Finite elasticity

For reference radii $a_0,b_0$, current inner radius $a$ and uniform axial stretch $z>0$, incompressibility integrates to $r^2=a^2+(R^2-a_0^2)/z$. The principal stretches are $r_{,R}=1/(z\lambda )$, $\lambda =r/R$ and $z$. Write $\hat{W}(\lambda ,z)=W((z\lambda {)}^{-1},\lambda ,z)$ and $\mathcal{E}=2\pi {\int }_{a_0}^{b_0}\hat{W}RdR$ per reference length. Virtual work gives $p={\mathcal{E}}_{,a}/(2\pi az)$ and the total axial wall load $N_{\mathrm{wall}}={\mathcal{E}}_{,z}$, both derivatives holding the other displayed parameter fixed. For pressurized closed ends the independently applied axial load is $N_{\mathrm{ext}}=N_{\mathrm{wall}}-\pi a^{2}p$. The distinction follows from the axial pressure power $\pi a^{2}p\dot{z}$.

For an equilibrium with $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$ and dead loads, the constrained second variation on ${\varphi }_{,F}:\delta F=0$ isIndeed, a constraint-preserving perturbation path has ${\varphi }_{,F}:{\delta }^{2}F=-{\varphi }_{,FF}[\delta F,\delta F]$, and stationarity of the total potential supplies the multiplier term. Linearized kinetic energy plus $\mathcal{Q}/2$ is conserved. Positivity gives the usual energy-norm linear stability; control in a specified displacement norm additionally requires the corresponding coercivity and treatment of rigid-motion modes.

For axial stretch $\lambda$ and twist per reference length $\alpha$, the deformation in local cylindrical orthonormal bases hasThe principal stretches follow from the eigenvalues of $F^{T}F$. For a cylinder of reference radius $A$, stored energy per reference length is $\mathcal{E}=2\pi {\int }_0^{A}W(F)RdR$. Reversible power $M\dot{\alpha }+N\dot{\lambda }=\dot{\mathcal{E}}$ gives torque $M={\mathcal{E}}_{,\alpha }$ and axial force $N={\mathcal{E}}_{,\lambda }$. The relevant derivative holds the reference twist fixed, not the twist per current length.

For axial stretch $\lambda$ and twist $\alpha$ per reference length, an incompressible neo-Hookean solid has current radius $a=R/\sqrt{\lambda }$. In local cylindrical orthonormal bases the deformation gradient is $A=\left(\begin{array}{ccc}{\lambda }^{-1/2}& 0& 0\\ 0& {\lambda }^{-1/2}& \alpha r\\ 0& 0& \lambda \end{array}\right)$. The Cauchy stress tensor is $\mu A{A}^T+qI$. Radial equilibrium gives $q^{\prime }=\mu {\alpha }^{2}r$, and a traction-free curved surface gives $q=-\mu /\lambda +\mu {\alpha }^2(r^2-a^2)/2$. The nominal end traction is the displayed vector, measured per reference area. Its moment about the axis is $\int (x_1{t}_2-x_2{t}_1)d{S}_0=\mu \alpha \int r^{2}d{S}_0=\mu \pi R^4\alpha /(2\lambda )$. This calculation uses current moment arms with reference-area tractions.

In cylindrical orthonormal bases the deformation gradient is $F=\left(\begin{array}{ccc}\rho /r& 0& 0\\ 0& r/\rho & \alpha r\\ 0& 0& 1\end{array}\right)$ and has unit determinant. One principal stretch is $\rho /r$; the squares of the other two are the roots of $z^2-[(r/\rho {)}^2+{\alpha }^2{r}^2+1]z+(r/\rho {)}^2=0$. The reference-volume energy per height is $E=2\pi \int \rho Wd\rho$. Virtual work gives end torque $M={\partial }_{\alpha }E$ and inner pressure $p=(2\pi a{)}^{-1}{\partial }_{a}E$ at fixed reference radii and height.

Here $b^2=b_0^2+a^2-a_0^2$. Substitute the neo-Hookean solid energy into the twist and inflation of an incompressible tube work identity. Differentiating with respect to $a$ gives the integrand $\mu \rho [{\rho }^{-2}-{\rho }^2/({\rho }^2+c{)}^2+{\alpha }^2]$, with $c=a^2-a_0^2$. Integrating between the two reference radii proves the displayed pressure when the outer surface is traction free.

