Finite Element Analysis MEEN 5330 Dustin Grant

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Transcript Finite Element Analysis MEEN 5330 Dustin Grant

Finite Element Analysis

MEEN 5330 Dustin Grant Kamlesh Borgaonkar Varsha Maddela Rupakkumar Patel Sandeep Yarlagadda

Introduction  What is finite element analysis, FEA?

 What is FEA used for?

 1D Rod Elements, 2D Trusses

Basic Concepts

f

T

  Loads

P i

ji

,

j

f

~

i

 0  Equilibrium  Boundary conditions

Development of Theory  Rayleigh-Ritz Method  Total potential energy equation  Galerkin’s Method

1D Rod Elements  To understand and solve 2D and 3D problems we must understand basic of 1D problems.

 Analysis of 1D rod elements can be done using Rayleigh Ritz and Galerkin’s method  To solve FEA problems same are modified in the Potential-Energy approach and Galerkin’s approach

1D Rod Elements  Loading consists of three types : body force f , traction force T, point load P i  Body force: distributed force , acting on every elemental volume of body i.e. self weight of body.

 Traction force: distributed force , acting on surface of body i.e. frictional resistance, viscous drag and surface shear  Point load: a force acting on any single point of element

1D Rod Elements  Element strain energy Element -1 Element-2

U e

 1 2 

q T

[

k e

] 

q

 Element stiffness matrix [

k e

] 

E e A e l e

   1 1  1 1    Load vectors  Element body load vector  Element traction-force vector

f

e

A e l e f

2

T

e

Tl e

2   1 1   1 1 

Example 1D Rod Elements Example 1 Problem statement: (Problem 3.1 from Chandrupatla and Belegunda’s book) Consider the bar in Fig.1, determine the following by hand calculation: 1) Displacement at point P 2) Strain and stress 3) Element stiffness matrix 4) strain energy in element Given:

E

  6

psi A e

 1.2

in

2

q

1  0.02

in q

2  0.025

in

Solution: 1) Displacement (q) at point P We have   (

x

2 2 

x

1 ) (

x x

1   2 Now linear shape functions N 1 (  ) and N 2 (  ) are given by

N

1  1   2  0.375

And

N

2  1   2  0.625

2D Truss  2 DOF  Transformations  Modified Stiffness Matrix  Methods of Solving

2D Truss  Transformation Matrix  Direction Cosines

l e

 

x

2 

x

1 [

L

]   

l

0

m

0

l

0 0

m

 

y

2 

y

1  2

l

 cos  

x

2 

x

1

l e m

 sin  

y

2 

y

1

l e

2D Truss  Element Stiffness Matrix [

k e

] 

E e A e l e

        

l

2

lm l

2

lm lm m

2 

lm

m

2 

l

2 

lm l

2

lm

 

m lm m

2

lm

2       

Methods of Solving  Elimination Approach  Eliminate Constraints  Penalty Approach  Will not discuss Today

Elimination Method  Set defection at the constraint to equal zero

Elimination Method  Modified Equation  DOF’s 1,2,4,7,8 equal to zero

2D Truss  Element Stresses  

E e

 

l l e

m l m

q

  Element Reaction Forces 

R

  

Q

2D Truss  Development of Tables  Coordinate Table  Connectivity Table  Direction Cosines Table

2D Truss  Coordinate Table

2D Truss  Connectivity Table

2D Truss  Direction Cosines Table

l e

 

x

2 

x

1   2

y

2 

y

1  2

l

 cos  

x

2

l e

x

1

m

 sin  

y

2

l e

y

1

Example 2D Truss

MATLAB Program TRUSS2D.M

3D Truss Stiffness Matrix  3D Transformation Matrix  Direction Cosines [

L

]   

l

0

m

0

n

0

l

0 0

m

0

n

 

l e

 

x

2 

x

1 

y

2 

y

1  

z

2 

z

1  2

l

 cos  

x

2 

x

1

l e m

 cos  

y

2 

y

1

l e n

 cos  

z

2 

z

1

l e

3D Truss Stiffness Matrix  3D Stiffness Matrix [

k e

] 

E e A e l e

           

l ln

2

lm l ln

2

lm lm m

2

mn

lm

m

2 

mn ln mn n

2 

ln

mn

n

2 

l

2 

lm

ln l

2

lm ln

lm

m

2 

mn lm m

2

mn

 

ln mn

   

n n ln mn

2 2      

Conclusion  Good at Hand Calculations, Powerful when applied to computers  Only limitations are the computer limitations

References

Homework