MESF593 Finite Element Methods

1
MESF593 Finite Element Methods
HW 3 Solutions
2
Prob. 1 (20)
P
(E, L)
Radius c
Radius 2c
Linear Model
Quadratic Model
A tapered bar with circular cross-section is
given as shown above. If two finite element
models (with 4 linear elements and 2 quadratic
elements, respectively) are used to simulate this
problem, compare the displacement results at each
node and the strain results in each element with
respect to the analytical exact solutions.
3
Prob. 1 Solution
IV
I
II
III
P
(E, L)
Radius 2c
Radius c
Linear Model
1
2
5
4
2
3
3
Quadratic Model
For the linear model, the strain in each element
will be constant. For the quadratic model, the
strain in each element will be linear. For
comparison purpose, choose blue line locations
for strain and red line locations for
displacement.
For exact solution -----
4
Prob. 1 Solution
u10
I
II
III
IV
Linear Model
P
1
2
4
3
5
A1(2c)2p
A5c2pA
A3(1.5c)2p
A2(1.75c)2p
A4(1.25c)2p
For a linear bar element with a length of L/4 and
taper cross-section as shown above
For the whole model
5
Prob. 1 Solution
6
Prob. 1 Solution
u10
I
II
Quadratic Model
P
2
4
1
5
3
A1(2c)2p
A5c2pA
A3(1.5c)2p
A2(1.75c)2p
A4(1.25c)2p
For a quadratic bar element with a length of L/2
and taper cross-section as shown above
7
Prob. 1 Solution
8
Prob. 1 Solution
9
Prob. 1 Solution
10
Prob. 2 (25)
Use the finite element method to solve for all
reaction forces and moment at the boundaries of
the beam given above.
11
Prob. 2 Solution
F
2L
I
II
L
1
2
3
EI
MFL
2EI
I
II
12
Prob. 2 Solution
13
Prob. 3 (30)
v4
A plane frame structure is given above. Plot the
deformed shape of the frame and find the value
(in terms of given symbols) of the vertical
deflection at the mid points indicated by the red
dots. (assume the axial deformation is
negligible i.e., use beam elements only)
14
Prob. 3 Solution
Due to the symmetry, we may take a half of the
frame as shown to the left for analysis. Nodes 7
and 8 are added. From observation, the essential
boundary conditions are u1 0, v1 1, q1
0 u7 0, q7 0 u8 0, q8 0
v
u
15
Prob. 3 Solution
Similarly, for Element II -----
M3
X3
3
2
X2
M2
For Element I -----
M2
M2
-X2
X2
2
2
1
1
-X1
X1
M1
M1
16
Prob. 3 Solution
For Element IV -----
For Element III -----
For Element IV -----
M2
M7
7
2
M3
M8
Y7
Y2
Y8
Y3
17
Prob. 3 Solution
For Element I -----
For Element II -----
For Element III -----
For Element IV -----
18
Prob. 3 Solution
Assembly the 4 elements together -----
19
Prob. 3 Solution
Assembly the 4 elements together -----
20
Prob. 3 Solution
Apply boundary conditions -----
21
Prob. 3 Solution
Discussion on boundary conditions -----
???
In the previous assembly equations, there is no
involvement of vertical support (in force or
displacement) at Node 1. As a result, rigid
body motion in the vertical direction will be
induced, causing invalid solution. Therefore,
further boundary conditions need to be enforced.
22
Prob. 3 Solution
Apply further boundary conditions -----

• The further boundary conditions are u2 v2 u3
  v3 0. The arguments for these are
• Bending and axial deformation are independent in
  the small formation theory.
• The current problem assumes negligible axial
  deformation.

Behave like hinge joints
23
Prob. 3 Solution
Reduced assembly equations -----
The deformed shape is
24
Prob. 4 (25)
Po 100 MPa
Po 100 MPa
2b 200 mm
2a 400 mm
(Youngs modulus is E 200 GPa and Poisson Ratio
is n 0.3)
A unit thickness (1 mm) plate is given above. The
boundary and loading conditions are specified as
shown. Use four plane stress elements to find the
displacement of the corner node (marked in red).
25
Prob. 4 Solution
where
26
Prob. 4 Solution
For all 4 elements, the stiffness matrix will
look the same ------
The principle of matrix assembly will be the same
as Prob. 4. But the assembly of the current 4
stiffness matrices will be too big (a 18x18
matrix). Therefore, in this case, working with
numerical values is better.
27
Prob. 4 Solution
Po 100 MPa
Po 100 MPa
6
9
5
III
2b 200 mm
3
8
4
I
II
7
1
2
2a 400 mm
III
II
28
Prob. 4 Solution
IV
III
9
6
5
III
3
8
4
I
II
1
7
2
29
Prob. 4 Solution
For Element I ------
30
Prob. 4 Solution
For Element II ------
31
Prob. 4 Solution
For Element III ------
32
Prob. 4 Solution
For Element IV ------
33
Prob. 4 Solution
0
0
0
0
0
0
0
0
0
0
0
0
0
0
(F6y and F7x are for reference only. They should
not be used because their corresponding EBCs v60
and u70 are specified.)
34
Prob. 4 Solution
After assembly, remove columns with zero
displacement and rows with unknown force------
The numerical values of stiffness matrix of each
element ------
35
Prob. 4 Solution
After substitutions of numbers------