Australian Bolt Capacity — AS 4100 Shear and Tension Capacity Tables
Complete reference for bolt shear and tension capacity in Australian steel design per AS 4100:2020 Clause 9.3.2.1. Factored capacity tables for Grade 8.8 and Grade 10.9 bolts from M12 to M36, including threads-in-shear-plane (N-grade) and threads-excluded (X-grade) conditions. Includes worked calculation examples.
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AS 4100 Bolt Capacity — Design Formulas
AS 4100:2020 Clause 9.3.2.1 defines the nominal capacity of a bolt in shear and tension. The design capacity is the nominal capacity multiplied by the capacity factor phi = 0.80.
Nominal Shear Capacity — Vfn
The nominal shear capacity of a single bolt is the lesser of:
(a) Bolt shear capacity: Vfn = 0.62 x fuf x kr x At(n) (threaded portion in shear plane) Vfn = 0.62 x fuf x kr x Ab (non-threaded portion in shear plane)
(b) Bolt thread stripping capacity (if threads extend into shear plane): Vfn = 0.372 x fuf x kr x At(n)
Where:
- fuf = minimum tensile strength of the bolt material
- kr = reduction factor for bolt length (Clause 9.3.2.2)
- kr = 1.0 if the bolt length < 12 x bolt diameter
- kr = 0.85 if the bolt length >= 12 x bolt diameter
- kr = 0.75 for bolts in connections with packers thicker than 6 mm
- At(n) = tensile stress area of threaded portion (mm2)
- Ab = nominal cross-sectional area of bolt body (pi x d2/4)
Simplified for design:
For threads included in the shear plane (N-grade): phi Vfn = 0.80 x 0.372 x fuf x kr x At = 0.298 x fuf x kr x At
For threads excluded from the shear plane (X-grade): phi Vfn = 0.80 x 0.62 x fuf x kr x Ab = 0.496 x fuf x kr x Ab
Nominal Tension Capacity — Ntf
The nominal tension capacity of a bolt is:
Ntf = fuf x At(n) x kr
Factored: phi Ntf = 0.80 x fuf x At(n) x kr
Grade 8.8 Bolt Capacity Tables (fuf = 830 MPa)
All values are factored (phi = 0.80) and assume kr = 1.0 (bolt length < 12d).
| Bolt Size | At (mm2) | Ab (mm2) | phi Vfn — N (kN) | phi Vfn — X (kN) | phi Ntf (kN) |
|---|---|---|---|---|---|
| M12 | 84.3 | 113 | 20.9 | 46.5 | 56.0 |
| M16 | 157 | 201 | 38.8 | 83.3 | 104.2 |
| M20 | 245 | 314 | 60.6 | 129.9 | 162.7 |
| M22 | 303 | 380 | 75.0 | 157.9 | 201.2 |
| M24 | 353 | 452 | 87.3 | 187.7 | 234.4 |
| M27 | 459 | 573 | 113.5 | 237.5 | 304.8 |
| M30 | 561 | 707 | 138.7 | 293.3 | 372.5 |
| M36 | 817 | 1018 | 202.0 | 421.7 | 542.5 |
Grade 8.8 — kr Reduction for Long Bolts
For bolts longer than 12 x bolt diameter (e.g., M20 bolts longer than 240 mm), kr = 0.85:
| Bolt Size | phi Vfn — N (kr = 0.85) | phi Vfn — X (kr = 0.85) | phi Ntf (kr = 0.85) |
|---|---|---|---|
| M20 | 50.5 kN | 108.4 kN | 138.3 kN |
| M24 | 72.8 kN | 157.2 kN | 199.2 kN |
| M30 | 117.9 kN | 249.3 kN | 316.6 kN |
Grade 10.9 Bolt Capacity Tables (fuf = 1040 MPa)
All values are factored (phi = 0.80) and assume kr = 1.0 (bolt length < 12d).
| Bolt Size | At (mm2) | Ab (mm2) | phi Vfn — N (kN) | phi Vfn — X (kN) | phi Ntf (kN) |
|---|---|---|---|---|---|
| M12 | 84.3 | 113 | 26.1 | 57.7 | 70.1 |
| M16 | 157 | 201 | 48.6 | 104.0 | 130.6 |
| M20 | 245 | 314 | 75.9 | 162.6 | 203.8 |
| M22 | 303 | 380 | 93.9 | 197.6 | 252.1 |
| M24 | 353 | 452 | 109.4 | 234.9 | 293.7 |
| M27 | 459 | 573 | 142.2 | 297.5 | 381.9 |
| M30 | 561 | 707 | 173.8 | 367.5 | 466.7 |
| M36 | 817 | 1018 | 253.1 | 528.5 | 679.7 |
Combined Shear and Tension
When a bolt is subject to combined shear and tension, AS 4100 Clause 9.3.4 specifies the interaction check:
For bolts where the threads are in the shear plane (N-grade): (Vf* / phi Vfn)2 + (Ntf* / phi Ntf)2 <= 1.0
For bolts where the threads are excluded from the shear plane (X-grade): (Vf* / phi Vfn)2 + (Ntf* / phi Ntf)2 <= 1.0
The interaction is quadratic — combined loading reduces the available capacity of each component. At Vf* = 0.5 x phi Vfn, the allowable tension reduces to phi Ntf x sqrt(1 - 0.52) = 0.866 x phi Ntf.
