External Force and Displacement Boundary Value Problems

Synonyms

Point heat source; Thermal stress; Uniform heat flux

Overview

The theory of thermoelasticity, as a classical problem in the field of solid mechanics, has been developed for a long time since the nineteenth century. The thermal stress problem still remains attracting research interest due to some new engineering applications such as in the very large-scale integration (VLSI) systems, new materials, nuclear engineering, and the field of aeronautics in recent years. Heating or nonuniform temperature fields make failures or reduce the service life of materials and structures. The investigations of the related problems have brought a considerable number of research works, both theoretical and experimental, treating various aspects of thermal stresses encountered in practice.

The following problems are solved: uniform heat flux problems for external force and displacement boundary value and heat point source problems for respective boundary values.

Scientific Fundamentals

Rational...

Appendix: Rational Mapping Function in Example 1

Equation 12 can be expressed in the following form and further expanded to a power series [1, 6, 12]:

$$ {{(1 - \zeta )}^{\alpha }} = \sum\limits_{{k = 0}}^{\infty } {\frac{{\alpha (\alpha - 1)\ldots (\alpha - k + 1)}}{{k!}}} {{( - 1)}^k}{{\zeta}^k} = 1 - {{a}_1}\zeta - {{a}_2}{{\zeta}^2} - {{\alpha}_3}{{\zeta}^3} + \ldots $$

(50)

This term represents the shape with a projection where coefficient $\alpha$ ($0 < \alpha < 1$) is determined by the angle of the projection.

Consider the following fractional expression and the expansion of the power series corresponding to (50):

$$\begin{array}{ll}1 + \sum\limits_{{j = 1}}^n {\left[ {\frac{{ - {{A}_j}}}{{1 - {{\alpha}_j}\zeta }} + {{A}_j}} \right]} = 1 \cr - \sum\limits_{{j = 1}}^n ( {{A}_j}{{\alpha}_j}\zeta + {{A}_j}\alpha_j^2{{\zeta}^2} + {{A}_j}\alpha_j^3{{\zeta}^3} + \ldots ) \end{array}$$

(51)

where $\left| {{{\alpha}_j}} \right| < 1$. The signs in front of ${{A}_j}$ is determined from the condition that the signs of the coefficients in (50) and (51) are identical. The 2n unknowns ${{A}_j},\;{ }{{\alpha}_j}$ are determined by approximately equating the corresponding coefficients of the terms with the same power in the series of (52) and (53), respectively,

$$ \sum\limits_{{j = 1}}^n {{{A}_j}\alpha_j^k} = {{a}_k} $$

(52)

The selection of 2n numbers of ${{a}_k}$ depends on the convergence of the expanded power series, which is treated separately for the cases of rapidly and slowly convergent terms. Here, considering the slow convergence of the series in (50), the monotonous decrease of its coefficients and the convenience of the numerical calculation ${{a}_k}$ are selected up to high order of power for $\zeta$ in the following way:

$$ \begin{array}{lll} \mathrm{{1^{st}} \; \rm{block}}\;\;\;\;\;\;\;\;{{a}_1},\quad{ }{{a}_2},\quad{ }{{a}_3},\quad{ }{{a}_4},\quad{ }{{a}_5},\quad{ }{{a}_6},\quad{ }{{a}_7},\quad{ }{{a}_8} \cr {\rm{2^{nd}} \; \mathrm{block}}\;\;\;\;\;\;\;{{a}_{{2M}}},\quad{{a}_{{3M}}}{, }\quad{{a}_{{4M}}},{ }\quad{{a}_{{5M}}} \cr {\rm{3^{rd}} \; \rm{block}}\;\;\;\;\;\;\;{{a}_{{2{{M}^2}}}},\quad{ }{{a}_{{3{{M}^2}}}}{, }\quad{{a}_{{4{{M}^2}}}},\quad{{a}_{{5{{M}^2}}}} \cr {\rm{4^{th}} \; \rm{block}}\;\;\;\;\;\;\;{{a}_{{2{{M}^3}}}},{ }\quad{{a}_{{3{{M}^3}}}}{, }\quad{{a}_{{4{{M}^3}}}},{ }\quad{{a}_{{5{{M}^3}}}}\; \cr {\rm{5^{th}} \; \rm{block}}\;\;\;\;\;\;\;{{a}_{{2{{M}^4}}}},{ }\quad{{a}_{{3{{M}^4}}}}{, }\quad{{a}_{{4{{M}^4}}}},{ }\quad{{a}_{{5{{M}^4}}}}\; \cr {\rm{6^{th}} \; \rm{block}}\;\;\;\;\;\;\;{{a}_{{2{{M}^5}}}},{ }\quad{{a}_{{3{{M}^5}}}}{, }\quad{{a}_{{4{{M}^5}}}},{ }\quad{{a}_{{5{{M}^5}}}} \end{array} $$

