Using (7.54), show that $\mathbf{T}=T_{\alpha \beta} \beta^{\mathbf{i} \alpha} \otimes \mathbf{i}_\beta-p \mathbf{i}_3 \otimes \mathbf{i}_3$, where
$$
T_{11}=-p-k \sin 2 \theta, \quad T_{22}=-p+k \sin 2 \theta, \quad \text { and } \quad T_{12}=T_{21}=k \cos 2 \theta .
$$
Equilibrium without body force, i.e., $T_{i j j}=0$, thus requires that $p_{, 3}=0$ and $T_{\alpha \beta, \beta}=0$, the latter yielding
$$
p_{, 1}+2 k\left(\theta_{, 1} \cos 2 \theta+\theta_{, 2} \sin 2 \theta\right)=0 \text { and } \quad p_{, 2}+2 k\left(\theta_{, 1} \sin 2 \theta-\theta_{, 2} \cos 2 \theta\right)=0
$$
Note that if, at a particular point, we orient the basis $\left\{\mathbf{i}_\alpha\right\}$ such that $\mathbf{i}_1=\mathbf{t}$ and $\mathbf{i}_2=\mathbf{s}$, then $\theta$ vanishes at that point, and these equations reduce to
$$
\begin{aligned}
& 0=p_1+2 k \theta_{, 1}=\mathbf{i}_1 \cdot \operatorname{grad}(p+2 k \theta)=\mathbf{t} \cdot \operatorname{grad}(p+2 k \theta), \quad \text { and, similarly, } \\
& 0=p_{12}-2 k \theta_{, 2}=\mathbf{i}_2 \cdot \operatorname{grad}(p-2 k \theta)=\mathbf{s} \cdot \operatorname{grad}(p-2 k \theta),
\end{aligned}
$$
which, of course, are just the Prandtl-Hencky equations.
With the Cartesian form of the equations in hand, the obvious thing to do when searching for simple solutions is to assume that $p$ or $\theta$ depends on just one of the coordinates. Let's consider the possibility that $\theta$ depends only on $y_2$; Eqs. (7.66) reduce to
$$
p_{, 1}=k(\cos 2 \theta)^{\prime} \quad \text { and } \quad p_{, 2}=k(\sin 2 \theta)^{\prime}
$$
where $(\cdot)^{\prime}=d(\cdot) / d y_2$. Assuming $p$ to be twice differentiable, we can eliminate it as follows:
$$
k(\cos 2 \theta)^{\prime \prime}=p, 12=p_{, 21}=k(\sin 2 \theta)_{, 21}=0 .
$$
Then $\cos 2 \theta$ is a linear function of $y_2$. One such solution, due to Prandtl, is
$$
\cos 2 \theta=-y_2 / h, \quad \sin 2 \theta=-\sqrt{1-\left(y_2 / h\right)^2},
$$
where $h$ is a positive constant. This is meaningful provided that $\left|y_2\right| \leq h$, and from (7.68) we have
$$
p=-k\left(y_1 / h\right)+f\left(y_2\right), \text { where } f^{\prime}=k(\sin 2 \theta)^{\prime} .
$$
Thus,
$$
p=p_0-k\left[y_1 / h+\sqrt{1-\left(y_2 / h\right)^2}\right]
$$
where $p_0$ is a constant, and the equilibrium stress field follows from the solution to Problem 7.10. Our solution is adapted from Prager's version of Prandtl's solution.
To obtain the velocity field, we invoke (7.25) with
$$
\tau=k \sin 2 \theta\left(\mathbf{i}_2 \otimes \mathbf{i}_2-\mathbf{i}_1 \otimes \mathbf{i}_1\right)+k \cos 2 \theta\left(\mathbf{i}_1 \otimes \mathbf{i}_2+\mathbf{i}_2 \otimes \mathbf{i}_1\right)
$$
Then,
$$
D=\alpha\left(i_1 \otimes i_1-i_2 \otimes i_2\right)+\beta\left(\mathbf{i}_1 \otimes \mathbf{i}_2+\mathbf{i}_2 \otimes \mathbf{i}_1\right)
$$
where
$$
\alpha=-\lambda k \sin 2 \theta \text { and } \beta=\lambda k \cos 2 \theta
$$
Because $\lambda \geq 0$ we then have $\lambda k=\sqrt{\alpha^2+\beta^2}$ and
$$
\alpha / \sqrt{\alpha^2+\beta^2}=-\sin 2 \theta, \quad \beta / \sqrt{\alpha^2+\beta^2}=\cos 2 \theta,
$$
Dxford University Press (2023).pdf
Open with
$$
\alpha=\sqrt{1-\left(y_2 / h\right)^2} \sqrt{\alpha^2+\beta^2} \text { and } \beta=-\left(y_2 / h\right) \sqrt{\alpha^2+\beta^2} \text {. }
$$
It follows from the expression for $\mathbf{D}$ that the components of the velocity field satisfy
$$
v_{1,1}=-v_{2,2}=\alpha \text { and } \quad \frac{1}{2}\left(v_{1,2}+v_{2,1}\right)=\beta .
