Me6603 sd by easy engineering.net

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2. SYLLABUS
ME 6603 FINITE ELEMENT ANALYSIS L T P C
3 0 0 3
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 9
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.
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)
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3. TABLE OF CONTENTS
S.NO TABLE OF CONTENTS
PAGE..
NO
a. Aim and Objective of the subject 4
b. Detailed Lesson Plan 5
c. Unit I- Introduction -Part A 8
d. Unit I- Introduction -Part B 10
e. Unit II- One-dimensional problems -Part A 37
f. Unit II- One-dimensional problems -Part B 39
g. Unit III- Two dimensional scalar variable problems -Part A 66
h. Unit III- Two dimensional scalar variable problems -Part B 68
i. Unit IV- Two Dimensional Vector Variable Problems -Part A 95
j. Unit IV- Two Dimensional Vector Variable Problems -Part B 96
k. Unit V- Isoparametric Formulation - Part A 117
l. Unit V- Isoparametric Formulation - Part B 120
m. Question bank 141

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ME 6603 FINITE ELEMENT ANALYSIS
AIM
 The goal is to understand the fundamentals of the finite element method for the
analysis of engineering problems arising in solids and structures.
 The course will emphasize the solution to real life problems using the finite
element method underscoring the importance of the choice of the proper
mathematical model, discretization techniques and element selection criteria.
OBJECTIVES:
1. To apply knowledge of mathematics, science and engineering to the analysis of simple
structures using the finite element method.
2. To analyze and interpret the results.
3. To identify, formulate, and solve engineering problems using the finite element
method.
4. To perform steady-state and transient heat transfer analysis including the effects of
conduction, convection, and radiation.
5. To perform modal analysis of a part to determine its natural frequencies, and analyze
harmonically-forced vibrations.

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SCAD GROUP OF INSTITUTIONS
Department of Mechanical Engineering
Detailed Lesson Plan
Name of the Subject& Code: ME 6603 FINITE ELEMENT ANALYSIS
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)*
S.No
Unit
No
Topic / Portions to be Covered
Hours
Required
/ Planned
Cumulative
Hrs
Books
Referred
1 1 Historical Background 1 1 T1,R1
2 1 Mathematical modeling of field problems in
Engineering
1 2 T1,R1
3 1 Governing Equations 1 3 T1,R1
4 1 Discrete and continuous models 1 4 T1,R1
5 1
Boundary, Initial and Eigen Value problems
1 5 T1,R1
6 1
Weighted Residual Methods concept
1 6 T1,R1
7 1
Weighted Residual Methods-Problems
1 7 T1,R1
8 1 Variational Formulation of Boundary Value
Problems
1 8 T1,R1
9 1
Ritz Technique concept
1 9 T1,R1
10 1
Ritz Technique -Problems
1 10 T1,R1
11 1
Basic concepts of the Finite Element Method.
1 11 T1,R1
12 2 One Dimensional Second Order Equations 1 12 T1,R1

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13 2 Discretization – Element types 1 13 T1,R1
14 2
Derivation of Shape functions and Stiffness
matrices and force vectors (Linear)
1 14
T1,R1
15 2
Derivation of Shape functions (Higher order
Elements)
1 15
T1,R1
16 2
Derivation of Stiffness matrices and force
vectors(Higher order Elements)
1 16
T1,R1
17 2
Solution of problems from solid mechanics
and heat transfer
1 17
T1,R1
18 2 Solution of problems from solid mechanics 1 18 T1,R1
19 2
Longitudinal vibration frequencies and mode
shapes
1 19
T1,R1
20 2 Fourth Order Beam Equation 1 20 T1,R1
21 2 Transverse deflections of beams. 1 21 T1,R1
22 2 Transverse Natural frequencies of beams. 1 22 T1,R1
23 3
Second Order 2D Equations involving Scalar
Variable Functions
1 23 T1,R1
24 3
Variational formulation -Finite Element
formulation
1 24
T1,R1
25 3
Triangular elements – Shape functions and
element matrices and vectors.
1 25
T1,R1
26 3 Application to Field Problems 1 26 T1,R1
27 3 Thermal problems 1 27 T1,R1
28 3 Torsion of Non circular shafts 1 28 T1,R1
29 3 Quadrilateral elements 1 29 T1,R1
30 3 Higher Order Elements concept 1 30 T1,R1
31 3 Higher Order Elements problems 1 31 T1,R1
32 4 Equations of elasticity 1 32 T1,R1

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33 4 Plane stress condition 1 33 T1,R1
34 4 plane strain conditions 1 34 T1,R1
35 4 Axisymmetric problems 1 35 T1,R1
36 4 Body forces in axisymmetric 1 36 T1,R1
37 4 temperature effects in axisymmetric 1 37 T1,R1
38 4 Stress calculations 1 38 T1,R1
39 4 Plate and shell elements 1 39 T1,R1
40 5 Natural co-ordinate systems 1 40 T1,R1
41 5 Isoparametric elements 1 41 T1,R1
42 5
Shape functions for iso parametric elements –
One and two dimensions
1 42
T1,R1
43 5 Serendipity elements 1 43 T1,R1
44 5
Numerical integration and application to
plane stress problems
1 44
T1,R1
45 5 Matrix solution techniques 1 45 T1,R1
46 5 Solutions Techniques to Dynamic problems 1 46 T1,R1
47 5 Introduction to Analysis Software 1 47 T1,R1

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UNIT-1 INTRODUCTION
Part- A
1. Distinguish one Dimensional bar element and Beam Element (May/June 2011)
1D bar element: Displacement is considered.
1D beam element: Displacement and slope is considered
2. What do you mean by Boundary value problem?
The solution of differential equation is obtained for physical problems, which satisfies some
specified conditions known as boundary conditions.
The differential equation together with these boundary conditions, subjected to a boundary
value problem.
Examples: Boundary value problem.
d2
y/dx2
- a(x) dy/dx – b(x)y –c(x) = 0 with boundary conditions, y(m) = S and y(n) = T.
3. What do you mean by weak formulation? State its advantages. (April/May 2015), (May/June
2013)
A weak form is a weighted integral statement of a differential equation in which the
differentiation is distributed among the dependent variable and the weight function and also
includes the natural boundary conditions of the problem.
 A much wider choice of trial functions can be used.
 The weak form can be developed for any higher order differential equation.
 Natural boundary conditions are directly applied in the differential equation.
 The trial solution satisfies the essential boundary conditions.
4. Why are polynomial types of interpolation functions preferred over trigonometric functions?
(May/June 2013)
Polynomial functions are preferred over trigonometric functions due to the following
reasons:
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.
5. What do you mean by elements & Nodes?(May/June 2014)
In a continuum, the field variables are infinite. Finite element procedure reduces such
unknowns to a finite number by dividing the solution region into small parts called Elements. The
common points between two adjacent elements in which the field variables are expressed are called
Nodes.
6. What is Ritz method?(May/June 2014)
It is integral approach method which is useful for solving complex structural problem,
encountered in finite element analysis. This method is possible only if a suitable function is
available. In Ritz method approximating functions satisfying the boundary conditions are used to
get the solutions

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7. Distinguish Natural & Essential boundary condition (May/June 2009)
There are two types of boundary conditions.
They are:
1. Primary boundary condition (or) Essential boundary condition
The boundary condition, which in terms of field variable, is known as primary
boundary condition.
2. Secondary boundary condition or natural boundary conditions
The boundary conditions, which are in the differential form of field variables, are
known as secondary boundary condition.
Example: A bar is subjected to axial load as shown in fig.
In this problem, displacement u at node 1 = 0, that is primary boundary condition.
EA du/dx = P, that is secondary boundary condition.
8. Compare Ritz method with nodal approximation method.(Nov/Dec 2014), (Nov/Dec 2012)
Similarity:
(i) Both methods use approximating functions as trial solution
(ii) Both methods take linear combinations of trial functions.
(iii) In both methods completeness condition of the function should be satisfied
(iv) In both methods solution is sought by making a functional stationary.
Difference
(i) Rayleigh-Ritz method assumes trial functions over entire structure, while finite element method
uses trial functions only over an element.
(ii) The assumed functions in Rayleigh-Ritz method have to satisfy boundary conditions over entire
structure while in finite element analysis, they have to satisfy continuity conditions at nodes and
sometimes along the boundaries of the element. However completeness condition should be
satisfied in both methods.
9. What do you mean by elements & Nodes?
In a continuum, the field variables are infinite. Finite element procedure reduces such
unknowns to a finite number by dividing the solution region into small parts called Elements. The
common points between two adjacent elements in which the field variables are expressed are called
Nodes.
10. State the discretization error. How it can be reduced? (April /May 2015)
Splitting of continuum in to smallest elements is known as discretization. In some context
like structure having boundary layer the exact connectivity can’t be achieved. It means that it may

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not resemble the original structure. Now there is an error developed in calculation. Such type of
error is discretization error.
To Reduce Error:
(i) Discretization error can be minimized by reducing the finite element (or) discretization
element.
(ii) By introducing finite element it has a curved member.
11. What are the various considerations to be taken in Discretization process?
(i) Types of Elements.
(ii) Size of Elements.
(iii) Location of Nodes.
(iv) Number of Elements.
12. State the principleofminimum potential energy. (Nov/Dec 2010)
Amongallthedisplacementequationsthatsatisfiedinternalcompatibilityandthe
boundaryconditionthosethatalsosatisfytheequationofequilibriummakethe potential energya
minimum is astable system.
PART-B
1. The following differential equation is available for a physical phenomenon. 𝑨𝑬 =
𝒅𝟐𝒖
𝒅𝒙𝟐
+
𝒂𝒙 = 𝟎, The boundary conditions are u(0) = 0, 𝑨𝑬 =
𝒅𝒖
𝒅𝒙 𝒙=𝑳
= 𝟎 By using Galerkin’s
technique, find the solution of the above differential equation.
Given Data:
Differential equ. 𝐴𝐸 =
𝑑2𝑢
𝑑𝑥2
+ 𝑎𝑥 = 0
Boundary Conditions 𝑢 0 = 0, 𝐴𝐸 =
𝑑2𝑢
𝑑𝑥2
+ 𝑎𝑥 = 0
To Find:
u(x) by using galerkin’s technique
Formula used
𝑤𝑖 𝑅 𝑑𝑥 = 0
𝐿
0
Solution:
Assume a trial function
Let 𝑢 𝑥 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2
+ 𝑎3𝑥3
…….. (1)
Apply first boundary condition
i.e) at x=0, u(x) = 0
1 ⟹ 0 = 𝑎0 + 0 + 0 + 0
𝑎0 = 0

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Apply first boundary condition i.e at x = L, 𝐴𝐸 =
𝑑𝑢
𝑑𝑥
= 0
⟹
𝑑𝑢
𝑑𝑥
= 0+𝑎1 + 2𝑎2𝑥 + 3𝑎3𝐿2
⟹ 0 = 𝑎1 + 2𝑎2𝐿 + 3𝑎3𝐿2
⟹ 𝑎1 = −(2𝑎2𝐿 + 3𝑎3𝐿2
)
sub 𝑎0 and 𝑎1 in value in equation (1)
𝑢 𝑥 = 0 + − 2𝑎2𝐿 + 3𝑎3𝐿2
𝑥 + 𝑎2𝑥2
+ 𝑎3𝑥3
= −2𝑎2𝐿𝑥 − 3𝑎3𝐿2
𝑎2𝑥 + 𝑎2𝑥2
+ 𝑎3𝑥3
= 𝑎2 𝑥2
− 2𝐿𝑥 + 𝑎3(𝑥3
− 3𝐿2
𝑥) ……… (2)
We Know That
Residual, 𝑅 = 𝐴𝐸
𝑑2𝑢
𝑑𝑥2
+ 𝑎𝑥 ………. (3)
(2) ⟹
𝑑𝑢
𝑑𝑥
= 𝑎2 2𝑥 − 2𝐿 + 𝑎3(3𝑥2
− 3𝐿2
)
𝑑2
𝑢
𝑑𝑥2
= 𝑎2 2 + 𝑎3(6𝑥)
𝑑2
𝑢
𝑑𝑥2
= 2𝑎2 + 6𝑎3𝑥
Sub
𝑑2𝑢
𝑑𝑥2
value in equation (3)
3 ⟹ 𝑅 = 𝐴𝐸 2𝑎2 + 6𝑎3𝑥 + 𝑎𝑥
Residual, 𝑅 = 𝐴𝐸 2𝑎2 + 6𝑎3𝑥 + 𝑎𝑥 ……… (4)
From Galerkn’s technique
𝑤𝑖 𝑅 𝑑𝑥 = 0
𝐿
0
. . … … . . . (5)
from equation (2) we know that
𝑤1 = 𝑥2
− 2𝐿𝑥
𝑤2 = 𝑥3
− 3𝐿2
𝑥
sub w1, w2 and R value in equation (5)
5 ⟹ 𝑥2
− 2𝐿𝑥
𝐿
0
𝐴𝐸 2𝑎2 + 6𝑎3𝑥 + 𝑎𝑥 𝑑𝑥 = 0 … … … … … (6)
𝑥3
− 3𝐿2
𝑥
𝐿
0
𝐴𝐸 2𝑎2 + 6𝑎3𝑥 + 𝑎𝑥 𝑑𝑥 = 0 … … … … … (7)
6 ⟹ 𝑥2
− 2𝐿𝑥
𝐿
0
𝐴𝐸 2𝑎2 + 6𝑎3𝑥 + 𝑎𝑥 𝑑𝑥 = 0

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𝑥2
− 2𝐿𝑥
𝐿
0
2𝑎2𝐴𝐸 + 6𝑎3𝐴𝐸𝑥 + 𝑎𝑥 𝑑𝑥 = 0
2𝑎2𝐴𝐸𝑥2
+ 6𝑎3𝐴𝐸𝑥3
+ 𝑎𝑥3
− 4𝑎2𝐴𝐸𝐿𝑥 − 12𝑎3𝐴𝐸𝐿𝑥2
− 2𝑎𝐿𝑥2
𝐿
0
= 0
⟹ [2𝑎2𝐴𝐸
𝑥3
3
+ 6𝑎3𝐴𝐸
𝑥4
4
+ 𝑎
𝑥4
4
− 4𝑎2𝐴𝐸𝐿
𝑥2
2
− 12𝑎3𝐴𝐸𝐿
𝑥3
3
− 2𝑎𝐿
𝑥3
3
]0
𝐿
= 0
⟹ 2𝑎2𝐴𝐸
𝐿3
3
+ 6𝑎3𝐴𝐸
𝐿4
4
+ 𝑎
𝐿4
4
− 4𝑎2𝐴𝐸
𝐿3
2
− 12𝑎3𝐴𝐸
𝐿4
3
− 2𝑎
𝐿4
3
= 0
⟹
2
3
𝑎2𝐴𝐸𝐿3
+
3
2
𝑎3𝐴𝐸 𝐿4
+ 𝑎
𝐿4
4
− 2𝑎2𝐴𝐸𝐿3
− 4𝑎3𝐴𝐸𝐿4
−
2
3
𝑎𝐿4
= 0
⟹ 𝐴𝐸𝑎2𝐿3
2
3
− 2 + 𝑎3𝐴𝐸 𝐿4
3
2
− 4 + 𝑎
𝐿4
4
−
2
3
𝑎2𝐿4
= 0
⟹
−4
3
𝐴𝐸𝐿3
𝑎2 −
5
2
𝐴𝐸𝐿4
𝑎3 =
2
3
−
1
4
𝑎𝐿4
−
4
3
𝐴𝐸𝐿3
𝑎2 −
5
2
𝐴𝐸𝐿4
𝑎3 =
5
12
𝑎𝐿4
−4
3
𝐴𝐸𝐿3
𝑎2 −
5
2
𝐴𝐸𝐿4
𝑎3 = −
5
12
𝑎𝐿4
… … … . 8
Equation (7)
⟹ (𝑥3
− 3𝐿2
𝑥)
𝐿
0
𝐴𝐸 2𝑎2 + 6𝑎3𝑥 + 𝑎𝑥 𝑑𝑥 = 0
⟹ (𝑥3
− 3𝐿2
𝑥)
𝐿
0
2𝑎2𝐴𝐸 + 6𝑎3𝐴𝐸𝑥 + 𝑎𝑥 𝑑𝑥 = 0
⟹ 2𝐴𝐸𝑎2𝑥3
+ 6𝐴𝐸𝑎3𝑥4
+ 𝑎𝑥4
− 6𝐴𝐸𝑎2𝐿2
𝑥 − 18𝐴𝐸𝑎3𝐿2
𝑥2
− 3𝑎𝐿2
𝑥2
𝑑𝑥 = 0
𝐿
0
⟹ 2𝐴𝐸𝑎2
𝑥4
4
+ 6𝐴𝐸𝑎3
𝑥5
5
+ 𝑎
𝑥5
5
𝑥2
2
𝑥3
3
− 3𝑎𝐿2
𝑥3
3 0
𝐿
= 0
⟹
1
2
𝐴𝐸𝑎2𝑥4
+
6
5
𝐴𝐸𝑎3𝑥5
+
1
5
𝑎𝑥5
𝑥2
𝑥3
− 𝑎𝐿2
𝑥3
0
𝐿
= 0
⟹
1
2
𝐴𝐸𝑎2𝐿4
+
6
5
𝐴𝐸𝑎3𝐿5
+
1
5
𝑎𝐿5
(𝐿2
) − 6𝐴𝐸𝑎3𝐿2
(𝐿3
) − 𝑎𝐿2
(𝐿3
) = 0
⟹
1
2
𝐴𝐸𝑎2𝐿4
+
6
5
𝐴𝐸𝑎3𝐿5
+
1
5
𝑎𝐿5
− 𝑎𝐿5
= 0
1
2
− 3 + 𝐴𝐸𝑎3𝐿5
6
5
− 6 + 𝑎𝐿5
+
1
5
− 1 = 0
5
2
−
24
5
𝐴𝐸𝑎3𝐿5
=
4
5
𝑎𝐿5
5
2
𝐴𝐸𝑎2𝐿4
+
24
5
𝐴𝐸𝑎3𝐿5
= −
4
5
𝑎𝐿5
… … … … . 9

