# Numerical Study of a Thermally Stressed State of a Rod

Authors: Anarbay Kudaykulov; Erkin Arinov; Nurlybek Ispulov; Abdul Qadir; Kalamkas Begaliyeva
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

This paper considers a new method based on the law of energy conservation for the study of thermo-stress-strain state of a rod of limited length with simultaneous presence of local heat fluxes, heat exchanges, and thermal insulation. The method allows determining the field of temperature distribution and the three components of deformations and stresses, as well as the magnitude of the rod elongation and the resulting axial force with an accuracy of satisfying the energy conservation laws. For specific initial data, all the sought-for ones are determined numerically with high accuracy. We found that all solutions satisfy the laws of energy conservation.


---

## Body

## 1. Introduction

This work is devoted to the study of the thermo-stress-strain state of a rod of constant cross section and of limited length. In this case, a heat flux of constant intensity is supplied to a closed local surface. The rest of the side surface of the rod is fully thermally insulated. Through the cross-sectional area of the two ends of the rod, convective heat exchange with the environment occurs. In this case, the coefficients of heat transfer and the temperature of the surrounding medium at the two ends of the rod are different. To determine the temperature field, the energy conservation law is used in the form of the total heat energy functional taking into account the existing dissimilar types of heat sources, physical and mechanical properties of the rod material, and its geometrical dimensions. Using this, the law of temperature distribution along the length of the rod under study is constructed. They determine the magnitude of the elongation of the rod, in the case of pinching one end. If both ends of the rod are clamped, then the magnitude of the resulting axial force is determined. The laws of distribution of thermoelastic, temperature, and elastic components of deformations and stresses are also determined depending on the presence of local heat flux, thermal insulation of heat exchanges, the geometry of the rod, and the physic-mechanical properties of the material of the rod. To determine the displacement field, the energy conservation law is used in the form of the potential energy functional of elastic deformation taking into account the presence of a temperature field. Further, the displacement field is determined taking into account the actual operating conditions. The developed program allows varying the values of the source data.In [1], the fundamental laws of the theory of thermoelasticity for deformable solids are given. In [2, 3], the results of a numerical study of a thermo-stress-deformed state of a rod under the action of laser beams are presented. In this case, the finite element method was used. The application of the finite element method is given in [4]. In [5], the exact solution of a two-dimensional definition problem in an ideally elastic-plastic cylindrical rod with a given uniform internal temperature is given. The dependence of voltage on temperature in a rod of limited length is given in [6]. In this work, the temporary factor is also taken into account. In [7], different statements of problems of thermal elasticity are given for structural elements under temperature effects. In this case, the theoretical basis of the method is focused on the Maysel formula of uncoupled thermoelasticity. In [8], on the basis of the small parameter method, the problem of determining the stress-strain state of an elastic-plastic pipe in the presence of temperature is considered. It uses the terms of Mises. In [9], the fundamentals of the theory of thermoelasticity and methods for their implementation in solving specific applied problems are presented. This takes into account the power and temperature factors. In the works [10, 11] methods and computational algorithms for numerical solution of the class of applied problems of mechanics are presented.In contrast to the above works, this paper uses the fundamental energy conservation laws in combination with the constructed quadratic spline functions to solve a particular applied problem.


## 2. Formulation of the Problem and Methods

We consider a horizontal bar of limited length L [cm], and a constant cross sectionF
c
m
2. The horizontal axisox we will direct from the left to right coincides with the axis of the rod. The side surfaces of the sections (0
≤
x
≤
l
1) and l
2
≤
x
≤
L the core are fully thermally insulated. A heat flux of constant intensity q
[
W
/
c
m
2
] is supplied on the lateral surface of the rod section l
1
≤
x
≤
l
2. Through the cross-sectional area of the left and right ends of the rod, heat exchange with the environment takes place.Heat transfer coefficients atx = 0 and atx = L
[
W
/
c
m
2
]. The temperature of the environments of these areas are T
E
n
v
1 and T
E
n
v
2 [K
o], respectively. The design scheme of the problem is shown in Figure 1.Figure 1
The design scheme of the problem.