(53)

28 of ${{a}_k}$ (in this case) was selected in six blocks of which the particular terms are defined by M. Equation 54 is solved by the iteration. In this case, the number of the fractional expressions is n = 14. The number of blocks may be increased or decreased due to the convergence of the series in (52). The value of M is related to the interval of ${{a}_k}$. $M = 8$ or its vicinity 7 or 9 is selected due to our experience for the good agreement between (50) and (51) and the good convergence of the numerical iteration. That more numbers of ${{a}_k}$ are adopted in the 1st block than others is to make the coefficients of low power agree well.

The nonlinear equations (52) are solved by the numerical iteration because it cannot be solved analytically in the following way: chiefly, ${{A}_1},{ }{{\alpha}_1}$ and ${{A}_2},{ }{{\alpha}_2}$ of ${{A}_j},{ }{{\alpha}_j}$ are determined from the 6th block and ${{A}_3},{ }{{\alpha}_3},{ }{{A}_4},{ }{{\alpha}_4}$ from the 5th block and so forth. The calculation is started from the 6th block. All initial values of ${{A}_j},{ }{{\alpha}_j}$ except ${{A}_1},{ }{{\alpha}_1}$ and ${{A}_2},{ }{{\alpha}_2}$ which will be determined from the 6th block can be treated as zero in the first iterative calculation. By subtracting the influence of ${{A}_3},\;{ }{{\alpha}_3},\;{ }{{A}_4},{ }\;{{\alpha}_4}$ in the 5th block, i.e., ${{A}_3}\alpha_3^{{2{{M}^5}}} + {{A}_4}\alpha_4^{{2{{M}^5}}}$, ${{A}_3}\alpha_3^{{3{{M}^5}}} + {{A}_4}\alpha_4^{{3{{M}^5}}}$, ${{A}_3}\alpha_3^{{4{{M}^5}}} + {{A}_4}\alpha_4^{{4{{M}^5}}}$, and ${{A}_3}\alpha_3^{{5{{M}^5}}} + {{A}_4}\alpha_4^{{5{{M}^5}}}$ from the 4 numbers of ${{a}_{{2{{M}^5}}}},\;{ }{{a}_{{3{{M}^5}}}}{, }\;{{a}_{{4{{M}^5}}}},\;{ }{{a}_{{5{{M}^5}}}}$ in the 6th block, ${{A}_1},{ }{{\alpha}_1}$ and ${{A}_2},{ }{{\alpha}_2}$ can be obtained by solving the following equations expressed in terms of $a{{_2^{,}}_{{{{M}^5}}}},\;{ }\;a{{_3^{,}}_{{{{M}^5}}}}{, }\;a{{_4^{,}}_{{{{M}^5}}}},\;{ }a{{_5^{,}}_{{{{M}^5}}}}$:

