Vibration of Orthotropic Rectangular Plates Under the Action of Moving Distributed Masses and Restin

International Journal of Advanced Engineering Research and Science (IJAERS) Peer-Reviewed Journal ISSN: 2349-6495(P) | 2456-1908(O) Vol-8, Issue-6; Jun, 2021 Journal Home Page Available: https://ijaers.com/ Article DOI:https://dx.doi.org/10.22161/ijaers.86.4

Vibration of Orthotropic Rectangular Plates Under the Action of Moving Distributed Masses and Resting on a Variable Elastic Pasternak Foundation with Clamped End Conditions Awodola T.O, Adeoye A.S Department of Mathematical Sciences, Federal University of Technology, Akure, Nigeria

Received: 15 Mar 2021; Received in revised form: 22 Apr 2021; Accepted: 25 May 2021; Available online: 11 Jun 2021 ©2021 The Author(s). Published by AI Publication.

This is an open access article under the CC BY license (https://creativecommons.org/licenses/by/4.0/).

Keywords— Variable bi-parametric foundation, orthotropic, foundation modulus, critical speed, flexural rigidity, shear modulus resonance, modified frequency, clamped end conditions.

I.

Abstract— This work investigates the vibration of orthotropic rectangular plate resting on a variable elastic Pasternak foundation under the action of moving distributed masses. The governing equation is a fourth order partial differential equation with variable and singular co-efficients. The solutions to the problem are obtained by transforming the fourth order partial differential equation for the problem to a set of coupled second order ordinary differential equations using the technique of Shadnam et al[11] which are then simplified using modified asymptotic method of Struble. The closed form solution is analyzed, resonance conditions are obtained and the results are presented in plotted curves for both cases of moving distributed mass and moving distributed force.

INTRODUCTION

The problems related to thin structural bodies (rods, beams, plates,and shells) with other bodies have widespread application in various fields of science and technology. The physical phenomena involved in the impact event include structural responses, contact effects and wave propagation.

The problems associated with these are always topical issues in the field of applied mechanics. Since these problems belong to the problems related to dynamic contact interaction, their solution is connected with cumbersome mathematical tasks. To this end , several researchers had worked and some are still working on the dynamic behavior of orthotropic rectangular plates. Ambartsumian [1] examined the five fundamental differential equations describing the equilibrium of an orthotropic plate with a cylindrical anisotropy for the case when all radial planes

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crossing the axis of anisotropy are the planes of elastic symmetry. Sveklo [2] suggested the contact theory for two anisotropic bodies under compression according to which the contact pressure is distributed over an elliptical contact region. The same structural effects are also true of the concrete slab in a composite girder bridge, but the steel orthotropic deck is considerably lighter, and therefore allows longer span bridges to be more efficiently designed.

Awodola [3] studied the effect of plate parameters on the vibrations under moving masses of elastically supported plate resting on bi-parametric foundation with stiffness variation.Szekrenyes [4] investigated the interface fracture in orthotropic composite plates using second order shear deformation theory. Yan [5] proposed elastic orthotropic models and used these in the nonlinear analysis of concrete structures subjected to monotonic or pseudo dynamic

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International Journal of Advanced Engineering Research and Science, 8(6)-2021

loading. Since these models can appropriately describe the strain softening behavior of concrete beyond the peak stress and show good agreement with the strength envelope obtained from experimental results Hu and Yao [6] studied the vibration solutions of rectangular orthotropic plates by symplectic geometry method. In the same vien, Alshaya, Hunt and Rowlands [7] examined stresses and strains in thick perforated orthotropic plates. Gbadeyan and Dada [8] found the natural frequency of rectangular plates traversed by moving concentrated masses. Awodola and Adeoye [9] investigated the behavior of simply supported orthotropic rectangular plate by applying the technique of variable separable. Adeoye and Awodola [10] studied the dynamic behavior of orthotropic rectangular plate with clamped-clamped boundary conditions by making use of the technique of Shadnam Due to inability of researchers to ∂4

𝐷𝑥

∂𝑥 4

∂4

[

∂𝑥 2 ∂𝑡 2

𝑊(𝑥, 𝑦, 𝑡) +

12𝑥 + 3𝑥 2 ) − ∑𝑁 𝑟=1 ∂2 ∂𝑥 2

∂4

𝑊(𝑥, 𝑦, 𝑡) + 2𝐵

∂ ∂𝑥

∂𝑦 2 ∂𝑡 2

II.

GOVERNING EQUATION

The dynamic transverse displacement 𝑊(𝑥, 𝑦, 𝑡) of orthotropic rectangular plates when it is resting on a bi-parametric elastic foundation and traversed by distributed mass 𝑀𝑟 moving with constant velocity 𝑐𝑟 along a straight line parallel to the x-axis issuing from point y = s on the y-axis with flexural rigidities 𝐷𝑥 and 𝐷𝑦 is governed by the fourth order partial differential equation given as ∂4

𝑊(𝑥, 𝑦, 𝑡) + 𝐷𝑦

∂𝑥 2 ∂𝑦 2 ∂4

solve orthotropic plates problems by analytical methods, this work aims at solving the governing equation by analytical solution and also considers the effect of the flexural rigidities in both x and y directions.

∂𝑦 4

∂2

𝑊(𝑥, 𝑦, 𝑡) + 𝜇

∂𝑡 2

𝑊(𝑥, 𝑦, 𝑡) − 𝜌ℎ𝑅0

𝑊(𝑥, 𝑦, 𝑡)] + 𝐾0 (4𝑥 − 3𝑥 2 + 𝑥 3 )𝑊(𝑥, 𝑦, 𝑡) + 𝑆0 (−13 +

𝑊(𝑥, 𝑦, 𝑡) − 𝑆0 (12 − 13𝑥 + 6𝑥 2 + 𝑥 3 )[

∂2

∂𝑥 2

∂2

[𝑀𝑟 𝑔𝐻(𝑥 − 𝑐𝑡)𝐻(𝑦 − 𝑠) − 𝑀𝑟 (

∂𝑡 2

𝑊(𝑥, 𝑦, 𝑡) +

𝑊(𝑥, 𝑦, 𝑡) + 2𝑐𝑟

∂2 ∂𝑥 ∂𝑡

∂2 ∂𝑦 2

𝑊(𝑥, 𝑦, 𝑡)]

𝑊(𝑥, 𝑦, 𝑡) +

(2.1)

𝑐𝑟2

𝑊(𝑥, 𝑦, 𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠)] = 0

where 𝐷𝑥 and 𝐷𝑦 are the flexural rigidities of the plate along x and y axes respectively. 𝐸𝑥 ℎ3

𝐷𝑥 =

12(1−𝜈𝑥 𝜈𝑦 )

, 𝐷𝑦 =

𝐸𝑦 ℎ3 12(1−𝜈𝑥 𝜈𝑦 )

, 𝐵 = 𝐷𝑥 𝐷𝑦 +

𝐺𝑜 ℎ3

(2.2)

6

𝐸𝑥 and 𝐸𝑦 are the Young's moduli along x and y axes respectively, 𝐺𝑜 is the rigidity modulus, 𝜈𝑥 and 𝜈𝑦 are Poisson's ratios for the material such that 𝐸𝑥 𝜈𝑦 = 𝐸𝑦 𝜈𝑥 , 𝜌 is the mass density per unit volume of the plate, h is the plate thickness, t is the time, x and y are the spatial coordinates in x and y directions respectively, 𝑅𝑜 is the rotatory inertia correction factor, 𝐾𝑜 is the foundation constant, 𝑆𝑜 shear modulus and g is the acceleration due to gravity, H(.) is the Heaviside function.

Rewriting equation (2.1), one obtains 𝜇

∂2 ∂𝑡 2

∂4

𝑊(𝑥, 𝑦, 𝑡) + 𝜇𝜔𝑛2 𝑊(𝑥, 𝑦, 𝑡) = 𝜌ℎ𝑅0 [

𝑊(𝑥, 𝑦, 𝑡) +

∂𝑥 2 ∂𝑡 2

∂4 ∂𝑥 2 ∂𝑦 2

𝑊(𝑥, 𝑦, 𝑡) − 𝐷𝑦

∂4 ∂𝑦 4

𝑊(𝑥, 𝑦, 𝑡) − 𝐷𝑥

∂4 ∂𝑥 4

∂𝑥 2

𝑊(𝑥, 𝑦, 𝑡) +

+2𝑐𝑟

∂2 ∂𝑥 ∂𝑡

∂2 ∂𝑦 2

𝑊(𝑥, 𝑦, 𝑡)] − 2𝐵

𝑊(𝑥, 𝑦, 𝑡) + 𝜇𝜔𝑛2 𝑊(𝑥, 𝑦, 𝑡) − 𝐾0 (4𝑥 −

3𝑥 2 + 𝑥 3 )𝑊(𝑥, 𝑦, 𝑡) + 𝐺𝑜 (−13 + 12𝑥 + 3𝑥 2 ) ∂2

∂4 ∂𝑦 2 ∂𝑡 2

∂ ∂𝑥

𝑊(𝑥, 𝑦, 𝑡) − 𝐺0 (12 − 13𝑥 + 6𝑥 2 + 𝑥 3 )[

𝑊(𝑥, 𝑦, 𝑡)] + ∑𝑁 𝑟=1 [𝑀𝑟 𝑔𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠) − 𝑀𝑟 (

∂2

∂𝑡 2

𝑊(𝑥, 𝑦, 𝑡) + 𝑐𝑟2

∂2 ∂𝑥 2

(2.3)

𝑊(𝑥, 𝑦, 𝑡)

𝑊(𝑥, 𝑦, 𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠)]

Simplifying equation (2.3) further, one obtains ∂2 ∂𝑡 2

𝑊(𝑥, 𝑦, 𝑡) + 𝜔𝑛2 𝑊(𝑥, 𝑦, 𝑡) = ∑𝑁 𝑟=1 [𝑅0 ( ∂4

∂𝑥 2 ∂𝑦 2

𝑊(𝑥, 𝑦, 𝑡) −

+𝑥 3 )𝑊(𝑥, 𝑦, 𝑡) + 𝑊(𝑥, 𝑦, 𝑡) + ∂2 ∂𝑥 ∂𝑡

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∂4

∂𝑥 2 ∂𝑡 2

∂2 ∂𝑦 2

𝐷𝑦 ∂4

𝐺𝑜 𝜇

𝜇 ∂𝑦 4

𝑊(𝑥, 𝑦, 𝑡) −

𝜇 ∂𝑥 4

(−13 + 12𝑥 + 3𝑥 2 ) 𝑀𝑟

𝑊(𝑥, 𝑦, 𝑡)) + ∑𝑁 𝑟=1 (

𝑊(𝑥, 𝑦, 𝑡) + 𝑐𝑟2

𝐷𝑥 ∂4

∂2 ∂𝑥 2

𝜇

∂ ∂𝑥

𝑊(𝑥, 𝑦, 𝑡) +

∂4 ∂𝑦 2 ∂𝑡 2

𝑊(𝑥, 𝑦, 𝑡)) −

𝑊(𝑥, 𝑦, 𝑡) + 𝜔𝑛2 𝑊(𝑥, 𝑦, 𝑡) −

𝑊(𝑥, 𝑦, 𝑡) −

𝐺0 𝜇

𝐾0 𝜇

2𝐵 𝜇

(4𝑥 − 3𝑥 2

(12 − 13𝑥 + 6𝑥 2 + 𝑥 3 )(

𝑔𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠) −

𝑀𝑟 𝜇

(

∂2 ∂𝑡 2

∂2 ∂𝑥 2

(2.4)

𝑊(𝑥, 𝑦, 𝑡) + 2𝑐𝑟

𝑊(𝑥, 𝑦, 𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠))]

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where 𝜔𝑛2 is the natural frequencies, 𝑛 = 1,2,3, . ..

