Design Problem

Given:

• Column section: W360x262 (US equivalent W14x176)
• Column height: 4.5 m (base to beam centreline)
• End conditions: Fixed base, pinned top (K_x = K_y = 1.0 for conservative design)
• Steel grade: CSA G40.21 350W
• Factored axial load: C_f = 4,500 kN (from NBCC ULS combination)
• Factored moment about major axis: M_fx = 120 kN·m (from frame action under wind)
• Factored moment about minor axis: M_fy = 0 kN·m
• The column is part of a braced frame (non-sway)

Step 1 — Section Properties (W360x262)

Section Classification Check (Cl. 11.2)

Flange: b = (b_f - t_w) / 2 = (374 - 21.6) / 2 = 176.2 mm b/t = 176.2 / 31.8 = 5.54 Class 1 limit (compression): 145/√350 = 7.75 5.54 < 7.75 → Class 1 flange ✓

Web: h = d - 2 × t_f = 368 - 2 × 31.8 = 304.4 mm h/w = 304.4 / 21.6 = 14.1 Class 1 limit (compression): 335/√350 = 17.9 14.1 < 17.9 → Class 1 web ✓

Overall: Class 1 section — full plastic resistance can be developed.

Step 2 — Slenderness Ratio

Check both axes:

Major axis: (K_x × L) / r_x = 1.0 × 4,500 / 178 = 25.3 Minor axis: (K_y × L) / r_y = 1.0 × 4,500 / 90.9 = 49.5 ← governs

The minor axis governs because r_y = 90.9 mm < r_x = 178 mm. This is typical for W-shape columns — the weak axis controls unless the effective lengths differ significantly.

Maximum slenderness check (Cl. 10.2.3):

K×L/r = 49.5 < 200 ✓ — within the code limit.

Step 3 — Non-Dimensional Slenderness λ

λ = (K×L/r) × sqrt(F_y / (π² × E))

= 49.5 × sqrt(350 / (π² × 200,000))

= 49.5 × sqrt(350 / 1,973,900)

= 49.5 × sqrt(1.773 × 10⁻⁴)

= 49.5 × 0.01332

= 0.659

Step 4 — Compression Resistance C_r (Cl. 13.3)

C_r = φ × A × F_y × (1 + λ^(2n))^(-1/n)

Where φ = 0.90, n = 1.34

Compute λ^(2n):

λ^(2 × 1.34) = λ^(2.68)

ln(0.659^(2.68)) = 2.68 × ln(0.659) = 2.68 × (-0.4169) = -1.117

0.659^(2.68) = e^(-1.117) = 0.327

Compute (1 + λ^(2n))^(-1/n):

1 + 0.327 = 1.327

(1.327)^(-1/1.34) = (1.327)^(-0.7463)

ln(1.327^(-0.7463)) = -0.7463 × ln(1.327) = -0.7463 × 0.2833 = -0.2114

1.327^(-0.7463) = e^(-0.2114) = 0.810

Compute C_r:

C_r = 0.90 × 33,400 × 350 × 0.810 / 1,000

= 0.90 × 33,400 × 283.5 / 1,000

= 0.90 × 9,469 / 1,000

= 8,522 kN

Step 5 — Axial Load Check

C_f / C_r = 4,500 / 8,522 = 0.528 ✓

The column is at 53% capacity for pure axial load. This leaves reserve for the coexisting moment.

Step 6 — Moment Resistance (Flexure)

Plastic moment capacity:

M_px = Z_x × F_y = 4,680 × 10³ × 350 / 10⁶ = 1,638 kN·m

Flexural resistance (Class 1 section, laterally braced at column ends):

M_rx = φ × M_px = 0.90 × 1,638 = 1,474 kN·m

Note: For a column in a braced frame, the member is braced at each floor level by the beam-to-column connections and the floor diaphragm. The unbraced length for LTB is the storey height (4.5 m). For this W360x262 section with its stocky proportions, LTB does not govern — the full plastic moment is attainable.

Step 7 — Beam-Column Interaction Check (Cl. 13.8)

Step 7a — Elastic buckling load C_ex:

C_ex = π² × E × I_x / (K_x × L)²

= π² × 200,000 × 1,060 × 10⁶ / (1.0 × 4,500)²

= π² × 200,000 × 1,060 × 10⁶ / 20.25 × 10⁶

= 1.9739 × 10⁵ × 1,060 / 20.25 × 10⁰

= 1.032 × 10¹¹ / 10³... let me recompute carefully.

