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2. SYLLABUS
ME8692
FINITE ELEMENT ANALYSIS
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 vectorsAssembly 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.
9
UNIT III TWO DIMENSIONAL SCALAR VARIABLE PROBLEMS
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.
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TEXT BOOK:
1. Reddy. J.N., “An Introduction to the Finite Element Method”, 3rd Edition, Tata McGrawHill, 2005
2. Seshu, P, “Text Book of Finite Element Analysis”, Prentice-Hall of India Pvt. Ltd., New
Delhi,2007.
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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
TABLE OF CONTENTS
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
Aim and Objective of the subject
Detailed Lesson Plan
Unit I- Introduction -Part A
Unit I- Introduction -Part B
Unit II- One-dimensional problems -Part A
Unit II- One-dimensional problems -Part B
Unit III- Two dimensional scalar variable problems -Part A
Unit III- Two dimensional scalar variable problems -Part B
Unit IV- Two Dimensional Vector Variable Problems -Part A
Unit IV- Two Dimensional Vector Variable Problems -Part B
Unit V- Isoparametric Formulation - Part A
Unit V- Isoparametric Formulation - Part B
Question bank
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95
96
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ME8692
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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Department of Mechanical Engineering
Detailed Lesson Plan
Name of the Subject& Code: ME8692 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)*
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1
Historical Background
2
1
3
1
Mathematical modeling of field problems in
Engineering
Governing Equations
4
1
Discrete and continuous models
5
1
Boundary, Initial and Eigen Value problems
6
1
Weighted Residual Methods concept
7
1
Weighted Residual Methods-Problems
8
1
Variational Formulation of Boundary Value
Problems
9
1
Ritz Technique concept
10
1
Ritz Technique -Problems
11
1
Basic concepts of the Finite Element Method.
12
2
One Dimensional Second Order Equations
A
S.No
Unit
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Topic / Portions to be Covered
Hours
Cumulative
Required
Hrs
/ Planned
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Books
Referred
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1
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1
6
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7
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1
8
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1
9
T1,R1
1
10
T1,R1
1
11
T1,R1
1
12
T1,R1
1
1
1
3
4
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13
2
Discretization – Element types
1
13
14
2
Derivation of Shape functions and Stiffness
matrices and force vectors (Linear)
1
14
15
2
Derivation of Shape functions (Higher order
Elements)
1
15
16
2
Derivation of Stiffness matrices and force
vectors(Higher order Elements)
1
16
17
2
Solution of problems from solid mechanics
and heat transfer
1
17
18
2
Solution of problems from solid mechanics
1
18
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T1,R1
T1,R1
T1,R1
20
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Fourth Order Beam Equation
21
2
Transverse deflections of beams.
22
2
Transverse Natural frequencies of beams.
1
22
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3
Second Order 2D Equations involving Scalar
Variable Functions
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3
Variational formulation -Finite Element
formulation
25
3
Triangular elements – Shape functions and
element matrices and vectors.
26
3
Application to Field Problems
1
26
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3
Thermal problems
1
27
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3
Torsion of Non circular shafts
1
28
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3
Quadrilateral elements
1
29
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3
Higher Order Elements concept
1
30
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3
Higher Order Elements problems
1
31
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4
Equations of elasticity
1
32
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Longitudinal vibration frequencies and mode
shapes
1
19
1
20
T1,R1
1
21
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33
4
Plane stress condition
1
33
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4
plane strain conditions
1
34
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4
Axisymmetric problems
1
35
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4
Body forces in axisymmetric
1
36
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4
temperature effects in axisymmetric
1
37
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4
Stress calculations
1
38
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4
Plate and shell elements
1
39
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5
Natural co-ordinate systems
1
40
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5
Isoparametric elements
1
41
5
Shape functions for iso parametric elements –
One and two dimensions
1
42
1
43
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Serendipity elements
44
5
Numerical integration and application to
plane stress problems
1
44
45
5
Matrix solution techniques
1
45
46
5
Solutions Techniques to Dynamic problems
47
5
Introduction to Analysis Software
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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.
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d y/dx - a(x) dy/dx – b(x)y –c(x) = 0 with boundary conditions, y(m) = S and y(n) = T.
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3. What do you mean by weak formulation? State its advantages. (April/May 2015), (May/June
2013)
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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.
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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:
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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.
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In this problem, displacement u at node 1 = 0, that is primary boundary condition.
EA du/dx = P, that is secondary boundary condition.
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8. Compare Ritz method with nodal approximation method.(Nov/Dec 2014), (Nov/Dec 2012)
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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.
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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
minimum is astable system.
PART-B
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𝒅𝟐 𝒖
The following differential equation is available for a physical phenomenon. 𝑨𝑬 = 𝒅𝒙𝟐 +
D
1.
𝒅𝒖
𝒂𝒙 = 𝟎, The boundary conditions are u(0) = 0, 𝑨𝑬 = 𝒅𝒙
𝒙=𝑳
= 𝟎 By using Galerkin’s
A
technique, find the solution of the above differential equation.
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Given Data:
𝑑2𝑢
Differential equ. 𝐴𝐸 = 𝑑𝑥 2 + 𝑎𝑥 = 0
Boundary Conditions 𝑢 0 = 0,
𝑑2𝑢
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𝐴𝐸 = 𝑑𝑥 2 + 𝑎𝑥 = 0
To Find:
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u(x) by using galerkin’s technique
Formula used
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𝐿
𝑤𝑖 𝑅 𝑑𝑥 = 0
0
Solution:
Assume a trial function
Let 𝑢 𝑥 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3
…….. (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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𝑑𝑢
Apply first boundary condition i.e at x = L, 𝐴𝐸 = 𝑑𝑥 = 0
⟹
𝑑𝑢
= 0+𝑎1 + 2𝑎2 𝑥 + 3𝑎3 𝐿2
𝑑𝑥
⟹ 0 = 𝑎1 + 2𝑎2 𝐿 + 3𝑎3 𝐿2
⟹ 𝑎1 = −(2𝑎2 𝐿 + 3𝑎3 𝐿2 )
sub 𝑎0 and 𝑎1 in value in equation (1)
𝑢 𝑥 = 0 + − 2𝑎2 𝐿 + 3𝑎3 𝐿2 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3
= −2𝑎2 𝐿𝑥 − 3𝑎3 𝐿2 𝑎2 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3
= 𝑎2 𝑥 2 − 2𝐿𝑥 + 𝑎3 (𝑥 3 − 3𝐿2 𝑥)
……… (2)
We Know That
𝑑2𝑢
Residual, 𝑅 = 𝐴𝐸 𝑑𝑥 2 + 𝑎𝑥
………. (3)
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𝑑𝑢
= 𝑎2 2𝑥 − 2𝐿 + 𝑎3 (3𝑥 2 − 3𝐿2 )
𝑑𝑥
(2) ⟹
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𝑑2 𝑢
= 𝑎2 2 + 𝑎3 (6𝑥)
𝑑𝑥 2
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3 ⟹ 𝑅 = 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥
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𝑑2𝑢
Sub 𝑑𝑥 2 value in equation (3)
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𝑑2 𝑢
= 2𝑎2 + 6𝑎3 𝑥
𝑑𝑥 2
Residual, 𝑅 = 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥
From Galerkn’s technique
𝐿
𝑤𝑖 𝑅 𝑑𝑥 = 0
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……… (4)
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. . … … . . . (5)
0
from equation (2) we know that
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𝑤1 = 𝑥 2 − 2𝐿𝑥
𝑤2 = 𝑥 3 − 3𝐿2 𝑥
sub w1, w2 and R value in equation (5)
𝐿
𝑥 2 − 2𝐿𝑥 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0
5 ⟹
… … … … … (6)
0
𝐿
𝑥 3 − 3𝐿2 𝑥 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0
… … … … … (7)
0
𝐿
𝑥 2 − 2𝐿𝑥 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0
6 ⟹
0
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𝐿
𝑥 2 − 2𝐿𝑥 2𝑎2 𝐴𝐸 + 6𝑎3 𝐴𝐸𝑥 + 𝑎𝑥 𝑑𝑥 = 0
0
𝐿
2𝑎2 𝐴𝐸𝑥 2 + 6𝑎3 𝐴𝐸𝑥 3 + 𝑎𝑥 3 − 4𝑎2 𝐴𝐸𝐿𝑥 − 12𝑎3 𝐴𝐸𝐿𝑥 2 − 2𝑎𝐿𝑥 2 = 0
0
⟹ [2𝑎2 𝐴𝐸
𝑥3
𝑥4
𝑥4
𝑥2
𝑥3
𝑥3
+ 6𝑎3 𝐴𝐸
+ 𝑎 − 4𝑎2 𝐴𝐸𝐿 − 12𝑎3 𝐴𝐸𝐿 − 2𝑎𝐿 ]𝐿0 = 0
3
4
4
2
3
3
𝐿3
𝐿4
𝐿4
𝐿3
𝐿4
𝐿4
⟹ 2𝑎2 𝐴𝐸 + 6𝑎3 𝐴𝐸
+ 𝑎 − 4𝑎2 𝐴𝐸 − 12𝑎3 𝐴𝐸 − 2𝑎 = 0
3
4
4
2
3
3
2
𝐿4
3
2
⟹ 3 𝑎2 𝐴𝐸𝐿3 + 2 𝑎3 𝐴𝐸 𝐿4 + 𝑎 4 − 2𝑎2 𝐴𝐸𝐿3 − 4𝑎3 𝐴𝐸𝐿4 − 3 𝑎𝐿4 = 0
2
3
𝐿4 2
− 2 + 𝑎3 𝐴𝐸 𝐿4 − 4 + 𝑎 − 𝑎2 𝐿4 = 0
3
2
4 3
−4
5
2 1
4
5
5 4
⟹
𝐴𝐸𝐿3 𝑎2 − 𝐴𝐸𝐿4 𝑎3 = − 𝑎𝐿4 − 𝐴𝐸𝐿3 𝑎2 − 𝐴𝐸𝐿4 𝑎3 =
𝑎𝐿
3
2
3 4
3
2
12
−4
5
5
𝐴𝐸𝐿3 𝑎2 − 𝐴𝐸𝐿4 𝑎3 = − 𝑎𝐿4
………. 8
3
2
12
⟹ 𝐴𝐸𝑎2 𝐿3
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𝐿
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Equation (7)
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𝐿
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0
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(𝑥 3 − 3𝐿2 𝑥) 𝐴𝐸 2𝑎2 + 6𝑎3 𝑥 + 𝑎𝑥 𝑑𝑥 = 0
⟹
(𝑥 3 − 3𝐿2 𝑥) 2𝑎2 𝐴𝐸 + 6𝑎3 𝐴𝐸𝑥 + 𝑎𝑥 𝑑𝑥 = 0
⟹
0
𝐿
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2𝐴𝐸𝑎2 𝑥 3 + 6𝐴𝐸𝑎3 𝑥 4 + 𝑎𝑥 4 − 6𝐴𝐸𝑎2 𝐿2 𝑥 − 18𝐴𝐸𝑎3 𝐿2 𝑥 2 − 3𝑎𝐿2 𝑥 2 𝑑𝑥 = 0
⟹
0
𝐿
𝑥4
𝑥5
𝑥5
𝑥2
𝑥3
𝑥3
⟹ 2𝐴𝐸𝑎2 + 6𝐴𝐸𝑎3 + 𝑎 − 6𝐴𝐸𝑎2 𝐿2 − 18𝐴𝐸𝑎3 𝐿2 − 3𝑎𝐿2
=0
4
5
5
2
3
3 0
t
𝐿
1
6
1 5
4
5
2 2
2 3
2 3
⟹ 𝐴𝐸𝑎2 𝑥 + 𝐴𝐸𝑎3 𝑥 + 𝑎𝑥 − 3𝐴𝐸𝑎2 𝐿 𝑥 − 6𝐴𝐸𝑎3 𝐿 𝑥 − 𝑎𝐿 𝑥
=0
2
5
5
0
1
6
1
𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 + 𝑎𝐿5 − 3𝐴𝐸𝑎2 𝐿2 (𝐿2 ) − 6𝐴𝐸𝑎3 𝐿2 (𝐿3 ) − 𝑎𝐿2 (𝐿3 ) = 0
2
5
5
1
6
1
⟹ 𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 + 𝑎𝐿5 − 3𝐴𝐸𝑎2 𝐿4 − 6𝐴𝐸𝑎3 𝐿5 − 𝑎𝐿5 = 0
2
5
5
1
6
1
⟹ 𝐴𝐸𝑎2 𝐿4 − 3 + 𝐴𝐸𝑎3 𝐿5 − 6 + 𝑎𝐿5 + − 1 = 0
2
5
5
5
24
4
⟹ 𝐴𝐸𝑎2 𝐿4
− 𝐴𝐸𝑎3 𝐿5 = 𝑎𝐿5
2
5
5
5
24
4
𝐴𝐸𝑎2 𝐿4 + 𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5
…………. 9
2
5
5
⟹
12
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Solving Equation (8) and (9)
4
5
5
5
24
4
Equation (8) ⟹ 3 𝐴𝐸𝑎2 𝐿3 + 2 𝐴𝐸𝑎3 𝐿4 = − 12 𝑎𝐿4
Equation (9) ⟹ 2 𝐴𝐸𝑎2 𝐿4 + 5 𝐴𝐸𝑎3 𝐿5 = − 5 𝑎𝐿5
5
4
Multiplying Equation (8) 2 𝐿 and Equation (9) by 3
20
25
25
𝐴𝐸𝑎2 𝐿4 +
𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5
6
4
24
20
25
16
𝐴𝐸𝑎2 𝐿4 +
𝐴𝐸𝑎3 𝐿5 = − 𝑎𝐿5
6
4
15
Subtracting
25 96
16 25
−
𝐴𝐸𝑎3 𝐿5 =
−
𝑎𝐿5
4 15
15 24
375 − 384
384 − 375
𝐴𝐸𝑎3 𝐿5 =
𝑎𝐿5
60
360
−9
9
⟹
𝐴𝐸𝑎3 𝐿5 =
𝑎𝐿5
60
360
ww
w.E
asy
En
gi
⟹ −0.15𝐴𝐸𝑎3 = 0.025𝑎
D
𝑎
𝐴𝐸
𝑎3 = −
SC
Substituting a3 value in Equation (8)
𝑎
6𝐴𝐸
A
𝑎3 = −0.1666
4
5
−𝑎 4 −5 4
𝐴𝐸𝑎2 𝐿3 + 𝐴𝐸
𝐿 =
𝑎𝐿
3
2
6𝐴𝐸
12
4
−5 4 5
−𝑎
𝐴𝐸𝑎2 𝐿3 =
𝑎𝐿 − 𝐴𝐸𝐿4 =
3
12
2
6𝐴𝐸
4
−5 4 5
𝐴𝐸𝑎2 𝐿3 =
𝑎𝐿 + 𝐴𝐸𝐿4
3
12
2
4
𝐴𝐸𝑎2 𝐿3 = 0
3
… … … . (10)
nee
rin
g
.ne
t
𝑎2 = 0
Sub a2 and a3 value in equation (2)
−𝑎
⟹ 𝑢 𝑥 = 0𝑥 𝑥2 − 2𝐿𝑥 +
6𝐴𝐸
𝑎
⟹𝑢 𝑥 =
3𝐿2 𝑥 − 𝑥 3
6𝐴𝐸
𝑥 3 − 3𝐿2 𝑥 = 0
Result:
𝑢 𝑥 =
𝑎
3𝐿2 𝑥 − 𝑥 3
6𝐴𝐸
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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
ww
Formula used
𝑑4 𝑦
𝐸𝐼 4 − 𝜔 = 0,
𝑑𝑥
0≤𝑥≤𝑙
w.E
asy
En
gi
𝑙
Sub-domain collocation method = 0 𝑅𝑑𝑥 = 0
𝑙
D
Point Collocation Method R = 0
A
Least Square Method 𝐼 = 0 𝑅 2 𝑑𝑥 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
Solution:
nee
SC
The differential equation governing the deflection of beam subjected to uniformly
distributed load is given by
𝑑4 𝑦
𝐸𝐼 4 − 𝜔 = 0,
𝑑𝑥
0≤𝑥≤𝑙
… … … . (1)
rin
g
.ne
The boundary conditions are Y=0 at x=0 and x = l, where y is the deflection.
𝑑4 𝑦
𝐸𝐼 4 = 0,
𝑑𝑥
𝑎𝑡 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 𝑙
t
Where
𝑑4𝑦
𝐸𝐼 𝑑𝑥 4 = 𝑀,
(Bending moment)
E → Young’s Modules
I → Moment of Inertia of the Beam.
𝜋𝑥
Let us select the trial function for deflection as 𝑌 = 𝑎𝑠𝑖𝑛 𝑙 ……. (2)
Hence it satisfies the boundary conditions
⟹
𝑑𝑦
𝜋
𝜋𝑥
= 𝑎 . cos
𝑑𝑥
𝑙
𝑙
𝑑2 𝑦
𝜋2
𝜋𝑥
⟹ 2 = −𝑎 2 . sin
𝑑𝑥
𝑙
𝑙
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𝑑3 𝑦
𝜋3
𝜋𝑥
⟹ 3 = −𝑎 3 . cos
𝑑𝑥
𝑙
𝑙
𝑑4 𝑦
𝜋4
𝜋𝑥
⟹ 4 = 𝑎 4 . sin
𝑑𝑥
𝑙
𝑙
Substituting the Equation (3) in the governing Equation (1)
𝜋4
𝜋𝑥
𝐸𝐼 𝑎 4 . sin
−𝜔 = 0
𝑙
𝑙
𝜋4
𝜋𝑥
Take, Residual 𝑅 = 𝐸𝐼𝑎 𝑙 4 . sin 𝑙 − 𝜔
a) Point Collocation Method:
In this method, the residuals are set to zero.
𝜋4
𝜋𝑥
⟹ 𝑅 = 𝐸𝐼𝑎 4 . sin
−𝜔 =0
𝑙
𝑙
ww
𝜋4
𝜋𝑥
𝐸𝐼𝑎 4 . sin
=𝜔
𝑙
𝑙
w.E
asy
En
gi
𝑙
To get maximum deflection, take 𝑘 = 2 (𝑖. 𝑒. 𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑜𝑓 𝑏𝑒𝑎𝑚)
𝜋4
𝜋 𝑙
4
𝜔𝑙 4
𝜋 4 𝐸𝐼
SC
𝑎=
A
𝜋
𝐸𝐼𝑎 4 = 𝜔
𝑙
D
𝐸𝐼𝑎 𝑙 4 . sin 𝑙 2 = 𝜔
Sub “a” value in trial function equation (2)
𝑌=
𝜔𝑙 4
𝜋𝑥
.
sin
𝜋 4 𝐸𝐼
𝑙
𝐴𝑡 𝑥 =
nee
[∵ sin
𝜋
= 1]
𝑙
rin
g
.ne
𝑙
𝜔𝑙 4
𝜋 𝑙
⟹ 𝑌max = 4 . sin
2
𝜋 𝐸𝐼
2 2
𝜔 𝑙4
𝑌max = 𝜋 4 𝐸𝐼
[∵ sin
𝜔𝑙 4
𝑌max =
97.4𝐸𝐼
t
𝜋
= 1]
2
b) Sub-domain collocation method:
In this method, the integral of the residual over the sub-domain is set to zero.
𝑙
𝑅𝑑𝑥 = 0
0
Sub R value
𝜋4
𝜋𝑥
⟹ 𝑎𝐸𝐼 4 sin
− 𝜔 𝑑𝑥 = 0
𝑙
𝑙
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𝑙
𝜋𝑥
−cos
𝑙 −𝜔 𝑥 = 0
𝜋
𝑙
0
4
𝜋
⟹ 𝑎𝐸𝐼 4
𝑙
𝑙
𝜋4
𝜋𝑥 𝑙
⟹ 𝑎𝐸𝐼 4 −cos
−𝜔𝑥 = 0
𝑙
𝑙 𝑢
0
∵ cos 𝜋 = −1
,
𝑐𝑜𝑠0 = 1
𝜋3
⟹ −𝑎𝐸𝐼 𝑙 3 cos𝜋 − 𝑐𝑜𝑠0 𝜔 𝑙 = 0
𝜋3
−𝑎𝐸𝐼 3 −1 − 1 = 𝜔 𝑙
𝑙
𝜔𝑙 4
𝜔𝑙 4
⟹ −𝑎 = 3 =
2𝜋 𝐸𝐼 62𝐸𝐼
Sub “a” value in the trial function equation (2)
𝜔𝑙 4
𝜋𝑥
𝑌=
. sin
62𝐸𝐼
𝑙
ww
𝑙
𝜔𝑙 4
𝜋 𝑙
𝐴𝑡 𝑥 = , 𝑌𝑚𝑎𝑥 =
. sin ( )
2
62𝐸𝐼
𝑙 2
w.E
asy
En
gi
𝜔𝑙 4
62𝐸𝐼
D
𝑌𝑚𝑎𝑥 =
𝑙
𝑅 2 𝑑𝑥 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝐼=
0
𝑙
𝜋4
𝜋𝑥
(𝑎𝐸𝐼 4 . 𝑠𝑖𝑛
− 𝜔)2 𝑑𝑥
𝑙
𝑙
𝐼=
0
𝑙
0
rin
g
.ne
2 2
𝜋8 1
2𝜋𝑥
= [𝑎 𝐸 𝐼 8 𝑥 𝑠𝑖𝑛
𝑙 2
𝑙
2
nee
𝜋8
𝜋𝑥
𝜋4
𝜋𝑥
2
2
[𝑎 𝐸 𝐼 8 . 𝑠𝑖𝑛
− 𝜔 − 2𝑎𝐸𝐼𝜔 4 . 𝑠𝑖𝑛 ]𝑑𝑥
𝑙
𝑙
𝑙
𝑙
2
=
SC
In this method the functional
A
c) Least Square Method:
2 2
= 𝑎2 𝐸 2 𝐼 2
t
𝑙
𝜋4
𝜋𝑥 𝑙 𝑙
2
+ 𝜔 − 2𝑎𝐸𝐼𝜔 4 . [−𝑐𝑜𝑠
]]
2𝜋
𝑙
𝑙 𝜋 0
𝜋8 1
𝑙
𝑙
−
𝑠𝑖𝑛2𝜋 − 𝑠𝑖𝑛0
𝑙8 2
2𝜋
𝜋4 𝑙
+ 𝜔2 𝑙 + 2𝑎𝐸𝐼𝜔 4 . [−𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠0]
𝑙 𝜋
∵ 𝑠𝑖𝑛2𝜋 = 0; 𝑠𝑖𝑛0 = 0; 𝑐𝑜𝑠𝜋 = 0; 𝑐𝑜𝑠0 = 1
𝐼 = 𝑎2 𝐸 2 𝐼 2
𝐼=
𝜋8 𝑙
𝜋3
2
+
𝜔
𝑙
+
2𝑎𝐸𝐼𝜔
. (−1 − 1)
𝑙2 2
𝑙3
𝑎2 𝐸 2 𝐼 2 𝜋 8
𝜋3
2
+
𝜔
𝑙
−
4𝑎𝐸𝐼𝜔
2𝑙 7
𝑙3
𝜕𝜋
Now, 𝜕𝑎 = 0
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𝑎2 𝐸 2 𝐼 2 𝜋 8
𝜋3
⟹
= 4𝐸𝐼𝜔 3
2𝑙 7
𝑙
𝑎2 𝐸 2 𝐼 2 𝜋 8
𝜋3
= 4𝐸𝐼𝜔 3
𝑙7
𝑙
4𝐸𝐼𝜔𝑙 5
𝑎= 5
𝜋 𝐸𝐼
Hence the trial Function
4𝜔𝑙 4
𝜋𝑥
𝑌 = 5 . sin
𝜋 𝐸𝐼
𝑙
𝑙
𝐴𝑡 𝑥 = 2 , max 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝜋
[∵ 𝑠𝑖𝑛 2 = 1]
4𝜔𝑙 4
𝜋 𝑙
𝑌𝑚𝑎𝑥 = 5 𝑠𝑖𝑛 ( )
𝜋 𝐸𝐼
2 2
𝜔𝑙 4
𝑌𝑚𝑎𝑥 =
76.5 𝐸𝐼
ww
d) Galerkin’s Method:
w.E
asy
En
gi
𝑙
D
In this method
⟹
𝑎𝑠𝑖𝑛
0
𝑙
𝜋𝑥
𝑙
𝑎𝐸𝐼
𝜋4
𝜋𝑥
𝑠𝑖𝑛
−𝜔
4
𝑙
𝑙
𝜋4
𝜋𝑥
𝜋𝑥
𝑎 𝐸𝐼 4 𝑠𝑖𝑛2
− 𝑎𝜔𝑠𝑖𝑛
𝑑𝑥 = 0
𝑙
𝑙
𝑙
2
⟹
0
𝑙
𝑑𝑥 = 0
SC
0
𝑙
A
𝑌. 𝑅 𝑑𝑥 = 0
nee
rin
g
.ne
𝜋4 1
2𝜋𝑥
𝜋𝑥
𝑎 𝐸𝐼 4 [ (1 − 𝑐𝑜𝑠
) − 𝑎𝜔𝑠𝑖𝑛
𝑑𝑥 = 0
𝑙 2
𝑙
𝑙
2
⟹
0
𝜋4 1
⟹ 𝑎 𝐸𝐼 4 [ 1 −
𝑙 2
2
1
2𝜋𝑥
𝑥−
𝑠𝑖𝑛2
2𝜋
𝑙
t
𝑙
𝑙
𝜋𝑥
+ 𝑎𝜔 𝑐𝑜𝑠
=0
𝜋
𝑙 0
𝜋4 𝑙
𝑙
𝑎 𝐸𝐼 4
− 2𝑎𝜔
=0
𝑙 2
𝜋
2
2𝜔𝑙 2𝑙 3
∴𝑎=
.
𝜋 𝐸𝐼𝜋 4
4𝜔𝑙 3
𝑎= 5
𝜋 𝐸𝐼
Hence the trial Function
4𝜔𝑙 4
𝜋𝑥
𝑌 = 5 . sin
𝜋 𝐸𝐼
𝑙
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𝑙
𝜋
𝐴𝑡 𝑥 = 2 , max 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
[∵ 𝑠𝑖𝑛 2 = 1]
4𝜔𝑙 4
𝜋 𝑙
𝑌𝑚𝑎𝑥 = 5 𝑠𝑖𝑛 ( )
𝜋 𝐸𝐼
2 2
4𝜔𝑙 4
𝑌𝑚𝑎𝑥 = 5
𝜋 𝐸𝐼
𝜔𝑙 4
𝑌𝑚𝑎𝑥 =
76.5 𝐸𝐼
Verification,
We know that simply supported beam is subjected to uniformly distributed load, maximum
deflection is,
𝑌𝑚𝑎𝑥 =
3) i)
ww
5 𝜔𝑙 4
384 𝐸𝐼
= 0.01
𝜔𝑙 4
𝐸𝐼
w.E
asy
En
gi
What is constitutive relationship? Express the constitutive relations for a linear
elastic isotropic material including initial stress and strain.
