Australian Bolt Group Capacity — AS 4100 Clause 9.3 Eccentric Load

Complete reference for bolt group capacity under eccentric loading per AS 4100:2020. Covers the elastic vector method for preliminary design, the instantaneous centre of rotation (ICR) method for rigorous analysis, prying action in tension bolts, combined shear and tension, and practical bolt pattern design.

Related pages: AS 4100 Bolt Bearing & Tearout | Bolt Spacing & Edge Distance | Bolt Torque Chart | Bolted Connection Calculator

Bolted Connection Design Philosophy — AS 4100 Clause 9.3

AS 4100 treats bolted connections through three limit states:

1. Bolt shear or tension failure (Clause 9.3.2) — the bolt shank or threaded portion fails.
2. Bearing and tearout of the connected ply (Clause 9.3.3) — the plate material around the bolt hole deforms or fractures.
3. Block shear rupture (for coped beams or gusset plates) — a block of material tears out along shear and tension planes.

This page focuses on limit state 1 (bolt group capacity) for the most common case of eccentrically loaded shear connections.

Elastic Vector Method

The elastic (vector) method assumes the bolt group rotates as a rigid body about its centroid. Each bolt resists the applied load with a force proportional to its distance from the centroid. This method is conservative and appropriate for all preliminary designs.

Procedure:

1. Determine the bolt group centroid (by symmetry for regular patterns).
2. Compute the polar moment of inertia: J = sum(r_i^2) for all bolts, where r_i is the distance from each bolt to the centroid.
3. Direct shear per bolt (equal distribution): V_d = P / n (in direction of applied load).
4. Torsional shear on bolt i: V_t_i = P x e x r_i / J (perpendicular to the radius from centroid).
5. Resultant on critical bolt: V_res = sqrt(V_d_x + V_t_x)^2 + (V_d_y + V_t_y)^2) — vector addition.

Check: V_res <= phi_Vfn (bolt shear capacity, Clause 9.3.2.3)

Elastic Method Example — 6-Bolt Bracket

Consider a 2-row by 3-bolt pattern (bolt spacing 70 mm horizontal, 75 mm vertical). Load P = 200 kN at eccentricity e = 250 mm from the bolt group centroid. Bolts are M20 Grade 8.8 (phi_Vfn = 107 kN, shank in shear plane).

Centroid at geometric centre. Bolt coordinates relative to centroid: Row 1 (y = -37.5), Row 2 (y = +37.5). Columns at x = -70, 0, +70.

Polar moment J = sum(x_i^2 + y_i^2) for 6 bolts: J = 2 x [(-70)^2 + (-37.5)^2 + 0^2 + (-37.5)^2 + 70^2 + (-37.5)^2] = 2 x [4900 + 1406 + 0 + 1406 + 4900 + 1406] = 2 x 14018 = 28,036 mm^2... wait, let me recalculate correctly.

J = 6 x (37.5)^2 + 4 x (70)^2 = 6 x 1406.25 + 4 x 4900 = 8437.5 + 19600 = 28,038 mm^2.

Direct shear per bolt: V_d = 200/6 = 33.3 kN (vertical, in direction of applied load).

Torsional shear on corner bolt (x = +/-70, y = +/-37.5, r = sqrt(70^2 + 37.5^2) = sqrt(4900 + 1406) = sqrt(6306) = 79.4 mm):

V_t = P x e x r_corner / J = 200 x 250 x 79.4 / 28038 = 141.6 kN.

Vector addition: V_d acts vertically at 33.3 kN. V_t acts perpendicular to r (which is at angle tan^-1(37.5/70) = 28.2 degrees from horizontal, so V_t components are V_tx = 141.6 x sin(28.2) = 66.9 kN, V_ty = 141.6 x cos(28.2) = 124.9 kN).

Resultant: V_res = sqrt(66.9^2 + (33.3 + 124.9)^2) = sqrt(4476 + 25027) = sqrt(29503) = 171.8 kN.

171.8 kN > 107 kN (phi_Vfn for M20). This bolt group FAILS the elastic check at 200 kN with e = 250 mm.

