Stress analysis for long thermoelastic rods with mixed boundary conditions

Abstract

A hybrid method involving boundary analysis and boundary collocation is used to obtain an approximate solution for a plane problem of uncoupled thermoelasticity with mixed thermal and mechanical boundary conditions in a square domain with one curved side. The unknown functions in the cross-section are obtained in the form of series expansions in Cartesian harmonics. A boundary analysis reveals the singular behavior of the solution at the transition points. In order to simulate the weak discontinuities of the temperature function and the discontinuities of stress, these expansions are enriched with proper harmonic functions with a singular behavior at the transition points. The results are discussed, and the functions of practical interest are represented on the boundary and also inside the domain. The locations where possible debonding of the fixed part of the boundary may take place are noted.

Appendices

Appendix 1: Boundary representation of harmonic functions

For an arbitrary point $(x,y) \in D$, the boundary integral representation of harmonic functions reads:

$$\begin{aligned} f(x,y)= \frac{1}{2 \pi }\oint _C \left( f \, \frac{\partial \ln R}{\partial n'} - \ln R \, \frac{\partial f}{\partial n'} \right) \, \mathrm{d}s'. \end{aligned}$$

Here R is the distance between the point (x, y) and the current integration point. When the point (x, y) tends to a boundary point, this relation transforms into an integral equation for the boundary values of the function f. Using integration by parts, this may be rewritten as

$$\begin{aligned} f(x,y) = \frac{1}{\pi }\oint _C\left( f(x',y')\frac{\partial \ln R}{\partial n'} +\big (f^{c}(x',y')-f^{c}(x,y)\big ) \frac{\partial \ln R}{\partial \tau '}\right) \, \mathrm{d}s', \end{aligned}$$

where $(x',y')$ is the current integration point.

This last equation contains removable singularities, which can be treated as explained in [67].

Appendix 2: The first and second derivatives of harmonic functions with respect to x and y on the boundary

For a general harmonic function f in the cross section, the following formulas are used:

$$\begin{aligned} \frac{\partial f}{\partial x}\bigg |_{i}&=\frac{1}{\omega _{i}^{2}}\left( \dot{f}^{c}_{i}\,\dot{y}_i+\dot{f}_{i}\,\dot{x}_i\right) , \\ \frac{\partial f}{\partial y}\bigg |_{i}&=\frac{1}{\omega _{i}^{2}}\left( \dot{f}_{i}\,\dot{y}_i-\dot{f}^{c}_{i}\,\dot{x}_i\right) , \\ \frac{\partial ^{2} f}{\partial y^{2}}\bigg |_{i}&=\frac{\alpha _{i}}{\omega ^{4}_{i}}\,\ddot{f}_{i}-\frac{\beta _{i}}{\omega ^{4}_{i}}\,\ddot{f}^{c}_{i}+\frac{\varrho _{i}}{\omega ^{6}_{i}}\,\dot{f}_{i}+\frac{\rho _{i}}{\omega ^{6}_{i}}\,\dot{f}^{c}_{i}, \\ \frac{\partial ^{2} f}{\partial x\, \partial y }\bigg |_{i}&=\frac{\beta _{i}}{\omega ^{4}_{i}}\,\ddot{f}_{i}+\frac{\alpha _{i}}{\omega ^{4}_{i}}\,\ddot{f}^{c}_{i}-\frac{\rho _{i}}{\omega ^{6}_{i}}\,\dot{f}_{i}-\frac{\varrho _{i}}{\omega ^{6}_{i}}\,\dot{f}^{c}_{i}, \\ \frac{\partial ^{2} f}{\partial x^{2}}\bigg |_{i}&=-\frac{\alpha _{i}}{\omega ^{4}_{i}}\,\ddot{f}_{i}+\frac{\beta _{i}}{\omega ^{4}_{i}}\,\ddot{f}^{c}_{i}+\frac{\varrho _{i}}{\omega ^{6}_{i}}\,\dot{f}_{i}-\frac{\rho _{i}}{\omega ^{6}_{i}}\,\dot{f}^{c}_{i}, \end{aligned}$$

where f stands for any one of the used harmonic functions, and

$$\begin{aligned} \rho _{i}&=\alpha _{i}\,\gamma _{i}+\beta _{i}\,\delta _{i},\\ \varrho _{i}&=\alpha _{i}\,\delta _{i}-\beta _{i}\,\gamma _{i}, \end{aligned}$$

with

$$\begin{aligned} \alpha _{i}&=\dot{y}^{2}_{i}-\dot{x}^{2}_{i},\\ \beta _{i}&=2\,\dot{x}_{i}\,\dot{y}_{i},\\ \gamma _{i}&=\dot{x}_{i}\,\ddot{y}_{i}-\dot{y}_{i}\,\ddot{x}_{i},\\ \delta _{i}&=\dot{x}_{i}\,\ddot{x}_{i}+\dot{y}_{i}\,\ddot{y}_{i}. \end{aligned}$$

