Problem 2: Dampened and Undamped System
On my honor, I have neither given nor received unauthorized aid in doing this assignment.
Spring-damper-body arrangement for R3.2 [ 1] Spring-damper-body arrangement as shown. Two separate forces applied to masses.
M
=
[
m
1
0
0
m
2
]
{\displaystyle M={\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}}
d
=
[
d
1
d
2
]
{\displaystyle d={\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}}
C
=
[
C
1
+
C
2
−
C
2
−
C
2
C
2
+
C
3
]
{\displaystyle C={\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}}
K
=
[
(
k
1
+
k
2
)
−
k
2
−
k
2
(
k
2
+
k
3
)
]
{\displaystyle K={\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}}
1.Draw the FBDs of all components in the spring-mass-damper system on p.53-13, with the known disp dofs being at the left and right supports:
d
3
=
0
{\displaystyle d_{3}=0}
d
4
=
0
{\displaystyle d_{4}=0}
2. Derive (1) p.53-12, and (1)-(3) p. 53-13.
Dampened:Spring-damper-body free body Undamped:Spring-body free body The free body diagrams for a dampened system and undamped system can be seen to the right.
Analyzing the forces on mass one, we obtain:
m
1
d
1
″
+
k
1
d
1
+
C
1
d
1
′
−
k
2
d
2
+
k
2
d
1
−
C
2
d
2
′
+
C
2
d
1
′
=
F
1
{\displaystyle \displaystyle m_{1}d''_{1}+k_{1}d_{1}+C_{1}d'_{1}-k_{2}d_{2}+k_{2}d_{1}-C_{2}d'_{2}+C_{2}d'_{1}=F_{1}}
m
1
d
1
″
+
C
1
d
1
′
+
C
2
(
d
1
′
−
d
2
′
)
+
k
1
d
1
+
k
2
(
d
1
−
d
2
)
=
F
1
{\displaystyle \displaystyle m_{1}d''_{1}+C_{1}d'_{1}+C_{2}(d'_{1}-d'_{2})+k_{1}d_{1}+k_{2}(d_{1}-d_{2})=F_{1}}
Analyzing the forces on mass two, we obtain:
m
2
d
2
″
+
C
2
(
d
2
′
−
d
1
′
)
+
k
2
(
d
2
−
d
1
)
+
C
3
d
2
′
+
k
3
d
2
=
F
2
{\displaystyle \displaystyle m_{2}d''_{2}+C_{2}(d'_{2}-d'_{1})+k_{2}(d_{2}-d_{1})+C_{3}d'_{2}+k_{3}d_{2}=F_{2}}
m
2
d
2
″
+
C
2
(
d
2
′
−
d
1
′
)
+
C
3
d
2
′
+
k
2
(
d
2
−
d
1
)
+
k
3
d
2
=
F
2
{\displaystyle \displaystyle m_{2}d''_{2}+C_{2}(d'_{2}-d'_{1})+C_{3}d'_{2}+k_{2}(d_{2}-d_{1})+k_{3}d_{2}=F_{2}}
[
m
1
0
0
m
2
]
[
d
1
″
d
2
″
]
+
[
C
1
+
C
2
−
C
2
−
C
2
C
2
+
C
3
]
[
(
k
1
+
k
2
)
−
k
2
−
k
2
(
k
2
+
k
3
)
]
[
d
1
′
d
2
′
]
+
[
d
1
d
2
]
{\displaystyle {\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}{\begin{bmatrix}d''_{1}\\d''_{2}\\\end{bmatrix}}+{\begin{bmatrix}C_{1}+C_{2}&-C_{2}\\-C_{2}&C_{2}+C_{3}\\\end{bmatrix}}{\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}{\begin{bmatrix}d'_{1}\\d'_{2}\\\end{bmatrix}}+{\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}}
Notice how this form matches each matrix given in the problem statement. Plugging in the original yields and proves the following equation of motion for an undamped system:
M
d
″
+
C
d
′
+
k
d
=
0
{\displaystyle Md''+Cd'+kd=0}
Analyzing the forces on mass one, we obtain:
m
1
d
1
″
+
k
1
d
1
−
k
2
d
2
+
k
2
d
1
=
F
1
{\displaystyle \displaystyle m_{1}d''_{1}+k_{1}d_{1}-k_{2}d_{2}+k_{2}d_{1}=F_{1}}
m
1
d
1
″
−
k
2
d
2
+
(
k
1
+
k
2
)
d
1
=
F
1
{\displaystyle \displaystyle m_{1}d''_{1}-k_{2}d_{2}+(k_{1}+k_{2})d_{1}=F_{1}}
Analyzing the forces on mass two, we obtain:
m
2
d
2
″
+
k
2
d
2
−
k
2
d
1
+
k
3
d
2
=
F
2
{\displaystyle \displaystyle m_{2}d''_{2}+k_{2}d_{2}-k_{2}d_{1}+k_{3}d_{2}=F_{2}}
m
2
d
2
″
+
(
k
2
+
k
3
)
d
2
−
k
2
d
1
=
F
2
{\displaystyle \displaystyle m_{2}d''_{2}+(k_{2}+k_{3})d_{2}-k_{2}d_{1}=F_{2}}
Plugging equations into matrix form we obtain:
[
m
1
0
0
m
2
]
[
d
1
″
d
2
″
]
+
[
(
k
1
+
k
2
)
−
k
2
−
k
2
(
k
2
+
k
3
)
]
[
d
1
d
2
]
{\displaystyle {\begin{bmatrix}m_{1}&0\\0&m_{2}\\\end{bmatrix}}{\begin{bmatrix}d''_{1}\\d''_{2}\\\end{bmatrix}}+{\begin{bmatrix}(k_{1}+k_{2})&-k_{2}\\-k_{2}&(k_{2}+k_{3})\\\end{bmatrix}}{\begin{bmatrix}d_{1}\\d_{2}\\\end{bmatrix}}}
Notice how this form matches each matrix given in the problem statement. Plugging in the original yields and proves the following equation of motion for an undamped system:
M
d
″
+
k
d
=
0
{\displaystyle Md''+kd=0}