## 3. The Solution to the Problem Using the Energy Conservation Law

The rod under consideration is discretized by elements of length L [cm]. Within the length of one discrete element, we approximate the temperature field by a second-order polynomial(1)
T
x
=
a
x
2
+
b
x
+
c
,
0
≤
x
≤
lwhere a, b, c are constants, the values of which are still unknown. The law of temperature distribution within the length of one discrete element is shown in Figure 2.Figure 2
The law of temperature distribution along the length of one discrete element.In the local coordinate system0
≤
x
≤
l, we fix three nodes with coordinates x
i
=
0
;
x
j
=
l
/
2
;
x
k
=
l.The temperature values in these sections will be denoted, respectively, by the following.(2)
T
x
=
x
i
=
0
=
T
i
;
T
x
=
x
j
=
l
2
=
T
j
;
T
x
=
x
k
=
l
=
T
kThen substituting (2) into (1), we get the following system.(3)
a
∙
0
+
b
∙
0
+
c
=
T
i
a
∙
l
2
2
+
b
∙
l
2
+
c
=
T
j
a
∙
l
2
+
b
∙
l
+
c
=
T
kSolving the system, we define the following.(4)
с
=
T
i
;
b
=
4
T
j
-
3
T
i
-
T
k
l
;
a
=
2
T
k
+
2
T
i
-
4
T
j
l
2Substituting (4) into (1), we have the following.(5)
T
x
=
2
T
k
+
2
T
i
-
4
T
j
l
2
∙
x
2
+
4
T
j
-
3
T
i
-
T
k
l
∙
x
+
T
i
=
2
x
2
-
3
l
x
+
l
2
l
2
T
i
+
4
l
x
-
4
x
2
l
2
T
j
+
2
x
2
-
l
x
l
2
T
k
,
0
≤
x
≤
lHere we introduce the following notations.(6)
φ
i
x
=
2
x
2
-
3
l
x
+
l
2
l
2
;
φ
j
x
=
4
l
x
-
4
x
2
l
2
;
φ
k
x
=
2
x
2
-
l
x
l
2
,
0
≤
x
≤
lThese functions are called quadratic spline functions in the local coordinate system. They have the following properties.(7)
φ
i
x
=
1
,
a
t
x
=
0
0
,
a
t
x
=
l
2
0
,
a
t
x
=
l
;
φ
j
x
=
0
,
a
t
x
=
0
1
,
a
t
x
=
l
2
0
,
a
t
x
=
l
;
φ
k
x
=
0
,
a
t
x
=
0
0
,
a
t
x
=
l
2
1
,
a
t
x
=
l
;
(8)
φ
i
x
+
φ
j
x
+
φ
k
x
=
1
;
0
≤
x
≤
lThe temperature gradient within the length of one discrete element in the local coordinate system has the following form.(9)
∂
T
∂
x
=
∂
φ
i
x
∂
x
T
i
+
∂
φ
j
x
∂
x
T
j
+
∂
φ
k
x
∂
x
T
k
=
4
x
-
3
l
l
2
T
i
+
4
l
-
8
x
l
2
T
j
+
4
x
-
l
l
2
T
k
,
0
≤
x
≤
lHere the following should be noted.(10)
∂
φ
i
x
∂
x
T
i
+
∂
φ
j
x
∂
x
T
j
+
∂
φ
k
x
∂
x
T
k
=