Bearing Capacity of Connected Parts
AS 4100 Clause 9.3.2.4 defines the bearing capacity of the connected plate at a bolt hole:
phi Vb = phi x 3.2 x df x tp x fup
Where:
- phi = 0.90 (for plate bearing)
- df = bolt diameter (mm)
- tp = plate thickness (mm)
- fup = minimum tensile strength of the plate (MPa)
For Grade 300PLUS plate (fup = 430 MPa), the bearing capacity per mm of plate thickness:
| Bolt Size | phi Vb per mm tp (kN/mm) |
|---|---|
| M12 | 14.9 |
| M16 | 19.8 |
| M20 | 24.8 |
| M24 | 29.7 |
| M30 | 37.1 |
| M36 | 44.6 |
Note: The bearing capacity is also limited by the edge distance per Clause 9.3.2.5. For bolts near an edge, reduced bearing capacities apply.
Worked Example 1: Simple Shear Connection
Problem: A beam-to-column connection uses 4 x M20 Grade 8.8 bolts in single shear. The threads are included in the shear plane (N-grade). Determine the shear capacity of the bolt group.
Solution:
- From the Grade 8.8 table: phi Vfn (M20, N) = 60.6 kN per bolt
- Number of bolts: 4
- Bolt group shear capacity: 4 x 60.6 = 242.4 kN
Check bearing: The connection plate is 10 mm thick Grade 300PLUS (fup = 430 MPa). phi Vb per bolt = 3.2 x 20 x 10 x 430 / 1000 x 0.90 = 247.7 kN per bolt (governs)
Since bearing capacity (247.7 kN per bolt) exceeds bolt shear capacity (60.6 kN per bolt), shear governs.
Worked Example 2: Tension Connection
Problem: A tension splice uses 6 x M24 Grade 8.8 bolts. Determine the tension capacity of the bolt group.
Solution:
- From the Grade 8.8 table: phi Ntf (M24) = 234.4 kN per bolt
- Number of bolts: 6
- Bolt group tension capacity: 6 x 234.4 = 1406.4 kN
Worked Example 3: Combined Shear and Tension
Problem: A bracket connection has a single M20 Grade 8.8 bolt (N-grade) subject to Vf* = 40 kN shear and Ntf* = 80 kN tension. Check adequacy.
Solution:
- From Grade 8.8 table: phi Vfn (M20, N) = 60.6 kN, phi Ntf = 162.7 kN
- Interaction: (40 / 60.6)2 + (80 / 162.7)2 = 0.436 + 0.242 = 0.678
- 0.678 < 1.0 — the bolt is adequate
Design Resources
- Australian Steel Design Guide — AS 4100 overview
- Australian Bolt Grades — Grade 8.8 and 10.9 properties
- Australian Bolt Hole Sizes — AS 4100 hole dimensions
- Australian Bolt Pretension — Slip-critical requirements
- AS 4100 Base Plate Design — Base plate reference
- Bolted Connections Calculator
- Beam Capacity Calculator
- Column Capacity Calculator
Frequently Asked Questions
How is bolt shear capacity calculated per AS 4100? Bolt shear capacity per AS 4100 Clause 9.3.2.1: Vfn = 0.62 x fuf x kr x A (threads excluded) or Vfn = 0.372 x fuf x kr x A (threads included). The factored capacity is phi Vfn with phi = 0.80. The area A is the body area (Ab) when the shear plane passes through the non-threaded portion, or the tensile stress area (At) when it passes through the threaded portion.
What is the difference between N-grade and X-grade bolt capacity? N-grade means the shear plane passes through the threaded portion of the bolt. X-grade means the shear plane passes through the non-threaded bolt body. For AS 4100, N-grade Vfn uses 0.372 x fuf x At, while X-grade Vfn uses 0.62 x fuf x Ab. For M20 Grade 8.8, X-grade capacity is approximately 2.14 times N-grade capacity (129.9 kN vs 60.6 kN). Where possible, detailing threads out of the shear plane significantly increases connection capacity.
What is the bolt tension capacity per AS 4100? Bolt tension capacity per AS 4100 Clause 9.3.2.1: Ntf = fuf x At x kr. For M20 Grade 8.8: Ntf = 830 x 245 / 1000 = 203.4 kN nominal. Factored: phi Ntf = 0.80 x 203.4 = 162.7 kN. The tension area used is the tensile stress area (At), not the body area, because the threaded portion is always present in the tension load path.
When does the bolt length reduction factor kr apply? The kr factor applies when the bolt grip length (total thickness of connected plies) exceeds 12 times the bolt diameter per AS 4100 Clause 9.3.2.2. For M20 bolts, kr = 0.85 when grip > 240 mm. For M24 bolts, kr = 0.85 when grip > 288 mm. Additionally, kr = 0.75 for any bolt with packers thicker than 6 mm, regardless of bolt length.
How is combined shear and tension checked in AS 4100? AS 4100 Clause 9.3.4 specifies a quadratic interaction: (Vf*/phi Vfn)2 + (Ntf*/phi Ntf)2 <= 1.0. This is a circular interaction curve (not linear), meaning that combined loading reduces capacity faster than simple addition would suggest. For bolts with significant combined load, the interaction typically governs over either pure shear or pure tension alone.
Educational reference only. All design values must be verified against the current edition of AS 4100:2020 and the project specification. This information does not constitute professional engineering advice. Always consult a qualified structural engineer for design decisions.