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Solving Equation (8) and (9)
Equation (8) ⟹
4
3
𝐴𝐸𝑎2𝐿3
+
5
2
𝐴𝐸𝑎3𝐿4
= −
5
12
𝑎𝐿4
Equation (9) ⟹
5
2
𝐴𝐸𝑎2𝐿4
+
24
5
𝐴𝐸𝑎3𝐿5
= −
4
5
𝑎𝐿5
Multiplying Equation (8)
5
2
𝐿 and Equation (9) by
4
3
20
6
𝐴𝐸𝑎2𝐿4
+
25
4
𝐴𝐸𝑎3𝐿5
= −
25
24
𝑎𝐿5
20
6
𝐴𝐸𝑎2𝐿4
+
25
4
𝐴𝐸𝑎3𝐿5
= −
16
15
𝑎𝐿5
Subtracting
25
4
−
96
15
𝐴𝐸𝑎3𝐿5
=
16
15
−
25
24
𝑎𝐿5
375 − 384
60
𝐴𝐸𝑎3𝐿5
=
384 − 375
360
𝑎𝐿5
⟹
−9
60
𝐴𝐸𝑎3𝐿5
=
9
360
𝑎𝐿5
⟹ −0.15𝐴𝐸𝑎3 = 0.025𝑎
𝑎3 = −0.1666
𝑎
𝐴𝐸
𝑎3 = −
𝑎
6𝐴𝐸
… … … . (10)
Substituting a3 value in Equation (8)
4
3
𝐴𝐸𝑎2𝐿3
+
5
2
𝐴𝐸
−𝑎
6𝐴𝐸
𝐿4
=
−5
12
𝑎𝐿4
4
3
𝐴𝐸𝑎2𝐿3
=
−5
12
𝑎𝐿4
−
5
2
𝐴𝐸𝐿4
=
−𝑎
6𝐴𝐸
4
3
𝐴𝐸𝑎2𝐿3
=
−5
12
𝑎𝐿4
+
5
2
𝐴𝐸𝐿4
4
3
𝐴𝐸𝑎2𝐿3
= 0
𝑎2 = 0
Sub a2 and a3 value in equation (2)
⟹ 𝑢 𝑥 = 0𝑥 𝑥2 − 2𝐿𝑥 +
−𝑎
6𝐴𝐸
𝑥3
− 3𝐿2
𝑥 = 0
⟹ 𝑢 𝑥 =
𝑎
6𝐴𝐸
3𝐿2
𝑥 − 𝑥3
Result:
𝑢 𝑥 =
𝑎
6𝐴𝐸
3𝐿2
𝑥 − 𝑥3

S
C
A
D
14
2. Find the deflection at the centre of a simply supported beam of span length “l” subjected
to uniformly distributed load throughout its length as shown in figure using (a) point
collocation method, (b) sub-domain method, (c) Least squares method, and (d) Galerkin’s
method. (Nov/Dec 2014)
Given data
Length (L) = 𝑙
UDL = 𝜔 𝑁/𝑚
To find
Deflection
Formula used
𝐸𝐼
𝑑4
𝑦
𝑑𝑥4
− 𝜔 = 0, 0 ≤ 𝑥 ≤ 𝑙
Point Collocation Method R = 0
Sub-domain collocation method = 𝑅𝑑𝑥 = 0
𝑙
0
Least Square Method 𝐼 = 𝑅2
𝑑𝑥 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑙
0
Solution:
The differential equation governing the deflection of beam subjected to uniformly
distributed load is given by
𝐸𝐼
𝑑4
𝑦
𝑑𝑥4
− 𝜔 = 0, 0 ≤ 𝑥 ≤ 𝑙 … … … . (1)
The boundary conditions are Y=0 at x=0 and x = l, where y is the deflection.
𝐸𝐼
𝑑4
𝑦
𝑑𝑥4
= 0, 𝑎𝑡 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 𝑙
Where
𝐸𝐼
𝑑4𝑦
𝑑𝑥4
= 𝑀, (Bending moment)
E → Young’s Modules
I → Moment of Inertia of the Beam.
Let us select the trial function for deflection as 𝑌 = 𝑎𝑠𝑖𝑛
𝜋𝑥
𝑙
……. (2)
Hence it satisfies the boundary conditions
⟹
𝑑𝑦
𝑑𝑥
= 𝑎
𝜋
𝑙
. cos
𝜋𝑥
𝑙
⟹
𝑑2
𝑦
𝑑𝑥2
= −𝑎
𝜋2
𝑙2
. sin
𝜋𝑥
𝑙

S
C
A
D
15
⟹
𝑑3
𝑦
𝑑𝑥3
= −𝑎
𝜋3
𝑙3
. cos
𝜋𝑥
𝑙
⟹
𝑑4
𝑦
𝑑𝑥4
= 𝑎
𝜋4
𝑙4
. sin
𝜋𝑥
𝑙
Substituting the Equation (3) in the governing Equation (1)
𝐸𝐼 𝑎
𝜋4
𝑙4
. sin
𝜋𝑥
𝑙
− 𝜔 = 0
Take, Residual 𝑅 = 𝐸𝐼𝑎
𝜋4
𝑙4
. sin
𝜋𝑥
𝑙
− 𝜔
a) Point Collocation Method:
In this method, the residuals are set to zero.
⟹ 𝑅 = 𝐸𝐼𝑎
𝜋4
𝑙4
. sin
𝜋𝑥
𝑙
− 𝜔 = 0
𝐸𝐼𝑎
𝜋4
𝑙4
. sin
𝜋𝑥
𝑙
= 𝜔
To get maximum deflection, take 𝑘 =
𝑙
2
(𝑖. 𝑒. 𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑜𝑓 𝑏𝑒𝑎𝑚)
𝐸𝐼𝑎
𝜋4
𝑙4
. sin
𝜋
𝑙
𝑙
2
= 𝜔
𝐸𝐼𝑎
𝜋4
𝑙4
= 𝜔
𝑎 =
𝜔𝑙4
𝜋4𝐸𝐼
Sub “a” value in trial function equation (2)
𝑌 =
𝜔𝑙4
𝜋4𝐸𝐼
. sin
𝜋𝑥
𝑙
𝐴𝑡 𝑥 =
𝑙
2
⟹ 𝑌max =
𝜔𝑙4
𝜋4𝐸𝐼
. sin
𝜋
2
𝑙
2
𝑌max =
𝜔𝑙4
𝜋4𝐸𝐼
𝑌max =
𝜔𝑙4
97.4𝐸𝐼
b) Sub-domain collocation method:
In this method, the integral of the residual over the sub-domain is set to zero.
𝑅𝑑𝑥 = 0
𝑙
0
Sub R value
⟹ 𝑎𝐸𝐼
𝜋4
𝑙4
sin
𝜋𝑥
𝑙
− 𝜔 𝑑𝑥 = 0
[∵ sin
𝜋
𝑙
= 1]
[∵ sin
𝜋
2
= 1]

S
C
A
D
16
⟹ 𝑎𝐸𝐼
𝜋4
𝑙4
−cos
𝜋𝑥
𝑙
𝜋
𝑙
−𝜔 𝑥
0
𝑙
= 0
⟹ 𝑎𝐸𝐼
𝜋4
𝑙4
−cos
𝜋𝑥
𝑙
𝑙
𝑢
− 𝜔 𝑥
0
𝑙
= 0
⟹ −𝑎𝐸𝐼
𝜋3
𝑙3 cos𝜋 − 𝑐𝑜𝑠0 𝜔 𝑙 = 0
−𝑎𝐸𝐼
𝜋3
𝑙3
−1 − 1 = 𝜔 𝑙
⟹ −𝑎 =
𝜔𝑙4
2𝜋3𝐸𝐼
=
𝜔𝑙4
62𝐸𝐼
Sub “a” value in the trial function equation (2)
𝑌 =
𝜔𝑙4
62𝐸𝐼
. sin
𝜋𝑥
𝑙
𝐴𝑡 𝑥 =
𝑙
2
, 𝑌𝑚𝑎𝑥 =
𝜔𝑙4
62𝐸𝐼
. sin
𝜋
𝑙
(
𝑙
2
)
𝑌𝑚𝑎𝑥 =
𝜔𝑙4
62𝐸𝐼
c) Least Square Method:
In this method the functional
𝐼 = 𝑅2
𝑑𝑥 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑙
0
𝐼 = (𝑎𝐸𝐼
𝜋4
𝑙4
. 𝑠𝑖𝑛
𝜋𝑥
𝑙
− 𝜔)2
𝑑𝑥
𝑙
0
= [𝑎2
𝐸2
𝐼2
𝜋8
𝑙8
. 𝑠𝑖𝑛2
𝜋𝑥
𝑙
− 𝜔2
− 2𝑎𝐸𝐼𝜔
𝜋4
𝑙4
. 𝑠𝑖𝑛
𝜋𝑥
𝑙
]𝑑𝑥
𝑙
0
= [𝑎2
𝐸2
𝐼2
𝜋8
𝑙8
1
2
𝑥 𝑠𝑖𝑛
2𝜋𝑥
𝑙
𝑙
2𝜋
+ 𝜔2
− 2𝑎𝐸𝐼𝜔
𝜋4
𝑙4
. [−𝑐𝑜𝑠
𝜋𝑥
𝑙
𝑙
𝜋
]]0
𝑙
= 𝑎2
𝐸2
𝐼2
𝜋8
𝑙8
1
2
𝑙 −
𝑙
2𝜋
𝑠𝑖𝑛2𝜋 − 𝑠𝑖𝑛0 + 𝜔2
𝑙 + 2𝑎𝐸𝐼𝜔
𝜋4
𝑙4
.
𝑙
𝜋
[−𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠0]
∵ 𝑠𝑖𝑛2𝜋 = 0; 𝑠𝑖𝑛0 = 0; 𝑐𝑜𝑠𝜋 = 0; 𝑐𝑜𝑠0 = 1
𝐼 = 𝑎2
𝐸2
𝐼2
𝜋8
𝑙2
𝑙
2
+ 𝜔2
𝑙 + 2𝑎𝐸𝐼𝜔
𝜋3
𝑙3
. (−1 − 1)
𝐼 =
𝑎2
𝐸2
𝐼2
𝜋8
2𝑙7
+ 𝜔2
𝑙 − 4𝑎𝐸𝐼𝜔
𝜋3
𝑙3
Now,
𝜕𝜋
𝜕𝑎
= 0
∵ cos 𝜋 = −1
𝑐𝑜𝑠0 = 1
,

S
C
A
D
17
⟹
𝑎2
𝐸2
𝐼2
𝜋8
2𝑙7
= 4𝐸𝐼𝜔
𝜋3
𝑙3
𝑎2
𝐸2
𝐼2
𝜋8
𝑙7
= 4𝐸𝐼𝜔
𝜋3
𝑙3
𝑎 =
4𝐸𝐼𝜔𝑙5
𝜋5𝐸𝐼
Hence the trial Function
𝑌 =
4𝜔𝑙4
𝜋5𝐸𝐼
. sin
𝜋𝑥
𝑙
𝐴𝑡 𝑥 =
𝑙
2
, max 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 [∵ 𝑠𝑖𝑛
𝜋
2
= 1]
𝑌𝑚𝑎𝑥 =
4𝜔𝑙4
𝜋5𝐸𝐼
𝑠𝑖𝑛
𝜋
2
(
𝑙
2
)
𝑌𝑚𝑎𝑥 =
𝜔𝑙4
76.5 𝐸𝐼
d) Galerkin’s Method:
In this method
𝑌. 𝑅 𝑑𝑥 = 0
𝑙
0
⟹ 𝑎𝑠𝑖𝑛
𝜋𝑥
𝑙
𝑎𝐸𝐼
𝜋4
𝑙4
𝑠𝑖𝑛
𝜋𝑥
𝑙
− 𝜔 𝑑𝑥 = 0
𝑙
0
⟹ 𝑎2
𝐸𝐼
𝜋4
𝑙4
𝑠𝑖𝑛2
𝜋𝑥
𝑙
− 𝑎𝜔𝑠𝑖𝑛
𝜋𝑥
𝑙
𝑑𝑥 = 0
𝑙
0
⟹ 𝑎2
𝐸𝐼
𝜋4
𝑙4
[
1
2
(1 − 𝑐𝑜𝑠
2𝜋𝑥
𝑙
) − 𝑎𝜔𝑠𝑖𝑛
𝜋𝑥
𝑙
𝑑𝑥 = 0
𝑙
0
⟹ 𝑎2
𝐸𝐼
𝜋4
𝑙4
[
1
2
1 − 𝑥 −
1
2𝜋
𝑠𝑖𝑛2
2𝜋𝑥
𝑙
+ 𝑎𝜔
𝑙
𝜋
𝑐𝑜𝑠
𝜋𝑥
𝑙 0
𝑙
= 0
𝑎2
𝐸𝐼
𝜋4
𝑙4
𝑙
2
− 2𝑎𝜔
𝑙
𝜋
= 0
∴ 𝑎 =
2𝜔𝑙
𝜋
.
2𝑙3
𝐸𝐼𝜋4
𝑎 =
4𝜔𝑙3
𝜋5𝐸𝐼
Hence the trial Function
𝑌 =
4𝜔𝑙4
𝜋5𝐸𝐼
. sin
𝜋𝑥
𝑙

S
C
A
D
18
𝐴𝑡 𝑥 =
𝑙
2
, max 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 [∵ 𝑠𝑖𝑛
𝜋
2
= 1]
𝑌𝑚𝑎𝑥 =
4𝜔𝑙4
𝜋5𝐸𝐼
𝑠𝑖𝑛
𝜋
2
(
𝑙
2
)
𝑌𝑚𝑎𝑥 =
4𝜔𝑙4
𝜋5 𝐸𝐼
𝑌𝑚𝑎𝑥 =
𝜔𝑙4
76.5 𝐸𝐼
Verification,
We know that simply supported beam is subjected to uniformly distributed load, maximum
deflection is,
𝑌𝑚𝑎𝑥 =
5
384
𝜔𝑙4
𝐸𝐼
= 0.01
𝜔𝑙4
𝐸𝐼
3) i) What is constitutive relationship? Express the constitutive relations for a linear
elastic isotropic material including initial stress and strain. (4)
[Nov/Dec 2009]
Solution:
It is the relationship between components of stresses in the members of a structure or in a
solid body and components of strains. The structure or solids bodies under consideration are made
of elastic material that obeys Hooke’s law.
𝜎 = 𝐷 {𝑒}
Where
[D] is a stress – strain relationship matrix or constitute matrix.
The constitutive relations for a linear elastic isotropic material is
𝜎𝑥
𝜎𝑦
𝜎𝑧
𝛿𝑥𝑦
𝛿𝑦𝑧
𝛿𝑧𝑥
=
𝐸
1 + 𝑣 1 − 2𝑣
(1 − 𝑣) 0 0
𝑣 (1 − 𝑣) 0
𝑣
0
0
0
𝑣
0
0
0
(1 − 𝑣)
0
0
0
0 0 0
0 0 0
0
1 − 2𝑣
2
0
0
0
0
1 − 2𝑣
2
0
0
0
0
1 − 2𝑣
2
𝑒𝑥
𝑒𝑦
𝑒𝑧
𝑣𝑥𝑦
𝑣𝑦𝑧
𝑣𝑧𝑥
ii) Consider the differential equation
𝒅𝟐𝒚
𝒅𝒙𝟐
+ 𝟒𝟎𝟎𝒙𝟐
= 𝟎 for 𝟎 ≤ 𝒙 ≤ 𝟏 subject to boundary
conditions Y(0) = 0, Y(1) = 0. The functions corresponding to this problem, to be eternized
is given by 𝑰 = −𝟎. 𝟓
𝒅𝒚
𝒅𝒙
𝟐
+ 𝟒𝟎𝟎𝒙𝟐
𝒀
𝒍
𝟎
. Find the solution of the problem using Ray
Light Ritz method by considering a two term solution as 𝒀 𝒙 = 𝒄𝟏𝒙 𝟏 − 𝒙 + 𝒄𝟐𝒙𝟐
(𝟏 −
𝒙) (12)

S
C
A
D
19
Given data
Differential equation =
𝑑2𝑦
𝑑𝑥2
+ 400𝑥2
= 0 for 0 ≤ 𝑥 ≤ 1
Boundary conditions Y(0) = 0, Y(1) = 0
𝐼 = −0.5
𝑑𝑦
𝑑𝑥
2
+ 400𝑥2
𝑌
𝑙
0
𝑌 𝑥 = 𝑐1𝑥 1 − 𝑥 + 𝑐2𝑥2
(1 − 𝑥)
To find:
Rayleigh- Ritz method
Formula used
𝜕𝐼
𝜕𝑐1
= 0
𝜕𝐼
𝜕𝑐2
= 0
Solution:
𝑌 𝑥 = 𝑐1𝑥 1 − 𝑥 + 𝑐2𝑥2
(1 − 𝑥)
𝑌 𝑥 = 𝑐1𝑥 𝑥 − 𝑥2
+ 𝑐2(𝑥2
− 𝑥3
)
𝑑𝑦
𝑑𝑥
= 𝑐1 1 − 2𝑥 + 𝑐2(2𝑥 − 3𝑥2
)
= 𝑐1 1 − 2𝑥 + 𝑐2𝑥(2 − 3𝑥)
𝑑𝑦
𝑑𝑥
2
= 𝑐1 1 − 2𝑥 + 𝑐2𝑥(2 − 3𝑥)2 2
= 𝑐1
2
1 − 4𝑥 + 4𝑥2
+ 𝑐2
2
𝑥2
4 − 12𝑥 + 9𝑥2
+ 2𝑐1𝑐2𝑥 1 − 2𝑥 (2 − 3𝑥)
= 𝑐1
2
1 − 4𝑥 + 4𝑥2
+ 𝑐2
2
𝑥2
4 − 12𝑥 + 9𝑥2
+ 2𝑐1𝑐2𝑥(2 − 3𝑥 − 4𝑥 + 6𝑥2
)
𝑑𝑦
𝑑𝑥
2
= 𝑐1
2
1 − 4𝑥 + 4𝑥2
+ 𝑐2
2
𝑥2
4 − 12𝑥 + 9𝑥2
+ 2𝑐1𝑐2𝑥(2 − 7𝑥 + 6𝑥2
)
We know that
𝐼 = [−0.5
𝑑𝑦
𝑑𝑥
2
+ 400𝑥2
𝑦
𝑙
0
] =
−1
2
𝑑𝑦
𝑑𝑥
2
+ 400 𝑥2
𝑦
𝑙
0
𝑙
0
= 𝑐1
2
1 − 4𝑥 + 4𝑥2
+ 𝑐2
2
𝑥2
4 − 12𝑥 + 9𝑥2
+ 2𝑐1𝑐2𝑥 2 − 7𝑥 + 6𝑥2
𝑙
0
+ 400[ 𝑥2
𝑐1𝑥 1 − 𝑥 + 𝑐2𝑥2
1 − 𝑥
𝑙
0
By Solving

S
C
A
D
20
𝐼 =
−1
2
𝑐1
2
3
+
2
15
𝑐2
2
+
1
3
𝑐1𝑐2 + 400
𝑐1
20
+
𝑐2
30
𝐼 =
−1
6
𝑐1
2
−
1
15
𝑐2
2
−
1
6
𝑐1𝑐2 + 20𝑐1 +
40
3
𝑐2
𝜕𝐼
𝜕𝑐1
= 0
⟹
−1
6
× 2𝑐1 −
1
6
𝑐2 + 20 = 0
⟹
−1
3
× 𝑐1 −
1
6
𝑐2 + 20 = 0 … … … . . (1)
Similarly,
𝜕𝐼
𝜕𝑐2
= 0
⟹
−2
15
𝑐2 −
1
6
𝑐1 +
40
3
= 0 … … … . . (2)
By Solving (1) and (2)
𝑐1 =
80
3
; 𝑐1 =
200
3
We know that
𝑌 = 𝑐1𝑥 1 − 𝑥 + 𝑐2𝑥2
(1 − 𝑥)
𝑌 =
80
3
𝑥 1 − 𝑥 +
200
3
𝑥2
1 − 𝑥
4) Consider a 1mm diameter, 50m long aluminum pin-fin as shown in figure used to
enhance the heat transfer from a surface wall maintained at 300C. Calculate the
temperature distribution in a pin-fin by using Rayleigh – Ritz method. Take, 𝒌 =
𝟐𝟎𝟎𝒘
𝒎𝐂 for aluminum h= 𝟐𝟎𝟎𝒘
𝒎𝟐𝐂
, 𝑻∞ = 𝟑𝟎𝐂.
𝒌
𝒅𝟐𝑻
𝒅𝒙𝟐 =
𝑷𝒉
𝑨
(𝑻 − 𝑻∞) , 𝑻 𝟎 = 𝑻𝒘 = 𝟑𝟎𝟎𝐂, 𝒒𝑳 = 𝑲𝑨
𝒅𝑻
𝒅𝒙
𝑳 = 𝟎 (insulated tip)

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Given Data:
The governing differential equation
𝑘
𝑑2
𝑇
𝑑𝑥2
=
𝑃𝑕
𝐴
(𝑇 − 𝑇∞)
Diameter d = 1mm = 1x10-3
m
Length L = 50mm = 50x10-3
m
Thermal K = 200𝑤
𝑚C
Conductivity Heat transfer co-efficient h = 200𝑤
𝑚C
Fluid Temp 𝑇∞ = 30C.
Boundary Conditions 𝑇 0 = 𝑇𝑤 = 300C
𝑞𝐿 = 𝐾𝐴
𝑑𝑇
𝑑𝑥
𝐿 = 0
To Find:
Ritz Parameters
Formula used
𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
Solution:
The equivalent functional representation is given by,
𝜋 = 𝑢 − 𝑣
𝜋 =
1
2
𝐾
𝑑𝑇
𝑑𝑥
2
𝑑𝑥 +
1
2
𝑃𝑕
𝐴
𝑇 − 𝑇∞
2
𝑑𝑥 − 𝑞𝐿𝑇𝐿 … … … … . (1)
𝐿
0
𝐿
0
𝜋 =
1
2
𝐾
𝑑𝑇
𝑑𝑥
2
𝑑𝑥 +
1
2
𝑃𝑕
𝐴
𝑇 − 𝑇∞
2
𝑑𝑥 … … … … . . 2
𝐿
0
𝐿
0
∵ 𝑞𝐿 = 0
Assume a trial function
Let
𝑇 𝑥 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2
… … … … … . . (3)
Apply boundary condition
at x = 0, T(x) = 300
300 = 𝑎0 + 𝑎1(0) + 𝑎2(0)2
𝑎0 = 300
Substituting 𝑎0 value in equation (3)
𝑇 𝑥 = 300 + 𝑎1𝑥 + 𝑎2𝑥2
… … … … … . . 4

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⟹
𝑑𝑇
𝑑𝑥
= 𝑎1 + 2𝑎2𝑥 … … … … … … (5)
Substitute the equation (4), (5) in (2)
𝜋 =
1
2
𝑘
𝑙
0
(𝑎1 + 2𝑎2𝑥)2
𝑑𝑥 +
1
2
𝑃𝑕
𝐴
270 + 𝑎1 + 𝑎2𝑥2 2
𝑑𝑥.
𝑙
0
[∵ 𝑎 + 𝑏 2
= 𝑎2
+ 𝑏2
+ 2𝑎𝑏; 𝑎 + 𝑏 + 𝑐 2
= 𝑎2
+ 𝑏2
+ 𝑐2
+ 2𝑎𝑏 + 2𝑏𝑐 + 2𝑐𝑎
𝜋 =
𝑘
2
(𝑎1
2
+ 4𝑎2
2
𝑥2
+ 4𝑎1
𝑙
0
𝑎2𝑥) +
𝑃𝑕
2𝐴
2702
+ 𝑎1
2
𝑥2
𝑙
0
+ 𝑎2
2
𝑥4
+ 540𝑎1𝑥 + 2𝑎1𝑥3
+ 540𝑎2𝑥2
𝑑𝑥
𝜋 =
𝑘
2
(𝑎1
2
𝑥 +
4𝑎2
2
𝑥3
3
+
4𝑎1𝑎2𝑥2
2 0
50𝑥10−3
+
𝑃𝑕
2𝐴
72900𝑘 +
𝑎1
2
𝑥3
3
+
𝑎2
2
𝑥5
5
+
540𝑎1𝑥2
2
+
2𝑎1𝑎2𝑥4
4
+
540𝑎2𝑥3
3 0
50𝑥10−3
[∵ 𝑙 = 50𝑥10−3
]
𝜋 =
𝑘
2
(50 × 10−3
)𝑎1
2
+
4𝑎2
2
(50 × 10−3
)3
3
+
4𝑎1𝑎2(50 × 10−3
)2
2
+
𝑃𝑕
2𝐴
72900𝑘 +
𝑎1
2
(50 × 10−3
)3
3
+
𝑎2
2
(50 × 10−3
)5
5
𝜋 =
200
2
50 × 10−3
𝑎1
2
+ 1.666 × 10−4
𝑎2
2
+ 50 × 10−3
𝑎1𝑎2 +
𝜋 × 10−3
× 20
2 ×
𝜋
2
× 10−3 2
= 364.5 + 4.166 × 10−5
𝑎1
2
+ 6.25 × 10−8
𝑎2
2
+ 0.675𝑎1 + 3.125 × 10−6
𝑎1𝑎2 + 0.0225𝑎2
𝜋 = 5𝑎1
2
+ 0.0166𝑎2
2
+ 0.5𝑎1𝑎2 + 14.58 × 10−7
+ 1.66912
+ 2.5 × 10−3
𝑎2
2
+ 2700 𝑎1
+ 0.125 𝑎1𝑎2 + 900𝑎2]
𝜋 = 6.66𝑎1
2
+ 0.0191𝑎2
2
+ 0.625𝑎1𝑎2 + 2700𝑎1 + 900𝑎2 + 14.58 × 107
Apply
𝜕𝜋
𝜕𝑎2
= 0
⟹ 13.32𝑎1 + 0.625𝑎2 + 27000 = 0
13.32𝑎1 + 0.625𝑎2 = − + 27000 … … … … … (6)
⟹ 0.625𝑎1 + 0.382𝑎2 + 900 = 0
0.625𝑎1 + 0.382𝑎2 = −900 … … … … . . (7)
Solve the equation (6) and (7)
13.32𝑎1 + 0.625𝑎2 = − + 27000 … … … … … (6)
0.625𝑎1 + 0.382𝑎2 = −900 … … … … . . 7
(6) x 0.625

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8.325𝑎1 + 0.3906𝑎2 = −16875 … … … … . . 8
(7) x -13.32
−8.325𝑎1 − 0.5088𝑎2 = 11988 … … … … . . 9
−0.1182𝑎2 = −4887
𝑎2 = 41345
Sub 𝑎2value in equation (6)
13.32𝑎1 + 0.625(41345) = − + 27000
𝑎1 = −3967.01
Sub 𝑎0, 𝑎1and 𝑎2values in equation (3)
𝑇 = 300 − 3697.01𝑥 + 41345𝑥2
5) Explain briefly about General steps of the finite element analysis.
[Nov/Dec 2014]
Step: 1
Discretization of structure
The art of sub dividing a structure into a convenient number of smaller element is known as
discretization.
Smaller elements are classified as
i) One dimensional element
ii) Two dimensional element
iii) Three dimensional element
iv) Axisymmetric element
(i) One dimensional element:-
a. A bar and beam elements are considered as one dimensional element has two nodes,
one at each end as shown.
(ii) Two Dimensional element:-
Triangular and Rectangular elements are considered as 2D element. These elements
are loaded by forces in their own plane.
1 2
3
1 2
3
4
2
1

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iii) Three dimensional element:-
The most common 3D elements are tetrahedral and lexahendral (Brick) elements. These
elements are used for three dimensional stress analysis problems.
iv) Axisymmetric element:-
The axisymmetric element is developed by relating a triangle or quadrilateral about a fixed
axis located in the plane of the element through 3600
. When the geometry and loading of the
problems are axisymmetric these elements are used.
The stress-strain relationship is given by,
𝜎 = 𝐸𝑒
Where, 𝜎 = Stress in 𝑥 direction
𝐸 = Modulus of elasticity
Step 2:- Numbering of nodes and Elements:-
The nodes and elements should be numbered after discretization process. The numbering
process is most important since if decide the size of the stiffness matrix and it leads the reduction of
memory requirement . While numbering the nodes, the following condition should be satisfied.
{Maximum number node} – {Minimum number node} = minimum

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Step 3:
Selection of a displacement function or a Interpolation function:-
It involves choosing a displacement function within each element. Polynomial of linear,
quadratic and cubic form are frequently used as displacement Function because they are simple to
work within finite element formulation. 𝑑 𝑥 .
The polynomial type of interpolation functions are mostly used due to the following
reasons.
1. It is easy to formulate and computerize the finite element equations.
2. It is easy to perform differentiation or Intigration.
3. The accuracy of the result can be improved by increasing the order of the polynomial.

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Step – 4:-
Define the material behavior by using strain – Displacemnt and stress. Strain
relationship:
Strain – displacement and stress – strain relationship and necessary for deriving the equatins
for each finite element.
In case of the dimensional deformation, the strain – displacement relationship is given by,
𝑒 =
𝑑𝑢
𝑑𝑥
Where, 𝑢 → displacement field variable 𝑥 direction 𝑒 → strain.
Step – 5
Deviation of equation is in matrix form as
𝑓1 𝑘11, 𝑘12, 𝑘13 … . . 𝑘1𝑛 𝑢1
𝑓2 𝑘21, 𝑘22, 𝑘23 … . . 𝑘2𝑛 𝑢2
𝑓3 𝑘31, 𝑘32, 𝑘33 … . . 𝑘3𝑛 𝑢3
𝑓
4 𝑘𝑛1, 𝑘42, 𝑘43 … . . 𝑘4𝑛 𝑢𝑛
In compact matrix form as.
Where,
𝑒 is a element, {𝐹} is the vector of element modal forces, [𝑘] is the element stiffness
matrix and the equation can be derived by any one of the following methods.
(i) Direct equilibrium method.
(ii) Variational method.
(iii) Weighted Residual method.
Step (6):-
Assemble the element equations to obtain the global or total equations.
The individual element equations obtained in step 𝑠 are added together by using a
method of super position i.e. direction stiffness method. The final assembled or global equation
which is in the form of
𝑓 = 𝑘 {𝑢}
Where, 𝐹 → Global Force Vector
𝐾 → Global Stiffness matrix
{𝑢} → Global displacement vector.
Step (7):-
Applying boundary conditions:
.
.
.
.
.
.
.
.
.

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The global stiffness matrix [𝑘] is a singular matrix because its determinant is equal
to zero. In order to remove the singularity problem certain boundary conditions are applied so that
the structure remains in place instead of moving as a rigid body.
Step (8):-
Solution for the unknown displacement formed in step (6) simultaneous algebraic
equations matrix form as follows.
Deviation of equation is in matrix form as
𝑓1 𝑘11, 𝑘12, 𝑘13 … . . 𝑘1𝑛 𝑢1
𝑓2 𝑘21, 𝑘22, 𝑘23 … . . 𝑘2𝑛 𝑢2
𝑓3 𝑘31, 𝑘32, 𝑘33 … . . 𝑘3𝑛 𝑢3
𝑓3 𝑘41, 𝑘42, 𝑘43 … . . 𝑘4𝑛 𝑢4
𝑓
4 𝑘𝑛1, 𝑘42, 𝑘43 … . . 𝑘4𝑛 𝑢𝑛
These equation can be solved and unknown displacement {𝑢} calculated by using
Gauss elimination.
Step (9):-
Computation of the element strains and stresses from the modal displacements 𝒖 :
In structural stress analysis problem. Stress and strain are important factors from the
solution of displacement vector {𝑢}, stress and strain value can be calculated. In case of 1D the
strain displacement can strain.
𝑒 =
𝑑
𝑢
= 𝑢2 − 𝑢1
Where, 𝑢1 and 𝑢2 are displacement at model 1 and 2
𝑥1 − 𝑥2 = Actual length of the element from that we can find the strain value,
By knowing the strain, stress value can be calculated by using the relation.
Stress 𝜎 = 𝐸𝑒
Where, 𝐸 → young’s modulus
𝑒 → strain
Step – 10
Interpret the result (Post processing)
.
.
.
.
.
.
.
.
.

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Analysis and Evaluation of the solution result is referred to as post-processing. Post processor
computer programs help the user to interpret the results by displaying them in graphical form.
6) Explain in detail about Boundary value, Initial Value problems.
The objective of most analysis is to determine unknown functions called dependent
variables, that are governed by a set of differential equations posed in a given domain. Ω and some
conditions on the boundary Γ of the domain. Often, a domin not including its boundary is called an
open domain. A domain boundary is called an open domain. A domain Ω with its boundary Γ is
called a closed domain.
Boundary value problems:- Steady state heat transfer : In a fin and axial deformation of a bar
shown in fig. Find 𝑢(𝑥) that satisfies the second – order differential equation and boundary
conditions.
−𝑑
𝑑𝑥
𝑎
𝑑𝑢
𝑑𝑥
+ 𝑐𝑢 = 𝑓 for 0 < 𝑥 < 𝐿
𝑢 𝑜 = 𝑢0, 𝑎
𝑑𝑢
𝑑𝑥 𝑥=𝐿
= 𝑞0
i) Bending of elastic beams under Transverse load : find 𝑢 𝑥 that satisfies the fourth order
differential equation and boundary conditions.
𝑑2
𝑑𝑥 2
𝑏
𝑑2𝑢
𝑑𝑥2
+ 𝑐𝑢 = 𝐹 for 0 < 𝑥 < −𝐿
𝑢 𝑜 = 𝑢0,
𝑑𝑢
𝑑𝑥 𝑥=0
= 𝑑0
𝑑
𝑑𝑥
𝑏
𝑑2𝑢
𝑑𝑥2
𝑥=𝐿
= 𝑚0 . 𝑏
𝑑2𝑢
𝑑𝑥2
0
= 𝓋0
Initial value problems:-
i) A general first order equation:-
Find 𝑢 𝑡 that satisfies the first-order differential equation and initial condition.
Equation and initial condition:-
𝑎
𝑑𝑢
𝑑𝑡
+ 𝑐𝑢 = 𝐹 for 0 < 𝑡 ≤ 𝑇
𝑢 0 = 𝑢0.
ii) A general second order equation:-
Find 𝑢 𝑡 that satisfies the second – order differential equation and initial conditions:-
x
x = 0 Ω = (o, L) x=L

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𝑎
𝑑𝑢
𝑑𝑡
+ 𝑏
𝑑2𝑢
𝑑𝑡 2
+ 𝑐𝑢 = 𝐹 for 0 < 𝑡 ≤ 𝑇
𝑢 𝑜 = 𝑢0, 𝑏
𝑑𝑢
𝑑𝑡 𝑡=0
= 𝑣0
Eigen value problems:-
(i) Axial vibration of a bar:
Find 𝑢 𝑥 and 𝑙 that satisfy the differential equation and boundary conditions.
−𝑑
𝑑𝑥
𝑎
𝑑𝑢
𝑑𝑥
− 𝜆𝑢 = 0 for 𝑜 < 𝑥 < 𝐿
𝑢 𝑜 = 0, 𝑎
𝑑𝑢
𝑑𝑥 𝑥=𝐿
= 0
(ii) Transverse vibration of a membrane:-
Find 𝑢 (𝑥, 𝑦) and 𝜆 that satisfy the partial differential equation and
boundary condition.
−
𝑑
𝑑𝑥
𝑎1
𝑑𝑢
𝑑𝑥
+
𝑑
𝑑𝑦
𝑎2
𝑑𝑢
𝑑𝑦
− 𝜆𝑢 = 0 in Ω
𝑢 = 0 on Γq
The values of 𝜆 are called cigen values and the associated functions 𝑢 are called cigen functions.
b) A simple pendulum consists of a bob of mass 𝒎(𝒌𝒈)attached to one end of a rod of
length 𝒍(𝒎) and the other end is pivoted to fixed point 𝟎.
Soln:-
𝐹 =
𝑑
𝑑𝑡
𝑚𝑣 = 𝑚𝑎
𝐹𝑥 = 𝑚.
𝑑𝑣𝑥
𝑑𝑡
−𝑚𝑔 sin 𝜃 = 𝑚𝑙
𝑑2
𝑄
𝑑𝑡2
or
𝑑2
𝑄
𝑑𝑡2
+
𝑔
𝑙
sin 𝑄 = 0
𝑑2
𝑄
𝑑𝑡2
+
𝑠
𝑙
𝑄 = 0
𝑑𝑄
𝑑𝑡
+ (𝑜) = 𝑈0.

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𝑄 𝑡 = 𝐴𝑠 𝑖𝑛 𝜆𝑡 + 𝐵 cos 𝜆 𝑡.
Where,
𝜆 =
𝑠
𝑙
and 𝐴 and 𝐵 are constant to be determined using the initial condition we
obtain.
𝐴 −
𝜈0
𝜆
, 𝐵 = 𝜃0
the solution to be linear problem is
𝜃 𝑡 =
𝜈0
𝜆
𝑆𝑖𝑛 ∧ 𝑡 + 0. 𝐶𝑜𝑠 𝜆𝑡
for zero initial velocity and non zero initial position 𝜃0 , we have.
𝜃 𝑡 = 𝜃0 cos 𝜆𝑡.
7) A simply supported beam subjected to uniformly distributed load over entire span and
it is subject to a point load at the centre of the span. Calculate the bending moment
and deflection at imdspan by using Rayleish – Ritz method. (Nov/Dec 2008).
Given data:-
To Find:
1. Deflection and Bending moment at mid span.
2. Compare with exact solutions.
Formula used
Solution:
We know that,
Deflection, y = a1 sin
πx
l
+ a2 sin
3πx
l 1
2

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Total potential energy of the beam is given by,
π = U − H
Where, U – Strain Energy.
H – Work done by external force.
The strain energy, U of the beam due to bending is given by,
U =
EI
2
d2y
dx2
2
dx
1
0
dy
dx
= a1 cos
πx
l
×
π
l
+ a2 cos
3πx
l
×
3π
l
dy
dx
=
a1πx
l
cos
πx
l
+
a23πx
l
cos
3πx
l
d2y
dx2 = −
a1π
l
sin
πx
l
×
π
l
−
a23π
l
sin
3πx
l
×
3π
l
d2y
dx2 = −
a1π2
l2 sin
πx
l
− 9
a2π2
l2 sin
3πx
l
Substituting
d2y
dx2
value in equation (3),
U =
EI
2
−
a1π2
l2 sin
πx
l
− 9
a2π2
l2 sin
3πx
l
2
dx
l
0
=
EI
2
a1π2
l2 sin
πx
l
+ 9
a2π2
l2 sin
3πx
l
2
dx
l
0
=
EI
2
π4
l4
a1
2
sin2 πx
l
+ 81a2
2
sin2 3πx
l
+ 2 a1 sin
πx
l
.9 a2 sin
3πx
l
dx
l
0
[∴ a + b 2
= a2
+ b2
+ 2ab]
U =
EI
2
π4
l4
a1
2
sin2 πx
l
+ 81a2
2
sin2 3πx
l
+ 18 a1a2 sin
πx
l
. sin
3πx
l
dx
l
0
a1
2
sin2 πx
l
dx =
𝑙
0
a1
2 1
2
1 − cos
2πx
l
l
0
dx ∴ sin2
x =
1−cos 2x
2
= a1
2 1
2
1 − cos
2πx
l
l
0
dx
=
a1
2
2
dx
𝑙
0
− cos
2πx
l
1
0
dx
2
2
3
2
4
2
5
2

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=
𝑎1
2
2
𝑥 0
𝑙
−
sin
2𝜋𝑥
𝑙
2𝜋
𝑙 0
𝑙
=
𝑎1
2
2
𝑙 − 0 −
1
2𝜋
sin
2𝜋𝑙
𝑙
− sin 0
=
𝑎1
2
2
𝑙 −
1
2𝜋
0 − 0 =
𝑎1
2 𝑙
2
∴ sin 2𝜋 = 0; sin 0 = 0
Similarly,
81 a2
2
sin2 3πx
l
dx =
𝑙
0
81a2
2 1
2
1 − cos
6πx
l
𝑙
0
dx ∴ sin2
x =
1−cos 2x
2
= 81a2
2 1
2
1 − cos
6πx
l
𝑙
0
dx
=
81a2
2
2
dx
𝑙
0
− cos
6πx
l
𝑙
0
dx
=
81𝑎2
2
2
𝑥 0
𝑙
−
sin
6𝜋𝑥
𝑙
6𝜋
𝑙 0
𝑙
=
81𝑎2
2
2
𝑙 − 0 −
1
6𝜋
sin
6𝜋𝑙
𝑙
− sin 0
=
81𝑎2
2
2
𝑙 −
1
6𝜋
0 − 0 =
𝑎1
2 𝑙
2
∴ sin 6𝜋 = 0; sin 0 = 0
18 a1a2 sin
πx
l
. sin
3πx
l
dx =
𝑙
0
18 a1a2 sin
πx
l
. sin
3πx
l
𝑙
0
dx
= 18 a1a2 sin
3πx
l
. sin
πx
l
𝑙
0
dx
= 18 a1a2
1
2
cos
2πx
l
− cos
4πx
l
𝑙
0
dx
∴ sin 𝐴 sin 𝐵 =
cos 𝐴−𝐵 −cos 𝐴+𝐵
2
=
18 a1a2
2
cos
2πx
l
dx
𝑙
0
− cos
4πx
l
𝑙
0
dx
a1
2
sin2
πx
l
dx =
𝑙
0
𝑎1
2
𝑙
2
81a2
2
sin2
3πx
l
dx =
𝑙
0
81𝑎2
2
𝑙
2
6
2
7
2

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=
18 a1a2
2
sin
2𝜋𝑥
𝑙
2𝜋
𝑙 0
𝑙
−
sin
4𝜋𝑥
𝑙
4𝜋
𝑙 0
𝑙
= 9 a1a2 0 − 0 = 0 ∴ sin 2𝜋 = 0; sin 4𝜋 = 0; sin 0 = 0
Substitute (6), (7) and (8) in equation (5),
U =
EI
2
π4
l4
𝑎1
2 𝑙
2
+
81𝑎2
2 𝑙
2
+ 0
U =
EI
4
π4 𝑙
l4
𝑎1
2
+ 81𝑎2
2
Work done by external forces,
𝐻 = 𝜔 𝑦 𝑑𝑥 + 𝑊 𝑦𝑚𝑎𝑥
𝑙
0
𝜔 𝑦 𝑑𝑥
𝑙
0
=
2𝜔𝑙
𝜋
𝑎1 +
𝑎2
3
We know that, 𝑦 = 𝑎1 sin
𝜋𝑥
𝑙
+ 𝑎2 sin
3𝜋𝑥
𝑙
In the span, deflection is maximum at 𝑥 =
1
2
𝑦𝑚𝑎𝑥 = 𝑎1 sin
𝜋 ×
1
2
𝑙
+ 𝑎2 sin
3𝜋×
1
2
𝑙
= 𝑎1 sin
𝜋
2
+ 𝑎2 sin
3𝜋
2
∴ sin
𝜋
2
= 1; sin
3𝜋
2
= −1
𝑦𝑚𝑎𝑥 = 𝑎1 − 𝑎2
Substitute (11) and (12) values in equation (8),
H =
2𝜔𝑙
𝜋
𝑎1 +
𝑎2
3
+ 𝑊 (𝑎1 − 𝑎2)
18 a1a2 sin
πx
l
. sin
3πx
l
dx =
𝑙
0
0
Strain Energy, U =
𝐸𝐼𝜋4
4𝑙3
𝑎1
2
+ 81𝑎2
2
8
2
9
2
10
11
12
13

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Substituting U and H values in equation (2), we get
𝜋 =
𝐸𝐼𝜋4
4𝑙3
𝑎1
2
+ 81𝑎2
2
−
2𝜔𝑙
𝜋
𝑎1 +
𝑎2
3
+ 𝑊 (𝑎1 − 𝑎2)
𝜋 =
𝐸𝐼𝜋4
4𝑙3
𝑎1
2
+ 81𝑎2
2
−
2𝜔𝑙
𝜋
𝑎1 +
𝑎2
3
− 𝑊 (𝑎1 − 𝑎2)
For stationary value of 𝜋, the following conditions must be satisfied.
𝜕𝜋
𝜕𝑎1
= 0and
𝜕𝜋
𝜕𝑎2
= 0
𝜕𝜋
𝜕𝑎1
=
𝐸𝐼𝜋4
4𝑙3
2𝑎1 −
2𝜔𝑙
𝜋
− 𝑊 = 0
𝐸𝐼𝜋4
2𝑙3
𝑎1 −
2𝜔𝑙
𝜋
− 𝑊 = 0
𝐸𝐼𝜋4
2𝑙3
𝑎1 =
2𝜔𝑙
𝜋
+ 𝑊
𝜕𝜋
𝜕𝑎2
=
𝐸𝐼𝜋4
4𝑙3
162𝑎2 −
2𝜔𝑙
𝜋
1
3
+ 𝑊 = 0
Similarly,
𝐸𝐼𝜋4
4𝑙3 162𝑎1 −
2𝜔𝑙
𝜋
+ 𝑊 = 0
𝐸𝐼𝜋4
2𝑙3 162𝑎1 =
2𝜔𝑙
𝜋
− 𝑊
From equation (12), we know that,
Maximum deflection, 𝑦𝑚𝑎𝑥 = 𝑎1 − 𝑎2
𝑦𝑚𝑎𝑥 =
2𝑙3
𝐸𝐼𝜋4
2𝜔𝑙
𝜋
+ 𝑊 −
2𝑙3
81𝐸𝐼𝜋4
2𝜔𝑙
3𝜋
− 𝑊
𝑎1 =
2𝑙3
𝐸𝐼𝜋4
2𝜔𝑙
𝜋
+ 𝑊
𝑎2 =
2𝑙3
81𝐸𝐼𝜋4
2𝜔𝑙
3𝜋
− 𝑊
14
15
16

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𝑦𝑚𝑎𝑥 =
4𝜔𝑙4
𝐸𝐼𝜋5
+
2𝑊𝑙3
𝐸𝐼𝜋4
−
4𝜔𝑙4
243𝐸𝐼𝜋5
+
2𝑊𝑙3
81𝐸𝐼𝜋4
𝑦𝑚𝑎𝑥 = 0.0130
𝜔𝑙4
𝐸𝐼
+ 0.0207
𝑊𝑙3
𝐸𝐼
We know that, simply supported beam subjected to uniformly distributed load, maximum deflection
is, 𝑦𝑚𝑎𝑥 =
5
384
𝜔𝑙4
𝐸𝐼
Simply supported beam subjected to point load at centre, maximum deflection is,
𝑦𝑚𝑎𝑥 =
𝜔𝑙3
48𝐸𝐼
So, total deflection, 𝑦𝑚𝑎𝑥 =
5
384
𝜔𝑙4
𝐸𝐼
+
𝜔𝑙3
48𝐸𝐼
From equations (17) and (18), we know that, exact solution and solution obtained by using
Rayleigh-Ritz method are same.
Bending Moment at Mid span
We know that,
Bending moment, M = EI
d2y
dx2
From equation (9), we know that,
d2y
dx2
= −
𝑎1𝜋2
𝑙2
sin
𝜋𝑥
𝑙
+
𝑎2 9𝜋2
𝑙2
sin
3𝜋𝑥
𝑙
Substitute 𝑎1 and 𝑎2 values from equation (15) and (16),
d2y
dx2
= −
2𝑙3
𝐸𝐼𝜋4
2𝜔𝑙
𝜋
+ 𝑊 ×
𝜋2
𝑙2
sin
𝜋𝑥
𝑙
+
2𝑙3
81𝐸𝐼𝜋4
2𝜔𝑙
3𝜋
− 𝑊 ×
9𝜋2
𝑙2
sin
3𝜋𝑥
𝑙
Maximum bending occurs at 𝑥 =
𝑙
2
= −
2𝑙3
𝐸𝐼𝜋4
2𝜔𝑙
𝜋
+ 𝑊 ×
𝜋2
𝑙2
sin
𝜋 ×
1
2
𝑙
+
2𝑙3
81𝐸𝐼𝜋4
2𝜔𝑙
3𝜋
− 𝑊 ×
9𝜋2
𝑙2
sin
3𝜋 ×
1
2
𝑙
𝑦𝑚𝑎𝑥 = 0.0130
𝜔𝑙4
𝐸𝐼
+ 0.0208
𝑊𝑙3
𝐸𝐼
18
19
17

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= −
2𝑙3
𝐸𝐼𝜋4
2𝜔𝑙
𝜋
+ 𝑊 ×
𝜋2
𝑙2
(1) +
2𝑙3
81𝐸𝐼𝜋4
2𝜔𝑙
3𝜋
− 𝑊 ×
9𝜋2
𝑙2
(−1)
∴ sin
𝜋
2
= 1; sin
3𝜋
2
= −1
= −
2𝑙
𝐸𝐼𝜋2
2𝜔𝑙
𝜋
+ 𝑊 −
2𝑙
9𝐸𝐼𝜋2
2𝜔𝑙
3𝜋
− 𝑊
= −
4𝜔𝑙2
𝐸𝐼𝜋3 +
2𝑊𝑙
𝐸𝐼𝜋2 −
4𝜔𝑙2
27𝐸𝐼𝜋3 +
2𝑊𝑙
9𝐸𝐼𝜋2
= −
3.8518𝜔𝑙2
𝐸𝐼𝜋3 +
2.222𝑊𝑙
𝐸𝐼𝜋2
Substitute
d2y
dx2
value in bending moment equation,
Mcentre = EI
d2y
dx2
= −𝐸𝐼 0.124
𝜔𝑙2
𝐸𝐼
+ 0.225
𝑊𝑙
𝐸𝐼
Mcentre = − 0.124 𝜔𝑙2
+ 0.225 𝑊𝑙
(∴Negative sign indicates downward deflection)
We know that, simply supported beam subjected to uniformly distributed load,
maximum bending moment is,
Mcentre =
𝜔𝑙2
8
Simply supported beam subjected to point load at centre, maximum bending moment
is,
Mcentre =
𝑊𝑙
4
Total bending moment, Mcentre =
𝜔𝑙2
8
+
𝑊𝑙
4
Mcentre = 0.125 𝜔𝑙2
+ 0.25 𝑊𝑙
From equation (20) and (21), we know that, exact solution and solution obtained by
using Rayleigh-Ritz method are almost same. In order to get accurate results, more terms in Fourier
series should be taken.
d2
y
dx2
= − 0.124
𝜔𝑙2
𝐸𝐼
+ 0.225
𝑊𝑙
𝐸𝐼
20
21

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UNIT – II ONE DIMENSIONAL PROBLEMS
PART - A
1. What is truss?(May/June 2014)
A truss is an assemblage of bars with pin joints and a frame is an assemblage of beam elements. Truss
can able to transmit load and it can deform only along its length. Loads are acting only at the joints.
2. 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.
3. What is natural co-ordinate?(Nov/Dec 2014), (April/May 2011)
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 inassembling of
stiffness matrices.
4. Define shape function. State its characteristics (May/June 2014), (Nov/Dec 2014), (Nov/Dec 2012)
In finite element method, field variables within an element are generally expressed by the
following approximate relation:
u (x,y) = N1(x,y) u1+N2 (x,y) u2+ N3(x,y) u3
Where u,1 u2, u3 are the values of the field variable at the nodes and N1 N2 N3 are interpolation
function. N1 N2 N3 is called shape functions because they are used to express the geometry or shape
of the element.
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.
5. 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 results can be improved by increasing the order of the Polynomial.
3. It is easy to formulate and computerize the finite element equations.
6. Write the governing equation for 1D Transverse and longitudinal vibration of the bar at one end
and give the boundary conditions. (April/May 2015)
The governing equation for free vibration of abeam is given by,
𝐸𝐼
𝜕4
𝑣
𝜕𝑥4
+ 𝜌𝐴
𝜕2
𝑣
𝜕𝑡2
= 0
Where,
E – Young’s modulus of the material.
I – Moment of inertia
Ρ – Density of the material.
A – Cross sectional area of the section of beam.
The governing equation for 1D longitudinal vibration of the bar at one end is given by
d2
U
dx2
AE + ρAUω2
= 0
Where,
U – axial deformation of the bar (m)
ρ – Density of the material of the bar (kg/m3
)

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ω – Natural frequency of vibration of the bar
A – Area of cross section of the bar (m2
)
7. Express the convections matrix for 1D bar element. (April/May 2015)
hPL
6
[
2 1
1 2
]
Convection stiffness matrix for 1D bar element:
hPTaL
2
1
1
Convection force matrix for 1D bar element:
Where,
h- Convection heat transfer coefficient (w/m2k)
P – Perimeter of the element (m)
L – Length of the element (m)
Ta – Ambient temperature (k)
8. State the properties of a stiffness matrix.(April/May 2015), (Nov/Dec 2012)
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.
9. Show the transformation for mapping x-coordinate system into a natural coordinate system for
a linear bar element and a quadratic bar element.(Nov/Dec 2012)
For example consider mapping of a rectangular parent element into a quadrilateral element
The shape functions of this element are
To get this mapping we define the coordinate of point P as,
10. Define dynamic analysis.(May/June 2014)
When the inertia effect due to the mass of the components is also considered in addition to the
externally applied load, then the analysis is called dynamic analysis.

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11. What are the types of boundary conditions used in one dimensional heat transfer problems?
(i) Imposed temperature
(ii) Imposed heat flux
(iii) Convection through an end node.
12. What are the difference between boundary value problem and initial value problem?
(i) The solution of differential equation obtained for physical problems which satisfies some
specified conditions known as boundary conditions.
(ii) If the solution of differential equation is obtained together with initial conditions then it is
known as initial value problem.
(iii)If the solution of differential equation is obtained together with boundary conditions then it is
known as boundary value problem.
PART -B
1. For the beam and loading shown in fig. calculate the nodal displacements.
Take [E] =210 GPa =210×109
𝑵 𝒎𝟐
, [I] = 6×10-6
m4
NOV / DEC 2013
Given data
Young’s modulus [E] =210 GPa =210×109
𝑁 𝑚2
Moment of inertia [I] = 6×10-6
m4
Length [L]1 = 1m
Length [L]2 = 1m
W=12 𝑘𝑁 𝑚 =12×103
𝑁 𝑚
F = 6KN
To find
 Deflection
Formula used
f(x)
−𝑙
2
−𝑙2
12
−𝑙
2
𝑙2
12
+
𝐹1
𝑀1
𝐹2
𝑀2
=
𝐸𝐼
𝑙3
12 6𝑙
6𝑙 4𝑙2
– 12 6𝑙
– 6𝑙 2𝑙2
– 12 – 6𝑙
6𝑙 2𝑙2
12 – 6𝑙
– 6𝑙 4𝑙2
𝑢1
𝜃1
𝑢2
𝜃2
Solution
For element 1
M1,θ1
M1,θ1
1 2
𝑣1,F1 𝑣2,F2
6 KN
2 m
1 m
12 𝐾𝑁 𝑚
6 KN

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f(x)
−𝑙
2
−𝑙2
12
−𝑙
2
𝑙2
12
+
𝐹1
𝑀1
𝐹2
𝑀2
=
𝐸𝐼
𝑙3
12 6𝑙
6𝑙 4𝑙2
– 12 6𝑙
– 6𝑙 2𝑙2
– 12 – 6𝑙
6𝑙 2𝑙2
12 – 6𝑙
– 6𝑙 4𝑙2
𝑢1
𝜃1
𝑢2
𝜃2
Applying boundary conditions
F1=0N ; F2=-6KN=-6×103
N; f(x)=0
M1=M2=0; u1=0; θ1=0; u2≠0; θ2≠0
103
×
0
0
−6
0
=
210×109×6×10−6
13
12 6
6 4
– 12 6
– 6 2
– 12 – 6
6 2
12 – 6
– 6 4
𝑢1
𝜃1
𝑢2
𝜃2
=1.26×106
12 6
6 4
−12 6
−6 2
−12 −6
6 2
12 −6
−6 4
0
0
𝑢2
0
For element 2
f(x)
−𝑙
2
−𝑙2
12
−𝑙
2
𝑙2
12
+
𝐹2
𝑀2
𝐹3
𝑀3
=
𝐸𝐼
𝑙3
12 6𝑙
6𝑙 4𝑙2
– 12 6𝑙
– 6𝑙 2𝑙2
– 12 – 6𝑙
6𝑙 2𝑙2
12 – 6𝑙
– 6𝑙 4𝑙2
𝑢2
𝜃2
𝑢3
𝜃3
f(x) = -12 𝑘𝑁
𝑚 =12×103 𝑁
𝑚; F2=F3=0=M2=M;
u2≠0; θ2≠0; u3=θ3=0
103
×
−6
−1
−6
1
+
0
0
0
0
= 1.26×106
×
12 6 − 12 6
6 4 − 6 2
−12 6 12 − 6
6 4 − 6 4
𝑢2
𝜃2
0
0
103
×
−6
−1
−6
1
= 1.26×106
×
12 6 − 12 6
6 4 − 6 2
−12 6 12 − 6
6 4 − 6 4
𝑢2
𝜃2
0
0
12 𝐾𝑁 𝑚 M3,θ3
M2,θ2
2 3
𝑣2,F2 𝑣3,F3

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Assembling global matrix
103
×
0
0
−12
−1
−6
1
= 1.26×106
×
12
6
−12
6
0
0
6
4
−6
2
0
0
−12
−6
24
0
−12
6
−6
2
0
8
−6
2
0
0
−12
−6
12
−6
0
0
6
2
−6
4
0
0
𝑢2
𝜃2
0
0
Solving matrix
-12×103
=1.26×106
×24u2=0; u2=-3.96×10-4
m
-1×103
=1.26×106
×8θ2=0; θ2=-9.92rad
Result
θ2=-9.92rad
u2=-3.96×10-4
m
2. Determine the axial vibration of a steel bar shown in fig. Take [E] =2.1×105
𝑵 𝒎𝒎𝟐
, [ρ] = 7800 𝒌𝒈 𝒎𝟑
NOV/DEC 2014
Given data
A1=1200mm2
; A2=900mm2
l1 =300mm; l2=400mm
Young’s modulus [E] =2.1×105
𝑁 𝑚𝑚2
Density [ρ] = 7800 𝐾𝑔 𝑚3
=7.8×10-6
𝐾𝑔 𝑚𝑚3
To find
 Stiffness matrix
 Mass matrix
 Natural frequency
 Mode shape
Formula used
General equation for free vibration of bar 𝑘 − 𝑚𝜆 {u}= 0
Stiffness matrix [k] =
𝐴𝐸
𝑙
1 – 1
– 1 1
Consistent mass matrix [m] =
𝜌𝐴𝐿
6
2 1
1 2
Lumped mass matrix [m] =
𝜌𝐴𝐿
2
1 0
0 1
Mode shape 𝑘 − 𝑚𝜆 U1 = 0 ; Normalization 𝑈1
𝑇
M U1 = 1
Solution
For element 1 u2
u1
300mm
1200mm2
400mm
300mm
1200mm2
900mm2

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𝐴𝐸
𝑙
1 – 1
– 1 1
[k1] =
𝐴1𝐸1
𝑙1
1 – 1
– 1 1
; =
1200×2.1×105
300
1 −1
−1 1
=8.4×105 1 – 1
– 1 1
; =105 8.4 – 8.4
– 8.4 8.4
𝜌𝐴𝐿
6
2 1
1 2
;
[m1] =
𝜌𝐴1𝐿1
6
2 1
1 2
=
1200×300×7.8×10−6
6
2 1
1 2
= 0.468×
2 1
1 2
[m1] =
0.936 0.468
0.468 0.936
For element 2
𝐴𝐸
𝑙
1 – 1
– 1 1
[k2] =
𝐴2𝐸2
𝑙2
1 – 1
– 1 1
;
=
900×2.1×105
400
1 −1
−1 1
= 4.73×105 1 −1
−1 1
[k2] = 105 4.73 – 4.73
– 4.73 4.73
;
𝜌𝐴𝐿
6
2 1
1 2
;
[m2] =
𝜌𝐴2𝐿2
6
2 1
1 2
=
900×400×7.8×10−6
6
2 1
1 2
= 0.468
2 1
1 2
[m2] =
0.936 0.468
0.468 0.936
Assembling global matrix
Stiffness matrix [k] = 105
8.4
−8.4
0
−8.4
13.13
−4.73
0
−4.73
4.73
0.936
0.468
0
0.468
1.87
0.468
0
0.468
0.936
u3
u2
400mm
900 mm2

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General equation for free vibration of bar 𝑘 − 𝑚𝜆 {u} = 0
105
8.4
−8.4
0
−8.4
13.13
−4.73
0
−4.73
4.73
- λ
0.936
0.468
0
0.468
1.87
0.468
0
0.468
0.936
=0
105 13.13
−4.73
−4.73
4.73
– λ
1.87
0.468
0.468
0.936
=0
13.13 × 105
− 1.87𝜆
−4.73 × 105
− 0.468𝜆
−4.73 × 105
− 0.468𝜆
4.73 × 105
− 0.936𝜆
= 0
[(13.13×105
-1.87λ)( 4.73 × 105
− 0.936𝜆) – (−4.73 × 105
− 0.468𝜆)( −4.73 × 105
− 0.468𝜆)] =0
6.2×1011
– 1.23× 106
λ – 8.84×10 5
λ + 1.75×λ2
-2.24×1011
-2.21×105
λ -2.21×105
λ – 0.22 λ2
=0
1.53λ2
-2.55×105
λ+3.96×1011
=0
Solving above equation
𝜆1 = 1.49×106
𝜆2 = 1.73×105
= 0.173×106
To find mode shape
𝑘 − 𝑚𝜆 {𝑢} = 0 ;
𝜆1 = 0.173×106
105 13.13
−4.73
−4.73
4.73
– 0.173×106 1.87
0.468
0.468
0.936
𝑢2
𝑢3
= 0
0.99 × 106
−0.55 × 106
−0.55 × 106
0.31 × 106
𝑢2
𝑢3
= 0
0.99×106
u2 – 0.55× 106
u3 =0
- 0.55×106
u2 + 0.31×106
u3 =0
u3 = 1.77u2
𝑘 − 𝑚𝜆 {𝑢} = 0
𝜆2 = 1.49×106
105 13.13
−4.73
−4.73
4.73
– 1.49×106 1.87
0.468
0.468
0.936
𝑢2
𝑢3
= 0
−1.48 × 106
−1.17 × 106
−1.17 × 106
−0.924 × 106
𝑢2
𝑢3
= 0
-1.482×106
u2 – 1.17× 106
u3 =0
- 1.17×106
u2 -0.924×106
u3 =0
𝑢3 =-1.26u2
Normalization 𝑈1
𝑇
M U1 = 1
Normalization of 𝜆1

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𝑢2 1.77𝑢2
1.87
0.46
0.468
0.936
𝑢2
1.77𝑢2
=1
𝑢2 1.77𝑢2
𝑢2
1.77𝑢2
= 1
2.7𝑢2
2
+ 3.79𝑢2
2
=1
𝑢2
2
=
1
6.4
; 𝑢2 = 0.392
𝑢3=1.78𝑢2; 𝑢3 = 0.698
Normalization of 𝜆2
𝑈2
𝑇
M U2 = 1
𝑢2 −1.26𝑢2
1.87
0.46
0.468
0.936
𝑢2
−1.26𝑢2
=1
1.28𝑢2 −0.707𝑢2
𝑢2
−1.256𝑢2
= 1
1.28𝑢2
2
+ 0.88𝑢2
2
=1
𝑢2
2
= 0.46; 𝑢3=-1.268𝑢2
𝑢3 = -0.84
Result
Mode shape
1
u3=-0.698
u2=0.678
u2=0.392
u3=0.698
u1=0
u1=0
Mode 2
Mode 1
2 3

