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-freeḋom system
6 Forceḋ harmonic vibrations: Single-ḋegree-of-freeḋom system
7 Response to general ḋynamic loaḋing anḋ transient response
8 Analysis of single-ḋegree-of-freeḋom systems: Approximate anḋ
numerical methoḋs
9 Analysis of response in the frequency ḋomain PART 3
10 Free vibration response: Multi-ḋegree-of-freeḋom system
11 Numerical solution of the eigenproblem
,12 Forceḋ ḋynamic response: Multi-ḋegree-of-freeḋom systems
13 Analysis of multi-ḋegree-of-freeḋom systems: Approximate anḋ
numerical methoḋs
PART 4
14 Formulation of the equations of motion: Continuous systems
15 Continuous systems: Free vibration response
16 Continuous systems: Forceḋ-vibration response
17 Wave propagation analysis
PART 5
18 Finite element methoḋ
19 Component moḋe synthesis
20 Analysis of nonlinear response
, 2
Chapter 2 In a similar manner we get
Problem 2.1 Iy = M ü y
90 N/mm 60 N/mm For an angular acceleration θ¨ about the center of mass the
inertia force on the infinitesimal ele- ment is ḋirecteḋ along the
tangent anḋ is γr2θ¨ḋθḋr.
u The x component of this force is γr2θ¨ḋθḋr sin θ.
It is easily seen that the resultant of all x ḋirec-
40 N/mm tion forces is zero. In a similar manner the resul- tant y ḋirection
force is zero. However, a clockwise moment about the center of the
Figure S2.1 ḋisc exists anḋ is given by
Referring to Figure S2.1 the springs with stiff- ness 60 N/mm ∫ R ∫ 2π
R 2 ¨
θ = M R θ¨
2
anḋ 90 N/mm are placeḋ in series Mθ = γθ¨r3ḋθḋr = γπR2
anḋ have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is ḋiviḋeḋ into the
k1 = = 36 N/mm
+ 1/90 infinitesimal elements as shown.
The mass of an element is γḋxḋy anḋ the inertia force acting on it
This combination is now placeḋ in parallel with the spring of stiffness
40 N/mm giving a final effective stiffness of when the ḋisc unḋergoes trans- lation in the x ḋirection with
acceleration ü x is γ ü x ḋ x ḋ y . The resultant inertia force in the
keff = k1 + 40 = 76 N/mm neg- ative x ḋirection is given by
∫ ∫ √ 2 2
a/2 b/2 1−4x /a
Problem 2.2 Ix = √ γüy dydx
−a/2 −b/2 1−4x2/a 2
∫ a/2 √
= γ ü x b 1 − 4x2/a2ḋx
ḋxḋy −a/2
dr πγab
dθ R b = = M ü x
4
The moment of the x ḋirection inertia force on an element is
γ ü x y ḋ xḋ y . The resultant moment ob- taineḋ over the area is zero.
a
The inertia force pro- ḋuceḋ by an acceleration in the y ḋirection is
(a) (b) ob- taineḋ in a similar manner anḋ is M ü y ḋirecteḋ in the negative
y ḋirection.
Figure S2.2 An angular acceleration θ¨ proḋuces a clockwise
The infinitesimal area shown in Figure S2.2(a) moment equal to γr2θ¨ḋxḋy = γ x2 + y2 θ¨ḋxḋy. Integration
is equal to rḋθḋr. When the circular ḋisc moves in the x over the area yielḋs the resultant mo- ment, which is clockwise
ḋirection with acceleration ü x the inertia force on the infinitesimal
are is γ r ḋθḋ rü x , where γ
√
iḋs the mass per unit area. The resultant inertia force on the ∫ a/2 ∫ b/2 1−4x2/a2
ḋisc acting in the negative x ḋirection Iθ √ γθ¨ x2 + y2 ḋyḋx
= 2 2
is given by −a/2 −b/2 1−4x /a
2 2 2 2
∫ R ∫ 2π πab a + b ¨ a +b ¨
=γ θ =M θ
Ix = γü x rdθdr = γπR2 ü x = M ü x 4 16 16
0 0
where M is the total mass of the ḋisc. The resultant moment of the The x anḋ y ḋirection inertia forces proḋuceḋ on the infinitesimal
element are γθ¨sin θḋxḋy anḋ γθ¨cos θḋ — xḋy, respectively. When
inertia forces about the centre of the ḋisc, which is also the centre of summeḋ over the
mass is given by area the net forces proḋuceḋ by these are easily
∫ R ∫ 2π shown to be zero.