0To construct resolving systems of equations taking into account the natural boundary conditions, we discretize the rod under study by three elemental lengths. For the first discrete element, the total energy functional with the thermal insulation of the side surface is as follows.(11)
J
1
=
∫
S
x
=
0
h
1
2
T
-
T
o
c
1
2
d
s
+
∫
V
1
K
x
x
2
∂
T
∂
x
2
d
v
,
0
≤
x
≤
l
1Here the first integral over the cross-sectional area of the left end takes place only for points of this surface. In the local coordinate system, taking into account the global node indexing, (11) can be rewritten in the following form:(12)
J
1
=
F
x
=
0
h
1
2
T
-
T
o
c
1
2
+
F
K
x
x
2
∫
0
l
4
x
-
3
l
1
l
1
2
T
1
+
4
l
1
-
8
x
l
1
2
T
2
+
4
x
-
l
1
l
1
2
T
3
2
d
x
=
F
h
1
2
T
1
-
T
o
c
1
2
+
F
K
x
x
6
l
1
7
T
1
2
-
16
T
1
T
2
+
2
T
1
T
3
-
16
T
2
T
3
+
16
T
2
2
+
7
T
3
2where h
1 is the heat transfer coefficient.Here it must be said that the lateral surface of the first discrete element is fully thermally insulated. In expression (12), it should be noted that the sum of the coefficients before the nodal values of temperature will be zero. For example, in the first bracket (1-1) = 0, and in the second bracket [7-16 + 2-16 + 16 + 7] = 0. Now in Figure 1 we consider the second discrete element. This item is internal. But on the side surface of this element summed thermal constant intensity q
[
W
/
c
m
2
], the total heat energy functional will have the following form.(13)
J
2
=
∫
V
2
K
x
x
2
∂
T
∂
x
2
d
v
+
∫
S
n
б
n
q
T
d
s
=
F
K
x
x
2
∫
0
l
4
x
-
3
l
2
l
2
2
T
3
+
4
l
2
-
8
x
l
2
2
T
4
+
4
x
-
l
2
l
2
2
T
5
2
d
x
+
q
P
∫
0
l
2
x
2
-
3
l
2
x
+
l
2
2
l
2
2
T
3
+
4
l
2
x
-
4
x
2
l
2
2
T
4
+
2
x
2
-
l
2
x
l
2
2
T
5
d
x
=
F
K
x
x
6
l
2
7
T
3
2
-
16
T
3
T
4
+
2
T
3
T
5
-
16
T
4
T
5
+
16
T
4
2
+
7
T
5
2
+
q
π
r
l
2
3
T
3
+
4
T
4
+
T
5
,
l
1
≤
x
≤
l
2HereV
2 is the volume of the second discrete element;P = 2πk is the perimeter of the cross section. Finally go to the last third discrete element. The side surface of this element is fully insulated, but through the cross-sectional area of the right end there is a heat exchange with its environment. In this case, the heat transfer coefficient is h
2, and the ambient temperature is T
E
n
v
2. The length of this item is l
