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-freedom system
6 Forced harmonic vibrations: Single-degree-of-freedom
system
7 Response to general dynamic loading and transient response
8 Analysis of single-degree-of-freedom systems: Approximate
and numerical methods
9 Analysis of response in the frequency domain
PART 3
10 Free vibration response: Multi-degree-of-freedom system
11 Numerical solution of the eigenproblem
,12 Forced dynamic response: Multi-degree-of-freedom
systems
13 Analysis of multi-degree-of-freedom systems: Approximate
and numerical methods
PART 4
14 Formulation of the equations of motion: Continuous
systems
15 Continuous systems: Free vibration response
16 Continuous systems: Forced-vibration response
17 Wave propagation analysis
PART 5
18 Finite element method
19 Component mode synthesis
20 Analysis of nonlinear response
, 2
Chapter 2 In a similar manner we get
Problem 2.1 Iy = M üy
For an angular acceleration θ̈ about the center
90 N/mm 60 N/mm of mass the inertia force on the infinitesimal ele-
ment is directed along the tangent and is γr2 θ̈dθdr.
u
The x component of this force is γr2 θ̈dθdr sin θ.
It is easily seen that the resultant of all x direc-
tion forces is zero. In a similar manner the resul-
40 N/mm tant y direction force is zero. However, a clockwise
moment about the center of the disc exists and is
Figure S2.1 given by
Referring to Figure S2.1 the springs with stiff- R 2π
R2 R2
ness 60 N/mm and 90 N/mm are placed in series Mθ = γ θ̈r3 dθdr = γπR2 θ̈ = M θ̈
and have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is
k1 = = 36 N/mm
1/60 + 1/90 divided into the infinitesimal elements as shown.
The mass of an element is γdxdy and the inertia
This combination is now placed in parallel with the
force acting on it when the disc undergoes trans-
spring of stiffness 40 N/mm giving a final effective
lation in the x direction with acceleration üx is
stiffness of
γ üx dxdy. The resultant inertia force in the neg-
keff = k1 + 40 = 76 N/mm ative x direction is given by
√
a/2 b/2 1−4x2 /a2
Problem 2.2 Ix = √ γ üy dydx
−a/2 −b/2 1−4x2 /a2
a/2
= γ üx b 1 − 4x2 /a2 dx
dxdy −a/2
dr
dθ R πγab
b = = M üx
4
The moment of the x direction inertia force on an
element is γ üx ydxdy. The resultant moment ob-
a tained over the area is zero. The inertia force pro-
duced by an acceleration in the y direction is ob-
(a) (b)
tained in a similar manner and is M üy directed in
Figure S2.2 the negative y direction.
An angular acceleration θ̈ produces
a clockwise
The infinitesimal area shown in Figure S2.2(a) moment equal to γr2 θ̈dxdy = γ x2 + y 2 θ̈dxdy.
is equal to rdθdr. When the circular disc moves Integration over the area yields the resultant mo-
in the x direction with acceleration üx the inertia ment, which is clockwise
force on the infinitesimal are is γrdθdrüx , where γ
ids the mass per unit area. The resultant inertia √
a/2 b/2 1−4x2 /a2
force on the disc acting in the negative x direction Iθ = √ γ θ̈ x2 + y 2 dydx
is given by −a/2 −b/2 1−4x2 /a2
2 2
R 2π πab a + b a2 + b2
=γ θ̈ = M θ̈
Ix = γ üx rdθdr = γπR2 üx = M üx 4 16 16
0 0
The x and y direction inertia forces produced on
where M is the total mass of the disc. The resultant
the infinitesimal element are −γ θ̈ sin θdxdy and
moment of the inertia forces about the centre of the
γ θ̈ cos θdxdy, respectively. When summed over the
disc, which is also the centre of mass is given by
area the net forces produced by these are easily
R 2π
shown to be zero.
Mx = γ üx r2 sin θdθdr = 0
0 0 @Seismicisolation
@Seismicisolation
, 3
Problem 2.3 The displacement at location A, shown in Figure
S2.4, due to a unit load is given by
L3b L3s L2 Ls
∆A = + + b
3EIb 3EIs GJs
3
12 × 768 363 × 64
= 6
+
3 × 30 × 10 3 × 30 × 106 × π
2
12 × 36 × 32
+
12 × 106 × π
= 0.01475 + 0.01056 + 0.0044
= 0.0297 in.
