Total strain energy formulae; derive shear energy

Total strain energy of a structural system

For a member of length $L$ with axial force $N$, bending moment $M$, shear force $V$ and torque $T$, the total strain energy is the sum:

$$U = \int_0^L \frac{N^2}{2AE}\,dx + \int_0^L \frac{M^2}{2EI}\,dx + \int_0^L \frac{\alpha\,V^2}{2GA}\,dx + \int_0^L \frac{T^2}{2GJ}\,dx$$

Here $A$ = area, $I$ = moment of inertia, $J$ = polar moment of inertia, $E$ and $G$ = Young's and shear moduli, $\alpha$ = shear form factor ($6/5$ for rectangle, $10/9$ for solid circle). For a structure, the sum is taken over all members.

Derivation of energy due to shear force

Consider an element of length $dx$ with shear force $V$. Shear stress $\tau$ at a level $y$ of the section (area element $dA$):

$$\tau = \frac{V A'\bar y}{Ib}$$

Energy per unit volume $= \dfrac{\tau^2}{2G}$, so for the element

$$dU = \int_A \frac{\tau^2}{2G}\,dA\,dx$$

If the shear stress were uniform, $\tau = V/A$, giving $dU = \dfrac{V^2}{2GA}dx$. To account for the actual non-uniform distribution a shear form factor is defined as

$$\alpha = \frac{A}{I^2}\int_A \left(\frac{A'\bar y}{b}\right)^2 dA$$

Then

$$dU = \frac{\alpha V^2}{2GA}\,dx \quad\Rightarrow\quad U_s = \int_0^L \frac{\alpha V^2}{2GA}\,dx$$

For a rectangular section $\tau = \dfrac{V}{2I}\left(\dfrac{d^2}{4}-y^2\right)$, which gives $\alpha = 1.2$.