Dynamics cheat sheet

⎡ ⎤ ⎧ ⎫ ⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ m 1 0 ⎨x ⎦
′′ ⎬
1
0 m
⎩
2 x ′′ ⎭ + ⎣ k 1 + k 2 −k 2
⎨x ⎦ 1 ⎬ ⎨
2 −k 2 k
⎩
2 x
⎭ = f 1 (t) ⎬
⎩
2 f 2 (t)
⎭
⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
⎣ 1 0 ⎨x ⎦
′′ ⎬
1
0 3
⎩
x ′′ ⎭ + ⎣ 3 −2 ⎨x ⎦ 1 ⎬ ⎨
−2 2
⎩
x
⎭ = 0 ⎬
⎩
2 sin (5t)
⎭
2
In this example m 11 = 1, m 12 = 0, m 21 = 0, m 22 = 3 and k 11 = 3, k 12 = −2, k 21 = −2, k 22 = 2 and f 1 (t) = 0
and f 2 (t) = sin (5t)
step 2 is now applied which solves the eigenvalue problem in order to find the two natural frequencies
Let ω 2 = λ hence
The solution is λ 1 = 3. 475 and λ 2 = 0.192, therefore
And
det ( [K] − ω 2 [M] ) = 0
⎛⎡
⎤ ⎡ ⎤⎞
det ⎝⎣ 3 −2 ⎦ − ω 2 ⎣ 1 0 ⎦⎠ = 0
−2 2 0 3
⎡
⎤
det ⎣ 3 − ω2 −2
⎦ = 0
−2 2 − 3ω 2
(
3 − ω
2 ) ( 2 − 3ω 2) − (−2) (−2) = 0
3ω 4 − 11ω 2 + 2 = 0
3λ 2 − 11λ + 2 = 0
ω 1 = √ 3.475 = 1.864
ω 2 = √ 0.192 = 0.438
step 3 is now applied which finds 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 eigen vectors ϕ 1 , ϕ 2
⎛⎡
⎤ ⎡ ⎤⎞
⎧ ⎫ ⎧ ⎫
⎝⎣ 3 −2 ⎦ − ω1
2 ⎣ 1 0 ⎨ϕ ⎦⎠
11 ⎬ ⎨
−2 2 0 3
⎩
ϕ
⎭ = 0⎬
⎩
21 0
⎭
For ω 1 = 1. 864
⎛⎡
⎤ ⎡ ⎤⎞
⎧ ⎫ ⎧ ⎫
⎝⎣ 3 −2 ⎦ − 1.864 2 ⎣ 1 0 ⎨ϕ ⎦⎠
11 ⎬ ⎨
−2 2
0 3
⎩
ϕ
⎭ = 0⎬
⎩
21 0
⎭
⎡
⎤ ⎧ ⎫ ⎧ ⎫
⎣ −0.475 −2 ⎨ 1 ⎬ ⎨
⎦
−2 −8.424
⎩
ϕ
⎭ = 0⎬
⎩
21 0
⎭
This gives one equation to solve for ϕ 21 (the first row equation is only used)
−0.475 − 2ϕ 21 = 0
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