Basics of Finite Element Analysis(FEA)

1. Introduction to FEA
By-
Kishan Sharma

2. Definition
The finite element method is a computational scheme to solve field problems in
engineering and science. The technique has very wide application, and has been
used on problems involving stress analysis, fluid mechanics, heat transfer,
diffusion, vibrations, electrical and magnetic fields, etc. The fundamental concept
involves dividing the body under study into a finite number of pieces (subdomains)
called elements.

3. Static vs. Dynamic FEM
Static analysis is good for engineering, to find just the end result.
Dynamic analysis is good for simulation, to find all intermediate steps.

4. Phases in FEA
 Preprocessing
o Understanding the Problem (Thermal , Structural, Dynamic etc)
o Element Selection (Solid i.e 3D , Shell i.e 2D , Beams i.e 1D etc)
o Deciding the Boundary Conditions (Constraints , Connections etc..)
o Load Application (Point , Surface , Body loads etc..)
 Solution
o Program derives the governing matrix equations from the model &
solves for the primary quantities.
 Post processing of Results (Deflection , Stress , Strain etc..)
o Validity check of the solution.
o Report Preparation.
o Observation and Conclusion from the Analysis. (MS Calcs, Design ok)
o Suggestion and Recommendation for Design Changes, if required.

5. 5
Idealization of
geometry (if necessary)
CAD geometry Simplified
geometry
CAD FEA Pre-
processing
Basic Steps In The Finite Element Analysis
Restraints
Material
properties
Type of
analysis Loads
MATHEMATICAL
MODEL

6. 6
FEA model FEA results
Discretization Numerical solver
FEA Pre-
processing
FEA
Solution
FEA Post-
processing
MATHEMATICAL
MODEL
Basic Steps In The Finite Element Analysis

8. Difference b/w 1D , 2D and 3D elements

9. Types of Materials
• Isotropic Materials- A material having identical values of a property
in all directions. Ex: Glass and metals.
• Anisotropic Materials- Properties such as Young's Modulus, change
with direction along the object. Ex: wood and composites.
• Orthotropic Materials- They have material properties that differ
along three mutually-orthogonal twofold axes of rotational symmetry.
They are a subset of anisotropic materials, because their properties
change when measured from different directions. Ex: wood.
• Homogeneous Material- A material of uniform composition
throughout that cannot be mechanically separated into
different materials. Ex: Certain types of plastics, ceramics, glass,
metals, alloys, paper, board, resins, and coatings.

10. Common Alloys used in Aircrafts
 Aluminium Alloys
1. Al 2024 Composition-
Due to its high strength and fatigue resistance, 2024 is widely used in aircraft structures,
especially wing and fuselage structures under tension. Al 2024 is known as the “aircraft
alloy” in machining rod.
a. Al 2024-T351 –
T351 temper 2024 plate has an ultimate tensile strength of 470 MPa (68 ksi) and yield
strength of 280 MPa (41 ksi). It has elongation of 20%.
b. Al 2024-T3 –
T3 temper 2024 sheet has an ultimate tensile strength of 400–430 MPa (58–62 ksi) and yield
strength of at least 270–280 MPa (39–40 ksi). It has an elongation of 10-15%.
Element Content (%)
Aluminium / Aluminum, Al 93.50
Copper, Cu 4.4
Magnesium, Mg 1.5
Manganese, Mn 0.6

11. 2. Al 5052 Composition-
Al 5052 is an aluminium alloy, primarily alloyed with magnesium and chromium.
Aluminium / Aluminum 5052 alloy is used in the manufacture of Milk crates, Hydraulic
tubes, Appliances, Kitchen cabinets, Small boats, home freezers, Aircraft tube, Fencing etc.
This alloy is also used widely in sheet metal work and in sheet metal parts.
3. Al 7050 Composition-
Element Content (%)
Aluminum, Al 89
Copper, Cu 2.3
Magnesium, Mg 2.3
Zinc, Zn 6.2
Zirconium, Zr 0.12

