Chapters 2 - 20 Covered
fg fg fg fg
SOLUTIONS
,Table of Contents fg fg
PART 1 fg
2 Formulation of the equations of motion: Single-degree-of- fg fg fg fg fg fg
freedom systems
fg fg
3 Formulation of the equations of motion: Multi-degree-of- fg fg fg fg fg fg
freedom systems
fg fg
4 Principles of analytical mechanics fg fg fg
PART 2
fg fg
5 Free vibrationresponse: Single-degree-of-freedom system
fg fg fg fg
6 Forced harmonic vibrations: Single-degree-of-freedom
fg fg fg
system
fg
7 Response to general dynamic loading and transient response
fg fg fg fg fg fg fg
8 Analysis of single-degree-of-freedom systems: Approximate
fg fg fg fg
and numerical methods
fg fg fg
9 Analysis of response in the frequency domain
fg fg fg fg fg fg
PART 3
fg fg
10 Free vibration response: Multi-degree-of-freedom system
fg fg fg fg
11 Numerical solution of the eigenproblem fg fg fg fg
,12 Forced dynamic response: Multi-degree-of-freedom
fg fg fg
systems
fg
13 Analysis of multi-degree-of-freedom systems: Approximate
fg fg fg fg
and numerical methods
fg fg fg
PART 4 fg
14 Formulation of the equations of motion: Continuous fg fg fg fg fg fg
systems
fg
15 Continuous systems: Free vibration response
fg fg fg fg
16 Continuous systems: Forced-vibration response
fg fg fg
17 Wave propagation analysis
fg fg
PART 5
fg fg
18 Finite element method
fg fg
19 Component mode synthesis fg fg
20 Analysis of nonlinear response
fg fg fg
, 2
Chapter In a similar manner we get
fg fg fg fg fg
2 f g
Iy = M u¨ y
Problem fg fg g
f
2.1
f g
For an angular acceleration θ ¨ about the
fg fg fg fg
fg
fg
center of mass the inertia force on the
fg fg fg fg fg fg fg fg
90 60
infinitesimal ele- ment is directed along the
fg fg fg fg fg fg fg
tangent and is γr θ¨dθdr.
2
N/mm N/mm fg fg fg fg
The x component of this force is γr θ¨dθdr
2
fg f g f g f g f g f g f g
u
sin θ.
fg fg
It is easily seen that the resultant of
f g f g f g f g f g f g f g
all x direc-
f g f g f g
40 tion forces is zero. In a similar manner the
fg fg fg f g fg fg fg fg
N/mm resul- tant y direction force is zero. However, a
fg fg fg fg fg fg fg fg fg
Figure f g S2.1 clockwise moment about the center of the disc
fg fg fg fg fg fg fg fg
exists and is given by
fg fg fg fg fg
Referring to Figure S2.1 the springs with fg fg fg fg fg fg ∫ fg
R fg∫ R2
stiff- ness 60 N/mm and 90 N/mm are γθ¨r3dθdr = γπR2 θ¨ = R2
θ¨
f g
fg fg f g f g f g f g f g f g Mθ = fg
fg
2π fg fg fg fg
placed in series
f g f g f g M fg
and have an effective stiffness given
fg fg fg fg fg 0 0 2 2
by
fg
1 The elliptical plate shown in Figure S2.2(c) is
k1 = fg = 36 fg
fg fg fg fg fg fg fg
1/60+ N/mm
fg
f divided into the infinitesimal elements
g f g f g f g f g
1/90
f g
as shown.
f g f g
The mass of an element is γdxdy and the
fg fg fg fg fg fg fg fg
This combination is now placed in parallel with
fg fg fg fg fg fg fg
inertia force acting on it when the disc
fg fg fg fg fg fg fg fg
the spring of stiffness 40 N/mm giving a final
fg fg fg fg fg fg fg fg fg
undergoes trans- lation in the x direction with
fg fg fg fg fg fg fg fg
effective stiffness of
fg fg fg
acceleration ü x is γ ü x dxdy. The resultant
fg fg fg fg f g fg
keff = k1 + 40 = 76 N/mm
fg fg fg fg fg fg fg inertia force in the neg- ative x direction is
fg fg fg fg fg fg fg fg fg
given by
fg fg
∫ ∫ √
Problem fg
a/2 f g
fg
b/2 1−4x2/a2
2.2
f g Ix f g √ γüy dydx
=
fg −a/2 −b/2 1−4x2/a 2
∫ √
f g gf
a/2
= γ ü x fg b 1 fg−fg4x2/a2dx
dxdy −a/2
dr
dθ πγab
R b = = M u¨ x
fg g
f
4
The moment of the x direction inertia force on
fg fg fg fg fg fg fg fg
an element is γüx ydxdy. The resultant moment
fg fg fg fg f g fg fg
a ob- tained over the area is zero. The inertia
fg fg fg fg fg fg fg f g fg
force pro- duced by an acceleration in the y
fg fg fg fg fg fg fg fg fg
(a) (b)
direction is ob- tained in a similar manner and
fg fg fg fg fg fg fg fg fg
Figure is M ü y directed in the negative y direction.
