CSA S16 Beam Design — Worked Example

Quick Reference: This worked example designs a W610x125 beam in Grade 350W for a 8 m simple span under NBCC 2020 factored loads. Checks include flexure (Cl. 13.5/13.6), shear (Cl. 13.4), deflection (L/360 live load), and lateral bracing requirements (Cl. 13.6.5).

Comprehensive worked examples are essential for understanding CSA S16 beam design. This guide takes you through a complete beam design from loads to final section selection, with all intermediate calculations shown.

Design Problem

Given:

Objective: Select an adequate W-shape beam and verify all limit states.

Step 1 — Factored Loads Per NBCC 2020

Unfactored loads:

Dead load: w_D = 3.0 kPa × 4.0 m = 12.0 kN/m Self-weight: Estimate W610x125 → 1.25 kN/m (W610x125 = 125 kg/m × 9.81 m/s² / 1,000) Total dead: w_D = 12.0 + 1.25 = 13.25 kN/m

Live load: w_L = 4.0 kPa × 4.0 m = 16.0 kN/m

Factored load (ULS):

Governing combination: NBCC Comb. 2 — 1.25D + 1.5L

w_f = 1.25 × 13.25 + 1.5 × 16.0 = 16.56 + 24.0 = 40.56 kN/m

Maximum factored moment:

M_f = w_f × L² / 8 = 40.56 × 8.0² / 8 = 40.56 × 64 / 8 = 324.5 kN·m

Maximum factored shear:

V_f = w_f × L / 2 = 40.56 × 8.0 / 2 = 162.2 kN

Step 2 — Select Trial Section

Required plastic modulus (assuming Class 1 or 2 section):

Z_x_req = M_f / (φ × F_y) = 324.5 × 10⁶ / (0.90 × 350) = 1.030 × 10⁶ mm³

Try W610x125 (US equivalent W24x84):

Property Value
d 612 mm
b_f 229 mm
t_f 19.6 mm
t_w 11.9 mm
Z_x 3,330 × 10³ mm³
S_x 2,970 × 10³ mm³
I_x 908 × 10⁶ mm⁴
r_y 49.8 mm
Mass 125 kg/m → 1.23 kN/m

Actual self-weight = 1.23 kN/m vs estimated 1.25 kN/m — close enough. No iteration needed.

Step 3 — Section Classification (Cl. 11.2, Table 2)

Flange classification (for flexure):

b = b_f / 2 = 229 / 2 = 114.5 mm (half-flange for I-section in flexure) b/t = 114.5 / 19.6 = 5.84

Limits for F_y = 350 MPa:

Web classification (for flexure):

h = d - 2 × t_f = 612 - 2 × 19.6 = 572.8 mm h/w = 572.8 / 11.9 = 48.1

Limits for F_y = 350 MPa:

Overall classification:

Both flange and web are Class 1. The section is Class 1 and can develop the full plastic moment with adequate rotation capacity.

Step 4 — Flexural Resistance (Cl. 13.5, 13.6)

Plastic moment capacity:

M_p = Z_x × F_y = 3,330 × 10³ × 350 / 10⁶ = 1,165.5 kN·m

φ × M_p = 0.90 × 1,165.5 = 1,049.0 kN·m

Check lateral-torsional buckling:

First, determine if LTB governs. The limiting unbraced length L_u:

r_t = radius of gyration of compression flange + 1/3 web in compression For doubly symmetric I-section: r_t ≈ r_y for simplification

L_u = 0.68 × sqrt(E/F_y) × r_y ≈ 1.1 × r_y × sqrt(E/F_y)

L_u = 1.1 × 49.8 × sqrt(200,000 / 350) = 1.1 × 49.8 × 23.9 = 1,309 mm = 1.31 m

Actual unbraced length L_b = 2.67 m > L_u = 1.31 m → LTB check required.

Elastic LTB moment (Cl. 13.6.5):

For W610x125:

L_b = 2,670 mm

M_u = ω₂ × (π/L_b) × sqrt(E × I_y × G × J + (π × E / L_b)² × I_y × C_w)

ω₂ = 1.14 (uniformly distributed load over the span, simple supports — from Table 3-2, CISC Handbook)

First term: E × I_y × G × J = 200,000 × 39.5 × 10⁶ × 77,000 × 1,230 × 10³ = 200,000 × 39.5 × 10⁶ × 77,000 × 1,230 × 10³ = 200,000 × 77,000 × 39.5 × 1,230 × 10⁹ = 1.54 × 10¹⁰ × 4.859 × 10⁴ × 10⁹ = 7.48 × 10²³ N²·mm⁶

Second term: (π × E / L_b)² × I_y × C_w = (π × 200,000 / 2,670)² × 39.5 × 10⁶ × 2,990 × 10⁹ = (235.5)² × 39.5 × 2,990 × 10¹⁵ = 55,460 × 1.181 × 10²⁰ = 6.55 × 10²⁴ N²·mm⁶

