Solution Manual for Structural Dynamics: Theory and Computation (6th Edition) by Mario Paz and William Leigh

1.1


If the weight w is displaced by amount, y, the beam and the springs will exert a total force
on the mass of
3𝐸𝐼
𝑃=# + 2𝑘, 𝑢
𝐿(

The beam and springs act in parallel. The equivalent stiffness is:
𝑃 3𝐸𝐼
𝑘 = =# + 2𝑘,
.
𝑢 𝐿(

Natural frequency:

𝑘 𝑔 3𝐸𝐼
𝜔=0 =0 # + 2𝑘,
𝑚 𝑊 𝐿(

Natural period:

2𝜋 𝑊 𝐿
𝑇= = 2𝜋𝐿0 # ,
𝜔 𝑔 3𝐸𝐼 + 2𝑘𝐿 (



1.2


Stiffness:

3𝐸𝐼 3 × 109
𝑘. = # + 2𝑘, = + 2 × 1000 = 4,300 𝑙𝑏/𝑖𝑛.
𝐿( 100(


Natural frequency:
0𝑘 4300 × 386
𝜔= =0 = 23.52 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑚 3000

Free vibration response of undamped oscillator:
𝑢O
𝑢(𝑡) = 𝑢O𝑐𝑜𝑠𝜔𝑡 + 𝑠𝑖𝑛𝜔𝑡
𝜔


1

, 𝑢(𝑡) = −𝑢O𝜔𝑠𝑖𝑛𝜔𝑡 + 𝑢O𝑐𝑜𝑠𝜔𝑡

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

𝑢O = 1 𝑖𝑛. , 𝑢O = 20 𝑖𝑛./𝑠𝑒𝑐:
20
𝑢(1) = 1 ∙ 𝑐𝑜𝑠(23.5 ∙ 1) + 𝑠𝑖𝑛(23.5 ∙ 1) = −0.89 𝑖𝑛.
23.5
𝑢(1) = −1 ∙ 23.5𝑠𝑖𝑛(23.5 ∙ 1) + 20𝑐𝑜𝑠(23.5 ∙ 1) = 22.66 𝑖𝑛./𝑠𝑒𝑐




1.3

The stiffness of the beam is


12𝐸𝐼W 3𝐸(2𝐼X) 12 ∙ (30 ∙ 10Z) ∙ 170.9 3 ∙ (30 ∙ 10Z) ∙ 82.5
𝑘=V ( + Y= + = 25,577 𝑙𝑏/𝑖𝑛.
𝐿 𝐿( 144( 144(


Natural frequency:

1 𝑘 1 25,577 × 386
0
𝑓= = 0 = 2.24 𝑐𝑝𝑠
2𝜋 𝑚 2𝜋 50,000




1.4

a) Infinitely rigid horizontal member

Stiffness:

12𝐸𝐼 12 ∙ (30 ∙ 10Z) ∙ 171
𝑘 = 2# ( , = = 21,100 𝑙𝑏/𝑖𝑛.
𝐿 (12 ∙ 15)(


Natural frequency:



2

, 0𝑘 21,100 × 386
𝜔= =0 = 18.05 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑚 25,000
𝜔
𝑓= = 2.87 𝑐𝑝𝑠
2𝜋

b) Flexible horizontal member consisting of W18X30

Compute the stiffness by moment distribution method. Displace the frame horizontally by
one inch and determine the stiffness of the frame as the sum of the shear forces in both
columns. Take advantage of the symmetry by modifying the stiffness of horizontal member
by factor 3/2.


Distribution factors
4𝐸𝐼 4𝐸
𝑘 = = ∙ 171 → 171 → 0.1244
^_
𝐿 𝐿
4𝐸𝐼 3 4𝐸 3
𝑘 = = ∙ ∙ 802 → 1203 → 0.8756
^a
𝐿 2 𝐿 2


Fixed end moments:
0.1244 0.8756
C
950 B
6𝐸𝐼 6 ∙ (30 ∙ 10Z) ∙ 170.9
𝑀 =𝑀 = = -118 -832
^_ _^ 𝐿X (12 ∙ 15)X -832
832
= 950 (𝑘 − 𝑖𝑛. )

Shear force:
832 + 891
𝑘= ∙ 2 = 19.14 𝑘𝑖𝑝/𝑖𝑛.
180 950 A D
-59
Natural frequency:
891

1 𝑘 1 19,140 × 386
0 = 0
𝑓= = 2.74 𝑐𝑝𝑠
2𝜋 𝑚 2𝜋 25,000



Note:
1. Assuming a flexible girder decreases the natural frequency by only


3

, 2.74
1− = 0.05 = 5%
2.87

2. Adding the weight of the girder and half of the weight of the two columns decrease the
natural frequency by:
15
𝑊 = 25,000 + 50 ∙ 15 + 2 ∙ 33 ∙ = 26,2456 𝑙𝑏
2

1 𝑘 1 19,140 × 386
0 = 0
𝑓= = 2.67 𝑐𝑝𝑠
2𝜋 𝑚 2𝜋 26,245

2.67
1− = 0.05 = 2.5%
2.74

1.5



Stiffness coefficient:

12𝐸𝐼 192 ∙ 𝐸𝐼
𝑘 = 2# (,=
(2⁄𝐿) (𝐿)(


Natural frequency:
𝑘
192 ∙ 𝐸𝐼𝑔
𝜔=0 =0
𝑚 (𝐿)(𝑊
𝜔
𝑓= 4 3 ∙ 𝐸𝐼𝑔
= 0
2𝜋 𝜋 (𝐿)(𝑊



1.6
𝑘 192 ∙ 10e ∙ 386
𝜔=0 =0 = 92.61 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑚 (120)( ∙ 5,000
𝑢O 15
𝑢(𝑡 = 2) = 𝑢 𝑐𝑜𝑠𝜔𝑡 + 𝑠𝑖𝑛𝜔𝑡 = 0.5 cos(92.61 ∙ 2) + 𝑠𝑖𝑛(92.61 ∙ 2)
O
𝜔 92.61

𝑢(𝑡 = 2) = −0.474 𝑖𝑛.



4