0.500
w_flange = 1.625 - 0.283 - 0.500 = 0.842 in
w/t_flange = 0.842 / 0.0566 = 14.9

For an unstiffened element with a lip, the plate buckling coefficient k depends on the lip stiffener adequacy per AISI S100-16 Section B4. For the 600S162-54, the 0.500-inch lip provides sufficient stiffening to treat the flange as partially stiffened (k ≈ 0.43 for the flange-lip assembly).

λ_flange = (1.052 / √0.43) × 14.9 × 0.0369 = 1.605 × 14.9 × 0.0369 = 0.882 > 0.673
ρ_flange = (1 - 0.22/0.882) / 0.882 = (1 - 0.249) / 0.882 = 0.851
be_flange = 0.851 × 0.842 = 0.717 in

The effective area Ae for axial compression accounts for the reduced web and flange widths:

Ae = Ag - (h - be_web) × t - 2 × (w_flange - be_flange) × t
Ae = 0.565 - (5.717 - 2.59) × 0.0566 - 2 × (0.842 - 0.717) × 0.0566
Ae = 0.565 - 3.127 × 0.0566 - 2 × 0.125 × 0.0566
Ae = 0.565 - 0.177 - 0.014 = 0.374 in²

Step 4: Distortional buckling check

Distortional buckling involves rotation of the flange-lip assembly about the flange-web junction. Per AISI S100-16 Section E4, the elastic distortional buckling stress Fd is computed using the analytical method (S100 Appendix 1, Section 2.3.3) or the simplified method.

For a 600S162-54 stud with 50 ksi steel, the critical elastic distortional buckling stress (simplified method) is approximately:

Fd = 35.2 ksi (for this section, obtained from AISI S100 Direct Strength Method tables or CUFSM analysis)

The nominal distortional buckling strength:

λd = √(Fy / Fd) = √(50 / 35.2) = √1.42 = 1.19
Since λd ≤ 1.5:
Pnd = (0.658^(λd²)) × Ag × Fy = 0.658^1.42 × 0.565 × 50
Pnd = 0.554 × 0.565 × 50 = 15.6 kips

Step 5: Nominal axial strength

The governing nominal axial strength is the minimum of the three buckling modes:

Pne (global) = Ae × Fn_x = 0.374 × 40.1 = 15.0 kips = 15,000 lb
Pnl (local)  = Ae × Fy (already captured in effective width reduction)
Pnd (distortional) = 15.6 kips = 15,600 lb

Global buckling with local interaction governs: Pn = 15.0 kips.

Per AISI S100-16 Section E2, φc = 0.85 for compression members:

φc Pn = 0.85 × 15.0 = 12.8 kips = 12,800 lb

Check LC1: Pu1 = 4,320 lb ≤ 12,800 lb → OK (utilization = 0.338)

Step 6: Bending strength (wind load)

Per AISI S100-16 Section F3, the nominal flexural strength for the strong axis is controlled by local buckling of the compression flange and web.

The effective section modulus Se for the wind moment case uses a stress gradient on the web (compression on one edge, tension on the other). For the wind case with f = Fy = 50 ksi:

Web in bending: k = 23.9 (stiffened element with stress gradient, ψ = -1.0)
λ_web_bend = (1.052 / √23.9) × 101.0 × 0.0369 = 0.215 × 101.0 × 0.0369 = 0.801 > 0.673
ρ_web_bend = (1 - 0.22/0.801) / 0.801 = 0.726 / 0.801 = 0.906

The compression flange in bending is unstiffened (edge stiffened by the lip):

λ_flange_bend = 1.605 × 14.9 × 0.0369 = 0.882 (same as compression case)
ρ_flange_bend = 0.851

The effective section modulus Se (computed about the shifted neutral axis after effective width reductions) is approximately:

Se = 0.487 in³ (iteration converged)
Mn = Se × Fy = 0.487 × 50 = 24.4 kip-in

Per AISI S100-16 Section F3, φb = 0.90 for laterally braced flexural members:

φb Mn = 0.90 × 24.4 = 21.9 kip-in

Check bending alone: Mu_wind = 4,990 lb-in = 5.0 kip-in ≤ 21.9 kip-in → OK (utilization = 0.228)

Step 7: Combined axial compression and bending

Per AISI S100-16 Section H1.2, the interaction equation for LRFD is:

Pu/(φc Pn) + Mu/(φb Mn) ≤ 1.0  [when Pu/(φc Pn) < 0.15, alternate equation applies]

For the governing load combination LC2 (axial = 3,240 lb, moment = 4,990 lb-in):

