Plug Weld Shear⚠ unverified
Mechanical / Welds · Compute the shear stress in a plug weld
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Applied load on the weld |
| d | d | m | 1.0 | Diameter of each plug weld |
| n | n | — | 1.0 | Number of plug welds, dimensionless |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | τ | Pa | Shear stress in the plug weld, in pascals (Pa). Returns 0.0 if ``d`` is not positive |
The science & history
Understanding the Parameters
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Plug diameter $d$ — the effective shear area is the hole's cross-section at the faying (contact) plane, $\pi d^2/4$. Codes cap the maximum hole size relative to plate thickness so the weld can fill completely; too large a plug leaves an unfused centre.
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Number of plugs $n$ — load is assumed to divide equally among the plugs. That is exact only when the group is symmetric about the load line; an eccentric group also develops torsional (secondary) shear that this formula omits (see Eccentric Weld Secondary Shear).
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Shear stress $\tau$ — averaged over the faying area. Real plug welds concentrate stress at the fusion boundary and can trap slag/porosity, so allowable values are conservative and inspection is limited (the weld interior is hidden).
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Failure plane — a plug weld is checked on the faying-surface area (the shear plane between plates), not on a $45^\circ$ throat; there is no $0.707$ factor.
Derivation (Approaching a Proof)
Each plug is a solid cylinder of weld metal spanning the interface. Under a shear load parallel to the plates, the fused disc at the faying plane is the surface that must shear. One plug presents a circular shear area
$$A_1 = \frac{\pi d^2}{4}.$$
With $n$ plugs sharing the load equally, the total resisting area is $A = n\,\pi d^2/4$, and the average shear stress is force over area:
$$\tau = \frac{F}{A} = \frac{F}{n\,\pi d^2/4} = \frac{4F}{n\,\pi d^2}.$$
Compare against the allowable weld shear stress to size the plug diameter or count.
Dimensional check. $\dfrac{F}{n\,\pi d^2/4} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$ — a stress ($n$ and $\pi/4$ are dimensionless).
History and Development
Plug and slot welds entered the codes as the welded successors to rivets and bolts for shear transfer across overlapped plates — for example stitching a cover plate to a flange, or a doubler over a wide panel where the interior is unreachable by a fillet. AWS D1.1 and AISC fix the geometry (hole size versus thickness, minimum spacing, fill requirements) precisely because the fused volume is buried and cannot be visually inspected, so the rules trade strength for reliability.
Related Concepts: Slot Weld Strength, Fillet Weld Capacity, Butt Weld Strength, Rivet Shear Stress, Shear Stress
Notes: Checked on the faying-plane circular area $\pi d^2/4$ per plug (no $0.707$ throat factor). Equal load-sharing assumes a symmetric group; eccentric groups add torsional shear (see Eccentric Weld Secondary Shear). Returns $0$ for non-positive $d$.