FEM 1d
FEM 1d
FEM 1d
formulation of 1D
problem
The following are the steps to be followed for carrying out the finite element analysis.
EA
Fx L L u1 EA 1 1 u
1
Fx2 EA
1
L 1
L
EA u2
1 u2
EA L
Element stiffness matrix by variational formulation
1
T
U 2 v
d v
We have the relation
Strain =
B B du s N
where d
is called as the strain displacement relationship.
Stress can be represented in terms of the strain as
where D
D
is called as the constituent relationship matrix.
Using the above relationship in the strain energy equation we get
21
B u D B u
T
U v
Applying the variational principle we haved v
U
F
u So we get the element stiffness matrix as
F B
k B
T
T
We know
D B dv
Fu
DBdv
that
v
k u
v
4 Stiffness of Truss Member
1. Element Stiffness of a truss member
𝜕𝑢
𝜀𝑥 = = 𝜕 𝑁1 𝜕𝑁2
Displacement function:
u x
0 𝜕𝑥 𝜕𝑥 𝑢, + 𝜕𝑥 𝑢 2
𝑢1
x 1 = 𝜕𝑁 1 𝜕𝑁2
𝑢2
𝜕𝑥 𝜕𝑥
x 0
1 1 𝑢1
= 𝐵
Applying boundary conditions:
At x= 0, u(0)= u1 𝑢2
v
1
Where, d N
B dx L L
1
So, the stiffness matrix will be:
k B D B d v
T
v
1
B
L T
Adx AE L L 1 1 AE 1 1
dx
0 0
1 L L 1 1
EB L
L
Thus, the stiffness matrix,
AE
k 1
1 L 1 1
4.2 Generalized Stiffness Matrix of a Plane Truss Member
Consider a member making an angle ‘’ with X axis as shown in Fig 6 below. By
resolving the forces along local X and Y direction, the following relations are
obtained.
Fx 1 Fx 1 c o s Fy 1 s i n
F x 2 c o s F y 2 sin
Fx 2
Fy 1 Fx 1 si n Fy 1 co s
F x 2 sin F y 2 c o s
Fy2
Where,
Fx1 Fx 2
are the axial forces along the member axis X
and
. Similarly,
Fy1 and are the forces perpendicular to the member axis
Fy 2
X
X ,u
Y, v
Y,v
X, u
sin 0
Fx1 0 Fx1
Fy1
cos sin cos 0 0 F
y1
F x 2 0 0
sin Fx 2
cos
Fy 2 0 cos Fy 2
0
One can express the as:
sin
above equation in short
F T F
Where, [T] is called transformation matrix that relates the global axis and member axis.
Similarly, the relations of nodal displacements between two coordinate systems may expressed
as:
d T d
Again, the equation stated in 16 can be generalized and expressed with respect to the member axis
as:
Fx1 1 0 u1
Fy1 AE 0 0 v
1
F x 2 0 0 u 2
Fy 2 0 v 2
1 0
F K d
F T K T
0
T d L 1
K T K T 0
T
0
0
sin sin
cos 1 0 1 0 cos
AE 0 0 0 cos
K sin
cos 0 sin
cos cos sin
sin 0 0 1 0
sin L 1
cos 0
0 0 sin
cos
3 x x
x2
2 0 dx
2
;
3 3
Now, applying boundary conditions one can find the following expressions from the above relations: At
x=0:
1 1
V 1 1 0 0 0 2 2
1 0 1 0 0
3
3
4
4
;
1
1
At x=L: 2 0 1 2 L 3L 2
2
V2 1 L L2 L3 2
3
3
4
4
Thus combining above expressions one can write:
0 00
V 1 1 1
1 0 0 2 A
0
1
2 3
3
V 2
2
1
0 L L L 2 4
1 2L 3 L1
0 0 V 1
1
1
2 0 0 2
So, 0
2
L3
3 1 V 2
2
0 0
1 L 0 V1
L3 2 3
1 1
2 L L 2
2L 1 2 1L V 2
L230 1 2L 3L
4 2
L2 3
L 2
L
1 0
0
0 1
0
Therefore,
;
1 0 0
0 V
;
0 1 0 V
1
2 0 3 1
V x 1 x x 2
x 3 3 2
2
1
N
1 N 2 N 3 N 4
V 2
1
l 1 l V 2 2
l 3 l 2 2
1
2 1
l
2 l
Where, 3 l 3 l2 2
N x x2 x
3
N1 1 2 x 2 x3
2
L L2
L2 L3
3x2 2x3 x 2 x3
N3 2
and N 4 L L2
L
N is called shape function which interpolates the beam displacement in terms of its
nodal
L3
displacements.
Now, the strain displacement relationship [B] can be expressed from the following expressions:
0
d v
2
1
0 0 2 6 x B B A 1
dx 2
d 2
Where, 4
1 0 0 0 V 1
B 0 0 2 6x; A 0 2
1 2
L3
; d V2
0
0 1 0 3L2 2
d 2v
L L EIBA
1
From the moment curvature relationship, we can M EI EI
write: dx 2
Strain Energy,
L
1 1
M d x
T
2L d
U 0 2 L
EI T
A
1 T
B T B A 1
= 2 d
d d x
0
Thus, F U E I L A 1
B B A
T T 1
0
d
d d x
L
So, the stiffness matrix will be:
B B A d x
K E I A
1 T T 1
0 L
B d x A
A
1 T T 1
EI B
0
00
L
B Bdx 0 6xdx
T L
Now 0 2
0 0 2
6x
0 0 0 0
0
L
0 0 0
dx
0 0 4 12x
2
0
0 0 12x 36x
1 3 2
0
So, 0 00 L 2
L 3 0 0 0
0 0
0 0 2L L12 0 0 0 0
K
E I A1 T
0
0 00
2
A 1 E I 3 A 1
0 0 2 0 0 4L 6 L
2
0 0 4L 6L
L2 L3 0 1 2 L3
1 0 6L2
12L 3 0 1
0 0 6L 2 0 L L2
0 12 6 12 6
1 0 L3 L 2
L 3
L2
0 0 0 6
0
6
4 6 6
0
EI0 0 2 0 L2
2
L L
2
L2 1 E I L1 2
L2
6 12 6
0
1
6l 23
3 2 3 2
6L 2
L L
4
L
02 3 2
L2 6
3 2 3 1
0 0 0 6 L L 2 L
L
0 0 2
L2 12 6L 12 6L
1
2 6L 2L2
Thus, the element stiffness of a beam member k EI 4L 2
6L
is:
L L L L
2
L3 12 6L 12
6L
6L 2L2 6L
4L
Dimension less number . Value is varies from -1 to
+1
Variation of natural
coordinate
𝜉 = 𝑎 + 𝑏x
We can write
2
𝜉 = 𝑥 2−𝑥 [x − x1)-1
1
Variation
of 𝜉
N2 variation
We can write the shape function
𝑁 𝜉 = 𝑎 + 𝑏𝜉
2
𝜉= 𝑥 − 𝑥1 − 1
𝜘2 − 𝜘1
Strain displacement relation
u= 𝑁𝑢𝑒
1 𝑢
𝜀 1 1
le 𝑢2
𝜀 = 𝐵𝑢
𝜎 = 𝐸ε = 𝐸𝐵u
Third step is finding stiffness matrix of individual
elements
Next step is assembly which gives global stiffness matrix
Now determine global load vector
displacement