The initial conditions, without any loss of generality, is taken as 𝑊(𝑥, 𝑦, 𝑡) = 0 =

III.

∂ ∂𝑡

𝑊(𝑥, 𝑦, 𝑡)

(2.5)

ANALYTICAL APPROXIMATE SOLUTION

In order to solve equation (2.4), one applies technique of Shadnam et al which requires that the deflection of the plates be in series form as 𝑊(𝑥, 𝑦, 𝑡) = ∑𝑁 𝑛=1 Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡)

(3.1)

where Ψ𝑛 (𝑥, 𝑦) = Ψ𝑗𝑚 (𝑥)Ψℎ𝑚 (𝑦) Ψ𝑗𝑚 𝑥 = sin𝜁𝑗𝑚 𝑥 + 𝐴𝑗𝑚 cos𝜁𝑗𝑚 𝑥 + 𝐵𝑗𝑚 sinh𝜁𝑗𝑚 𝑥 + 𝐶𝑗𝑚 cosh𝜁𝑗𝑚 𝑥 Ψℎ𝑚 𝑦 = sin𝜑ℎ𝑚 𝑦 + 𝐴ℎ𝑚 cos𝜑ℎ𝑚 𝑦 + 𝐵ℎ𝑚 sinh𝜑ℎ𝑚 𝑦 + 𝐶ℎ𝑚 cosh𝜑ℎ𝑚 𝑦

𝜁𝑗𝑚 =

𝜙𝑗𝑚

, 𝜑ℎ𝑚 =

𝐿𝑥

𝜙ℎ𝑚 𝐿𝑦

The right hand side of equation (2.4), taken into account equation (3.1), written in the form of series takes the form ∑∞ 𝑛=1 [𝑅0 (

∂2 ∂𝑥 2

Ψ𝑛 (𝑥, 𝑦)𝑄̈𝑛 (𝑡) +

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) − 𝑄𝑛 (𝑡) + 𝑄𝑛 (𝑡) + 2𝑐

∂ ∂𝑥

𝐺𝑜 𝜇

𝐷𝑥 ∂4 𝜇 ∂𝑥 4

∂4

∂𝑦 2

∂ ∂𝑥

Ψ𝑛 (𝑥, 𝑦)𝑄̇𝑛 (𝑡) + 𝑐𝑟2

∂𝑥 2

𝜔𝑛2 Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡)

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) −

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡)) + ∑𝑁 𝑟=1 ( ∂2

∂4

𝜇 ∂𝑥 2 ∂𝑦 2

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) +

(−13 + 12𝑥 + 3𝑥 2 )

∂2

2𝐵 Ψ𝑛 (𝑥, 𝑦)𝑄̈𝑛 (𝑡)) −

∂𝑦 2

𝑀𝑟 𝜇

𝐺0 𝜇

−

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) −

𝐾0 𝜇

2

𝐷𝑦 ∂4 𝜇 ∂𝑦 4

3

(4𝑥 − 3𝑥 + 𝑥 )Ψ𝑛 (𝑥, 𝑦)

(12 − 13𝑥 + 6𝑥 2 + 𝑥 3 )(

𝑔𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠) −

𝑀𝑟 𝜇

∂2 ∂𝑥 2

Ψ𝑛 (𝑥, 𝑦)

(3.2)

(Ψ𝑛 (𝑥, 𝑦)𝑄̈𝑛 (𝑡) +

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠))] = ∑𝑁 𝑛=1 Ψ𝑛 (𝑥, 𝑦)Θ𝑛 (𝑡)

Multiplying both sides of equation (3.2) by Ψ𝑚 (𝑥, 𝑦) and integrating on area A of the plate and considering the orthogonality of Ψ𝑛 (𝑥, 𝑦), one obtains Θ𝑛 (𝑡) = 𝑄𝑛 (𝑡) −

1 𝜃∗

∂2

𝐷𝑦 ∂4 𝜇 ∂𝑦 4

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) −

3𝑥 2 + 𝑥 3 )Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) + +6𝑥 2 + 𝑥 3 )(

∂2

∂𝑥 2

−

𝑀𝑟 𝜇

∂4

̈ ̈ ∑∞ 𝑛=1 ∫𝐴 [𝑅0 (∂𝑥 2 Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) + ∂𝑦 2 Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡)) − 𝐺𝑜 𝜇

𝐷𝑥 ∂4 𝜇 ∂𝑥 4

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) +

(Ψ𝑛 (𝑥, 𝑦)𝑄̈𝑛 (𝑡) + 2𝑐

∂ ∂𝑥

∂𝑦 2

𝜇 ∂𝑥 2 ∂𝑦 2

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) + 𝜔𝑛2 Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) −

(−13 + 12𝑥 + 3𝑥 2 ) ∂2

∂4

2𝐵

∂ ∂𝑥

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡) −

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡)) + ∑𝑁 𝑟=1 (

Ψ𝑛 (𝑥, 𝑦)𝑄̇𝑛 (𝑡) + 𝑐𝑟2

∂2 ∂𝑥 2

𝑀𝑟 𝜇

𝐺0 𝜇

Ψ𝑛 (𝑥, 𝑦)

𝐾0 𝜇

(4𝑥 −

(12 − 13𝑥

𝑔𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠)

(3.3)

Ψ𝑛 (𝑥, 𝑦)𝑄𝑛 (𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠)

)]Ψ𝑚 (𝑥, 𝑦)𝑑𝐴 and zero when 𝑛 ≠ 𝑚 where 𝜃 ∗ = ∫𝐴 Ψ𝑛2 (𝑥, 𝑦)𝑑𝐴

(3.4)

Making use of equation (3.3) and taking into account equation (3.2), equation (2.4) can be written as

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International Journal of Advanced Engineering Research and Science, 8(6)-2021

Ψ𝑛 (𝑥,𝑦)

Ψ𝑛 (𝑥, 𝑦)[𝜔𝑛2 𝑄𝑛 (𝑡) + 𝑄̈𝑛 (𝑡)] =

2𝐵 ∂2 Ψ𝑞 (𝑥,𝑦)

Ψ𝑚 (𝑥, 𝑦)𝑄̈𝑞 (𝑡)) − Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) − Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) −

∂Φ𝑞 (𝑥,𝑦) ∂𝑥

∂𝑥 2 ∂𝑦 2

𝜇 𝐾0

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) −

∂2 Ψ𝑞 (𝑥,𝑦) ∂𝑥 2

Ψ𝑚 (𝑥, 𝑦)𝑄̈𝑞 (𝑡) +

𝐷𝑦 ∂4 Ψ𝑞 (𝑥,𝑦) ∂𝑦 4

𝜇

∂2 Ψ𝑞 (𝑥,𝑦)

𝐺𝑜

(12 − 13𝑥 + 6𝑥 2 + 𝑥 3 )(

𝜇

𝜇

∑∞ 𝑞=1 ∫𝐴 [𝑅0 (

∂𝑥 2

𝑔Ψ𝑚 (𝑥, 𝑦)𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠) −

Φ𝑚 (𝑥, 𝑦)𝑄̇𝑞 (𝑡) + 𝑐𝑟2

∂2 Φ𝑞 (𝑥,𝑦) ∂𝑥 2

𝐺𝑜 𝜇

𝜇

∂𝑦 2 𝐷𝑥 ∂4 Ψ𝑞 (𝑥,𝑦) ∂𝑥 4

𝜇

(−13 + 12𝑥 + 3𝑥 2 )

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) +

𝑀𝑟

∂2 Ψ𝑞 (𝑥,𝑦)

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) +

(4𝑥 − 3𝑥 2 + 𝑥 3 )Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) +

𝜇

𝑀𝑟

𝑄𝑞 (𝑡)) + ∑𝑁 𝑟=1 (

𝜃∗

∂2 Ψ𝑞 (𝑥,𝑦) ∂𝑦 2

∂Ψ𝑞 (𝑥,𝑦) ∂𝑥

(3.5)

Ψ𝑚 (𝑥, 𝑦)

(Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑄̈𝑞 (𝑡) + 2𝑐𝑟

Φ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠))]𝑑𝐴

On further simplification of equation (3.5), one obtains 𝑄̈𝑛 (𝑡) + 𝜔𝑛2 𝑄𝑛 (𝑡) = −

2𝐵 ∂2 Ψ𝑞 (𝑥,𝑦) 𝜇

∂𝑥 2 ∂𝑦 2

1 𝜃∗

∂2 Ψ𝑞 (𝑥,𝑦)

∑∞ 𝑞=1 ∫𝐴 [𝑅0 (

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) −

∂𝑥 2

𝐷𝑦 ∂4 Ψ𝑞 (𝑥,𝑦) ∂𝑦 4 𝐾0

𝜇

𝑄𝑞 (𝑡) + 𝜔𝑞2 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑄𝑛 (𝑡) − +12𝑥 + 3𝑥 2 ) +

∂2 Ψ𝑞 (𝑥,𝑦) ∂𝑦 2

∂Ψ𝑞 (𝑥,𝑦) ∂𝑥

𝜇

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡)) + ∑𝑁 𝑟=1 (

𝑀𝑟 𝜇

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) −

𝜇

∂𝑦 2

Ψ𝑚 (𝑥, 𝑦)𝑄̈𝑞 (𝑡))

𝐷𝑥 ∂4 Ψ𝑞 (𝑥,𝑦) 𝜇

∂𝑥 4

(12 − 13𝑥 + 6𝑥 2 + 𝑥 3 )(

Ψ𝑚 (𝑥, 𝑦)

∂2 Ψ𝑞 (𝑥,𝑦) ∂𝑥 2

𝑔Ψ𝑚 (𝑥, 𝑦)𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠) − 2 2 ∂ Φ𝑞 (𝑥,𝑦) 𝑟 ∂𝑥 2

