On my honor, I have neither given nor received unauthorized aid in doing this assignment.
Mode Shape for system. Mode slope increases and crosses into the positive region 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}}}
k
1
=
1
,
k
2
=
3
,
k
3
=
3
{\displaystyle k_{1}=1,k_{2}=3,k_{3}=3}
K
=
[
3
−
2
−
2
5
]
{\displaystyle K={\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}}
[
K
−
{\displaystyle [K-}
I
]
x
=
{\displaystyle I]x=}
[
3
−
2
−
2
5
]
{\displaystyle {\begin{bmatrix}3&-2\\-2&5\end{bmatrix}}}
−
{\displaystyle -}
[
1
0
0
1
]
{\displaystyle {\begin{bmatrix}1&0\\0&1\end{bmatrix}}}
)
{\displaystyle )}
{
x
1
x
2
}
{\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}
=
{\displaystyle =}
{
0
0
}
{\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}
d
e
t
{
3
−
γ
−
2
−
2
5
−
γ
}
=
γ
2
−
8
γ
+
11
=
0
{\displaystyle det{\begin{Bmatrix}3-\gamma &-2\\-2&5-\gamma \end{Bmatrix}}=\gamma ^{2}-8\gamma +11=0}
Find the eigenvector
x
2
{\displaystyle x_{2}}
γ
2
{\displaystyle \gamma _{2}}
Eigenvalues are found
γ
1
=
4
+
5
>
0
{\displaystyle \gamma _{1}=4+{\sqrt {5}}>0}
γ
2
=
4
−
5
>
0
{\displaystyle \gamma _{2}=4-{\sqrt {5}}>0}
γ
2
{\displaystyle \gamma _{2}}
γ
2
=
4
+
5
{\displaystyle \gamma _{2}=4+{\sqrt {5}}}
[
K
−
γ
2
I
]
x
=
{\displaystyle [K-\gamma _{2}I]x=}
[
−
1
−
5
−
2
−
2
1
−
5
]
{\displaystyle {\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}}
{
x
1
x
2
}
{\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}
=
{\displaystyle =}
{
0
0
}
{\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}
x
2
=
1
{\displaystyle x_{2}=1}
(
−
1
−
5
)
x
1
−
(
2
)
x
2
=
0
{\displaystyle (-1-{\sqrt {5}})x_{1}-(2)x_{2}=0}
x
1
=
2
−
1
−
5
{\displaystyle x_{1}={\frac {2}{-1-{\sqrt {5}}}}}
Eigenvectors are orthogonal to each other:
EDU>> x= [-.8507;-.5257];
EDU>> y= [-.5257;.8507];
EDU>> transpose(y)*x
ans = 0
On my honor, I have neither given nor received unauthorized aid in doing this assignment.
Use same given values as in problem 4.1
Mode Shape for system. Notice plot is the same even with different initial conditions Find the eigenvectors for
γ
1
{\displaystyle \gamma _{1}}
γ
2
{\displaystyle \gamma _{2}}
x
1
=
1
{\displaystyle x_{1}=1}
We find the eigenvectors from
γ
1
{\displaystyle \gamma _{1}}
γ
1
=
4
−
5
{\displaystyle \gamma _{1}=4-{\sqrt {5}}}
[
K
−
γ
1
I
]
x
=
{\displaystyle [K-\gamma _{1}I]x=}
[
−
1
+
5
−
2
−
2
1
+
5
]
{\displaystyle {\begin{bmatrix}-1+{\sqrt {5}}&-2\\-2&1+{\sqrt {5}}\end{bmatrix}}}
