How does the elastic method calculate bolt forces in an eccentrically loaded group?

The elastic method assumes the connecting plate is rigid and rotates about the bolt group centroid. Direct shear V* is distributed equally among n bolts. The eccentric moment M* = V* x e is resisted by bolt forces proportional to distance from centroid: V_moment,i = M* x ri / Σr². The resultant per bolt is the vector sum. The most heavily loaded bolt is furthest from the centroid in the direction of rotation.

What is the shear capacity of an M20 Grade 8.8/S bolt per AS 4100?

Per AS 4100 Clause 9.3.2.1: φVf = φ x 0.62 x fuf x Ac where φ = 0.80, fuf = 830 MPa, Ac = 225 mm² (core area). φVf = 92.6 kN with threads in the shear plane. With shank in shear plane (Ao = 314 mm²), capacity increases to 129 kN — a 40% gain.

When should the instantaneous centre method be used instead of the elastic method?

The elastic method is conservative (assumes linear behaviour to failure). The IC method accounts for bolt ductility and load redistribution, yielding 15-35% higher group capacity. Use elastic for preliminary sizing and small eccentricities (e 0.90. Both methods are acceptable per AS 4100.

AS 4100 Bolt Group Design Worked Example — Eccentric Load, Elastic Method, M20 8.8/S

Bolt group design under eccentric loading is a fundamental calculation for steel connections. A bracket, fin plate, or beam end plate carries shear that does not pass through the bolt group centroid, producing an in-plane moment that loads the outer bolts more heavily than the inner ones. This page walks through a complete AS 4100 bolt group design using the elastic vector method, with six M20 Grade 8.8/S bolts, 200 mm eccentricity, and full bearing and tearout checks.

PRELIMINARY — NOT FOR CONSTRUCTION. This is an educational worked example. All designs must be independently verified by a licensed Professional Engineer before use in any project.

Design Brief

Parameter	Value
Connection type	Bracket plate, single shear, bearing-type
Applied shear	V* = 120 kN (factored, downward)
Eccentricity	e = 200 mm (horizontal from centroid to load line)
Bolts	6 x M20 Grade 8.8/S (fuf = 830 MPa)
Bolt layout	2 columns x 3 rows, 70 mm pitch x 80 mm gauge
Plate material	Grade 300, tp = 12 mm (fup = 440 MPa)
Edge distance	ae = 35 mm (minimum 30 mm per Cl 9.6.2)
Thread condition	Threads in shear plane assumed (conservative)
Capacity factors	0.80 (bolt shear), 0.90 (bearing)

Step 1 — Bolt Group Geometry and Centroid

Six bolts in a 2 x 3 grid. Centroid at the geometric centre: x-bar = 40 mm from each column, y-bar = 70 mm from bottom row.

Distances from centroid to each bolt:

Bolt 1 (bottom-left):  dx = -40, dy = -70, r1 = 80.6 mm, r1^2 = 6,500
Bolt 2 (middle-left):  dx = -40, dy = 0,   r2 = 40.0 mm, r2^2 = 1,600
Bolt 3 (top-left):     dx = -40, dy = +70, r3 = 80.6 mm, r3^2 = 6,500
Bolt 4 (bottom-right): dx = +40, dy = -70, r4 = 80.6 mm, r4^2 = 6,500
Bolt 5 (middle-right): dx = +40, dy = 0,   r5 = 40.0 mm, r5^2 = 1,600
Bolt 6 (top-right):    dx = +40, dy = +70, r6 = 80.6 mm, r6^2 = 6,500

Sum of r^2 = 4 x 6,500 + 2 x 1,600 = 26,000 + 3,200 = 29,200 mm2

Step 2 — Direct Shear Per Bolt

V_dir = V* / n = 120 / 6 = 20.0 kN per bolt (downward)

Each bolt carries 20 kN vertically downward from direct shear.

Step 3 — Moment-Induced Shear Per Bolt

The eccentric moment about the centroid:

M* = V* x e = 120 x 200 = 24,000 kNmm

Shear force on bolt i due to moment (perpendicular to radius vector):

V_moment,i = M* x ri / sum(r^2)

For corner bolts (r = 80.6 mm): V_moment = 24,000 x 80.6 / 29,200 = 66.2 kN
For middle bolts (r = 40.0 mm): V_moment = 24,000 x 40.0 / 29,200 = 32.9 kN

Step 4 — Vector Resultant for Critical Bolt

For Bolt 1 (bottom-left, r = 80.6 mm), resolving moment force components and adding direct shear:

Moment force acts perpendicular to radius vector from centroid: angle approximately 150 degrees from horizontal.

