ME6603- Finite-Element-Analysis - III YEAR-VI SEMESTER.pdf

:

1

SUBJECT

CODE

ME 6603

2

SUBJECT

NAME

FINITE

ELEMENT

ANALYSIS

3 YEAR/SEM

III / VI

4 COURSE

CODE C314

ME 6603 -FINITE ELEMENT ANALYSIS

.

PREREQUISITE:

• Heat and Mass transfer

• Basic Mathematics

• Engineering Mechanics

• Strength of Materials

COURSE OBJECTIVES:

• To introduce the concepts of Mathematical Modeling of Engineering Problems. • To appreciate the use of FEM to a range of Engineering Problems.

COURSE OUTCOMES:

Upon completion of this course the student will be able to:

CO314.1 Acquire the concept and purpose of Finite element methods.

CO314.2 Understand the methods of FEA by using different types of elements.

CO314.3 Solve engineering problems by mathematical differential equations.

CO314.4 Implement the methodology of FEA for different applications.

CO314.5 Apply the concept of finite element analysis for solving different kind of

engineering problems.

CO-PO MAPPING

CO/P PO PO PO PO PO PO PO PO PO PO1 PO1 PO1

O 1 2 3 4 5 6 7 8 9 0 1 2

CO314.1 3 3 2 1 1

CO314.2 3 3 2 2 1

C314.3 2 3 1 2

C314.4 3 2 1 1

C315.5 2 1 2 2 1

SYLLABUS

UNIT I INTRODUCTION 9 Historical Background – Mathematical Modeling of field problems in Engineering – Governing Equations – Discrete and continuous models – Boundary, Initial and Eigen Value problems– Weighted Residual Methods – Variational Formulation of Boundary Value Problems – RitzTechnique – Basic concepts of the Finite Element Method.

UNIT II ONE-DIMENSIONAL PROBLEMS 9 One Dimensional Second Order Equations – Discretization – Element types- Linear and Higher order Elements – Derivation of Shape functions and Stiffness matrices and force vectors- Assembly of Matrices - Solution of problems from solid mechanics and heat transfer. Longitudinal vibration frequencies and mode shapes. Fourth Order Beam Equation –Transverse deflections and Natural frequencies of beams.

UNIT III TWO DIMENSIONAL SCALAR VARIABLE PROBLEMS 9 Second Order 2D Equations involving Scalar Variable Functions – Variational formulation –Finite Element formulation – Triangular elements – Shape functions and element matrices and vectors. Application to Field Problems - Thermal problems – Torsion of Non circular shafts –Quadrilateral elements – Higher Order Elements.

UNIT IV TWO DIMENSIONAL VECTOR VARIABLE PROBLEMS 9

Equations of elasticity – Plane stress, plane strain and axisymmetric problems – Body forces and temperature effects – Stress calculations - Plate and shell elements.

UNIT V ISOPARAMETRIC FORMULATION

Natural co-ordinate systems – Isoparametric elements – Shape functions for iso parametric elements – One and two dimensions – Serendipity elements – Numerical integration and application to plane stress problems - Matrix solution techniques – Solutions Techniques to Dynamic problems – Introduction to Analysis Software.

TOTAL: 45 PERIODS CONTENT BEYOND SYLLABUS

• Three dimensional problems • Meshing of complex shaper using analysis software • Application of theoretical knowledge in software like ANSYS for solving structural

problems.

LEARNINGRESOURCES:

TEXT BOOK: 1. Reddy. J.N., “An Introduction to the Finite Element Method”, 3rd Edition, Tata

McGraw-Hill, 2005 2. Seshu, P, “Text Book of Finite Element Analysis”, Prentice-Hall of India Pvt.

Ltd., New Delhi, 2007.

REFERENCES: 1. Rao, S.S., “The Finite Element Method in Engineering”, 3rd Edition, Butterworth

Heinemann, 2004 2. Logan, D.L., “A first course in Finite Element Method”, Thomson Asia Pvt. Ltd.,

2002 3. Robert D. Cook, David S. Malkus, Michael E. Plesha, Robert J. Witt, “Concepts

and Applications of Finite Element Analysis”, 4th Edition, Wiley Student Edition,

2002. 4. Chandrupatla & Belagundu, “Introduction to Finite Elements in Engineering”,

3rd Edition, Prentice Hall College Div, 1990 5. Bhatti Asghar M, "Fundamental Finite Element Analysis and

Applications", John Wiley & Sons, 2005 (Indian Reprint 2013)*

ADDITIONAL RESOURCES : 1. NPTEL TUTORIALS (Internal Server) 2. FEA Books (PDF Formats) 3. Online Objective Question

Unit 1

ONE-DIMENSIONAL PROBLEMS

Part –A

1. Name the variational methods. [CO1 – L2 – May 14] 1. Ritz method 2. Rayleigh –Ritz method

2. Name the weighted residual methods. [CO1 – L2 – May 14] 1. Point collocation method 2. Sub-domain collocation method 3. Least squares method 4. Galerkin‟s method.

3. What is Raleigh – Ritz method? . [CO1 – L2 – Dec 10]

Rayleigh-Ritz method is an integral method which is useful for solving complex structural

problems, encountered in finite element analysis. This method is possible only if a

suitable functional is available.

4. What is meant by discretisation and assembly? . [CO1 – L2 – May 07]

The art of sub dividing the structure into a convenient number of smaller components is

known as discretisation. The smaller components are put together and this process of

combining all the elements together is known as assemblage.

5. What is aspects ratio? . [CO1 – L2 – Dec 11]

Aspect ratio is the ratio of the largest dimension of the element to the smallest dimension

of the element. In many cases, if the aspect ratio increases the inaccuracy of the solution

increases. The aspect ratio should be close to unity as for as possible.

6. What is meant by finite element analysis? . [CO1 – L2 ]

A finite element method is a numerical method for solving problems of engineering and

mathematical physics. In this method, instead of solving the problem for the entire body

in one operation, we formulate equations for each element and combine them to obtain

the solution for the whole body.

