UNIT 1
1.What is meant by finite element?
A small units having definite shape of geometry and nodes is called finite element.
2. What is meant by node or joint?
Each kind of finite element has a specific structural shape and is inter- connected with
the adjacent element by nodal point or nodes. At the nodes, degrees of freedom are
located. The forces will act only at nodes at any others place in the element.
3. What is the basic of finite element method?
Discretization is the basis of finite element method. The art of subdividing a structure
in to convenient number of smaller components is known as discretization.
4. What are the types of boundary conditions?
I. Primary boundary conditions
II. Secondary boundary conditions
5. State the methods of engineering analysis?
I. Experimental methods
II. Analytical methods
III. Numerical methods or approximate methods
6. What are the types of element?
I. 1D element
II. 2D element
III. 3D element
7. State the three phases of finite element method.
1. Preprocessing
2. Analysis
3. Post Processing
8. What is structural problem?
Displacement at each nodal point is obtained. By these displacements solution stress and
strain in each element can be calculated.
9. What is non structural problem?
Temperature or fluid pressure at each nodal point is obtained. By using these values
properties such as heat flow fluid flow for each element can be calculated.
10. What are the methods are generally associated with the finite element analysis?
1. Force method
2. Displacement or stiffness method.
11. Explain stiffness method.
Displacement or stiffness method, displacement of the nodes is considered as the
unknown of the problem. Among them two approaches, displacement method is
desirable.
12. What is meant by post processing?
Analysis and evaluation of the solution result is referred to as post processing. Postprocessor
computer program help the user to interpret the result by displaying them in graphical form.
13. Name the variation methods.
1. Ritz method.
2. Ray-Leigh Ritz method.
14. What is meant by degrees of freedom?
When the force or reaction act at nodal point node is subjected to deformation. The
deformation includes displacement rotation, and or strains. These are collectively known as degrees
of freedom
15. What is meant by discretization and assemblage?
The art of subdividing a structure in to convenient number of smaller components is known
as discretization. These smaller components are then put together. The process of uniting the
various elements together is called assemblage.
16. What is Rayleigh-Ritz method?
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.
17. What is Aspect ratio?
It is defined as the ratio of the largest dimension of the element to the smallest dimension. In
many cases, as the aspect ratio increases the in accuracy of the solution increases. The conclusion of
many researches is that the aspect ratio
18. What is truss element?
The truss elements are the part of a truss structure linked together by point joint which
transmits only axial force to the element.
19. What are the h and p versions of finite element method?
It is used to improve the accuracy of the finite element method. In h version, the order of
polynomial approximation for all elements is kept constant and the numbers of elements are
increased. In p version, the numbers of elements are maintained constant and the order of
polynomial approximation of element is increased.
20. Name the weighted residual method
1. Point collocation method
2. Sub domain collocation method
3. Least squares method
4. Galerkins method.
21. Distinguish ID bar element and Beam Element (May/June 2011)
1D bar element: Displacement is considered.
1D beam element: Displacement and slope is considered
22. 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.
d2y/dx
2 - a(x) dy/dx β b(x)y βc(x) = 0 with boundary conditions, y(m) = S and y(n) = T.
23. What do you mean by weak formulation? State its advantages. (April/May 2015), (May/June
2013)
A weak form is a weighted integral statement of a differential equation in which the
differentiation is distributed among the dependent variable and the weight function and also
includes the natural boundary conditions of the problem.
A much wider choice of trial functions can be used.
The weak form can be developed for any higher order differential equation.
Natural boundary conditions are directly applied in the differential equation.
The trial solution satisfies the essential boundary conditions.
24. Why are polynomial types of interpolation functions preferred over trigonometric functions?
(May/June 2013)
Polynomial functions are preferred over trigonometric functions due to the following
reasons:
1. It is easy to formulate and computerize the finite element equations
2. It is easy to perform differentiation or integration
3. The accuracy of the results can be improved by increasing the order of the polynomial.
25. 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.
26. 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.
27. Distinguish Natural & Essential boundary condition (May/June 2009)
There are two types of boundary conditions.
They are:
1. Primary boundary condition (or) Essential boundary condition
The boundary condition, which in terms of field variable, is known as primary
boundary condition.
2. Secondary boundary condition or natural boundary conditions
The boundary conditions, which are in the differential form of field variables, are
known as secondary boundary condition.
Example: A bar is subjected to axial load as shown in fig.
In this problem, displacement u at node 1 = 0, that is primary boundary condition.
EA du/dx = P, that is secondary boundary condition.
28. Compare Ritz method with nodal approximation method.(Nov/Dec 2014), (Nov/Dec 2012)
Similarity:
(i) Both methods use approximating functions as trial solution
(ii) Both methods take linear combinations of trial functions.
(iii) In both methods completeness condition of the function should be satisfied
(iv) In both methods solution is sought by making a functional stationary.
Difference
(i) Rayleigh-Ritz method assumes trial functions over entire structure, while finite element method
uses trial functions only over an element.
(ii) The assumed functions in Rayleigh-Ritz method have to satisfy boundary conditions over entire
structure while in finite element analysis, they have to satisfy continuity conditions at nodes and
sometimes along the boundaries of the element. However completeness condition should be
satisfied in both methods.
29. 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.
30. 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 canβt be achieved. It means that it may
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.
31. 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.
32. State the principleofminimum potential energy. (Nov/Dec 2010)
Amongallthedisplacementequationsthatsatisfiedinternalcompatibilityandthe
boundaryconditionthosethatalsosatisfytheequationofequilibriummakethe potential energy minimum
is a stable system.
UNIT-1
PART-B
1. The following differential equation is available for a physical phenomenon. π¨π¬ = π ππ
π ππ+
ππ = π, The boundary conditions are u(0) = 0, π¨π¬ = π π
π π π=π³
= π By using Galerkinβs
technique, find the solution of the above differential equation.
