ritz method(app soln)

18/04/2023

CE502: Finite Element Analysis

By

Dr. A. ChakrabortyDepartment of Civil Engineering

Indian Institute of Technology Guwahati, India

1

RITZ METHOD

18/04/2023

2

' '

( ) ( )

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.

18/04/2023

3

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:

18/04/2023

4

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

18/04/2023

5

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)

18/04/2023

6

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

18/04/2023

7

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

18/04/2023

8

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

18/04/2023

9

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

18/04/2023

10

Thank You!!!