Axisymmetric deformation in transversely isotropic magneto-thermoelastic solid with Green–Naghdi III due to inclined load

Abstract

The axisymmetric problem in two-dimensional transversely isotropic magneto-thermoelastic (TIMT) solid due to inclined load with Green–Naghdi (GN)-III theory and two temperature (2T) has been studied. The Laplace and Hankel transform has been used to get the expressions of temperature distribution, displacement, and stress components with the horizontal distance in the physical domain. The effect of Green–Naghdi theories of type I, II, and III theories of thermoelasticity has been studied graphically on the resulting quantities. A special case for the magneto-thermoelastic isotropic medium has also been studied.

Introduction

The study of deformation in a thermoelastic medium is one of the wide and dynamic domains of continuum dynamics. It is well known that all the rotating large bodies have angular velocity, as well as magnetism; therefore, the thermoelastic interactions in a rotating medium under magnetic field are of importance. The study of thermoelasticity is beneficial to analyze the deformation field such as geothermal engineering, advanced aircraft structure design, thermal power plants, composite engineering, geology, high-energy particle accelerators, and many developing technologies.

Eubanks and Sternberg (1954) discussed the axisymmetric issue of elasticity concept for a transversely isotropy medium. Vendhan and Archer (1978) electrostatically analyzed transversely isotropic (TI) finite stress-free cylinders with lateral surfaces using displacement potential. Green and Naghdi (1992, 1993) dealt with the linear and the nonlinear theories of the thermoelastic body with and without energy dissipation. Three new thermoelastic theories were proposed by them, based on entropy equality. Their theories are known as GN-I, GN-II, and GN-III theories of thermoelasticity. On linearization, type I becomes the classical heat equation, whereas on linearization, type II as well as type III theories predicts the finite speed of thermal wave. Savruk (1994) discussed the axisymmetric deformation of a TI body containing cracks.

Tarn et al. (2009) analyzed the axisymmetric and stress dispersal in a TI roundabout barrel-shaped body utilizing Hamiltonian variational definition through Legendre’s change. Liang and Wu (2012) discussed the axisymmetric deformation of one TI cylinder with the Lure method. Mahmoud (2012) considered the impact of relaxation times, the rotation, and the initial stress on Rayleigh waves. Shi et al. (2016) presented the thermomagnetoelectroelastic field in a heterogeneous annular multi-ferric composite plate with thermal loadings which is uniformly distributed on the boundaries. Kumar et al. (2016a, 2016b) studied the conflicts of thermomechanical sources in a TI homogeneous thermoelastic rotating medium with magnetic effect as well as two temperature applied to the thermoelasticity GN-III theories. Li et al. (2016) presented a set of axisymmetric solutions of the thermoelastic field in a heterogeneous circular plate which could either simply reinforced or fastened persuaded by the external thermal load.

Including these, various researchers dealt with various theory of thermoelasticity such as Marin (Marin 1997a, 1997b; Marin 1998; Marin 1999), Marin (Marin 2008; Marin 1997a, 1997b), Ezzat et al. (2012), Atwa (2014), Marin et al. (2013), Marin (2016), Marin and Baleanu (2016), Bijarnia and Singh (2016), Sharma et al. (Sharma et al. 2015a, 2015b; 2016; 2017), Ezzat et al. (2017), Lata (2018), Lata et al. (2016), Marin and Öchsner (2017), Othman and Marin (2017), Ezzat et al. (2017), Chauthale and Khobragade (2017), Kumar et al. (2017), Lata and Kaur (2018), Shahani and Torki (2018), Lata and Kaur (2019a, 2019b, 2019c, 2019d), Kaur and Lata (2019a, 2019b), Bhatti and Lu (2019a, 2019b), and Marin et al. (2019).

Despite these, very less work has been done in thermomechanical interactions in TIMT rotating solid with GN-III theory, with two temperature in the axisymmetric medium. Remembering these contemplations, analytic expressions for the displacement components, stress components, and temperature distribution in two-dimensional homogeneous, TIMT rotating solids with GN-III theories, with two temperature have been derived.

