Solutions Manual for
Structural Dynamics: Concepts and Applications
Henry R. Busby
George H. Staab
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
the figure.
Figure 1.72 One-degree-of-freedom systems
 t 
as indicated in
Solutions
(a)
(b)
spring force =  3EI / L3  u
spring force =  48EI / L  u
3
mu 
mu 
48EI
u  P(t )
L3
(c)
3EI
u  P(t )
L3
(d)
spring force =  3EI / L3  u   3EI / L2   (t )
spring force =  3EI / L3   u  a 
mu 
3EI
u  a   0
L3
3EI
3EI
mu  3 u  3 a(t )
L
L
mu 
(e)
3EI
3EI
u  2  (t )
3
L
L
(f)
spring force =  EA / L  u
EA
mu 
u  P(t )
L
spring force = 2  3EI / L3  u   6 EI / L3  u
mu 
6 EI
u  P(t )
L3
(g)
(h)
For m: u " 
spring force = 2  EI / L  u   6 EI / L  u
3
mu 
6EI
6EI
u  3 a (t )
3
L
L
3
P   L
 L
 3  L   
6EI   2 
2
muL3
3EI
 5PL3
 
 48EI
5PL3 muL3
L3
For P: u  u ' u " 


 5P  16mu 
48EI 3EI
48EI
2
For P: u ' 
16mu 
48EI
u  5P(t )
L3
3
mu 
3EI
5
u  P( t )
3
L
16
(i)
spring force =  48EI / L3  u
mu 
48EI
48EI
u  3 a(t )
3
L
L
1.2 Find the natural frequency 0 of each of the systems in Figures 1.72 (a) … (i) (write the general expression for
0 ) and calculate values using E = 30 x 106 psi, I = 80 in4, A = 10 in2, L = 100 in., and the weight of the mass
500 lbs.
Solutions
m
500
 1.294  1.3
32.2(12)
(a) o 
48EI
48(30 106 )(80)

 297.7 rad/sec
mL3
1.3(100)3
(b) o 
3EI
3(30 106 )(80)

 74.4 rad/sec
mL3
1.3(100)3
(c) o 
3EI
 74.4 rad/sec
mL3
(d) o 
3EI
 74.4 rad/sec
mL3
(e) o 
EA
(30 106 )(10)

 1519 rad/sec
mL
1.3(100)
(f) o 
6 EI
6(30 106 )(80)

 105.3 rad/sec
3
mL
1.3(100)3
(g) o 
6 EI
 105.3 rad/sec
mL3
(h) o 
3EI
 74.4 rad/sec
mL3
(i) o 
48EI
 297.7 rad/sec
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
(a) U * 
 
u* 
1
1



2
3
P *  m(i )  48EI / L  115, 200  1.3 2 
(b) U * 
 
u* 
1
1



P *  m(i ) 2  3EI / L3   7200  1.3 2 
(c) U * 
 
u* 
3EI / L3
1



2
3
2
P *  m(i )  3EI / L   7200  1.3 
(d) U * 
  720,000 
u* 
3EI / L2


2
3
P *  m(i )  3EI / L   7200  1.3 2 
(e) U * 
 
u* 
1
1




2
2
P *  m(i )  EA / L   3,000,000  1.3 
(f) U * 
 
u* 
1
1



P *  m(i ) 2  6 EI / L3  14, 400  1.3 2 
(g) U * 
 
u* 
6 EI / L3
14, 400



P *  m(i ) 2  6 EI / L3  14, 400  1.3 2 
(h) U * 
 
u* 
5 / 16
0.3125



2
3
2
P *  m(i )  3EI / L   72,000  1.3 
(i) U * 
 
u* 
48EI / L3
115,000



2
3
2
P *  m(i )  48EI / L  115,000  1.3 
U*
Plots for cases (a),(f),(g),(h)
0,005
0,004
0,003
0,002
0,001
0
-0,001 0
-0,002
-0,003
-0,004
-0,005
-0,006
a
100
200
300
400
500
f,g
h
Plots for cases (b), (c)
b,c
0,015
U*
0,01
0,005
b,c
0
0
-0,005
100
200
300
400
500
Plot for case (e)
e
0,00005
0
U*
-0,00005 0
500
1000
1500
2000
-0,0001
e
-0,00015
-0,0002
-0,00025
-0,0003
Plot for case (i)
i
200
100
U*
0
0
100
200
300
400
i
500
-100
-200
-300
Plot for case (d)
d
10000
8000
U*
6000
4000
d
2000
0
-2000 0
-4000
100
200
300
400
500
(B) For an inelastic material we can use the same solution if we replace E with E *  E   iE  , where
E   0.03E  and E   30 106 psi.

   48E ' I  m 2   i  48E " I  
 

3
   L3
u* 
1
  L 
(a) U * 



2
2 
P *  m(i ) 2  48 I ( E ' iE ")    48E ' I
 48E " I  
2

m


3
 

3
L

   L3
  L  


1

U* 
2
2
  48E ' I
 48E " I 
2

m


 

3
  L3
  L 
 

 
 

 
 

1
115, 200  1.3    3456 
2 2
2





48E " I 




3456

1 
L3
   tan 1 
   tan 
2 
48E ' I
115, 200  1.3 

 m 2 
 L3



1

(b) U *  
2
2
  3E ' I
 3E " I 
2

m


  3 
  L3
  L 
 

 
 

 
 

1
 7200  1.3    216 
2 2
2





3E " I 

  L3

216


   tan 
  tan 1 

2 
3E ' I
7200

1.3

2


 3  m 
 L

1
3E ' I
3E " I

i 3
3
u* 
L
L

(c) U * 
a *  m(i ) 2  3E ' I  i 3E " I

L3
L3
   3E ' I
   L3

 
 
3 E " I    3E ' I
 3E " I  
 m 2   i 3  
 
L3   L3
L 


2
2
 3E ' I
 3E '' I 
2

 3  m    3 

 L
  L 
i
2
 3E ' I  3 E ' I

 3E " I 
 3E " I
2
2

m


 3  3
  3   i  3 m  
 L  L
  L 
 L

U* 

2
2
 3E ' I
 3E '' I 
2



m


 3
  3 


 L
  L 

2 2
2
  3E ' I  3E ' I  m 2    3E " I     3E " I m 2 
  3 
 3

  L3  L3
  L    L


U* 
2
2
 3E ' I
 3E '' I 
2

 3  m    3 

 L
  L 


2 2
2
2 2
 7200  7200  1.3    216     216(1.3) 
U* 
2

 7200  1.3 2    2162















3E " I




 3 m 2
216(1.3 2 )



1 
L
  tan 1 

tan

2
2 
2 
2 2
 3E ' I  3E ' I  m 2    3E '' I  
 7200  7200  1.3    216  
  3  
 L3  L3
  L  
 3E ' I 3E ' I
  2  3
u *   L  L
(d) U * 

* 


2
2 
  3E " I   3E " I    3E " I

 m 2    2   3     2 m 2  
  L   L   L
 

2
2
 3E ' I
 3E '' I 
2

 3  m    3 

 L
  L 

2

2
2 2
 720,000  7200  1.3    21,600  216    21,600(1.3) 
U* 
2
2
7200  1.3 2    216 










3E " I


 2 m 2


L
  tan 1 

2
 3E ' I  3E ' I  m 2    3E '' I   3E '' I  
  2  3 
 L2  L3
  L  L 


21,600(1.3 2 )


 tan 1 

2
2
 720,000  7200  1.3   21,600  216  
(e)


u* 
1
U* 

2
2
P*   E ' A
 E" A
2

m


 

  L
  L 
 

 
 

 
 

1
 3,000,000  1.3    90,000
2 2
90, 000

2 
3,
000,
000

1.3




  tan 1 


1

(f) U *  
2
2
  6E ' I
 6E " I 
2

m


  3 
  L3
  L 
 

 
 

 
 

1
14, 400  1.3    432 
6E " I 

  L3

432


  tan 
 tan 1 

2 
6E ' I
14,
400

1.3

2


 3  m 
 L

1
2 2
2





2






2 2
2
  6 E ' I  6 E ' I  m 2    6 E " I     6 E " I m 2 
  3 
 3

  L3  L3
  L    L

(g) U *   
2
2
 6E ' I
 6 E '' I 
2

 3  m    3 

 L
  L 


2 2
2
2 2
 14, 400 14, 400  1.3    432     432(1.3) 
U* 
2

14, 400  1.3 2    4322




  tan 1 
 3E ' I
 L3













3E " I



m 2
3
432(1.3 2 )


1 
L

tan

2
2 
2 
2 2
3
E
'
I
3
E
''
I


 
2
14, 400 14, 400  1.3    432  
 3  m    3  
 L
  L  



5 / 16

(h) U *  
2
2
  3E ' I
 3E " I 
2

m


  3 
  L3
  L 
 

 
 

 
 

0.3125
 7200  1.3    216 
2 2
2





3E " I 

  L3

216

1 
  tan 
  tan 
2 
3E ' I
7200

1.3

2


 3  m 
 L

1
(i)



U* 



 48E ' I

3
 L
2
2
2
 48E ' I
 48 E " I    48 E " I
2
2

m



m



 



3
3
3
 L
  L    L

2
2
 48E ' I
  48E '' I 
 m 2   


3
3
 L
  L 

2 2
2
2 2
 115, 200 115, 200  1.3    3456     3456(1.3) 
U* 
2

115, 200  1.3 2   3456 2















48E " I
2





m

3456(1.3 2 )




L3
  tan 1 
 tan 1 

2
2 
2
2
2
 48E ' I  48E ' I  m 2    48E '' I  
115, 200 115, 200  1.3   3456 
 

3
 L3  L3
  L  
Plots for cases (a), (b), (e), (f), (h)
0,005
0,0045
U*
0,004
0,0035
a
0,003
b
0,0025
e
0,002
f
0,0015
h
0,001
0,0005
0
0
100
200
300
400
500
Plots for cases (c), (g, (i)
35
30
U*
25
20
c
15
g
i
10
5
0
0
100
200
300
400
500
Plot for case (d)
d
3500
3000
U*
2500
2000
d
1500
1000
500
0
0
100
200
300
400
500
1.4 Find the frequency response functions of the maximum bending moments M *  i  for systems (a), (b), (c), (d),
(f), (g), (h), and (i) of problem 1.1; plot values of M * / P * or M * / a* or M * / Q* , as appropriate, versus the
frequency  . Find the frequency response of the axial force, N *  i  , in the system (e); plot N * / P * versus
the frequency  for 0    500 being careful to avoid the singularity. Assume EI  6.25 106 N-m2 ,
EA  1.25 106 N , L  2.5 m , and m  200 kg .
Solutions
(a) M  Fd 
1 48EI  L   12 EI 
u   
u
2 L3  2   L2 
 12 EI 
 12 EI 
M *   2  u*   2 U * P *
 L 
 L 


 
M *  12 EI 
12 EI 
1
75,000,000

  2 U *  2 

2
48
EI
P*  L 
L 
 m 2  19, 200,000  200 
 L3

(a)
4000
3000
2000
(a)
1000
0
-1000
0
100
200
300
400
3EI
 3EI 
 3EI 
uL   2  u
M *   2 U * P *
3
L
 L 
 L 


 
M *  3EI 
1
3,000,000

 3EI  
  2 U *   2  

2
3
EI
P*  L 
 L 
 m 2  1, 200,000  200 
 L3

(b) M  Fd 
500
(b)
200
150
100
(b)
50
0
0
100
200
300
400
500
3EI
 3EI 
 3EI 
 3EI 
u  a  L   2  u  a 
M *   2   u * a *   2  U * a * a *
3 
L
 L 
 L 
 L 
3EI




2
3
  3EI   m 2
 3,000,000  200 
M *  3EI 
 3EI  
L

 1   2  

 U * 1   2   3EI
2
3EI
a *  L2 
 L 
 m 2   L   3  m 2  1, 200,000  200
3
 L

 L

(c) M  Fd 
(c)
300000000
200000000
(c)
100000000
0
0
(d) M  Fd 
100
200
3EI
 3EI 
u  QL  L   2   u  QL 
3 
L
 L 
300
400
500
 3EI 
 3EI 
M *   2   u * Q * L    2  U * Q * Q * L 
 L 
 L 


3EI


  3EI   mL 2 
2
M *  3EI 
 3EI  
L

 L   2  

 U *  L    2   3EI
Q *  L2 
2
 L  
  L   3EI  m 2 
 m 
3

3
  L

 L



 3,000,000  2.5(3,000,000  200 2 )
1, 200,000  200 2
(d)
100
80
60
(d)
40
20
0
0
(e) N  F 
EA
u
L
100
N* 
200
300
400
500
EA
 EA 
u*  
U * P *
L
 L 



N *  EA 
1
500,000
 EA  


U *  
  EA
2
P*  L 
 L 
 m 2  500,000  200
 L

(e)
14
12
10
8
6
4
2
0
(e)
0
100
200
300
400
3EI
 3EI 
 3EI 
 3EI 
uL   2  u
M *   2  u*   2 U * P *
L3
L
L




 L 


 
M *  3EI 
1
3,000,000

 3EI  


 U *   2   6 EI
2
P *  L2 
2,
400,000

200

2
 L 

 m  
 L3

(f) M  Fd 
500
(f)
200
150
100
(f)
50
0
0
100
200
300
400
500
 3EI 
 3EI 
M *   2   u * a *   2  U * a * a *
 L 
 L 
6
EI




