Application Of Mathematics In Mechanical Engineering

International Journal of Trend in Scientific Research and Development (IJTSRD)Volume 5 Issue 2, January-FebruaryFebruary 2021 Available Online: www.ijtsrd.com e-ISSN: 2456 – 6470Application off Mathematics inn Mechanical EngineeringLeena M. Bhoyar, Prerna M. Parkhi, Sana AnjumAssistant Professor, Department ofo Mathematics, J D College off Engineering, Nagpur, Maharashtra,Maharashtra IndiaHow to cite this paper:paper Leena M. Bhoyar Prerna M. Parkhi Sana ing"PublishedinInternationalJournal of Trend inScientific ResearchIJTSRD38348andDevelopment(ijtsrd), ISSN: 245624566470, Volume-5Volume Issue-2, apers/ijtsrd38348.pdfABSTRACTApplied Mathematics have been successfully used in the development ofscience and technology in 20th –21st21st century. In Mechanical Engineers, anapplication of Mathematics gives mechanical engineers convenient accessto the essential problem solvingng tools that they use. In this paper, we willdiscuss some examples of applications of mathematics in MechanicalEngineering. We conclude that the role of mathematics in engineeringremains a vital problem, and find out that mathematics should be afundamentalental concern in the design and practice of engineering.KEYWORDS: Matrices, Laplace transform, Partial differential equationCopyright 202120by author(s) andInternational Journal of Trend inScientific Research and DevelopmentJournal. This is an Open Access articledistributed underthe terms of theCreative CommonsAttribution License (CC BY ONIn this paper, several examples of applications ofmathematics in mechanical engineering are discussed.discussedMathematics occupies a unique role in the MechanicalEngineering and represents a strategic key in thedevelopment of the technology. In this paper we elaboratesome topics such as Matrices, Laplace transform, Partialdifferential equation for Mechanical Engineering.EngineeringSOME OF THE MATHEMATICAL TOOLS THAT AREUSED IN MECHANICAL ENGINEERINGMatrices,Laplace transformPartial differential equationdesign a robot without the use of matrices. All the jointvariables for forward/inverse kinematics and dynamicsproblems of the subject are noting down by matrix.Similarly, many concepts of matrices are used in FiniteElement Analysis (FEA) and Finite Element Methods(FEM) for solving problems, just like CAD does. Mainlyeigen value concept of matrices is used here.These are some applications of matrices in mechanicalengineering. Now we discuss a one example of matrixrepresentingenting stress for calculation of principal stress.INTRODUCTION TO THE STRESS TENSORMatricesMatrices: A rectangular arrangement of number, symbols,or expressions in rows and columns is known as matrix.Matrices play an important role in mechanical engineeringsyllabus. Some subjects mention below in which we willapply matrix knowledge –In Engineering Materialaterial Sciences (Miller indices) matrixplay an important role for defining crystal latticegeometries. In Strength of materials Strain matrix, stressmatrix and the moment of inertia tensors are used forsolving problems. We will also find application of matricesin analogous subjects like Design of Machine Elements,Design of Mechanical Systems. MATLAB stands for “MatrixLaboratory”. Matrix is basic building block of MATLAB.Computer -aidedaided Designing (CAD) cannot exist withoutmatrices. In Robotics Engineeringeering it is impossible to@ IJTSRD Unique Paper ID – IJTSRD3834838348 Volume – 5 Issue – 2 January-FebruaryFebruary 2021Page 142

