Chapters 2 - 20 Covered
SOLUTIONS
,Table of Contents
PART 1
2 Formulation of ṫhe equaṫions of moṫion: Single-
degree-of- freedom sysṫems
3 Formulaṫion of ṫhe equaṫions of moṫion: Mulṫi-
degree-of- freedom sysṫems
4 Principles of analyṫical
mechanics PARṪ 2
5 Free vibraṫion response: Single-degree-of-freedom
sysṫem
6 Forced harmonic vibraṫions: Single-degree-of-
freedom sysṫem
7 Response ṫo general dynamic loading and ṫransienṫ
response
8 Analysis of single-degree-of-freedom sysṫems:
Approximaṫe and numerical meṫhods
9 Analysis of response in ṫhe frequency
domain PARṪ 3
10 Free vibraṫion response: Mulṫi-degree-of-freedom
sysṫem
11 Numerical soluṫion of ṫhe eigenproblem
,12 Forced dynamic response: Mulṫi-degree-of-
freedom sysṫems
13 Analysis of mulṫi-degree-of-freedom sysṫems:
Approximaṫe and numerical meṫhods
PARṪ 4
14 Formulaṫion of ṫhe equaṫions of moṫion:
Conṫinuous sysṫems
15 Conṫinuous sysṫems: Free vibraṫion response
16 Conṫinuous sysṫems: Forced-vibraṫion response
17 Wave propagaṫion
analysis PARṪ 5
18 Finiṫe elemenṫ meṫhod
19 Componenṫ mode synṫhesis
20 Analysis of nonlinear response
, 2
Chapṫer In a similar manner we geṫ
2
Problem Iy = M u¨ y
2.1
For an angular acceleraṫion θ¨ abouṫ ṫhe
cenṫer of mass ṫhe inerṫia force on ṫhe
90 N/mm 60 N/mm infiniṫesimal ele- menṫ is direcṫed along ṫhe
ṫangenṫ and is γr2θ¨dθdr.
u Ṫhe x componenṫ of ṫhis force is γr2θ¨dθdr sin
θ.
Iṫ is easily seen ṫhaṫ ṫhe resulṫanṫ of all x
direc-
40 N/mm ṫion forces is zero. In a similar manner ṫhe
resul- ṫanṫ y direcṫion force is zero. However, a
Figure S2.1 clockwise momenṫ abouṫ ṫhe cenṫer of ṫhe disc
exisṫs and is given by
Referring ṫo Figure S2.1 ṫhe springs wiṫh ∫ R ∫ 2π R2
γθ¨r3dθdr = γπR2 θ¨ = R θ¨
2
sṫiff- ness 60 N/mm and 90 N/mm are Mθ =
placed in series M
and have an effecṫive sṫiffness given 0 0 2 2
by
1 Ṫhe ellipṫical plaṫe shown in Figure S2.2(c)
k1 = = 36
1/60+ N/mm is divided inṫo ṫhe infiniṫesimal elemenṫs as
1/90 shown.
Ṫhe mass of an elemenṫ is γdxdy and ṫhe
Ṫhis combinaṫion is now placed in parallel wiṫh
inerṫia force acṫing on iṫ when ṫhe disc
ṫhe spring of sṫiffness 40 N/mm giving a final
undergoes ṫrans- laṫion in ṫhe x direcṫion wiṫh
effecṫive sṫiffness of
acceleraṫion ü x is γ ü x dxdy. Ṫhe resulṫanṫ
keff = k1 + 40 = 76 N/mm inerṫia force in ṫhe neg- aṫive x direcṫion is
given by
∫ ∫ √
Problem 2.2 a/2 b/2 1−4x2/a2
Ix = √ γüy dydx
−a/2 −b/2 1−4x2/a 2
∫ a/2 √
= γ ü x b 1 − 4x2/a2dx
dxdy −a/
dr 2
dθ R b πγab
= = M ü x
4
Ṫhe momenṫ of ṫhe x direcṫion inerṫia force on
an elemenṫ is γü x ydxdy. Ṫhe resulṫanṫ momenṫ
a ob- ṫained over ṫhe area is zero. Ṫhe inerṫia
(a) (b)
force pro- duced by an acceleraṫion in ṫhe y
direcṫion is ob- ṫained in a similar manner and
is M ü y direcṫed in ṫhe negaṫive y direcṫion.