7. What are the types of boundary condition? . [CO1 – L2 – May 11]

There are two types of boundary condition. They are: 1. Primary boundary condition 2. Secondary boundary condition

8. What are the methods generally associated with the finite element analysis?

. [CO1 – L1] Force method and Displacement or stiffness method are the two methods.

9. Expalin force method. . [CO1 – L2 – may 15]

In force method, internal forces are considered as unknowns of the problem. In

displacement or stiffness method, the displacements are considered as unknowns of the

problem. Among the two methods , displacement method is desirable.

10. Why polynomial type of interpolation functions are mostly used due to

the following reasons: . [CO1 – L2 – dec 14] 1. It is easy to formulate and computerize the finite element equations 2. It is easy to perform differentiation or integration 3. The accuracy of the results can be improved by increasing the order of the polynomial.

11. What are ‘h’ and ‘p’ versions of finite element method? [CO1 – L2 – Dec 03]

In „h‟ version, the order of the polynomial approximation for all elements is kept

constant and the number of elements increased.

In „p‟ version the number of elements is maintained constant and the order of

polynomial approximation of element is increased.

12. Name any four FEA software’s. [CO1 – L2 – May08] 1. Ansys 2. NASTRAN 3. COSMOS 4. NISA 5. ABAQUS

13. Differentiate between global and local axes. [CO1 – H2 –may 13]

Local axes are established in an element, they change with change in orientation of

the element. The direction differs from element to element.

Global axes are defined for the entire system. They have the same direction for all

the elements even though the elements are differently oriented.

13. List the Advantages of Finite Element Method . [CO1 – H2 – may 15] 1. FEM can handle irregular geometry in a convenient manner. 2. Handles general load conditions without difficulty 3. Non – homogeneous materials can be handled easily. 4. Higher order elements may be implemented.

14. List the Disadvantages of Finite Element Method 1. It requires a digital computer and fairly extensive 2. It requires longer execution time compared with FEM. 3. Output result will vary considerably.

15. What are the Applications of Finite Element Analysis? [AU – Dec 06]

Structural Problems:

1. Stress analysis including truss and frame analysis 2. Stress concentration problems typically associated with holes, fillets or

other changes in geometry in a body. 3. Buckling Analysis: Example: Connecting rod subjected to axial compression. 4. Vibration Analysis: Example: A beam subjected to different types of

loading. Non - Structural Problems: 1. Heat Transfer analysis:

Example: Steady state thermal analysis on composite cylinder. 2. Fluid flow analysis:

Example: Fluid flow through pipes. 16. State the methods of engineering analysis. [CO1 – L2 – dec 14]

1. Experimental Method 2. Analytical Method 3. Numerical Method (or) Approximate Method

17. What is meant by node or joint? [CO1 – L2 – May12]

Each kind of finite element has a specific structural shear and is interconnected

with the adjacent element by node points (or) nodes. At the nodes, degrees of freedom

are located. The forces will act at nodes and not at any other place in the element.

18. What is Structural and non-structural problems? [CO1 – L1 – dec 13]

Structural Problems In structural problems, displacement at each nodal point is obtained. By using these

displacement solutions, stress and strain in each element can be calculated. Non-

Structural Problems: In Non-Structural problems, temperature or fluid pressure at each nodal point is

obtained. By using these values, properties such as heat flow, fluid flow, etc.,

Part –B

1. List and briefly describe the general steps of the finite element method

and Historical Background . [CO1 – L2 – may 13]

Finite Element Analysis





3. The following differential equation is available for a physical phenomenon

AE d2y/dx2 +q0 =0 with the boundary condition y(0)=0, x=L, dy/dx=0, find

the value of f(x) using the weighted residual method. . [CO1 – H2 – May 14]



4. Find the solution for the following differential equation. d2y/dx2+50=0,

0<x<10 Trial function is y=a1x(10-x) Boundary conditions are y(0)=0

Y(10)=0

Find the value of the parameter a1 by the following methods (i)point collocations, (ii)subdomain collocation, (iii)least square method, (iv)galerkins [CO1 –



5. Determine the expression for the deflection and bending moment in a

simply supported beam subjected to uniformly distributed load over the entire span. Find the deflection and moment at midspan and compare with

exact solution using Rayleigh Ritz method Use y = a1sin (πx/l) + a2 sin (3πx/l).[CO1 – H2 – Nov/dec 2008)



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Finite Element Analysi




6. A beam AB of span ‘l’ simply supported at the ends and carrying a

concentrated load ‘W at the centre ‘C’ as shown in figure. Determine the

deflection at the mid span by using Rayleigh-Ritz method and compare with

exact solution. Use a suitable one term trigonometric trial function.

[CO1 – H2 – May 13]


Finite Element Analysi






7. A simply supported beam is subjected to uniformly distributed load over

entire span and it is subjected to a point load at the centre of the span.

Calculate the deflection at the mid span using Rayleigh-Ritz method and

compare with exact solution. [CO1 – H2 – DEC 14]




Finite Element Ana

8. The differential equation of a physical phenomenon is given by

d2y/dx2+ 500x2=0 ;0<x<1

By using the trial function, Y=a1(x-x3)+a2(x-x

5), calculate the value of the

parameter a1 and a2 by the following methods. (i)point collocation

(ii)subdomain method

(iii)least square


(iv)galerkins




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57 Finite Element Analysis

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UNIT II

ONE-DIMENSIONAL PROBLEMS

PART A

1. What are the types loading acting on a structure? [CO1-L1-MAY/JUNE 2015]

There are three type of loading acting on a structure. They are,

1. Body force (f) 2. Traction force (T) 3. Point load (P)

2. Define body force. [CO1-L1-NOV/DEC 2011]

A body force is distributed force acting on every elemental volume of the

body. Unit: force per unit volume

3. Define traction force. [CO1-L1-MAY/JUNE 2015]

Traction force is defined as distributed force acting on the surface of the body. Unit:

Force per unit area Examples: Frictional resistance, viscous

4. What is a point load? [CO1-L2-NOV/DEC 2010]

Point load is load acting at a particular point which causes displacement.

