Download 7. IITD 2012 Theory of Vibration

CEL-707
SOIL DYNAMICS AND GEOTECHNICAL EARTHQUAKE ENGINEERING
ASSIGNMENTASSIGNMENT-1
THEORY OF VIBRATIONS
SUBMITTED BY:
SWARUP DAS
ENTRY NO. 2012VST9412
M.TECH , PCR - IIT DELHI, (JULY-NOV’ 2012)
UNDER THE GUIDANCE OF DR. R. AYOTHIRAMAN, CIVIL ENGG.
TOTAL SHEETS - 1+8
Assignment-1
CEL-707
CEL-707
Soil Dynamics and Geotechnical Earthquake Engineering
ASSIGNMENT-1
Submitted by: Swarup Das
Dated: 06.09.2012
Problem No. 1 : A harmonic motion has amplitude of 0.2 cm and a period of 0.15 sec.
Determine the following:
a) maximum velocity
b) maximum acceleration
Solution:
The harmonic motion can be demonstrated by a mass suspended from a weight less spring
as shown in the Fig. 1.1. When the mass(m) is displaced from its rest(equilibrium) position
and released, it will oscillate up and down. Harmonic motionis is often represented as the
projection on a straight line of a point that is moving on a circle at a constant angular speed
ω, the displacement z can be written as
z = Asinωt
Where, A is the amplitude of oscillation, usually measured mm from the equilibrium position
of the mass and τ is the period of harmonic motion, usually measured in second.
Because the motion repeats itself in 2π radians, we have the relationship ω = 2π / t.
Period of harmonic motion
Amplitude of oscillation
Circular frequency of the motion
Velocity of the motion, (Using dot-notation for the derivative)
Acceleration of the motion,
Fig. 1.1
τ =
A=
ω = 2 π/τ =
v = z - dot = ω Acos ω t
0.15 sec.
0.2 cm
41.89 rad/sec.
v max. = z - dot max = ω Acos(0) =
83.78 mm/sec
a = z - double dot = ω 2 Asin ω t
a max. = z - double dot = ω 2 Asin( π /2) =
Swarup Das
m
3509 mm/sec^2
Sheet 1 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 2 : A mass-spring-dashpot system (W=356N) is subjected to a sinusoidal force.
The amplitude of forcing is F0 = 40N and the damping ratio is 0.05.
a) What will be the total spring force?
Solution: The system can be demonstrated by the Fig. 1.2.
c
k
W
F = F0 sinωt
where, ω is the frequency of excitation.
Weight of the system
Mass of the system
Harmonic exciting force
Stiffness of the spring (Spring constant)
Damping factor i.e, the ratio of actual damping c to critical
damping cc
Actual viscous damping of the system
Critical viscous damping of the system
The frequency of the excitation
The natural circular frequency of the system
At resonance, the frequency ratio,
The amplitude of oscillation,
W=
m=
F0 =
356 N
36.29 kg
40 N
k=
ξ
= c/c c =
0.05
c = c c ξ = 2m ω n ξ =
c c =2m ω n =
=ω =
= ωn =
r = ω/ωn =
1.05
20.00
20.00
1
Z = (F0/k) / [(1-r2)2 + (2ξr)2]1/2 =
i.e, at resonance, Z = (F0/k) / 2ξ =
Therefore, the spring force at resonance
i.e, kZ = F0 / 2ξ =
400 N
Problem No. 3 : Consider a simple spring-mass-dashpot system with parameters, m=4kg, k=1.6x103 N/m and the two cases of
damping : (i) c=80N-sec/n )ii) c = 320N-sec/m.
a) Study the nature of the free response in each case.
k
c
m
Solution:
Mass of the system
Stiffness of the spring (Spring constant)
Undamped natural circular frequency of the system
Case-i) Actual viscous damping of the system
Damping factor i.e, the ratio of actual damping c to critical
damping cc
m=
k=
ω n = sqrt(k/m) =
c = c c ξ = 2m ω n ξ =
ξ
= c/c c =
4 kg
1600 N/m
20 radians/sec.
