Strain energy due to shear in rectangular beam

Strain energy due to shear is the energy stored because of the shear deformation of the beam. For a rectangular section the shear stress varies parabolically, so the energy is found by integrating over the section.

Derivation

Take a beam of rectangular section (width $b$, depth $d$) carrying shear force $V$ at a section. Shear stress at a distance $y$ from the neutral axis:

$$\tau = \frac{V A\bar y}{I b} = \frac{V}{2I}\left(\frac{d^2}{4} - y^2\right)$$

Strain energy per unit volume $= \dfrac{\tau^2}{2G}$. For an element of length $dx$, with area element $dA = b\,dy$:

$$dU = \int_{-d/2}^{d/2} \frac{\tau^2}{2G}\, b\,dy\,dx$$

Substituting $\tau$:

$$\begin{aligned} dU &= \frac{V^2 b\,dx}{8GI^2}\int_{-d/2}^{d/2}\left(\frac{d^2}{4}-y^2\right)^2 dy \\ &= \frac{V^2 b\,dx}{8GI^2}\cdot\frac{d^5}{30} \end{aligned}$$

Using $I = \dfrac{bd^3}{12}$, so $I^2 = \dfrac{b^2d^6}{144}$:

$$dU = \frac{V^2 b\, d^5 \times 144}{8G\,b^2 d^6\times 30}\,dx = \frac{3V^2}{5\,G\,bd}\,dx = \frac{6}{5}\cdot\frac{V^2}{2GA}\,dx$$

where $A = bd$. Hence the shear strain energy for the whole beam is

$$U_s = \int_0^L \frac{6}{5}\,\frac{V^2}{2GA}\,dx = \int_0^L \frac{3V^2}{5GA}\,dx$$

The factor $\dfrac{6}{5} = 1.2$ is the form factor (shape factor) for shear $\alpha$ of a rectangle. If shear were uniform, $dU = \dfrac{V^2}{2GA}dx$.

Example (check): simply supported beam with central load $P$: $V = \pm P/2$,

$$U_s = \frac{6}{5}\times\frac{(P/2)^2}{2GA}\times L = \frac{3P^2L}{20\,GA}$$

The deflection under the load (from $\delta = \partial U/\partial P$) is $\delta_s = \dfrac{3PL}{10GA}$.

Result: $U_s = \dfrac{6}{5}\displaystyle\int_0^L \dfrac{V^2}{2GA}\,dx$ for a rectangular section.