Given Data:
Differential equ. π΄πΈ = π2π’
ππ₯2 + ππ₯ = 0
Boundary Conditions π’ 0 = 0, π΄πΈ = π2π’
ππ₯2 + ππ₯ = 0
To Find:
u(x) by using galerkinβs technique
Formula used
π€π π ππ₯ = 0
πΏ
0
Solution:
Assume a trial function
Let π’ π₯ = π0 + π1π₯ + π2π₯2 + π3π₯3 . β¦β¦.. (1)
Apply first boundary condition
i.e) at x=0, u(x) = 0
1 βΉ 0 = π0 + 0 + 0 + 0
π0 = 0
Apply first boundary condition i.e at x = L, π΄πΈ =ππ’
ππ₯= 0
βΉππ’
ππ₯= 0+π1 + 2π2π₯ + 3π3πΏ2
βΉ 0 = π1 + 2π2πΏ + 3π3πΏ2
βΉ π1 = β(2π2πΏ + 3π3πΏ2)
sub π0 and π1 in value in equation (1)
π’ π₯ = 0 + β 2π2πΏ + 3π3πΏ2 π₯ + π2π₯2 + π3π₯3
= β2π2πΏπ₯ β 3π3πΏ2π2π₯ + π2π₯2 + π3π₯3
= π2 π₯2 β 2πΏπ₯ + π3(π₯3 β 3πΏ2π₯) β¦β¦β¦ (2)
We Know That
Residual, π = π΄πΈπ2π’
ππ₯2+ ππ₯ β¦β¦β¦. (3)
(2) βΉππ’
ππ₯= π2 2π₯ β 2πΏ + π3(3π₯2 β 3πΏ2)
π2π’
ππ₯2= π2 2 + π3(6π₯)
π2π’
ππ₯2= 2π2 + 6π3π₯
Sub π2π’
ππ₯2 value in equation (3)
3 βΉ π = π΄πΈ 2π2 + 6π3π₯ + ππ₯
Residual, π = π΄πΈ 2π2 + 6π3π₯ + ππ₯ β¦β¦β¦ (4)
From Galerknβs technique
π€π π ππ₯ = 0
πΏ
0
. . β¦β¦ . . . (5)
from equation (2) we know that
π€1 = π₯2 β 2πΏπ₯
π€2 = π₯3 β 3πΏ2π₯
sub w1, w2 and R value in equation (5)
5 βΉ π₯2 β 2πΏπ₯
πΏ
0
π΄πΈ 2π2 + 6π3π₯ + ππ₯ ππ₯ = 0 β¦β¦ β¦ β¦ β¦ (6)
π₯3 β 3πΏ2π₯
πΏ
0
π΄πΈ 2π2 + 6π3π₯ + ππ₯ β¦ β¦ β¦ β¦ β¦ (7)
6 βΉ π₯2 β 2πΏπ₯
πΏ
0
π΄πΈ 2π2 + 6π3π₯ + ππ₯ ππ₯ = 0
π₯2 β 2πΏπ₯
πΏ
0
2π2π΄πΈ + 6π3π΄πΈπ₯ + ππ₯ ππ₯ = 0
2π2π΄πΈπ₯2 + 6π3π΄πΈπ₯3 + ππ₯3 β 4π2π΄πΈπΏπ₯ β 12π3π΄πΈπΏπ₯2 β 2ππΏπ₯2
πΏ
0
= 0
βΉ [2π2π΄πΈπ₯3
3+ 6π3π΄πΈ
π₯4
4+ π
π₯4
4β 4π2π΄πΈπΏ
π₯2
2β 12π3π΄πΈπΏ
π₯3
3β 2ππΏ
π₯3
3]0πΏ = 0
βΉ 2π2π΄πΈπΏ3
3+ 6π3π΄πΈ
πΏ4
4+ π
πΏ4
4β 4π2π΄πΈ
πΏ3
2β 12π3π΄πΈ
πΏ4
3β 2π
πΏ4
3= 0
βΉ2
3π2π΄πΈπΏ3 +
3
2π3π΄πΈ πΏ4 + π
πΏ4
4β 2π2π΄πΈπΏ3 β 4π3π΄πΈπΏ4 β
2
3ππΏ4 = 0
βΉ π΄πΈπ2πΏ3 2
3β 2 + π3π΄πΈ πΏ4
3
2β 4 + π
πΏ4
4β
2
3π2πΏ4 = 0
βΉβ4
3π΄πΈπΏ3π2 β
5
2π΄πΈπΏ4π3 =
2
3β
1
4 ππΏ4 β
4
3π΄πΈπΏ3π2 β
5
2π΄πΈπΏ4π3 =
5
12ππΏ4
β4
3π΄πΈπΏ3π2 β
5
2π΄πΈπΏ4π3 = β
5
12ππΏ4 β¦ β¦ β¦ . 8
Equation (7)
βΉ (π₯3 β 3πΏ2π₯)
πΏ
0
π΄πΈ 2π2 + 6π3π₯ + ππ₯ ππ₯ = 0
βΉ (π₯3 β 3πΏ2π₯)
πΏ
0
2π2π΄πΈ + 6π3π΄πΈπ₯ + ππ₯ ππ₯ = 0
βΉ 2π΄πΈπ2π₯3 + 6π΄πΈπ3π₯4 + ππ₯4 β 6π΄πΈπ2πΏ2π₯ β 18π΄πΈπ3πΏ2π₯2 β 3ππΏ2π₯2 ππ₯ = 0
πΏ
0
βΉ 2π΄πΈπ2
π₯4
4+ 6π΄πΈπ3
π₯5
5+ π
π₯5
5β 6π΄πΈπ2πΏ2
π₯2
2β 18π΄πΈπ3πΏ2
π₯3
3β 3ππΏ2
π₯3
3
0
πΏ
= 0
βΉ 1
2π΄πΈπ2π₯4 +
6
5π΄πΈπ3π₯5 +
1
5ππ₯5 β 3π΄πΈπ2πΏ2π₯2 β 6π΄πΈπ3πΏ2π₯3 β ππΏ2π₯3
0
πΏ
= 0
βΉ1
2π΄πΈπ2πΏ4 +
6
5π΄πΈπ3πΏ5 +
1
