We consider a transversely isotropic thin circular plate of thickness 2b occupying space D defined by 0 ≤ r ≤ ∞,−h ≤ z ≤ h. Thin plates are usually characterised by the ratio a/h (the ratio between the length of a side, a, and the thickness of the material, h, falling between the values of 8 and 80 as mentioned by Ventsel and Krauthammer (2001). Let the plate be subjected to axisymmetric heat supply thermal shock load applied into its inner boundary having initially undisturbed state at a uniform temperature T0. We use plane cylindrical coordinates (r, θ, z) with the centre of the plate as the origin (Fig. 1).
As the problem considered is plane axisymmetric, (u, v, w, and φ) are independent of θ. We restrict our analysis to two-dimension problem with \( \overrightarrow{u}=\left(u,0,w\right) \), also applying the transformation:
$$ {x}^{\prime }=x\cos \phi +y\sin \phi, {y}^{\prime }=-x\sin \phi +y\cos \phi, \kern0.5em {z}^{\prime }=z. $$
where ϕ is the angle of rotation in x-y plane, on the set of Eqs. (1)–(3) to derive the equations for TIT solid with two temperatures and with energy dissipation, to obtain:
$$ {C}_{11}\left(\frac{\partial^2u}{{\partial r}^2}+\frac{1}{r}\frac{\partial u}{\partial r}-\frac{1}{r^2}u\right)+{C}_{13}\left(\frac{\partial^2w}{\partial r\partial z}\right)+{C}_{44}\frac{\partial^2u}{{\partial z}^2}+{C}_{44}\left(\frac{\partial^2w}{\partial r\partial z}\right)-{\beta}_1\frac{\partial }{\partial r}\left\{\varphi -{a}_1\left(\frac{\partial^2\varphi }{{\partial r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{{\partial z}^2}\right\}=\rho \left(\frac{\partial^2u}{\partial {t}^2}-{\varOmega}^2u+2\varOmega \frac{\partial w}{\partial t}\right), $$
(8)
$$ \left({C}_{11}+{C}_{44}\right)\left(\frac{\partial^2u}{\partial r\partial z}+\frac{1}{r}\frac{\partial u}{\partial z}\right)+{C}_{44}\left(\frac{\partial^2w}{{\partial r}^2}+\frac{1}{r}\frac{\partial w}{\partial r}\right)+{C}_{33}\frac{\partial^2w}{{\partial z}^2}-{\beta}_3\frac{\partial }{\partial z}\left\{\varphi -{a}_1\left(\frac{\partial^2\varphi }{{\partial r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{{\partial z}^2}\right\}=\rho \left(\frac{\partial^2w}{\partial {t}^2}-{\varOmega}^2w-2\varOmega \frac{\partial u}{\partial t}\right), $$
(9)
$$ \left({K}_1^{\ast }+{K}_1\frac{\partial }{\partial t}\right)\left(\frac{\partial^2\varphi }{{\partial r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)+\left({K}_3^{\ast }+{K}_3\frac{\partial }{\partial t}\right)\frac{\partial^2\varphi }{{\partial z}^2}={T}_0\left(1+{\tau}_0\frac{\partial }{\partial t}\right)\left({\beta}_1\frac{\partial \dot{u}}{\partial r}+{\beta}_3\frac{\partial \dot{w}}{\partial z}\right)+\rho {C}_E\left(1+{\tau}_0\frac{\partial }{\partial t}\right)\left\{\dot{\varphi}-{a}_1\left(\frac{\partial^2\dot{\varphi}}{{\partial r}^2}+\frac{1}{r}\frac{\partial \dot{\varphi}}{\partial r}\right)-{a}_3\frac{\partial^2\dot{\varphi}}{{\partial z}^2}\right\}. $$
(10)
In above equations, we use the contracting subscript notations (1 → 11, 2 → 22, 3 → 33, 5 → 23, 4 → 13, 6 → 12) to relate Cijkl to Cmn. Also a1 and a3 are two temperature parameters.