A unit reference fibre direction $m_0$ retains its length under this constraint on the deformation gradient. Its reaction contributes an undetermined tension along the current fibre direction, independently of the pressure reaction enforcing incompressibility.

For two distinct reference fibre directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0)$, with $0<\theta <\pi /2$, consider the volume-preserving deformation with right stretch tensor $U=\mathrm{diag}({\lambda }^{-1/2}\alpha ,{\lambda }^{-1/2}{\alpha }^{-1},\lambda )$, with positive $\lambda ,\alpha$. The inextensible fibre constraint gives ${\alpha }^2{\cos }^2\theta +{\alpha }^{-2}{\sin }^2\theta =\lambda$. Setting $b={\alpha }^2$ gives a quadratic with the displayed roots. There are two distinct positive branches for $\lambda >\sin 2\theta$. The smallest possible axial stretch is $\lambda =\sin 2\theta$, at which ${\alpha }^2=\tan \theta$ and $(U{l}_{+})\cdot (U{l}_{-})=0$. Since both image fibres have unit length, their mutual area cannot exceed one; their area is $\sin 2\theta /\lambda$, which gives the same lower bound. Coincident or antiparallel fibres at the angle endpoints do not give the two-family orthogonality conclusion.

Let $e=(\cos \theta ,\sin \theta )$ be the current fibre direction and $n=(-\sin \theta ,\cos \theta )$. Inextensibility fixes the first column of the deformation gradient to $e$, and incompressibility fixes the second to $\gamma e+n$. Commuting mixed derivatives gives ${\gamma }_{,1}={\theta }_{,1}$ and ${\theta }_{,2}=\gamma {\theta }_{,1}$. Thus $\gamma =\theta +f(X_2)$, and curves tangent to $(-\gamma ,1)$ map to straight transverse curves with constant $\theta$.

Write the Cauchy stress tensor in the local fibre frame as $\sigma =se\otimes e+h(e\otimes n+n\otimes e)+tn\otimes n$. The compatible geometry has $\nabla \theta =\kappa e$, $\nabla \cdot e=0$ and $\nabla \cdot n=-\kappa$. With zero body force, force balance becomes $e\cdot \nabla s=2\kappa h-n\cdot \nabla h$ and $n\cdot \nabla t-\kappa t=-e\cdot \nabla h-\kappa s$. For specified shear stress $h$, the first equation determines $s$ along fibres, then the second determines $t$ along transverse curves. Pressure and fibre tension supply these two normal constraint reactions. This is a local stress construction; boundary tractions must be chosen consistently with its integration data.

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 a plane material-coordinate elastic wave with $u=(0,w(X_1,t),0)$, set $q=w_{,X_1}$. The longitudinal equation is $0={\partial }_{X_1}{P}_{11}(1,q,0)$. Arbitrary nonuniform pure transverse profiles thus require the displayed constitutive compatibility, which does not follow from isotropy alone. If it holds and ${\tau }^{\prime }(q)>0$, the shear equation closes as ${\rho }_0{w}_{tt}={\partial }_{X_1}\tau (q)$ with speed $c(q{)}^2={\tau }^{\prime }(q)/{\rho }_0$.

An explicit objective isotropic example is $\mathcal{W}=\mu (S-1)/2+h(S-1{)}^2/4+(\lambda +\mu )(J-1{)}^2/2$, with $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and $\lambda ,\mu >0$, $h\ge 0$. On $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, $S-1=(a-1{)}^2+q^2$ and $J=a$, so $P_{11}=0$ at $a=1$ while the shear nominal stress is the displayed cubic. The natural-state quadratic energy has Lamé parameters $\lambda ,\mu$, and the shear speed is $\sqrt{(\mu +3h{q}^2)/{\rho }_0}$.

This objective isotropic constitutive model uses the Green-Lagrange strain tensor $E=(A^{\mathsf{T}}A-I)/2$. It reproduces linear elasticity at small strain. Under simple shear $A=[(1,q,0{)}^{\mathsf{T}},e_2,e_3]$, its longitudinal nominal stress is $(\lambda +2\mu )q^2/2$. Nonuniform finite shear therefore generates longitudinal acceleration, even though the corresponding infinitesimal transverse mode is uncoupled.

