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Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

Differentiate the pulled-back Kirchhoff stress tensor $T=F^T\tau F$, using the velocity gradient convention $L=\dot{F}{F}^{-1}$:Thus the lower-convected stress pull-back givesFor an explicit objective time derivative check, superpose $x^{\ast }=Q(t)x+c(t)$, with $Q^{T}Q=I$ and $\Omega =\dot{Q}{Q}^T$ skew. Then ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\Omega$, whereasThe additional terms in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$ are $-\Omega {\tau }^{\ast }+{\tau }^{\ast }\Omega$. They cancel the extra material-derivative terms, proving $\boxed{{\mathcal{D}}_l{\tau }^{\ast }=Q({\mathcal{D}}_l\tau )Q^T}$. The translation has no effect on a stress or velocity gradient.

For the incompressible Oldroyd-A model, avoid using the same symbol for a stress tensor and its relaxation time by referring to $\tau >0$ below only as the scalar relaxation time. Define $S={\sigma }^d-2{\mu }_{r}D$. The constitutive equation becomesBy the pull-back identity, $A=F^{T}SF$ satisfiesAn integrating factor givesPush forward by $F(t{)}^{-T}$ and $F(t{)}^{-1}$. For a convergent, fading remote-past history with the transported homogeneous term tending to zero as $t_0\to -\infty$, this proves the fading-memory representation of the Oldroyd-A model:For arbitrary finite-time initial data retain $e^{-(t-t_0)/\tau }F(t{)}^{-T}A(t_0)F(t{)}^{-1}$. Thus the printed integral-only expression incorporates an initial-memory condition; it is not an unconditional replacement for all initial-value solutions.

For time-dependent simple shear,Put $\Delta \gamma =\gamma (s)-\gamma (t)$. The relative deformation is $F(s)F(t{)}^{-1}=\left(\begin{array}{cc}1& \Delta \gamma \\ 0& 1\end{array}\right)$, so the entire transported integrand simplifies toConsequently the explicit nonpressure stresses areSymmetry gives ${\sigma }_{21}^d={\sigma }_{12}^d$, and the unshown third-direction nonpressure components vanish. The superscript $d$ here labels the paper's nonpressure constitutive stress; it need not have zero trace. Subtracting $pI$ supplies the total Cauchy stress tensor.

For constant shear rate $s_0$, use $\gamma (s)-\gamma (t)=-s_0(t-s)$ and the integrals ${\int }_0^{\infty }{e}^{-a/\tau }da=\tau$ and ${\int }_0^{\infty }a{e}^{-a/\tau }da={\tau }^2$. This gives the steady simple shear of an Oldroyd-A fluid resultand thereforeThe incompressibility pressure cancels from this normal-stress difference. With the usual $\mu \ge {\mu }_r\ge 0$, it is nonnegative, while the second normal-stress difference equals its negative.

Use both material isotropy and material frame indifference. The polar decomposition in continuum mechanics writes $F=VR$, where $V$ is the left stretch tensor. Isotropy removes the reference rotation: $\sigma (F)=\sigma (V)$. Choose an orthonormal eigenbasis of $V$. A half-turn about any principal axis leaves $V$ fixed but reverses appropriate off-diagonal entries of $\sigma (V)$. Frame indifference therefore makes those entries zero. If a principal stretch is repeated, rotations in its eigenspace additionally force the stress to be isotropic there. Thus the principal stresses share the stretch axes: this proves coaxiality of isotropic elastic stress with $V$ and with $B=V^2=F{F}^T$.

For simple shear, the left Cauchy-Green deformation tensor isIts principal axes make an angle $\vartheta$ with the coordinate axes satisfying $2\cot (2\vartheta )=\gamma$ when $\gamma \ne 0$. If the two principal stresses are $s_1,s_2$, rotating their diagonal matrix back givesConsequently the universal simple-shear normal-stress identity isEquivalently, it follows immediately from the off-diagonal entry of $\sigma B-B\sigma =0$. At $\gamma =0$, the undeformed material has isotropic stress, so the identity holds there too. No particular elastic constitutive equation has been used.

The inextensible fibre constraint gives $|{\mathbf{x}}_{,1}|=1$. Write the fibre direction and its transverse unit direction asThen $x_{1,1}=\cos \theta$ and $x_{2,1}=\sin \theta$. In plane strain, incompressibility gives $\det F=1$. Decomposing the second column of the deformation gradient along $e,n$ therefore gives ${\mathbf{x}}_{,2}=\gamma e+n$, where $\gamma ={\mathbf{x}}_{,2}\cdot e$. HenceThis is a local rotation followed by simple shear.

The compatibility of the deformation map requires ${\mathbf{x}}_{,12}={\mathbf{x}}_{,21}$. Since $e_{,j}={\theta }_{,j}n$ and $n_{,j}=-{\theta }_{,j}e$, this isTaking components gives the compatibility relations for incompressible inextensible-fibre plane strain:Along a reference curve,Thus $\theta$ is constant when $d{X}_2/d{X}_1=-1/\gamma$. The equivalent condition $d{X}_1+\gamma d{X}_2=0$ also covers $\gamma =0$. Its current tangent is proportional to $F(-\gamma ,1{)}^T=n$, so these transverse curves are straight wherever the deformation is regular.

To construct the stress with zero body force, specify the local shear stress $h$ as the material's function or functional of $\gamma$ under simple shear. The normal stresses are constraint reactions: incompressibility supplies pressure, and the inextensible fibre constraint supplies fibre tension. WriteThe compatible geometry has $\nabla \theta =\kappa e$ with $\kappa ={\theta }_{,1}$, so $\nabla \cdot e=0$, $\nabla \cdot n=-\kappa$, $(e\cdot \nabla )e=\kappa n$ and $(n\cdot \nabla )e=0$. Resolving $\nabla \cdot \sigma =0$ along the two directions givesThe triangular equilibrium construction for inextensible-fibre plane strain now determines $s$ by integration along fibres and then $t$ along transverse curves. Thus any regular compatible deformation has a local zero-body-force stress construction once the shear law and suitable boundary integration data are supplied. There is no additional constitutive normal-stress law to specify: the two normal components are fixed by force balance and boundary tractions. This does not assert that arbitrary separately prescribed boundary tractions can accompany an arbitrary deformation.

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.