For axisymmetric problem following Lata and Kaur (2019b), the constitutive relations are:
$$ {t}_{rr}={c}_{11}{e}_{rr}+{c}_{12}{e}_{\theta \theta}+{c}_{13}{e}_{zz}-{\beta}_1T, $$
$$ {t}_{zr}=2{c}_{44}{e}_{rz}, $$
$$ {t}_{zz}={c}_{13}{e}_{rr}+{c}_{13}{e}_{\theta \theta}+{c}_{33}{e}_{zz}-{\beta}_3T, $$
$$ {t}_{\theta \theta}={c}_{12}{e}_{rr}+{c}_{11}{e}_{\theta \theta}+{c}_{13}{e}_{zz}-{\beta}_1T, $$
where
$$ {e}_{rz}=\frac{1}{2}\left(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial r}\right), $$
$$ {e}_{rr}=\frac{\partial u}{\partial r}, $$
$$ {e}_{\theta \theta}=\frac{u}{r}, $$
$$ {e}_{zz}=\frac{\partial w}{\partial z}, $$
$$ T=\varphi -{a}_1\left(\frac{\partial^2\varphi }{{\partial r}^2}+\frac{1}{r}\frac{\partial \varphi }{\partial r}\right)-{a}_3\frac{\partial^2\varphi }{{\partial z}^2}, $$
$$ {\beta}_{ij}={\beta}_i{\delta}_{ij},\kern0.5em {K}_{ij}={K}_i{\delta}_{ij,} $$
$$ {\beta}_1=\left({c}_{11}+{c}_{12}\right){\alpha}_1+{c}_{13}{\alpha}_3, $$
$$ {\beta}_3=2{c}_{13}{\alpha}_1+{c}_{33}{\alpha}_3. $$
To facilitate the solution, the following dimensionless quantities are introduced:
$$ {r}^{\prime }=\frac{r}{L},\kern0.75em {z}^{\prime }=\frac{z}{L},{t}^{\prime }=\frac{c_1}{L}t,{u}^{\prime }=\frac{{\rho c}_1^2}{L{\beta}_1{T}_0}u,{w}^{\prime }=\frac{{\rho c}_1^2}{L{\beta}_1{T}_0}w,{T}^{\prime }=\frac{T}{T_0},{t}_{zr}^{\prime }=\frac{t_{zr}}{\beta_1{T}_0},{t}_{zz}^{\prime }=\frac{t_{zz}}{\beta_1{T}_0},{\varOmega}^{\prime }=\frac{L}{C_1}\varOmega, {\varphi}^{\prime }=\frac{\varphi }{T_0},{a}_1^{\prime }=\frac{a_1}{L^2},{a}_3^{\prime }=\frac{a_3}{L^2}, $$
(11)
where \( {c}_1^2=\frac{c_{11}}{\rho } \), and L is a constant of dimension of length.
Using the dimensionless quantities defined by Eq. (11) in Eqs. (8)–(10) and after that suppressing the primes and applying the Laplace and Hankel transforms defined by:
$$ {f}^{\ast}\left(r,z,s\right)=\underset{0}{\overset{\infty }{\int }}f\left(r,z,t\right){e}^{- st} dt, $$
(12)
$$ \overset{\sim }{f}\left(\xi, z,s\right)=\underset{0}{\overset{\infty }{\int }}{f}^{\ast}\left(r,z,s\right)r{J}_n\left( r\xi \right) dr. $$
(13)
on the resulting quantities, we obtain:
$$ \left(-{\xi}^2-{s}^2+{\varOmega}^2+{\delta}_2{D}^2\right)\overset{\sim }{u}+\left[{\xi \delta}_1D-2\varOmega s\right]\overset{\sim }{w}+\left(\xi \left(1-{a}_3{D}^2\Big)+{a}_1{\xi}^3\right)\right)\overset{\sim }{\varphi }=0, $$
(14)
$$ \left({\delta}_1\xi D+2\varOmega s\right)\overset{\sim }{u}+\left({\delta}_3{D}^2-{\delta}_2{\xi}^2-{s}^2+{\varOmega}^2\right)\overset{\sim }{w}-\left(\frac{\beta_3}{\beta_1}D\left[\left(1-{a}_3{D}^2\right)+{\xi}^2{a}_1\right]\right)\overset{\sim }{\varphi }=0, $$
(15)