5. What are the basic steps involved in the finite element modeling.

[CO1-L2-NOV/DEC 2012]

Finite element modeling consists of the following: 1. Discretisation of the structure 2.

Numbering of the nodes.

6. What are the classifications of the coordinates? [CO1-L2-MAY/JUNE 2009]

The co-ordinates are generally classified as , 1. Global co-coordinates 2. Local

Co-ordinates 3. Natural co-ordinates

7. What is natural co-ordinates? [CO1-L1-NOV/DEC 2014]


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A natural co-ordinate system is used to define any point inside the element by a set of

dimensionless numbers, whose magnitude never exceeds unity. This system is useful

in assembling of stiffness matrices.

8. Define shape function. [CO1-L1-MAY/JUNE 2011]

In finite element method, field variables within an element are generally expressed by

the following approximate relation: values of + N2 + N3 are the the field variable at the

nodes and N1 N2 N3 N4 are interpolation function. N1 N2 N3 N4 are called shape

functions because they are used to express the geometry or shape of the element.

9. What are the characteristics of shape function? [CO1-L1-MAY/JUNE 2012]

The characteristics of the shape functions are follows: 1. The shape function has

unit value at one nodal point and zero value at the other nodes. 2. The sum of the

shape function is equal to one.

10. Why polynomials are generally used as shape function?

[CO1-L1-MAY/JUNE 2010]

Polynomials are generally used as shape functions due to the following reasons: 1.

Differentiation and integration of polynomials are quite easy. 2. The accuracy of the

results can be improved by increasing the order of the polynomial. 3. It is easy to

formulate and computerize the finite element equations.

11.Give the expression for element stiffness matrix. [CO1-L1-MAY/JUNE 2011]

Stiffness matrix [K] = [B] T [D][B]d v

Where, [B] matrix is a strain displacement matrix [D] matrix is stress, strain

relationship matrix

12.Write down the expression of stiffness matrix for one dimensional element bar

element. [CO1-L1-MAY/JUNE 2014]

AE 1 Stiffness matrix [K] = l 1 Where, A is the area of the bar element E is the

young‟s modulus of the bar element L is the length of the bar element

13. State the properties of a stiffness matrix. [CO1-L1-MAY/JUNE 2013]


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The properties of the stiffness matrix [K] are, 1. It is a symmetric matrix 2. The sum of

the elements in any column must be equal to zero. 3. It is an unstable element, so the

determinant is equal to zero.

14.Write down the general finite element equation. [CO1-L1-MAY/JUNE 2011]

General finite element equation is, {F} = [K] {u} Where, {F} is a force vector [K] is

the stiffness matrix {u} is the degrees of freedom

15. state the assumptions made in the case of truss element.


The following assumptions are made in the case of truss element, 1. All the members

are pin jointed. 2. The truss is loaded only at the joints 3. The self weight of the

members are neglected unless stated.

16. Write down the expression of shape function N and displacement u for one

dimensional bar element. [CO1-L1-NOV/DEC 2011]

For on dimensional bar element, Displacement function, u = N1 u1 + N2 u 2 l x

Where, Shape function, N1 = l x N2 = l

17. State the principle of minimum potential energy. [CO1-L1-MAY/JUNE 2010]

total strain energy U and the potential energy of the external forces, (W)

18. Distinguish between essential boundary condition and natural boundary

condition. [CO1-L1-NOV/DEC 2011]

There are two type of boundary conditions. They are, 1. Primary boundary condition

(or) essential boundary condition: The boundary condition which in terms of the field

variables is known as primary boundary condition 2. Secondary boundary condition or

natural boundary condition: The boundary conditions which are in the differential form of

field variables is known as secondary boundary condition.

19. What are the difference between boundary value problem and initial value

problem? [CO1-L1-MAY/JUNE 2010]

The solution of differential equation obtained for physical problems which satisfies some

specified conditions known as boundary conditions l 2 Stiffness matrix, [K] = A E e e lm l 2 l

e l m lm l 2 lm m 2 lm m2 lm l 2 lm m 2 lm m 2 If the solution of differential equation is

obtained If the solution of differential equation is obtained together with initial conditions


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then it is known as initial value problem. If the solution of differential equation is obtained

together with boundary conditions then it is known as boundary value problem.

PART-B

1. What is one dimensional element? Explain their types?

[CO1-H1-MAY/JUNE 2010]

Bar and beam elements are considered as One Dimensional elements. These elements are often used to model trusses and frame structures.

Bar, Beam and Truss Bar is a member which resists only axial loads. A beam can resist axial, lateral and twisting loads. A truss is an assemblage of bars with pin joints and a frame is an assemblage of beam elements?

Stress, Strain and Displacement Stress is denoted in the form of vector by the variable x as σx, Strain is denoted in the form of vector by the variable x as ex, Displacement is denoted in the form of vector by the variable x as ux.

Types of Loading (1) Body force (f) It is a distributed force acting on every elemental volume of the body. Unit is Force / Unit volume. Ex: Self weight due to gravity. (2) Traction (T) It is a distributed force acting on the surface of the body. Unit is Force / Unit area. But for one dimensional problem, unit is Force / Unit length. Ex: Frictional resistance, viscous drag and Surface shear. (3) Point load (P) It is a force acting at a particular point which causes displacement.

Finite Element Modeling It has two processes. (1) Discretization of structure (2) Numbering of nodes.


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(A) Global co – ordinates, (B) Local co – ordinates and (C) Natural co –ordinates.

Natural Co – Ordinate (ε) Integration of polynomial terms in natural co – ordinates for two dimensional elements can be performed by using the formula,

Shape function N1N2N3 are usually denoted as shape function. In one dimensional problem, the displacement

i ui =N1 u1 For two noded bar element, the displacement at any point within the element is given by,

i ui =N1 u1 + N2 u2 For three noded triangular element, the displacement at any point within the element is given by,

i ui =N1 u1 + N2 u2 + N3 u3

i vi =N1 v1 + N2 v2 + N3 v3

(a) First derivatives should be finite within an element; (b) Displacement should be continuous across the element boundary

Polynomial Shape function Polynomials are used as shape function due to the following reasons, (1) Differentiation and integration of polynomials are quite easy. (2) It is easy to formulate and computerize the finite element equations. Properties of Stiffness Matrix

1. It is a symmetric matrix,

2. The sum of elements in any column must be equal to zero, 3. It is an unstable element. So the determinant is equal to zero.

