All Chapters Covered
SOLUTIONS
,Table of Contents
1. Single-Degree-of-Freedom Sỵstems
2. Random Vibrations
3. Dỵnamic Response of SDOF Sỵstems Using Numerical Methods
4. Sỵstems with Several Degrees of Freedom
5. Equations of Motion of Continuous Sỵstems
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
, Chapter 1
1.1 Write the equations of motion for the one-degree-of-freedom sỵstems 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 sỵstems
@@SSee
isim
smiciicsio
solala
tio
tionn
, Solutions
(a) (b)
(
spring force = 3EI / L3 u )
(
spring force = 48EI / L3 u ) 3EI
mu + u = P(t)
48EI L3
mu + 3 u = P(t)
L
(c) (d)
( ) (
spring force = 3EI / L3 u − 3EI / L2 (t))
(
spring force = 3EI / L 3
)(u − a) mu +
3EI
u=
3EI
(t)
L3 L2
3EI
mu + 3 (u − a ) = 0
L
3EI 3EI
mu + u= a(t)
3
L L3
(e) (f)
spring force = (EA / L)u
EA ( ) (
spring force = 2 3EI / L3 u = 6EI / L3 u )
mu + u = P(t) 6EI
@@SSee
isim
smiciicsio
solala
tio
tionn
SOLUTIONS
,Table of Contents
1. Single-Degree-of-Freedom Sỵstems
2. Random Vibrations
3. Dỵnamic Response of SDOF Sỵstems Using Numerical Methods
4. Sỵstems with Several Degrees of Freedom
5. Equations of Motion of Continuous Sỵstems
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
, Chapter 1
1.1 Write the equations of motion for the one-degree-of-freedom sỵstems 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 sỵstems
@@SSee
isim
smiciicsio
solala
tio
tionn
, Solutions
(a) (b)
(
spring force = 3EI / L3 u )
(
spring force = 48EI / L3 u ) 3EI
mu + u = P(t)
48EI L3
mu + 3 u = P(t)
L
(c) (d)
( ) (
spring force = 3EI / L3 u − 3EI / L2 (t))
(
spring force = 3EI / L 3
)(u − a) mu +
3EI
u=
3EI
(t)
L3 L2
3EI
mu + 3 (u − a ) = 0
L
3EI 3EI
mu + u= a(t)
3
L L3
(e) (f)
spring force = (EA / L)u
EA ( ) (
spring force = 2 3EI / L3 u = 6EI / L3 u )
mu + u = P(t) 6EI
@@SSee
isim
smiciicsio
solala
tio
tionn
