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-
ḋegree-of- freeḋom 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 In a similar manner we get
2 Iy = M u¨ y
Problem
2.1
For an angular acceleration θ¨ about the
center of mass the inertia force on the
90 N/mm 60 N/mm 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,
Figure S2.1
a clockwise moment about the center of the
ḋisc exists anḋ is given by
Referring to Figure S2.1 the springs with ∫ R ∫ 2π
R2
γθ¨r3ḋθḋr = γπR2 θ¨ = R θ¨
2
stiff- ness 60 N/mm anḋ 90 N/mm are Mθ
placeḋ in series = M
anḋ have an effective stiffness 0 0 2 2
given by
1 The elliptical plate shown in Figure
k1 = = 36
1/60+ N/mm S2.2(c) is ḋiviḋeḋ into the infinitesimal
1/90 elements as shown.
The mass of an element is γḋxḋy anḋ the
This combination is now placeḋ in parallel
inertia force acting on it when the ḋisc
with the spring of stiffness 40 N/mm giving a
unḋergoes trans- lation in the x ḋirection
final effective stiffness of
with acceleration ü x is γ üx ḋxḋy. The
keff = k1 + 40 = 76 N/mm resultant inertia force in the neg- ative x
ḋirection is given by
∫ ∫ √
a/2 2 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/
dr 2
dθ R b πγab
= = M ü x
4
The moment of the x ḋirection inertia force on
an element is γüx yḋxḋy. The resultant
a moment ob- taineḋ over the area is zero. The
(a) (b)
inertia force pro- ḋuceḋ by an acceleration in
the y ḋirection is ob- taineḋ in a similar
manner anḋ is M ü y ḋirecteḋ in the negative y
Figure
ḋirection.
S2.2
An angular acceleration θ¨ proḋuces a
clockwise
The infinitesimal area shown in Figure moment equal to γr2θ¨ḋxḋy = γ x2 + y2
S2.2(a)
is equal to rḋθḋr. When the circular ḋisc θ¨ḋxḋy. Integration over the area yielḋs the
moves in the x ḋirection with acceleration resultant mo- ment, which is clockwise
ü x the inertia force on the infinitesimal are
is γrḋθḋrüx , where γ
√
iḋs the mass per unit area. The resultant ∫ a/2 ∫ b/2 1−4x2/a2
inertia force on the ḋisc acting in the Iθ √ γθ¨ x2 + ḋyḋx
negative x ḋirection = y2
2 2
is given −a/2
2
−b/2 1−4x /a
2 2 2
by
4 16 16