4. BEM solution for displacement field

The equilibrium Eqs. (1) for anisotropic plate structures can be written as follows [47]

$$C\_{ijkl}\frac{d^4w}{d\mathfrak{x}^4} - T\frac{d^2w}{d\mathfrak{x}^2} = p + \frac{T\_1}{r} \tag{33}$$

where

$$T\_2 = -\frac{pr}{2} \tag{34}$$

$$T\_1 = \mathbf{C}\_{ijkl} T\_2 - \mathbf{C}\_{ijkl} h \frac{w}{r} \tag{35}$$

By using (34) and (35), we can write (33) in the following form

$$C\_{ijkl}\frac{d^4w}{d\kappa^4} + \frac{pr}{2}\frac{d^2w}{d\kappa^2} + \frac{C\_{ijkl}h}{r^2}w = p\left(1 - \frac{C\_{ijkl}}{2}\right) \tag{36}$$

where

$$\mathbf{A} = \frac{\mathbf{C}\_{ijkl}h}{r^2} \tag{37}$$

$$\mathbf{B} = p \left( \mathbf{1} - \frac{\mathbf{C}\_{ijkl}}{2} \right) \tag{38}$$

By using Eqs. (37) and (38), we can write (36) as follows

$$C\_{ijkl}\frac{d^4w}{d\mathfrak{x}^4} - T\frac{d^2w}{d\mathfrak{x}^2} + \text{A}w = \mathbf{B} \tag{39}$$

where

$$\beta = \frac{T\_2}{2\sqrt{C\_{ijkl}k}}, 0 < \beta^2 < 1 \tag{40}$$

The general solution of (39) can be obtained as

$$\mathbf{w}(\mathbf{x}) = \mathbf{C}\_1 \mathbf{c}h \mathbf{\hat{x}} \cos \gamma \mathbf{x} + \mathbf{C}\_2 ch \mathbf{\hat{x}} \sin \gamma \mathbf{x} + \mathbf{C}\_3 sh \mathbf{\hat{x}} \cos \gamma \mathbf{x} + \mathbf{C}\_4 sh \mathbf{\hat{x}} \sin \gamma \mathbf{x} + w\_{\text{part}} \tag{41}$$

where

$$\delta = a\sqrt{1+\beta}; \gamma = a\sqrt{1-\beta}; a = \sqrt[4]{\frac{k}{4\mathcal{C}\_{ijkl}}}, \beta = \frac{T\_2}{2\sqrt{\mathcal{C}\_{ijkl}k}} \tag{42}$$

and the particular solution can be determined as p ¼ constant as follows

$$w\_{\text{part}} = \frac{pr^2}{\mathcal{C}\_{ijkl}h} \left(1 - \frac{\mathcal{C}\_{ijkl}}{2}\right) \tag{43}$$

Thus, Eq. (41) can be written as

$$w(\mathbf{x}) = \frac{pr^2}{\mathbf{C}\_{ijkl}h} \left(\mathbf{1} - \frac{\mathbf{C}\_{ijkl}}{2}\right) + \mathbf{C}\_1 ch \delta \mathbf{x} \cos \gamma \mathbf{x} + \mathbf{C}\_4 sh \delta \mathbf{x} \sin \gamma \mathbf{x} \tag{44}$$

By implementing the following boundary conditions.

$$\text{at } \mathbf{x} = \pm \frac{l}{2} \frac{dw}{d\mathbf{x}} = \mathbf{0} \tag{45}$$

$$\text{at } \mathbf{x} = \frac{l}{2}\boldsymbol{w} = \frac{2pr^2}{\mathbf{C}\_{ijkl}h} \frac{d^3w}{d\boldsymbol{x}^3} \tag{46}$$

we can write the unknown C<sup>1</sup> and C<sup>4</sup> as follows

$$\mathbf{C}\_{1} = -\frac{2pr^{2}}{C\_{ijkl}h} \left( 1 - \frac{C\_{ijkl}}{2} \right) \frac{u\_1 ch u\_1 \sin u\_2 + u\_2 sh u\_1 \cos u\_2}{u\_2 sh 2u\_1 + u\_1 \sin 2u\_2} \mathbf{e}\_1 \tag{47}$$

$$\mathbf{C}\_{4} = -\frac{2pr^{2}}{\mathbf{C}\_{ijkl}h} \left( 1 - \frac{\mathbf{C}\_{ijkl}}{2} \right) \frac{u\_{2}ch u\_{1} \sin u\_{2} - u\_{1}sh u\_{1} \cos u\_{2}}{u\_{2}sh 2u\_{1} + u\_{1} \sin 2u\_{2}} \mathbf{e}\_{1} \tag{48}$$

