Shear deflection negligible vs bending

Strain energy is the energy stored in an elastic body due to deformation produced by loads. It equals the work done by the gradually applied loads: $U = \tfrac12 P\Delta$. For a beam,

$$U = \int \frac{M^2}{2EI}\,dx + \int \frac{\alpha V^2}{2GA}\,dx$$

with the first term due to bending and the second due to shear.

Comparison of bending and shear deflections

Take a simply supported beam of span $L$, depth $d$ and rectangular section with a central load $P$. Here $M = Px/2$ (for $x \le L/2$) and $V = P/2$.

Bending energy:

$$U_b = \frac{P^2L^3}{96EI}\ \Rightarrow\ \delta_b = \frac{\partial U_b}{\partial P} = \frac{PL^3}{48EI}$$

Shear energy:

$$U_s = \frac{\alpha (P/2)^2 L}{2GA} = \frac{\alpha P^2L}{8GA}\ \Rightarrow\ \delta_s = \frac{\partial U_s}{\partial P} = \frac{\alpha PL}{4GA}$$

Ratio:

$$\frac{\delta_s}{\delta_b} = \frac{\alpha PL}{4GA}\times\frac{48EI}{PL^3} = 12\,\alpha\,\frac{E}{G}\,\frac{I}{AL^2}$$

For a rectangular section $I/A = d^2/12$, so

$$\frac{\delta_s}{\delta_b} = \alpha\,\frac{E}{G}\left(\frac{d}{L}\right)^2$$

Substituting $\alpha = 1.2$ and $E/G = 2.4$:

$$\frac{\delta_s}{\delta_b} = 1.2\times 2.4\times\left(\frac{d}{L}\right)^2 = 2.88\left(\frac{d}{L}\right)^2$$

For an ordinary beam, $L/d$ is 10 to 20, so

Hence the shear deflection is only about 1 to 3% of the bending deflection and can be neglected in an ordinary beam. It becomes important only for deep beams ($L/d < 5$ or so).