Canadian Bolt Group Capacity — CSA S16 Eccentric Load Design

Complete reference for the analysis of eccentrically loaded bolt groups in CSA S16-19 steel design. Covers the elastic vector method for preliminary design, the instantaneous centre (IC) method for rigorous capacity analysis, C-value tables from the CISC Handbook, and bolt deformation-based load distribution models.

Related pages: CSA Bolt Bearing & Tearout | CSA Bolt Spacing | CSA Bolt Capacity Tables | Bolted Connection Calculator

Eccentric Bolt Group Philosophy

When a bolt group is subjected to an eccentric load (the line of action of the resultant force does not pass through the bolt group centroid), the bolts resist not only the direct shear but also a torsional moment equal to P x e. The distribution of forces among the bolts depends on the assumed rotation behaviour of the group.

CSA S16-19 does not prescribe a single analysis method. The CISC Handbook of Steel Construction recommends two approaches of increasing accuracy:

1. Elastic vector method — conservative, simple, suitable for all preliminary designs
2. Instantaneous centre (IC) method — rigorous, accounts for non-linear bolt behaviour and ductile load redistribution

Both methods are accepted under CSA S16 provided the bolt capacities per Clause 13.12 are satisfied.

Elastic Vector Method

The elastic method assumes the bolt group rotates as a rigid body about its centroid. Each bolt experiences a direct shear component (equal for all bolts) and a torsional shear component proportional to its distance from the centroid.

Direct shear per bolt: V_dir = P / n (in the direction of the applied load)

Torsional shear on bolt i: V_tor_i = P x e x r_i / J

Where J = sum(r_i^2) is the polar moment of inertia of the bolt group about its centroid (units: mm^2 for bolts treated as point elements).

Resultant on critical bolt: V_res = vector sum of V_dir and V_tor. The critical bolt is typically the one farthest from the centroid where the torsional component is largest.

Check: V_res <= Vr (factored shear resistance per bolt, Clause 13.12.3)

Elastic Method — 8-Bolt Bracket Example

Consider an 8-bolt bracket: 2 vertical rows (gauge = 80 mm) x 4 bolts per row (spacing = 70 mm). Total height = 210 mm. Load P = 300 kN at eccentricity e = 250 mm from the centroid.

Bolt group centroid: at geometric centre (by symmetry). r_max (corner bolt) = sqrt((80/2)^2 + (105)^2) = sqrt(1600 + 11025) = sqrt(12625) = 112.4 mm.

J = sum(r_i^2) = 4 x (40^2 + 105^2) + 4 x (40^2 + 35^2) = 4 x (1600 + 11025) + 4 x (1600 + 1225) = 4 x 12625 + 4 x 2825 = 50500 + 11300 = 61,800 mm^2.

Direct shear per bolt: V_dir = 300/8 = 37.5 kN (downward).

Torsional shear on corner bolt: V_tor = 300 x 250 x 112.4 / 61800 = 136.4 kN. V_tor acts perpendicular to r (at angle tan^-1(105/40) = 69.1 degrees from horizontal).

V_tor_x = 136.4 x cos(69.1) = 48.7 kN (horizontal) V_tor_y = 136.4 x sin(69.1) = 127.5 kN (downward)

Resultant: V_res = sqrt(48.7^2 + (37.5 + 127.5)^2) = sqrt(2372 + 27225) = sqrt(29597) = 172.1 kN.

For a M20 A325M bolt (AX, threads excluded): Vr = 100.0 kN. 172.1 > 100.0 — FAILS elastic check.

Instantaneous Centre (IC) Method

The IC method is based on the experimentally observed behaviour of eccentrically loaded bolt groups: the group does NOT rotate about its centroid. Instead, the instantaneous centre of rotation shifts away from the load line, and individual bolt forces follow a non-linear load-deformation relationship.