∂Φ (𝑥,𝑦) 𝑄̈𝑞 (𝑡) + 2𝑐𝑟 𝑞 Φ𝑚 (𝑥, 𝑦)𝑄̇𝑞 (𝑡) + 𝑐 ∂𝑥

∂2 Ψ𝑞 (𝑥,𝑦)

(4𝑥 − 3𝑥 2 + 𝑥 3 )Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) +

𝐺𝑜

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) −

Ψ𝑚 (𝑥, 𝑦)𝑄̈𝑞 (𝑡) +

𝑀𝑟 𝜇

𝐺𝑜 𝜇

(−13 (3.6)

Ψ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡) (Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)

Φ𝑚 (𝑥, 𝑦)𝑄𝑞 (𝑡))𝐻(𝑥 − 𝑐𝑟 𝑡)𝐻(𝑦 − 𝑠))]𝑑𝐴

The system of equations in equation (3.6) is a set of coupled ordinary differential equations where 𝐻(𝑥 − 𝑐𝑟 𝑡) and 𝐻(𝑦 − 𝑠) are the Heaviside functions which are defined as 1,𝑓𝑜𝑟 𝑥≥𝑐𝑟 𝑡 𝑥<𝑐𝑟 𝑡

𝐻(𝑥 − 𝑐𝑟 𝑡) = {0,𝑓𝑜𝑟

1,𝑓𝑜𝑟 𝑦≥𝑠 𝑦<𝑠

,

𝐻(𝑦 − 𝑠) = {0,𝑓𝑜𝑟

(3.7)

With the properties (𝑖)

𝑑 𝑑𝑥

𝑑

[𝐻(𝑥 − 𝑐𝑟 𝑡)] = 𝛿(𝑥 − 𝑐𝑟 𝑡),

𝑓(𝑥),𝑓𝑜𝑟 𝑥≥𝑐𝑟 𝑡 𝑥<𝑐𝑟 𝑡

(𝑖𝑖)𝑓(𝑥)𝐻(𝑥 − 𝑐𝑟 𝑡) = {0,𝑓𝑜𝑟

𝑑𝑦

,

[𝐻(𝑦 − 𝑠)] = 𝛿(𝑦 − 𝑠)

(3.8)

𝑓(𝑦),𝑓𝑜𝑟 𝑦≥𝑠 𝑦<𝑠

𝑓(𝑦)𝐻(𝑦 − 𝑠) = {0,𝑓𝑜𝑟

(3.9)

Using the Fourier series representation, the Heaviside functions take the form 1

1

sin(2𝑛+1)𝜋(𝑥−𝑐𝑟𝑡)

4

𝜋

2𝑛+1

𝐻(𝑥 − 𝑐𝑟 𝑡) = + ∑𝑁 𝑟=1

1

1

sin(2𝑛+1)𝜋(𝑦−𝑠)

4

𝜋

2𝑛+1

𝐻(𝑦 − 𝑠) = + ∑𝑁 𝑟=1

,0 < 𝑥 < 1

(3.10)

,0 < 𝑦 < 1

(3.11)

On putting equations (3.7) to (3.11) into equation (3.6) and simplifying, one obtains 𝑄̈𝑛 (𝑡) + 𝜔𝑛2 𝑄𝑛 (𝑡) −

1 𝜃∗

𝐷𝑦 2𝐵 𝐷 ̈ ∑∞ 𝑇2 𝑄𝑞 (𝑡) − 𝑇3 𝑄𝑞 (𝑡) − 𝑥 𝑇4 𝑄𝑞 (𝑡) + (𝜔𝑞2 𝐹4∗ − 𝑞=1 [𝑅0 𝑇1 𝑄𝑞 (𝑡) −

𝜇 𝜇 𝜇 𝐺0 𝑀𝑟 1 cos(2𝑗+1)𝜋𝑐𝑟 𝑡 ∞ ∗ 𝑁 ∗ 𝐹 )𝑄𝑞 (𝑡) + (𝑇6 + 𝑇7 )𝑄𝑞 (𝑡) − ∑𝑟=1 ((𝑇8 + 2 (∑𝑗=1 𝐸1 − ∑∞ 𝑗=1 𝜇 5 𝜇 𝜇 𝜋 2𝑗+1 sin(2𝑗+1)𝜋𝑐𝑟 𝑡 1 ∗ cos(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ cos(2𝑘+1)𝜋𝑠 ∗ sin(2𝑘+1)𝜋𝑠 )(∑∞ − ∑∞ ) + (∑∞ 𝑘=1 𝐸3 𝑘=1 𝐸4 𝑗=1 𝐸5 2𝑗+1 2𝑘+1 2𝑘+1 4𝜋 2𝑗+1 𝐾0

1 ∗ cos(2𝑘+1)𝜋𝑠 ∗ sin(2𝑘+1)𝜋𝑠 (∑∞ − ∑∞ ))𝑄̈𝑞 (𝑡) + 𝑘=1 𝐸7 𝑘=1 𝐸8 4𝜋 2𝑘+1 2𝑘+1 1 ∗ cos(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ cos(2𝑘+1)𝜋𝑠 (∑∞ − ∑∞ )(∑∞ 𝑗=1 𝐸9 𝑗=1 𝐸10 𝑘=1 𝐸11 𝜋2 2𝑗+1 2𝑗+1 2𝑘+1

∗ − ∑∞ 𝑗=1 𝐸6

2𝑐𝑟 (𝑇9 +

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sin(2𝑗+1)𝜋𝑐𝑟 𝑡

𝐸2∗

2𝑗+1

)+

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International Journal of Advanced Engineering Research and Science, 8(6)-2021 ∗ sin(2𝑗+1)𝜋𝑐𝑟 𝑡 − ∑∞ ) 𝑗=1 𝐸14 2𝑗+1 2𝑗+1 sin(2𝑘+1)𝜋𝑠 1 ∗ ∗ ∑∞ ))𝑄̇𝑞 (𝑡) + 𝑐𝑟2 (𝑇10 + 2 (∑∞ 𝑘=1 𝐸16 𝑗=1 𝐸17 2𝑘+1 𝜋 cos(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ cos(2𝑘+1)𝜋𝑠 ∗ sin(2𝑘+1)𝜋𝑠 − ∑∞ )(∑∞ − ∑∞ 𝑗=1 𝐸18 𝑘=1 𝐸19 𝑘=1 𝐸20 2𝑗+1 2𝑗+1 2𝑘+1 2𝑘+1 1 1 ∞ ∞ ∞ ∗ cos(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑟 𝑡 ∗ cos(2𝑘+1)𝜋𝑠 ∑ ) + (∑𝑗=1 𝐸21 − 𝑗=1 𝐸22 ) + (∑𝑘=1 𝐸23 4𝜋 2𝑗+1 2𝑗+1 4𝜋 2𝑘+1 ∗ − ∑∞ 𝑘=1 𝐸12

sin(2𝑘+1)𝜋𝑠

1

)+

2𝑘+1 4𝜋 1 ∞ ∗ cos(2𝑘+1)𝜋𝑠 + (∑𝑘=1 𝐸15 − 4𝜋 2𝑘+1

∗ − ∑∞ 𝑘=1 𝐸24

sin(2𝑘+1)𝜋𝑠 2𝑘+1

∗ (∑∞ 𝑗=1 𝐸13

cos(2𝑗+1)𝜋𝑐𝑟 𝑡

𝑁 ))𝑄𝑞 (𝑡))] = ∑∞ 𝑞=1 ∑𝑟=1

𝑀𝑟 𝑔 𝜇𝜃 ∗

(3.12)

Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

which is the transformed equation governing the problem of an orthotropic rectangular plate resting on bi-parametric elastic foundation. where 𝑇1 = ∫𝐴 [

∂2

∂𝑥 2

Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦) +

𝑇2 = ∫𝐴

∂2

[

∂2

∂𝑥 2 ∂𝑥 2

∂4

𝑇3 = ∫𝐴

∂𝑦 4

𝑇4 = ∫𝐴

∂𝑥 4

∂4

∂2 ∂𝑦 2

Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)]𝑑𝐴

Ψ𝑞 (𝑥, 𝑦)]Ψ𝑚 (𝑥, 𝑦)𝑑𝐴

(3.14)

[Ψ𝑞 (𝑥, 𝑦)]Ψ𝑚 (𝑥, 𝑦)𝑑𝐴

(3.15)

[Ψ𝑞 (𝑥, 𝑦)]Ψ𝑚 (𝑥, 𝑦)𝑑𝐴

(3.16)

𝐹4∗ = ∫𝐴 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑑𝐴 𝑇5 = 4𝑈1 − 3𝑈2 + 𝑈3 ,

(3.17)

𝑇6 = −13𝐴1 + 12𝐴2 + 3𝐴3

(3.18)

𝑇7 = 12𝑓1 − 13𝑓2 + 6𝑓3 + 𝑓4 + 12𝑓5 − 13𝑓6 + 6𝑓7 + 𝑓8

(3.19)

𝑇8 =

1 ∫ 16 𝐴

Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑑𝐴

(3.20)

𝐸1∗ = ∫𝐴 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)sin(2𝑗 + 1)𝜋𝑥𝑑𝐴

(3.21)

𝐸2∗ = ∫𝐴 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)cos(2𝑗 + 1)𝜋𝑥𝑑𝐴

(3.22)

𝐸3∗ = ∫𝐴 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)sin(2𝑘 + 1)𝜋𝑦𝑑𝐴

(3.23)

𝐸4∗ = ∫𝐴 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)cos(2𝑘 + 1)𝜋𝑦𝑑𝐴

(3.24)

𝐸5∗ = 𝐸1∗ , 𝐸6∗ = 𝐸2∗ , 𝐸7∗ = 𝐸3∗ , 𝐸8∗ = 𝐸4∗

(3.25)

𝑇9 =

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(3.13)

1 ∂ Ψ (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)𝑑𝐴 ∫ 16 𝐴 ∂𝑥 𝑞

(3.26)

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021

∂

𝐸9∗ = ∫𝐴

∂𝑥

∗ 𝐸10 = ∫𝐴

∂𝑥

∗ 𝐸11 = ∫𝐴

∂𝑥

(Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)sin(2𝑗 + 1)𝜋𝑥𝑑𝐴

(3.27)

(Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)cos(2𝑗 + 1)𝜋𝑥𝑑𝐴

(3.28)

(Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)sin(2𝑘 + 1)𝜋𝑦𝑑𝐴

(3.29)

Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)cos(2𝑘 + 1)𝜋𝑦𝑑𝐴

(3.30)

∗ ∗ ∗ ∗ ∗ ∗ ∗ 𝐸13 = 𝐸9∗ , 𝐸14 = 𝐸10 , 𝐸15 = 𝐸11 , 𝐸16 = 𝐸12

(3.31)

