Dynamics cheat sheet

are the natural frequencies of the system. Since there are two degrees of freedom, there will be two natural
frequencies ω 1 , ω 2 for the system.
det ( [K] − ω 2 [M] ) = 0
⎛⎡
⎤ ⎡ ⎤⎞
det ⎝⎣ k 11 k 12
⎦ − ω 2 ⎣ m 11 m 12
⎦⎠ = 0
k 21 k 22 m 21 m 22
⎡
⎤
det ⎣ k 11 − ω 2 m 11 k 12 − ω 2 m 12
⎦ = 0
k 21 − ω 2 m 21 k 22 − ω 2 m 22
(
k11 − ω 2 ) (
m 11 k22 − ω 2 ) (
m 22 − k12 − ω 2 ) (
m 12 k21 − ω 2 )
m 21 = 0
ω 4 (m 11 m 22 − m 12 m 21 ) + ω 2 (−k 11 m 22 + k 12 m 21 + k 21 m 12 − k 22 m 11 ) + k 11 k 22 − k 12 k 21 = 0
The above is a polynomial in ω 4 . Let ω 2 = λ it becomes
λ 2 (m 11 m 22 − m 12 m 21 ) + λ (−k 11 m 22 + k 12 m 21 + k 21 m 12 − k 22 m 11 ) + k 11 k 22 − k 12 k 21 = 0
This quadratic polynomial in λ which is now solved using the quadratic formula. Then the positive square
root of each λ root to obtain ω 1 and ω 2 which are the roots of the original eigenvalue problem. Assuming from
now that these roots are ω 1 and ω 2 the next step is to obtain the non-normalized shape vectors ϕ 1 , ϕ 2 also
called the eigenvectors associated with ω 1 and ω 2
1.3 Step 3, finding the non-normalized eigenvectors
For each natural frequency ω 1 and ω 2 the corresponding shape function is found by solving the following two
sets of equations for the vectors ϕ 1 , ϕ 2
⎡
⎡
⎧
⎫
⎣ k 11 k 12
k 21
⎤
⎦ − ω1
2
k 22
⎣ m 11 m 12
m 21
⎤ ⎫ ⎧
⎨ϕ ⎦ 11 ⎬ ⎨
m
⎩
22 ϕ
⎭ = 0⎬
⎩
21 0
⎭
and
⎡ ⎤
⎣ k 11 k 12
⎦ − ω2
2
k 21 k 22
⎡
⎣ m 11 m 12
m 21
⎤ ⎧ ⎫ ⎧ ⎫
⎨ϕ ⎦ 12 ⎬ ⎨
m
⎩
22 ϕ
⎭ = 0⎬
⎩
22 0
⎭
For ω 1 , let ϕ 11 = 1 and solve for
⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ k 11 k 12
⎦ − ω1
2 ⎣ m 11 m 12
⎨ 1 ⎬ ⎨
⎦
k 21 k 22 m 21 m
⎩
22 ϕ
⎭ = 0⎬
⎩
21 0
⎭
⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ k 11 − ω1 2m 11 k 12 − ω1 2m ⎨
12 1 ⎬ ⎨
⎦
k 21 − ω1 2m 21 k 22 − ω1 2m ⎩
22 ϕ
⎭ = 0⎬
⎩
21 0
⎭
Which gives one equation now to solve for ϕ 21 (the first row equation is only used)
(
k11 − ω 2 1m 11
)
+ ϕ21
(
k12 − ω 2 1m 12
)
= 0
Hence
ϕ 21 = − ( k 11 − ω1 2m )
11
(
k12 − ω1 2m 12)
5