C_ex = π² × E × I_x / (K_x × L)²

= π² × 200,000 × 1,060 × 10⁶ / (4,500)²

= 9.8696 × 200,000 × 1.060 × 10⁹ / 20.25 × 10⁶

= (9.8696 × 200,000 / 20.25) × (1.060 × 10⁹ / 10⁶)

= (97,537) × 1,060

= 103,389 kN

Step 7b — Factor U_1x (Cl. 13.8.4):

U_1x = ω_1 / (1 - C_f / C_ex)

where ω_1 = 0.6 for Class 1 and 2 sections per CSA S16 Cl. 13.8.4.

U_1x = 0.6 / (1 - 4,500 / 103,389) = 0.6 / (1 - 0.0435) = 0.6 / 0.9565 = 0.627

U_1x is less than 1.0. The amplified moment is 0.627 × 120 = 75.2 kN·m — the P-Delta effect is modest because the column is stocky and C_f/C_ex is small (4.4%).

Step 7c — Cross-section strength check (Cl. 13.8.2):

C_f / C_r + 0.85 × U_1x × M_fx / M_rx + 0.85 × U_1y × M_fy / M_ry ≤ 1.0

= 0.528 + 0.85 × 0.627 × 120 / 1,474 + 0.85 × 0 × ... / ...

= 0.528 + 0.85 × 0.627 × 0.0814

= 0.528 + 0.85 × 0.0510

= 0.528 + 0.0434

= 0.571 ✓ — Cross-section strength is adequate.

Step 7d — Member stability check (Cl. 13.8.3):

C_f / C_r + U_1x × M_fx / M_rx + U_1y × M_fy / M_ry ≤ 1.0

= 0.528 + 0.627 × 0.0814

= 0.528 + 0.0510

= 0.579 ✓ — Member stability is adequate.

Step 8 — Column Design Summary

Selected section: W360x262 (Grade 350W) — adequate for the combined axial load and moment.

Alternative Section Check — W360x196

For comparison, let us check if a lighter section could work:

W360x196 (US equivalent W14x132):

• A = 24,900 mm²
• r_y = 81.8 mm
• Z_x = 3,280 × 10³ mm³
• I_x = 694 × 10⁶ mm⁴

Slenderness:

K×L/r_y = 1.0 × 4,500 / 81.8 = 55.0

Non-dimensional slenderness:

λ = 55.0 × 0.01332 = 0.733

Compression resistance:

λ^(2.68) = 0.733^(2.68) = e^(2.68 × ln(0.733)) = e^(2.68 × (-0.3105)) = e^(-0.832) = 0.435

(1 + 0.435)^(-0.7463) = 1.435^(-0.7463) = e^(-0.7463 × ln(1.435)) = e^(-0.7463 × 0.3615) = e^(-0.2698) = 0.764

C_r = 0.90 × 24,900 × 350 × 0.764 / 1,000 = 5,994 kN

Interaction check:

C_f / C_r = 4,500 / 5,994 = 0.751

C_ex = π² × 200,000 × 694 × 10⁶ / (4,500)² = 67,640 kN

U_1x = 0.6 / (1 - 4,500 / 67,640) = 0.6 / 0.9335 = 0.643

M_rx = 0.90 × 3,280 × 10³ × 350 / 10⁶ = 1,033 kN·m

Cl. 13.8.3: 0.751 + 0.643 × 120 / 1,033 = 0.751 + 0.0747 = 0.826 ✓

W360x196 would work at 83% utilisation. The original W360x262 (58% utilisation) was conservatively chosen — W360x196 is a more economical option.

Base Plate Design Summary

The column transfers its load to a concrete footing through a base plate. For C_f = 4,500 kN on 30 MPa concrete:

Required base plate area: A_req = C_f / (0.85 × φ_c × f'_c) = 4,500 × 10³ / (0.85 × 0.65 × 30) = 271,493 mm²

Try 600 mm × 500 mm plate: A = 300,000 mm² ✓

Base plate thickness per CSA S16 Cl. 26.3: t_p = m × sqrt(2 × C_f / (0.90 × F_y × B × N))

Where m = (N - 0.95 × d) / 2 = (500 - 0.95 × 368) / 2 = 75.2 mm

t_p = 75.2 × sqrt(2 × 4,500 × 10³ / (0.90 × 350 × 600 × 500)) = 75.2 × sqrt(9.0 × 10⁶ / 94.5 × 10⁶) = 75.2 × 0.309 = 23.2 mm → use 25 mm plate

Four 25 mm diameter anchor rods at 300 mm spacing, embedded 400 mm into the footing.

