The direct shear $\vec P/n$ is equal on every bolt and parallel to the load. The torsional shear is proportional to the distance $r_i$ from the centroid and perpendicular to the radius (tangent $\hat t_i$). The bolt with the largest resultant $|\vec F_i|$ is the critical bolt. $J=\sum r_j^2$ is the polar moment; $M=Pe$ is the moment about the centroid. This simplified model assumes elastic behaviour and rigid rotation; slip, bearing, and code factors need separate checks.
What is an eccentric bolt group?
When a load P does not pass through the centroid of a bolt group, the connection carries not only shear but also a moment M = P·e about the centroid. This "eccentric shear" gives each bolt two parts: (1) a direct shear P/n shared equally by all bolts, and (2) a torsional shear P·e·rᵢ/J that resists the rotation. Because these are vectors, the resultant Rᵢ on each bolt is a vector sum — not a plain addition. This is the elastic vector method.
In the live view, three arrows leave each bolt — direct shear (cyan), torsional shear (orange), and resultant (yellow) — and they stretch in real time as the load and eccentricity change. The bolt with the largest resultant (red) is the critical bolt; the design is governed by keeping that single bolt below the allowable.
Physics model and key equations
With rᵢ the distance from the centroid to bolt i, the polar moment is $J=\sum_j r_j^2$. The direct shear is $\vec P/n$ parallel to the load; the torsional shear has magnitude $M r_i/J$ and acts perpendicular to the radius (tangent). The resultant is the vector sum:
Here θᵢ is the angle between the two components. The resultant is largest where they line up — so the farthest bolt is not always critical; alignment of the directions matters, which is exactly why this tool uses a vector sum.
Learn the eccentric bolt group by dialogue
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An eccentric bolt group is basically a bolted joint under a moment, right? Why isn't "the farthest bolt is the most loaded" good enough?
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Good question. The torsional shear P·e·rᵢ/J does grow with distance rᵢ. But every bolt also carries the direct shear P/n in the load direction. The resultant peaks on the bolt where those two arrows line up. That's why the red bolt in the figure often sits a bit off from directly under the load — you need to add the vectors.
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So direction matters. When I raise eccentricity e, the orange arrows keep getting longer.
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Right — the torsional part scales with M=P·e, so it grows linearly with e. Set e=0 and the orange vanishes: every bolt sees only the same cyan direct shear (pure shear). Push e up and the orange overtakes the cyan, so the critical resultant climbs fast. On a steel bracket connection, taking the arm too long is exactly how this eccentricity overloads the bolts.
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Does adding bolts or spreading the layout help? Increasing "Layout scale r" seems to lower the resultant.
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It does. J=Σrᵢ² grows with the square of the spread, so widening the layout lowers the torsional shear M·rᵢ/J. More bolts n also lowers the direct shear P/n. Compare the "single row" and "3×3" presets: at the same bolt count, a layout that is tall in the eccentricity direction resists the moment better. That's the usual fix in practice.
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What is the eccentricity map for?
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It colours the critical-bolt utilization over V (vertical) and e (horizontal). The white cross is the current operating point. Green means margin; red means over the allowable. In design you check whether a small increase in e jumps you into the red. A single-point calc can look safe yet flip to the unsafe side once e drifts due to fit-up tolerance. Confirm final numbers against code values (JIS, AISC) and the real material.
Real-world applications
Steel beam-end bracket and gusset connections: a load on an offset arm puts the bolt group under moment.
Machine frame and bracket fastenings: a cantilever load applies eccentric shear to the support bolts.
First-pass checks for crane/conveyor rail brackets and knee-brace (strut) pipe-support connections.
Common misconceptions and cautions
"Farthest bolt = critical bolt" is wrong. The critical bolt is the one where the direct and torsional shears align. This tool computes the vector sum on every bolt.
Do not add the components scalar-wise. Direct and torsional shear are vectors, so the maximum bolt force is combined with a square root (largest when codirectional).
This elastic vector method is a first-pass estimate assuming rigid rotation and elastic behaviour. Slip resistance (slip-critical high-strength bolts), bearing, plate yielding, and code-specific factors need separate checks.
FAQ
For each bolt the direct shear P/n and the torsional shear P·e·rᵢ/J are summed as vectors; the bolt with the largest resultant |Fᵢ| is the critical bolt. It is not necessarily the farthest bolt — the position where the two components align governs. This tool computes all bolts and highlights it in red.
It is the sum of squared distances from the centroid, J=Σrᵢ². It is the denominator of the torsional shear M·rᵢ/J; spreading the layout outward increases it with the square of the distance and lowers the torsional stress. It is the measure of resistance to the eccentric moment.
The torsional shear scales with M=P·e, so reducing e (moving the load toward the centroid) is the most effective. If that is impractical, widen the layout scale r to increase J, or add bolts n. Use the eccentricity map to confirm the operating point stays in the green region.
The elastic vector method is a first-pass estimate assuming rigid rotation and elastic behaviour. It does not cover slip in slip-critical bolts, bearing, plate yielding, or code-specific correction factors. Final decisions require code values (JIS, AISC, etc.), measured data, detailed analysis, and vendor limits.
How to Use
Enter applied shear force V in kN and eccentricity distance e in mm from bolt group centroid.
Input bolt circle radius r in mm and total number of bolts n in the group.
Simulator calculates direct shear per bolt (V/n), moment arm effect (M·c/J), resultant force, and utilization ratio against bolt tensile capacity.
Worked Example
A W14x90 beam connection with 8 Grade 8.8 M20 bolts arranged in circle r=80mm receives V=120kN at e=150mm from centroid. Direct shear: 120/8=15kN per bolt. Polar moment J=2·π·80²=40,212mm⁴. Moment M=120×150=18,000N·mm. Maximum eccentric component=18,000×80/40,212=35.7kN. Resultant on critical bolt=√(15²+35.7²)=38.6kN. With M20 tensile area=245mm², utilization=38,600/(245×640)=24.6%.
Practical Notes
Eccentric loads on unstiffened angles or clip-angle connections create rotation; position critical bolt diametrically opposite load application for maximum force.
Increase bolt circle radius or number of bolts to reduce eccentric stress concentration; r=100mm versus r=60mm reduces moment arm effect by 40%.
Grade 8.8 bolts (640 MPa tension) tolerate higher utilization than Grade 5.6 (300 MPa); verify material for high-eccentricity industrial connections.