The vibration of a nanobeam subjected to constant magnetic field and ramp-type heat under non-Fourier heat conduction law based on the Lord-Shulman model

Abstract

In this work, the usual Euler–Bernoulli nanobeam has been modeled in the context of the Lord-Shulman thermoelastic theorem which contains the non-Fourier heat conduction law. The nanobeam has been subjected to a constant magnetic field and ramp-type thermal loading. The Laplace transform definition has been applied to the governing equations and the solutions have been obtained by using a direct approach. The inversions of the Laplace transform have been calculated numerically by using the Tzou approximation method. The solutions have been applied to a nanobeam made of silicon nitride. The distributions of the temperature increment, lateral deflection, strain, stress, and strain-energy density have been represented in figures with different values of the magnetic field intensity and ramp-time heat parameter. The value of the magnetic field intensity and ramp-time heat parameter have significant effects on all the studied functions, and they could be used as tuners to control the energy which has been generated through the nanobeam.

Introduction

It has been shown that the theory of coupled thermoelasticity, a kind of heat conduction, solves a number of problems.^1,2 It is consisting of two partial differential equations based on Fourier’s law of heat conduction, the equation of motion and the rule of conservation of energy.^3–6 The propagation velocity of heat waves is infinitely increased by this kind of heat conduction. A extended theory of thermoelasticity mechanical theory with the relaxation time of isotropic objects was presented out by Lord and Shulman. ⁷ According to this theory, Fourier’s law is replaced with Cattaneo’s law or a non-law Fourier’s of heat conduction. The heat equation removes the paradox of infinite propagation velocity since it is a hyperbola. ⁸

Several scientists have worked on micro/nano electric devices in recent years. Micro/nanoelectromechanical beam resonators have a wide range of uses, including actuators, beams, sensors, pumps, resonators, and motors. They are also important for physical applications.^9–12

Studying the vibration of micro/nano-beam resonators is important. Alghamdi ³ used the generalization theory of thermoelasticity with a dual-phase lag to study vibration damping caused by the cavity of thermoelastic beam resonators (DPL). Thin, homogeneous, isotropic, thermoelastic micro/nanoscale beam cavity resonators’ lateral oscillations were investigated by Sharma and Grover. ¹³ Sun and Saka ¹⁴ investigated thermoelastic disk resonators’ vibration dampening outside the plane of the microplate. They changed Lifshitz and Roukes’ ¹⁵ formula for thermoelastic damping by introducing a new coefficient. Several scientists have looked at heat transfer techniques and nanobeam vibration.^16–20 Al-Lehaibi and Youssef ¹⁶ examined the gold nanobeam’s thermoelastic vibration due to thermal shock. Kidawa-Kukla ¹⁷ studied the internal and external damping effects on the lateral oscillations of the beam caused by the mobile heat source using the characteristics of Green’s function. On the vibration of a rectangular, simply supported nanobeam, Boley ¹⁸ examined the effects of thermal shock spread along the span. Manolis and Beskos ¹⁹ studied the thermoelastic dynamic response of the nanobeam exposed to thermal loads using the numerical technique. Using the Laplace transform method, Al-Huniti et al. ²⁰ examined the displacements, stresses, and dynamic behavior of a rod heated by a moving laser beam. Alghamdi ²¹ examined the thermoelastic vibration of a micro/nano-beam that was exposed to a moving heat source.

The standard Euler-Bernoulli nanobeam has been modeled in this study within the framework of the non-Fourier heat conduction law found in the Lord-Shulman thermoelastic theorem. A ramp-type thermal loading and a constant magnetic field will both be applied to the nanobeam. The governing equations will be subjected to the Laplace transform definition, and a direct technique will be used to arrive at the solutions. Using the Tzou approximation approach, the inversions of the Laplace transform will be numerically determined.

It must be noted that applying a constant magnetic field to a nanobeam is a novel application which has not been executed before, thus the results of this work will be new results.