2
3
  3EI   m 2
  3,000,000  200  
M *  3EI 
 3EI  
L
  2  U * 1*   2  
 1   2  



2
a*  L 
 L   6 EI  m 2   L   6 EI  m 2   2, 400,000  200 
 L3

 L3

(g) M  Fd 
3EI
 3EI 
u  a  L   2  u  a 
3 
L
 L 
(g)
400000000
300000000
200000000
(g)
100000000
0
0
200
300
400
 3EI 
 3EI 
M *   2  u*   2 U * P *
 L 
 L 


M *  3EI 
 3EI   5 / 16   0.3125  3,000,000 
  2 U *   2  

2 
P*  L 
 L   3EI  m 2   1, 200,000  200 
 L3

(h) M  Fd 
3EI
 3EI 
uL   2  u
3
L
 L 
100
500
(h)
60
40
(h)
20
0
0
100
200
300
400
500
1 48EI
L
12EI
12EI
12EI
u  a    2  u  a  M *   2  u * a *   2  U * a * a *
2 L3
2  L 
L


 L 
48EI




2
3
  12 EI  
 12,000,000  200  
M *  12 EI 
200 2
 12 EI  
L
  2  U * 1   2  
 1   2  




2
a*  L 
 L   48EI  m 2   L   48EI  200 2  19, 200,000  200 
 L3

 L3

(i) M  Fd 
(i)
1,5E+10
1E+10
5E+09
(i)
0
0
100
200
300
400
500
-5E+09
1.5 Using Laplace transformations, find the transfer functions for the systems given in Problem 1.1. Determine the
transient responses. Assume:
(A) Elastic material.
(B) Kelvin model, with E = 30 x 106 psi,   60 106 psi.sec.,
(C) Maxwell model, with E = 30 x 106 psi,   18 108 psi.sec.,
(D) Standard solid model, with E0  30 106 psi., E1  60 106 psi.
  60 106 psi.sec.
Solutions
(A) Elastic material – no damping
(a) U ( s) 
u 
1

 2
. For the transient response
3
P  ms  48EI / L 
U (t )  L
U (s)  L
-1
-1
1/ m


 2
3
 s  48EI / mL 
 48EI
sin 
3
48EI
 mL
m
mL3
1

 t

Or
U (t ) 
(b) U ( s) 
U (s)  L
-1
-1
1/ m
1


sin o t
 2
3
 s  3EI / mL  mo
o2 
3EI
mL3
U (s)  L
-1

-1
3EI / mL3 
1


sin o t
 2
3

 s  3EI / mL 
 mo
o2 
3EI
mL3
 3EI / L2 
 2
. For the transient response
  ms  3EI / L3 
U (s)  L
-1

-1
3EI / mL2 
EI
 2 sin ot
 2
3
 s  3EI / mL  mL o
o2 
3EI
mL3
u 
1

 2
. For the transient response
P  ms  EA / L 
U (t )  L
(f) U ( s) 
mL3
u
U (t )  L
(e) U ( s) 
48EI
u  3EI / L3 

 . For the transient response
P  ms 2  3EI / L3 
U (t )  L
(d) U ( s) 
o 
u 
1

 2
. For the transient response
3
P  ms  3EI / L 
U (t )  L
(c) U ( s) 
1
sin o t
mo
U (s)  L
-1
-1
1/ m
1

sin o t
 2

 s  EA / mL  mo
u 
1


. For the transient response
P  ms 2  6EI / L3 
o2 
EA
mL
(g) U ( s ) 
o2 
6EI
mL3
U (s)  L
-1

-1
6EI / mL3 

 sin ot
 2
3

 s  6EI / mL 

o2 
6EI
mL3
u 
5 /16

 2
. For the transient response
3
P  ms  3EI / L 
U (t )  L
(i) U ( s ) 
1/ m
1


sin ot
 2
3
m
o
 s  6EI / mL 
-1
u  6 EI / L3 

. For the transient response
a  ms 2  6 EI / L3 
U (t )  L
(h) U ( s) 
-1
U (s)  L
U (t )  L
U (s)  L
-1
-1
5 /16m
5


sin ot
 2
3
 s  3EI / mL  16mo
o2 
3EI
mL3
u  48EI / L3 

. For the transient response
a  ms 2  48EI / L3 
U (t )  L
U (s)  L
-1

-1
48EI / mL3 

 sin ot
 2
3
 s  48EI / mL 


o2 
48EI
mL3
(B) For the Kelvin model, replace E by E ( s )  E   s with E = 30 x 106 psi,   60 106 psi.sec.,

 

 
1/ m
1/ m
(a) U ( s )  

48EI 48 Is   2 48 I
48EI
2
s 
 s 

s
mL3
mL3  
mL3
mL3


1/ m

-1
U (t )  L 
48 I
48EI
2
s 
s

mL3
mL3

1
U (t ) 
m
48EI  24 I 


mL3  mL3 
2
e




L



24 I
mL3










1/ m
-1

2 
2
24 I 
48EI  24 I  

  s  mL3    mL3   mL3   

 

2
 48EI
 24 I  
sin 

t


 mL3  mL3  


(C) For the Maxwell model, replace E by E ( s ) 
 Es
with E = 30 x 106 psi,   18 108 psi.sec.,
E s




1 / m  E s
1 / m  E   s 
1/ m

(a) U ( s )  

48
EI

s
 2 48EI   ES    s 3  Es 2 

E
48EI 
s  s2  s 
3
 s  mL3  E   s  
mL

mL3 




(b) U ( s) 
u 
1

. For the transient response
 2
P  ms  3EI / L3 
U (t )  L
(c) U ( s) 
o2 
3EI
mL3
U (s)  L
-1

-1
3EI / mL3 
1


sin o t
 2
3
m
o
s

3
EI
/
mL




o2 
3EI
mL3
 3EI / L2 
 2
. For the transient response
  ms  3EI / L3 
U (s)  L
-1

-1
3EI / mL2 
EI
 2 sin ot
 2
3
 s  3EI / mL  mL o
o2 
3EI
mL3
u 
1


. For the transient response
P  ms 2  EA / L 
U (t )  L
(f) U ( s) 
1/ m
1

sin o t
 2

 s  3EI / mL3  mo
u
U (t )  L
(e) U ( s) 
-1
u  3EI / L3 

 . For the transient response
P  ms 2  3EI / L3 
U (t )  L
(d) U ( s) 
U (s)  L
-1
U (s)  L
-1
-1
1/ m
1

sin o t
 2

m
o
 s  EA / mL 
o2 
EA
mL
u 
1

 2
. For the transient response
3
P  ms  6EI / L 
U (t )  L
U (s)  L
-1
-1
1/ m
1


sin ot
 2
3
 s  6EI / mL  mo
o2 
6EI
mL3
(g) U ( s ) 
u  6 EI / L3 

. For the transient response
a  ms 2  6 EI / L3 
U (t )  L
(h) U ( s) 

-1
6EI / mL3 

 sin ot
 2
3
 s  6EI / mL 


o2 
6EI
mL3
u 
5 /16

. For the transient response

P  ms 2  3EI / L3 
U (t )  L
(i) U ( s ) 
U (s)  L
-1
U (s)  L
-1
-1
5 /16m
5


sin ot
 2
3
16
m
o
 s  3EI / mL 
o2 
3EI
mL3
u  48EI / L3 

. For the transient response
a  ms 2  48EI / L3 
U (t )  L
U (s)  L
-1

-1
48EI / mL3 

 sin ot
 2
3
s

48
EI
/
mL




o2 
48EI
mL3
1.6 Find the particular solution (steady state) for the displacement if the loading is
P(t )  P0 sin t ,
P(t )  P0 cos t
a (t )  a0 sin t ,
a (t )  a0 cos t
 (t )  0 sin  ,  (t )   0 cos 
Consider the elastic and inelastic materials as in Problem 1.3, viscoelastic materials as in Problem 1.5.
Solution
(a) For P(t )  P0 sin t , u(t )  P0 Im U *eit 
Elastic material: U *(i ) eit 
cos t
sin t
i
 48EI
 48EI
2
2
 3  m   3  m 
 L
  L

u (t ) 
Inelastic material:
Po sin t
 48EI
2
 3  m 
L


U *(i ) eit
 48E ' I

 48E ' I

 m 2  cos t
 m 2  sin t


3
3
L
L






2
2
2
2
 48E ' I
 48E " I   48E ' I
 48E " I 
2
2
 m   
 m   

 

3
3
3
3
 L
  L   L
  L 
  48E ' I

48E " I

 m 2  sin t 
cos t 

3
3


L
L

 i 

2
2
48
E
'
I
48
E
"
I




2


 m   


3
3


 L
  L 
  48E ' I

48E " I

 m 2  sin t 
cos t 

3
  L3

L

u(t )  Po 

2
2
 48E ' I
 48E " I 
2


 m   


3
3


 L
  L 
Viscoelastic material: Kelvin model - E *(i )  E  i   30  106  1  2i   E ' iE
  144  107 I


288  107 I
 m 2  sin t 
cos t 

3
3
L
L



u(t )  Po  

2
2
7
7
 144  10 I
 288  10 I 


2

m



 

3
3


L
L

 



For P(t )  P0 cos t , u(t )  P0 Re U *eit 
Elastic material:
u (t ) 
Po cos t
 48EI
2
 3  m 
L


  48E ' I

48E " I

 m 2  cos t 
sin t 

3
3


L
L

Inelastic material: u(t )  Po  

2
2
48
E
'
I
48
E
"
I



2



m



 

3
3


 L
  L 
Viscoelastic material: Kelvin model - E *(i )  E  i   30  106  1  2i   E ' iE
  144  107 I


288  107 I
 m 2  cos t 
sin t 

3
3
L
L



u(t )  Po  

2
2
7
7
 144  10 I
 288  10 I 


2

m



 

3
3


L
L

 



1.7 Find the solution for the displacement u(t) if the loading is as in Problem 1.6 for t > 0, and the initial conditions
are
u u 0
for t  0
Solution
From Problem 1.6, the particular solution for an elastic material is
up 
Po sin t
 48EI
2
 3  m 
 L

o2 
The natural frequency is
up 
48EI
, so
mL3
 Po / m  sin t

2
o
 2 
The corresponding solution is
u  A sin o t  B cos o t 
 Po / m  sin t

2
o
 2 
With initial conditions u(0)  u(0)  0
u(0)  A(0)  B(1)  0  B  0
u(0)  o A(1)  o B (0) 
 Po / m  

2
o
 2 
0 
A
Po / mo
2
2
o  

Therefore
u
 Po / m 



 sin t    sin o t 
    
 o 

2
o
2
For inelastic or viscoelastic materials the solution is the same with the new particular solution used.
1.8 Find the particular solution (steady state) for the displacement for the loading shown in Figure 1.74
Figure 1.73 Loading sequence
Consider systems as in Problem 1.1, with various types of material. Calculate the first three terms of the
Fourier series involved.
Solution
f ( x) 
ao  
 n x 
 n x  
  an cos 
  bn sin 

2 n 1 
p


 p 
1 p
2 P T /2
f ( x )dx 
dt  P


p
p
T 0
Where ao 
T /2
 n x 
1 p
2 P T /2
P
 2n t 
 2n t 
an   f ( x ) cos 
cos 
sin 
dt 
 0
dx 


p
0
p
T
n
 T 
 T 0
 p 
T /2
 n x 
1 p
2 P T /2  2n t 
P
 2n t 
f ( x ) sin 
sin 
cos 
dt 

dx 



p
0
p
T
n
 T 
 T 0
 p 
P
2P

n  1,3,5,...
 cos n  1 
n
n
bn 
Therefore
f (t ) 

P
2P
2n t
 
sin
2 n 1,3,5 n
T
P (t ) 
P 4  1
2n t 
1   sin

2   n 1,3,5 n
T 
Using the first three terms
P( t ) 
Since  
2
T
P 4
2 t 4
6 t 
1  sin
 sin

2 
T
3
T 
P (t ) 
P 2P
2P

sin t 
sin 3 t
2 
3
As in Problem 6, the steady-state displacement is
P / 2  2P 
 2P 
it
3it

 ImU *(i )e   
 ImU *(3i )e 
K
 
 3 
u(t ) 
u (t ) 
PL3

96 EI
2 P sin t
2 P sin 3t

 48EI
 48EI
2
  3  m  3  3  9m 2 
 L

 L

For inelastic and viscoelastic materials proceed as in Problem 6
Chapter 2
2.1 What are the dimensions of the correlation function R x and the spectral density S x if the random function X (t )
is
(A) Displacement (in.)
(B) Acceleration ( in./sec2 )
(C) Force (lb.)
(D) Stress (psi)
Solution
R x   
1 T
x(t ) x(t   )d
T T
S x ( ) 
2
1 *
1
xT  i  
2T
2T



Part
Function
R x  
Sx ( )
A
Displacement (in.)
in.2
in.2  sec
B
Acceleration ( in./sec2 )
in.2 / sec4
in.2 / sec3
C
Force (lb.)
lb.2
lb.2  sec
D
Stress (psi)
lb.2 / in.4
lb.2  sec/ in.4
xT (t )eit dt
2
2.2. What are the dimensions of the probability function Fx and the probability density function f x if the random
function X (t ) is
(A) Displacement (in.)
(B) Acceleration ( in./sec2 )
(C) Force (lb.)
(D) Stress (psi)
Solution
Fx  


f ( x)dx
Part
Function
A
fx 
2
2
1
e  ( x  x ) /2 c
2 c
Displacement (in.)
Fx
Dimensionless
fx
1/ in
B
Acceleration ( in./sec2 )
Dimensionless
sec2 / in.
C
Force (lb.)
Dimensionless
1/lbs.
D
Stress (psi)
Dimensionless
in.2 /lbs
2.3. Define a stationary random process.
Solution
A stationary random process is defined by the correlations function of a random variable being constant regardless
of the time at which it is evaluated, so long as the period over which each evaluation is taken is unchanged in
duration.
2.4. Consider a stationary ergodic process X (t ) ; express E  X  , E  X 2  , and R x   as the time averages.
Solution
For a stationary ergodic process X (t ) we have in terms of time averages
1 T
x(t )dt
2T T
1 T 2
E  X 2 (t )  lim
x (t )dt
T  2T T
E  X (t )  lim
T 
R x    E  X (t ) X (t   )  lim
T 
1
2T