$$ \eqalign{ {{A}_1}\alpha_1^{{2{{M}^5}}} + {{A}_2}\alpha_2^{{2{{M}^5}}} = {{a}_{{2{{M}^5}}}} - ({{A}_3}\alpha_3^{{2{{M}^5}}} + {{A}_4}\alpha_4^{{2{{M}^5}}}) \equiv a_{{2{{M}^5}}}^{,} \cr {{A}_1}\alpha_1^{{3{{M}^5}}} + {{A}_2}\alpha_2^{{3{{M}^5}}} = {{a}_{{3{{M}^5}}}} - ({{A}_3}\alpha_3^{{3{{M}^5}}} + {{A}_4}\alpha_4^{{3{{M}^5}}}) \equiv a_{{3{{M}^5}}}^{,} \cr {{A}_1}\alpha_1^{{4{{M}^5}}} + {{A}_2}\alpha_2^{{4{{M}^5}}} = {{a}_{{4{{M}^5}}}} - ({{A}_3}\alpha_3^{{4{{M}^5}}} + {{A}_4}\alpha_4^{{4{{M}^5}}}) \equiv a_{{4{{M}^5}}}^{,} \cr {{A}_1}\alpha_1^{{5{{M}^5}}} + {{A}_2}\alpha_2^{{5{{M}^5}}} = {{a}_{{5{{M}^5}}}} - ({{A}_3}\alpha_3^{{5{{M}^5}}} + {{A}_4}\alpha_4^{{5{{M}^5}}}) \equiv a_{{5{{M}^5}}}^{,} \cr } $$

(54)

$\alpha_1^{{{{M}^5}}},{}\alpha_2^{{{{M}^5}}}$ are obtained by solving the linear equations in two variables and a square equation. After that, ${{A}_1},{ }{{A}_2}$ are obtained from the 1st two equations in (54). Next, consider the 5th block to determine ${{A}_3},{ }{{\alpha}_3},{ }{{A}_4},{ }{{\alpha}_4}$. By subtracting the effect of ${{A}_5},{ }{{\alpha}_5},{ }{{A}_6},{ }{{\alpha}_6}$ and ${{A}_1},{ }{{\alpha}_1},{ }{{A}_2},{ }{{\alpha}_2}$ in the 4th and 6th block, respectively, which corresponds to ${{a}_{{2{{M}^4}}}}$ et al., from the 4 numbers of ${{a}_{{2{{M}^4}}}}$ et al. in the 5th block, ${{A}_3},{}\alpha_3^{{{{M}^4}}},{{A}_4},{}\alpha_4^{{{{M}^4}}}$ can be obtained by solving four equations similar equation (52) to the 6th block. In case of treating the 4th block, the effect of the 3rd, 5th, and 6th blocks needs to be considered. The influence of the blocks in lower order on the blocks in higher order is small because $\left| {{{\alpha}_j}} \right| < 1$. For example, the values of $\alpha_9^{{{{M}^5}}}$and $\alpha_{{10}}^{{{{M}^5}}}$ obtained from the 2nd block or the values of $\alpha_{{11}}^{{{{M}^5}}},{}\alpha_{{12}}^{{{{M}^5}}},{}\alpha_{{13}}^{{{{M}^5}}},{}\alpha_{{14}}^{{{{M}^5}}}$ from the 1st block have almost no effect on the blocks in high order in case $M = 8$. The 1st block, having 8 elements, may be determined by solving the linear equations in 4 variables and an equation of 4th power. Thus, after solving the 1st block, return to the 6th block and do the same iterative procedure till the variation becomes small enough. Finally, we have the values of ${{A}_j},{ }{{\alpha}_j}$, which should satisfy the condition $\left| {{{\alpha}_j}} \right| < 1$.

From the above statement, we know the number of elements in each block in (53) does not have to be 4. If it is 6, in order to determine the fractional expressions in each block, the linear equations in 3 variables instead of 2 and a cube equation instead of the square one must be solved. And it does not matter to use 4 elements in the 1st block in (53). The selection depends on the needed accuracy, the interval of ${{a}_j}$, and the value of M. The term of (53) must present in the mapping function in case that there is a projection in the configuration to be analyzed. The index $\alpha$ depends on the angle of the projection. Accordingly, e.g., $\alpha = 0.5$ when the angle of the projection is 90°. It is not necessary to compute the values of ${{A}_{{j{,}}}}{ }{{\alpha}_j}$ every time in the analysis once they have been obtained for the particular angles.