θ 1 lb.
u Equivalent spring stiffness = 0.0297 = 33.67 in.
2
10
Mass = 386.4 = 0.02588 lb.s
in.
2uk mü T Mü
The equation of motion is given by
0.02588ü + 33.67u = 0
I0θ̈ or
ü + 1300u = 0
T
Figure S2.3 Problem 2.5
As the center of the pulley moves a distance u,
the spring is stretched by 2u and the pulley rotates
M(L1 ⫹ R)θ̈
through u/R. Referring to the free body diagrams
shown in Figure S2.3, the conditions of equilibrium
are
2uk × 2R + müR + I0 θ̈ − T R = 0 B
I0θ̈
T + M ü = 0
Substituting I0 = mR2 /2 and eliminating T gives
3m
M+ ü + 4ku = 0 θ
2
cd θ̇
Problem 2.4 kL2θ
Figure S2.5
3⬘0⬙ Taking moments about B shown in Figure
Light, flexible
S2.5, and noting that I0 = M R2 /2
Light, flexible
1⬘0⬙
2 M R2
Wheel M (L1 + R) θ̈ + θ̈ + cd2 θ̇ + kL22 θ = 0
2
u
or
Figure S2.4
3 2
3 M L21 + R + 2L1 R θ̈ + cd2 θ̇ + kL22 θ = 0
1 1 1 2
Ib = ×1× =
12 4 768
π π
Is = , Js =
64 32 @Seismicisolation
@Seismicisolation
, 4
Problem 2.6 Problem 2.8
ug(t ) u
d Mb θ̈ kLθ
dx
x
mL2
θ̈
3 θ
A1 A2
cdθ̇
Mbθ̈ Input Output
MR 2
θ̈
2
L
Figure S2.8
Figure S2.6
Taking moments about A shown in Figure S2.6 dξ 1
= , A(ξ) = A1 + (A2 − A1 ) ξ
mL2 dx L
+ M b θ̈ + cd2 θ̇ + kL2 θ = 0
2
dψ 1
3 = (2 − 2ξ) , m(ξ) = ρA(ξ)
dx L
When the flywheel is braked, an additional rota-
tional inertia term will exist. Moment equilibrium where ρ is the mass density.
gives L
m∗ = m(x) {ψ(x)} dx
2
mL2 2 R2
+M b + θ̈ + cd2 θ̇ + kL2 θ = 0
0
3 2 1
= ρL {A1 + (A2 − A1 ) ξ}
0
2
Problem 2.7 4ξ + ξ 4 − 4ξ 3 dξ
Boundary conditions ρL
= (5A1 + 11A2 )
30
ψ = 0 L 2
dψ(x)
dψ x=0 k∗ = EA(x) dx
= 0 0 dx
dx
E 1
= {A1 + (A2 − A1 ) ξ}
ψ = 0 L 0
x=L
d2 ψ 4 + 4ξ 2 − 8ξ dξ
EI 2 = 0
dx E 1
where = A1 + A2
L 3
dψ dψ 1 1 3 L
= × = 8ξ − 15ξ 2 + 6ξ p∗ = − m(x)üg ψ(x)dx
dx dξ L L
0
2
d ψ 1 1
= 2 24ξ 2 − 30ξ + 6
dx 2 L =− ρLüg {A1 + (A2 − A1 ) ξ} 2ξ − ξ 2 dξ
0
All four boundary conditions are satisfied
1 5
L = −ρLüg A1 + A2
2 4 12
m∗ = m {ψ(x)} dx
0
1
= mL
2
{ψ(ξ)} dξ = 0.03016mL Problem 2.9
0
L 2 Refer to Figure S2.9 where the inertia force
d2 ψ(x) 2¨ut
k =∗
EI dx on an infinitesimal element is shown as m̄ ∂∂x 2 dx.
