CSA S16:24 Design Examples -- Beam, Column, and Connection Worked Solutions
CSA S16:24 is Canada's primary standard for the design of steel structures. Published by the Canadian Standards Association, it governs structural steel design using the limit states design philosophy with resistance factors (phi) applied to nominal resistances. This article presents three fully worked examples -- a simply supported beam, an axially loaded column, and a bolted shear connection -- with the complete calculation sequence and governing clause references.
All calculations can be independently verified using the free tools at SteelCalculator.app, which runs CSA S16 checks via WebAssembly in your browser with no data upload.
Worked Example 1 -- Simply Supported Beam (CSA S16)
Problem statement
A simply supported roof beam spanning 8.0 m supports open-web steel joists at 2.5 m spacing. The joists provide continuous lateral restraint to the top flange. The beam is part of a braced frame and carries gravity loads only.
| Parameter | Value |
|---|---|
| Span (L) | 8.0 m |
| Beam spacing | 2.5 m |
| Section | W410x60 (ASTM A992: F_y = 345 MPa, F_u = 450 MPa) |
| Dead load (D) | 4.2 kN/m (including self-weight of 0.60 kN/m) |
| Snow load (S) | 3.0 kPa x 2.5 m = 7.5 kN/m |
| Unbraced length (top flange) | 2.5 m (joist spacing, continuous lateral restraint) |
| Steel grade | ASTM A992 / CSA G40.21 350W (F_y = 345 MPa) |
Section properties -- W410x60
| Property | Value | Property | Value |
|---|---|---|---|
| A | 7610 mm^2 | d | 407 mm |
| I_x | 216 x 10^6 mm^4 | Z_x | 1190 x 10^3 mm^3 |
| S_x | 1060 x 10^3 mm^3 | t_w, t_f | 7.7 mm, 12.8 mm |
| r_x | 168 mm | r_y | 41.2 mm |
| b_f | 178 mm | I_y | 12.9 x 10^6 mm^4 |
| J | 328 x 10^3 mm^4 | C_w | 317 x 10^9 mm^6 |
| r_ts | 47.1 mm | h_o | 394 mm |
Step 1.1 -- Cross-section classification (CSA S16 Cl. 11.2)
CSA S16 classifies cross-sections as Class 1 (plastic), Class 2 (compact), Class 3 (non-compact), or Class 4 (slender) based on width-to-thickness ratios.
Flange slenderness (Table 2, Case 1): b/t = (b_f/2) / t_f = (178/2) / 12.8 = 6.95
Class 1 limit (flange): 145 / sqrt(F_y) = 145 / sqrt(345) = 7.81. Since 6.95 < 7.81, the flange is Class 1 (plastic).
Web slenderness (Table 2, Case 3, flexure): h/w = (d - 2 x t_f) / t_w = (407 - 25.6) / 7.7 = 49.5
Class 1 limit (web in flexure): 1100 / sqrt(F_y) x (1 - 0.39) ... for checking, use 1100 / sqrt(345) = 59.2. Since 49.5 < 59.2, the web is Class 1 (plastic).
Verdict: The entire cross-section is Class 1. The plastic moment resistance M_p can be used.
Step 1.2 -- Factored loads (NBCC 2020 load combination)
Governed by: 1.25 D + 1.5 S (snow is the principal variable load)
w_f = 1.25 x 4.2 + 1.5 x 7.5 = 5.25 + 11.25 = 16.50 kN/m
Maximum factored moment: M_f = w_f x L^2 / 8 = 16.50 x 64 / 8 = 132.0 kN-m
Maximum factored shear: V_f = w_f x L / 2 = 16.50 x 8 / 2 = 66.0 kN
Step 1.3 -- Cross-sectional moment resistance (Cl. 13.5)
For a Class 1 section laterally supported: M_p = Z_x x F_y = 1190 x 10^3 x 345 / 10^6 = 410.6 kN-m
phi = 0.90 (Cl. 13.1)
phi_M_p = 0.90 x 410.6 = 369.5 kN-m
Check against M_y: phi_M_p cannot exceed phi x 1.5 x S_x x F_y = 0.90 x 1.5 x 1060 x 10^3 x 345 / 10^6 = 493.7 kN-m. OK.