Basic equations

The field equations with and without energy dissipation, without body forces and heat sources for an anisotropic thermoelastic medium following Lata and Kaur (2019d), are:

$$ {t}_{ij}={C}_{ij kl}{e}_{kl}-{\beta}_{ij}T, $$

(1)

$$ {K}_{ij}{\varphi}_{, ij}+{K}_{ij}^{\ast }{\overset{.}{\varphi}}_{, ij}={\beta}_{ij}{T}_0{\ddot{\mathrm{e}}}_{ij}+\rho {C}_E\ddot{T}. $$

(2)

and the equation of motion for a medium rotating uniformly and Lorentz force is

$$ {t}_{ij,j}+{F}_i=\rho \left\{{\ddot{u}}_i+\right(\varOmega \times {\left(\varOmega \times u\right)}_i+{\left(2\varOmega \times \overset{.}{u}\right)}_i\Big\}, $$

(3)

where

Ω = Ωn, n is a unit vector signifying the direction of the rotating axis.

$$ {F}_i={\mu}_0\left(\overrightarrow{j}\times {\overrightarrow{H}}_0\right), $$

$$ T=\varphi -{a}_{ij}{\varphi}_{, ij}, $$

(4)

$$ {\beta}_{ij}={C}_{ij kl}{\alpha}_{ij}, $$

(5)

$$ {e}_{ij}=\frac{1}{2}\left({u}_{i,j}+{u}_{j,i}\right).i=1,2,3 $$

(6)

$$ {\beta}_{ij}={\beta}_i{\delta}_{ij},{K}_{ij}={K}_i{\delta}_{ij},{K}_{ij}^{\ast }={K}_i^{\ast }{\delta}_{ij},i\; is\kern0.17em not\kern0.17em summed $$

$$ Here\;{C}_{ijkl}\; having\kern0.17em symmetry\;\left({C}_{ijkl}={C}_{klij}={C}_{jikl}={C}_{ijlk}\right). $$

Method and formulation of the problem

Consider a TIMT homogeneous medium with an initial temperature T₀. We consider a cylindrical polar coordinate system (r, θ, z) with symmetry about the z-axis. For a plane axisymmetric problem, v = 0, and u, w, and φ are independent of θ. Additionally, we take

$$ \varOmega =\left(0,\varOmega, 0\right). $$

and

$$ {J}_2=0. $$

The density components J₁and J₃ are given as

$$ {J}_1=-{\varepsilon}_0{\mu}_0{H}_0\frac{\partial^2w}{\partial {t}^2}, $$

(7)

$$ {J}_3={\varepsilon}_0{\mu}_0{H}_0\frac{\partial^2u}{\partial {t}^2}. $$

(8)

Using the appropriate transformation following Slaughter (2002) on Eqs. (1)–(3) to determine the conditions for TI thermoelastic solid with 2T and with and without energy dissipation, we get

$$ {C}_{11}\left(\frac{\partial^2u}{\partial {r}^2}+\frac{1}{r}\frac{\partial u}{\partial r}-\frac{1}{r^2}u\right)+{C}_{13}\left(\frac{\partial^2w}{\partial r\partial z}\right)+{C}_{44}\frac{\partial^2u}{\partial {z}^2}+{C}_{44}\left(\frac{\partial^2w}{\partial r\partial z}\right)-{\beta}_1\frac{\partial }{\partial r}\left\{\varphi -{a}_1\left(\frac{\partial^2\varphi }{\partial {r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right\}-{\mu}_0{J}_3{H}_0=\rho \left(\frac{\partial^2u}{\partial {t}^2}-{\varOmega}^2u+2\varOmega \frac{\partial w}{\partial t}\right), $$

(9)

$$ \left({C}_{11}+{C}_{44}\right)\left(\frac{\partial^2u}{\partial r\partial z}+\frac{1}{r}\frac{\partial u}{\partial z}\right)+{C}_{44}\left(\frac{\partial^2w}{\partial {r}^2}+\frac{1}{r}\frac{\partial w}{\partial r}\right)+{C}_{33}\frac{\partial^2w}{\partial {z}^2}-{\beta}_3\frac{\partial }{\partial z}\left\{\varphi -{a}_1\left(\frac{\partial^2\varphi }{\partial {r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right\}+{\mu}_0{J}_1{H}_0=\rho \left(\frac{\partial^2w}{\partial {t}^2}-{\varOmega}^2w-2\varOmega \frac{\partial u}{\partial t}\right), $$

(10)

$$ \left({K}_1+{K}_1^{\ast}\frac{\partial }{\partial t}\right)\left(\frac{\partial^2\varphi }{\partial {r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)+\left({K}_3+{K}_3^{\ast}\frac{\partial }{\partial t}\right)\frac{\partial^2\varphi }{\partial {z}^2}={T}_0\frac{\partial^2}{\partial {t}^2}\left({\beta}_1\frac{\partial u}{\partial r}+{\beta}_3\frac{\partial w}{\partial z}\right)+\rho {C}_E\frac{\partial^2}{\partial {t}^2}\left\{\varphi -{a}_1\left(\frac{\partial^2\varphi }{\partial {r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{\partial {z}^2}\right\}. $$