3. The total heat energy functional for the third discrete element will be as follows:(14)
J
3
=
∫
V
3
K
x
x
2
∂
T
∂
x
2
d
v
+
∫
S
x
=
l
1
+
l
2
+
l
3
h
2
2
T
-
T
o
c
2
2
d
s
=
F
K
x
x
2
∫
0
l
4
x
-
3
l
3
l
3
2
T
5
+
4
l
3
-
8
x
l
3
2
T
6
+
4
x
-
l
3
l
3
2
T
7
2
d
x
+
F
h
2
2
T
7
-
T
o
c
2
2
=
F
K
x
x
6
l
3
7
T
5
2
-
16
T
5
T
6
+
2
T
5
T
7
-
16
T
6
T
7
+
16
T
6
2
+
7
T
7
2
+
F
h
2
2
T
7
-
T
o
c
2
2
,
l
2
≤
x
≤
Lwhere V
3  is the volume of the third discrete element andF is the cross-sectional area of the rod; then the total heat energy functional for the rod under study has the following form.(15)
J
=
J
1
+
J
2
+
J
3
=
F
h
1
2
T
1
-
T
o
c
1
2
+
F
K
x
x
6
l
1
×
7
T
1
2
-
16
T
1
T
2
+
2
T
1
T
3
-
16
T
2
T
3
+
16
T
2
2
+
7
T
3
2
+
F
K
x
x
6
l
2
×
F
K
x
x
6
l
2
·
7
T
3
2
-
16
T
3
T
4
+
2
T
3
T
5
-
16
T
4
T
5
+
16
T
4
2
+
7
T
5
2
+
q
π
r
l
2
3
T
3
+
4
T
4
+
T
5
+
F
K
x
x
6
l
3
7
T
5
2
-
16
T
5
T
6
+
2
T
5
T
7
-
16
T
6
T
7
+
16
T
6
2
+
7
T
7
2
+
F
h
2
2
T
7
-
T
o
c
2
2To construct resolving systems of linear algebraic equations for the nodal values of temperatures, the functionalJ is minimized byT1, T2, …, T7.(16)
1
∂
J
∂
T
1
=
0
;
⇒
F
h
1
T
1
-
T
E
n
v
1
+
F
K
x
x
6
l
1
14
T
1
-
16
T
2
+
2
T
3
=
0
;
2
∂
J
∂
T
2
=
0
;
⇒
F
K
x
x
6
l
1
-
16
T
1
+
32
T
2
-
16
T
3
=
0
;
3
∂
J
∂
T
3
=
0
;
⇒
F
K
x
x
2
T
1
+
16
T
2
+
14
T
3
6
l
1
+
14
T
3
-
16
T
4
+
2
T
5
6
l
2
+
q
π
r
l
2
3
=
0
;
4
∂
J
∂
T
4
=
0
;
⇒
F
K
x
x
6
l
2
-
16
T
3
+
32
T
4
-
16
T
5
+
4
q
π
r
l
2
3
=
0
;
5
∂
J
∂
T
5
=
0
;
⇒
F
K
x
x
2
T
3
-
16
T
4
+
14
T
5
6
l
2
+
14
T
5
-
16
T
6
+
2
T
7
6
l
3
+
q
π
r
l
2
3
=
0
;
6
∂
J
∂
T
6
=
0
;
⇒
F
K
x
x
6
l
3
-
16
T
5
+
32
T
6
-
16
T
7
=
0
;
7
∂
J
∂
T
7
=
0
;
⇒
F
K
x
x
6
l
3
2
T
5
-
16
T
6
+
14
T
7
+
F
h
2
T
7
-
T
E
n
v
2
=
0
.Solving this system, the node values of temperaturesT
1
,
T
2
,
…
,
T
7 are calculated. They are used to construct the law of temperature distribution along the length of three discrete sections of the rod.(17)
T
I
x
=
2
x
2
-
3
l
1
x
+
l
1
2
l
1
2
T
1
+
4
l
1
x
-
4
x
2
l
1
2
T
2
+
2
x
2
-
l
1
x
l
1
2
T
3
,
0
≤
x
≤
l
1
T
I
I
x
=
2
x
2
-
3
l
2
x
+
l
2
2
l
2
2
T
3
+
4
l
2
x
-
4
x
2
l
2
2
T
4
+
2
x
2
-
l
2
x
l
2
2
T
5
,
0
≤
x
≤
l
2
T
I
I
I
x
=
2
x
2
-
3
l
3
x
+
l
3
2
l
3
2