80
0.5
The damping is less than critical damping, i.e ξ<1. Such a system is called underdamped. Thus, the under
damped natural circular frequency in viscously damping vibration equal ωnd = ωn(1-ξ2)1/2
Case-ii) Actual viscous damping of the system
Damping factor i.e, the ratio of actual damping c to critical
damping cc
c = c c ξ = 2m ω n ξ =
320
ξ
2.0
= c/c c =
The damping is more than critical damping, i.e ξ>1. Thus, z decreases as the t increases but it never
changes sign. Such a system is called Overdamped or non-oscillatory . If an innitial displacement is given
to the system, the mass is pulled back by the springs and dampers absorb all the energy by the time the
mass returns to the initial position.
Swarup Das
Sheet 2 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 4 : A new concrete hall is to be protected from the ground vibrations from an adjacent highway by mounting the hall
on rubber blocks, The predominant frequency of the sinusoidal vibration is 40Hz and a motion transmissibility of 0.1 is to be
attained at that frequency.
a) Calculate the static deflection required in the rubber blocks. Assuming that these act as linear undamped spring
Solution: F = F0 sinωt
By assuming linear undamped spring,
ξ = c/c c =
2
2 2
T R = [ F T /F 0 ] = SQRT [ (1+(2 ξ r) / {(1-r ) +(2 ξ r) 2 } ] =
Transmissibility
2 2
(1-r ) =
(1-r 2 ) =
i. e, r =
or, r =
r = ω/ωn =
The frequency ratio,
The frequency of the excitation
f=
The circular frequency of the excitation
ω =2 π f =
ω n = ω /r =
Undamped natural circular frequency of the system
Also, ω n = sqrt(k/m) = sqrt(W/ δ stat / m) = sqrt(mg / m δ stat )= sqrt(g / δ stat )
Therefore, δ stat = g / ω n 2 =
0
0.1
100
10
3.32
3.00
3.32
40 Hz(cps)
251 rad/sec
75.78 radians/sec.
0.17 cm
Problem No. 5 : An electrical motor used as a mechanical drive is mounted at the centre of a light table. The weight of the motor
added to the effective weight of the table is 363N. The armature of the rotating part weights 910N and has an eccentricity =10mm.
The table is observed to deflect 3.2mm when the motor is mounted in a free vibration amplitude decreases to 0.2 of the innitial
value after 6 consecutive cycles. The operating speed of the motor is 900rpm.
a) Calculate the amplitide of vibration of the system at operating frequency.
Solution:
W1 =
Weight of the rotating part
910 N
Mass of the rotating part
m1 =
92.76 kg
Effective weight of the table
Mass of the Table
W2 =
363
37.00 kg
129.77 kg
m2 =
M = (m1 + m2)=
Total mass of the motor-table system
Z1 =
3.2 mm
Z6 = 0.2Z1 =
0.64 mm
Max. amplitude in 1st cycle
Max. amplitude in 6th cycle
Frequency of the motor(frequency of the excitation)
The circular frequency of the excitation
Damping factor i.e, the ratio of actual damping c to critical
damping cc
Static deflection
Undamped natural circular frequency of the system
The frequency ratio,
The eccentricity of the rotating part
The amplitude of the vibtation
Swarup Das
f=
ω
ξ
=2 π f =
900 rpm
94.25 rad/sec
= (1/2 π n) ln(z 1 /z 6 ) =
0.0427
δ stat =
ω n = sqrt(g / δ stat ) =
r = ω/ωn =
3.20
55.37
1.70
10
5.72
e=
Z = (m1e/M)r2 / [(1-r2)2 + (2ξr)2]1/2 =
mm
rad/sec.
mm
mm
Sheet 3 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 6 : A shock absorber is to be designed so that its overshoot is 10 percent of the initial displacement when released.
a) Determinne ξ1.
b) If ξ is made equal to 0.5ξ1. what will be the overshoot.
