Solutions Manual for Structural Dynamics Concepts and Applications 1st Edition by H. R. Busby PDF | Step-by-Step Worked Solutions for Vibration Analysis, Dynamic Loading, Damping, and Structural Response | Includes Complete Answers to All Chapter Problems

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All Chapters Covered




SOLUTIONS

,Table of Contents
1. Single-Degree-of-Freedom Systems

2. Random Vibrations

3. Dynamic Response of SDOF Systems Using Numerical Methods

4. Systems with Several Degrees of Freedom

5. Equations of Motion of Continuous Systems

6. Vibration of Strings and Bars

7. Beam Vibrations

8. Continuous Beams and Frames

9. Vibrations of Plates

10. Vibration of Shells

11. Finite Elements and Time Integration Numerical Techniques

12. Shock Spectra

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Chapter 1



1.1 Write the equations of motion for the one-degree-of-freedom systems shown in Figures1.72 (a) … (i). Assume
that the loading is in the form of a force P(t), a given displacement a(t), or a given rotation t as
indicated in the figure.




Figure 1.72 One-degree-of-freedom systems

,Solutions


(a) (b)




spring force = 3EI / L3 u
3
spring force = 48EI / L u 3EI
mu u P(t)
48EI L3
mu u P(t)
L3


(c) (d)




spring force = 3EI / L3 u 3EI / L2 (t)
3EI 3EI
spring force = 3EI / L3 u mu u (t)
a
L3 L2
3EI
mu u a
L3
0
3EI 3EI
mu u a(t)
L3 L3



(e) (f)




spring force = EA / L u
EA spring force = 2 3EI / L3 u 6EI / L3 u
mu u P(t) 6EI
L mu u P(t)
L3

,(g) (h)




muL3
3 For m: u"
spring force = 2 EI / L u 6EI / 3EI
3
L u 2 3
6EI 6EI
mu u a(t)
3 3 P L L 5PL3
L L L
For P: u ' 3 2 48EI
6EI 2
5PL3
For P: u muL3
u ' u"  5P 16mu
48EI 3EI 48EI

48EI 3EI 5
16mu u 5P(t) mu u P(t)
3
L 16
L3


(i)




spring force = 48EI / L3 u
48EI 48EI
mu u a(t)
3
L L3




1.2 Find the natural frequency 0 of each of the systems in Figures 1.72 (a) … (i) (write the general expression for
6 4 2
0 ) and calculate values using E = 30 x 10 psi, I = 80 in , A = 10 in , L = 100 in., and the weight of the mass

500 lbs.



Solutions

, 500
m 1.294 1.3
32.2(12)

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48EI 48(30 106 )(80)
(a) o 297.7 rad/sec
mL3 1.3(100)3
3EI 3(30106 )(80)
(b) o 74.4 rad/sec
mL3 1.3(100)3

(c) o 74.4 rad/sec
mL3

(d) o 74.4 rad/sec
mL3
EA (30106 )(10)
o 1519 rad/sec
(e) mL 1.3(100)
6EI 6(30 106 )(80)
o 105.3 rad/sec
(f) mL3 1.3(100)3

o 105.3 rad/sec
(g) mL3

o 74.4 rad/sec
(h) mL3

o 297.7 rad/sec
(i) mL3




1.3 Find the frequency response functions U * i for the output u(t) and inputs as indicated in Figures 1.72 (a) …
(i) , assuming:

(A) An elastic material (without damping)
(B) An inelastic material with the complex modulus E* E iE

For part (A), plot U * versus the frequency for 0 500 , except for case e, where 0 2000 . For part
*
(B) , plot the absolute values of the frequency functions U versus the frequency for 0 500 . Assume
the same numerical data as given in Problem 1.2, with E 0.03E and E 30 106 psi.



Solutions

(A) Elastic material – no damping
u* 1 1
(a) U*
P *  m(i )2 48EI / L3 115, 200 1.3 2
u* 1 1
(b) U*
P*  m(i )2 3EI / L3 7200 1.3 2

u* 3EI / L3 1
(c) U* 2 3 2
P*  )
m(i 3EI / L  7200 1.3

, u* 3EI / L2 720, 000
(d) U* 2 3 2
P*  )
m(i 3EI / L  7200 1.3
(e) U* u* 1 1

P *  m(i )2 EA / L 3, 000, 000 1.3 2


(f) U* u* 1 1

P*  m(i )2 6EI / L3 14, 400 1.3 2

u* 6EI / L3 14, 400
(g) U* 2 3 2
P*  )
m(i 6EI / L  14, 400 1.3
u*
(h) U* 0.3125


P*  m(i )2 3EI / L3 72, 000 1.3 2

u* 48EI / L3 115, 000
(i) U* 2 3 2
P*  )
m(i 48EI / L  115, 000 1.3


Plots for cases (a),(f),(g),(h)


0.005
0.004
0.003
0.002
0.001 a
0
U*




f,g
-0.001 0 100 200 300 400 500
-0.002 h
-0.003
-0.004
-0.005
-0.006


Plots for cases (b), (c)



b,c
0.015


0.01
U*




0.005 b,c


0
0 100 200 300 400 500
-0.005

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Plot for case (e)



e
0.00005
0
-0.00005 0 500 1000 1500 2000
-0.0001
U*




e
-0.00015
-0.0002
-0.00025
-0.0003


Plot for case (i)



i
200

100

0
U*




0 100 200 300 400 500 i
-100

-200

-300

Plot for case (d)


d
10000
8000
6000
4000
U*




d
2000
0
0 100 200 300 400 500
-2000
-4000

, (B) For an inelastic material we can use the same solution if we replace E with E* E iE , where
6
E 0.03E and E 30 10 psi.

48E ' I 2 48E " I
u* 1 m i
L3 L3
(a) U* 48I 2 2
P* 2 48E ' I 2
48E " I
m(i ) m
L3 (E ' iE ")  3 3 
L
L

1
U* 1

115, 200 1.3   3456
2
 48E ' I  m2    48E " I  2 2

 L3   
   L 
3



1
48E "
tan 3456
I 3 tan 1
48E ' I L 2 115, 200 1.3 2

m
L3



(b) U * 1 1

7200 1.3   216
2
 3E ' I  m   3E " I  2 2

   
   
3E "
216
tan 1 I 3 tan 1
3E ' I L 2 7200 1.3 2

m
L3



3E ' I i3E 3" I 3E ' I 3E " I 3E ' I 2 3E " I
u* i m i
3
3 L33 L3
L L L L
(c) U*
a* 3E 3' I 3E " I 2 2

 i 3E ' m 2 3E '' I
m(i ) 2 3 I 
L L L3
3
3E ' I 3E ' I 2 3E " I 3E "LI
2
m i m 2


U3 *
L L3
L3 L3
2
3E ' I 3E '' I 2
m 2
 
L3 L3


  3E " I  
2
 3E " I
 m2    m2 
U* 2 3E '' I
2
3E ' I
m 2
 
L3 L3


7200 7200 1.3   216   216(1.3)
U*  