Formulation of the problem

Although challenging but creating a beam with a rectangular cross-section is the simplest when compared to other cross-sections. Consider small flexural deflections of a thin elastic beam of length $\ell (0\le x\le \ell )$ , width $b\left(-\frac{b}{2}\le y\le \frac{b}{2}\right)$ , and thickness $h(-\frac{h}{2}\le z\le \frac{h}{2})$ as in Figure 1, for which the $x,y,z$ axes are defined along the longitudinal, width, and thickness directions of the beam, respectively. In equilibrium, the beam is unstrained, unstressed, with no damping mechanism present, and at temperature $T_0$ everywhere.^{4,6,9,13,21–24}

Figure 1.

Nanobeam subjected to constant magnetic field.

In the present study, the usual Euler–Bernoulli assumption is adopted, that is, any plane cross-section, initially perpendicular to the axis of the beam remains plane and perpendicular to the neutral surface during bending. Thus, the displacement components $(u,v,w)$ are given by:

$$\begin{array}{c}\multicolumn{1}{c}{u(x,y,z,t)=-z\frac{\partial w(x,t)}{\partial x},v(x,y,z,t)=0,\mathrm{and}}\\ w(x,y,z,t)=w(x,t),\hfill \end{array}$$

(1)

and the volumetric strain is given by:

$$e=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=-z\frac{{\partial }^{2}w}{\partial x^2}.$$

(2)

Hence, the differential equation of thermally induced lateral vibration of the beam may be expressed in the form^{4,6,9,13,21–24}:

$$(\lambda +2\mu )I\frac{{\partial }^{4}w}{\partial x^4}+\rho A\frac{{\partial }^{2}w}{\partial t^2}+\beta \frac{{\partial }^2{M}_T}{\partial x^2}=F_x(x,t),$$

(3)

where $\lambda ,\mu$ are the elastic constant, $I=h^{3}b/12$ is the inertial moment about the x-axis, $\rho$ the density of the beam, $\beta ={\alpha }_T(3\lambda +2\mu )$ is the coefficient of linear thermal expansion, $w(x,t)$ is the lateral deflection, x is the distance along the length of the beam, $A=\mathrm{hb}$ is the cross-section area, and t the time and $F(x,t)$ is the external body force. $M_T$ is the thermal moment which is defined as:

$$M_T(x,z,t)=b\underset{-h/2}{\overset{h/2}{\int }}\theta (x,z,t)z\mathrm{dz},$$

(4)

and the stress-strain relation is given by:

$${\sigma }_{\mathrm{xx}}=(\lambda +2\mu )e-\beta \theta ,$$

(5)

where $\theta (x,z,t)=(T-T_0)$ is the temperature increment of the resonator, which $T(x,z,t)$ is the temperature distribution, and $T_0$ is the environmental temperature.

According to the Lord-Shulman model (L-S) with one-relaxation time, the non-Fourier heat conduction equation has the following form^{4,6,9,13,21–24}:

$$K\frac{{\partial }^2\theta }{\partial x^2}=(\frac{\partial }{\partial t}+{\tau }_o\frac{{\partial }^2}{\partial t^2})(\rho C_E\theta +\beta T_{0}e),$$

(6)

where $C_E$ is the specific heat at constant volume, ${\tau }_0$ the thermal relaxation time, $K$ the thermal conductivity, and the dot above it means partial derivative concerning time.

A constant magnetic field with components $\overrightarrow{H}(0,0,H_0)$ permeates the medium in the absence of an external electric field $\overrightarrow{E}=0$ .^25–27

The external force is given by:

$$\stackrel{\rightharpoonup }{F}=\left(F_x,F_y,F_z\right)=\left(\stackrel{\rightharpoonup }{J}\wedge \stackrel{\rightharpoonup }{B}\right),$$

(7)

where $\stackrel{\rightharpoonup }{B}$ is the magnetic induction vector given by^25–27:

$$\stackrel{\rightharpoonup }{B}={\mu }_o\stackrel{\rightharpoonup }{H},$$

(8)

where ${\mu }_0$ is the permeability of the vacuum

and $\stackrel{\rightharpoonup }{J}$ is the conduction current density given by Ohm’s law as follows^25–27:

$$\overrightarrow{J}=j_o\left[\stackrel{\rightharpoonup }{E}+\frac{\partial \stackrel{\rightharpoonup }{u}}{\partial t}\wedge \stackrel{\rightharpoonup }{B}\right]=j_o\frac{\partial \stackrel{\rightharpoonup }{u}}{\partial t}\wedge \stackrel{\rightharpoonup }{B},$$

(9)

where $j_0$ is known as the electrical conductivity of the medium and $\stackrel{\rightharpoonup }{E}=0$ .