dx2
0
The total displacement ut = u + ug , where u is
1 2
EI d2 ψ(ξ) EI the displacement relative to the ground and ug is
= dξ = 7.2
L3 dξ 2 L3 the displacement of the ground. The inertia force
0 @Seismicisolation
@Seismicisolation
, 5
thus becomes m̄ (z̈(t)ψ(x) + üg ) dx. The first term Equation of motion:
within the parentheses produces the generalized
mass m∗ while the second term produces the gen- mL π 4 EI πvt
z̈ + z = F sin
eralized force p∗ as follows 2 2L3 L
L
33
∗
m = m̄[ψ(x)]2 dx = m̄L
0 140
L
3
p∗ = −m̄üg ψ(x)dx = − m̄Lüg
0 8 F
The generalized stiffness is given by
L v
′′ 2 3EI
∗
k = EI ψ (x) dx = 3
0 L x
u
The geometric stiffness produced by the axial force
L
S(x) = (L − x)m̄g is obtained from
L ′ 2 3 Figure S2.10
kG = S(x) ψ (x) dx = m̄g
0 8
The equation of motion including the gravity effect Problem 2.11
is
The mass matrix M and the stiffness matrix
33 3EI 3 3
m̄Lz̈ + − m̄g z = − m̄Lg K are given by
140 L3 8 8
2 0 0
x
M = 0 2 0 kip s2 /in.
0 0 1
1000 −500 0
K = −500 750 −250 kips/in.
0 −250 250
s(x)
2ut
—d dx With ψ T = [ 1 2 3 ], the generalized mass
m dx M
dt2 d2ut
⫽z̈(t) ψ(x)⫹üg m and generalized stiffness k ∗ are obtained from
∗
d2
ψ(x) ∂2u
M ⫽ EI 2 ⫽EI z(t) ψ″(x) m∗ = ψ T Mψ = 19 kip s2 /in.
x ∂t
—
S(x) ⫽(L ⫺x) m g k ∗ = ψ T Kψ = 1250 kips/in.
The generalized force is obtained from
ug 1
p∗ = −ψ T M 1 üg = −1170 sin(6πt) kips
Figure S2.9 1
Problem 2.10 The equation of motion is given by
u = zψ(x) 19z̈ + 1250z = −1170 sin(6πt)
πx d2 ψ π2 πx
ψ = sin , 2
= − 2 sin
L dx L L Problem 2.12
L
∗ 2 πx mL
m = m sin dx = Taking moments about A shown in Figure
0 L 2
L S2.12
π4 πx π 4 EI
k∗ = EI 4 sin2 dx = 3
ml2 θ̈ + 2ka2 θ + mgl sin θ = 0
0 L L 2L
L For small vibrations sin θ ≈ θ, hence
πx
p∗ = F δ (x − vt) sin dx
0 L
ml2 θ̈ + 2ka2 + mgl θ = 0
πvt
= F sin
L @Seismicisolation
@Seismicisolation
, 6
A
θ
2 ka θ
S
ml θ̈
mg
Figure S2.12
@Seismicisolation
@Seismicisolation
, 7
Chapter 3 Collectively, the four equations can be expressed in
the matrix form
Problem 3.1
M∗ ü + C∗ u̇ + K∗ u = 0
uT = [ q1 aq2 q3 q4 ]
q1 W
q2
w, r g
W r 2
M∗ =
g a2
w
a a g
w
g
q3 q4
2 0 −1 −1
ks cs ks cs
0 2 1 −1
w w C∗ = cS
−1 1 1 0
kT kT
−1 −1 0 1
2 0 −1 −1
0 2 1 −1
K∗ = kS
−1 1 1 + kkTS 0
cs(q· 3 ⫺ q· 1 ⫹ q· 2a) cs(q· 4 ⫺q· 1 ⫺q· 2a) −1 −1 0 1 + kkTS
ks(q3 ⫺ q1 ⫹ q2a)
ks(q4 ⫺q1 ⫺ q2a)
Problem 3.2
