Eccentric Load Angle⚠ unverified
Mechanical / Welds · Compute the combined weld stress for an eccentric load applied at an angle
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Applied load on the weld |
| theta | θ | — | 1.0 | Angle of the applied load measured from the weld axis, in degrees |
| L | L | m | 1.0 | Length of the weld |
| h | h | m | 1.0 | Weld leg size |
| e | e | m | 1.0 | Eccentricity of the load from the weld group centroid |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | τ | Pa | Combined weld stress, in pascals (Pa) |
The science & history
Understanding the Parameters
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Load angle $\theta$ — resolves the load into an axial component $F\cos\theta$ (direct shear along the weld) and a transverse component $F\sin\theta$ (which bends the weld about its own length). At $\theta = 0$ the load is pure direct shear; at $\theta = 90^\circ$ it is pure bending.
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Direct term $\cos\theta/(0.707\,h\,L)$ — the axial component divided by the throat area $0.707\,h\,L$: a uniform shear along the weld.
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Bending term $6\,e\,\sin\theta/(h\,L^2)$ — the transverse component $F\sin\theta$ at eccentricity $e$ makes a moment $M = F e\sin\theta$; dividing by a line-weld section modulus $\sim hL^2/6$ gives the peak bending stress at the weld end.
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Eccentricity $e$ — only the transverse load has a lever arm here, so eccentricity amplifies the bending term alone; a load applied through the weld ($e = 0$) leaves only direct shear.
Derivation (Approaching a Proof)
Resolve $F$ at angle $\theta$ from the weld axis into two components and superpose their stresses on a single vertical weld of length $L$ and throat $t = 0.707\,h$:
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Axial component $F\cos\theta$ acts along the weld and is carried in direct shear over the throat area $A = 0.707\,h\,L$: $$\tau_{\text{direct}} = \frac{F\cos\theta}{0.707\,h\,L}.$$
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Transverse component $F\sin\theta$ acts across the weld at eccentricity $e$, creating a bending moment $M = F e\sin\theta$. Treating the weld as a line of length $L$, its section modulus is $S = tL^2/6$; the registry drops the throat factor and uses $S \approx hL^2/6$, giving $$\tau_{\text{bend}} = \frac{M}{S} = \frac{F e\sin\theta}{hL^2/6} = \frac{6\,F\,e\,\sin\theta}{hL^2}.$$
Adding the two (a scalar sum here, taking them as co-directed for the worst case):
$$\tau = \frac{F\cos\theta}{0.707\,h\,L} + \frac{6\,F\,e\,\sin\theta}{hL^2} = F\left(\frac{\cos\theta}{0.707\,h\,L} + \frac{6\,e\,\sin\theta}{hL^2}\right).$$
Dimensional check. Direct term: $\dfrac{\text{N}}{\text{m}\cdot\text{m}} = \text{Pa}$. Bending term: $\dfrac{\text{N}\cdot\text{m}}{\text{m}\cdot\text{m}^2} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$. Both terms are stresses, so $\tau$ is in Pa.
History and Development
Combining direct and bending stresses by superposition is the everyday tool of eccentric-connection design, predating welding itself (it is how riveted brackets were checked). For welds, the AISC/Shigley "line method" treats the weld as a one-dimensional line, computes its section properties per unit throat, and adds direct and moment-induced stresses — exactly the two terms above. Resolving the load by angle simply generalises the pure-shear and pure-bending cases into one expression for an obliquely applied load.
Related Concepts: Eccentric Weld Primary Shear, Eccentric Weld Secondary Shear, Weld Bending Stress, Eccentric Weld Load, Fillet Weld Size, Section Modulus
Notes: Supply $\theta$ in radians (trig functions). Bending term drops the $0.707$ throat factor — inconsistent with the direct term; approximate combined stress. Scalar (worst-case co-directed) sum.