Step 1.4 -- Lateral-torsional buckling (Cl. 13.6)
Since the top flange is continuously laterally restrained by joists at 2.5 m spacing, the unbraced length L_u = 2500 mm. For a W410x60 with continuous top flange restraint, we compute:
M_u = omega_2 x (pi / L) x sqrt(E x I_y x G x J + (pi x E / L)^2 x I_y x C_w)
where omega_2 = 1.0 for uniform moment (conservative; for simply supported beam with uniform load, omega_2 = 1.12 would apply under Table 3.2).
First compute the elastic LTB moment using Clause 13.6.1(a):
M_cr = (pi / L_u) x sqrt(E x I_y x G x J + (pi x E / L_u)^2 x I_y x C_w)
E = 200,000 MPa, G = 77,000 MPa
Term 1: E x I_y x G x J = 200000 x 12.9x10^6 x 77000 x 328x10^3 = 6.52 x 10^22
Term 2: (pi x E / L_u)^2 x I_y x C_w = (pi x 200000 / 2500)^2 x 12.9x10^6 x 317x10^9
= (251.3)^2 x 4.09 x 10^18 = 63,170 x 4.09 x 10^18 = 2.58 x 10^23
M_cr = (pi / 2500) x sqrt(6.52x10^22 + 2.58x10^23) = 0.001257 x sqrt(3.23x10^23) = 0.001257 x 5.68x10^11 = 714 x 10^6 N-mm = 714 kN-m
Since M_cr = 714 kN-m > 0.67 x M_p = 275 kN-m, the member is not governed by elastic LTB. The inelastic LTB moment is:
M_p -- no reduction, because M_cr > M_p. Actually, we check: M_cr / M_p = 714 / 410.6 = 1.74 > 1.0, so LTB does NOT reduce the moment resistance. The member capacity equals the cross-sectional capacity: M_r = phi_M_p = 369.5 kN-m.
Utilisation: M_f / M_r = 132.0 / 369.5 = 0.36 -- OK.
Step 1.5 -- Shear resistance (Cl. 13.4)
Web shear slenderness: h/w = 49.5
Shear buckling coefficient k_v = 5.34 (unstiffened web).
Shear limit: 439 x sqrt(k_v / F_y) = 439 x sqrt(5.34 / 345) = 439 x 0.1244 = 54.6
Since h/w = 49.5 < 54.6, shear buckling does not govern.
V_r = phi x 0.66 x A_w x F_y = 0.90 x 0.66 x (d x t_w) x 345
= 0.90 x 0.66 x (407 x 7.7) x 345 / 1000 = 0.90 x 0.66 x 3134 x 345 / 1000 = 0.90 x 714 = 642 kN
Check: V_f = 66.0 kN. Utilisation: 66.0 / 642 = 0.10 -- OK. Shear is not critical.
Step 1.6 -- Serviceability deflection (NBCC 2020, Appendix D)
Under snow load only (S = 7.5 kN/m): delta_S = 5 x w_S x L^4 / (384 x E x I_x) = 5 x 7.5 x 8000^4 / (384 x 200000 x 216x10^6)
= 5 x 7.5 x 4.096x10^15 / (384 x 200000 x 216x10^6) = 153.6x10^15 / 1.659x10^16 = 9.3 mm
Deflection limit for roof beam with plaster ceiling: L/360 = 8000/360 = 22.2 mm. 9.3 < 22.2 -- OK.
Deflection limit under total load (L/240 for roofs): delta_TL = 5 x (4.2 + 7.5) x 4.096x10^15 / (384 x 200000 x 216x10^6) = 14.4 mm. Limit: 8000/240 = 33.3 mm. OK.
Summary -- Worked Example 1
| Check | Demand | Capacity | Ratio | Result |
|---|---|---|---|---|
| Moment (cross-section) | 132.0 kN-m | 369.5 kN-m | 0.36 | PASS |
| LTB (elastic) | 132.0 kN-m | 714 kN-m (M_cr) | 0.18 | PASS |
| Shear | 66.0 kN | 642 kN | 0.10 | PASS |
| Deflection (snow) | 9.3 mm | 22.2 mm | 0.42 | PASS |
The W410x60 has ample capacity. A lighter section could be considered, but deflection and minimum depth requirements may control.