[Nov/Dec 2009]
D
A
Solution:
(4)
It is the relationship between components of stresses in the members of a structure or in a
nee
SC
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
rin
g
[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
(1 − 𝑣)
𝑣
𝐸
𝑣
=
1 + 𝑣 1 − 2𝑣
0
0
0
0
(1 − 𝑣)
𝑣
0
0
0
0
0
0
0
0
(1 − 𝑣)1 − 2𝑣
0
2
0
0
0
0
0
0
0
0
1 − 2𝑣
2
0
.ne
0
𝑒𝑥
0
𝑒𝑦
0
𝑒𝑧
0
𝑣𝑥𝑦
0
1 − 2𝑣 𝑣𝑦𝑧
𝑣𝑧𝑥
2
t
𝒅𝟐 𝒚
ii) Consider the differential equation 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
𝒍
𝒅𝒚 𝟐
is given by 𝑰 = 𝟎 −𝟎. 𝟓 𝒅𝒙
+ 𝟒𝟎𝟎𝒙𝟐 𝒀 . Find the solution of the problem using Ray
Light Ritz method by considering a two term solution as 𝒀 𝒙 = 𝒄𝟏 𝒙 𝟏 − 𝒙 + 𝒄𝟐 𝒙𝟐 (𝟏 −
𝒙)
(12)
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Downloaded From : www.EasyEngineering.net
Given data
𝑑2𝑦
Differential equation = 𝑑𝑥 2 + 400𝑥 2 = 0 for 0 ≤ 𝑥 ≤ 1
Boundary conditions Y(0) = 0, Y(1) = 0
𝑑𝑦 2
𝑙
𝐼 = 0 −0.5 𝑑𝑥
+ 400𝑥 2 𝑌
𝑌 𝑥 = 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 (1 − 𝑥)
To find:
Rayleigh- Ritz method
Formula used
𝜕𝐼
=0
ww
𝜕𝐼
=0
𝜕𝑐2
Solution:
w.E
asy
En
gi
𝑌 𝑥 = 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 (1 − 𝑥)
A
𝑌 𝑥 = 𝑐1 𝑥 𝑥 − 𝑥 2 + 𝑐2 (𝑥 2 − 𝑥 3 )
D
𝜕𝑐1
𝑑𝑦
= 𝑐1 1 − 2𝑥 + 𝑐2 (2𝑥 − 3𝑥 2 )
𝑑𝑥
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= 𝑐1 1 − 2𝑥 + 𝑐2 𝑥(2 − 3𝑥)
𝑑𝑦 2
= 𝑐1 1 − 2𝑥 + 𝑐2 𝑥(2 − 3𝑥)2 2
𝑑𝑥
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= 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥 1 − 2𝑥 (2 − 3𝑥)
= 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥(2 − 3𝑥 − 4𝑥 + 6𝑥 2 )
𝑑𝑦
𝑑𝑥
2
t
= 𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥(2 − 7𝑥 + 6𝑥 2 )
We know that
𝑙
𝐼=
0
𝑑𝑦 2
−1
[−0.5
+ 400𝑥 2 𝑦] =
𝑑𝑥
2
𝑙
0
𝑑𝑦 2
+ 400
𝑑𝑥
𝑙
𝑥2 𝑦
0
𝑙
𝑐12 1 − 4𝑥 + 4𝑥 2 + 𝑐22 𝑥 2 4 − 12𝑥 + 9𝑥 2 + 2𝑐1 𝑐2 𝑥 2 − 7𝑥 + 6𝑥 2
=
0
𝑙
+ 400[ 𝑥 2 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 1 − 𝑥
0
By Solving
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−1 𝑐12
2
1
𝑐1 𝑐2
𝐼=
+ 𝑐22 + 𝑐1 𝑐2 + 400
+
2 3 15
3
20 30
𝐼=
−1 2
1 2 1
40
𝑐1 −
𝑐2 − 𝑐1 𝑐2 + 20𝑐1 + 𝑐2
6
15
6
3
𝜕𝐼
=0
𝜕𝑐1
−1
1
× 2𝑐1 − 𝑐2 + 20 = 0
6
6
−1
1
⟹
× 𝑐1 − 𝑐2 + 20 = 0
3
6
⟹
… … … . . (1)
Similarly,
𝜕𝐼
=0
𝜕𝑐2
⟹
−2
1
40
𝑐2 − 𝑐1 +
=0
15
6
3
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… … … . . (2)
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By Solving (1) and (2)
We know that
𝑌 = 𝑐1 𝑥 1 − 𝑥 + 𝑐2 𝑥 2 (1 − 𝑥)
4)
80
200 2
𝑥 1−𝑥 +
𝑥 1−𝑥
3
3
SC
𝑌=
D
80
200
; 𝑐1 =
3
3
A
𝑐1 =
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Consider a 1mm diameter, 50m long aluminum pin-fin as shown in figure used to
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enhance the heat transfer from a surface wall maintained at 300C. Calculate the
t
temperature distribution in a pin-fin by using Rayleigh – Ritz method. Take, 𝒌 =
𝟐𝟎𝟎𝒘
𝒅𝟐 𝑻
𝟐𝟎𝟎𝒘
𝒎𝐂 for aluminum h=
, 𝑻 = 𝟑𝟎𝐂.
𝒎𝟐 𝐂 ∞
𝑷𝒉
𝒅𝑻
𝒌 𝒅𝒙𝟐 = 𝑨 (𝑻 − 𝑻∞ ) , 𝑻 𝟎 = 𝑻𝒘 = 𝟑𝟎𝟎𝐂, 𝒒𝑳 = 𝑲𝑨 𝒅𝒙 𝑳 = 𝟎 (insulated tip)
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Given Data:
The governing differential equation
𝑘
𝑑2 𝑇 𝑃
=
(𝑇 − 𝑇∞ )
𝑑𝑥 2
𝐴
d = 1mm = 1x10-3m
Diameter
Length L = 50mm = 50x10-3m
K = 200𝑤 𝑚C
Thermal
Conductivity Heat transfer co-efficient h = 200𝑤 𝑚C
Fluid Temp 𝑇∞ = 30C.
Boundary Conditions 𝑇 0 = 𝑇𝑤 = 300C
𝑑𝑇
𝑞𝐿 = 𝐾𝐴 𝑑𝑥 𝐿 = 0
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To Find:
Ritz Parameters
Formula used
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Solution:
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𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝜋 =𝑢−𝑣
𝐿
𝜋=
0
𝐿
𝜋=
0
1 𝑑𝑇 2
𝐾
𝑑𝑥 +
2 𝑑𝑥
1 𝑑𝑇 2
𝐾
𝑑𝑥 +
2 𝑑𝑥
𝐿
0
𝐿
0
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𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
A
The equivalent functional representation is given by,
1 𝑃
𝑇 − 𝑇∞ 2 𝑑𝑥 − 𝑞𝐿 𝑇𝐿
2 𝐴
1 𝑃
𝑇 − 𝑇∞ 2 𝑑𝑥
2 𝐴
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… … … … . (1)
………….. 2
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t
∵ 𝑞𝐿 = 0
Assume a trial function
Let
𝑇 𝑥 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2
… … … … … . . (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 + 𝑎1 𝑥 + 𝑎2 𝑥 2
…………….. 4
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⟹
𝑑𝑇
= 𝑎1 + 2𝑎2 𝑥
𝑑𝑥
… … … … … … (5)
Substitute the equation (4), (5) in (2)
𝑙
𝜋=
0
1
𝑘 (𝑎1 + 2𝑎2 𝑥)2 𝑑𝑥 +
2
𝑙
0
1 𝑃
270 + 𝑎1 + 𝑎2 𝑥 2 2 𝑑𝑥.
2 𝐴
[∵ 𝑎 + 𝑏 2 = 𝑎2 + 𝑏 2 + 2𝑎𝑏; 𝑎 + 𝑏 + 𝑐 2 = 𝑎2 + 𝑏 2 + 𝑐 2 + 2𝑎𝑏 + 2𝑏𝑐 + 2𝑐𝑎
𝑙
𝑘
𝑃
𝜋=
(𝑎12 + 4𝑎22 𝑥 2 + 4𝑎1 𝑎2 𝑥) +
2
2𝐴
0
𝑙
2702 + 𝑎1 2 𝑥 2 + 𝑎2 2 𝑥 4 + 540𝑎1 𝑥 + 2𝑎1 𝑥 3 + 540𝑎2 𝑥 2 𝑑𝑥
0
50𝑥10 −3
𝑘
4𝑎22 𝑥 3 4𝑎1 𝑎2 𝑥 2
𝜋 = (𝑎12 𝑥 +
+
2
3
2
0
50𝑥10 −3
𝑃
𝑎1 2 𝑥 3 𝑎2 2 𝑥 5 540𝑎1 𝑥 2 2𝑎1 𝑎2 𝑥 4 540𝑎2 𝑥 3
+
72900𝑘 +
+
+
+
+
2𝐴
3
5
2
4
3
0
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[∵ 𝑙 = 50𝑥10−3 ]
𝑃
𝑎1 2 (50 × 10−3 )3 𝑎2 2 (50 × 10−3 )5
72900𝑘 +
+
2𝐴
3
5
A
+
D
𝑘
4𝑎22 (50 × 10−3 )3 4𝑎1 𝑎2 (50 × 10−3 )2
𝜋 = (50 × 10−3 )𝑎12 +
+
2
3
2
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𝜋 × 10−3 × 20
50 × 10−3 𝑎12 + 1.666 × 10−4 𝑎22 + 50 × 10−3 𝑎1 𝑎2 +
𝜋
2
2 × 2 × 10−3 2
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𝜋=
rin
g
= 364.5 + 4.166 × 10−5 𝑎12 + 6.25 × 10−8 𝑎22 + 0.675𝑎1 + 3.125 × 10−6 𝑎1 𝑎2 + 0.0225𝑎2
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𝜋 = 5𝑎12 + 0.0166𝑎22 + 0.5𝑎1 𝑎2 + 14.58 × 10−7 + 1.66912 + 2.5 × 10−3 𝑎22 + 2700 𝑎1
+ 0.125 𝑎1 𝑎2 + 900𝑎2 ]
𝜋 = 6.66𝑎12 + 0.0191𝑎22 + 0.625𝑎1 𝑎2 + 2700𝑎1 + 900𝑎2 + 14.58 × 107
t
𝜕𝜋
Apply 𝜕𝑎 = 0
2
⟹ 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
0.625𝑎1 + 0.382𝑎2 = −900
… … … … … (6)
………….. 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 𝑎2 value in equation (6)
13.32𝑎1 + 0.625(41345) = − + 27000
𝑎1 = −3967.01
Sub 𝑎0 , 𝑎1 and 𝑎2 values in equation (3)
𝑇 = 300 − 3697.01𝑥 + 41345𝑥 2
5) Explain briefly about General steps of the finite element analysis.
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Step: 1
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[Nov/Dec 2014]
D
Discretization of structure
discretization.
A
The art of sub dividing a structure into a convenient number of smaller element is known as
Smaller elements are classified as
(i)
One dimensional element
ii)
Two dimensional element
iii)
Three dimensional element
iv)
Axisymmetric element
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i)
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One dimensional element:-
t
a. A bar and beam elements are considered as one dimensional element has two nodes,
one at each end as shown.
1
(ii)
2
Two Dimensional element:Triangular and Rectangular elements are considered as 2D element. These elements
are loaded by forces in their own plane.
3
1
4
3
1
2
2
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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.
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A
D
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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
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process is most important since if decide the size of the stiffness matrix and it leads the reduction of
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memory requirement . While numbering the nodes, the following condition should be satisfied.
{Maximum number node} – {Minimum number node} = minimum
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D
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A
Step 3:
Selection of a displacement function or a Interpolation function:-
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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 .
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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,
𝑑𝑢
𝑒 = 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
ww
𝑘21 ,
𝑘22 ,
𝑘23 … . . 𝑘2𝑛
𝑢2
𝑓3
𝑘31 ,
𝑘32 ,
𝑘33 … . . 𝑘3𝑛
𝑢3
𝑓4
𝑘𝑛1 ,
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𝑘42 ,
𝑘43 … . . 𝑘4𝑛
A
In compact matrix form as.
Where,
.
.
.
𝑢𝑛
.
.
.
D
.
.
.
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𝑒 is a element, {𝐹} is the vector of element modal forces, [𝑘] is the element stiffness
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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.
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Step (6):-
t
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}
𝐹 → Global Force Vector
Where,
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
.
.
.
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𝑓4
𝑘𝑛1 ,
𝑘42 ,
.
.
.
𝑘43 … . . 𝑘4𝑛
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𝑢𝑛
.
.
.
These equation can be solved and unknown displacement {𝑢} calculated by using
A
Step (9):-
D
Gauss elimination.
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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
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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
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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
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conditions.
𝑑𝑥
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𝑑𝑢
𝑎 𝑑𝑥 + 𝑐𝑢 = 𝑓 for 0 < 𝑥 < 𝐿
𝑑𝑢
𝑢 𝑜 = 𝑢0 , 𝑎 𝑑𝑥
= 𝑞0
Bending of elastic beams under Transverse load : find 𝑢 𝑥 that satisfies the fourth order
A
i)
𝑥=𝐿
D
−𝑑
differential equation and boundary conditions.
𝑑𝑥 2
𝑑2𝑢
𝑏
𝑑𝑥 2
𝑢 𝑜 = 𝑢0 ,
𝑑2𝑢
𝑑
𝑏 𝑑𝑥 2
𝑑𝑥
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+ 𝑐𝑢 = 𝐹 for 0 < 𝑥 < −𝐿
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𝑑2
𝑥=𝐿
= 𝑚0 .
𝑑2𝑢
𝑏 𝑑𝑥 2
0
Ω = (o, L)
x=0
𝑑𝑢
= 𝑑0
𝑑𝑥 𝑥=0
= 𝓋0
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x=L
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t
x
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:𝑑𝑢
𝑎 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:28
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𝑑2𝑢
𝑑𝑢
𝑎 𝑑𝑡 + 𝑏 𝑑𝑡 2 + 𝑐𝑢 = 𝐹 for 0 < 𝑡 ≤ 𝑇
𝑑𝑢
𝑢 𝑜 = 𝑢0 , 𝑏 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
𝑑𝑢
𝑎 𝑑𝑥 − 𝜆𝑢 = 0 for 𝑜 < 𝑥 < 𝐿
𝑢 𝑜 = 0, 𝑎
(ii)
𝑑𝑢
=0
𝑑𝑥 𝑥=𝐿
Transverse vibration of a membrane:Find 𝑢 (𝑥, 𝑦) and 𝜆 that satisfy the partial differential equation and
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boundary condition.
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𝑑
𝑑𝑢
𝑑
𝑑𝑢
− 𝑑𝑥 𝑎1 𝑑𝑥 + 𝑑𝑦 𝑎2 𝑑𝑦 − 𝜆𝑢 = 0 in Ω
D
𝑢 = 0 on Γq
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A simple pendulum consists of a bob of mass 𝒎(𝒌𝒈)attached to one end of a rod of
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b)
A
The values of 𝜆 are called cigen values and the associated functions 𝑢 are called cigen functions.
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length 𝒍(𝒎) and the other end is pivoted to fixed point 𝟎.
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Soln:𝑑
𝐹 = 𝑑𝑡 𝑚𝑣 = 𝑚𝑎
𝑑𝑣
𝐹𝑥 = 𝑚. 𝑑𝑡𝑥
−𝑚𝑔 sin 𝜃 = 𝑚𝑙
t
𝑑2 𝑄
𝑑𝑡 2
or
𝑑2 𝑄 𝑔
+ sin 𝑄 = 0
𝑑𝑡 2
𝑙
𝑑2 𝑄 𝑠
+ 𝑄=0
𝑑𝑡 2
𝑙
𝑑𝑄
+ (𝑜) = 𝑈0.
𝑑𝑡
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𝑄 𝑡 = 𝐴𝑠 𝑖𝑛 𝜆𝑡 + 𝐵 cos 𝜆 𝑡.
Where,
𝑠
𝜆=
and 𝐴 and 𝐵 are constant to be determined using the initial condition we
𝑙
obtain.
𝜈
𝐴 − 𝜆0 , 𝐵 = 𝜃0
the solution to be linear problem is
𝜈
𝜃 𝑡 = 𝜆0 𝑆𝑖𝑛 ∧ 𝑡 + 0. 𝐶𝑜𝑠 𝜆𝑡
for zero initial velocity and non zero initial position 𝜃0 , we have.
𝜃 𝑡 = 𝜃0 cos 𝜆𝑡.
7)
ww
A simply supported beam subjected to uniformly distributed load over entire span and
w.E
asy
En
gi
it is subject to a point load at the centre of the span. Calculate the bending moment
SC
A
Given data:-
D
and deflection at imdspan by using Rayleish – Ritz method. (Nov/Dec 2008).
nee
rin
g
.ne
t
To Find:
1. Deflection and Bending moment at mid span.
2. Compare with exact solutions.
Formula used
𝜋 = 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 − 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
Solution:
We know that,
πx
3πx
Deflection, y = a1 sin l + a2 sin l
1
2
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Total potential energy of the beam is given by,
π=U−H
2
2
Where, U – Strain Energy.
H – Work done by external force.
The strain energy, U of the beam due to bending is given by,
1 d2y
0 dx 2
EI
U= 2
2
3
dx
2
πx
dy
ww
a 1 πx
=
dx
πx
a 2 3πx
cos l +
l
+ a2 cos
l
l
3πx
3π
×
l
l
3πx
cos
l
w.E
asy
En
gi
d2y
dx 2
a1 π
=−
πx
a1 π2
=−
dx 2
π
sin l × l −
l
d2y
d2y
π
= a1 cos l ×
l2
EI
= 2
−
π2
l a1
0
l2
EI π 4
l
0
= 2 l4
a1 π2
l2
3πx
sin l ×
− 9 2l 2 sin
SC
l
0
EI
l
a π2
πx
sin l
Substituting dx 2 value in equation (3),
U= 2
a 2 3π
πx
π2
2
3πx
a21 sin2 l + 81a22 sin2
3πx
l
2
nee
3πx
sin l + 9 2l 2 sin l
πx
4
l
− 9 2l 2 sin l
a
l
3πx
a π2
πx
sin l
3π
D
dx
A
dy
2
dx
dx
rin
g
πx
.ne
3πx
+ 2 a1 sin l .9 a2 sin l
dx
t
[∴ a + b 2 = a2 + b2 + 2ab]
EI π 4
U = 2 l4
l
0
πx
a21 sin2 l + 81a22 sin2
𝑙 2
πx
a sin2 l dx
0 1
1
l1
= a21 0 2
l
1 − cos l
2πx
a2
l
πx
+ 18 a1 a2 sin l . sin
2πx
= a21 2 0 1 − cos l
= 21
3πx
𝑙
dx −
0
dx
3πx
dx
l
2
∴ sin x =
5
1−cos 2x
2
2
dx
1
2πx
cos l dx
0
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=
=
=
𝑙
0
a21 sin2
𝑎 12
2
𝑎 12
𝑙
2𝜋𝑥
𝑙
2𝜋
𝑙
sin
𝑥 𝑙0 −
1
0
2𝜋𝑙
𝑙 − 0 − 2𝜋 sin 𝑙 − sin 0
2
𝑎 12
1
𝑙 − 2𝜋 0 − 0
2
𝑎 12 𝑙
=
∴ sin 2𝜋 = 0; sin 0 = 0
2
πx
𝑎12 𝑙
dx =
l
2
6
2
Similarly,
𝑙
3πx
81 a22 sin2 l dx
0
ww
1
𝑙1
= 81a22 0 2 1 − cos
𝑙
6πx
6πx
∴ sin2 x =
dx
l
w.E
asy
En
gi
= 81a22 2 0 1 − cos
=
=
𝑙
0
81a22 sin2
81𝑎 22
2
𝑙
dx −
0
𝑙
6πx
cos l dx
0
𝑥 𝑙0 −
sin
6𝜋𝑥
𝑙
6𝜋
𝑙
81𝑎 22
2
81𝑎 22
2
D
2
𝑙
A
=
81a 22
2
dx
l
SC
=
1−cos 2x
1
nee
0
6𝜋𝑙
𝑙 − 0 − 6𝜋 sin 𝑙 − sin 0
1
𝑙 − 6𝜋 0 − 0
=
rin
g
.ne
𝑎 12 𝑙
∴ sin 6𝜋 = 0; sin 0 = 0
2
3πx
81𝑎22 𝑙
dx =
l
2
t
7
2
𝑙
πx
3πx
18 a1 a2 sin l . sin l dx
0
𝑙
πx
= 18 a1 a2 0 sin l . sin
𝑙
3πx
3πx
l
dx
πx
= 18 a1 a2 0 sin l . sin l dx
𝑙1
= 18 a1 a2 0 2 cos
2πx
l
− cos
4πx
l
dx
∴ sin 𝐴 sin 𝐵 =
=
18 a 1 a 2
2
𝑙
2πx
cos l dx −
0
cos 𝐴−𝐵 −cos 𝐴+𝐵
2
𝑙
4πx
cos l dx
0
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=
2
𝑙
𝑙
4𝜋𝑥
𝑙
4𝜋
𝑙
sin
−
0
0
= 9 a1 a 2 0 − 0 = 0
∴ sin 2𝜋 = 0; sin 4𝜋 = 0; sin 0 = 0
πx
3πx
. sin
dx = 0
l
l
8
𝑙
0
2𝜋𝑥
𝑙
2𝜋
𝑙
sin
18 a 1 a 2
18 a1 a2 sin
2
Substitute (6), (7) and (8) in equation (5),
EI π 4 𝑎 12 𝑙
81𝑎 22 𝑙
+
+0
2
2
U = 2 l4
ww
EI π 4 𝑙
U = 4 l 4 𝑎12 + 81𝑎22
w.E
asy
En
gi
𝐸𝐼𝜋 4 2
𝑎 + 81𝑎22
4𝑙 3 1
𝑙
𝜔 𝑦 𝑑𝑥 + 𝑊 𝑦𝑚𝑎𝑥
0
𝜔 𝑦 𝑑𝑥 =
0
SC
𝐻=
𝑙
A
Work done by external forces,
9
D
Strain Energy, U =
2𝜔𝑙
𝑎2
𝑎1 +
𝜋
3
𝜋𝑥
2
nee
rin
g
11
3𝜋𝑥
𝑦 = 𝑎1 sin 𝑙 + 𝑎2 sin 𝑙
We know that,
10
.ne
t
1
In the span, deflection is maximum at 𝑥 = 2
𝑦𝑚𝑎𝑥 = 𝑎1 sin
𝜋×
1
2
𝑙
𝜋
+ 𝑎2 sin
3𝜋×
1
2
𝑙
3𝜋
𝜋
= 𝑎1 sin 2 + 𝑎2 sin 2
3𝜋
∴ sin 2 = 1; sin 2 = −1
𝑦𝑚𝑎𝑥 = 𝑎1 − 𝑎2
12
Substitute (11) and (12) values in equation (8),
2𝜔𝑙
H= 𝜋
𝑎
𝑎1 + 32 + 𝑊 (𝑎1 − 𝑎2 )
13
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Substituting U and H values in equation (2), we get
𝜋=
𝜋=
𝐸𝐼𝜋 4
𝑎12 + 81𝑎22 −
4𝑙 3
𝐸𝐼𝜋 4
2𝜔𝑙
𝜋
2𝜔𝑙
𝑎12 + 81𝑎22 − 𝜋
4𝑙 3
𝑎
𝑎1 + 32 + 𝑊 (𝑎1 − 𝑎2 )
𝑎
𝑎1 + 32 − 𝑊 (𝑎1 − 𝑎2 )
14
For stationary value of 𝜋, the following conditions must be satisfied.
𝜕𝜋
𝜕𝜋
= 0and𝜕𝑎 = 0
𝜕𝑎 1
2
𝐸𝐼𝜋 4
𝜕𝜋
𝜕𝑎 1
𝐸𝐼𝜋 4
2𝑙 3
2𝜔𝑙
= 4𝑙 3 2𝑎1 − 𝜋 − 𝑊 = 0
2𝜔𝑙
𝑎1 − 𝜋 − 𝑊 = 0
ww
𝐸𝐼𝜋 4
2𝜔𝑙
𝑎
=
+𝑊
1
2𝑙 3
𝜋
w.E
asy
En
gi
2𝑙 3 2𝜔𝑙
+𝑊
𝐸𝐼𝜋 4 𝜋
15
D
A
𝑎1 =
𝜕𝜋
𝐸𝐼𝜋 4
2𝜔𝑙 1
=
162𝑎2 −
+𝑊 =0
3
𝜕𝑎2
4𝑙
𝜋 3
SC
Similarly,
𝐸𝐼𝜋 4
4𝑙 3
nee
2𝜔𝑙
.ne
162𝑎1 − 𝜋 + 𝑊 = 0
𝐸𝐼𝜋 4
2𝑙 3
2𝜔𝑙
162𝑎1 = 𝜋 − 𝑊
𝑎2 =
rin
g
2𝑙 3
2𝜔𝑙
−𝑊
81𝐸𝐼𝜋 4 3𝜋
t
16
From equation (12), we know that,
Maximum deflection, 𝑦𝑚𝑎𝑥 = 𝑎1 − 𝑎2
2𝑙 3
𝑦𝑚𝑎𝑥 = 𝐸𝐼𝜋 4
2𝜔𝑙
2𝑙 3
+ 𝑊 − 81𝐸𝐼𝜋 4
𝜋
2𝜔𝑙
3𝜋
−𝑊
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4𝜔 𝑙 4
2𝑊𝑙 3
4𝜔 𝑙 4
2𝑊𝑙 3
𝑦𝑚𝑎𝑥 = 𝐸𝐼𝜋 5 + 𝐸𝐼𝜋 4 − 243𝐸𝐼𝜋 5 + 81𝐸𝐼𝜋 4
𝜔 𝑙4
𝑊𝑙 3
𝑦𝑚𝑎𝑥 = 0.0130 𝐸𝐼 + 0.0207 𝐸𝐼
17
We know that, simply supported beam subjected to uniformly distributed load, maximum deflection
5 𝜔 𝑙4
𝑦𝑚𝑎𝑥 = 384 𝐸𝐼
is,
Simply supported beam subjected to point load at centre, maximum deflection is,
𝜔 𝑙3
𝑦𝑚𝑎𝑥 = 48𝐸𝐼
5 𝜔 𝑙4
ww
𝜔 𝑙3
𝑦𝑚𝑎𝑥 = 384 𝐸𝐼 + 48𝐸𝐼
So, total deflection,
𝑦𝑚𝑎𝑥 = 0.0130
𝜔𝑙 4
𝑊𝑙 3
+ 0.0208
𝐸𝐼
𝐸𝐼
w.E
asy
En
gi
18
D
From equations (17) and (18), we know that, exact solution and solution obtained by using
A
Rayleigh-Ritz method are same.
We know that,
SC
Bending Moment at Mid span
d2y
nee
Bending moment, M = EI dx 2
rin
g
19
.ne
From equation (9), we know that,
d2y
dx 2
= −
𝑎1 𝜋 2
𝑙2
𝜋𝑥
sin 𝑙 +
𝑎 2 9𝜋 2
𝑙2
3𝜋𝑥
sin 𝑙
t
Substitute 𝑎1 and 𝑎2 values from equation (15) and (16),
d2y
dx 2
2𝑙 3
= − 𝐸𝐼𝜋 4
𝜋2
2𝜔𝑙
𝜋𝑥
2𝑙 3
+ 𝑊 × 𝑙 2 sin 𝑙 + 81𝐸𝐼𝜋 4
𝜋
2𝜔𝑙
9𝜋 2
3𝜋𝑥
− 𝑊 × 𝑙 2 sin 𝑙
3𝜋
𝑙
Maximum bending occurs at 𝑥 = 2
1
1
𝜋×2
3𝜋 × 2
2𝑙 3 2𝜔𝑙
𝜋2
2𝑙 3
2𝜔𝑙
9𝜋 2
= −
+ 𝑊 × 2 sin
+
− 𝑊 × 2 sin
𝐸𝐼𝜋 4 𝜋
𝑙
𝑙
81𝐸𝐼𝜋 4 3𝜋
𝑙
𝑙
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2𝑙 3 2𝜔𝑙
𝜋2
2𝑙 3
2𝜔𝑙
9𝜋 2
= −
+ 𝑊 × 2 (1) +
− 𝑊 × 2 (−1)
𝐸𝐼𝜋 4 𝜋
𝑙
81𝐸𝐼𝜋 4 3𝜋
𝑙
𝜋
3𝜋
∴ sin 2 = 1; sin 2 = −1
= −
2𝑙 2𝜔𝑙
2𝑙
2𝜔𝑙
+
𝑊
−
−𝑊
𝐸𝐼𝜋 2 𝜋
9𝐸𝐼𝜋 2 3𝜋
4𝜔𝑙 2
4𝜔 𝑙 2
2𝑊𝑙
2𝑊𝑙
= − 𝐸𝐼𝜋 3 + 𝐸𝐼𝜋 2 − 27𝐸𝐼𝜋 3 + 9𝐸𝐼𝜋 2
=−
𝐸𝐼𝜋 3
+
2.222𝑊𝑙
𝐸𝐼𝜋 2
d2 y
𝜔𝑙 2
𝑊𝑙
=
−
0.124
+
0.225
dx 2
𝐸𝐼
𝐸𝐼
ww
d2y
3.8518 𝜔𝑙 2
w.E
asy
En
gi
Substitute dx 2 value in bending moment equation,
𝜔 𝑙2
𝑊𝑙
D
d2y
Mcentre = EI dx 2 = −𝐸𝐼 0.124 𝐸𝐼 + 0.225 𝐸𝐼
A
Mcentre = − 0.124 𝜔𝑙 2 + 0.225 𝑊𝑙
20
nee
SC
(∴Negative sign indicates downward deflection)
rin
g
We know that, simply supported beam subjected to uniformly distributed load,
maximum bending moment is,
Mcentre =
.ne
𝜔 𝑙2
8
t
Simply supported beam subjected to point load at centre, maximum bending moment
is,
Mcentre =
𝑊𝑙
4
Total bending moment, Mcentre =
𝜔 𝑙2
8
𝑊𝑙
+ 4
Mcentre = 0.125 𝜔𝑙 2 + 0.25 𝑊𝑙
21
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.
