>[!info] Work done on linear elastic bodies >- Work done by a force: $W_{P}(e_{1})=\int_{0} ^{e_{1}}P(e)de$ where $e$ is elongation > - $W_{P}=\frac{1}{2}Pe$ >- Work done by a torque: $W_{T}(\phi_{1})=\int_{0} ^{\phi_{1}}T(\phi )d\phi$ > - $W_{T}=\frac{1}{2}T\phi$ >- Work done by a moment: $W_{M}(\theta_{1})=\int_{0}^{\theta_{1}}M(\theta)d\theta$ > - $W_{M}=\frac{1}{2}M\theta$ # Elastic strain energy Under combined loading, each point in the body may experience a different state of stress, thus a different amount of elastic energy will be stored at different differential volumes in the body $ U=\int_{V}\bar{u}dV $ Energy strain density: $ \bar{u}=\frac{1}{2}[\sigma_{x}(\varepsilon _{x}-\alpha\Delta T)+\sigma_{y}(\varepsilon_{y}-\alpha\Delta T)+\sigma_{z}(\varepsilon_{z}-\alpha\Delta T)+\tau_{xy}\gamma_{xy}+\tau_{xz}\gamma_{xz}+\tau_{yz}\gamma_{yz}] $ **Goal**: use strain energy density to determine an expression for the energy stored in an elastic body under axial/torsional/bending loads > [!info] Form factor > $f_{s}$ is a form factor for shear that depends on the shape of the cross section > $ > f_{s}=\frac{A(x)}{I(x)^2} \int_{A} \frac{Q(x,y)^2}{t(y)^2}dA > $ > > | Section | $f_{s}$ | | --------- | ------- | | Rectangle | $6 /5$ | | Circle | $10 /9$ | | Thin tube | 2 | >[!info] Work and elastic strain energy >![[Pasted image 20261019165231.png|center|500]] >- **Work done by the force:** $W_{ext}=\frac{1}{2}P_{A}e_{A}$ > - $U=\frac{1}{2} \int_{0} ^L \frac{P(x)^2}{E(x)A(x)}dx$ > - $U=\frac{P^2L}{2EA}$ >- **Work done by the torque:** $W_{ext}=\frac{1}{2}T_{A}\phi_{A}$ > - $U=\frac{1}{2}\int_{0}^L \frac{T(x)^2}{G(x)I_{p}(x)}dx$ > - $U=\frac{T^2L}{2GI_{p}}$ >- **Work done by the moment:** $W_{ext}=\frac{1}{2}M_{A}\theta _A =\frac{1}{2}P_{C}\Delta_{C}$ > - $U_{\sigma}=\frac{1}{2} \int_{0}^L \frac{M(x)^2}{E(x)I(x)}dx$ > - $U_{\tau}=\frac{1}{2}\int_{0}^L \frac{f_{s}V(x)^2}{G(x)A(x)}dx$ > - $U_{\tau}\ll U_{\sigma}$ # Castigliano's Second Theorem Consider a determinant linearly elastic deformable body acted upon by $N_{P}$ forces $P_{i},$ $N_{M}$ moments $M_{i}$, and $N_{T}$ torques $T_{i}.$ Among all possible equilibrium configurations of the body, the actual configuration is the one for which: $ \Delta_{i}=\frac{ \partial U }{ \partial P_{i} }\qquad i=1,2,\dots,N_{P} $ $ \theta_{i}=\frac{ \partial U }{ \partial M_{i} }\qquad i=1,2,\dots,N_{M} $ $ \phi_{i}=\frac{ \partial U }{ \partial T_{i} }\qquad i=1,2,\dots,N_{T} $