$$
Accordingly, $\alpha$ and $\beta$ are not arbitrary. Assuming the velocity components to be thrice differentiable, we derive the compatibility condition
$$
\alpha_{, 11}-\alpha_{, 22}+2 \beta_{, 12}=0
$$
which, with reference to $(7.77)$, is easily seen to be solved if
$$
\sqrt{\alpha^2+\beta^2}=\frac{C}{h}\left(\sqrt{1-\left(y_2 / h\right)^2}\right)^{-1} ; \quad C=\text { const. }
$$
For we then have
$$
\alpha=\frac{C}{h} \quad \text { and } \quad \beta=-\frac{C}{h} \frac{y_2}{h}\left(\sqrt{1-\left(y_2 / h\right)^2}\right)^{-1}
$$
a constant and a function of $y_2$ alone, respectively.
The first and second of Eqs. (7.78) integrate to give
$$
v_1=\frac{C}{h} y_1+f\left(y_2\right) \text { and } v_2=-\frac{C}{h} y_2+g\left(y_1\right),
$$
where $f$ and $g$ are functions to be determined. These yield $v_{1,2}=f^{\prime}\left(y_2\right)$ and $v_{2,1}=g^{\prime}\left(y_1\right)$, which combine with the third equality in (7.78) and the second in (7.81) to give $g^{\prime \prime}\left(y_1\right)=0$ and
$$
v_2=-\frac{C}{h} y_2+D y_1+E,
$$
where $D$ and $E$ are constants. Then, $v_{1,2}=2 \beta-D$, yielding
$$
v_1=C\left[y_1 / h+2 \sqrt{1-\left(y_2 / h\right)^2}\right]-D y_2+F
$$
where $F$ is another constant.
Clearly $E$ and $F$ are rigid-body translational velocities. Further, the terms involving $D$ contribute a fixed tensor to the vorticity $\mathbf{W}$, and therefore represent a rigid-body rotation. Suppressing the rigid-body terms, we thus have the velocity field
$$
v_1=C\left[y_1 / h+2 \sqrt{1-\left(y_2 / h\right)^2}\right], \quad v_2=-C y_2 / h,
$$
valid in the region defined by $\left|y_2\right| \leq h$.
Evidently $v_2=\mp C$ at $y_2= \pm h$, and so this solution simulates the squeezing, assuming $C>0$, of a layer of thickness $2 h$ between rigid walls approaching each other at the relative speed $2 C$. The solution predicts that $T_{12}=\mp k$ at $y_2= \pm h$ but does not satisfy the noslip conditions $v_1=0$ there. We also note that $\left|D_{12}\right|(=|\beta|) \rightarrow \infty$ as $\left|y_2\right| \rightarrow h$. Thus, if the material is viscoplastic, then the viscosity cannot be neglected when $y_2$ is close to $\pm h$. In this case the present solution is valid outside a thin boundary layer adjoining the walls in which Bingham's model may be used to meet a no-slip condition. See Prager's book for a discussion of the relevant boundary-layer theory. Further discussion of the present solution is given in the books by Nadai and Hill, the latter including corrections to accommodate finite-width strips.
Corresponding slip lines may be determined, if desired, by using $(7.70)_1$ to obtain
$$
d y_2 / d y_1=2 h(\sin 2 \theta) d \theta / d y_1
$$
Then, on the trajectory with unit tangent $\mathbf{t}$, where $d y_2 / d y_1=\tan \theta$, and on that with unit tangent s, where $d y_2 / d y_1=-\cot \theta$, we have
$$
y_1=h(\sin 2 \theta+2 \theta)+\text { const. } \quad \text { and } \quad y_1=h(\sin 2 \theta-2 \theta)+\text { const. }
$$
respectively. These, together with $(7.70)_1$, are the parametric equations of a pair of orthogonal cycloids.