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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,
𝜕4 𝑣
𝜕2 𝑣
𝐸𝐼 4 + 𝜌𝐴 2 = 0
𝜕𝑥
𝜕𝑡
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.
SC
A
D
ww
w .E
asy
En
gin
eer
ing
.ne
t
The governing equation for 1D longitudinal vibration of the bar at one end is given by
d2 U
AE + ρAUω2 = 0
dx 2
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 1
1
2
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.
D
ww
w .E
asy
En
SC
A
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
gin
eer
ing
.ne
t
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
ww
w .E
asy
En
12 𝐾𝑁 𝑚
6 KN
D
1m
A
2m
Given data
gin
eer
SC
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
ing
.ne
t
−𝑙
2
−𝑙 2
f(x)
12
−𝑙
2
𝑙2
𝐹1
𝑀
+ 1 =
𝐹2
𝑀2
𝐸𝐼
𝑙3
12
6𝑙
6𝑙
4𝑙 2
– 12
6𝑙
– 6𝑙
2𝑙 2
– 12 6𝑙
– 6𝑙 2𝑙 2
12 – 6𝑙
– 6𝑙
12
𝑢1
𝜃1
𝑢2
𝜃2
4𝑙 2
M1,θ1
6 KN
M1,θ1
Solution
1
For element 1
𝑣1, F1
2
𝑣2 ,F2
39
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Downloaded From : www.EasyEngineering.net
−𝑙
2
−𝑙 2
f(x)
12
−𝑙
2
𝑙2
𝐹1
𝑀1
+
=
𝐹2
𝑀2
𝐸𝐼
𝑙3
12
6𝑙
6𝑙
4𝑙 2
– 12
6𝑙
– 6𝑙
2𝑙 2
– 12 6𝑙
– 6𝑙 2𝑙 2
12 – 6𝑙
– 6𝑙 4𝑙 2
𝑢1
𝜃1
𝑢2
𝜃2
12
Applying boundary conditions
F1=0N ;
F2=-6KN=-6×103 N;
M1=M2=0; u1=0;
θ1=0; u2≠0;
0
103× 0 =
−6
0
f(x)=0
θ2≠0
– 12 6
–6 2
210×10 9 ×6×10 −6
3
1
– 12 – 6 12 – 6
6
2 –6 4
12
6 −12
6
4 −6
=1.26×106
−12 −6 12
6
2 −6
12
6
6
4
6
2
−6
4
0
0
𝑢2
0
ww
w .E
asy
En
𝑢1
𝜃1
𝑢2
𝜃2
For element 2
12
−𝑙
2
𝑙2
12
𝐸𝐼
𝑙3
12
6𝑙
6𝑙
4𝑙 2
– 12
6𝑙
– 6𝑙
2𝑙 2
– 12 6𝑙
– 6𝑙 2𝑙 2
12 – 6𝑙
– 6𝑙 4𝑙 2
A
f(x)
𝐹2
𝑀2
+
=
𝐹3
𝑀3
SC
2
−𝑙 2
𝑢2
𝜃2
𝑢3
𝜃3
gin
Applying boundary conditions
f(x) = -12 𝑘𝑁 𝑚 =12×103 𝑁 𝑚;
M3,θ3
2
D
−𝑙
12 𝐾𝑁 𝑚
M2,θ2
3
𝑣3 ,F3
𝑣2, F2
eer
ing
F2=F3=0=M2=M;
u2≠0; θ2≠0; u3=θ3=0
0
12
−6
6
103 × −1 + 0 = 1.26×106×
−6
0
−12
1
0
6
6 − 12 6
4 −6
2
6
12 − 6
4 −6
4
𝑢2
𝜃2
0
0
12
−6
6
103 × −1 = 1.26×106×
−6
−12
1
6
6 − 12 6
4 −6
2
6
12 − 6
4 −6
4
𝑢2
𝜃2
0
0
.ne
t
40
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Downloaded From : www.EasyEngineering.net
Assembling global matrix
12
0
6
0
−12
3
6 −12
10 ×
= 1.26×10 ×
6
−1
−6
0
1
0
6
4
−6
2
0
0
Solving matrix
-12×103=1.26×106×24u2=0;
-1×103=1.26×106×8θ2=0;
Result
θ2=-9.92rad
u2=-3.96×10-4m
−12
−6
24
0
−12
6
0
0
6
2
−6
4
0
0
𝑢2
𝜃2
0
0
u2=-3.96×10-4m
θ2=-9.92rad
ww
w .E
asy
En
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
1200mm2
900mm2
300mm
400mm
D
2.
0
0
−12
−6
12
−6
−6
2
0
8
−6
2
SC
A
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
1 –1
𝐴𝐸
Stiffness matrix
[k] = 𝑙
–1
1
𝜌𝐴𝐿 2 1
Consistent mass matrix [m] = 6
1 2
𝜌𝐴𝐿 1 0
Lumped mass matrix [m] = 2
0 1
Mode shape 𝑘 − 𝑚𝜆 U1 = 0 ;
Normalization 𝑈1𝑇 M U1 = 1
Solution
For element 1
gin
u1
eer
ing
1200mm2
.ne
t
u2
300mm
41
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Downloaded From : www.EasyEngineering.net
𝐴𝐸
Stiffness matrix [k] =
[k1] =
=8.4×105
1 –1
–1
1
1 –1
;
–1
1
=
1
–1
=105
𝐴1 𝐸1
𝑙1
𝑙
1200 ×2.1×10 5
–1
;
1
Consistent mass matrix [m] =
𝜌𝐴1 𝐿1
[m1] =
6
=
2
1
300
6
1200 ×300×7.8×10 −6
6
8.4 – 8.4
– 8.4
8.4
2 1
;
1 2
𝜌𝐴𝐿
1
2
1 −1
−1
1
2 1
1 2
2 1
1 2
0.936 0.468
0.468 0.936
= 0.468×
ww
w .E
asy
En
[m1]
=
u2
For element 2
𝑙
D
1 –1
–1
1
𝐴𝐸
Stiffness matrix [k] =
400mm
A
1 –1
;
𝑙2
–1
1
900×2.1×10 5
1 −1
=
400
−1
1
1 −1
5
= 4.73×10
−1
1
4.73 – 4.73
[k2] = 105
;
– 4.73
4.73
𝐴2 𝐸2
SC
[k2] =
Consistent mass matrix [m] =
[m2] =
𝜌𝐴2 𝐿2
=
6
2
1
u3
900 mm2
𝜌𝐴𝐿
6
1
2
900×400×7.8×10 −6
6
2 1
1 2
0.936 0.468
0.468 0.936
gin
eer
ing
2 1
;
1 2
2
1
.ne
t
1
2
= 0.468
[m2]
=
Assembling global matrix
8.4
Stiffness matrix [k] = 105 −8.4
0
0.936
Consistent mass matrix [m] = 0.468
0
−8.4
13.13
−4.73
0.468
1.87
0.468
0
−4.73
4.73
0
0.468
0.936
42
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Downloaded From : www.EasyEngineering.net
General equation for free vibration of bar 𝑘 − 𝑚𝜆 {u} = 0
8.4 −8.4
0
0.936 0.468
0
105 −8.4 13.13 −4.73 - λ 0.468 1.87 0.468 =0
0
−4.73
4.73
0
0.468 0.936
13.13
−4.73
105
−4.73
1.87
–λ
4.73
0.468
13.13 × 105 − 1.87𝜆
−4.73 × 105 − 0.468𝜆
0.468
=0
0.936
−4.73 × 105 − 0.468𝜆 = 0
4.73 × 105 − 0.936𝜆
[(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
ww
w .E
asy
En
Solving above equation
𝜆1 = 1.49×106
𝜆2 = 1.73×105 = 0.173×106
𝑘 − 𝑚𝜆 {𝑢} = 0
;
SC
13.13
−4.73
A
𝜆1 = 0.173×106
105
D
To find mode shape
gin
−4.73
1.87
– 0.173×106
4.73
0.468
0.99 × 106
−0.55 × 106
−0.55 × 106
0.31 × 106
6
6
𝑢2
𝑢3 = 0
eer
0.468
0.936
0.99×10 u2 – 0.55× 10 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
1.87
– 1.49×106
4.73
0.468
−1.48 × 106
−1.17 × 106
−1.17 × 106
−0.924 × 106
0.468
0.936
𝑢2
𝑢3 = 0
ing
.ne
t
𝑢2
𝑢3 = 0
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 𝑈1𝑇 M U1 = 1
Normalization of 𝜆1
43
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Downloaded From : www.EasyEngineering.net
𝑢2
𝑢2
1.77𝑢2 =1
1.87 0.468
0.46 0.936
1.77𝑢2
𝑢2
𝑢2
1.77𝑢2 = 1
1.77𝑢2
2.7𝑢22 + 3.79𝑢22 =1
𝑢22 = 6.4 ;
1
𝑢2 = 0.392
𝑢3 =1.78𝑢2 ;
𝑢3 = 0.698
Normalization of 𝜆2
𝑈2𝑇 M U2 = 1
𝑢2
−1.26𝑢2
1.87 0.468
0.46 0.936
1.28𝑢2
−0.707𝑢2
𝑢2
−1.26𝑢2 =1
ww
w .E
asy
En
1.28𝑢22 + 0.88𝑢22 =1
𝑢22 = 0.46;
A
𝑢3 =-1.268𝑢2
D
𝑢2
−1.256𝑢2 = 1
Result
Mode shape
SC
𝑢3 = -0.84
gin
eer
2
1
3
u2=0.392
Mode 1
ing
u3=0.698
.ne
t
u1=0
u2=0.678
u1=0
Mode 2
u3=-0.698
44
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Downloaded From : www.EasyEngineering.net
3.
Consider the simply supported beam shown in fig. let the length L=1m,
E=2×1011𝑵 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
L
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
SC
Formula used
A
Lumped mass matrix
Consistent mass matrix
Natural frequency
D
ww
w .E
asy
En
gin
eer
General equation for free vibration of beam 𝑘 − 𝜔2 𝑚 {u} = 0
– 12 6𝑙
– 6𝑙 2𝑙 2
𝐸𝐼
Stiffness matrix[k] = 𝑙 3
– 12 – 6𝑙 12 – 6𝑙
6𝑙 2𝑙 2 – 6𝑙 4𝑙 2
156 22𝑙
22𝑙 4𝑙 2
𝜌𝐴𝐿
Consistent mass matrix [m] = 420
54
13𝑙
−13𝑙 −3𝑙 2
1 0 0 0
𝜌𝐴𝑙 0 0 0 0
Lumped mass matrix [m] = 2
0 0 1 0
0 0 0 0
12
6𝑙
6𝑙
4𝑙 2
ing
54 −13𝑙
13𝑙 −3𝑙 2
156 – 22𝑙
−22𝑙 4𝑙 2
.ne
t
Solution
For element 1
45
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Downloaded From : www.EasyEngineering.net
6𝑙1
4𝑙12
−6𝑙1
2𝑙12
12
6𝑙1
𝐸1 𝐼
Stiffness matrix[k]1 = 3
𝑙 1 −12
6𝑙1
−12
−6𝑙1
12
−6𝑙1
6𝑙1
2𝑙12
−6𝑙1
4𝑙12
θ1
θ2
1
2
0.5 m
𝑣1
12
6 × 0.5
−12
6 × 0.5
2×10 11 ×100×−12
=
0.53
12
3
[k]1 =160×
−12
3
6 × 0.5
4 × 0.52
−6 × 0.5
2 × 0.52
−12
−3
12
−3
3
1
−3
0.5
2
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
76518 ×3×10 −3 ×0.5
=
SC
A
2
[m]1
57.38
= 0
0
0
0
0
0
0
=
420
156
22 × 0.5
54
−13 × 0.5
42.63
[m]1 = 3
14.74
−1.77
3
0.27
1.77
−0.20
0
0
1
0
0
0
0
0
gin
156
22𝑙1
𝜌𝐴 𝑙
Consistent mass matrix [m]1 = 4201
54
−13𝑙1
76518 ×3×10 −3 ×0.5
6 × 0.5
2 × 0.52
−6 × 0.5
4 × 0.52
0
0
0
0
D
Lumped mass matrix [m]1 =
1
0
0
0
−12
−6 × 0.5
12
−6 × 0.5
3
0.5
−3
1
ww
w .E
asy
En
𝜌𝐴 𝑙 1
𝑣2
22𝑙1
4𝑙12
13𝑙1
−3𝑙12
22 × 0.5
4 × 0.52
13 × 0.5
−3 × 0.52
eer
ing
54
13𝑙1
156
−22𝑙1
−13𝑙1
−3𝑙12
−22𝑙1
4𝑙12
54
13 × 0.5
156
−22 × 0.5
.ne
t
−13 × 0.5
−3 × 0.52
−22 × 0.5
4 × 0.52
14.74 −1.77
1.77 −0.20
42.63
−3
−3
0.27
46
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Downloaded From : www.EasyEngineering.net
For element 2
6𝑙2
4𝑙22
−6𝑙2
2𝑙22
12
6𝑙2
𝐸𝐼
Stiffness matrix[k]2 = 𝑙 3
−12
2
6𝑙2
6𝑙2
2𝑙22
−6𝑙2
4𝑙22
−12
−6𝑙2
12
−6𝑙2
θ2
θ3
2
3
0.5 m
𝑣2
12
6 × 0.5
−12
6 × 0.5
2×10 11 ×100×−12
=
0.53
12
3
[k]2 = 160×
−12
3
3
1
−3
0.5
−12
−3
12
−3
1
0
0
0
0
0
1
0
ww
w .E
asy
En
2
57.38
= 0
0
0
0
0
0
0
[m]2
1
0
0
0
=
420
42.63
[m]2 = 3
14.74
−1.77
156
22 × 0.5
54
−13 × 0.5
3
0.27
1.77
−0.20
0
0
0
0
0
0
1
0
0
0
0
0
57.38 0
0
0
156
22𝑙2
𝜌𝐴 𝑙
Consistent mass matrix [m]2 = 4202
54
−13𝑙2
76518 ×3×10 −3 ×0.5
6 × 0.5
2 × 0.52
−6 × 0.5
4 × 0.52
3
0.5
−3
1
0
0
0
0
A
76518 ×3×10 −3 ×0.5
SC
=
2
0
0
0
0
−12
−6 × 0.5
12
−6 × 0.5
D
Lumped mass matrix [m]2 =
𝜌𝐴 𝑙 2
6 × 0.5
4 × 0.52
−6 × 0.5
2 × 0.52
𝑣3
0
0
0
0
gin
22𝑙2
4𝑙22
13𝑙2
−3𝑙22
22 × 0.5
4 × 0.52
13 × 0.5
−3 × 0.52
eer
54
13𝑙2
156
−22𝑙2
ing
−13𝑙2
−3𝑙22
−22𝑙2
4𝑙22
54
13 × 0.5
156
−22 × 0.5
.ne
t
−13 × 0.5
−3 × 0.52
−22 × 0.5
4 × 0.52
14.74 −1.77
1.77 −0.20
42.63
−3
−3
0.27
Global matrix
47
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Downloaded From : www.EasyEngineering.net
12
3
−12
Stiffness matrix [k] =160×
3
0
0
3
1
−3
0.5
0
0
−12
−3
24
0
−12
3
57.38
0
0
Lumped mass matrix [m]=
0
0
0
0
0
0
0
0
0
0
0
114.77
0
0
0
42.63
3
14.74
Consistent mass matrix[m]=
−1.77
0
0
Frequency for lumped mass matrix
𝑘 − 𝜔2 𝑚 {u} = 0
3
0.27
1.77
−0.2
0
0
−12
−3
24
0
−12
3
3
0.5
0
2
−3
0.5
0
0
−12
−3
12
−3
Applying boundary conditions
𝑣1 =0=𝜃1 ;
𝑣2 ≠0;
𝜃2 ≠0
12
3 −12
3
0
3
1
−3 0.5
0
−12 −3
24
0 −12
160 ×
3 0.5
0
2
−3
0
0 −12 −3
12
0
0
3 0.5
−3
160 ×
24
0
0
114.7
− 𝜔2
2
0
3840 − 𝜔2 × 114.7
0−0
0
0
0
0
3
0.5
−3
1
0
0
0
0
0
0
0
0
0
0
57.38
0
0
0
0
0
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
114.77
0
0
0
0
0
0
0
0
0
0
57.38
0
0
3
0
2
−𝜔
0.5
0
−3
0
1
0
0
0
0
0
0
0
0
57.38
0
0
3
0
2
−𝜔
0.5
0
−3
0
1
0
𝑣3 =0=𝜃3 ;
0
0
0
0
0 114.77
0
0
0
0
0
0
A
3
1
−3
0.5
0
0
SC
12
3
−12
160 ×
3
0
0
0
0
−12
−3
12
−3
D
ww
w .E
asy
En
3
0.5
0
2
−3
0.5
gin
eer
0
0
−1.77
−0.2
−3
0.27
0
0
0
0
57.38
0
ing
0
0
0
0
0
0
0
0
0
0
0
0
𝑣1
𝜃1
𝑣2
𝜃2 =0
𝑣3
𝜃3
.ne
t
0
0
0
0
57.38
0
0
0
0
0
0
0
0
0
𝑣2
𝜃2 =0
0
0
𝑣2
𝜃2 = 0
0−0
=0
320 − 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
48
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Downloaded From : www.EasyEngineering.net
𝑘 − 𝜔2 𝑚 {u} = 0
12
3
−12
160 ×
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
42.63
0
3
3
14.74
− 𝜔2
0.5
−1.77
0
−3
0
1
Applying boundary conditions
𝑣1 =0=𝜃1 ;
𝑣2 ≠0;
𝜃2 ≠0
12
3
−12
160 ×
3
0
0
0
42.63
0
3
3
14.74
2
−𝜔
0.5
−1.77
0
−3
0
1
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
24
0
0
2
− 𝜔2
3840 − 85.26ω2
0−0
85.26
0
0
0.5
𝑣2
𝜃2
−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 =0
𝑣3
𝜃3
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
𝑣3 =0=𝜃3 ;
14.74
1.77
85.26
0
14.74
−1.77
3
0.27
1.77
−0.2
0
0
ww
w .E
asy
En
160 ×
14.74
1.77
85.26
0
14.74
−1.77
3
0.27
1.77
−0.2
0
0
−1.77
−0.2
0
0.5
1.77
−0.2
=0
D
0−0
=0
320 − 0.5ω2
2
SC
Take λ = 𝜔
A
(3840 − 85.26𝜔2 ) 320 − 0.5𝜔2 = 0
1.23×106-1920𝜔2 -27283.2𝜔2 +42.63𝜔4 =0
42.63 λ2 -29203.3 λ+1.23×106 =0
𝜆=
gin
eer
29203 .3 ± 29203 .32 −4×42.63×1.23×10 6
2×42.63
29203 .3 ±25359 .28
=
2𝑎
.ne
t
85.26
29203 .3+25359 .28
𝜆1 =
−𝑏± 𝑏 2 −4𝑎𝑐
ing
ax2 +bx+c=0; x =
85.26
29203 .3−25359 .28
𝜆2 =
;
𝜆1 =639.95;
85.26
𝜆2 =45.08
λ = 𝜔2
𝜔1 = λ1 ;
𝜔 2 = 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 = 2×105𝑴𝑵 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
80mm
150m
m
P
300m
m
40m
Given data
Mass density ρ = 7800𝑘𝑔 𝑚3 m
= 7800 × 9.81=76518 𝑁 𝑚3
= 7.65 × 10-5 𝑁 𝑚𝑚3
Young’s modulus E = 2×105𝑀𝑁 𝑚2 ;
= 2×105 × 106 𝑁 𝑚2
= 2×105 𝑁 𝑚𝑚2
D
ww
w .E
asy
En
A
Point load P = 10 KN
To find
gin
{F} =[K] {u}
SC
Displacement at each node
Reaction force at the support
Formula used
𝐴𝐸
Stiffness matrix [k] = 𝑙
𝜌𝐴𝑙
Force vector 𝐹 = 2
1 – 1 𝑢1
𝐹1
𝐴𝐸
= 𝑙
𝐹2
–1
1 𝑢2
1
1
1 – 1 𝑢1
–1
1 𝑢2
eer
ing
.ne
t
{R} =[K] {u} -{F}
Solution
The given taper bar is considered as stepped bar as shown in fig.
W1=80mm
W1=80mm
P
150mm
150m
m
2
300m
m
1
10KN
150mm
3
W3=40
mm
50
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Downloaded From : www.EasyEngineering.net
W1 = 80mm
W2 =
𝑊1 +𝑊3
2
=
80+40
= 60 mm
2
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
ww
w .E
asy
En
= W3 × t3 = 40 × 10 =400mm2
Average area of element 1
2
Average area of element 2
For element 1
𝐴1 + 𝐴2
2
𝐴2 + 𝐴3
A
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 2 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 3
2
SC
Ā2 =
=
1
–1
Ā 𝐸
Stiffness matrix [k]1 = 1𝑙 1
1
700 ×2×10 5
=
150
=
=
2
2
= 700mm2
1
−1
=
600+400
2
= 500mm2
eer
– 1 𝑢1
1 𝑢2
u1,F1
ing
−1 𝑢1
1 𝑢2
−4.67 𝑢1
4.67 𝑢2
4.67
= 2× 10
−4.67
𝜌 Ā1 𝑙 1
800+600
gin
2
5
Force vector 𝐹 1 =
=
D
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 1 +𝐴𝑟𝑒𝑎 𝑜𝑓 𝑛𝑜𝑑𝑒 2
Ā1 =
.ne
t
150mm
u2,F2
10KN
1
1
7.65×10 −5 ×700×150
2
1
1
=
4.017
4.017
u2,F2
For element 2
Ā 𝐸
Stiffness matrix [k]2 = 2𝑙 2
2
=
– 1 𝑢2
1 𝑢3
1
–1
500 ×2×10 5
150
= 2× 105
1
−1
3.33
−3.33
10KN
−1 𝑢2
1 𝑢3
150mm
u3,F3
−3.33 𝑢2
3.33 𝑢3
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Force vector 𝐹 2 =
=
𝜌 Ā2 𝑙 2
2
1
1
7.65×10 −5 ×500×150
2
=
2.869
2.869
−4.66
7.99
−3.33
0
−3.33
3.33
1
1
Global matrix
4.66
Stiffness matrix [k] = 2×105 × −4.66
0
4.017
Force vector 𝐹 = 6.88
2.87
Finite element equation
{F} =[K] {u}
ww
w .E
asy
En
𝐹1
4.66
𝐹2 = 2×105 × −4.66
𝐹3
0
−4.66
7.99
−3.33
0
−3.33
3.33
𝑢1
𝑢2
𝑢3
D
Applying boundary conditions
𝑢1 = 0; 𝑢2 ≠ 0; 𝑢3 ≠ 0; 𝐹2 = 10 × 103 N
0
−3.33
3.33
A
−4.66
7.99
−3.33
SC
𝐹1
4.66
𝐹2 = 2×105 × −4.66
𝐹3
0
gin
4.017
4.66
5
3
=
2×10
×
6.88 + 10 × 10
−4.66
2.87
0
10006.88
7.99
= 2×105
2.86
−3.33
𝑢1
𝑢2
𝑢3
eer
−4.66
7.99
−3.33
−3.33 𝑢2
3.33 𝑢3
2×105 (7.99𝑢2 − 3.33𝑢3 ) = 10006.88
0
−3.33
3.33
0
𝑢2
𝑢3
ing
.ne
t
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
52
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Reaction force
{R} =[K] {u} -{F}
𝑅1
4.66
𝑅2 = 2×105 × −4.66
𝑅3
0
−4.66
7.99
−3.33
𝑢1
𝐹1
0
−3.33 𝑢2 - 𝐹2
𝐹3
3.33 𝑢3
𝑅1
4.66
5
𝑅2 = 2×10 × −4.66
𝑅3
0
−4.66
7.99
−3.33
0
−3.33
3.33
𝑢1
4.017
0.01074 - 10006.88
0.01074
2.87
0 − 0.05 + 0
4.017
=2×105 0 + 0.086 − 0.036 - 10006.88
0 − 0.036 + 0.036
2.87
4.017
−0.05
0.05 - 10006.88
2.87
0
=
SC
Result
−10004.017
−6.88
−2.86
A
−10000
4.017
= 10000 - 10006.88
0
2.87
𝑅1
−10004.017
𝑅2 =
−6.88
𝑅3
−2.86
5.
D
ww
w .E
asy
En
= 2×105
gin
eer
ing
.ne
t
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 1000OC and outside
of the insulation is exposed to atmospheric air at 30oc with heat transfer coefficient of 35 W/m2k. Calculate the nodal temperature. NOV/DEC 2014
Given Data:Thickness of the wall, l1 = 0.6m
Conduction
Convection
Conduction
Thermal conductivity of the wall K1= 1.2W/mk
h
Thickness of the insulation l2 = 0.06m
T1
T3
T2
Thermal Conductivity of the wall K2 = 0.3W/mk
Inner surface temperature T1= 1000oC+273
Wall
Insulation
𝑙1
𝑙2
𝑇∞
= 1273 K
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Atmospheric air temperature
T2 = 30 +273
= 303 K
Heat transfer co-efficient at outer side h = 35W/m2k
To find
Nodal temperature T2 and T3
Formula used
1D Heat conduction
𝐴𝑘 1
𝐹1
=
𝐹2
𝑙 –1
– 1 𝑇1
1 𝑇2
1D Heat conduction with free end convection
𝐴𝑘
[K]= 𝑙
1 –1
0
+ hA
0
–1
1
0
1
ww
w .E
asy
En
Solution
Conduction
A
SC
k1 A1 1 −1 T1
f1
=
f2
l1 −1 1 T2
For unit area: A1 = 1m2
1.2 1
−1 T1
= 0.6
−1 1 T2
f1
2 −2 T1
=
f2
−2 2 T2
D
For element 1
T1
gin
eer
For element (2)
A2 K 2 1 −1
0 0 T2
0
+ hA
= h T2 A
T
0 1
1
l2 −1 1
3
T
1 X 0.3 1
0 0
−1
0
2
+ 35 × 1
=35×303×1×
0.06 −1
T3
0 1
1
1
T1
0 0
5 −5
0
+
=
0 35 T2
−5 5
10.605 × 103
5 −5 T1
0
=
T
−5 5
10.605 × 103
2
T2
ing
L1
.ne
t
Convection
T3
h T∞
Conduction
T2
L2
Assembling finite element equation
f1
2 −2 0 T1
f2 = −2 7 −5 T2
f3
0 −5 40 T3
Applying boundary conditions
f1 = 0
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f2 = 0
f3 = 10.605 x 103
2 −2 0 T1
0
−2 7 −5 T2 =
0
T
10.605 × 103
0 −5 40
3
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 T1
0
T
0 7 −5 2 =
0
10.605 × 103
0 −5 40 T3
Step – II
The first row of the force matrix is replaced by the known temperature at node 1
ww
w .E
asy
En
1 0
0 T1
1273
0 7 −5 T2 =
0
10.605 × 103
0 −5 40 T3
SC
A
D
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 T1
1273
0 7 −5 T2 =
2546
T
10.605
× 103
0 −5 40
3
⟹ 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
gin
eer
ing
.ne
t
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
55
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Downloaded From : www.EasyEngineering.net
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.
𝓍
1
2
𝑢1
𝑢2
ww
w .E
asy
En
𝑙
Fig Two node bar element
SC
A
D
Since the element has got two degrees of freedom, it will have two generalized co-ordinates.
𝑢 = 𝑎0 + 𝑎1 𝑥
Where, 𝑎0 and 𝑎1 are global or generalized co – ordinates.
Writing the equation in matrix form,
𝑎0
𝑢 = 1𝑥 𝑎
1
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
1 0
𝑢2 = 1 𝑙
gin
eer
ing
.ne
t
𝑎0
𝑎1
𝑢∗
𝐶
𝐴
∗
Where, 𝑢 ⟶ Degree of freedom.
𝐶 ⟶ Connectivity matrix.
𝐴 ⟶ Generalized or global co-ordinates matrix.
