Weld Torsional Shear⚠ unverified
Mechanical / Welds · Compute the torsional shear stress in a weld group
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Applied load on the weld group |
| e | e | m | 1.0 | Eccentricity of the load from the weld group centroid |
| r | r | m | 1.0 | Radial distance from the weld group centroid to the point of interest |
| J | J | m^4 | 1.0 | Polar moment of inertia of the weld group |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | τ | Pa | Torsional shear stress, in pascals (Pa). Returns 0.0 if ``J`` is not positive |
The science & history
Understanding the Parameters
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Couple $M = Fe$ — an eccentric load is equivalent to a centroidal force plus this twisting moment; the moment is what the torsional-shear term resists. Reducing $e$ (moving the load line toward the weld centroid) is the most effective way to cut this stress.
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Radial distance $r$ — the stress rises linearly with distance from the centroid, so the corners of a weld group — largest $r$ — are the critical points.
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Polar moment $J$ — the group's twist stiffness, $J = I_x + I_y$ from Weld Group Polar Moment. Spreading the welds out raises $J$ and lowers the torsional stress everywhere.
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Direction — $\tau$ acts perpendicular to the radius to the point, mirroring shaft torsion; this direction is what the vector sum with the primary (load-parallel) shear needs.
Derivation (Approaching a Proof)
Take the eccentric load $F$ at eccentricity $e$ and replace it by a centroidal force plus a couple $M = Fe$. The couple twists the planar weld group in its own plane. Treating the throat area as a thin section, the in-plane torsional shear at a point distance $r$ from the centroid follows the torsion formula $\tau = M\rho/J$ with $\rho \to r$:
$$\tau = \frac{M\,r}{J} = \frac{F\,e\,r}{J}, \qquad J = \int r^2\,dA = I_x + I_y.$$
The centroidal force is carried by the direct-shear term $\tau' = F/A$; the total design stress is the vector sum $\lvert\boldsymbol{\tau'}+\boldsymbol{\tau}\rvert$ at the worst point.
Dimensional check. $\dfrac{F\,e\,r}{J} = \dfrac{\text{N}\cdot\text{m}\cdot\text{m}}{\text{m}^4} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$ — a stress, as required.
History and Development
Weld-group torsional shear is Coulomb's shaft-torsion result ($\tau = T\rho/J$, 1784) applied to a planar pattern of weld metal — the same transplant used for eccentrically loaded bolt and rivet groups in 19th- and 20th-century structural practice. Codified in AISC and Shigley as the "elastic method", it reduces the group to a line weld, computes $J$ per unit throat, and locates the peak by vector addition. Its clarity made it the standard hand method before the more efficient (ductility-based) instantaneous-centre approach appeared in modern design tables.
Related Concepts: Eccentric Weld Secondary Shear, Eccentric Weld Primary Shear, Eccentric Load Angle, Weld Group Polar Moment, Weld Group Moment Of Inertia, Torsional Shear Stress
Notes: Identical to Eccentric Weld Secondary Shear. Perpendicular to $r$; combine vectorially with direct shear $F/A$. Max at largest-$r$ point. $J = I_x + I_y$ from Weld Group Polar Moment. Returns $0$ for non-positive $J$.