Chapters 2 - 20 Covered
SOLUTIONS
,Table of Contents
PART 1
2 Formulation of the equations of motion: Single-degree-of-
freedom systems
3 Formulation of the equations of motion: Multi-degree-of-
freedom systems
4 Principles of analytical mechanics
PART 2
5 Free vibration response: Single-degree-of-freedom system
6 Forced harmonic vibrations: Single-degree-of-freedom
system
7 Response to general dynamic loading and transient response
8 Analysis of single-degree-of-freedom systems: Approximate
and numerical methods
9 Analysis of response in the frequency domain
PART 3
10 Free vibration response: Multi-degree-of-freedom system
11 Numerical solution of the eigenproblem
,12 Forced dynamic response: Multi-degree-of-freedom
systems
13 Analysis of multi-degree-of-freedom systems: Approximate
and numerical methods
PART 4
14 Formulation of the equations of motion: Continuous
systems
15 Continuous systems: Free vibration response
16 Continuous systems: Forced-vibration response
17 Wave propagation analysis
PART 5
18 Finite element method
19 Component mode synthesis
20 Analysis of nonlinear response
, 2
Chapter 2 In a similar manner we get
Problem 2.1 Iy = M üy
For an angular acceleration θ̈ about the center
90 N/mm 60 N/mm of mass the inertia force on the infinitesimal ele-
ment is directed along the tangent and is γr2 θ̈dθdr.
u
The x component of this force is γr2 θ̈dθdr sin θ.
It is easily seen that the resultant of all x direc-
tion forces is zero. In a similar manner the resul-
40 N/mm tant y direction force is zero. However, a clockwise
moment about the center of the disc exists and is
Figure S2.1 given by
Referring to Figure S2.1 the springs with stiff- R 2π
R2 R2
ness 60 N/mm and 90 N/mm are placed in series Mθ = γ θ̈r3 dθdr = γπR2 θ̈ = M θ̈
and have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is
k1 = = 36 N/mm
1/60 + 1/90 divided into the infinitesimal elements as shown.
The mass of an element is γdxdy and the inertia
This combination is now placed in parallel with the
force acting on it when the disc undergoes trans-
spring of stiffness 40 N/mm giving a final effective
lation in the x direction with acceleration üx is
stiffness of
γ üx dxdy. The resultant inertia force in the neg-
keff = k1 + 40 = 76 N/mm ative x direction is given by
√
a/2 b/2 1−4x2 /a2
Problem 2.2 Ix = √ γ üy dydx
−a/2 −b/2 1−4x2 /a2
a/2
= γ üx b 1 − 4x2 /a2 dx
dxdy −a/2
dr
dθ R πγab
b = = M üx
4
The moment of the x direction inertia force on an
element is γ üx ydxdy. The resultant moment ob-
a tained over the area is zero. The inertia force pro-
duced by an acceleration in the y direction is ob-
(a) (b)
tained in a similar manner and is M üy directed in
Figure S2.2 the negative y direction.
An angular acceleration θ̈ produces
a clockwise
The infinitesimal area shown in Figure S2.2(a) moment equal to γr2 θ̈dxdy = γ x2 + y 2 θ̈dxdy.
is equal to rdθdr. When the circular disc moves Integration over the area yields the resultant mo-
in the x direction with acceleration üx the inertia ment, which is clockwise
force on the infinitesimal are is γrdθdrüx , where γ
ids the mass per unit area. The resultant inertia √
a/2 b/2 1−4x2 /a2
force on the disc acting in the negative x direction Iθ = √ γ θ̈ x2 + y 2 dydx
is given by −a/2 −b/2 1−4x2 /a2
2 2
R 2π πab a + b a2 + b2
=γ θ̈ = M θ̈
Ix = γ üx rdθdr = γπR2 üx = M üx 4 16 16
0 0
The x and y direction inertia forces produced on
where M is the total mass of the disc. The resultant
the infinitesimal element are −γ θ̈ sin θdxdy and
moment of the inertia forces about the centre of the
γ θ̈ cos θdxdy, respectively. When summed over the
disc, which is also the centre of mass is given by
area the net forces produced by these are easily
R 2π
shown to be zero.