Instantaneous Centre of Rotation (ICR) Method

The ICR method recognises that the bolt group does not rotate about its centroid under eccentric load. Instead, the instantaneous centre shifts toward the load line, and bolt forces follow a non-linear load-deformation relationship. This provides a more accurate capacity assessment, typically 15-35% higher than the elastic method.

The ICR location is found by iterating until the three equilibrium equations are satisfied: sum of horizontal bolt forces = 0, sum of vertical bolt forces = P, and sum of bolt moments about the ICR = P x e_ICR (where e_ICR is the eccentricity of P from the ICR).

Bolt load-deformation model (Clause 9.3.8 Commentary): R_i = R_ult x (1 - e^(-mu x Delta_i))^lambda

Where typical values from calibration studies are mu = 10.0 and lambda = 0.55 for Grade 8.8 bolts in bearing-type connections with standard holes.

Deformation: Delta_i = r_i x theta, where r_i is the distance from bolt i to the ICR (not the centroid) and theta is the rotation of the group.

The ICR method is computationally intensive for hand calculation but is standard in commercial software (SteelCalculator.app, Limcon, RAM Connection).

C-Values for Rapid Design

The C-value approach pre-computes ICR results and presents them as a multiplier on single-bolt capacity. For a bolt group with a given geometry, the eccentric load capacity is:

P_design = C x phi_Vfn

C-values for 4-bolt rectangular patterns (M20 8.8/S, Grade 300 plate):

A 4-bolt group with e = 200 mm and s = 75 mm can carry P = 1.62 x 107 = 173 kN.

Prying Action in Tension Bolts — Clause 9.3.4

For bolts in tension (end plates, tension splices, column base plates with uplift), prying action amplifies the bolt force beyond the applied tension. Prying occurs when the connected plate bends, creating a lever effect that magnifies bolt tension.

Prying force: Q = T* x (b' - d_b/2) / (a' - d_b/2)

Where a' and b' define the bolt geometry relative to the plate edge and stiffener/beam flange. The bolt must be designed for T* + Q, not just T*.

Mitigation:

• Increase plate thickness (reduces bending, Q approaches zero for very thick plates)
• Reduce the distance from bolt centre to the plate edge (smaller lever arm)
• Provide stiffeners to eliminate the prying mechanism entirely
• Use pre-tensioned bolts (HSFG) which clamp the plates together and reduce prying

Worked Example — 8-Bolt Column Connection

Problem: An 8-bolt connection (4 rows x 2 bolts per row, bolt spacing 75 mm vertical, 100 mm horizontal gauge) connects a 310UC137 column to a base plate. The connection carries V* = 380 kN shear at an eccentricity e = 150 mm from the bolt group centroid (due to beam framing eccentricity). M20 Grade 8.8 bolts, threads excluded from shear plane (AX).

Step 1 — Bolt group layout: Rows at y = -112.5, -37.5, +37.5, +112.5 mm (75 mm spacing, total 225 mm height). Columns at x = +/-50 mm (gauge = 100 mm). 8 bolts total.

Step 2 — Elastic method: Bolt coordinates: for each of 4 rows, two bolts at (-50, y) and (+50, y). J = sum(x^2 + y^2) = 8 x (50^2) + 2 x (112.5^2 + 37.5^2 + 37.5^2 + 112.5^2) = 8 x 2500 + 2 x (12656 + 1406 + 1406 + 12656) = 20000 + 2 x 28124 = 20000 + 56248 = 76,248 mm^2.

Direct shear per bolt: V_d = 380/8 = 47.5 kN.

Critical bolt (corner, at x = 50, y = 112.5, r = sqrt(50^2 + 112.5^2) = sqrt(2500 + 12656) = sqrt(15156) = 123.1 mm):

V_t = 380 x 150 x 123.1 / 76248 = 92.0 kN.

Vector components: V_t at angle tan^-1(112.5/50) = 66 degrees from horizontal. V_tx = 92.0 x cos(66) = 37.4 kN. V_ty = 92.0 x sin(66) = 84.0 kN.

Resultant: V_res = sqrt((37.4)^2 + (47.5 + 84.0)^2) = sqrt(1399 + 17318) = sqrt(18717) = 136.8 kN.