5ππΏ5 β 3π΄πΈπ2πΏ2(πΏ2) β 6π΄πΈπ3πΏ2(πΏ3) β ππΏ2(πΏ3) = 0
βΉ1
2π΄πΈπ2πΏ4 +
6
5π΄πΈπ3πΏ5 +
1
5ππΏ5 β 3π΄πΈπ2πΏ4 β 6π΄πΈπ3πΏ5 β ππΏ5 = 0
βΉ π΄πΈπ2πΏ4 1
2β 3 + π΄πΈπ3πΏ5
6
5β 6 + ππΏ5 +
1
5β 1 = 0
βΉ π΄πΈπ2πΏ4 5
2 β
24
5π΄πΈπ3πΏ5 =
4
5ππΏ5
5
2 π΄πΈπ2πΏ4 +
24
5π΄πΈπ3πΏ5 = β
4
5ππΏ5 β¦ β¦ β¦β¦ . 9
Solving Equation (8) and (9)
Equation (8) βΉ4
3π΄πΈπ2πΏ3 +
5
2π΄πΈπ3πΏ4 = β
5
12ππΏ4
Equation (9) βΉ5
2π΄πΈπ2πΏ4 +
24
5π΄πΈπ3πΏ5 = β
4
5ππΏ5
Multiplying Equation (8) 5
2πΏ and Equation (9) by
4
3
20
6π΄πΈπ2πΏ4 +
25
4π΄πΈπ3πΏ5 = β
25
24ππΏ5
20
6π΄πΈπ2πΏ4 +
25
4π΄πΈπ3πΏ5 = β
16
15ππΏ5
Subtracting
25
4β
96
15 π΄πΈπ3πΏ5 =
16
15β
25
24 ππΏ5
375 β 384
60 π΄πΈπ3πΏ5 =
384 β 375
360 ππΏ5
βΉβ9
60π΄πΈπ3πΏ5 =
9
360ππΏ5
βΉ β0.15π΄πΈπ3 = 0.025π
π3 = β0.1666π
π΄πΈ
π3 = βπ
6π΄πΈ β¦ β¦ β¦ . (10)
Substituting a3 value in Equation (8)
4
3π΄πΈπ2πΏ3 +
5
2π΄πΈ
βπ
6π΄πΈ πΏ4 =
β5
12ππΏ4
4
3π΄πΈπ2πΏ3 =
β5
12ππΏ4 β
5
2π΄πΈπΏ4 =
βπ
6π΄πΈ
4
3π΄πΈπ2πΏ3 =
β5
12ππΏ4 +
5
2π΄πΈπΏ4
4
3π΄πΈπ2πΏ3 = 0
π2 = 0
Sub a2 and a3 value in equation (2)
βΉ π’ π₯ = 0π₯ π₯2 β 2πΏπ₯ + βπ
6π΄πΈ π₯3 β 3πΏ2π₯ = 0
βΉ π’ π₯ = π
6π΄πΈ 3πΏ2π₯ β π₯3
Result:
π’ π₯ = π
6π΄πΈ 3πΏ2π₯ β π₯3
2. Find the deflection at the centre of a simply supported beam of span length βlβ 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) Galerkinβs
method. (Nov/Dec 2014)
Given data
Length (L) = π
UDL = π/π
To find
Deflection
Formula used
πΈπΌπ4π¦
ππ₯4β π = 0, 0 β€ π₯ β€ π
Point Collocation Method R = 0
Sub-domain collocation method = π ππ₯ = 0π
0
Least Square Method πΌ = π 2ππ₯ ππ ππππππ’ππ
0
Solution:
The differential equation governing the deflection of beam subjected to uniformly
distributed load is given by
πΈπΌπ4π¦
ππ₯4β π = 0, 0 β€ π₯ β€ π β¦β¦ β¦ . (1)
The boundary conditions are Y=0 at x=0 and x = l, where y is the deflection.
πΈπΌπ4π¦
ππ₯4= 0, ππ‘ π₯ = 0 πππ π₯ = π
Where
πΈπΌπ4π¦
ππ₯4 = π, (Bending moment)
E β Youngβs Modules
I β Moment of Inertia of the Beam.
Let us select the trial function for deflection as π = ππ ππππ₯
π β¦β¦. (2)
Hence it satisfies the boundary conditions
βΉππ¦
ππ₯= π
π
π. cos
ππ₯
π
βΉπ2π¦
ππ₯2= βπ
π2
π2. sin
ππ₯
π
βΉπ3π¦
ππ₯3= βπ
π3
π3. cos
ππ₯
π
βΉπ4π¦
ππ₯4= π
π4
π4. sin
ππ₯
π
Substituting the Equation (3) in the governing Equation (1)
πΈπΌ ππ4
π4. sin
ππ₯
π β π = 0
Take, Residual π = πΈπΌππ4
π4 . sinππ₯
πβ π
a) Point Collocation Method:
In this method, the residuals are set to zero.