$$ {\delta}_4\left(s+{\tau}_0{s}^2\right)\xi \overset{\sim }{u}-{\delta}_5\left(s+{\tau}_0{s}^2\right)D\overset{\sim }{w}+\left(-{\delta}_6\left(s+{\tau}_0{s}^2\right)\left(1+{\xi}^2{a}_1\right)-\left({K}_1^{\ast }+\frac{C_1}{L}{K}_1s\right){\xi}^2+{D}^2\left({K}_3^{\ast }+\frac{C_1}{L}{K}_3s+{a}_3{\delta}_6\left(s+{\tau}_0{s}^2\right)\right)\right)\overset{\sim }{\varphi }=0, $$
(16)
where
$$ {\delta}_1=\frac{c_{13}+{c}_{44}}{c_{11}},\kern0.5em {\delta}_2=\frac{c_{44}}{c_{11}},\kern0.5em {\delta}_3=\frac{c_{33}}{c_{11}},\kern0.5em {\delta}_4=\frac{\beta_1^2{T}_0}{\rho },\kern0.75em {\delta}_5=\frac{\beta_1{\beta}_3{T}_0}{\rho },\kern0.5em {\delta}_6=\rho {C}_E{C}_1^2, $$
and
$$ D\equiv \frac{d}{dz}, $$
$$ \overset{\sim }{t_{zz}}=\frac{C_{13}}{C_{11}}\xi \overset{\sim }{u}+{\delta}_3D\overset{\sim }{w}-\frac{\upbeta_3}{\upbeta_1}\left(1+{a}_1{\xi}^2-{a}_3{D}^2\right)\overset{\sim }{\varphi }, $$
(17)
$$ \overset{\sim }{t_{rz}}={\delta}_2D\overset{\sim }{u}-\xi {\delta}_2\overset{\sim }{w}, $$
(18)
$$ \overset{\sim }{t_{rr}}=-\xi \overset{\sim }{u}+\frac{C_{12}\xi }{C_{11}}\overset{\sim }{u}+\frac{C_{13}}{C_{11}}D\overset{\sim }{w}-\left(1+{a}_1{\xi}^2-{a}_3{D}^2\right)\overset{\sim }{\varphi }. $$
(19)
We will obtain the non-trivial solution of Eqs. (14)–(16) if the determinant of the coefficient\( \overset{\sim }{\ u} \), \( \overset{\sim }{w} \), and \( \overset{\sim }{\varphi\ } \)vanishes, which yields to the following characteristic equation:
$$ {AD}^6+{BD}^4+{CD}^2+E=0, $$
(20)
where
$$ \mathrm{A}={\updelta}_2{\updelta}_3{\upzeta}_{12}-{\updelta}_2{\upzeta}_{10}{\upzeta}_8, $$
$$ \mathrm{B}={\upzeta}_1{\upzeta}_{12}{\updelta}_3-{\zeta}_1{\zeta}_{10}{\upzeta}_8+{\updelta}_2{\updelta}_3{\upzeta}_{11}+{\updelta}_2{\upzeta}_{12}{\upzeta}_6-{\updelta}_2{\upzeta}_{10}{\upzeta}_7-{\zeta}_2^2{\upzeta}_{12}-{\zeta}_2{\zeta}_9{\upzeta}_8+{\zeta}_5{\zeta}_{10}{\upzeta}_2-{\upzeta}_5{\upzeta}_9{\updelta}_3, $$
$$ C={\updelta}_3{\upzeta}_1{\upzeta}_{11}+{\updelta}_2{\upzeta}_6{\upzeta}_{11}+{\upzeta}_1{\upzeta}_6{\upzeta}_{12}-{\upzeta}_1{\upzeta}_{10}{\upzeta}_7-{\zeta}_2^2{\upzeta}_{11}+{\upzeta}_2{\upzeta}_7{\upzeta}_9-{\upzeta}_5{\upzeta}_6{\upzeta}_9+{\upzeta}_4{\upzeta}_2{\upzeta}_{10}-{\updelta}_3{\upzeta}_4{\upzeta}_9+{\zeta}_3^2{\upzeta}_{12}, $$
$$ E={\upzeta}_6{\upzeta}_1{\upzeta}_{11}-{\upzeta}_4{\upzeta}_6{\upzeta}_9. $$
$$ {\upzeta}_1=-{\xi}^2-{\mathrm{s}}^2+{\Omega}^2, $$
$$ {\upzeta}_2={\delta}_1\xi, $$
$$ {\upzeta}_3=2\Omega s, $$
$$ {\upzeta}_4=\upxi \left(1+{\mathrm{a}}_1{\upxi}^2\right), $$
$$ {\upzeta}_5={a}_3\xi, $$
$$ {\upzeta}_6=-{\delta}_2{\xi}^2-{s}^2+{\Omega}^2, $$
$$ {\upzeta}_7=-\frac{\upbeta_3}{\upbeta_1}\left(1+{\mathrm{a}}_1{\upxi}^2\right), $$