The components of the electromagnetic induction vector are given by:

$$B_x=B_y=0,B_z={\mu }_o{H}_o.$$

(10)

Hence, the components $\stackrel{\rightharpoonup }{F}=\stackrel{\rightharpoonup }{J}\wedge \stackrel{\rightharpoonup }{B}$ are given^25–27:

$$F_x=-j_o{\mu }_o^2{H}_o^2\frac{\partial u}{\partial t}=j_o{\mu }_o^2{H}_o^{2}z\frac{\partial {}^{2}w(x,t)}{\partial x\partial t},F_y=F_z=0.$$

(11)

Then, we have:

$$\begin{array}{c}(\lambda +2\mu )I\frac{{\partial }^{4}w}{\partial x^4}+\rho A\frac{{\partial }^{2}w}{\partial t^2}+\beta b\frac{{\partial }^2}{\partial x^2}\underset{-h/2}{\overset{h/2}{\int }}\theta (x,z,t)z\mathrm{dz}\hfill \\ =j_o{\mu }_o^2{H}_o^{2}z\frac{\partial {}^{2}w(x,t)}{\partial x\partial t},\hfill \end{array}$$

(12)

and

$$\frac{{\partial }^2\theta }{\partial x^2}+\frac{{\partial }^2\theta }{\partial z^2}=(\frac{\partial }{\partial t}+{\tau }_o\frac{{\partial }^2}{\partial t^2})(\frac{\rho C_E}{K}\theta -\frac{\beta T_0}{K}z\frac{{\partial }^{2}w(x,t)}{\partial x^2}).$$

(13)

where there is no heat flow across the upper and lower surfaces of the beam, so that $\frac{\partial \theta }{\partial z}$ at $z=\pm h/2$ for a very thin beam and assuming the temperature varies in terms of a $\sin \left(\mathrm{pz}\right)$ function along the thickness direction, where $p=\pi /h$ , gives:

$$\theta (x,z,t)=\vartheta (x,t)\sin \left(\mathrm{pz}\right).$$

(14)

Then, we have

$$\begin{array}{c}\multicolumn{1}{c}{\frac{{\partial }^{4}w}{\partial x^4}+\frac{\rho A}{(\lambda +2\mu )I}\frac{{\partial }^{2}w}{\partial t^2}+\frac{b\beta }{(\lambda +2\mu )I}\frac{{\partial }^2\vartheta }{\partial x^2}\underset{-h/2}{\overset{h/2}{\int }}z\sin \left(\mathrm{pz}\right)}\\ \mathrm{dz}=\frac{j_o{\mu }_o^2{H}_o^{2}z}{(\lambda +2\mu )I}\frac{\partial {}^{2}w}{\partial x\partial t},\hfill \end{array}$$

(15)

which gives after executing the integration the following form:

$$\begin{array}{c}\multicolumn{1}{c}{\frac{{\partial }^{4}w}{\partial x^4}+\frac{\rho A}{(\lambda +2\mu )I}\frac{{\partial }^{2}w}{\partial t^2}+\frac{2h^{2}b\beta }{{\pi }^2(\lambda +2\mu )I}\frac{{\partial }^2\vartheta }{\partial x^2}}\\ =\frac{j_o{\mu }_o^2{H}_o^{2}z}{(\lambda +2\mu )I}\frac{\partial {}^{2}w}{\partial x\partial t}.\hfill \end{array}$$

(16)

Hence, we have

$$\begin{array}{c}\multicolumn{1}{c}{\frac{{\partial }^{4}w}{\partial x^4}+\frac{12\rho }{(\lambda +2\mu )h^2}\frac{{\partial }^{2}w}{\partial t^2}+\frac{24\beta }{{\pi }^2(\lambda +2\mu )h}\frac{{\partial }^2\vartheta }{\partial x^2}}\\ =\frac{12j_o{\mu }_o^2{H}_o^{2}z}{(\lambda +2\mu )b{h}^3}\frac{\partial {}^{2}w}{\partial x\partial t}.\hfill \end{array}$$

(17)