(11)

Constitutive relations are

$$ {\displaystyle \begin{array}{c}{t}_{rr}={c}_{11}{e}_{rr}+{c}_{12}{e}_{\theta \theta}+{c}_{13}{e}_{zz}-{\beta}_1T,\\ {}{t}_{zr}=2{c}_{44}{e}_{rz},\\ {}{t}_{zz}={c}_{13}{e}_{rr}+{c}_{13}{e}_{\theta \theta}+{c}_{33}{e}_{zz}-{\beta}_3T,\\ {}{t}_{\theta \theta}={c}_{12}{e}_{rr}+{c}_{11}{e}_{\theta \theta}+{c}_{13}{e}_{zz}-{\beta}_3T,\end{array}} $$

(12)

where

$$ {\displaystyle \begin{array}{c}{e}_{rz}=\frac{1}{2}\left(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial r}\right),\\ {}{e}_{rr}=\frac{\partial u}{\partial r},\\ {}{e}_{\theta \theta}=\frac{u}{r},\\ {}{e}_{zz}=\frac{\partial w}{\partial z},\\ {}T=\varphi -{a}_1\left(\frac{\partial^2\varphi }{\partial {r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{\partial {z}^2},\\ {}{\beta}_1=\left({c}_{11}+{c}_{12}\right){\alpha}_1+{c}_{13}{\alpha}_3,\\ {}{\beta}_3=2{c}_{13}{\alpha}_1+{c}_{33}{\alpha}_3.\end{array}} $$

We consider that a primary medium is at rest. Therefore, the preliminary and symmetry conditions are assumed as

$$ {\displaystyle \begin{array}{c}u\left(r,z,0\right)=0=\overset{.}{u}\left(r,z,0\right),\\ {}\ w\left(r,z,0\right)=0=\overset{.}{w}\left(r,z,0\right),\\ {}\varphi \left(r,z,0\right)=0=\overset{.}{\varphi}\left(r,z,0\right)\ \mathrm{for}\ z\ge 0,-\infty <r<\infty, \\ {}u\left(r,z,t\right)=w\left(r,z,t\right)=\varphi \left(r,z,t\right)=0\ \mathrm{for}\ t>0\ \mathrm{when}\ z\to \infty .\end{array}} $$

To simplify the solution, the following dimensionless quantities are introduced

$$ {r}^{\hbox{'}}=\frac{r}{L},{z}^{\hbox{'}}=\frac{z}{L},{t}^{\hbox{'}}=\frac{c_1}{L}t,{u}^{\hbox{'}}=\frac{\rho {c}_1^2}{L{\beta}_1{T}_0}u,{w}^{\hbox{'}}=\frac{\rho {c}_1^2}{L{\beta}_1{T}_0}w,{T}^{\hbox{'}}=\frac{T}{T_0},{t}_{zr}^{\hbox{'}}=\frac{t_{zr}}{\beta_1{T}_0},{t}_{zz}^{\hbox{'}}=\frac{t_{zz}}{\beta_1{T}_0},{\varphi}^{\hbox{'}}=\frac{\varphi }{T_0},{a}_1^{\hbox{'}}=\frac{a_1}{L^2},{a}_3^{\hbox{'}}=\frac{a_3}{L^2},{h}^{\hbox{'}}=\frac{h}{H_0},{\varOmega}^{\hbox{'}}=\frac{L}{C_1}\Omega . $$

(13)

Applying the dimensionless quantities introduced in (13) on Eqs. (9)–(11) and subsequently suppressing the primes and using the following Laplace and Hankel transforms

$$ \overline{f}\left(r,z,s\right)=\underset{0}{\overset{\infty }{\int }}f\left(r,z,t\right){e}^{- st} dt, $$

(14)

$$ \tilde{f}\left(\xi, z,s\right)={\int}_0^{\infty}\overline{f}\left(r,z,s\right)r{J}_n\left( r\xi \right) dr. $$

(15)

on the resulting quantities, we obtain

$$ \left(-{\xi}^2+{\delta}_2{D}^2-{s}^2{\delta}_7+{\varOmega}^2\right)\tilde{u}+\left({\delta}_1 D\xi -2\varOmega \mathrm{s}\right)\tilde{w}+\left(-\xi \left(1+{a}_1{\xi}^2\right)+{a}_3\xi {D}^2\right)\tilde{\varphi}=0, $$

