First plug given values into the stiffness matrix:
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
=
[
(
10
+
20
)
−
20
−
20
(
20
+
15
)
]
{\displaystyle K={\begin{bmatrix}(10+20)&-20\\-20&(20+15)\\\end{bmatrix}}}
K
=
[
30
−
20
−
20
35
]
{\displaystyle K={\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}}
Next plug stiffness matrix in equation below:
[
K
−
γ
I
]
x
=
0
{\displaystyle [K-\gamma _{I}]x=0}
Take the determinant:
d
e
t
{
30
−
γ
−
20
−
20
35
−
γ
}
=
[
(
30
−
γ
)
(
35
−
γ
)
]
−
(
−
20
)
(
−
20
)
=
0
{\displaystyle det{\begin{Bmatrix}30-\gamma &-20\\-20&35-\gamma \end{Bmatrix}}=[(30-\gamma )(35-\gamma )]-(-20)(-20)=0}
Simplifying:
1050
−
30
γ
−
35
γ
+
γ
2
−
400
{\displaystyle 1050-30\gamma -35\gamma +\gamma ^{2}-400}
γ
2
−
65
γ
+
650
=
0
{\displaystyle \gamma ^{2}-65\gamma +650=0}
Using the Quadratic formula to solve:
γ
1
=
65
+
65
2
−
(
4
)
(
650
)
2
{\displaystyle \gamma _{1}={\frac {65+{\sqrt {65^{2}-(4)(650)}}}{2}}}
γ
2
=
65
−
65
2
−
(
4
)
(
650
)
2
{\displaystyle \gamma _{2}={\frac {65-{\sqrt {65^{2}-(4)(650)}}}{2}}}
Final gamma values:
γ
1
=
52.66
{\displaystyle \gamma _{1}=52.66}
γ
2
=
12.34
{\displaystyle \gamma _{2}=12.34}
To find the eigenvectors, first we:
[
K
−
γ
I
]
x
=
[
[
30
−
20
−
20
35
]
−
γ
∗
[
1
0
0
1
]
[
]
x
1
x
2
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-\gamma *{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}{\begin{bmatrix}]x_{1}\\x_{2}\\\end{bmatrix}}=0}
To solve, lets set
x
1
{\displaystyle x_{1}}
equal to 1:
[
K
−
γ
I
]
x
=
[
[
30
−
20
−
20
35
]
−
52.66
∗
[
1
0
0
1
]
]
[
1
x
2
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}
[
K
−
γ
I
]
x
=
[
[
30
−
20
−
20
35
]
+
[
−
52.66
0
0
−
52.66
]
]
[
1
x
2
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}+{\begin{bmatrix}-52.66&0\\0&-52.66\\\end{bmatrix}}]{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}
[
K
−
γ
I
]
x
=
[
[
30
−
52.66
−
20
−
20
35
−
52.66
]
[
1
x
2
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30-52.66&-20\\-20&35-52.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}
[
K
−
γ
I
]
x
=
[
[
−
22.66
−
20
−
20
−
17.66
]
[
1
x
2
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}1\\x_{2}\\\end{bmatrix}}=0}
Multiplying out the matrices, we obtain:
−
20
−
17.66
x
2
=
0
{\displaystyle -20-17.66x_{2}=0}
x
2
=
1.1325
{\displaystyle x_{2}=1.1325}
Using the same process we find
x
1
{\displaystyle x_{1}}
:
[
K
−
γ
I
]
x
=
[
[
30
−
20
−
20
35
]
−
52.66
∗
[
1
0
0
1
]
]
[
x
1
1
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}30&-20\\-20&35\\\end{bmatrix}}-52.66*{\begin{bmatrix}1&0\\0&1\\\end{bmatrix}}]{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0}
[
K
−
γ
I
]
x
=
[
[
−
22.66
−
20
−
20
−
17.66
]
[
x
1
1
]
=
0
{\displaystyle [K-\gamma _{I}]x=[{\begin{bmatrix}-22.66&-20\\-20&-17.66\\\end{bmatrix}}{\begin{bmatrix}x_{1}\\1\\\end{bmatrix}}=0}
Multiplying out the matrices, we obtain:
−
22.66
x
1
−
20
=
0
{\displaystyle -22.66x_{1}-20=0}
x
1
=
0.8826
{\displaystyle x_{1}=0.8826}