## 3 Evaluation of Disturbing Factors
Let us consider \(\ddot{u}_{z}(x,\ t)\) in a more general way:  
\[\ddot{u}_{z}(x,\ t)=\frac{\partial^{2}u_{z}(x,\ t)}{\partial x^{2}}\ddot{u}_{ x}(x,\ t)+2\frac{\partial^{2}u_{z}(x,\ t)}{\partial x\,\partial t}\dot{u}_{z}(x,\ t)+\frac{\partial^{2}u_{z}(x,\ t)}{\partial t^{2}} \tag{9}\]  
Taking into account (8), we have  
\[ \begin{array}{l}\ddot{u}_{z}(x,\ t)=\Big{[}12\frac{At}{t+a}\big{(}x^{2}-2 lx +l^{2}\big{)}+\frac{\partial^{2}u_{z0}(x,\ 0)}{\partial x^{2}}\Big{]}\ddot{u}_{x}(x,\ t)+\\ +8\frac{a}{(t+a)^{2}}\big{(}x^{3}-3 lx^{2}+3l^{2}x\big{)}\dot{u}_{z}(x,\ t)- \\ -2\frac{a}{(t+a)^{3}}\big{(}x^{4}-4 lx^{3}+6l^{2}x^{2}\big{)},\ 0\leq x\leq l,\ t>0.\end{array}  \tag{10}\]Analysis of Expression (10) shows that acceleration (\(\ddot{u}_{z}(x,\ t)\)) is not necessarily opposite to the deflection (\(u_{z}(x,\ t)\)) itself, as in the case when \(\ddot{u}_{z}(x,\ t)\approx\frac{\partial^{2}u_{z}(x,\ t)}{\partial t^{2}}\). It is more general. In this case, the inertia force (7) depends on the initial deflection, because the first term of the right part (10) contains the second derivative of the initial deflection function (\(u_{z0}=u_{z0}(x,\ 0)\)). Let us use the approximation of the temperature field proposed in [54]:  
\[T(x,\ z,\ t)=Cz\frac{t}{t+a}-Mx+T_{0},\ 0\leq x\leq l,\ 0\leq z\leq h,\ t>0 \tag{11}\]  
where \(C\) and \(M\) are some positive constants [54]. Then, considering \(z=h/2\) (the median surface of the plate), we obtain expressions for \(\dot{u}_{x}(x,\ t)\) and \(\ddot{u}_{x}(x,\ t)\):  
\[\dot{u}_{x}(x,\ t)=\frac{\partial u_{x}}{\partial x}\frac{dx}{dt}+\frac{ \partial u_{x}}{\partial t} \tag{12}\]  
We believe that \(\dot{u}_{x}(x,\ t)\equiv\frac{dx}{dt}\). Therefore, we have  
\[\dot{u}_{x}(x,\ t)=\frac{1}{1-\frac{\partial u_{x}}{\partial x}}\frac{ \partial u_{x}}{\partial t} \tag{13}\]  
Considering Expressions (6) and (11), we obtain  
\[\frac{\partial u_{x}}{\partial x}=\alpha\left(C\frac{t}{t+a}\frac{h}{2}-Mx\right) \tag{14}\]  
\[\frac{\partial u_{x}}{\partial t}=\alpha\ C\frac{a}{(t+a)^{2}}\frac{h}{2}x \tag{15}\]  
Substituting (14) and (15) into (13), we get  
\[\dot{u}_{x}(x,\ t)=\frac{\alpha Cahx}{\left[2-\alpha\left(C\frac{t}{t+a}h-2 Mx \right)\right]\left(t+a\right)^{2}} \tag{16}\]  
The second total derivative has the following form:  
\[\ddot{u}_{x}(x,\ t)=\frac{\partial\dot{u}_{x}}{\partial x}\frac{dx}{dt}+\frac {\partial\dot{u}_{x}}{\partial t} \tag{17}\]  
Then  
\[\frac{\partial\dot{u}_{x}}{\partial x}=\frac{\alpha\ Cahx}{\left(t+a\right)^{ 2}}\frac{2-\alpha\ C\frac{t}{t+a}h}{\left[2-\left(\alpha\ C\frac{t}{t+a}h-2 Mx \right)\right]^{2}} \tag{18}\]  
\[\frac{\partial\dot{u}_{x}}{\partial t}=\frac{\alpha\ Cahx}{\left(t+a\right)^{ 3}\left[2-\left(\alpha\ C\frac{t}{t+a}h-2 Mx\right)\right]}\left(\frac{\alpha \ Cah}{\left(t+a\right)\left[2-\left(\alpha\ C\frac{t}{t+a}h-2 Mx\right) \right]^{2}}-2\right) \tag{19}\]  
Substituting (18) and (19) into (17), we get  
\[ \begin{array}{l}\ddot{u}_{x}=\frac{\alpha\ Cahx}{\left(t+a\right)^{3}\left[2 -\left(\alpha\ C\frac{t}{t+a}h-2 Mx\right)\right]}\left\{\frac{\alpha\ Cah}{ \left(t+a\right)\left[2-\left(\alpha\ C\frac{t}{t+a}h-2 Mx\right)\right]} \right.\\ \cdot\left[\frac{2-\alpha\ C\frac{t}{t+a}h}{2-\left(\alpha\ C\frac{t}{t+a}h-2 Mx \right)}+1\right]-2\right\}.\end{array}  \tag{20}\]  
It is possible to obtain estimations of the disturbing effect of temperature shock on the small spacecraft with the symmetrical scheme of flexible elements by substituting (16) and (20) into (10) at first, and after that (10) into (7).