Rayleigh Ritz Method

An Approximation Method for a Mathematical Formulation

1. prepared by
R.Sakthivel Murugan,
Assistant Professor, KCET

2. •Engineering Analysis
•Analytical Methods (or) Theoretical Analysis
•Numerical Methods (or) Approximate Methods

3. Functional Approximation
Finite Difference Method (FDM)
Finite Element Method (FEM)

4. -Rayleigh Ritz Method
-Weighted Residual Method

5. Potential Energy is the capacity to do work.
Total Potential Energy = Internal Potential
Energy
+External
Potential Energy.
Principle of minimum potential energy

6. It is an integral approach method
Useful for solving Structural Mechanics
Problems.
It is also known as Variational Approach.

7. Potential Energy ,
Total Potential Energy
π = Strain Energy – Work done by external forces
= U- H
ò
=P
2
1
)''','','(
x
x
dxyyyf

8. It should satisfy the geometric boundary
condition.
It should have at least one Rayleigh Ritz
parameter.
It should represented as either polynomial or
trigonometrical.

9. Polynomial  Bar Element
Trignometric  Beam Element
.......
3
3
2
210 ++++= xaxaxaay
.....
3
sinsin
1 +
P
+
P
=
l
x
l
x
ay

11. •Step 1
–Setting an approximation Function
•Step 2
–Determine Strain Energy, U
•Step 3
–Determine Work Done by External Force , H
•Step 4
–Total Potential Energy, π= U-H
•Step 5
–To find Ritz Parameter by Partial Differentiation (step 4 result)
•Step 6
–Determine deflection for beam element
–Determine displacement for bar element
•Step 7
–Determine Bending Moment for beam element
–Determine Stress for bar element

12. A simply supported beam subjected to
uniformly distributed load over entire span.
Determine the bending moment and
deflection at mid span by using Rayleigh Ritz
method.

13. Step 1
Setting approximation function for beam

14. Step 2
Strain Energy
Solving this we get ,

15. Step 3
Work done by External Forces
Solving this we get,
dxyH
l
ò
=
0
w

16. Step 4
Total Potential Energy , π= U- H

17. Step 5
To find Ritz Parameter by Partial
Differentiation
Solving this we get ,
&
So,

18. Step 6
Maximum Deflection
Sub x = l/2 in y 

19. Step 7
Maximum Bending Moment
Solving this we get,
M
max= -0.124ω
2
2
2
dx
yd
EIM=

20. Thank you