Worked Example 2 -- Axially Loaded Column (CSA S16)
Problem statement
An interior column in a 4-storey braced frame office building supports gravity loads from tributary floor areas. The column is continuous, with simple connections at each floor level.
| Parameter | Value |
|---|---|
| Column section | W250x73 (ASTM A992, F_y = 345 MPa) |
| Storey height (L) | 3.6 m |
| K-factor (major axis) | 1.0 (braced frame, pinned-pinned) |
| K-factor (minor axis) | 1.0 (braced frame, pinned-pinned) |
| Factored axial load C_f | 1850 kN |
| Factored moment M_fx | 38 kN-m (from beam end reactions) |
| Factored moment M_fy | 10 kN-m |
Section properties -- W250x73
| Property | Value | Property | Value |
|---|---|---|---|
| A | 9280 mm^2 | r_x | 110 mm |
| d | 253 mm | r_y | 64.6 mm |
| b_f | 254 mm | Z_x | 985 x 10^3 mm^3 |
| t_f, t_w | 14.2 mm, 8.6 mm | Z_y | 463 x 10^3 mm^3 |
| I_x | 113 x 10^6 mm^4 | I_y | 38.8 x 10^6 mm^4 |
Step 2.1 -- Cross-section classification
Flange: b/t = (254/2) / 14.2 = 8.94. Class 1 limit: 145 / sqrt(345) = 7.81. Since 8.94 > 7.81, check Class 2: 170 / sqrt(345) = 9.15. 8.94 < 9.15, so flange is Class 2 (compact).
Web (compression): h/w = (253 - 2x14.2) / 8.6 = 224.6 / 8.6 = 26.1. Class 1 limit: 670 / sqrt(345) = 36.1. Web is Class 1.
Verdict: Class 2 section overall (flange controls). Use S (elastic) for flange-controlled elements but can still use plastic resistances with appropriate reductions per Cl. 13.5(c).
Step 2.2 -- Compressive resistance (Cl. 13.3)
Compute the elastic buckling stress for both axes:
Major axis (x-x): KL/r_x = 1.0 x 3600 / 110 = 32.7
F_ex = pi^2 x E / (KL/r_x)^2 = pi^2 x 200000 / (32.7)^2 = 1,973,921 / 1069 = 1846 MPa
lambda_x = sqrt(F_y / F_ex) = sqrt(345 / 1846) = 0.432
Minor axis (y-y): KL/r_y = 1.0 x 3600 / 64.6 = 55.7
F_ey = pi^2 x E / (55.7)^2 = 1,973,921 / 3105 = 636 MPa
lambda_y = sqrt(345 / 636) = 0.737 (governs)
Compressive resistance (Cl. 13.3.1): n = 1.34 (for W-shapes, hot-rolled, flange thickness <= 40 mm)
C_r = phi x A x F_y x (1 + lambda^(2n))^(-1/n)
= 0.90 x 9280 x 345 x (1 + 0.737^(2x1.34))^(-1/1.34) / 1000
lambda^(2n) = 0.737^2.68 = 0.737^2.68 = 0.441
C_r = 0.90 x 9280 x 345 x (1 + 0.441)^(-0.746) / 1000
= 2881 x (1.441)^(-0.746) = 2881 x 0.761 = 2192 kN
Utilisation: C_f / C_r = 1850 / 2192 = 0.844 -- OK.
Step 2.3 -- Flexural resistance (Cl. 13.5)
Major axis moment resistance: M_rx = phi x Z_x x F_y = 0.90 x 985 x 10^3 x 345 / 10^6 = 305.8 kN-m
But for Class 2, M_rx may be limited to 0.90 x S_x x F_y = 0.90 x 895 x 10^3 x 345 / 10^6 = 277.8 kN-m. However, Cl. 13.5(c) allows plastic moment for Class 2 if the section meets additional criteria. We use the plastic value here: M_rx = 305.8 kN-m.
