AS 4100 Bolt Design Guide — Shear, Bearing & Tearout (Clause 9.3)

Complete AS 4100:2020 bolted connection design reference: bolt shear capacity (V_f), ply bearing capacity (V_b), tearout limit state, edge distance rules, bolt spacing requirements, and bolt group analysis. This guide covers Grade 8.8 and 10.9 structural bolts to AS/NZS 1252 with worked examples for a beam-to-column end plate connection using M20 bolts.

This page covers the full bolted connection workflow under AS 4100 Clause 9. The Steel Calculator WASM engine performs bolt group checks automatically for AS 4100 (Australia), AISC 360 (USA), EN 1993 (Europe), CSA S16 (Canada), and IS 800 (India).

PRELIMINARY — NOT FOR CONSTRUCTION. All results are for educational and reference use only. Must be independently verified by a Chartered Professional Engineer (CPEng) or RPEQ before use in any project.

Bolt Grades to AS/NZS 1252

Australian structural bolts are specified to AS/NZS 1252.1:2016. Two grades dominate:

The suffix TF (Tensioned, Fully threaded) and TB (Tensioned, Bearing) indicate the bolt assembly type. Grade 8.8 is the default choice for structural steel in Australia — it provides a good balance of strength and ductility.

AS 4100 vs AISC 360: Australian Grade 8.8 (f_uf = 830 MPa) is equivalent to ASTM A325 (f_uf = 830 MPa for diameters up to M24) in the US system. Grade 10.9 (f_uf = 1,040 MPa) is equivalent to ASTM A490.

AS 4100 Bolt Shear Capacity (Clause 9.3.2.1)

The design shear capacity of a single bolt depends on whether threads are in the shear plane.

Threads in Shear Plane (conservative)

The nominal shear capacity per bolt when threads intercept the shear plane:

V_f = 0.62 x f_uf x (n_n x A_c + n_x x A_o)

where:

• f_uf = 830 MPa (Grade 8.8 minimum tensile strength)
• n_n = number of shear planes with threads in them
• n_x = number of shear planes without threads (shank in shear plane)
• A_c = core area (tensile stress area through the threaded portion)
• A_o = nominal shank area

For a single shear plane with threads in the plane (most common case):

V_f = 0.62 x 830 x A_c

M20 Grade 8.8 Bolt — Example Calculation

M20 bolt properties (coarse thread, M20 x 2.5 pitch):

• Nominal diameter d_f = 20 mm
• A_c = 225 mm^2 (core area / tensile stress area for M20 coarse thread)
• A_o = pi x 20^2 / 4 = 314 mm^2

Threads in shear plane (n_n = 1, n_x = 0):

V_f = 0.62 x 830 x (1 x 225 + 0 x 314) = 0.62 x 830 x 225 = 115.8 kN

Design capacity: phi x V_f = 0.80 x 115.8 = 92.6 kN

Shank in shear plane (n_n = 0, n_x = 1):

V_f = 0.62 x 830 x (0 x 225 + 1 x 314) = 0.62 x 830 x 314 = 161.5 kN

Design capacity: phi x V_f = 0.80 x 161.5 = 129.2 kN

There is a 40% increase in shear capacity when the bolt shank (not the threaded portion) is in the shear plane. For this reason, structural details should specify the grip length such that threads fall outside the shear plane wherever possible.

Bolt Shear Capacity Comparison Table

All values for Grade 8.8 single shear plane, phi = 0.80. Note: M30 bolts use a coarser thread pitch (M30 x 3.5) resulting in a larger core area relative to the nominal area compared to M20.

Ply Bearing Capacity (AS 4100 Clause 9.3.2.4)

When a bolt bears against the hole wall in the connected ply, the bearing capacity V_b must be checked. This is frequently the governing limit state for thin connecting plates.

V_b = 3.2 x d_f x t_p x f_up (for standard holes, a_e >= 2 x d_f)

where:

• d_f = bolt diameter (20 mm for M20)
• t_p = ply thickness (mm)
• f_up = ply ultimate tensile strength (MPa)

For Grade 300 plate (f_up = 440 MPa):

For Grade 250 plate (f_up = 410 MPa), multiply these values by 410/440 = 0.932.

The bearing capacity typically exceeds the bolt shear capacity for plates 8mm and thicker, making bolt shear the governing limit state for standard connection details.