T
T
x(t ) x(t   )dt
2.5. Prove the relations between the spectral density S   and the correlation function R   :
R   
1
2


 S  e d
i


S     R  ei d

Solution
1
2T
R x    E  X (t ) X (t   )  lim
T 




T
T
x(t ) x(t   )dt


1  i
1 
e d  x (t )x(t   )dt 
d  x (t )x(t   )e i ( t  )eit dt






2T
2T

1 
it
 i ( t  )

x( t) e dt  x( t   )e
d

2T 
R  ei d 

Since X *(i)   x(t )eit dt




R  ei d 
and

X *(i)   x(t   )e i ( t  )d

1
X *(i) X *( i)
2T
Take the limit of both sides as T  

lim  R  ei d  lim
T  


T 
1
2
X *(i) X *( i)
2T
lim R  ei d  S ()
 T 

S ()   R  ei d

Thus, S () 
F R( )
R( ) 
and
R( ) 
1
2



F
S ()
1
S   ei d
t
2.6. Assuming that the input-output relation is y(t )    (t   ) x( )d  . Show that

R y    



 
     R x      d  d and S y     *(i ) S x  
Solution
Given y (t ) we know that the correlation function is
2
R y    lim
T 
1
2T

T
T
y(t ) y(t   )dt  lim
T 
1
2T

T
T




dt  x(t   ) ( )d   x(t     ) ( )d
where

y  t     (t   )x( )d    x(t   ) ( )d    x(t   ) ( )d 
t
t


y t     
t 



x(t     ) ( )d   x(t     ) ( )d



R y    lim  d   d ( ) ( )
T  
lim
T 
1
2T

T
T

1 
x(t   ) x(t     )dt
2T 
x(t   )x(t     )dt  R x      
t    t1
and t      t2 so t2  t1 =    
therefore
R y    



 
     R x      d  d



  
S ( )   R  e i d  

 

     R x       e i d d  d






  d  d   d e i      R x      






  d  d   d ei e i e i (     )     R x       e i






     e i d      ei d   R x       e  i (     ) d
  *(i ) *(i )Sx ( )
therefore
S y     *(i ) S x  
2
2.7. Derive the expression for the average number of crossings of the value x   assuming that the joint
distribution function f ( ,  ) of X (t ) and X ( t ) is
(A) two-dimensional normal
(B) arbitrary
Solution
(a)
The joint distribution function f ( ,  ) of X (t ) and X ( t ) is given by
 2
2 
 2  2 
1
2C 2C
f ( ,  ) 
e  1 2 
2 C1C2
where C12  E  X 2  and C22  E  X 2 
The probability that   X    d and   X    d  is f ( ,  )d d  . The duration of the passage
from  to   d is d /  . Thus the passages form  to   d at a velocity  per unit time is
f ( ,  )d d   f ( ,  )d d 

   f ( ,  )d  
d
d

The passages from  to   d at any rate is equal to the sum of the passages at each value of  , or
N  


1
 f ( ,  )d  
2




C1C2
e
 2
2 
 2  2 
 2 C1 2 C2 
d
2.8. The spectral density of the loading (input) is given in the form shown in Figure 2.29
Figure 2.29Loading input
Find the spectral density’s of the outputs (displacement, bending moment, etc.) of the systems described in
problem 1(a) of Chapter 1.
Solution
From Problem 3(a) of Chapter 1 we know that
1
1


U*  
or  *(i ) 
2 
115, 200  1.3 2
115, 200  1.3 
S y   *(i ) Sx
2
At   100
 *(i100) 
1
 9.785 106  9.8 106
115, 200  1.3(100)2
 *(i100)   9.8 106   9.6 1011 so S y   9.6 1011  S x
2
2
Also, the mean square value of u( x, t ) is
E (u 2 ) 
1
2



S x ( )d  
1
2

100
101
(1)d  
1
2

1001
100
(1)d  
1

2.9. A random function X (t ) has its spectral density as in Problem 2.8. Find the mean square value of X (t ) . Find
the average number of crossings of the values x  0 , x  1 , and x  10 .
Solution
From Problem 8 we have
E (u 2 ) 
1

For x  0
1/ 2
  2

1  0  S x ( )d 
N0 
   S ( )d 
 0 x

For x  1

1 10, 000 100

rad/sec

1

N  N0 e

2
2C12
N1 
100
N1 
100

where C12  E (u 2 ) 
e (1)
2
/2

100

1

e / 2 rad/sec
For x  10

e (10)
2
/2

100

e50 rad/sec
Chapter 3
3.1 A 4 lb cart is attached to a 12 lb/in spring starts from rest and is subjected to an applied force F(t) as illustrated
in Figure 3.15. The equation of motion for this system is mx  kx  F (t ) . Using Euler finite differences solve the
differential equation above and note the position, velocity and acceleration at various time intervals up to 0.2
seconds. Assume a time interval of t  T /10 .
Figure 3.15 Spring-mass system with an applied force.
Solution
The mass of the cart is m 
4 lb
 0.124 lb-s2 /ft  0.01035 lb-s2 /in . The equation of motion is therefore
32.2 ft/s2
0.01035x  12x  F (t ) or x  96.6F (t ) 1159.4 x .The time period is T  2
time interval is t 
m
0.01035
 2
 0.1845 s . The
k
12
T 0.1845

 0.01845  0.02 . Since the cart starts from rest, x1  x1  0 and the acceleration is
10
10
x1  96.6(10)  1159.4(0)  966 .
Since x2  x1  x1t 
1
2
x1  t  and x2  x1t we have, at t  0.02
2
1
2
x2  0  0  (966)  0.02  0.1932 and x2  966(0.02)  19.32
2
x2  96.6( F2 )  1159.4x2  96.6(12)  1159.4(0.1932)  935.2
At t  2t  0.04 : x3  x2  x2 t 
1
1
2
2
x2  t   0.1932  19.32(0.02)  (935.2)  0.02   0.7664
2
2
x3  2 x2 t  x1  2(935.2)(0.02)  0  37.4
x3  96.6( F3 )  1159.4x3  96.6(14)  1159.4(0.7664)  463.8
Following the same procedures as above we construct the table below
t (s)
F (t )
0
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
10
12
14
16
18
20
18
16
15
15
15
Position
0
0.193
0.766
1.608
2.301
2..608
2.404
1.813
0.877
0.137
-0.084
Velocity
0
19.32
37.408
37.859
24.676
0.689
-19.035
-41.253
-41.322
-23.969
10.283
Acceleration
966
935.2
463.5
-318.3
-929.2
-1092.8
-1048.6
-577.2
432.1
1290.1
1546.8
3.2. Repeat problem 1 using t  T  20  0.01 . Compare this solution to that of problem 3.1 by plotting the
displacements, velocities, and accelerations for both cases.
Solution
Since the cart starts from rest, x1  x1  0 and the acceleration is x1  96.6(10)  1159.4(0)  966 .
Since x2  x1  x1t 
1
2
x1  t  and x2  x1t we have, at t  0.01
2
1
2
x2  0  0  (966)  0.01  0.0483 and x2  966(0.01)  9.66
2
x2  96.6( F2 )  1159.4x2  96.6(11)  1159.4(0.0483)  1006.6
At t  2t  0.02 : x3  x2  x2 t 
1
1
2
2
x2  t   0.0483  9.66(0.01)  (1006.6)  0.01  0.1952
2
2
x3  2 x2 t  x1  2(1006.6)(0.01)  0  20.13
x3  96.6( F3 )  1159.4x3  96.6(12)  1159.4(0.19523)  932.8
velocity
displacement
This continues in the same manner and the plots below can be generated using an appropriate computer code.
3
2,5
2
1,5
1
0,5
0
-0,5 0
50
40
30
20
10
0
-10 0
-20
-30
-40
-50
Dt=0.02
Dt=0.01
0,05
0,1
time
0,15
0,2
Dt=0.02
0,05
0,1
time
0,15
0,2
Dt=0.01
2000
1500
acceleration
1000
500
Dt=0.02
0
-500
0
0,05
0,1
0,15
0,2
Dt=0.01
-1000
-1500
time
3.3. A spring mass system is governed by the equation 0.2 x  32 x  5  100t , where 0  t  1 s . Assume the system
starts from rest, the time interval is t  T /10 , and determine the system response using the Euler finite
difference method for 0  t  0.5 s.
Solution
T 0.5
m
0.2
 2
 0.4967  0.5 s .The time interval is t  
 0.05 . The equation
k
32
10 10
of motion is written as x  25  500t  160 x .
The time period is T  2
Since the system starts from rest, x1  x1  0 and the acceleration at t  0 is x1  25 .
Since x2  x1  x1t 
1
2
x1  t  and x2  x1t we have, at t  0.05
2
1
2
x2  0  0  (25)  0.05  0.03125 and x2  25(0.05)  1.25
2
x2  25  500t  160 x2  25  500(0.05)  160(0.03125)  45
At t  2t  0.10 : x3  x2  x2 t 
1
1
2
2
x2  t   0.03125  1.25(0.05)  (45)  0.05  0.15
2
2
x3  2 x2 t  x1  2(45)(0.05)  0  4.5
x3  25  500(0.1)  160 x3  25  500(0.1)  160(0.15)  51
Following the same procedures as above we construct the table below
t (s)
0
0.05
0.1
0.15
Position
0
0.03125
0.15
0.43875
Velocity
0
1.25
4.5
6.35
Acceleration
25
45
51
29.8
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.7935
1.165
1.427
1.552
1.533
1.475
1.418
7.48
6.154
3.839
0.818
-1.001
-1.208
0.679
-1.96
-36.41
-53.36
-48.4
-20.27
16.79
48.1
3.4. Repeat problem 3.3 using t  T  20  0.025 . Compare this solution to that of problem 3.3 by plotting the
displacements, velocities, and accelerations for both cases.
Solution
Since the system starts from rest, x1  x1  0 and the acceleration at t  0 is x1  25 .
Since x2  x1  x1t 
1
2
x1  t  and x2  x1t we have, at t  0.05
2
1
2
x2  0  0  (25)  0.025  0.007813 and x2  25(0.025)  0.625
2
x2  25  500t  160 x2  25  500(0.025)  160(0.007813)  36.25
At t  2t  0.05 : x3  x2  x2 t 
1
1
2
2
x2  t   0.007813  0.625(0.025)  (36.25) 0.025  0.03477
2
2
x3  2 x2 t  x1  2(36.25)(0.025)  0  1.8125
x3  25  500(0.05)  160 x3  25  500(0.05)  160(0.03477)  44.437
This continues in the same manner and the plots below can be generated using an appropriate computer code.
8
7
6
velocity
5
4
3
Dt=0/05
2
Dt=0.25
1
0
-1 0
0,1
0,2
displacement
-2
0,3
0,4
0,5
time
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
Dt=0/05
Dt=0.25
0
0,1
0,2
0,3
0,4
0,5
time
60
acceleration
40
20
Dt=0/05
0
-20
0
0,1
0,2
0,3
0,4
0,5
Dt=0.25
-40
-60
time
3.5. A dynamic system is governed by the equation u  10u  250u  F (t ) , where F (t ) is shown in Figure 3.16.
Assume the system starts from rest, the time interval is t  T /10 , and plot the position and velocity using the
Euler finite difference method for 0  t  0.5 s.
Figure 3.16. Forcing function for a dynamic system.
Solution
From the given equation we can assume m  1 so the time period is T  2
interval is t 
m
1
 2
 0.397  0.4 s . The time
k
250
T 0.4