βΉ π = πΈπΌππ4
π4. sin
ππ₯
πβ π = 0
πΈπΌππ4
π4. sin
ππ₯
π= π
To get maximum deflection, take π =π
2(π. π. ππ‘ πππ ππ ππ ππππ)
πΈπΌππ4
π4 . sinπ
π π
2 = π
πΈπΌππ4
π4= π
π =ππ4
π4πΈπΌ
Sub βaβ value in trial function equation (2)
π =ππ4
π4πΈπΌ. sin
ππ₯
π
π΄π‘ π₯ =π
2βΉ πmax =
ππ4
π4πΈπΌ. sin
π
2 π
2
πmax = ππ4
π4πΈπΌ
πmax =
ππ4
97.4πΈπΌ
b) Sub-domain collocation method:
In this method, the integral of the residual over the sub-domain is set to zero.
π ππ₯ = 0π
0
Sub R value
βΉ ππΈπΌπ4
π4sin
ππ₯
πβ π ππ₯ = 0
βΉ ππΈπΌπ4
π4 βcos
ππ₯π
ππ
βπ π₯
0
π
= 0
[β΅ sinπ
π= 1]
[β΅ sinπ
2= 1]
βΉ ππΈπΌπ4
π4 βcos
ππ₯
π
π
π’ β π π₯
0
π
= 0
βΉ βππΈπΌπ3
π3 cosπ β πππ 0 π π = 0
βππΈπΌπ3
π3 β1 β 1 = π π
βΉ βπ =ππ4
2π3πΈπΌ=
ππ4
62πΈπΌ
Sub βaβ value in the trial function equation (2)
π =ππ4
62πΈπΌ. sin
ππ₯
π
π΄π‘ π₯ =π
2, ππππ₯ =
ππ4
62πΈπΌ. sin
π
π(π
2)
ππππ₯ =ππ4
62πΈπΌ
c) Least Square Method:
In this method the functional
πΌ = π 2ππ₯ ππ ππππππ’π
π
0
πΌ = (ππΈπΌπ4
π4. π ππ
ππ₯
πβ π)2ππ₯
π
0
= [π2πΈ2πΌ2π8
π8. π ππ2
ππ₯
πβ π2 β 2ππΈπΌπ
π4
π4. π ππ
ππ₯
π]ππ₯
π
0
= [π2πΈ2πΌ2π8
π8 1
2π₯ π ππ
2ππ₯
π
π
2π + π2 β 2ππΈπΌπ
π4
π4. [βπππ
ππ₯
π π
π ]]0
π
= π2πΈ2πΌ2π8
π8 1
2π β
π
2π π ππ2π β π ππ0 + π2π + 2ππΈπΌπ
π4
π4.π
π[βπππ π β πππ 0]
β΅ π ππ2π = 0; π ππ0 = 0; πππ π = 0; πππ 0 = 1
πΌ = π2πΈ2πΌ2π8
π2
π
2+ π2π + 2ππΈπΌπ
π3
π3. (β1 β 1)
πΌ =π2πΈ2πΌ2π8
2π7+ π2π β 4ππΈπΌπ
π3
π3
Now, ππ
ππ= 0
βΉπ2πΈ2πΌ2π8
2π7= 4πΈπΌπ
π3
π3
β΅ cos π = β1
πππ 0 = 1 ,
π2πΈ2πΌ2π8
π7= 4πΈπΌπ
π3
π3
π =4πΈπΌππ5
π5πΈπΌ
Hence the trial Function
π =4ππ4
π5πΈπΌ. sin
ππ₯
π
π΄π‘ π₯ =π
2, max πππππππ‘πππ [β΅ π ππ
π
2= 1]
ππππ₯ =4ππ4
π5πΈπΌ π ππ
π
2(π
2)
ππππ₯ =ππ4
76.5 πΈπΌ
d) Galerkinβs Method:
In this method
π. π ππ₯ = 0
π
0
βΉ ππ ππππ₯
π ππΈπΌ
π4
π4π ππ
ππ₯
πβ π ππ₯ = 0
π
0
βΉ π2πΈπΌπ4
π4π ππ2
ππ₯
πβ πππ ππ
ππ₯
π ππ₯ = 0
π
0
βΉ π2πΈπΌπ4
π4[1
2(1 β πππ
2ππ₯
π) β πππ ππ
ππ₯
π ππ₯ = 0
π
0
βΉ π2πΈπΌπ4
π4[1
2 1 β π₯ β
1
2π π ππ2
2ππ₯
π + ππ
π
ππππ
ππ₯
π
0
π
= 0
π2πΈπΌπ4
π4 π
2 β 2ππ
π
π = 0
β΄ π =2ππ
π.
2π3
πΈπΌπ4
π =4ππ3
π5πΈπΌ
Hence the trial Function
π =4ππ4
π5πΈπΌ. sin
ππ₯
π
π΄π‘ π₯ =π
2, max πππππππ‘πππ [β΅ π ππ
π
2= 1]
ππππ₯ =4ππ4
π5πΈπΌ π ππ
π
2(π
2)
ππππ₯ =4ππ4
π5 πΈπΌ
ππππ₯ =ππ4
76.5 πΈπΌ
Verification,
We know that simply supported beam is subjected to uniformly distributed load, maximum
deflection is,
ππππ₯ =5
384
ππ4
πΈπΌ
= 0.01ππ4
πΈπΌ
3) i) What is constitutive relationship? Express the constitutive relations for a liner
elastic isotropic material including initial stress and strain.
[Nov/Dec 2009]
Solution:
It is the relationship between components of stresses in the members of a structure or in a
solid body and components of strains. The structure or solids bodies under consideration are made
of elastic material that obeys Hookeβs law.
π = π· {π}
Where
[D] is a stress β strain relationship matrix or constitute matrix.