(16)

$$ \left({\delta}_1 D\xi +2\varOmega s\right)\tilde{u}+\left({\delta}_3{D}^2-{\delta}_2{\xi}^2-{s}^2{\delta}_7+{\varOmega}^2\right)\tilde{w}-\left(\frac{\beta_3}{\beta_1}D\left[\left(1+{\xi}^2{a}_1\right)-{a}_3{D}^2\right]\right)\tilde{\varphi}=0, $$

(17)

$$ -{\delta}_6{s}^2\xi \tilde{u}-\frac{\beta_3}{\beta_1}{\delta}_6{s}^2D\tilde{w}+\left(-{\delta}_8{s}^2\left(1+{\xi}^2{a}_1-{a}_3{D}^2\right)-{\xi}^2\left({K}_1+{\delta}_4s\right)+{D}^2\left({K}_3+{\delta}_5s\right)\right)\tilde{\varphi}=0. $$

(18)

where

$$ {\delta}_1=\frac{c_{13}+{c}_{44}}{c_{11}},{\delta}_2=\frac{c_{44}}{c_{11}},{\delta}_3=\frac{c_{33}}{c_{11}},{\delta}_4=\frac{K_1^{\ast }{C}_1}{L},{\delta}_5=\frac{K_3^{\ast }{C}_1}{L},{\delta}_6=-\frac{T_0{\beta}_1^2}{\rho },{\delta}_7=\frac{\varepsilon_0{\mu}_0^2{H}_0^2}{\rho }+1,{\delta}_8=-\rho {C}_E{C}_1^2. $$

The non-trivial solution of (16)–(18) by eliminating$\overset{\sim }{\ u}$, $\overset{\sim }{w}$, and $\overset{\sim }{\varphi }$ yields

$$ A{D}^6+B{D}^4+C{D}^2+E=0 $$

(19)

where

$$ {\displaystyle \begin{array}{c}A={\delta}_2{\delta}_3{\zeta}_{11}-{\zeta}_9{\updelta}_2{\zeta}_7,\\ {}B={\delta}_2{\zeta}_3{\zeta}_{11}+{\delta}_3{\zeta}_1{\zeta}_{11}+{\delta}_2{\delta}_3{\zeta}_{10}-{\delta}_2{\zeta}_7{\zeta}_{10}-{\zeta}_7{\zeta}_1{\zeta}_9-{\zeta}_2^2{\zeta}_{11}+{\zeta}_2{\zeta}_4{\zeta}_9+{\zeta}_8{\zeta}_7{\zeta}_2-{\delta}_3{\zeta}_4{\zeta}_8,\\ {}C={\zeta}_1{\zeta}_3{\zeta}_{11}+{\delta}_2{\zeta}_{10}{\zeta}_5+{\delta}_3{\zeta}_3{\zeta}_{10}-{\zeta}_9{\zeta}_1{\zeta}_6-{\zeta}_2^2{\zeta}_{10}+{\zeta}_2{\zeta}_6{\zeta}_8+{\zeta}_3{\zeta}_2{\zeta}_9-{\zeta}_3{\zeta}_8{\delta}_3-{\zeta}_4{\zeta}_8{\zeta}_5+4{\varOmega}^2{s}^2{\zeta}_{11},\\ {}E={\zeta}_5{\zeta}_1{\zeta}_{10}-{\zeta}_8{\zeta}_5{\zeta}_3+4{\varOmega}^2{s}^2{\zeta}_{10}.\\ {}{\zeta}_1=-{\xi}^2-{s}^2{\delta}_7+{\varOmega}^2,\\ {}{\zeta}_2={\delta}_1\xi, \\ {}{\zeta}_3=-\xi \left({a}_1{\xi}^2+1\right),\\ {}{\zeta}_4={a}_3\xi, \\ {}{\zeta}_5=-{\delta}_2{\xi}^2-{s}^2{\delta}_7+{\varOmega}^2,\\ {}{\zeta}_6=-\frac{\beta_3}{\beta_1}\left(1+{a}_1{\xi}^2\right),\\ {}{\zeta}_7={a}_3\frac{\beta_3}{\beta_1},\\ {}{\zeta}_8=-\xi {\delta}_6{s}^2,\\ {}{\zeta}_9=-\frac{\beta_3}{\beta_1}{\delta}_6{s}^2,\\ {}{\zeta}_{10}=-{\delta}_8{s}^2\left(1+{a}_1{\xi}^2\right)-{\xi}^2\left({K}_1+{\delta}_4s\right),\\ {}{\zeta}_{11}=\left({K}_3+{\delta}_5s\right)+{\delta}_8{s}^2{\mathrm{a}}_3.\end{array}} $$

The solutions of Eq. (19) can be written as

$$ \hat{u}\left(\xi, z,s\right)={\sum}_{j=1}^3{A}_j{e}^{-{\lambda}_jz}, $$

(20)

$$ \hat{w}\left(\xi, z,s\right)={\sum}_{j=1}^3{d}_j{A}_j{e}^{-{\lambda}_jz}, $$

(21)

$$ \hat{\varphi}\left(\xi, z,s\right)=\sum \limits_{j=1}^3{l}_j{A}_j{e}^{-{\lambda}_jz}, $$

(22)

where A_j being arbitrary constants, ±λ_j represents the roots of Eq. (19) and d_j and l_j are given by

$$ {d}_j=\frac{{\delta}_{2}{\zeta}_{11}{\lambda}^{4}_{j}+(\zeta_{11}\zeta_{1}-\zeta_{4}\zeta_{8}+\delta_{2}\zeta_{10})\lambda^{2}_{j}+{\zeta}_{1}{\zeta}_{10}-{\zeta}_{8}{\zeta}_{3}} {(\delta_{3}\zeta_{11}-\zeta_{7}\zeta_{9})\lambda^{4}_{j}+(\delta_{3}\zeta_{10}+\zeta_{5}\zeta_{11}-\zeta_{9}\zeta_{6})\lambda^{2}_{j}+ \zeta_{5}\zeta_{10}}, $$

$$ {l}_j=\frac{{\delta}_{2}{\delta}_{3}{\lambda}^{4}_{j}+(\delta_{2}\zeta_{5}+\zeta_{1}\delta_{3}-\zeta^{2}_{2})\lambda^{2}_{j}+{\zeta}_{1}{\zeta}_{5}+4\Omega^{2}s^{2}} {(\delta_{3}\zeta_{11}-\zeta_{7}\zeta_{9})\lambda^{4}_{j}+(\delta_{3}\zeta_{10}+\zeta_{5}\zeta_{11}-\zeta_{9}\zeta_{6})\lambda^{2}_{j}+ \zeta_{5}\zeta_{10}}, $$

$$ \tilde{t_{zz}}=\sum {A}_j\left(\xi, s\right){\eta}_j{e}^{-{\lambda}_jz}, $$

(23)

$$ \tilde{t_{rz}}=\sum {A}_j\left(\xi, s\right){\mu}_j{e}^{-{\lambda}_jz}, $$

(24)

$$ \tilde{t_{rr}}=\sum {A}_j\left(\xi, s\right){Q}_j{e}^{-{\lambda}_jz}, $$

(25)

where

$$ {\eta}_j={\delta}_{11}\xi -{\delta}_3{\lambda}_j{d}_j-\frac{\beta_3}{\beta_1}\left(1+{a}_1{\xi}^2\right){\mathrm{l}}_{\mathrm{j}}+\frac{\beta_3}{\beta_1}{\mathrm{a}}_3{l}_j{\lambda}_j^2, $$

(26)

$$ {\mu}_j={\delta}_2\left(-{\lambda}_j+\xi {d}_j\right), $$

$$ {Q}_j=\left({\delta}_{10}+1\right)\xi -{\delta}_{11}{\lambda}_j{d}_j-{l}_j\left(1+{a}_1{\xi}^2\right)+{\mathrm{a}}_3{l}_j{\lambda}_{\mathrm{j}}^2,i,j=1,2,3. $$

Boundary conditions

The boundary conditions when normal force and tangential load are applied to the half-space (z = 0) are

$$ {t}_{zz}\left(r,z,t\right)=-{F}_1{\psi}_1(r)H(t), $$

(27)

$$ {t}_{rz}\left(r,z,t\right)=-{F}_2{\psi}_2(r)H(t), $$

(28)

$$ \frac{\partial \varphi \left(r,z,t\right)}{\partial z}+ h\varphi \left(r,z,t\right)=0. $$

