Cold-Formed Steel Floor Joist Design: AISI S100-16 Worked Example (1200S250-97)

Cold-formed steel floor joists carry superimposed dead and live loads over spans typically ranging from 10 to 30 feet. The 1200S250-97 section — a 12-inch-deep lipped channel with 2.5-inch flanges and 97-mil design thickness — is one of the workhorse CFS joist sections for mid-span residential and light commercial floors. Its deeper web and thicker steel provide substantially more flexural capacity than the 8-inch and 10-inch C-sections commonly used in wall framing.

This worked example demonstrates a complete flexural design check per AISI S100-16 for a simply supported 1200S250-97 floor joist at 16 inches on center spanning 18 feet. You will work through section property derivation, local buckling effective widths for flexure, distortional buckling check, lateral-torsional buckling assessment, shear strength, web crippling at supports, and live-load deflection verification.

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 — 1200S250-97 (gross)

The 1200S250-97 is a lipped C-section with a nominal 12-inch web depth, 2.5-inch flange, 0.625-inch lip, and 0.1017-inch design thickness (97 mil base steel, 0.0970 in base metal). Gross properties are computed using the AISI S100 centerline method.

Step 1: Factored loads (LRFD)

Tributary width: 16/12 = 1.333 ft
Superimposed dead: w_D_sup = 15 psf × 1.333 ft = 20.0 lb/ft
Self-weight: w_D_self = 2.3 lb/ft
Total dead load: w_D = 22.3 lb/ft
Live load: w_L = 40 psf × 1.333 ft = 53.3 lb/ft

LRFD load combination (ASCE 7-22):

wu = 1.2 w_D + 1.6 w_L = 1.2 × 22.3 + 1.6 × 53.3
wu = 26.8 + 85.3 = 112.1 lb/ft = 9.34 lb/in

Mu = wu × L² / 8 = 9.34 × 216² / 8 = 9.34 × 46,656 / 8
Mu = 435,754 / 8 = 54,469 lb-in = 54.5 kip-in

Step 2: Flexural strength — local buckling effective section

Per AISI S100-16 Appendix 1, the effective section modulus Se is computed for the stress distribution on the cross-section at yield. The compression flange and the compression portion of the web are reduced for local buckling.

Compression flange (unstiffened element under uniform compression):

The flat width (centerline) of the compression flange:

w_f = B - t - R - d_centerline = 2.50 - 0.1017 - 0.1526 - 0.625 = 1.621 in
w_f/t = 1.621 / 0.1017 = 15.94

For a flange with a 0.625-inch lip providing edge stiffening, the plate buckling coefficient depends on the lip stiffener adequacy per AISI S100 Section B4. The lip slenderness:

d_flat = 0.625 - 0.1526 = 0.472 in
d_flat/t = 0.472 / 0.1017 = 4.64

Per Table B4-1, for a lip with d/t ≤ 14 and Is/Ia ≥ 1.0, the flange is adequately stiffened and k = 4.0 (stiffened element behavior). Compute the lip stiffener moment of inertia:

Is = d³ × t / 12 = 0.472³ × 0.1017 / 12 = 0.105 × 0.1017 / 12 = 0.000890 in⁴
Ia = 399 × t⁴ × [(w_f/t)/S - √(ku/4)]³ (iterative; for w_f/t = 15.94, Ia ≈ 0.000520 in⁴)
Is / Ia = 0.000890 / 0.000520 = 1.71 > 1.0 → flange is stiffened, k = 4.0

With k = 4.0:

λ_flange = (1.052 / √4.0) × (w_f/t) × √(f/E) = 0.526 × 15.94 × √(50/29,500)
λ_flange = 0.526 × 15.94 × 0.0412 = 0.345 < 0.673 → fully effective (ρ = 1.0)

Since λ < 0.673, the compression flange is fully effective. No width reduction needed.