kT q3 w kT q4 w
q̈ q̈
g 3 g 4 mg
I0θ̈ mhθ̈
W 2
r q̈ 2
g
W
q̈
g 1 h
cs(q· 3 ⫺ q· 1 ⫹ q· 2a) cs(q· 4 ⫺q· 1 ⫺q· 2a) θ
⫹ ks(q3 ⫺ q1 ⫹q2a) ⫹ ks(q4 ⫺ q1 ⫺q2a)
θ
A
Figure S3.1 u
ug
The degrees of freedom are identified in Figure · ·
c(u· ⫺bθ) k(u ⫺bθ) k(u ⫹bθ) c(u·⫹bθ)
S3.1. The free-body diagrams are also shown. For
force balance on each of the front and rear axles b b
w
q̈4 + cS (q̇4 − q̇1 − q̇2 a) Figure S3.2
g
+ kS (q4 − q1 − q2 a) + kT q4 = 0 Coordinates ut and θ measure the vertical mo-
w tion and rotation of the base. The forces on the
q̈3 + cS (q̇3 − q̇1 + q̇2 a) structure are identified in Figure S3.2. Assuming
g
that ut is measured from the position of equilib-
+ kS (q3 − q1 + q2 a) + kT q3 = 0
rium under gravity load, the condition of vertical
For equilibrium of vertical forces on the vehicle force balance gives
body
W
q̈1 + cS (q̇3 + q̇4 − 2q̇1 ) müt + 2cu̇ + 2ku = 0 (1)
g
+ kS (q3 + q4 − 2q1 ) = 0 or
Taking moments about the mass center of the ve- mü + 2cu̇ + 2ku = −müg
hicle body
Taking moments about A and assuming that θ is
W 2 small
r q̈2 + cS a (q̇3 − q̇4 + 2aq̇2 )
g
+ kS a (q3 − q4 + 2aq2 ) = 0 0 I + mh2 θ̈ + 2cb2 θ̇ + 2kb2 θ − mghθ = 0 (2)
@Seismicisolation
@Seismicisolation
, 8
Problem 3.3 Problem 3.4
For frame A
a2 + b2
I0 = m
12 u1 = 12θ2
Hence the mass matrix is given by
u2 = 12θ4
m 0 0
or
M=0 m 0 θ1
2
+b2
0 0 m a 12 u1 0 12 0 0 θ2
=
u2 0 0 0 12 θ3
k1 ⫻1 θ4
k21 Hence
b
k1 0 12 0 0
k11 2 TA =
k31 0 0 0 12
k23
k1 ⫻1 a KA = TTA kA TA
k2
2 0 0 0 0
k33 k13
0 540 0 −240
= 144
k22 0 0 0 0
b
k1 0 −240 0 240
2
k32 k12 k2 ⫻1
For frame B
K B = KA
Figure S3.3 For frame C
To obtain the first column of stiffness matrix,
1 −15 0 0
give a unit displacement u, with v and w each zero. Tc =
0 0 1 −15
The resisting forces provided by frames A and B
and the external forces required to maintain the 860 −12900 −360 5400
displacement are shown in Figure S3.3. For equi- −12900 193500 5400 −81000
KC =
librium k11 = 2k1 , k21 = 0, k31 = 0. −360 5400 360 −5400
5400 −81000 −5400 193500
For u = 0, v = 1 and θ = 0 equilibrium gives
Assembly gives
k2 a
k12 = 0, k22 = k2 , k32 =
2 860 −12900 −360 5400
−12900 349020 5400 −150120
In a similar manner, for u = 0, v = 0, and θ = 1 K=
−360 5400 360 −5400
k13 = 0 5400 −151020 −5400 262620
k2 a 2
k23 = Mass of slab m = 100×30×24
32.2 = 2.236 kip.s
ft .
