AS 4100 Worked Examples -- Beam, Column, and Bolt Group Design
AS 4100:2020 is Australia's primary standard for structural steel design. This article presents three fully worked examples -- a simply supported beam, an axially loaded column, and a bolted connection -- with the complete calculation sequence shown at every step. Each example includes the governing clause reference, intermediate values, and a final utilisation ratio.
All calculations can be independently verified using the free tools at SteelCalculator.app, which runs AS 4100 checks via WebAssembly in your browser with no data upload.
Worked Example 1 -- Simply Supported Beam (AS 4100)
Problem statement
A simply supported floor beam spanning 9.0 m supports composite steel decking at 3.0 m spacing. The beam is unshored during construction and acts compositely with the slab in the final condition. We design the steel beam for the construction stage only (no composite action), which is the governing case for the steel section.
| Parameter | Value |
|---|---|
| Span (L) | 9.0 m |
| Beam spacing | 3.0 m |
| Section | 410UB53.7 (Grade 300) |
| Dead load (G) | 3.5 kN/m (deck + beam self-weight) |
| Live load (Q) | 4.0 kPa x 3.0 m = 12.0 kN/m |
| Unbraced length (top flange) | 3.0 m (deck attached at 450 mm centres) |
| Steel grade | AS/NZS 3679.1 Grade 300 (f_y = 300 MPa, f_u = 440 MPa) |
Section properties -- 410UB53.7
| Property | Value | Property | Value |
|---|---|---|---|
| A_g | 6850 mm^2 | d | 406 mm |
| I_x | 186 x 10^6 mm^4 | Z_ex | 918 x 10^3 mm^3 |
| S_x | 1050 x 10^3 mm^3 | t_w, t_f | 7.8 mm, 10.9 mm |
| r_x | 165 mm | r_y | 29.7 mm |
| k_f | 0.93 | I_y | 6.03 x 10^6 mm^4 |
Step 1.1 -- Section classification (AS 4100 Cl. 5.2)
Flange slenderness: lambda_ef = (b_f - t_w) / (2 x t_f) x sqrt(f_y/250) = (178 - 7.8) / (2 x 10.9) x sqrt(300/250) = 7.81 x 1.095 = 8.55
Flange yield limit lambda_ey = 16 (Table 5.2, flat element, one supported edge). Since 8.55 < 16, flange is fully effective. Flange plasticity limit lambda_ep = 10. Since 8.55 < 10, flange can reach yield in compression without local buckling.
Web slenderness: lambda_ew = (d - 2 x t_f) / t_w x sqrt(f_y/250) = (406 - 21.8) / 7.8 x 1.095 = 49.3 x 1.095 = 53.9
Web yield limit lambda_ey = 115 (both edges supported). Since 53.9 < 115, web is fully effective. Web plasticity limit lambda_ep = 82. Since 53.9 < 82, web is compact.
Verdict: The section is compact. Form factor k_f = 0.93 (per published section properties; the small reduction is from the web-flange junction radius).
Step 1.2 -- Section moment capacity (Cl. 5.2.3)
M_sx = f_y x Z_ex = 300 x 1050 x 10^3 / 10^6 = 315 kN-m
But check the plastic capacity limit: M_sx <= 1.5 x f_y x S_x / 10^6 = 1.5 x 300 x 918 x 10^3 / 10^6 = 413 kN-m. OK, M_sx = 315 kN-m.
phi_Msx = 0.90 x 315 = 283.5 kN-m
Step 1.3 -- Member moment capacity (Cl. 5.6)
For a beam with the top flange continuously restrained by decking, the effective length for lateral-torsional buckling (LTB) is the segment between deck attachments:
L_e = k_t x k_l x k_r x L = 1.0 x 1.0 x 1.0 x 3000 = 3000 mm (top flange restrained)
For a compact section, the reference buckling moment is: M_oa = sqrt((pi^2 x E x I_y / L_e^2) x (G x J + pi^2 x E x I_w / L_e^2))
Using section properties (J = 218 x 10^3 mm^4, I_w = 0.284 x 10^12 mm^6 from published data):
M_oa = sqrt((pi^2 x 200000 x 6.03x10^6 / 3000^2) x (80000 x 218x10^3 + pi^2 x 200000 x 0.284x10^12 / 3000^2)) / 10^6
First term: pi^2 x 200000 x 6.03x10^6 / 9x10^6 = 1.323 x 10^6 N
Torsion term: 80000 x 218x10^3 = 17.44 x 10^9 N-mm^2
Warping term: pi^2 x 200000 x 0.284x10^12 / 9x10^6 = 62.3 x 10^9 N-mm^2
M_oa = sqrt(1.323x10^6 x (17.44 + 62.3)x10^9) / 10^6 = sqrt(105.5x10^15) / 10^6 = 325 x 10^6 / 10^6 = 325 kN-m
Slenderness reduction factor: alpha_s = 0.6 x (sqrt((M_oa/M_sx)^2 + 3) - M_oa/M_sx) = 0.6 x (sqrt((325/315)^2 + 3) - 325/315) = 0.6 x (sqrt(1.064 + 3) - 1.032) = 0.6 x (2.016 - 1.032) = 0.590
Moment modification factor alpha_m = 1.0 (simply supported, uniform load).