Tearout Limit State (AS 4100 Clause 9.3.2.4 — Edge Distance Reduction)

When the edge distance a_e from the bolt centre to the plate edge is less than 2 x d_f, the bearing capacity reduces linearly:

V_b_reduced = V_b x (a_e / (2 x d_f)) for a_e < 2 x d_f

For an M20 bolt with a_e = 30 mm in 10 mm Grade 300 plate:

a_e / (2 x d_f) = 30 / 40 = 0.75

V_b_reduced = 281.6 x 0.75 = 211.2 kN phi x V_b_reduced = 0.90 x 211.2 = 190.1 kN

At a_e = 25 mm (the absolute minimum for M20 with sheared edges):

a_e / (2 x d_f) = 25 / 40 = 0.625 V_b_reduced = 281.6 x 0.625 = 176.0 kN phi x V_b_reduced = 158.4 kN — still above the bolt shear capacity of 92.6 kN.

This means that for Grade 300 plate 10mm thick, edge distance tearout is unlikely to govern provided the minimum AS 4100 edge distances are observed.

Edge Distance and Bolt Spacing Rules (AS 4100 Clause 9.6)

AS 4100 prescribes both minimum and maximum edge distances and bolt spacing. These limits prevent local plate failure and ensure adequate access for tightening.

Minimum Edge Distance (Table 9.6.2)

From bolt centre to plate edge:

Minimum Bolt Spacing (Pitch)

Minimum centre-to-centre distance s_g = 2.5 x d_f:

• M20: 50 mm
• M24: 60 mm

Maximum Edge Distance and Spacing

Maximum edge distance: 12 x t_p but not greater than 150 mm. Maximum bolt spacing: 15 x t_p (non-stressed direction) or 32 x t_p (stressed direction), with an upper limit of 200 mm to prevent moisture ingress between plies.

Worked Example: Beam End Plate Connection

Problem Statement

A 310UB40.4 beam is connected to a column flange via a 10mm Grade 300 end plate with 4 x M20 Grade 8.8 bolts in a 2 x 2 pattern. The design shear force at the connection is V* = 95 kN.

Connection Geometry

• End plate: 160 mm wide x 250 mm deep x 10 mm thick, Grade 300
• Bolt pattern: 2 columns x 2 rows
• Gauge (horizontal spacing): g = 90 mm
• Pitch (vertical spacing): p = 70 mm
• Edge distance from top/bottom: a_e_top = 90 mm
• Edge distance from sides: a_e_side = 35 mm
• Bolt holes: 22 mm diameter (standard clearance for M20)

Step 1 — Check Minimum Edge Distance

From bolt centre to plate edge: a_e_side = 35 mm > 30 mm (1.5 x d_f for sheared edge) OK. a_e_top = 90 mm > 30 mm OK.

Step 2 — Bolt Shear Capacity (Per Bolt)

Threads in shear plane (conservative for end plate connections):

phi x V_f = 0.80 x 0.62 x 830 x 225 = 92.6 kN per bolt per shear plane.

With a single shear plane (end plate to column flange), V*_bolt = 95 / 4 = 23.75 kN < 92.6 kN per bolt. Shear is not governing.

Step 3 — Ply Bearing Capacity

The 10mm Grade 300 end plate is the critical ply (f_up = 440 MPa):

V_b = 3.2 x 20 x 10 x 440 = 281.6 kN per bolt phi x V_b = 0.90 x 281.6 = 253.4 kN per bolt

Since a_e_side = 35 mm > 40 mm (2 x d_f)? No — 35 mm < 40 mm, so an edge distance reduction applies.

For the outer bolts: a_e = 35 mm. a_e / (2 x d_f) = 35/40 = 0.875.

V_b_reduced = 281.6 x 0.875 = 246.4 kN phi x V_b_reduced = 0.90 x 246.4 = 221.8 kN per bolt

Inner bolts (not near any edge): full V_b = 281.6 kN applies.

23.75 kN << 221.8 kN — bearing is not governing.

Step 4 — Bolt Group Capacity Under Pure Shear

For a symmetrically loaded bolt group under pure shear (no eccentricity):

V*_group = 95 kN Total capacity = 4 bolts x 92.6 kN = 370.4 kN

D/C = 95 / 370.4 = 0.256 — the connection has significant reserve capacity.

Step 5 — Eccentric Shear Check (if applicable)

If the beam reaction is applied at some distance from the bolt group centroid (e.g., an extended end plate or seated connection), the eccentric moment must be considered. Assume a shear load applied at 60 mm below the bolt group centroid:

M* = V* x e = 95 x 0.060 = 5.7 kN.m

Using the elastic vector method, the bolt forces are:

Direct shear per bolt: V_dir = 95 / 4 = 23.75 kN (vertical)

Torsional component: I_p = sum of (x_i^2 + y_i^2) for all bolts

For bolts at (+/-45 mm, +/-35 mm) from centroid: x_i^2 + y_i^2 = 45^2 + 35^2 = 2,025 + 1,225 = 3,250 mm^2 per bolt Total I_p = 4 x 3,250 = 13,000 mm^2

r_max = sqrt(45^2 + 35^2) = 57.0 mm

V_torsion = M* x r_max / I_p = 5.7 x 10^6 x 57.0 / 13,000 x 10^3 = 25.0 kN

The vector sum of direct shear and torsional component (at 90 degrees for a corner bolt):

V_resultant = sqrt(23.75^2 + 25.0^2) = 34.5 kN

D/C = 34.5 / 92.6 = 0.373 — still well within capacity.