 0.04 . Within the first 2 seconds, the force is expressed as F (t )  5  75t . Between 0.2
10 10
and 0.3 seconds F (t )  20  100t then remains constant for 0.1 sec until the last segment where the force is
F (t )  10  50t .
Since the system starts from rest, u1  u1  0 and the acceleration at t  0 is u1  5 .
1
2
Since u2  u1  u1t  u1  t  and u2  u1t we have, at t  0.04 , F (t )  8
2
1
2
u2  0  0  (5)  0.04  0.004 and u2  5(0.04)  0.20
2
u2  8  10(0.2)  250(0.004)  5
At t  2t  0.08 : F (t )  11 and
1
1
2
2
u3  u2  u2 t  u2  t   0.004  0.2(0.04)  (5)  0.04   0.016
2
2
u3  2u2 t  u1  2(5)(0.04)  0  0.4
u2  11  10(0.4)  250(0.016)  3
Following the same procedure we end up with the plot below
1,5
1
0,5
0
-0,5 0
0,1
0,2
0,3
0,4
0,5
Position
Velocity
-1
-1,5
-2
-2,5
Time (s)
3.6. Solve problem 3.1 using the Runge-Kutta method with h  0.02 . Plot position, velocity, and acceleration as a
function of time using the Runge-Kutta method and compare it to the solutions obtained in problem 1.
Solution
From problem 1 we have x  96.6F (t )  1159.4 x , y  x , y  x  96.6F (t )  1159.4x , h  t  0.02
F (t )  10  100t . For the Runge-Kutta method we follow the procedure outlined in the table below
t
x (position)
y  x (velocity)
f  y  x (acceleration)
T1  ti
X 1  xi
Y1  yi
F1  f T1 , X1 ,Y1 
h
2
h
X 3  xi  Y2
2
X 4  xi  Y3h
h
2
h
Y3  yi  F2
2
Y4  yi  F3h
F2  f T2 , X 2 ,Y2 
h
2
h
T3  ti 
2
T4  ti  h
T2  ti 
X 2  xi  Y1
Y2  yi  F1
F3  f T3 , X 3 ,Y3 
F4  f T4 , X 4 ,Y4 
For subsequent steps we use
xi 1  xi 
h
Y1  2Y2  2Y3  Y4 
6
and
yi 1  yi 
h
 F1  2F2  2F3  F4 
6
As a result, we can generate a simple computer code to establish the required parameters. The first several steps of
the solution are as presented below
step
1
t
0
0.01
0.01
0.02
F
10
11
11
12
x
0
0
0.0966
0.10626
y
0
9.66
10.626
9.50602
f
966
1062.6
950.602
1036.002
2
0.02
12
0.166927
20.09469
965.6651
0.03
0.03
0.04
0.04
3
13
13
14
14
0.367874
0.46444
0.734678
0.736306
29.75134
28.38756
34.44125
35.29306
829.2873
717.3281
500.6144
498.727
Following this procedure and incorporating the results form problem 1we can generate the curves shown
displacement
3
2,5
2
1,5
Euler
1
Runge-Kutta
0,5
0
0
0,05
0,1
0,15
0,2
time
60
velocity
40
20
Euler
0
-20
0
0,05
0,1
0,15
0,2
Runge-Kutta
-40
-60
time
2000
acceleration
1500
1000
500
Euler
0
-500 0
0,05
0,1
0,15
0,2
Runge-Kutta
-1000
-1500
time
3.7. Solve problem 3.3 using the Runge-Kutta method with h  0.05 . Plot position, velocity, and acceleration as a
function of time using the Runge-Kutta method and compare it to the solutions obtained in problem 3.
Solution
From problem 3 we have x  25  500t  160 x , y  x , h  t  0.05 . For the Runge-Kutta method we follow the
procedure outlined in the table below
t
x (position)
y  x (velocity)
f  y  x (acceleration)
T1  ti
X 1  xi
Y1  yi
F1  f T1 , X1 ,Y1 
h
2
h
X 3  xi  Y2
2
X 4  xi  Y3h
h
2
h
Y3  yi  F2
2
Y4  yi  F3h
F2  f T2 , X 2 ,Y2 
h
2
h
T3  ti 
2
T4  ti  h
T2  ti 
X 2  xi  Y1
Y2  yi  F1
F3  f T3 , X 3 ,Y3 
F4  f T4 , X 4 ,Y4 
For subsequent steps we use
xi 1  xi 
h
Y1  2Y2  2Y3  Y4 
6
h
 F1  2F2  2F3  F4 
6
and
yi 1  yi 
As a result, we can generate a simple
computer code to establish the required
parameters. The first several steps of
the solution are as presented in this
table.
step
1
t
0
0.025
0.025
0.05
0.05
0.075
0.075
0.1
0.1
2
Following this procedure and
incorporating the results form problem
1we can generate the curves shown
displacement
3
x
0
0
0.015625
0.046875
0.048438
0.092708
0.119115
0.196563
0.192545
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
y
0
0.625
0.9375
2.6875
1.770833
2.827083
2.9625
3.942917
4.004306
f
25.00
37.50
35.00
42.50
42.25
47.67
43.44
43.55
44.19
Euler
Runge-Kutta
0
0,1
0,2
0,3
time
0,4
0,5
8
velocity
6
4
Euler
2
Runge-Kutta
0
0
0,1
0,2
-2
0,3
0,4
0,5
time
60,00
acceleration
40,00
20,00
Euler
0,00
0
0,1
0,2
0,3
-20,00
0,4
0,5
Runge-Kutta
-40,00
-60,00
time
3.8. A beam of mass m is supported by two identical square steel columns of length L. The beam is acted on by a
6 EI
time varying force P(t ) . The equation of motion for this system is mu  3 u  P(t ) . The dimension of each
L
column is a  a , and the overall dimensions of the system are such that the equation of motion can be written as
1800u  3.7 109 a4u  50t . The application of this system is such that the horizontal displacement, u, must not
exceed 25 mm. We have available column sizes of a  25 mm , 38 mm , 50 mm and want to approximate the
size of the column. Use the Newmark Beta method to determine the displacement of this system and approximate
the required column size so that the allowable horizontal displacement is not exceeded. Assume h  0.2 ,
  1/ 2 ,   1/ 6 , and that the system starts from rest with the applied load duration being 6 seconds.
Solution
Using the equation of motion, 1800u  3.7 109 a4u  50t , and following Table 3.6 we have
9 4
uo (0)  uo (0)  uo (0)  0 . With   1/ 2 ,   1/ 6 , m  1800 ,   0 , k  3.7 10 a , h  0.2, and p(t )  50t we
determine
1

kˆ 
m
  k  150m  15  k  150(1800)  15(0)  3.7  109 a 4  270,000  3.7 109 a 4
2
h
h
a
1

m
  150m  15  270,000,
2
h
h
b


1
m    1  30m  2  54,000,
h


 1

 

c
 1 m  h 
 1  2m  0.1  3600
 2

 2

pˆ i 1  pi 1  aui  bui  cui  pi 1  270,000ui  54,000ui  3600ui
ui 1 
pˆ i 1 pi 1  270,000ui  54,000ui  3600ui

270,000  3.7 109 a 4
kˆ
ui 1 
 1

1
1
u  ui  
ui  
 1 ui  150  ui 1  ui   30ui  2ui
2  i 1

h
2

h


ui 1  ui  1    hui   hui 1  ui  0.1 ui  ui1 
Horizontal Displacement
(m)
Developing a simple computer code we can generate a plot of the displacements as shown below
0,300
0,250
0,200
0,150
0,100
0,050
0,000
a = 25 mm
a = 38 mm
a = 50 mm
0
1
2
3
4
5
6
time
Based on this we can see that a  50 mm is an appropriate solution.
3.9. Assume a single degree of freedom system starts from rest and is defined by 6u  3.6u  54u  360t . Use the
Newmark Beta method to determine the response (displacement, velocity, and acceleration) of this system.
Assume h  t  0.1T ,   1/ 2 and   1/ 6 . Plot the responses above 0  t  5 sec.
Solution
The natural angular frequency is   k / m  3 and the period is T  2 /   2 / 3  2.094 . As a result,
h  t  0.1T  0.2 . Following Table 3.6 we have uo (0)  uo (0)  uo (0)  0 . Next, using   1/ 2 ,
  1/ 6 , m  6 ,   3.6, k  54 , h  0.2, and p(t )  360t we determine
1

kˆ 
m
  k  150m  15  k  150(6)  15(3.6)  54  1008
2

h
h
a
1
m
 h2

  150m  15  954,
h
b
1
h


m    1  30m  2  187.2,



 1

 

c
 1 m  h 
 1  2m  0.1  12.36
 2

 2

pˆ i 1  pi 1  aui  bui  cui  pi 1  954ui  187.2ui  12.36ui
ui 1 
ui 1 
pˆ i 1 pi 1  954ui  187.2ui  12.36ui

1008
kˆ
1
 h2
 ui 1  ui  
1
h
 1

ui  
 1 ui  150  ui 1  ui   30ui  2ui
2



ui 1  ui  1    hui   hui 1  ui  0.1 ui  ui1 
For i  0 , pˆ i 1  pi 1  p1  360(0.2)  72 , u1 
72
 0.0714 , u1  150  0.0714  0  2(0)  30(0)  10.714 , and
1008
u1  0  0.1 0  1.0714  0.10714 . Subsequently, the equations above can be programed into an appropriate
computer code and the plot below can be generated.
Position, Velocity,
Accekleration
35,000
25,000
15,000
Position
Acceleration
5,000
Velocity
-5,000 0
-15,000
1
2
3
4
5
time
3.10. Work Problem 3.3 using the Newmark Beta method with   1/ 2 ,   1/ 6 , and h  t  0.1T . Compare the
results by plotting the displacement for both cases.
Solution
From problem 3.3 we had 0.2 x  32 x  5  100t , T  0.5 , so h  0.05 . Following Table 3.6 we have
uo (0)  uo (0)  0 and uo (0)  5 / 0.2  25 . Next, using   1/ 2 ,
  1/ 6 , m  0.2 ,   0, k  32 , h  0.05, and p(t)  100t  5 we determine
1

kˆ 
m
  k  2400m  60  k  2400(0.2)  15(0)  32  512
2

h
h
a
1
 h2
m



1
  2400m  60  480, b 
m    1  120m  2  24,
h
h



 1

 

c
 1 m  h 
 1  2m  0.1  0.4
 2

 2

pˆ i 1  pi 1  aui  bui  cui  pi 1  480ui  24ui  0.4ui
ui 1 
ui 1 
pˆ i 1 pi 1  480ui  24ui  0.4ui

512
kˆ
1
h
2
 ui 1  ui  
 1

1
ui  
 1 ui  2400  ui 1  ui   120ui  2ui
h
 2

ui 1  ui  1    hui   hui 1  ui  0.025 ui  ui1 
Following the procedures of problem 8 and comparing those displacements with the ones from problem 3 we
generate the plot below.
displacement
2
1,5
1
Euler
Newmark-Beta
0,5
0
0
0,1
0,2
0,3
0,4
0,5
time
Chapter 4
4.1. Write the equations of motion for the systems shown in Figure 4.21 (a)...(e). Write the elements of the mass
and stiffness (or flexibility) matrices (no damping). Write the frequency equation of each system and solve the
equation.
Figure 4.21 Multiple degree of freedom systems.
Solution
(a) The deformation will resemble that
shown. Assume no axial deformation of
the columns. The horizontal members
will experience the forces shown
(related to the spring rates k1 and k2 ,
defined as
k1  k2  k 
12 EI
L3
The equations of motion are
m2 x2  2k2  x2  x1   P(t )
m1 x1  2k1  x1  a(t )   2k2  x2  x1   0
Since k1  k2  k 
12 EI
,
L3
m1 x1  4kx1  2kx2  2ka(t )
m2 x2  2kx1  2kx2  P(t )
In matrix notation
2k   x1  2ka (t ) 
 

2k   x2   P (t ) 
 m1 0   x1   4k
 0 m   x    2k

2 2

The mass and stiffness matrices are
0
m
M  1

 0 m2 
 4k
and K  
 2k
2k 
2k 
Assume x1  A sin t , x2  B sin t , a(t )  P(t )  0 so that we have
x1   A 2 sin t and x2   B 2 sin t
m1 A 2 sin t  4kA sin t  2kB sin t  0
m2 B 2 sin t  2 Ak sin t  2 Bk sin t  0
Thus;
2k
 2k  m  
 4k  m12 
2k
2
2
 0  4k 2   4k  m1 2  2k  m2 2 
Therefore the frequency equation is given by
m1m24  2k2  m1  2m2   4k 2  0
Solving for  2 gives
2k  m1  2m2   4k 2  m1  2m2   16m1m2 k 2
2
 
2


2 
(b)
2m1m2
2k  m1  2m2   4m12 k 2  16m1m2 k 2  16m22 k 2  16m1m2 k 2
2m1m2
2k  m1  2m2   4k 2  m12  4m12 
k
m1m2
2m1m2
m  2m  
1
2
m12  4m22

Because of the loading, the system will undergo a vertical
displacement as well as a rotation. The two equations governing
these are
Y  m
d 2v
Pa
d 2
and

M


I
z
dt 2
2
dt 2
Where
Y  k11V  k12 and M  k21V  k22
 k11 k12  V   Py 
k
    
 21 k22      M 
Apply a unit displacement in the y direction of V  1
k11V  Py and k21V  M  0  k12
 3EI 
k11V  Py  2  3 V
 b 
 k11 
6EI
b3
Apply a unit rotation of   1 so that k12  Py  0 and k22  M . Assuming small angles
sin    

a/2

2
a Fb3
1    
a
2 3EI
 a  3EI
k22  M  F    3 a 2
 2  4b
So Y 
 k22 
 F
3EI
a
2b 3
3EI 2
a
4b3
6EI
3EI
v and M  3 a 2 and the equations of motion are
3
b
4b
mv 
6 EI
v0
b3
m 0 
M 

 0 Iz 
and
I z 
3EIa 2
Pa

3
4b
2
 6 EI
 b3
and K  
 0




3EIa 2 
4b3 
0
Assume v  Asin t and   B sin t , so that v   A2 sin t and   B2 sin t results in
12 
6 EI
mb3

1 
6 EI
mb3
22 
3EIa 2
4 Ib3

2 
3Ea 2
4b 3
(c)
This is a 5 degree of freedom system with 3
displacements (u, v, and w) and 2 rotations (  and  ) as
illustrated in the figure. We have
d 2u
dt 2
d 2v
-Y  m 2
dt
-X  m
- M   P (t )
- M  I w
- Z  P (t )  m
 k11
k
 21
 k31

 k41
 k51
k12
k22
k32
k42
k52
L
d 2
 Iv 2
2
dt
d 2
dt 2
d 2w
dt 2
k13 k14
k23 k24
k33 k34
k43 k44
k53 k54
k15   u   X 
k25   v   Y 
    
k35   w    Z 

k45      M  
   
k55     M  
Only the major diagonal terms are non-zero
Spring pairs 2 and 3 are in parallel and spring pair 1 is in series so the equivalent springs are
k1 eq  K1  k1 / 2 , k2  eq  K 2  2k2 , and k3 eq  K 3  2k3
For rotation about the v and w axes we get k44  k3  L / 2 ( L) so that k44  k3 L2 / 2 and
k55  k2  L / 2 ( L) so that k55  k2 L2 / 2
 k1 / 2 0
 0
2k 2

