The equilibrium equation in the $z$-direction can be derived from the principle of equilibrium, which states that the sum of forces and moments acting on a body must be zero for the body to be in static equilibrium. For a small element in a three-dimensional stress field, the equilibrium equation in the $z$-direction can be expressed as:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$,

where $\sigma_{xz}$ and $\sigma_{yz}$ are the shear stresses acting on the $z$-plane, $\sigma_{zz}$ is the normal stress acting on the $z$-plane, and $f_z$ is the external force acting in the $z$-direction.

Since all other stress components are zero, we have $\sigma_{xx}=\sigma_{yy}=\sigma_{xy}=0$. Thus, the equilibrium equation reduces to:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Since $\sigma_{zy}=\sigma_{yz}$, we can write:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Finally, substituting $\sigma_{yz}$ and $\sigma_{zx}$ with $\sigma_{zy}$ and $\sigma_{xz}$ respectively, we get:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Therefore, the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{zy}$ and $\sigma_{xz}$ with all other stress components equal to zero is:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$

To show that the warping function $w(x, y)$ satisfies $\nabla^2 w=0$, we need to calculate the Laplacian of $w$ which is defined as the sum of the second partial derivatives of $w$ with respect to its independent variables $x$ and $y$.

$\nabla^2 w=\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}$

Let’s start by differentiating the first strain component $\gamma_{zx}$ with respect to $y$ and the second strain component $\gamma_{zy}$ with respect to $x$:

$\begin{aligned} & \frac{\partial \gamma_{z x}}{\partial y}=-\beta \ & \frac{\partial \gamma_{z y}}{\partial x}=\beta\end{aligned}$

Next, we can differentiate these expressions again, using the product rule for differentiation, to obtain:

$\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=0 \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=0\end{aligned}$

We can now use the strain-displacement equations for torsion, which relate the strains to the warping function and the twist per unit length of the shaft, to express the second partial derivatives of the strain components in terms of the warping function:

$\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial \gamma_{z x}}{\partial x}\right)=\frac{\partial}{\partial y}\left(\frac{\partial w}{\partial x}\right)=\frac{\partial^2 w}{\partial x \partial y} \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial \gamma_{z y}}{\partial y}\right)=\frac{\partial}{\partial x}\left(\frac{\partial w}{\partial y}\right)=\frac{\partial^2 w}{\partial y \partial x}\end{aligned}$

Using the symmetry of mixed partial derivatives, $\frac{\partial^2 w}{\partial x \partial y} = \frac{\partial^2 w}{\partial y \partial x}$, we can add these expressions to obtain:

$\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}=0$

which is the Laplace equation for $w$. Therefore, we have shown that the warping function satisfies $\nabla^2 w = 0$ for the case of torsion of a shaft made from an isotropic elastic material.

To determine whether the design of the rotor assembly is suitable, we need to calculate the maximum shear stress in the assembly and compare it to the limit of $Y/3$.

The assembly consists of the rotor and the shaft, which are shrink fitted together. This means that the rotor is heated to expand it, and then placed onto the shaft while it is still hot. As the rotor cools and contracts, it tightly grips the shaft. This creates a state of residual stress in both the rotor and the shaft.

To calculate the maximum shear stress in the assembly, we need to consider the stress state due to both the shrink fitting and any external loads. Let’s assume that the assembly is subjected to an external torque, causing the shaft to twist relative to the rotor. This creates a shear stress in the radial direction of the shaft.

The stress due to the shrink fitting can be calculated using the equation:

$\Delta \sigma = \frac{E \alpha \Delta T}{1-\nu}$

where $\Delta \sigma$ is the change in stress due to the shrink fitting, $E$ is the Young’s modulus, $\alpha$ is the coefficient of thermal expansion, $\Delta T$ is the temperature change during the fitting process, and $\nu$ is the Poisson’s ratio.

Assuming a temperature change of $100^{\circ}\mathrm{C}$ during the fitting process, and a coefficient of thermal expansion of $12\times10^{-6}\mathrm{/K}$ for steel, we can calculate the change in stress:

$\Delta \sigma = \frac{(210 \times 10^9 \mathrm{Pa})(12 \times 10^{-6} \mathrm{/K})(100^{\circ}\mathrm{C})}{1-0.3} \approx 2.52 \mathrm{MPa}$

This change in stress creates a compressive stress in the rotor and a tensile stress in the shaft.

Now let’s consider the external torque. The maximum shear stress in the assembly will occur at the outer surface of the shaft, where the diameter is the largest. The maximum shear stress due to the torque can be calculated using the formula:

$\tau = \frac{T r}{J}$

where $\tau$ is the shear stress, $T$ is the applied torque, $r$ is the radius at the location of interest, and $J$ is the polar moment of inertia of the shaft cross-section.

The polar moment of inertia of a solid cylinder is $J=\frac{\pi}{2}r^4$. Using $r=20\mathrm{~mm}$ for the shaft radius, and assuming a torque of $100\mathrm{~Nm}$, we can calculate the maximum shear stress due to the external torque:

$\tau = \frac{(100\mathrm{~Nm})(20\mathrm{~mm})}{\frac{\pi}{2}(20\mathrm{~mm})^4} \approx 8.03 \mathrm{MPa}$

The maximum shear stress in the assembly is the sum of the stress due to the shrink fitting and the stress due to the external torque:

$\tau_{max} = \Delta \sigma + \tau \approx 10.55 \mathrm{MPa}$

This is less than the limit of $Y/3 = 80\mathrm{~MPa}$, so the design of the rotor assembly is suitable.