Eccentric Weld Load⚠ unverified
Mechanical / Joints · Compute the additional bending stress from an eccentric weld load
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Applied load on the weld |
| e | e | m | 1.0 | Eccentricity of the load from the weld centroid |
| L | L | m | 1.0 | Length of the weld |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | σ | Pa | Additional bending stress from eccentricity, in pascals (Pa) |
The science & history
Understanding the Parameters
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Eccentricity $e$ — the moment arm; the bending moment $M = F e$ grows linearly with it. Reducing $e$ (bringing the load line toward the weld) is the most effective way to cut this stress.
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Weld length $L$ — enters as $L^2$: a taller weld resists eccentric bending far better (section modulus $\propto L^2$). Height dominates, exactly as for a beam.
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Line section modulus $L^2/6$ — the weld is idealised as a line of length $L$; its unit second moment is $L^3/12$ and unit section modulus $S_u = (L^3/12)/(L/2) = L^2/6$ (units m²). Dividing $M$ by $S_u$ gives a per-unit-throat force, which becomes a stress after dividing by the throat.
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Combine with direct shear — this bending term adds (vectorially) to the direct shear $F/A$; the resultant governs the weld.
Derivation (Approaching a Proof)
Idealise the fillet as a line of length $L$ (the line method). Its second moment of area per unit throat about the horizontal centroidal axis is the rectangle result with unit width:
$$I_u = \frac{L^3}{12}, \qquad S_u = \frac{I_u}{L/2} = \frac{L^2}{6}.$$
An eccentric load $F$ at arm $e$ applies moment $M = F e$. The line-method flexure result is
$$\sigma_u = \frac{M}{S_u} = \frac{F e}{L^2/6} = \frac{6 F e}{L^2},$$
which carries units of force per unit length (N/m) — it is stress $\times$ throat. Dividing by the effective throat $t = 0.707\,h$ converts it to an actual stress:
$$\sigma = \frac{\sigma_u}{t} = \frac{6 F e}{0.707\,h\,L^2}.$$
The registry stops at $\sigma_u = 6Fe/L^2$ and labels it Pa; treat it as the per-unit-throat quantity to be divided by the throat.
Dimensional check. $\dfrac{F e}{L^2/6} = \dfrac{\text{N}\cdot\text{m}}{\text{m}^2} = \dfrac{\text{N}} {\text{m}}$ — force per unit length, confirming the note. Dividing by throat $t$ (m) yields $\text{N}/\text{m}^2 = \text{Pa}$.
History and Development
Omer Blodgett's line method (Design of Welded Structures, Lincoln Electric, 1966) treats a weld as a line, tabulates unit section properties ($S_u$, $J_u$) for standard patterns, and defers the throat until the end — so the designer works in convenient force-per-length units and converts to leg size once. This eccentric-bending term $6Fe/L^2$ is the single-vertical-weld case of that method, combined with direct and torsional shear for the full eccentric weld-group analysis.
Related Concepts: Weld Bending Stress, Eccentric Load Angle, Eccentric Weld Primary Shear, Eccentric Weld Secondary Shear, Section Modulus, Throat Thickness
Notes: Line-method result is N/m (force per unit throat), not Pa — divide by throat $0.707\,h$ for actual stress. $S_u = L^2/6$ (line section modulus). Combine with direct shear for the resultant. Height $L$ dominates ($\propto L^2$).