0
K    0

0
 0
 0
0
0
0
0 
0
0
0 

2k 3
0
0 

0 k3 L2 / 2
0 
0
0
k2 L2 / 2
mu  (k1 / 2)u  0
mv  2k2 v  0
mw  2k3 w  P(t )
I v  k3 ( L2 / 2)  P(t )(L/ 2)
I w  k2 ( L2 / 2)  0
m 0 0 0
0 m 0 0

m   0 0 m 0

 0 0 0 Iv
 0 0 0 0
0
0

0

0
I w 
Assume u  Asin t , v  B sin t , w  C sin t ,   D sin t ,   E sin t
mA12 sin t  (k1 / 2) A sin t  0
1  k1 / 2m

mB sin t  2k2 Bsin t  0

2  2k2 / m
mC sin t  2k3C sin t  0

3  2k3 / m
2
2
2
3
 I v D42 sin t  (k3 L2 / 2) D sin t  0

4  k3 L2 / 2 I v
 I w E52 sin t  ( k2 L2 / 2) E sin t  0

4  k2 L2 / 2 I w
(d)
- X  P1 (t )  m
 k11 k12
k
k
 21 22
 k31 k32
d 2u
dt 2
, - Y  P2 (t )  m
, - Z  P3 (t )  m
d 2w
dt 2
k13   u   X 
   
k23   v    Y 

  
k33  
w  Z 
For each rod we know  
k11 
d 2v
dt 2
EA1
L1
k22 
,
PL
P AE
 k 
AE

L
EA2
L2
,
k33 
EA3
L3
k12  k21  k13  k31  k23  k 32  0
 EA1

 L1

K    0


 0

mu 
0
EA2
L2
0

0 


0 

EA3 

L3 
m 0 0 
m   0 m 0 
 0 0 m 
EA
EA1
EA
 P1 (t ) , mv  2  P2 (t ) , mw  3  P3 (t )
L1
L2
L3
Setting P1 (t )  P2 (t )  P3 (t )  0 and assuming u  Asin t , v  B sin t , w  C sin t
EA1
A sin t  0
L1
mA12 sin t 
12 
EA1
mL1
1 
EA1
mL1
mB12 sin t 
EA2
EA2
B sin t  0  2 
L2
mL2
mC32 sin t 
EA3
EA3
Csin t  0  3 
L3
mL3
(e)
d 2u
d 2v
, - Y  P2 (t )  m 2
2
dt
dt
d 2
- M  P1 (t ) a  I 2
dt
- X  P1 (t )  m
 k11 k12
k
k
 21 22
 k31 k32
k13  u   X 
   
k23   v    Y 

  
k33  
   M 
Let u  1 , v    0 and get
k11u  X , k21u  Y , k31u  M
Each structural member will carry an axial load
and will have a stiffness. Looking at one member,
we note its elongation to be

PL1
AE
 P
AE
AE

u cos 
L1
L1
Resolving this into components: Px 
Px' 
AE
AE
u cos2  , Py 
u cos  sin  . For the other member we find
L1
L1
AE
AE
u cos2  , Py' 
u cos  sin  .
L2
L2
 cos2  cos2  
k11u  2 Px  2 Px'  2 AEu 


L2 
 L1
 cos2  cos2  
k11  2 AE 


L2 
 L1
k21v  Y  Py  Py'  Py  Py'  0
k21  0
 2a cos2  2a cos2  b cos  sin  b cos  sin  
k31u  M  2 Px a  2 Px' a  Py' b  Py b  AEu 




L1
L2
L2
L1


  2a cos2  2a cos2  
 cos  sin  cos  sin 
k31  AE  2a 


  b
L1
L2
L2
L1


 
 

 
Next, let v  1 , then k12 v  X , k22v  Y , k32v  M . As before, k21  k12  0 . As
before, we note
PL1
AE
AE
AE

v sin 
L1
L1
AE
AE
Resolving this into components: Py * 
v sin 2  , Px * 
v cos  sin 
L1
L1
AE
AE
For the other member we find Py' * 
v sin 2  , Px' * 
v cos  sin 
L2
L2

 P
 sin 2  sin 2  
k22 v  Y  2 Py * 2 Py' *  2 AEv 


L2 
 L1
 sin 2  sin 2  
k22  2 AE 


L2 
 L1
k32 v  M  0  k32  0
Next, let   1 , then k13  X , k23  Y , k33  M . From the rotation of the mass we see
that for small rotations
sin     v / b  v  b
  u / a  u  a
The forces due to horizontal ( u  a ) and vertical ( v  b )
displacements are as shown.
k13  k31 , k23  Y  0  k23  k32  0
k33  a  2  Px  Px'   b  Py'  Py   b  Py  Py' 
 b  2  Py *  Py' *  a  2  Px' *  Px *
 2a  Px  Px'   2a  Px' *  Px *  2bPy'  2bPy
 2b  Py *  Py' *
 AE

AE
k33  M  2a 
u cos2  
u cos2  
L1
 L2

 AE

AE
 2a 
v cos  sin  
v cos  sin  
L1
 L2

AE
AE
 2b
u cos  sin   2b
u cos  sin 
L2
L1
 AE

AE
 2b 
v sin 2  
v sin 2  
L2
 L1

With u  a and v  b this becomes
 AE

 AE

AE
AE
AE
k33  2a 
a cos2  
a cos2    2a 
b cos  sin  
b cos  sin    2b
a cos  sin 
L
L
L
L
L2
1
1
 2

 2

 2b
 AE

AE
AE
a cos  sin   2b 
b sin 2  
b sin 2  
L1
L
L
2
 1


 AE

 AE

AE
AE
AE
k33   2a 2 
cos2  
cos 2    2ab 
cos  sin  
cos  sin    2ab
cos  sin 
L1
L1
L2

 L2

 L2

 2ab
 AE 2
AE
AE 2  
cos  sin   2b2 
sin  
sin   
L1
L
L2
 1
 
 AE

 AE

 AE 2
AE
AE
AE 2 
k33  2a 2 
cos2  
cos2    4ab 
cos  sin  
cos  sin    2b2 
sin  
sin  
L
L
L
L
L
L2
1
1
 2

 2

 1







 cos 2  cos 2  

2 AE 

0

L2 

 L1



 sin 2  sin 2  
K

0
2 AE 

  

L1
L2 




2
2

  2a cos  2a cos  
AE
2
a





L1
L2
 


0

 cos  sin  cos  sin   
  b



L2
L1

 


2
2
  2a cos  2a cos  
AE 2a 


L1
L2

 
 cos  sin  cos  sin 
 b

L2
L1

 

 
0
 cos 2  cos 2  

AE 2a 2 


L1 

 L2
 cos  sin  cos  sin 
 4ab 

L2
L1

 sin 2 
sin 2   
 2b 2 


L2  
 L1
The equations of motion are
mu  k11u  k13  P1 (t) , mv  k22 v  P2 (t ) , I  k13u  k33   P1 (t )a
























Assume P1 (t )  P2 (t )  0 and u  A sin t , v  B sin t ,   C sin t
(1)
 mA 2 sin t  k11 A sin t  k13C sin t  0
(2)
 mB 2 sin t  k22 B sin t  0
(3)
 IC 2 sin t  k13 A sin t  k33C sin t  0
From (1) and (3)
k
 m 2  A  k13C  0 and  k33  I  2  C  k13  0
11
For a non-trivial solution
k
11
 m 2 
k13
k
k13
33
 I
2

 0   k11  m 2  k33  I  2   k132
mI4   k33m  k11I  2  k11k33  k132  0
 
2

 k33m  k11 I    k33m  k11I 
2
 4mI (k11k 33  k132 )
2mI
 k33m  k11 I  
k m  2k11k 33mI  k112 I 2  4mIk132  k 33m  k 11I 
2
33
2
2
2mI
 m 0 0
m   0 m 0
 0 0 I 
4.2. Find two normalized natural modes for the system in problem 4.1(a). Assume m  1 . Write the elements i( r )
of the modal matrix  .
Solution
From Problem 4.1 we have
A1 2k  m2 2

A2
2k
For 1 we have
 2k  m2 2  (1)
A1(1)  
 A2 or
2k


A1(1)  XA2(1)
For 1 we have A1(2)  YA2(2) . In addition we have i( r )   Ai( r )
For 1 ; 1(1)   A1(1) , 2(1)   A2(1)
For 2 ; 1(2)   A1(2) , 2(2)   A2(2)
1(1)   XA2(1)  X 2(1) ; 2(1)  1(1) / X
1(2)   A2(2)  YA2(2)  Y 2(2) ; 2(2)  1(2) / Y
For normalized modes, we have
 m 
 j(s)   rs m
(r)
ij i
i
j
m 
  mi 2i( r )2(s)    rs m
( r ) (s)
i1 i
1
i
m111( r )1(s)  m121( r )2(s)  m212( r )1(s)  m222(r )2(s)   rs m
Since m12  m21  0 , m11  m1 , m22  m2 , and m  1
m11(1)1(1)  m22(1)2(1)  1
m11(2)1(2)  m22(2)2(2)  1
Substituting and solving for 1(1) , 2(1) , 1(2) , and 2(2) gives
m1 

(1) 2
1
 m2
(1) 2
1
X2
1
 X 2 m1  m2 

 1
X2


 
(1) 2
1
X
1(1) 
2(1) 
X m1  m2
2
m1 1(2)   m2
2
 
 
(2) 2
1
 
(2) 2
1
Y2
1
X m1  m2
2
1
 Y 2 m1  m2 

 1
Y2


Y
1(2) 
Y m1  m2
2
The modal matrix is
1(1) 1(2) 
(1)
(2) 
2 2 
   
2(2) 
1
Y m1  m2
2
4.3. Write the uncoupled equations of motion for forced vibrations of system of problem 4.1 (a)
Solution
Assume q1  g1 (t )1(1)  g2 (t )1(2) and q2  g1 (t )2(1)  g 2 (t )2(2) , then
g1  12 g1 
f1
f
and g2  22 g2  2 , where m  1 and
m
m
f1  p11(1)  p22(1) and f 2  p11(2)  p22(2)
Where p1  p1 (t )  P(t ) and p2  p2 (t )  2ka(t )
The solution is given by
gr  Ar cos r t  Br sin r t  gr (t )
Where
t
t
0
0
gr (t )   Gr (t  t ') f r (t ')dt '  
1
sin r (t  t ') f r (t ')dt '
mr
Therefore
qi   Ari( r ) cos r t   Bri( r ) sin r t   gr (t )i( r )
r
r
r
4.4. Assume that the material of the structure in problem 4.1(a) is an arbitrary viscoelastic material (all of the
elements of each structure are made from the same material). Write the uncoupled equations of motion for forced
vibrations using:
(A) Laplace transformations
(B) Fourier transformations
Solution
(a) L
Kij   Kij ( s)
, L
 pi (t )  pi (s)
Assume q1  g1 (t )1(1)  g2 (t )1(2) and q2  g1 (t )2(1)  g 2 (t )2(2)
Then q1  g11(1)  g 21(2) and q2  g12(1)  g 22(2)
Then and g 2  22 g 2  f 2
And
g1 
s
g2 
s
2
1
 f1 ( s )
1
 f 2 ( s)
 12 ( s ) 
 22 ( s ) 
2
Where
f1 ( s )  P1 ( s )1(1)  P2 ( s )2(1)
f 2 ( s )  P1 ( s )1(2)  P2 ( s )2(2)
P(t )
And from the system P1( s)  L
; P2 ( s)  2kL
a(t)
From Problem 4.2 we have 1(1) , 1(2) , 2(1) , 2(2) , 12 , and 22 . Substitute these values to obtain f1 ( s ) and f 2 ( s ) ,
then knowing f ( s ) , g (s) can be calculated and then
qi ( s )  g( s )i( r )  gi( r ) ( s )
qi (s)
1
q(t )  L
(b)
m g  k g
ij
i
ij
j
i
 Pi (t )
j
 mij 2 g *j   kij g *j  Pi*
j
j
For a viscoelastic material replace kij* with kij
 mij 2 g *j   kij g *j  Pi *
j
j
*
* (r)
Assume q j   gr j
r
 mij 2 g r* j( r )   kij g r* j( r )  Pi *
j
r
j
 mij g  
2
i
j
* (s)
r i
r
(r )
j
  kij g r*i( s ) j( r )   Pi *i( s )
r
i
j
r
i
 m rs g   m rs g   P 
2
*
r
r
r
*
* (s)
Let f s   Pi i
i

*2
s
*
r
  2  g r*  f s*
*
r
* (s )
i i
i
k
 kij"   kij*
'
ij
Pi (t )  Pi*
Assume q1*  g1*1(1)  g 2*2(1) and q2*  g1*1(2)  g 2*2(2) . Thus
g1* 
f1*
f 2*
*
and
g

2
1*2   2
2*2   2
Where f1*  P1*1(1)  P2*2(1) and f 2*  P1*1(2)  P2*2(2) . For the system
P1*  P1 (t ) and P2*  2kij* a (t )
Knowing 1(1) , 1(2) , 2(1) , and 2(2) we find f1*and f 2* to obtain g1* and g 2*
4.5. For the system shown in Figure 4.21(a), assume a structural rigidity of EI  117 103 N-m2 and that the
masses are related through m1  nm2 , where 0.2  n  5 . If m2  225 kg and the distance between masses is
L  3 m , plot the frequencies as a function of n.
Solution
From the solution of problem 4.1(a) we had
2 
2k  m1  2m2   4k 2  m12  4m12 
2m1m2
or
12 
With k 
k
m1m2
12 117 103 
(3)3
12 
22 
k
nm22
k
nm22
 m  2m  
1
2
m12  4m22