T
5
+
4
l
3
x
-
4
x
2
l
3
2
T
6
+
2
x
2
-
l
3
x
l
3
2
T
7
,
0
≤
x
≤
l
3If the coefficient of thermal expansion of the material of the rodα
1
/
K
0 is a constant value, then the value of thermal elongation of the rod Δ
l
T
[
c
m
] in case of pinching with one end is determined in accordance with the theory of thermal physics [11]. (18)
Δ
l
T
=
∫
0
L
α
·
T
x
d
x
.
(19)
w
h
e
r
e
T
x
=
T
I
x
+
T
I
I
x
+
T
I
I
I
xIf both ends of the rod are clamped, then it cannot lengthen, but there is an axial compressive force R [kG], which is determined by the conditions of compatibility deformation.Then substituting (19) in (18), we get the following.(20)
Δ
l
T
=
α
l
1
6
T
1
+
2
l
1
3
T
2
+
l
1
6
T
3
+
l
2
6
T
3
+
2
l
2
3
T
4
+
l
2
6
T
5
+
l
3
6
T
5
+
2
l
3
3
T
6
+
l
3
6
T
7
=
α
6
l
1
T
1
+
4
l
1
T
2
+
l
1
+
l
2
T
3
+
4
l
2
T
4
+
l
2
+
l
3
T
5
+
4
l
3
T
6
+
l
3
T
7If both ends of the rod are clamped, then it cannot lengthen, but there is an axial compressive force R [kG], which is determined by the conditions of compatibility deformation. The essence of this approach is as follows. First, consider the horizontal rod clamped by the left end (Figure 3).Figure 3
The left end of the rod is pinned under the influence of compressive force R [kG].This rod is under the influence of the compressive force R which is applied at the right free end. Then, according to Hooke’s law, it is reduced by the value ofΔ
l
R
(21)
Δ
l
R
=
R
∙
L
E
∙
Fwhere L
=
l
1
+
l
2
+
l
3 is the total length of the test rod, E
k
G
/
c
m modulus of elasticity of the material of the rod, and F
[
c
m
2
] rod area.If both ends of the rod under study are rigidly clamped, then naturally it does not lengthen or shorten; i.e.,(22)
Δ
l
R
+
Δ
l
T
=
0
.Then, taking into account (21), we will determine the value of the axial compressive force R that arises in the clamped rod by the two ends of the rod under study.(23)
R
=
-
E
∙
F
∙
Δ
l
T
L
=
-
E
∙
F
∙
α
6
L
l
1
T
1
+
4
l
1
T
2
+
l
1
+
l
2
T
3
+
4
l
2
T
4
+
l
2
+
l
3
T
5
+
4
l
3
T
6
+
l
3
T
7In this case, the distribution field of the thermoelastic component of stressσ
[
k
G
/
c
m
2
] also arises, which is determined in accordance with Hooke’s law [1].In this case, the distribution field of the thermoelastic component of stressσ
