Finite Element Method involves breaking structures down into small discrete elements. This document discusses 1D and 2D beam elements. 1) 1D bar elements have axial displacement degrees of freedom at nodes. The element stiffness matrix relates nodal forces and displacements. Systems of multiple elements are assembled into a global stiffness matrix. 2) 2D beam elements have transverse displacement and slope degrees of freedom at nodes. Interpolation functions define displacement across the element. Integrating the strain energy derives the element stiffness matrix relating nodal forces and displacements. 3) Examples demonstrate assembling systems of springs and bars, computing displacements and stresses, and derive the complete stiffness matrix for a beam element including axial, shear and

1. Finite Element Method
By Sohail Iqbal

2. Google Classroom

3. 1D elements & computations
procedure
• 1D elements include
– Straight bar loaded axially
– Straight beam loaded laterally
– Bar conducts heat or electricity etc
• Degree of freedom (DOF) is number of parameters that may vary
independently
• Total DOF in structure = number of nodes * DOF at each node
• Order of structural stiffness = Total DOF in structure * number of
nodes in structure
• i.e spring element with 2 nodes has order of structural stiffness of
2*2
3

4. Bar element
4
u1 and u2 axial displacement at nodes
σ internal axial stress
F1 and F2 nodal forces

5. Bar element
• F1= AE/L(u1-u2)
• F2= AE/L(u2-u1) ……………(1)
• where k= AE/L
• In matrix form can be written
…………..(2)
AE/L 1 -1 u1 F1
-1 1 u2 = F2 ………..(3)
5

6. Bar element
Or [K]{d}= {r}…….……….(4)
[k] is the stiffness of linear spring
6

7. Bar element
• Consider 2 uniform elastic bars attached end
to end
7

8. Bar element
• F1= k1(u1-u2)
• F2= -k1u1+ (k1+k2)u2- k2u3 ………………(1)
• F3= k2(u3-u2)
• In matrix form can be written
………(2)
• Where [K] is the global stiffness matrix or
structural stiffness
8

9. Bar element
1 -1 0 u1 F1
AE/L -1 2 1 u2 = F2
0 -1 1 u3 F3 …………..(3)
Where K= AE/L
{qi}= [Ki]{ui}………………….(4)
9

10. Spring Element
• Consider a linear elastic spring
10

11. Spring Element
When spring is not deformed net spring
deformation is given
δ= u2 − u1………………………(1)
As a result of axial force
f = kδ = k(u2 − u1)……………..(2)
f1 + f2 = 0 or f1 = −f2…………..(3)
From 2 & 3
11

12. Spring Element
f1 = −k(u2 − u1)
f2 = k(u2 − u1) …………..(4)
In matrix form
k −k u1 f1
−k k u2 = f2 ……………..(5)
Or [k]{u} = { f }………………………(6)
12

13. Spring Element
Where [K]= k −k
−k k
Nodal displacements can be found
u1 f1
u2 = [k]−1 f2 ……….…………(7)
13

14. System assembly in global coordinates
• Process of finding nodal equilibrium equations
• Take individual stiffness components and put
them together to obtain the system equation
• Consider a system of 2 linear spring elements
connected
• Has 2 different spring constants k1 & k2
• Find the system stiffness matrix [k]?
14

15. System assembly in global coordinates
15

16. System assembly in global coordinates
• Assume system is in equilibrium & Using equation 5
…………….……..(8)
• Displacement compatibility conditions relate
element displacements to system displacements
16

17. System assembly in global coordinates
……..…..(9)
Put eqn 9 in 8 we get
…………………(10a)
And
…………………..(10b)
• Element 1 is not connected to node 3 &
element 2 is not connected to node 1
17

18. System assembly in global coordinates
• Expand both matrix equations into 3*3
……………..(11)
………………(12)
• Adding equations 11 & 12 gives us
18

19. System assembly in global coordinates
• ……..………(13)
• Equilibrium conditions of node 1, 2 & 3 figure
c d & e gives us
………………..(14)
Substitute 14 in to 13 we get
19

20. System assembly in global coordinates
………………(15)
Which is the form [K]{U} = {F}
System stiffness matrix is
k1 -k1 0
[K]= -k1 k1+k2 -k2 ………………..……(16)
0 -k2 k2
20

21. Example 1
• Consider 2 element spring
• Node 1 is attached to fixed support, stiffness of
element 1 is 50 lb/in & 75 lb/in for element 2.
find displacements of free nodes. Where F2= F3=
75 lb
21
k1 k2
F1x F2x F3x
x

22. Example 1
• Using equation 15 and given values we get
50 -50 0 0 F1
-50 125 -75 U2 = 75
0 -75 75 U3 75
-50U2= F1 constraint equation
From 2nd and 3rd rows we get
22

23. Example 1
125 -75 U2 = 75
-75 75 U3 75
U2 125 -75 -1 75
U3 = -75 75 75
U2 = 0.02 0.02 75 3
U3 0.02 0.033333 75 = 4
U2= 3 U3= 4
23

24. Example 2
• Assembly of 2 bar elements made of different
materials compute nodal displacements,
stresses and the reaction force
24

