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
∫=Π
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
sinsin1 +
Π
+
Π
=
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
∫=
0
ω

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,
Mmax= -0.124ω2
2
2
dx
yd
EIM =

20. Thank you