22 
k
m1m2
 m  2m  
1
2
m12  4m22
 52,000 , and m1  nm2
n  2 m  m
2
2
n  2 m  m
2
2

n2  4 

n2  4 
k
nm2
k
nm2
n  2 
n  2 

n2  4 

n2  4 

231
n  2  n2  4
n

231
n  2  n2  4
n



omega
Plotting these two equations results in
6000
5000
4000
3000
2000
1000
0
Omega 1
Omega 2
0
1
2
3
4
5
n
4.6. For the system shown in Figure 4.21 (b) , assume the beam is made from aluminum ( E  10  106 psi ) and has
dimensions such that I  2 in 4 . The weight of the rigid body is 300 lb. The beam segments are related by b  na
, where 1  n  5 and a  24" . Plot the frequencies as a function of n.
Solution
From the solution of problem 4.1 (b) 1 
6 EI
mb3
and 2 
6 Ea 2
4b 3
With E  10 106 , I  2 , EI  20 106 , b  na , a  24 , and m  300 / 32.2(12)  0.776 these become
6  20 106 
6 EI
1 

=105.7 1/ n3
3
mb3
0.776  24n 
2 
6 Ea 2
6(10 106 )(24) 2

=790.6 1/ n3
4b3
4(24n)3
1000
Omega
800
600
400
Omega 1
200
Omega 2
0
0
1
2
3
4
5
n
4.7. For the system shown in Figure 4.21 (c), assume the rigid bar is a cylinder made from an aluminum alloy (
E  70 GPa ) and has a weight of 30 N/m. The length of the bar is between 1 and 4 meters. The horizontal spring
is relatively soft k1  880 N/m while springs 2 and 3 are stiffer with k2  k3  3520 N/m . Plot the frequencies as
a function of L.
Solution
From the solution of problem 4.1(c) we have 5 equations:
mu  (k1 / 2)u  0
,
mv  2k2 v  0 ,
mw  2k3w  P(t )
I v  k3 ( L / 2)  P(t )(L/ 2) , I w  k2 ( L / 2)  0
2
2
The mass matrix is
m 0 0 0
0 m 0 0

m   0 0 m 0

 0 0 0 Iv
 0 0 0 0
0
0

0

0
I w 
Assuming u  Asin t , v  B sin t , w  C sin t ,   D sin t ,   E sin t we get
mA12 sin t  (k1 / 2) A sin t  0
1  k1 / 2m

mB22 sin t  2k2 Bsin t  0

2  2k2 / m
mC32 sin t  2k3C sin t  0

3  2k3 / m
 I v D42 sin t  (k3 L2 / 2) D sin t  0

4  k3 L2 / 2 I v
 I w E52 sin t  ( k2 L2 / 2) E sin t  0

5  k2 L2 / 2 I w
Using the information from the problem statement
m
2
30L
mL2  3.06L  L
 3.06L , I v  I w 

 0.255L3 . Therefore
9.81
12
12
1  k1 / 2m  880 / 2(3.06L)  11.99 1/ L  12 1/ L
2  3  2(3520) / 3.06L  47.97 1/ L  48 1/ L
Since I v  I w 4  5  3520 L2 / 2(0.255L3 )  83.1 1 / L  83 1 / L
Plotting the results gives
Omega
90
80
70
60
50
40
30
20
10
0
Omega 1
Omega 2,3
Omega 4,5
0
1
2
3
4
L (m)
4.8. For the system shown in Figure 4.21 (d) , assume m  180 kg and the three identical supporting rods are
aluminum ( E  70 GPa ) . Each rod has a length which is related to its diameter by L  nd , where 5  n  30 .
Three possible rod diameters are being considered; d  12.5 , 25 , 37.5 mm . Plot the frequencies as a function
of n.
Solution
From problem 1(d) we had 1 
EA3
EA1
EA2
, 2 
, and 3 
.
mL2
mL3
mL1
Since all rods are identical, we
need only consider one frequency. With E  70 109 , L  nd , and m  180 , the frequency become

E d 2 / 4

m  nd 
 70 10   d
9
2
4(180)  nd 
 17, 477
d
n
Omega
Plotting this equation using the data given yields the figure shown
1600
1400
1200
1000
800
600
400
200
0
d=12.5 mm
d=25 mm
d=37.5 mm
0
10
20
n
30
4.9. For the system shown in Figure 4.21 (e), assume the box weighs 200 lb and the three supporting rods are all 1”
in diameter aluminum with AE  7.9 106 lb . For the crate we approximate I  0.863 in4 . The box is a
rectangle with dimensions a  1 ft, b  2 ft The lengths of the rods are L1  5 ft and L2  10 ft . The box is
suspended between 2 and 4 feet above the ground. Plot the frequencies as a function height above the ground.
Solution
From the solution of problem 1(e) we had
2 
 k33m  k11I  
2
k33
m 2  2k11k 33mI  k112 I 2  4mIk132  k 33m  k11I 
2
2mI
h
 L (1  cos  ) 
Where   sin 1   and   sin 1  1
 and m  200 / 32.2(12)  0.5175
L2
L


 1
 cos2  cos2  
 sin 2  sin 2  
k11  2 AE 


 , k22  2 AE 

L2 
L2 
 L1
 L1
  cos2  cos2    cos  sin  cos  sin 

k31  AE 2a 


  b
L2  
L2
L1

  L1





Omega
 AE

 AE

 AE 2
AE
AE
AE 2 
k33  2a 2 
cos2  
cos2    4ab 
cos  sin  
cos  sin    2b2 
sin  
sin  
L
L
L
L
L
L2
1
1
 2

 2

 1

18000
16000
14000
12000
10000
8000
6000
4000
2000
0
Omega 1
Omega 2
2
2,5
3
3,5
h
4
4,5
5
Chapter 5
5.1 A plate with the dimension shown in Figure 5.21 is placed between two rigid walls and subjected to an axial
compressive force F. Determine the displacement equations (u, v, and w) of the plate in the x, y, and zdirections.
Figure 5.21 Plate fixed between two rigid walls.
Solution
F
and the rigid walls produce a compressive stress  y    o .
ht
  xz   yz  0 . From the axial displacement of the plate and the
The applied force is modeled an axial stress  x  
The remaining stress components are  z   xy
fixed walls we note that  x  

L
and  y  0 . In addition, the shear stresses are all zero (  xy   xz   yz  0 ).
From Hook’s law
x  

L
y  0 

1
1 F

 x  y     ( o ) 
E
E  ht




1
 y  x
E


  y   x   o  
  1  2
z  
F
ht
FL
 Eht

 F
 
  F

 x  y      o    o  

E
E  ht
 E  ht
 1  L
Integrating the strains gives
u

L

x  1  2
F
x
 Eht
, v0 , w
 
1  L
z
 F
z
1  Eht
5.2 An elastic material ( E  70 GPa,   0.33 ) fills a cavity in a rigid block. The dimensions of the cavity are
a  75 mm, b  125 mm, L  300 mm . A rigid cap is placed on the material and a force Po is applied to the
center of the cap as illustrated in Figure 5.22. The elastic material is compressed an amount  . Determine
general expressions for the applied force Po as well as the net forces that exist on the x and z faces of the material
 Px , Pz  . In addition, determine the explicit force in each direction if the axial strain is 0.01%.
Figure 5.22 Elastic material compressed by an applied force.
Solution
From the axial displacement of the cap and the fixed walls we note that  x  0 ,  y  

L
, and  z  0 . In addition,
Po
and
ab
compressive normal stresses  x and  z will exist on other two faces of the material. We can express the strainthe shear stresses are all zero (  xy   xz   yz  0 ). The normal stress in the y-direction will be  y  
stress relationship in matrix form as
 x 
1
  1
 y    
  E  
 z


1

From this we find  x   z 


L

L
 0 
   x 

 11
 
 


or
  y 
     

L
E



 
1  
 z 
0





1 
y  

1

 
   x 
 P 
    o 
 ab

1  



z


 Po
. From the second row of the matrix we have
1  ab

1  Po
 Po  Po  2 2 
1

  2

E  ab
1   ab  Eab  1   

Po
1

  x    z 
E
ab


Po  1    2 2  Po  1  2 1    




Eab  1  
1 

 Eab 

L

Po 
E 1   ab 
1  2 1    L
Px   x a  L    and Pz   z b  L   
Px  
  E 1     

 a L  
1    1  2 1    L 
Pz  
  E 1     

bL  
1    1  2 1    L 
From the information provided, E  70  109 ,   0.33 , a  75 mm , b  125 mm , L  300 mm ,
 / L  0.0001    0.03 mm  0.0003 m
Po 
70  109 1  0.33 (0.075)(0.125)
 0.0001  97.2 kN
1  0.661  0.33
Px  
Pz  
9

0.33  70  10 1  0.33
 0.0001 0.075 0.3  0.0003   114.8 kN

1  0.33  1  0.661  0.33

9

0.33  70  10 1  0.33
 0.0001 0.125 0.30  0.0003   191.3 kN

1  0.33  1  0.661  0.33

Po  97.2 kN
Px  114.8 kN
Pz  191.3 kN
5.3 Consider a cantilevered beam of length L subjected to a uniform load q applied over its entire length. Assume
 ( x) is approximated by the static deflection curve for the beam. In addition, assume the beam dimensions are
such that I  bh3 /12 and A  bh . Use Rayleigh’s method to determine the fundamental frequency for this
beam. Assume the beam is aluminum and the height can be 25, 50, or 75 mm and the length is between 1 and 3
meters. Plot the frequency as a function of beam length for each height.
Solution
The static deflection curve is w( x) 


q
x 4  4 Lx3  6 L2 x 2 .
24 EI

We therefore approximate ( x)  a1 x 4  4Lx3  6L2 x 2

L
L
 144 5 
2U ()   EI ('' )2 dx   EIa12 (12 x 2  24Lx  12 L2 ) 2dx  EIa12 
L 
0
0
 5

L
L
0
0
2K ' ()   A ()2 dx   A a12 (x 4  4Lx3  6L2 x 2 ) 2 dx  A a12 (2.31L9 )
12
R
U ( )
K ( )
'


EIa12 144 / 5L5
A a12 (2.31L9 )
  12.467 EI
A L4
EI 3.53 bh3 I
1.019h E
 2

A
12
bh


L
L2
3.53
2
L
Frequency


1.019h E
450
400
350
300
250
200
150
100
50
0
L2

h=25 mm
h=50 mm
h=75 mm
0
0,5
1
1,5
2
2,5
3
Beam Length (m)
5.4 The static deflection curve of a uniform beam with a distributed weight q fixed to rigid walls at both ends is


w( x )   q / 24 EI  x 4  2 Lx 3  L2 x 2 . Assume  ( x) is approximated by this deflection curve and use
Rayleigh’s method to determine the fundamental frequency for this beam.
Solution

 '( x )  a  4 x

 2L x 
We approximate ( x)  a1 x 4  2 Lx 3  L2 x 2
1
3
 6Lx 2

 "( x )  a1 12 x 2  12 Lx  2 L2
2



2
L
L
4 
2U ()   EI ('' )2 dx   EIa12 12 x 2  12 Lx  2 L2 dx  EIa12  L5 
0
0
5 
L
L
0
0

2 K ' ()   A ()2 dx   A a12 x 4  2 Lx3  L2 x 2
12
R
U ( )
K ( )
'


EIa12 0.8 L5

A a12 (0.00158L9 )


2
dx  A a12 (0.00158L9 )
506.3EI
A L4
1 
22.5
2
L
EI
A
5.5 The tapered circular shaft shown in Figure 5.23 is subjected to a longitudinal vibration, for which the EI term in
the strain energy is replaced by EA and  " is replaced by du / dx . Using Rayleigh’s method, approximate the
x
lowest natural frequency of this shaft. Assume the displacement is approximated by u( x)  A sin
.
2L
Figure 5.23 Tapered circular shaft.
Solution
The radius of the shaft is a function of x, resulting in the area being a function of x. Therefore
x 

r( x )  r 1  
 2L 
L
x 

 A( x )   r 2  1  
 2L 
2U ( )   EA( x)(du / dx) dx  
2
0
L
0


2
x  
 x
E r 1    cos  dx
2L 
 2L   2L
2
2
E 3r 2
4 L2
E 3r 2
4 L2
2
L
0
x
 x x2
x
2x
 cos2
 2 cos2
 cos
dx
2L L
2L 4L
2L 

x L
 x 1  x 2 Lx  x L2
x
sin
  
sin
 2 cos
 

L L  4 2
L 2
L 
 2 2
L
1  x 3  x 2 L L3   xL2
 x 
 2  
 3  sin  2 cos  
L 
L  
4 L  6  2  
0
 L 1  L2 L2  1  L3 2 L3  
    2   2   2 
 2 L  4   4 L  6   

E 3r 2
4 L2

E 3r 2  1  1 1  1  1 2  
Er 2





2.6533




4 L  2  4  2  4  6  2  
L
L
L
x 
x

2 K ' ()   A (u)2 dx    r 2 1   sin 2
dx
0
0
2L
 2L 
2
L
 x x 2  x x2
x
  r 2   sin 2
 sin
 2 sin2
dx
0
2L L
2L 4L
2L 

 x L  x 1  x 2 2 Lx  x 4 L2
x
  r 2   sin
  
sin
 2 cos 
L L  4 4
L 8
L 
 2 
1  x 3  x 2 L L3   xL2
 x 
 2  
 3  sin  2 cos

L  
4 L  6  2   L 
L
0
 L 1  L2 L2  1  L3 2 L3  
  r 2     2   2   2  
 2 L  4   4 L  6   
 1  1 1  1  1 2 
  L r 2     2     2    0.7567  Lr 2
 2  4   4  6  
12  R 
U ( ) 2.6533Er 2 / L
E