25. Example 2
1 -1 = 4*15*106 1 -1 = 3*106 1 -1
K1= A1E1/L1 -1 1 20 -1 1 -1 1
1 -1 = 2.25*10*106 1 -1 = 1.125*106 1 -1
K2= A2E2/L2 -1 1 20 -1 1 -1 1
u1 u2 u3
3 -3 0 u1
[K]= 1*106 -3 4.125 -1.125 u2
0 -1.125 1.125 u3
25

26. Example 2
• {qi}= [Ki]{ui}
R1 3 -3 0 u1
0 = 1*106 -3 4.125 -1.125 u2
2*104 0 -1.125 1.125 u3
2*104 0 -3 3 u3
0 = 1*106 -1.125 4.125 -3 u2
R1 1.125 -1.125 0 u1
26

27. Example 2
qf Kff Kfs Uf Us =0
qs = Ksf Kss Us Uf = [Kff]-1{qf}
U2 1 0 -3 -1 2*104
U3 = 1*106 -1.125 4.125 0
U2 -1.2222 -.8889 2*104 = -0.0244
U3 = 1/1*106 -0.3333 0 0 -0.0066
U2= -0.0244’’ U3= -0.0066’’
27

28. Example 2
{qs}= [Ksf]{Uf} + [Kss]{Us}
-.0244
{qs}= [Ksf]{Uf} = 1*106[1.125 -1.125] -.0066
{R1}= 1*106[-0.0200]
{R1}= {-20000} lb
28

29. Example 2
• Stresses
σ1 = E1(u2-u1)/L1 = 15*106 (-0.0244)/20
σ1 = -18300 psi
σ2 = E2(u3-u2)/L2 = 10*106 (-0.0066+0.0244)/20
σ2= 8900 psi
29

30. 2D Beam Element
Has 2 nodes
2 DOF at each node
Nodal variables are
transverse displacements v1
& v2 at the nodes & the
slopes (rotations) θ1 & θ2
30

31. Beam Element
• v(x ) = f (v1, v2, θ1, θ2, x )
• Boundary conditions:
• v(x = x1) = v1 v(x = x2) = v2
• dv/dx│x= x1 = θ1 dv/dx│x= x2 = θ2 ……..…..(1)
• Assume the system coordinate x1 = 0 and x2= L
• assume the displacement function in the
form v(x ) = a0 + a1x + a2x2 + a3x 3 .............(2)
31

32. Beam Element
• Equations 1 & 2 gives
• v(x = 0) = v1 = a0
• v(x = L) = v2 = a0 + a1L + a2L2 + a3L3
• dv/dx│x= 0 = θ1 = a1 ...........(3)
• dv/dx│x= L = θ2 = a1 + 2a2L + 3a3L2
• Solving simultaneously 3 reduces to
32

33. Beam Element
• a0 = v1 a1 = θ1
• a2 = 3/L2(v2 – v1)- 1/L(2θ1+ θ2) ………...…(4)
• a3 = 2/L3(v1- v2) + 1/L2(θ1+ θ2)
• Substitute 4 in 2 & arranging coefficients of
the nodal variables gives us
……..………………..(5)
33

34. Beam Element
• Which is the form
• v(x ) = N1(x)v1 + N2(x)θ1 + N3(x)v2 + N4(x )θ2
................(6a)
or in matrix from
v1
v(x) = [N1 N2 N3 N4] θ1 = [N]{δ}..............(6b)
v2
θ2
34

35. Beam Element
• N1 N2 N3 N4 are interpolation functions that
describe the distribution of displacement
Equations 5 & 6a
• N1(x) = (1-3x2/L2 +2x3/L3) N1″(x) = -6/L2+12x/L3
• N2(x) = (x-2x2/L+x3/L2) N2″(x) = -4/L+6x/L2
• N3(x) = (3x2/L2-2x3/L3) N3″(x) = 6/L2-12x/L3
• N4(x) = (-x2/L+x3/L2) N4″(x) = -2/L+6x/L2
……..(7)
35

36. Beam Element
• Strain energy of bending for any constant
cross-section beam
…………….……(8)
Substitute 8 in 6a we get
……….(9)
36

37. Beam Element
• Transverse force at node 1 can be found from
equation 9 w.r.t nodal displacement v1
……….(10)
• w.r.t rotational displacement gives us
……….(11)
37

38. Beam Element
• For node 2
……….….(12)
……………(13)
• Equation 10-13 algebraically relate the 4 nodal
displacement values to the 4 applied nodal
forces which is of the form
38

39. Beam Element
…………….(14)
Standard form of element stiffness coefficients is
Kij = EI0∫LNi’’(x)Nj’’(x)dx…………………..(15)
i= row j= column
K11= EI0∫ L N1″(x)N1″(x)dx………………….(16)
39

40. Beam Element
• Equation 7 & 16 gives
K11= EI0∫ L (-6/L2+12x/L3)2dx = EI0∫ L(36/L4-
144x/L5+144x2/L6)dx
=EI[36x/L4-144x2/2L5+144x3/3L6]0
L
=EI/L4(36L-144L2/2L+144L3/3L2)
= EI/L4(36L-72L+48L)
= EI/L4(12L)
K11=12EI/L3
40

41. Beam Element
• Complete stiffness matrix for the beam
element in equation 14 becomes
……………..(17)
41

42. Frame

44. General transformed global stiffness matrix for a
beam element that includes axial force, shear
force, and bending moment effects as follows: