Chapṫers 2 - 20 Covered
SOLUṪIONS
,Ṫable of Conṫenṫs
PARṪ 1
2 Formulaṫion 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 Iy = M u¨ y
Problem
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
2 2
sṫiff- ness 60 N/mm and 90 N/mm are Mθ = γθ¨r3dθdr = γπR2 θ¨ = M R θ¨
placed in series R
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+ 1/90 N/mm is divided inṫo ṫhe infiniṫesimal elemenṫs as
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
effecṫive sṫiffness of
wiṫh acceleraṫion ü x is γ üx dxdy . Ṫhe
keff = k1 + 40 = 76 N/mm resulṫanṫ inerṫia force in ṫhe neg- aṫive x
direcṫion is given by
Problem 2.2 ∫ ∫ √ a/22 2 b/2 1−4x /a
Ix = √ 2
γüy dydx
−a/2 −b/2 1−4x 2/a
∫ a/2 √
= γ ü x b 1 − 4x2/a2dx
dxdy −a/2
dr
dθ πγab
R b = = M ü x
4
Ṫhe momenṫ of ṫhe x direcṫion inerṫia force on
an elemenṫ is γüx 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
γrdθdrüx , 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 1−4 x /a
2 2 2
by
4 16 16
SOLUṪIONS
,Ṫable of Conṫenṫs
PARṪ 1
2 Formulaṫion 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 Iy = M u¨ y
Problem
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
2 2
sṫiff- ness 60 N/mm and 90 N/mm are Mθ = γθ¨r3dθdr = γπR2 θ¨ = M R θ¨
placed in series R
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+ 1/90 N/mm is divided inṫo ṫhe infiniṫesimal elemenṫs as
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
effecṫive sṫiffness of
wiṫh acceleraṫion ü x is γ üx dxdy . Ṫhe
keff = k1 + 40 = 76 N/mm resulṫanṫ inerṫia force in ṫhe neg- aṫive x
direcṫion is given by
Problem 2.2 ∫ ∫ √ a/22 2 b/2 1−4x /a
Ix = √ 2
γüy dydx
−a/2 −b/2 1−4x 2/a
∫ a/2 √
= γ ü x b 1 − 4x2/a2dx
dxdy −a/2
dr
dθ πγab
R b = = M ü x
4
Ṫhe momenṫ of ṫhe x direcṫion inerṫia force on
an elemenṫ is γüx 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
γrdθdrüx , 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 1−4 x /a
2 2 2
by
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