Combined: 7.48 × 10²³ + 6.55 × 10²⁴ = 7.30 × 10²⁴

sqrt(7.30 × 10²⁴) = 2.70 × 10¹² N·mm³

M_u = 1.14 × (π / 2,670) × 2.70 × 10¹² = 1.14 × 0.001177 × 2.70 × 10¹² = 3.62 × 10⁹ N·mm = 3,620 kN·m

Check LTB resistance:

0.67 × M_p = 0.67 × 1,165.5 = 780.9 kN·m

M_u = 3,620 kN·m > 780.9 kN·m → inelastic LTB region applies:

M_r = 1.15 × φ × M_p × (1 - 0.28 × M_p / M_u)

= 1.15 × 1,049.0 × (1 - 0.28 × 1,165.5 / 3,620) = 1,206.4 × (1 - 0.0902) = 1,206.4 × 0.9098 = 1,097.4 kN·m

BUT M_r cannot exceed φ × M_p = 1,049.0 kN·m

Therefore: M_r = 1,049.0 kN·m

Check:

M_f / M_r = 324.5 / 1,049.0 = 0.309 ✓ — Flexure is adequate with significant reserve.

Step 5 — Shear Resistance (Cl. 13.4)

Web slenderness for shear:

h/w = 48.1 Limit for shear buckling: 439 × sqrt(k_v / F_y) where k_v = 5.34 (unstiffened web)

439 × sqrt(5.34 / 350) = 439 × 0.1235 = 54.2

48.1 < 54.2 → Shear buckling does not govern. Use basic shear equation.

Shear resistance:

V_r = φ × 0.60 × F_y × d × t_w

= 0.90 × 0.60 × 350 × 612 × 11.9 / 1,000 = 0.90 × 0.60 × 350 × 7,283 / 1,000 = 1,377 kN

Check:

V_f / V_r = 162.2 / 1,377 = 0.118 ✓ — Shear is not critical for this W-shape under uniform loading.

Step 6 — Deflection Check (Serviceability)

Live load deflection (governs per NBCC):

w_L = 16.0 kN/m (unfactored)

δ_LL = 5 × w_L × L⁴ / (384 × E × I_x)

= 5 × 16.0 × (8,000)⁴ / (384 × 200,000 × 908 × 10⁶)

= 5 × 16.0 × 4.096 × 10¹⁵ / (384 × 200,000 × 908 × 10⁶)

= 3.277 × 10¹⁷ / 6.973 × 10¹³

= 4,700 mm⁴ · kN/m² / mm⁴... Let me recalculate carefully.

= 5 × 16.0 × 4,096 × 10¹² / (384 × 200,000 × 908 × 10⁶)

= 5 × 16.0 × 4.096 × 10¹⁵¹⁴... let me be precise:

5 × 16.0 = 80 80 × 8,000⁴ = 80 × 4.096 × 10¹⁵ = 3.277 × 10¹⁷

384 × E × I = 384 × 200,000 × 908 × 10⁶ = 384 × 1.816 × 10¹⁴ = 6.973 × 10¹⁶

δ_LL = 3.277 × 10¹⁷ / 6.973 × 10¹⁶ = 4.70 mm × (kN/m → N/mm)...

Actually, careful with units: w = 16.0 kN/m = 16.0 N/mm L = 8,000 mm E = 200,000 N/mm² I = 908 × 10⁶ mm⁴

δ_LL = 5 × 16.0 × 8,000⁴ / (384 × 200,000 × 908 × 10⁶)

5 × 16 = 80 80 × 8,000⁴ = 80 × 4.096 × 10¹⁵ = 3.277 × 10¹⁷ N·mm³

384 × 200,000 = 76,800,000 76,800,000 × 908 × 10⁶ = 6.973 × 10¹⁶ N·mm²

δ_LL = 3.277 × 10¹⁷ / 6.973 × 10¹⁶ = 4.70 mm

Allowable live load deflection:

L / 360 = 8,000 / 360 = 22.2 mm

δ_LL = 4.70 mm < 22.2 mm ✓ — Live load deflection is well within limits.

Total load deflection (SLS-1):

w_total = 13.25 + 16.0 = 29.25 kN/m (unfactored)

δ_total = δ_LL × (29.25 / 16.0) = 4.70 × 1.828 = 8.59 mm

L / 240 = 8,000 / 240 = 33.3 mm

8.59 mm < 33.3 mm ✓ — Total load deflection is adequate.

Step 7 — Lateral Bracing Requirements (Cl. 13.6.2)

Bracing spacing:

Maximum bracing spacing for Class 1 sections (Cl. 13.6.2):

L_b_max = 0.86 × r_y × sqrt(E / F_y) × (M_p / M_f) for segments with moment gradient

For the end segment (simple support to first brace at 2.67 m): Moment at brace = w × x × (L - x) / 2 where x = 2.67 m M_at_brace = 40.56 × 2.67 × (8.0 - 2.67) / 2 = 40.56 × 2.67 × 5.33 / 2 = 288.7 kN·m

M_p / M_at_brace = 1,165.5 / 288.7 = 4.04

L_b_max = 0.86 × 49.8 × sqrt(200,000 / 350) × 4.04 = 0.86 × 49.8 × 23.9 × 4.04 = 4,134 mm = 4.13 m

The actual bracing spacing (2.67 m) is less than the maximum (4.13 m) → OK.

