Shape functions

ManojShukla 983 views 12 slides Aug 06, 2020
Slide 1
Slide 1 of 12
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10
Slide 11
11
Slide 12
12

About This Presentation

1 d formulation


Slide Content

Adrian Egger
Method of Finite Elements I:
Shape Functions

Why shape functions?
Discretization leads to solution in the nodes, but no information concerning the
space in between
Shape functions required to approximate quantities between nodes
Underlying assumption of how quantities are distributed in an element
(stiffness, mass, element loads; displacements, strains, stress, internal forces, etc.)
Geometry transformation
3/24/2015 2Adrian Egger | FEM I | FS 2015
?

What can shape functions be used for?
1.Used to interpolate between nodes
i.e. discrete nodal quantities continuous across element
??????�= �
??????(�)??????
??????=�
1(�)??????
1+�
2(�)??????
2
3/24/2015 3Adrian Egger | FEM I | FS 2015

What can shape functions be used for?
2.Used to discretize continuous quantities to nodal DOF
i.e. continuous across element discrete nodal quantities
3/24/2015 4
??????
??????�??????�??????�??????
��????????????�
=
??????=0
??????
??????
??????
??????(�)??????�=
??????=0
??????
�
01
1
�
11
1
�
02
1
�
12
1
??????(�)??????�=
??????�
2
??????�
2
12
??????�
2

??????�
2
12
??????
1
�
1
??????
2
�
2
Adrian Egger | FEM I | FS 2015

Alternative way to derive loading vector
Recap: We calculate the solution in the nodes
“What is the influence of element loading in the nodes”
We must fix the element such that reaction forces develop
in the nodal DOF we are interested in!
Equivalent to solving differential equation
??????��
????????????
=−??????
3/24/2015 5Adrian Egger | FEM I | FS 2015

How to derive shape functions
Interpolation functions are generally assumed!
(within certain parameters and restrictions)
Minimal amount of continuity / differentiability
Etc.
Wish to implement this repetitive task as easily as possible,
i.e. computer implementation using highly optimized
numerical schemes, and thus natural coordinates (r,s,t) are
introduced ranging from -1 < r,s,t< 1.
3/24/2015 6Adrian Egger | FEM I | FS 2015

Derivation of shape functions:
Bar element (I)
1.Find a relationship for r(x). We choose -1 < r < 1.
2.Choose an appropriate shape function polynomial
3.Evaluate A at each DOF by substituting values of “r”
3/24/2015 7Adrian Egger | FEM I | FS 2015

Derivation of shape functions:
Bar element (II)
4.Reorder the previous equation
5.Substitute into previous equation
6.Extract shape functions (as a function of “r”)
3/24/2015 8Adrian Egger | FEM I | FS 2015

Derivation of shape functions:
Beam element (I)
1.Find a relationship for r(x). We choose 0 < r < 1.
2.Choose an appropriate shape function polynomial
3/24/2015 9Adrian Egger | FEM I | FS 2015

Derivation of shape functions:
Beam element (II)
3.Find an expression linking displacements and rotations
4.Evaluate A at each DOF by substituting values of “r”
3/24/2015 10Adrian Egger | FEM I | FS 2015

Derivation of shape functions:
Beam element (III)
4.Reorder the previous equation
5.Substitute into previous equation
6.Extract shape functions (as a function of “r”)
3/24/2015 11Adrian Egger | FEM I | FS 2015

Questions
3/24/2015 12Adrian Egger | FEM I | FS 2015
Tags