[
k
G
/
c
m
2
], which is determined in accordance with Hooke’s law [1], also appears.(24)
σ
x
=
R
F
=
c
o
n
s
tThen, according to Hooke’s law, the field distribution of the thermoelastic component of deformation has the following form.(25)
ε
x
=
σ
x
E
=
c
o
n
s
tThe field distribution of the temperature component of the strain and stress is determined on the basis of the general theory of thermoelasticity.(26)
Ɛ
T
x
=
-
α
T
x
,
(27)
σ
T
x
=
E
·
ε
T
x
=
-
α
E
T
x
,From these relations, in accordance with the theory of thermoelasticity, the field distribution of elastic components of deformations and stresses is determined.(28)
ε
x
x
=
ε
-
ε
T
=
σ
x
E
+
α
·
T
x
,
(29)
σ
x
x
=
E
·
ε
x
x
=
σ
x
-
σ
T
x
,The displacement field along the length of one discrete element of length l [cm] is approximated by a complete second-order polynomial.(30)
U
x
=
φ
i
x
U
i
+
φ
j
x
U
j
+
φ
k
x
U
k
,
0
≤
x
≤
lThe gradient of displacement is determined from here(31)
∂
u
∂
x
=
4
x
-
3
l
l
2
U
i
+
4
l
-
8
x
l
2
U
j
+
4
x
-
l
l
2
U
k
,
0
≤
x
≤
lwhere U
i
=
U
(
x
=
0
)
;
U
j
=
U
(
x
=
l
/
2
)
;
U
k
=
U
(
x
=
l
). The potential energy functional of elastic deformation in the presence of a temperature field has the following form [1].(32)
П
=
∫
V
σ
x
x
2
ε
x
x
d
v
-
∫
V
α
E
·
T
x
·
ε
x
x
d
vwhere ε
x
(
x
)
=
∂
U
/
∂
x
=
4
x
-
3
l
/
l
2
U
i
+
4
l
-
8
x
/
l
2
U
j
+
4
x
-
l
/
l
2
U
k is the elastic component of deformation; σ
x
(
x
)
=
E
ε
x
(
x
)
=
E
(
∂
U
/
∂
x
)
=
E
4
x
-
3
l
/
l
2
U
i
+
4
l
-
8
x
/
l
2
U
j
+
4
x
-
l
/
l
2
U
k is the elastic component of stress.Given these relations, the expression of the potential energy of other strains for the rod under study can be rewritten as follows.(33)
П
=
∫
V
1
σ
x
x
2
ε
x
x
d
v
-
∫
V
1
α
·
E
·
T
x
·
ε
x
x
d
v
+
∫
V
2
σ
x
x
2
ε
x
x
d
v
-
∫
V
2
α
·
E
·
T
x
·
ε
x
x
d
v
+
∫
V
3
σ
x
x
2
ε
x
x
d
v
-
∫
V
3
α
·
E
·
T
x
·
ε
x
x
d
v
=
E
F
2
∫
0
l
1
4
x
-
3
l
l
2
U
1
+
4
l
-
8
x
l
2
U
2
+
4
x
-
l
l
2
U
3
2
d
x
-
α
E
F
∫
0
l
1
2
x
2
-
3
l
x
+
l
l
2
T
1
+
4
l
-
4
x
2
l
2
T
2
+
2
x
2
-
l
x
l
2
T
3
×
4
x
-
3
l
l
2
U
1
+
4
l
-
8
x
l
2
U
2
+
4
x
-
l
l
2
U
3
d
x
+
E
F
2
∫
0
l
2
4
x
-
3
l
l
2
U