{
x
1
x
2
}
{\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}
=
{\displaystyle =}
{
0
0
}
{\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}
x
1
=
1
{\displaystyle x_{1}=1}
(
−
1
−
5
)
x
1
−
(
2
)
x
2
=
0
{\displaystyle (-1-{\sqrt {5}})x_{1}-(2)x_{2}=0}
x
2
=
−
1
+
5
2
{\displaystyle x_{2}={\frac {-1+{\sqrt {5}}}{2}}}
γ
2
{\displaystyle \gamma _{2}}
γ
2
=
4
+
5
{\displaystyle \gamma _{2}=4+{\sqrt {5}}}
[
K
−
γ
2
I
]
x
=
{\displaystyle [K-\gamma _{2}I]x=}
[
−
1
−
5
−
2
−
2
1
−
5
]
{\displaystyle {\begin{bmatrix}-1-{\sqrt {5}}&-2\\-2&1-{\sqrt {5}}\end{bmatrix}}}
{
x
1
x
2
}
{\displaystyle {\begin{Bmatrix}x_{1}\\x_{2}\end{Bmatrix}}}
=
{\displaystyle =}
{
0
0
}
{\displaystyle {\begin{Bmatrix}0\\0\end{Bmatrix}}}
x
1
=
1
{\displaystyle x_{1}=1}
(
−
1
−
5
)
x
1
−
(
2
)
x
2
=
0
{\displaystyle (-1-{\sqrt {5}})x_{1}-(2)x_{2}=0}
x
2
=
−
1
−
5
2
{\displaystyle x_{2}={\frac {-1-{\sqrt {5}}}{2}}}
On my honor, I have neither given nor received unauthorized aid in doing this assignment.
Free body diagram of an element
P
1
(
e
)
=
(
f
1
(
e
)
)
2
+
(
f
2
(
e
)
)
2
{\displaystyle P_{1}^{(e)}={\sqrt {(f_{1}^{(e)})^{2}+(f_{2}^{(e)})^{2}}}}
P
2
(
e
)
=
(
f
3
(
e
)
)
2
+
(
f
4
(
e
)
)
2
{\displaystyle P_{2}^{(e)}={\sqrt {(f_{3}^{(e)})^{2}+(f_{4}^{(e)})^{2}}}}
P
(
e
)
=
T
(
e
)
f
(
e
)
{\displaystyle \mathbf {P} ^{(e)}=\mathbf {T} ^{(e)}\mathbf {f} ^{(e)}}
{
P
1
(
e
)
P
2
(
e
)
}
=
[
l
(
e
)
m
(
e
)
0
0
0
0
l
(
e
)
m
(
e
)
]
{
f
1
(
e
)
f
2
(
e
)
f
3
(
e
)
f
4
(
e
)
}
{\displaystyle {\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}{\begin{Bmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\end{Bmatrix}}}
Discuss computational efficiency of each method.
Reconcile analytically using both algebra and geometry
P
1
(
e
)
=
(
f
1
(
e
)
)
2
+
(
f
2
(
e
)
)
2
{\displaystyle P_{1}^{(e)}={\sqrt {(f_{1}^{(e)})^{2}+(f_{2}^{(e)})^{2}}}}
P
2
(
e
)
=
(
f
3
(
e
)
)
2
+
(
f
4
(
e
)
)
2
{\displaystyle P_{2}^{(e)}={\sqrt {(f_{3}^{(e)})^{2}+(f_{4}^{(e)})^{2}}}}
The first method uses the Pythagorean Theorem, which is also a distance formula, to find the axial member forces from the nodal forces. This method only requires the two nodal forces on a node to find an axial member force. To use this method, the nodal forces are defined, and then put into the distance formula. Additional axial forces are found by defining additional nodal forces. The formula must be repeated each time to give each axial member force.