∂

∂

∗ 𝐸12 = ∫𝐴

∂ ∂𝑥

𝑇10 = ∂2

∗ 𝐸17 = ∫𝐴

∂𝑥 2

∗ 𝐸18 = ∫𝐴

∂𝑥 2

∂2

1 ∂2 (Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)𝑑𝐴 ∫ 16 𝐴 ∂𝑥 2

(3.32)

(Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)sin(2𝑗 + 1)𝜋𝑥𝑑𝐴

(3.33)

(Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)cos(2𝑗 + 1)𝜋𝑥𝑑𝐴

(3.34)

∗ 𝐸19 = ∫𝐴 Ψ𝑞 (𝑥, 𝑦)Ψ𝑚 (𝑥, 𝑦)sin(2𝑘 + 1)𝜋𝑦𝑑𝐴

∗ 𝐸20 = ∫𝐴

∂2 ∂𝑥 2

(3.35)

(Ψ𝑞 (𝑥, 𝑦))Ψ𝑚 (𝑥, 𝑦)cos(2𝑘 + 1)𝜋𝑦𝑑𝐴

(3.36)

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 𝐸21 = 𝐸17 , 𝐸22 = 𝐸18 , 𝐸23 = 𝐸19 , 𝐸24 = 𝐸20

(3.37)

Ψ𝑚 (𝑥, 𝑦) is assumed to be the products of functions Ψ𝑝𝑚 (𝑥)Ψ𝑏𝑚 (𝑦) which are the beam functions in the directions of x and y axes respectively. That is Ψ𝑚 (𝑥, 𝑦) = Ψ𝑝𝑚 (𝑥)Ψ𝑏𝑚 (𝑦)

(3.38)

where Φ𝑚 (𝑥) = sin

Φ𝑚 (𝑦) = sin

Γ𝑚 𝑥 𝐿𝑥

Γ𝑚 𝑦 𝐿𝑦

+ 𝐴𝑚 cos

+ 𝐴𝑚 cos

Γ𝑚 𝑥 𝐿𝑥

Γ𝑚 𝑦 𝐿𝑦

+ 𝐵𝑚 sinh

+ 𝐵𝑚 sinh

Γ𝑚 𝑥 𝐿𝑥

Γ𝑚 𝑦 𝐿𝑦

+ 𝐶𝑚 cosh

+ 𝐶𝑚 cosh

Γ𝑚 𝑥 𝐿𝑥

Γ𝑚 𝑦 𝐿𝑦

(3.39)

(3.40)

where 𝐴𝑝𝑚 , 𝐵𝑝𝑚 , 𝐶𝑝𝑚 , 𝐴𝑏𝑚 , 𝐵𝑏𝑚 and 𝐶𝑏𝑚 are constants determined by the boundary conditions. And Ψ𝑝𝑚 and Ψ𝑏𝑚 are called the mode frequencies where 𝜆𝑝𝑚 =

𝜉𝑝𝑚 𝐿𝑥

,

𝜆𝑏𝑚 =

𝜉𝑏𝑚 𝐿𝑦

(3.41)

Considering a unit mass, equation (3.12) can be re-written as

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021

𝑄̈𝑛 (𝑡) + 𝜔𝑛2 𝑄𝑛 (𝑡) −

1 𝜃∗

𝐷𝑦 2𝐵 𝐷 ̈ ∑∞ 𝑇2 𝑄𝑞 (𝑡) − 𝑇3 𝑄𝑞 (𝑡) − 𝑥 𝑇4 𝑄𝑞 (𝑡) + 𝑞=1 [𝑅0 𝑇1 𝑄𝑞 (𝑡) − 𝜇

𝜇 ∗ cos(2𝑗+1)𝜋𝑐𝑡 − 𝑇5 )𝑄𝑞 (𝑡) + (𝑇6 + 𝑇7 )𝑄𝑞 (𝑡) − 𝛼𝜌((𝑇8 + 𝐸1 𝜇 𝜇 2𝑗+1 sin(2𝑗+1)𝜋𝑐𝑡 cos(2𝑘+1)𝜋𝑠 sin(2𝑘+1)𝜋𝑠 1 ∞ ∞ ∞ ∗ ∗ ∗ − ∑𝑗=1 𝐸2 )(∑𝑘=1 𝐸3 − ∑𝑘=1 𝐸4 )+ ( 2𝑗+1 2𝑘+1 2𝑘+1 4𝜋 cos(2𝑗+1)𝜋𝑐𝑡 sin(2𝑗+1)𝜋𝑐𝑡 1 cos(2𝑘+1)𝜋𝑠 ∞ ∞ ∗ ∞ ∗ ∗ ∑𝑗=1 𝐸5 − ∑𝑗=1 𝐸6 ) + (∑𝑘=1 𝐸7 − 2𝑗+1 2𝑗+1 4𝜋 2𝑘+1 𝐾0

(𝜔𝑞2 𝐹4∗

∗ ∑∞ 𝑘=1 𝐸8

sin(2𝑘+1)𝜋𝑠 2𝑘+1

sin(2𝑗+1)𝜋𝑐𝑡 2𝑗+1

cos(2𝑗+1)𝜋𝑐𝑡

2𝑘+1

(∑∞ 𝑗=1 𝜋2

))𝑄̈𝑞 (𝑡) + 2𝑐(𝑇9 +

∗ )(∑∞ 𝑘=1 𝐸11

2𝑗+1 sin(2𝑘+1)𝜋𝑠

𝜇 1

𝐺0

cos(2𝑘+1)𝜋𝑠

2𝑗+1 1

cos(2𝑘+1)𝜋𝑠

∗ − ∑∞ 𝑗=1 𝐸22

2𝑘+1

−

sin(2𝑗+1)𝜋𝑐𝑡 2𝑗+1

1 4𝜋

∗ (∑∞ 𝑘=1 𝐸15

cos(2𝑘+1)𝜋𝑠 2𝑘+1

∗ − ∑∞ 𝑘=1 𝐸16

∗ sin(2𝑗+1)𝜋𝑐𝑡 − ∑∞ 𝑗=1 𝐸18 2𝑗+1 2𝑗+1 sin(2𝑘+1)𝜋𝑠 1 cos(2𝑗+1)𝜋𝑐𝑡 ∞ ∞ ∗ ∗ ∑𝑘=1 𝐵20 ) + (∑𝑗=1 𝐸21 2𝑘+1 4𝜋 2𝑗+1

)+

𝑁 )𝑄𝑞 (𝑡))] = ∑∞ 𝑞=1 ∑𝑟=1

)+

∗ (∑∞ 𝑗=1 𝐸9

cos(2𝑗+1)𝜋𝑐𝑡

∗ − ∑∞ 𝑗=1 𝐸10 2𝑗+1 1 ∗ sin(2𝑘+1)𝜋𝑠 ∗ ∑∞ ) + (∑∞ 𝑘=1 𝐸12 𝑗=1 𝐸13 2𝑘+1 4𝜋

sin(2𝑗+1)𝜋𝑐𝑡

∗ − ∑∞ 𝑗=1 𝐸14

))𝑄̇𝑞 (𝑡) + 𝑐 2 (𝑇10 +

∗ )(∑∞ 𝑘=1 𝐸19

−

2𝑘+1

1

𝜋2

𝜋2

1 4𝜋

𝑀𝑔 𝜇𝜃 ∗

∗ (∑∞ 𝑗=1 𝐸17

∗ (∑∞ 𝑘=1 𝐸23

cos(2𝑗+1)𝜋𝑐𝑡

cos(2𝑘+1)𝜋𝑠 2𝑘+1

∗ − ∑∞ 𝑘=1 𝐸24

sin(2𝑘+1)𝜋𝑠 2𝑘+1

)

Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

(3.42)

equation (3.42) is the fundamental equation of the problem. where 𝑀

𝛼=

𝜇𝜌

, 𝜌 = 𝐿𝑥 𝐿𝑦

(3.43)

Ψ𝑚 (𝑐𝑡) = sin𝜒𝑚 𝑡 + 𝐴𝑚 cos𝜒𝑚 𝑡 + 𝐵𝑚 sinh𝜒𝑚 𝑡 + 𝐶𝑚 cosh𝜒𝑚 𝑡

(3.44)

Ψ𝑚 (𝑠) = sin𝜈𝑚 + 𝐴𝑚 cos𝜈𝑚 + 𝐵𝑚 sinh𝜈𝑚 + 𝐶𝑚 cosh𝜈𝑚

(3.45)

𝜒𝑚 =

𝜙𝑚 𝑐 𝐿𝑥

, 𝜈𝑚 =

𝜙𝑚 𝑠

(3.46)

𝐿𝑦

3.1 Orthotropic Rectangular Plate Traversed by a Moving Force In moving force, we account for only the load being transferred to the structure. In this case, the inertia effect is negligible.

Setting 𝜛 = 0 in the fundamental equation (3.42), one obtains ∗

𝐹 𝑄̈𝑛 (𝑡) + (1 − 4∗)𝜔𝑛2 𝑄𝑛 (𝑡) − 𝜃

1 𝜇𝜃 ∗

(𝜇𝑅0 𝑇1 𝑄̈𝑛 (𝑡) − 2𝐵𝑇2 𝑄𝑛 (𝑡) − 𝐷𝑦 𝑇3 𝑄𝑛 (𝑡) − 𝐷𝑥 𝑇4 𝑄𝑛 (𝑡)

−𝐾𝑜 𝑇5 𝑄𝑛 (𝑡) + 𝐺0 (𝑇6 + 𝑇7 )𝑄𝑛 (𝑡)) −

1 𝜇𝜃 ∗

̈ ∑∞ 𝑞=1 [𝜇𝑅0 𝑇1 𝑄𝑞 (𝑡) − 2𝐵𝑇2 𝑄𝑞 (𝑡) − 𝐷𝑦 𝑇3 𝑄𝑞 (𝑡)

𝑁 −𝐷𝑥 𝐹4∗ 𝑄𝑞 (𝑡) + (𝜇𝜔𝑞2 𝐹4∗ − 𝐾𝑜 𝑇5 )𝑄𝑞 (𝑡) + 𝐺0 (𝑇6 + 𝑇7 )𝑄𝑞 (𝑡)] = ∑∞ 𝑞=1 ∑𝑟=1

𝑀𝑔 𝜇Θ∗

(3.47)

Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

which can further be simplified as 𝑄̈𝑛 (𝑡) + 𝜒𝑛2 𝑄𝑛 (𝑡) − 𝜏 ∗ (𝜇𝑅0 𝑇1 𝑄̈𝑛 (𝑡) − 2𝐵𝑇2 𝑄𝑛 (𝑡) − 𝐷𝑦 𝑇3 𝑄𝑛 (𝑡) − 𝐷𝑥 𝑇4 𝑄𝑛 (𝑡) − 𝐾𝑜 𝑇5 ̈ 𝑄𝑛 (𝑡) + 𝐺0 (𝑇6 + 𝑇7 )𝑄𝑛 (𝑡)) − 𝜏 ∗ ∑∞ 𝑞=1 [𝜇𝑅0 𝑇1 𝑄𝑞 (𝑡) − 2𝐵𝑇2 𝑄𝑞 (𝑡) − 𝐷𝑦 𝑇3 𝑄𝑞 (𝑡) − 𝐷𝑥 𝑇4

(3.48)

𝑁 ∗ 𝑄𝑞 (𝑡) + (𝜇𝜔𝑞2 𝐹4∗ − 𝐾𝑜 𝑇5 )𝑄𝑞 (𝑡) + 𝐺0 (𝑇6 + 𝑇7 )𝑄𝑞 (𝑡)] = ∑∞ 𝑞=1 ∑𝑟=1 𝑀𝑔𝜏 Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

On expanding, re-arranging and simplifying equation (4.48), one obtains 𝑄̈𝑛 (𝑡) +

2 −𝜏 ∗ 𝐽∗ ) (𝜒𝑛 6

[1−𝜏∗ 𝜇𝑅0 𝑇1 ]

𝑄𝑛 (𝑡) +

𝜏 [1−𝜏∗ 𝜇𝑅0 𝑇1 ]

̈ ∑∞ 𝑞=1,𝑞≠𝑛 (𝜇𝑅0 𝑇1 𝑄𝑞 (𝑡) − 2𝐵𝑇2 𝑄𝑞 (𝑡) − 𝐷𝑦 𝑇3 𝑄𝑞 (𝑡)

−𝐷𝑥 𝑇4 𝑄𝑞 (𝑡) + (𝜇𝜔𝑞2 𝐹4∗ − 𝐾𝑜 𝑇5 )𝑄𝑞 (𝑡) + 𝐺0 (𝑇6 + 𝑇7 )𝑄𝑞 (𝑡)) =

𝜏∗ 𝑀𝑔 [1−𝜏∗ 𝜇𝑅0 𝐹1∗ ]

(3.49) Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

where

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Awodola T.O et al.

𝜏∗ =

International Journal of Advanced Engineering Research and Science, 8(6)-2021

1 𝜇𝜃 ∗

, 𝜒𝑛2 = (1 −

𝐹4∗ 𝜃∗

For any arbitrary ratio Υ, defined as 𝜏 ∗ =

)𝜔𝑛2 , 𝐽6∗ = −2𝐵𝑇2 − 𝐷𝑦 𝑇3 − 𝐷𝑥 𝑇4 − 𝐾0 𝑇5 + 𝐺0 (𝑇6 + 𝑇7 ) Υ 1+Υ

(3.50)

, one obtains

Υ=

𝜏∗ 1−𝜏∗

= 𝜏 ∗ + 𝑜(𝜏 ∗2 )+. ..

(3.51)

For only 𝑜(𝜏 ∗ ), one obtains 𝜏∗ = Υ

(3.52)

Applying binomial expansion, 1 1−Υ𝜇𝑅0 𝐹1∗

= 1 + Υ𝜇𝑅0 𝐹1∗ + 𝑜(Υ 2 )+. ..

(3.53)

On putting equation (4.53) into equation (4.49), one obtains 𝑄̈𝑛 (𝑡) + (𝜒𝑛2 − Υ𝐽6∗ )(1 + Υ𝜇𝑅0 𝑇1 + 𝑜(Υ 2 )+. . . )𝑄𝑛 (𝑡) + Υ(1 + 𝜏 ∗ 𝜇𝑅0 𝑇1 + 𝑜(Υ 2 )+. . . ) ∗ ∗ ̈ ∗ ∗ ∗ 2 ∑∞ 𝑞=1,𝑞≠𝑛 (𝜇𝑅0 𝐹1 𝑍𝑞 (𝑡) − 2𝐵𝐹2 𝑄𝑞 (𝑡) − 𝐷𝑦 𝐹3 𝑄𝑞 (𝑡) − 𝐷𝑥 𝐹4 𝑄𝑞 (𝑡) + (𝜇𝜔𝑞 − 𝐾0 )𝐹5 𝑄𝑞 (𝑡) + 𝐺0 (𝐹6∗

+

𝐹7∗ )𝑄𝑞 (𝑡))

(3.54)

2

= (1 + Υ𝜇𝑅0 𝑇1 + 𝑜(Υ )+. . . )𝑀𝑔Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

Retaining only 𝑜(Υ), equation (4.54) becomes ∗ ̈ ∗ ∗ 𝑄̈𝑛 (𝑡) + (𝜒𝑛2 (1 + Υ𝜇𝑅0 𝑇1 ) − Υ𝐽6∗ )𝑄𝑛 (𝑡) + Υ ∑∞ 𝑞=1,𝑞≠𝑛 (𝜇𝑅0 𝐹1 𝑄𝑞 (𝑡) − 2𝐵𝐹2 𝑄𝑞 (𝑡) − 𝐷𝑦 𝐹3 𝑄𝑞 (𝑡) − 𝐷𝑥 𝐹4∗ 𝑄𝑞 (𝑡) + (𝜇𝜔𝑞2 − 𝐾0 )𝐹5∗ 𝑄𝑞 (𝑡) + 𝐺0 (𝐹6∗ + 𝐹7∗ )𝑄𝑞 (𝑡)) = Υ𝑀𝑔Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

(3.55)

which is simplified further as ∗ ̈ ∗ ∗ ∗ 𝑄̈𝑛 (𝑡) + 𝐽7∗ 𝑄𝑛 (𝑡) + Υ ∑∞ 𝑞=1,𝑞≠𝑛 (𝜇𝑅0 𝐹1 𝑄𝑞 (𝑡) − 2𝐵𝐹2 𝑄𝑞 (𝑡) − 𝐷𝑦 𝐹3 𝑄𝑞 (𝑡) − 𝐷𝑥 𝐹4 𝑄𝑞 (𝑡) + ∗ 2 ∗ ∗ (𝜇𝜔𝑞 − 𝐾0 )𝐹5 𝑄𝑞 (𝑡) + 𝐺0 (𝐹6 + 𝐹7 )𝑄𝑞 (𝑡)) = Υ𝑀𝑔Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

(3.56)

where 𝐽7∗ = 𝜒𝑛2 (1 + Υ𝜇𝑅0 𝑇1 ) − Υ𝐽6∗

(3.57)

Using Struble's technique, the solution to the homogeneous part of part of equation (3.57) is assumed to take the form 𝑄𝑛 (𝑡) = 𝜑(𝑛, 𝑡)cos(𝜒𝑛 𝑡 − 𝜌(𝑛, 𝑡))+. ..

(3.58)

𝜑(𝑛, 𝑡) = 𝜀𝑛

(3.59)

where

and 2 −𝐽∗ 𝜒𝑛 7

𝜙(𝑛, 𝑡) = (

2𝜒𝑛

)𝑡 + 𝜄𝑛

(3.60)

On putting equations (3.59) and (3.60) into equation (3.58), one obtains 𝑄𝑛 (𝑡) = 𝜀𝑛 cos(𝜒𝑛 𝑡 − (

2 −𝐽 ∗ 𝜑(𝑛,𝑡) 𝜒𝑛 7

)𝑡 − 𝜄𝑛 )

2𝜒𝑛

(3.61)

On further simplification, one obtains 𝑄𝑛 (𝑡) = 𝜀𝑛 cos(𝜐𝑛 𝑡 − 𝜄𝑛 )

(3.62)

where 𝜐𝑛 = 𝜒𝑛 − (

2 −𝐽∗ 𝜑(𝑛,𝑡) 𝜒𝑛 7

2𝜒𝑛

)

(3.63)

is the modified frequency for moving force problem for orthotropic rectangular plate resting on variable elastic bi-parametric foundation.

Using equation (3.63), the homogeneous part of equation (3.56) 𝑄̈𝑛 (𝑡) + 𝜐𝑛2 𝑄𝑛 (𝑡) = 0

(3.64)

𝑄̈𝑛 (𝑡) + 𝜐𝑛2 𝑄𝑛 (𝑡) = Υ𝑀𝑔Ψ𝑚 (𝑐𝑡)Ψ𝑚 (𝑠)

(3.65)

Hence, the entire equation (3.56) gives

Re-writing equation (3.65), one obtains

www.ijaers.com

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021 𝑄̈𝑛 (𝑡) + 𝜐𝑛2 𝑄𝑛 (𝑡) = Υ𝑀𝑔Ψ𝑚 (𝑠)[sin𝜒𝑚 (𝑡) + 𝐴𝑚 cos𝜒𝑚 𝑡 + 𝐵𝑚 sinh𝜒𝑚 𝑡 +𝐶𝑚 cosh𝜒𝑚 𝑡]

(3.66)

To obtain the solution to equation (3.66), one makes use of Laplace transformation techniques to obtain 𝑄𝑛 (𝑡) =

𝑀𝑔ΥΦ𝑚 (𝑠) 4 −𝜐4 ) 𝜐𝑛 (𝜒𝑚 𝑛

2 2 [(𝜒𝑚 + 𝜐𝑛2 )(𝜒𝑚 sin𝜐𝑛 𝑡 − 𝜐𝑛 sin𝜒𝑚 𝑡) − 𝐴𝑚 𝜐𝑛 (𝜒𝑚 + 𝜐𝑛2 )

2 2 (cos𝜒𝑚 𝑡 − cos𝜐𝑛 𝑡) − 𝐵𝑚 (𝜒𝑚 − 𝜐𝑛2 )(𝛼𝑚 sin𝜐𝑛 𝑡 − 𝜐𝑛 sinh𝜒𝑚 𝑡) + 𝐶𝑚 𝜐𝑛 (𝜒𝑚 − 𝜐𝑛2 ) (cosh𝜒𝑚 𝑡 − cos𝜐𝑛 𝑡)]

(3.67)

which on inversion yields 𝑀𝑔ΥΦ𝑚 (𝑠) 2 2 2 4 −𝜐4 ) [(𝜒𝑚 + 𝜐𝑛 )(𝜒𝑚 sin𝜐𝑛 𝑡 − 𝜐𝑛 sin𝜒𝑚 𝑡) − 𝐴𝑚 𝜐𝑛 (𝜒𝑚 𝜐𝑛 (𝜒𝑚 𝑛 2 2 +𝜐𝑛2 )(cos𝜒𝑚 𝑡 − cos𝜐𝑛 𝑡) − 𝐵𝑚 (𝜒𝑚 − 𝜐𝑛2 )(𝜒𝑚 sin𝜐𝑛 𝑡 − 𝜐𝑛 sinh𝜒𝑚 𝑡) + 𝐶𝑚 𝜐𝑛 (𝜒𝑚 𝜙 𝜙 𝜙 𝜙𝑗𝑚 𝑗𝑚 𝑗𝑚 𝑗𝑚 −𝜐𝑛2 )(cosh𝜒𝑚 𝑡 − cos𝜐𝑛 𝑡)](sin 𝑥 + 𝐴𝑗𝑚 cos 𝑥 + 𝐵𝑗𝑚 sinh 𝑥 + 𝐶𝑗𝑚 cosh 𝑥) 𝐿 𝐿 𝐿 𝐿 ∞ 𝑊(𝑥, 𝑦, 𝑡) = ∑∞ 𝑗𝑚=1 ∑ℎ𝑚=1

𝑥

(sin

𝜙ℎ𝑚 𝐿𝑦

𝑦 + 𝐴ℎ𝑚 cos

𝜙ℎ𝑚 𝐿𝑦

𝑥

𝑦 + 𝐵ℎ𝑚 sinh

𝜙ℎ𝑚 𝐿𝑦

𝑥

𝑦 + 𝐶ℎ𝑚 cosh

𝜙ℎ𝑚 𝐿𝑦

(3.68)

𝑥

𝑦)

which is the transverse displacement response to a moving force of orthotropic rectangular plate resting on variable elastic bi-parametric foundation.