Problem (I set) 1. A two noded truss element is shown in figure. The nodal displacements are u1

= 5 mm and u2 = 8 mm. Calculate the displacement at x = ¼, 1/3 and ½.

Problem (II set) 1. Consider a three bar truss as shown in figure. It is given that E = 2 x 105 N/mm2. Calculate (a) Nodal displacement, (b) Stress in each member and (c) Reactions at the support. Take Area of element 1 = 2000 mm2, Area of element 2 = 2500 mm2, Area of element 3 = 2500 mm2.


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`

2. Classification of beams and explain briefly [CO2-H3-NOV/DEC 2010]

1. Cantilever beam, 2. Simply Supported beam, 3. Over hanging beam, 4. Fixed beam and 5. Continuous beam.

1. Point or Concentrated Load, 2. Uniformly Distributed Load and 3. Uniformly

Problem (III set) 1. A fixed beam of length 2L m carries a uniformly distributed load of w (N/m) which runs over a length of L m from the fixed end. Calculate the rotation at Point B.

LINEAR STATIC ANALYSIS (BAR ELEMENT)


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Most structural analysis problems can be treated as linear static problems, based on the following assumptions 1. Small deformations (loading pattern is not changed due to the deformed shape) 2. Elastic materials (no plasticity or failures) 3. Static loads (the load is applied to the structure in a slow or steady fashion) Linear analysis can providemost of the information about the behavior of a structure, and can be a good approximation for many analyses. It is also the bases of nonlinear analysis in most of the cases.

BEAM ELEMENT A beam element is defined as a long, slender member (one dimension is much larger than the other two) that is subjected to vertical loads and moments, which produce vertical displacements and rotations. The degrees of freedom for a beam element are a vertical displacement and a rotation at each node, as opposed to only an horizontal displacement at each node for a truss element. Degrees of Freedom Degrees of freedom are defined as the number of independent coordinates necessary to specify the configuration of a system. The degrees of freedom for a general situation consists of three translations in the x, y, and z directions and three rotations about the x, y, and z axes. A one-dimensional beam element has four degrees of freedom, which include, a vertical displacement and a rotation at each node.

Assumptions Nodal Forces and Moments Forces and moments can only be applied at the nodes of the beam element, not between the nodes. The nodal forces and moments, , are related to the nodal displacements and rotations, through the element stiffness matrix, . Constant Load The loads that are applied to the beam element are assumed to be static and not to vary over the time period being considered, this assumption is only valid if the rate of change of the force is much less than the applied force (F >> dF/dt). If the loads vary significantly, (if the variation in load is not much less than the applied force) then the problem must be considered as dynamic. Weightless Member The weight (W) of the beam is neglected, if it is much less than the total resultant forces (F) acting on the beam. If the weight of the beam is not neglected, then its effects must


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be represented as vertical forces acting at the nodes, by dividing up the weight and lumping it at the nodes, proportionally according to it's placement along the beam. Prismatic Member The beam element is assumed to have a constant cross-section, which means that the cross-sectional area and the moment of inertia will both be constant (i.e., the beam element is a prismatic member). If a beam is stepped, then it must be divided up into sections of constant cross-section, in order to obtain an exact solution. If a beam is tapered, then the beam can be approximated by using many small beam elements, each having the same crosssection as the middle of the tapered length it is approximating. The more sections that are used to approximate a tapered beam, the more accurate the solution will be. The moment of inertia is a geometric property of a beam element, which describes the beams resistance to bending and is assumed to be constant through the length of the element. The moment of inertia can be different along different axes if the beam element is not

symmetric, we use the moment of inertia (I) of the axis about which the bending of the beam occurs

Where (Iz) refers to the moment of inertia, resisting bending about the "z" axis and (Iy) about the "y" axis. The Beam Element is a Slender Member A beam is assumed to be a slender member, when it's length (L) is more than 5 times as long as either of it's cross-sectional dimensions (d) resulting in (d/L<.2). A beam must be slender, in order for the beam equations to apply, that were used to derive our FEM equations.


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3. What are the Beams different Bends without Twisting? [CO1-H1-NOV/DEC 2009]

It is assumed that the cross-section of the beam is symmetric about the plane of bending (x-y plane in this case) and will undergo symmetric bending (where no twisting of the beam occurs during the bending process). If the beam is not symmetric about this plane, then the beam will twist during bending and the situation will no longer be one-dimensional and must be approached as an unsymmetric bending problem (where the beam twists while bending) in order to obtain a correct solution. Cross Section Remains Plane When a beam element bends, it is assumed that it will deflect uniformly, thus the cross section will move uniformly and remain plane to the beam centerline. In other words, plane sections remain plane and normal to the x axis before and after bending. Axially Rigid The one-dimensional beam element is assumed to be axially rigid, meaning that there will be no axial displacement (u) along the beams centriodal axis. This implies that forces will only be applied perpendicular to the beams centriodal axis. The one-dimensional beam element can be used only when the degrees of freedom are limited to vertical displacements (perpendicular to the beams centriodal axis) and rotations in one plane. If axial displacements are present then a one-dimensional bar element must be superimposed with the onedimensional beam element in order to obtain a valid solution. Homogenous Material A beam element has the same material composition throughout and therefore the same mechanical properties at every position in the material. Therefore, the modulus of elasticity E is constant throughout the beam element. A member in which the material properties varies from one point to the next in the member is called inhomogenous (non-homogenous). If a beam is composed of different types of materials, then it must be divide up into elements that are each of a single homogeneous material, otherwise the solution will not be exact.