𝑎0
1 0 −1 𝑢1
=
𝑎1
𝑢2
1 𝑙
= 𝑙−0
1 −0
−1 1
𝑎12 −1
𝑎22
=
1
𝑎11
𝑁𝑜𝑡𝑒: 𝑎
21
𝑢1
𝑢2
1
𝑎22
× −𝑎
21
𝑎11 𝑎22 − 𝑎12 𝑎21
−𝑎12
𝑎11
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𝑎0
1
𝑙
0 𝑢1
𝑎1 = 𝑙 −1 1 𝑢2
𝑎0
Substitute 𝑎 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
1
1
𝑙
0 𝑢1
𝑢 = 1 𝑥 𝑙
−1 1 𝑢2
1
𝑙
0 𝑢1
= 𝑙 1 𝑥
−1 1 𝑢2
𝑢1
1
= 𝑙 1−𝑥
0+𝑥 𝑢
2
∵ 𝑀𝑎𝑡𝑟𝑖𝑥 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 1 × 2 2 × 2 = 1 × 2
𝑢1
1− 𝑥 𝑥
𝑢 = 𝑙 𝑙
𝑢2
𝑢1
𝑢 = 𝑁1 𝑁2 𝑢
2
Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2
𝑙− 𝑥
Where, Shape function, 𝑁1 =
ww
w .E
asy
En
𝑙
𝑥
; 𝑠𝑎𝑝𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑁2 = 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.
A
1
D
𝓍
gin
SC
𝑢1
Stiffness matrix, 𝐾 =
𝑣
𝑙− 𝑥
𝑙
𝑢2
eer
𝑙
ing
B T 𝐷 𝐵 𝑑𝑣
Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2
Shape function, 𝑁1 =
2
𝑥
; 𝑠𝑎𝑝𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑁2 = 𝑙
Strain displacement matrix,[B] =
𝑑𝑁1
𝑑𝑁2
𝑑𝑥
−1
1
= 𝑙
.ne
t
𝑑𝑥
𝑙
−1
𝑙
1
[B]T=
𝑙
One dimensional problem [D] = [E] = young’s modulus
−1
[K] =
𝒍
𝟎
𝑙
1
×𝐸×
−1
1
𝑙
𝑙
𝑑𝑣
𝑙
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1
𝑙 𝑙2
= 0 −1
𝑙2
−1
1
𝑙 𝑙2
0 −1
𝑙2
−1
=
𝑙2
1
× 𝐸 × 𝑑𝑣
𝑙2
𝑙2
1
× 𝐸 × A × dx
𝑙2
1
−1
𝑙2
= AE −1
𝑙2
1
𝑙2
𝑙2
1
−1
𝑙2
= AE −1
𝑙2
1
𝑙2
𝑙2
×
1
−1
𝑙2
1
𝑙2
𝑙2
𝐴𝐸𝑙
1
−1
−1
1
𝐴𝐸
1 –1
–1
1
ww
w .E
asy
En
A
[K] = 𝑙
1
−1
𝑙2
= AE −1
𝑙2
1
𝑙2
𝑙2
𝑥 𝑙0
D
= 𝑙2
𝑙
𝑑𝑥
0
(𝑙 − 0)
𝑙2
−1
= AE 𝑙
[dv = A×dx
gin
SC
Finite element equation for finite element analysis
{F} =[K] {u}
eer
𝐴𝐸 1 – 1 𝑢1
𝐹1
=
𝐹2
1 𝑢2
𝑙 –1
ing
.ne
t
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
x
Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2
Where; 𝑁1 =
𝑙− 𝑥
𝑙
𝑥
; 𝑁2 = 𝑙 ;
[N] =
𝑙− 𝑥
𝑥
𝑙
𝑙
xb
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𝑙− 𝑥
[N]T =
𝑙
𝑥
𝑙
Substitute Xb and [N]T values
𝐹 𝑒=
𝑙
0
𝑙− 𝑥
𝑙
𝑥
𝑙
= ρA
= ρA
ρA dx
𝑥−
𝑥2
2𝑙
𝑥2
2𝑙
𝑙− 𝑥
𝑙
𝑥
𝑙
𝑙
= ρA 0
𝑙
= ρA
𝑙−
𝑙2
2𝑙
0
dx
𝑙2
2𝑙
= ρA
1
1
DERIVATION OF SHAPE FUNCTION AN STIFFNESS MATRIX FOR ONEDIMENSIONAL QUADRATIC BAR ELEMENT: May / June 2012
D
7.
𝜌𝐴𝑙
2
𝑙
2
𝑙
2
𝑙
2
𝑙
2
ww
w .E
asy
En
Force vector {F} =
𝑙−
𝜐1
1
SC
𝓍
A
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.
3
𝑙
gin
eer
ing
.ne
t
2 𝜐2
𝜐3
2
𝑙
Fig. (i). Quadratic bar element
Since the element has got three nodal displacements, it will have three generalized
coordinates.
u = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2
Where, 𝑎0 , 𝑎1 𝑎𝑛𝑑 𝑎2 are global or generalized coordinates. Writing the equation is matrix
form,
59
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𝑈 = 1𝑥 𝑥
𝑎0
𝑎1
𝑎2
2
At node 1, u = u1 , 𝑥 = 0
At node 2,
u = u2 , 𝑥 = 1
1
At node 3, u = u3 , 𝑥 = 2
Substitute the above values in equation.
u1 = 𝑎0
u2 = 𝑎0 + 𝑎1 𝑙 + 𝑎2 𝑙 2
𝑙
u3 = 𝑎0 + 𝑎1
2
ww
w .E
asy
En
𝑙 2
+ 𝑎2
2
Substitute the equation we get
u2 = 𝑢1 + 𝑎1 𝑙 + 𝑎2 𝑙 2
𝑎 𝑙2
𝑎 𝑙
D
u3 = 𝑢1 + 21 + 24
A
u2 − u1 = 𝑎1 𝑙 + 𝑎2 𝑙 2
gin
𝑎 𝑙2
𝑎 𝑙
SC
u3 − 𝑢1 = 21 + 24
Arranging the equation in matrix form,
𝑙
u2 − u1
=
𝑙
u3 − 𝑢1
𝑙2
2
4
𝑙
𝑙2
⇒
a1
a2
=
eer
𝑙
𝑙2
2
4
−1
u2 − u1
u3 − 𝑢1
𝑙2
= 𝑙3
1
𝑙3
−
4
2
𝑎11
𝑁𝑜𝑡𝑒 ∵ 𝑎
21
⇒
ing
a1
a2
𝑙2
4
−𝑙
−𝑙 2
𝑙
2
1
𝑎12 −1
𝑎22
=
X
𝑎22
−𝑎21
𝑎11 𝑎22 − 𝑎12 𝑎21
a1
a2 =
−4
⇒ 𝑎1 = 𝑙 3
𝑙2
1
−𝑙 3
4
𝑙2
4
4
−𝑙
2
−𝑙 2
𝑙
.ne
t
u2 − u1
u3 − 𝑢1
−𝑎12
𝑎11
u2 − u1
u3 − 𝑢1
u2 − u1 −𝑙 2 u3 − 𝑢1
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−4
−𝑙
⇒ 𝑎2 = 𝑙 3
Equation
u2 − u1 + 𝑙 u3 − 𝑢1
2
𝑙2 𝑢2
−4
𝑎1 = 𝑙 3
𝑙2 𝑢1
−
4
4
−4𝑙 2 𝑢 2
=
+
4𝑙 3
− 𝑢2
=
𝑙
𝑙
4𝑙 2 𝑢 1
+
4𝑙 3
𝑢
4 𝑢3
+ 𝑙1 +
−3 𝑢 1
𝑎1 =
−𝑙 2 𝑢3 + 𝑙 2 𝑢1
𝑢
−
𝑙3
4𝑙 2 𝑢 1
𝑙3
4 𝑢1
−
𝑙
− 𝑙2 +
4𝑙 2 𝑢 3
𝑙
4 𝑢3
𝑙
Equation
−4
−𝑙𝑢 2
𝑎2 = 𝑙 3
2
4𝑙 𝑢 2
𝑙
− 2 𝑢1 + 𝑙𝑢3 − 𝑙𝑢1
4𝑙
4𝑙
4𝑙
2 𝑙3
+ 2 𝑙 3 𝑢1 − 𝑙 3 𝑢3 + 𝑙 3 𝑢1
2𝑢 2
2
ww
w .E
asy
En
=
=
𝑙2
4
4
− 𝑙 2 𝑢1 − 𝑙 2 𝑢3 + 𝑙 2 𝑢1
2
2𝑢
4
D
𝑎2 = 𝑙 2 𝑢1 + 𝑙 2 2 − 𝑙 2 𝑢3
Arranging the equation in matrix form,
0
0
𝑢1
𝑢2
𝑢3
A
1
𝑎0
−3
𝑎1 = 𝑙
2
𝑎2
gin
−1
SC
𝑙
2
𝑙2
𝑙2
4
𝑙
−4
𝑙2
Substitution the equation
𝑢 = 1 𝑥
𝑢 =
𝑢 = 𝑁1
𝑥2
−3
−1
0
0
𝑙
2
𝑙
2
𝑙
−4
𝑙2
𝑙2
𝑙2
3
1− 𝑙 𝑥+
𝑁2
𝑁3
1
4
2 𝑥2
−𝑥
𝑙2
𝑙
eer
ing
𝑢1
𝑢2
𝑢3
2 𝑥2
+ 𝑙2
4𝑥
𝑙
4 𝑥2
− 𝑙2
.ne
t
𝑢1
𝑢2
𝑢3
𝑢1
𝑢2
𝑢3
𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3
Where, shape function,
𝑁1 = 1 −
3𝑥
2𝑥 2
+ 𝑙2
𝑙
𝑁2 =
−𝑥
2𝑥 2
𝑁3 =
4𝑥
4𝑥 2
+ 𝑙2
𝑙
− 𝑙2
𝑙
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STIFFNESS MATRIX FOR ONE-DIMENSIONAL QUADRATIC BAR ELEMENT:
1
1
𝑙
𝜐1
2
2 3 𝜐2
2
2
𝑙
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.
ww
w .E
asy
En
We know that,
𝐵 𝑇 𝐷 𝐵 𝑑𝑣
D
Stiffness matrix, 𝑘 =
𝑣
A
In one dimensional quadratic bar element,
gin
Displacement function, 𝑢 = 𝑁1 𝑢1 + 𝑁2 𝑢2 + 𝑁3 𝑢3
SC
𝜐1
Where, 𝑁1 = 1 −
𝑁3 =
+ 𝑙2
𝑙
2𝑥 2
−𝑥
𝑁2 =
eer
2𝑥 2
3𝑥
+ 𝑙2
𝑙
ing
4𝑥 2
4𝑥
− 𝑙2
𝑙
We know that,
𝑑 𝑁1 𝑑 𝑁2 𝑑 𝑁3
Strain – Displacement matrix, 𝐵 =
⟹
𝑑 𝑁1
⟹
𝑑 𝑁2
⟹
𝑑 𝑁3
𝑑𝑥
𝑑𝑥
.ne
t
𝑑𝑥
𝑑𝑥
=
−3
4𝑥
=
−1
𝑙
+ 2
4
8𝑥
𝑑𝑥
+ 𝑙2
𝑙
4𝑥
𝑙
𝑑𝑥
= 𝑙 + 𝑙2
−3
4𝑥
Substitute the equation
𝐵 =
𝑙
+ 𝑙2
−1
𝑙
4𝑥
+ 𝑙2
4
𝑙
8𝑥
− 𝑙2
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−3
𝑙
−1
𝐵𝑇=
𝑙
4
4𝑥
+ 𝑙2
4𝑥
+ 𝑙2
8𝑥
+ 𝑙2
𝑙
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 4𝑥
+ 2
𝑙
𝑙
−1 4𝑥
+ 2
𝑙
𝑙
4 8𝑥
−
𝑙 𝑙2
𝑙
⟹=
0
−3 4𝑥
+ 2
𝑙
𝑙
−1 4𝑥
+ 2
𝑙
𝑙
ww
w .E
asy
En
⟹ 𝑘 = 𝐸𝐴
𝑙
−3
𝑙
0
4𝑥
−3
4𝑥
𝑙
−1
4𝑥
4
+ 𝑙2
+ 𝑙2
𝑙
−3
8𝑥
−1
4𝑥
𝑙
−1
4𝑥
4
+ 𝑙2
𝑙
−1
32𝑥
𝑙4
64𝑥 2
𝑙
4𝑥
𝑙4
16𝑥 2
𝑙
𝑙
4𝑥
𝑙
4𝑥
𝑙4
16𝑥 2
16𝑥
𝑙4
32𝑥 2
𝑙2
24𝑥
16𝑥
+ 𝑙3 + 𝑙3 −
𝑙4
𝑙2
9𝑥 12𝑥 2 12𝑥 2 16𝑥 3
−
−
+
𝑙2
2 𝑙3
2 𝑙3
3 𝑙4
2
2
3𝑥 12𝑥
4𝑥
16𝑥 3
−
−
+
𝑙2
2 𝑙3
2 𝑙3
3 𝑙4
−12 24𝑥 2 16𝑥 2 32𝑥 2
+
+
−
𝑙2
2 𝑙3
2 𝑙3
3 𝑙4
⟹ 𝑘 = 𝐸𝐴
− 3+
3
− 𝑙3 − 𝑙3 +
8𝑥
+ 𝑙3 + 𝑙3 −
eer
𝑙2
16
𝑙4
𝑙2
3𝑥 12𝑥 2 4𝑥 2 16𝑥 3
−
− 3+
𝑙2
2 𝑙3
2𝑙
3 𝑙4
2
2
𝑥 4𝑥
4𝑥
16𝑥 2
−
−
+
𝑙2 2 𝑙3 2 𝑙3
3 𝑙4
−4 8𝑥 2 16𝑥 2 32𝑥 2
+ 3+
−
𝑙2
2𝑙
2 𝑙3
3 𝑙4
9 6 6 16
− − +
𝑙 𝑙 𝑙 3𝑙
3 6 2 16
− − +
𝑙
𝑙 𝑙 3𝑙
−12 12 8 32
+
+ −
𝑙
𝑙
𝑙 3𝑙
+
2
+
3
−
3
ing
+ 𝑙3 + 𝑙3 −
32𝑥
8𝑥
− 𝑙2
8𝑥
− 𝑙2
𝑙
8𝑥
𝑑𝑥
.ne
t
− 𝑙3 − 𝑙3 +
𝑙4
−12 24𝑥 2 16𝑥 2 32𝑥 3
+
+
−
𝑙2
2 𝑙3
2 𝑙3
3 𝑙4
2
2
−4 8𝑥
16𝑥
32𝑥 2
+
+
−
𝑙2
2 𝑙3
2 𝑙3
3 𝑙4
16 32𝑥 2 32𝑥 64𝑥 2
−
− 3+
𝑙2
2 𝑙3
2𝑙
3 𝑙4
3 6 2 16
− − +
𝑙
𝑙 𝑙 3𝑙
1 2 4 16
− − +
𝑙 𝑙 𝑙
𝑙
−4 4 8 32
+ + −
𝑙2
𝑙 𝑙 3𝑙
𝑑𝑥
− 𝑙2
𝑙
𝑙4
32𝑥 2
𝑙
12𝑥
𝑙2
−4
4
− 𝑙2
𝑙
𝑙
16𝑥
𝑙
𝑙4
32𝑥 2
− 𝑙2
𝑙
𝑙
8𝑥
16𝑥 2
− 𝑙3 − 𝑙3 +
8𝑥
+ 𝑙2
𝑙
𝑙
−4
4𝑥
𝑙2
−12
4𝑥
𝑙
4
32𝑥 2
12𝑥
1
4
4
16𝑥
3
−
2
8𝑥
4𝑥
+ 𝑙2
24𝑥
16𝑥 2
+
3
4𝑥
𝑙
−1
−12
12𝑥
−
3
−3
gin
12𝑥
−
2
4𝑥
+ 𝑙2
+ 𝑙2
𝑙
+ 𝑙2
𝑙
A
− 𝑙2
𝑙
× E 𝑑𝑣
4𝑥
+ 𝑙2
9
3
= 𝐸𝐴
4𝑥
𝑙
−1
SC
𝑙
0
−3
+ 𝑙2
𝑙
+ 𝑙2
𝑙
⟹ 𝑘 = 𝐸𝐴
4𝑥
+ 𝑙2
D
−3
4 8𝑥
−
𝑙 𝑙2
𝑑𝑥
−12 12 8 32
+
+ −
𝑙
𝑙
𝑙 3𝑙
−4 4 8 32
+ + −
𝑙
𝑙 𝑙 3𝑙
16 16 16 64
−
−
+
𝑙
𝑙
𝑙
3𝑙
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7
1 −8
3𝑙 3𝑙 3𝑙
1
7 −8
⟹ 𝑘 = 𝐸𝐴
3𝑙 3𝑙 3𝑙
−8 −8 16
3𝑙 3𝑙 3𝑙
1 −8
𝐸𝐴 7
⟹ 𝑘 =
1
7 −8
3𝑙
−8 −8 16
LOAD VECTOR FOR ONE DIMENSIONAL QUADRATIC BAR ELEMENT:
`We know that, general force vector is,
𝐹 =
𝑙
𝑁𝑇
0
Xb
ww
w .E
asy
En
1−
𝑁1
Where, 𝑁 𝑇 = 𝑁2 =
𝑁3
3𝑥
2𝑥 2
−𝑥
𝑙2
2𝑥 2
+
𝑙
𝑙
4𝑥
𝑙
+
𝑙2
4𝑥 2
−
𝑙2
D
Due to self weight, Xb = ρ A 𝑑𝑥
A
Substitute the equation,
gin
SC
1−
𝐹 =
𝑙
0
3𝑥
2𝑥 2
−𝑥
𝑙2
2𝑥 2
+
𝑙
𝑙
4𝑥
𝑙
−
𝑙2
ρ A 𝑑𝑥
2𝑥 3
−𝑥 2
3 𝑙2
2𝑥 3
+
2𝑙
2𝑙
4𝑥 2
2𝑙
1−
= ρA
𝑙−
=ρA
eer
𝑙2
4𝑥 2
3𝑥 2
𝑥−
𝐹 =ρA
+
+
−
1
3 𝑙2
4𝑥 3
3 𝑙2
2𝑙 3
3 𝑙2
2 𝑙3
3 𝑙2
4𝑙 3
3 𝑙2
3𝑙
2𝑙
2
4𝑙
2
+
+
−
.ne
t
0
3𝑙 2
+
2𝑙
−𝑙 2
+
2𝑙
4 𝑙2
−
2𝑙
2
−𝑙
ing
3
2𝑙
3
4𝑙
3
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=ρA
0.166 𝑙
0.166 𝑙
0.166 𝑙
0.166
= ρ A 𝑙 0.166
0.166
𝐹 =ρA𝑙
1
6
1
6
2
3
𝐹1
𝐹2 = ρ A 𝑙
𝐹3
1
6
1
6
2
3
SC
A
D
ww
w .E
asy
En
gin
eer
ing
.ne
t
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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.
ww
w.E
asy
En
Cn – Governing differential equations is polynomial.
3. How do we specify two dimensional elements? (May/June 2014)
SC
A
D
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)
gin
Ten noded triangular elements are known as Quadratic strain element (QST).
eer
i ng
.ne
t
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 𝑑2 ∅ 1 𝑑2 ∅
+
+ 2𝜃 = 0
𝐺 𝑑𝑥 2 𝐺 𝑑𝑦 2
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)
ww
w.E
asy
En
𝑞1
0
𝑟1
0
𝑟1
𝑞1
𝑞2 0 𝑞3 0 𝑞 = 𝑦 − 𝑦 𝑞 = 𝑦 − 𝑦
0 𝑟2 0 𝑟3 𝑟1 = 𝑥 2 − 𝑥 3 𝑟2 = 𝑥 3 − 𝑥 1
3
2
2
1
3
𝑟2 𝑞2 𝑟3 𝑞3 1
𝑝1 = 𝑥2 𝑦3 − 𝑥3 𝑦2
𝑝2 = 𝑥3 𝑦1 − 𝑥1 𝑦3
𝑞3 = 𝑦1 − 𝑦2
𝑟3 = 𝑥2 − 𝑥1
𝑝3 = 𝑥1 𝑦2 − 𝑥2 𝑦1
SC
A
D
1
[B]=
2𝐴
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.
gin
eer
10. Evaluate the following area integrals for the three noded triangular element
𝛼 ! 𝛽! 𝛾!
𝑋 2𝐴
𝛼+ 𝛽+ 𝛾+2
i ng
𝑁𝑖 𝑁𝑗2 𝑁𝑘3 𝑑𝐴. (May/June 2013), (Nov/Dec 2012)
We know that,
𝛽 𝛾
𝐿𝛼𝑖 𝐿2 𝐿𝑘 𝑑𝐴 =
1! 2! 3!
𝑋 2𝐴
(1+ 2+ 3+2)!
Here, α = 1, β = 2, γ = 3
𝑁𝑖 𝑁𝑗2 𝑁𝑘3 𝑑𝐴 =
1𝑋2𝑋1𝑋3𝑋2𝑋1
𝑋 2𝐴
(8𝑋7𝑋6𝑋5𝑋4𝑋3𝑋2𝑋1)
=
1! 2! 3!
𝑋 2𝐴
(8)!
.ne
t
𝐴
=1680 𝑁𝑖 𝑁𝑗2 𝑁𝑘3 𝑑𝐴
11. Write the strain displacement relation for CST element.
𝑒𝑋
1 𝑞1
𝑒𝑌 =
0
2𝐴 𝑟
𝛾𝑥𝑦
1
0
𝑟1
𝑞1
𝑞2
0
𝑟2
0
𝑟2
𝑞2
𝑞3
0
𝑟3
0
𝑟3
𝑞3
𝑢1
𝑣1
𝑢2
𝑣2
𝑢3
𝑣3
67
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Downloaded From : www.EasyEngineering.net
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)
1
𝑁1 = 2𝐴 𝑝1 + 𝑞1 𝑥 + 𝑟1 𝑦
1
1
; 𝑁2 = 2𝐴 𝑝2 + 𝑞2 𝑥 + 𝑟2 𝑦 ; 𝑁3 = 2𝐴 𝑝3 + 𝑞3 𝑥 + 𝑟3 𝑦 ;
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=420C, T2=540C, T3= 560C, & T4=
460C. Also determine 3 points on the 500C contour line.
ww
w.E
asy
En
Given:
ϕi= 420C
m (5,5) 460C
k(8,5) 560C
SC
A
D
ϕj= 540C
ϕk=560C
ϕm=460C
2b=3
2a=2
b=3/2
gin
i (5,3) 460C
a=1
To find:
1. Temperature at point (2,1),ϕ
2. Three points on 500C.
eer
j(8,3) 540C
i ng
.ne
t
Formula used:
s
t s t
Ni= 1 1 1 1
2b 2a 3 2
t s
t
s
Nj= 1 1
2b 2a 3 2
st
Nk=
4ab
st
= st
4 3 1 6
2
s t s
t
Nm= 1 1
2a 2b 2 3
68
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Downloaded From : www.EasyEngineering.net
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.
2 1 1
Ni= 1 1 =
3 2 6
2 1 1
Nj= 1 =
3 2 3
ww
w.E
asy
En
1 2 1
Nm = 1 =
2 3 6
SC
A
D
2 1 1
Nk=
=
6 3
1
1
1
1
ϕ = 42 54 56 46 .
6
3
3
6
ϕ = 51.40C
0
gin
The x,y coordinates of 50 C contour line are
m (5,5) 460C
𝜙 𝑗 −𝜙 𝑖
=
eer
𝑥 𝑗 −𝑥
𝑥 𝑗 −𝑥 𝑖
𝑦 𝑗 −𝑦
i ng
=
𝑦 𝑗 −𝑦 𝑖
k(8,5) 560C
.ne
t
j(8,3) 540C
i
460C (5,3)
i,j
𝜙 𝑗 −𝜙
500C
54 50 8 x 3 y
54 42 8 5 3 3
(1)
(2)
(3)
Equating(1),(2)
equating (1),(3)
69
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Downloaded From : www.EasyEngineering.net
4 8 x
12
3
4 3 y
12
0
x 7cm
m,k
𝜙 𝑘 −𝜙
𝜙 𝑘 −𝜙 𝑚
y 3cm
=
𝑥 𝑘 −𝑥
𝑥 𝑘 −𝑥 𝑚
=
𝑦 𝑘 −𝑦
𝑦 𝑘 −𝑦𝑚
56 50 8 x 5 y
56 46 8 5 5 5
(1)
(2)
(3)
Equating (1),(2)
equating (1),(3)
6 8 x
10
3
;
x 6.2cm
;
6 5 y
10
0
ww
w.E
asy
En
y=4
[lower point yi=3, upper point ym=5]
SC
A
D
Third point
y 5cm
Centre line between the sides i,j&k,m
Local coordinates
t = y-yi= 4-3 = 1
gin
ϕ = Niϕi + Njϕj + Nkϕk + Nmϕm
50=
s 1
s 1
1 1 42 1 54
3 2
3 2
eer
i ng
1 s
s 1
56 1 46
2 3
6
.ne
t
s
s
1 21 93 9.33s 231
3
3
50=
21 73 9s 9.33s 23 7.66s
s 1.63cm
(6.2,5)
s x xj
1.63 5 x
(6.7,4)
x 6.7cm
y 4cm
500C
(7,3)
70
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Downloaded From : www.EasyEngineering.net
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.
ww
w.E
asy
En
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.
SC
A
D
Given:
Nodal Displacements:
U1=2.0mm;
U2=0.5mm;
U3=3.0mm;
v1=1.0mm;
gin
X1= 20mm
y1=30mm
X2= 80mm
y2=30mm
X3=50mm
y3=120mm
v2=0.0mm;
v3=1.0mm.
eer
i ng
.ne
t
Young’s modulus, E= 210 GPa =210x109 Pa
= 210x109N/m2 = 210x103 N/mm2
=2.1x 105 N/mm2
Poisson’s ratio,
Thickness,
ν=0.25
t= 10mm
71
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Downloaded From : www.EasyEngineering.net
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 =
x y
2
ww
w.E
asy
En
Minimum normal stress, σmin = σ2 =
Solution: we know that
x y
2 xy
2
2
2 xy
x y
1 x1
1
Area of the element, A= 1 x 2
2
1 x3
=
2
2
SC
A
D
principle angle, tan 2θp=
x y
x y
2 xy
2
gin
y1
1 20 30
1
y 2 1 80 30
2
1 50 120
y3
eer
i ng
1
x[ 1x(80x120-50x30)-20(120-30)+30(50-80)]
2
1
=
x [8100-1800-900]
2
A=2700 mm2
.ne
t
….. (1)
We know that,
Strain Displacement matrix,
q1
1
0
[B]=
2A
r1
Where,
0 q 2 0 q3 0
r1 0 r 2 0 r 3
q1 r 2 q 2 r 3 q3
…… (2)
q1 = y2 – y3 = 30-120 = -90
q2= y3 – y1 = 120- 30 = 90
q3= y1- y2 = 30 – 30 = 0
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Downloaded From : www.EasyEngineering.net
r1= x3- x2 = 50-80 = -30
r2= x1- x3 = 20-50 = -30
r3= x2- x1 = 80-20 = 60
Substitute the above values in equation no. (2),
0
90
0
90
1
0
30
0
30
[B] =
2A
30 90 30 90
0
60
0
0
0
60
Substitute Area, A value,
0
90
0
90
1
0
30
0
30
[B] =
2 2700
30 90 30 90
60
3
0
3 0
30
0 1 0 1
=
2 2700
1 3 1 3
0
2
0
ww
w.E
asy
En
0
0
SC
A
D
2
0
60
0
0
0
3
0
3 0
[B] = 5.555 x 10 0 1 0 1
1 3 1 3
-3
We know that
0
0
0
2
0
gin
2
eer
Stress strain relationship matrix [D] for plane stress problem is,
[D]=
1 v
0
E
v 1
0
1 v2
1 v
0 0
2
………(3)
i ng
.ne
t
1
0.25
0
2.1 10
0.25
1
0
=
1 (0.25) 2
1 0.25
0
0
2
5
0.25
0
1
2.1 10 5
=
0.25
1
0
0.9375
0
0
0.375
73
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Downloaded From : www.EasyEngineering.net
4 1 0
2.1x10 5 0.25
=
1 4 0
0.9375
0 0 1.5
4 1 0
= 56 10 1 4 0
0 0 1.5
…. (4)
3
3
0
3 0
4 1 0
-3
[D] [B] = 56 10 1 4 0 x 5.555 x 10 0 1 0 1
1 3 1 3
0 0 1.5
3
3
0
4 1 0 3 0
= 311.08 x 1 4 0 0 1 0 1
0 0 1.5 1 3 1 3
ww
w.E
asy
En
Stress { σ} = [D] [B] {u}
2
0 1 0
040
000
000
0 0 4.5
0 0 1.5
0 0 4.5
003
1
4
0
0
4.5
3
2
8
0
gin
u1
v
1
u 2
= [D] [B]
v2
u 3
v3
1
12
12
4
3
= 311.08 3
1.5 4.5 1.5
1
4
0
0
4.5
3
0
2
0
0
2
0
0
0
12 0 0
3 0 0
1
12
12
4
3
=311.08 x 3
1.5 4.5 1.5
We know that
2
0 1 0
040
SC
A
D
12 0 0
= 311.08 3 0 0
0 0 1.5
0
0
eer
i ng
020
0 8 0
0 0 0
.ne
t
2
1
2
0.5
8 X
0
0
3
1
74
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Downloaded From : www.EasyEngineering.net
(12 2) (1 1) (12 0.5) (1 0) (0 3) (2 1)
= 311.08
(3 2) (4 1) (3 0.5) (4 0) (0 3) (8 1)
(1.5 2) (4.5 1) (1.5 0.5) (4.5 0) (3 3) (0 1)
17
{σ} =311.08 0.5
0.75
x 5288.36
y = 155.54
233.31
z
Normal stress, σx = 5288.36 N/mm2
Normal stress, σy = 155.54 N/mm2
Shear stress, xy = 233.31 N/mm2
ww
w.E
asy
En
We know that,
x y
2 xy ….. (7)
2
2
SC
A
D
Maximum normal stress, σmax = σ1 =
x y
2
gin
= 5288.36 155.54 5288.36 155.54 (233.31) 2
2
2
σ1 = -144.956 N/mm2
Minimum normal stress, σmin = σ2 =
x y
2
2
eer
2
5288.36 155.54
5288.36 155.54
2
(233.31)
2
2
2
=
i ng
x y
2 xy …… (8)
2
.ne
t
σ2 = -5298.9N/mm2
We know that principle angle, tan 2θp=
2 xy
x y
2 xy
tan 2θp = tan-1
x y
2 233.31
= tan-1
5288.36 155.54
2θp=-5.1940
75
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Downloaded From : www.EasyEngineering.net
θp = -2.590
Result: 1. Element stress
a) Normal stress, σx= 5288.36 N/mm2
b) Normal stress, σy= 155.54 N/mm2
c) Shear stress, xy = 233.31 N/mm2
d) Maximum normal stress, σ1= -144.956 N/mm2
e) Minimum normal stress, σ2= -5298.9N/mm2
2. Principle angle,θp= -2.590
3.