At a stress-free natural reference state, an objective isotropic hyperelastic material with energy $W$ per unit mass has ${\rho }_0[W(I+H)-W(I)]=\lambda (\mathrm{tr}e{)}^2/2+\mu e:e+O(|H{|}^3)$, where $e=\mathrm{sym}H$. Twice differentiating gives the displayed Lamé parameters and the usual elastic stiffness tensor. A prestressed reference requires initial-stress corrections.

At leading thin-shell order, a spherical balloon with reference radius $a_0$, thickness $h_0$ and current radius $a=\lambda a_0$ has reference elastic energy $4\pi a_0^2{h}_{0}W(\lambda ,\lambda ,{\lambda }^{-2})$. Equating its differential to pressure work $p4\pi a^{2}da$ gives the displayed pressure difference. Its thickness becomes $h_0{\lambda }^{-2}$ by incompressibility. The same result follows from the spherical membrane force balance $p=2h{\sigma }_t/a$, with current tangential Cauchy stress tensor component ${\sigma }_t=\lambda N$ and the equibiaxial nominal tension of an incompressible sheet. Thin-shell validity requires $h/a\ll 1$.

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 static base state with zero incremental body force, homogeneous incremental tractions or homogeneous Dirichlet boundary conditions eliminate the boundary term in integration by parts. The displayed Rayleigh quotient then gives real squared frequencies because of major symmetry. Positive moduli give strictly positive squared frequencies for modes with nonzero displacement gradient. Pure dead loading leaves constant translations at zero; fixed displacement conditions remove those modes.

Strong ellipticity requires the displayed inequality for every nonzero displacement vector $m$ and material direction $\nu$. Equivalently, every acoustic tensor is positive-definite. It is weaker than positivity of the moduli on all matrices. A quadratic minor null Lagrangian vanishes on rank-one matrices, so adding such a term does not change the acoustic tensor or this condition.

The acoustic tensor determines the plane waves of a locally uniform incremental elastic medium. For a unit material direction $\nu$, its eigenvalues are ${\rho }_0{c}^2$ and its eigenvectors are displacement polarizations. Major symmetry of the incremental elastic moduli makes $Q$ a real symmetric matrix. Positivity on every rank-one matrix $m\otimes \nu$ makes all its wave speeds real and nonzero.

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 $\det A=1$.

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.

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.

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.

For the volume-preserving family $F=\mathrm{diag}(\lambda ,{\lambda }^{-1/2},{\lambda }^{-1/2})$, equal dead nominal tension $T$ gives potentialIts derivative factors as ${\Pi }^{\prime }=(1-{\lambda }^{-3/2})[\mu (\lambda +{\lambda }^{-1/2})-T]$. The nontrivial branch has a fold at $\lambda =2^{-2/3}$, $T/\mu =32^{-2/3}$, and meets the trivial branch at $T/\mu =2$. Stability in this one-parameter family is determined by ${\Pi }^{\prime \prime }$, not by branch count alone. At $T/\mu =2$ there is only one nontrivial positive solution, $\lambda =(3-\sqrt{5})/2$; the other branch value is $\lambda =1$.

In three-dimensional isotropic linear elasticity, decompose the symmetric strain as $e=pI+d$ with $p=\mathrm{tr}(e)/3$ and $\mathrm{tr}d=0$. The stored energy is $E=\mu {\sum }_{ij}{d}_{ij}^2+\frac{3}{2}(3\lambda +2\mu )p^2$. Its nonnegativity for every strain is equivalent to the displayed inequalities: the traceless and purely volumetric test strains prove necessity, and the sum of squares proves sufficiency.

Material isotropy makes the elastic response independent of a rotation of reference material directions: $\sigma (FR)=\sigma (F)$ for proper rotations $R$. Combined with material frame indifference, it yields coaxiality of isotropic elastic stress.

The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

For simple shear, $B=\left(\begin{array}{cc}1+{\gamma }^2& \gamma \\ \gamma & 1\end{array}\right)$. The off-diagonal entry of $\sigma B-B\sigma$ is $\gamma ({\sigma }_{11}-{\sigma }_{22})-{\gamma }^2{\sigma }_{12}$. Coaxiality of isotropic elastic stress therefore proves the identity for $\gamma \ne 0$. At $\gamma =0$, material isotropy makes the stress isotropic, proving the same identity.