fg fg fg fg fg fg fg fg fg
S2.2 f g An angular acceleration θ¨ produces a clockwise
fg fg fg
fg
fg fg
fg
The infinitesimal area shown in Figure
fg fg fg fg fg
moment equal to γr θ¨dxdy = γ x + y
2 2
fg
2
fg fg fg fg f g fg fg
fg
S2.2(a) fg
is equal to rdθdr. When the circular disc
fg fg fg f g fg fg fg
f gθ¨dxdy. Integration over the area yields thefg fg fg fg fg fg
moves in the x direction with acceleration
fg f g fg fg fg fg fg
resultant mo- ment, which is clockwise
fg fg fg fg fg fg
ü x the inertia force on the infinitesimal are is
fg fg fg fg fg fg fg fg fg
γrdθdrü x , where γ
√
fg fg fg
ids the mass per unit area.
f g f g The f g f g f g f g
∫ a/2 ∫ b/2 1−4x2/a2
f gresultant inertia force on the disc acting in
f g fg fg fg fg fg fg
Iθ √ γθ¨ x2 + fg dydx
the negative x direction 2
fg fg fg fg
=
fg
2 f g 2
f g
y fg
is given
fg −a/2 f g f g −b/2 f g f g 1−4x f g /a
2 2 2 2
by
fg
∫ R ∫ fg
fg
fg
2π
4 16 16
γü x rdθdr = γ πR 2 ü x = fg fg fg
fg fg fg fg
SOLUTIONS
,Table of Contents fg fg
PART 1 fg
2 Formulation of the equations of motion: Single-degree-of- fg fg fg fg fg fg
freedom systems
fg fg
3 Formulation of the equations of motion: Multi-degree-of- fg fg fg fg fg fg
freedom systems
fg fg
4 Principles of analytical mechanics fg fg fg
PART 2
fg fg
5 Free vibrationresponse: Single-degree-of-freedom system
fg fg fg fg
6 Forced harmonic vibrations: Single-degree-of-freedom
fg fg fg
system
fg
7 Response to general dynamic loading and transient response
fg fg fg fg fg fg fg
8 Analysis of single-degree-of-freedom systems: Approximate
fg fg fg fg
and numerical methods
fg fg fg
9 Analysis of response in the frequency domain
fg fg fg fg fg fg
PART 3
fg fg
10 Free vibration response: Multi-degree-of-freedom system
fg fg fg fg
11 Numerical solution of the eigenproblem fg fg fg fg
,12 Forced dynamic response: Multi-degree-of-freedom
fg fg fg
systems
fg
13 Analysis of multi-degree-of-freedom systems: Approximate
fg fg fg fg
and numerical methods
fg fg fg
PART 4 fg
14 Formulation of the equations of motion: Continuous fg fg fg fg fg fg
systems
fg
15 Continuous systems: Free vibration response
fg fg fg fg
16 Continuous systems: Forced-vibration response
fg fg fg
17 Wave propagation analysis
fg fg
PART 5
fg fg
18 Finite element method
fg fg
19 Component mode synthesis fg fg
20 Analysis of nonlinear response
fg fg fg
, 2
Chapter In a similar manner we get
fg fg fg fg fg
2 f g
Iy = M u¨ y
Problem fg fg g
f
2.1
f g
For an angular acceleration θ ¨ about the
fg fg fg fg
fg
fg
center of mass the inertia force on the
fg fg fg fg fg fg fg fg
90 60
infinitesimal ele- ment is directed along the
fg fg fg fg fg fg fg
tangent and is γr θ¨dθdr.
