1. AERSP 301 Finite Element Method
Jose Palacios
July 2008

2. Today HW 5 has been uploaded (Stationary Principle, MOS FOS)
Short assignment
Due Thursday
HW’s 4 due Tuesday
I will upload solutions Wednesday evening
HW’s 3 & 4 due no later than Friday
Solutions will be uploaded Friday evening
No Class on Thursday
Friday during class, Exam Review
Exam
Time
Equation Sheet
Start Finite Element Method - Bars

3. Axial deformations in bars
We were able to write expressions for the strain energy, U, and external work, W, for a bar under axial loading.
For the 1-DOF and 2-DOF spring problems, the next step was to apply the stationary principle.
For the 1-DOF system, gave the equilibrium equation.
Likewise, for the 2-DOF system,
gave the (system of two)
equilibrium equations.
As in HW #5, Problem 2

4. Axial deformations in bars
For the bar, though being a continuum structure – it has an infinite number of degrees of freedom.
Displacement at every x will vary, u = u(x).
Second issue – if you were applying the stationary principle, differentiating is a problem because the expressions for U and W have integrals.
The finite element method represents continuum structures (with an infinite number of degrees of freedom) into a system with a finite number of degrees of freedom.
Represents structure in an approximate sense
The structure is subdivided (“discretized”) into a number of segments (or finite elements)

5. Axial deformations in bars
Variation of displacement (u) within each element is assumed to be very simple, such that U and W may be easily integrated across the element length
U and W are related to displacements at a finite number of points which are the end-points of the element
Solution of the problem is then reduced to the system of an algebraic system for a finite set of “end point” or NODAL values.

6. Axial deformations in bars
Entire Bar:
Displacements (u’s) must be continuous along length of bar (no jumps)

7. Axial deformations in bars
When the elements of the bar are reassembled:
This GLOBAL system has 4 DOFs (q1…q4)
Called GLOBAL degrees of freedom = Global nodal displacements
Each element has 2 DOFs:
Called LOCAL DOFs or Element DOFs or Element nodal displacements

8. Axial deformations in bars
Every local DOF corresponds to a nodal displacement
Looking closely at a single element:
u1, u2: local DOFs
x1, x2: element nodal coordinates
l = x2 – x1 = length of element

9. Axial deformations in bars
Global x-coordinate system is difficult to use so we transform to a non-dimensional local coordinate system (s-coordinate) such that:
At x = x1, s = 0
At x = x2, s = 1
Transformation expressed as
where (1-s) and (s) are mapping functions and dx = lds where l is the length scaling for mapping

10. Axial deformations in bars
For individual element, express the displacements within the elements (in terms of values of nodes)
For this, assume some displacement profile or shape within the element (in terms of element nodal displacements u1 and u2)
An obvious choice is a linear profile wherein
Such that at
And at

11. Axial deformations in bars
This is the expression for the axial displacement, u, within an element (at every point within the element) – in terms of local coordinates, s, and displacements at end points (nodes)
u1, u2:
Define N1(s) = 1 – s , and N2(s) = s  displacement shape functions
Shape functions are always in the local coordinate system
Recall that
The mapping functions are identical to the shape functions for this problem (this is not always the case)
where M1 and M2 are mapping functions

12. Axial deformations in bars – Element Stiffness Matrix
Recall , where
 – Total Potential
U – Strain Energy
W – External Work Potential
Look at U (Strain Energy)
For the bar in extension:
Then discretize the bar into a number of elements:

13. Axial deformations in bars – Element Stiffness Matrix
Since strain energy, U, is a scalar quantity, the integral
may be broken up
Strain energy may be calculated for each individual element and summed to obtain the total strain energy of the entire structure:
Consider a single element:
Mapping function
Also Recall:
Assumed displacement within the element with shape fns, N1(s) & N2(s)

14. Axial deformations in bars – Element Stiffness Matrix
Then
This partial derivative can be written in two forms:
Introducing into the strain energy for the ith element:
How? Recall:
EA a function of s?
Are u1 and u2 functions of s?

15. Axial deformations in bars – Element Stiffness Matrix
Look at
Each term in the matrix may
be integrated individually.

16. Axial deformations in bars – Element Stiffness Matrix
Individual terms are:
This yields: or
ELEMENT STIFFNESS MATRIX
What if an element moves through space without elongation (Rigid Body Translation)?
Can an element represent a constant strain condition?

17. Axial deformations in bars – Element Load Vector
Now look at the External Work Potential, W:
Recall
Consider the integral term::
Discretize the bar using the finite element method
Since Work is also a scalar and can be calculated for individual elements and added.

18. Axial deformations in bars – Element Load Vector
Look at the contribution from a single element (putting integral in local coordinate system):

19. Axial deformations in bars – Element Load Vector
Or in vector form:
This is the contribution to the External Work Potential of the distributed force, f, over an element.
What is the physical implication of representing:
Distributed load is represented as two individual forces (one at each node)

20. Axial deformations in bars – Element Load Vector
Now consider a constant distributed force, f = c
Half of total load, cl, is placed at each node. What about other loading conditions?
Could integrate exactly across element length, if convenient.
Or, could use linear interpolation across the element
Try these.

21. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
So far, have obtained expressions for strain energy, Ui, and external work potential, Wi, for any individual element.
Now add the contributions from individual elements together to obtain the strain energy, U, and external work potential, W, for the entire structure.
This process is called ASSEMBLY.
Look at the External Work Potential, W, due to the distributed force, f.

22. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
The external work potential for element #1, W1, can be written as:
Similarly, the external work potential for element #2, W2, can be written as:

23. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
And, the external work potential for element #3, W3, can be written as:
The total external work potential due to the distributed force, f, is:
Element #1
Element #2
Element #3

24. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
The external work potential due to the tip load, P, is W = q4P
Similarly, there is a reaction force, R, acting at the root, x = 0 (because the bar is clamped at the end). The external work potential due to this reaction force, R, is W = q1R

25. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
The total External Work Potential, W, is obtained by adding each load vector
What if there were additional concentrated loads acting at some other locations on the bar?
(obtained from elemental load vectors and contributions of concentrated loads)
Global Load Vector

26. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
Now, look at the strain energy, U, of the structure.
The strain energy for element #1, U1, can be written as:

27. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
Similarly, the strain energy for element #2, U2, can be written as:
And, the strain energy for element #3, U3, can be written as:

28. Axial deformations in bars Global Load Vector and Global Stiffness Matrix
The total strain energy of the structure:
Global Stiffness Matrix
(obtained from 2x2 elemental stiffness matrices)