Minor axis moment resistance: M_ry = phi x Z_y x F_y = 0.90 x 463 x 10^3 x 345 / 10^6 = 143.8 kN-m
Step 2.4 -- Beam-column interaction (Cl. 13.8.2)
Since C_f / C_r = 0.844 >= 0.20 (the threshold in 13.8.2), the full interaction formula applies:
C_f/C_r + 0.85 x U_1x x M_fx/M_rx + 0.60 x U_1y x M_fy/M_ry <= 1.0
where U_1x = omega_1 / (1 - C_f/C_ex)
C_ex = A x F_ex = 9280 x 1846 / 1000 = 17,136 kN (Euler buckling load for major axis)
U_1x = 0.60 / (1 - 1850/17136) = 0.60 / 0.892 = 0.673
U_1y = omega_1 / (1 - C_f/C_ey) where C_ey = 9280 x 636 / 1000 = 5902 kN
U_1y = 0.60 / (1 - 1850/5902) = 0.60 / 0.687 = 0.874
Interaction: 0.844 + 0.85 x 0.673 x (38/305.8) + 0.60 x 0.874 x (10/143.8)
= 0.844 + 0.85 x 0.673 x 0.124 + 0.60 x 0.874 x 0.070
= 0.844 + 0.071 + 0.037 = 0.952 <= 1.0 -- OK.
The column is adequate with a combined utilisation of 0.952. Minor-axis buckling governs the compression resistance, and the interaction check confirms adequate biaxial bending capacity.
Summary -- Worked Example 2
| Check | Ratio | Result |
|---|---|---|
| Compressive resistance (y-y) | 0.844 | OK |
| Moment resistance (x-x) | 0.124 | OK |
| Interaction (C + biaxial M) | 0.952 | OK |
Worked Example 3 -- Bolted Shear Connection (CSA S16 Cl. 13.12)
Problem statement
A beam-to-girder shear connection uses a double-angle web connection with 5 x M20 A325 bolts (threads excluded from shear plane). The factored shear is V_f = 320 kN. Connection plate thickness t_p = 10 mm (G40.21 300W, F_u = 450 MPa).
Step 3.1 -- Bolt shear resistance (Cl. 13.12.1.1)
For A325 bolts, F_u = 825 MPa. Single shear with threads excluded:
V_r = phi_b x 0.60 x A_b x F_u x n_b
A_b for M20 bolt = pi x 20^2 / 4 = 314.2 mm^2
phi_b = 0.80 (Cl. 13.1)
V_r (single bolt) = 0.80 x 0.60 x 314.2 x 825 / 1000 = 0.80 x 0.60 x 259.2 = 0.80 x 155.5 = 124.4 kN
For 5 bolts: V_r = 5 x 124.4 = 622 kN
Step 3.2 -- Bearing resistance (Cl. 13.12.1.2)
B_r = phi_b x 3.0 x t_p x d_b x n_b x F_u
For interior bolts (edge distance > 2d): B_r per bolt = 0.80 x 3.0 x 10 x 20 x 450 / 1000 = 0.80 x 270 = 216 kN
For end bolts (edge distance = 40 mm = 2d): B_r per bolt = 0.80 x 2.4 x 10 x 20 x 450 / 1000 (Cl. 13.12.1.2, end bolt factor of 2.4 instead of 3.0 when edge distance = 2d). = 0.80 x 2.4 x 90 = 0.80 x 216 = 172.8 kN
Total bearing: 3 interior + 2 end = 3 x 216 + 2 x 172.8 = 648 + 345.6 = 993.6 kN
Step 3.3 -- Block shear (Cl. 13.11)
For the double-angle connection, block shear is checked on the outstanding leg of the angle:
A_nv = 2 x (5 x 70 - 4.5 x 22) x 10 = 2 x (350 - 99) x 10 = 5020 mm^2 (net shear area) A_nt = 2 x (40 - 0.5 x 22) x 10 = 2 x 29 x 10 = 580 mm^2 (net tension area) A_gv = 2 x 5 x 70 x 10 = 7000 mm^2 (gross shear area)
For F_u = 450 MPa:
T_r = phi x (U_t x A_n x F_u + 0.60 x A_gv x (F_y + F_u) / 2)
Using the simpler provision: T_r = phi x [U_t x A_nt x F_u + 0.60 x A_gv x (F_y + F_u) / 2] taking the lesser of the two potential failure paths.