2
b2 a2 Mass of moment of inertia
k33 = 2k1 + k2
4 4 a2 + b2 302 + 242
I0 = m = 2.236 ×
On assembly, the following stiffness matrix is ob- 12 12
tained = 275.03 kip.ft.s2
2k1 0 0 Hence
k2 a
K= 0 k2 2
k2 a k1 b 2
k2 a2 2.236 0 0 0
0 2 + 4
2 0 275.03 0 0
M=
0 0 2.236 0
0 0 0 275.03
30 × 12
p1 = × p0 sin Ωt = 0.360p0 sin Ωt, p2 = 0
1000
@Seismicisolation
@Seismicisolation
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-freedom system
6 Forced harmonic vibrations: Single-degree-of-freedom
system
7 Response to general dynamic loading and transient response
8 Analysis of single-degree-of-freedom systems: Approximate
and numerical methods
9 Analysis of response in the frequency domain
PART 3
10 Free vibration response: Multi-degree-of-freedom system
11 Numerical solution of the eigenproblem
,12 Forced dynamic response: Multi-degree-of-freedom
systems
13 Analysis of multi-degree-of-freedom systems: Approximate
and numerical methods
PART 4
14 Formulation of the equations of motion: Continuous
systems
15 Continuous systems: Free vibration response
16 Continuous systems: Forced-vibration response
17 Wave propagation analysis
PART 5
18 Finite element method
19 Component mode synthesis
20 Analysis of nonlinear response
, 2
Chapter 2 In a similar manner we get
Problem 2.1 Iy = M üy
For an angular acceleration θ̈ about the center
90 N/mm 60 N/mm of mass the inertia force on the infinitesimal ele-
ment is directed along the tangent and is γr2 θ̈dθdr.
u
The x component of this force is γr2 θ̈dθdr sin θ.
It is easily seen that the resultant of all x direc-
tion forces is zero. In a similar manner the resul-
40 N/mm tant y direction force is zero. However, a clockwise
moment about the center of the disc exists and is
Figure S2.1 given by
Referring to Figure S2.1 the springs with stiff- R 2π
R2 R2
ness 60 N/mm and 90 N/mm are placed in series Mθ = γ θ̈r3 dθdr = γπR2 θ̈ = M θ̈
and have an effective stiffness given by 0 0 2 2
1 The elliptical plate shown in Figure S2.2(c) is
k1 = = 36 N/mm
1/60 + 1/90 divided into the infinitesimal elements as shown.
The mass of an element is γdxdy and the inertia
This combination is now placed in parallel with the
force acting on it when the disc undergoes trans-
spring of stiffness 40 N/mm giving a final effective
lation in the x direction with acceleration üx is
stiffness of
γ üx dxdy. The resultant inertia force in the neg-
keff = k1 + 40 = 76 N/mm ative x direction is given by
√
a/2 b/2 1−4x2 /a2
Problem 2.2 Ix = √ γ üy dydx
−a/2 −b/2 1−4x2 /a2
a/2
= γ üx b 1 − 4x2 /a2 dx
dxdy −a/2
dr
dθ R πγab
b = = M üx
4
The moment of the x direction inertia force on an
element is γ üx ydxdy. The resultant moment ob-
a tained over the area is zero. The inertia force pro-
duced by an acceleration in the y direction is ob-
(a) (b)
tained in a similar manner and is M üy directed in
Figure S2.2 the negative y direction.
An angular acceleration θ̈ produces
a clockwise
The infinitesimal area shown in Figure S2.2(a) moment equal to γr2 θ̈dxdy = γ x2 + y 2 θ̈dxdy.
is equal to rdθdr. When the circular disc moves Integration over the area yields the resultant mo-
in the x direction with acceleration üx the inertia ment, which is clockwise
force on the infinitesimal are is γrdθdrüx , where γ
ids the mass per unit area. The resultant inertia √
a/2 b/2 1−4x2 /a2
force on the disc acting in the negative x direction Iθ = √ γ θ̈ x2 + y 2 dydx
is given by −a/2 −b/2 1−4x2 /a2
2 2
R 2π πab a + b a2 + b2
=γ θ̈ = M θ̈
Ix = γ üx rdθdr = γπR2 üx = M üx 4 16 16
0 0
The x and y direction inertia forces produced on
where M is the total mass of the disc. The resultant
the infinitesimal element are −γ θ̈ sin θdxdy and
moment of the inertia forces about the centre of the
γ θ̈ cos θdxdy, respectively. When summed over the
disc, which is also the centre of mass is given by
area the net forces produced by these are easily
R 2π
shown to be zero.