Web element (stiffened with stress gradient):

The web flat height (centerline):

h = D - 2(t + R) = 12.00 - 2 × (0.1017 + 0.1526) = 12.00 - 0.509 = 11.491 in
h/t = 11.491 / 0.1017 = 113.0

For a stiffened element with stress gradient (web in bending, ψ = -1.0 at first yield), the plate buckling coefficient is:

k_web = 4 + 2 × (1 - ψ)³ + 2 × (1 - ψ) = 4 + 2 × (1 - (-1.0))³ + 2 × (1 - (-1.0))
k_web = 4 + 2 × 8 + 2 × 2 = 4 + 16 + 4 = 24.0 (for ψ = -1.0 per Table B2.3-2)

For bending, use k = 23.9 per AISI S100 Table B2.3-2 for ψ = -1.0:

f = Fy = 50 ksi (stress at extreme compression fiber at first yield)
λ_web = (1.052 / √23.9) × 113.0 × 0.0412 = 0.215 × 113.0 × 0.0412
λ_web = 0.215 × 4.66 = 1.00 > 0.673 → ineffective portion

Effective web depth factor:

ρ_web = (1 - 0.22/λ) / λ = (1 - 0.22/1.00) / 1.00 = 0.78

The effective width of the compression portion of the web (from the extreme compression fiber to the neutral axis, approximately h/2 for first yield):

be_comp = ρ_web × (h/2) = 0.78 × 5.746 = 4.48 in
ineffective_comp = 5.746 - 4.48 = 1.27 in

The tension portion of the web is fully effective. After shifting the neutral axis to balance effective areas, the effective section modulus converges to:

Se = 5.62 in³ (after 3 iterations)
Mn_local = Se × Fy = 5.62 × 50 = 281 kip-in

Step 3: Distortional buckling check

Per AISI S100-16 Section F3.2, the distortional buckling moment capacity uses the elastic distortional buckling stress Fd for the compression flange-lip assembly. For the 1200S250-97 section in bending, computed via CUFSM or the simplified analytical method:

Fd = 38.1 ksi (critical elastic distortional buckling stress)
λd = √(Fy / Fd) = √(50 / 38.1) = √1.31 = 1.146

Since λd ≤ 1.673 for the strength curve (Table F3.2-1):

Mnd = My × (1 - 0.22/λd) / λd

Wait — this uses the AISI global curve. Actually, for the DSM approach per AISI S100 Appendix 1 Section 2.3.2:

Mnd = My for λd ≤ 0.673
Mnd = (1 - 0.22 × (Mcr_d/My)^(-0.5)) × (Mcr_d/My)^(-0.5) × My for λd > 0.673

Using the simplified explicit expression for the DSM distortional buckling curve:

Mcr_d = Sxc × Fd = 6.10 × 38.1 = 232.4 kip-in
λd = √(My / Mcr_d) = √(6.10 × 50 / 232.4) = √(305 / 232.4) = √1.312 = 1.146
Mnd = My × (1 - 0.22 × (1/λd)) / λd = 305 × (1 - 0.22/1.146) / 1.146
Mnd = 305 × (1 - 0.192) / 1.146 = 305 × 0.808 / 1.146 = 215 kip-in

Step 4: Lateral-torsional buckling

The top flange is continuously braced by the plywood subfloor screwed at 12 in o.c. Per AISI S100-16 Section F3.1.2, for a C-section with one flange through-fastened to sheathing, LTB is considered prevented. The nominal flexural strength for LTB:

Lb = 0 (continuously braced, unbraced length for twist is zero)
Mn_LTB = ∞ (does not govern)

Verification: Per the SSMA Technical Guide and AISI S240, a screw-attached structural sheathing (plywood or OSB) at 12 in o.c. maximum provides continuous lateral restraint. The torsional restraint from the sheathing diaphragm is adequate for the full flexural capacity.

Step 5: Nominal flexural strength

Per AISI S100-16 Section F3.1:

Mn = min(Mn_local, Mnd, Mn_LTB) = min(281, 215, ∞) = 215 kip-in
φb = 0.90 (LRFD for laterally braced flexural members)
φb Mn = 0.90 × 215 = 194 kip-in

Check flexure: Mu = 54.5 kip-in ≤ 194 kip-in → OK (utilization = 0.281)

The distortional buckling limit state controls. This is typical for 12-inch-deep CFS joists where the compression flange-lip assembly governs the ultimate capacity before local web buckling fully develops.

Step 6: Shear strength

Per AISI S100-16 Section G2, the nominal shear strength Vn of a web without transverse stiffeners is governed by web shear buckling:

h/t = 113.0
Ekv check: (Ek_v / Fy)^(0.5) = (29,500 × 5.34 / 50)^(0.5)
              = (3,150)^(0.5) = 56.1
h/t = 113.0 > 56.1 → elastic shear buckling

The shear buckling coefficient for an unstiffened web is k_v = 5.34.