However, for compactness in design, the actual bracing spacing of 2.67 m already satisfies L_u (1.31 m) check completed above, and the LTB check confirmed the section is adequate.

Step 8 — Final Check Summary

Check Demand Capacity Ratio Status
Flexure (M_r) 324.5 kN·m 1,049.0 kN·m 0.31 ✓
Shear (V_r) 162.2 kN 1,377 kN 0.12 ✓
Live load deflection 4.70 mm 22.2 mm 0.21 ✓
Total deflection 8.59 mm 33.3 mm 0.26 ✓
LTB (M_r) 324.5 kN·m 1,049.0 kN·m 0.31 ✓

Selected section: W610x125 (Grade 350W) with intermediate bracing at 2.67 m spacing.

Design Optimisation

The W610x125 is conservatively sized (31% flexural utilisation). A lighter section could be considered if deflection does not govern:

Try W530x92 (US equivalent W21x62):

Deflection check: I_x for W530x92 = 552 × 10⁶ mm⁴ δ_LL = 4.70 × (908 / 552) = 7.73 mm < 22.2 mm ✓

LTB check (L_b = 2.67 m): r_y = 50.1 mm (similar to W610x125) LTB resistance likely adequate for this short unbraced length.

W530x92 would be adequate with 45% flexural utilisation. The W610x125 was selected to demonstrate the full design procedure for a heavier section, but W530x92 would be more economical in practice.

Related Pages

Frequently Asked Questions

When should intermediate lateral bracing be provided for a steel beam?

Intermediate lateral bracing is required when the unbraced length L_b exceeds the limiting unbraced length L_u. For a W610x125 in 350W, L_u ≈ 1.31 m. Without intermediate bracing, the full 8.0 m span would require an LTB check with ω₂ = 1.14 and L = 8.0 m, resulting in a much lower M_r. For this example, bracing at 2.67 m (one-third points) ensures the beam can develop its full plastic moment. In practice, steel deck with puddle welds at ~300 mm spacing provides continuous lateral restraint to the top (compression) flange, eliminating the need for separate bracing. Unbraced beams (such as crane girders or exposed roof beams) require explicit LTB verification.

How do I check combined bending and shear per CSA S16?

CSA S16 Clause 14.6 requires interaction checks when V_f exceeds 60% of V_r. The interaction equation is: M_f / M_r + 0.6 × V_f / V_r ≤ 1.33 for V_f / V_r > 0.6. In this worked example, V_f / V_r = 0.118, far below the 0.6 threshold. Shear rarely governs for standard W-shape beams under uniform loading — the interaction check is typically only relevant for beams with heavy concentrated loads near supports (transfer beams, crane girders, and deep plate girders with thin webs). For plate girders (h/w > 440/sqrt(F_y)), the shear resistance may be governed by post-buckling tension field action per Cl. 14.4, and combined bending and shear must always be checked per Cl. 14.5.

What live load deflection limit does NBCC 2020 specify for floors?

NBCC 2020 does not prescribe specific deflection limits — these are established by the engineer of record based on occupancy requirements and industry practice. The commonly used L/360 for live load deflection of floor beams comes from the CSA S16 Commentary (Appendix D) and industry standards. More stringent limits may apply: L/480 for floors with brittle finishes (terrazzo, stone tile, thin-set ceramic), L/600 for vibration-sensitive occupancies (laboratories, operating rooms, gymnasia). The Canadian Sheet Steel Building Institute (CSSBI) provides additional guidance for steel deck floor vibration criteria. For this office beam example, L/360 is appropriate for the 325 Pa partition load allowance.

How is the moment gradient factor ω₂ determined for Canadian beam design?

CSA S16 references ω₂ from Table 3-2 of the CISC Handbook of Steel Construction. The factor accounts for the shape of the bending moment diagram along the unbraced segment: ω₂ = 1.00 for uniform moment (worst case), 1.14 for uniformly distributed load on a simple span, 1.35 for a central point load, and up to 1.75 for double curvature bending (equal and opposite end moments). The ω₂ factor in CSA S16 is conceptually identical to AISC's C_b factor but uses slightly different reference values. Unlike AISC, CSA S16 provides ω₂ in a single table rather than a formula, making it simpler to use. For segments with linear moment gradient, ω₂ can be calculated as 1.75 + 1.05β + 0.3β² where β = |M₁|/|M₂| ≤ 1.0 at the segment end with the smaller-to-larger moment ratio.

This page is for educational reference. Beam design per CSA S16-19 and NBCC 2020. Verify section properties against current CISC Handbook of Steel Construction. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent P.Eng. verification.