# Chapter 15: Vibrations¶

## Example 1, Page 535¶

In [1]:
import math

#Variable declaration
W=.3*2240#lb
l=36#in
D=3.#in
k=15#in
A=math.pi*(D/2)**2
E=30*10**6#youngs modulus
C=12*10**6
g=32.2#ft/s^2

#Calculations
d=W*l/(A*E)
Fl=187.8/(d)**(1./2)
I=math.pi*(d/2)**4
d1=W*(l**3)*64/(3*E*math.pi*(3**4))
Ft=187.8/(d1)**(1./2)
j=math.pi*3**4./32
q=C*j/l
Ftor=(1/(2*math.pi))*(q*g*12/(W*k**2))**(1./2)
F1=Ftor*60

#Results
print "a) Frequency of Longitudinal vibrations = %.f per min\nb) Frequency of the transverse vibrations = %.f per min"\
"\nc) Frequency of the torsional vibration = %.f per min"%(Fl,Ft,F1)

a) Frequency of Longitudinal vibrations = 17583 per min
b) Frequency of the transverse vibrations = 634 per min
c) Frequency of the torsional vibration = 786 per min


## Example 2, Page 536¶

In [18]:
import math

#Variable declaration
l1=3#ft
l2=2.#ft
l=l1+l2#ft
W=.5*2240#lb
k=20#in
d=2.#in

#Calculations
Wa=2*W/5
E=30*10**6
d1=(Wa*l1*12)/(math.pi*E)
N1=187.8/math.sqrt(d1)

I=math.pi*(d)**4./64
d2=W*(l1*12)**3*(l2*12)**3/(3*E*(l*12)**3*I)
N2=187.8/(d2)**(1./2)
C=12*10**6#given
g=32.2#given
J=math.pi*d**4/32
q=C*J*((1./(l1*12))+(1./(l2*12)))
n=(1./(2*math.pi))*(q*g*12/(W*k**2))**(1./2)
N3=n*60

#Results
print "a)Longitudinal vibration = %.f per min\nb)Transverse Vibration = %.f per min\nc)Torsional Vibration = %.f per min"%(N1,N2,N3)

a)Longitudinal vibration = 14356 per min
b)Transverse Vibration = 863 per min
c)Torsional Vibration = 321 per min


## Example 3, Page 547¶

In [19]:
#Variable declaration
l=10#ft
d=4#in
E=30*10**6#youngs modulus
d1=0.0882#inches; maximum deflection as shown in the figure

#Calculations
N=207./(d1)**(1./2)#From 15.20

#Result
print "Frequency of natural transverse vibration = %.f per min"%N

Frequency of natural transverse vibration = 697 per min


## Example 4, Page 552¶

In [21]:
import math

#Variable declaration
m=50.#lb
k=100#lb/in
g=32.2#ft/s

#Calculations
d=m/k#static deflection
n=(1/(2*math.pi))*(g*12/d)**(1./2)
#part 2
b=g*12./d
a=(b/20.79)**(1./2)
nd=(1./(2*math.pi))*((b-(a/2)**2))**(1./2)
A=nd/n

#Results
print "Frequency of free vibrations = %.3f per sec\nFrequency of damped vibrations = %.3f per sec"\
"\nThe ratio of the frequencies of damped and free vibrationsis %.3f"%(n,nd,A)

Frequency of free vibrations = 4.424 per sec
Frequency of damped vibrations = 4.398 per sec
The ratio of the frequencies of damped and free vibrationsis 0.994


## Example 5, Page 553¶

In [22]:
import math

#Variable declaration
#damping torque is directly proposrtional to the angular velocity
C=12*10**6#Modulus of rigidity
l=3#ft
d=1#in
g=32.2#ft/s**2
I=500#lb ft^2 ; moment of inertia

#Calculations
J=math.pi*d**4/32
q=C*J/(l*12)
n=(1./(2*math.pi))*(q*g*12/(I*12**2))**(1./2)
#part 2
b1=(q*g*12/(I*12**2))
a1=(b1/10.15)**(1./2)#by reducing equation 15.28
nd=(1./(2*math.pi))*(b1-(a1/2)**2)**(1./2)
A=nd/n

#Results
print "The frequency of natural vibration = %.2f per sec\nThe frequency of damped vibration = %.2f per sec"\
"\nThe ratio nd/n = %.3f"%(n,nd,A)

The frequency of natural vibration = 2.11 per sec
The frequency of damped vibration = 2.08 per sec
The ratio nd/n = 0.988


## Example 6, Page 560¶

In [23]:
#Variable declaration
m=20.#lb
k=50#lb/in
F=30.#lb
w=50#sec^-1
g=32.2#ft/s^2

#Calculations
d=m/k
x=F/k#extension of the spring
b=g*12./d
a=(b/30.02)**(1./2)#from equation 15.28
D=1/((1-w**2/b)**2+a**2*w**2/b**2)**(1./2)
Af=D*x#amplitude of forced vibration
D=(b/a**2)**(1./2)#At resonance
A=D*x#amplitude at resonance

#Results
print "Amplitude of forced vibrations = %.3f in\nAmplitude of the forced vibrations at resonance = %.2f in"%(Af,A)