For simplicity, we will consider that $\frac{\partial {}^{2}w}{\partial x\partial t}\approx \frac{1}{\ell }\frac{\partial w}{\partial t}$ , then we have:

$$\begin{array}{c}\multicolumn{1}{c}{\frac{{\partial }^{4}w}{\partial x^4}+\frac{12\rho }{(\lambda +2\mu )h^2}\frac{{\partial }^{2}w}{\partial t^2}+\frac{24\gamma }{{\pi }^2(\lambda +2\mu )h}\frac{{\partial }^2\vartheta }{\partial x^2}}\\ =\frac{12j_o{\mu }_o^2{H}_o^{2}z}{(\lambda +2\mu )b{h}^3\ell }\frac{\partial w}{\partial t},\hfill \end{array}$$

(18)

and the heat conduction equation as:

$$\begin{array}{c}(\frac{{\partial }^2\vartheta }{\partial x^2}-p^2\vartheta )\sin \left(\mathrm{pz}\right)=(\frac{\partial }{\partial t}+{\tau }_o\frac{{\partial }^2}{\partial t^2})\hfill \\ (\frac{\rho C_E}{K}\vartheta \sin \left(\mathrm{pz}\right)-\frac{\gamma T_0}{K}z\frac{{\partial }^{2}w}{\partial x^2}).\hfill \end{array}$$

(19)

Hence, we have:

$$\frac{{\partial }^2\vartheta }{\partial x^2}-p^2\vartheta =(\frac{\partial }{\partial t}+{\tau }_o\frac{{\partial }^2}{\partial t^2})(\frac{\rho C_E}{K}\vartheta -\frac{\gamma T_{0}z}{K\sin \left(\mathrm{pz}\right)}\frac{{\partial }^{2}w}{\partial x^2}).$$

(20)

The following non-dimensional variables will be applied to the above governing equations^{4,6,22,24–27}:

$$\begin{array}{c}(x\prime ,w\prime ,h\prime )=\epsilon c_o(x,w,h),(t\prime ,{\tau \prime }_o)=\epsilon c_o^2(t,{\tau }_o),\hfill \\ \vartheta \prime =\frac{\vartheta }{T_0},\sigma \prime =\frac{\sigma }{\lambda +2\mu },\hfill \end{array}$$

(21)

where $c_o^2=\frac{(\lambda +2\mu )}{\rho }$ and $\epsilon =\frac{\rho C_E}{K}$ .

Hence, we have:

$$\frac{{\partial }^{4}w}{\partial x^4}+{\alpha }_1\frac{{\partial }^{2}w}{\partial t^2}+{\alpha }_2\frac{{\partial }^2\vartheta }{\partial x^2}-{\alpha }_3\frac{\partial w}{\partial t}=0,$$

(22)

$$\frac{{\partial }^2\vartheta }{\partial x^2}-{\alpha }_4\vartheta =(\frac{\partial }{\partial t}+{\tau }_o\frac{{\partial }^2}{\partial t^2})(\vartheta -{\alpha }_5\frac{{\partial }^{2}w}{\partial x^2}),$$

(23)

and

$${\sigma }_{\mathrm{xx}}=e-\alpha \theta ,$$

(24)

where ${\alpha }_1=\frac{12}{h^2},{\alpha }_2=\frac{24\beta T_o}{{\pi }^{2}h(\lambda +2\mu )},{\alpha }_3=\frac{12j_o{\mu }_o^2{H}_o^2\epsilon c_o^{2}z}{(\lambda +2\mu )b{h}^3\ell },{\alpha }_4=p^2,{\alpha }_5=\frac{\beta z}{\epsilon K\sin \left(\mathrm{pz}\right)},$ $\alpha =\frac{\beta T_0}{(\lambda +2\mu )}$ .

