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7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
1/23
Remember from last lecture, single element (2 nodes)matrix can be written as follows
Element e
1e 2e
u1e
u2e
F1e F2
e
ke
2
e
2
k ue
1
u-k
e2
k u
e
e1uk
e
eF=+
e1
F=
1 2
1
2
2
e
2
ke ue
1
u-k
e2
ke u
e
e1uk
e
eF=+
e1
F=
1 2
1
2
ke - ke
ke- ke
The stiffness matrix from:
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Lets write the individual matrix for each element
1
2
3
4
P applied load
u
1
u
2
u
3
u
4
R1 unknown
reaction force
k2
k3
ele. 1
ele. 2
ele. 3
k1
5
R1
K1(u2 u1)
k4 ele. 4
K1(u2 u1)
K2(u3 u2)K2(u3 u2)
K3(u4 u3)
K3(u4 u3)
K4(u5 u4)
K4(u5 u4)
P
Node 1
Node 2
Node 3
Node 4
Node 5
ee
ee
kk
kk
11
11Element 1
Element (e) Nodes
1 1, 2
2 2, 3
3 3, 4
4 4, 5
00000
00000
00000
00000
00000
1 2 3 4 5
1
23
4
5
This is a plainmatrix
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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ee
ee
kk
kk
11
11Element 1
00000
00000
00000
000
000
1 2 3 4 5
1
2
3
4
5
1
2
3
4
P applied load
u
1
u
2
u3
u4
R1 unknown
reaction force
k2
k3
ele. 1
ele. 2
ele. 3
k1
5
R1
K1(u2 u1)
k4 ele. 4
K1(u2 u1)
K2(u3 u2)K2(u3 u2)
K3(u4 u3)
K3(u4 u3)
K4(u5 u4)
K4(u5 u4)
P
Node 1
Node 2
Node 3
Node 4
Node 5
Element 1 is elated to nodes 1 & 2So we can place this matrix in the
plain global matrix at theintersection of nodes 1 & 2
00000
00000
00000
00000
00000
1 2 3 4 5
1
2
3
4
5
ee
ee
kk
kk
11
11
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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ee
ee
kk
kk
22
22Element 2
00000
00000
000
00
000
1 2 3 4 5
1
2
3
4
5
1
2
3
4
P applied load
u
1
u
2
u3
u4
R1 unknown
reaction force
k2
k3
ele. 1
ele. 2
ele. 3
k1
5
R1
K1(u2 u1)
k4 ele. 4
K1(u2 u1)
K2(u3 u2)K2(u3 u2)
K3(u4 u3)
K3(u4 u3)
K4(u5 u4)
K4(u5 u4)
P
Node 1
Node 2
Node 3
Node 4
Node 5
Element 2 is elated to nodes 2 & 3So we can place this matrix in the
plain global matrix at theintersection of nodes 2 & 3
00000
00000
00000
00000
00000
1 2 3 4 5
1
2
3
4
5
ee
ee
kk
kk
22
22
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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ee
ee
kk
kk
33
33Element 3
00000
000
00
00
000
1 2 3 4 5
1
2
3
4
5
1
2
3
4
P applied load
u
1
u
2
u3
u4
R1 unknown
reaction force
k2
k3
ele. 1
ele. 2
ele. 3
k1
5
R1
K1(u2 u1)
k4 ele. 4
K1(u2 u1)
K2(u3 u2)K2(u3 u2)
K3(u4 u3)
K3(u4 u3)
K4(u5 u4)
K4(u5 u4)
P
Node 1
Node 2
Node 3
Node 4
Node 5
Element 3 is elated to nodes 3 & 4So we can place this matrix in the
plain global matrix at theintersection of nodes 3 & 4
00000
00000
00000
00000
00000
1 2 3 4 5
1
2
3
4
5
ee
ee
kk
kk
33
33
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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ee
ee
kk
kk
44
44Element 4
000
00
00
00
000
1 2 3 4 5
1
2
3
4
5
1
2
3
4
P applied load
u
1
u
2
u3
u4
R1 unknown
reaction force
k2
k3
ele. 1
ele. 2
ele. 3
k1
5
R1
K1(u2 u1)
k4 ele. 4
K1(u2 u1)
K2(u3 u2)K2(u3 u2)
K3(u4 u3)
K3(u4 u3)
K4(u5 u4)
K4(u5 u4)
P
Node 1
Node 2
Node 3
Node 4
Node 5
Element 4 is elated to nodes 4 & 5So we can place this matrix in the
plain global matrix at theintersection of nodes 4 & 5
00000
00000
00000
00000
00000
1 2 3 4 5
1
2
3
4
5
ee
ee
kk
kk
44
44
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
7/23
The final global matrix of the system is obtained simply byassembling or adding the individual element matrices
000
00
00
00
000
1 2 3 4 5
1
2
3
4
5ee
ee
kk
kk
44
44
ee
ee
kk
kk
33
33
ee
ee
kkkk
22
22
ee
ee
kk
kk
11
11
The overlappingK will be addedin the system
global matrix
The assembly of the above matrix gives the global matrix of the system
+
+
+
44
4433
3322
2211
11
000
00
00
00
000
kk
kkkk
kkkk
kkkk
kkThis method givesthe same matrixwhich we havedeveloped fromindividual equilibrium
equation for eachnode
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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To calculate the stiffness of any element
k = stiffness
E = Modules of elasticity
A = cross sectional area ( width x thickness )
l= length of the element
1
2
3
4
A
B
D
C
F
a
a
a
b c bwall
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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For the suspended plate which is fixed to the roof as shown in Fig. belowa. Model the plate in suitable FE form.b. Calculate the stiffness of each element, ke
c. Write the element nodes table.d. Set the global stiffness matrix of the model.e. Solve for the elongation in each element
E = Modulus of Elasticity = 100 GPa
All dimensions in mmt = thickness of the plate = 1mm
a = 20 mm
a a a a a
200
100
10 kN
Example 1
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Model the plate in suitable FE form.
