2.092/2.093
FINITE ELEMENT OF SOLIDS AND FLUIDS I
FALL 2009
Homework 1- solution
Instructor:
TA:
Prof. K. J. Bathe
Seounghyun Ham
Assigned:
Due:
Session 3
Session 5
Problem 1 (30 points):
The truss geometry is shown below. Bar mumbers are circled. Joint numbers are placed
adjacent to their respective joints.
U4
U2
U3
3
U8 2
4
4
3 U6
a
5
U7
1
U1
U5
2
1
a
a
For a linear static analysis, we have:
K8×8U8×1=R8×1
a) The K matrix is calculated column by column. The ith column of the stiffness
matrix represents the external force vector required to give the structure unit
displacement about the ith degree of freedom and zero displacement about all other
degree of freedom. Take a look at one example how to construct it.
Calculate column 5:
Imposing the following displacement pattern:
U5=1, U1=U2=U3=U4=U6=U7=U8=0
1
The resulting external force vector under this set of displacement conditions is equal
to the 5th column of the stiffness matrix K. In this case, the truss bar 1 changes length
by 1, the truss bar 5 shrinks by 1 , and all other truss bars are fixed in length.
2
Hence, the bar axial forces are as follows.
N1 =
AE
AE
, N 2 = N 3 = N 4 = 0, N 5 = −
a
2a
Positive and negative signs of the axial forces imply tension and compression,
respectively. Hence the reaction forces, the entries of the 5th column, are obtained
from the equilibrium equations at the joints.
K15 =
K 25 = −
AE
AE 1
AE
AE
, K 55 =
(
+
1), K 65 =
, K 75 = −
, K 35 =
K 45 =
K85 =
0
a 2 2
a
2 2a
2 2a
After assembling all columns, the following K is determined:
1
⎡ 1
⎢ 2 2 +1 2 2
⎢
1
⎢ 1
⎢
2 2
2 2
⎢
⎢
0
⎢
-1
⎢
⎢
0
0
EA
⎢
K =
⎢
1
1
a
⎢
−
−
⎢ 2 2
2 2
⎢
1
⎢− 1
−
⎢ 2 2
2 2
⎢
⎢ 0
0
⎢
⎢
⎢ 0
0
⎢⎣
⎤
⎥
2 2
2 2
⎥
1
1
⎥
−
−
0
0
0
0 ⎥
2 2
2 2
⎥
1
1
1
1 ⎥
+1
−
−
0
0
2 2
2 2
2 2
2 2⎥
⎥
1
1
1
1 ⎥
+1
−
−
0
-1
2 2
2 2
2 2
2 2⎥
⎥
1
1
0
0
+1
−1
0 ⎥
⎥
2 2
2 2
⎥
1
1
0
-1
+1
0
0 ⎥
⎥
2 2
2 2
⎥
1
1
1
1 ⎥
−
-1
0
+1
−
2 2 ⎥
2 2
2 2
2 2
⎥
1
1
1
1 ⎥
−
0
0
−
2 2
2 2 ⎥⎦
2 2
2 2
-1
0
−
1
−
1
0
0
b) Since U1 = U 2 = U 4 = U 7 = U8 = 0 , we can reduce K8×8U8×1=R8×1 to KaaUa=Ra
as follows:
2
⎡ 1
⎤
0
0 ⎥
⎢ 2 2 +1
⎢
⎥ ⎡U 3 ⎤ ⎡ R3 ⎤ ⎡ 0 ⎤
1
1 ⎥⎢ ⎥ ⎢ ⎥ ⎢
EA ⎢
0
+1
U 5 ⎥ = ⎢ R5 ⎥ = ⎢ 6 ×104 ⎥
⎢
⎥
⎢
⎥ N.
