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On an antiplane shear $F=I+e_3\otimes (p,q,0)$, the first strain invariant satisfy $I_1=I_2=3+p^2+q^2$, $I_3=1$. Restricting an isotropic objective strain energy density to this family determines only derivatives tangent to the family. It fixes $S_{31}=2w^{\prime }(p^2+q^2)p$ and $S_{32}=2w^{\prime }(p^2+q^2)q$, but does not specify the whole Cauchy stress tensor. For example the neo-Hookean solid energy $W=\mu (|F{|}^2-3)/2-\mu \log \det F$ gives $S=\mu (B-I)$ on this family. Even though the reference state is stress free, this is not generally a scalar multiple of $B$.

The bulk modulus relates hydrostatic compressive pressure to fractional volume compression. In isotropic three-dimensional linear elasticity, the spherical part of the stress tensor is $\kappa \mathrm{tr}(\epsilon )I$, so uniform dilation has strain energy density $\kappa (\mathrm{tr}\epsilon {)}^2/2$. Positive bulk modulus and positive shear modulus make the isotropic elastic stiffness tensor positive on nonzero symmetric strains.

For strain energy density $W=\mu [({\lambda }_r^2+{\lambda }_{\theta }^2+{\lambda }_{\varphi }^2{)}^{\alpha }-3^{\alpha }]/2$ with $\mu >0$ and $\alpha >0$, the critical loading for elastic cavitation is finite exactly in the displayed range. To prove this, put $S={\lambda }^{-4}+2{\lambda }^2$ in the reduced strain energy density. The critical-pressure integrand is $2\mu \alpha S^{\alpha -1}({\lambda }^{-2}+{\lambda }^{-5})$. It is finite at $\lambda =1$ and asymptotic to $\mu \alpha 2^{\alpha }{\lambda }^{2\alpha -4}$ at infinity. The comparison test for the resulting improper integral gives $2\alpha -4<-1$.

For a regular material constraint $F(A)=0$, admissible rates of the deformation gradient lie in the kernel of $F_{,A}$. The difference between the nominal stress and the derivative of the strain energy density annihilates that tangent space by virtual work, so it is a Lagrange multiplier times $F_{,A}$. For incompressibility, the derivative of the determinant is its cofactor matrix, and the material-first reaction is $q{A}^{-1}$ when $\det A=1$.

The effective elastic energy is the minimum averaged strain energy density under prescribed affine boundary displacement. For differentiable stable elastic response without body force, varying the prescribed mean strain gives $D{W}^{\mathrm{eff}}(E)=⟨\sigma ⟩$. The result follows by integration by parts, because the interior variation with zero boundary data does no work against an equilibrated stress tensor.

A thin isotropic hyperelastic material subject to incompressibility, with equal in-plane principal stretches $\lambda$ has transverse stretch ${\lambda }^{-2}$. With unloaded transverse faces, $N_{33}=W_3+q{\lambda }^2=0$. Hence $N_{11}=N_{22}=W_1-{\lambda }^{-3}{W}_3$, using isotropy to set $W_1=W_2$. The derivative along the constrained path is $W_1+W_2-2{\lambda }^{-3}{W}_3$, proving the displayed formula. The two equal nominal tensions do work $2Nd\lambda$ per reference volume, exactly the change in strain energy density.

The first principal invariant of the left Cauchy-Green deformation tensor is the sum of the squares of the principal stretches. Since $B=F{F}^T$, it also equals ${\sum }_{i,I}{F}_{iI}^2$. Its derivative with respect to the deformation gradient is $2F$. For a strain energy density $W=w(I_1)$ with incompressibility, the Cauchy stress tensor is $T=2w^{\prime }(I_1)B-pI$.

A hyperelastic material is homogeneous when its strain energy density $W(F)$ has no explicit dependence on the reference configuration position. Its first Piola-Kirchhoff stress tensor is $T=DW(F)$. Homogeneity of the material does not require the deformation gradient to be spatially constant.

A hyperelastic material derives its elastic stress from a strain energy density. In reference-volume units its first Piola-Kirchhoff stress tensor is the derivative of that density with respect to the deformation gradient.

For unconstrained hyperelastic material, the Hessian of the strain energy density gives $\delta N_{\alpha i}=c_{\alpha i\beta j}\delta A_{j\beta }$. Commuting the second derivatives gives major symmetry $c_{\alpha i\beta j}=c_{\beta j\alpha i}$. These material-coordinate moduli linearize about a finite base deformation; they need not have the minor symmetries of a small-strain elastic stiffness tensor. Material constraints require their additional reaction increments and admissibility equations.

For a material isotropy and material frame indifference strain energy density $W(F)=h(I_1,I_2,I_3)$, take $B=F{F}^T$, $I_1=\mathrm{tr}B$, $I_2=((\mathrm{tr}B{)}^2-\mathrm{tr}{B}^2)/2$, $I_3=\det B=J^2$, $J=\det F>0$. The chain rule and $d(\det F)=J{F}^{-T}:dF$ give $DW=2h_{1}F+2h_2(I_{1}I-B)F+2I_3{h}_3{F}^{-T}$. Multiplying by $F^T/J$ yields the displayed Cauchy stress tensor. Material isotropy allows an identity term and a quadratic term in $B$, as well as the term proportional to $B$.