Solution:
Damping factor i.e, the ratio of actual damping c to critical
damping cc
ξ
or,
ξ
= (1/2 π n) ln(z 1 /z 2 ) =
2
0.3665
0.3440
0.183
3.16
=> z 2 = 0.316 * z1
or, z 2 = 0.334 * z1
= sqrt(1- ξ )/2 π ln(z 1 /z 2 ) =
if, ξ = 0.366 /2, i.e, ξ =
i.e, 0.183 = (1/2 π) ln(z 1 /z 2 ) => z 1 /z 2 =
Problem No. 7 : A machine mass is 1.95kg vibrates in a viscous medium. Determine the damping coefficient when a harmonic
exciting force 24.46N results in resonant amplitude of 1.27 cm with a period of 0.2sec. If the same system in excited by harminic
force of frequency 4cps, what will be the percentage increase in the amplitude of forced vibration when the dash-pot is removed ?
Solution: Mass of the machine
m=
τn =
Period of motion
The natural circular frequency of the system
ω n =2 π / τ n =
k = mωn
Stiffness of the spring (Spring constant)
2
=
F0 =
Harmonic exciting force
The amplitude of oscillation,
c
31.42 rad/sec
1925 N/m
24.46 N
Z = (F 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2
i.e, at resonance, r = 1, Z r=1 = (F 0 /k) / 2 ξ =
i.e, at resonance, ξ = (F 0 /k) / 2Z =
i.e, ξ =
f=
ω =2 π f =
r = ω/ωn =
When the dashpot is removed,
Harminic force of frequency
The circular frequency of the excitation
The frequency ratio,
The amplitude of oscillation,
1.27 cm
0.50
0
4 Hz(cps)
25.13 rad/sec
0.80
Z = (F 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2
i.e, when ξ = 0, Z ξ =0 = (F 0 /k) / (1-r 2 ) =
Therefore, increase in percentage amplitude of forced vibration
Swarup Das
1.95Wkg
0.2 sec
(Z ξ =0 - Z r=1 ) / Z r=1 ] *100 -100=
3.53 cm
77.98 % increased
Sheet 4 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 8 : A sensitive instrument with mass 113kg is to be installed at a location where the acceleration is 15.24cm/sec^2 at
a frequency 20 Hz. It is proposed to mount the instrument on a rubber pad with the following properties:
k = 2802N/cm and ξ = 0.10. What acceleration is transmitted to the instrument ?
Solution: Mass of the instrument
Harminic force of frequency
Period of motion
The circular frequency of the excitation
m=
f=
τ =1/f =
ω =2 π f =
2
k = mωn =
Stiffness of the spring (Spring constant)
The natural circular frequency of the system
ω n = sqrt(k/m) =
r = ω/ωn =
The frequency ratio,
Damping factor i.e, the ratio of c to cc
ξ = c/c c =
2
2 2
2
T R = [ F T /F 0 ] = SQRT [ (1+(2 ξ r) / {(1-r ) +(2 ξ r) } ] =
Transmissibility
2
a = z - double dot = ω Asin ω t
Acceleration of the location
= a * TR =
Acceleration transmitted to the instrument,
113
20
0.05
125.66
kg
Hz(cps)
sec
rad/sec
2802 N/cm
49.80 rad/sec
2.52
0.1
0.20
15.24 cm/sec^2
3.0 cm/sec^2
Problem No. 9 : A body weight 60kg is suspended from a spring, which deflects 1.2cm under the load. It is subjected to a
damping effect adjusted to value 0.2 times that required for critical damping. Find the natural frequency of the undamped, damped
vibration and in the later case, determine the ratio of successing amplitudes. If the body is subjected to a harmonic force with a
maximum value of 250N at a frequency equal to twice of its undamped natural frequency, determine the amplitude of forces
vibrations and the phase differentce with respect to the disturbing forces.
Solution:
m=
δ stat =
ω nd = ω n sqrt(1- ξ ) =
60
1.2
0.2
28.59
28.01
Z 1 / Z 2 = exp power (2 πξ /sqrt(1- ξ 2 )=
3.61
r = ω/ωn =
2
Mass of the body
Static deflection
Damping factor i.e, the ratio of c to cc
The natural circular frequency of the system
Damped natural frequency
Ratio of successive amplitudes
ξ
2
The frequency ratio, ω = 2ωn (Given)
Harmonic exciting force
Stiffness of the spring (Spring constant)
The amplitude of oscillation,
Phase differece between amplitude and disturbing force
Swarup Das
= c/c c =
ω n = sqrt(g / δ stat ) =
kg
cm
rad/sec
rad/sec
F0 =
250 N
2
49050 N/m
k = mωn
=
Z = (F 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2 =
φ = tan -1 [2 ξ r / (1-r 2 )] =
1.70
-14.93
o
Sheet 5 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 10 : A machine foundation supported by a soil layer can be approximated as a mass-spring-dashpot system. The
weight of foundation with machine is 800kN. The stiffness constant k is 200x10^3 kN/m and damping constant c is 2340kN-sec/m
a) Determine critical damping coefficient, damping ratio, logarithmic decrement and damped natural frequency.