Example 1
1
2
3
Element 1 Element 2 Element 3
10kN
Element(e) Nodes
1 1, 2
2 1, 2
3
2, 34
1, 2
To calculate the stiffness
Element 4
k = stiffness
E = Modules of elasticity
A = cross sectional area ( width x thickness )
l= length of the element
Since elements 1, 2, & 3 have the same area. Module ofelasticity & length, hence their stiffness are equalK1 = K2 = K3
E = 100 GPa
t = thickness of the plate = 1mm
a = 20 mm
a a a a a
200
100
10 kN
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Individual matrix of each element
ee
ee
kk
kk
11
11
ee
ee
kk
kk
22
22
ee
ee
kk
kk
33
33
Element 1 Element 2 Element 3 Element(e) Nodes
1 1, 2
2 1, 2
3
2, 34
1, 2
000
000
000
1 2 3
1
2
3
0
0
ee
ee
kk
kk
11
11
ee
ee
kk
kk
22
22
ee
ee
kk
kk
33
33
ee
ee
kk
kk
44
44
Element 4
ee
ee
kk
kk
44
44
+++
++
44
44321321
321321
0
0
kk
kkkkkkkk
kkkkkk
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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=
550
5116
066
107G
K
+++
++
44
44321321
321321
0
0
kkkkkkkkkk
kkkkkk
Substitute the above k values into the developed global matrix gives
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Remember from last lecture
So we can write the matrix as
}F{}u{]K[ GG =
=
10000
0
0
550
5116
066
10
3
2
1
7
u
u
u
Zero because its a fixed point
=
10000
0
0
550
5116
066
10
3
2
1
7
u
u
u
In any matrix if there zero then the columnand row of the zero will be cancelled
=
10000
0
55
51110
3
27
u
u
U1=0 because its a fixed poin
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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=
10000
0
55
511
103
27
u
u We get from this matrix
two equations
Two equations with two unknowns can be easily solved and get
U2=0.167mm
U3=0.367mm
U1=0
Multiplying the matrices (refer to matrix lecture slides) gives
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Example 2:
The beam shown in the figure below is
cantilevered at one end and is pulled by a force of
100 N at the other end. Calculate the totalelongation of the beam.
38
E1 E2 100 N
m0275.0
m0425.0
2
1
=
=
h
h
h1h2
0.1m 0.1m
E= 200000 Pa
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Ex 2: Solution
Use 2 uniform cross section elements.
39
E1 E2 100 N
m0275.0
m0425.0
2
1
=
=
h
h
h1h2
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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E1 E2 100 N
100 N
k1 k2Node 1 Node 2 Node 3
R1
Two Elements Stiffness Matrix
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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K=
By using the method to assemble the global matrix fromindividual elements matrix. The resulted global stiffness
matrix for the system is:
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Ex 2: Solution
Since this is a uniaxial 1-D problem, the elementscan be simplified as 2 springs in series.
42
100 N
k1 k2Node 1 Node 2 Node 3
R1
with element stiffness,ki = E * Ai / L
Where Ai=2h (cross section area is square)
k1 = [(200000*0.04252)/0.1] = 3612.5 N/mk2 = [(200000*0.0275
2)/0.1] = 1512.5 N/m
E1 E2 100 N
m0275.0
m0425.0
=
=
h
h
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Ex 1: Solution
43
K= =
Global Stiffness Matrix**:
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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44
Ex 2: Solution
Arranging the problem in matrix form (K*x = F) yield;
A known solution, u1 = 0. Thus the problem can be
reduced to (by eliminating first column & first row):
=
=
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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45
Option 1 solution: Using the inverse matrix method, x = K * F ;-1
Ex 1: Solution
= =
=
-1
Refer to previous matrix
lecture slides
7/27/2019 Computer Aided Engineering- FE Lecture 4 Slides
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Option 2 solution: Use two equations solution
=
Two equations with two unknowns can be easily solved and get
U2=0.027
U3=0.093
U1=0