Vn = (0.905 × E × k_v × t³) / h = (0.905 × 29,500 × 5.34 × 0.1017³) / 11.491
Vn = (0.905 × 29,500 × 5.34 × 0.001052) / 11.491
Vn = 149.7 / 11.491 = 13.0 kips

Per AISI S100-16 Section G2, φv = 0.95:

φv Vn = 0.95 × 13.0 = 12.4 kips

Vu = wu × L / 2 = 112.1 × 18 / 2 = 1,009 lb = 1.01 kips
Vu = 1.01 kips ≤ 12.4 kips → OK (utilization = 0.081)

Step 7: Web crippling at supports

Per AISI S100-16 Section F4.1, web crippling is checked for the end reaction. For a C-section with the load applied to one flange and reacted by the opposite flange (EOF condition):

The web crippling strength depends on the bearing length, web slenderness, and inside bend radius. For an EOF condition with bearing length N = 3.5 in:

h/t = 113.0, R/t = 0.1526/0.1017 = 1.50, N/t = 3.5/0.1017 = 34.4, N/h = 3.5/11.491 = 0.305

Pn_crippling = C × t² × Fy × sin(θ) × (1 - Cr × √(R/t)) × (1 + Cl × √(N/t)) × (1 - Cw × √(h/t))

Using the coefficients from AISI S100 Table F4.1-1 for EOF end-one-flange loading of C-sections:

C = 4.0, Cr = 0.14, Cl = 0.35, Cw = 0.02

Pn_cr = 4.0 × 0.1017² × 50 × 1.0 × (1 - 0.14 × √1.50) × (1 + 0.35 × √34.4) × (1 - 0.02 × √113.0)
       = 4.0 × 0.01034 × 50 × (1 - 0.171) × (1 + 0.35 × 5.86) × (1 - 0.02 × 10.63)
       = 4.0 × 0.01034 × 50 × 0.829 × (1 + 2.05) × (1 - 0.213)
       = 4.0 × 0.01034 × 50 × 0.829 × 3.05 × 0.787
       = 4.0 × 0.01034 × 50 × 1.988
       = 4.11 kips

Per AISI S100-16, φw = 0.75 for web crippling:

φw Pn_cr = 0.75 × 4.11 = 3.08 kips

End reaction check: Ru = Vu = 1.01 kips ≤ 3.08 kips → OK (utilization = 0.328)

Step 8: Deflection — serviceability (live load only)

Per AISI S240-15, floor joist live load deflection is limited to L/360:

w_L = 53.3 lb/ft = 4.44 lb/in

For a simply supported joist with uniformly distributed load:
Δ_LL = 5 × w_L × L⁴ / (384 × E × Ieff)

The effective moment of inertia accounts for partial yielding at service load. At service live load, the extreme fiber stress is:

M_LL = w_L × L² / 8 = 4.44 × 216² / 8 = 4.44 × 46,656 / 8 = 25,895 lb-in
f_LL = M_LL / Sxg = 25,895 / 6.10 = 4,245 psi = 4.25 ksi

Since f_LL << Fy, the section is fully effective at service load: Ieff = Ixg = 36.6 in⁴.

Δ_LL = 5 × 4.44 × 216⁴ / (384 × 29,500,000 × 36.6)
Δ_LL = 5 × 4.44 × 2,176,782,336 / (384 × 29,500,000 × 36.6)
Δ_LL = 48,312,456,000 / 414,643,200,000 = 0.117 in

Deflection limit: L/360 = 216/360 = 0.600 in > 0.117 in → OK (utilization = 0.194)

Total load deflection for floor performance (L/240):

w_total = w_D + w_L = 22.3 + 53.3 = 75.6 lb/ft = 6.30 lb/in
Δ_total = Δ_LL × (6.30/4.44) = 0.117 × 1.419 = 0.166 in
L/240 = 216/240 = 0.900 in > 0.166 in → OK

Step 9: Floor vibration (qualitative assessment)

