Comment on the paper “On thermoelastic problem of a thermosensitive functionally graded rectangular plate with instantaneous point heat source, V. R. Manthena, G. D. Kedar, Journal of Thermal Stresses, 2019, Vol. 42, No. 7, 849-862”

This article refers to:

On thermoelastic problem of a thermosensitive functionally graded rectangular plate with instantaneous point heat source

First error

In the above paper the Equation (2)(2) $$k(x,T)\frac{\partial T}{\partial x}-{\epsilon }_2(T-T_0)=f(y,z,t)\hspace{0.5em}\text{at}\hspace{0.5em}x=a$$(2) are as follows (1) $$k(x,T)\frac{\partial T}{\partial x}-{\epsilon }_1(T-T_0)=0\hspace{0.5em}\text{at}\hspace{0.5em}x=0$$(1) (2) $$k(x,T)\frac{\partial T}{\partial x}-{\epsilon }_2(T-T_0)=f(y,z,t)\hspace{0.5em}\text{at}\hspace{0.5em}x=a$$(2) where $k(\mathrm{kgKelvi}{\mathrm{n}}^{-1}\mathrm{m}{\hspace{0.17em}\text{sec}}^{-3})$ is the thermal conductivity, $T(\mathrm{Kelvin})$ is the temperature and $x(m)$ is the Cartesian coordinate. In a physics equation all terms must have the same units and from Equations (1)(1) $$k(x,T)\frac{\partial T}{\partial x}-{\epsilon }_1(T-T_0)=0\hspace{0.5em}\text{at}\hspace{0.5em}x=0$$(1) and (2)(2) $$k(x,T)\frac{\partial T}{\partial x}-{\epsilon }_2(T-T_0)=f(y,z,t)\hspace{0.5em}\text{at}\hspace{0.5em}x=a$$(2) it is found that the units of ${\epsilon }_1$ and ${\epsilon }_2$ are $\mathrm{kgKelvi}{\mathrm{n}}^{-1}{\hspace{0.17em}\text{sec}}^{-3}.$

In equation (10) in [1] the following temperature is defined (3) $$\Theta (T)=\underset{0}{\overset{T}{\int }}k(x,T)dT$$(3)

From Equation (3)(3) $$\Theta (T)=\underset{0}{\overset{T}{\int }}k(x,T)dT$$(3) it is found that the units of $\Theta (T)$ are $\mathrm{kgm}{\hspace{0.17em}\text{sec}}^{-3}.$ The boundary conditions (12) in [1] are (4) $$\frac{\partial \Theta }{\partial x}-{\epsilon }_1\Theta =0\hspace{0.5em}\text{at}\hspace{0.5em}x=0$$(4) (5) $$\frac{\partial \Theta }{\partial x}+{\epsilon }_2\Theta =f(y,z,t)\hspace{0.5em}\text{at}\hspace{0.5em}x=a$$(5)

The units of the term $\frac{\partial \Theta }{\partial x}$ are $\mathrm{k}\mathrm{g}{\hspace{0.17em}\text{sec}}^{-3}$ whereas the units of the terms ${\epsilon }_1\Theta$ and ${\epsilon }_2\Theta$ are $\mathrm{k}{\mathrm{g}}^2\mathrm{Kelvi}{\mathrm{n}}^{-1}\mathrm{m}{\hspace{0.17em}\text{sec}\hspace{0.17em}}^{-6}.$ For that reason the Equations (4)(4) $$\frac{\partial \Theta }{\partial x}-{\epsilon }_1\Theta =0\hspace{0.5em}\text{at}\hspace{0.5em}x=0$$(4) and (5)(5) $$\frac{\partial \Theta }{\partial x}+{\epsilon }_2\Theta =f(y,z,t)\hspace{0.5em}\text{at}\hspace{0.5em}x=a$$(5) are wrong.

Second error

Below equation (14) in [1] appear the terms $\text{sin}\hspace{0.17em}{\beta }_{m}x$ and $\text{cos}\hspace{0.17em}{\beta }_{m}x.$ From these terms it is clear that the units of ${\beta }_m$ are ${\mathrm{m}}^{-1}(\mathrm{lengt}{\mathrm{h}}^{-1})$ in order that the quantity ${\beta }_{m}x$ is dimensionless (The trigonometric function of a dimensional quantity is meaningless).

In equation (21) in [1] appear the terms (6) $$E_1=\frac{A_4}{(2\omega +2A_6)}$$(6) (7) $$E_2=\frac{A_4}{(2\omega -2A_6)}$$(7) (8) $$E_3=\frac{A_4\omega }{A_6^2-{\omega }^2}+A_5$$(8) where $\omega (\mathrm{se}{\mathrm{c}}^{-1})$ is the frequency and $A_6={\beta }_m^2+\mathrm{otherterms}.$ The units of $A_6$ are ${\mathrm{m}}^{-2}(\mathrm{lengt}{\mathrm{h}}^{-2}).$ In Physics it is not allowed to add quantities with different units. Therefore the terms $(2\omega +2A_6),$ $(2\omega -2A_6),$ $(A_6^2-{\omega }^2)$ are wrong, the terms $E_1,E_2,E_3$ are also wrong and consequently the equation (21) in [1] is wrong.

Third error

In paragraph “Numerical results and discussion” in [1] the following dimensionless parameters are presented (9) $$\left({\overline{M}}_x,{\overline{M}}_y,{\overline{M}}_{xy}\right)=\frac{\left(M_x,M_y,M_{xy}\right)}{E_0{b}^3}$$(9)

From equation (8) in [1] it is found that the units of $M_x,M_y,M_{xy}$ are $\mathrm{kgmse}{\mathrm{c}}^{-2}.$ $E_0(\mathrm{kg}{\mathrm{m}}^{-1}\mathrm{se}{\mathrm{c}}^{-2})$ is the Young’s modulus and $b=4\mathrm{cm}$ is the width of the plate. From equation (9) it is found that the units of $({\overline{M}}_x,{\overline{M}}_y,{\overline{M}}_{xy})$ are ${\mathrm{m}}^{-1}.$ This means that $({\overline{M}}_x,{\overline{M}}_y,{\overline{M}}_{xy})$ are dimensional and wrong.

Probably a correct form of equation (9)(9) $$\left({\overline{M}}_x,{\overline{M}}_y,{\overline{M}}_{xy}\right)=\frac{\left(M_x,M_y,M_{xy}\right)}{E_0{b}^3}$$(9) is (10) $$\left({\overline{M}}_x,{\overline{M}}_y,{\overline{M}}_{xy}\right)=\frac{\left(M_x,M_y,M_{xy}\right)}{E_0{b}^2}$$(10)

In addition the symbol $\omega$ has been used either as deflection with units $\mathrm{meters}(\mathrm{length})$ or as frequency with units ${\hspace{0.17em}\text{sec}\hspace{0.17em}}^{-1}(\mathrm{tim}{\mathrm{e}}^{-1})$ and this causes confusion in the paper.