Appendix 3: Treating the stress singularities

Fig. 17

The function h(x)

Fig. 18

Green’s function technique

In order to simulate the singular behaviours of stresses at the two boundary points of contact, we have proceeded according to the following scheme:

• 1. Our goal is to introduce a harmonic function that has weak singularities at a boundary point, more precisely a harmonic function that has discontinuous second derivatives at a boundary point.

• 2. Let h(x) be a function of the real variable x defined on the real axis, such that it is continuous with a continuous derivative everywhere, but has a finite jump in the second derivative at the point $x=0$. Let this function be of the form:

  $$\begin{aligned} h(x)= {\left\{ \begin{array}{ll} 0 &{} x\le 0,\\ \\ \dfrac{1}{2}\,x^2\,\mathrm{e}^{-x}&{} x>0. \end{array}\right. } \end{aligned}$$

  The exponential function is needed here to make the function integrable on the real axis. The graph of this function is shown in Fig. 17.

• 3. The boundary integral representation of harmonic functions applied to a function f which is harmonic in a simply connected domain D bounded by a smooth surface S yields

  $$\begin{aligned} f(x, y) = \oint _s \left[ f \frac{\partial }{\partial n'}{\left( \frac{1}{2\pi } \ln R \right) } - \left( \frac{1}{2\pi } \ln R \right) \frac{\partial }{\partial n'} {f} \right] \, \mathrm{d}s', \end{aligned}$$

  (22)

  where (x, y) is an arbitrary point in D, $n'$ denotes the outward unit normal to S at the integration point, and R is the distance function between the point (x, y) and the integration point.

  If D is the upper half-plane $(\xi , \eta ), \, \eta > 0$, the previous equation takes the form:

  $$\begin{aligned} f(x, y) = - \int _{-\infty }^{+\infty } \left[ f \frac{\partial }{\partial \eta }{\left( \frac{1}{2\pi } \ln R \right) } - \left( \frac{1}{2\pi } \ln R \right) \frac{\partial }{\partial \eta } {f} \right] \, \mathrm{d} \xi . \end{aligned}$$

  (23)

  Let $R_1$ be the distance shown in Fig. 18. The function $\ln R_1$ is harmonic on the upper half-plane. A well-known theorem of the Theory of Potential, applied to the harmonic functions f and $\ln R_1$, yields

  $$\begin{aligned} 0 = \int _{-\infty }^{+\infty } \left[ f \frac{\partial }{\partial \eta }{\left( \frac{1}{2\pi } \ln R_1 \right) } - \left( \frac{1}{2\pi } \ln R_1 \right) \frac{\partial }{\partial \eta } {f} \right] \, \mathrm{d} \xi . \end{aligned}$$

  (24)

  Adding and noting that $R = R_1$ on the $\xi$-axis, one gets

  $$\begin{aligned} f(x,y)= & {} - \int _{-\infty }^{+\infty } \left[ f \frac{\partial }{\partial \eta }{\left( \frac{1}{2\pi } \ln R - \frac{1}{2\pi } \ln R_1 \right) } \right] \, \mathrm{d} \xi \\= & {} \int _{-\infty }^{+\infty } \left[ f \frac{\partial }{\partial \eta }{\left( \frac{1}{2\pi } \ln \frac{1}{R} - \frac{1}{2\pi } \ln \frac{1}{R_1} \right) } \right] \, \mathrm{d} \xi . \end{aligned}$$

  The function under the normal derivative in the last equation is just Green’s function for the Dirichlet problem in the upper half-plane:

  $$\begin{aligned} G(\xi , \eta ; x, y)= & {} \frac{1}{2\pi } \ln \frac{1}{R} - \frac{1}{2\pi } \ln \frac{1}{R_1} \\= & {} \frac{1}{2\pi } \ln \frac{R_1}{R}. \end{aligned}$$