 3.506 2
'
2
K ( ) 0.7567 Lr
L
1 
1.87 E
L 
5.6 Assume that ( x)  a1x2  a2 x3 for a uniform cantilever beam. Determine the first two natural frequencies
using the Ritz method.
Solution
d
d 2
 2a1 x  3a2 x 2 and
 2a1  6a2 x
dx
dx 2
L
L
U11   EI (2)(2)dx  4EIL
U12  U 21   EI (2)(6 x)dx  6EIL2
0
0
L
U 22   EI (6 x)(6 x)dx  12EIL3
0
mL5
5
L
K11   m( x 2 )( x 2 )dx 
0
L
K 22   m( x3 )( x3 )dx 
0
L
K12  K 21   m( x 2 )( x3 )dx 
0
7
mL
7
The frequency equation is
4 EIL 
6 EIL 
2
 2 mL5
5
 2 mL6
6
6 EIL2 
12 EIL 
3
 2 mL6
6
 2 mL7
7
0
mL6
6
2
2
7
2
6

 2 mL5  
3  mL  
2  mL 
 4 EIL 
  12 EIL 
   6 EIL 
 0
5 
7  
6 

104
1
1


48E 2 I 2 L4 
EI  2 mL8   4 m2 L12   36E 2 I 2 L4  2 EI  2 mL8   4 m2 L12   0
35
35
36


12 E 2 I 2 L4  0.971EI  2mL8  0.000794 4m 2 L12  0
2
 EI  2
 EI 
  15,120  4   0
4 
mL


 mL 
 4  1223.46 
EI
mL4
1  3.53
2  34.8
EI
mL4
5.7 For the fixed-fixed beam described in problem 5.4, we can assume the mode shape to be expressed as




w( x)  a11 ( x)  a2  2 ( x) , where 1 ( x)  x 4  2 Lx3  L2 x 2 and  2 ( x)  x5  3L2 x3  2 L3 x 2 . Using these
expressions for 1 ( x) and  2 ( x) , approximate the lowest two natural frequencies using Ritz’s method.
Solution
L
Uij   EI ("i )("j )dx
0
L
Kij    A(i )( j )dx
0
1'  4x3  6Lx2  2L2 x
1"  12x2 12Lx  2L2
'2  5x4  9L2 x2  4L3 x
"2  20x3 18L2 x  4L3
L


2
U11   EI 12 x 2  12 Lx  2 L2 dx  0.8EIL5
0
L



U12  U 21   EI 12 x2  12Lx  2L2 20 x3  18L2 x  4L3 dx  2EIL6
0

  A  x

L
U 22   EI 20 x3  18L2 x  4 L3
0
K11
L
4
0
L

2
dx  5.142EIL7
2
 2Lx3  L2 x 2 dx  A (0.001587 L9 )

K12  K21   A x4  2Lx3  L2 x2
0
L


 x
5

 3L2 x3  2 L3 x2 dx  A (0.003968L10 )
2
K 22   A x5  3L2 x3  2L3 x 2 dx  A (0.0099567 L11 )
0

0.8EIL5   2 A (0.001587L9 )
2EIL6   2 A (0.003968L10 )
2EIL6   2 A (0.003968L10 )
5.142EIL7   2 A (0.0099567L11 )
0.8EI   2 A (0.001587 L4 )
2EIL   2 A (0.003968L5 )
2EIL   2 A (0.003968L5 )
5.142EIL2   2 A (0.0099567 L6 )
0
 0.8EI  0.001587
2
A L4
 5.142EIL  0.0099567
2
2
 
A L6  2EIL  0.003968 2 A L5

2
0
2
 EI  2
 EI 
  4612.7 
  2065454.5 
0
4 
4 
 A L 
 A L 
4
1 
22.4
EI
A
2
L
2 
64.1 EI
L2 A
5.8 Use the Ritz method to determine the two lowest frequencies for shaft in problem 5.5 assuming 1  sin
and  2  sin
x
2L
3 x
.
2L
Solution
L
L
Uij   EA(i' )('j )dx
Kij    A(i )( j )dx
0
x 

r( x )  r 1  
 2L 
0
x 

 A( x )   r 2  1  
 2L 
2
1' 

2L
cos
x
2L
 '2 
3
3 x
cos
2L
2L
2
2
3 2
L
L E r 
x   
x
x
x2
E r 2

2 x
2 x
2 x
U11   E r 2 1 
cos
dx

cos

cos

cos
dx

0.9017


 

0 4L2  2L L
0
2L 
2 L 4 L2
2 L 
L
 2L   2L

L
x  
 x  3
3 x 

U12  U 21   E r 2 1 
cos
cos
 

 dx
0
2 L  2 L
2L 
 2L   2L
2

r 
x
3 x x
x
3 x x 2
x
3 x 
E r 2
cos
cos

cos
cos

cos
cos
dx

0.6094


2L
2L L
2L
2 L 4 L2
2L
2 L 
L
4 L2 
L 3E
0
3 2
3 2
L
L 9 E r 
x   3
3 x 
x
x2
3 x 

2 3 x
2 3 x
U 22   E r 2 1 
cos
dx

cos

cos

cos 2

 dx
 


2
2

0
0
2L 
2L L
2L 4L
2 L 
4L 
 2L   2L
2
2
 6.664
E r 2
L

L
L
x   x
 x x 2  x x2
x

2
K11    r 2 1 
sin
dx    r 2  sin 2
 sin
 2 sin 2
 dx  0.2157  r L



0
0
2
L
2
L
2
L
L
2
L
2
L
4L

 



2
2
L
x    x 
3 x 

K12  K 21    r 2 1 
  sin
 sin
 dx
0
2
L
2
L
2L 

 

2
 x
L
3 x x
x
3 x x 2
x
3 x 
2
   r 2  sin
sin
 sin
sin
 2 sin
sin
 dx  0.0697  r L
0
2
L
2
L
L
2
L
2
L
2
L
2
L
4L


2
2

L
L
x  
3 x 
3 x x 2 3 x x 2
3 x 

2
K 22    r 2 1 
sin
dx    r 2  sin 2
 sin
 2 sin 2
 dx  0.2832  r L



0
0
2
L
2
L
2
L
L
2
L
2
L
4L

 



E r 2
 0.2157  r 2 L 2
L
E r 2
0.6094
 0.0697  r 2 L 2
L
0.9017
E r 2
 0.0697 r 2 L 2
L
0
E r 2
2
2
6.664
 0.2832 r L
L
0.6094
2



E r 2
E r 2
E r 2
2
2 
 0.2157 r 2 L 2 
6.664

0.2832

r
L


0.6094
 0.0697 r 2 L 2   0
 0.9017





L
L
L


 

 E  2
 E 
  100.25  2   0
2 
 L 
 L 
 4  28.59 
1 
2.02 E
L

2 
L
L
0
0
4.95 E
L

5.9 For a torsion bar Uij   GJ (i / x)( j / x)dx and Kij    J (i )( j )dx . The 1” diameter, 48” long
solid aluminum bar in Figure 5.24 is fixed to a wall at one end and has a torsional spring with a spring rate of
k is attached at one end. Assume 1  x3  3L2 x and 2  x 2  2Lx . Use the Ritz method to determine the
two lowest frequency and plot the results as a function of k for 500  k  2000 .
Figure 5.24 Torsion rod with a torsional spring at the end.
Solution
For this shaft J 
 (0.5)4
2
 0.0982 , G  3  106 , GJ  0.2946  106 ,   0.10 ,  J  0.00982
L
     j 
Uij   GJ  i  
 dx  k i ( L) j ( L)
0
  x   x 
L
Kij    J ( i )( j )dx
0
1
 3x 2  3L2
x
1  x3  3L2 x ,
L

U11   GJ 3x 2  3L2
0

2

dx  k 2 L3

2
 2 x  2L
x
and 2  x 2  2Lx ,

2

 4.8GJL5  4k L6   4.8GJ  4k L  L5

 
U12  U21   GJ 3x2  3L2  2 x  2 L dx  k 2 L3  L2   2.5GJ  2k L L4
L
0
 
U 22   GJ  2 x  2 L  dx  k  L2
L
2
0
2
 1.333GJ  k L  L3
L
K11    J ( x3  3L2 x)2 dx  1.9428 JL7
0
L
K12  K21    J ( x3  3L2 x)( x2  2 Lx)dx  1.0167 JL6
0
L
K22    J ( x 2  2Lx)2 dx  0.5333 JL5
0
 4.8GJ  4k L  L2  1.9428 JL4 2  2.5GJ  2k L  L  1.016 JL3 2 3
 L  0
 2.5GJ  2k L  L  1.016 JL3 2 1.333GJ  k L   0.5333 JL2 2
 4.8GJ  4k L  L  1.9428 JL  1.333GJ  k L   0.5333 JL  
  2.5GJ  2k L  L  1.016 JL    0
2
4
2
2
2
3 2 2


2


 2 
 G 
 G 
 kG  
 k 

3.1478


38.5

34.34
0










2
2
2 3
  JL  
 L 
 L 
  JL  




 4  17.87 


Using an appropriate computer code, the graph below is produced.
350
lowest frequency
300
250
200
150
100
50
0
0
5000
10000
torsional spring rate
15000
20000
Chapter 6
6.1. A string is fixed to a mass and a spring at point A and fixed to a wall at point B as shown in Figure 6.21 (a) and
a free body diagram of the mass is shown in Figure 6.21 (b). Use a free body diagram approach to develop an
expression which allows you to determine the natural frequencies. Approximate the first two natural frequencies
by assuming the cable length is l  1 , the spring rate is k A  25 , the mass is m  0.005 , and T /   10,000
Figure 6.21 String with a spring and mass at one end.
Solution
Using the free body diagram we sum forces vertically and get
mvA (0, t )  k Ava (0, t )  T sin
Assuming small angles so that sin     dy / dx 
mvA (0, t )  k Av A (0, t )  T
v A
(0, t ) we can write this as
x
v A
(0, t )
x
Using vA  X ( x)  A cos t  B sin t  and X  x   C cos
x
c
 D sin
x
c
we determine
v A (0, t )   2 X (0)  A cos t  B sin t    2C  A cos t  B sin t 
k Av A (0, t )  k A X (0)  k AC
v A
X

(0, t ) 
(0, t )  D
x
x
c
Therefore
m2C  A cos t  B sin t   k AC  A cos t  B sin t   D
or
k
A

 m C   D
2

c

k
D  C
At end B we have vB (l , t )  0  X (l )  C cos
l
c
A

 m 2 c

 D sin
l
c
or

c
 A cost  B sin t 
C cos
Letting  

c
l
c
k
C
A

 m 2 c

sin
l
c
0
and 2   2c2 this becomes




k A  m 2 c 2
C cos  l 
sin  l   0





The non-trivial solution to this is
tan  l 

k A  m 2 c 2
Since c  T /  this becomes
tan  l 

T 
k A  m 2  

Using the data from the problem statement, this becomes tan  

25  50 2
. Solving results in
1  0.6059 , 2  2.5355
Since    / c and c  T /   100 we get
1  60.59 , 2  253.55
6.2. Assume a fixed-fixed string of length l is subjected to an applied load. Determine an expression for the steadystate forced vibration response if the applied load is;
(A) A concentrated force p( x, t )  P(t )  x  xo   Po  x  xo 
(B) A constant velocity (v) moving load
 P(t )  x  vt  0  vt  l
p ( x, t )  
0
vt  l

Solution
(A) The steady-state forced vibration response is given in Equation (6.27) without the complementary solution as
v ( x, t ) 
1
n x  l
nc (t   ) 
sin
d 
 0 Pn ( )sin
c n 1 n
l 
l

2


for n  1, 2,3,...
Where Pn (t ) is given by Equation (6.24) as
Pn (t )  0l p( x, t )sin
n xo
n x
n x
dx  0l P(t )  x  xo  sin
dx  P(t )sin
l
l
l
Since P(t )  Po , we can write P(t )  Po H (t ) , where H (t ) is the Heaviside function defined by
0 t  0
H (t )  
1 t  0
Therefore
2 Po  1
n xo
n x  l
nc (t   ) 
sin
d 
 sin
 0 H (t ) sin
c n 1 n
l 
l
l

v ( x, t ) 


2 Po l
c 
2
2

n 1
n xo
1
n x 
n ct 
sin
sin
 1  cos

l
l 
l 
n2
(B) The steady-state response is given by
v ( x, t ) 
1
n x  l
nc (t   ) 
sin
d 
 0 Pn ( )sin
c n 1 n
l 
l


2

for n  1, 2,3,...
Where Pn (t ) is given by Equation (6.24) as
Pn (t )  0l p( x, t )sin
n x
n x
n vt
dx  0l P(t )  x  vt  sin
dx  P(t )sin
l
l
l
Therefore
v ( x, t ) 
2
c
1
n x  l
n vt
nc (t   ) 
sin
sin
d 
 0 P( )sin
n
l 
l
l



n 1
Using P(t )  Po H (t )
v ( x, t ) 
2 Po  1
n x  l
n vt
nc (t   ) 
sin
d 
 sin
 0 H ( ) sin
c n 1 n
l 
l
l