Member capacity: phi_Mbx = phi x alpha_m x alpha_s x M_sx <= phi_Msx
phi_Mbx = 0.90 x 1.0 x 0.590 x 315 = 167.3 kN-m. This is less than phi_Msx = 283.5 kN-m, so LTB governs -- even with deck restraint at 3 m spacing.
Step 1.4 -- Demand and utilisation
Factored load: w* = 1.2 x 3.5 + 1.5 x 12.0 = 4.20 + 18.00 = 22.20 kN/m
Maximum moment: M* = w* x L^2 / 8 = 22.20 x 81 / 8 = 224.8 kN-m
Utilisation: M*/phi_Mbx = 224.8 / 167.3 = 1.34 -- FAILS.
The 410UB53.7 is inadequate for the 9.0 m span with 3.0 m deck spacing. Options:
- Reduce beam spacing to 2.4 m (reduces live load to 9.6 kN/m; M* drops to 156 kN-m; utilisation = 0.93 -- OK).
- Use 460UB67.1 (increased I_y and stronger LTB resistance; phi_Mbx approx 245 kN-m; utilisation = 0.92 -- OK).
- Provide intermediate fly braces to reduce L_e further (L_e = 1.5 m gives alpha_s = 0.88; phi_Mbx = 249 kN-m; utilisation = 0.90 -- OK).
Step 1.5 -- Shear capacity (Cl. 5.11)
Web slenderness: d_p/t_w = (406 - 2x10.9)/7.8 = 384.2/7.8 = 49.3
Shear yield limit: 82 x sqrt(250/f_y) = 82 x sqrt(250/300) = 74.9. Since 49.3 < 74.9, shear buckling does not govern.
V_v = 0.6 x f_y x A_w = 0.6 x 300 x (406 x 7.8) / 1000 = 570 kN phi_Vv = 0.90 x 570 = 513 kN
Maximum shear: V* = w* x L / 2 = 22.20 x 9 / 2 = 99.9 kN. Utilisation = 99.9/513 = 0.19 -- OK. Shear does not govern.
Step 1.6 -- Deflection (serviceability, Appendix B)
Under live load (Q = 12.0 kN/m): delta_LL = 5 x w_LL x L^4 / (384 x E x I_x) = 5 x 12.0 x 9000^4 / (384 x 200000 x 186x10^6) = 5 x 12.0 x 6.56x10^15 / (384 x 200000 x 186x10^6) = 27.6 mm
Limit: L/360 = 9000/360 = 25.0 mm. 27.6 > 25.0 -- deflection also fails.
Under total load: delta_TL = 34.2 mm. Limit: L/250 = 36.0 mm. OK.
Deflection governs live load, LTB governs the strength check. Both must be satisfied.
Summary -- Worked Example 1
| Check | Demand | Capacity | Ratio | Result |
|---|---|---|---|---|
| Moment (LTB) | 224.8 kN-m | 167.3 kN-m | 1.34 | FAIL |
| Shear | 99.9 kN | 513 kN | 0.19 | PASS |
| Live load deflection | 27.6 mm | 25.0 mm | 1.10 | FAIL |
The 410UB53.7 is undersized. Increasing to 460UB67.1 or reducing beam spacing resolves both issues.
Worked Example 2 -- Axially Loaded Column (AS 4100)
Problem statement
An internal column in a multi-storey braced frame supports gravity loads from floors above. The column is continuous through two storeys (storey height 4.0 m) with simple (nominally pinned) connections to the beams at each floor level.
| Parameter | Value |
|---|---|
| Column section | 310UC118 (Grade 300) |
| Storey height (L) | 4.0 m |
| Effective length factor (k_e) | 1.0 (braced frame, pinned ends) |
| Design axial load N* | 2450 kN (after 1.2G + 1.5Q) |
| Bending moment M*_x | 45 kN-m (nominal beam end moment) |
| Bending moment M*_y | 12 kN-m (minor axis) |
| Steel grade | Grade 300 (f_y = 300 MPa, f_u = 440 MPa) |
Section properties -- 310UC118
| Property | Value | Property | Value |
|---|---|---|---|
| A_g | 15000 mm^2 | d, b_f | 315 mm, 307 mm |
| I_x | 275 x 10^6 mm^4 | I_y | 90.4 x 10^6 mm^4 |
| r_x | 135 mm | r_y | 77.5 mm |
| t_f, t_w | 18.7 mm, 11.9 mm | k_f | 0.96 |
| Z_ex, Z_ey | 1750, 590 x 10^3 mm^3 | S_x, S_y | 1990, 903 x 10^3 mm^3 |
Step 2.1 -- Section classification
Flange: lambda_ef = ((307-11.9)/(2x18.7)) x sqrt(300/250) = 7.89 x 1.095 = 8.64. lambda_ey = 16. Flange is fully effective.