12. Aluminium / Aluminum 7050 alloy is a heat treatable alloy. It has high toughness, strong
mechanical strength, and good stress corrosion cracking resistance.
Aluminium / Aluminum 7050 alloy is mainly used in manufacturing aircraft and other
structures.
a. Al 7050-T76511 -
Solution heat-treated, stress-relieved by controlled stretching and then artificially overaged
in order to achieve a good exfoliation corrosion resistance. The products receive no further
straightening after stretching, except that minor straightening is allowed after stretching to
comply with standard tolerances.
4. Al 7075 Composition-
Element Content (%)
Aluminum, Al 90
Zinc, Zn 5.6
Magnesium, Mg 2.5
Copper, Cu 1.6
Chromium, Cr 0.23

13. Aluminum alloy 7075 is an aluminum alloy, with zinc as the primary alloying element. It is
strong, with a strength comparable to many steels, and has good fatigue strength and
average machinability, but has less resistance to corrosion than many other Al alloys.
Aluminum 7075 alloy is mainly used in manufacturing aircraft and other aerospace
applications.
a. Al 7075-T6 -
T6 temper 7075 has an ultimate tensile strength of 510–540 MPa (74,000–78,000 psi) and
yield strength of at least 430–480 MPa (63,000–69,000 psi). It has a failure elongation of 5–
11%.
b. Al 7075-T 6511 –
Solution heat-treated, stress-relieved by controlled stretching (permanent set 1% to 3% for
extruded rod, bar, shapes and tube, 0.5% to 3% for drawn tube) and then artificially aged.
Minor straightening is allowed after stretching to comply with standard tolerances

14. Meshing( Discretization )
Discretization is the method of approximating the differential equations
by a system of algebraic equations for the variables at some set of discrete
locations in space and time.
Continuous domain Discretized domain

15. Concept of Discretization (Meshing)
Concept of FEM is all about Discretization (Meshing) i.e. Dividing a big
structure/component into small discrete Blocks (Nodes and Element concept)
But why we do Meshing ???
Ans: Ex: (1) No. of Points =∞
DoF per point = 6
Total No of Equations to be solved=∞ * 6 = ∞
(2) No. of Points =8
DoF per point = 6
Total No of Equations to be solved=8 * 6 = 48
*From Infinite to Finite…Hence the Term “Finite Element Method”*

16. Basic Steps in the Finite Element Method
Time Independent Problems
 Domain Discretization
 Select Element Type (Shape and Approximation)
 Derive Element Equations.
 Assemble Element Equations to Form Global System.
[K]{U} = {F}
where, [K] = Stiffness or Property Matrix
{U} = Nodal Displacement Vector
{F} = Nodal Force Vector
 Incorporate Boundary and Initial Conditions.
 Solve Assembled System of Equations for Unknown Nodal
Displacements and Secondary Unknowns of Stress and Strain Values.

17. Fundamental Concept

18. Derivation of Stiffness matrix for a Single Spring Element

19. Derivation of a Global Stiffness Matrix

20. In general, we will have a stiffness matrix
of the form (assume for now that we do
not know k11, k12, etc)











333231
232221
131211
kkk
kkk
kkk
K
The finite element force-
displacement relations:































3
2
1
3
2
1
333231
232221
131211
F
F
F
d
d
d
kkk
kkk
kkk
k1
k2F1x F2x F3x
x
1 2 3
Element 1 Element 2
d1x d2x
d3x
Physical Significance of Stiffness Matrix

21. The first equation is,
1313212111 Fdkdkdk 
Force equilibrium
equation at node 1
What if d1=1, d2=0, d3=0 ?
313
212
111
kF
kF
kF


 Force along node 1 due to unit displacement at node 1
Force along node 2 due to unit displacement at node 1
Force along node 3 due to unit displacement at node 1
While nodes 2 and 3 are held fixed
Similarly we obtain the physical significance of the other entries of the
global stiffness matrix.
Columns of the global stiffness matrix

22. Steps in solving a problem
Step 1: Write down the node-element connectivity table linking local and
global displacements
Step 2: Write down the stiffness matrix of each element
Step 3: Assemble the element stiffness matrices to form the global stiffness
matrix for the entire structure using the node element connectivity table
Step 4: Incorporate appropriate boundary conditions
Step 5: Solve resulting set of reduced equations for the unknown
displacements
Step 6: Compute the unknown nodal forces

