Past exam of the solid mechanics course of the University of Oxford / 2011 / 1

A homogeneous hyperelastic material has a strain energy density per unit reference volume $W(F)$ depending on the deformation gradient, with no explicit dependence on the reference configuration. For orientation-preserving gradients $F\in G{L}^{+}(3)$, its first Piola-Kirchhoff stress tensor is the derivative:Equivalently the infinitesimal stored-energy change is $DW(F)[H]=T:H$. Reference and spatial indices are distinguished here by upper- and lowercase index letters.

The material symmetry group may be written $G_W=\{H\in G{L}^{+}(3):W(FH)=W(F)\text{ for every }F\in G{L}^{+}(3)\}$. A volume-preserving relabelling convention restricts the candidate $H$ to $SL(3)$; the material isotropy criterion is the same. The material has material isotropy when $SO(3)\subseteq G_W$, meaning $W(FQ)=W(F)$ for every proper rotation $Q$. If only rotational symmetries are included in the definition of the group, material isotropy means that group equals $SO(3)$. This right-rotation invariance is distinct from material frame indifference, the left-rotation invariance $W(QF)=W(F)$.

The polar decomposition in continuum mechanics gives the unique factorizationswhere $U,V$ are positive-definite symmetric matrices. Their common positive eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$ are the principal stretches, equivalently the singular values of $F$. They describe the local stretch magnitudes after separating out a rigid rotation.

Define the invariants by the characteristic polynomialBecause $B=F{F}^T$ has positive eigenvalues ${\lambda }_j^2$, its invariants are the elementary symmetric functions in those eigenvalues:The Frobenius norm gives $|F{|}^2=\mathrm{tr}(F^{T}F)={\sum }_j{\lambda }_j^2$. Write $J=\det F>0$. Since $\mathrm{cof}F=J{F}^{-T}$, its singular values are $J/{\lambda }_j$, soFinally $\det B=\det F\det F^T=J^2$. ThereforeThese identities also give $I_B=\mathrm{tr}B$ and $I{I}_B=((\mathrm{tr}B{)}^2-\mathrm{tr}(B^2))/2$.

Take a singular value decomposition $F=Q_{1}D{Q}_2^T$, with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$ positive and $Q_1,Q_2\in SO(3)$. Such a choice is possible because $\det F>0$. Left material frame indifference removes $Q_1$, and right material isotropy removes $Q_2$, giving $W(F)=W(D)$.

Moreover $W(D)$ is unchanged when the positive diagonal entries are permuted. Every permutation can be implemented by a signed permutation matrix of determinant one; the signs disappear in conjugating a diagonal matrix. Thus the energy depends only on the unordered principal stretches.

The three invariants determine precisely this unordered triple: the roots of $t^3-I_B{t}^2+I{I}_{B}t-II{I}_B$ are ${\lambda }_1^2,{\lambda }_2^2,{\lambda }_3^2$, and their positive square roots give the stretches. Consequently, on the set of admissible invariant triples, define $h(I_B,I{I}_B,II{I}_B)$ to be $W(D)$. The preceding argument proves that this is well defined and thatNot every positive triple is realizable by three positive roots. Since the requested statement only asserts existence of a function on $(0,\infty {)}^3$, extend $h$ arbitrarily on the nonrealizable triples; no choice there affects the formula for a deformation in continuum mechanics. Smooth invariant representations, where needed for differentiating the constitutive equation, are understood locally on the realizable set.

Retain the material frame indifference assumption used in part (c), in addition to material isotropy. Put $p=u_{,1}$, $q=u_{,2}$ and $s=p^2+q^2$. For this antiplane shear,The invariant identities give $I_B=I{I}_B=3+s$ and $II{I}_B=1$. Hence the restriction of the strain energy density isAntiplane energy does not determine the full stress: the restriction gives one scalar energy along antiplane gradients. It does not determine the full stress derivative with respect to every entry of $F$.

Indeed, using the isotropic hyperelastic stress invariant representation and differentiating the energy, with $h_j$ denoting the corresponding partial derivatives, givesUsing $S=J^{-1}T{F}^T$, the general objective isotropic Cauchy stress tensor isAll derivatives are evaluated at the current invariant triple. The extra terms generally remain nonzero on antiplane gradients, so the printed assertion that the full stress must be $\beta B$ does not follow.

A stress-free, objective isotropic counterexample is the compressible neo-Hookean solid $W(F)=\frac{\mu }{2}(|F{|}^2-3)-\mu \log J$, with $\mu >0$. Its derivative is $T=\mu F-\mu F^{-T}$, so on antiplane gradientsTake $u(x_1,x_2)=x_1$. Then $B_{22}=1$ but $S_{22}=0$, whereas $B_{13}=1$ and $S_{13}=\mu$. The diagonal entry would force $\beta =0$, contradicting the shear entry. Thus this is not a matter of choosing the right scalar coefficient.

For the additional one-invariant constitutive assumption $W(F)=\hat{w}(I_B)$, the stronger formula does hold, withThis is a valid special case, rather than a consequence of material isotropy alone. For the general restricted energy, the shear entries are still fixed: $S_{31}=2w^{\prime }(s)u_{,1}$ and $S_{32}=2w^{\prime }(s)u_{,2}$, since $w^{\prime }=h_1+h_2$ along $(3+s,3+s,1)$. The full normal stress needs the constitutive information that the printed claim omits. If material isotropy alone is interpreted without material frame indifference, even the first reduction to $|Du{|}^2$ needs that additional material frame indifference assumption.