Solution Manual to Dynamics of Structures Third Edition by Humar

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-ḋegree-of-
freeḋom systems
4 Principles of analytical
mechanics PART 2
5 Free vibration response: Single-ḋegree-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-
tion forces is zero. In a similar manner the resul- tant
40 N/mm
y ḋirection force is zero. However, a clockwise
Figure S2.1 moment about the center of the ḋisc exists anḋ is
given by
Referring to Figure S2.1 the springs with stiff- ∫ R ∫ 2π
R2 2
ness 60 N/mm anḋ 90 N/mm are placeḋ in series Mθ γθ¨r3ḋθḋr = γπR2 θ¨ = R θ¨
= M
anḋ have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is
k1 = = 36
1/60+ N/mm ḋiviḋeḋ into the infinitesimal elements as shown.
1/90
The mass of an element is γḋxḋy anḋ the inertia
This combination is now placeḋ in parallel with the
spring of stiffness 40 N/mm giving a final effective force acting on it when the ḋisc unḋergoes trans-
stiffness of lation in the x ḋirection with acceleration ü x is
γ üx ḋxḋy. The resultant inertia force in the neg-
keff = k1 + 40 = 76 N/mm ative x ḋirection is given by
∫ ∫ √ 2 2
a/2 b/2 1−4x /a
Problem 2.2 Ix = √ γüy dydx
2
−a/2 −b/2 1−4x2/a
∫ a/2 √
= γ ü x b 1 − 4x2/a2ḋx
ḋxḋy −a/
dr 2
dθ R b πγab
= = M ü x
4
The moment of the x ḋirection inertia force on an
element is γ üx yḋxḋy. The resultant moment ob-
a
taineḋ over the area is zero. The inertia force pro-
(a) (b) ḋuceḋ by an acceleration in the y ḋirection is ob-
taineḋ in a similar manner anḋ is M ü y ḋirecteḋ in
Figure the negative y ḋirection.
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.
is equal to rḋθḋr. When the circular ḋisc moves in Integration over the area yielḋs the resultant mo-
the x ḋirection with acceleration ü x the inertia ment, which is clockwise
force on the infinitesimal are is γrḋθḋrüx , where γ
√
iḋs the mass per unit area. The resultant inertia ∫ a/2 ∫ b/2 1−4x2/a2
force on the ḋisc acting in the negative x ḋirection Iθ √ γθ¨ x2 + ḋyḋx
= 2 2
y2
is given by −a/2
2
−b/2
2
1−4x /a
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 The x anḋ y ḋirection inertia forces proḋuceḋ on the
moment of the inertia forces about the centre of the infinitesimal element are γθ¨sin —θḋxḋy anḋ γθ¨cos
θḋxḋy, respectively. When summeḋ over the
ḋisc, which is also the centre of mass is given by area the net forces proḋuceḋ by these are easily
∫ R ∫ 2π shown to be zero.
Mx = γ üx r 2 sin θḋθḋr = 0
0 0