(29)

where ψ₁(r) and ψ₂(r) are the vertical and the tangential load applied on along the r-axis and

$$ H(t)=\Big\{{\displaystyle \begin{array}{c}1t>0\\ {}0t<0\end{array}} $$

$$ {\displaystyle \begin{array}{c}\sum {A}_j\left(\xi, s\right){\eta}_j=-{F}_1{\psi}_1\left(\xi \right),\\ {}\sum {A}_j\left(\xi, s\right){\mu}_j=-{F}_2{\psi}_2\left(\xi \right),\\ {}\sum {A}_j\left(\xi, s\right){P}_j=0.\ \mathrm{where},{P}_j={l}_j\left(-{\lambda}_j+h\right).\end{array}} $$

Solving Eqs. (27)–(29) with the aid of (20)–(25), we obtain

$$ \tilde{u}=\frac{F_1\tilde{\psi_1}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{\varLambda}_{1j}{\theta}_j\right]+\frac{F_2\tilde{\psi_2}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{\varLambda}_{2j}{\theta}_j\right], $$

(30)

$$ \tilde{w}=\frac{F_1\tilde{\psi_1}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{d}_j{\varLambda}_{1j}{\theta}_j\right]+\frac{F_2\tilde{\psi_2}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{d}_j{\varLambda}_{2j}{\theta}_j\right], $$

(31)

$$ \tilde{\varphi}=\frac{F_1\tilde{\psi_1}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{l}_j{\varLambda}_{1j}{\theta}_j\right]+\frac{F_2\tilde{\psi_2}\left(\xi \right)}{\varLambda}\left[\sum \limits_{i=1}^3{l}_j{\varLambda}_{2j}{\theta}_j\right], $$

(32)

$$ \tilde{t_{rr}}=\frac{F_1\tilde{\psi_1}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{Q}_j{\varLambda}_{1j}{\theta}_j\right]+\frac{F_2\tilde{\psi_2}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{Q}_j{\varLambda}_{2j}{\theta}_j\right], $$

(33)

$$ \tilde{t_{zr}}=\frac{F_1\tilde{\psi_1}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{\mu}_j{\varLambda}_{1j}{\theta}_j\right]+\frac{F_2\tilde{\psi_2}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{\mu}_j{\varLambda}_{2j}{\theta}_j\right], $$

(34)

$$ \tilde{t_{zz}}=\frac{F_1\tilde{\psi_1}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{\eta}_j{\varLambda}_{1j}{\theta}_j\right]+\frac{F_2\tilde{\psi_2}\left(\xi \right)}{\varLambda}\left[\sum \limits_{j=1}^3{\eta}_j{\varLambda}_{2j}{\theta}_j\right], $$

(35)

where

$$ {\displaystyle \begin{array}{l}{\varLambda}_{11}=-{\mu}_2{P}_3+{P}_2{\mu}_3,\\ {}{\varLambda}_{12}={\mu}_1{P}_3-{P}_1{\mu}_3,\\ {}{\varLambda}_{13}=-{\mu}_1{P}_2+{P}_1{\mu}_2,\\ {}{\varLambda}_{21}={\eta}_2{P}_3-{P}_2{\eta}_3\\ {}{\varLambda}_{22}=-{\eta}_1{P}_3+{P}_1{\eta}_3\\ {}{\varLambda}_{23}={\eta}_1{P}_2-{P}_1{\eta}_2\\ {}\varLambda =-{\eta}_1{\varLambda}_{11}-{\eta}_2{\varLambda}_{12}-{\eta}_3{\varLambda}_{13},j=1,2,3.\end{array}} $$

Special cases

Concentrated normal force (CNF)

The CNF applied on the half-space is taken as

$$ {\psi}_1(r)=\frac{\delta (r)}{2\pi r},{\psi}_2(r)=\frac{\delta (r)}{2\pi r}. $$

(36)

Applying Hankel transform, we get

$$ {\hat{\psi}}_1\left(\xi \right)=\frac{1}{2\pi \xi},{\hat{\psi}}_2\left(\xi \right)=\frac{1}{2\pi \xi}. $$

(37)

The solution of Eqs. (30)–(35) with CNF is obtained using (37).

Uniformly distributed force (UDF)

Let a uniform force F₁/constant temperature F₂ be applied over a uniform circular region of radius a. We obtained the solution with UDF applied on the half-space by taking

$$ {\psi}_1(r)={\psi}_2(r)=\frac{H\left(a-r\right)}{\pi {a}^2}, $$

(38)

where H(a − r) is a Heaviside function. The Hankel transforms of ψ₁(r) and ψ₂(r)are given by

$$ {\hat{\psi}}_1\left(\xi \right)={\hat{\psi}}_2\left(\xi \right)=\left\{\frac{J_1\left(\xi a\right)}{2\pi a\xi}\right\},\xi \ne 0. $$

(39)

The solution of Eqs. (30)–(35) with UDF is obtained using (39).