3.2 Orthotropic Rectangular Plate Traversed by a Moving Mass In moving mass problem, the moving load is assumed rigid, and the weight and as well as inertia forces are transferred to the moving load. That is the inertia effect is not negligible. Thus 𝜛 ≠ 0 and so it is required to solve the entire equation (3.42).To solve the equation,one employs analytical approximate method. This method is known as an approximate analytical method of Struble. The homogeneous part of equation (3.42) shall be replaced by a free system operator defined by the modified frequency 𝜐𝑛 . Thus, the entire equation becomes ∗ 𝑄̈𝑛 (𝑡) + 𝜐𝑛2 𝑄𝑛 (𝑡) + 𝛼𝜌∗ ∑∞ 𝑞=1 [(𝐹8 + sin(2𝑗+1)𝜋𝑐𝑡

∗ )(∑∞ 𝑘=1 𝐸3

2𝑗+1 cos(2𝑗+1)𝜋𝑐𝑡 2𝑗+1 sin(2𝑘+1)𝜋𝑠 2𝑘+1

−

cos(2𝑘+1)𝜋𝑠

cos(2𝑘+1)𝜋𝑠 2𝑘+1

−

cos(2𝑗+1)𝜋𝑐𝑡

2𝑗+1 sin(2𝑘+1)𝜋𝑠 ∞ ∗ ∑𝑘=1 𝐸4 ) 2𝑘+1

+

+

∗ − ∑∞ 𝑗=1 𝐸2 1

4𝜋 1 ∞ ∗ cos(2𝑘+1)𝜋𝑠 (∑𝑘=1 𝐸7 4𝜋 2𝑘+1

∗ (∑∞ 𝑗=1 𝐸5 ∗ − ∑∞ 𝑘=1 𝐸8

𝜋2

∗ − ∑∞ 𝑗=1 𝐸22

))𝑄𝑞 (𝑡)] = where 𝜌∗ =

2𝑗+1

)+

1

∗ (∑∞ 𝑗=1 𝐸9

cos(2𝑗+1)𝜋𝑐𝑡

∗ sin(2𝑘+1)𝜋𝑠 − ∑∞ 𝑘=1 𝐸16 2𝑘+1 2𝑘+1 1 cos(2𝑗+1)𝜋𝑐𝑡 sin(2𝑗+1)𝜋𝑐𝑡 ∞ ∞ 2 ∗ ∗ ∗ ))𝑄̇𝑞 (𝑡) + 𝑐 (𝐹10 + 2 (∑𝑗=1 𝐸17 − ∑𝑗=1 𝐸18 ) 𝜋 2𝑗+1 2𝑗+1 1 ∗ cos(2𝑘+1)𝜋𝑠 ∗ sin(2𝑘+1)𝜋𝑠 ∗ cos(2𝑗+1)𝜋𝑐𝑡 (∑∞ − ∑∞ ) + (∑∞ 𝑘=1 𝐸19 𝑘=1 𝐸20 𝑗=1 𝐸21 2𝑘+1 2𝑘+1 4𝜋 2𝑗+1 ∗ − ∑∞ 𝑗=1 𝐸14

sin(2𝑗+1)𝜋𝑐𝑡

1

∗ (∑∞ 𝑗=1 𝐸1

∗ sin(2𝑗+1)𝜋𝑐𝑡 − ∑∞ 𝑗=1 𝐸10 2𝑗+1 2𝑗+1 sin(2𝑘+1)𝜋𝑠 1 cos(2𝑗+1)𝜋𝑐𝑡 ∞ ∞ ∗ ∗ ∑𝑗=1 𝐸12 ) + (∑𝑗=1 𝐸13 2𝑘+1 4𝜋 2𝑗+1

))𝑄̈𝑞 (𝑡) + 2𝑐(𝐹9∗ +

∗ )(∑∞ 𝑘=1 𝐸11

−

2𝑘+1 sin(2𝑗+1)𝜋𝑐𝑡 ∞ ∗ ∑𝑗=1 𝐸6 ) 2𝑗+1

1 𝜋2

4𝜋

∗ (∑∞ 𝑘=1 𝐸15

sin(2𝑗+1)𝜋𝑐𝑡

1 ) + (∑∞ 𝑘=1 2𝑗+1 4𝜋 𝑔𝛼 ∞ ∑𝑞=1 ∗ Φ𝑚 (𝑐𝑡)Φ𝑚 (𝑠) 𝜃

∗ 𝐸23

cos(2𝑘+1)𝜋𝑠

cos(2𝑘+1)𝜋𝑠 2𝑘+1

∗ − ∑∞ 𝑘=1 𝐸24

(3.69)

sin(2𝑘+1)𝜋𝑠 2𝑘+1

𝜌 𝜃∗

On expanding,simplifying and rearranging equation (4.88), one obtains 𝑄̈𝑛 (𝑡) + 2𝑐𝛼𝜌∗ (𝐹9∗ +

1

∗ (∑∞ 𝑗=1 𝐸9

cos(2𝑗+1)𝜋𝑐𝑡

∗ (∑∞ 𝑗=1 𝐸1

cos(2𝑗+1)𝜋𝑐𝑡

∗ − ∑∞ 𝑗=1 𝐸10

sin(2𝑗+1)𝜋𝑐𝑡

∗ )(∑∞ 𝑘=1 𝐸11 2𝑗+1 cos(2𝑘+1)𝜋𝑐𝑡 1 ∗ sin(2𝑘+1)𝜋𝑐𝑡 ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ − ∑∞ ) + (∑∞ − ∑∞ 𝑘=1 𝐸12 𝑗=1 𝐸13 𝑗=1 𝐸14 2𝑘+1 2𝑘+1 4𝜋 2𝑗+1 sin(2𝑗+1)𝜋𝑐𝑡 ∗ cos(2𝑘+1)𝜋𝑐𝑡 ∗ sin(2𝑘+1)𝜋𝑐𝑡 ) + (∑∞ − ∑∞ ))𝑄̇𝑛 (𝑡) + 𝑘=1 𝐸15 𝑘=1 𝐸16 2𝑗+1 2𝑘+1 2𝑘+1

(𝜐𝑛2 (1 − 𝛼𝜌∗ (𝐹8∗ +

𝜋2

1

𝜋2 ∞ ∗ cos(2𝑘+1)𝜋𝑐𝑡 (∑𝑘=1 𝐸3 2𝑘+1

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−

2𝑗+1

∗ − ∑∞ 𝑗=1 𝐸2

2𝑗+1 sin(2𝑘+1)𝜋𝑐𝑡 ∞ ∗ ∑𝑘=1 𝐸4 ) 2𝑘+1

+

sin(2𝑗+1)𝜋𝑐𝑡

)

2𝑗+1 1 cos(2𝑗+1)𝜋𝑐𝑡 ∞ (∑𝑗=1 𝐸5∗ 4𝜋 2𝑗+1

−

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∗ ∑∞ 𝑗=1 𝐸6

sin(2𝑗+1)𝜋𝑐𝑡 2𝑗+1 1

∗ 𝑐 2 𝛼𝜌∗ (𝐹10 + cos(2𝑘+1)𝜋𝑐𝑡

∗ ) + (∑∞ 𝑘=1 𝐸7

cos(2𝑘+1)𝜋𝑐𝑡

∗ − ∑∞ 𝑘=1 𝐸8

2𝑘+1 ∞ ∗ cos(2𝑗+1)𝜋𝑐𝑡 (∑𝑗=1 𝐸17 − ∑∞ 𝑗=1 𝜋2 2𝑗+1

sin(2𝑘+1)𝜋𝑐𝑡

)))) +

2𝑘+1 ∗ sin(2𝑗+1)𝜋𝑐𝑡 𝐸18 )(∑∞ 𝑘=1 2𝑗+1

∗ 𝐸19

1 ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ ) + (∑∞ − ∑∞ 𝑗=1 𝐸21 𝑗=1 𝐸22 2𝑘+1 4𝜋 2𝑗+1 ∗ cos(2𝑘+1)𝜋𝑐𝑡 ∗ sin(2𝑘+1)𝜋𝑐𝑡 ) + (∑∞ − ∑∞ ))𝑄𝑛 (𝑡) 𝑘=1 𝐸23 𝑘=1 𝐸24 2𝑗+1 2𝑘+1 2𝑘+1 1 ∞ ∗ ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑡 +𝛼𝜌∗ ∑∞ − ∑∞ )(∑∞ 𝑞=1,𝑞≠𝑛 [(𝐹8 + 𝜋2 (∑𝑗=1 𝐸1 𝑗=1 𝐸2 𝑘=1 2𝑗+1 2𝑗+1 2𝑘+1 sin(2𝑗+1)𝜋𝑐𝑡

cos(2𝑘+1)𝜋𝑐𝑡

∗ − ∑∞ 𝑘=1 𝐸20

sin(2𝑘+1)𝜋𝑐𝑡

𝐸3∗

1 ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ ) + (∑∞ − ∑∞ 𝑗=1 𝐸5 𝑗=1 𝐸6 2𝑘+1 4𝜋 2𝑗+1 ∗ cos(2𝑘+1)𝜋𝑐𝑡 ∗ sin(2𝑘+1)𝜋𝑐𝑡 ) + (∑∞ − ∑∞ ))𝑄̈𝑞 (𝑡) + 𝑘=1 𝐸7 𝑘=1 𝐸8 2𝑗+1 2𝑘+1 2𝑘+1 1 ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑡 ∗ cos(2𝑘+1)𝜋𝑐𝑡 2𝑐(𝐹9∗ + 2 (∑∞ − ∑∞ )(∑∞ 𝑗=1 𝐸9 𝑗=1 𝐸10 𝑘=1 𝐸11 𝜋 2𝑗+1 2𝑗+1 2𝑘+1 2𝑘+1 sin(2𝑗+1)𝜋𝑐𝑡