136.8 kN > 107 kN (phi_Vfn M20 AX). Elastic method FAILS.

Step 3 — ICR method (C-value approach): From tabulated values (interpolated for 8-bolt 2x4 pattern, e = 150 mm, gauge = 100 mm, spacing = 75 mm): C ~ 4.2.

P_ICR = C x phi_Vfn = 4.2 x 107 = 449 kN > 380 kN. OK.

The ICR method shows the connection is adequate with a ratio of 380/449 = 0.85. The elastic method is conservative by about 28% here.

Step 4 — Check alternate failure modes:

• Bolt bearing on plate: phi_Vb = phi x 3.2 x d_f x t_p x f_up = 0.90 x 3.2 x 20 x 20 x 440 / 1000 = 507 kN per bolt >> 136.8 kN. OK.
• Tearout: Edge distance = 40 mm. Not governing for this geometry.
• Block shear: Check path around bolt group. With 20 mm plate Grade 300, typically not governing for 8 bolts.

Final design: 8-M20 Grade 8.8/S bolts, 20 mm base plate Grade 300, 75 mm bolt spacing, 100 mm gauge, standard 22 mm holes.

Block Shear Failure — Clause 9.3.6

Block shear is a fracture mode where a block of plate material tears out along a shear path parallel to the load and a tension path perpendicular to it. This is critical for connections where the bolt group is close to the plate edge (coped beam webs, gusset plates).

Nominal block shear capacity: The lesser of shear yield + tension fracture, and shear fracture + tension yield, where the shear and tension areas are defined by the bolt group perimeter. The capacity factor phi = 0.75 (fracture limit state).

In practice, block shear governs when the edge distance from the bolt group to the plate edge is less than about 1.5 times the bolt spacing. Providing adequate edge distance (> 1.5 x bolt diameter) usually eliminates block shear as a governing limit state.

Practical Bolt Pattern Design

• Minimum bolt spacing: 2.5 x bolt diameter (Clause 9.3.5), typically 50 mm for M20 bolts.
• Preferred spacing: 3 x bolt diameter (75 mm for M20) — provides wrench clearance and reduces bearing interaction.
• Minimum edge distance: 1.5 x bolt diameter (30 mm for M20) for sheared edges; 1.25 x bolt diameter for rolled edges.
• Gauge distance for column connections: Match the column section gauge where possible. For UC sections, gauges of 100-150 mm are typical.
• Avoid single-bolt connections in primary members — redundancy is a fundamental connection design principle.

Frequently Asked Questions

When should I use the ICR method instead of the elastic method for bolt groups? The ICR method should be used for production design when the elastic method fails (i.e., V_res > phi_Vfn), when the eccentricity is large (e > 3 x bolt spacing), or when the connection is critical and every kN of capacity matters. The elastic method is always acceptable for preliminary sizing and for configurations where the result is clearly conservative. A 15-35% capacity gain from ICR can eliminate a bolt row, saving fabrication cost.

How does prying affect bolt group design in end plate connections? Prying amplifies bolt tension by 10-30% for typical end plate geometries. The prying force Q must be added to the direct tension T* before checking bolt tension capacity. Thick end plates (tp > bolt diameter / 2) essentially eliminate prying. AS 4100 Clause 9.3.4 provides the prying model, which is a simplified T-stub approach. For moment end plates, the top row of bolts in the tension zone experiences both direct tension and prying and is typically the critical bolt.

What is the capacity of a single M20 Grade 8.8 bolt in shear? phi_Vfn = 0.80 x 0.62 x 830 x 245 / 1000 = 100.8 kN (threads in shear plane, N category). If the shear plane passes through the shank (threads excluded, X category): phi_Vfn = 0.80 x 0.80 x 830 x 314 / 1000 = 166.7 kN per bolt, where 314 mm^2 is the shank area. Always specify whether threads are in or out of the shear plane on fabrication drawings.

This page is for educational reference. Bolt group analysis per AS 4100:2020 Clause 9.3. Verify C-values and bolt capacities against current ASI design guides. All structural designs must be independently verified by a licensed Professional Engineer or Structural Engineer. Results are PRELIMINARY — NOT FOR CONSTRUCTION.