# Primary Seal Deformation in Multipane Glazing Units
## 4 Derivation of Analytical Expressions for Deformation Mode D  
Deformation mode D considers opening or closing the PIB layer between the glass pane and spacer (Figure 5). Relative displacements in the normal direction cause the thinning or thickening of the PIB layer. As shown in Figure 1, the deformation pattern of the PIB cross-section includes normal strains \(\varepsilon_{xx}(x,y,z)\), \(\varepsilon_{yy}(x,y,z)\), and shear strains \(\gamma_{xy}(x,y,z)\). Determination of all these strain field distributions will be based on the following analytical approach. The first question defines the shape of the PIB when the distance between both surfaces (glass and spacer) increases/decreases. If the PIB is considered to be a liquid medi  
Figure 5: Thinning of the primary sealant layer: (**a**) front view of the unit; (**b**) cross-section before expansion; (**c**) cross-section after expansion. by [PERSON] et al. [28], who examined the capillary liquid bridge between solid surfaces. Liquid takes a form that minimizes its outer surface area due to surface tension, \(q_{surf}\), which holds a curved differential patch of a doubly-curved liquid surface in equilibrium. Both principal radii of the curvature are related to the Young-Laplace equation, which also considers the pressure difference between the liquid pressure just below the surface and the gas pressure just above the surface, \(\Delta p\). \[\frac{1}{R_{y}}+\frac{1}{R_{z}}=\frac{\Delta p}{q_{surf}} \tag{9}\]  
In the absence of gravitational and electrostatic forces, the pressure gradient inside a fluid is zero. For plane strain conditions, one of the principal radii is infinite, meaning that a cross-sectional curve is a circular arc (Figure 6). For the sake of brevity, let us assume an initially rectangular domain of PIB with dimensions \(2a\times 2b\), which is stretched by glass panes in the \(x\)-direction. Due to symmetry, only a quarter of the domain is presented in Figure 7. All points of the PIB at \(x=a\) are attached on a glass pane, hence their displacements follow the pane, i.e., they move horizontally by \(\Delta a\). The free surface at \(y=b\) forms a circular arc \(Y(X,b)\) in a deformed state. We also assume that all points at the same height \(y\) also form a circular arc \(Y(X,y)\) with radius \(R(y)\), which follows the equation:  
\[X^{2}+\left(Y(X,y)-B(y)\right)^{2}=R(y)^{2} \tag{10}\]  
where \(B(y)\) determines the location of the circle centre for each arc along \(y\) and is determined from the condition that each circle must lead through the point \(Y(a+\Delta a,y)=y\). Each circular arc is then defined as  
\[Y(X,y)=y-\sqrt{R(y)^{2}-X^{2}}+\sqrt{R(y)^{2}-\left(a+\Delta a\right)^{2}} \tag{11}\]  
where \(X\) (location in a deformed state) is assumed to be linearly scaled from the initial domain as:  
\[X=X(x)=x\frac{a+\Delta a}{a} \tag{12}\]  
Figure 6: The capillary surface of the plane strain state [28]. If we consider the incompressibility of the PIB, the radius \(R(y)\) is determined from the volume preservation, i.e., the initial volume (or area under plane strain conditions) \(V_{0}(y)=ay\) is the same as the deformed volume (area) \(V(y)\), defined as:  
\[\left\{ \begin{array}{l}V(y)=\int\limits_{0}^{a+\Delta a}Y(X,y)dX=(a+\Delta a )y+\frac{(a+\Delta a)^{2}}{2K^{2}}\big{(}K\sqrt{1-K^{2}}-\text{Arcsin}(K)\big{)} \\ K=K(y)=\frac{a+\Delta a}{R(y)}\end{array} \right. \tag{13}\]  
Considering \(V_{0}(y)=V(y)\), it follows an implicit nonlinear equation for \(K\):  
\[y\Delta a+\frac{(a+\Delta a)^{2}}{2K^{2}}\big{(}K\sqrt{1-K^{2}}-\text{Arcsin}( K)\big{)}=0 \tag{14}\]  
However, \(K\) cannot be extracted explicitly, but since \(K\ll 1\), it is admissible to perform a Taylor series expansion for some functions in the above expression.