The constitutive relations for a linear elastic isotropic material is
ππ₯
ππ¦
ππ§
πΏπ₯π¦
πΏπ¦π§
πΏπ§π₯
=πΈ
1 + π£ 1 β 2π£
(1 β π£) 0 0
π£ (1 β π£) 0π£000
π£000
(1 β π£)000
0 0 00 0 00
1 β 2π£
200
00
1 β 2π£
20
000
1 β 2π£
2
ππ₯
ππ¦
ππ§
π£π₯π¦
π£π¦π§
π£π§π₯
ii) Consider the differential equation π ππ
π ππ + πππππ = π for π β€ π β€ π subject to boundary
conditions Y(0) = 0, Y(1) = 0. The functions corresponding to this problem, to be eternized
is given by π° = βπ. π π π
π π π
+ ππππππ π
π. Find the solution of the problem using Ray
Light Ritz method by considering a two term solution as π π = πππ π β π + ππππ(π β
π)
Given data
Differential equation = π2π¦
ππ₯2+ 400π₯2 = 0 for 0 β€ π₯ β€ 1
Boundary conditions Y(0) = 0, Y(1) = 0
πΌ = β0.5 ππ¦
ππ₯
2
+ 400π₯2π π
0
π π₯ = π1π₯ 1 β π₯ + π2π₯2(1 β π₯)
To find:
Ray Light Ritz method
Formula used
ππΌ
ππ1= 0
ππΌ
ππ2= 0
Solution:
π π₯ = π1π₯ 1 β π₯ + π2π₯2(1 β π₯)
π π₯ = π1π₯ π₯ β π₯2 + π2(π₯2 β π₯3)
ππ¦
ππ₯ = π1 1 β 2π₯ + π2(2π₯ β 3π₯2)
= π1 1 β 2π₯ + π2π₯(2 β 3π₯)
ππ¦
ππ₯
2
= π1 1 β 2π₯ + π2π₯(2 β 3π₯)2 2
= π12 1 β 4π₯ + 4π₯2 + π2
2π₯2 4 β 12π₯ + 9π₯2 + 2π1π2π₯ 1 β 2π₯ (2 β 3π₯)
= π12 1 β 4π₯ + 4π₯2 + π2
2π₯2 4 β 12π₯ + 9π₯2 + 2π1π2π₯(2 β 3π₯ β 4π₯ + 6π₯2)
ππ¦
ππ₯
2
= π12 1 β 4π₯ + 4π₯2 + π2
2π₯2 4 β 12π₯ + 9π₯2 + 2π1π2π₯(2 β 7π₯ + 6π₯2)
We know that
πΌ = [β0.5 ππ¦
ππ₯
2
+ 400π₯2π¦
π
0
] =β1
2
ππ¦
ππ₯
2
+ 400 π₯2π¦
π
0
π
0
= π12 1 β 4π₯ + 4π₯2 + π2
2π₯2 4 β 12π₯ + 9π₯2 + 2π1π2π₯ 2 β 7π₯ + 6π₯2
π
0
+ 400[ π₯2 π1π₯ 1 β π₯ + π2π₯2 1 β π₯
π
0
By Solving
πΌ =β1
2 π1
2
3+
2
15π2
2 +1
3π1π2 + 400
π1
20+
π2
30
πΌ =β1
6π1
2 β1
15π2
2 β1
6π1π2 + 20π1 +
40
3π2
ππΌ
ππ1= 0
βΉβ1
6Γ 2π1 β
1
6π2 + 20 = 0
βΉβ1
3Γ π1 β
1
6π2 + 20 = 0 β¦β¦ β¦ . . (1)
Similarly,
ππΌ
ππ2= 0
βΉβ2
15π2 β
1
6π1 +
40
3= 0 β¦ β¦ β¦ . . (2)
By Solving (1) and (2)
π1 =80
3; π1 =
200
3
We know that
π = π1π₯ 1 β π₯ + π2π₯2(1 β π₯)
π = 80
3π₯ 1 β π₯ +
200
3π₯2 1 β π₯
4) Consider a 1mm diameter, 50m long aluminum pin-fin as shown in figure used to
enhance the heat transfer from a surface wall maintained at 300C. Calculate the
temperature distribution in a pin-fin by using Rayleigh β Ritz method. Take, π =
ππππππ for aluminum h= ππππ
πππ , π»β = πππ.
ππ ππ»
π ππ =π·π
π¨(π» β π»β) , π» π = π»π = ππππ, ππ³ = π²π¨
π π»
π π π³ = π (insulated tip)
Given Data:
The governing differential equation
ππ2π
ππ₯2=
ππ
π΄(π β πβ)
Diameter d = 1mm = 1x10-3
m
Length L = 50mm = 50x10-3
m
Thermal K = 200π€πC
Conductivity Heat transfer co-efficient h = 200π€πC
Fluid Temp πβ = 30C.
Boundary Conditions π 0 = ππ€ = 300C
ππΏ = πΎπ΄ππ
ππ₯ πΏ = 0
To Find:
Ritz Parameters
Formula used
π = π π‘ππππ ππππππ¦ β π€πππ ππππ
Solution:
The equivalent functional representation is given by,
π = π π‘ππππ ππππππ¦ β π€πππ ππππ
π = π’ β π£
π = 1
2πΎ
ππ
ππ₯
2
ππ₯ + 1
2
ππ
π΄ π β πβ 2ππ₯ β ππΏππΏ β¦ β¦ β¦ β¦ . (1)
πΏ
0
πΏ
0
π = 1
2πΎ
ππ
ππ₯
2
ππ₯ + 1
2
ππ
π΄ π β πβ 2ππ₯ β¦ β¦ β¦ β¦ . . 2
πΏ
0
πΏ
0
β΅ ππΏ = 0
Assume a trial function
Let
π π₯ = π0 + π1π₯ + π2π₯2 β¦ β¦ β¦ β¦ β¦ . . (3)
Apply boundary condition
at x = 0, T(x) = 300
300 = π0 + π1(0) + π2(0)2
π0 = 300
Substituting π0 value in equation (3)
π π₯ = 300 + π1π₯ + π2π₯2 β¦β¦ β¦ β¦ β¦ . . 4
βΉππ
ππ₯= π1 + 2π2π₯ β¦ β¦ β¦β¦ β¦ β¦ (5)
Substitute the equation (4), (5) in (2)
π = 1
2 π
π
0
(π1 + 2π2π₯)2ππ₯ + 1
2
ππ
π΄ 270 + π1 + π2π₯2 2ππ₯.