∗ − ∑∞ 𝑘=1 𝐸4

sin(2𝑘+1)𝜋𝑐𝑡

1 ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑡 ) + (∑∞ − ∑∞ ) 𝑗=1 𝐸13 𝑗=1 𝐸14 2𝑘+1 4𝜋 2𝑗+1 2𝑗+1 cos(2𝑘+1)𝜋𝑐𝑡 sin(2𝑘+1)𝜋𝑐𝑡 1 ∗ ∗ ∗ +(∑∞ − ∑∞ ))𝑄̇𝑞 (𝑡) + 𝑐 2 (𝐹10 + 2 (∑∞ 𝑘=1 𝐸15 𝑘=1 𝐸16 𝑗=1 2𝑘+1 2𝑘+1 𝜋 cos(2𝑗+1)𝜋𝑐𝑡 sin(2𝑗+1)𝜋𝑐𝑡 cos(2𝑘+1)𝜋𝑐𝑡 ∗ ∗ ∗ − ∑∞ )(∑∞ − ∑∞ 𝑗=1 𝐸18 𝑘=1 𝐸19 𝑘=1 𝐸20 2𝑗+1 2𝑗+1 2𝑘+1 ∗ − ∑∞ 𝑘=1 𝐸12

sin(2𝑘+1)𝜋𝑐𝑡

sin(2𝑘+1)𝜋𝑐𝑡

∗ 𝐸17

1 1 ∗ cos(2𝑗+1)𝜋𝑐𝑡 ∗ sin(2𝑗+1)𝜋𝑐𝑡 (∑∞ − ∑∞ )+ 𝑗=1 𝐸21 𝑗=1 𝐸22 4𝜋 2𝑗+1 2𝑗+1 4𝜋 𝑔𝛼 ∗ cos(2𝑘+1)𝜋𝑐𝑡 ∗ sin(2𝑘+1)𝜋𝑐𝑡 (∑∞ − ∑∞ ))𝑄𝑞 (𝑡)] = ∗ Φ𝑚 (𝑐𝑡)Φ𝑚 (𝑠) 𝑘=1 𝐸23 𝑘=1 𝐸24 2𝑘+1 2𝑘+1 𝜃 2𝑘+1

)+

(3.70)

Applying modified asymptotic method of Struble, the solution to equation (3.70) takes the form 𝑄𝑛 (𝑡) = 𝜓(𝑛, 𝑡)cos(𝜐𝑛 𝑡 − 𝐺(𝑛, 𝑡)) + 𝜐𝑛 𝑄1 (𝑡)+. ..

(3.71)

where 𝜓(𝑛, 𝑡) = 𝛽𝑒 −𝛼𝜌

∗𝐽 𝑡 8

(3.72)

and 𝐹(𝑛, 𝑡) =

1 2𝜐𝑛

1

[𝜐𝑛2 𝜛𝜑 ∗ (𝐹9∗ +

∗ −𝑐 2 𝜛𝜑 ∗ (𝐹10 +

1 4𝜋

4𝜋

∗ (∑∞ 𝑘=1 𝐸7

∗ (∑∞ 𝑘=1 𝐸23

cos(2𝑘+1)𝜋𝑠 2𝑘+1

cos(2𝑘+1)𝜋𝑐𝑡 2𝑘+1

−

∗ − ∑∞ 𝑘=1 𝐸8

sin(2𝑘+1)𝜋𝑠

2𝑘+1 sin(2𝑘+1)𝜋𝑐𝑡 ∞ ∗ ∑𝑘=1 𝐸24 ))]𝑡 2𝑘+1

)) (3.73)

+ 𝜏𝑛

On putting equations (3.72) and (3.73) into equation (3.71), one obtains 𝑄𝑛 (𝑡) = 𝑍𝑒 −𝛼𝜌 sin(2𝑘+1)𝜋𝑠 2𝑘+1

∗𝐽 𝑡 8

cos[𝜐𝑛 𝑡 −

1

1 ∗ cos(2𝑘+1)𝜋𝑠 ∗ (∑∞ − ∑∞ 𝑘=1 𝐸7 𝑘=1 𝐸8 4𝜋 2𝑘+1 ∗ cos(2𝑘+1)𝜋𝑐𝑡 ∗ sin(2𝑘+1)𝜋𝑐𝑡 (∑∞ − ∑∞ ) 𝑘=1 𝐸23 𝑘=1 𝐸24 4𝜋 2𝑘+1 2𝑘+1

2𝜐𝑛 1

∗ )) − 𝑐 2 𝛼𝜌∗ (𝐹10 +

(𝜐𝑛2 𝛼𝜌∗ (𝐹8∗ +

(3.74)

))𝑡 − 𝜏𝑛 ] On further simplifications, one obtains 𝑄𝑛 (𝑡) = 𝑍𝑒 −𝛼𝜌

∗𝐽 𝑡 8

cos[𝛽𝑛 𝑡 − 𝜀𝑛 ]

(3.75)

where 𝛽𝑛 = 𝜐𝑛 −

1 2𝜐𝑛

(𝜐𝑛2 𝛼𝜌∗ (𝐹8∗ +

∗ −𝑐 2 𝛼𝜌∗ (𝐹10 +

1 4𝜋

∗ (∑∞ 𝑘=1 𝐸7

1

cos(2𝑘+1)𝜋𝑠

∗ cos(2𝑘+1)𝜋𝑐𝑡 (∑∞ 𝑘=1 𝐸23 4𝜋 2𝑘+1

2𝑘+1

−

∗ − ∑∞ 𝑘=1 𝐸8

sin(2𝑘+1)𝜋𝑠

2𝑘+1 sin(2𝑘+1)𝜋𝑐𝑡 ∞ ∗ ∑𝑘=1 𝐸24 ))) 2𝑘+1

)) (3.76)

is the modified frequency representing the frequency of the free system.

Using the same argument used to solve equation (3.56), the homogeneous part of equation (3.70) becomes 𝑄̈𝑛 (𝑡) + 𝛽𝑛2 𝑄𝑛 (𝑡) = 0

(3.77)

𝑔𝛼 𝑄̈𝑛 (𝑡) + 𝛽𝑛2 𝑄𝑛 (𝑡) = ∗ Φ𝑚 (𝑐𝑡)Φ𝑚 (𝑠)

(3.78)

Hence, the entire equation becomes 𝜃

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Rewriting equation (3.78), one obtains 𝑔𝛼 𝑄̈𝑛 (𝑡) + 𝛽𝑛2 𝑄𝑛 (𝑡) = ∗ Φ𝑚 (𝑠)[sin𝜒𝑚 𝑡 + 𝐴𝑚 cos𝜒𝑚 𝑡 + 𝐵𝑚 sinh𝜒𝑚 𝑡 + 𝐶𝑚 cosh𝜒𝑚 𝑡]

(3.79)

𝜃

Following the procedures applied to solve equation (3.66), one obtains 𝑄𝑛 (𝑡) =

𝑔𝛼Φ𝑚 (𝑠) 4 −𝛽 4 ) 𝜃 ∗ 𝛽𝑛 (𝜒𝑚 𝑛

2 2 [(𝜒𝑚 + 𝛽𝑛2 )(𝜒𝑚 sin𝛽𝑛 𝑡 − 𝛽𝑛 sin𝜒𝑚 𝑡) − 𝐴𝑚 𝛽𝑛 (𝜒𝑚 + 𝛽𝑛2 )

(3.80)

2 2 (cos𝜒𝑚 𝑡 − cos𝛽𝑛 𝑡) − 𝐵𝑚 (𝜒𝑚 − 𝛽𝑛2 )(𝜒𝑚 sin𝛽𝑛 𝑡 − 𝛽𝑛 sinh𝜒𝑚 𝑡) + 𝐶𝑚 𝛽𝑛 (𝜒𝑚 − 𝛽𝑛2 ) (cosh𝜒𝑚 𝑡 − cos𝛽𝑛 𝑡)]

which on inversion yields 𝑔𝛼Φ𝑚 (𝑠) 2 2 2 4 −𝛽 4 ) [(𝜒𝑚 + 𝛽𝑛 )(𝜒𝑚 sin𝛽𝑛 𝑡 − 𝛽𝑛 sin𝜒𝑚 𝑡) − 𝐴𝑚 𝛽𝑛 (𝜒𝑚 𝜃 ∗ 𝛽𝑛 (𝜒𝑚 𝑛 2 2 (cos𝜒𝑚 𝑡 − cos𝛽𝑛 𝑡) − 𝐵𝑚 (𝜒𝑚 − 𝛽𝑛2 )(𝜒𝑚 sin𝛽𝑛 𝑡 − 𝛽𝑛 sinh𝜒𝑚 𝑡) + 𝐶𝑚 𝛽𝑛 (𝜒𝑚 − 𝛽𝑛2 ) 𝜙𝑗𝑚 𝜙𝑗𝑚 𝜙𝑗𝑚 𝜙𝑗𝑚 (cosh𝜒𝑚 𝑡 − cos𝛽𝑛 𝑡)](sin 𝑥 + 𝐴𝑗𝑚 cos 𝑥 + 𝐵𝑗𝑚 sinh 𝑥 + 𝐶𝑗𝑚 cosh 𝑥) 𝐿 𝐿 𝐿 𝐿

∞ 𝑊(𝑥, 𝑦, 𝑡) = ∑∞ 𝑗𝑚=1 ∑ℎ𝑚=1

𝑥

(sin

𝜙ℎ𝑚 𝐿𝑦

𝑦 + 𝐴ℎ𝑚 cos

𝜙ℎ𝑚 𝐿𝑦

𝑥

𝑦 + 𝐵ℎ𝑚 sinh

𝜙ℎ𝑚 𝐿𝑦

𝑥

𝑦 + 𝐶ℎ𝑚 cosh

𝜙ℎ𝑚 𝐿𝑦

+ 𝛽𝑛2 ) (3.81)

𝑥

𝑦)

which is the transverse displacement response to a moving mass of an orthotropic rectangular plate resting on variable elastic bi-parametric foundation.

