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. For exact solution -----
Prob. #1 Solution I II
III IV
(E, L) P
Radius: 2c
Radius: c
Linear Model 1 2 2 3 3 4 5
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 u1=0 Linear Model PII
III IV
Linear Model P 1 2 3 4 5
A1=(2c)2p
A2=(1.75c)2p
A3=(1.5c)2p
A4=(1.25c)2p
A5=c2p=A
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 x u1=0 Quadratic Model PII
Quadratic Model P 1 2 3 4 5
A1=(2c)2p
A2=(1.75c)2p
A3=(1.5c)2p
A4=(1.25c)2p
A5=c2p=A
For a quadratic bar element with a length of L/2 and taper cross-section as shown above 1 3 2 x
x=-1
x=1

7. Prob. #1 Solution

8. Prob. #1 Solution

9. Prob. #1 Solution

10. Prob. #2 (25%) F 2L
M=FL L
2EI EI
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 I 2L II L 1 2 3 EI
M=FL
2EI I II

12. Prob. #2 Solution

13. Prob. #3 (30%)
P (N/m) 1 L 2L EI
12EI 2 3 5 6 4 v3
x = x1
q(x)
Hint------
x = x2 v1 v2
L = x2-x1 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 I II IV
III 1 2 7 3 8
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 q2 v2
x = x1
q(x)
For a general beam ---
x = x2 v1 q1
L = x2-x1

15. Prob. #3 Solution Similarly, for Element II ----- For Element I -----IV
III 1 2 7 3 8
Similarly, for Element II ----- M3 X3 3 u v 2 X2 M2
For Element I ----- M2 M2
-X2 X2 2 2 1
-X1 1 X1 M1 M1

16. Prob. #3 Solution For Element IV ----- For Element III -----1 2 7 3 8 u v
For Element III -----
For Element IV ----- 3 8 M2 2 7 M7 M3 M8 Y2 Y7 Y3 Y8

17. Prob. #3 Solution For Element I ----- For Element II -----1 2 M1 M2 X2 X1 2 3 M2 M3 X3 X2
For Element III -----
For Element IV ----- 2 7 M2 Y2 M7 Y7 M8 Y8 M3 Y3 3 8

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 ----- ??? 1 2 7 3II IV
III 1 2 7 3 8 u v
???
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. (Young’s modulus is E = 200 GPa and Poisson Ratio is n = 0.3)
Prob. #4 (25%)
Po = 100 MPa
Po = 100 MPa
2b = 200 mm
2a = 400 mm
(Young’s 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 2b = 200 mm 2a = 400 mm IV6 IV 5 9
III
2b = 200 mm 4 3 8 I II 1 2 7
2a = 400 mm
III II

28. Prob. #4 Solution IV
III 6 IV 5 9
III 4 3 8 I II 1 2 7

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
(F6y and F7x are for reference only. They should not be used because their corresponding EBCs v6=0 and u7=0 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------