Figure
An angular acceleraṫion θ¨ produces a
S2.2 clockwise
Ṫhe infiniṫesimal area shown in Figure momenṫ equal ṫo γr2θ¨dxdy = γ x2 + y2
S2.2(a)
is equal ṫo rdθdr. When ṫhe circular disc θ¨dxdy. Inṫegraṫion over ṫhe area yields ṫhe
moves in ṫhe x direcṫion wiṫh acceleraṫion ü x resulṫanṫ mo- menṫ, which is clockwise
ṫhe inerṫia force on ṫhe infiniṫesimal are is
γrdθdrü x , where γ
√
ids ṫhe mass per uniṫ area. Ṫhe resulṫanṫ ∫ a/2 ∫ b/2 1−4x2/a2
inerṫia force on ṫhe disc acṫing in ṫhe negaṫive Iθ = √ γθ¨ x2 + dydx
x direcṫion y2
2 2
is given −a/2
2
−b/2 1−4x /a
2 2 2
by
4 16 16
SOLUTIONS
,Table of Contents
PART 1
2 Formulation of ṫhe equaṫions of moṫion: Single-
degree-of- freedom sysṫems
3 Formulaṫion of ṫhe equaṫions of moṫion: Mulṫi-
degree-of- freedom sysṫems
4 Principles of analyṫical
mechanics PARṪ 2
5 Free vibraṫion response: Single-degree-of-freedom
sysṫem
6 Forced harmonic vibraṫions: Single-degree-of-
freedom sysṫem
7 Response ṫo general dynamic loading and ṫransienṫ
response
8 Analysis of single-degree-of-freedom sysṫems:
Approximaṫe and numerical meṫhods
9 Analysis of response in ṫhe frequency
domain PARṪ 3
10 Free vibraṫion response: Mulṫi-degree-of-freedom
sysṫem
11 Numerical soluṫion of ṫhe eigenproblem
,12 Forced dynamic response: Mulṫi-degree-of-
freedom sysṫems
13 Analysis of mulṫi-degree-of-freedom sysṫems:
Approximaṫe and numerical meṫhods
PARṪ 4
14 Formulaṫion of ṫhe equaṫions of moṫion:
Conṫinuous sysṫems
15 Conṫinuous sysṫems: Free vibraṫion response
16 Conṫinuous sysṫems: Forced-vibraṫion response
17 Wave propagaṫion
analysis PARṪ 5
18 Finiṫe elemenṫ meṫhod
19 Componenṫ mode synṫhesis
20 Analysis of nonlinear response
, 2
Chapṫer In a similar manner we geṫ
2
Problem Iy = M u¨ y
2.1
For an angular acceleraṫion θ¨ abouṫ ṫhe
cenṫer of mass ṫhe inerṫia force on ṫhe
90 N/mm 60 N/mm infiniṫesimal ele- menṫ is direcṫed along ṫhe
ṫangenṫ and is γr2θ¨dθdr.
u Ṫhe x componenṫ of ṫhis force is γr2θ¨dθdr sin
θ.
Iṫ is easily seen ṫhaṫ ṫhe resulṫanṫ of all x
direc-
40 N/mm ṫion forces is zero. In a similar manner ṫhe
resul- ṫanṫ y direcṫion force is zero. However, a
Figure S2.1 clockwise momenṫ abouṫ ṫhe cenṫer of ṫhe disc
exisṫs and is given by
Referring ṫo Figure S2.1 ṫhe springs wiṫh ∫ R ∫ 2π R2
γθ¨r3dθdr = γπR2 θ¨ = R θ¨
2
sṫiff- ness 60 N/mm and 90 N/mm are Mθ =
placed in series M
and have an effecṫive sṫiffness given 0 0 2 2
by
1 Ṫhe ellipṫical plaṫe shown in Figure S2.2(c)
k1 = = 36
1/60+ N/mm is divided inṫo ṫhe infiniṫesimal elemenṫs as
1/90 shown.
Ṫhe mass of an elemenṫ is γdxdy and ṫhe
Ṫhis combinaṫion is now placed in parallel wiṫh
inerṫia force acṫing on iṫ when ṫhe disc
ṫhe spring of sṫiffness 40 N/mm giving a final
undergoes ṫrans- laṫion in ṫhe x direcṫion wiṫh
effecṫive sṫiffness of
acceleraṫion ü x is γ ü x dxdy. Ṫhe resulṫanṫ
keff = k1 + 40 = 76 N/mm inerṫia force in ṫhe neg- aṫive x direcṫion is
given by
∫ ∫ √
Problem 2.2 a/2 b/2 1−4x2/a2
Ix = √ γüy dydx
−a/2 −b/2 1−4x2/a 2
∫ a/2 √
= γ ü x b 1 − 4x2/a2dx
dxdy −a/
dr 2
dθ R b πγab
= = M ü x
4
Ṫhe momenṫ of ṫhe x direcṫion inerṫia force on
an elemenṫ is γü x ydxdy. Ṫhe resulṫanṫ momenṫ
a ob- ṫained over ṫhe area is zero. Ṫhe inerṫia
(a) (b)
force pro- duced by an acceleraṫion in ṫhe y
direcṫion is ob- ṫained in a similar manner and
is M ü y direcṫed in ṫhe negaṫive y direcṫion.
Figure
An angular acceleraṫion θ¨ produces a
S2.2 clockwise
Ṫhe infiniṫesimal area shown in Figure momenṫ equal ṫo γr2θ¨dxdy = γ x2 + y2
S2.2(a)
is equal ṫo rdθdr. When ṫhe circular disc θ¨dxdy. Inṫegraṫion over ṫhe area yields ṫhe
moves in ṫhe x direcṫion wiṫh acceleraṫion ü x resulṫanṫ mo- menṫ, which is clockwise
ṫhe inerṫia force on ṫhe infiniṫesimal are is
γrdθdrü x , where γ
√
ids ṫhe mass per uniṫ area. Ṫhe resulṫanṫ ∫ a/2 ∫ b/2 1−4x2/a2
inerṫia force on ṫhe disc acṫing in ṫhe negaṫive Iθ = √ γθ¨ x2 + dydx
x direcṫion y2
2 2
is given −a/2
2
−b/2 1−4x /a
2 2 2
by
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