Elasticity/Principle of minimum complementary energy

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Complementary strain energy

The complementary strain energy density is given by

U^{c}({\boldsymbol {\sigma }})=\int _{0}^{\boldsymbol {\sigma }}\varepsilon _{ij}~d\sigma _{ij}

For linear elastic materials

U^{c}={\frac {1}{2}}{\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}=U

Principle of minimum complementary energy

Let ${\mathcal {A}}$ be the set of all statically admissible states of stress.

Let the complementary energy functional be

\Pi ^{c}[{\boldsymbol {\sigma }}]=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }})~dV-\int _{\partial {\mathcal {B}}^{u}}\mathbf {t} \bullet {\widehat {\mathbf {u} }}~dA

Then the Principle of Stationary Complementary Energy states that:

Among all stress fields ${\boldsymbol {\sigma }}$ in ${\mathcal {A}}$ , the functional $\Pi ^{c}$ is rendered stationary only by actual stress fields which satisfy compatibility and the displacement BCs.

The Principle of Minimum Complementary Energy states that:

For linear elastic materials, the complementary energy functional is rendered an absolute minimum by the actual stress field.

Note that the complementary energy corresponding to the actual stress field is the negative of the potential energy corresponding to the actual displacement field.

Proof

Let $[\mathbf {u} ,{\boldsymbol {\varepsilon }},{\boldsymbol {\sigma }}]$ be a solution of a the mixed boundary value problem of linear elasticity.

Let ${\tilde {\boldsymbol {\sigma }}}\in {\mathcal {A}}$ .

Define

{\boldsymbol {\sigma }}^{'}={\tilde {\boldsymbol {\sigma }}}-{\boldsymbol {\sigma }}

Then ${\boldsymbol {\sigma }}^{'}$ satisfies equilibrium and the traction BCs, i.e.,

{\begin{aligned}{\boldsymbol {\nabla }}\bullet {{\boldsymbol {\sigma }}^{'}}=0&~~\forall \mathbf {x} \in {\mathcal {B}}\\\mathbf {t} ^{'}={\widehat {\mathbf {n} }}{}\bullet {\boldsymbol {\sigma }}^{'}=0&~~\forall \mathbf {x} \in \partial {\mathcal {B}}^{t}\end{aligned}}

Since

{\boldsymbol {\varepsilon }}={\text{S}}:{\boldsymbol {\sigma }}

and

\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }}^{'}+{\boldsymbol {\sigma }})~dV=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }}^{'})~dV+\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }})~dV+\int _{\mathcal {B}}{\boldsymbol {\sigma }}^{'}:({\text{S}}:{\boldsymbol {\sigma }})~dV

we have,

{\text{(1)}}\qquad \int _{\mathcal {B}}\left[U^{c}({\tilde {\boldsymbol {\sigma }}})-U^{c}({\boldsymbol {\sigma }})\right]~dV=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }}^{'})~dV+\int _{\mathcal {B}}{\boldsymbol {\sigma }}^{'}:{\boldsymbol {\varepsilon }}~dV

We can also show that

\int _{\partial {\mathcal {B}}}\mathbf {t} \bullet \mathbf {u} ~dA=\int _{\mathcal {B}}\mathbf {u} \bullet ({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})~dV+{\frac {1}{2}}\int _{\mathcal {B}}{\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})~dV

Therefore,

{\text{(2)}}\qquad \int _{\partial {\mathcal {B}}^{u}}\mathbf {t} ^{'}\bullet \mathbf {u} ~dA=\int _{\mathcal {B}}{\boldsymbol {\sigma }}^{'}:{\boldsymbol {\varepsilon }}~dV

Now,

\Pi ^{c}[{\boldsymbol {\sigma }}]=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }})~dV-\int _{\partial {\mathcal {B}}^{u}}\mathbf {t} \bullet {\widehat {\mathbf {u} }}~dA

Hence,

\Pi ^{c}[{\tilde {\boldsymbol {\sigma }}}]=\int _{\mathcal {B}}U^{c}({\tilde {\boldsymbol {\sigma }}})~dV-\int _{\partial {\mathcal {B}}^{u}}(\mathbf {t} +\mathbf {t'} )\bullet {\widehat {\mathbf {u} }}~dA

Therefore,

{\text{(3)}}\qquad \Pi ^{c}[{\tilde {\boldsymbol {\sigma }}}]-\Pi ^{c}[{\boldsymbol {\sigma }}]=\int _{\mathcal {B}}\left[U^{c}({\tilde {\boldsymbol {\sigma }}})-U^{c}({\boldsymbol {\sigma }})\right]~dV-\int _{\partial {\mathcal {B}}^{u}}\mathbf {t} ^{'}\bullet {\widehat {\mathbf {u} }}~dA

From equations (1), (2), and (3), we have,

\Pi ^{c}[{\tilde {\boldsymbol {\sigma }}}]-\Pi ^{c}[{\boldsymbol {\sigma }}]=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }}^{'})~dV+\int _{\partial {\mathcal {B}}^{u}}\mathbf {t} ^{'}\bullet \mathbf {u} ~dA-\int _{\partial {\mathcal {B}}^{u}}\mathbf {t} ^{'}\bullet {\widehat {\mathbf {u} }}~dA

Since $\mathbf {u} ={\widehat {\mathbf {u} }}$ on $\partial {\mathcal {B}}^{u}$ , we have,

\Pi ^{c}[{\tilde {\boldsymbol {\sigma }}}]-\Pi ^{c}[{\boldsymbol {\sigma }}]=\int _{\mathcal {B}}U^{c}({\boldsymbol {\sigma }}^{'})~dV>0

Hence, proved.