Per AISI S240-15 Commentary, CFS floor joists at spans exceeding 15 ft should be evaluated for footfall vibration. For the 1200S250-97 at 18-ft span, the fundamental frequency is:

f_n ≈ 1.57 × √(E × I × g / (w × L⁴))
where w = total dead + 0.25 × live (for vibration mass)
w = (22.3 + 0.25 × 53.3) / 12 = (22.3 + 13.3) / 12 = 2.97 lb/in
f_n ≈ 1.57 × √(29.5 × 10⁶ × 36.6 × 386 / (2.97 × 216⁴))
f_n ≈ 1.57 × √(4.17 × 10¹¹ / (2.97 × 2.18 × 10⁹))
f_n ≈ 1.57 × √(4.17 × 10¹¹ / 6.47 × 10⁹) = 1.57 × √64.4 = 1.57 × 8.03 = 12.6 Hz

f_n = 12.6 Hz > 8 Hz (AISI S240 recommended minimum). The joist floor is expected to perform adequately for residential footfall vibration.

Results summary

Key takeaways

• Distortional buckling controls CFS joist flexure. Even at 97 mil thickness, the 12-inch-deep joist's compression flange-lip assembly buckles before the web local buckling limit state develops. The distortional mode is the most common flexural limit state for deeper CFS joists.

• Shear is rarely critical in CFS joists. The 0.081 shear utilization here is typical. CFS joists are flexure-governed; shear only controls in short-span, heavily loaded conditions or when large web openings are present.

• Web crippling is the most sensitive detail. The end bearing condition depends critically on bearing length N. Increasing the track flange width from 3.5 to 5.0 inches nearly doubles the crippling capacity. Always confirm the actual bearing length from the construction drawings.

• Sheathing attachment provides both bracing and diaphragm action. Screwing the subfloor to the top flange at 12 in o.c. prevents lateral-torsional buckling entirely and contributes to floor diaphragm stiffness. Missing screws or overdriven fasteners can compromise both functions.

• Vibration governs long spans. At 18 ft, the 1200S250-97 satisfies the 8 Hz minimum. For spans over 20 ft with 97-mil joists, consider deeper sections (14 in), closer spacing (12 in o.c.), or a concrete topping to add mass and increase the fundamental frequency.

FAQ

When should I use 1200S250-97 instead of 1200S200-68?

The 1200S250-97 provides approximately 65% more moment capacity and 80% more web crippling capacity than the 1200S200-68 (68 mil). Use the 97-mil section when: (1) the span exceeds 16 ft under residential loading, (2) heavier floor finishes (tile, stone) add substantial dead load, (3) vibration performance is critical, or (4) bearing length is limited to under 4 inches. The 68-mil section is adequate for spans under 14 ft with carpet or vinyl flooring.

What floor sheathing is required for CFS joists?

Per AISI S240-15, the minimum floor sheathing is 23/32-inch APA-rated tongue-and-groove plywood or OSB screwed to each joist flange at 12 in o.c. maximum. Screws must be #8 minimum with sufficient penetration through the steel. Glued-and-screwed assemblies provide superior vibration performance and are recommended for spans exceeding 16 ft.

How do I handle web openings in CFS joists?

Web openings for mechanical services should be located in the middle third of the span where shear is low. AISI S100-16 Section G5 provides the web opening provisions. For a 1200S250-97 joist, rectangular openings up to 6 in x 12 in may be punched in the web centerline without reinforcement if the opening edge is at least 2 in from the flange and the opening length does not exceed 24 in. Larger openings require stiffeners per the SSMA Technical Guide.

Is the AISI S100 Direct Strength Method better for joist design?

The Direct Strength Method (DSM) typically provides 5–15% higher capacities than the Effective Width Method for CFS joists, particularly for sections governed by distortional buckling. DSM uses elastic buckling analysis results (from CUFSM) with separate design curves for local, distortional, and global buckling. Many manufacturers publish DSM-based capacity tables for their proprietary joist sections. Always confirm whether your project specification permits DSM before using it.

What fire protection is needed for CFS floor joists?

Per IBC Section 721.6, CFS floor assemblies require fire-resistance-rated construction for multi-family and commercial buildings. A typical 1-hour assembly uses 5/8-inch Type X gypsum board on the ceiling with resilient channels, mineral wool insulation in the joist cavity, and 3/4-inch plywood subfloor above. The steel temperature rise under fire exposure may require reduced design strength; refer to AISI S100-16 Appendix 4 for elevated-temperature design provisions.