Steel Framed Walls — Steel Plate Shear Walls and CFS Stud Walls

Steel framed walls serve two distinct roles in construction: as lateral force-resisting elements (steel plate shear walls, SPSW) and as non-structural or lightly-structural partitions (cold-formed steel stud walls). This reference covers both systems, their design provisions, and the practical engineering checks required for each.

Steel plate shear walls (SPSW) per AISC 341

A steel plate shear wall consists of a thin steel infill plate bounded by boundary columns (called vertical boundary elements, VBE) and boundary beams (horizontal boundary elements, HBE). Under lateral load, the thin plate buckles and develops a diagonal tension field that resists story shear.

Design principles

The infill plate is designed to yield in tension along the diagonal. The capacity per AISC 341-22 Section F5 is:

Vn = 0.42 * Fy * tw * Lcf * sin(2 * alpha)

where:

Worked example — SPSW infill plate sizing

Given: 3-story building, SDC D. Story shear at Level 2: V_u = 480 kips. Bay width = 20 ft (L_cf = 19 ft = 228 in. after deducting column flanges). Story height = 13 ft. A36 infill plate. R = 7, Omega_0 = 2 (SPSW system factors per ASCE 7 Table 12.2-1).

Step 1 — Required plate thickness: Assuming alpha = 42 degrees: sin(2 * 42) = sin(84) = 0.995

phi _ V_n = phi _ 0.42 _ Fy _ tw * Lcf * sin(2 _ alpha) 480 = 0.90 _ 0.42 _ 36 _ t*w * 228 _ 0.995 480 = 0.90 _ 3,445 _ t_w 480 = 3,100 * t_w t_w = 480 / 3,100 = 0.155 in.

Use 3/16 in. (0.1875 in.) plate. This is close to the practical minimum for welded plate construction.

Step 2 — VBE (column) design: The boundary columns must resist the horizontal component of the tension field plus frame action. Per AISC 341 Section F5.4b, the VBE must be designed for the forces corresponding to full yielding of the infill plates above and below, using Ry * Fy (expected yield = 1.5 * 36 = 54 ksi for A36).

Step 3 — HBE (beam) design: The boundary beams must transfer the tension field forces from the plate to the columns. They experience both the uniform pull from the tension field (acting like a beam on elastic foundation) and the frame moments from lateral drift.

Cold-formed steel (CFS) stud walls

CFS stud walls use light-gage steel framing (typically 20-12 gage, 33-50 ksi) for interior partitions, exterior curtain walls, and load-bearing walls in low-rise construction.

Design per AISI S100 and S240

CFS studs act as columns subject to axial load and/or beam-columns when wind or seismic out-of-plane loading is present. Key limit states include:

Worked example — CFS exterior stud wall

Given: 600S162-54 stud (6 in. deep, 1.625 in. flange, 54 mil = 0.054 in., Fy = 50 ksi). Stud height = 12 ft. Studs at 16 in. o.c. Wind pressure = 25 psf. Axial dead load from parapet = 200 lb per stud. Bridging at mid-height and third points.

Step 1 — Section properties (from AISI S100 tables): A_g = 0.840 in.^2, I_x = 5.91 in.^4, S_x = 1.97 in.^3, r_x = 2.65 in., r_y = 0.727 in.

Step 2 — Axial capacity (weak axis governs with bridging at third points): Unbraced length for weak axis: L_y = 12/3 = 4 ft = 48 in. (bridging at third points) KL/r_y = 1.0 * 48 / 0.727 = 66.0

F*e = pi^2 * E / (KL/r)^2 = pi^2 _ 29500 / 66^2 = 291,100 / 4,356 = 66.8 ksi lambda_c = sqrt(Fy / F_e) = sqrt(50 / 66.8) = 0.865

Since lambda*c <= 1.5: F_n = 0.658^(lambda_c^2) * Fy = 0.658^0.748 _ 50 = 0.718 * 50 = 35.9 ksi

phi _ P_n = 0.85 _ 35.9 * 0.840 = 25.6 kips >> 0.200 kips (axial demand trivially low)

Step 3 — Flexural capacity (strong axis, wind loading): M*u = w * L^2 / 8 = (25 _ 16/12) / 1000 _ 144^2 / 8 = 0.0333 _ 20,736 / 8 = 86.3 kip-in. Wait — recalculate:

w = 25 psf _ (16/12 ft) = 33.3 lb/ft = 0.0333 kip/ft M_u = 0.0333 _ 12^2 / 8 = 0.0333 * 144 / 8 = 0.600 kip-ft = 7.2 kip-in.

phi _ M_n = 0.90 _ Fy _ S_e (effective section modulus for local buckling check) For this stud with Fy = 50 ksi, S_e approximately = 1.85 in.^3 (slight reduction from local buckling) phi _ M*n = 0.90 * 50 _ 1.85 = 83.3 kip-in. >> 7.2 kip-in. (OK)

Step 4 — Combined axial + bending interaction (AISI S100 Section C5.2.2): Pu / (phi * Pn) + M_u / (phi * M_n) = 0.200/25.6 + 7.2/83.3 = 0.008 + 0.086 = 0.094 < 1.0 (OK — stud is adequate with significant reserve)

Code comparison

Aspect AISC 341 (SPSW) / AISI S100 (CFS) EN 1998-1 / EN 1993-1-3 AS 4100 / AS 4600 CSA S16 / CSA S136
SPSW provisions AISC 341 Section F5 EN 1998-1 Section 7 (limited) Not widely codified CSA S16 Clause 27.9
SPSW R factor 7.0 q approximately 5-6 Not defined (rare use) Rd * Ro = 5.0 * 1.6
CFS stud design AISI S100-22 EN 1993-1-3 AS/NZS 4600 CSA S136-16
Distortional buckling AISI S100 Appendix 2 EN 1993-1-3 Section 5.5 AS 4600 Section 3 CSA S136 Section C
CFS seismic AISI S400 EN 1998-1 (general) No specific standard AISI S400 (adopted)

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Disclaimer

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.