Calculate the element stiffness matrix and the temperature force vector for the plane stress
element as shown in figure. The element experiences a 20°C increase in temperature, Assume
coefficient of thermal expansion is 6 x 10-6/°C. Take Young’s modulus E = 2 X
105N/mm2,possion ratio
v=0.25,Thickness t=
5mm.
Given data:
X1 = 0;
Y1 = 0
X2 = 2;
Y2 = 0
X3 = 1;
Y3 = 3
E = 2 X 105N/mm2
SC
A
D
ww
w.E
asy
En
gin
V = 0.25
t= 5mm
eer
i ng
.ne
t
ΔT = 10°C
α = 6 x 10-6/°C
To find: 1. Element stiffness matrix [K]
2. The temperature force vector [F]
Formula used:
Stiffness matrix [K] = [B] T [D] A t
Temperature force vector, {F} = [B] T [D] {eo} A t
Solution:
We know that, stiffness matrix [K] = [B] T [D] A t
Where A = Area of the element
76
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Downloaded From : www.EasyEngineering.net
1 X 1 Y1
1
1 X 2 Y2 =
2
1 X 3 Y3
1
=
2
=
1 0 0
1 2 0
1 1 3
1
[1(6-0)-0+0]; A= 3 mm2.
2
q1
1
0
Strain –Displacement matrix [B] =
2A
r1
0
r1
q2
0
0
r2
q3
0
q1
r2
q2
r3
Where, q1 = y2 – y3 = 0-3 = -3;
r1 = x3 – x2 = 1-2 = -1
q2 = y3 – y1 = 3-0 = 3;
r2 = x1 – x3 = 0-1 = -1
q3 = y1 – y2 = 0-0 = 0;
r3 = x2 – x1 = 2-0 = 2
0
r3
q3
Substitute the above values in [B] matrix equation
ww
w.E
asy
En
3
0
3 0
0 1 0 1
1 3 1 3
1
Substitute “A” value, [B] =
23
[B] = 0.1667
0
2
0
0
0
2
SC
A
D
1
[B] =
2A
3
0
3 0
0 1 0 1
1 3 1 3
3
0
3 0
0 1 0 1
1 3 1 3
0
0
gin
0
2
0
2
0
2
0
0
0
2
eer
i ng
.ne
t
We know that, stress-strain relationship matrix [D] for plane stress problem is
E
[D] = 1V 2
2 x105 0.25
=
0.9375
We know [B] = 0.1667
1 v
0
v 1
0
1 v
0 0
2
4 1 0
1 4 0
0 0 1.5
2105
= 10.252
1
0.25
0
0.25
1
0
1 0.25
0
0
2
; [D] = 53.33 x 10
3
0
3 0
0 1 0 1
1 3 1 3
0
0
2
3
4 1 0
1 4 0
0 0 1.5
0
2
0
77
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Downloaded From : www.EasyEngineering.net
[B]T = 0.1667
T
[B] [D] = 0.1667
3 0 1
0 1 3
3
0 1
0 1 3
0
0
2
2
0
0
3 0 1
0 1 3
3
0 1
0 1 3
0
0
2
2
0
0
x 53.33 x 10
12
1
12
1
0
2
3
4
3
4
3
4
3
4
1.5
4.5
1.5
4.5
3
0
3
4 1 0
1 4 0
0 0 1.5
1.5
4.5
1.5
4.5
3
0
ww
w.E
asy
En
[B]T [D] = 8.890 X 103
0
8
SC
A
D
= 0.1667 X 53.33 X 103
12
1
12
1
0
2
0
8
gin
eer
i ng
12
1
12
1
0
2
3
4
3
4
37.5
7.5
34.5
[B]T [D] [B] = 1.482 X 103
1.5
3
6
7.5
17.5
1.5
9.5
34.5
1.5
37.5
7.5
1.5
9.5
7.5
17.5
3
9
3
9
9
8
3
6
9
8
6
0
T
3
[B] [D] [B] == 8.890 X 10
0
8
.ne
t
1.5
4.5
3
0
3 0
1.5
x 0.1667 0 1 0 1
4.5
1 3 1 3
3
0
0
0
2
0
2
0
6
8
6
8
0
16
Substitute [B]T [D] [B] and A, t values in stiffness matrix
78
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Downloaded From : www.EasyEngineering.net
Stiffness matrix [K] = [B] T [D] A t
37.5
7.5
34.5
1.5
3
6
7.5
17.5
1.5
9.5
34.5
1.5
37.5
7.5
1.5
9.5
7.5
17.5
3
9
3
9
9
8
3
6
9
8
6
0
37.5
7.5
34.5
[K] = 22.23 X 103
1.5
3
6
7.5
17.5
1.5
9.5
34.5
1.5
37.5
7.5
1.5
9.5
7.5
17.5
3
9
3
9
9
8
3
6
9
8
6
0
[K] = 1.482 X 103
ww
w.E
asy
En
6
8
6
x3x5
8
0
16
6
8
6
8
0
16
We know that, for plane stress problem, Initial strain {eo} =
SC
A
D
6 x10 6 x10
60
-6
6
{eo} = 6 x10 x10 = 1 x 10 60
60
0
gin
We know that, Temperature force vector, {F} = [B] T [D] {eo} A t
{F} = 8.890 x 103
12
1
12
1
0
2
3
4
3
4
0
8
1.5
4.5
1.5
4.5
3
0
x 1 x 10-6
eer
60
60
60
i ng
xAxt
.ne
t
Substitute “A” and “t” values
3
= 8.890 x 10
x 1 x 10
-6
x3x5
12
1
12
1
0
2
3
4
3
4
0
8
1.5
4.5
1.5
x
4.5
3
0
60
60
60
79
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Downloaded From : www.EasyEngineering.net
(12 x60) (3 x60) 0
(1x60) (4 x60) 0
(12 x60) (3 x60) 0
= 0.1335 (1x60) (4 x60) 0
000
(2 x60) (8 x60) 0
900
300
900
300
0
600
= 0.1335
120.15
40.05
120.15
{F} =
40.05
0
80.10
ww
w.E
asy
En
Result:
7.5
17.5
1.5
9.5
34.5
1.5
37.5
7.5
SC
A
D
37.5
7.5
34.5
1.5
3
6
Stiffness matrix [K] = 22.23 X 103
Temperature force vector, {F} =
4.
gin
9
8
120.15
40.05
120.15
40.05
0
80.10
3
6
1.5
9.5
7.5
17.5
3
9
3
9
9
8
6
0
eer
i ng
6
8
6
8
0
16
.ne
t
A thin plate is subjected to surface traction as shown in figure. Calculate the global stiffness
matrix.
fig (i)
Take Young’s modulus E = 2 X 105N/mm2, possion ratio v=0.30, Thickness t=25mm.Assume plane
stress condition.
80
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Downloaded From : www.EasyEngineering.net
Given data:
E = 2 X 105N/mm2;
Breath b =250mm
V = 0.25;
length l =500mm
t= 25mm;
tensile surface traction T= 0.4 N/mm2
1
“T” is converted into nodal force F = 2 T A = ½ x T x (b x t)
1
= 2 x 0.4 x 250 x 25
ww
w.E
asy
En
F = 1250 N
Fig (ii) Discretized plate
To find:
Formula used:
SC
A
D
Global stiffness matrix [K].
Global Stiffness matrix [K]1 = [B] T [D][B] A t
Solution:
gin
eer
i ng
.ne
t
Fig (iii)
For element (1) - Nodal displacements are u1, v1, u3, v3 and u4 v4
Fig (iv)
Take node 1 as origin;
81
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Downloaded From : www.EasyEngineering.net
For node 1: X1= 0, Y1=0; For node 3: X2=500, Y2=250; For node 4: X3= 0, Y3=250;
We know that, stiffness matrix [K]1 = [B] T [D][B] A t
1 X 1 Y1
1 X 2 Y2
1
Where A =Area of the triangular element =
2
=
1 X3
1 0
0
1 500 250
1
=
2
Y3
1
0
250
1
x 1 (500x250 -0) = 62500mm2
2
A = 62.5 X 103 mm2
1
Strain –Displacement matrix [B] =
2A
q1
0
0
r1
q2
0
0
r2
q3
0
0
r3
r1
q1
r2
q2
r3
q3
ww
w.E
asy
En
Where, q1 = y2 – y3 = 250-250 = 0
r1 = x3 – x2 = 0-500 = -500
q2 = y3 – y1 = 250-0 = 250
r2 = x1 – x3 = 0-0 = 0
q3 = y1 – y2 = 0-250 =-250
r3 = x2 – x1 = 500-0 = 500
1
[B] =
2A
0
0
500
SC
A
D
Substitute the above values in [B] matrix equation
0
500
250
0
0
0
0
0
250
250
0
0
500
500
250
gin
0
0
1
Substitute “A” value, [B] =
0
500
2 62.5 103
500
0
[B] =
250
2 62.5 103
0 0
0 2
2 0
1
0
0
0
1
0
0
1
2
eer
250
0
0
0
250
0
0
250
500
0
2
1
i ng
0
500
250
.ne
t
We know that, stress-strain relationship matrix [D] for plane stress problem is
E
[D] = 1V 2
2105
= 0.91
1 v
0
v 1
0
1 v
0 0
2
21 0
5
= 1 ( 0.3) 2
1 0.3
0
0.3 1
0
1 0.3
0
0
2
0
1 0.3
0.3 1
0
0
0 0.35
82
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Downloaded From : www.EasyEngineering.net
[D][B] =
= 439.56
0
1 0.3
0.3 1
0
0
0 0.35
2 x105
0.91
0
0
0.7
0.6
2
1
0.3
0
0
1
0.3
0
0
0.35
0.7
250
We know that, [B] =
2 62.5 103
0 0
0 2
2 0
1
0
0
0
1
0
0
1
2
0
0
1
0
0
1
2
0
2
1
0
2
1
0 2
0
0 2 0
1
0
0
0
1
0
1 0
2
2 1
0
SC
A
D
0 2
0
0 2 0
0
1
0
0
[B]T [D] [B] = 2 x 10-3 x 439.56
x 0
0
1
0
0.7
1 0
2
2 1
0
0.6
2
1
0.3
0
0
1
0.3
0
0
0.35
0.7
gin
= 0.8791
1
0
0.6
2
0.35
ww
w.E
asy
En
[B]T = 2 x 10-3
0 0
0 2
2 0
250
x
2 x62.5 x10 3
1.4
0
0
0.7
1.4
0.7
0
4
0.6
0
0
0.6
1
0
0.7
0
0
0.35
1.4
0.6
1
0.7
0.6
4
1
0.6
0.7
0.35
2.4
1.3
eer
0.7
4
0.6
0.35
1.3
4.35
i ng
0.6
2
0.35
.ne
t
Substitute [B]T [D] [B] and A, t values in stiffness matrix
Stiffness matrix [K]1 = [B] T [D] A t
1.4
0
0
Stiffness matrix [K]1 =0.8791
0.7
1.4
0.7
0
4
0.6
0
0
0.6
1
0
0.7
0
0
0.35
1.4
0.6
1
0.7
0.6
4
1
0.6
0.7
0.35
2.4
1.3
0.7
4
0.6
3
x 6.25x 10 x25
0.35
1.3
4.35
83
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Downloaded From : www.EasyEngineering.net
1.4
0
0
0.7
1.4
0.7
=1373.59 x 103
U1
1923.026
0
0
[K]1 =1x103
961.513
1923.026
961.513
0
4
0.6
0
0
0.6
1
0
0.7
0
0
0.35
1.4
0.6
1
0.7
0.6
4
1
0.6
0.7
0.35
2.4
1.3
v1
u3
v3
0
5494.36
824.154
0
0
824.154
1373.59
0
824.154
5494.36
1373.59
824.154
u4
v4
961.513
0
0
480.7565
1923.026
961.513 u1
824.154
5494.36 v1
1373.59
824.154 u3
961.513
480.7565 v3
961.513
3296.616 1785.667 u 4
480.7565 1785.667 5975.1165 v4
SC
A
D
ww
w.E
asy
En
For element (2):
0.7
4
0.6
0.35
1.3
4.35
gin
eer
i ng
fig(v)
Nodal displacements are u1, v1, u3, v3 and u4 v4
.ne
t
Take node 1 as origin; For node 1: X1= 0, Y1=0; For node2: X2=500, Y2=0; For node 3: X3= 500, Y3=250;
We know that, stiffness matrix [K]2 = [B] T [D][B] A t
1
Where A =Area of the triangular element =
2
=
1 X 1 Y1
1 X 2 Y2
1 X3
Y3
1
=
2
1 0
1 500
0
0
1 500 250
1
x 1 (500x250 -0) = 62500mm2
2
A = 62.5 x 103 mm2
84
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Downloaded From : www.EasyEngineering.net
q1
1
0
Strain –Displacement matrix [B] =
2A
r1
0
r1
q2
0
0
r2
q3
0
q1
r2
q2
r3
0
r3
q3
Where, q1 = y2 – y3 = 0-250 = -250;
r1 = x3 – x2 = 500-500 = 0
q2 = y3 – y1 = 250-0 = 250;
r2 = x1 – x3 = 0-500 = -500
q3 = y1 – y2 = 0-0 =0;
r3 = x2 – x1 = 500-0 = 500
Substitute the above values in [B] matrix equation
1
[B] =
2A
0
250
0
250
0
0
0
500
0
250 500 250
ww
w.E
asy
En
0
500
0
0
0
500
1
250
0
Substitute “A” value, [B] =
2 62.5 103
0
0
0
1
0
0
2
0
0
1 2
1
2
0
2
0
E
[D] = 1V 2
1
2 x105
= 0.91 0.3
0
5
2 x10
[D][B] = 0.91
= 439.56
1
0.3
0
SC
A
D
We know that, stress-strain relationship matrix [D] for plane stress problem is
1 v
0
v 1
0
1 v
0 0
2
0.3
1
0
5
2 x10
= 1 ( 0.3) 2
gin
0
0
0.35
0
1 0.3
250
0.3 1
x
0
2 x62.5 x10 3
0
0 0.35
0
0
1
0.3
0.6
2
0
0
0.35
0.7
0.35
0.7
250
We know that, [B] =
2 62.5 103
1 0.3
0
0.3 1
0
1 0.3
0
0
2
eer
i ng
1 0 1 0
0 0 0 2
0 1 2 1
0
0
2
.ne
t
0
2
0
0.6
2
0
1 0 1 0
0 0 0 2
0 1 2 1
0
0
2
0
2
0
85
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Downloaded From : www.EasyEngineering.net
[B]T = 2 x 10-3
0
1 0
0
0 1
1
0 2
0 2 1
0
0
2
2
0
0
0
1 0
0
0 1
1
1
0 2
T
-3
[B] [D] [B] = 2 x 10 x 439.56
0.3
0 2 1 0
0
0
2
2
0
0
1
0
1
0.6
0
0.6
0
0.35
0.7
0.35
1
0.7
2.4
1.3
0.6
0.35
1.3
4.35
0
0.7
1.4
0.7
0.7
0
1.4
0.6
0.7
4
1.4
0
= 0.8791
1
0.3
0.6
2
0
0
0.35
0.7
0.35
0.7
0.6
2
0
0.6
0
0.6
4
0
4
SC
A
D
ww
w.E
asy
En
0
0
Substitute [B]T [D] [B] and A, t values in stiffness matrix
gin
Stiffness matrix [K]1 = [B] T [D] A t
Stiffness matrix [K]1 =0.8791
1
0
1
0.6
0
0.6
=1373.59 x 103
u1
eer
i ng
1
0
1
0.6
0
0.6
0
0.35
0.7
0.35
1
0.7
2.4
1.3
0.6
0.35
1.3
4.35
0
0.7
1.4
0.7
0.7
0
1.4
0.6
0.7
4
1.4
0
0
0.35
0.7
0.35
1
0.7
2.4
1.3
0.6
0.35
1.3
4.35
0
0.7
1.4
0.7
0.7
0
1.4
0.6
0.7
4
1.4
0
0.6
0
0.6
4
0
4
v1
u3
v3
0.6
0
0.6
3
x 6.25x 10 x25
4
0
4
u4
.ne
t
v4
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Downloaded From : www.EasyEngineering.net
0
1373.59
824.154
0
480.7565
961.513
480.7565 961.513
961.513
3296.616 1785.667 1923.026
480.7565 1785.667 5975.1165
961.513
1373.59
0
1373.59
[K]2 =1x103
824.154
0
824.154
961.513
0
1923.026
824.154
961.513
5494.36
824.154 u1
v
0
1
824.154 u3
5494.36 v3
u4
0
5494.36 v4
1923.026
0
Global stiffness matrix [K]:
Assemble the stiffness matrix equations [K]1 & [K]2 = 1 x 103 x
u1
v1
1923.026+
0+0
u2
v2
-1373.59
824.154
u3
v3
u4
v4
-961.513+
-1923.026
961.513
u1
0+0
824.154
-5494.36
v1
-1923.026
824.154
0
0
u2
0+0
1373.59
0+0
-824.154
5494.36+
961.513
-480.7565
-824.154+
ww
w.E
asy
En
480.7565
-961.513
-1373.59
961.513
3296.616
824.154
-480.7565
-1785.667 5975.116
961.513
-5494.36
0
0
v2
0+0
-824.154+
0+
1373.59+
0+0
-1373.59+
824.154+
u3
-961.513
-1923.026 961.513
1923.026
0
0
0+0
0+
0+
0+0
961.513+
-480.7565
824.154
-5494.36
0
+0
961.513
-1785.667
u4
5975.116
v4
-824.154
-1923.026
824.154
0
961.513
-5494.36
0
0+
SC
A
D
-961.513+
-1785.667
gin
480.7565
+
5494.36
0
-1373.59
0
824.154
eer
3296.616
-480.7565
-1785.667
u3
v3
u4
Global stiffness matrix [K] = 1 x 103 x
u1
v1
u2
v2
v3
i ng
.ne
t
v4
3296.616
0
-1373.59
824.154
0
-1785.667
-1923.026
961.513
u1
0
5975.1165
961.513
-480.7565
-1785.667
0
824.154
-5494.36
v1
-1373.59
961.513
9296.616
-1785.667
-1923.026
824.154
0
0
u2
824.154
-480.7565
-1785.667 5975.116
5
961.513
-5494.36
0
0
v2
0
-1785.667
-1923.026 961.513
9296.616
0
-1373.59
824.154
u3
-1785.667
0
824.154
-5494.36
0
5975.116
5
961.513
-480.7565
v3
-1923.026
824.154
0
0
-1373.59
961.513
3296.616
-1785.667
u4
961.513
-5494.36
0
0
824.154
-480.7565
-1785.667
5975.116
v4
87
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Downloaded From : www.EasyEngineering.net
5. Derive the Shape function for the six noded triangular elements.
Fig. A six noded triangular element
Consider a six-noded triangular element is shown in figure. It belongs to the serendipity family of
elements. It consists of six nodes, which are located on the boundary.
We know that, shape function N1=1 at node 1 and 0 at all other nodes. The natural coordinates of the
nodes are indicated in the figure. By following our earlier procedure, the shape functions can be obtained as,
ww
w.E
asy
En
At node 1: (Coordinates L1 =1, L2 =0, L3 =0)
Shape function N1=1 at node 1
N1=0 at all other nodes,
Substitute L1= 1 and N1 =1
1
); where C is constant.
2
SC
A
D
N1 has to be in the form of
N1 = C L1 (L1 -
gin
N1 = C x 1 (1 C=2
1
)
2
Substitute C value in the above equation
1
N1 = 2 L1 (L1 - )
2
eer
N1 = L1 (2L1 -1)
At node 2: (Coordinates L1 =0, L2 =1, L3 =0)
i ng
.ne
t
Shape function N2=1 at node 2
N2=0 at all other nodes,
N2 has to be in the form of
N2 = C L2 (L2 -
Substitute L2= 1 and N2 =1
1
); where C is constant.
2
N2 = C x 1 (1 -
1
)
2
C=2
Substitute C value in the above equation
N2= 2 L2 (L2 -
1
)
2
N2 = L2 (2L2 -1)
88
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Downloaded From : www.EasyEngineering.net
At node 3: (Coordinates L1 =0, L2 =0, L3 =1)
Shape function N3=1 at node 3
N3=0 at all other nodes,
N3 has to be in the form of
N3 = C L3 (L3 -
Substitute L3= 1 and N3 =1
1
); where C is constant.
2
N3 = C x 1 (1 -
1
)
2
C=2
Substitute C value in the above equation
N3= 2 L3 (L3 -
1
)
2
N3 = L3 (2L3 -1)
Now, we define N4, N5 and N6 at the mid-points.
ww
w.E
asy
En
At node 4: (Coordinates L1 =
1
1
, L2 = , L3 =0)
2
2
SC
A
D
Shape function N4=1 at node 4
N4=0 at all other nodes,
N4 has to be in the form of
Substitute L4=
1
1
and L2 =
2
2
N4 = C L1L2;
where C is constant.
N4 = C x
C=4
gin
1 1
x
2 2
Substitute C value in the above equation
eer
N4 = 4L1 L2
At node 5: (Coordinates L1 =0, L2 =
1
1
, L3 = )
2
2
Shape function N5=1 at node 5
i ng
.ne
t
N5=0 at all other nodes,
N5 has to be in the form of
Substitute L2=
N5 = C L2L3;
1
1
and L3 =
2
2
N5= C x
where C is constant.
1 1
x
2 2
C=4
Substitute C value in the above equation
N5 = 4L2 L3
At node 6: (Coordinates L1 =
1
1
, L2 =0, L3 = )
2
2
Shape function N6 =1 at node 6
N6=0 at all other nodes,
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Downloaded From : www.EasyEngineering.net
N6 has to be in the form of
Substitute L1=
N6 = C L1L3;
1
1
and L3 =
2
2
N6= C x
where C is constant.
1 1
x
2 2
C=4
Substitute C value in the above equation
N6 = 4L1 L3
Shape functions are,
N1 = L1 (2L1 -1)
N2 = L2 (2L2 -1)
N3 = L3 (2L3 -1)
N4 = 4L1 L2
N5 = 4L2 L3
ww
w.E
asy
En
N6 = 4L1 L3
6. Derive the Shape function for the Constant Strain Triangular element (CST).
SC
A
D
We begin this section with the development of the shape function for a basic two dimensional
finite element, called constant stain triangular element (CST).
We consider the CST element because its derivation is the simplest among the available two
dimensional elements.
gin
eer
i ng
.ne
t
Fig. Three noded CST elements.
Consider a typical CST element with nodes 1, 2 and 3 as shown in fig. let the nodal
displacements to be u1, u2, u3, v1, v2, v3.
u1
u
2
u3
Displacement u
v1
v2
v3
90
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Downloaded From : www.EasyEngineering.net
Since the CST element has gat two degrees of freedom at each node (u, v), the total degree of
freedom are 6. Hence it has 6 generalised coordinates.
Let,
u a1 a2 x a3 y
… (3.1)
v a 4 a5 x a 6 y
… (3.2)
Where a1, a2, a3, a4, a5, and a6 are globalised coordinates
u1 a1 a2 x1 a3 y1
u 2 a1 a2 x2 a3 y 2
u3 a1 a2 x3 a3 y3
Write the above equations in matrix form,
u1 1 x1
u 2 1 x 2
u 1 x
3
3
y1 a1
y 2 a 2
y 3 a3
a1 1 x1
a 2 1 x 2
a 1 x
3
3
y1
y 2
y 3
1 x1
Let D = 1 x 2
1 x3
y1
y 2
y 3
We know, D-1 =
CT
D
ww
w.E
asy
En
u1
u 2
u
3
SC
A
D
1
gin
Find the co-factor of matrix D.
C11 =
x2
y2
x3
y3
C12 =
1 y2
C13 =
1 x2
C21 =
x1
y1
x3
y3
1 y3
1 x3
eer
… (3.3)
i ng
.ne
t
… (3.4)
( x 2 y 3 x3 y 2 )
( y 3 y 2 ) y 2 y 3
( x3 x 2 )
( x1 y3 x3 y1 ) x3 y1 x1 y3
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Downloaded From : www.EasyEngineering.net
C22 =
1 y1
C23 =
1 x1
1 y3
1 x3
( y3 y1 )
( x3 x1 ) x1 x3
x1
x2
y1
x1 y 2 x2 y1
y2
C32 =
1 y1
( y 2 y1 ) y1 y 2
C33 =
1 x1
( x2 x1 )
1 x2
C31 =
1 y2
ww
w.E
asy
En
x 2 y 3 x3 y 2 y 2 y 3 x3 x 2
x3 y1 x1 y3 y3 y1 x1 x3
C = x1 y 2 x 2 y1 y1 y 2 x 2 x1
SC
A
D
x2 y3 x3 y 2 x3 y1 x1 y3 x1 y 2 x2 y1
y 2 y3
y3 y1
y1 y 2
T
C = x3 x 2
x1 x3
x2 x1
1 x1
y1
We know that, D= 1 x 2
y2
1 x3
y3
gin
D = 1 ( x2 y3 x3 y 2 ) x1 y3 y 2 y1 x3 x2
Substitute CT and D value in equation (3.4)
D-1 =
1
( x2 y3 x3 y 2 ) x1 y3 y 2 y1 x3 x2
eer
…(3.5)
i ng
.ne
t
…(3.6)
x2 y3 x3 y 2 x3 y1 x1 y3 x1 y2 x2 y1
y 2 y3
y3 y1
y1 y2
x3 x 2
x1 x3
x2 x1
Substitute D-1 value in equation (3.3)
a1 1 x1
a 2 1 x 2
a 1 x
3
3
y1
y 2
y 3
1
u1
u 2
u
3
92
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a1
x2 y3 x3 y2
1
a
y2 y3
2 =
a ( x2 y3 x3 y2 ) x1 y3 y 2 y1 x3 x2
x3 x2
3
x3 y1 x1 y3 x1 y2 x2 y1 u1
y3 y1
y1 y2 u2
x1 x3
x2 x1 u3 ..(3.7)
The area of the triangle can be expressed as a function of the x,y coordinate of the nodes 1,2 and 3.
