Solution Manual for Structural Dynamics Theory and Computation Sixth Edition by Paz

1

, If the weight w is displaced by amount, y, the beam and the springs will exert a total force on the
g g g g g g g g g g g g g g g g g g g g




mass of
g g




3𝐸𝐼 g



𝑃=( + 2𝑘)𝑢 g g g g




𝐿3

The beam and springs act in parallel. The equivalent stiffness is:
g g g g g g g g g g




𝑃 3𝐸𝐼 g g



𝑘 = =( + 2𝑘)
g g g g




𝑒
𝑢 𝐿3

Natural frequency: g




𝑘 𝑔 3𝐸𝐼
𝜔=√ =√ ( +2𝑘)
g g


g g g g




𝑚 𝑊 𝐿3

Natural period: g




2𝜋 g 𝑊 g 𝐿
𝑇=
g g = 2𝜋𝐿√ g ( )
𝜔 𝑔 3𝐸𝐼 + 2𝑘𝐿 3



1.2


Stiffness:

3𝐸𝐼 3 × 108 g g




𝑘𝑒 = ( + 2𝑘) = + 2 × 1000 = 4,300 𝑙𝑏/𝑖𝑛.
𝐿3 1003
g g g g g g g g g g g
g g




Natural frequency: g




√𝑘 g 4300 × 386 g g




𝜔= g
=√ = 23.52 𝑟𝑎𝑑/𝑠𝑒𝑐
3000
g g g



𝑚 g




2

,Free vibration response of undamped oscillator:
g g g g g




𝑢˙O
𝑢(𝑡) = 𝑢O𝑐𝑜𝑠𝜔𝑡+ 𝑠𝑖𝑛𝜔𝑡
g


g g
g g



𝜔




3

, 𝑢˙(𝑡) = −𝑢O𝜔𝑠𝑖𝑛𝜔𝑡 + 𝑢˙O𝑐𝑜𝑠𝜔𝑡
g g g g g




Displacement and velocity at 𝑡 = 1𝑠𝑒𝑐 with the initial values
g g g g g g g g g g




𝑢O = 1𝑖𝑛. , 𝑢˙O = 20 𝑖𝑛./𝑠𝑒𝑐:
g g g g g g g g




20
𝑢(1) = 1 ∙ 𝑐𝑜𝑠(23.5 ∙ 1) +
g

𝑠𝑖𝑛(23.5 ∙ 1) = −0.89 𝑖𝑛.
g g g g g g g




23.5
g g g g g




𝑢˙(1) = −1 ∙ 23.5𝑠𝑖𝑛(23.5 ∙ 1) + 20𝑐𝑜𝑠(23.5 ∙ 1) = 22.66 𝑖𝑛./𝑠𝑒𝑐
g g g g g g g g g g g g g g




1.3

The stiffness of the beam is
g g g g g




12𝐸𝐼1 3𝐸(2𝐼2) 12 ∙ (30 ∙ 106) ∙ 170.9
g g g g g g 3 ∙ (30 ∙ 106) ∙ 82.5
g g g g g g




𝑘 =V + Y= + = 25,577 𝑙𝑏/𝑖𝑛.
𝐿3 𝐿3 1443 1443
g g g g g g g




Natural frequency:g




𝑘 1
1 25,577 × 386
√
g g g




𝑓= = √ = 2.24 𝑐𝑝𝑠
2𝜋 𝑚 2𝜋 50,000
g g g g g


g g




1.4

a) Infinitelyrigidhorizontalmember g g g




Stiffness:
g




12𝐸𝐼 12 ∙ (30 ∙ 106) ∙ 171 g g g g g g




𝑘 = 2( )= = 21,100 𝑙𝑏/𝑖𝑛.
𝐿3 (12 ∙ 15)3
g g g g g g g
g g
g g




Natural frequency: g




4