International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470Here diagonal elements are normal stresses and offdiagonal elements are shear stresses. On stress elementin 3D, the normal and shear stresses can be compiled intoa 3 3 matrix is called a stress tensor. From ourobservation, it is possible to find a set of threeprincipal stresse for a given system. We know that, theshear stresses always become zero when principalstresses are acting. So in view of stress tensor, accordingto mathematical terms, this is the processof digonalization of matrix in which the eigen value ofgiven matrix play the role of principal stresses.ExampleThe state of plane stress at a point is represented by thestress element below. Find the principal stresses andangles at which the principal stresses act.# 10& #2.867634.867910%00On solving above matrix, first eigenvector is given by0.28671For22.8679 %00%2022.86791012# 10& #34.86792.867910# 10& #22.867900On solving above matrix, second eigenvector is given by3.48671Now we calculate angles at which the principal stressesact, but before that we can check whether eigenvectors arecorrect or not. For this we will digonalized given matrix asfallowD B-1AB0.265 0.265 2010Let1012We consider the matrix form AX X, where I is identitymatrix and become eigen value, X is eigenvector suchthat (A- I)X 0.To find the value of we consider Characteristic equationwhich is given by det(A- I) 0201001012λ8 340 0λ 14.8679, 22.8679So the principal stresses are 14.8679 and -22.8679 asdiscuss above. By using eigen values, we can calculateeigenvectors. These eigen vector are use for calculatingthe angle at which the principal stresses act.# Let X # be the eigen vector such that (A(A I)X 0# 20100%& #101200014.8679 %@ IJTSRD 2014.86791014.860022.86201010120.286713.48671Here B is model matrix (matrix combination of alleigenvector) and D is digonalized form of matrix ANow we calculate the eigen value of matrix A.For0.9240.075912Now compile this unit eigenvector into rotation matrix Rsuch that determinant R 10.96120.2756R 0.2756 0.9612Determinant R (0.9612) (0.9612) (0.2756) (0.2756) 1R * ,,./-,./* ,-D’ RTAR # 10& #14.867914Unique Paper ID – IJTSRD3834838348These imply that the calculated eigenvector are correct.Now to calculate angle, we must calculate unit eigenvector.0.28673.48670.96120.2756 and 201010 0.961212 0.27560.27560.9612So θ 16 , as we found earlier for one of the principalangles. Using the rotation angle of 16 , the matrix A(representing the original stress state of the element) canbe transformed to matrix D’ (representing the principalstress state).Volume – 5 Issue – 2 January-FebruaryFebruary 2021Page 143

International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470The quantity of heat Q2 flowing out of the section at adistance x x will be u Q2 kA x x δx Per secondThe amount of heat retained by the slab with thickness δxis, therefore, u u Q1 Q2 KA x x δx x x Persecond . (1)In this way we can calculate the principal stresses andangles at which the principal stresses act by matrixmethod.Partial differential equationsA large number of problems in fluid mechanics, solidmechanics, heat transfer, electromagnetic theory andother areasas of physics and engineering science aremodelled as Initial Value Problems and boundary valueproblems consisting of partial differential equations. Inthis paper, some of most important partial differentialequations of one dimensional heat equation havehav beenderived and solved.Partial differential equations are used for heatconduction analysis.Second order differential equation is used to findmaxima and minima of function of several variables.Partial differential equation help to provide shape andinterior, exterior design of machine.Partial differential equation is used to calculate heatflow in one and two dimensions.One dimensional heat flowLet us consider a conduction of heat along a bar whoseboth sides are insulated. Also the loss of heat from thesides of the bar by conduction or radiation is negligible.One end of the bar is taken as origin