Chapter 2:Moment Of Inertia

Problem no 2.1,Page no.29

In [10]:
import math

#Initilization of Variables
 
#Rectangle-1
b_1=10 #cm #width of Rectangle-1
d_1=2 #cm #breadth of Rectangle-1
a_1=40 #cm**2 #Area of Rectangle-1
y_1=1 #cm #Distance of centroid-1

#Rectangle-2
b_2=2 #cm #width of Rectangle-2
d_2=10 #cm #breadth of Rectangle-2
a_2=20 #cm**2 #Area of rectangle-2
y_2=7 #cm #Distance of centroid-2

#Rectangle-3
b_3=20 #cm #width of Rectangle-3
d_3=2 #cm  #breadth of Rectangle-3
a_3=20 #cm**2 #Area of rectangle-3
y_3=13 #cm #Distance of centroid-3



#Calculation
Y_bar=((a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1) #cm #centre of gravity of section

Y_1=4.5 #cm #Distance of centroid of rectangle 1 to C.G
Y_2=1.5 #cm #Distance of centroid of rectangle 2 to C.G
Y_3=7.5 #cm #Distance of centroid of rectangle 3 to C.G

I_x_x_1=b_1*d_1**3*12**-1+a_1*Y_1**2 #moment of inertia of rectangle 1 about centroidal x-x axis of the section
I_x_x_2=b_2*d_2**3*12**-1+a_2*Y_2**2 #moment of inertia of rectangle 2  about centroidal x-x axis of the section
I_x_x_3=b_3*d_3**3*12**-1+a_3*Y_3**2 #moment of inertia of rectangle 3 about centroidal x-x axis of the section
I_x_x=I_x_x_1+I_x_x_2+I_x_x_3 #cm**4 

#Result
print"Moment of Inertia of the section is",round(I_x_x,2),"cm**4"

Moment of Inertia of the section is 2166.67 cm**4

Problem no 2.2,Page no.31

In [11]:
import math

#Initilization of Variables

#Rectangle-1
b_1=2 #cm #width of Rectangle-1
d_1=12 #cm #breadth of Rectangle-1
a_1=24 #cm**2 #Area of Rectangle-1
y_1=6 #cm #Distance of centroid-1

#Rectangle-2
b_2=6 #cm #width of Rectangle-2
d_2=2 #cm #breadth of Rectangle-2
a_2=12 #cm**2 #Area of rectangle-2
y_2=1 #cm #Distance of centroid-2

#Rectangle-3
b_3=2 #cm #width of Rectangle-3
d_3=12 #cm  #breadth of Rectangle-3
a_3=24 #cm**2 #Area of rectangle-3
y_3=6 #cm #Distance of centroid-3

#Calculation
Y_bar=((a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1) #cm #centre of gravity of section

Y_1=6 #cm #Distance of centroid of rectangle 1 to base 
Y_2=1 #cm #Distance of centroid of rectangle 2 to base
Y_3=6 #cm #Distance of centroid of rectangle 3 to base

I_x_x_1=b_1*d_1**3*12**-1+a_1*Y_1**2 #moment of inertia of rectangle 1 about centroidal x-x axis of the section
I_x_x_2=b_2*d_2**3*12**-1+a_2*Y_2**2 #moment of inertia of rectangle 2  about centroidal x-x axis of the section
I_x_x_3=b_3*d_3**3*12**-1+a_3*Y_3**2 #moment of inertia of rectangle 3 about centroidal x-x axis of the section
I_x_x=I_x_x_1+I_x_x_2+I_x_x_3 #cm**4 


#Result
print"Moment of Inertia of the section is",round(I_x_x,2),"cm**4"

Moment of Inertia of the section is 2320.0 cm**4

Problem no2.3,Page no.32

In [12]:
import math

#Initilization of Variables

#Rectangle-1
b_1=12 #cm #width of Rectangle-1
d_1=2 #cm #breadth of Rectangle-1
a_1=24 #cm**2 #Area of Rectangle-1
y_1=1 #cm #Distance of centroid-1

#Rectangle-2
b_2=2 #cm #width of Rectangle-2
d_2=6 #cm #breadthof Rectangle-2
a_2=12 #cm**2 #Area of rectangle-2
y_2=5 #cm #Distance of centroid-2

#Rectangle-3
b_3=5 #cm #width of Rectangle-3
d_3=2 #cm  #breadth of Rectangle-3
a_3=10 #cm**2 #Area of rectangle-3
y_3=9 #cm #Distance of centroid-3

#Calculation
Y_bar=((a_1*y_1+a_2*y_2+a_3*y_3)*(a_1+a_2+a_3)**-1) #cm #centre of gravity of section

Y_1=2.78 #cm #Distance of centroid of rectangle 1 to C.G 
Y_2=1.22 #cm #Distance of centroid of rectangle 2 to C.G
Y_3=5.22 #cm #Distance of centroid of rectangle 3 to C.G 

I_x_x_1=b_1*d_1**3*12**-1+a_1*Y_1**2 #moment of inertia of rectangle 1 about centroidal x-x axis of the section
I_x_x_2=b_2*d_2**3*12**-1+a_2*Y_2**2 #moment of inertia of rectangle 2  about centroidal x-x axis of the section
I_x_x_3=b_3*d_3**3*12**-1+a_3*Y_3**2 #moment of inertia of rectangle 3 about centroidal x-x axis of the section
I_x_x=I_x_x_1+I_x_x_2+I_x_x_3 #cm**4 