Amplitude of forced vibrations = 0.372 in
Amplitude of the forced vibrations at resonance = 3.29 in


## Example 7, Page 563¶

In [26]:
import math

#Variable declaration
e=1./30
n=1200.#rpm
w=math.pi*n/30
m=3.#lb
g=32.2#ft/s^2
stroke=3.5#in

#Calculations
r=stroke/2
k=(1+1./e)**(1./2)#nf/n=k
d=(k/187.7)**2
W=200.#lb ; given
s=W/d#combined stiffness
p=1./14.1#As a^2/b=1/198
T=((1+p**2*k**2/((1-k**2)**2+p**2*k**2)))**(1./2)#actual value of transmissibility
F=(m/g)*w**2*r/12#maximum unbalanced force transmitted on the machine
Fmax=F*T#maximum force transmitted to the foundation
#case b
E=((1+p**2)/(p**2))**(1./2)
Nreso=215.5#rpm
Fub=F*(Nreso/n)**2
Ftmax=E*Fub
D=E#dynamic magnifier
deln=Fub/152#static deflection
A=deln*D

#Results
print "a) Maximum force transmitted at 1200 rpm = %.1f lb\nb) The amplitude of the forced vibrations of the machine at"\
"resonance = %.3f in\n   Force transmitted = %.f lb"%(Fmax,A,Fub)

a) Maximum force transmitted at 1200 rpm = 214.6 lb
b) The amplitude of the forced vibrations of the machine atresonance = 0.643 in
Force transmitted = 7 lb


## Example 8, Page 570¶

In [27]:
import math

#Variable declaration
l1=11#in
l2=10#in
l3=15#in
l4=4#in
l5=10#in
d1=3#in
d2=5#in
d3=3.5#in
d4=7#in
d5=5#in
I1=1500#lb ft^2
I2=1000#lb ft^2
leq=3#in from 15.49
g=32.2#ft/s^2
C=12*10**6

#Calculations
J=math.pi*leq**4./32
l=l1+l2*(leq/d2)**4+l3*(leq/d3)**4+l4*(leq/d4)**4+l5*(leq/d5)**4
la=I2*l/(I1+I2)
qa=C*J/la
n=(1./(2*math.pi))*(qa*g*12/(I1*12**2))**(1./2)

#Results
print "The frequency of the natural torsional oscillation of the system = %.1f per sec"%n

The frequency of the natural torsional oscillation of the system = 23.8 per sec


## Example 9, Page 572¶

In [28]:
import math

#Variable declaration
Ia=2.5#ton ft^2
Ib=7.5#ton ft^2
Ic=3.#ton ft^2
g=32.2#ft/s^2
AB=9.5#ft
BC=25#ft
d=8.5#in
C=11.8*10**6#lb/in^2

#Calculations
k=Ic/Ia#la/lc=k
lc1=(25.6+(25.6**2-4*114.1)**(1./2))/2#from 1 and 2 , reducing using quadratic formula
lc2=(25.6-(25.6**2-4*114.1)**(1./2))/2#from 1 and 2 , reducing using quadratic formula
la1=lc1*k
la2=lc2*k
J=math.pi*d**4/32
q=C*J/(lc1*12)#torsional stiffness
IC=Ic*2240*12**2/(g*12)#moment of inertia
nc=(1./(2*math.pi))*(q/IC)**(1./2)#fundamental frequency of vibration
a1=nc*60
a=math.floor(a1)
n=16*(lc1/lc2)**(1./2)
b=n*60

#Results
print "Fundamental frequency of vibration = %.f per min\nTwo node frequency = %.f per min"%(a,b)

Fundamental frequency of vibration = 961 per min
Two node frequency = 1784 per min


## Example 11, Page 587¶

In [29]:
import math

#Variable declaration
g=32.3#ft/s^2
l2=25.5#in
d1=2.75#in
d2=3.5#in
C=12*10**6#modulus of rigidity
G=1/0.6#given speed ratio
Ib=54.#lb in^2
Ic=850.#lb in^2
Id=50000.#lb in^2

#Calculations
Id1=Id/G**2#15.62
Ia=1500#lb in^2
la=Id1/(Id1+Ia)*66.5
J=math.pi*d1**4/32
q=C*J/la#torsional stiffness
n=(1/(2*math.pi))*(q*g*12/Ia)**(1./2)
nf=n*60#for minutes
#case b)
Ib1=Ib+Ic/(G**2)
a=63.15#in; distance of the node from rotor A (given)
b=3.661#in; distance of the node from rotor A (given)
N1=n*(la/a)**(1./2)
N2=n*(la/b)**(1./2)
N1f=N1*60#for minutes
N2f=N2*60#for minutes

#Results
print "a) The frequency of torsional vibrations n = %.1f per sec or %.f per min\nb) The fundamental frquency = %.1f per sec"\
"or %.f per min\n   and the two node frequency = %.f per sec or %.f per min"%(n,nf,N1,N1f,N2,N2f)

a) The frequency of torsional vibrations n = 84.8 per sec or 5086 per min
b) The fundamental frquency = 83.6 per secor 5014 per min
and the two node frequency = 347 per sec or 20824 per min