$$ {\upzeta}_8=\frac{\upbeta_3}{\upbeta_1}{\mathrm{a}}_3, $$
$$ {\upzeta}_9={\delta}_4\left(s+{\tau}_0{s}^2\right)\xi, $$
$$ {\upzeta}_{10}=-{\delta}_5\left(s+{\tau}_0{s}^2\right), $$
$$ {\zeta}_{11}=-{\delta}_6\left(s+{\tau}_0{s}^2\right)\left(1+{\mathrm{a}}_1{\upxi}^2\right)-{\xi}^2\left({K}_1^{\ast }+\frac{C_1}{L}{K}_1s\right), $$
$$ {\zeta}_{12}=\left({K}_3^{\ast }+\frac{C_1}{L}{K}_3s+{a}_3{\delta}_6\left(s+{\tau}_0{s}^2\right)\right). $$
The solutions of Eq. (20) can be written in the form by making use of the radiation condition that \( \overset{\sim }{\ u} \), \( \overset{\sim }{w} \), \( \overset{\sim }{\varphi\ }\to 0 \) as z → ∞ :
$$ \overset{\sim }{u}=\sum {A}_i\left(\xi, s\right)\cosh \left({q}_iz\right), $$
(21)
$$ \overset{\sim }{w}=\sum {d}_i{A}_i\left(\xi, s\right)\cosh \left({q}_iz\right), $$
(22)
$$ \overset{\sim }{\varphi }=\sum {l}_i{A}_i\left(\xi, s\right)\cosh \left({q}_iz\right), $$
(23)
where Ai, being arbitrary constants,±qi are the roots of Eq. (20), and
$$ {d}_i=\frac{\left({\zeta}_2{\zeta}_{12}-{\zeta}_8{\zeta}_9\right){q}_i^3+{\zeta}_3{\zeta}_{12}{q}_i^2+\left({\zeta}_2{\zeta}_{11}-{\zeta}_7{\zeta}_9\right){q}_i+{\zeta}_3{\zeta}_{11}}{\left(-{\zeta}_8{\zeta}_{10}+{\delta}_3{\upzeta}_{12}\right){q}_i^4+\left({\delta}_3{\zeta}_{11}+{\zeta}_6{\zeta}_{12}-{\zeta}_7{\zeta}_{10}\right){q}_i^2+{\zeta}_6{\zeta}_{11}}, $$
(24)
$$ {l}_i=\frac{\left(-{\zeta}_9{\delta}_3+{\zeta}_1{\zeta}_{10}\right){q}_i^2+{\zeta}_3{\zeta}_{10}{q}_i-{\zeta}_8{\zeta}_9}{\left(-{\zeta}_8{\zeta}_{10}+{\delta}_3{\upzeta}_{12}\right){q}_i^4+\left({\delta}_3{\zeta}_{11}+{\zeta}_6{\zeta}_{12}-{\zeta}_7{\zeta}_{10}\right){q}_i^2+{\zeta}_6{\zeta}_{11}}, $$
(25)
where i = 1, 2, 3.
Also, using Eqs. (21)–(23) in expressions (17)–(19), we obtain stress components as:
$$ \overset{\sim }{t_{zz}}=\sum {A}_i\left(\xi, s\right){\eta}_i\cosh \left({q}_iz\right)+\sum {\mu}_i{A}_i\left(\xi, s\right)\sinh \left({q}_iz\right), $$
(26)
$$ \overset{\sim }{t_{rz}}=\sum {A}_i\left(\xi, s\right){M}_i\cosh \left({q}_iz\right)+\sum {N}_i{A}_i\left(\xi, s\right)\sinh \left({q}_iz\right), $$
(27)
$$ {t}_{rr}=\sum {A}_i\left(\xi, s\right){R}_i\cosh \left({q}_iz\right)+\sum {S}_i{A}_i\left(\xi, s\right)\sinh \left({q}_iz\right), $$
(28)
where
$$ {\eta}_i=\frac{C_{13}}{C_{11}}\xi -\frac{\upbeta_3}{\upbeta_1}{l}_i\left(1+{a}_1{\xi}^2-{\mathrm{a}}_3{\mathrm{q}}_{\mathrm{i}}^2\right), $$
$$ {R}_i=-\xi +\frac{C_{12}\xi }{C_{11}}-\left(1+{a}_1{\xi}^2-{a}_3{\mathrm{q}}_{\mathrm{i}}^2\right), $$
$$ {S}_i=\frac{C_{13}}{C_{11}}{d}_i{q}_i, $$
$$ {\mu}_i={\delta}_3{d}_i{q}_i, $$
$$ {M}_i={\delta}_2{d}_i\xi, $$
$$ {N}_i={\delta}_2{q}_i,i=1,2,3. $$