π
0
[β΅ π + π 2 = π2 + π2 + 2ππ; π + π + π 2 = π2 + π2 + π2 + 2ππ + 2ππ + 2ππ
π =π
2 (π1
2 + 4π22π₯2 + 4π1
π
0
π2π₯) +ππ
2π΄ 2702 + π1
2π₯2
π
0
+ π22π₯4 + 540π1π₯ + 2π1π₯3 + 540π2π₯2 ππ₯
π =π
2 (π1
2π₯ +4π2
2π₯3
3+
4π1π2π₯2
2
0
50π₯10β3
+ππ
2π΄ 72900π +
π12π₯3
3+
π22π₯5
5+
540π1π₯2
2+
2π1π2π₯4
4+
540π2π₯3
3
0
50π₯10β3
[β΅ π = 50π₯10β3]
π =π
2(50 Γ 10β3)π1
2 +4π2
2(50 Γ 10β3)3
3+
4π1π2(50 Γ 10β3)2
2
+ππ
2π΄ 72900π +
π12(50 Γ 10β3)3
3+
π22(50 Γ 10β3)5
5
π =200
2 50 Γ 10β3π1
2 + 1.666 Γ 10β4π22 + 50 Γ 10β3π1π2 +
π Γ 10β3 Γ 20
2 Γπ2 Γ 10β3 2
= 364.5 + 4.166 Γ 10β5π12 + 6.25 Γ 10β8π2
2 + 0.675π1 + 3.125 Γ 10β6π1π2 + 0.0225π2
π = 5π12 + 0.0166π2
2 + 0.5π1π2 + 14.58 Γ 10β7 + 1.66912 + 2.5 Γ 10β3π22 + 2700 π1
+ 0.125 π1π2 + 900π2]
π = 6.66π12 + 0.0191π2
2 + 0.625π1π2 + 2700π1 + 900π2 + 14.58 Γ 107
Apply ππ
ππ2= 0
βΉ 13.32π1 + 0.625π2 + 27000 = 0
13.32π1 + 0.625π2 = β + 27000 β¦ β¦ β¦β¦ β¦ (6)
βΉ 0.625π1 + 0.382π2 + 900 = 0
0.625π1 + 0.382π2 = β900 β¦ β¦ β¦ β¦ . . (7)
Solve the equation (6) and (7)
13.32π1 + 0.625π2 = β + 27000 β¦ β¦ β¦β¦ β¦ (6)
0.625π1 + 0.382π2 = β900 β¦ β¦ β¦ β¦ . . 7
(6) x 0.625
8.325π1 + 0.3906π2 = β16875 β¦ β¦ β¦ β¦ . . 8
(7) x -13.32
β8.325π1 β 0.5088π2 = 11988 β¦ β¦ β¦ β¦ . . 9
β0.1182π2 = β4887
π2 = 41345
Sub π2value in equation (6)
13.32π1 + 0.625(41345) = β + 27000
π1 = β3967.01
Sub π0, π1and π2values in equation (3)
π = 300 β 3697.01π₯ + 41345π₯2
6) Explain in detain about Boundary value, Initial Value problems.
(ME ENG.DESIGN Nov/DEC 2011)
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. Ξ© and some
conditions on the boundary Ξ 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 Ξ© with its boundary Ξ is
called a closed domain.
Boundary value problems:- Steady state leat transfer : In a fin and axial deformation of a bar
shown in fig. Find π’(π₯) that satisfies the second β order differential equation and boundary
conditions.
βπ
ππ₯ π
ππ’
ππ₯ + ππ’ = π for 0 < π₯ < πΏ
π’ π = π’0, π ππ’
ππ₯
π₯=πΏ= π0
i) Bending of elastic beams under Transverse load : find π’ π₯ that satisfies the fourth order
differential equation and boundary conditions.
π2
ππ₯ 2 π π2π’
ππ₯2 + ππ’ = πΉ for 0 < π₯ < βπΏ
π’ π = π’0, ππ’
ππ₯
π₯=0= π0
π
ππ₯ π
π2π’
ππ₯2 π₯=πΏ
= π0 . ππ2π’
ππ₯2 0
= π0
x
x = 0 Ξ© = (o, L) x=L
Initial value problems:-
i) A general first order equation:-
Find π’ π‘ that satisfies the first-order differential equation and initial condition.
Equation and initial condition:-
π ππ’
ππ‘+ ππ’ = πΉ for 0 < π‘ β€ π
π’ 0 = π’0.
ii) A general second order equation:-
Find π’ π‘ that satisfies the second β order differential equation and initial conditions:-
π ππ’
ππ‘+ π
π2π’
ππ‘ 2 + ππ’ = πΉ for 0 < π‘ β€ π
π’ π = π’0, πππ’
ππ‘ π‘=0
= π£0
Eigen value problems:-
(i) Axial vibration of a bar:
Find π’ π₯ and π that satisfy the differential equation and boundary conditions.
βπ
ππ₯ π
ππ’
ππ₯ β ππ’ = 0 for π < π₯ < πΏ
π’ π = 0, πππ’
ππ₯ π₯=πΏ
= 0
(ii) Transverse vibration of a membrane:-
Find π’ (π₯, π¦) and π that satisfy the partial differential equation and
boundary condition.