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3. Consider the simply supported beam shown in fig. let the length L=1m,
E=2×1011
𝑵 𝒎𝟐
, area of cross section A=30cm2
, moment of inertia I=100mm4
,
density[ρ] = 7800𝒌𝒈 𝒎𝟑
. Determine the natural frequency using two types of
mass matrix. Lumped mass matrix and consistent mass matrix. APRIL / MAY 2011
Given data
Length = 1m
Young’s modulus E=2×1011 𝑁
𝑚2
Area A=30cm2
= 3×10-3
m2
Moment of inertia I=100mm4
= 100×10-12
m4
Density[ρ] = 7800 kg/m3
=76518 𝑁
𝑚3
To find
 Lumped mass matrix
 Consistent mass matrix
 Natural frequency
Formula used
General equation for free vibration of beam 𝑘 − 𝜔2
𝑚 {u} = 0
Stiffness matrix[k] =
𝐸𝐼
𝑙3
12 6𝑙
6𝑙 4𝑙2
– 12 6𝑙
– 6𝑙 2𝑙2
– 12 – 6𝑙
6𝑙 2𝑙2
12 – 6𝑙
– 6𝑙 4𝑙2
𝜌𝐴𝐿
420
156 22𝑙
22𝑙 4𝑙2
54 −13𝑙
13𝑙 −3𝑙2
54 13𝑙
−13𝑙 −3𝑙2
156 – 22𝑙
−22𝑙 4𝑙2
Lumped mass matrix [m] =
𝜌𝐴𝑙
2
1 0
0 0
0 0
0 0
0 0
0 0
1 0
0 0
Solution
For element 1
L

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Stiffness matrix[k]1 =
𝐸1𝐼
𝑙1
3
12
6𝑙1
−12
6𝑙1
6𝑙1
4𝑙1
2
−6𝑙1
2𝑙1
2
−12
−6𝑙1
12
−6𝑙1
6𝑙1
2𝑙1
2
−6𝑙1
4𝑙1
2
=
2×1011 ×100×−12
0.53
12
6 × 0.5
−12
6 × 0.5
6 × 0.5
4 × 0.52
−6 × 0.5
2 × 0.52
−12
−6 × 0.5
12
−6 × 0.5
6 × 0.5
2 × 0.52
−6 × 0.5
4 × 0.52
[k]1 =160×
12
3
−12
3
3
1
−3
0.5
−12
−3
12
−3
3
0.5
−3
1
Lumped mass matrix [m]1 =
𝜌𝐴𝑙1
2
1 0
0 0
0 0
0 0
0 0
0 0
1 0
0 0
=
76518×3×10−3×0.5
2
1 0
0 0
0 0
0 0
0 0
0 0
1 0
0 0
[m]1 =
57.38
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
Consistent mass matrix [m]1 =
𝜌𝐴𝑙1
420
156
22𝑙1
54
−13𝑙1
22𝑙1
4𝑙1
2
13𝑙1
−3𝑙1
2
54
13𝑙1
156
−22𝑙1
−13𝑙1
−3𝑙1
2
−22𝑙1
4𝑙1
2
=
76518×3×10−3×0.5
420
156
22 × 0.5
54
−13 × 0.5
22 × 0.5
4 × 0.52
13 × 0.5
−3 × 0.52
54
13 × 0.5
156
−22 × 0.5
−13 × 0.5
−3 × 0.52
−22 × 0.5
4 × 0.52
[m]1 =
42.63
3
14.74
−1.77
3
0.27
1.77
−0.20
14.74
1.77
42.63
−3
−1.77
−0.20
−3
0.27
𝑣1 𝑣2
0.5 m
θ2
θ1
1 2

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For element 2
Stiffness matrix[k]2 =
𝐸𝐼
𝑙2
3
12
6𝑙2
−12
6𝑙2
6𝑙2
4𝑙2
2
−6𝑙2
2𝑙2
2
−12
−6𝑙2
12
−6𝑙2
6𝑙2
2𝑙2
2
−6𝑙2
4𝑙2
2
=
2×1011 ×100×−12
0.53
12
6 × 0.5
−12
6 × 0.5
6 × 0.5
4 × 0.52
−6 × 0.5
2 × 0.52
−12
−6 × 0.5
12
−6 × 0.5
6 × 0.5
2 × 0.52
−6 × 0.5
4 × 0.52
[k]2 = 160×
12
3
−12
3
3
1
−3
0.5
−12
−3
12
−3
3
0.5
−3
1
Lumped mass matrix [m]2 =
𝜌𝐴𝑙2
2
1 0
0 0
0 0
0 0
0 0
0 0
1 0
0 0
=
76518×3×10−3×0.5
2
1 0
0 0
0 0
0 0
0 0
0 0
1 0
0 0
[m]2 =
57.38
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
Consistent mass matrix [m]2 =
𝜌𝐴𝑙2
420
156
22𝑙2
54
−13𝑙2
22𝑙2
4𝑙2
2
13𝑙2
−3𝑙2
2
54
13𝑙2
156
−22𝑙2
−13𝑙2
−3𝑙2
2
−22𝑙2
4𝑙2
2
=
76518×3×10−3×0.5
420
156
22 × 0.5
54
−13 × 0.5
22 × 0.5
4 × 0.52
13 × 0.5
−3 × 0.52
54
13 × 0.5
156
−22 × 0.5
−13 × 0.5
−3 × 0.52
−22 × 0.5
4 × 0.52
[m]2 =
42.63
3
14.74
−1.77
3
0.27
1.77
−0.20
14.74
1.77
42.63
−3
−1.77
−0.20
−3
0.27
Global matrix
𝑣2 𝑣3
0.5 m
θ3
θ2
2 3

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Stiffness matrix [k] =160×
12
3
−12
3
0
0
3
1
−3
0.5
0
0
−12
−3
24
0
−12
3
3
0.5
0
2
−3
0.5
0
0
−12
−3
12
−3
0
0
3
0.5
−3
1
Lumped mass matrix [m]=
57.38
0
0
0
0
0
0
0
0
0
0
0
0
0
114.77
0
0
0
0
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
0
0
Consistent mass matrix[m]=
42.63
3
14.74
−1.77
0
0
3
0.27
1.77
−0.2
0
0
14.74
1.77
85.26
0
14.74
−1.77
−1.77
−0.2
0
0.5
1.77
−0.2
0
0
14.74
1.77
42.63
−3
0
0
−1.77
−0.2
−3
0.27
Frequency for lumped mass matrix
𝑘 − 𝜔2
𝑚 {u} = 0
160 ×
12
3
−12
3
0
0
3
1
−3
0.5
0
0
−12
−3
24
0
−12
3
3
0.5
0
2
−3
0.5
0
0
−12
−3
12
−3
0
0
3
0.5
−3
1
− 𝜔2
57.38
0
0
0
0
0
0
0
0
0
0
0
0
0
114.77
0
0
0
0
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
0
0
𝑣1
𝜃1
𝑣2
𝜃2
𝑣3
𝜃3
=0
𝑣1=0=𝜃1; 𝑣2≠0; 𝜃2≠0 𝑣3=0=𝜃3;
160 ×
12
3
−12
3
0
0
3
1
−3
0.5
0
0
−12
−3
24
0
−12
3
3
0.5
0
2
−3
0.5
0
0
−12
−3
12
−3
0
0
3
0.5
−3
1
− 𝜔2
57.38
0
0
0
0
0
0
0
0
0
0
0
0
0
114.77
0
0
0
0
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
0
0
0
0
𝑣2
𝜃2
0
0
=0
160 ×
24
0
0
2
− 𝜔2 114.7
0
0
0
𝑣2
𝜃2
= 0
3840 − 𝜔2
× 114.7
0 − 0
0 − 0
320 − 0
= 0
{(3840 − 𝜔2
× 114.7) × ( 320 − 0)-0-0} =0
1228800-36704𝜔2
= 0
𝜔2
= 33.47
𝜔 = 5.78 𝑟𝑎𝑑
𝑠
Frequency for consistent mass matrix

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𝑘 − 𝜔2
𝑚 {u} = 0
160 ×
12
3
−12
3
0
0
3
1
−3
0.5
0
0
−12
−3
24
0
−12
3
3
0.5
0
2
−3
0.5
0
0
−12
−3
12
−3
0
0
3
0.5
−3
1
− 𝜔2
42.63
3
14.74
−1.77
0
0
3
0.27
1.77
−0.2
0
0
14.74
1.77
85.26
0
14.74
−1.77
−1.77
−0.2
0
0.5
1.77
−0.2
0
0
14.74
1.77
42.63
−3
0
0
−1.77
−0.2
−3
0.27
𝑣1
𝜃1
𝑣2
𝜃2
𝑣3
𝜃3
=0
𝑣1=0=𝜃1; 𝑣2≠0; 𝜃2≠0 𝑣3=0=𝜃3;
160 ×
12
3
−12
3
0
0
3
1
−3
0.5
0
0
−12
−3
24
0
−12
3
3
0.5
0
2
−3
0.5
0
0
−12
−3
12
−3
0
0
3
0.5
−3
1
− 𝜔2
42.63
3
14.74
−1.77
0
0
3
0.27
1.77
−0.2
0
0
14.74
1.77
85.26
0
14.74
−1.77
−1.77
−0.2
0
0.5
1.77
−0.2
0
0
14.74
1.77
42.63
−3
0
0
−1.77
−0.2
−3
0.27
0
0
𝑣2
𝜃2
0
0
=0
160 ×
24
0
0
2
− 𝜔2 85.26
0
0
0.5
𝑣2
𝜃2
= 0
3840 − 85.26ω2
0 − 0
0 − 0
320 − 0.5ω2 =0
(3840 − 85.26𝜔2
) 320 − 0.5𝜔2
= 0
1.23×106
-1920𝜔2
-27283.2𝜔2
+42.63𝜔4
=0
Take λ = 𝜔2
42.63 λ2
-29203.3 λ+1.23×106
=0 ax2
+bx+c=0; x =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
𝜆 =
29203.3 ± 29203.32−4×42.63×1.23×106
2×42.63
=
29203.3 ±25359.28
85.26
𝜆1 =
29203.3+25359.28
85.26
; 𝜆2 =
29203.3−25359.28
85.26
𝜆1 =639.95; 𝜆2=45.08
λ = 𝜔2
𝜔1 = λ1 ; 𝜔2 = λ2
𝜔1= 639.95 𝜔2 = 45.08
𝜔1= 25.3 𝑟𝑎𝑑 𝑠 𝜔2= 6.7 𝑟𝑎𝑑 𝑠

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4. For a tapered plate of uniform thickness t = 10mm as shown in fig. find the
displacements at the nodes by forming in to two element model. The bar has mass
density ρ = 7800𝑲𝒈 𝒎𝟑
Young’s modulus E = 2×105
𝑴𝑵 𝒎𝟐
. In addition to self
weight the plate is subjected to a point load p = 10KN at its centre. Also
determine the reaction force at the support. Nov/Dec 2006
Given data
Mass density ρ = 7800𝑘𝑔 𝑚3
= 7800 × 9.81=76518 𝑁 𝑚3
= 7.65 × 10-5
𝑁 𝑚𝑚3
Young’s modulus E = 2×105
𝑀𝑁 𝑚2
;
= 2×105
× 106
𝑁 𝑚2
= 2×105
𝑁 𝑚𝑚2
Point load P = 10 KN
To find
 Displacement at each node
 Reaction force at the support
Formula used
{F} =[K] {u}
𝐴𝐸
𝑙
1 – 1
– 1 1
𝑢1
𝑢2
Force vector 𝐹 =
𝜌𝐴𝑙
2
1
1
𝐹1
𝐹2
=
𝐴𝐸
𝑙
1 – 1
– 1 1
𝑢1
𝑢2
{R} =[K] {u} -{F}
Solution
The given taper bar is considered as stepped bar as shown in fig.
1
40m
m
P
80mm
150m
m
300m
m
W1=80mm
W3=40
mm
P
W1=80mm
150m
m
300m
m
150mm
150mm
3
10KN
2

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W1 = 80mm
W2 =
𝑊1+𝑊3
2
=
80+40
2
= 60 mm
W3 = 40mm
Area at node 1 A1 = Width × thickness
=W1 × t1
= 80 × 10 = 800mm2
Area at node 2; A2 = Width × thickness
=W2 × t2 = 60 × 10 =600mm2
Area at node 1 A1 = Width × thickness
= W3 × t3 = 40 × 10 =400mm2
Average area of element 1
Ā1 =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 1 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 2
2
=
𝐴1 + 𝐴2
2
=
800+600
2
= 700mm2
Average area of element 2
Ā2 =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 2 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 3
2
=
𝐴2 + 𝐴3
2
=
600+400
2
= 500mm2
For element 1
Stiffness matrix [k]1 =
Ā1𝐸1
𝑙1
1 – 1
– 1 1
𝑢1
𝑢2
=
700 ×2×105
150
1
−1
−1
1
𝑢1
𝑢2
= 2× 105 4.67
−4.67
−4.67
4.67
𝑢1
𝑢2
Force vector 𝐹 1 =
𝜌Ā1𝑙1
2
1
1
=
7.65×10−5×700×150
2
1
1
=
4.017
4.017
For element 2
Stiffness matrix [k]2 =
Ā2𝐸2
𝑙2
1 – 1
– 1 1
𝑢2
𝑢3
=
500 ×2×105
150
1
−1
−1
1
𝑢2
𝑢3
= 2× 105 3.33
−3.33
−3.33
3.33
𝑢2
𝑢3
150mm
10KN
u2,F2
u1,F1
150mm
u2,F2
u3,F3
10KN

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Force vector 𝐹 2 =
𝜌Ā2𝑙2
2
1
1
=
7.65×10−5×500×150
2
1
1
=
2.869
2.869
Global matrix
Stiffness matrix [k] = 2×105
×
4.66
−4.66
0
−4.66
7.99
−3.33
0
−3.33
3.33
Force vector 𝐹 =
4.017
6.88
2.87
Finite element equation
{F} =[K] {u}
𝐹1
𝐹2
𝐹3
= 2×105
×
4.66
−4.66
0
−4.66
7.99
−3.33
0
−3.33
3.33
𝑢1
𝑢2
𝑢3
𝑢1 = 0; 𝑢2 ≠ 0; 𝑢3 ≠ 0; 𝐹2 = 10 × 103
N
𝐹1
𝐹2
𝐹3
= 2×105
×
4.66
−4.66
0
−4.66
7.99
−3.33
0
−3.33
3.33
𝑢1
𝑢2
𝑢3
4.017
6.88 + 10 × 103
2.87
= 2×105
×
4.66
−4.66
0
−4.66
7.99
−3.33
0
−3.33
3.33
0
𝑢2
𝑢3
10006.88
2.86
= 2×105 7.99
−3.33
−3.33
3.33
𝑢2
𝑢3
2×105
(7.99𝑢2 − 3.33𝑢3) = 10006.88
2×105
(-3.33𝑢2 + 3.33𝑢3) = 2.86
Solving above equation
2×105
(4.66 𝑢2) = 10009.74
𝑢2 = 0.01074 mm
2×105
(-3.33×0.01074+3.33𝑢3) = 2.86
666000𝑢3 = 2.86 + 7152.88
𝑢3 = 0.01074

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Reaction force
{R} =[K] {u} -{F}
𝑅1
𝑅2
𝑅3
= 2×105
×
4.66
−4.66
0
−4.66
7.99
−3.33
0
−3.33
3.33
𝑢1
𝑢2
𝑢3
-
𝐹1
𝐹2
𝐹3
𝑅1
𝑅2
𝑅3
= 2×105
×
4.66
−4.66
0
−4.66
7.99
−3.33
0
−3.33
3.33
𝑢1
0.01074
0.01074
-
4.017
10006.88
2.87
=2×105
0 − 0.05 + 0
0 + 0.086 − 0.036
0 − 0.036 + 0.036
-
4.017
10006.88
2.87
= 2×105
−0.05
0.05
0
-
4.017
10006.88
2.87
=
−10000
10000
0
-
4.017
10006.88
2.87
=
−10004.017
−6.88
−2.86
Result
𝑅1
𝑅2
𝑅3
=
−10004.017
−6.88
−2.86
5. A wall of 0.6m thickness having thermal conductivity of 1.2 W/mk. The wall is to
be insulated with a material of thickness 0.06m having an average thermal
conductivity of 0.3 W/mk. The inner surface temperature in 1000O
C and outside
of the insulation is exposed to atmospheric air at 30o
c with heat transfer co-
efficient of 35 W/m2
k. Calculate the nodal temperature. NOV/DEC 2014
Given Data:-
Thickness of the wall, l1 = 0.6m
Thermal conductivity of the wall K1= 1.2W/mk
Thickness of the insulation l2 = 0.06m
Thermal Conductivity of the wall K2 = 0.3W/mk
Inner surface temperature T1= 1000o
C+273
= 1273 K
𝑙1 𝑙2
𝑇∞
h
Convection
T3
T1
Conduction Conduction
T2
Wall Insulation

