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    Steel / Bolt groups

    Eccentric bolt group

    When the load does not pass through the centroid of a bolt group, the group twists as well as translates. This vector-sums the direct and torsional shear on every bolt and reports the worst one — the check a hand calculation gets wrong most often.

    Bolts4Eccentricity200 mmVerdictPass
    Critical bolt
    45.1 kN
    at (-50, -50)
    Utilisation
    75%
    capacity 60 kN/bolt
    Direct shear
    12.5 kN
    P/4
    Torsional
    35.4 kN
    M·r/Σr²
    Group response
    Elastic method about the group centroid.
    Group response
    QuantityValue
    Bolts4
    Polar moment Σ(x² + y²)2.000e+4 mm²
    Direct shear per bolt12.50 kN
    Torsional component at critical bolt35.36 kN
    Resultant at critical bolt45.07 kN
    Bolt capacity60 kN

    Elastic method: the group rotates about its centroid, which is conservative.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Bolt force against eccentricity
    Torsion grows linearly with the lever arm.
    Bolt force against eccentricity0.00017.9335.8653.7971.7289.640 mm0 mm50 mm50 mm100 mm100 mm150 mm150 mm200 mm200 mm300 mm300 mm400 mm400 mmMax bolt force (kN)
    Canonical equations
    The exact relationships implemented by this engine.

    f_d = P / n · Shared equally by the group.

    f_t = M·r / Σ(x² + y²) · Perpendicular to the radius.

    f = √(f_dx + f_tx)² + (f_dy + f_ty)² · Vector sum, not arithmetic sum.

    engineering methodology

    A clear path from inputs to results.

    When a load does not pass through the centroid of a bolt group, the group rotates as well as translates. The bolt furthest from the centroid sees the sum of both, and that vector sum — not an arithmetic addition — is what the connection must carry.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Set the grid: columns, rows and spacings.
    2. Enter the design load and its eccentricity from the centroid.
    3. Read the critical bolt force and compare it with the single-bolt capacity.

    The formula

    fd=Pn,ft=MrΣ(x2+y2),f=(fdx+ftx)2+(fdy+fty)2f_d = \frac{P}{n}, \qquad f_t = \frac{M r}{\Sigma(x^2 + y^2)}, \qquad f = \sqrt{(f_{dx}+f_{tx})^2 + (f_{dy}+f_{ty})^2}

    Reference: AISC Manual Part 7, elastic method (conservative against the instantaneous centre method).

    Worked example

    2 × 2 group, 100 mm spacing, 50 kN at 200 mm

    Inputs
    • Four bolts, capacity 60 kN each.
    Calculation path
    1. 1

      Σr² = 20,000 mm². Direct 12.5 kN. M = 10,000 kN·mm, torsional at the critical bolt 35.4 kN. Vector sum 45.1 kN.

    Result

    45.1 kN on the critical bolt, 75% of its 60 kN capacity.

    Interpreting the result

    Torsion grows linearly with the lever arm, so halving the eccentricity roughly halves the torsional component but leaves the direct shear unchanged.

    Assumptions & limitations

    • Elastic method: the plate is rigid and bolts share load by stiffness.
    • Load applied in the plane of the group.
    • Bolt capacities come from the connection check, not from this page.
    • The instantaneous centre method would give 10–20% more capacity.

    Frequently asked questions

    Why not add the components arithmetically?

    They are perpendicular vectors; adding magnitudes overstates the force by up to 40%.

    Is the elastic method always conservative?

    For a ductile group at ultimate it can be slightly conservative or slightly unconservative near the centre; the IC method is the refinement.

    Which bolt is critical?

    Usually the one furthest from the centroid in the direction that combines both components — the page names it.

    Does the plate have to be rigid?

    The elastic method assumes it; a flexible plate redistributes load and needs a different analysis.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.