3
+
4
l
-
8
x
l
2
U
4
+
4
x
-
l
l
2
U
5
2
d
x
-
α
E
F
∫
0
l
2
2
x
2
-
3
l
x
+
l
l
2
T
3
+
4
l
-
4
x
2
l
2
T
4
+
2
x
2
-
l
x
l
2
T
5
×
4
x
-
3
l
l
2
U
3
+
4
l
-
8
x
l
2
U
4
+
4
x
-
l
l
2
U
5
d
x
+
E
F
2
∫
0
l
3
4
x
-
3
l
l
2
U
5
+
4
l
-
8
x
l
2
U
6
+
4
x
-
l
l
2
U
7
2
d
x
-
α
E
F
∫
0
l
3
2
x
2
-
3
l
x
+
l
l
2
T
5
+
4
l
-
4
x
2
l
2
T
6
+
2
x
2
-
l
x
l
2
T
7
×
4
x
-
3
l
l
2
U
5
+
4
l
-
8
x
l
2
U
6
+
4
x
-
l
l
2
U
7
d
x
=
E
F
6
l
1
7
U
1
2
+
16
U
2
2
+
7
U
3
2
-
16
U
1
U
2
+
2
U
1
U
3
-
16
U
2
U
3
-
α
E
F
-
1
2
T
1
U
1
+
2
3
T
1
U
2
-
1
6
T
1
U
3
-
2
3
T
2
U
1
+
2
3
T
2
U
3
+
1
6
T
3
U
1
-
2
3
T
3
U
2
+
E
F
6
l
2
7
U
3
2
+
16
U
4
2
+
7
U
5
2
-
16
U
3
U
4
+
2
U
3
U
5
-
16
U
4
U
5
-
α
E
F
2
3
T
3
U
4
-
1
6
T
3
U
5
-
2
3
T
4
U
3
+
2
3
T
4
U
5
+
1
6
T
5
U
3
-
2
3
T
5
U
4
+
E
F
6
l
3
7
U
5
2
+
16
U
6
2
+
7
U
7
2
-
16
U
5
U
6
+
2
U
5
U
7
-
16
U
6
U
7
-
α
E
F
2
3
T
5
U
6
-
1
6
T
5
U
7
-
2
3
T
6
U
5
+
2
3
T
6
U
7
+
1
6
T
7
U
5
-
2
3
T
7
U
6
+
1
2
T
7
U
7Due to the pinch of both ends of the rod,U
1
=
U
7
=
0.By minimizing this functional on the nodal displacementsU
2
,
U
3
,
…
,
U
6, a resolving system of linear algebraic equations is constructed for the desired quantities, taking into account the simultaneous presence of heterogeneous heat sources and thermal insulation(34)
1
∂
П
∂
U
2
=
0
;
⇒
E
F
6
l
1
32
U
2
-
16
U
3
-
α
E
F
2
3
T
1
-
2
3
T
3
=
0
;
2
∂
П
∂
U
3
=
0
;
⇒
E
F
14
U
3
-
16
U
2
16
l
1
+
14
U
3
-
16
U
4
+
2
U
5
16
l
2
-
α
E
F
-
1
6
T
1
+
2
3
T
2
-
2
3
T
4
+
1
6
T
5
=
0
;
3
∂
П
∂
U
4
=
0
;
⇒
E
F
6
l
2
32
U
4
-
16
U
3
-
16
U
5
-
α
E
F
2
3
T
3
-
2
3
T
5
=
0
;
4
∂
П
∂
U
5
=
0
;
⇒
E
F
14
U
5
+
2
U
3
-
16
U
4
16
l
2
+
14
U
5
-
16
U
6
+
2
U
7
16
l
7
-
α
E
F
-
1
6
T
3
+
2
3
T
4
-
2
3
T
6
+
1
6
T
7
=
0
;
5
∂
П
∂
U
6
=
0
;
⇒
E
F
6
l
3
32
U
6
-
16
U
5
-
16
U
7
-
α
E
F
2
3
T
5
-
2
3
T
7
=
0
.By solving the latter system, the valuesU
2
,
U
3
,
…
,
U
6 are determined. According to them, the displacement field is constructed within three discrete elements.(35)
U
I
x
=
4
l
x
-
4
x
2
l
2
·
U
2
+
2
x
2
-
l
x
l
2
·
U
3
U
I
I
x
=
2
x
2
-
3
l
x
+
l
2
l
2
·
U
3
4
l
x
-
4
x
2
l
2
·
U
4
+
2
x
2
-
l
x
l
2
·
U
5
U
I
I
I
x
=
2
x
2
-
3
l
x
+
l
2
l
2
·
U
5
4
l
x
-
4
x
2
l
2
·
U
6