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Atmospheric air temperature T2 = 30 +273
= 303 K
Heat transfer co-efficient at outer side h = 35W/m2
k
To find
Nodal temperature T2 and T3
Formula used
1D Heat conduction
𝐹1
𝐹2
=
𝐴𝑘
𝑙
1 – 1
– 1 1
𝑇1
𝑇2
1D Heat conduction with free end convection
[K]=
𝐴𝑘
𝑙
1 – 1
– 1 1
+ hA
0 0
0 1
Solution
For element 1
f1
f2
=
k1A1
l1
1 −1
−1 1
T1
T2
For unit area: A1 = 1m2
=
1.2
0.6
1 −1
−1 1
T1
T2
f1
f2
=
2 −2
−2 2
T1
T2
For element (2)
A2K2
l2
1 −1
−1 1
+ hA
0 0
0 1
T2
T3
= h T2A
0
1
1 X 0.3
0.06
1 −1
−1 1
+ 35 × 1
0 0
0 1
T2
T3
=35×303×1×
0
1
5 −5
−5 5
+
0 0
0 35
T1
T2
= 0
10.605 × 103
5 −5
−5 5
T1
T2
= 0
10.605 × 103
Assembling finite element equation
f1
f2
f3
=
2 −2 0
−2 7 −5
0 −5 40
T1
T2
T3
f1 = 0
Convection
h T∞
L2
Conduction
T2 T3
L1
Conduction
T1 T2

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f2 = 0
f3 = 10.605 x 103
2 −2 0
−2 7 −5
0 −5 40
T1
T2
T3
=
0
0
10.605 × 103
Step (1)
The first row and first column of the stiffness matrix K have been set equal to 0
except for the main diagonal.
1 0 0
0 7 −5
0 −5 40
T1
T2
T3
=
0
0
10.605 × 103
Step – II
The first row of the force matrix is replaced by the known temperature at node 1
1 0 0
0 7 −5
0 −5 40
T1
T2
T3
=
1273
0
10.605 × 103
Step – III
The second row first column of stiffness K value is multiplied by known
temperature at node 1 -2 × 1273 = -2546. This value positive digit 2546 has been
added to the second row of the force matrix.
1 0 0
0 7 −5
0 −5 40
T1
T2
T3
=
1273
2546
10.605 × 103
⟹ 7 T2 − 5 T3 = 2546
−5 T2 + 40 T3 = 10.605 × 103
Solving above Eqn ×8 56 T2 − 40T3 = 20.368 × 103
5 T2 − 40T3 = 10.605 × 103
51 T2 = 30973
T2 = 607.31 K
7 × 607.31 -5 T3 = 2546
4251.19 - 5 T3 = 2546
-−5 T3 = −1705
T3 = 341.03 K
Result
Nodal Temp T1 = 1273 K
T2 = 607.31K
T3 = 341.03 K

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7. Derivation of the displacement function u and shape function N for one dimensional
linear bar element. OR
Derive the shape function, stiffness matrix and load vector for one dimensional bar
element. May / June 2013
Consider a bar with element with nodes 1 and 2 as shown in Fig. 𝜐1 and 𝜐2 are the
displacement at the respective nodes. 𝜐1 And 𝜐2 is degree of freedom of this bar element.
Fig Two node bar element
Since the element has got two degrees of freedom, it will have two generalized co-ordinates.
𝑢 = 𝑎0 + 𝑎1𝑥
Where, 𝑎0 and 𝑎1 are global or generalized co – ordinates.
Writing the equation in matrix form,
𝑢 = 1 𝑥
𝑎0
𝑎1
At node 1, 𝑢 = 𝑢1, 𝑥 = 0
At node 1, 𝑢 = 𝑢2, 𝑥 = 1
Substitute the above values ion equation,
𝑢1 = 𝑎0
𝑢2 = 𝑎0 + 𝑎1 𝑙
Arranging the equation in matrix form,
𝑢1
𝑢2
=
1 0
1 𝑙
𝑎0
𝑎1
𝑢∗
𝐶 𝐴
Where, 𝑢∗
⟶ Degree of freedom.
𝐶 ⟶ Connectivity matrix.
𝐴 ⟶ Generalized or global co-ordinates matrix.
𝑎0
𝑎1
=
1 0
1 𝑙
−1 𝑢1
𝑢2
=
1
𝑙−0
1 −0
−1 1
𝑢1
𝑢2
𝑁𝑜𝑡𝑒:
𝑎11 𝑎12
𝑎21 𝑎22
−1
=
1
𝑎11 𝑎22 − 𝑎12𝑎21
×
𝑎22 −𝑎12
−𝑎21 𝑎11
𝓍
1 2
𝑢1 𝑢2
𝑙

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𝑎0
𝑎1
=
1
𝑙
𝑙 0
−1 1
𝑢1
𝑢2
Substitute
𝑎0
𝑎1
𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑢 = 1 𝑥
1
𝑙
𝑙 0
−1 1
𝑢1
𝑢2
=
1
𝑙
1 𝑥
𝑙 0
−1 1
𝑢1
𝑢2
=
1
𝑙
1 − 𝑥 0 + 𝑥
𝑢1
𝑢2
∵ 𝑀𝑎𝑡𝑟𝑖𝑥 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 1 × 2 2 × 2 = 1 × 2
𝑢 =
1− 𝑥
𝑙
𝑥
𝑙
𝑢1
𝑢2
𝑢 = 𝑁1 𝑁2
𝑢1
𝑢2
Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2
Where, Shape function, 𝑁1 =
𝑙− 𝑥
𝑙
; 𝑠𝑕𝑎𝑝𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑁2 =
𝑥
𝑙
Stiffness matrix for one dimensional linear bar element
Consider a bar with element with nodes 1 and 2 as shown in Fig. 𝜐1 and 𝜐2 are the
displacement at the respective nodes. 𝜐1 And 𝜐2 is degree of freedom of this bar element.
Stiffness matrix, 𝐾 = B T
𝐷 𝐵 𝑑𝑣
𝑣
Shape function, 𝑁1 =
𝑙− 𝑥
𝑙
; 𝑠𝑕𝑎𝑝𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑁2 =
𝑥
𝑙
Strain displacement matrix,[B] =
𝑑𝑁1
𝑑𝑥
𝑑𝑁2
𝑑𝑥
=
−1
𝑙
1
𝑙
[B]T
=
−1
𝑙
1
𝑙
One dimensional problem [D] = [E] = young’s modulus
[K] =
−1
𝑙
1
𝑙
× 𝐸 ×
−1
𝑙
1
𝑙
𝑑𝑣
𝒍
𝟎
𝓍
1 2
𝑢1 𝑢2
𝑙

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=
1
𝑙2
−1
𝑙2
−1
𝑙2
1
𝑙2
𝑙
0
× 𝐸 × 𝑑𝑣 [dv = A×dx
=
1
𝑙2
−1
𝑙2
−1
𝑙2
1
𝑙2
𝑙
0
× 𝐸 × A × dx
= AE
1
𝑙2
−1
𝑙2
−1
𝑙2
1
𝑙2
× 𝑑𝑥
𝑙
0
= AE
1
𝑙2
−1
𝑙2
−1
𝑙2
1
𝑙2
𝑥 0
𝑙
= AE
1
𝑙2
−1
𝑙2
−1
𝑙2
1
𝑙2
(𝑙 − 0)
= AE 𝑙
1
𝑙2
−1
𝑙2
−1
𝑙2
1
𝑙2
=
𝐴𝐸𝑙
𝑙2
1
−1
−1
1
[K] =
𝐴𝐸
𝑙
1 – 1
– 1 1
Finite element equation for finite element analysis
{F} =[K] {u}
𝐹1
𝐹2
=
𝐴𝐸
𝑙
1 – 1
– 1 1
𝑢1
𝑢2
Load vector [F]
Consider a vertically hanging bar of length𝑙, uniform cross section A, density ρ and young’s
modulus E. this bar is subjected to self weight Xb
The element nodal force vector
𝐹 𝑒 = 𝑁 𝑇
Xb
Self weight due to loading force Xb = ρAdx
Where; 𝑁1 =
𝑙− 𝑥
𝑙
; 𝑁2 =
𝑥
𝑙
;
[N] =
𝑙− 𝑥
𝑙
𝑥
𝑙
x
xb

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[N]T
=
𝑙− 𝑥
𝑙
𝑥
𝑙
Substitute Xb and [N]T
values
𝐹 𝑒 =
𝑙− 𝑥
𝑙
𝑥
𝑙
𝑙
0
ρA dx = ρA
𝑙− 𝑥
𝑙
𝑥
𝑙
𝑙
0
dx
= ρA
𝑥 −
𝑥2
2𝑙
𝑥2
2𝑙 0
𝑙
= ρA
𝑙 −
𝑙2
2𝑙
𝑙2
2𝑙
= ρA
𝑙 −
𝑙
2
𝑙
2
= ρA
𝑙
2
𝑙
2
Force vector {F} =
𝜌𝐴𝑙
2
1
1
7. DERIVATION OF SHAPE FUNCTION AN STIFFNESS MATRIX FOR ONE-
DIMENSIONAL QUADRATIC BAR ELEMENT: May / June 2012
Consider a quadratic bar element with nodes 1,2 and 3 as shown in
Fig.(i), 𝑢1, 𝑢2 𝑎𝑛𝑑 𝑢3 are the displacement at the respective nodes. So, 𝑢1, 𝑢2 𝑎𝑛𝑑 𝑢3 are
considered as degree of freedom of this quadratic bar element.
Fig. (i). Quadratic bar element
Since the element has got three nodal displacements, it will have three generalized
coordinates.
u = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2
Where, 𝑎0, 𝑎1 𝑎𝑛𝑑 𝑎2 are global or generalized coordinates. Writing the equation is matrix
form,
𝓍
𝜐1 1 2 𝜐2
𝑙
2
𝑙
3 𝜐3

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𝑈 = 1 𝑥 𝑥2
𝑎0
𝑎1
𝑎2
At node 1, u = u1, 𝑥 = 0
At node 2, u = u2, 𝑥 = 1
At node 3, u = u3, 𝑥 =
1
2
Substitute the above values in equation.
u1 = 𝑎0
u2 = 𝑎0 + 𝑎1 𝑙 + 𝑎2 𝑙2
u3 = 𝑎0 + 𝑎1
𝑙
2
+ 𝑎2
𝑙
2
2
Substitute the equation we get
u2 = 𝑢1 + 𝑎1 𝑙 + 𝑎2 𝑙2
u3 = 𝑢1 +
𝑎1 𝑙
2
+
𝑎2 𝑙2
4
u2 − u1 = 𝑎1 𝑙 + 𝑎2 𝑙2
u3 − 𝑢1 =
𝑎1 𝑙
2
+
𝑎2 𝑙2
4
u2 − u1
u3 − 𝑢1
=
𝑙 𝑙2
𝑙
2
𝑙2
4
a1
a2
⇒
a1
a2
=
𝑙 𝑙2
𝑙
2
𝑙2
4
−1
u2 − u1
u3 − 𝑢1
=
1
𝑙3
4
−
𝑙3
2
𝑙2
4
−𝑙2
−𝑙
2
𝑙
u2 − u1
u3 − 𝑢1
𝑁𝑜𝑡𝑒 ∵
𝑎11 𝑎12
𝑎21 𝑎22
−1
=
1
𝑎11𝑎22 − 𝑎12𝑎21
X
𝑎22 −𝑎12
−𝑎21 𝑎11
⇒
a1
a2
=
1
−𝑙3
4
𝑙2
4
−𝑙2
−𝑙
2
𝑙
u2 − u1
u3 − 𝑢1
⇒ 𝑎1 =
−4
𝑙3
𝑙2
4
u2 − u1 −𝑙2
u3 − 𝑢1

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⇒ 𝑎2 =
−4
𝑙3
−𝑙
2
u2 − u1 + 𝑙 u3 − 𝑢1
Equation 𝑎1 =
−4
𝑙3
𝑙2 𝑢2
4
−
𝑙2 𝑢1
4
−𝑙2
𝑢3 + 𝑙2
𝑢1
=
−4𝑙2 𝑢2
4𝑙3
+
4𝑙2 𝑢1
4𝑙3
+
4𝑙2 𝑢3
𝑙3
−
4𝑙2 𝑢1
𝑙3
=
− 𝑢2
𝑙
+
𝑢1
𝑙
+
4 𝑢3
𝑙
−
4 𝑢1
𝑙
𝑎1 =
−3 𝑢1
𝑙
−
𝑢2
𝑙
+
4 𝑢3
𝑙
Equation
𝑎2 =
−4
𝑙3
−𝑙𝑢2
2
−
𝑙
2
𝑢1 + 𝑙𝑢3 − 𝑙𝑢1
=
4𝑙 𝑢2
2 𝑙3
+
4𝑙
2 𝑙3
𝑢1 −
4𝑙
𝑙3
𝑢3 +
4𝑙
𝑙3
𝑢1
=
2𝑢2
𝑙2
−
2
𝑙2
𝑢1 −
4
𝑙2
𝑢3 +
4
𝑙2
𝑢1
𝑎2 =
2
𝑙2 𝑢1 +
2𝑢2
𝑙2 −
4
𝑙2 𝑢3
𝑎0
𝑎1
𝑎2
=
1 0 0
−3
𝑙
−1
𝑙
4
𝑙
2
𝑙2
2
𝑙2
−4
𝑙2
𝑢1
𝑢2
𝑢3
Substitution the equation
𝑢 = 1 𝑥 𝑥2
1 0 0
−3
𝑙
−1
𝑙
4
𝑙
2
𝑙2
2
𝑙2
−4
𝑙2
𝑢1
𝑢2
𝑢3
𝑢 = 1 −
3
𝑙
𝑥 +
2 𝑥2
𝑙2
−𝑥
𝑙
+
2 𝑥2
𝑙2
4𝑥
𝑙
−
4 𝑥2
𝑙2
𝑢1
𝑢2
𝑢3
𝑢 = 𝑁1 𝑁2 𝑁3
𝑢1
𝑢2
𝑢3
𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3
Where, shape function,
𝑁1 = 1 −
3𝑥
𝑙
+
2𝑥2
𝑙2
𝑁2 =
−𝑥
𝑙
+
2𝑥2
𝑙2
𝑁3 =
4𝑥
𝑙
−
4𝑥2
𝑙2

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STIFFNESS MATRIX FOR ONE-DIMENSIONAL QUADRATIC BAR ELEMENT:
Fig. A bar element with three nodes
Consider a one dimensional quadratic bar element with nodes 1,2, and 3 as shown in Fig. 2.
Let 𝑢1, 𝑢2 𝑎𝑛𝑑 𝑢3 be the nodal displacement parameters or otherwise known as degree of
freedom.
We know that,
Stiffness matrix, 𝑘 = 𝐵 𝑇
𝐷 𝐵 𝑑𝑣
𝑣
In one dimensional quadratic bar element,
Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3
Where, 𝑁1 = 1 −
3𝑥
𝑙
+
2𝑥2
𝑙2
𝑁2 =
−𝑥
𝑙
+
2𝑥2
𝑙2
𝑁3 =
4𝑥
𝑙
−
4𝑥2
𝑙2
We know that,
Strain – Displacement matrix, 𝐵 =
𝑑 𝑁1
𝑑𝑥
𝑑 𝑁2
𝑑𝑥
𝑑 𝑁3
𝑑𝑥
⟹
𝑑 𝑁1
𝑑𝑥
=
−3
𝑙
+
4 𝑥
𝑙2
⟹
𝑑 𝑁2
𝑑𝑥
=
−1
𝑙
+
4 𝑥
𝑙2
⟹
𝑑 𝑁3
𝑑𝑥
=
4
𝑙
+
8 𝑥
𝑙2
Substitute the equation
𝐵 =
−3
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
𝜐1 1 2 3 𝜐2
𝑙
2
𝑙
1 2 𝜐1 2