and direction of heatflow is along positive x-axis.axis. The temperature u at anypoint of the bar is depend upon the distance x of the pointfrom one end and the time t. The temperature of all pointsof any cross-section is the same.The rate of increase of heat in the slab is sρAδx u, t . (2)Where s is thee specific heat and p is the density of thematerial of the bar.From (1) and (2) we have therefore232323,0 1#5 6 76 72423:24; 23E24 2 ?6 @ 72F 32 F@AB@9 ?289C6 @ 7@D, GHIJI E:; is known as the thermal diffusivity of the material of thebar.Solution of heat EquationThe heat equation isKLK LE . .1KMK#where the symbols have got their usual meanings.Let L O M P # , . .22Then2324AndK LK#PQRQ4S POS#Taking the above substitutions in (1), we obtainSOS PPEOSMS#1 SO1S P5 ,TU 3 E O SMP S#Hence, the quantity of heat Q1 flowing into the section at adistance x will be u Q1 KA x x Per secondThe negative sign on RHS is because u decreases as xincreases,@ IJTSRD Unique Paper ID – IJTSRD3834838348 The solutions will now be found under the following threecases:Case I: When K 0, we have from (3)1S P1 SO0,0P S#E O SM SOSM0,S PS#Volume – 5 Issue – 2 0January-FebruaryFebruary 2021Page 144

International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470L #, ML #, MAfter integration we get O E , T/S P E # EWL #, MUsing this in equation (2), we getL E E # EW .Which is a solution of (1).Case II: When K m2, i.e. K is 0, we have from equation(3)1 SO1S PX ,XE O SMP S# SOOX E SM,S PS#X E M log O OE] I F a F4X P0log E] , . ., P E. b.0Ec I X0Ed I C f. g.S PX PS#X E M n oEp ,X E SM,A.E.X0 C F a F 4 O Ep I, X.0, sXE I C6 } F F7 a 4EW sin 6/ #7 72 ‚ I C6 ƒ } F F7 a 4sin 6/ #7 82 ƒ } ‚ sin 6 7 ‚ 1 ‚c‚ sin 63}‚7‚W‚W sin 6‚]W}‚d7‚] sin 6 ]}07From (8), we have thereforec} F F}F F #5 #L ‚ I C ] a 4 ,./‚c I C6 7 a 4 ,./22From (2) we haveF FL Ep I C a 4 Ev cos X# Ew sin X# . Ewhich is a solution of equation (1).LAmong these solutions, we have to choose that solutionwhich is consistent with physical nature of problem. Sinceu decreases as t increases, the only suitable solution ofheat equation (1) is solution (C).ExampleExample: Find the temperature in a bar of length 2 unitswhose ends are kept at zero temperature and lateral}surface insulated if the initial temperature is ,./c}From (6) we haveUsing (5) in (8), we obtain #5 #/ #sin3 sin ‚ sin 67222X C.F. I u Ev * ,s# Ew ,./s# I x Ev * ,X# Ew ,./X# Ev * ,X# Ew ,./X# ( f. g. 03 ,./Using condition (4) in (2), we obtainF F0 E I C a 4 EW sin 2X sin 2X 0 ,.// 2X / X / /2L0 s. tFrom (2), we have thereforeF FL E I C a 4 EW sin X# 6The general solution iswhich is a solution of equation (1)Case III: When K - m2, i.e. K 0, we have from (3)1 SO1S PX ,XE O SMP S#SOO log O0 . 32 . 45 #3 sinTM M 0 . 52Using the condition (3) in (2), we obtainF F0 E I C a 4 E E0LFrom (2), we have thereforeF Fj@lj@ LE] I a 4 ah i 8ak i . m 0 TM #0 TM # #sin2.}F F4IC ] aThe solution of equation (1) consistent with physicalnature of problem is given byF FL E I C a 4 E cos X# EW sin X# 2 #2c} F F7 a 43I C6,./5 #2which is the required temperature.Laplace TransformLaplace Transform: The Laplace Transform is thetransform to time domain (t) into complex domain (s).Laplace Transform plays an important role in engineeringsystem. The concept of Laplace Transform is applied in thearea of science and technology such as to find the transferfunction in mechanical system.The Laplace Transform of f(t) is defined and denoted as † ‡ M ˆSolution: One dimensional heat equation isKLK LE . 1KMK#,./‰x I C;4 ‡ M SM Where f(t) is the function of t and t 0. Provided thatintegral exist, s is the parameter which may be real andcomplex.Find the transfer function of the system shown below?Where@ IJTSRD Unique Paper ID – IJTSRD38348 Volume – 5 Issue – 2 January-February 2021Page 145