1 x1
1
1 x2
2
A= 1 x3
A
y1
y 2
y 3
1
( x2 y3 x3 y 2 ) x1 y3 y 2 y1 x3 x2
2
2 A ( x2 y3 x3 y 2 ) x1 y3 y 2 y1 x3 x2
... (3.8)
ww
w.E
asy
En
Substitute 2A value in equation (3.7),
a1
p1
1
q1
a 2 =
2A
a3
r1
x3 y1 x1 y3 x1 y2 x2 y1 u1
y3 y1
y1 y2 u2
u
x1 x3
x2 x1
3
SC
A
D
a1
x2 y3 x3 y2
1
y 2 y3
a 2 =
2A
a3
x3 x2
p 2 p3 u1
q 2 q3 u 2
r 2 r 3 u3
gin
eer
...(3.9)
p1 x2 y3 x3 y2 p 2 x3 y1 x1 y3 p3 x1 y2 x2 y1
q1 y2 y3
Where,
r1 x3 x2
q 2 y3 y1
q3 y1 y2
r 2 x1 x3
r 3 x2 x1
From eq (3.1) we know that
u= 1 x
i ng
.ne
t
a1
y a 2
a3
a1
Sub a 2 values from Eq (3.10)
a3
u= 1 x
p1
1
q1
y
2A
r1
p 2 p3 u1
q 2 q3 u 2
r 2 r 3 u3
93
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Downloaded From : www.EasyEngineering.net
1
1 x
2A
p1 p2 p3 u1
y q1 q 2 q3 u 2
r1 r 2 r 3 u3
1
p1 q1 x r1 y
2A
u=
u!
p3 q3 x r3 y u2
u
3
p2 q2 x r2 y
p1 q1 x r1 y
2A
u!
p3 q3 x r3 y
u2
2A
u3
p2 q2 x r2 y
2A
ww
w.E
asy
En
The above equation is in the form of
u=
V =
Similarly,
Where shape function , N1=
N2=
N3=
N2
u1
N 3 u 2
u
3
SC
A
D
u = N1
N1
N2
v1
N 3 v2
v
3
p1 q1 x r1 y
2A
p1 q1 x r1 y
2A
p3 q3 x r3 y
2A
gin
… (3.11)
… (3.12)
eer
i ng
.ne
t
Assembling the equations (3.11) and (3.12) in matrix form
Displacement matrix u =
u ( x, y ) N1
v ( x, y ) 0
0
N1
N2
0
0
N2
N3
N3
u1
v
1
0
u
2
0 v2
u3
v3
… (313)
94
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Downloaded From : www.EasyEngineering.net
UNIT IV – TWO DIMENSIONAL VECTOR VARIABLE PROBLEMS
PART - A
1. What is meant by axisymmetric field problem? Give example.(April/May 2010)
In some of the three dimensional solids like flywheel, turbine, discs etc, the material is
symmetric with respect to their axes. Hence the stress developed is also symmetric. Such solids are
known as axisymmetric solids. Due to this condition, three dimensional solids can be treated as two
dimensional elements.
2. List the required conditions for a problem assumed to be axisymmetric. (April/May 2011)
The condition to be axisymmetric is as follows:
Problem domain must be symmetric about the axis of revolution.
All boundary conditions must be symmetric about the axis of revolution.
All loading conditions must be symmetric about the axis of revolution.
3. What is Plane stress and Plane strain condition? (April/May 2015), (May/June 2013)
ww
Plane stress - A state of plane stress is said to exist when the elastic body is very thin and there is
no load applied in the coordinate direction parallel to the thickness.
w.E
asy
En
gi
Example: A ring press-fitted on a shaft in a plane stress problem. In plane stress problem σz,
τyz, τzx are zero.
SC
A
D
Plane strain – A state of plane strain is said to exist when the strain at the plane perpendicular
to the plane of application of load is constant.
4. What are the forces acting on shell elements? Give its applications
The two forces in which the shell element is subjected to are:
Bending moments
Membrane forces
nee
rin
g
Shell elements can be employed in the analysis of the following structures,
Example:
Sea shell, egg shell (the wonder of the nature);
Containers, pipes, tanks;
Car bodies;
Roofs, buildings (the Superdome), etc.
.ne
t
5. Write the constitutive relations for axisymmetric problems.
6. Define body force.
A body force is distributed force acting on every elemental volume of the body.
Unit: force per unit volume.
7. Write the governing equation for 2D bending of plates.
95
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Downloaded From : www.EasyEngineering.net
8. Write the stress strain relationship for plane stress problems.
ww
9. Differentiate material non linearity and geometric non linearity. (Nov/Dec 2012)
Material Non linearity
Geometric non linearity
(i) The stress – strain relation for the
(i) The Strain – Displacement relations
material may not be linear.
are not linear.
(ii) Basic non-linear relations are
(ii) Need consideration of actual strain
time dependent complex constitutive
displacement relations rather than linear
relations
strain displacement.
SC
A
D
w.E
asy
En
gi
nee
10. Write the equilibrium equations for two dimensional elements. (Nov/Dec 2012)
In elasticity theory, the stresses in the structure must satisfythe following equilibrium
equations,
rin
g
.ne
wherefx and fy are body forces (such as gravity forces) per unit volume.
t
PART - B
1. For the axe symmetric element shown in fig .Determine the element stresses. Take E=
2.1 x 105 N/mm2 𝝂 = 0.25. The co-ordinates shown in fig are in mm. The nodal
displacements are u1=0.05 mm, u2=0.02 mm, u3=0.0 mm, 𝝎𝟏 = 𝟎. 𝟎𝟑 𝒎𝒎, 𝝎𝟐 =
𝟎. 𝟎𝟐 𝒎𝒎, 𝝎𝟑 = 𝟎. 𝟎 𝒎𝒎.
Z
(0,0) 1
3 (30,50)
2 (60,0)
96
r
Downloaded From : www.EasyEngineering.net
Downloaded From : www.EasyEngineering.net
Given data:
r1 = 0 mm
z1=0 mm
u1=0.05 mm
𝜔1 = 0.03 𝑚𝑚
r2 = 60 mm
z2=0 mm
u2=0.02 mm
𝜔2 = 0.02 𝑚𝑚
r3 = 30 mm
z3=50 mm
u3=0.0 mm
𝜔3 = 0.0 𝑚𝑚
E= 2.1 x 105 N/mm2, 𝜈 = 0.25
To find
i. Radial stress 𝜎𝑟
ii. Circumferential stress 𝜎𝜃
iii. Longitudinal stress 𝜎𝑧
iv. Shear stress 𝜏𝑟𝑧
ww
Formula used
{σ}
w.E
asy
En
gi
Solution:
SC
A
D
𝜎𝑟
𝜎𝜃
𝜎𝑧
𝜏𝑟𝑧
= [D] [B] {u}
𝑢1
𝑤1
𝑢2
= [D] [B] 𝑤
2
𝑢3
𝑤3
nee
{σ} = [D] [B] {u}
D = Stress - Strain relationship matrix
𝐸
D= 1+𝜈 1−2𝜈
=
1−𝜈
𝜈
𝜈
0
𝜈
1−𝜈
𝜈
0
2.1 x 105
1 + 0.25 1 − 2 × 0.25
3
1
[D] = 336 × 103 × 0.25
1
0
𝜈
𝜈
1−𝜈
0
0
0
0
rin
g
.ne
1− 2𝜈
2
1 − 0.25
0.25
0.25
0.25
1 − 0.25
0.25
0.25
0.25
1 − 0.25
0
1
3
1
0
1
1
3
0
0
0
0
0
1
0
3
1
= 84 × 103
1
0
1
3
1
0
1
1
3
0
t
0
0
0
1 − 2 × 0.25
2
0
0
0
1
[B] =Strain displacement relationship matrix or gradient matrix
[B] =
1
2𝐴
𝛽1
𝛾₁𝑧
+
𝛽₁
+
𝑟
𝑟
0
𝛾1
𝛼₁
0
0
𝛾1
𝛽1
𝛽2
𝛾2 𝑧
+
𝛽
+
2
𝑟
𝑟
0
𝛾2
𝛼2
0
0
𝛾2
𝛽2
𝛽3
𝛾3 𝑧
+
𝛽
+
3
𝑟
𝑟
0
𝛾3
𝛼3
0
0
𝛾3
𝛽3
97
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𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 = 60 × 50 − 30 × 0 = 3000𝑚𝑚2
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 = 30 × 0 − 0 × 50 = 0
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 = 0 × 0 − 60 × 0 = 0
𝛽1 = 𝑧2 − 𝑧3 = 0 − 50 = −50 ;
𝛾1 = 𝑟3 − 𝑟2 = 30 − 60 = −30 ;
𝛽2 = 𝑦3 − 𝑦1 = 50 − 0 = 50 ;
𝛾2 = 𝑟1 − 𝑟3 = 0 − 30 = −30 ;
𝑟=
𝑟1 + 𝑟2 + 𝑟3 0 + 60 + 30
=
= 30 𝑚𝑚
3
3
𝑧=
𝑧1 + 𝑧2 + 𝑧3 0 + 0 + 50
=
= 16.67 𝑚𝑚
3
3
𝛽3 = 𝑦1 − 𝑦2 = 0 − 0 = 0
𝛾3 = 𝑟2 − 𝑟1 = 60 − 0 = 60
𝛼₁
𝛾1 𝑧
3000
(−30 × 16.67)
+ 𝛽1 +
=
+ −50 +
= 33.33 𝑚𝑚
𝑟
𝑟
30
30
ww
𝛼2
𝛾2 𝑧
(−30 × 16.67)
+ 𝛽2 +
= 0 + 50 +
= 33.33 𝑚𝑚
𝑟
𝑟
30
w.E
asy
En
gi
SC
A
D
𝛼3
𝛾3 𝑧
60 × 16.67
+ 𝛽3 +
= 0+0+
= 33.33 𝑚𝑚
𝑟
𝑟
30
1 𝑟1
𝐴 = 2 1 𝑟2
1 𝑟3
1
1
𝑧1
1
1
𝑧2 = 1
2
𝑧3
1
0
0
60 0
30 50
nee
= 2 [1 3000 − 0 − 0 50 − 0 + 0 30 − 60 ]=1500 𝑚𝑚2
[B] =
−50
0
1
33.33
0
2 × 1500
0
−30
−30 −50
rin
g
0
0
0
50
0
33.33 0
33.33
−30
0
60
0
50
60
0
−30
3 1 1 0
1 3 1 0
[D] [B] = 84 × 103
× 3.34 ×10-4
1 1 3 0
0 0 0 1
0
0
0
−50
0
50
0
33.33 0
33.33
0
33.33
−30
0
60
0
−30
0
50
60
0
−30 −50 −30
−116.67
49.99
= 28
−16.67
−30
−30
−30
−90
−50
𝜎𝑟
−116.67 −30
𝜎𝜃
49.99
−30
= 28
𝜎𝑧
−16.67 −90
𝜏𝑟𝑧
−30
−50
183.33 −30
149.99 −30
83.33 −90
−30
50
183.33
149.99
83.33
−30
.ne
t
33.33 60
99.99 60
33.33 180
60
0
−30 33.33 60
−30 99.99 60
−90 33.33 180
50
60
0
0.05
0.03
0.02
0.02
0
0
98
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𝜎𝑟
−3.66
−102.65
𝜎𝜃
4
112
= 28
=
𝜎𝑧
−3.66
−102.65
𝜏𝑟𝑧
−2.6
−72.8
Results
Radial stress 𝜎𝑟 = −102.65 N/mm2
Circumferential stress 𝜎𝜃 = 112 N/mm2
Longitudinal stress 𝜎𝑧 = −102.65 N/mm2
Shear stress 𝜏𝑟𝑧 = −72.8 N/mm2
2. Calculate the element stiffness matrix and the thermal force vector for the
axisymmetric triangular element shown in figure. The element experiences a 15 0c
increase in temperature. The co-ordinates are in mm. Take α= 10 x 10-6/0c ; E= 2x 105
N/mm2 , 𝝂 = 0.25
Z
3 (9,10)
ww
w.E
asy
En
gi
2 (8,7)
SC
A
D
(6,7) 1
Given data:
r1 = 6 mm
z1=7 mm
r2 = 8 mm
z2=7 mm
r3 = 9 mm
z3=10 mm
5
2
E= 2 × 10 N/mm , 𝜈 = 0.25, α= 10 × 10-6/0c
To find
Thermal force vector {F}t
Formula used
[K]=[𝐵]T D B 2πr A
nee
r
rin
g
.ne
{F}= 𝐵 T D 𝑒𝑡 2πr A
Solution:
[B] =Strain displacement relationship matrix or gradient matrix
0
𝛽3
𝛽1
0
𝛽2
𝛼₁
𝛾₁𝑧
𝛼2
𝛾2 𝑧
𝛼3
𝛾 𝑧
+ 𝛽₁ +
0
+ 𝛽2 +
0
+ 𝛽3 + 3
1
𝑟
𝑟
𝑟
𝑟
𝑟
[B] = 2𝐴 𝑟
0
𝛾1
0
𝛾2
0
𝛾1
𝛽1
𝛾2
𝛽2
𝛾3
t
0
0
𝛾3
𝛽3
𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2 = 8 × 10 − 9 × 7 = 17𝑚𝑚2
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3 = 9 × 7 − 6 × 10 = 3𝑚𝑚2
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1 = 6 × 7 − 8 × 7 = 13𝑚𝑚2
𝛽1 = 𝑧2 − 𝑧3 = 7 − 10 = −3𝑚𝑚 ;
𝛾1 = 𝑟3 − 𝑟2 = 9 − 8 = 1𝑚𝑚 ;
𝛽2 = 𝑦3 − 𝑦1 = 10 − 7 = 3 ;
𝛾2 = 𝑟1 − 𝑟3 = 6 − 9 = −3 ;
𝛽3 = 𝑦1 − 𝑦2 = 7 − 7 = 0
𝛾3 = 𝑟2 − 𝑟1 = 8 − 6 = 2
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𝑟=
𝑟1 + 𝑟2 + 𝑟3 6 + 8 + 9
=
= 7.67 𝑚𝑚
3
3
𝑧=
𝑧1 + 𝑧2 + 𝑧3 7 + 7 + 10
=
= 8 𝑚𝑚
3
3
𝛼₁
𝛾1 𝑧
17
(1 × 8)
+ 𝛽1 +
=
+ −3 +
= 0.26 𝑚𝑚
𝑟
𝑟
7.67
7.67
𝛼2
𝛾2 𝑧
3
(−3 × 8)
+ 𝛽2 +
=
+3+
= 0.26 𝑚𝑚
𝑟
𝑟
7.67
7.67
𝛼3
𝛾3 𝑧 −14
2 × 8
+ 𝛽3 +
=
+0+
= 0.26 𝑚𝑚
𝑟
𝑟
7.67
7.67
1 𝑟1
1
𝐴 = 2 1 𝑟2
1 𝑟3
𝑧1
1
1
𝑧2 = 1
2
𝑧3
1
ww
[D]
w.E
asy
En
gi
−3
1
0.26
2×3
0
1
𝐸
= 1+𝜈 1−2𝜈
0
0
1
−3
3
0
0.26 0
0
−3
−3
3
1−𝜈
𝜈
𝜈
0
𝜈
1−𝜈
𝜈
0
0
0.26
0
2
0
0
;
2
0
SC
A
D
[B] =
6 7
1
8 7 = 2 [1 80 × 63 − 6 10 − 7 + 7(9 − 8)=3 𝑚𝑚2
9 10
2 × 10 5
= 1+ 0.25 1−2× 0.25
1 − 0.25
0.25
0.25
0
3
1
= 320 × 105 × 0.25
1
0
−3 0.26
0
0
[B]T[D] = 0.167 3 0.26
0
0
0 0.26
0
0
𝜈
𝜈
1−𝜈
0
1
3
1
0
1
1
3
0
0
0
0
nee
0
1
1 −3
0 −3
−3 3
0
2
0
2
1− 2𝜈
2
rin
g
0.25
0.25
1 − 0.25
0.25
0.25
1 − 0.25
0
0
0
0
0
1
3
1
0
1
1
3
0
.ne
1− 2× 0.25
2
0
0
0
1
0
1
3
1 −3
0 −3 × 8×104 1
−3 3
1
0
2
0
0
2
−8.7 −2.2
1
1
= 13.36×103 9.26 3.78
−3
−3
0.26 0.78
2
2
−3 0.26
0
0
[B]T = 0.167 3 0.26
0
0
0 0.26
0
0
t
0
0
0
1
−2.7 1
3
−3
3.26 −3
−9
3
0.26 2
6
0
100
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−8.7 −2.2
1
1
T
3 9.26
3.78
[B] [D][B] = 13.36×10
−3
−3
0.26 0.78
2
2
−2.7 1
−3 0.26
3
−3
0
0
3.26 −3 × 0.167 3 0.26
−9
3
0
0
0.26 2
0 0.26
6
0
0
0
0
1
1 −3
0 −3
−3 3
0
2
0
2
1.42 −5.4
26.63
−5.7 −29.79 11.21
−5.7
−18
−5.7
12
12.26
6
−5.01
3 −29.79 12.26
−18.78
37.76
6.5
[K]= 321.27 × 10
36
11.21
−18 −18.78
5.2
−18
5.2
1.42
−5.7
−5.01
4.2
0.52
−18
−5.4
6
6.5
0.52
12
Thermal force vector {F}= 𝐵 T D 𝑒𝑡 2πr A
𝑒𝑡 =
𝛼∆𝑡
10 × 10−6 × 15
150
−6
𝛼∆𝑡
10
×
10
×
15
-6 150
=
=10
0
0
0
−6
𝛼∆𝑡
150
10 × 10 × 15
ww
−8.7 −2.2
1
1
{F}= [B]T[D] 𝑒𝑡 2πr A = 13.36×103 9.26 3.78
−3
−3
0.26 0.78
2
2
SC
A
D
w.E
asy
En
gi
−2.7 1
3
−3
150
3.26 −3 × 10-6 150 × 2π × 7.67 × 3
−9
3
0
0.26 2
150
6
0
−1493.46
−150
1506.54
= 1.927
−450
456.54
600
𝐹1𝑢
−2878.25
𝐹1𝑤
−289.08
𝐹2𝑢
2903.45
Thermal force vector {F} =
=
𝐹2𝑤
−867.25
879.86
𝐹3𝑢
1156.34
𝐹3𝑤
3.
nee
rin
g
.ne
t
DERIVE THE EXPRESSION FOR STRESS – STRAIN RELATIONSHIP FOR A
2D- ELEMENT?
EQUATION OF ELASTICITY
1. Stress – strain relationship matrix for a two dimensional element
Consider a three dimensional body as shown in fig. which is subjected to a stress σx σy
and σz
101
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Y
σy
σz
σx
σ
x
x
σz
ww
Hook’sZlaw
σy
σ = Ee
w.E
asy
En
gi
𝜎
SC
A
D
e=𝐸
The stress in the x direction produces a positive strain in x direction as shown in fig.
𝜎
ex = 𝐸𝑥
nee
The positive stress in the y direction produces a negative strain in the x direction
ey =
−𝜈𝜎 𝑦
𝐸
rin
g
The positive stress in the z direction produces a negative strain in the x direction
ez =
ex =
−𝜈𝜎 𝑧
𝐸
𝜈𝜎 𝑦
𝜎𝑥
𝜈𝜎
− 𝐸 − 𝐸𝑧
𝐸
.ne
t
𝜎𝑦 𝜈𝜎
𝜈𝜎
ey = − 𝐸 𝑥 + 𝐸 − 𝐸 𝑧
ez = −
𝜈𝜎 𝑥
𝐸
−
𝜈𝜎 𝑦
𝐸
𝜎
+ 𝑧
𝐸
Solving 3 equations
𝐸
e𝑥 1−𝑣 +𝑣 𝑒𝑦 +𝑉 𝑒2
𝐸
v e𝑥 1−𝑣 − 𝑒𝑦 +𝑉 𝑒2
𝐸
v e𝑥+𝑣 𝑒𝑦 + 1−𝑣 𝑒2
𝜎𝑥 = 1+𝑣 1−2𝑣
𝜎𝐽 = 1+𝑣 1−2𝑣
𝜎2 = 1+𝑣 1−2𝑣
The shear stress and shear strain relationship
𝜏 = 𝐺𝛾 where, 𝜏 - Shear Stress
102
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Downloaded From : www.EasyEngineering.net
𝛾 – Shear Strain
G – Modular of rigidity
𝜏 𝑥𝑦 = G𝛾𝑥𝑦
𝜏 𝑦𝑧 = G𝛾𝑦𝑧
𝜏 𝑧𝑥
= G𝛾𝑧𝑥
𝐸
G
Modular of rigidity = 2 1+𝑣
𝐸
𝐸
1−2𝑣
𝜏 𝑥𝑦 = 2 1+𝑣 𝛾𝑥𝑦 ; 𝜏 𝑥𝑦 = 2 1+𝑣 1−2𝑣 𝛾 𝑦 2
𝜏 𝑦𝑧 =
=
2
1+𝑣 1−2𝑣
𝛾𝑥𝑧 ;𝜏 𝑦𝑧 =
1−2𝑣
=
𝐸
1−2𝑣
1+𝑣 1−2𝑣
2
1−v
v
v
1− v
v
v
0
0
v
v
1− v
0
w.E
asy
En
gi
𝐸
1+𝑣 1−2𝑣
𝜎
= 𝐷
0
0
0
0
0
0
𝐸
1−v
v
v
1− v
v
v
0
0
v
v
1− v
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1−2𝑣
0
0
0
1−2𝑣
2
2
nee
𝑒
D- in a stress strain relation ship matrix
𝐷 = 1+𝑣 1−2𝑣
𝛾𝑦𝑧
𝛾𝑧𝑥
2
SC
A
D
𝜎𝑥
𝜎𝑦
𝜎𝑧
𝜎𝑥𝑦
𝜎𝑦𝑧
𝜎𝑧𝑥
1−2𝑣
𝐸
𝜏 𝑧𝑥
ww
𝐸
1+𝑣 1−2𝑣
2
0
0
0
0
0
0
0
0
1−2𝑣
0
0
0
2
2
2
rin
g
0
0
0
0
1−2𝑣
1−2𝑣
𝑒𝑥
𝑒𝑦
𝑒𝑧
𝛾𝑥𝑦
𝛾𝑦𝑧
𝛾𝑧𝑥
.ne
t
1−2𝑣
2
Where E – Yours Modules
V – Poisson Ratio
(i) PLANE STRESS CONDITION:Plane stress is defined to be a state of stress in which the normal stress 𝜎 and shear
stress 𝜏 cleared perpendicular to the plane are assumed to be zero.
Normal stress 𝜎𝑧 = 0; Shear Stress 𝜏𝑥𝑧 + 𝜏𝑦𝑧 = 0
𝜎𝑧 = 𝜏 𝑥𝑧 = 𝜏 𝑦𝑧 = 0
103
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𝜎𝑦
𝜎
𝜎𝑦
𝜎
𝑒𝑥 = 𝐸𝑥 - v 𝐸 ; 𝑒𝑦 = -v 𝐸𝑥 + 𝐸
𝜎𝑦
𝜎
= 𝐸𝑥 -v 𝐸
𝑒𝑥
𝜎𝑥
v𝑒𝑦 = −𝑣 2
𝜎𝑦
𝐸
+𝑣 𝐸
𝜎
𝑣 2 𝜎𝑥
𝑒𝑥 + v𝑒𝑦 = 𝐸𝑥 -
𝐸
𝜎𝑥
𝑒𝑥 + v𝑒𝑦 = 𝐸 - 1 − 𝑣 2
𝐸
𝜎𝑥 = 1−𝑣 2
𝑒𝑥 + 𝑣 𝑒𝑦
𝜎𝑦
𝜎
v 𝑒𝑥 = v 𝐸𝑥 -V 2 𝐸
𝜎𝑦
𝜎
𝑒𝑦 = -v 𝐸𝑥 + 𝐸
𝜎𝑦
𝜎𝑦
v 𝑒𝑥 + 𝑒𝑦 = -V 2 𝐸 + 𝐸
ww
𝜎𝑦
v 𝑒𝑥 + 𝑒𝑦 = 𝐸 1 − 𝑣 2
w.E
asy
En
gi
𝐸
𝑣𝑒𝑥 + 𝑒𝑦
SC
A
D
𝜎𝑦 = 1−𝑣 2
Share Stress 𝜏 𝑥𝑧 = G 𝛾𝑥𝑧
Where G
𝛾𝑥𝑦
Modular of rigidity =
nee
Share Strain
V – Poisson ratio
𝐸
𝜏𝑥𝑦 = 2 1+𝑣 𝛾𝑥𝑦
𝐸
𝜏𝑥𝑦 = 1+𝑣 1−𝑣 ×
𝐸
𝜏𝑥𝑦 = 1−𝑣 2 ×
1−𝑣
𝐸
1−𝑣
2
𝛾𝑥𝑦
2 1+𝑣
rin
g
.ne
× 𝛾𝑥𝑦
2
t
Above equation matrix form
𝜎𝑥
𝜎𝑦
𝜏𝑥𝑦
𝐸
= 1−𝑣
1
𝑣
0
𝑣
1
0
0
0
1−𝑣
2
𝑒𝑥
𝑒𝑦
𝜏𝑥𝑦
Two dimensional stress strain relationship matrix for phase stress location.
𝐸
𝐷 =
1−𝑣
1
𝑣
𝑣
1
0
0
0
0
1−𝑣
2
104
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(ii) PLANE STRAIN CONDITION
Plane strain is defined to be a state of strain in which the strain normal to the xy plane
and the shear strain are assumed to be zero.
Normal strain 𝑒𝑧 =0
Shear Stress 𝛾𝑥𝑧 = 0 =𝛾𝑦𝑧
𝜎𝑥
𝜎𝑦
𝜎𝑧
𝜎𝑥𝑦
𝜎𝑦𝑧
𝜎𝑧𝑥
𝐸
= 1+𝑣 1−2𝑣
1−v
v
v
1− v
v
v
0
0
v
v
1− v
0
0
0
0
0
0
0
𝑒𝑧 =0 ; 𝛾𝑥0 =𝛾𝑦𝑧 =0
Sub in above matrix.
𝜎𝑥
𝐸
𝜎𝑦 =
1+𝑣 1−2𝑣
𝛾𝑥𝑦
1−𝑣
𝑣
0
ww
𝑣
1−𝑣
0
0
0
1−2𝑣
0
0
0
0
0
0
0
0
0
0
0
0
1−2𝑣
0
0
0
1−2𝑣
2
2
2
𝑒𝑥
𝑒𝑦
𝛾𝑥𝑦
w.E
asy
En
gi
2
1−2𝑣
𝑒𝑥
𝑒𝑦
𝑒𝑧
𝛾𝑥𝑦
𝛾𝑦𝑧
𝛾𝑧𝑥
Stress Strain relationship matrix for phase strain condition.
1+𝑣 1−2𝑣
1−𝑣
𝑣
0
𝑣
1−𝑣
0
0
0
SC
A
D
𝐷 =
𝐸
1−2𝑣
2
nee
4. A long hollow cylinder of inside diameter 100 mm and outside diameter 140 mm is
subjected to an internal pressure of 4 N/mm2 as shown in figure.(i) By using two
elements on the 15 mm length shown in figure. (ii) Calculate the displacements at the
F1
inner radiusTake E=2×105 N/mm2. V=0.3.