Isotropic Material A beam element has the same mechanical and physical properties in all directions, i.e., they are independent of direction. For instance, cutting out three tensile test specimens,

one in the x-direction, one in the y-direction and the other oriented 45 degrees in the x-y plane,

a tension test on each specimen, will result in the same value for the modulus of


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elasticity (E), yield strength and ultimate strength . Most metals are considered isotropic. In contrast fibrous materials, such as wood, typically have properties that are directionaly dependant and are generally considered anisotropic (not isotropic). The Proportional Limit is not Exceeded It is assumed that the beam element is initially straight and unstressed. It is also assumed that the material does not yield, therefore the beam will be straight after the load is released. These assumptions mean that the beam must be made of an elastic material, one which will return to it's original size and shape when all loads are removed, if not stressed past the materials elastic or proportional limit. It is also assumed that the beam is not stressed past the proportional limit, at which point the beam will take a permanent set and will not fully return to it's original size and shape, when all loads are removed. Below the proportional limit an elastic material is in the linear elastic range, where the strain ( ) varies linearly with the applied load and the stress ( ) varies linearly according to: , where E is the modulus of elasticity. Rigid Body Modes for the One-Dimensional Beam Element Rigid body motion occurs when forces and/or moments are applied to an unrestrained mesh (body), resulting in motion that occurs without any deformations in the entire mesh (body). Since no strains (deformations) occur during rigid body motion, there can be no stresses developed in the mesh. In order to obtain a unique FEM solution, rigid body motion must be constrained. If rigid body motion is not constrained, then a singular system of equations will result, since the determinate of the mesh stiffness matrix is equal to zero (i.e., ). There are two rigid body modes for the one-dimensional beam element, a translation (displacement) only and a rotation only. These two rigid body modes can occur at the same time resulting in a displacement and a rotation simultaneously. In order to eliminate rigid body motion in a 1-D beam element (body), one must prescribe at least two nodal degrees of freedom (DOF), either two displacements or a displacement and a rotation. A DOF can be equal to zero or a non-zero known value, as long as the element is restrained from rigid body motion (deformation can take place when forces and moments are applied) . For simplicity we will introduce the rigid body modes using a mesh composed of a single element. If only translational rigid body motion occurs, then the displacement at local node I will be equal to the displacement at local node J. Since the displacements are equal there is no strain developed in the element and the applied nodal forces cause the element to move in a rigid (non-deflected) vertical motion (which can be either up as shown below or it can be in the downward direction depending on the direction of the applied forces).

This rigid body mode can be suppressed by prescribing a vertical nodal displacement. If

rotational rigid body motion occurs, then the rotation at local node I will be equal to

the rotation at local node J (i.e., in magnitude and direction). In this situation the nodal forces


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and/or moments applied to the element, cause the element to rotate as a rigid body (either clockwise as shown below or counterclockwise depending on the direction of the applied forces and/or moments).

This rigid body mode can be suppressed by prescribing a nodal translation or rotation. If translational and rotational rigid body motion occurs simultaneously then:


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4. DERIVE THE 1-D 2-NODED CUBIC BEAM ELEMENT MATRICES [CO1-H1-MAY/JUNE 2009]

A single 1-d 2-noded cubic beam element has two nodes, with two degrees of freedom at each node (one vertical displacement and one rotation or slope). There is a total of 4 dof and the displacement polynomial function assumed should have 4 terms, so we choose a cubic polynomial for the vertical deflection. Slope is a derivative of the vertical deflections.


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5. FORMULATE THE DEVELOPMENT OF ELEMENT EQUATION

[CO2-H2-NOV/DEC 2009]


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6. EXPLAIN ABOUT THE BEAM ELEMENT AND FORMULATE THE STIFFNESS MATRIX

[CO2-H1-MAY/DEC 2009]

A beam is a long, slender structural member generally subjected to transverse loading that produces significant bending effects as opposed to twisting or axial effects. An elemental length of a long beam subjected to arbitrary loading is considered for analysis. For this elemental beam length L, we assign two points of interest, i.e., the ends of the beam, which become the nodes of the beam element. The bending deformation is measured as a transverse (vertical) displacement and a rotation (slope). Hence, for each node, we have a vertical displacement and a rotation (slope) – two degrees of freedom at each node. For a single 2-noded beam element, we have a total of 4 degrees of freedom. The associated “forces” are shear force and bending moment at each node.

8 Finite Element Analysis

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The stiffness term ij k indicates the force (or moment) required at i to produce a unit deflection (or rotation) at j, while all other degrees of freedom are kept zero. Sign conventions followed Upward forces are positive and upward displacements are positive. Counter-clockwise moments are positive and counter-clockwise rotations are positive. Formulae required – cantilever beam subjected to concentrated load and moment.


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7. Find the slopes at the supports and support reaction forces and support

reaction moments for the beam shown in Figure. Take E=210 GPa, I = 2×10-4 m4



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8. Given that E=210 GPa and I=4×10-4m4, cross section of the beam is constant. Determine the deflection and slope at point C. calculate the reaction forces and moments.



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UNIT III

TWO DIMENSIONAL SCALAR VARIABLE PROBLEMS

PART - A

1. Write a displacement function equation for CST element. [CO1-L2-NOV/DEC 2012]

Displacement function

{

( )

} [ ] ( )

{ }

where, N1, N2, N3 are shape functions.

2. Define path line and streamline? [CO1-L2-NOV/DEC 2012]

A path line is defined as locus of points through which a fluid particle of fixed identity passes as it moves in space.

A streamline is an imaginary line that connects a series of ponts in space at a given instant in such a manner that all particles falling on the line at that instant have velocities whose vectors are tangent to the line.

3. How do you define two dimensional elements? [CO1-L2-NOV/DEC 2012]

Two dimensional elements are define by three or more nodes in a two dimensional plane. The basic element useful for two dimensional analysis is the triangular element.

4. What is CST element? [CO1-L2-NOV/DEC 2012]

Three noded triangular elements are known as CST. It has six

unknown displacement degrees of freedom (u1, v1, u2, v2, u3, v3). The element is called CST because it has a constant strain throughout it.


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5. What is LST element? [CO1-L2-NOV/DEC 2012]

Six nodded triangular elements are known as LST. It has twelve unknown displacement degrees of freedom. The displacement function for the elements are quadratic instead of linear as in the CST.