Bolt load-deformation model (Kulak, Fisher, and Struik — University of Toronto / CISC research): R_i = R_ult x (1 - exp(-10 x Delta_i))^0.55

Where:

• R_ult = ultimate shear strength of a single bolt
• Delta_i = bolt deformation at bolt i = r_i x theta (distance from IC x rotation)
• 10 and 0.55 are empirically calibrated coefficients for A325 and A490 bolts in standard holes

IC solution procedure:

1. Assume IC location — initially at coordinates (x_0, y_0) relative to the bolt group centroid
2. Calculate r_i for each bolt from the IC (not the centroid)
3. Assume a rotation theta (scale factor)
4. Compute each bolt force R_i from the load-deformation model
5. Check three equilibrium equations:
   • Sum of horizontal bolt force components = 0
   • Sum of vertical bolt force components = P
   • Sum of bolt moments about the IC = P x e_IC (where e_IC is the distance from the IC to the load line)
6. Iterate the IC location (x_0, y_0) until all three equations are satisfied

This is computationally intensive by hand and is typically performed with software (SteelCalculator.app calculates IC bolt group capacities using the Kulak-Fisher-Struik model).

C-Value Tables — CISC Handbook Part 5

For hand calculations, the CISC Handbook provides pre-computed C-values that encapsulate the IC method results. The C-value is the multiplier on single-bolt capacity:

P_ult = C x Vr_per_bolt

Enter the CISC tables with: number of bolts, vertical spacing, horizontal gauge, eccentricity e.

Example C-values for single vertical row of bolts, 75 mm spacing, e measured from bolt line:

For the 8-bolt bracket above: e = 250 mm, s = 70 mm (interpolate between 200 and 300 mm). C ~ 1.72. With M20 AX, Vr = 100.0 kN: P_ICR = 1.72 x 100 = 172 kN. But P = 300 kN — still fails even with IC method. The bracket requires more bolts or a wider pattern.

Improved design: Add a second vertical row (gauge = 80 mm) — now 2 x 4 = 8 bolts in a rectangular pattern. For e = 250 mm with 2 rows at 80 mm, 4 bolts per row at 70 mm: C ~ 3.0 (from CISC Table 5-4). P_ICR = 3.0 x 100 = 300 kN. Exactly at the limit.

Or use A490M bolts: Vr per M20 A490M AX = 1.27 x 100 = 127 kN. C = 1.72 (single row): P_ICR = 1.72 x 127 = 218 kN — still below 300.

Conclusion: For 300 kN at e = 250 mm, use 8 bolts in a 2x4 rectangular pattern with A325M M20 bolts. The double-row pattern provides the torsional stiffness necessary to resist the eccentric moment.

Connection Ductility and the IC Method Advantage

The IC method gives higher capacities than the elastic method (typically 15-35% more) because it accounts for ductile load redistribution. As the most heavily loaded bolt reaches its ultimate capacity, it continues to deform (ductile behaviour) rather than fracturing immediately, allowing neighbouring bolts to pick up additional load. This redistribution is possible because:

• A325 and A490 bolts have sufficient ductility (elongation >= 14% per ASTM F3125)
• Standard holes (2 mm oversize) provide deformation capacity
• Bearing-type connections can accommodate the 2-4 mm deformations associated with full bolt group capacity

For connections in seismic force-resisting systems, the IC method is the default in Canadian practice because it represents actual connection behaviour under large deformations.

Combined Shear and Tension on Bolts

Bolts in end-plate moment connections, column base plates with uplift, and bracing connections experience both shear and tension simultaneously. CSA S16 Clause 13.12.4 provides the interaction equation:

(Vf/Vr)^2 + (Tf/Tr)^2 <= 1.0

This elliptical interaction is more realistic than linear interaction because the shear and tension failure mechanisms are partially independent. A bolt at 80% of its shear capacity can still carry approximately 60% of its tension capacity.