Block Shear (AS 4100 Clause 9.2.3)

Block shear is a tearing failure where a block of plate material rips out around the bolt group. Two potential failure paths are checked:

Failure Path 1: Tension rupture + shear yield

phi x V_b1 = phi x (0.60 x f_ui x A_nt + 0.60 x f_yi x A_gv)

where A_nt = net area in tension (perpendicular to load) A_gv = gross area in shear (parallel to load)

Failure Path 2: Shear rupture + tension yield

phi x V_b2 = phi x (0.60 x f_ui x A_nv + f_yi x A_gt)

phi = 0.75 for block shear (AS 4100 Table 3.4)

For the end plate example, block shear rarely governs for standard beam end plates — the end plate is sized for the beam depth, providing ample edge distance in the load direction.

Bolt Capacity Summary Table — AS 4100 vs International

The most notable difference is the capacity factor: AS 4100 uses phi = 0.80 for bolts (vs AISC 0.75), making AS 4100 bolt design about 7% less conservative than AISC for an identical bolt size and grade.

Frequently Asked Questions

What is the design shear capacity of a single M20 Grade 8.8 bolt per AS 4100?

Per AS 4100 Clause 9.3.2.1, the design shear capacity of a single Grade 8.8 M20 bolt with threads in the shear plane is phi-Vf = 0.80 x 0.62 x fuf x Ac x nn = 0.80 x 0.62 x 830 x 225 x 1 = 92.6 kN. For a bolt with the shank (unthreaded portion) in the shear plane, the capacity is 129 kN. The capacity factor phi = 0.80 for structural bolts per AS 4100 Table 3.4.

How does AS 4100 calculate bolt bearing capacity at a ply?

AS 4100 Clause 9.3.2.4 calculates the bearing capacity Vb = 3.2 x df x tp x fup where df is the bolt diameter, tp is the ply thickness, and fup is the ply tensile strength. A reduction factor applies when the edge distance ae is less than 2 x df. For an M20 bolt in 10 mm Grade 300 plate (fup = 440 MPa) with ae = 30 mm: Vb = 3.2 x 20 x 10 x 440 = 281.6 kN, with phi-Vb = 0.90 x 281.6 = 253.4 kN.

What edge distance does AS 4100 require for bolted connections?

AS 4100 Clause 9.6.2 specifies minimum edge distances: 1.5 x df for sheared or hand-flame-cut edges and 1.25 x df for rolled, machined, or sawn edges. For an M20 bolt, this means 30 mm minimum from the bolt centre to the plate edge for sheared edges. Minimum bolt spacing (pitch) is 2.5 x df = 50 mm for M20. These limits prevent premature edge tearing and ensure adequate load distribution in the bolt group.

How does AS 4100 handle bolt groups under eccentric shear?

AS 4100 Clause 9.3.4 permits two analysis methods for bolt groups under eccentric shear: the elastic (vector) method and the instantaneous centre of rotation (IC) method. The elastic method superposes direct shear and torsional components — it is conservative for ductile Grade 8.8 bolts. The IC method accounts for bolt deformation capacity and provides 10-30% more capacity by allowing the bolt group to rotate about an instantaneous centre, but requires iterative solution.

Try it now: Check your bolt design with our free Australian Bolt Capacity calculator

Related Pages

• AS 4100 Beam Design — 310UB40.4 Worked Example — Flexure, shear, LTB
• AS 4100 Column Design — 200UC46 Worked Example — Axial compression and buckling
• AS 4100 Fillet Weld Design — Clause 9.7 — Weld capacity SP/GP method
• AS 4100 Load Combinations — AS 1170.0 — ULS/SLS/STB combinations
• Australia AS 4100 Steel Design Guide — Complete standards reference
• Bolt Shear vs Bearing — Design Guide — Comparative bolt failure modes
• Bolt Group Capacity Calculator — Free bolt group analysis tool
• Block Shear Design — Multi-Code Reference — Block shear per AISC & AS 4100

This page is for educational reference. All resistance formulae are per AS 4100:2020 with AS/NZS 1252 bolt properties. Verify the applicable edition of the National Construction Code for your project jurisdiction. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent review by a registered structural engineer (CPEng/RPEQ).