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𝐵 𝑇
=
−3
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
4
𝑙
+
8𝑥
𝑙2
In one dimensional problems,
𝐷 = 𝐸 = 𝐸 = 𝑌𝑜𝑢𝑛𝑔′
𝑠𝑀𝑜𝑑𝑢𝑙𝑢𝑠
Substitute 𝐵 𝐵 𝑇
𝑎𝑛𝑑 𝐷 values in stiffness matrix equation 𝐿𝑖𝑚𝑖𝑡 𝑖𝑠 0 𝑡𝑜 𝑙 .
⟹ =
−3
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
𝑙
0
−3
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
× E 𝑑𝑣
⟹ 𝑘 = 𝐸𝐴
−3
𝑙
+
4𝑥
𝑙2
−3
𝑙
+
4𝑥
𝑙2
−3
𝑙
+
4𝑥
𝑙2
−3
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
−3
𝑙
+
4𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
−3
𝑙
+
4𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
−1
𝑙
+
4𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
4
𝑙
−
8𝑥
𝑙2
𝑙
0
𝑑𝑥
⟹ 𝑘 = 𝐸𝐴
9
𝑙2
−
12𝑥
𝑙3
−
12𝑥
𝑙3
+
16𝑥2
𝑙4
3
𝑙2
−
12𝑥
𝑙3
−
4𝑥
𝑙3
+
16𝑥2
𝑙4
−12
𝑙2
+
24𝑥
𝑙3
+
16𝑥
𝑙3
−
32𝑥2
𝑙4
3
𝑙2
−
12𝑥
𝑙3
−
4𝑥
𝑙3
+
16𝑥2
𝑙4
1
𝑙2
−
4𝑥
𝑙3
−
4𝑥
𝑙3
+
16𝑥2
𝑙4
−4
𝑙2
+
8𝑥
𝑙3
+
16𝑥
𝑙3
−
32𝑥2
𝑙4
−12
𝑙2
+
24𝑥
𝑙3
+
16𝑥
𝑙3
−
32𝑥2
𝑙4
−4
𝑙2
+
8𝑥
𝑙3
+
16𝑥
𝑙3
−
32𝑥2
𝑙4
16
𝑙2
−
32𝑥
𝑙3
−
32𝑥
𝑙3
+
64𝑥2
𝑙4
𝑙
0
𝑑𝑥
= 𝐸𝐴
9𝑥
𝑙2
−
12𝑥2
2 𝑙3
−
12𝑥2
2 𝑙3
+
16𝑥3
3 𝑙4
3𝑥
𝑙2
−
12𝑥2
2 𝑙3
−
4𝑥2
2 𝑙3
+
16𝑥3
3 𝑙4
−12
𝑙2
+
24𝑥2
2 𝑙3
+
16𝑥2
2 𝑙3
−
32𝑥3
3 𝑙4
3𝑥
𝑙2
−
12𝑥2
2 𝑙3
−
4𝑥2
2 𝑙3
+
16𝑥3
3 𝑙4
𝑥
𝑙2
−
4𝑥2
2 𝑙3
−
4𝑥2
2 𝑙3
+
16𝑥2
3 𝑙4
−4
𝑙2
+
8𝑥2
2 𝑙3
+
16𝑥2
2 𝑙3
−
32𝑥2
3 𝑙4
−12
𝑙2
+
24𝑥2
2 𝑙3
+
16𝑥2
2 𝑙3
−
32𝑥2
3 𝑙4
−4
𝑙2
+
8𝑥2
2 𝑙3
+
16𝑥2
2 𝑙3
−
32𝑥2
3 𝑙4
16
𝑙2
−
32𝑥2
2 𝑙3
−
32𝑥
2 𝑙3
+
64𝑥2
3 𝑙4
𝑑𝑥
⟹ 𝑘 = 𝐸𝐴
9
𝑙
−
6
𝑙
−
6
𝑙
+
16
3 𝑙
3
𝑙
−
6
𝑙
−
2
𝑙
+
16
3 𝑙
−12
𝑙
+
12
𝑙
+
8
𝑙
−
32
3 𝑙
3
𝑙
−
6
𝑙
−
2
𝑙
+
16
3 𝑙
1
𝑙
−
2
𝑙
−
4
𝑙
+
16
𝑙
−4
𝑙
+
4
𝑙
+
8
𝑙
−
32
3 𝑙
−12
𝑙
+
12
𝑙
+
8
𝑙
−
32
3 𝑙
−4
𝑙2
+
4
𝑙
+
8
𝑙
−
32
3 𝑙
16
𝑙
−
16
𝑙
−
16
𝑙
+
64
3 𝑙

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⟹ 𝑘 = 𝐸𝐴
7
3 𝑙
1
3 𝑙
−8
3 𝑙
1
3 𝑙
7
3 𝑙
−8
3 𝑙
−8
3 𝑙
−8
3 𝑙
16
3 𝑙
⟹ 𝑘 =
𝐸𝐴
3 𝑙
7 1 −8
1 7 −8
−8 −8 16
LOAD VECTOR FOR ONE DIMENSIONAL QUADRATIC BAR ELEMENT:
`We know that, general force vector is,
𝐹 = 𝑁 𝑇
Xb
𝑙
0
Where, 𝑁 𝑇
=
𝑁1
𝑁2
𝑁3
=
1 −
3𝑥
𝑙
+
2𝑥2
𝑙2
−𝑥
𝑙
+
2𝑥2
𝑙2
4𝑥
𝑙
−
4𝑥2
𝑙2
Due to self weight, Xb = ρ A 𝑑𝑥
Substitute the equation,
𝐹 =
1 −
3𝑥
𝑙
+
2𝑥2
𝑙2
−𝑥
𝑙
+
2𝑥2
𝑙2
4𝑥
𝑙
−
4𝑥2
𝑙2
ρ A 𝑑𝑥
𝑙
0
𝐹 = ρ A
𝑥 −
3𝑥2
2 𝑙
+
2𝑥3
3 𝑙2
−𝑥2
2 𝑙
+
2𝑥3
3 𝑙2
4𝑥2
2 𝑙
−
4𝑥3
3 𝑙2
0
1
= ρ A
1 −
3𝑙2
2 𝑙
+
2𝑙3
3 𝑙2
−𝑙2
2 𝑙
+
2 𝑙3
3 𝑙2
4 𝑙2
2 𝑙
−
4𝑙3
3 𝑙2
= ρ A
𝑙 −
3 𝑙
2
+
2𝑙
3
−𝑙
2
+
2 𝑙
3
4𝑙
2
−
4𝑙
3

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= ρ A
0.166 𝑙
0.166 𝑙
0.166 𝑙
= ρ A 𝑙
0.166
0.166
0.166
𝐹 = ρ A 𝑙
1
6
1
6
2
3
𝐹1
𝐹2
𝐹3
= ρ A 𝑙
1
6
1
6
2
3

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UNIT-III TWO DIMENSIONAL SCALAR VARIABLE PROBLEMS
PART- A
1. Differentiate CST and LST elements. (Nov/Dec 2014)
Three nodded triangular element is known as constant strain triangular element. It has 6unknown degrees
of freedom called u1, v1, u2, v2, u3, v3. The element is called CST because it has constant strain throughout
it.
Six nodded triangular element is known as Linear Strain Triangular element. It has 12unknown
displacement degrees of freedom. The displacement function for the element are quadratic instead of linear
as in the CST.
2. What do you mean by the terms: C0
, C1
and Cn
continuity?
C0
– Governing differential equation is quasiharmonic, ø has to be continuous.
C1
– Governing differential equation is biharmonic, øas well as derivative has to be continuous inside
and between the elements.
Cn
– Governing differential equations is polynomial.
3. How do we specify two dimensional elements? (May/June 2014)
Two dimensional elements are defined by three or more nodes in two dimensional plane (i.e x and y
plane). The basic element useful for two dimensional analysis is a triangular element.
4. What is QST element?(May/June 2014)
Ten noded triangular elements are known as Quadratic strain element (QST).
5. Write the governing differential equation for two dimensional heat transfer.
The governing differential equation for two dimensional heat transfer is given by,
6. Write the governing differential equation for shaft with non-circular cross-section subjected to
torsion.
The governing differential equation is given by,

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1
𝐺
𝑑2
∅
𝑑𝑥2
+
1
𝐺
𝑑2
∅
𝑑𝑦2
+ 2𝜃 = 0
Where,
Ø – Field variable
 - Angle of twist per unit length (rad/m)
G – Modulus of rigidity or shear modulus (N/m2
)
7. What is geometric isotropy?(May/June 2013)
An additional consideration in the selection of polynomial shape function for the displacement
model is that the pattern should be independent of the orientation of the local coordinate system. This
property is known as Geometric Isotropy, Spatial Isotropy or Geometric Invariance.
8.Write the strain displacement matrix of CST element.(Nov/Dec 2012),(April/May 2011)
9. Why higher order elements are preferred?
Higher order elements are preferred to,
(i) Represent the curved boundaries
(ii) Reduce the number of elements when compared with straight edge elements to model geometry.
10. Evaluate the following area integrals for the three noded triangular element
𝛼! 𝛽! 𝛾!
𝛼+ 𝛽+ 𝛾+2
𝑋 2𝐴 𝑁𝑖 𝑁𝑗
2
𝑁𝑘
3
𝑑𝐴. (May/June 2013), (Nov/Dec 2012)
We know that,
𝐿𝑖
𝛼
𝐿2
𝛽
𝐿𝑘
𝛾
𝑑𝐴 =
1! 2! 3!
(1+ 2+ 3+2)!
𝑋 2𝐴
Here, α = 1, β = 2, γ = 3
𝑁𝑖 𝑁𝑗
2
𝑁𝑘
3
𝑑𝐴 =
1𝑋2𝑋1𝑋3𝑋2𝑋1
(8𝑋7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1)
𝑋 2𝐴 =
1! 2! 3!
(8)!
𝑋 2𝐴
=
𝐴
1680
𝑁𝑖 𝑁𝑗
2
𝑁𝑘
3
𝑑𝐴
11. Write the strain displacement relation for CST element.
𝑒𝑋
𝑒𝑌
𝛾𝑥𝑦
=
1
2𝐴
𝑞1 0 𝑞2
0 𝑟1 0
𝑟1 𝑞1 𝑟2
0 𝑞3 0
𝑟2 0 𝑟3
𝑞2 𝑟3 𝑞3
𝑢1
𝑣1
𝑢2
𝑣2
𝑢3
𝑣3
𝑝1 = 𝑥2𝑦3 − 𝑥3𝑦2 𝑝2 = 𝑥3𝑦1 − 𝑥1𝑦3 𝑝3 = 𝑥1𝑦2 − 𝑥2𝑦1
[B]=
1
2𝐴
𝑞1 0 𝑞2
0 𝑟1 0
𝑟1 𝑞1 𝑟2
0 𝑞3 0
𝑟2 0 𝑟3
𝑞2 𝑟3 𝑞3
𝑞1 = 𝑦2 − 𝑦3 𝑞2 = 𝑦3 − 𝑦1 𝑞3 = 𝑦1 − 𝑦2
𝑟1 = 𝑥3 − 𝑥2 𝑟2 = 𝑥1 − 𝑥3 𝑟3 = 𝑥2 − 𝑥1

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12. List out the two theories for calculating the shear stress in a solid non circular shaft subjected to
torsion.
The two theories which helps in evaluating the shear stresses in a solid non circular shaft is proposed
by,
(i) St. Venant called as St.Venant theory
(ii) Prandtl called as Prandtl’s theory.
13. Write down the shape functions associated with three noded linear triangular element (April/May
2015)
𝑁1 =
1
2𝐴
𝑝1 + 𝑞1𝑥 + 𝑟1𝑦 ; 𝑁2 =
1
2𝐴
𝑝2 + 𝑞2𝑥 + 𝑟2𝑦 ; 𝑁3 =
1
2𝐴
𝑝3 + 𝑞3𝑥 + 𝑟3𝑦 ;
PART - B
1. For a four Noded rectangular element shown in fig. determine the temperature at the
point (7, 4). The nodal values of temperature are T1=420
C, T2=540
C, T3= 560
C, & T4=
460
C. Also determine 3 points on the 500
C contour line.
Given:
ϕi= 420
C m (5,5) 460
C k(8,5) 560
C
ϕj= 540
C
ϕk=560
C
ϕm=460
C
2b=3 2a=2
b=3/2 a=1
To find:
1. Temperature at point (2,1),ϕ
2. Three points on 500
C.
Formula used:
Ni= 













a
t
b
s
2
1
2
1  













2
1
3
1
t
s
Nj= 












a
t
b
s
2
1
2
 












2
1
3
t
s
Nk= 





ab
st
4














 1
2
3
4
st
= 





6
st
Nm= 












b
s
a
t
2
1
2
 












3
1
2
s
t
j(8,3) 540
C
i (5,3) 460
C

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69
Solution:
The point (7,4) in global coordinate (x,y) is changed in the local coordinate (s,t)
S= x-xi  7-5=2
t= y-yi  4-3=1
the temperature at point (2,1) in local coordinate as
ϕ = Niϕi + Njϕj + Nkϕk + Nmϕm.
Ni= 













2
1
1
3
2
1 =
6
1
Nj= 












2
1
1
3
2
=
3
1
Nk= 




 
6
1
2
=
3
1
Nm = 












3
2
1
2
1
=
6
1
ϕ = 46
6
1
56
3
1
54
3
1
42
6
1






 .
ϕ = 51.40
C
The x,y coordinates of 500
C contour line are
𝜙𝑗 −𝜙
𝜙𝑗 −𝜙𝑖
=
𝑥𝑗 −𝑥
𝑥𝑗 −𝑥𝑖
=
𝑦𝑗 −𝑦
𝑦𝑗 −𝑦𝑖
m (5,5) 460
C k(8,5) 560
C
i j(8,3) 540
C
460
C (5,3) 500
C
i,j
3
3
3
5
8
8
42
54
50
54







 y
x
(1) (2) (3)
Equating(1),(2) equating (1),(3)

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C
A
D
70
3
8
12
4 x


0
3
12
4 y


cm
x 7
 cm
y 3

m,k
𝜙𝑘−𝜙
𝜙𝑘−𝜙𝑚
=
𝑥𝑘−𝑥
𝑥𝑘−𝑥𝑚
=
𝑦𝑘−𝑦
𝑦𝑘−𝑦𝑚
5
5
5
5
8
8
46
56
50
56







 y
x
(1) (2) (3)
Equating (1),(2) equating (1),(3)
3
8
10
6 x

 ;
0
5
10
6 y


cm
x 2
.
6
 ; cm
y 5

Third point y=4 [lower point yi=3, upper point ym=5]
Centre line between the sides i,j&k,m
Local coordinates
t = y-yi= 4-3 = 1
ϕ = Niϕi + Njϕj + Nkϕk + Nmϕm
50= 54
2
1
1
3
42
2
1
1
3
1 





















s
s
46
3
1
2
1
56
6
1














  s
s

















3
1
23
33
.
9
93
21
3
1
s
s
s
50= s
s
s 66
.
7
23
33
.
9
9
73
21 




cm
s 63
.
1
 (6.2,5)
j
x
x
s 

x

5
63
.
1 (6.7,4)
cm
y
cm
x
4
7
.
6


500
C (7,3)

S
C
A
D
S
C
A
D
71
2. For the plane stress element shown in Fig, the nodal displacements are:
[Anna University, May 2002]
U1=2.0mm; v1=1.0mm;
U2=0.5mm; v2=0.0mm;
U3=3.0mm; v3=1.0mm.
Determine the element stresses σx, σy, σ1, and σ2 and the principal angle θp, let E=210 GPA,
ν= 0.25 and t=10 mm. All coordinates are in millimetre.
Given:
Nodal Displacements: U1=2.0mm; v1=1.0mm;
U2=0.5mm; v2=0.0mm;
U3=3.0mm; v3=1.0mm.
X1= 20mm y1=30mm
X2= 80mm y2=30mm
X3=50mm y3=120mm
Young’s modulus, E= 210 GPa =210x109
Pa
= 210x109
N/m2
= 210x103
N/mm2
=2.1x 105
N/mm2
Poisson’s ratio, ν=0.25
Thickness, t= 10mm

S
C
A
D
S
C
A
D
72
To find: 1. Element stress
a) Normal stress, σx
b) Normal stress, σy
c) Shear stress, xy

d) Maximum normal stress, σ1
e) Minimum normal stress, σ2
2. Principle angle,θp
Formula used:
 Stress {σ} = [D] [B] {u}
 Maximum normal stress, σmax = σ1 = xy
y
x
y
x 2
2
2
2













 


 Minimum normal stress, σmin = σ2 = xy
y
x
y
x 2
2
2
2













 


 principle angle, tan 2θp=
y
x
xy




2
Solution: we know that
Area of the element, A=





















120
50
1
30
80
1
30
20
1
2
1
3
3
1
2
2
1
1
1
1
2
1
y
x
y
x
y
x
=
2
1
x[ 1x(80x120-50x30)-20(120-30)+30(50-80)]
=
2
1
x [8100-1800-900]
A=2700 mm2
….. (1)
We know that,
Strain Displacement matrix,
[B]=










3
3
2
2
1
1
3
0
2
0
1
0
0
3
0
2
0
1
2
1
q
r
q
r
q
r
r
r
r
q
q
q
A
…… (2)
Where, q1 = y2 – y3 = 30-120 = -90
q2= y3 – y1 = 120- 30 = 90
q3= y1- y2 = 30 – 30 = 0

Me6603 sd by easy engineering.net

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