Related Pages

• Canada CSA S16 Steel Design Guide — Full CSA S16 design reference
• CSA S16 Column Buckling — Euler & Fcr Curves — Column buckling theory
• CSA S16 Beam Design — Worked Example — Step-by-step beam design example
• CSA S16 Beam Design — Flexure, LTB & Shear — Detailed beam design guide
• Canadian Steel Beam Sizes — W Shapes, HSS — Complete section tables
• Canadian Steel Grades — G40.21 300W to 480W — Material properties
• CSA S16 Load Combinations — NBCC ULS & SLS — Canadian load combination guide
• CSA S16 Base Plate Design — Step-by-Step Guide — Anchorage and bearing per CSA S16
• BEAM Capacity Calculator — Free multi-code beam calculator

Frequently Asked Questions

How is the effective length factor K determined for columns in a braced frame?

For a braced frame (non-sway), K = 1.0 is conservative. The actual K depends on end restraint from beams and columns at each joint, calculated from the alignment chart where G = (ΣI_c/L_c) / (ΣI_g/L_g). For typical braced frames, K ranges from 0.75 to 0.90. Using K = 1.0 underestimates capacity by 10-20% compared to refined values. For unbraced (sway) frames, K can exceed 2.0 for slender frames. CSA S16 Cl. 13.3.4 permits the use of K = 1.0 for braced frames without further refinement, which is common in design practice. For this worked example (K=1.0), the design is conservative.

What is the P-Delta effect and when must it be considered for column design?

P-Delta (or second-order) effects refer to the additional moment caused by the axial load acting through the lateral deflection of the column. CSA S16 Cl. 27.10 requires P-Delta effects to be considered when the inter-storey drift exceeds 0.01 × h_s. For this column, the P-Delta amplification is captured by the U_1 factor in the beam-column interaction check. The elastic buckling load C_ex = 103,389 kN is much larger than C_f = 4,500 kN, so U_1 = 0.627 is actually less than 1.0 — P-Delta effects are negligible for this stocky column. In general, P-Delta is more significant for slender columns (K×L/r > 80) and unbraced frames where lateral displacements accumulate.

How does CSA S16 handle biaxial bending in column design?

For biaxial bending (moment about both axes), CSA S16 Cl. 13.8.2 (cross-section strength) and Cl. 13.8.3 (member stability) include both M_fx/M_rx and M_fy/M_ry terms with the same 0.85 factor for the cross-section check. A common simplification is to use the 1.0 interaction equation: C_f/C_r + U_1x × M_fx/M_rx + U_1y × M_fy/M_ry ≤ 1.0. For square HSS columns with similar moment about both axes, the biaxial interaction is significant. For W-shape columns, the weak-axis moment (M_ry) is typically much smaller than the strong-axis moment (M_rx), so minor axis bending may still govern if the eccentricity is large.

What is the minimum base plate thickness for a CSA S16 column?

CSA S16 Cl. 26.3 specifies that base plate thickness is determined by the cantilever bending of the plate projecting beyond the column footprint. The critical section is at a distance 0.8 × b_f from the plate edge for the flange cantilever and 0.95 × d from the plate edge for the web cantilever. The thickness is the larger of: t_p = m × sqrt(2 × C_f / (φ × F_y × B × N)) for axial load, or the anchor rod tension case if uplift governs. Minimum thickness in practice is typically 12 mm for light columns and 20-40 mm for heavy columns. Anchor rods are designed per CSA S16 Cl. 26.4 for tension, shear, or combined loading with φ = 0.67 for tension and φ = 0.55 for shear in anchor rods.

This page is for educational reference. Column design per CSA S16:24. Verify section properties against current CISC Handbook of Steel Construction. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent P.Eng. verification.

Design Resources

Calculator tools

• Column Capacity Calculator
• Steel Buckling Calculator
• Steel Column Base Design Calculator

Design guides

• Column Capacity Worked Example
• Column Buckling Guide
• Column Buckling Calculator Guide
• EN 1993-1-1 Column Buckling Worked Example