Problem formulation in the Laplace transform domain

The Laplace transform will be applied which is defined as follows:

$$\overline{f}(x,s)=\underset{0}{\overset{\infty }{\int }}{e}^{-st}f(x,t)\mathrm{dt},$$

(25)

where the inversion of the Laplace transform may be calculated numerically by the following iteration:

$$\begin{array}{c}\multicolumn{1}{c}{f(x,t)=L^{-1}\left[\overline{f}(x,s)\right]\approx \frac{e^{\kappa t}}{2t}\overline{f}(x,\kappa )}\\ +\frac{e^{\kappa t}}{t}\mathrm{Re}\sum _{n=1}^N{(-1)}^n\overline{f}(x,\frac{\kappa t+in\pi }{t}),\hfill \end{array}$$

(26)

where “Re” denotes the real part, while “i” defines the unit imaginary number. Numerous numerical tests have been conducted to determine if the value of “ $\kappa$ ” may meet the relation $\kappa t\approx 4.7$ .^1,2

Then, we obtain:

$$\frac{d^4\overline{w}}{d{x}^4}+({\alpha }_1{s}^2-{\alpha }_{3}s)\overline{w}+{\alpha }_2\frac{d^2\overline{\vartheta }}{d{x}^2}=0,$$

(27)

$$\frac{d^2\overline{\vartheta }}{d{x}^2}-({\alpha }_4+s+{\tau }_o{s}^2)\overline{\vartheta }=-{\alpha }_5(s+{\tau }_o{s}^2)\frac{d^2\overline{w}}{d{x}^2},$$

(28)

and

$${\overline{\sigma }}_{\mathrm{xx}}=\overline{e}-\alpha \overline{\theta }.$$

(29)

While applying the Laplace transforms, we used the following initial conditions:

$$w(x,0)=\vartheta (x,0)=\frac{\partial \vartheta (x,0)}{\partial x}=\frac{\partial w(x,0)}{\partial x}=0.$$

(30)

After re-writing the equations (27) and (28), we have the following system of ordinary differential equations:

$$[D^4+({\alpha }_1{s}^2-{\alpha }_{3}s)]\overline{w}+{\alpha }_2{D}^2\overline{\vartheta }=0,$$

(31)

and

$${\alpha }_5(s+{\tau }_o{s}^2)D^2\overline{w}+[D^2-({\alpha }_4+s+{\tau }_o{s}^2)]\overline{\vartheta }=0,$$

(32)

where $D^r=\frac{d^r}{d{x}^r}$ .

The above system gives the following characteristic equation:

$$[D^6-L{D}^4+M{D}^2-N]\{\overline{\vartheta },\overline{w}\}=0,$$

(33)

where $L={\alpha }_4+s+{\tau }_o{s}^2+{\alpha }_2{\alpha }_5(s+{\tau }_o{s}^2),M={\alpha }_1{s}^2-{\alpha }_{3}s,N=({\alpha }_1{s}^2-{\alpha }_{3}s)({\alpha }_4+s+{\tau }_o{s}^2)$ .

The general solutions of the equation (33) when the second end of the nanobeam $x=\ell$ is simply supported with zero temperature increment will take the forms:

$$\overline{\vartheta }(x,s)=\sum _{i=1}^3{A}_i\sinh \left(k_i(\ell -x)\right),$$

(34)

and

$$\overline{w}(x,s)=\sum _{i=1}^3{A}_i{w}_i\sinh \left(k_i(\ell -x)\right),$$

(35)

where $A_i\mathrm{and}{w}_i$ are some parameters to be determined, while the parameters $\pm k_1,\pm k_2\mathrm{and}\pm k_3$ denote the solutions of the following characteristic equation:

$$k^6-L{k}^4+M{k}^2-N=0.$$

(36)

To obtain $w_i$ , we will substitute from equations (34), and (35) in the equation (32), hence, we have:

$$w_i=\frac{({\alpha }_4+s+{\tau }_o{s}^2)-k_i^2}{{\alpha }_5(s+{\tau }_o{s}^2)k_i^2}.$$

(37)

Then, we obtain:

$$\overline{w}(x,s)=\sum _{i=1}^3\frac{{\alpha }_6-k_i^2}{{\alpha }_5(s+{\tau }_o{s}^2)k_i^2}{A}_i\sinh \left(k_i(\ell -x)\right),$$

(38)

$$\mathrm{where}{\alpha }_6={\alpha }_4+s+{\tau }_o{s}^2$$

To obtain $A_1,A_2,\mathrm{and}{A}_3$ , certain boundary conditions will be used, thus, we assume the end of the nanobeam $x=0$ is loaded by a thermal function $g\left(t\right)$ and is simply supported.