a
2 2
2 2
⎢
⎥ ⎢⎣U 6 ⎥⎦ ⎢⎣ R6 ⎥⎦ ⎢⎣ 0 ⎥⎦
1
1
⎢
⎥
+ 1⎥
⎢⎣
0
2 2
2 2 ⎦
The solution is
0
⎡U 3 ⎤
⎡
⎤
⎢U ⎥ = a ⎢ 4.76 × 104 ⎥ ×10 4
⎢ 5 ⎥ EA ⎢
⎥
⎢−1.24
⎢U
×104 ⎥⎦
⎣ 6 ⎥⎦
⎣
c) Since we know the values of U 3 , U 5 , and U 6 , we can calculate the reaction forces
from KbaUa=Rb,
⎡
⎢ −1
⎢
R
⎡ 1⎤ ⎢ 0
⎢R ⎥ ⎢
⎢ 2⎥ ⎢ 1
⎢R4 ⎥ = ⎢
⎢ ⎥ ⎢ 2 2
⎢ R7 ⎥ ⎢ 1
⎢
⎣⎢ R8 ⎥⎦ ⎢ − 2 2
⎢
⎢− 1
⎢⎣ 2 2
−
1
2 2
1
−
2 2
0
−1
0
1 ⎤
2 2⎥
⎥
1 ⎥
⎡ −1.24 × 104 ⎤
⎡ R1 ⎤
−
⎥
⎢
⎥
2 2
⎢ ⎥
4
0
⎥⎡
⎤
⎢ −1.24 ×10 ⎥
⎢ R2 ⎥
−1 ⎥⎥ ⎢ 4.76 × 104 ⎥ N = ⎢ 1.24 ×104 ⎥ N = ⎢ R4 ⎥
⎢
⎥
⎢
⎢ ⎥
4⎥
⎥ ⎢⎣ −1.24 ×10 4 ⎥⎦
⎢ −4.76 ×10 ⎥
⎢ R7 ⎥
⎢
⎥
0 ⎥
⎢⎣ R8 ⎥⎦
0
⎣
⎦
⎥
⎥
0 ⎥
⎥⎦
−
The undeformed and deformed meshes with applied boundary conditions and loads
are plotted on the next page.
3
To calculate all internal forces, let’s draw the equilibrium diagram for each joint.
R=1.2426×104 N
P=6×105 N
R
R
2
3.8284R
3.8284R
R
R
4
3
1.4142R
5
1.4142R
1
3.8284R
P=4.8284R
Therefore, the internal forces are
Element 1: tension 4.7572×104 N
Element 2: no force
4
Element 3: tension 1.2426×104 N
Element 4: no force
Element 5: Compression 1.7573×104 N
⎡ R1 ⎤ ⎡ −1.24 ×104 ⎤
⎢R ⎥ ⎢
4 ⎥
⎢ 2 ⎥ ⎢ −1.24 ×10 ⎥
⎢ R3 ⎥ ⎢
⎥
0
⎢ ⎥ ⎢
4 ⎥
R4
1.24 ×10 ⎥
and Reactions: ⎢ ⎥ = ⎢
⎢ R ⎥ ⎢ 6×105 ⎥ N
⎢ 5⎥ ⎢
⎥
0
⎢ R6 ⎥ ⎢
⎥
⎢ R ⎥ ⎢ −4.76 ×104 ⎥
⎢ 7⎥ ⎢
⎥
⎢R
0
⎦⎥
⎣ 8 ⎥⎦ ⎢⎣
d) We can make sure that element 3 and joint 3 are in equilibrium explicitly by the
diagram below.
1.2424×104N
1.2424×104N
1.2424×104N
1.2424×104N
5
External forces acting on structure.
12.4kN
12.4kN
12.4kN
60kN
12.4kN
∑F
x
= −12.4kN − 47.6kN + 60kN = 0
∑F
y
= 12.4kN −12.4kN = 0
∑M
at 4
�
�
�
= −12.4kN × a − 47.6kN × a + 60kN × a = 0
Therefore, the structure is in equilibrium.
Question: Why does the joint 3 not move horizontally? Assume you have not
calculated the detailed solution given above. Give your answer and a physical reason.
6
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2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
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