All 18 Chapters Covered
SOLUTION MANUAL
, CHAPTER 1
Problem 1.1
If ke is the effective stiffness,
fS = keu
u
k1
k1 u
fS fS
k2 u
k2
Equilibrium of forces: =fS ( k1 + k2 ) u
Effective stiffness: k=e fS =
u k1 + k2
Equation of motion: mu&& + keu = p( t )
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This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
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1
,Problem 1.2
If ke is the effective stiffness,
fS = keu (a)
u
k1 k2
fS
If the elongations of the two springs are u1 and u2 ,
=
u u1 + u2 (b)
Because the force in each spring is fS ,
fS = k1u1 fS = k2u2 (c)
Solving for u1 and u2 and substituting in Eq. (b) gives
fS fS f 1 1 1
= + S ⇒ = + ⇒
ke k1 k2 ke k1 k2
k1 k2
ke =
k1 + k2
Equation of motion: mu&& + keu =
p( t )
© 2020 Pearson Education Ltd. All rights reserved.
This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education Ltd.
2
, Problem 1.3
k1 k3
m Fig. 1.3(a)
k2
⇓
k 1+ k 2 k3
m Fig. 1.3(b)
⇓
u
ke
m Fig. 1.3(c)
This problem can be solved either by starting from the
definition of stiffness or by using the results of Problems
P1.1 and P1.2. We adopt the latter approach to illustrate
the procedure of reducing a system with several springs to
a single equivalent spring.
First, using Problem 1.1, the parallel arrangement of
k1 and k2 is replaced by a single spring, as shown in
Fig. 1.3(b). Second, using the result of Problem 1.2, the
series arrangement of springs in Fig. 1.3(b) is replaced by
a single spring, as shown in Fig. 1.3(c):
1 1 1
= +
ke k1 + k2 k3
Therefore the effective stiffness is
( k1 + k2 ) k3
ke =
k1 + k2 + k3
The equation of motion is mu&& + keu =
p( t ) .
© 2020 Pearson Education Ltd. All rights reserved.
This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education Ltd.
3
SOLUTION MANUAL
, CHAPTER 1
Problem 1.1
If ke is the effective stiffness,
fS = keu
u
k1
k1 u
fS fS
k2 u
k2
Equilibrium of forces: =fS ( k1 + k2 ) u
Effective stiffness: k=e fS =
u k1 + k2
Equation of motion: mu&& + keu = p( t )
© 2020 Pearson Education Ltd. All rights reserved.
This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education Ltd.
1
,Problem 1.2
If ke is the effective stiffness,
fS = keu (a)
u
k1 k2
fS
If the elongations of the two springs are u1 and u2 ,
=
u u1 + u2 (b)
Because the force in each spring is fS ,
fS = k1u1 fS = k2u2 (c)
Solving for u1 and u2 and substituting in Eq. (b) gives
fS fS f 1 1 1
= + S ⇒ = + ⇒
ke k1 k2 ke k1 k2
k1 k2
ke =
k1 + k2
Equation of motion: mu&& + keu =
p( t )
© 2020 Pearson Education Ltd. All rights reserved.
This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education Ltd.
2
, Problem 1.3
k1 k3
m Fig. 1.3(a)
k2
⇓
k 1+ k 2 k3
m Fig. 1.3(b)
⇓
u
ke
m Fig. 1.3(c)
This problem can be solved either by starting from the
definition of stiffness or by using the results of Problems
P1.1 and P1.2. We adopt the latter approach to illustrate
the procedure of reducing a system with several springs to
a single equivalent spring.
First, using Problem 1.1, the parallel arrangement of
k1 and k2 is replaced by a single spring, as shown in
Fig. 1.3(b). Second, using the result of Problem 1.2, the
series arrangement of springs in Fig. 1.3(b) is replaced by
a single spring, as shown in Fig. 1.3(c):
1 1 1
= +
ke k1 + k2 k3
Therefore the effective stiffness is
( k1 + k2 ) k3
ke =
k1 + k2 + k3
The equation of motion is mu&& + keu =
p( t ) .
© 2020 Pearson Education Ltd. All rights reserved.
This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education Ltd.
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