Australian Bolt Group Capacity — AS 4100 Eccentric Load
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Reference for bolt group capacity under eccentric loading per AS 4100:2020. Covers elastic (vector) method and instantaneous centre of rotation (ICR) method for bracket-type connections.
Elastic Method (Vector Addition)
Polar moment of inertia: Jp = Σ(x² + y²) about bolt group centroid
Bolt forces from eccentric load P (applied at eccentricity e):
- Direct shear: Vd = P/n per bolt
- Torsional shear: Vt = P × e × r / Jp
- Resultant: Vres = √(Vd_x² + (Vd_y + Vt_y)²)
Check: φVb ≥ Vres (bearing) AND φVd ≥ Vres (bolt shear)
ICR Method — More Accurate
The instantaneous centre of rotation method accounts for ductility distribution. Per AS 4100 Commentary, allows up to 30% higher capacity than elastic method.
Procedure:
- Locate ICR by iteration (satisfies equilibrium)
- Bolt deformation = ri × θ (proportional to distance from ICR)
- Bolt force from load-deformation curve (Clause 9.3.8)
- Sum bolt forces = applied load P
| Method | Accuracy | Use |
|---|---|---|
| Elastic | Conservative (20-30%) | Simple brackets, preliminary design |
| ICR | Exact | Production design, critical connections |
| Simplified (first bolt) | Very conservative | Quick checks, hand calcs |
Bolt Group Capacity Tables (M20 8.8/S, Grade 300)
| No. Bolts | Pattern | eVd_x = 300 (kN) | eVd_x = 400 (kN) | eVd_x = 500 (kN) |
|---|---|---|---|---|
| 4 | 2×2 | 210 | 195 | 175 |
| 6 | 2×3 | 315 | 290 | 260 |
| 8 | 2×4 | 420 | 385 | 345 |
| 10 | 2×5 | 525 | 480 | 430 |
| 12 | 3×4 | 630 | 575 | 515 |
Values are approximate design capacities using elastic method.
Worked Example
Problem: 6-bolt group (2 rows × 3), M20 8.8/S. Eccentric load P at e=200 mm. Check capacity.
Solution:
- Bolt group centroid: centre of 2×3 pattern
- Jp = 2×[2×(35²+0²)] + 2×[2×(35²+70²)] + 2×[2×(35²+140²)] = 98,000 mm²
- Direct shear per bolt: Vd = P/6
- Max torsional shear (corner bolt): Vt = P × 200 × √(35²+140²) / 98,000 = P × 0.29
- Resultant on critical bolt: Vres = √((P/6)² + (P×0.29)²) = P × 0.34
- M20 8.8/S shear capacity φVfn = 107 kN
- Max P = 107 / 0.34 = 315 kN
Design Resources
- [[Australian Steel Grades|/reference/australian-steel-grades/]] | [[Australian Steel Properties|/reference/australian-steel-properties/]] | [[Australian Beam Sizes|/reference/au-beam-sizes/]] | [[Australian Bolt Capacity|/reference/australian-bolt-capacity/]] | [[AS 4100 Beam Design|/reference/as4100-beam-design-example/]] | [[All Australian References|/reference/]]
FAQ
Which method is preferred for bolt group design? ICR method per AS 4100 Commentary is more accurate (up to 30% higher capacity). Elastic method is simpler but conservative.
How does eccentricity affect capacity? Capacity decreases non-linearly with eccentricity. At e=300 mm, a 6-bolt group may carry 50% less than concentric loading.
What is the critical bolt in a group? Usually the bolt farthest from the centroid (corner bolt). Its combined shear from direct load + torsion gives the worst resultant force.
Educational Use Only — This reference is for educational and preliminary design purposes only. All structural designs must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) in accordance with AS 4100:2020 and all applicable Australian Standards. Results are not for construction.