Bolt Group Capacity — Eccentrically Loaded Bolt Groups

When the line of action of a force does not pass through the centroid of a bolt group, the bolts must resist both direct shear and a torsional moment. The instantaneous center of rotation (ICR) method is the most accurate approach. AISC Manual Tables 7-6 through 7-14 provide tabulated C coefficients based on this method.

Concentric loading (simple case)

For bolt groups loaded through the centroid, each bolt carries an equal share: phi*Rn_group = n * phi*Rn_bolt, where n = number of bolts and Rn_bolt is the lesser of bolt shear and bearing/tearout at each location (AISC 360-22 Sections J3.6 and J3.10). Always check bearing/tearout at edge bolts and block shear for the pattern.

Eccentric loading — elastic method

The elastic method assumes linear force-deformation and finds the critical (most distant) bolt. Each bolt carries direct shear r_v = P/n plus torsional force r_m,i = Md_i / sum(d_i^2), where M = Pe (eccentricity moment) and d_i = distance from bolt to centroid. The resultant r_max = vector sum of direct and torsional components on the critical bolt.

Instantaneous center of rotation (ICR) method

The ICR method uses a nonlinear bolt force-deformation curve: R_i = R_ult*(1 - e^(-10*delta_i))^0.55 (AISC Manual Equation 7-1). It finds the instantaneous center by iteration. Each bolt's deformation is proportional to its distance from the IC. This method typically gives 15-30% higher capacity than the elastic method because it accounts for load redistribution among bolts.

Using AISC Manual tables (C coefficients)

AISC Manual Tables 7-6 through 7-14 tabulate coefficient C for standard patterns:

phiRn_group = C * phiRn_bolt

C represents the effective number of bolts. For example, Table 7-7 (single vertical line), 4 bolts at 3" spacing with e = 6": C approximately 2.10, so phiRn_group = 2.10 * 17.9 = 37.6 kips.

Worked example — 6-bolt bracket (2 columns x 3 rows)

Given: 6-bolt bracket (2 columns, 3 rows), 3/4" A325-N bolts in single shear, phi*rn = 17.9 kips/bolt. Vertical load P = 60 kips applied at eccentricity e = 8" from bolt group centroid. Bolt spacing: 3" pitch (vertical), 5.5" gage (horizontal).

Step 1 — Bolt coordinates relative to centroid: Row 1 (top): (-2.75, 3.0) and (2.75, 3.0). Row 2 (middle): (-2.75, 0) and (2.75, 0). Row 3 (bottom): (-2.75, -3.0) and (2.75, -3.0).

Step 2 — Polar moment of the bolt group: sum(d*i^2) = 6 * 2.75^2 + 4 _ 3.0^2 = 6 _ 7.5625 + 4 _ 9.0 = 45.375 + 36.0 = 81.375 in^2.

Wait -- using Ip = sum(x*i^2 + y_i^2): each bolt has x = +/-2.75 and y = +3, 0, -3. Ip = 6 * 2.75^2 + 2 _ 3^2 + 2 _ 0^2 + 2 _ 3^2 = 45.375 + 18 + 0 + 18 = 81.375 in^2.

Step 3 — Moment: M = P _ e = 60 _ 8 = 480 kip-in.

Step 4 — Forces on critical bolt (top-right, farthest from centroid): d*crit = sqrt(2.75^2 + 3.0^2) = sqrt(7.5625 + 9.0) = sqrt(16.5625) = 4.07 in. Direct shear (vertical): r_vy = P/n = 60/6 = 10.0 kips (downward). Torsional shear: r_mx = M * y / Ip = 480 _ 3.0 / 81.375 = 17.69 kips (horizontal, to the right). r_my = M _ x / Ip = 480 _ 2.75 / 81.375 = 16.22 kips (vertical, upward on right bolts for clockwise moment -- but direction depends on sign convention; the torsional vertical component is downward on the critical bolt for this load direction).

For a vertical load at eccentricity to the right: the moment is clockwise. On the top-right bolt, the torsional component acts perpendicular to the radius, which has components: horizontal = +My/Ip = +17.69 kips (rightward) and vertical = -Mx/Ip = -16.22 kips (downward).

Total vertical: 10.0 + 16.22 = 26.22 kips. Total horizontal: 17.69 kips. Resultant: r_max = sqrt(26.22^2 + 17.69^2) = sqrt(687.5 + 312.9) = sqrt(1000.4) = 31.6 kips.

Step 5 — Check: phirn = 17.9 kips < 31.6 kips. FAILS by elastic method. Need stronger bolts (7/8" A325-N: phirn = 24.4 kips -- still fails) or more bolts.

