Dynamic Responses - Mechanics of Structures: Variational and Computational Methods - page 622

1 / 2 ,s

Division of the mode shapes by ( M s ) = 1 , 2 , leads to scaled mode shapes

0 and φ 2 = 0

. .

1
√ m

459701 1

√ m

888074

φ 1 =

(4)

0 . 627963 − 0 . 325057

Then

0

1
√ m

. 459701 0 . 888074

Φ = (5)

0 . 627963 − 0 . 325057

Step 3. Compute the generalized mass and loading matrix. For the scaled mode shapes, the
generalized mass matrix is a unit matrix, and the load matrix is

0 t

P 0
√ m

. 627963

Φ T P P = = ≥ 0 (6)

− 0 . 325057

Steps 4, 5, and 6. Obtain the uncoupled equations of motion. The i th ( i = 1 , 2 ) equation is

2
i q i

+ ω = ( )

q i P i t (7)

and the initial conditions are calculated from Eq. (10.85) as q 0 .

Step 7. Compute the modal response due to the applied loading. From Eq. (10.87) with
ζ

( 0 ) = ˙ q ( 0 ) =

= 0 ,

i

t

P 0
√ m

1
ω

q 1 ( t ) = 0 . 627963 sin ω ( t − τ) d τ

1

1

0

P 0

= 0 . 627963

1 √ m (

1 − cos ω

1 t

) (8)

2

ω

P 0

q 2 ( t ) =− 0 . 325057

2 √ m (

1 − cos ω

2 t

)

2

ω

Step 8. From Eq. (10.90), the physical displacement response is computed as V = Φq . Thus,

0

P 0
m

1
ω

1
ω

V 1 = u 1 . 459701 × 0 . 62727963

1 (

1 − cos ω 1 t ) − 0 . 888074 × 0 . 325057

2 (

1 − cos ω 2 t )

796266 k 538188 k

0

455295 1

m t

122 1

m t

P 0
k

= . − cos 0 . − 0 . − cos 1 . (9)

and

0

u 2 P 0
m

1
ω

1
ω

627963 2 325057 2

V 2 = .

1 (

1 − cos ω

1 t

) + 0 .

2 (

1 − cos ω

2 t

)

796266 k 538188 k

0

621945 1

m t

44658 1

m t (10)

P 0
k

= . − cos 0 . + 0 . − cos 1 .