Elasticity/Energy methods example 4

Application of the Principle of Virtual Work
The virtual work done by the external applied forces in moving through the virtual displacement $$\delta w(x)\,$$ is given by

\delta W_{\text{ext}} = \int_0^L q~\delta w~dx + P~\delta w(L) $$ The work done by the internal forces are,
 * $$\begin{align}

\delta W_{\text{int}} & = \int_{\mathcal R} \delta U~dV \\ & = \int_0^L\int_{\mathcal S}\delta\left(\frac{1}{2}\sigma_{ij}\varepsilon_{ij}\right)~dA~dx\\ & = \int_0^L\int_{\mathcal S}\sigma_{ij} \delta\varepsilon_{ij}~dA~dx \end{align}$$ From beam theory, the displacement field at a point in the beam is given by

u = -z\frac{dw}{dx} ; v = 0 ; w = w(x) $$ The strains are, neglecting Poisson effects,

\varepsilon_{xx} = \frac{\partial u}{\partial x} = -z\frac{d^2 w}{dx^2} ; \varepsilon_{yy} = 0 ; \varepsilon_{zz} = 0 $$ and the corresponding stresses are

\sigma_{xx} = E\varepsilon_{xx} = -Ez\frac{d^2 w}{dx^2} ; \sigma_{yy} = 0 ; \sigma_{zz} = 0 $$ If we also neglect the shear strains and stresses, we get
 * $$\begin{align}

\delta W_{\text{int}} = & = \int_0^L\int_{\mathcal S}\sigma_{xx} \delta\varepsilon_{xx}~dA~dx \\ & = \int_0^L\int_{\mathcal S} E~\varepsilon_{xx} \delta\varepsilon_{xx}~dA~dx \\ & = \int_0^L E\frac{d^2w}{dx^2}\frac{d^2(\delta w)}{dx^2} \left(\int_{\mathcal S} z^2~dA\right)~dx \\ & = \int_0^L E~I\frac{d^2w}{dx^2}\frac{d^2(\delta w)}{dx^2} dx \end{align}$$ Therefore, from the principle of virtual work,

\delta W = \int_0^L \left(E~I\frac{d^2w}{dx^2}\frac{d^2(\delta w)}{dx^2}    + q~\delta w\right)~dx + P~\delta w(L) = 0 $$ Integrating by parts and after some manipulation, we get,
 * $$\begin{align}

0 &= \int_0^L \left[\frac{d^2}{dx^2}\left(E~I\frac{d^2w}{dx^2}\right) + q + P\delta(L-x)\right]\delta w~dx + \left.\left[\left(E~I\frac{d^2w}{dx^2}\right)\delta\left(\frac{dw}{dx}\right) + \frac{d}{dx}\left(E~I\frac{d^2w}{dx^2}\right)\delta w\right]\right|_0^L \end{align}$$ where $$\delta(L-x)$$ is the Dirac delta function,

\int_{-\infty}^{\infty} \delta(L-x)~f(x)~dx = f(L) $$ The Euler equation for the beam is, therefore,

\frac{d^2}{dx^2}\left(E~I\frac{d^2w}{dx^2}\right) + q + P\delta(L-x) = 0 $$ and the boundary conditions are
 * $$\begin{align}

E~I\frac{d^2w}{dx^2}(L) & = 0 \\ \left.\frac{d}{dx}\left(E~I\frac{d^2w}{dx^2}\right)\right|_{x=L} & = 0 \end{align}$$

Application of the Hellinger-Prange-Reissner variational principle
The governing equations of the cantilever beam can be written as

Kinematics


\kappa = \frac{d^2 w}{dx^2} ; w(0) = 0 ; \left.\frac{dw}{dx}\right|_{x=0} = 0 $$

Constitutive Equation


M = EI\kappa $$

Equilibrium (kinetics)


\frac{d^2M}{dx^2} + q + P\delta(L-x) = 0 ; M(L) = 0 ; \left.\frac{dM}{dx}\right|_{x=L} = 0 $$

Recall that the Hellinger-Prange-Reissner functional is given by

{\mathcal H}[s] = \int_{\mathcal{B}} U^c(\boldsymbol{\sigma}) - \int_{\mathcal{B}} \boldsymbol{\sigma}:\boldsymbol{\varepsilon}~dV - \int_{\mathcal{B}} \mathbf{f}\bullet\mathbf{u}~dV + \int_{\partial{\mathcal{B}}^{u}} \mathbf{t}\bullet(\mathbf{u}-\widehat{\mathbf{u}})~dA + \int_{\partial{\mathcal{B}}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{u}~dA $$

If we apply the strain-displacement constraints using the Lagrange multipliers $$\boldsymbol{\lambda}\,$$ and the displacement boundary conditions using the Lagrange multipliers $$\boldsymbol{\mu}\,$$, we get a modified functional

\bar{\mathcal H}[\mathbf{u},\boldsymbol{\varepsilon},\boldsymbol{\lambda},\boldsymbol{\mu}] = \int_{\mathcal{B}} \left[U(\boldsymbol{\varepsilon}) + \boldsymbol{\lambda}:[\frac{1}{2}(\boldsymbol{\nabla}\mathbf{u}-\boldsymbol{\nabla}\mathbf{u}^T) - \boldsymbol{\varepsilon}] - \mathbf{f}\bullet\mathbf{u}\right]~dV - \int_{\partial{\mathcal{B}}^{u}} \boldsymbol{\mu}\bullet(\mathbf{u}-\widehat{\mathbf{u}})~dA - \int_{\partial{\mathcal{B}}^{t}} \widehat{\mathbf{t}}\bullet\mathbf{u}~dA $$

For the cantilevered beam, the above functional becomes
 * $$\begin{align}

\bar{\mathcal H}[w,\kappa,\lambda,\mu_1,\mu_2] = & \int_0^L \left[\frac{EI}{2}\kappa^2 + \left(\frac{d^2w}{dx^2}-\kappa\right)\lambda + [q + P\delta(L-x)]~w\right]~dx \\ & - M(L)\frac{dw}{dx}(L) - \frac{dM}{dx}(L)~w(L) + \mu_1[w(0) - 0] + \mu_2[\frac{dw}{dx}(0) - 0] \end{align}$$

Taking the first variation of the functional, we can easily derive the Euler equations and the associated BCs.
 * $$\begin{align}

\delta\kappa : & EI\kappa - \lambda = 0 \\ \delta w : & \frac{d^2\lambda}{dx^2} + q + P\delta(L-x) = 0 \\ \delta\lambda : & \frac{d^2 w}{dx^2} -\kappa = 0 \end{align}$$ and
 * $$\begin{align}

\frac{d}{dx}(\delta w) : & \lambda(0) = \mu_2, \lambda(L) = M(L)\\ \delta w : & \frac{d\lambda}{dx}(0) = -\mu_1 , \frac{d\lambda}{dx}(L) = \frac{dM}{dx}(L)\\ \delta\mu_1 : & w(0) = 0 \\ \delta\mu_1 : & \frac{dw}{dx}(0) = 0 \end{align}$$

The same process can be used to derive Euler equations using the Hu-Washizu variational principle.