4
Z
1
rin
g
.ne
1
t
15 mm
Element
Cylinder
Axis of the hallow Cylinder
Element
2
2
3
F2
50 mm
100 mm
70 mm
140 mm
105
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Downloaded From : www.EasyEngineering.net
Given data:
Inner diameter, de= 100mm
Inner radius re= 50 mm
Outer diameter De=140 mm
Outer radius Re=70mm
Internal pressure P=4N/mm2
Length le=15mm
Young’s modulus E=2×105 N/mm2
Poison’s ratio v= 0.3
To Find
ww
W1
u1, w1, u2, w2, u3, w3, u4, w4
w.E
asy
En
gi
Solution
𝑈
For element (1)
(Nodal displacements u1, w1, u2, w2, u4, w4)
Co ordinates
Axis of the hallow cylinder
𝐹 =𝐾
At node 1
U1
Element
(r1 Z1)
U4
(r3 Z3 )
1
SC
A
D
Formula used
W4
15 mm
nee
Z
r1=50mm
W2
rin
g
2 (r2 Z2 )
50mm
z1=15mm
70 mm
At node 2
r
r1=50mm
U2
.ne
t
z1=0mm
At node 3
r1=70mm
z1=15mm
We know that, 𝑤ℎ𝑒𝑟𝑒 𝑟 =
𝑟1 +𝑟2 +𝑟3
3
=
50+50+70
3
r = 56.6667mm
𝑧 +𝑧2 +𝑧3
𝑧= 1
3
=
15+0+15
3
;
z= 10 mm
1
Area of the triangle element = × 𝐵𝑟𝑒𝑎𝑑𝑡ℎ × 𝐻𝑒𝑖𝑔ℎ𝑡
2
106
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Downloaded From : www.EasyEngineering.net
1
= 2 × 20 × 15 ;
A = 150 mm
We know that,
Stiffness matrix for axisymmetric triangular element (1),
𝐾 1 =2 𝜋 rA 𝐵 T 𝐷 B
1−𝜈
𝜈
𝜈
0
𝐸
Stress strain relationship matrix 𝐷 = 1+𝜈 1−2𝜈
2𝑋10 5
Stress strain relationship matrix 𝐷 = 1+0.3 1−(2×0.3)
ww
2×10 5
𝜈
1−𝜈
𝜈
0
0.5
0
0
0
1− 2𝜈
2
1 − 0.3
0.3
0.3
0.3
1 − 0.3
0.3
𝜈
𝜈
1 − 0.3
0
0
0
0
0
0
1−(2×0.3)
2
0.7 0.3 0.3 0
0.3 0.7 0.3 0
𝜈
𝜈 0.7 0
0
0
0 0.2
w.E
asy
En
gi
=
𝜈
𝜈
1−𝜈
0
SC
A
D
0.7 0.3 0.3 0
0.3 0.7 0.3 0
= 384.6153×103
𝜈
𝜈 0.7 0
0
0
0 0.2
nee
We know that , strain-Displacement matrix
1
B = 2𝐴
𝛽1
𝛾₁𝑧
+
𝛽₁
+
𝑟
𝑟
0
𝛾1
𝛼₁
0
0
𝛾1
𝛽1
𝛽2
𝛾2 𝑧
+
𝛽
+
2
𝑟
𝑟
0
𝛾2
𝛼2
𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2
0
0
𝛾2
𝛽2
𝛽3
𝛾3 𝑧
+
𝛽
+
3
𝑟
𝑟
0
𝛾3
𝛼3
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3
𝛼2 = 70 × 15 − 50 × 15
𝛼1 = 750 𝑚𝑚2
𝛼2 = 300𝑚𝑚2
.ne
𝛼3 = −750𝑚𝑚2
𝛽2 = 𝑦3 − 𝑦1
𝛾2 = 𝑟1 − 𝑟3
𝛽3 = 𝑦1 − 𝑦2
𝛾3 = 𝑟2 − 𝑟1
𝛽1 = 0 − 15
𝛾1 = 70 − 50
𝛽2 = 15 − 15
𝛾2 = 50 − 70
𝛽3 = 15 − 0
𝛾3 = 50 − 50
𝛽1 = −15𝑚𝑚
𝛾1 = 20𝑚𝑚
𝛽2 = 0
𝛾2 = −20𝑚𝑚
𝛽3 = 15𝑚𝑚
𝛾3 = 0
𝛾₁𝑧
𝛼2
+ 𝛽2 + 2𝑟
=
+ 𝛽3 + 3𝑟
= 56.6667 + 15 + 0
+ 𝛽₁ + 𝑟
𝑟
𝛾 𝑧
𝛾 𝑧
=
750
20×10
+ (−15) + 56.6667
56.6667
300
56.6667
−750
(−20×10)
+ 0 + 56.6667
t
𝛼3 = 50 × 0 − 50 × 15
𝛽1 = 𝑧2 − 𝑧3
𝛾1 = 𝑟3 − 𝑟2
𝛼₁
𝑟
rin
g
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1
𝛼1 = 50 × 15 − 70 × 0
𝑟
𝛼3
0
0
𝛾3
𝛽3
=1.7647 mm
= 1.7647 mm
=1.7647mm
107
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Downloaded From : www.EasyEngineering.net
𝛼₁
𝛾₁𝑧 𝛼
𝛾 𝑧
𝛼3
Substitute𝛽1, 𝛽2 , 𝛽3, 𝑟 + 𝛽₁ + 𝑟 , 𝑟2 + 𝛽2 + 2𝑟 ,
𝑟
𝛾 𝑧
+ 𝛽3 + 3𝑟 , 𝛾1 , 𝛾2, 𝛾3 and A values in
equations no 5, we get,
−15
1
1.7647
B = 2×150
0
20
0
0
20
−15
−15
1.7647
B =3.333 × 10−3
0
20
0
1.7647
0
−20
0
0
−20
0
15
1.7647
0
0
0
0
0
15
0
0
20
−15
0
1.7647
0
−20
0
0
−20
0
15
1.7647
0
0
−15 1.7647
0
0
0
1.7647
B T=3.333 × 10−3 0
0
15 1.7647
0
0
ww
0
0
0
15
0
20
20 −15
0
−20
−20
0
0
0
0
15
w.E
asy
En
gi
0.7 0.3 0.3 0
0.7 0.3 0
D B = 384.6153×10
×
0.3 0.3 0.7 0
0
0
0 0.2
−15
0
1.7647
0
3.33310−3
0
20
20
−15
SC
A
D
3 0.3
−9.9706
0.3
D B = 1.282×103
0.3
0
6 0.5294
0.7
0.3
0.3
0.7
0
0
0
1.7647
0
−20
0
0
−20
0
nee
−6 11.0294 0
−6
5.7353 0
−14 5.0294 0
0
0
0
15
1.7647
0
0
0
0
0
15
rin
g
.ne
−9.9706 6 0.5294 −6 11.0294 0
0.3
0.7
0.3
−6
5.7353 0 X3.33 10-3
D B B T =1.282×103
𝜈
𝜈
0.7
−14 5.0294 0
0
0
0
0
0
0
0
20
−15 1.7647
20 −15
0
0
0 1.7647
0
−20
0
0
−20
0
15 1.7647
0
0
0
0
0
15
223.798 −139.4118 −85.7611
−139.412
325
70.588
−85.7612
70.588
82.18
D B B 𝑇 = 4.2733
79.412
−280
−10.588
−155.3202 100.5882 10.1210
60
−45
−60
t
79.4118 −155.32 60
−280
100.588 −45
−10.588 10.1211 −60
280
−100.588
0
−100.588 175.5621
0
0
0
45
108
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Downloaded From : www.EasyEngineering.net
Substitute D B B 𝑇 value in equ no 4
𝐾 1 = 2 𝜋 ×56.6667×150×4.2733
223.798 −139.4118 −85.7611
−139.412
325
70.588
−85.7612
70.588
82.18
×
79.412
−280
−10.588
−155.3202 100.5882 10.1210
60
−45
−60
79.4118 −155.32 60
−280
100.588 −45
−10.588 10.1211 −60
280
−100.588
0
−100.588 175.5621
0
0
0
45
223.798 −139.4118 −85.7611
−139.412
325
70.588
70.588
82.18
𝐾 1 =228224.6× −85.7612
79.412
−280
−10.588
−155.3202 100.5882 10.1210
60
−45
−60
79.4118 −155.32 60
−280
100.588 −45
−10.588 10.1211 −60
280
−100.588
0
−100.588 175.5621
0
0
0
45
u1
w1
u2
ww
51.076 −31.817
−31.817 74.173
𝐾 1= −19.573 16.110
18.124
−63.903
−35.448
22.597
13.693
10.270
w2
−19.573
16.110
18.755
−2.416
2.310
−13.693
u4
w4
18.124 −35.448 13.693
−63.903 22.597 −10.270
−2.416
2.310
−13.693
63.903 −22.597
0
−22.597
40.068
0
0
0
10.270
SC
A
D
w.E
asy
En
gi
For element (2) (Nodal displacements, u2, w2, u3, w3, u4, w4)
Co ordinates
At node 2
nee
r1=50mm
Z
z1=0mm
W4
(r3,z3)
rin
g
U4
4
.ne
At node 3
t
15 mm
r1=70mm
z1=0mm
Element
At node 4
r1=70mm
W2
z1=15mm
U2
𝑟1 +𝑟2 +𝑟3
We know that, 𝑤ℎ𝑒𝑟𝑒 𝑟 =
2
3
=
2
3
(r,z1)
W3
U3
(r2,z2)
50+70+70
3
50mm
r = 63.3333mm,
𝑧 +𝑧2 +𝑧3
𝑧= 1
3
=
0+0+15
3
;z= 5 mm
70 mm
109
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Downloaded From : www.EasyEngineering.net
1
Area of the triangle element = 2 × 𝐵𝑟𝑒𝑎𝑑𝑡ℎ × 𝐻𝑒𝑖𝑔ℎ𝑡
1
= 2 × 20 × 15
A = 150 mm
We know that,
Stiffness matrix for axisymmetric triangular element (2),
𝐾 2 =2 𝜋 rA 𝐵 T 𝐷 B
1−𝜈
𝜈
𝜈
0
𝐸
Stress strain relationship matrix 𝐷 = 1+𝜈 1−2𝜈
ww
2𝑋10 5
Stress strain relationship matrix 𝐷 = 1+0.3 1−(2×0.3)
w.E
asy
En
gi
= 0.5
𝜈
𝜈
1−𝜈
0
1− 2𝜈
nee
0.7 0.3 0.3 0
0.3 0.7 0.3 0
=384.6153×103
𝜈
𝜈 0.7 0
0
0
0 0.2
rin
g
B=
1
2𝐴
0
𝛾1
0
0
0
1−(2×0.3)
2
.ne
We know that, strain-Displacement matrix
𝛽1
𝛾₁𝑧
+ 𝛽₁ + 𝑟
𝑟
2
0.7 0.3 0.3 0
0.3 0.7 0.3 0
𝜈
𝜈 0.7 0
0
0
0 0.2
0.7 0.3 0.3 0
0.3 0.7 0.3 0
=384.6153×103
𝜈
𝜈 0.7 0
0
0
0 0.2
𝛼₁
0
0
0
1 − 0.3
0.3
0.3
0.3
1 − 0.3
0.3
𝜈
𝜈
1 − 0.3
0
0
0
SC
A
D
2×10 5
𝜈
1−𝜈
𝜈
0
0
0
𝛾1
𝛽1
𝛽2
𝛾 𝑧
+ 𝛽2 + 2𝑟
𝑟
𝛼2
0
𝛾2
𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2
𝛼1 = 70 × 15 − 70 × 0
𝛽3
𝛾 𝑧
+ 𝛽3 + 3𝑟
𝑟
𝛼3
𝛾2
𝛽2
0
𝛾3
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3
𝛼2 = −750𝑚𝑚2
𝛽2 = 𝑦3 − 𝑦1
𝛾2 = 𝑟1 − 𝑟3
0
0
𝛾3
𝛽3
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1
𝛼2 = 70 × 0 − 50 × 15
𝛼1 = 1050 𝑚𝑚2
𝛽1 = 𝑧2 − 𝑧3
𝛾1 = 𝑟3 − 𝑟2
0
0
t
𝛼3 = 50 × 0 − 70 × 0
𝛼3 = 0
𝛽3 = 𝑦1 − 𝑦2
𝛾3 = 𝑟2 − 𝑟1
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𝛼₁
𝑟
𝛾₁𝑧
+ 𝛽₁ + 𝑟
𝛼2
𝛾 𝑧
=
𝛽1 = 0 − 15
𝛾1 = 70 − 70
𝛽2 = 15 − 0
𝛾2 = 50 − 70
𝛽3 = 0 − 0
𝛾3 = 70 − 50
𝛽1 = −15𝑚𝑚
𝛾1 = 0
𝛽2 = 15𝑚𝑚
𝛾2 = −20𝑚𝑚
𝛽3 = 0
𝛾3 = −20𝑚𝑚
1050
63.333
−750
+ (−15) + 0
(−20×5)
𝑟
+ 𝛽2 + 2𝑟
=
𝑟
+ 𝛽3 + 3𝑟
= 0 + 0 + 63.333
𝛼3
𝛾 𝑧
63.333
𝛼₁
=1.579 mm
+ 15 + 63.333
(20×5)
= 1.579 mm
=1.579mm
𝛾₁𝑧 𝛼
𝛾 𝑧
Substitute 𝛽1, 𝛽2 , 𝛽3, 𝑟 + 𝛽₁ + 𝑟 , 𝑟2 + 𝛽2 + 2𝑟 ,
𝛼3
𝑟
𝛾 𝑧
+ 𝛽3 + 3𝑟 , 𝛾1 , 𝛾2, 𝛾3 and A values in
equations no 10, we get,
−15
1
1.579
B = 2×150
0
0
ww
0
0
0
−15
0
15
0
1.579
−20
0
15
−20
0
1.579
0
20
0
0
20
0
w.E
asy
En
gi
15
1.579
0
−20
0
0
0
1.579
−20
0
15
20
SC
A
D
−15
0
1.579
0
B =3.333 × 10−3
0
0
0
−15
0
0
20
0
nee
D B = 384.6153×103
0.7 0.3 0.3 0
−15
0.3 0.7 0.3 0
1.579
× 3.333 × 10−3
0.3 0.3 0.7 0
0
0
0
0 0.2
0
−10.0263
3 −3.3947
D B = 1.282×10
−4.0263
0
0 10.9737
0
5.6053
0
4.9737
−3
−4
0
0
0
−15
0
15
0
1.579
−20
0
15
−20
rin
g
−6 0.4737 6
−6
1.1053 6
−14 0.4737 14
3
4
0
0
1.579
0
20
.ne
0
0
20
0
t
We know that
−15
0
1.579
0
B =3.333 × 10−3
0
0
0
−15
15
1.579
0
−20
−15 1.579
0
0
T
−3 15 1.579
B =3.333 × 10
0
0
0 1.579
0
0
0
0
0
1.579
−20
0
15
20
0
0
20
0
0
0
0 −15
0
−20
−20 15
0
20
20
0
111
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Downloaded From : www.EasyEngineering.net
−10.0263
−3.3947
D B B T =1.282×103
−4.0263
0
0
0
0
−3
10.9737
5.6053
4.9737
−4
−15 1.579
0
0
15 1.579
0
0
0 1.579
0
0
−6 0.4737 6
−6
1.1053 6
3.333 × 10−3
−14 0.4737 14
3
4
0
0
0
0 −15
0
−20
−20 15
0
20
20
0
145.034
0 −155.755
0
45
60
−155.755 60 253.456
D B B T =4.2733
80.526
−45
−159.474
−5.360
−60
−71.149
−80.526
0
99.474
80.526
−5.360 −80.526
−45
−60
0
−159.474 −71.149 99.474
325 50.256
−280
50.526 81.745
9.474
−280
9.474
280
ww
w.E
asy
En
gi
SC
A
D
Substitute D B B 𝑇 value in equ no 8
145.034
0 −155.755
0
45
60
−155.755
60
253.456
𝐾 2 =2 𝜋 ×63.333×150×4.2733×
80.526
−45
−159.474
−5.360
−60
−71.149
−80.526
0
99.474
80.526
−5.360 −80.526
−45
−60
0
−159.474 −71.149 99.474
325 50.256
−280
50.526 81.745
9.474
−280
9.474
280
nee
145.034
0 −155.755
0
45
60
−155.755 60 253.456
𝐾 2 =255.074X103
80.526
−45
−159.474
−5.360
−60
−71.149
−80.526
0
99.474
36.994
0
−39.729
0
11.478 15.304
−39.729 15.304 64.650
𝐾 2 =106
20.540
−11.478
−40.678
−1.367
−15.304
−18.148
−20.540
0
25.373
rin
g
80.526
−5.360 −80.526
−45
−60
0
−159.474 −71.149 99.474
325 50.256
−280
50.526 81.745
9.474
−280
9.474
280
.ne
t
20.540
−1.367 −20.540
−11.478 −15.304
0
−40.678 −18.148 25.373
82.899 12.877
−71.421
12.877
20.851
2.417
−71.421
2.417
71.421
112
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Downloaded From : www.EasyEngineering.net
Assemble the equations.
Global stiffness matrix, [ K ] =
51.076
+0
-31.817
+0
-19.573
+0
-31.817
+0
74.173
+0
16.110
+0
18.124
+0
0
-63.903
+0
0
ww
0
-35.448
+0
13.693+
0
18.124+0 0
63.903+0
-2.416
+
0
-2.416
63.903
+0
+11.478
0+
0+
(-39.729) 15.304
20.540+0 11.478+0
2.310-22.597
1.367
-15.304
-13.693
0+0
-20.540
0
-35.448+0
13.693+0
0
0
22.957+0
-10.270+0
-39.729
+0
20.540
+0
2.3101.367
-13.693
-20.540
-22.957
-15.304
0+
(-18.148)
12.887+0
0+0
40.068+20
.851
0+2.417
0+2.417
0+15.304 0
-11.478
0+64.650 0+
(-40.678)
82.899+0
40.678+0
0
0+12.887
-18.148
0+25.373 0-71.421
w.E
asy
En
gi
22.597+0
-10.270+0
SC
A
D
0
-19.573
+0
16.110
+0
18.755+
36.994
nee
Global stiffness matrix, [ K ] =
51.076
-31.817
-19.573
-31.817
74.173
16.110
18.124
0
0
-35.448
13.693
-63.903
0
0
22.597
-10.270
0+25.373
-71.421+0
10.270
+71.421
-19.573
16.110
55.749
18.124
-63.903
-2.416
0
0
-39.729
0
0
20.540
-35.448
22.957
0.943
13.693
-10.270
-34.233
-2.416
(-39.729)
20.540
0.943
-34.233
75.381
15.304
-11.478
-38.261
0
15.304
64.650
-40.678
18.148
25.373
-11.478
-40.678
82.899
12.887
71.421
-38.261
-18.148
12.887
60.919
2.417
0
25.373
-71.421
2.417
81.691
rin
g
.ne
t
We know that
𝐹 =𝐾
𝑈
51.076
−31.817 −19.573
𝐹1𝑢
−31.817
74.173
16.110
𝐹2𝑢
−19.5573 16.110
55.759
𝐹3𝑢
18.124 −63.903 −2.416
6
=10
𝐹4𝑢
0
0
−39.729
0
0
20.540
𝐹5𝑢
−35.448
22.957
0.943
𝐹6𝑢
13.693
−10.270
−34.233
18.124
0
0
−63.903
0
0
−2.416 −39.729 20.540
75.381 15.304 −11.478
15.304 64.650 −40.678
−11.478 −40.678 82.899
−38.261 −18.148 12.887
0
25.373 −71.421
−35.448 13.693
22.957 −10.270
0.943
−34.233
−38.261
0
−18.148 25.373
12.887 −71.421
60.919
2.417
2.417
81.691
𝑢1
𝑤1
𝑢2
𝑤2
𝑢3
𝑤3
𝑢4
𝑤4
113
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Downloaded From : www.EasyEngineering.net
Forces we know that
F1u = F2u =
2 𝑃𝜋 𝑟 𝑒𝑙 𝑒
2
=
2× 𝜋×50×15×4
2
= 9424.77 N
The remaining forces are zero F1w, F2w, F3u, F3w, F4w, are zero.
Displacements
1. Node 1 is moving in r direction. u1
0 but w1 =0
2. Node 2 is moving in r direction. u2
0 but w2 =0
3. Node 3 & 4 are fixed. So u3, w3 u4 and w4 are zero.
Substitute nodal force and nodal displacements values in eqn 12
18.124
0
0
51.076
−31.817 −19.573
−35.448 13.693
−63.903
0
0
−31.817
74.173
16.110
22.957 −10.270
9424.77
−2.416 −39.729 20.540
−19.5573 16.110
55.759
0.943
−34.233
0
9424.77
18.124 −63.903 −2.416
−38.261
0
75.381 15.304 −11.478
=106
×
0
0
0
−39.729
15.304 64.650 −40.678
−18.148 25.373
0
12.887 −71.421
0
0
20.540 −11.478 −40.678 82.899
0
−38.261 −18.148 12.887
60.919
2.417
−35.448
22.957
0.943
0
25.373 −71.421
2.417
81.691
13.693
−10.270
−34.233
ww
SC
A
D
w.E
asy
En
gi
𝑢1
0
𝑢2
0
0
0
0
0
Delete second row, second column, fourth row, fourth column, fifth row, fifth column,
sixth row, sixth column, seventh row, seventh column, and eighth row and eight column of the
above matrix. Hence the Equation reduces to
nee
9424.77
51.706
=106
9424.77
−19.5573
−19.5573
55.759
X
𝑢1
𝑢2
rin
g
9424.77 = 106 (51.706u1-19.573u2)
.ne
9424.77 = 106 (-19.573u1-55.749u2)
Above equations we solving and we get
u1 =2.88×10-4mm
t
u2 =2.70×10-4mm
RESULTS
DISPLACEMENTS
u1 =2.88×10-4mm
w1=0
u2 =2.70×10-4mm
w2=0
u3 =0
w3=0
u4 =0
w4=0
114
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Downloaded From : www.EasyEngineering.net
5. DERIVE
THE
EXPRESSION
FOR
STRAIN-DISPLACEMENT
RELATIONSHIP FOR AXISYMMETRIC ELEMENT.
Shape function are given below
U = N1u1+N2u2+N3u3 --------------------------- 1
W = N1w1+N2w2 +N3w3 --------------------- 2
𝜕𝑢
Radial strain er = 𝜕𝑟
Eqn 1 d.w.r to “r “
𝜕𝑢
𝜕𝑁
𝜕𝑁
𝜕𝑁
er = 𝜕𝑟 = 𝜕𝑟1 𝑢1 + 𝜕𝑟2 𝑢2 + 𝜕𝑟3 𝑢3 ------------------- 3
𝑢
ww
Circumferential strain e Ɵ = 𝑟
𝑁
𝑁
𝑁
w.E
asy
En
gi
e Ɵ = 𝑟1 𝑢1 + 𝑟2 𝑢2 + 𝑟3 𝑢3 --------- 4
𝜕𝑤
Longitudinal strain ez =
SC
A
D
𝜕𝑧
𝜕𝑁1
ez =
𝜕𝑧
𝜕𝑢
𝜕𝑁
𝜕𝑁
𝑤1 + 𝜕𝑧2 𝑤2 + 𝜕𝑧3 𝑤3 ---------- 5
nee
𝜕𝑤
Shear strain ϒ rz = 𝜕𝑧 + 𝜕𝑟
𝜕𝑁1
𝜕𝑁2
𝜕𝑁3
𝜕𝑁1
𝜕𝑁2
rin
g
𝜕𝑁
.ne
ϒ rz = 𝜕𝑧 𝑢1 + 𝜕𝑧 𝑢2 + 𝜕𝑧 𝑢3 + 𝜕𝑟 𝑤1 + 𝜕𝑟 𝑤2 + 𝜕𝑟3 𝑤3 ------ 6
Arranging equation 3, 4, 5 & 6 in matrix form
𝜕𝑁1
𝑒𝑟
𝑒𝜃
=
𝑒𝑧
𝛾𝑟𝑧
0
𝜕𝑟
𝑁1
0
𝑟
𝜕𝑁1
0
𝜕𝑁1
𝜕𝑧
𝜕𝑁1
𝜕𝑧
𝜕𝑟
𝜕𝑁2
𝜕𝑟
𝑁2
𝑟
0
0
0
𝜕𝑁2
𝜕𝑁3
𝜕𝑟
𝑁3
0
0
𝑟
𝜕𝑁3
0
𝜕𝑁2
𝜕𝑧
𝜕𝑁2
𝜕𝑁3
𝜕𝑧
𝜕𝑁3
𝜕𝑧
𝜕𝑟
𝜕𝑧
𝜕𝑟
𝑢1
𝑤1
𝑢2
𝑤2 ------------- 7
𝑢3
𝑤3
t
Shape function
1
𝑁1 = 2𝐴 𝛼1 + 𝛽1 𝑟 + 𝛾1 𝑧
1
𝑁2 = 2𝐴 𝛼2 + 𝛽2 𝑟 + 𝛾2 𝑧
;
;
1
𝑁3 = 2𝐴 𝛼3 + 𝛽3 𝑟 + 𝛾3 𝑧 ;
115
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𝜕𝑁1
𝛽1
=
𝜕𝑟
2𝐴
𝑁1
1 ∝1
𝛾1 𝑧
=
+ 𝛽1 +
𝑟
2𝐴 𝑟
𝑟
𝜕𝑁1
𝛾1
=
𝜕𝑧
2𝐴
𝜕𝑁2
𝛽2
=
𝜕𝑟
2𝐴
𝑁2
1 ∝2
𝛾2 𝑧
=
+ 𝛽2 +
𝑟
2𝐴 𝑟
𝑟
𝜕𝑁2
𝛾2
=
𝜕𝑧
2𝐴
ww
𝜕𝑁3
𝛽3
=
𝜕𝑟
2𝐴
w.E
asy
En
gi
𝜕𝑁3
𝛾3
=
𝜕𝑧
2𝐴
SC
A
D
𝑁3
1 ∝3
𝛾3 𝑧
=
+ 𝛽3 +
𝑟
2𝐴 𝑟
𝑟
nee
Above values substitute in eqn 7
𝛽1
𝑒𝑟
𝑒𝜃
=
𝑒𝑧
𝛾𝑟𝑧
𝛼₁
𝛾₁𝑧
1
+ 𝛽₁ +
𝑟
𝑟
2𝐴
0
{e} =
[B]{u}
𝛾1
𝛽1 = 𝑧2 − 𝑧3
𝛾1 = 𝑟3 − 𝑟2
𝛼1 = 𝑟2 𝑧3 − 𝑟3 𝑧2
1
[B] = 2𝐴
𝛽1
𝛾₁𝑧
+ 𝛽₁ + 𝑟
𝑟
0
𝛾1
𝛼₁
0
0
𝛾1
𝛽1
0
0
𝛾1
𝛽1
𝛼2
𝛾2 𝑧
+ 𝛽2 +
𝑟
𝑟
0
𝛾2
𝛽2 = 𝑧3 − 𝑧1
𝛾2 = 𝑟1 − 𝑟3
𝛼2 = 𝑟3 𝑧1 − 𝑟1 𝑧3
𝛽2
𝛾 𝑧
+ 𝛽2 + 2𝑟
𝑟
0
𝛾2
𝛼2
𝛽2
0
0
𝛾2
𝛽2
rin
g
0
0
𝛾2
𝛽2
0
.ne
𝛼3
𝛾3 𝑧
+ 𝛽3 +
𝑟
𝑟
0
𝛾3
0
t
𝛾3
𝛽3
𝑢1
𝑤1
𝑢2
𝑤2
𝑢3
𝑤3
𝛽3 = 𝑧1 − 𝑧2
𝛾3 = 𝑟2 − 𝑟1
𝛼3 = 𝑟1 𝑧2 − 𝑟2 𝑧1
𝑏3
𝛾 𝑧
+ 𝛽3 + 3𝑟
𝑟
0
𝛾3
𝛼3
𝑏3
0
0
𝛾3
𝛽3
116
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Downloaded From : www.EasyEngineering.net
UNIT V
ISOPARAMETRIC FORMULATION
PART A
1. What do you mean by uniqueness of mapping?
It is absolutely necessary that a point in parent element represents only one point in the isoperimetric
element. Some times, due to violent distortion it is possible to obtain undesirable situation of
nonuniqueness. Some of such situations are shown in Fig. If this requirement is violated determinant of
Jacobiam matrix (to be explained latter) becomes negative. If this happens coordinate transformation fails
and hence the program is to be terminated and mapping is corrected.
ww
w.E
asy
E
Non Uniqueness of Mapping
2. What do you mean by iso parametric element?(April/May 2011)
SC
A
D
If the shape functions defining the boundary and displacements are the same, the element is called
as isoparametric element and all the eight nodes are used in defining the geometry and displacement.
ngi
n
eer
in
3. What do you mean by super parametric element?
The element in which more number of nodes are used to define geometry compared to the number of
nodes used to define displacement are known as superparametric element.
g.n
e
t
4. What do you mean by sub parametric element?
The fig shows subparametric element in which less number of nodes are used to define geometry
compared to the number of nodes used for defining the displacements. Such elements can be used
advantageously in case of geometry being simple but stress gradient high.
117
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5. What do you mean by iso parametric formulation?(April/May 2011)
The principal concept of isoparametric finite element formulation is to express the element
coordinates and element displacements in the form of interpolations using the natural coordinate system of
the element. These isoparametric elements of simple shapes expressed in natural coordinate system,
known as master elements, are the transformed shapes of some arbitrary curves sided actual elements
expressed in Cartesian coordinate system.
6. What is a Jacobian matrix of transformation?(April/May 2011)
ww
w.E
asy
E
It‟s the transformation between two different co-ordinate system. This transformation is
used to evaluate the integral expression involving „x‟ interms of expressions involving ε.