## 4. Analysis of the Obtained Results

For the practical application of the above method and algorithm, we take the following initial data:(36)
L
=
3
c
m
;
r
=
1
c
m
;
α
=
0
,
0000125
1
/
К
0
;
E
=
2
·
1
0
6
k
G
/
c
m
2
;
К
x
x
=
100
w
a
t
t
/
c
m
·
K
0
;
h
1
=
h
2
=
10
w
a
t
t
/
c
m
2
·
K
0
;
T
a
x
i
s
1
=
T
a
x
i
s
2
=
4
0
0
K
;
q
=
-
500
w
a
t
t
/
c
m
2
;with these initial data, the resulting solution is shown in Figures4–7.Figure 4
Temperature dependence T along the length of the rod.Figure 5
Dependencies deformations along the length of the rod. (1) Distribution law of temperature component of deformationε
T
(
x
); (2) law distribution of thermoelastic component of deformation ε
(
x
); (3) law distribution of elastic component of deformation ε
x
(
x
).Figure 6
Stress Dependencies along the length of the rod. (1) Law distribution of temperature component of stressesσ
T
(
x
); (2) law distribution of thermoelastic component of stresses σ
(
x
); (3) law distribution of elastic component of stresses σ
x
(
x
).Figure 7
Dependence of displacement along the length of the rod.For the numerical solution of the problem under consideration, the rod under study was discretized by n = 24—discrete elements of the same lengthl
=
L
/
n
=
30
/
24
=
1
,
25
c
m. The side surface of the first and last 8 elements is fully insulated. The heat flux q
=
-
500
w
a
t
t
/
c
m
2 is supplied to the lateral surface of the middle 8 elements.From Figure4, it is seen that, due to the symmetry of the problem under consideration, at the ends of the rod the temperature values will be T (x = 0) = T (x = L) = 5400 K. At that time, the highest temperature occurs in the middle of the rod, T
x
x
=
L
/
2
=
116
5
0
K
. It can also be seen from Figure 4 that the temperature distribution field along the length of the rod under study has a parabolic character.The laws of distribution of the three component deformations are shown in Figure5. Here, the distribution field of the temperature component of deformation Ɛ
T
x has a compressive character along the entire length of the rod under study. It has a parabolic shape with a bulge down. Its value at the ends of the investigated rod will be Ɛ
T
x
=
0
=
Ɛ
T
x
=
L
=
-
0.0675
. At the time in the middle of the rod, it has Ɛ
T
x
=
L
/
2
=
-
0.014562
. This shows that in the middle of the rod the temperature component of the deformation will be 2.157 times greater than that at the end of the rod. At that time, the value of the thermoelastic component of deformations Ɛ
x will be constant along the length of the rod and Ɛ
x
=
-
0.0112
. The behavior of the elastic component of deformation will be of alternating sign. For example, in the middle part of the rod, where the heat flux is supplied, the elastic component of deformation Ɛ
x has a tensile character. Amplitude Ɛ
x
x
=
L
/
2
=
-
0.033
. At that time, the first and last 1/3 of the rod will experience compressive ε
x
x. It should be noted that at the ends of the rod Ɛ
x
x
=
0
=
Ɛ
x
x
=
L
=
-
0.0045
. In general, the distribution field Ɛ
x
x has a parabolic character, convex at the top. In the relevant generalized Hooke’s law, the behavior of the stress components will be appropriate for the corresponding component strain (Figure 6).The values of the thermoelastic component of stressσ
(
x
) along the entire length of the rod will be constant σ
x
=
-
22528
k
G
/
c
m
2
. However, the law of distribution of the temperature component of voltage σ
T
x will have a parabolic character, bulge down, and it will have a compressive character along the entire length of the rod. The values of σ
T
x at the ends of the rod will be σ
T
x
=
0
=
σ
T
x
=
L
=
-
13500
k
G
/
c
m
2. At that time in the middle of the rod σ
T
(
x
=
L
/
2
)
=
-
29125
k
G
/
c
m
2. This shows that in the middle of the bar the value of σ
T will be 2.157 times more than that at its ends. The elastic component of stress σ
x
(
x
) in the first and last 1/3 of the rod behaves as compressive, and in the middle 1/3 as tensile. With this(37)
σ
x
x
=
0
=
σ
x
x
=
L
=
-
9027
k
G
/
c
m
2
,
σ
x
x
=
L
2
=
6597
k
G
/
c
m
2
.The distribution law of the displacement of the rod sections is not shown in Figure7. From this figure, it is clear that all sections of the left half of the rod are moved against the direction of the axisox, and sections of the right half are shifted in the direction of the axis of the oh. This process is due to the presence on the surface of the middle 1/3 of the investigated rod summed heat flux. At that time, the amplitude of the displacement of the section whose coordinate x =7.5 cm will be equal to U=(x=7.5)= - 0.01627 cm. Similarly, we have that U=(x=22.5)=0.01627 cm. This shows that the process under study is strictly symmetric about the middle of the rod.


## 5. Conclusion

The developed method based on the fundamental laws of energy conservation, a computational algorithm, and an application program in Python allows automating the construction of resolving systems of equations taking into account the natural boundary conditions for rods of limited length under the influence of dissimilar types of heat sources. The developed system also allows determining the laws of temperature distribution, all components of deformations and stresses, and displacements. In this regard, it can be stated that the developed method, algorithm, and Python programs are relatively universal in the sense of studying the steady thermo-stress-strain state of the supporting core elements of strategic equipment under the influence of various local heat sources. In this case, the obtained numerical results will differ in high accuracy, since these results satisfy the fundamental laws of energy conservation.



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*Source: 8986010-2019-06-02.xml*