Web: lambda_ew = ((315-2x18.7)/11.9) x 1.095 = 23.3 x 1.095 = 25.5. lambda_ey = 45. Web is fully effective. lambda_ep = 35. Web is compact.
Section is non-slender (compact) for both axes.
Step 2.2 -- Section capacity (Cl. 8.2)
Compression: N_s = k_f x A_g x f_y = 0.96 x 15000 x 300 / 1000 = 4320 kN. phi_Ns = 0.90 x 4320 = 3888 kN.
Moment (major axis): phi_Msx = 0.90 x 300 x 1990x10^3 / 10^6 = 537 kN-m.
Moment (minor axis): phi_Msy = 0.90 x 300 x 903x10^3 / 10^6 = 244 kN-m.
Step 2.3 -- Member capacity (Cl. 8.4)
Effective lengths: L_ex = L_ey = 1.0 x 4000 = 4000 mm (braced frame, pinned ends).
Modified slenderness: lambda_nx = (L_ex/r_x) x sqrt(k_f) x sqrt(f_y/250) = (4000/135) x sqrt(0.96) x 1.095 = 29.6 x 0.98 x 1.095 = 31.8 lambda_ny = (4000/77.5) x 0.98 x 1.095 = 51.6 x 0.98 x 1.095 = 55.4
Compression member capacity: alpha_a (Cl. 6.3.3) = 2100 x (lambda_n - 13.5) / (lambda_n^2 - 15.3 x lambda_n + 2050)
For lambda_nx = 31.8: alpha_ax = 2100 x (31.8-13.5) / (31.8^2 - 15.3x31.8 + 2050) = 2100 x 18.3 / (1011.2 - 486.5 + 2050) = 38430 / 2574.7 = 0.927
For lambda_ny = 55.4: alpha_ay = 2100 x (55.4-13.5) / (55.4^2 - 15.3x55.4 + 2050) = 2100 x 41.9 / (3069.2 - 847.6 + 2050) = 87990 / 4271.6 = 0.817
phi_Ncx = 0.90 x alpha_ax x N_s = 0.90 x 0.927 x 4320 = 3604 kN phi_Ncy = 0.90 x 0.817 x 4320 = 3176 kN (minor axis governs)
Moment member capacity (major axis): M_ox = sqrt((pi^2 x E x I_y / L_e^2) x (G x J + pi^2 x E x I_w / L_e^2))
Using published values for 310UC118 (J = 1620x10^3 mm^4, I_w = 1.98x10^12 mm^6):
M_ox = sqrt((pi^2 x 200000 x 90.4x10^6 / 4000^2) x (80000 x 1620x10^3 + pi^2 x 200000 x 1.98x10^12 / 4000^2)) / 10^6
= sqrt((1.115x10^7) x (129.6x10^9 + 244.2x10^9)) / 10^6 = sqrt(4.168x10^18) / 10^6 = 2042 x 10^6 / 10^6 = 2042 kN-m
alpha_sx = 0.6 x (sqrt((2042/537)^2 + 3) - 2042/537) = 0.6 x (sqrt(14.46 + 3) - 3.80) = 0.6 x (4.18 - 3.80) = 0.226
alpha_mx = 1.13 (member with end moments only, M1/M2 = 0.5). Per Table 5.6.2: alpha_m = 1.75 + 1.05 x 0.5 + 0.3 x 0.25 = 2.35, capped at 2.5.
phi_Mbx = 0.90 x 2.35 x 0.226 x 537 = 256 kN-m. (This is less than phi_Msx = 537, so LTB reduces the member moment capacity significantly.)
Step 2.4 -- Interaction check (Cl. 8.4.5.1)
N*/phi_Nc = 2450/3176 = 0.771
Since N*/phi_Nc > 0.15, the full interaction equation applies:
M*_x/phi_Mbx x (1 - N*/phi_Ncx)/(1 - N*/N_ombx) + M*_y/phi_Msy <= 1.0
The denominator term N_ombx (Euler buckling load for the column buckling about the major axis, which governs in-plane strength) = pi^2 x E x I_x / L_ex^2 = pi^2 x 200000 x 275x10^6 / (4000)^2 = 33,950 kN.
