ritz method(app soln)
DESCRIPTION
Ritz Method(App Soln)TRANSCRIPT
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CE502: Finite Element Analysis
By
Dr. A. ChakrabortyDepartment of Civil Engineering
Indian Institute of Technology Guwahati, India
1
RITZ METHOD
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' '
( ) ( )
1 =
2T T T T
i iiv v s
Objective Find disp u of the elastic body
Strain Energy U Work Potential WP
U WP
dv u fdv u Tds u P
Walter Ritz – Swiss theoritical physicist(1878-1909). He developed the approximate solution using Principle of Minimum Potential Energy.
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2 2 2 21 1 2 2 3 3 4 4 1 1 3 3
1 1 1 1
2 2 2 2k k k k Fq F q
Principle of Min Pot Eng – For conservative systems, of all the kinematically admissible displacement field, those corresponding to equilibrium, extremize the total potential energy. If the extremum conditionis min, the equilibrium state is stable.
Example 11
32
2
1
433F
2q1q
3q
1F
Sol:
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2 2 2 21 1 2 2 2 3 3 2 4 3 1 1 3 3
1 11 1
1 1 2 3 3 2
3 3 4 33
1 2 3
1 1 1 1 ( ) ( )
2 2 2 2
0 , 1, 2,3
0
. 0
0
, , .
i
k q q k q k q q k q Fq F q
iq
q Fk k
K q F k k k k k q
k k k Fq
Solve for q q q
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Rayleigh-Ritz Method:For a continua, the total potential energy can be used for finding approximate solution. Ritz method involved the construction of an assumed disp field.
𝛱
1 2 3
(x, y, z) 1
(x, y, z) j 1 m
(x, y, z) k 1 n
( , , ,..........)
0 .
x i i
y j j
z k k
i
U a i to l
U a l to n m l
U a m to
a a a
form the set of algebraic eqnq
Coeff. (unknown) Polynomial or Trigonometric function(known)
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2
1
21 2 3
02
1
1 2 3
23
1 3
3
12
2
,
0 |
0
2 4 0
( 2 x x )
2 ( 1 )
xx
duEA dx u
dx
Let u a a x a x
It must satisfy u
a
a a a
u a
u a
dua x
dx
Example 2
1
1.5 1
-1
+
-
0.75 1
Exact
App
Exact
App
21 1
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22 23 3
0
23 3
33
3
1 3
2
14 ( 1 ) 2( )
2
4 2
38
0 2 03
0.75
0.75
0.75(2 ) Ans.
1.5(1 x)
a x dx a
a a
aa
a
u a
u x x
duE
dx
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Example 3
w
PL/2
u
L
22
20
1 2
0 0
22 2
1 2
0
1 2
1
2
3, w= sin sin
w 0 | , M 0 |
3 3sin sin
2
3 sin sin
2 2
L
c
x xx L x L
L
d wEI dx Pw
dx
x xLet
L LBC
EI x xdx
L L L L
P
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4 42 2
1 2 1 2
4
11
4
22
43 3
1 24 4
3
4
3. . ( )
2 2 2
0 2 . 02 2
3 0 2 . 0
2 2
2 2 1 ,
3
2sin
EI L LP
L L
EI LP
L
EI LP
L
P L P L
EI EI
PL xw
EI L
4
3
4 4 4
3 3
1 3sin
3
2 1 1 1 ...........
3 5
0.02148
c
c c
x
L
PLIf n w
EI
PL PLExact w w
EI EI
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Thank You!!!