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.

AISC Manual Tables 7-1 through 7-11 overview

The AISC Manual 16th Edition organizes bolt group design data across several table series. Understanding which table to use for each design scenario is essential for efficient connection design:

Table 7-1: Single-bolt shear and tension strengths. Provides phiRn for each bolt grade, diameter, and thread condition (N vs X). This is the starting point for all bolt group calculations — the C coefficient from the eccentric bolt group tables is multiplied by the single-bolt capacity from Table 7-1.

Table 7-2: Single-bolt bearing and tearout strengths. Provides phiRn for bearing on the connected material at bolt holes. Values are tabulated by bolt diameter, material Fy/Fu, hole type (standard, oversized, short-slot, long-slot), and edge distance (1.5d or 3.0d). When bearing controls over bolt shear, use the bearing value as phiRn_bolt in the C-coefficient equation.

Tables 7-3 through 7-5: Single bolts and small groups under special loading. These tables cover prying action, filler considerations, and combined shear-tension interaction for single bolts.

Tables 7-6 through 7-14: Eccentric bolt group C coefficients. These are the core tables for eccentric loading design. Each table covers a specific bolt pattern geometry:

Table 7-15: Coefficients C for bolt groups subject to inclined loads. For loads not aligned with the bolt pattern axes, this table provides C coefficients for specific load angles.

Using the tables efficiently: Enter the table with the number of bolts per line, bolt spacing, and eccentricity. Read the coefficient C. Then compute phiRn_group = C x phiRn_bolt. If the result exceeds the applied load, the bolt group is adequate. If not, increase bolt count, bolt diameter, or bolt grade.

Instantaneous center of rotation method — detailed procedure

The instantaneous center (ICR) method provides the most accurate prediction of eccentric bolt group capacity. It is based on the observation that at ultimate load, the bolt group rotates about a single point (the instantaneous center) and every bolt reaches the same deformation relative to its distance from that center. The method requires iterative computation:

Step 1 — Assume an initial IC location. For a vertical load applied at eccentricity e to the right of the bolt group centroid, the IC is to the left of the centroid at some distance x_IC. Initial guess: x_IC = -e/3.

Step 2 — Compute distances and deformations. For each bolt i at position (xi, yi) relative to the assumed IC: di = sqrt(xi^2 + yi^2). The deformation at each bolt is proportional to its distance: delta_i = delta_max x di / d_max, where delta_max = 0.34 in (the deformation at which a bolt reaches its ultimate strength).

Step 3 — Compute bolt forces. Each bolt force is: Ri = Rult x (1 - e^(-10 x delta_i))^0.55, where Rult is the single-bolt ultimate shear capacity. This is the AISC Manual Equation 7-1 nonlinear force-deformation curve.

Step 4 — Check equilibrium. Three equilibrium equations must be satisfied simultaneously: sum of horizontal forces = 0, sum of vertical forces = 0, and sum of moments about the IC = P x (e + x_IC). If the applied load P is the unknown, solve for P from the moment equilibrium equation and verify force equilibrium.

Step 5 — Iterate. Adjust the IC location and repeat until all three equilibrium equations converge (typically to within 0.1%). The converged P value is the nominal group capacity. The design capacity is phiPn = 0.75 x P.

The ICR method is always equal to or less conservative than the elastic method. The typical range of capacity increase is 15 to 30%, with larger gains for bolt groups with more rows and moderate eccentricities. For very large eccentricities (e greater than 2 x bolt group depth), the two methods converge toward similar results because most of the capacity is used by the few bolts farthest from the IC.

Elastic method worked example — 4-bolt single-line bracket

Given: A single vertical line of 4 bolts (3/4 in A325-N, phiRn = 17.9 kips per bolt) supports a vertical load P = 30 kips at eccentricity e = 5 in from the bolt line. Bolt spacing = 3 in vertical pitch.

Step 1 — Bolt coordinates relative to group centroid: Bolts at y = +4.5, +1.5, -1.5, -4.5 in (4 bolts at 3 in pitch, centroid at midheight). All bolts have x = 0 (single line).

Step 2 — Polar moment of inertia: Ip = sum(yi^2) = 4.5^2 + 1.5^2 + 1.5^2 + 4.5^2 = 20.25 + 2.25 + 2.25 + 20.25 = 45.0 in^2.

Step 3 — Applied moment: M = P x e = 30 x 5 = 150 kip-in.

Step 4 — Forces on critical bolt (top or bottom bolt, farthest from centroid): Torsional shear (horizontal): r_mx = M x y_crit / Ip = 150 x 4.5 / 45 = 15.0 kips (horizontal). Direct shear (vertical): r_vy = P / n = 30 / 4 = 7.5 kips (vertical). Resultant: r_max = sqrt(15.0^2 + 7.5^2) = sqrt(225 + 56.25) = sqrt(281.25) = 16.8 kips.

Step 5 — Check capacity: phiRn = 17.9 kips per bolt. r_max = 16.8 kips less than 17.9 kips. OK.

Step 6 — Compare with AISC Table 7-6 (ICR method): From Table 7-6, 4 bolts at 3 in pitch, e = 5 in: C = 2.30. phiRn_group = 2.30 x 17.9 = 41.2 kips. This is 37% higher than the elastic method capacity of 30 kips x (17.9/16.8) = 32.0 kips. The ICR method confirms the connection is adequate with significant margin.

Bolt group coefficient reference tables

The following abbreviated tables provide C coefficients for common bolt patterns and eccentricities. These are extracted from AISC Manual Tables 7-6 and 7-7 for preliminary design use:

Single vertical line (Table 7-6 type), 3 in pitch

Two vertical lines, 3 in pitch, 5.5 in gage (Table 7-8 type)

How to use these tables: Find your bolt count and eccentricity. Multiply C by the single-bolt capacity (phiRn from Table 7-1). The result is the design capacity of the bolt group. For example, 4 bolts in a single line at 3 in pitch with e = 8 in, 3/4 in A325-N: phiRn_group = 1.77 x 17.9 = 31.7 kips.

Important: These C coefficients assume the single-bolt limit state is bolt shear. If bearing/tearout at an edge bolt controls with a lower capacity, use that lower value: phiRn_group = C x phiRn_bearing.

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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Related references

• Bolt Capacity Table
• Bolt Spacing Requirements
• Bolt Bearing and Tearout
• Bolt Grades
• Gusset Plate Design
• How to Verify Calculations

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.

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