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Misrimal Navajee Munoth Jain Engineering College
Department of Mechanical Engineering
ME 6603 – FINITE ELEMENT ANALYSIS
Assignment-1
B.E. Mechanical 3rd Year
Due Date:02.01.2018
1. Solve the differential equation for the physical problem expressed as with the boundary
d2y
500x 2 0, 0 x 1
conditions as
dx2
y(0)=0 and y(1)=0 using a two-term trial function by (i) point collocation method, (ii)
subdomain collocation method, (iii) Least square method and (iv) Galerkin’s method.
d2y
300x 2 0, 0 x 1 with
2. Consider the differential equation for a problem as
dx2
the boundary conditions y(0)=0 and y(1)=0. Find the solution by using trial function as
y a1x(1 x3 ) . Use (i) point collocation method, (ii) subdomain collocation method,
(iii) Least square method and (iv) Galerkin’s method.
3. Solve the differential equation for the physical problem expressed as with the boundary
d2y
100 0, 0 x 10
conditions as
dx2
y(0)=0 and y(10)=0 using (i) point collocation method, (ii) subdomain collocation
method, (iii) Least square method and (iv) Galerkin’s method
4. A simply supported beam (span L and flexural rigidity EI) subjected to a point load at
the centre of the span. Calculate the deflection at the mid-span using Galerkin’s
weighted residual method and compare with the exact solution. Assume trial solution
x
as y aSin .
L
5. A simply supported beam (span L and flexural rigidity EI) subjected to a point load at
the centre of the span. Calculate the deflection at the mid-span using Rayleigh-Ritz
method and compare with the exact solution.
6. A cantilever beam is subjected to a uniformly distributed load throughout its length.
Find the deflection at free end using Rayleigh-Ritz method.
7. A cantilever beam of length L is loaded with a point load at the free end. Find the
maximum deflection and maximum bending moment using Rayleigh-Ritz method
x
using the function y a 1 Cos . Given EI is constant.
2 L
d2y
400x 2 0, 0 x 1 subject to the
2
dx
boundary conditions y(0)=0 and y(1)=0. The functional corresponding to the problem
2
1
dy
2
I
0
.
5
400
x
y
to be extremised is given by
dx . Find the solution using
0 dx
Rayleigh-Ritz method by considering a two-term trial solution as
y( x) a1x(1 x) a2 x(1 x3 )
8. Consider the differential equation

Misrimal Navajee Munoth Jain Engineering College
Department of Mechanical Engineering
Formulae for ME 6603 - Finite Element Analysis (6th Semester)
Unit -1 Introduction
Weighted Residual methods:
1. Point collocation method
The residual(R) is forced to zero at discrete number of points. The number of points is
equal to the number of undetermined parameters in the assumed trial solution.
2. Sub-domain method
The average residual in the sub-domain is forced to zero. The number of sub-domains is
equal to the number of undetermined parameters in the assumed trial solution.
3. Least-square method
The integral of the weighted square of the residual over the domain is minimized.
max
R
I
2
dx
min
Suppose if there are two undetermined coefficients C1 and C2, then
I
R
R
dx 0
C1 min C1
max
I
R
R
dx 0
C2 min C2
max
4. Galerkin Weighted residual method
Galerkin introduced the idea of letting w(x) to be the same as trial functions
max
w ( x) R( x)dx 0
i
min
Rayleigh-Ritz method:
1. Assume a trial solution (generally polynomial)
u ( x) c1 c2 x c3 x 2 ...
2. Evaluate the total potential
U V
Where U-Strain energy and V-Work done by external forces
3. Setup and solve the system of equations
0 , where i=1,2,3,…No. of undetermined parameters in trial solution
ci
Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering
Page 1

Unit-2 One Dimensional problems
1. The general finite element equation is
{F e } [ K e ]{u e }
Where{Fe}-nodal force vector
[Ke]-stiffness matrix
{ue}-nodal field variable vector
2. For linear bar element
AE 1 1
[K e ]
L 1 1
3. The field variable in terms of shape functions (interpolation functions)
u( x) N1 ( x)u1 N 2 ( x)u2 N3 ( x)u3 ...
4. Strain in the element,
u1
dN 3
dN 2
du dN1
u 2
dx dx
dx
dx
u3
Bu
Where B-Strain displacement matrix
5. The element stiffness matrix
L
K B T DBdx , for 1-D case D=E, Young’s modulus
e
0
6. Element stresses
D
7. Body force vector
{ f e } [ N T qAdx , where q-body force per unit length
8. For Truss element
l2
lm
l 2 lm
m 2 lm m 2
AE lm
e
[K ]
Le l 2 lm l 2
lm
2
lm
m 2
lm m
x x
y y1
Sin
where l 2 1 Cos and m 2
Le
Le
9. One-Dimensional heat transfer
k 1 1 PhL 2 1 T1
Ph L 2 Q0
q
L 1 1 6 A 1 2 T 0 A T L Q
2
c
c
2 l
Where Q0 and Ql represent the heat flux at the ends of the element (nodes)
10. Plane beam element
6 L 12 6 L
12
6 L 4 L2 6 L 2 L2
EI
[K e ] 3
L 12 6 L 12 6 L
2
2
6L 2L 6L 4L
Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering
Page 2

11. Dynamic analysis
K 2 m u F
Where [K]-stiffness matrix
[m]-mass matrix
ω – natural frequency
12. Longitudinal vibration
2u
2u
Equation of motion is AE 2 A 2
x
t
AL 2 1
Consistent mass matrix is m
6 1 2
AL 1 0
Lumped mass matrix is m
2 0 1
13. Transverse vibration
4v
2v
Equation of motion is EI 4 A 2 0
dx
t
156
4 L2
AL 22 L
Consistent mass matrix is m
13L
420 54
2
13L 3L
1
AL 0
Lumped mass matrix is m
2 0
0
0
0
0
0
0
0
1
0
symmetric
156
22 L
4 L2
0
0
0
0
Unit-3 Two-Dimensional Scalar Variable Problems
3 ( x3,y3)
1. For a three noded triangular element
Field variable, u(x,y) is given as
u1
v
1
u N1 0 N 2 0 N 3 0 u 2
1
2
v 0 N1 0 N 2 0 N 3 v 2
(x1,y1)
(x2,y2)
u3
v3
3 x 3
1 x 1
2 x 2
Where N1 1
, N2 2
and N 3 3
2A
2A
2A
1 x2 y3 x3 y2 , 2 x3 y1 x1 y3 and 3 x1 y2 x2 y1
1 y2 y3 , 2 y3 y1 and 3 y1 y2
1 x3 x2 , 2 x1 x3 and 3 x2 x1
Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering
Page 3

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