Hence, we have:

$$w(0,t)=\frac{{\partial }^{2}w(0,t)}{\partial x^2}=0,\vartheta (0,t)={\vartheta }_{0}g\left(t\right),$$

(39)

where ${\vartheta }_0$ is constant.

Applying the Laplace transform gives that:

$$\overline{w}{|}_{x=0}={\frac{d^2\overline{w}}{d{x}^2}|}_{x=0}=0,$$

(40)

and

$${\overline{\vartheta }|}_{x=0}={\vartheta }_0\overline{g}\left(s\right).$$

(41)

Hence, the following system of linear equations has been obtained:

$$\sum _{i=1}^3{A}_i\sinh \left(k_i\ell \right)={\vartheta }_0\overline{g}\left(s\right),$$

(42)

$$\sum _{i=1}^3\frac{{\alpha }_6-k_i^2}{k_i^2}{A}_i\sinh \left(k_i\ell \right)=0,$$

(43)

and

$$\sum _{i=1}^3({\alpha }_6-k_i^2)A_i\sinh \left(k_i\ell \right)=0.$$

(44)

After solving the above system, we get:

$A_1=\frac{{\vartheta }_0\overline{g}\left(s\right)k_1^2({\alpha }_6-k_2^2)({\alpha }_6-k_3^2)}{{\alpha }_6\sinh \left(k_1\ell \right)(k_1^2-k_2^2)(k_1^2-k_3^2)}$ , $A_2=\frac{{\vartheta }_0\overline{g}\left(s\right)k_2^2({\alpha }_6-k_1^2)({\alpha }_6-k_3^2)}{{\alpha }_6\sinh \left(k_2\ell \right)(k_2^2-k_1^2)(k_2^2-k_3^2)}$ , and $A_3=\frac{{\vartheta }_0\overline{g}\left(s\right)k_3^2({\alpha }_6-k_1^2)({\alpha }_6-k_2^2)}{{\alpha }_6\sinh \left(k_3\ell \right)(k_3^2-k_1^2)(k_3^2-k_2^2)}$ .

Thus, the solutions in the Laplace transform domain have the following forms:

The temperature increment function as:

$$\begin{array}{c}\overline{\theta }(x,z,s)=\frac{{\vartheta }_0\overline{g}\left(s\right)\sin \left(\mathrm{pz}\right)}{{\alpha }_6}\hfill \\ \left[\begin{array}{c}\frac{k_1^2({\alpha }_6-k_2^2)({\alpha }_6-k_3^2)}{\sinh \left(k_1\ell \right)(k_1^2-k_2^2)(k_1^2-k_3^2)}\sinh \left(k_1(\ell -x)\right)+\hfill \\ \multicolumn{1}{c}{\frac{k_2^2({\alpha }_6-k_1^2)({\alpha }_6-k_3^2)}{\sinh \left(k_2\ell \right)(k_2^2-k_1^2)(k_2^2-k_3^2)}\sinh \left(k_2(\ell -x)\right)+}\\ \multicolumn{1}{c}{\frac{k_3^2({\alpha }_6-k_1^2)({\alpha }_6-k_2^2)}{\sinh \left(k_3\ell \right)(k_3^2-k_1^2)(k_3^2-k_2^2)}\sinh \left(k_3(\ell -x)\right)}\end{array}\right].\hfill \end{array}$$

(45)

The lateral deflection (vibration) function as:

$$\begin{array}{c}\overline{w}(x,s)=\frac{{\vartheta }_0\overline{g}\left(s\right)({\alpha }_6-k_1^2)({\alpha }_6-k_2^2)({\alpha }_6-k_3^2)}{{\alpha }_5{\alpha }_6(s+{\tau }_o{s}^2)}\hfill \\ \left[\begin{array}{c}\frac{\sinh \left(k_1(\ell -x)\right)}{\sinh \left(k_1\ell \right)(k_1^2-k_2^2)(k_1^2-k_3^2)}+\hfill \\ \multicolumn{1}{c}{\frac{\sinh \left(k_2(\ell -x)\right)}{\sinh \left(k_2\ell \right)(k_2^2-k_1^2)(k_2^2-k_3^2)}+}\\ \multicolumn{1}{c}{\frac{\sinh \left(k_3(\ell -x)\right)}{\sinh \left(k_3\ell \right)(k_3^2-k_1^2)(k_3^2-k_2^2)}}\end{array}\right].\hfill \end{array}$$

