# Engineering of a FGM Interlayer to Reduce the Thermal Stresses Inside the PFCs
## 2 Analytical Methods for the FGM Engineering
### Hollow Cylinder with Thin Thickness and a Homogeneous Thermal Load  
The constitutive equations in cylindrical coordinates (\(r\),\(t\),\(z\)), for a 3D cylindrical hollow body having a small thickness \(\delta\) and made of a homogeneous and isotropic material, are obtained by considering the radial normal stress to be negligible compared to the other components of the stress tensor (\(\sigma_{r}\sim 0\)):  
\[ \begin{cases}\epsilon_{t}&=\frac{1}{c}(\sigma_{t}-\
u\sigma_{z})+\alpha(T-T_{ 0})\\ \epsilon_{z}&=\frac{1}{c}(\sigma_{z}-\
u\sigma_{t})+\alpha(T-T_{0})\end{cases}  \tag{2}\]  
where \(\epsilon_{i}\) are the total strain components in the principal directions and \(\sigma_{t}\) the principal stresses. The fact that \(\delta\) is small compared to the other dimensions of the body, together with the lack of bending due to the axisymmetric geometry and temperature field, implies that the \(\epsilon_{i}\) can be considered constant over the thickness. Moreover, since the temperature field is a function of the cylinder thickness only, the principal strains do not vary with \(t\) and \(z\). Additionally, in absence of external forces, the integral over the thickness of all the principal stresses is zero. Therefore, by integrating Equation (2) over the thickness one finds:  
\[\epsilon_{t}=\epsilon_{z}=\frac{1}{\delta}\int_{0}^{\delta}\alpha(T-T_{0})dy \tag{3}\]  
In this case, \(y=0\) at the inner radius and \(y=\delta\) at the outer radius of the hollow body. If the tangential strain can be considered constant along the thickness, the congruence equation coming from the cylindrical geometry ensures that also the radial strain is constant and equal to \(\epsilon_{r}=\frac{d}{dr}(\epsilon_{t}r)=\epsilon_{t}\). As a result, the total strain components of the body are all equal to the mean thermal strain, \(\epsilon_{T,m}=\frac{1}{\delta}\int_{0}^{\delta}\alpha(T-T_{0})dy\). This means that the cylinder, due to the thermal expansion generated by the temperature field \(T\), deforms into a homothetically bigger cylinder having all its linear dimensions increased by a ratio \((1+\epsilon_{T,m})\). In the special case where \(\alpha\) does not vary with the temperature, the cylinder deforms into the larger cylinder exactly the size it would have if heated uniformly to its average temperature. Therefore, the heat sink would experience such deformation if it were detached from the rest of the component. Using the same approach, one obtains that also the thin cylindrical armor would expand freely in this fashion, but with the substantial difference that, due to the dissimilar CTE and local temperature, the homotetic ratio will be different. This is in fact the reason why, once we bond the PFM to the heat sink, additional stresses are generated. However, if all the sub-components would expand in the same exact manner, this contribution to stresses would be eliminated. To achieve so, the mean thermal strain of the two bodies must be equal:  
\[ \begin{split}\epsilon_{T,m1}&=\epsilon_{T,m2}\\ \frac{1}{\delta_{1}}\int_{\delta_{2}}^{\delta_{1}+\delta_{2}} \alpha_{1}(T-T_{0})dy&=\frac{1}{\delta_{2}}\int_{0}^{\delta_{2}} \alpha_{2}(T-T_{0})dy\end{split}  \tag{4}\]  
where the subscript 1 refers to the PFM and 2 to the heat sink material. In the simplified case of linear thermoelasticity, for which \(\alpha_{i}\) are constants, this equation translates in a condition on the mean temperatures:\[T_{m1}=\frac{\alpha_{2}}{\alpha_{1}}T_{m2}+\left(1-\frac{\alpha_{2}}{\alpha_{1}} \right)T_{0} \tag{5}\]  
In other words, if the W would be hot enough to match the heat sink's expansion, no stresses due to the CTE mismatch will be generated. A straightforward solution to satisfy such condition is to increase the armor thickness above the minimum requirement coming from the sputtering erosion. One must in this case solve Equation (4) for \(\delta_{1}\). However, while reaching our objective, a thicker PFM would also increase the contribution to thermal stresses which are due to the temperature gradient.