Applications

We considered an inclined load (F₀/unit length) applied on a uniform circular region and its inclination with the z-axis is θ (see Fig. 1), we have

$$ {F}_1={F}_0\cos \theta\ \mathrm{and}\ {F}_2={F}_0\sin \theta $$

(40)

Fig. 1

Geometry of the problem

Using Eq. (40) in Eqs. (30)–(35) and with the aid of Eqs. (37) and (39), we obtain displacement components, stress components, and conductive temperature with uniformly distributed force and concentrated force on the surface of TIMT body with and without energy dissipation.

Particular cases

• a) If we take${K}_{ij}^{\ast}\ne 0$, Eq. (2) is GN-III theory, and thus we obtain the solution of (30)–(35) for TIMT solid with rotation and GN-III theory.

• b) Equation (2) becomes GN-II theory if we take ${K}_{ij}^{\ast }=0$, and we obtain the solution of (30)–(35) for TIMT solid with rotation and GN-II theory.

• c) If we take K_ij = 0, the equation of GN-III theory reduces to the GN-I theory, which is identical with the classical theory of thermoelasticity, and thus we obtain the solution of (30)–(35) for TIMT solid with rotation and GN-I theory.

• d) If ${C}_{11}={C}_{33}=\lambda +2\mu, {C}_{12}={C}_{13}=\lambda, {C}_{44}=\mu, \kern0.5em {\alpha}_1={\alpha}_3={\alpha}^{\prime },\kern0.5em {a}_1={a}_3=a,\kern0.5em {b}_1={b}_3=b,{K}_1={K}_3=K,{K}_1^{\ast }={K}_3^{\ast }={K}^{\ast }$, we obtain the solution of (30)–(35) for TIMT materials with rotation and with GN-III theory.

Inversion of the transformation

To obtain the solution to the problem in the physical domain following Sharma et al. (2015a, 2015b), invert the transforms in Eqs. (30)–(35) by inverting the Hankel transform using

$$ {f}^{\ast}\left(r,z,s\right)=\underset{0}{\overset{\infty }{\int }}\xi \overset{\sim }{f}\left(\xi, z,s\right){J}_n\left(\xi r\right) d\xi . $$

(41)

The integral in Eq. (41) is calculated using Romberg’s integration through adaptive step size as defined in Press et al. (1986).

Numerical results and discussion

For determining the theoretical results and influence of GN-I, GN-II, and GN-III theories of thermoelasticity, the physical data for cobalt material has been considered from Dhaliwal and Singh (1980) as

$$ {\displaystyle \begin{array}{l}{c}_{11}=3.07\times {10}^{11}\ \mathrm{N}\ {\mathrm{m}}^{-2},{c}_{33}=3.581\times {10}^{11}\mathrm{N}\ {\mathrm{m}}^{-2},{c}_{13}=1.027\times {10}^{10}\ \mathrm{N}\ {\mathrm{m}}^{-2},{c}_{44}=1.510\times \\ {}{10}^{11}\ \mathrm{N}\ {\mathrm{m}}^{-2},{\beta}_1=7.04\times {10}^6\mathrm{N}\ {\mathrm{m}}^{-2}\ {\deg}^{-1},{\beta}_3=6.90\times {10}^6\mathrm{N}\ {\mathrm{m}}^{-2}\ {\deg}^{-1},\rho =8.836\times {10}^3\mathrm{kg}\ {\mathrm{m}}^{-3},\\ {}\ {C}_E=4.27\times {10}^2{\mathrm{jkg}}^{-1}{\deg}^{-1},{K}_1=0.690\times {10}^2\mathrm{W}\ {\mathrm{m}}^{-1}\ {\deg}^{-1},{K}_3=0.690\times {10}^2\mathrm{W}\ {\mathrm{m}}^{-1}\ {\mathrm{K}}^{-1},{T}_0=298\ \mathrm{K},\\ {}\ {H}_0=1\ \mathrm{J}\ {\mathrm{m}}^{-1}{\mathrm{nb}}^{-1},\kern0.5em {\varepsilon}_0=8.838\times {10}^{-12}\mathrm{F}\ {\mathrm{m}}^{-1},\kern0.5em L=1,\kern0.5em \mathrm{and}\ \varOmega =0.5.\end{array}} $$

Using these values, the graphical illustrations of displacement components (u and w), conductive temperature φ, normal force stress t_zz, tangential stress t_zr, and radial stress t_rr,for a TIMT solid with GN-III theory and with 2T due to inclined load, have been illustrated. The numerical calculations have been obtained by developing a FORTRAN program using the above values for cobalt material.