Mx = γ üx r2 sin θdθdr = 0
0 0 @Seismicisolation
@Seismicisolation
, 3
Problem 2.3 The displacement at location A, shown in Figure
S2.4, due to a unit load is given by
L3b L3s L2 Ls
∆A = + + b
3EIb 3EIs GJs
3
12 × 768 363 × 64
= 6
+
3 × 30 × 10 3 × 30 × 106 × π
2
12 × 36 × 32
+
12 × 106 × π
= 0.01475 + 0.01056 + 0.0044
= 0.0297 in.
θ 1 lb.
u Equivalent spring stiffness = 0.0297 = 33.67 in.
2
10
Mass = 386.4 = 0.02588 lb.s
in.
2uk mü T Mü
The equation of motion is given by
0.02588ü + 33.67u = 0
I0θ̈ or
ü + 1300u = 0
T
Figure S2.3 Problem 2.5
As the center of the pulley moves a distance u,
the spring is stretched by 2u and the pulley rotates
M(L1 ⫹ R)θ̈
through u/R. Referring to the free body diagrams
shown in Figure S2.3, the conditions of equilibrium
are
2uk × 2R + müR + I0 θ̈ − T R = 0 B
I0θ̈
T + M ü = 0
Substituting I0 = mR2 /2 and eliminating T gives
3m
M+ ü + 4ku = 0 θ
2
cd θ̇
Problem 2.4 kL2θ
Figure S2.5
3⬘0⬙ Taking moments about B shown in Figure
Light, flexible
S2.5, and noting that I0 = M R2 /2
Light, flexible
1⬘0⬙
2 M R2
Wheel M (L1 + R) θ̈ + θ̈ + cd2 θ̇ + kL22 θ = 0
2
u
or
Figure S2.4
3 2
3 M L21 + R + 2L1 R θ̈ + cd2 θ̇ + kL22 θ = 0
1 1 1 2
Ib = ×1× =
12 4 768
π π
Is = , Js =
64 32 @Seismicisolation
@Seismicisolation
, 4
Problem 2.6 Problem 2.8
ug(t ) u
d Mb θ̈ kLθ
dx
x
mL2
θ̈
3 θ
A1 A2
cdθ̇
Mbθ̈ Input Output
MR 2
θ̈
2
L
Figure S2.8
Figure S2.6
Taking moments about A shown in Figure S2.6 dξ 1
= , A(ξ) = A1 + (A2 − A1 ) ξ
mL2 dx L
+ M b θ̈ + cd2 θ̇ + kL2 θ = 0
2
dψ 1
3 = (2 − 2ξ) , m(ξ) = ρA(ξ)
dx L
When the flywheel is braked, an additional rota-
tional inertia term will exist. Moment equilibrium where ρ is the mass density.
gives L
m∗ = m(x) {ψ(x)} dx
2
mL2 2 R2
+M b + θ̈ + cd2 θ̇ + kL2 θ = 0
0
3 2 1
= ρL {A1 + (A2 − A1 ) ξ}
0
2
Problem 2.7 4ξ + ξ 4 − 4ξ 3 dξ
Boundary conditions ρL
= (5A1 + 11A2 )
30
ψ = 0 L 2
dψ(x)
dψ x=0 k∗ = EA(x) dx
= 0 0 dx
dx
E 1
= {A1 + (A2 − A1 ) ξ}
ψ = 0 L 0
x=L
d2 ψ 4 + 4ξ 2 − 8ξ dξ
EI 2 = 0
dx E 1
where = A1 + A2
L 3
dψ dψ 1 1 3 L
= × = 8ξ − 15ξ 2 + 6ξ p∗ = − m(x)üg ψ(x)dx
dx dξ L L
0
2
d ψ 1 1
= 2 24ξ 2 − 30ξ + 6
dx 2 L =− ρLüg {A1 + (A2 − A1 ) ξ} 2ξ − ξ 2 dξ
0
All four boundary conditions are satisfied
1 5
L = −ρLüg A1 + A2
2 4 12
m∗ = m {ψ(x)} dx
0
1
= mL
2
{ψ(ξ)} dξ = 0.03016mL Problem 2.9
0
L 2 Refer to Figure S2.9 where the inertia force
d2 ψ(x) 2¨ut
k =∗
EI dx on an infinitesimal element is shown as m̄ ∂∂x 2 dx.