Mx = γ üx r2 sin θdθdr = 0
0 0 @Seismicisolation
@Seismicisolation
SOLUTIONS
,Table of Contents
PART 1
2 Formulation of the equations of motion: Single-degree-of-
freedom systems
3 Formulation of the equations of motion: Multi-degree-of-
freedom systems
4 Principles of analytical mechanics
PART 2
5 Free vibration response: Single-degree-of-freedom system
6 Forced harmonic vibrations: Single-degree-of-freedom
system
7 Response to general dynamic loading and transient response
8 Analysis of single-degree-of-freedom systems: Approximate
and numerical methods
9 Analysis of response in the frequency domain
PART 3
10 Free vibration response: Multi-degree-of-freedom system
11 Numerical solution of the eigenproblem
,12 Forced dynamic response: Multi-degree-of-freedom
systems
13 Analysis of multi-degree-of-freedom systems: Approximate
and numerical methods
PART 4
14 Formulation of the equations of motion: Continuous
systems
15 Continuous systems: Free vibration response
16 Continuous systems: Forced-vibration response
17 Wave propagation analysis
PART 5
18 Finite element method
19 Component mode synthesis
20 Analysis of nonlinear response
, 2
Chapter 2 In a similar manner we get
Problem 2.1 Iy = M üy
For an angular acceleration θ̈ about the center
90 N/mm 60 N/mm of mass the inertia force on the infinitesimal ele-
ment is directed along the tangent and is γr2 θ̈dθdr.
u
The x component of this force is γr2 θ̈dθdr sin θ.
It is easily seen that the resultant of all x direc-
tion forces is zero. In a similar manner the resul-
40 N/mm tant y direction force is zero. However, a clockwise
moment about the center of the disc exists and is
Figure S2.1 given by
Referring to Figure S2.1 the springs with stiff- R 2π
R2 R2
ness 60 N/mm and 90 N/mm are placed in series Mθ = γ θ̈r3 dθdr = γπR2 θ̈ = M θ̈
and have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is
k1 = = 36 N/mm
1/60 + 1/90 divided into the infinitesimal elements as shown.
The mass of an element is γdxdy and the inertia
This combination is now placed in parallel with the
force acting on it when the disc undergoes trans-
spring of stiffness 40 N/mm giving a final effective
lation in the x direction with acceleration üx is
stiffness of
γ üx dxdy. The resultant inertia force in the neg-
keff = k1 + 40 = 76 N/mm ative x direction is given by
√
a/2 b/2 1−4x2 /a2
Problem 2.2 Ix = √ γ üy dydx
−a/2 −b/2 1−4x2 /a2
a/2
= γ üx b 1 − 4x2 /a2 dx
dxdy −a/2
dr
dθ R πγab
b = = M üx
4
The moment of the x direction inertia force on an
element is γ üx ydxdy. The resultant moment ob-
a tained over the area is zero. The inertia force pro-
duced by an acceleration in the y direction is ob-
(a) (b)
tained in a similar manner and is M üy directed in
Figure S2.2 the negative y direction.
An angular acceleration θ̈ produces
a clockwise
The infinitesimal area shown in Figure S2.2(a) moment equal to γr2 θ̈dxdy = γ x2 + y 2 θ̈dxdy.
is equal to rdθdr. When the circular disc moves Integration over the area yields the resultant mo-
in the x direction with acceleration üx the inertia ment, which is clockwise
force on the infinitesimal are is γrdθdrüx , where γ
ids the mass per unit area. The resultant inertia √
a/2 b/2 1−4x2 /a2
force on the disc acting in the negative x direction Iθ = √ γ θ̈ x2 + y 2 dydx
is given by −a/2 −b/2 1−4x2 /a2
2 2
R 2π πab a + b a2 + b2
=γ θ̈ = M θ̈
Ix = γ üx rdθdr = γπR2 üx = M üx 4 16 16
0 0
The x and y direction inertia forces produced on
where M is the total mass of the disc. The resultant
the infinitesimal element are −γ θ̈ sin θdxdy and
moment of the inertia forces about the centre of the
γ θ̈ cos θdxdy, respectively. When summed over the
disc, which is also the centre of mass is given by
area the net forces produced by these are easily
R 2π
shown to be zero.
Mx = γ üx r2 sin θdθdr = 0
0 0 @Seismicisolation
@Seismicisolation