3.3 ILLUSTRATIVE EXAMPLES 3.3.1 Orthotropic Rectangular Plate Clamped at All Edges For an orthotropic plate clamped at all its edges, the boundary conditions are given by 𝑊(0, 𝑦, 𝑡) = 0, 𝑊(𝐿𝑥 , 𝑦, 𝑡) = 0

(3.82)

𝑊(𝑥, 0, 𝑡) = 0, 𝑊(𝑥, 𝐿𝑦 , 𝑡) = 0

(3.83)

∂𝑊(0,𝑦,𝑡)

(3.84)

=0

(3.85)

𝜉𝑝𝑚 (0) = 0, 𝜉𝑝𝑚 (𝐿𝑥 ) = 0

(3.86)

𝜉𝑞𝑚 (0) = 0, 𝜉𝑞𝑚 (𝐿𝑦 ) = 0

(3.87)

∂𝑊(𝑥,0,𝑡) ∂𝑦

= 0,

∂𝑊(𝐿𝑥 ,𝑦,𝑡)

=0

∂𝑥

= 0,

∂𝑥

∂𝑊(𝑥,𝐿𝑦 ,𝑡) ∂𝑦

Thus,for the normal modes

∂𝜉𝑝𝑚 (0) ∂𝑥

∂𝜉𝑞𝑚 (0) ∂𝑦

= 0,

= 0,

∂𝜉𝑝𝑚 (𝐿𝑥 ) ∂𝑥

∂𝜉𝑞𝑚 (𝐿𝑦 ) ∂𝑦

=0

(3.88)

=0

(3.89)

For simplicity, our initial conditions are of the form 𝑊(𝑥, 𝑦, 0) = 0 =

∂𝑊(𝑥,𝑦,𝑜) ∂𝑡

(3.90)

Using the boundary conditions in equations (3.86) to (3.89) and the initial conditions given by equation (3.90), it can be shown that 𝐴𝑝𝑚 =

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sinh𝜉𝑝𝑚 −sin𝜉𝑝𝑚 cos𝜉𝑝𝑚 −cosh𝜉𝑝𝑚

=

cos𝜉𝑝𝑚 −cosh𝜉𝑝𝑚 sin𝜉𝑝𝑚 +sinh𝜉𝑝𝑚

(3.91)

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𝐴𝑞𝑚 =

sinh𝜉𝑞𝑚 −sin𝜉𝑞𝑚 cos𝜉𝑞𝑚 −cosh𝜉𝑞𝑚

=

cos𝜉𝑞𝑚 −cosh𝜉𝑞𝑚 sin𝜉𝑞𝑚 +sinh𝜉𝑞𝑚

(3.92)

In the same vein, we have 𝐴𝑚 =

sinh𝜉𝑚 −sin𝜉𝑚 cos𝜉𝑚 −cosh𝜉𝑚

=

cos𝜉𝑚 −cosh𝜉𝑚 sin𝜉𝑚 +sinh𝜉𝑚

𝐵𝑝𝑚 = −1, 𝐵𝑞𝑚 = −1,

⇒ 𝐵𝑚 = −1

𝐶𝑝𝑚 = −𝐴𝑝𝑚 , 𝐶𝑞𝑚 = −𝐴𝑞𝑚 ,

⇒ 𝐶𝑚 = −𝐴𝑚

(3.93)

(3.94)

(3.95)

and from equation (3.93) , one obtains cos𝜉𝑚 cosh𝜉𝑚 = 1

(3.96)

which is termed the frequency equation for the dynamical problem, such that 𝜉1 = 4.73004, 𝜉2 = 7.85320, 𝜉3 = 10.9951

(3.97)

On using equations (3.91), (3.92), (3.93),(3.94),(3.95) and (3.97) in equations (3.68) and (3.81), one obtains the displacement response to a moving force and a moving mass of clamped orthotropic rectangular plate resting on bi-parametric condition respectively.

3.3.2 Graphs of the Clamped-clamped End Conditions Figures 1 and 2 display the effect of foundation modulus 𝐾𝑜 on the deflection profile of clamped-clamped orthotropic rectangular plate under the action of load moving at constant velocity in both cases of moving distributed forces and moving distributed masses respectively. The graphs show that the response amplitude decreases as the value of foundation modulus 𝐾𝑜 increases.

Fig.1: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐾𝑜 and Traversed by Moving Force

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Fig.2: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐾𝑜 and Traversed by Moving Mass Figures 3 and 4 display the effect of shear modulus 𝐺𝑜 on the deflection profile of clamped-clamped orthotropic rectangular plate under the action of load moving at constant velocity in both cases of moving distributed forces and moving distributed masses respectively. The graphs show that the response amplitude decreases as the value of shear modulus 𝐺𝑜 increases.

Fig.3: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐺𝑜 and Traversed by Moving Force

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021

Fig.4: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐺𝑜 and Traversed by Moving Mass Figures 5 and 6 display the effect of flexural rigidity of the plate along x-axis 𝐷𝑥 on the deflection profile of Clamped-clamped orthotropic rectangular plate under the action of load moving at constant velocity in both cases of moving distributed forces and moving distributed masses respectively. The graphs show that the response amplitude decreases as the value of flexural rigidity 𝐷𝑥 increases.

Fig.5: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐷𝑥 and Traversed by Moving Force

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021

Fig.6: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐷𝑥 and Traversed by Moving Mass Figures 7 and 8 display the effect of flexural rigidity of the plate along y-axis 𝐷𝑦 on the deflection profile of Clamped-clamped orthotropic rectangular plate under the action of load moving at constant velocity in both cases of moving distributed forces and moving distributed masses respectively. The graphs show that the response amplitude decreases as the value of flexural rigidity 𝐷𝑦 increases.

Fig.7: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐷𝑦 and Traversed by Moving Force

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International Journal of Advanced Engineering Research and Science, 8(6)-2021

Fig.8: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝐷𝑦 and Traversed by Moving Mass Figures 9 and 10 display the effect of rotatory inertia 𝑅𝑜 on the deflection profile of Clamped-clamped orthotropic rectangular plate under the action of load moving at constant velocity in both cases of moving distributed forces and moving distributed masses respectively. The graphs show that the response amplitude decreases as the value of rotatory inertia 𝑅𝑜 increases.

Fig.9: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝑅𝑜 and Traversed by Moving Force

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021

Fig.10: Displacement Profile of Clamped-clamped Orthotropic Rectangular Plate with Varying 𝑅𝑜 and Traversed by Moving Mass Figure 11 displays the comparison between moving force and moving mass for fixed values of 𝑅𝑜 , 𝐺𝑜 , 𝐾𝑜 , 𝐷𝑥 and 𝐷𝑦 .

Fig.11: Displacement Profile of Comparison between Moving Force and Moving Mass

IV.

CONCLUSION

In this research work, the problem of vibration of orthotropic rectangular plate under the action of moving masses and resting on a variable elastic Pasternak foundation with clamped end conditions has been studied.

The closed form solutions of the fourth order partial differential equations with variable and singular coefficients governing the orthotropic rectangular plates is obtained for both cases of moving force and moving mass

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using a solution technique that is based on the separation of variables which was used to remove the singularity in the governing fourth order partial differential equation and thereby reducing it to a sequence of coupled second order differential equations. The modified asymptotic method of Struble and Laplace transformation techniques are then employed to obtain the analytical solution to the two-dimensional dynamical problem.

The solutions are then analyzed. The analyses show that, for

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Awodola T.O et al.

International Journal of Advanced Engineering Research and Science, 8(6)-2021

the same natural frequency and the critical speed, the moving mass problem is smaller than that of the moving force problem. Resonance is reached earlier in the moving mass system than in the moving force problem. That is to say the moving force solution is not an upper bound for the accurate solution of the moving mass problem.

The results in plotted curves show that as foundation modulus 𝐾𝑜 and the shear modulus 𝐺𝑜 increase, the amplitudes of plates decrease for both cases of moving force and moving mass problems. As the rotatory inertia correction factor 𝑅𝑜 increases, the amplitudes of plates decrease for both cases of moving force and moving mass problems. As the flexural rigidities along both the x-axis 𝐷𝑥 and y-axis 𝐷𝑦 increase, the amplitudes of plates decrease for both cases of moving force and moving mass problems.

Loads.

Journal of Nigerian Mathematical Physics.5,185-200.

[9] Awodola,T.O and Adeoye A.S (2020): Behavior under Moving Distributed Masses of Simply Supported Orthotropic Rectangular Plate Resting on a Constant Elastic Bi-Parametric Foundation. Asian Research Journal of Mathematics, 16(8): 64 - 92, 2020, Article no. ARJOM. 57775, ISSN: 2456 - 477X. [10] Adeoye A.S and Awodola,T.O (2020): Dynamic Behavior of Moving Distributed Masses of Orthotropic Rectangular Plate with Clamped-clamped Boundary Conditions Resting on a Constant Elastic Bi-Parametric Foundation.

International Journal of Chemistry, Mathematics and Physics, Vol. 4, Jul-Aug. [11] Shadnam, M.R, Mofid M. and Akin J.E (2001): On the dynamic Response of Rectangular Plate with Moving Mass.

Thin-walled Structures,39, pp 797 - 806.

It can be shown further from the results that for fixed values of foundation modulus and shear modulus, rotatory inertia correction factor, flexural rigidities along both x-axis and y-axis,the amplitude for the moving mass problem is greater than that of the moving force problem which implies that resonance is reached earlier in moving mass problem than in moving force problem of simply supported orthotropic rectangular plates resting on bi-parametric foundation.

REFERENCES [1] Ambartsumian S. A. (1969): Theory of Anisotropic Plates (Engl. transl. by T. Cheron, ed. by J.E. Ashton from Russian edition by Nauka, Moscow), Technomic Publishing Company [2] Sveklo V. A.(1964): Boussinesq type problem for the anisotropic half-space, J. Appl. Math. Mech. 28, pp.

1099?1105.

[3] Awodola T.O (2015):On the Vibrations of Moving Mass of Elastically Supported Plate Resting on Bi-Parametric Foundation with Stiffness Variation, Journal of the Nigerian Association of Mathematical Physics, vol. 29, pp (65-80) [4] Szekrenyes, A.(2013): Interface Fracture in Orthotropic Composite Plates Using Second-Order Shear Deformation Theory, International Journal of Damage Mechanics; 22(8): 1161-1185 [5] Yan D. and Lin G. (2007): Dynamic behavior of concrete in biaxial compression, Magazine of Concrete Research, 59(1), 42-52.

[6] Hu, X.F,Yao W.A. (2012):Vibration Solutions of Rectangular Orthotropic Plates by Symplectic Geometry Method. pg. 211-221 [7] Alshaya, A. , John H., Rowlands R. (2016): Stresses and Strains in Thick Perforated Orthotropic Plates. Journal of Engineering Mechanics. vol. 142(11), 10p.

[8] Gbadeyan, J.A. and Dada, M.S. (2001): Dynamic Response of Plates on Pasternak Foundation to Distributed Moving

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