dx2
0
The total displacement ut = u + ug , where u is
1 2
EI d2 ψ(ξ) EI the displacement relative to the ground and ug is
= dξ = 7.2
L3 dξ 2 L3 the displacement of the ground. The inertia force
0 @Seismicisolation
@Seismicisolation
, 5
thus becomes m̄ (z̈(t)ψ(x) + üg ) dx. The first term Equation of motion:
within the parentheses produces the generalized
mass m∗ while the second term produces the gen- mL π 4 EI πvt
z̈ + z = F sin
eralized force p∗ as follows 2 2L3 L
L
33
∗
m = m̄[ψ(x)]2 dx = m̄L
0 140
L
3
p∗ = −m̄üg ψ(x)dx = − m̄Lüg
0 8 F
The generalized stiffness is given by
L v
′′ 2 3EI
∗
k = EI ψ (x) dx = 3
0 L x
u
The geometric stiffness produced by the axial force
L
S(x) = (L − x)m̄g is obtained from
L ′ 2 3 Figure S2.10
kG = S(x) ψ (x) dx = m̄g
0 8
The equation of motion including the gravity effect Problem 2.11
is
The mass matrix M and the stiffness matrix
33 3EI 3 3
m̄Lz̈ + − m̄g z = − m̄Lg K are given by
140 L3 8 8
2 0 0
x
M = 0 2 0 kip s2 /in.
0 0 1
1000 −500 0
K = −500 750 −250 kips/in.
0 −250 250
s(x)
2ut
—d dx With ψ T = [ 1 2 3 ], the generalized mass
m dx M
dt2 d2ut
⫽z̈(t) ψ(x)⫹üg m and generalized stiffness k ∗ are obtained from
∗
d2
ψ(x) ∂2u
M ⫽ EI 2 ⫽EI z(t) ψ″(x) m∗ = ψ T Mψ = 19 kip s2 /in.
x ∂t
—
S(x) ⫽(L ⫺x) m g k ∗ = ψ T Kψ = 1250 kips/in.
The generalized force is obtained from
ug 1
p∗ = −ψ T M 1 üg = −1170 sin(6πt) kips
Figure S2.9 1
Problem 2.10 The equation of motion is given by
u = zψ(x) 19z̈ + 1250z = −1170 sin(6πt)
πx d2 ψ π2 πx
ψ = sin , 2
= − 2 sin
L dx L L Problem 2.12
L
∗ 2 πx mL
m = m sin dx = Taking moments about A shown in Figure
0 L 2
L S2.12
π4 πx π 4 EI
k∗ = EI 4 sin2 dx = 3
ml2 θ̈ + 2ka2 θ + mgl sin θ = 0
0 L L 2L
L For small vibrations sin θ ≈ θ, hence
πx
p∗ = F δ (x − vt) sin dx
0 L
ml2 θ̈ + 2ka2 + mgl θ = 0
πvt
= F sin
L @Seismicisolation
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, 6
A
θ
2 ka θ
S
ml θ̈
mg
Figure S2.12
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@Seismicisolation
, 7
Chapter 3 Collectively, the four equations can be expressed in
the matrix form
Problem 3.1
M∗ ü + C∗ u̇ + K∗ u = 0
uT = [ q1 aq2 q3 q4 ]
q1 W
q2
w, r g
W r 2
M∗ =
g a2
w
a a g
w
g
q3 q4
2 0 −1 −1
ks cs ks cs
0 2 1 −1
w w C∗ = cS
−1 1 1 0
kT kT
−1 −1 0 1
2 0 −1 −1
0 2 1 −1
K∗ = kS
−1 1 1 + kkTS 0
cs(q· 3 ⫺ q· 1 ⫹ q· 2a) cs(q· 4 ⫺q· 1 ⫺q· 2a) −1 −1 0 1 + kkTS
ks(q3 ⫺ q1 ⫹ q2a)
ks(q4 ⫺q1 ⫺ q2a)
Problem 3.2
kT q3 w kT q4 w
q̈ q̈
g 3 g 4 mg
I0θ̈ mhθ̈
W 2
r q̈ 2
g
W
q̈
g 1 h
cs(q· 3 ⫺ q· 1 ⫹ q· 2a) cs(q· 4 ⫺q· 1 ⫺q· 2a) θ
⫹ ks(q3 ⫺ q1 ⫹q2a) ⫹ ks(q4 ⫺ q1 ⫺q2a)