Pu/(φc Pn) = 3.24 / 12.8 = 0.253 > 0.15 → use full interaction:
Pu/(φc Pn) + Mu/(φb Mn) ≤ 1.0
0.253 + 5.0/21.9 = 0.253 + 0.228 = 0.481 ≤ 1.0 → OK

Step 8: Serviceability — wall deflection under wind

Per AISI S240-15 (North American Standard for Cold-Formed Steel Structural Framing), wall deflection under wind is limited to L/360 for walls with flexible finishes:

Δ_wind = 5 × w_wind × L⁴ / (384 × E × Ieff)
w_wind = 33.3 lb/ft = 2.78 lb/in
Ieff ≈ 0.80 × Ixg = 0.80 × 2.64 = 2.11 in⁴ (accounting for reduced stiffness at service load)
Δ_wind = 5 × 2.78 × 120⁴ / (384 × 29,500,000 × 2.11)
Δ_wind = 5 × 2.78 × 207,360,000 / (384 × 29.5 × 10⁶ × 2.11)
Δ_wind = 2,882,304,000 / 23,893,776,000 = 0.121 in

Deflection limit: L/360 = 120/360 = 0.333 in > 0.121 in → OK

Results summary

Key takeaways

• Effective width governs CFS capacity. The local buckling reduction can remove 30-50% of the gross cross-section from the effective area calculation. Always verify that your effective width analysis converges — the stress-dependent iteration on λ can require 2-3 cycles for singly symmetric sections.

• Distortional buckling can control. Although global buckling governed for this stud, distortional buckling can be the critical mode for sections with longer lips or wider flanges. The simplified analytical method in AISI S100 gives conservative results; use CUFSM or the Direct Strength Method for optimization.

• Sheathing restraint matters. The wall sheathing (OSB both sides with screws at 12 in o.c.) provides continuous weak-axis bracing and torsional restraint. Without adequate sheathing attachment, the weak-axis unbraced length governs, dramatically reducing axial capacity.

• The interaction equation is linear for Pr/Pc ≥ 0.15. When the axial utilization is high, combined loading drives the design. When axial is low (<0.15), AISI S100 permits a more favorable interaction expression.

• Track-to-stud connection is a separate check. This example covered the stud body only. End reactions must be transferred through the track with adequate bearing and screw shear capacity per AISI S240.

FAQ

What does the 600S162-54 designation mean?

Per AISI S200-12 (North American Standard for Cold-Formed Steel Framing — General Provisions): 600 = nominal web depth in 1/100 inches (6.00 in), S = stud (lipped channel), 162 = nominal flange width in 1/100 inches (1.625 in), and 54 = base steel thickness in mils (0.054 in base metal, 0.0566 in design thickness after coating). The actual design thickness is always larger than the base metal thickness because the coating (typically G60 or G90 galvanizing) adds 0.0013 to 0.0026 in per side.

When does distortional buckling govern instead of local buckling?

Distortional buckling tends to govern when the lip stiffener is relatively short (d/B < 0.25), the flange is wide relative to thickness, or the member is moderately long (intermediate unbraced lengths). For 600S162 members with standard lip proportions (d ≈ 0.5 in), local buckling usually governs for stockier flanges (1.625 in) and distortional becomes critical for wider flanges (2.5 in, 3.5 in). Always verify both modes per AISI S100 Section E4.

How does sheathing attachment affect stud capacity?

Per AISI S240-15, continuous sheathing attached with screws at 12 in o.c. (maximum) provides full weak-axis bracing and partial torsional restraint. If the sheathing is only on one side or the screw spacing exceeds 12 in, the designer must check weak-axis buckling without full bracing and possibly account for torsional-flexural buckling per AISI S100 Section E3.

Can I use the Direct Strength Method instead of the Effective Width Method?

Yes. AISI S100-16 Appendix 1 (Section 2.3) permits the Direct Strength Method (DSM) as an alternative to the Effective Width Method. DSM uses gross section properties and applies separate strength curves for global, local, and distortional buckling based on elastic buckling analysis (typically from CUFSM software). DSM can produce less conservative results because it more accurately captures post-buckling reserve strength, particularly for distortional buckling.

What track thickness is required for this stud?

Per AISI S240-15 Section C2.1.1, the minimum track design thickness for a 54-mil stud is 33 mil (0.0346 in) for non-load-bearing walls and 43 mil (0.0451 in) for load-bearing walls. The track must be at least the same steel grade as the stud. Always check the end reaction bearing length and the track web crippling per AISI S100 Section F4.1.