#Result
print"Moment of Inertia of the section is",round(I_x_x,2),"cm**4"

Moment of Inertia of the section is 523.16 cm**4

Problem no2.4,Page no.33

In [23]:
import math

#Initilization of variables

D=10 #cm #diameter of circle
b=4 #cm #width of rectangle 
d=4 #cm #breadth of rectangle
Y=1 #cm #Distance of centroid of rectangle 1 to C.G
a=16 #cm**2 #area of rectangle

#Calculations

I_x_x_1=pi*64**-1*(D**4) #cm**4 #moment of inertia of circle about x-x axis
I_x_x_2=b*d**3*12**-1+a*Y**2 #cm**4 #moment of inertia of rectangle about x-x axis
I_x_x=I_x_x_1-I_x_x_2 #cm**4 #Total moment of inertia of the section

#Result
print"Total moment of inertia of the section is",round(I_x_x,2),"cm**4"

Total moment of inertia of the section is 453.54 cm**4

Problem no 2.5,Page no.33

In [39]:
import math

#Initilization of variables

b_1=10 #cm #Breadth of the triangle
h=9 #cm #Height of triangle
b_2=2 #cm #width of rectangle
d=3 #cm #Depth of rectangle

#Triangle ABC-1
a_1=45 #cm**2 #Area of triangle
y_1=3 #cm #C.G of triangle

#Rectanglar hole-2
a_2=6 #cm**2 #Area of rectangle
y_2=4.5 #cm #C.G of rectangle

#Calculations

#Using relations
Y_bar=((a_1*y_1-a_2*y_2)*(a_1-a_2)**-1) #cm

I_1=b_1*h**3*36**-1+a_1*(y_1-Y_bar)**2 #cm**4 #M.I of triangle ABC about x-x passing through C.G of section
I_2=b_2*d**3*12**-1+a_2*(y_2-Y_bar)**2 #cm**4 #M.I of rectangular hole about x-x passing through C.G of section
I=I_1-I_2 #cm**4 #M.I of whole section about x-x passing through the C.G 

I_3=b_1*h**3*12**-1 #cm**4 #M.I of triangle ABC about the base BC
I_4=b_2*d**3*12**-1+a_2*y_2**2 #cm**4 #M.I of Rectangular hole about the base BC

I_5=I_3-I_4 #cm**4 #M.I of the whole section about the base BC

#Result
print"M.I of whole section about x-x passing through the C.G",round(I,2),"cm**4"
print"M.I of the whole section about the base BC is",round(I_5,2),"cm**4"

M.I of whole section about x-x passing through the C.G 182.42 cm**4
M.I of the whole section about the base BC is 481.5 cm**4

Problem no2.6,Page no.34

In [30]:
import math

#Initilization of variables


#Notifications has been changed as per requirement


h=8 #cm #height of triangle
b=8 #cm #breadth of triangle or diameter semicircle
d=4 #cm #diameter of circle enclosed

#Calculations

I_1=b*h**3*12**-1 #cm #moment of inertia of the triangle ABC about the axis AB
I_2=pi*b**4*128**-1 #cm ##moment of inertia of the semicircle about the axis AB
I_3=pi*d**4*64**-1 #cm #moment of inertia of circle about the circle about the axis

I=I_1+I_2-I_3 #cm #Moment of Inertia of the shaded area about the axia AB

#Result
print"Moment of Inertia of the shaded area is",round(I,2),"cm"

Moment of Inertia of the shaded area is 429.3 cm

Problem no.2.12,Page no.38

In [50]:
import math

#Initilization of variables

#Rectangle
a_1=600 #cm**2 #Area of the Rectangle
y_1=15 #cm #C.G of Rectangle
b=20 #cm #width of rectangle
d=30 #cm #depth of rectangle
D=15 #cm #Diameter of circle

#Circle
a_2=176.7 #cm**2 #Area of the circle
y_2=20 #cm #C.G of the circle
 
#Calculation

Y_bar=((a_1*y_1-a_2*y_2)*(a_1-a_2)**-1) #cm #Distance of C.G From the AB
Y_bar_1=2.1 #cm
Y_bar_2=7.1 #cm

I_1=b*d**3*12**-1 #cm**4 #M.I of the rectangle about its C.G and parallel to x-x axis
I_2=I_1+a_1*Y_bar_1**2
I_3=pi*D**4*64**-1+a_2*Y_bar_2**2 #cm**4 #M.I of circular section about x-x axis

I=I_2-I_3 #cm**4 #M.I of the section about x-x axis

#Result
print"M.I of the section about x-x axis",round(I,2),"cm**4"

M.I of the section about x-x axis 36253.5 cm**4

Problem no.2.13,Page no.38

In [6]:
import math

#Initilization of variables

d=90 #cm #Diameter of grindstone
t=10 #cm #thickness of grindstone
rho=0.0026 #Kg/cm**3 #Density

#calculations

#M=Mass of grindstone=Volume *Density=Area*Thickness*Density
M=pi*4**-1*d**2*t*rho #Kg 
R=d*2**-1 #cm #radius
I_g=M*R**2*2**-1 #Kg*m**2

k=R*(2**0.5)**-1 #cm #Radius of gyration

#Result
print"Radius of gyration is",round(k,2),"cm"

Radius of gyration is 31.82 cm