θ
A
Figure S3.1 u
ug
The degrees of freedom are identified in Figure · ·
c(u· ⫺bθ) k(u ⫺bθ) k(u ⫹bθ) c(u·⫹bθ)
S3.1. The free-body diagrams are also shown. For
force balance on each of the front and rear axles b b
w
q̈4 + cS (q̇4 − q̇1 − q̇2 a) Figure S3.2
g
+ kS (q4 − q1 − q2 a) + kT q4 = 0 Coordinates ut and θ measure the vertical mo-
w tion and rotation of the base. The forces on the
q̈3 + cS (q̇3 − q̇1 + q̇2 a) structure are identified in Figure S3.2. Assuming
g
that ut is measured from the position of equilib-
+ kS (q3 − q1 + q2 a) + kT q3 = 0
rium under gravity load, the condition of vertical
For equilibrium of vertical forces on the vehicle force balance gives
body
W
q̈1 + cS (q̇3 + q̇4 − 2q̇1 ) müt + 2cu̇ + 2ku = 0 (1)
g
+ kS (q3 + q4 − 2q1 ) = 0 or
Taking moments about the mass center of the ve- mü + 2cu̇ + 2ku = −müg
hicle body
Taking moments about A and assuming that θ is
W 2 small
r q̈2 + cS a (q̇3 − q̇4 + 2aq̇2 )
g
+ kS a (q3 − q4 + 2aq2 ) = 0 0 I + mh2 θ̈ + 2cb2 θ̇ + 2kb2 θ − mghθ = 0 (2)
@Seismicisolation
@Seismicisolation
, 8
Problem 3.3 Problem 3.4
For frame A
a2 + b2
I0 = m
12 u1 = 12θ2
Hence the mass matrix is given by
u2 = 12θ4
m 0 0
or
M=0 m 0 θ1
2
+b2
0 0 m a 12 u1 0 12 0 0 θ2
=
u2 0 0 0 12 θ3
k1 ⫻1 θ4
k21 Hence
b
k1 0 12 0 0
k11 2 TA =
k31 0 0 0 12
k23
k1 ⫻1 a KA = TTA kA TA
k2
2 0 0 0 0
k33 k13
0 540 0 −240
= 144
k22 0 0 0 0
b
k1 0 −240 0 240
2
k32 k12 k2 ⫻1
For frame B
K B = KA
Figure S3.3 For frame C
To obtain the first column of stiffness matrix,
1 −15 0 0
give a unit displacement u, with v and w each zero. Tc =
0 0 1 −15
The resisting forces provided by frames A and B
and the external forces required to maintain the 860 −12900 −360 5400
displacement are shown in Figure S3.3. For equi- −12900 193500 5400 −81000
KC =
librium k11 = 2k1 , k21 = 0, k31 = 0. −360 5400 360 −5400
5400 −81000 −5400 193500
For u = 0, v = 1 and θ = 0 equilibrium gives
Assembly gives
k2 a
k12 = 0, k22 = k2 , k32 =
2 860 −12900 −360 5400
−12900 349020 5400 −150120
In a similar manner, for u = 0, v = 0, and θ = 1 K=
−360 5400 360 −5400
k13 = 0 5400 −151020 −5400 262620
k2 a 2
k23 = Mass of slab m = 100×30×24
32.2 = 2.236 kip.s
ft .
2
b2 a2 Mass of moment of inertia
k33 = 2k1 + k2
4 4 a2 + b2 302 + 242
I0 = m = 2.236 ×
On assembly, the following stiffness matrix is ob- 12 12
tained = 275.03 kip.ft.s2
2k1 0 0 Hence
k2 a
K= 0 k2 2
k2 a k1 b 2
k2 a2 2.236 0 0 0
0 2 + 4
2 0 275.03 0 0
M=
0 0 2.236 0
0 0 0 275.03
30 × 12
p1 = × p0 sin Ωt = 0.360p0 sin Ωt, p2 = 0
1000
@Seismicisolation
@Seismicisolation