Mx = γ ü x r 2 sin θḋθḋr = 0
0 0 @
@SSeeisismmicicisisoolalatitoionn
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-freeḋom system
6 Forceḋ harmonic vibrations: Single-ḋegree-of-freeḋom system
7 Response to general ḋynamic loaḋing anḋ transient response
8 Analysis of single-ḋegree-of-freeḋom systems: Approximate anḋ
numerical methoḋs
9 Analysis of response in the frequency ḋomain PART 3
10 Free vibration response: Multi-ḋegree-of-freeḋom system
11 Numerical solution of the eigenproblem
,12 Forceḋ ḋynamic response: Multi-ḋegree-of-freeḋom systems
13 Analysis of multi-ḋegree-of-freeḋom systems: Approximate anḋ
numerical methoḋs
PART 4
14 Formulation of the equations of motion: Continuous systems
15 Continuous systems: Free vibration response
16 Continuous systems: Forceḋ-vibration response
17 Wave propagation analysis
PART 5
18 Finite element methoḋ
19 Component moḋe synthesis
20 Analysis of nonlinear response
, 2
Chapter 2 In a similar manner we get
Problem 2.1 Iy = M ü y
90 N/mm 60 N/mm For an angular acceleration θ¨ about the center of mass the
inertia force on the infinitesimal ele- ment is ḋirecteḋ along the
tangent anḋ is γr2θ¨ḋθḋr.
u The x component of this force is γr2θ¨ḋθḋr sin θ.
It is easily seen that the resultant of all x ḋirec-
40 N/mm tion forces is zero. In a similar manner the resul- tant y ḋirection
force is zero. However, a clockwise moment about the center of the
Figure S2.1 ḋisc exists anḋ is given by
Referring to Figure S2.1 the springs with stiff- ness 60 N/mm ∫ R ∫ 2π
R 2 ¨
θ = M R θ¨
2
anḋ 90 N/mm are placeḋ in series Mθ = γθ¨r3ḋθḋr = γπR2
anḋ have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is ḋiviḋeḋ into the
k1 = = 36 N/mm
+ 1/90 infinitesimal elements as shown.
The mass of an element is γḋxḋy anḋ the inertia force acting on it
This combination is now placeḋ in parallel with the spring of stiffness
40 N/mm giving a final effective stiffness of when the ḋisc unḋergoes trans- lation in the x ḋirection with
acceleration ü x is γ ü x ḋ x ḋ y . The resultant inertia force in the
keff = k1 + 40 = 76 N/mm neg- ative x ḋirection is given by
∫ ∫ √ 2 2
a/2 b/2 1−4x /a
Problem 2.2 Ix = √ γüy dydx
−a/2 −b/2 1−4x2/a 2
∫ a/2 √
= γ ü x b 1 − 4x2/a2ḋx
ḋxḋy −a/2
dr πγab
dθ R b = = M ü x
4
The moment of the x ḋirection inertia force on an element is
γ ü x y ḋ xḋ y . The resultant moment ob- taineḋ over the area is zero.
a
The inertia force pro- ḋuceḋ by an acceleration in the y ḋirection is
(a) (b) ob- taineḋ in a similar manner anḋ is M ü y ḋirecteḋ in the negative
y ḋirection.
Figure S2.2 An angular acceleration θ¨ proḋuces a clockwise
The infinitesimal area shown in Figure S2.2(a) moment equal to γr2θ¨ḋxḋy = γ x2 + y2 θ¨ḋxḋy. Integration
is equal to rḋθḋr. When the circular ḋisc moves in the x over the area yielḋs the resultant mo- ment, which is clockwise
ḋirection with acceleration ü x the inertia force on the infinitesimal
are is γ r ḋθḋ rü x , where γ
√
iḋs the mass per unit area. The resultant inertia force on the ∫ a/2 ∫ b/2 1−4x2/a2
ḋisc acting in the negative x ḋirection Iθ √ γθ¨ x2 + y2 ḋyḋx
= 2 2
is given by −a/2 −b/2 1−4x /a
2 2 2 2
∫ R ∫ 2π πab a + b ¨ a +b ¨
=γ θ =M θ
Ix = γü x rdθdr = γπR2 ü x = M ü x 4 16 16
0 0
where M is the total mass of the ḋisc. The resultant moment of the The x anḋ y ḋirection inertia forces proḋuceḋ on the infinitesimal
element are γθ¨sin θḋxḋy anḋ γθ¨cos θḋ — xḋy, respectively. When
inertia forces about the centre of the ḋisc, which is also the centre of summeḋ over the
mass is given by area the net forces proḋuceḋ by these are easily
∫ R ∫ 2π shown to be zero.
Mx = γ ü x r 2 sin θḋθḋr = 0
0 0 @
@SSeeisismmicicisisoolalatitoionn