XB
1
f ( x)dx f ( )d
1
SC
A
D
xA
The differential element dx in the global co-ordinate system x is related to differential
element dε in natural co-ordinate system ε by
ngi
n
dx = dx/ dε . dε
dx = J . dε
𝐽
Jacobian matrix of transformation J =dx/ dε = 11
𝐽21
7. Differentiate the serendipity and langrangian elements
Serendipity elements
eer
in
𝐽12
𝐽22
langrangian elements
g.n
e
In discretized element
In discretized element, if nodes
If nodes lies on corner, then the
are present in both centre of element
element are known as serendipity
and corner are known as langrangian
elements.
elements.
t
8. Explain Gauss quadrature rule.(Nov/Dec 2012), (April/May 2011)
The idea of Gauss Quadrature is to select “n” Gauss points and “n” weight functions such that the
integral provides an exact answer for the polynomial f(x) as far as possible, Suppose if it is necessary to
evaluate the following integral using end point approximation then
1
I=
f ( x)dx
1
118
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The solution will be
1
f ( x)dx w f ( x ) w f ( x ) ......... w f ( x )
1
1
2
2
n
n
1
w1,w2,…………..…., wnare weighted function, x1,x2……………….., xnare Gauss points
9. What are the differences between implicit and explicit direct integration methods?
Implicit direct integration methods:
(i)
Implicit methods attempt to satisfy the differential equation at time „t‟ after the solution at time “t∆t”is found
(ii)
These methods require the solution of a set of linear equations at each time step.
(iii)
Normally larger time steps may be used.
(iv)
Implicit methods can be conditionally or unconditionally stable.
ww
w.E
asy
E
Explicit direct integration methods:
These methods do not involve the solution of a set of linear equations at each step.
SC
A
D
(i)
(ii)
Basically these methods use the differential equations at time „t‟ to predict a solution at time
“t+∆t”
ngi
n
(iii)
Normally smaller time steps may be used
(iv)
All explicit methods are conditionally stable with respect to size of time step.
eer
in
(v)
Explicit methods initially proposed for parabolic PDES and for stiff ODES with widely separated
time constants.
10. State the three phases of finite element method.
The three phases of FEM is given by,
(i)
Preprocessing
(ii)
Analysis
(iii)
Post Processing
g.n
e
t
11. List any three FEA software.(Nov/Dec 2014)
The following list represents FEA software as,
(i)
ANSYS
(ii)
NASTRAN
(iii)
COSMOS
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PART-B
1.
A four noded rectangular element is shown in Fig. Determine the following
1. jacobian matrix 2. Strain – Displacement matrix 3. Element Stresses.
SC
A
D
ww
w.E
asy
E
T
Take E = 2 10 N/mm ; v = 0.25 ; u = 0, 0, 0, 0.003, 0.004, 0.006, 0.004, 0, 0
5
2
Assume the plane Stress condition.
Given Data
ngi
n
Cartesian co – ordinates of the points 1,2,3 and 4
𝑥1 = 0;
𝑦1 = 0
𝑥2 = 2;
𝑥3 = 2;
𝑥4 = 0;
𝑦2 = 0
𝑦3 = 1
𝑦4 = 1
Young‟s modulus, E = 2 105 N/mm2
Poisson‟s ratio v = 0.25
0
0
0.003
0.004
Displacements, u =
0.006
0.004
0
0
Natural co-ordinates , ε = 0 , = 0
eer
in
g.n
e
ε=0;=0
t
To find:
1. Jacobian matrix, J
2. Strain – Displacement matrix [B]
3. Element Stress σ.
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Formulae used
J =
J22 −J12 0 0
1
1
0 −J21 J11
𝐵 = 𝐉 0
4
−J21 J11 J22 −J12
𝐽11 𝐽12
𝐽21 𝐽22
0
0
0
−(1 − )
(1 − )
(1 + ) 0 −(1 + )
0
0
0
0
−(1 − 𝜀)
−(1 + 𝜀)
(1 + 𝜀)
(1 − 𝜀)
−(1 − )
(1 − )
−(1 + )
0
0
0 (1 + )
0
−(1 − 𝜀)
−(1 + 𝜀) 0 (1 + 𝜀)
(1 − 𝜀)
0
0
0
Solution :Jacobian matrix for quadrilateral element is given by,
ww
w.E
asy
E
J =
𝐽11 𝐽12
𝐽21 𝐽22
1
4
SC
A
D
Where ,
1
J11 = 4 −(1 − )𝑥1 + (1 − )𝑥2 +(1 + )𝑥3 −(1 + )𝑥4
ngi
n
(1)
−(1 − )𝑦1 + (1 − )𝑦2 +(1 + )𝑦3 −(1 + )𝑦4
(2)
J21 = 4 −(1 − 𝜀)𝑥1 − (1 + 𝜀)𝑥2 +(1 + 𝜀)𝑥3 +(1 − 𝜀)𝑥4
(3)
J12 =
1
1
J22 = 4 −(1 − 𝜀)𝑦1 − (1 + 𝜀)𝑦2 +(1 + 𝜀)𝑦3 +(1 − 𝜀)𝑦4
eer
in
(4)
g.n
e
Substitute 𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑦1, 𝑦2, 𝑦3, 𝑦14, ε and values in equation (1), (2),(3) and (4)
1
(1) J11 = 4 0 + 2 + 2 − 0
t
𝐉𝟏𝟏 = 1
(2)
1
J12 = 4 0 + 0 + 1 − 1
J12 = 0
(3)
1
J21 = 4 0 − 2 + 2 − 0
J21 = 0
(4)
1
J22 = 4 −0 − 0 + 1 + 1
J22 = 0.5
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J =
𝐽11 𝐽12
𝐽21 𝐽22
J =
Jacobian matrix
1 0
0 0.5
(5)
J = 10.5- 0
J = 0.5
We Know that, Strain – Displacement matrix for quadrilateral element is,
1
𝐵 = 𝐉
J22 −J12 0 0
1
0
0 −J21 J11
4
−J21 J11 J22 −J12
0
0
0
−(1 − )
(1 − )
(1 + ) 0 −(1 + )
0
0
0
0
−(1 − 𝜀)
−(1 + 𝜀)
(1 + 𝜀)
(1 − 𝜀)
−(1 − )
(1 − )
−(1 + )
0
0
0 (1 + )
0
−(1 − 𝜀)
−(1 + 𝜀) 0 (1 + 𝜀)
(1 − 𝜀)
0
0
0
ww
w.E
asy
E
Substitute 𝐉𝟏𝟏 , 𝐉𝟏𝟐, 𝐉𝟐𝟏, 𝐉𝟐𝟐 𝐉 , 𝜺 𝐚𝐧𝐝 𝐯𝐚𝐥𝐮𝐞𝐬
−1 0 1 0 10−1 0
0
1 −1 0 −1 0 10 1 0
1 4
0 −1 0 1 01 0 −1
1
0 −1 0 −101 0 1
SC
A
D
1
𝐵 = 0.5
0.5 0 0
0 0 0
0 1 0.5
ngi
n
−0.5 0 0.5 0 0.5 0 −0.5 0
1
𝐵 = 0.54 0 −1 0 −1 0 1 0
1
−1 −0.5−10.5 1 0.5 1 −0.5
−1 0 1 0 10−1 0
0.5
= 0.54 0 −2 0 −202 0 2
−2−1−2 1 21 2 −1
eer
in
g.n
e
t
−1 0 1 0 10−1 0
𝐵 = 0.25 0 −2 0 −202 0 2
−2−1−2 1 21 2 −1
We know that,
Element stress, σ = 𝐃 𝑩 𝒖
For plane stress condition,
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Stress- strain relationship matrix,
D =
1𝑣 0
𝑣
1 0
1−𝑣 2 0 0 1−𝑣
𝐸
2
=
2 10 5
1 0.25 0
0
0.25 1
1−0.25
0
0
1− (0.25)2
2
1 0.25 0
= 213.33 103 0.25 1
0
0 0.375
0
41 0
= 213.33103 0.25 1 4 0
0 0 1.5
ww
w.E
asy
E
41 0
= 53.333103 1 4 0
0 0 1.5
SC
A
D
Substitute 𝐷 , 𝐵 and 𝑢
−1 0 1 0 10−1 0
41 0
σ = 53.333103 1 4 0 0.25 0 −2 0 −202 0 2
0 0 1.5
−2−1−2 1 21 2 −1
ngi
n
−4 2 4 −24 2 −4 2
= 53.333103 0.25 −1 −8 1 −81 8 −1 8
−3−1.5−31.531.5 3 −1.5
=13.33310
3
0
0
0.003
0.004
0.006
0.004
0
0
eer
in
0
0
0.003
0.004
0.006
0.004
0
0
g.n
e
t
0 + 0 + 4 × 0.003 + −2 × 0.004 + 4 × 0.006 + 2 × 0.004 + 0 + 0
0 + 0 + 1 + 0.003 + −8 × 0.004 + 1 × 0.006 + 8 × 0.004 + 0 + 0
0 + 0 + −3 × 0.003 + 1.5 × 0.004 + 3 × 0.006 + 1.5 × 0.004 + 0 + 0
0.036
𝜎 = 13.333103 0.009
0.021
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480
𝜎 = 120 N/m2
280
Result :
J = 0.5
480
𝜎 = 120 N/m2
280
For the isoparametric quadrilateral element shown in Fig. the Cartesian co-ordinate of
point P are (6,4). The loads 10KN and 12KN are acting in x and y direction on the point P.
Evaluate the nodal equivalent forces.
ww
w.E
asy
E
SC
A
D
2.
ngi
n
Givendata :
eer
in
Cartesian co- ordinates of point P,
X = 6;
y=4
g.n
e
t
The Cartesian co-ordinates of point 1,2,3 and 4 are
𝑥1 = 2;
𝑦1 = 1
𝑥2 = 8;
𝑦2 = 4
𝑥3 = 6;
𝑦3 = 6
𝑥4 = 3;
𝑦4 = 5
Loads ,F𝑥 = 10𝐾𝑁F𝑦 = 12𝐾𝑁
To find : Nodal equivalent forces for x and y directions,
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i,e., F1𝑥 , F2𝑥 , F3𝑥 , F4𝑥 , F1𝑦 , F2𝑦 , F3𝑦 , F4𝑦
Formulae Used
1
N1 = 4 (1-ε) (1-)
1
N2 = 4 (1+ ε) (1- )
1
N3 = 4 (1+ ε) (1+)
1
N4 = 4 (1-ε) (1+)
Fx
Element force vector, F e = N T F
y
ww
w.E
asy
E
solution:
Shape functions for quadrilateral elements are,
1
1
SC
A
D
N1 = 4 (1-ε)(1-)(1)
N2 = 4 (1+ ε) (1- )
1
N3 = (1+ ε) (1+)
4
1
N4 = 4 (1-ε) (1+)
ngi
n
Cartesian co-ordinates of the point,P(x,y)
(2)
(3)
eer
in
(4)
𝑥 = N1 𝑥1 +N2 𝑥2 + N3 𝑥3 + N4 𝑥4
g.n
e
(5)
𝑦 = N1 𝑦1 +N2 𝑦2 + N3 𝑦3 + N4 𝑦4
(6)
t
Substitute 𝑥,𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , 𝑁1 , 𝑁2 , 𝑁3 , 𝑎𝑛𝑑 𝑁4 values in equation.
1
6 = 4 [(1-ε) (1-) 2 +(1+ε) (1- )8 + (1+ ε) (1+)6 +(1 - ε) (1+)3]
24= [(1--ε+ε)2+(1-+ε-ε)8+(1++ε+ε)6+(1+-ε-ε)3]
24 = 19-+9ε-3ε
5 = -+9ε - 3ε
9ε - - 3ε = 5
(7)
Substitute 𝑦,𝑦1 , 𝑦2 , 𝑦3 , 𝑦4 , 𝑁1 , 𝑁2 , 𝑁3 , 𝑎𝑛𝑑 𝑁4 values in equation.
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1
4 = 4 [(1-ε) (1-) 1 +(1+ε) (1- )4 + (1+ ε) (1+)6 +(1 - ε) (1+)5]
16 = [1--ε+ε+4-4+4ε-4ε+6+6+6ε+6ε+5+5-5ε-5ε]
16= [16+6+4ε-2ε]
4ε + 6 - 2ε = 0
(8)
Equation (7) multiplied by 2 and equation (8) multiplied by (-3).
18ε - 2 - 6ε = 10
(9)
-12ε - 18 + 6ε = 0
ww
w.E
asy
E
6ε – 20 = 10
-20 = 10 - 6ε
20 = 6ε -10
6𝜀−10
SC
A
D
=
20
ngi
n
Substituting value in equation (7),
9ε – (0.3ε – 0.5) - 3ε (0.3ε – 0.5) = 5
= 0.3ε – 0.5
10.2ε – 0.9ε2 – 4.5 = 0
0.9ε2 - 10.2ε + 4.5 = 0
ε=
=
(10)
eer
in
10.2± (−10.2)2 −4 0.9 (4.5)
2(0.9)
(11)
g.n
e
t
10.2−9.372
1.8
ε = 0.46
Substitute ε and values in equation (1),(2),(3) and (4)
(1) N1 =
1
4
(1 - 0.46) (1+ 0.362)
N1 = 0.18387
(2)
N2 =
1
4
(1 + 0.46) (1+ 0.362)
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N2 = 0.49713
N3 =
(3)
1
(1 + 0.46) (1 - 0.362)
4
N3 = 0.23287
N4 =
(4)
1
(1 - 0.46) (1 - 0.362)
4
N3 = 0.08613
We know that,
Fx
Element force vector, F e = N T F
(12)
y
ww
w.E
asy
E
F1𝑥
F2𝑥
F3𝑥
F4x
𝑁1
𝑁2
=
𝑁3
𝑁4
F1𝑥
F2𝑥
F3𝑥
F4x
0.18387
0.49713
=
0.23287
0.08613
F1𝑥
F2𝑥
F3𝑥
F4x
1.8387
4.9713
=
KN
2.3287
0.8613
SC
A
D
F𝑥
ngi
n
10
eer
in
g.n
e
t
Similarly,
F1𝑦
F2𝑦
F3𝑦
F4y
𝑁1
𝑁2
=
𝑁3
𝑁4
F𝑦
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F1𝑦
F2𝑦
F3𝑦
F4y
0.18387
0.49713
=
0.23287
0.08613
F1𝑦
F2𝑦
F3𝑦
F4y
2.20644
5.96556
=
KN
2.79444
1.03356
12
Result:
Nodal forces for x directions,
F1𝑥
F2𝑥
F3𝑥
F4x
1.8387
4.9713
=
KN
2.3287
0.8613
ww
w.E
asy
E
Nodal forces for y directions,
4.
2.20644
5.96556
=
KN
2.79444
1.03356
SC
A
D
F1𝑦
F2𝑦
F3𝑦
F4y
ngi
n
Derive the shape function for the Eight Noded Rectangular Element
eer
in
Consider a eight noded rectangular element is shown in fig. It belongs to the
serendipity family of elements. It consists of eight nodes, which are located on the boundary.
g.n
e
We know that, shape function N1 = 1 at node 1 and 0 at all other nodes.
t
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N1=0 at all other nodes
N1 has to be in the form of N1 =C(1- ε)(1-)(1+ε+)
(1)
Where C is constant
Substitute ε = -1 and = -1 in equation (1)
N1 = C (1+1)(1+1)(-1)
1 = -4C
1
C =-4
ww
w.E
asy
E
Substitute C value in equation
1
N1= -4 (1+ ε) (1 +) (1+ε+)
(2)
SC
A
D
At node 2 :(Coordinates ε =1,= -1)
Shape Function N2 = 1 at node 2
ngi
n
N2 = 0 at all other nodes
N2has to be in the form of N2 =C(1 +ε)(1-)(1-ε+)
Substitute ε = 1 and = -1 in equation (3)
N2 = C (1+1) (1+1) (-1)
eer
in
1 = -4C
1
C =-4
(3)
g.n
e
t
Substitute C value in equation (3)
1
N2= -4 (1+ ε) (1 - ) (1- ε +)
(4)
At node 3 :(Coordinates ε =1,= 1)
Shape Function N3 = 1 at node 3
N3 = 0 at all other nodes
N3has to be in the form of N3 =C(1+ε)(1+)(1- ε - )
(5)
Substitute ε = 1 and = 1 in equation (5)
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N3 = C (1+1) (1+1) (-1)
1 = -4C
1
C =-4
Substitute C value in equation (5)
1
N3= − 4 (1+ ε) (1+ ) (1- ε - )
(6)
At node 4 :(Coordinates ε =- 1,= 1)
Shape Function N4 = 1 at node 4
N4 = 0 at all other nodes
ww
w.E
asy
E
N4 has to be in the form of N4 =C(1- ε)(1 + )(1+ε - )
(7)
Substitute ε = -1 and = 1 in equation (7)
SC
A
D
N4 = C (1+1) (1+1) (-1)
1 = -4C
1
C = −4
Substitute C value in equation (3)
1
ngi
n
N4= - 4 (1- ε) (1 + ) (1+ ε -)
Now , we define N5,N6,N7 and N8 at the mid points.
eer
in
At node 5 :(Coordinates ε = - 1,= - 1)
(8)
g.n
e
Shape Function N5 = 1 at node 5
t
N5 = 0 at all other nodes
N5has to be in the form of N5 =C(1- ε)(1 -)(1+ε )
N5 = C (1- ε2)(1 - )
(9)
Substitute ε = 0 and = -1 in equation (9)
N5 = C (1-0)(1+1)
1 = 2C
1
C=2
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Substitute C value in equation (9)
1
N5= 2 (1- ε2)(1 - )
(10)
At node 6 :(Coordinates ε = 1,= - 1)
Shape Function N6 = 1 at node 6
N6 = 0 at all other nodes
N6 has to be in the form of N6 =C (1+ε)(1 - )(1+ )
N6 = C (1 + ε)(1 - 2)
(11)
Substitute ε = 1 and = 0 in equation (11)
N6 = C (1+1) (1 - 0)
ww
w.E
asy
E
1 = 2C
1
C=2
Substitute C value in equation (11)
1
2
(1+ ε)(1 - 2)
SC
A
D
N6 =
At node 7 :(Coordinates ε = 1,= 1)
Shape Function N7 = 1 at node 7
ngi
n
(12)
eer
in
N7 = 0 at all other nodes
N7 has to be in the form of N7 =C (1+ε)(1 + )(1- ε )
N7 = C (1 – ε2)(1 + )
Substitute ε = 0 and = 1 in equation (12)
N7 = C (1-0) (1 + 1)
g.n
e
(13)
t
1 = 2C
1
C=2
Substitute C value in equation (13)
N7 =
1
2
(1 – ε2)(1 + )
(14)
At node 8 :(Coordinates ε = -1,= 1)
Shape Function N8 = 1 at node 8
N8 = 0 at all other nodes
N8 has to be in the form of N8 =C (1-ε)(1 + )(1- )
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N8 = C (1 – ε)(1 -2)
(15)
Substitute ε = -1 and = 0 in equation (15)
N8 = C (1+1) (1 - 0)
1 = 2C
1
C=2
Substitute C value in equation (15)
1
N8 =
2
(1 – ε)(1 -2)
(16)
Shape Functions are,
1
N1= - 4 (1+ ε) (1 +) (1+ε+)
ww
w.E
asy
E
1
N2 = -
4
(1+ ε) (1 - ) (1- ε + )
1
N3= − 4 (1+ ε) (1 + ) (1- ε - )
1
N5 =
N6 =
N7 =
N8 =
5.
SC
A
D
N4= - 4 (1- ε) (1 + ) (1+ ε -)
1
2
1
2
1
2
1
2
(1- ε2)(1 - )
(1+ ε)(1 - 2)
(1 – ε2)(1 + )
ngi
n
(1 – ε)(1 - )
2
eer
in
g.n
e
Derive the shape function for 4 noded rectangular parent element by using natural coordinate system and co-ordinate transformation
η
4 (-1,1)
η (+1)
3 (1,1)
ε
t
ε (+1)
ε (-1)
1(-1,-1)
η (-1)
2 (1,-1)
Consider a four noded rectangular element as shown in FIG. The parent element is defined in ε
and η co-ordinates i.e., natural co-ordinates ε is varying from -1 to 1 and η is also varying -1 to 1.
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We know that,
Shape function value is unity at its own node and its value is zero at other nodes.
At node 1: (co-ordinate ε = -1, η = -1)
Shape function N1 = 1 at node 1.
N1 = 0 at nodes 2, 3 and 4
N1has to be in the form of N1 = C (1 - ε) (1 -η)
(1)
Where, C is constant.
Substitute ε = -1 and η = -1 in equation (1)
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N1 = C (1+1)(1+1)
N1= 4C
1
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C=4
Substitute C value in equation (1)
1
N1 = 4(1 - ε) (1 -η)
ngi
n
(2)
N2 = 0 at nodes 1, 3 and 4
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N1has to be in the form of N2 = C (1 + ε) (1 -η)
(3)
At node 2: (co-ordinate ε = 1, η = -1)
Shape function N2 = 1 at node 2.
Where, C is constant.
g.n
e
t
Substitute ε = 1 and η = -1 in equation (3)
N2 = C (1+1) (1+1)
N2 = 4C
1
C=4
Substitute C value in equation (1)
1
(4)
N2 = 4(1 + ε) (1 -η)
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At node 3: (co-ordinate ε = 1, η = 1)
Shape function N3 = 1 at node 3.
N3 = 0 at nodes 1, 2 and 4
N1has to be in the form of N3 = C (1 + ε) (1 +η)
(5)
Where, C is constant.
Substitute ε = 1 and η = 1 in equation (5)
N3 = C (1+1)(1+1)
N3 = 4C
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1
C=4
Substitute C value in equation (1)
1
(6)
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N3 = 4(1 +ε) (1 + η)
At node 4: (co-ordinate ε = -1, η = 1)
Shape function N4 = 1 at node 4.
ngi
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N4 = 0 at nodes 1, 2 and 3
N1has to be in the form of N4 = C (1 - ε) (1 +η)
Where, C is constant.
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(7)
Substitute ε = -1 and η = 1 in equation (1)
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N4 = C (1+1) (1+1)
t
N4 = 4C
1
C=4
Substitute C value in equation (1)
1
(8)
N4 = 4(1 - ε) (1 +η)
Consider a point p with co-ordinate (ε ,η). If the displacement function u =
𝑢
represents the
𝑣
displacements components of a point located at (ε ,η) then,
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u = N1 𝑢1 +N2 𝑢2 +N3 𝑢3 +N4 𝑢4
v = N1 𝑣1 +N2 𝑣2 +N3 𝑣3 +N4 𝑣4
It can be written in matrix form as,
u=
𝑢
=
𝑣
𝑁1 0 𝑁2 0
0 𝑁1 0 𝑁2
𝑁3 0 𝑁4 0
0 𝑁3 0 𝑁4
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𝑢1
𝑣1
𝑢2
𝑣2
𝑢3
𝑣3
𝑢4
𝑣4
(9)
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In the isoparametric formulation i,e., for global system, the co-ordinates of the nodal points are
𝑥1 , 𝑦1 , 𝑥2 , 𝑦2 , 𝑥3 , 𝑦3 , and 𝑥4 , 𝑦4 . In order to get mapping the co-ordinate of point p is
defined as
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𝑥 = N1 𝑥1 +N2 𝑥2 +N3 𝑥3 +N4 𝑥4
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𝑦 = N1 𝑦1 +N2 𝑦2 +N3 𝑦3 +N4 𝑦4
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The above equation can be written in matrix form as,
𝑥
u= 𝑦 =
𝑁3 0 𝑁4 0
0 𝑁3 0 𝑁4
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(10)
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6.
𝑁1 0 𝑁2 0
0 𝑁1 0 𝑁2
𝑥1
𝑦1
𝑥2
𝑦2
𝑥3
𝑦3
𝑥4
𝑦4
For the isoparametric four noded quadrilateral element shown in fig. Determine the
Cartesian co-ordinates of point P which has local co-ordinatesε= 0.5 , η =0.5
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Given data
Natural co-ordinates of point P
ε= 0.5
η =0.5
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Cartesian co-ordinates of the point 1,2,3 and 4 P 𝑥, 𝑦
𝑥1 = 1;
𝑦1 = 1
𝑥2 = 5;
𝑦2 = 1
𝑥3 = 6;
𝑦3 = 6
𝑥4 = 1;
𝑦4 = 4
To find : Cartesian co-ordinates of the point P(x,y)
Formulae used:
Co -ordinate, 𝑥 = N1 𝑥1 +N2 𝑥2 +N3 𝑥3 +N4 𝑥4
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Co-ordinate, 𝑦 = N1 𝑦1 +N2 𝑦2 +N3 𝑦3 +N4 𝑦4
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Solution
Shape function for quadrilateral elements are,
1
ngi
n
N1 = 4(1 - ε) (1 -η)
1
N2 = (1 + ε) (1 -η)
4
1
N3 = 4(1 +ε) (1 + η)
1
N4 = 4(1 - ε) (1 +η)
Substitute ε and η values in the above equations,
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1
N1 = 4(1 – 0.5) (1 –0.5) = 0.0625
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N2 = 4(1 + 0.5) (1 –0.5) = 0.1875
1
N3 = 4(1 +0.5) (1 + 0.5) =0.5625
1
N4 = 4(1 – 0.5) (1 +0.5) = 0.1875
We know that,
Co-ordinate, 𝑥 = N1 𝑥1 +N2 𝑥2 +N3 𝑥3 +N4 𝑥4
= 0.0625×1+0.1875×5+0.5625×6+0.1875×1
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Downloaded From : www.EasyEngineering.net
𝑥 = 4.5625
Similarly,
Co-ordinate, 𝑦 = N1 𝑦1 +N2 𝑦2 +N3 𝑦3 +N4 𝑦4
= 0.0625×1+0.1875×1+0.5625×6+0.1875×4
y = 4.375
𝟏
𝟏
𝒙
𝟐
𝒆
+
𝒙
+
dx using Gaussian integration with one,
−𝟏
𝒙+𝟕
,two , three integration points and compare with exact solution
Evaluate the integral I =
Given:
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1
−1
I=
𝑒 𝑥 + 𝑥2 +
1
𝑥+7
dx
To Find:
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7.
Evaluate the integral by using Gaussian.
Formulae used:
1
I=
−1
ngi
n
𝑒 𝑥 + 𝑥2 +
1
𝑥+7
f 𝑥1 ,w1 f 𝑥1 ,
w1 f 𝑥1 + w2 f 𝑥2 + w3 f 𝑥3
Solution
1. point Gauss quadrature
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dx
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𝑥1 = 0; w1 = 2
f 𝑥 =
f 𝑥1 =
𝑒 𝑥 + 𝑥2 +
𝑒0 + 0 +
1
𝑥+7
1
0+7
f 𝑥1 = 1.1428
w1 f 𝑥1 = 2 ⨯1.1428
= 2.29
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Downloaded From : www.EasyEngineering.net
2. point Gauss quadrature
𝑥1 =
1
3
=0.5773;
1
𝑥2 = − 3= -0.5773;
w1 = w2 = 1
f 𝑥 =
𝑒 𝑥 + 𝑥2 +
1
𝑥+7
f 𝑥1 = 𝑒 0.5773 + 0.57732 +
1
0.5773 +7
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f 𝑥1 = 1.7812 + 0.33327 + 0.13197
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f 𝑥1 = 2.246
w1 f 𝑥1 = 1 ⨯2.246
= 2.246
f 𝑥2 = 𝑒 −0.5773 + (−0.5773)2 +
ngi
n
1
−0.5773 +7
= 0.5614 + 0.3332+0.15569
f 𝑥2 = 1.050
w2 f 𝑥2 = 1 ⨯1.050
= 1.050
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w1 f 𝑥1 + w2 f 𝑥2 = 2.246 + 1.050
= 3.29
3. point Gauss quadrature
𝑥1 =
3
5
=0.7745;
𝑥2 = 0:
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𝑥1 = −
3
5
= - 0.7745;
5
w1 = 9 = 0.5555;
8
w2 = 9 = 0.8888
5
w2 = 9 = 0.5555
1
f 𝑥 = 𝑒 𝑥 + 𝑥 2 + 𝑥+7
1
f 𝑥1 = 𝑒 0.7745 + 0.77452 + 0.7745 +7
f 𝑥1 = 2.1697 + 0.6 + 0.1286
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f 𝑥1 = 2.898
w1 f 𝑥1 = 0.55555⨯2.898
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= 1.610
1
f 𝑥2 = 1+ 7
f 𝑥2 = 1.050
ngi
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w2 f 𝑥2 = 0.888⨯1.143
= 1.0159
eer
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w1 f 𝑥1 + w2 f 𝑥2 + w3 f 𝑥3 = 1.160 + 1.0159 +0.6786
= 2.8545
Exact Solution
I=
1
−1
= 𝑒 𝑥 1−1 +
1
1
𝑒 𝑥 + 𝑥 2 + 𝑥+7 dx
𝑥3
1
3 −1
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+ ln(𝑥 + 7) 1−1
−1
= 𝑒 +1 − 𝑒 −1 + 3 − 3 + ln(1 + 7) − ln(−1 + 7)
2
= 2.7183 − 0.3678 + 3 + ln(8) − ln(6)
= 2.3505 +0.6666 + 2.0794 − 1.7917
= 3.0171 + 0.2877 = 3.3048
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UNIVERSITY QUESTION PAPERS
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