• i. The black line with a square symbol relates to ${K}_{ij}^{\ast}\ne 0$ for TIMT solid with rotation and GN-III theory.

• ii. The red line with a circle symbol relates to ${K}_{ij}^{\ast }=0$ for TIMT solid with rotation and GN-II theory.

• iii. The blue line with a triangle symbol relates to K_ij = 0 the GN theory of type I.

Case 1: Concentrated normal force

Figures 2, 3, 4, 5, 6, and 7 illustrate the deviations of u and w, conductive temperature φ, and t_rr, t_rz, and t_zz for a TIMT medium with concentrated normal force, with rotation, and due to inclined load. From the graph, we find that that the displacement component (u) and conductive temperature φ decreases while stress components ( t_rr, t_rz, and t_zz) and displacement component (w) show a sharp increase and then an oscillatory pattern with an amplitude difference.

Fig. 2

Deviations of u with x (with CNF)

Fig. 3

Deviations of w with x (with CNF)

Fig. 4

Deviations of φ with x (with CNF)

Fig. 5

Deviations of t_rr with x (with CNF)

Fig. 6

Deviations of t_rz with x (with CNF)

Fig. 7

Deviations of t_zz with x (with CNF)

Case II: Uniformly distributed force (UDF)

Figures 8, 9, 10, 11, 12, and 13 illustrate the deviations of u and w, conductive temperature φ, and t_rr, t_rz, and t_zz for a TIMT medium with UDF and with rotation, and due to inclined load. The displacement components (u), stress components ( t_rr, t_rz, and t_zz), and temperature φ decrease sharply and then show the small oscillatory pattern, while the displacement component (w) first increases during the initial range of distance near the loading surface and follows the small oscillatory pattern for the rest of the values of distance.

Fig. 8

Deviations of u with x (with UDF)

Fig. 9

Deviations of w with x (with UDF)

Fig. 10

Deviations of φ with x (with UDF)

Fig. 11

Deviations of t_rr with x (with UDF)

Fig. 12

Deviations of the t_rzwith x (with UDF)

Fig. 13

Deviations of t_zzwith x (with UDF)

Conclusions

In the above research, we conclude:

• The components of displacement, stress, and temperature distribution for TIMT solid with GN-III theory, with 2T with inclined load, are calculated numerically.

• The study motivates to consider magneto-thermoelastic materials as an inventive field of thermoelastic solids. The shape of curves demonstrates the effect of various GN theories and rotation on the body and fulfills the purpose of the study.

• The outcomes of this research are extremely helpful in the 2-D problem with dynamic response of inclined load in TIMT medium with rotation which is beneficial to detect the deformation field such as geothermal engineering, advanced aircraft structure design, thermal power plants, composite engineering, geology, high-energy particle accelerators, geophysics, auditory range, and geomagnetism. The proposed model is significant to different problems in thermoelasticity and thermodynamics.

Nomenclature

Symbol	Name of Symbol	SI Unit	Symbol	Name of Symbol	SI Unit
δ ij	Kronecker delta,		ω	Frequency	Hz
C ijkl	Elastic parameters,	Nm −2	β ij	Thermal elastic coupling tensor,	Nm −2 K −1
τ 0	Relaxation Time	s	Ω	Angular Velocity of the Solid	s −1
Fi	Components of Lorentz force	N	T	Absolute temperature,	K
${\overrightarrow{H}}_0$	Magnetic field intensity vector	Jm−1nb−1	e ij	Strain tensors,	Nm −2
φ	conductive temperature,	Wm −1 K −1	$\overrightarrow{j}$	Current Density Vector	Am −2
t ij	Stress tensors,	Nm −2	$\overrightarrow{u}$	Displacement Vector	m
μ 0	Magnetic permeability	Hm−1	T 0	Reference temperature,	K
u i	Components of displacement,	m	ε 0	Electric permeability	Fm−1
ρ	Medium density,	Kgm −3	δ(t)	Dirac’s delta function
C E	Specific heat,	JKg −1 K −1	K ij	Materialistic constant,	Wm −1 K −1
α ij	Linear thermal expansion coefficient,	K −1	${K}_{ij}^{\ast }$	Thermal conductivity,	Ns −2 K −1
H(t)	Heaviside unit step function		F1, F2	magnitude of applied forces	N