  One has

  $$\begin{aligned} \frac{\partial G}{\partial \eta }= & {} \frac{1}{2 \pi } \frac{\partial }{\partial \eta } \ln \frac{R_1}{R} \\= & {} \frac{1}{4 \pi } \frac{\partial }{\partial \eta } \ln \frac{(\xi - x)^2 + (\eta + y)^2}{(\xi - x)^2 + (\eta - y)^2} \\= & {} \frac{1}{2 \pi } \frac{(\xi - x)^2 + (\eta - y)^2}{(\xi - x)^2 + (\eta + y)^2} \frac{(\eta +y) \left[ (\xi - x)^2 + (\eta - y)^2 \right] -(\eta -y) \left[ (\xi - x)^2 + (\eta + y)^2 \right] }{\left[ (\xi - x)^2 + (\eta - y)^2 \right] ^2}. \end{aligned}$$

  Thus,

  $$\begin{aligned} \left. \frac{\partial G}{\partial \eta }\right| _{\eta =0} = \frac{1}{2 \pi } \frac{2y}{(\xi - x)^2 + y^2}. \end{aligned}$$

  and the function f takes the form:

  $$\begin{aligned} f(x, y)= & {} \frac{1}{2 \pi } \int _{-\infty }^{+\infty } h(\xi ) \, \frac{2\,y}{(\xi - x)^2 + y^2} \, \mathrm{d} \xi \end{aligned}$$

  (25)
  $$\begin{aligned}= & {} \frac{1}{2 \pi } \int _{0}^{+\infty } \xi ^2\,e^{-\xi }\, \frac{y}{(\xi - x)^2 + y^2} \, \mathrm{d} \xi . \end{aligned}$$

  (26)

Fig. 19

Singular stresses

• 4. In order to perform the integration, we write

  $$\begin{aligned} \frac{\xi ^2}{(\xi - x)^2 + y^2} = \frac{\xi ^2}{\xi ^2 - 2x \xi + x^2 + y^2} = \frac{\xi ^2}{(\xi + c_1)(\xi + c_2)} , \end{aligned}$$

  (27)

  where

  $$\begin{aligned} c_1 = -x + i y, \qquad c_2 = -x - i y = \bar{c_1} \end{aligned}$$

  (28)

  and “bar” denotes the conjugate. Expansion in partial fractions yields after some manipulations:

  $$\begin{aligned} \frac{\xi ^2}{(\xi - x)^2 + y^2} = 1 -\frac{c_1^2}{2 i y} \frac{1}{\xi + c_1}+ \frac{c_2^2}{2 i y} \frac{1}{\xi + c_2}. \end{aligned}$$

  (29)

  Hence,

  $$\begin{aligned} f(x, y) = \frac{1}{2 \pi }\, \int _{0}^{+\infty }\left[ y -\frac{c_1^2}{2 i}\, \frac{1}{\xi + c_1}+ \frac{c_2^2}{2 i}\, \frac{1}{\xi + c_2}\right] \,\mathrm{e}^{-\xi }\, \mathrm{d} \xi . \end{aligned}$$

  (30)

• 5. Next, introduce the integral exponential function $E_1(z)$ of the complex argument z by the formula ([69, p. 62]):

  $$\begin{aligned} E_1(z) = \int _{z}^{\infty } \frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t = \mathrm{e}^{- z} \int _{0}^{\infty } \frac{\mathrm{e}^{-t}}{t+z}\,\mathrm{d}t. \end{aligned}$$

  (31)

  One has

  $$\begin{aligned} E_1(z) = -\gamma - \ln (z) - \sum _{n=1}^{\infty } \frac{(-1)^n\,z^n}{n\,n!}, \end{aligned}$$

  (32)

  where $\gamma = 0.5772156649$ is Euler’s constant. Finally,

  $$\begin{aligned} f(x,y) = \frac{1}{2\pi } \left[ y + 2 R\mathrm{e}\left( \frac{i\,c_1^2}{2}\, \mathrm{e}^{c_1}\, E_1(c_1)\right) \right] . \end{aligned}$$

  Substitution for the exponential integral function allows one to compute the function f(x, y).

• 6. The obtained function will now be centered at each of the two boundary separation points in order to simulate the behavior of stresses there. The sum of the resulting two functions is now taken to replace the function $\psi$ in the above formulation, and will be denoted $\psi S$. The three-dimensional plots illustrated in Fig. 19 show the harmonic function with singular boundary behavior, and the singular stresses as calculated from it as the stress function.