β π
ππ₯ π1
ππ’
ππ₯ +
π
ππ¦ π2
ππ’
ππ¦ β ππ’ = 0 in Ξ©
π’ = 0 on Ξq
The values of π are called cigen values and the associated functions π’ are called cigen functions.
b) A simple pendulum consists of a bob of mass π(ππ)attached to one end of a rod of
length π(π) and the other end is pivoted to fixed point π.
Soln:-
πΉ = π
ππ‘ ππ£ = ππ
πΉπ₯ = π.ππ£π₯
ππ‘
βππ sin π = ππ π2π
ππ‘2
or
π2π
ππ‘2+
π
πsin π = 0
π2π
ππ‘2+
π
ππ = 0
ππ
ππ‘+ (π) = π0.
π π‘ = π΄π ππ ππ‘ + π΅ cos π π‘.
Where,
π = π
π and π΄ and π΅ are constant to be determined using the initial condition we
obtain.
π΄ β π0
π, π΅ = π0
the solution to be linear problem is
π π‘ = π0
π πππ β§ π‘ + 0. πΆππ ππ‘
for zero initial velocity and non zero initial position π0 , we have.
π π‘ = π0 cos ππ‘.
7) A simply supported beam subjected to uniformly distributed load over entire span and
it is subject to a point load at the centre of the span. Calculate the bending moment
and deflection at imdspan by using Rayleish β Ritz method. (Nov/Dec 2008).
Given data:-
To Find:
1. Deflection and Bending moment at mid span.
2. Compare with exact solutions.
Formula used
π = π π‘ππππ ππππππ¦ β π€πππ ππππ
Solution:
We know that,
Deflection, y = a1 sinΟx
l+ a2 sin
3Οx
l
Total potential energy of the beam is given by,
Ο = U β H
Where, U β Strain Energy.
H β Work done by external force.
The strain energy, U of the beam due to bending is given by,
U = EI
2
d2y
dx 2 2
dx1
0
dy
dx= a1 cos
Οx
l Γ
Ο
l + a2 cos
3Οx
l Γ
3Ο
l
dy
dx=
a1Οx
lcos
Οx
l +
a23Οx
lcos
3Οx
l
d2y
dx 2 = β a1Ο
lsin
Οx
l Γ
Ο
l β
a23Ο
lsin
3Οx
lΓ
3Ο
l
d2y
dx 2 = β a1Ο2
l2 sinΟx
l β 9
a2Ο2
l2 sin3Οx
l
Substituting d2y
dx 2 value in equation (3),
U = EI
2 β
a1Ο2
l2sin
Οx
l β 9
a2Ο2
l2sin
3Οx
l
2
dxl
0
= EI
2
a1Ο2
l2 sinΟx
l + 9
a2Ο2
l2 sin3Οx
l
2
dxl
0
= EI
2
Ο4
l4 a12 sin2 Οx
l + 81a2
2 sin2 3Οx
l+ 2 a1 sin
Οx
l .9 a2 sin
3Οx
l dx
l
0
[β΄ a + b 2 = a2 + b2 + 2ab]
=EI
2
Ο4
l4 a12 sin2 Οx
l + 81a2
2 sin2 3Οx
l+ 18 a1a2 sin
Οx
l . sin
3Οx
l dx
l
0
12
22
32
42
5
a12 sin2 Οx
ldx =
π
0a1
2 1
2 1 β cos
2Οx
l
l
0dx β΄ sin2 x =
1βcos 2x
2
= a12 1
2 1 β cos
2Οx
l
l
0dx
= a1
2
2 dx
π
0β cos
2Οx
l
1
0 dx
= π1
2
2 π₯ 0
π β sin
2ππ₯
π2π
π
0
π
= π1
2
2 π β 0 β
1
2π sin
2ππ
πβ sin 0
= π1
2
2 π β
1
2π 0 β 0 =
π12 π
2 β΄ sin 2π = 0; sin 0 = 0
Similarly,
81 a22 sin2 3Οx
ldx =
π
081a2
2 1
2 1 β cos
6Οx
l
π
0dx β΄ sin2 x =
1βcos 2x
2
= 81a22 1
2 1 β cos
6Οx
l
π
0dx
= 81a2
2
2 dx
π
0β cos
6Οx
l
π
0 dx
= 81π2
2
2 π₯ 0
π β sin
6ππ₯
π6π
π
0
π
= 81π2
2
2 π β 0 β
1
6π sin
6ππ
πβ sin 0
= 81π2
2
2 π β
1
6π 0 β 0 =
π12 π
2 β΄ sin 6π = 0; sin 0 = 0
18 a1a2 sinΟx
l . sin
3Οx
ldx =
π
018 a1a2 sin
Οx
l . sin
3Οx
l
π
0dx
= 18 a1a2 sin3Οx
l . sin
Οx
l
π
0dx
a12 sin2
Οx
l dx =
π
0
π12 π
2
81a22 sin2
3Οx
l dx =
π
0
81π22 π
2
62
72
= 18 a1a2 1
2 cos
2Οx
lβ cos
4Οx
l
π
0dx
β΄ sin π΄ sin π΅ =cos π΄βπ΅ βcos π΄+π΅
2
= 18 a1a2
2 cos
2Οx
ldx
π
0β cos
4Οx
l
π
0 dx
= 18 a1a2
2
sin2ππ₯
π2π
π
0
π
β sin
4ππ₯
π4π
π
0
π
= 9 a1a2 0 β 0 = 0 β΄ sin 2π = 0; sin 4π = 0; sin 0 = 0
Substitute (6), (7) and (8) in equation (5),
U =EI
2
Ο4
l4 π1
2 π
2+
81π22 π
2+ 0
U =EI
4
Ο4 π
l4 π1
2 + 81π22
Work done by external forces,
π» = π π¦ ππ₯ + π π¦πππ₯π
0
π π¦ ππ₯π
0
= 2ππ
π π1 +
π2
3
We know that, π¦ = π1 sinππ₯
π+ π2 sin
3ππ₯
π
In the span, deflection is maximum at π₯ =1
2
π¦πππ₯ = π1 sinπ Γ
1
2
π+ π2 sin
3πΓ1
2
π
= π1 sinπ
2+ π2 sin
3π
2 β΄ sin
π
2= 1; sin
3π
2= β1
18 a1a2 sinΟx
l . sin
3Οx
ldx =
π
0
0
Strain Energy, U =πΈπΌπ4
4π3 π1
2 + 81π22
82
92
10
11
12
π¦πππ₯ = π1 β π2
Substitute (11) and (12) values in equation (8),
H = 2ππ
π π1 +
π2
3 + π (π1 β π2)
Substituting U and H values in equation (2), we get
π = πΈπΌπ4
4π3 π1
2 + 81π22 β
2ππ
π π1 +
π2
3 + π (π1 β π2)
π = πΈπΌπ4
4π3 π1
2 + 81π22 β
2ππ
π π1 +
π2
3 β π (π1 β π2)
For stationary value of π, the following conditions must be satisfied.
