Shear Deflection⚠ unverified
Mechanical / Beams · Compute the additional beam deflection due to transverse shear
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| V | V | N | 1.0 | Transverse shear force |
| L | L | m | 1.0 | Length over which the shear acts |
| A | A | m**2 | 1.0 | Cross-sectional area |
| G | G | Pa | 1.0 | Shear modulus of the beam material |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | Δs | m | Shear deflection, in metres (m) |
The science & history
Understanding the Parameters
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Shear force $V$ — the transverse (cross-cutting) internal force. Shear deflection is proportional to $V$, and $V$ is largest near supports and point loads, so shear deformation is concentrated there.
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Length $L$ — the span over which the shear acts. Note shear deflection grows linearly with $L$, whereas bending deflection grows as $L^3$. This different scaling is exactly why shear matters only for short beams: as $L$ shrinks, $L^3$ bending shrinks far faster than $L^1$ shear, so shear's share rises.
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Cross-sectional area $A$ and shear modulus $G$ — the product $AG$ is the shear rigidity, the shear analogue of bending's $EI$. A larger area or stiffer-in-shear material reduces shear deflection.
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Shear form factor $f_s$ — a dimensionless correction (also written as $1/k$, the reciprocal of the Timoshenko shear coefficient) accounting for the fact that shear stress is not uniform over the section. It is $6/5$ for a rectangle, $10/9$ for a solid circle, and roughly $A/A_{\text{web}}$ for an I-beam (nearly all shear is carried by the web).
Derivation (Approaching a Proof)
Shear deflection is derived cleanly from energy. The strain energy stored in transverse shear over a length $L$ is
$$U_s = \int_0^L \frac{f_s\, V^2}{2\,A\,G}\,\mathrm{d}x,$$
where the form factor $f_s$ arises from integrating the actual parabolic shear-stress distribution $\tau(y)$ over the section rather than assuming uniform $\tau = V/A$ — the non-uniformity stores more energy than the average would, and $f_s > 1$ corrects for it.
By Castigliano's second theorem, the deflection at the load in the direction of a force equals the partial derivative of strain energy with respect to that force. For constant $V$ over the length,
$$\delta_s = \frac{\partial U_s}{\partial V} = \frac{f_s\, V\, L}{A\, G}.$$
The parallel with bending is exact in structure: bending deflection is $\propto M L^{\dots}/EI$, shear deflection is $\propto V L/(AG)$. Timoshenko beam theory combines the two, adding this shear flexibility to the Euler–Bernoulli bending flexibility so that plane sections may rotate relative to the axis.
Dimensional check. $[\delta_s] = \dfrac{\text{N}\cdot\text{m}}{\text{m}^2 \cdot \text{Pa}} = \dfrac{\text{N}\cdot\text{m}}{\text{m}^2 \cdot \text{N/m}^2} = \text{m}$. ✓
History and Development
Shear deformation of beams was formalised by Stephen Timoshenko in 1921–22, extending Euler–Bernoulli theory by adding transverse-shear and rotary-inertia effects; the resulting Timoshenko beam is standard for stubby beams, high-frequency vibration, and sandwich/composite construction. The shear coefficients ($f_s$ or Timoshenko's $k$) were later refined by Cowper (1966) from three-dimensional elasticity. The energy/Castigliano route used here is the classical hand-calculation method found in Roark and Timoshenko's Strength of Materials.
Related Concepts: Beam Bending Stress, Shear Stress, Beam Moment At X, Rectangular Moment of Inertia, Section Modulus
Notes: Add $\delta_s$ to the bending deflection for total deflection (Timoshenko). Negligible for slender beams ($L/h \gtrsim 10$); significant for deep beams, short spans, and low-$G$ materials (elastomers, honeycomb cores). Confirm the assumed $f_s$ (rectangle $= 1.2$) — see the registry note.