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Therefore the first shape vector is<br />
⎧ ⎫ ⎧<br />
⎨ϕ 11 ⎬ ⎪⎨<br />
ϕ 1 =<br />
⎩<br />
ϕ<br />
⎭ = 1<br />
⎪<br />
21<br />
⎩<br />
− ( k 11 −ω1 2m )<br />
( 11<br />
k12 −ω1 2m 12)<br />
⎫<br />
⎪⎬<br />
⎪⎭<br />
Similarly the second shape function is obtained. For ω 2 , let ϕ 12 = 1 and solve for<br />
⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ k 11 k 12<br />
⎦ − ω2<br />
2 ⎣ m 11 m 12<br />
⎨ 1 ⎬ ⎨<br />
⎦<br />
k 21 k 22 m 21 m<br />
⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
⎡<br />
⎤ ⎧ ⎫ ⎧ ⎫<br />
⎣ k 11 − ω2 2m 11 k 12 − ω2 2m ⎨<br />
12 1 ⎬ ⎨<br />
⎦<br />
k 21 − ω2 2m 21 k 22 − ω2 2m ⎩<br />
22 ϕ<br />
⎭ = 0⎬<br />
⎩<br />
22 0<br />
⎭<br />
Which gives one equation now to solve for ϕ 22 (the first row equation is only used)<br />
(<br />
k11 − ω 2 2m 11<br />
)<br />
+ ϕ22<br />
(<br />
k12 − ω 2 2m 12<br />
)<br />
= 0<br />
Hence<br />
ϕ 22 = − ( k 11 − ω2 2m )<br />
11<br />
(<br />
k12 − ω2 2m 12)<br />
Therefore the second shape vector is<br />
⎧ ⎫ ⎧<br />
⎨ϕ 12 ⎬ ⎪⎨<br />
ϕ 2 =<br />
⎩<br />
ϕ<br />
⎭ = 1<br />
⎪<br />
22<br />
⎩<br />
− ( k 11 −ω2 2m )<br />
( 11<br />
k12 −ω2 2m 12)<br />
Now that the two non-normalized shape vectors are found, the next step is to perform mass normalization<br />
1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors)<br />
Let<br />
µ 1 = ϕ T 1 [M] ϕ 1<br />
This results in a scalar value µ 1 , which is later used to normalize ϕ 1 . Similarly<br />
µ 2 = ϕ T 2 [M] ϕ 2<br />
⎫<br />
⎪⎬<br />
⎪⎭<br />
For example, to find µ 1<br />
⎧ ⎫T ⎡ ⎤ ⎧ ⎫<br />
⎨ϕ 11 ⎬<br />
µ 1 = ⎣ m 11 m 12<br />
⎨ϕ ⎦ 11 ⎬<br />
⎩<br />
ϕ<br />
⎭<br />
21 m 21 m<br />
⎩<br />
22 ϕ<br />
⎭<br />
21<br />
{ } ⎡ ⎤ ⎧ ⎫<br />
= ϕ 11 ϕ ⎣ m 11 m 12<br />
⎨ϕ ⎦ 11 ⎬<br />
21<br />
m<br />
⎩<br />
22 ϕ<br />
⎭<br />
21<br />
m 21<br />
}<br />
=<br />
{ϕ ⎧ ⎫<br />
⎨ϕ 11 ⎬<br />
11 m 11 + ϕ 21 m 21 ϕ 11 m 12 + ϕ 21 m 22<br />
⎩<br />
ϕ<br />
⎭<br />
21<br />
= ϕ 11 (ϕ 11 m 11 + ϕ 21 m 21 ) + ϕ 21 (ϕ 11 m 12 + ϕ 21 m 22 )<br />
6