Term for major axis moment: (45/256) x (1 - 2450/3604) / (1 - 2450/33950) = 0.176 x 0.320 / 0.928 = 0.061
Term for minor axis moment: 12/244 = 0.049
Total interaction: 0.771 + 0.061 + 0.049 = 0.881 <= 1.0 -- OK.
The column is adequate. Minor axis compression governs (lambda_ny = 55.4 vs lambda_nx = 31.8), which is expected for a UC section.
Summary -- Worked Example 2
| Check | Ratio | Result |
|---|---|---|
| Section compression | 0.63 | OK |
| Member compression (minor) | 0.77 | OK |
| Section moment (major) | 0.08 | OK |
| Interaction (comp + biaxial) | 0.88 | OK |
Worked Example 3 -- Bolted Connection (AS 4100 Cl. 9.2)
A beam-to-column shear connection using a web-side plate with 4 x M20 8.8/S bolts (snug-tight, threads in shear plane). The design shear is V* = 180 kN. This is a concise summary; for the detailed five-limit-state walkthrough, see our AS 4100 Bolt Group Design guide.
| Check | Formula | Capacity | Utilisation |
|---|---|---|---|
| Bolt shear (single) | phi x 0.62 x f_uf x A_c | 92.6 kN | -- |
| Bolt shear (group) | 4 x V_f | 370 kN | 0.49 |
| Ply bearing (group) | 4 x phi x 3.2 x d_f x t_p x f_u | 1014 kN | 0.18 |
| Tear-out (end bolts) | phi x a_e x t_p x f_u | 95.0 kN each | 0.47 |
| Block shear | phi x (0.6 x f_u x A_nv + f_u x A_nt) | 608 kN | 0.30 |
| Net section | phi x 0.85 x f_u x A_n | 424 kN | 0.42 |
Controlling mode: Bolt shear at 0.49 utilisation. The connection has adequate reserve capacity.
Key Takeaways Across All Three Examples
- Lateral-torsional buckling often governs beam design, even with deck restraint. Checking LTB is not optional -- always verify the unrestrained segment length between effective restraints.
- Column capacity is governed by minor-axis buckling for UC sections, where r_y << r_x. The interaction check (compression + biaxial bending) is mandatory when N*/phi_Nc > 0.15.
- Bolt groups must be checked for all limit states, not just bolt shear. In thin-plate connections, bearing or block shear often controls.
- Deflection can be more restrictive than strength. The beam example failed live-load deflection before the member capacity was exhausted.
- Always record phi factors and clause references. AS 4100 uses different phi factors for different limit states (0.90 for moment/shear, 0.80 for bolts, 0.75 for block shear). Missing the correct phi is one of the most common errors in steel design.
Educational reference only. All steel designs must be independently verified by a licensed Professional Engineer. Results are PRELIMINARY -- NOT FOR CONSTRUCTION. Consult AS 4100:2020 and the National Construction Code for authoritative requirements.
Frequently Asked Questions
What is the difference between AS 4100 and AISC 360?
AS 4100 and AISC 360 both use the limit states design philosophy with capacity reduction factors (phi factors), but they differ in several details. AS 4100 uses slightly different phi factors (e.g., 0.90 for beams vs AISC's 0.90 for flexure -- same in this case, but they diverge for connections). AS 4100 also uses the alpha_s formulation for LTB rather than the AISC C_b-based moment gradient approach. Section classifications (compact/non-compact/slender) have different lambda limits in each standard.
Is AS 4100 equivalent to Eurocode 3 (EN 1993)?
No, not directly. AS 4100 and EN 1993 follow different structural reliability frameworks (Australian vs European). While the underlying mechanics are similar, the partial factors, buckling curves, and connection design rules differ. EN 1993 uses multiple buckling curves (a0, a, b, c, d) while AS 4100 has a unified alpha_a/alpha_b formulation. Design outputs are usually comparable but not identical -- always use the code specified for your jurisdiction.
How do I verify my hand calculations?
SteelCalculator.app provides free browser-based AS 4100 member design, connection design, and load combination calculators. Enter your section, loads, and restraints to get independent verification. All computation runs in your browser via WebAssembly with no data upload. Use the Designer Hub member design page for a full AS 4100 beam and column check.
Can I use these examples for university assignments?
Yes, these worked examples are designed for educational use. They show the correct calculation sequence, intermediate values, and cross-checks at each step. However, always check the current edition of AS 4100 (2020 with amendments) and confirm any referenced clauses against the official standard.
What other AS 4100 design guides are available?
SteelCalculator.app has dedicated guides on AS 4100 bolt group design (with a full M20 8.8 worked example), AS 4100 fillet weld design, and AS 4100 beam sizes. See the blog index for all articles and the Designer Hub for interactive AS 4100 load combination and member design tools.