Worked Example — Beam-to-Column Shear Tab

Problem: A W460 beam-to-column shear tab connection with 6-M20 A325M bolts in a single vertical row at 75 mm centres. The beam reaction Vf = 250 kN is offset from the bolt line by e = 100 mm (from the bolt line to the column face, plus half the bolt spacing). Bolts are AX (threads excluded from shear plane). All plates are 350W.

Step 1 — Bolt shear capacity: Vr per M20 A325M AX = 100.0 kN.

Step 2 — Elastic method: V_dir = 250/6 = 41.7 kN per bolt. J = sum(y_i^2). For 6 bolts spanning 375 mm height (5 x 75 = 375): y_i from centroid: [+/-187.5, +/-112.5, +/-37.5] = [35156, 12656, 1406] each end. J = 2 x (35156 + 12656 + 1406) = 2 x 49218 = 98,436 mm^2.

r_max = 187.5 mm (top or bottom bolt). V_tor = P x e x r_max / J = 250 x 100 x 187.5 / 98436 = 47.6 kN.

Resultant: V_res = sqrt(41.7^2 + 47.6^2) = sqrt(1739 + 2266) = sqrt(4005) = 63.3 kN < 100.0 kN. OK.

Step 3 — IC method (C-value): From CISC Table 5-1, for 6 bolts, s = 75 mm, e = 100 mm: C ~ 2.38. P_ICR = 2.38 x 100 = 238 kN < 250 kN. FAILS IC check on capacity.

The elastic method passes but the IC method does not — this is unusual. Reviewing the C-value: at e = 100 mm with 6 bolts, C = 2.38 produces P_ICR = 238 kN vs demand of 250 kN. The ratio is 250/238 = 1.05, a marginal overstress.

Option A: Increase to 8 bolts (C ~ 2.54): P_ICR = 254 kN. OK. Option B: Increase spacing to 100 mm to improve the geometric efficiency. Option C: Accept the elastic method result (conservative check passed) — the elastic method is sufficient per CSA S16 for this non-seismic static connection.

Frequently Asked Questions

When should I use the CISC C-value tables vs the elastic method? Use the elastic method for: (a) preliminary sizing, (b) simple connections where the result is clearly adequate, and (c) when C-values are not readily available for the specific bolt pattern. Use C-values (IC method) for production design when the elastic method indicates marginal overstress and the connection geometry falls within the tabulated range. The IC method extracts an additional 15-35% capacity from the same bolt group.

What is the physical reason the IC method gives higher capacity? The elastic method assumes the bolt group rotates about its centroid, which produces a symmetric force distribution. In reality, the centre of rotation shifts toward the load line, reducing the lever arm on the far-side bolts and allowing a more uniform load distribution. Additionally, the most heavily loaded bolt deforms plastically rather than fracturing, redistributing load to less stressed bolts. This ductile redistribution cannot be captured by elastic analysis.

How does bolt spacing affect the bolt group capacity? Wider bolt spacing increases J (polar moment) proportionally to the square of the distance, improving torsional resistance. However, the minimum spacing is 2.7 x bolt diameter (54 mm for M20) per CSA S16 Clause 22.3, and the maximum for efficiency is 3-4 x bolt diameter (60-80 mm for M20). Spacing beyond 100 mm provides diminishing returns because the lever arm benefit is offset by reduced bearing interaction and block shear considerations.

Can the elastic method ever give unsafe results compared to the IC method? No. The elastic method is always conservative because it assumes all bolts remain elastic and the group rotates about its centroid. Both assumptions under-predict the true capacity. The IC method is the "true" capacity and the elastic method is a lower bound. If the elastic method fails, the IC method should be checked — it may pass. If the elastic method passes, the connection is safe.

This page is for educational reference. Bolt group analysis per CSA S16-19 Clause 13.12 and CISC Handbook Part 5. Verify C-values against the current edition of the CISC Handbook. All structural designs must be independently verified by a licensed Professional Engineer. Results are PRELIMINARY — NOT FOR CONSTRUCTION.