# Primary Seal Deformation in Multipane Glazing Units
## 4 Derivation of Analytical Expressions for Deformation Mode D
Fourth-order expansions would give \(K\sqrt{1-K^{2}}\approx K-\frac{K^{3}}{2}\) and \(\text{Arcsin}(K)\approx K+\frac{K^{3}}{6}\), which gives a linear expression for K and consequently it can be extracted as:  
\[K=\frac{3y\Delta a}{\left(a+\Delta a\right)^{2}} \tag{15}\]  
Which finally gives the linear approximation of the radius \(R(y)\) :  
\[R(y)=\frac{\left(a+\Delta a\right)^{3}}{3y\Delta a} \tag{16}\]  
Insertion of \(R(y)\) into Equation (11) gives the function of all circular arcs along with the height of domain \(Y(X,y)\), which is presented in Figure 7:  
\[Y(X,y)=y-\sqrt{\frac{\left(a+\Delta a\right)^{6}}{\left(3y\Delta a\right)^{2} }-X^{2}}+\sqrt{\frac{\left(a+\Delta a\right)^{6}}{\left(3y\Delta a\right)^{2} }-\left(a+\Delta a\right)^{2}} \tag{17}\]  
Figure 7: Initial and deformed PIB (a quarter of the domain). A point initially located in \((x,y)\) moves to the displaced coordinates \((X,Y)\), hence the vertical displacement is given by \(v(x,y)=Y(X(x),y)-y\), which finally gives the vertical component of the displacement field:  
\[v(x,y)=\sqrt{\frac{\left(a+\Delta a\right)^{6}}{\left(3y\Delta a\right)^{2}}- \left(a+\Delta a\right)^{2}}-\sqrt{\frac{\left(a+\Delta a\right)^{6}}{\left(3y \Delta a\right)^{2}}-\left(x\frac{a+\Delta a}{a}\right)^{2}} \tag{18}\]  
Considering \(\Delta a\ll a\), we can approximate the vertical displacement by means of a Taylor series expansion, which, after rearrangement, gives:  
\[v(x,y)=\frac{3y\Delta a}{2a^{3}}\left(x^{2}-a^{2}\right) \tag{19}\]  
Now strains \(\varepsilon_{yy}(x,y)\) can be calculated as:  
\[\varepsilon_{yy}(x,y)=\frac{\partial v(x,y)}{\partial y}=\frac{3\Delta a}{2a^ {3}}\left(x^{2}-a^{2}\right) \tag{20}\]  
The next task is to determine the horizontal displacement component \(u(x,y)\). For incompressible solids no volume change occurs, hence, according to small strain theory, the trace of a strain tensor is zero:  
\[\varepsilon_{xx}(x,y)+\varepsilon_{yy}(x,y)+\varepsilon_{zz}(x,y)=0 \tag{21}\]  
The elaborated case can be considered as a plane strain problem, where \(\varepsilon_{zz}(x,y)=0\), therefore the absolute values of the strains are the same for the whole domain \(\varepsilon_{xx}(x,y)=-\varepsilon_{yy}(x,y)\). Horizontal displacement is now calculated from the definition:  
\[K=\frac{3y\Delta a}{\left(a+\Delta a\right)^{2}} \tag{15}\]0  
Considering the symmetry boundary condition \(u(0,y)=\ \ 0\), it follows:  
\[K=\frac{3y\Delta a}{\left(a+\Delta a\right)^{2}} \tag{15}\]1  
All displacement and strain fields are presented in \[K=\frac{3y\Delta a}{\left(a+\Delta a\right)^{2}} \tag{15}\]2  
If the obtained solution is checked with regard to boundary conditions, it is evident that all conditions are fulfilled, except that the shear strains on the free (upper) edge \(Y(x,b)\) are not zero, as they should be, since the edge is traction-free. It is reasonable to assume that the solution is a  
\[K=\frac{3y\Delta a}{\left(a+\Delta a\right)^{2}} \tag{15}\]3 Displacement and strain field without correction. The equation system is linear and all seven unknowns, \(a_{1}\) to \(a_{7}\), can be obtained analytically. Using known functions, the upgraded displacement field can be defined as:  
\[u_{ALL}(x,y)=u(x,y)+\left\{ \begin{array}{ccc}f_{corr}(y)&;&y>f_{bound}(x)\\ 0&;&y\leq f_{bound}(x)\end{array} \right. \tag{25}\]  
Once \(u_{ALL}(x,y)\) is known, \(v_{ALL}(x,y)\) is derived analogically to Equation (22). Without the loss of generality, the solution can be converted into a more convenient form using non-dimensional variables \(\xi=\frac{x}{a}\), \(\eta=\frac{y}{b}\) and \(\varepsilon_{0}=\frac{\Delta a}{a}\). The final solution of the displacement field is:  
\[\left\{ \begin{array}{l}u_{ALL}(\xi,\eta)=\frac{a\xi_{0}\left(3-\xi^{2}\right) }{2}-\left\{\begin{array}{ccc}u_{corr}(\xi,\eta)&;&\eta>f_{bound}(\xi)\\ 0&;&\eta\leq f_{bound}(\xi)\end{array}\right.\\ \\ u_{corr}(\xi,\eta)=\frac{3\xi\varepsilon_{0}\left(a^{2}\left(1-\xi^{2} \right)-b^{2}\left(1-\eta\right)\right)^{2}}{2a^{3}\left(1-\xi^{2}\right)} \end{array} \right. \tag{26}\]  
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