where

$$\varepsilon\_1 = \frac{1}{1 + \frac{lh}{A} A\_1(u\_1, u\_2)}\tag{49}$$

$$A\_1(u\_1, u\_2) = \sqrt{1 - \beta^2} \frac{ch2u\_1 - \cos 2u\_2}{u\_2 sh2u\_1 + u\_1 \sin 2u\_2} \tag{50}$$

$$
\mu\_1 = \frac{\delta l}{2} = \mu\sqrt{1+\beta},
\mu\_2 = \frac{\gamma l}{2} = \mu\sqrt{1+\beta},
\mu = 0.6425\frac{1}{\sqrt{r h}}\tag{51}
$$

If we neglected the longitudinal forces influence on the bending of the circular cylindrical shell, we can write (39) in the following form

$$\mathbf{C}\_{ijkl}w^{IV} + kw = q \tag{52}$$

Now, the approximate solution has been reduced for solving problem of bending single span beam with the following compliance

A New Computerized Boundary Element Model for Three-Temperature Nonlinear Generalized… DOI: http://dx.doi.org/10.5772/intechopen.90053

$$k\_{II} = \frac{2r^2}{C\_{ijkl}A} \tag{53}$$

The deflection of the considered shell in the cross section and reference section, respectively, is as follows

$$w(\mathbf{0}) = \frac{pr^2}{C\_{ijkl}h} \left(\mathbf{1} - \frac{v}{2}\right) \left[\mathbf{1} - \frac{\mathcal{D}\_1(u)}{\mathbf{1} + B\_1}\right] \tag{54}$$

$$w\left(\frac{l}{2}\right) = \frac{pr^2}{C\_{ijkl}h} \left(1 - \frac{v}{2}\right) \frac{B\_1}{1 + B\_1} \tag{55}$$

Also, the bending moment in the cross section and reference section, respectively, is as follows

$$M\_1(\mathbf{0}) = -\frac{pl^2}{24} \left( \mathbf{1} - \frac{\nu}{2} \right) \left[ \mathbf{1} - \frac{\chi\_1(u)}{\mathbf{1} + B\_1} \right] \tag{56}$$

$$M\_1\left(\frac{l}{2}\right) = \frac{pl^2}{12}\left(1 - \frac{v}{2}\right)\frac{\chi\_2(u)}{1 + B\_1} \tag{57}$$

The Cauchy model with two-bed scheme can be described as follows

$$D v^{IY}(\mathbf{x}) + \frac{pr}{2} v''(\mathbf{x}) + \frac{C\_{ijkl}h}{r^2} v(\mathbf{x}) = p\left(1 - \frac{\mu}{2}\right) \tag{58}$$

$$
v(\mathbf{0}); \rho(\mathbf{0}) = \boldsymbol{\nu}'(\mathbf{0})\tag{59}$$

$$M(\mathbf{0}) = -Dv''(\mathbf{0}) - Tv(\mathbf{0})\tag{60}$$

$$Q(\mathbf{0}) = -Dv''''(\mathbf{0}) - Tv'(\mathbf{0})\tag{61}$$

where the characteristic equation of (58) can be defined as

$$\mathbf{C}\_{ijkl}\mathbf{k}^4 + \frac{pr}{2}\mathbf{k}^2 + \frac{\mathbf{C}\_{ijkl}h}{r^2} = \mathbf{0}, \mathbf{k}^2 = \mathbf{t} \tag{62}$$

$$\mathbf{C}\_{ijkl}t^2 + \frac{pr}{2}t + \frac{\mathbf{C}\_{ijkl}h}{r^2} = \mathbf{0} \tag{63}$$

which roots

$$k\_{1,2,3,4} = \pm \sqrt{\frac{-\frac{pr}{2} \pm \sqrt{\left(\frac{pr}{2}\right)^2 - 4C\_{ijkl}\frac{h}{r^2}}}}\tag{64}$$

$$t\_{1,2} = \frac{-\frac{pr}{2} \pm \sqrt{\left(\frac{pr}{2}\right)^2 - 4C\_{ijkl}{}^2 \frac{h}{r^2}}}{2C\_{ijkl}}\tag{65}$$

The systems (32) and (58) can be solved by using the algorithm of Fahmy [35] to obtain the three temperatures and displacements components. Then we can compute thermal stresses distributions along radial distance r. we refer the reader to recent references [48–51] for details of boundary element technique.