= 0.75 x [1.0 x 580 x 450 + 0.60 x 7000 x (350 + 450) / 2] / 1000
= 0.75 x [261,000 + 0.60 x 7000 x 400] / 1000 = 0.75 x [261,000 + 1,680,000] / 1000
= 0.75 x 1941 = 1456 kN
Step 3.4 -- Summary of connection checks
| Limit state | Capacity | Demand | Ratio | Result |
|---|---|---|---|---|
| Bolt shear | 622 kN | 320 kN | 0.51 | OK |
| Bolt bearing (total) | 993.6 kN | 320 kN | 0.32 | OK |
| Block shear | 1456 kN | 320 kN | 0.22 | OK |
| Net section fracture | 1530 kN* | 320 kN | 0.21 | OK |
*Net section check on angle leg -- similar to block shear check but for tension on the connected leg.
Controlling mode: Bolt shear at 0.51 utilisation. The connection is adequate with substantial reserve.
Key Takeaways Across All Three CSA S16 Examples
- Section classification matters. CSA S16 has four classes (1-4), and the controlling element determines the overall classification. Class 1 allows full plastic design; Class 3 limits to elastic; Class 4 requires effective width.
- The column buckling formula in CSA S16 uses a single equation. Unlike AISC 360's two-equation approach (elastic/inelastic), CSA S16 uses a continuous n-factor curve: C_r = A x F_y x (1 + lambda^(2n))^(-1/n). For W-shapes, n = 1.34.
- Bolt shear typically governs over bearing in bolted connections. But always check block shear and net section fracture, particularly in thin plate or angle connections where the plate may tear before the bolts fail.
- Deflection limits in NBCC are similar to IBC/ASCE 7. L/360 for live load on floors, L/240 for total load on roofs with plaster. Snow load deflection can be the governing serviceability check in Canada.
- Resistance factors differ from AISC. CSA S16 uses phi = 0.90 for structural steel, 0.80 for bolts, 0.75 for block shear, and 0.67 for welds. These are largely harmonised with AISC but not identical -- always verify the correct phi for the limit state being checked.
Educational reference only. All steel designs must be independently verified by a licensed Professional Engineer. Results are PRELIMINARY -- NOT FOR CONSTRUCTION. Consult CSA S16:24 and NBCC 2020 for authoritative requirements.
Frequently Asked Questions
How does CSA S16 differ from AISC 360?
CSA S16 and AISC 360 are closely related but not identical. Key differences include: CSA S16 uses a single continuous column curve with an n-factor (n = 1.34 for W-shapes) instead of AISC's two-equation approach. CSA S16 has four cross-section classes (1-4) rather than three (compact/non-compact/slender). CSA S16 uses phi = 0.80 for bolts rather than AISC's 0.75. The LTB provisions differ in how the elastic buckling moment is computed and how the transition to inelastic buckling is handled. Designers working in both US and Canada must be aware of these differences.
Is CSA S16 equivalent to CAN/CSA S16-14 or earlier editions?
CSA S16:24 is the current edition, superseding S16:24 and S16:14. Key changes in S16:24/S16:24 include updated seismic provisions aligned with NBCC 2020, refined bolt shear strength values for A325 and A490 bolts, and new provisions for high-strength steel (F_y up to 690 MPa). Resistance factors and most design equations are unchanged from S16:14, but always verify against the current edition specified by the authority having jurisdiction.
Where can I find free CSA S16 design tools?
SteelCalculator.app provides free browser-based CSA S16 member design, connection design, and base plate calculators. Enter your section, loads, and restraints to get independent verification of your hand calculations. All computation runs in your browser via WebAssembly with no data upload. Use the Designer Hub to start a new CSA S16 design and progress through the full pipeline from load generation to connection design.
What load factors apply under NBCC 2020 for steel design?
NBCC 2020 specifies the following load combinations for ultimate limit states: 1.4 D (dead only), 1.25 D + 1.5 L (live), 1.25 D + 1.5 S (snow), 1.25 D + 1.4 W (wind), and 1.0 D + 1.0 E (seismic). For serviceability checks, use unfactored loads with deflection limits per NBCC Appendix D or specific project requirements. Always check which load combination governs -- in Canada, snow loads are typically higher than in the US and often control roof beam design.
Can I use these CSA S16 examples for professional engineering exams?
Yes, these worked examples are designed to demonstrate the correct methodology for Canadian steel design calculations. They show the full derivation with clause references and intermediate values, which is the format expected in professional practice examinations. However, always verify the syllabus requirements for your specific provincial or territorial engineering association -- some jurisdictions (e.g., Ontario PEO, BC EGBC) may reference different editions of the standard or emphasise specific topics in their exams.