6. What is QST element? [CO1-L2-NOV/DEC 2012]

Ten nodded triangular elements are known as Quadratic strain triangle. It is also called as cubic displacement triangle.

7. What meant by plane stress analysis? [CO1-L2-NOV/DEC 2012]

Plane stress is defined to be a state of stress in which the normal stress and shear stress directed perpendicular to the plane are assumed to be zero.


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8. Define plane strain analysis. [CO1-L2-NOV/DEC 2012]

Plane strain is defined to be state of strain normal to the xy plane and the shear strains are assumed to be zero.

9. Write down the stiffness matrix equation for two dimensional CST element. [CO1-L2-NOV/DEC 2012]


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Stiffness matrix , - , - , -, -

, - - Strain displacement , - - Stress strain matrix

, - - Strain displacement matrix A - Area of the element

T - Thickness of the element

10. Write down the stress-strain relationship matrix for plane stress condition. [CO1-L2-NOV/DEC 2012]

For plane stress problems, stress-strain relationship matrix is, , - [ ]

Where, E = Young‟s modulus V = Poisson‟s ratio

11. Write down the stress-strain relationship matrix for plane strain condition. [CO1-L2-NOV/DEC 2012]

For plane strain problems, stress-strain relationship matrix is,

( )

, -

[

( )

]

( ) ( )

Where, E = Young‟s modulus V = Poisson‟s ratio

12. Write a strain-displacement matrix for CST element. [CO1-L2-NOV/DEC 2012]


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Strain-displacement matrix for CST element is,

, - [ ]

Where, A = Area of the element

; ; ;

; ; ;

13. Write down the expression for the shape function for a constant strain triangular element. [CO1-L2-NOV/DEC 2012]

For CST element,

Shape function,

Where,

;

;

;

;

; ;

14. State the assumption in the theory of pure torsion. [CO1-L2-NOV/DEC 2012]

• The material of the shaft is homogeneous, perfectly elastic and obey‟s Hooe‟s law.

• Twist is uniform along the length of the shaft.

• The stress does not exceed the limit of proportionality.

• Strain and deformation are small.

15. Write down the finite element equation for torsinal bar element. [CO1-L2-NOV/DEC 2012]


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[ [ ] [ ]] { } { }

PART – B

1. Derive the shape function for the constant strain triangular element. [CO3-H2-NOV/DEC 2012]

Figure shows a 2-D two-variable linear triangular element with three nodes and the two dof at each node. The nodes are placed at the corners of the triangle. The two variables (dof) are displacement in x-direction (u) and displacement in y-direction (v). Since each node has two dof, a single element has 6 dof. The nodal displacement vector is given by

* + { }

We select a linear displacement function for each dof as ( ) ( )


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where u x, y v x, y x, y the element.

The above two algebraic equations can also be written as

Using steps we had developed for the 2-D single-variable linear triangular element, we can write

and using the interpolation functions we had developed for the 2-D single-variable linear triangularelement, we can write


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Writing the above equations in matrix form

1. For the constant strain triangular element shown in figure. Assemble strain-

displacement matrix. Take t = 20 mm, E = 2 X 105 N / mm2. [CO3-H2-NOV/DEC 2012]


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Given data:

Young‟s modulus, E = 2 X 105 N / mm

2

Thickness, t = 20 mm

To find: Strain-displacement matrix [B]

Solution: We know that,

Strain-displacement matrix for CST element is, , - [ ] .... (1)

Where, A = Area of the element

Substitute the above values in eqation (1)


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, - [ ] .... (2)

Where, A = Area of the element | |

| | , ( ) ( ) ( )-

Substitute A value in equation (2) , - [ ] , - [ ]

Result :

Strain-displacement matrix , - [ ]

2. Determine the shape functions N1, N2 and N3 at the interior point P for the

triangular element shown in figure [CO3-H2-NOV/DEC 2012]


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Given data:

To find: Shape functions N1, N2 and N3 at the interior point P

Solution: We know that,

( ) ( ) …(1)

( ) ( ) …(2)

Substituting the co-ordinates values,

( ) ( ) …(3

( ) ( ) …(4)

Equation (3) becomes,

…(5)

Equation (4) becomes,


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…(6)

Solving equation (5) and (6)

Solving,

Substituting N1 value in equation (5) or equation (6), ( )

we know that, N1+N2+N3 = 1

0.4166+0.1111+N3 = 1

Solution:

Shape functions N1, N2 and N3 at the interior point P

3. Determine the stiffness matrix for the straight-sided triangular element of

thickness t = 1 mm, as shown. Use E = 70 GPa, n = 0.3 and assume a plane

stress condition. [CO3-H2-NOV/DEC 2012]


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Solution:


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4. Consider a thin plate having thickness t= 0.5 in. being modeled using two CST

elements, as shown. Assuming plane stress condition, (a) determine the

displacements of nodes 1 and 2, and (b) estimate the stresses in both

elements. [CO3-H2-NOV/DEC 2012]


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IV Unit

TWO DIMENSIONAL VECTOR VARIABLE PROBLEMS

Part A

1. What are the basic steps involved in the finite element modeling.

[CO1-L2-NOV/DEC 2012]

Finite element modeling consists of the following: 1. Discretisation of the structure 2.

Numbering of the nodes.

2. What are the classifications of the coordinates? [CO1-L2-MAY/JUNE 2009]

The co-ordinates are generally classified as , 1. Global co-coordinates 2. Local

Co-ordinates 3. Natural co-ordinates

3. What is natural co-ordinates? [CO1-L1-NOV/DEC 2014]

A natural co-ordinate system is used to define any point inside the element by a set of

dimensionless numbers, whose magnitude never exceeds unity. This system is useful

in assembling of stiffness matrices.

4. Define shape function. [CO1-L1-MAY/JUNE 2011]

In finite element method, field variables within an element are generally expressed by

the following approximate relation: values of + N2 + N3 are the the field variable at the

nodes and N1 N2 N3 N4 are interpolation function. N1 N2 N3 N4 are called shape

functions because they are used to express the geometry or shape of the element.