2P 
 o 
c n 1
 n x 
sin 

 l 
n c n c
n v 
 n v
sin
t
sin
t
l
l
l 
 l
2 
 n v   n c 
n 
 
 
 l   l  
2
6.3. For the bars shown in Figure 6.22, find the natural frequencies and modes of longitudinal vibrations and show
the shapes of the first five modes. All bars have a length L .
Figure 6.22 Bars with different end conditions
Solution
2
 2u
E  2u Eg  2u
2  u

a


t 2
x 2  x 2
 x 2
The equation of motion is
Assume
u( x, t )  X ( x)(t ) and substitute into the equation of motion, giving
X ( x) (t )  a 2 (t ) X ( x )
Since separation of variables was assumed, each must be equal to a constant, i.e.
 (t )
X ( x)
 a2
  2
 (t )
X ( x)
 (t )   2 (t )  0
D
2
X (x ) 
  2   (t )  0
2
a2
X (x )  0
D 2   2 where D  i

 (t )  C1eit  C2eit  C1  cos t  i sin t   C2  cost  i sin t 
  C1  C2  cost  i C1  C2  sin t  A cost  B sint
 2   2 
 D     X ( x )  0
a 

X ( x)  C cos

a
D  i
x  E sin
(a) Free-free ends: @ x  0 :   E

a

a
x
u
 X 
 0 and 
 0
x
 x  x 0
X
 



 C sin x  E cos x  0  E  0 so E  0
x
a
a
a
a
a
@x  L:   E
u
 X 
 0 and 
 0
x
 x  x  L
X
 


 
 L
 C sin x  E cos x  C sin (L)  0  C sin
0
x
a
a
a
a
a
a
a
a
For a non-trivial solution sin
L
a
 0 , so
Therefore the natural frequencies are
L
a
  x or  
an
L
where n  1,2,3,...
an n
=
L
L
n 
Eg

where n  1, 2,3,...
The modes are
X n ( x )  Cn cos
n x
a
 Cn cos
n x
L
u


cos
n 1,2,3,..
n x 
n at
n at 
 Bn sin
 An cos

L 
L
L 
The first 5 modes are shown below.
1
0,8
0,6
0,4
n=1
0,2
n=2
0
-0,2
n=3
0
0,2
0,4
0,6
0,8
1
n=4
n=5
-0,4
-0,6
-0,8
-1
(b) Cantilevered beam: u( x, t )  X ( x) (t ) ,  (t )  A cos t  B sin t , X ( x )  C cos
Boundary conditions:  u x 0  0
 X  x 0  0

a
E cos
L
a
0 
L
a

n
2
n  1,3,5,...
Therefore the natural frequencies are
n 
The modes are
an n Eg
=
2L 2L 
a
x  E sin
X (0)  0  C  0

L
 X 

  0  E cos
a
a
 x  x  L
 u 
  0
 x  x  L

where n  1,3,5,...
E is arbitrary

a
x
X n ( x )  En sin
n x
2a
 En sin
n x
2L


u
sin
n 1,3,5,..
n x 
n at
n at 
 Bn sin
 An cos

2L 
2L
2L 
The first 5 modes are shown below.
1
0,8
0,6
0,4
n=1
0,2
n=3
0
n=5
-0,2
0
0,2
0,4
0,6
0,8
1
n=7
n=9
-0,4
-0,6
-0,8
-1
(c) Fixed-fixed: u( x, t )  X ( x) (t ) ,  (t )  A cos t  B sin t , X ( x )  C cos
Boundary conditions:  u x 0  0
 X  x 0  0
L
a
L
0 
a
a
x  E sin

a
x
X (0)  0  C  0
u xL  0  u  x  L  0  E sin
E sin

L
a
 n
E is arbitrary
n  1, 2,3,...
Therefore the natural frequencies are
n 
an n
=
L
L
Eg

where n  1, 2,3,...
The modes are
X n ( x )  En sin
n x
a
 En sin
n x
L
u


n 1,2,3,..
sin
n x 
n at
n at 
 Bn sin
 An cos

L 
2L
2L 
The first 5 modes are shown below.
1
0,8
0,6
0,4
n=1
0,2
n=2
0
-0,2
n=3
0
0,2
0,4
0,6
0,8
1
n=4
n=5
-0,4
-0,6
-0,8
-1
6.4. For the bars in Figure 6.23 find the general expressions for the displacement u caused by arbitrary forces P(t )
acting on the bars. Use normal mode expansions.
Figure 6.23 End loaded bars with different end conditions
Solution
 2u 2  2u
1
a

p ( x, t )
t 2
x 2 A
The equation of motion is


n 1
n 1
(a) Free-free with external loading. Assume u( x, t )   f n (t ) X n (t ) and p( x, t )   Pn (t ) X n (t )
where

P (t ) 
L
0
n
p( x, t ) X n dx

L
0
2
X dx
n
Substituting into the equation of motion

f
n 1

n
1 
 Pn X n
A n 1
X n  a 2  f n X n" 
n 1
For free vibrations a 2 X n"   2 X n , resulting in

f
n 1

X n   f nn2 X n" 
n
n 1
1 
 Pn X n
A n 1
Since the series on the left equals the series on the right we have
f n  n2 f n 
Pn (t )
A
From Problem 1 for the free-free beam we had
X n ( x )  Cn cos
n x
L
where Cn  1, n 
n
L
Eg

where n  1, 2,3,...
Assume p( x, t )  P(t ) ( x  L)
Pn (t ) 

L
0
n x
dx
2 P(t ) cos n
L

L
L
2 n x
0 cos L dx
P(t )  x  L  cos
so
f n  n2 f n 
2 P(t ) cos n
A L
Taking the Laplace transform gives
s 2 f n  n2 f n 
2 P cos n
A L
s
2cos n
P
A L
2
 n2  f n 
f n (t )  L
 2  n
 f ( s)  A L 0 sinn  t    P   d
n
-1

 2  n
A n
t

sinn  t    P   d
Eg 0
or
f n (t ) 
Thus
 2  n
An
2cos n
P
A L  s 2  n2 
 fn 
g
E
0 sinn  t    P   d
t
t

u( x, t )  
 2 n
g
E
An
n 1

0 sinn  t    P   d  cos
t
n x 

L 
or
1
u( x, t ) 
A
g
E
   2 n 

n x  t

 cos
 0 sinn  t    P   d 
L 
 n 1 n 

(b) From Problem 1; X n ( x)  En sin
n x
n Eg
with En  1, n 
where n  1,3,5,...
2L
2L 
Assume p( x, t )  P(t ) ( x  L)
Pn (t ) 

L
0
n x
dx
n 1
2 P (t )
n
 2 P (t ) 
2L
2

sin


1




L
L
2
2 n x
 L 
0 sin 2 L dx
P(t )  x  L  sin
so
f n  n2 f n   1
u( x, t ) 
1
A
g
E
n 1
2
 2 P (t ) 


 A L 
   1n  n x  t


 sin
 0 sinn  t    P   d 
n 
2L 
 n 1

6.5. Find the frequency response functions of the displacement u for the bar and loadings shown in Problem 6.4.
Use natural mode expansions. Develop expressions for U *( x, ) if the material is elastic, Inelastic with
E*  E ' iE " , a Kelvin-type material, and a Maxwell-type material.
Solution
For a free-free elastic beam
 2u 2  2u g
a

P (t )
t 2
x 2 A
Assume P(t )  eit and u( x, t )  U * ( x, i )eit . Substituting into the equation of motion
(i ) 2 U * ( x, i )  a 2
 2U * ( x, i )
x 2
U * ( x, i)  C sin

a
x  D cos
 u 
Boundary conditions: E    0
 x  x 0

a
x
 U * 
E
 0
 x  x 0
 U * 

 
 


  C   cos x  D   sin x  0  C  

x
a
a
a
a




a

 x 0
P
 u 
E    E   

x
A
 xL
1
 U * 

 

x
EA

xL
 U * 
1
   L


   D   sin
a
EA
a
 x  x  L
 D
 

cos x 
a 
a
U ( x, i )  


 EA  sin  L 
a 

For an inelastic beam, replace E with E *(i) and a with a *(i) so that



 cos a *(i ) x 
a
*(
i

)
U * ( x, i )  


 E *(i ) A  sin  L 

a *(i ) 

For a Kelvin model, replace E *(i) with E  i so that



 cos a *(i ) x 
a
*(
i

)
U * ( x, i )  


  E  i  A  sin  L 

a *(i ) 

Ei
so that
i  E / 



cos
x

a *(i )
a *(i )
U * ( x, i )  


 Ei   sin  L 

 A
a *(i ) 
 i  E /   
(b) Fixed-free beam:
 2u 2  2u g
a

P (t )
t 2
x 2 A
 for P  1
a
1
 EA sin  L
a
*
For a Maxwell model, replace E *(i) with
 C0
Assume P(t )  eit and u( x, t )  U * ( x, i )eit . Substituting into the equation of motion
(i ) 2 U * ( x, i )  a 2
 2U * ( x, i )
x 2
U * ( x, i)  C sin
Boundary conditions:

a
x  D cos
 u  x 0  0

a
x
U *x0  0


 
 
 sin x  D   cos x  0  D  
a
a
a
a
a
U *x0  C 

P
 u 
E    E   

x
A
 xL
1
 U * 

 

x
EA

xL
 U * 
L 1
 


  C   cos
a
EA
a
 x  x  L
 C
 

sin x 
a 
a
U ( x, i )  


 EA  cos  L 
a 

For an inelastic beam, replace E with E *(i) and a with a *(i) so that



 sin a *(i ) x 
a
*(
i

)
U * ( x, i )  


 E *(i ) A  cos  L 

a *(i ) 

For a Kelvin model, replace E *(i) with E  i so that



sin
x
a *(i ) 
a *(i ) 
U ( x, i )  


  E  i  A  cos  L 

a *(i ) 

*
For a Maxwell model, replace E *(i) with
 for P  1
a
1
 EA cos  L
a
*
Ei
so that
i  E / 



 sin a *(i ) x 
a
*(
i

)
U * ( x, i )  


 Ei   cos  L 

 A
a *(i ) 
 i  E /   
 D0
6.6. A viscoelastic bar as shown in Figure 6.24 whose properties are specified by the complex modulus
E*  E  i , where E and  are constants, is subjected to an end load P(t )  Po cos t . Determine a
particular solution for the axial displacement u( x, t ) and the axial stress  ( x, t ) using
(A) The natural mode method
(B) The closed form solution.
Figure 6.24 Viscoelastic bar with an end load.
Solution
(A) A
 2u
t
2
 E* A
 2u
x
2
 p ( x, t )
 2u

For free vibrations and E being elastic
t
 2u
t 2

2

E*  2u
 x 2
E  2u
 x 2

p ( x, t )
A
0
Assume u( x, t )  X ( x)  A cos  t  B sin  t  and substitute into the equation of motion. The result is
 2 X 
E d2X
X  C cos
 dX 2

E
 x  D sin

E
x
The boundary conditions are
x  0:
u  0 X  0
Substituting, C  0 and D cos

cos
E

X n  Dn sin
E
L  0
x  L:

E

n x  Dn sin
 x  u / x  0  dX / dx  0
 L  0 . Since D  0 is the trivial solution we explorer

E
n x
2L
L
n
2
d2 Xn
dx 2


n 1
n 1
 n 
n E
2L 

 n2   X n ( x)
E
Assume u ( x, t )   X n ( x)n (t ) and p ( x, t )   X n ( x) f n (t ) , where
n  1,3,5,....
l
fn
 p( x, t ) X m dx
(t )  0
l
2
u( x, t )   X n ( x)n (t ) 
E*
0 X m dx
Substituting results in

n 1
Recalling that
d2 Xn
dx
2
 n2

E



n 1
d 2 Xn

dx 2
1 
 X n f n (t )
A n1
X n ( x)


E* 2 
1 
X
(
x
)

(
t
)

n X nn  

 n n
 X n fn (t )

E
n 1 
 A n 1

or
 E*
n (t )  
 E

n2  n 

f n (t )
A
For the loading condition given ( P(t )  Po cos  t )
n x
n 1
 ( x  L )dx 2 P cos  t
2 Po cos  t
n
2L
 o
sin
 (1) 2
n  1,3,5,...
L
L
2
L
2 n x
sin
dx
0
2L
L
f n (t ) 
Po cos  t  sin
0
 E*
n (t )  
 E

n2  n  (1)
n 1
2

2 Po cos  t
n  1,3,5,..
A L
Using the convolution integral gives for the particular solution
n  (1)
n 1
2
2 Po
A Ln
E*
E
L
0


E* 
 (t  t ')  cos t ' dt ' n  1,3,5,....
sin  n
E 





n 1
 2P
u ( x, t )   o
 A L

n x
(1) 2 sin
E  
2L

E  n 1
n

*
L

0 sin  n


E* 
 (t  t ')  cos t ' dt '
E 



(B) Assume u( x, t )  U * ( x, i)eit and p( x, t )  Po eit . This results in
(i ) 2 U *eit 
U *  C sin
E * d 2U *
 dx 2

E
*
eit  0
 x  D cos

E*
or
x
d 2U *
dx
2

 2
E
*
U*  0
The boundary conditions become
C0
D

E
*
 cos

E
*
x  0:
L
Po
*
E A
U *  0 and
 D
x  L:
Po
1
 A  E*
P
dU *
 *o . Substituting yields
dx
E A
1
cos

E*
L
So that



x
 cos
*
E


U* 
* 


 A  E cos
L

*
E


Po



x
 cos
*
E


u ( x, L ) 
* 


 A  E cos
L

*
E


Po eit