Let’s consider the solution of the problem with the following initial data:

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}l=90cm;\phantom{\rule{thinmathspace}{0ex}}a=-\frac{1}{15};\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}b=12cm;\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}q=-500\frac{Bm}{c{m}^{2}};}\\ {\displaystyle {k}_{xx}=100\frac{Bm}{cm{\cdot}^{\circ}C};\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}h=10\frac{Bm}{c{m}^{2}{\cdot}^{\circ}C};\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{T}_{oc}={20}^{\circ}C,}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{T}_{k}=T(x=l)={120}^{\circ}C;\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{T}_{j}=T(x=\frac{l}{2})=483,{46}^{\circ}C;}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{T}_{i}=T(x=0)=743,{077}^{\circ}C.}\end{array}$$

In this case, the temperature distribution along the length of the rod according to (13) has the following form:

$$\begin{array}{}{\displaystyle T(x,l,h,{k}_{xx},{T}_{oc},q,a,b)=-0,02564{x}^{2}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-4,6154x+743,077.}\end{array}$$(14)

Then:

$$\begin{array}{}{\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}T(x=0)={T}_{i}=743,{077}^{\circ}C;}\\ {\displaystyle T(x=\frac{l}{2})=T(x=45)={T}_{j}=483,{46}^{\circ}C;}\\ {\displaystyle \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}T(x=l)=T(x=90)={120}^{\circ}C.}\end{array}$$

Then the graph of the temperature distribution field along the length of the rod under the assumed initial data will look like Figure 2.

Figure 2 The field of temperature distribution along the length of the rod.

The figure shows that the distribution of temperatures has a slightly parabolic character. This is due to the variability of the cross-section of the rod along its length that varies according to the square law, and the radius which varies according to the linear law.

If one of the ends is firmly fixed and the other is free, due to the heat source it becomes longer. The value of elongation can be determined on the General laws of thermophysics:

$$\begin{array}{}{\displaystyle \mathit{\Delta}{l}_{T}=\underset{0}{\overset{l}{\int}}\alpha T(x)dx=\underset{0}{\overset{l}{\int}}\alpha (-0.02564{\text{x}}^{2}-4.6154x}\\ \phantom{\rule{2em}{0ex}}+743.077)dx=0.0000125(-0.02564{x}^{3}-4.6154{x}^{2}\\ {\displaystyle \phantom{\rule{2em}{0ex}}+743.077x)\underset{0}{\overset{l}{\int}}=0.524455}\end{array}$$

Here *α* = 0.0000125
$\begin{array}{}{\displaystyle (\frac{1}{{}^{\circ}C})}\end{array}$
- coefficient of thermal expansion of the material of the rod.

If both ends are firmly fixed, then axial compressive force in the rod due to thermal expansion occurs. Its value can be determined by using the compatibility conditions of deformation [13]. In our case, according to [13] the value of the axial force is determined by the next formula:

$$\begin{array}{}{\displaystyle R=-\frac{EF\mathit{\Delta}{l}_{T}}{l}=-\frac{E\mathit{\Delta}{l}_{T}}{l}\underset{0}{\overset{l}{\int}}F(x)dx=\frac{EF\mathit{\Delta}{l}_{T}}{l}.}\end{array}$$

Because *E* = 2 * 10^{6} *F*_{cp} = 9, then *R* = −3062209.933. In this case the field distribution of thermo-elastic stress in the cross-sections of the studied rod occurs, which is defined as:

$$\begin{array}{}{\displaystyle \sigma (x)=-\frac{=-3062209.933}{(-\frac{1}{15}x+12{)}^{2}},\phantom{\rule{1em}{0ex}}0\le x\le l=90cm,}\end{array}$$

The law of deformation component of thermoelastic distributiom is determined based on the laws of physics [13]:

$$\begin{array}{}{\displaystyle \epsilon (x)=\frac{\sigma (x)}{E},\phantom{\rule{1em}{0ex}}0\le x\le l=90cm,}\end{array}$$

Using the found temperature distribution law (14) we can calculate the distribution law of thermal component of deformation:

$$\begin{array}{}{\displaystyle {\epsilon}_{T}(x)=\alpha T(x)=\alpha (0.02564{x}^{2}+4.6154x-743.077)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}0\le x\le l=90cm,}\end{array}$$

Then according to the generalized Hooke’s law we can determine the distribution law of thermal component of stress:

$$\begin{array}{}{\displaystyle \phantom{\rule{thinmathspace}{0ex}}{\sigma}_{T}(x)=E{\epsilon}_{T}(x)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}=2\cdot {10}^{6}(0.02564{\text{x}}^{2}+4.6154\text{x}-743.077),}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\le x\le l=90cm,}\end{array}$$

Further, it’s possible to determine the distribution law of elastic component of deformation:

$$\begin{array}{}{\displaystyle {\epsilon}_{x}(x)=\epsilon (x)-{\epsilon}_{T}(x),\phantom{\rule{1em}{0ex}}0\le x\le l=90cm,}\end{array}$$

Further, based on generalized Hooke’s law we can determine the distribution law of elastic component of stress:

$$\begin{array}{}{\displaystyle 0\le x\le l=90cm,}\end{array}$$

Figure 3 shows the distribution of the three strain components along the length of the rod with variable cross-section, firmly fixed at both ends.

Figure 3 The distribution law of deformation: *1* - *ϵ*(*x*); *2* - *ϵ*_{T} (*x*) *u3* - *ϵ*_{x} (*x*).

The graph shows that the temperature and thermo - elastic components of deformation are compressive in nature throughout the length of the rod. When the elastic component of deformation on the length of the rod 0 ≤ *x* ≤ *l* = 67 *cm* has stretchable nature, on the length 67 ≤ *x* ≤ *l* = 90 *cm* – compressive nature. This process caused by the presence of a large heat flow
$\begin{array}{}{\displaystyle (q=-500\frac{Bm}{c{m}^{2}})}\end{array}$
on the left end of the rod, where the cross-sectional area at the left end is four times more than on the right end.

Figure 4 shows the distribution of three stress components along the length of the rod with variable cross-section area and firmly fixed at both ends.

Figure 4 The laws of stress distribution: *1* - *σ(x)*; *2* - *σ*_{T}(x) u 3 – *σ*_{x} (*x*).

From the figure it is evident that all stress components have the same character as the deformation.

Figure 5 shows the distribution of displacement of cross-sections of the rod.

Figure 5 The distribution law of displacement of cross-sections of the rod.

The figure shows that all cross sections move from left to right. It is also due to the large heat flux at the left end of the rod. The largest displacement is on the cross section of the rod with coordinates *x*=*51 cm*, as the displacement on both ends of the rod is equal to 0.

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