b) If the foundation subjected to a vertical force Pz = P0 sin ωt in which P0 = 25kN and ω = 100 rad/sec, determine the amplitide of
vertical vibration of foundation and the maximum dynamic forcetransmitted to the subgrade.
Solution:
a) Weight of foundation with machine
Mass of foundation with machine
W=
m1 =
800 kN
81.55 kg
Harmonic exciting force
P0 =
250 kN
k = mωn
Stiffness of the spring (Spring constant)
Actual viscous damping of the system
Critical viscous damping of the system
Damping factor i.e, the ratio of c to cc
The circular frequency of the excitation
The circular natural frequency of the system
Damped natural frequency of the system
=
c = c c ξ = 2m ω n ξ =
c c =2m ω n = 2sqrt(km) =
ξ = c/c c =
ω =2 π f =
ω n = sqrt(k/m) =
ω nd = ω n sqrt(1- ξ 2 ) =
=>
b) The frequency ratio,
The amplitude of oscillation,
Transmissibility
2
2
ln(z 1 /z 2 ) = 2 πξ / sqrt(1- ξ ) =
r = ω/ωn =
Z = (P 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2 =
T R = [ F T /F 0 ] = SQRT [ (1+(2 ξ r) 2 / {(1-r 2 ) 2 +(2 ξ r) 2 } ] =
Harmonic exciting force
Force transmitted to the subgrade
2.0.E+08 N/m
2.34.E+06
8.08.E+06
0.29
100.00
49.52
47.40
1.902
2.02
0.38
0.47
N-sec/m
N-sec/m
rad/sec
rad/sec
rad/sec
mm
P0 =
25 kN
TF = [ P0* TR ] =
11.69 kN
Problem No. 11 : A machine and its foundation weigh 200kN. The spring constant is 120MN/m and damping factor is 20%.
Steady state vibration of the foundation is caused by a disturbing force of Fd = F0 sin ωt, in which F0=50kN and the circular
frequency of 120 rad/sec. Determine
(a) Undamped natural frequency,
(b) Amplitude of vibration and
(c) Phase angle
Solution: Harmonic exciting force
The circular frequency of the excitation
Weight of machine and its foundation
Mass of machine and its foundation
Damping factor i.e, the ratio of c to cc
Stiffness of the spring (Spring constant)
a) The circular natural frequency of the system
b) The frequency ratio,
The amplitude of oscillation,
c) Phase differece between amplitude and disturbing force
Swarup Das
F0 =
ω
=2 π f =
W=
m1 =
ξ
= c/c c =
2
k = mωn =
ω n = sqrt(k/m) =
r = ω/ωn =
Z = (P 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2 =
φ = tan -1 [2 ξ r / (1-r 2 )] =
50 kN
120 rad/sec
200 kN
20.39 kg
0.20
1.2.E+08 N/m
76.72 rad/sec
1.56
0.26 mm
-23.40 o
Sheet 6 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 12 : A machine is supported on four steel springs for which damping can be neglected. The natural frequency of
vertical vibration of the machine-spring system is 3.33Hz. The machine genetares a vertical force F(t) = F0 sinωt, the amplitude of
the resulting steady state vertical vibration of the system is 5mm when the machine is operating at 0.33Hz and 0.6mm at 10Hz.
Calculate the amplitude of vertical bibration of the machine if the steel springs are replaced by rubber isolators, which provided the
same stiffness, but introduce damping(D) equivalent to 0.25 for the system. Comment on the effectiveness of isolators at both
frequencies of machine.