ππ
ππ1= 0and
ππ
ππ2= 0
ππ
ππ1=
πΈπΌπ4
4π3 2π1 β
2ππ
πβ π = 0
πΈπΌπ4
2π3 π1 β2ππ
πβ π = 0
πΈπΌπ4
2π3π1 =
2ππ
π+ π
ππ
ππ2=
πΈπΌπ4
4π3 162π2 β
2ππ
π
1
3 + π = 0
Similarly,
πΈπΌπ4
4π3 162π1 β
2ππ
π+ π = 0
πΈπΌπ4
2π3 162π1 =
2ππ
πβ π
From equation (12), we know that,
π1 =2π3
πΈπΌπ4
2ππ
π+ π
π2 =2π3
81πΈπΌπ4
2ππ
3πβ π
13
14
15
16
Maximum deflection, π¦πππ₯ = π1 β π2
π¦πππ₯ = 2π3
πΈπΌπ4 2ππ
π+ π β
2π3
81πΈπΌπ4 2ππ
3πβ π
π¦πππ₯ = 4ππ4
πΈπΌπ5 + 2ππ3
πΈπΌπ4 β 4ππ4
243πΈπΌπ5 + 2ππ3
81πΈπΌπ4
π¦πππ₯ = 0.0130 ππ4
πΈπΌ+ 0.0207
ππ3
πΈπΌ
We know that, simply supported beam subjected to uniformly distributed load, maximum deflection
is, π¦πππ₯ =5
384
ππ4
πΈπΌ
Simply supported beam subjected to point load at centre, maximum deflection is,
π¦πππ₯ =ππ3
48πΈπΌ
So, total deflection, π¦πππ₯ =5
384
ππ4
πΈπΌ+
ππ3
48πΈπΌ
From equations (17) and (18), we know that, exact solution and solution obtained by using
Rayleigh-Ritz method are same.
Bending Moment at Mid span
We know that,
Bending moment, M = EI d2y
dx 2
From equation (9), we know that,
d2y
dx 2 = β π1π2
π2 sinππ₯
π+
π2 9π2
π2 sin3ππ₯
π
Substitute π1 and π2 values from equation (15) and (16),
d2y
dx 2 = β 2π3
πΈπΌπ4 2ππ
π+ π Γ
π2
π2 sinππ₯
π+
2π3
81πΈπΌπ4 2ππ
3πβ π Γ
9π2
π2 sin3ππ₯
π
Maximum bending occurs at π₯ =π
2
π¦πππ₯ = 0.0130 ππ4
πΈπΌ+ 0.0208
ππ3
πΈπΌ
18
19
17
= β 2π3
πΈπΌπ4
2ππ
π+ π Γ
π2
π2sin
π Γ12
π+
2π3
81πΈπΌπ4
2ππ
3πβ π Γ
9π2
π2sin
3π Γ12
π
= β 2π3
πΈπΌπ4
2ππ
π+ π Γ
π2
π2 (1) +
2π3
81πΈπΌπ4
2ππ
3πβ π Γ
9π2
π2 (β1)
β΄ sinπ
2= 1; sin
3π
2= β1
= β 2π
πΈπΌπ2
2ππ
π+ π β
2π
9πΈπΌπ2
2ππ
3πβ π
= β 4ππ2
πΈπΌπ3 + 2ππ
πΈπΌπ2 β 4ππ2
27πΈπΌπ3 + 2ππ
9πΈπΌπ2
= β 3.8518ππ2
πΈπΌπ3 + 2.222ππ
πΈπΌπ2
Substitute d2y
dx 2 value in bending moment equation,
Mcentre = EId2y
dx 2 = βπΈπΌ 0.124ππ2
πΈπΌ+ 0.225
ππ
πΈπΌ
Mcentre = β 0.124 ππ2 + 0.225 ππ
(β΄Negative sign indicates downward deflection)
We know that, simply supported beam subjected to uniformly distributed load,
maximum bending moment is,
Mcentre = ππ2
8
Simply supported beam subjected to point load at centre, maximum bending moment
is,
Mcentre = ππ
4
Total bending moment, Mcentre = ππ2
8+
ππ
4
Mcentre = 0.125 ππ2 + 0.25 ππ
d2y
dx2= β 0.124
ππ2
πΈπΌ+ 0.225
ππ
πΈπΌ
20
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.