2
N/mm N/mm fg fg fg fg
The x component of this force is γr θ¨dθdr
2
fg f g f g f g f g f g f g
u
sin θ.
fg fg
It is easily seen that the resultant of
f g f g f g f g f g f g f g
all x direc-
f g f g f g
40 tion forces is zero. In a similar manner the
fg fg fg f g fg fg fg fg
N/mm resul- tant y direction force is zero. However, a
fg fg fg fg fg fg fg fg fg
Figure f g S2.1 clockwise moment about the center of the disc
fg fg fg fg fg fg fg fg
exists and is given by
fg fg fg fg fg
Referring to Figure S2.1 the springs with fg fg fg fg fg fg ∫ fg
R fg∫ R2
stiff- ness 60 N/mm and 90 N/mm are γθ¨r3dθdr = γπR2 θ¨ = R2
θ¨
f g
fg fg f g f g f g f g f g f g Mθ = fg
fg
2π fg fg fg fg
placed in series
f g f g f g M fg
and have an effective stiffness given
fg fg fg fg fg 0 0 2 2
by
fg
1 The elliptical plate shown in Figure S2.2(c) is
k1 = fg = 36 fg
fg fg fg fg fg fg fg
1/60+ N/mm
fg
f divided into the infinitesimal elements
g f g f g f g f g
1/90
f g
as shown.
f g f g
The mass of an element is γdxdy and the
fg fg fg fg fg fg fg fg
This combination is now placed in parallel with
fg fg fg fg fg fg fg
inertia force acting on it when the disc
fg fg fg fg fg fg fg fg
the spring of stiffness 40 N/mm giving a final
fg fg fg fg fg fg fg fg fg
undergoes trans- lation in the x direction with
fg fg fg fg fg fg fg fg
effective stiffness of
fg fg fg
acceleration ü x is γ ü x dxdy. The resultant
fg fg fg fg f g fg
keff = k1 + 40 = 76 N/mm
fg fg fg fg fg fg fg inertia force in the neg- ative x direction is
fg fg fg fg fg fg fg fg fg
given by
fg fg
∫ ∫ √
Problem fg
a/2 f g
fg
b/2 1−4x2/a2
2.2
f g Ix f g √ γüy dydx
=
fg −a/2 −b/2 1−4x2/a 2
∫ √
f g gf
a/2
= γ ü x fg b 1 fg−fg4x2/a2dx
dxdy −a/2
dr
dθ πγab
R b = = M u¨ x
fg g
f
4
The moment of the x direction inertia force on
fg fg fg fg fg fg fg fg
an element is γüx ydxdy. The resultant moment
fg fg fg fg f g fg fg
a ob- tained over the area is zero. The inertia
fg fg fg fg fg fg fg f g fg
force pro- duced by an acceleration in the y
fg fg fg fg fg fg fg fg fg
(a) (b)
direction is ob- tained in a similar manner and
fg fg fg fg fg fg fg fg fg
Figure is M ü y directed in the negative y direction.
fg fg fg fg fg fg fg fg fg
S2.2 f g An angular acceleration θ¨ produces a clockwise
fg fg fg
fg
fg fg
fg
The infinitesimal area shown in Figure
fg fg fg fg fg
moment equal to γr θ¨dxdy = γ x + y
2 2
fg
2
fg fg fg fg f g fg fg
fg
S2.2(a) fg
is equal to rdθdr. When the circular disc
fg fg fg f g fg fg fg
f gθ¨dxdy. Integration over the area yields thefg fg fg fg fg fg
moves in the x direction with acceleration
fg f g fg fg fg fg fg
resultant mo- ment, which is clockwise
fg fg fg fg fg fg
ü x the inertia force on the infinitesimal are is
fg fg fg fg fg fg fg fg fg
γrdθdrü x , where γ
√
fg fg fg
ids the mass per unit area.
f g f g The f g f g f g f g
∫ a/2 ∫ b/2 1−4x2/a2
f gresultant inertia force on the disc acting in
f g fg fg fg fg fg fg
Iθ √ γθ¨ x2 + fg dydx
the negative x direction 2
fg fg fg fg
=
fg
2 f g 2
f g
y fg
is given
fg −a/2 f g f g −b/2 f g f g 1−4x f g /a
2 2 2 2
by
fg
∫ R ∫ fg
fg
fg
2π
4 16 16
γü x rdθdr = γ πR 2 ü x = fg fg fg