Solution: The frequency of the excitation
The circular frequency of the excitation
The amplitude of oscillation,
The frequency of the excitation
The circular frequency of the excitation
The amplitude of oscillation,
The frequency of the excitation
The circular frequency of the excitation
Damping factor i.e, the ratio of c to cc
The frequency ratio,
ωn
f=
3.33 Hz(cps)
=2 π f =
20.92 rad/sec
Z1 =
f1=
ω1
=2 π f 1 =
Z2 =
f2=
ω 2 =2 π f 2 =
ξ = c/c c =
r1= ω1/ωn =
r2= ω2/ωn =
The amplitude of oscillation,
5.00
0.33
2.07
0.60
10.00
62.83
0
0.099
3.003
mm
Hz(cps)
rad/sec
mm
Hz(cps)
rad/sec
Z = (F 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2
i.e, Z = (F 0 /k) / (1-r 2 )
=> (F 0 /k) = Z 1 * (1-r 2 ) =
4.95 mm
-4.81 mm
Now, damping factor i.e, the ratio of c to cc
0.25
ξ = c/c c =
f1=
The frequency of the excitation
0.33 Hz(cps)
The circular frequency of the excitation
2.07 rad/sec
ω 1 =2 π f 1 =
2 2
2 1/2
Z 1 = (F 0 /k) / [(1-r 1 ) + (2 ξ r 1 ) ]
=
The amplitude of oscillation,
4.99 mm
Z 2 = (F 0 /k) / [(1-r 1 2 ) 2 + (2 ξ r 1 ) 2 ] 1/2 =
The amplitude of oscillation,
0.61 mm
Comment: Since the amplitude is more or less same as spring system. Therefore, effect of rubber isolator is same as
spring systems.
=> (F 0 /k) = Z 1 * (1-r 2 ) =
Problem No. 13 : A machine of 100kg mass is supported on springs of total stiffness 700kN/m, which has an unbalanced rotating
element that results in a disturbing force of 350N at a speed of 3000rpm. Assume the damping of system is 0.2. Determine
(a) Amplitude of vibration
(b) Transmissibility and
(c) Force transmitted to the foundation-soil system.
Solution: Harmonic exciting force
The frequency of the excitation
The circular frequency of the excitation
Mass of machine
Damping factor i.e, the ratio of c to cc
Stiffness of the spring (Spring constant)
F0 =
f=
ω
=2 π f =
m1 =
ξ
= c/c c =
k = mωn
2
=
ω n = sqrt(k/m) =
The circular natural frequency of the system
r = ω/ωn =
The frequency ratio,
Z = (P 0 /k) / [(1-r 2 ) 2 + (2 ξ r) 2 ] 1/2 =
a) The amplitude of oscillation,
2
2 2
2
b) Transmissibility
T R = [ F T /F 0 ] = SQRT [ (1+(2 ξ r) / {(1-r ) +(2 ξ r) } ] =
TF = [ F0* TR ] =
Force transmitted to the subgrade
Swarup Das
350 kN
3000 Hz(cps)
314 rad/sec
100 kg
0.20
7.0.E+05 N/m
83.67
3.75
0.038
13.69
47.90
rad/sec
mm
%
kN
Sheet 7 OF 8
9/6/2012
Assignment-1
CEL-707
Problem No. 14 : A two storied building is represented as a lumped mass system (fig.-1) in which m1 =0.5m2 and k1=2k2(The
stiffness given represents equivalent lateral stiffness of columns). Determine its natural frequencyes, amplitude of vibration and
plot normal mode shapes.
x1
z2
m1
k2
x2
m2
m2
z1
k1
m1
Mathematical
Representation
Fig. 1
The equation of motion can be writted as follows:
m1x1.. + k1(x1 - x2) = 0
..
=> m1x1 + k1x1 - k1x2 = 0
and
………….. (1)
m2x2.. + k1(x2 - x1) + k2x2 = 0
..
=> m2x2 + (k1+k2)x2 - k1x1 = 0
………….. (2)
From equation (1) & (2) we get,
k1-m1ωn2
-k1
-k1
k1+k2 -m2ωn2
=
0
Mode Shapes:
Mode-1
Swarup Das
Mode-2
Sheet 8 OF 8
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