For a homogeneous isotropic plate of thickness $h$, Young's modulus $E$ and Poisson's ratio $\nu$, the bending stiffness is $D=E{h}^3/[12(1-{\nu }^2)]$. Its quadratic bending energy isIt follows by integrating the strain energy density $\frac{1}{2}{\tau }_{ij}{e}_{ij}$ through the thickness: linear bending strains are proportional to $z$, and ${\int }_{-h/2}^{h/2}{z}^{2}dz=h^3/12$. Taking a first variation and integrating by parts twice gives $D{\Delta }^{2}w=0$ without applied transverse load. Clamped boundary variations have both zero value and zero normal derivative.

The symmetry group of a hyperelastic material consists of invertible reference transformations $H$ preserving its strain energy density: $W(FH)=W(F)$ for every admissible deformation gradient $F$. A convention restricting reference transformations to determinant one restricts $H$ to the special linear group. Material isotropy requires all proper rotations in this group. This right action differs from the left action in material frame indifference.

Let a convex strain energy density have derivative equal to the actual stress tensor. Under mixed displacement and traction data, any equilibrated stress ${\sigma }_0$ with the prescribed body force and boundary tractions represents the external work. For any compatible trial strain ${\epsilon }^{\prime }$, the convex supporting-plane inequality gives $W({\epsilon }^{\prime })-W(\epsilon )\ge \sigma :({\epsilon }^{\prime }-\epsilon )$. The integral of $(\sigma -{\sigma }_0):({\epsilon }^{\prime }-\epsilon )$ vanishes by integration by parts. Thus the actual strain minimizes $\int (W-{\sigma }_0:\epsilon )$.

An incompressible Mooney-Rivlin solid has strain energy density $W=C_1(I_1-3)+C_2(I_2-3)$, with right Cauchy-Green deformation tensor $C=F^{T}F$, $I_1=\mathrm{tr}C$ and $I_2=\mathrm{tr}{C}^{-1}$ when $\det C=1$. It extends the neo-Hookean solid by a second invariant. A convention $W={\mu }_1(I_1-3)/2-{\mu }_2(I_2-3)/2$ means $C_1={\mu }_1/2$, $C_2=-{\mu }_2/2$; the minus sign must be retained when comparing material constants.

An unconstrained natural state is a stress-free homogeneous deformation gradient. For $J>0$, its equivalent conditions are $DW(F)=0$ and $J^{-1}DW(F)F^T=0$. An interior differentiable minimum of the strain energy density is a natural state, although a stationary state need not minimize the energy. Constraints require allowing their reaction stress; the unconstrained equivalence must then be modified. Under strict Baker-Ericksen inequalities, an isotropic natural state is a dilation times a proper rotation.

For an incompressible neo-Hookean solid, a convenient strain energy density is $W=\mu (\mathrm{tr}C-3)/2$, where $C$ is the right Cauchy-Green deformation tensor. An additive constant in $W$ does not affect the stress. The shear modulus $\mu$ fixes the small-strain stiffness.

Choose zero energy in the unstrained state. The required strain energy density isIts variation is $\delta W=(\lambda \mathrm{tr}(e)I+2\mu e):\delta e$, giving the prescribed stress as the tensor gradient. This variational notation handles the symmetric off-diagonal components without a spurious factor of two.

Decompose $e=e^{\mathrm{dev}}+(\mathrm{tr}e)I/3$ into orthogonal deviatoric and spherical parts. ThenThe two squares vanish together only for $e=0$. Thus $\mu >0$ and $3\lambda +2\mu >0$ are sufficient for positive definiteness. They are necessary by testing a nonzero trace-free strain and a pure dilation separately. Equivalently both shear modulus $\mu$ and bulk modulus $K=\lambda +2\mu /3$ must be positive.

For a surface with unit normal $\mathbf{n}$, traction is $t_i={\tau }_{ij}{n}_j$. Thus ${\tau }_{ij}$ is the $i$th force-per-area component on a plane whose positive normal is ${\mathbf{e}}_j$. Diagonal entries are normal stresses and off-diagonal entries are shear stresses, with tensile normal stress positive in this convention.

The symmetric strain has six independent components and so does the symmetric stress. A general linear map between these two six-dimensional spaces therefore hasEquivalently choose the minor symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$. Components antisymmetric in the strain indices have no observable effect and are omitted from this representative.

A further distinction is important: if a strain energy density exists, $C_{ijkl}={\partial }^{2}W/(\partial e_{ij}\partial e_{kl})$ has the major symmetry $C_{ijkl}=C_{klij}$. Then the six-by-six map is symmetric and has $6\cdot 7/2=21$ independent coefficients. Thus 36 follows from the two stated tensor symmetries alone; 21 additionally uses the energy/reciprocity assumption, as in the variational setting later in the paper. These are the minor and major symmetries of linear elasticity.

The perturbation preserves the prescribed displacement because $\mathbf{w}=0$ on the boundary. Strain varies as $e_{ij}(\mathbf{u}+\alpha \mathbf{w})=e_{ij}(\mathbf{u})+\alpha (w_{i,j}+w_{j,i})/2$. Taylor expansion of the strain energy density givesSymmetry of the stress makes its contraction with the two symmetric terms equal, henceA minimizer must have zero first variation for both signs of $\alpha$. Integrating by parts and using the zero boundary variation yields ${\int }_B(-{\partial }_j{\tau }_{ij}-\rho g_i)w_{i}dV=0$. The fundamental lemma of the calculus of variations, applied to arbitrary compactly supported variations, givesThis is the static momentum equation. Fixed-displacement data remove the surface variation; prescribed tractions would instead appear through boundary work in the potential energy.