23. MNTamin, CSMLab
ke

Ae Ee
 
 1 1
le 1 1 
3. Element force vector
due to body force, fb
1
2
e e b
f e A l f  
  
1
4. Element force vector
due to traction force, T
1e
Te Tl  
  
2 1
 1q
le
 
 
E
1 1  
q2 
Some formulas used in 1-D problems
1. Stress-strain relation
2. Element stiffness matrix

24. Problem

25. Solution
1. Transform the given plate into 2 sections, each
having uniform cross-sectional area.
Note:
Area at midpoint is
Amid = 4.5 in2.
Average area of section 1 is
A1 = (6 + 4.5)/2 = 5.25 in2.
Average area of section 2 is
A2 = (4.5 + 3)/2 = 3.75 in2.
2. Model each section using 1-D
(line) element.

26. element 2:
6
12
k(2) 3.753010  1
  
1
1 1 
4. Assemble global stiffnessmatrix,
6
12
 5.25 
K
3010 5.25 3.75


5.25 0
9.00 
0 3.75 3.75 
3. Write the element stiffness matrix for each element
element 1:
6
12
1
k(1)

5.253010  1
1 1 
Note: The main diagonal must contain positive numbersonly!

27. 5. Write the element force vector for each element
a) Due to body force, fb = 0.2836lb/in3
element 1  (1)
12
fb
5.25120.2836 1
  
 
element 2  (2)
12
fb
3.75120.2836 1
  
 
Assemble global force vector due to body force,
15.3
2
bF 
5.25  8.9
120.2836 
9.00   
   
3.75  6.4
   

28. b) Due to traction force, T = 36 lb/ft
element 1
1
2
 3612     
T(1)
  12 
1
18   
1 1
element 2
1
2
 3612     
T(2)
 12 
1
18   
1 1
Assemble global force vector due to traction force,
1 18
   
FT 182  36
1 18
   

29. c) Due to concentrated load, P = 100 lb at node 2
 
 0 
FP 100
 0 
 
6. Assemble all element force vectors to form the globalforce
vector for the entire structure.
lb
 8.918 0  26.9
   
F 15.336100  151.3
   24.46.4180   

30. 12
0 3.75
 Q1   26.9
30106
    
 5.25
5.25 3.75 Q2   151.3
 Q  24.4
5.25 0
9.00
3.75   3   
7. Write system of linear equations (SLEs) for entire model
The SLEs can be written in condensed matrix form as
KQF
Expanding all terms and substituting values, we get
Note:
1. The global force term includes the unknown reaction force R1 at
the support. But it is ignored for now.
2. The SLEs have no solutions since the determinant of [K] = 0;
Physically, the structure moves around as a rigid body.

31. 8. Impose boundary conditions (BCs) on the global SLEs
There are 2 types of BCs:
a) Homogeneous = specified zero displacement;
b) Non-homogeneous = specified non-zero displacement.
In this example, homogeneous BC exists at node 1.
How to impose this BC on the global SLEs?
DELETE ROW AND COLUMN #1 OF THE SLEs!
12
0 3.75
 5.25 5.25 0 Q1   26.9
30106
    
5.25 3.75Q2   151.3
 Q  24.4
9.00
3.75   3   

32. 30106

2
   
 9.00 3.75 Q  151.3
12 3.75 3.75  Q3   24.4
Solve using Gaussian elimination method, yields
2
in
5
Q    1.33910
   5
Q3  1.59910 
9. Solve the reduced SLEs for the unknown nodal
displacements
The reduced SLEs are,
Quiz: Does the answers make sense? Explain…

33. 10. Estimate stresses in each elements
 1q
le
(e)  
 EBq E 
1
1 1  
q2 
element 1
0
12
1
 1  30106

1
1

 33.48 psi 
1.339105

element 2
12
5
11.33910
 2
 30106

1
1

 6.5psi 
1.599105

Recall,

34. 12
0
5
 5.25    26.9 R1
30106
  5.25 3.75 1.33910  151.3
24.4
  


   
5.25 0 0
9.00
3.75

 3.75 1.599 105
  
6
1
0
12
 
 
R 
3010
5.25 5.25 01.339105
  26.9334
1.599105 
 
R1  202.68lb
11. Compute the reaction force R1 at node1
We now include the reaction force term in the global SLEs.
From the 1st. equation we get,
We have,