(46)

The volumetric deformation function as:

$$\begin{array}{c}\overline{e}(x,z,s)=-\frac{z{\vartheta }_0\overline{g}\left(s\right)({\alpha }_6-k_1^2)({\alpha }_6-k_2^2)({\alpha }_6-k_3^2)}{{\alpha }_5{\alpha }_6(s+{\tau }_o{s}^2)}\hfill \\ \left[\begin{array}{c}\frac{k_1^2\sinh \left(k_1(\ell -x)\right)}{\sinh \left(k_1\ell \right)(k_1^2-k_2^2)(k_1^2-k_3^2)}+\hfill \\ \multicolumn{1}{c}{\frac{k_2^2\sinh \left(k_2(\ell -x)\right)}{\sinh \left(k_2\ell \right)(k_2^2-k_1^2)(k_2^2-k_3^2)}+}\\ \multicolumn{1}{c}{\frac{k_3^2\sinh \left(k_3(\ell -x)\right)}{\sinh \left(k_3\ell \right)(k_3^2-k_1^2)(k_3^2-k_2^2)}}\end{array}\right].\hfill \end{array}$$

(47)

Numerical results and discussion

To obtain the computational results, silicon nitride has been taken as the thermoelastic material for which we take the following values of the different physical constants^{4,6,16,21,22,27}:

$$\begin{array}{c}{\mathrm{\alpha }}_{\mathrm{T}}=2.71{\left(10\right)}^{-6}{\mathrm{K}}^{-1},\mathrm{\rho }=3200\mathrm{kg}{\mathrm{m}}^{-3},{\mathrm{T}}_0=293\mathrm{K},\hfill \\ {\mathrm{C}}_{\upsilon }=630{\mathrm{m}}^2{\mathrm{s}}^{-2}{\mathrm{K}}^{-1},\lambda =217\times {10}^9\mathrm{kg}{\mathrm{m}}^{-1}{s}^{-2},\hfill \\ \mathrm{\mu }=108\times {10}^9\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2},\mathrm{K}=43.5\mathrm{kg}\mathrm{m}{\mathrm{K}}^{-1}{\mathrm{s}}^{-3},\hfill \\ {\mathrm{\mu }}_0=4.0\mathrm{\pi }{10}^{-7}\mathrm{kg}{\mathrm{s}}^{-2}{\mathrm{A}}^{-2},{\mathrm{j}}_0={10}^{-8}{\mathrm{A}}^2{\mathrm{s}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{m}}^{-3},\hfill \\ {\mathrm{H}}_0={10}^9\mathrm{kg}{\mathrm{A}}^{-1}{\mathrm{s}}^{-2}.\hfill \end{array}$$

where $\mathrm{A}$ is the ampere unit.

The aspect ratios of the beam are fixed as $\ell /h=8$ and $b=h/2$ . For the nanoscale beam, we will take the range of the beam length $\ell$ in the nanometer scale and the original time t and the relaxation time ${\tau }_0$ in the picosecond scale, respectively. The figures were prepared by using the non-dimensional variables for beam length $\ell =1.0$ , ${\theta }_0=1.0$ , $z=h/3$ , ${\tau }_0=0.02$ , and $t=1.0$ .

Discussion

Two groups of figures show the numerical results of the problem; the first group represents the distributions of the temperature increment, vibration (lateral deflection), cubical deformation, stress, and strain-energy density with various values of the magnetic field when ( $t\ge t_0$ and $t<t_0$ ).

While the second group represents the distributions of the same functions when $t=t_0,H_0={10}^9$ .

Figures 2 and 7 represent the temperature increment with respect to x due to the various values of the magnetic field when ( $t\ge t_0$ and $t<t_0$ ) respectively. It is observed that the value of the magnetic field has a weak effect on the temperature increment, while the value of the ramp-time heat parameter has a significant effect on the temperature increment where ${\theta (0,t)|}_{t\ge t0}=0.7\mathrm{and}{\theta \left(0,t\right)|}_{t<t0}=0.6$ .

Figure 2.

The temperature increment distribution with various values of the magnetic field when $t\ge t_0$ .