5. What are the characteristics of shape function? [CO1-L1-MAY/JUNE 2012]

The characteristics of the shape functions are follows: 1. The shape function has

unit value at one nodal point and zero value at the other nodes. 2. The sum of the

shape function is equal to one.

6. Why polynomials are generally used as shape function?


Polynomials are generally used as shape functions due to the following reasons: 1.

Differentiation and integration of polynomials are quite easy. 2. The accuracy of the


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results can be improved by increasing the order of the polynomial. 3. It is easy

to formulate and computerize the finite element equations.

7.Give the expression for element stiffness matrix. [CO1-L1-MAY/JUNE 2011]

Stiffness matrix [K] = [B] T [D][B]d v

Where, [B] matrix is a strain displacement matrix [D] matrix is stress, strain

relationship matrix

8.Write down the expression of stiffness matrix for one dimensional element bar

element. [CO1-L1-MAY/JUNE 2014]

AE 1 Stiffness matrix [K] = l 1 Where, A is the area of the bar element E is the

young‟s modulus of the bar element L is the length of the bar element

9. State the properties of a stiffness matrix. [CO1-L1-MAY/JUNE 2013]

The properties of the stiffness matrix [K] are, 1. It is a symmetric matrix 2. The sum of

the elements in any column must be equal to zero. 3. It is an unstable element, so the

determinant is equal to zero.

10.Write down the general finite element equation. [CO1-L1-MAY/JUNE 2011]

General finite element equation is, {F} = [K] {u} Where, {F} is a force vector [K] is

the stiffness matrix {u} is the degrees of freedom

11. state the assumptions made in the case of truss element.


The following assumptions are made in the case of truss element, 1. All the members

are pin jointed. 2. The truss is loaded only at the joints 3. The self weight of the members

are neglected unless stated.

12. Write down the expression of shape function N and displacement u for one

dimensional bar element. [CO1-L1-NOV/DEC 2011]

For on dimensional bar element, Displacement function, u = N1 u1 + N2 u 2 l x

Where, Shape function, N1 = l x N2 = l

13. State the principle of minimum potential energy. [CO1-L1-MAY/JUNE 2010]

total strain energy U and the potential energy of the external forces, (W)


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Part B

1. Derive the shape function equation for the Axisymmetric element. [CO4 – H2 – Nov/Dec’10]


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2. Derive the Strain Displacement Matrix for the Axisymmetric element. [CO4 – H2 – Nov/Dec’11]


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3. Derive the Stress Strain Relationship matrix for the Axisymmetric

triangular element. [CO4 – H2 – Nov/Dec’13]


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4. The nodal co-ordinates for an axisymmetric triangular element are given below.

r1= 10 mm , r2= 30 mm , r3= 30 mm z1 = 10 mm, z2 = 10 mm, z3 = 40 mm.

Evaluate [B] Matrix for the element[CO4 – H2 – Nov/Dec’14]


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5. The nodal coordinates for an Axisymmetric triangular element are given below

r1= 20 mm , r2= 40 mm , r3= 30 mm z1 = 40 mm, z2 = 40 mm, z3 = 60 mm.

Evaluate [B] Matrix for the element[CO4 – H2 – Nov/Dec’15]


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6. For the element shown in fig, determine the stiffness matrix. Take E = 200Gpa

and v = 0.25.The co-ordinates shown in fig are in millimeters. [CO4 – H2 –

Nov/Dec’13]


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7. For the axisymmetric elements shown in fig, determine the stiffness matrix.

Let E = 2.1 x 105 N/mm2 and v = 0.25. The co-ordinates shown in fig are in

millimeters. [CO4 – H2 – Nov/Dec’12]


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8. For the axisymmetric elements shownin fig.Determine the element stresses.

Let E = 210 Gpa and v = 0.25. The co ordinates are in millimeters.

The nodal displacements are u1= 0.05 mm , u2= 0.02 mm , u3= 0 mm ,w1 = 0.03 mm

, w2 = 0.02 mm, w3 = 0 mm. [CO4 – H2 – Nov/Dec’14]


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UNIT V

ISOPARAMETRIC FORMULATION

PART A

1. Define Isoparametric element? [CO5 – L2 – MAY/JUN’11]

Isoparametric Elements. Isoparametric Formulation of the Bar Element. The term

isoparametric is derived from the use of the same shape functions (or

interpolation functions) [N] to define the element's geometric shape as are used

to define the displacements within the element.

2. Differentiate between Isoparametric, super parametric and sub-parametric

elements. [CO5 – L3 – MAY/JUN’13]

Isoparametric If the number of nodes used for defining the geometry is same as

number of nodes used for defining the displacements is known as isoparametric

element. Super Parametric: If the number nodes used for defining the geometry is more

than number of nodes used for defining the displacements is known as super

parametric element. Sub-parametric elements: If the number of nodes used for defining the geometry

is less than number of nodes used for defining the displacements known as

isoparametric element.

3. Define Isoparametric formulation? [CO5 – L2 – MAY/JUN’10]

Isoparametric formulations help us solve two problems. „ Help simplify the

definition of the approximate displacement field for more complex planar

elements 4-sided elements, elements with curved edges.„ Significantly reduce the

integration process by ensuring that we always have integrals in terms of natural

coordinates that are taken over fixed bounds

4. Explain the Jacobian transformation? [CO5 – L3 – MAY/JUN’12]

The Jacobian generalizes the gradient of a scalar-valued function of multiple


https://en.wikipedia.org/wiki/Gradient

https://en.wikipedia.org/wiki/Scalar_(mathematics)

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variables, which itself generalizes the derivative of a scalar-valued function of a

single variable. In other words, the Jacobian for a scalar-valued multivariate function

is the gradient and that of a scalar-valued function of single variable is simply its

derivative. The Jacobian can also be thought of as describing the amount of

"stretching", "rotating" or "transforming" that a transformation imposes locally.

