# In-Plane Structural Analysis of Coated Masonry Walls via a Homogenized Model
## 3 Stress Distribution in the Three-Layer Model
### Self-Stress State  
Also in this case, an interesting investigation concerns the distribution of the self-stress state \(\bar{\sigma}_{xe},\bar{\sigma}_{ye}\), resulting from the transverse-to-load behavior of the non-homogeneous cell under the normal forces \(N_{y}\) and \(N_{x}\), respectively.  
Let us consider the transverse dilatation along \(x\) (due to \(N_{y}\)), modeled by the spring system in Figure 3b, where \(\triangle/2\) is the unknown displacement at nodes. By denoting with \(F_{xe}\) the elastic force at the spring \(e\), the elastic problem is governed by:  
\[\begin{array}{l}F_{x1}=k_{x1}(\triangle-\bar{\varepsilon}_{x1}B),\\ F_{x2}=k_{x2}(\triangle-\bar{\varepsilon}_{x2}B)\end{array} \tag{10}\]  
which express the elastic law for the springs, and by:  
\[\begin{array}{l}F_{x1}+2F_{x2}=0\end{array} \tag{11}\]  
which expresses the equilibrium condition at the nodes. The resolution to this problem, specifically, the overall elongation of the assembly, is as follows:  
\[\begin{array}{l}\Delta=-\frac{(2\beta+1)(\alpha_{Ex}+2\beta\alpha_{xy})}{(2 \beta+\alpha_{Ex})(2\beta\alpha_{Ey}+1)}\frac{v_{xy}^{uw}}{E_{y}^{uw}}B\bar{ \sigma}_{y}\end{array} \tag{12}\]  
By substituting the solution Equation (12) in Equation (10), the spring forces are evaluated and, by letting \(\tilde{\sigma}_{x1}=\frac{F_{x1}}{hA}\), \(\tilde{\sigma}_{x2}=\frac{F_{x2}}{tA}\), it follows:  
\[\begin{array}{l}\tilde{\sigma}_{x1}=\frac{2\beta\alpha_{Ex}(2\beta+1)(1- \alpha_{xy})}{(2\alpha_{Ex}\beta+1)(2\alpha_{Ey}\beta+1)}\
u_{xy}^{uw}\frac{ E_{x}^{uw}}{E_{y}^{uw}}\bar{\sigma}_{y},\\ \\ \tilde{\sigma}_{x2}=\frac{\alpha_{Ex}(2\beta+1)(-1+\alpha_{xy})}{(2\alpha_{Ex} \beta+1)(2\alpha_{Ey}\beta+1)}\
u_{xy}^{uw}\frac{E_{x}^{uw}}{E_{y}^{uw}}\bar{ \sigma}_{y}\end{array} \tag{13}\]  
In the same way:  
\[\begin{array}{l}\tilde{\sigma}_{y1}=\frac{2\alpha_{Ey}\beta(2\beta+1)(1- \alpha_{xy})}{(2\alpha_{Ey}\beta+1)(2\alpha_{Ex}\beta+1)}\
u_{xy}^{uw}\tilde{ \sigma}_{x},\\ \\ \tilde{\sigma}_{y2}=\frac{\alpha_{Ey}(2\beta+1)(-1+\alpha_{xy})}{(2\alpha_{Ey} \beta+1)(2\alpha_{Ex}\beta+1)}\
u_{xy}^{uw}\bar{\sigma}_{x}\end{array} \tag{14}\]  
with the nondimensional coefficient \(\alpha_{xyx}=\alpha_{xyy}\frac{E_{y}^{uw}}{E_{x}^{uw}}\).  
It should be noticed that, from Equation (13), \(\frac{\tilde{\sigma}_{x2}}{\tilde{\sigma}_{x1}}=-\frac{1}{2\beta}\) follows; since \(\beta\) is small (because the wall is much thicker than the reinforcing coating), it turns out that \(\bar{\sigma}_{x1}\ll\bar{\sigma}_{x2}\), that is, the self-stress at the masonry is negligible. Conversely, the self-stress at the reinforcement can be significant. Similar considerations arise from Equation (14). Thus, for small values of \(\beta\), the self-stress expressions can be approximated as:  
\[\begin{split}\tilde{\sigma}_{x1}&\simeq 0,\\ \tilde{\sigma}_{x2}&\simeq\alpha_{Ex}(1-\alpha_{ vxy})v_{xy}^{uu}\frac{E_{x}^{uu}}{E_{y}^{uu}}\tilde{\sigma}_{y}\end{split} \tag{15}\]  
and  
\[\begin{split}\tilde{\sigma}_{y1}&\simeq 0,\\ \tilde{\sigma}_{y2}&\simeq\alpha_{Ey}(1-\alpha_{ vyx})v_{xy}^{uu}\tilde{\sigma}_{x}\end{split} \tag{16}\]  
Finally, the total stress is determined by the combination of self-stress induced by transverse elongation and the stress triggered by axial elongation:  
\[\begin{split}\tilde{\sigma}_{xe}&=\sigma_{xe}+\tilde {\sigma}_{xe},\\ \tilde{\sigma}_{ye}&=\sigma_{ye}+\tilde{\sigma}_{ye} \end{split} \tag{17}\]  
for \(e=1,2\) and where Equations (6), (7), (13) and (14) hold.