## 3 Evaluation of Disturbing Factors  
To assess the disturbing factors, let us use the studies conducted in [3; 22; 53]. Let the vector of displacements of the flexible element points under the temperature shock have the general form of \(\overset{\rightarrow}{u}\left(u_{x},\ u_{y},\ u_{z}\right)\). Studies [52; 25] show that only the deflections of points of the flexible element \(u_{z}=u_{z}(x,\ t)\) can be taken into account when evaluating perturbations, because the other components of the displacement vector are negligible compared with deflections. Thus, numerical modeling in [25] shows that the component of the displacement vector \(u_{y}=u_{y}(x,\ y,\ t)\) is more than two orders of magnitude smaller than the deflection. Let us consider the component of the displacement vector \(u_{x}=u_{x}(x,\ t)\). It is shown in [8; 22] that the points with coordinates \(l/2\leq x\leq l\) should move along the \(x\)-axis. Let us write down the equilibrium equation for these points [3; 22]:  
\[N=E\iint\limits_{S}\left\{\varepsilon_{xx}-\alpha[T(x,\ z,\ \ t)-T_{0}(x,\ z,\ \ 0)]\ \right\}\ dS \tag{2}\]  
where \(dS=dy\ dz\) is an infinitesimal element of the cross section of the flexible element; \(\varepsilon_{xx}\) is an element of the strain tensor; \(\alpha\) is the coefficient of thermal expansion; \(E\) is [PERSON]'s modulus; and \(N\) is the internal longitudinal force arising in the infinitesimal element of the cross section of the flexible element. With the free expansion of the flexible element:  
\[N=0,\ \frac{l}{2}\leq x\leq l \tag{3}\]  
Therefore, instead of (2), we have  
\[\varepsilon_{xx}=\alpha[T(x,\ \ z,\ \ t)-T_{0}(x,\ \ z,\ \ 0)],\ \ \frac{l}{2}\leq x\leq l,\ \ z=\frac{h}{2},\ \ t>0 \tag{4}\]  
Equation (4), as well as the second equation of System (1), is considered valid for the median surface of the plate. This is quite enough for a thin plate. We have a displacement estimation [55]:  
\[\frac{\partial u_{x}}{\partial x}=\alpha[T(x,\ \ z,\ \ t)-T_{0}(x,\ \ z,\ \ 0)],\ \ \frac{l}{2}\leq x\leq l,\ \ z=\frac{h}{2},\ \ t>0 \tag{5}\]  
After integrating (5), we have  
\[u_{x}=\int\alpha[T(x,\ \ z,\ \ t)-T_{0}(x,\ \ z,\ \ 0)]dx,\ \ \frac{l}{2}\leq x\leq l,\ \ z=\frac{h}{2},\ \ t>0 \tag{6}\]  
With a maximum temperature difference of 100 \(K\) and a coefficient of thermal expansion of \(\alpha=2.6\cdot 10^{-6}\ K^{-1}\), the maximum \(u_{x}\) values are about 0.13 mm. This is more than an order of magnitude lower than the deflection values (about 3 mm [25]). Therefore, it is sufficient to take into account only deflections that are much more significant than other components of the displacement vector to neutralize the temperature shock. It was noted in [3] that with the symmetrical formulation (symmetrical arrangement of flexible elements), only one disturbing factor from the temperature shock is unbalanced. This is the force of inertia in the direction of the \(z\)-axis [3]:  
\[\Phi_{2}(t)=-k\int\limits_{0}^{m_{1}}\ddot{u}_{z}(x,\ t)\ dm=-k\frac{m_{1}}{l} \int\limits_{0}^{l}\ddot{u}_{z}(x,\ \ t)\ dx \tag{7}\]  
where \(k\) is the number of flexible elements; \(m_{1}\) is the mass of one flexible element; and \(\ddot{u}_{z}(x,\ t)\) is the second total derivative of the deflection with respect to time. Let us use the expression obtained in [54] for deflections in the framework of the two-dimensional model of thermoelasticity (Expression (16) [54]):  
\[u_{z}(x,\ t)=\frac{At}{t+a}\Big{(}x^{4}-4 lx^{3}+6l^{2}x^{2}\Big{)}+u_{z0}(x, \ 0),\ 0\leq x\leq l,\ \ t>0 \tag{8}\]  
where \(a\) and \(A\) are some positive constants [54]. In [3], it was assumed that \(\ddot{u}_{z}(x,\ t)\approx\frac{\partial^{2}u_{z}(x,\ t)}{\partial t^{2}}\). In this case, the inertia force (7) does not depend on the initial deflection. It coincides with a similar force for the one-dimensional formulation of the thermoelasticity problem [22].