Step 6 — AISC Table comparison: From Table 7-8 (2 vertical lines, 3 bolts per line, 3" spacing, e = 8"): C approximately 3.07. phiRn*group = 3.07 * 17.9 = 54.9 kips < 60 kips. Still fails by ICR method, but the ICR method shows 54.9 vs. elastic effective capacity of 60/31.6 _ 17.9 = 34.0 kips -- ICR gives 62% higher capacity. Solution: add a 4th row (8 bolts) or increase to 7/8" bolts.

In-plane vs. out-of-plane eccentricity

In-plane: Load in the faying surface plane causes torsion about the centroid. Use ICR method or AISC tables. This is the case covered above.

Out-of-plane: Load perpendicular to the faying surface creates prying moment. Some bolts go into tension, compression is resisted by bearing. Use T-stub or prying action analysis per AISC Manual Part 9. Out-of-plane eccentricity is treated separately and is not combined with in-plane torsion via the C-coefficient method.

Multi-code comparison

AISC 360-22 (USA): ICR method per Manual Tables 7-6 through 7-14. Bolt shear capacity per Section J3.6 with phi = 0.75. Bearing/tearout per Section J3.10 with phi = 0.75. The C-coefficient method is unique to the AISC Manual and is the standard U.S. approach for eccentrically loaded bolt groups.

AS 4100-2020 (Australia): Clause 9.3.2.2 addresses bolt groups under combined shear and moment. The elastic analysis method is used to find the critical bolt force. AS 4100 does not include a C-coefficient approach; the polar moment method (elastic) is standard. phi = 0.8 for bolt shear (higher than AISC's 0.75 but with different load factors). Bolt shear capacity per Clause 9.2.2.1: Vf = phi _ 0.62 _ fuf * Ac (where Ac is the core area).

EN 1993-1-8 (Europe): Clause 3.12 covers bolt groups under eccentric loading. The elastic method distributes forces based on the polar second moment of area of the bolt group. No tabulated C-coefficients exist in Eurocode. Partial safety factor gamma*M2 = 1.25 for bolt shear. Bolt shear resistance per Clause 3.6.1: Fv,Rd = alpha_v * fub _ A / gamma_M2 (alpha_v = 0.6 for 8.8 bolts through the threaded area, 0.5 for 10.9).

CSA S16-19 (Canada): Clause 13.12 covers eccentrically loaded bolt groups. Both elastic and ICR methods are recognized. CSA S16 commentary references the AISC C-coefficient tables as an acceptable design aid. phib = 0.80 for bolt shear. Bolt shear resistance per Clause 13.12.1.2: Vr = 0.60 * phib * n _ m _ Ab * Fu.

Common mistakes

  1. Using only the elastic method for final design. The elastic method can underestimate capacity by 15-30%. Use AISC C-coefficient tables (ICR-based) for economic designs, especially when the elastic method shows a marginal failure.

  2. Forgetting individual bolt limit state checks. The group capacity from the C-coefficient table assumes each bolt reaches its single-bolt capacity. If bearing or tearout at an edge bolt is lower than bolt shear, use the reduced single-bolt capacity: phiRn_group = C * (min of phirn_shear, phirn_bearing).

  3. Ignoring eccentricity entirely. Treating all bolts as equally loaded (P/n) is only valid when the load passes through the bolt group centroid. Even small eccentricities (2-3 bolt diameters) can increase the critical bolt force by 50% or more.

  4. Mixing in-plane and out-of-plane eccentricity methods. In-plane eccentricity uses the ICR/torsion approach. Out-of-plane eccentricity uses prying action (T-stub model). These are separate analyses -- do not combine them by adding torsional shear to prying tension.

  5. Interpolating C-tables beyond their range. The AISC tables cover specific spacings and eccentricities. Extrapolating beyond the table limits (e.g., very large eccentricities or non-standard spacings) requires direct ICR analysis or finite element methods.

Frequently asked questions

What is the ICR method? It finds the point about which the bolt group rotates at failure using a nonlinear bolt force-deformation relationship (R_i = R_ult*(1-e^(-10*delta))^0.55). More accurate and less conservative than the elastic method.

How do I use the AISC bolt group tables? Match your pattern to Tables 7-6 through 7-14 (organized by number of bolt columns and rows), find C for your eccentricity and bolt count, then: phiRn_group = C * phiRn_bolt.

When does eccentricity matter most? When the lever arm is large relative to the bolt group dimensions and when fewer bolts provide less torsional resistance. A rule of thumb: if eccentricity exceeds twice the bolt group depth, capacity drops below 50% of the concentric value.

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Disclaimer

This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against AISC 360-22 Chapter J and the AISC Manual bolt group tables. The site operator disclaims liability for any loss arising from the use of this information.