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We are going to illustrate how stress and other relevant unknowns change when the material manifold is perturbed.

Equilibrium problem.

For the sake of simplicity we consider the equilibrium problem for an elastic bar¹¹By a bar we mean a body that is slender enough to be described as one dimensional, but such that the radius of gyration of the cross section of the bar is large enough compared with its length to prevent bending instabilities. clamped on one end, and subject to a traction on the opposite end.

The equilibrium state of the body is governed by the folliowing boundary-value problem:

		$\displaystyle\left.\begin{aligned} &\displaystyle\sigma^{\prime}+b=0\\ &\displaystyle\sigma=\psi^{\prime}(\lambda),\\ &\displaystyle\lambda=y^{\prime},\\ \end{aligned}\qquad\right\}\text{ in }(x_{0},x_{1})$		(1)
		$\displaystyle\left.\begin{aligned} &\displaystyle y(x_{0})=Y_{0},\\ &\displaystyle\sigma(x_{1})=F.\end{aligned}\right.$		(1)

Perturbed problem.

We suppose that the domain $(x_{0},x_{1})$ undergoes an transformation which changes its extreme points into $(x_{0,\varepsilon},x_{1,\varepsilon})$ . We ask how such perturbation affects the solution.²²To gain generality, we assume that the perturbation of the domain is accompanied by a perturbation of loads and constraints, but not of the free energy. Accordingly, we write the following perturbed system:

		$\displaystyle\left.\begin{aligned} &\displaystyle\sigma^{\prime}_{\varepsilon}% +b_{\varepsilon}=0\\ &\displaystyle\sigma_{\varepsilon}=\widehat{\sigma}(\lambda_{\varepsilon}),\\ &\displaystyle\lambda_{\varepsilon}=y^{\prime}_{\varepsilon},\\ \end{aligned}\qquad\right\}\text{ in }(x_{0,\varepsilon},x_{1,\varepsilon})$		(2)
		$\displaystyle\left.\begin{aligned} &\displaystyle y(x_{0,\varepsilon})=Y_{0,% \varepsilon},\\ &\displaystyle\sigma(x_{1,\varepsilon})=F_{0,\varepsilon}.\end{aligned}\right.$		(2)

In the above perturbation $\varepsilon$ is a parameter. We suppose that the perturbation is null for $\varepsilon=0$ , so that $x_{0}=x_{0,0}$ and $x_{1}=x_{1,0}$ .

Change of domain.

To cope with the domain depending on $\varepsilon$ we reformulate the perturbed problem in the initial domain $(x_{0},x_{1})$ . We do this by introducing a $\varepsilon$ -parametrized class of diffeomorphisms $\varphi_{\varepsilon}:(x_{0},x_{1})\to(x_{0,\varepsilon},x_{1,\varepsilon})$ such that

\varphi_{\varepsilon}(x_{0})=x_{0,\varepsilon}

and

\varphi_{\varepsilon}(x_{1})=x_{1,\varepsilon}

(3)

We defined the pullback of the traction as

\widetilde{\sigma}_{\varepsilon}(\widetilde{x})=\sigma_{\varepsilon}(\varphi_{% \varepsilon}(\widetilde{x})).

(4)

The equilibrium problem in the initial domain is

		$\displaystyle\left.\begin{aligned} &\displaystyle\widetilde{\sigma}^{\prime}_{% \varepsilon}+\varphi_{\varepsilon}^{\prime}\widetilde{b}_{\varepsilon}=0\\ &\displaystyle\widetilde{\sigma}_{\varepsilon}=\psi^{\prime}(\widetilde{% \lambda}_{\varepsilon}),\\ &\displaystyle\widetilde{\lambda}_{\varepsilon}\varphi_{\varepsilon}^{\prime}=% \widetilde{y}^{\prime}_{\varepsilon}-\varphi_{\varepsilon}^{\prime},\\ \end{aligned}\qquad\right\}\text{ in }(x_{0},x_{1})$		(5)
		$\displaystyle\left.\begin{aligned} &\displaystyle\widetilde{y}(x_{0})=Y_{0,% \varepsilon},\\ &\displaystyle\widetilde{\sigma}(x_{1})=F_{0,\varepsilon}.\end{aligned}\right.$		(5)

Perturbation analysis.

We introduce the perturbation of the stress $\delta\widetilde{\sigma}(\widetilde{x})=\left.\frac{\partial}{\partial% \varepsilon}\right|_{\varepsilon=0}\widetilde{\sigma}_{\varepsilon}(\widetilde% {x})$ . We define the other perturbations in a similar manner. By linearizing (5) we obtain

		$\displaystyle\left.\begin{aligned} &\displaystyle\delta\widetilde{\sigma}^{% \prime}+\delta\widetilde{b}+b\delta\varphi^{\prime}=0\\ &\displaystyle\delta\widetilde{\sigma}=EA\delta\widetilde{\lambda},\\ &\displaystyle\delta\widetilde{\lambda}+\lambda\delta\varphi^{\prime}=\delta% \widetilde{y}^{\prime}-\delta\varphi^{\prime},\\ \end{aligned}\qquad\right\}\text{ in }(x_{0},x_{1})$		(6)
		$\displaystyle\left.\begin{aligned} &\displaystyle\delta\widetilde{y}(x_{0})=% \delta Y_{0},\\ &\displaystyle\delta\widetilde{\sigma}(x_{1})=\delta F.\end{aligned}\right.$		(6)

We next define $\delta\sigma(x)=\left.\frac{\partial}{\partial\varepsilon}\right|_{\varepsilon% =0}\sigma_{\varepsilon}(x)$ . From this definition and from (4), it follows that

\delta\widetilde{\sigma}=\delta\sigma+\sigma^{\prime}\delta\varphi.

(7)

Similar relations hold for all other increments, such as $\delta b$ , etc.

Consider the first of (10). If we write this equation in terms of the increments $\delta\sigma$ , $\delta b$ instead of $\delta\widetilde{\sigma}$ , $\delta\widetilde{b}$ , making use of (7) and of the other relations we obtain

(\delta\sigma+\sigma^{\prime}\delta\varphi)^{\prime}+\delta b+b^{\prime}\delta% \varphi+b\delta\varphi^{\prime}=0.

(8)

Using the equation $\sigma^{\prime}+b=0$ one can check that most of the terms in (8) cancel out, and

\delta\sigma^{\prime}+\delta b=0.

(9)

		$\displaystyle\left.\begin{aligned} &\displaystyle\delta\sigma^{\prime}+\delta b% =0\\ &\displaystyle\delta\sigma=\psi^{\prime\prime}(\sigma)\delta\lambda,\\ &\displaystyle\delta\lambda=\delta y^{\prime},\\ \end{aligned}\qquad\right\}\text{ in }(x_{0},x_{1})$		(10)
		$\displaystyle\left.\begin{aligned} &\displaystyle\delta y(x_{0})+y^{\prime}(x_% {0})\delta x_{0}=\delta Y_{0},\\ &\displaystyle\delta\sigma(x_{1})+\sigma^{\prime}(x_{1})\delta x_{1}=\delta F.% \end{aligned}\right.$		(10)

For a similar result in the three-dimensional setting see here.