5. Write down the jacobian matrix for four noded quadrilateral element. [CO5 – L2 – MAY/JUN’10]

Jacobian matrix,[ J ] = J11 J12 J21 J22

Where, J11 = ¼[-(1- η)x1 + (1- η)x2 + (1+η)x3 - (1+ η)x4 ] J12 = ¼[-(1- η)y1 + (1- η)y2 + (1+η)y3 - (1+ η)y4 ] J21 = ¼[-(1- Є)x1 - (1+ Є)x2 + (1+ Є)x3 + (1- Є)x4 ] J22 = ¼[-(1- Є)y1 - (1+ Є)y2 + (1+ Є)y3 + (1- Є)y4 ]

Where, Є and η are natural co-ordinates. x1 x2 x3 x4 y1 y2 y3 y4 are Cartesian coordinates.

6. What is the purpose of isoparametric elements? [CO5 – L1 – MAY/JUN’13]

It is difficult to represent the curved boundaries by straight edges finite elements. A large number of finite elements may be used to obtain reasonable resemblance between original body and assemblage. In order to overcome this drawback, isoparametric elements are used i.e. for problems involving curved boundaries; a family of elements known as “isoparametic elements”.

7. Write down the shape function for 4 noded rectangular elements using natural co-ordinate system. [CO5 – L3 – MAY/JUN’12]

Shape functions: N1 = ¼ (1-Є) (1-η) N2 = ¼ (1+Є) (1-η) N3 = ¼ (1+Є) (1+η) N4 = ¼ (1-Є) (1+η)

Where, Є and η are natural co-ordinates.

8. What is the difference between natural co-ordinates and simple natural co-ordinate? [CO5 – L2 – MAY/JUN’113]

A natural co-ordinate is one whose value lies between zero and one. Examples: L2 = x/l; l =

(1-x/l) Area co-ordinates : L1 = A1/A ; L2 = A2/A ; L3 = A3/A. A simple natural co


https://en.wikipedia.org/wiki/Multivariate_function



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ordinates is one whose value lies between -1 to +1

9. Write down the Jacobian matrix for four noded Quadrilateral element. [CO5 – L2 – MAY/JUN’10]

10. Name a few FEA packages. [CO5 – L2 – MAY/JUN’13]

ABACUS, ADINA, ANSYS,SAP,NISA,COSMOS,PAFEC.

11. Write down the stiffness matrix equation for four noded isoparametric quadrilateral elements. [CO5 – L1 – MAY/JUN’12]

1 1 Stiffness matrix, [k] = t ∫ ∫ [B]T [D] [B] X |J| X δ Є X δ η

-1 -1 Where t = thickness of the element

|J|=Determinant of the jacobian Є and η are natural co-ordinates [B] Strain displacement matrix

[D] Stress strain relationship matrix.


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12. Write down the Gaussian quadrature expression for numerical integration. [CO5 – L2 – MAY/JUN’11]

Gaussian quadrature expression 1 n

∫ f(x) dx = ∑ wi f(xi) -1 i=1

Wi weight function F(xi) values of the function at pre-determined sampling points.

13. Define super parametric element. [CO5 – L3 – MAY/JUN’10]

If the number nodes used for defining the geometry is more than number of

nodes used for defining the displacements is known as super parametric element.

14. What are the types of Eigen value problems? [CO5 – L2 –

MAY/JUN’13]

A set of patch tests is first performed to assess the stability and consistency of the above hierarchic form. The set consists of one-, two-, four-, and eight-element patches. First, we perform a stability assessment by determining the number of zero eigenvalues for each patch.

15. What are the four basic sets of elasticity equations. [CO5 – L1 –

MAY/JUN’12]

An important class of anisotropic materials is one for which three planes of

symmetry exist and is called an orthotropic material. Here the principal axes are

also in rectangular cartesian coordinates. For orthotropic materials it is common to

define the elastic material parameters in terms of theirYoung‟s moduli,

Poisson

ratios and shear moduli.

16. Explain consistent load vector. [CO5 – L3 – MAY/JUN’11]

As part of the input data it is necessary to describe the element formulation to be used in forming the „stiffness‟ matrix and „load‟ vector of each problem. This may be provided either by user written modules (see below) or using the element library provided with the program.

17. What is meant by transverse vibration. [CO5 – L3 – MAY/JUN’10]


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For a properly formulated finite element procedure, the displacement interpolation functions ensure that compatibility is satisfied. The stress-strain law is also enforced; however, differential equilibrium is not satisfied at every point in the continuum. Rather, equilibrium is only satisfied at every node and on each element (see properties I and II above). As such, the finite element solution is approximate, and this approximation becomes more accurate as the mesh is refined, or the interpolation order of the element is increased.

PART B

1. For the isoparametric quadrilateral element shown in fig determine the local co-ordinates of the point P which has Cartesian co-ordinates (7,4). [CO5 – H2 – Nov/Dec’13]

Solution


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2. Evaluate the integeral I = 1 (2 + x + x 2 ) dx and compare with exact solution.

−1

[CO5 – H2 – Nov/Dec’14]

Given Integeral I = 1

(2 + x + x 2 ) dx −1


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3. Derive the shape functions for 4 noded rectangular parent element by using natural co-ordinates system and co-ordinate transformation. [CO5 – H2 – Nov/Dec’15]

Solution:


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4. Evaluate the integral by using Gaussian Quadrature 1 x 2 dx

−1

[CO5 – H2 – Nov/Dec’10]


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5. Derive the element stiffness matrix equation for 4 noded isoparametric quadrilateral element. [CO5 – H2 – Nov/Dec’11]


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=


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6. Evaluate the integral 1 ( x 4 − 3 x + 7) dx . [CO5 – H2 – Nov/Dec’11]

−1


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7. Evaluate the integral by applying 1 Cos x dx 3 point Gaussian 1 − x2

−1

quadrature. [CO5 – H2 – Nov/Dec’11]


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1 1 8. Evaluate I = 3e x + x 2 + dx using one point and two point Gauss

−

1 x + 2 Quadrature. [CO5 – H2 – Nov/Dec’12]


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PREPARED BY R.PRASANTH HTS1089 AND P.BALAJI HTS1340

ASSISITANT PROFESSOR/MECH


ME6603- Finite-Element-Analysis - III YEAR-VI SEMESTER.pdf

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