Figures 3 and 8 represent the vibration (lateral deflection) with respect to x due to the various values of the magnetic field when ( $t\ge t_0$ and $t<t_0$ ) respectively. It is observed that an increased magnetic value results in an increase in the absolute value of lateral deflection (vibration). Moreover, the ramp-time heat parameter has a significant effect on the absolute value of lateral deflection (vibration).

Figure 3.

The lateral deflection distribution with various values of the magnetic field when $t\ge t_0$ .

Figures 4 and 9 represent cubical deformation (strain) with respect to x due to the various values of the magnetic field when ( $t\ge t_0$ and $t<t_0$ ) respectively. It is observed that an increased magnetic value results in an increase in the strain’s absolute value (cubical deformation). Also, the ramp-time heat parameter has a significant effect on the absolute value of strain.

Figure 4.

The deformation distribution with various values of the magnetic field when $t\ge t_0$ .

Figures 5 and 10 represent stress with respect to x due to the various values of the magnetic field when ( $t\ge t_0$ and $t<t_0$ ) respectively. It is observed that an increased magnetic value results in an increase in the stress’s absolute value. In addition, the ramp-time heat parameter has a significant effect on the absolute value of stress.

Figure 5.

The stress distribution with various values of the magnetic field when $t\ge t_0$ .

Figures 6 and 11 represent strain energy density with respect to x due to the various values of the magnetic field when ( $t\ge t_0$ and $t<t_0$ ) respectively. It is observed that an increase in magnetic value results in an increase in the strain energy density’s value. Moreover, the ramp-time heat parameter has a significant effect on the strain energy density’s value.

Figure 6.

The strain-energy density distribution with various values of the magnetic field when $t\ge t_0$ .

Figure 7.

The temperature increment distribution with various values of the magnetic field when $t<t_0$ .

Figure 8.

The lateral deflection distribution with various values of the magnetic field when $t<t_0$ .

Figure 9.

The deformation distribution with various values of the magnetic field when $t<t_0$ .

Figure 10.

The stress distribution with various values of the magnetic field when $t<t_0$ .

Figure 11.

The strain-energy density distribution with various values of the magnetic field when $t<t_0$ .

Figure 12 represents the temperature increment with respect to x due to the various values of z when $t=t_0,H_0={10}^9$ . It is observed that a decrease in z values results in a decrease in the temperature increment.

Figure 12.

The temperature increment distribution with various values of z when $t=t_0,H_0={10}^9$ .

Figure 13, represent the vibration (lateral deflection) concerning × due to the various values of z when $t=t_0,H_0={10}^9$ . It is observed that a decrease in z values results in a decrease in the vibration’s absolute value.

Figure 13.

The lateral deflection distribution with various values of z when $t=t_0,H_0={10}^9$ .

Figure 14, represent strain with respect to x due to the various values of z when $t=t_0,H_0={10}^9$ . It is observed that a decrease in z values results in a decrease in the strain’s absolute value.

Figure 14.

The deformation distribution with various values of z when $t=t_0,H_0={10}^9$ .

Figure 15, represent stress with respect to x due to the various values of z when $t=t_0,H_0={10}^9$ . It is observed that a decrease in z values results in a decrease in the stress’s absolute value.

Figure 15.

The stress distribution with various values of z when $t=t_0,H_0={10}^9$ .

Figure 16, represent strain energy density function concerning x due to the various values of z when $t=t_0,H_0={10}^9$ . It is observed that a decrease in z values results in a decrease in the strain energy density’s value.

Figure 16.

The strain-energy density distribution with various values of z when $t=t_0,H_0={10}^9$ .

Conclusion

The distributions of the temperature increment, vibration (lateral deflection), cubical deformation, stress, and strain-energy density are significantly affected by the ramp-time heat parameter and the magnitude of the magnetic field, and these parameters might be utilized as tuners to regulate the energy produced by the nanobeam.

The effect of the magnetic field is very limited on the temperature increment. While increasing the value of the magnetic field causes an increase in the values of the absolute vibration (lateral deflection), cubical deformation, stress, and strain-energy density functions for all the values of the ramp-time heat parameter.

Moreover, the ramp-time heat parameter and the thickness of the beam have significant impacts on the temperature increment, vibration (lateral deflection), cubical deformation, stress, and strain-energy density functions.