P
(
e
)
=
T
(
e
)
f
(
e
)
{\displaystyle \mathbf {P} ^{(e)}=\mathbf {T} ^{(e)}\mathbf {f} ^{(e)}}
{
P
1
(
e
)
P
2
(
e
)
}
=
[
l
(
e
)
m
(
e
)
0
0
0
0
l
(
e
)
m
(
e
)
]
{
f
1
(
e
)
f
2
(
e
)
f
3
(
e
)
f
4
(
e
)
}
{\displaystyle {\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}}={\begin{bmatrix}l^{(e)}&m^{(e)}&0&0\\0&0&l^{(e)}&m^{(e)}\end{bmatrix}}{\begin{Bmatrix}f_{1}^{(e)}\\f_{2}^{(e)}\\f_{3}^{(e)}\\f_{4}^{(e)}\end{Bmatrix}}}
The second method uses the transformation matrix to add the projections of the nodal forces along the element. This method requires the two nodal forces and an angle. However, the angle can be applied to both ends of the node. To use this method, the transformation matrix,
T
(
e
)
{\displaystyle \mathbf {T} ^{(e)}}
l
(
e
)
=
cos
θ
{\displaystyle l^{(e)}=\cos {\theta }}
m
(
e
)
=
sin
θ
{\displaystyle m^{(e)}=\sin {\theta }}
f
(
e
)
{\displaystyle \mathbf {f} ^{(e)}}
The matrix multiplication performs the following operations:
P
1
=
l
(
e
)
f
1
(
e
)
+
m
(
e
)
f
2
(
e
)
+
0
+
0
{\displaystyle P_{1}=l^{(e)}f_{1}^{(e)}+m^{(e)}f_{2}^{(e)}+0+0}
P
1
=
f
1
(
e
)
cos
θ
+
f
2
(
e
)
sin
θ
{\displaystyle P_{1}=f_{1}^{(e)}\cos \theta +f_{2}^{(e)}\sin \theta }
P
2
=
0
+
0
+
l
(
e
)
f
3
(
e
)
+
m
(
e
)
f
4
(
e
)
{\displaystyle P_{2}=0+0+l^{(e)}f_{3}^{(e)}+m^{(e)}f_{4}^{(e)}}
P
2
=
f
3
(
e
)
cos
θ
+
f
4
(
e
)
sin
θ
{\displaystyle P_{2}=f_{3}^{(e)}\cos \theta +f_{4}^{(e)}\sin \theta }
Free body diagram of a node
The Pythagorean Theorem is used for the first method. To reconcile this method with the second method, geometry is used to define the nodal forces in terms of
P
1
{\displaystyle P_{1}}
f
1
(
e
)
=
P
1
(
e
)
cos
θ
{\displaystyle f_{1}^{(e)}=P_{1}^{(e)}\cos \theta }
f
2
(
e
)
=
P
1
(
e
)
sin
θ
{\displaystyle f_{2}^{(e)}=P_{1}^{(e)}\sin \theta }
P
1
(
e
)
=
(
f
1
(
e
)
)
2
+
(
f
2
(
e
)
)
2
{\displaystyle P_{1}^{(e)}={\sqrt {(f_{1}^{(e)})^{2}+(f_{2}^{(e)})^{2}}}}
P
1
(
e
)
=
(
P
1
(
e
)
cos
θ
)
2
+
(
P
1
(
e
)
sin
θ
)
2
{\displaystyle P_{1}^{(e)}={\sqrt {(P_{1}^{(e)}\cos \theta )^{2}+(P_{1}^{(e)}\sin \theta )^{2}}}}
P
1
(
e
)
=
(
P
1
(
e
)
)
2
(
cos
2
θ
+
sin
2
θ
)
{\displaystyle P_{1}^{(e)}={\sqrt {(P_{1}^{(e)})^{2}(\cos ^{2}\theta +\sin ^{2}\theta )}}}
P
1
(
e
)
=
(
P
1
(
e
)
)
2
(
1
)
{\displaystyle P_{1}^{(e)}={\sqrt {(P_{1}^{(e)})^{2}(1)}}}
P
1
(
e
)
=
(
P
1
(
e
)
)
2
{\displaystyle P_{1}^{(e)}={\sqrt {(P_{1}^{(e)})^{2}}}}
P
1
(
e
)
=
P
1
(
e
)
{\displaystyle P_{1}^{(e)}=P_{1}^{(e)}}
![{\displaystyle I]x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/659628c77794b974e5177670b1ca076cad3439d5)
![{\displaystyle [K-\gamma _{2}I]x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04ed7874295f7de914404fc662cec20c8f867145)
![{\displaystyle [K-\gamma _{1}I]x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96adbcb4d69ed6a2a42045cd879a8831625b7e66)