Vx_moment = 66.2 x cos(-30) = 57.3 kN (positive x)
Vy_moment = 66.2 x sin(-30) = -33.1 kN (downward)

Adding direct shear (0, -20):
Vx_total = 57.3 + 0 = 57.3 kN
Vy_total = -33.1 + (-20) = -53.1 kN

V_resultant = sqrt(57.3^2 + 53.1^2) = sqrt(3283 + 2820) = sqrt(6103) = 78.1 kN

The critical corner bolt sees 78.1 kN resultant shear.

Step 5 — Bolt Shear Capacity

Per AS 4100 Clause 9.3.2.1, threads in shear plane:

M20 Grade 8.8/S: fuf = 830 MPa, core area Ac = 225 mm2

Design shear capacity = 0.80 x 0.62 x 830 x 225 x 1 = 92.6 kN

Utilisation: 78.1 / 92.6 = 0.844 (84.4%) OK

If the shank is in the shear plane (grip length controlled), capacity increases to 129 kN, reducing utilisation to 60.5%.

Step 6 — Bearing and Edge Distance Checks

Bearing per AS 4100 Cl 9.3.2.4:

Vb = 3.2 x df x tp x fup = 3.2 x 20 x 12 x 440 = 337.9 kN
Design bearing = 0.90 x 337.9 = 304.1 kN
Utilisation: 78.1 / 304.1 = 0.257 (25.7%) OK

Tearout check (ae = 35 mm < 2 x df = 40 mm, reduced bearing applies):

Vb_tearout = ae x tp x fup = 35 x 12 x 440 = 184.8 kN
Design tearout capacity = 0.90 x 184.8 = 166.3 kN
Utilisation: 78.1 / 166.3 = 0.470 (47.0%) OK

Bolt shear governs at 84.4% utilisation.

Step 7 — Net Section and Block Shear

Net section tension check for the 120 mm wide bracket plate through two bolt holes:

Hole diameter: dh = 22 mm (standard hole)
Net width: 120 - 2 x 22 = 76 mm
Net area: An = 76 x 12 = 912 mm2

Design tension capacity = 0.90 x 0.85 x 1.0 x 912 x 440 = 307 kN
Tension demand (horizontal component) = 57.3 kN
Utilisation: 57.3 / 307 = 0.187 (18.7%) OK

Block shear combines tension on the net section with shear on gross area (failure path through two bolt holes):

Shear area gross: 2 x 175 x 12 = 4,200 mm2
Tension area net: 80 x 12 - 1 x 22 x 12 = 696 mm2

Block shear capacity = 0.90 x (0.6 x 300 x 4,200 + 300 x 696) = 0.90 x (756,000 + 208,800) = 868 kN
Utilisation: 57.3 / 868 = 0.066 (6.6%) OK

Results Summary

Check	Demand	Capacity	Utilisation	Status
Bolt shear (threads in plane)	78.1 kN	92.6 kN	0.844	OK
Bolt shear (shank in plane)	78.1 kN	129.0 kN	0.605	OK
Bearing at bolt hole	78.1 kN	304.1 kN	0.257	OK
Tearout	78.1 kN	166.3 kN	0.470	OK
Net section tension	57.3 kN	307.0 kN	0.187	OK
Block shear	57.3 kN	868.0 kN	0.066	OK

The bolt group is adequate. Bolt shear with threads in the shear plane governs at 84.4% utilisation.

Frequently Asked Questions

What is the difference between the elastic method and the instantaneous centre method?

The elastic method assumes bolt forces are proportional to distance from centroid, which is conservative. The instantaneous centre (IC) method accounts for bolt ductility and load redistribution, typically predicting 15-35% higher group capacity. Use the elastic method for preliminary sizing and small eccentricities. Use the IC method for final design when eccentricity is large. Both methods are acceptable per AS 4100.

When does bearing or tearout govern over bolt shear?

Bearing governs when the connected plate is thin (tp < 8 mm for M20). Tearout governs when edge distance is at the minimum (ae < 2 x df). For standard Australian practice (12 mm minimum plate thickness, 35 mm edge distance), bolt shear governs by a wide margin for Grade 8.8 bolts.

What changes if bolts are in double shear?

In double shear, the shear plane count doubles, giving 2 x 92.6 = 185.2 kN per bolt. Bearing and tearout are checked at each ply separately. Double-shear connections are standard for splice plates, beam-to-column web connections, and bracing gusset plates.

How do I handle bolts in combined shear and tension?

AS 4100 Clause 9.3.3 requires: (V*/phiVf)^2 + (N*/phiNtf)^2 <= 1.0. For the M20 8.8/S with V* = 78.1 kN (84.4% of shear capacity), the remaining tension capacity fraction is sqrt(1 - 0.844^2) = 0.537, meaning the bolt can carry up to 0.537 x 145 kN = 77.8 kN in simultaneous tension.

This worked example is for educational purposes only. Verify all bolt grades, plate materials, and code references against the current editions of AS 4100, AS/NZS 1252, and AS/NZS 3678 before use in any project. All connection designs must be independently checked and certified by a licensed Professional Engineer.