Chapter 12 Moment of Inertia

Example 12.7 Moment of Inertia of an area of a plane figure with respect to an axis in its plane

In [1]:
# Initilization of variables
A= 50 # cm**2 # area of the shaded portion
J_A=22.5*10**2 # cm**4 # polar moment of inertia of the shaded portion
d=6 # cm
# Calculations
J_c=J_A-(A*d**2) 
# substuting the value of I_x from eq'n 2 in eq'n 1 we get,
I_y=J_c/3 # cm**4 # M.O.I about Y-axis
# Now from eq'n 2,
I_x=2*I_y # cm**4 # M.O.I about X-axis
# Results
print('The centroidal moment of inertia about X-axis (I_x) is %d cm**4'%I_x)
print('The centroidal moment of inertia about Y-axis (I_y) is %d cm**4'%I_y)

The centroidal moment of inertia about X-axis (I_x) is 300 cm**4
The centroidal moment of inertia about Y-axis (I_y) is 150 cm**4

Example 12.8 Moment of Inertia of a Composite area or hollow section

In [2]:
import math
# Initilization of variables
b=20 # cm # width of the pate
d=30 # cm # depth of the plate
r=15 # cm # radius of the circular hole
h=20 # cm # distance between the centre of the circle & the x-axis
# Calculations
# (a) Location of the centroid of the composite area
A_1=b*d # cm**2 # area of the plate
y_1=d/2 # cm # y-coordinate of the centroid
A_2=(math.pi*r**2)/4 # cm**2 # area of the circle removed (negative)
y_2=h # cm # y-coordinate of the centroid
y_c=((A_1*y_1)-(A_2*y_2))/(A_1-A_2) # cm # from the bottom edge
# (b) Moment of Inertia of the composite area about the centroidal x-axis
# Area (A_1) M.I of area A_1 about x-axis
I_x1=(b*(d**3))/12 # cm**4
# M.I of the area A_1 about the centroidal x-axis of the composite area (By parallel-axis theorem)
OC_1=15 # cm # from the bottom edge
OC_2=20 # cm
OC=12.9 # cm # from the bottom edge
d_1=OC_1-OC # cm
d_2=OC_2-OC # cm 
I_X1=(I_x1)+(A_1*d_1**2) # cm**4
# Area(A_2) M.I of area A_2 about x-axis
I_x2=(math.pi*r**4)/64 # cm**2
# M.I of the area A_2 about the centroidal x-axis of the composite area (By parallel-axis theorem)
I_X2=(I_x2)+(A_2*d_2**2) # cm**4
# COMPOSITE AREA:M.O.I of the composite area about the centroidal x-axis
I_x=(I_X1)-(I_X2) # cm**4
# Results
print('The M.O.I of the composite area about the centroidal x-axis is %f cm**4'%I_x)
# There may be a small error in the answer due to a decimal point 

The M.O.I of the composite area about the centroidal x-axis is 36252.768805 cm**4

Example 12.9 Moment of Inertia of a Composite area or hollow section

In [3]:
from __future__ import division
import math
# Initilization of variables
b1=80 # mm # width of the flange pate
d1=20 # mm # depth of the flange plate
b2=40 # mm # width/thickness of the web
d2=60 # mm # depth of the web
# Calculations
# (a) Location of the centroid of the composite area
A_1=b1*d1 # mm**2 # area of the flange plate
y_1=d2+(d1/2) # mm # y-coordinate of the centroid
A_2=b2*d2 # mm**2 # area of the web
y_2=d2/2 # mm # y-coordinate of the centroid
y_c=((A_1*y_1)+(A_2*y_2))/(A_1+A_2) # mm # from the bottom edge
# (b) Moment of Inertia of the composite area about the centroidal x-axis
# Area (A_1) M.I of area A_1 about x-axis
I_x1=(b1*(d1**3))/12 # mm**4
# M.I of the area A_1 about the centroidal x-axis of the composite area (By parallel-axis theorem)
OC_1=70 # mm # from the bottom edge
OC_2=30 # mm # from the bottom edge
OC=y_c # mm # from the bottom edge
d_1=(d2-y_c)+(d1/2) # mm
d_2=y_c-OC_2 # mm 
I_X1=(I_x1)+(A_1*d_1**2) # mm**4
# Area(A_2) M.I of area A_2 about x-axis
I_x2=(b2*d2**3)/12 # mm**4
# M.I of the area A_2 about the centroidal x-axis of the composite area (By parallel-axis theorem)
I_X2=(I_x2)+(A_2*d_2**2) # mm**4
# COMPOSITE AREA:M.O.I of the composite area about the centroidal x-axis
I_x=(I_X1)+(I_X2) # mm**4
# Results
print('The M.O.I of the composite area about the centroidal x-axis is %f mm**4'%I_x)
# NOTE: The answer given in the text book is 2.31*10**3 insted of 2.31*10**6.

The M.O.I of the composite area about the centroidal x-axis is 2309333.000000 mm**4

Example 12.10 Moment of Inertia of a Composite area or hollow section

In [6]:
from __future__ import division
import math
# Initilization of variables
b1=120 # mm # width of the flange pate of L-section
d1=20 # mm # depth of the flange plate
b2=20 # mm # width/thickness of the web
d2=130 # mm # depth of the web
# Calculations
# (a) Location of the centroid of the composite area
A_1=b1*d1 # mm**2 # area of the flange plate
A_2=b2*d2 # mm**2 # area of the web
y_1=d2+(d1/2) # mm # y-coordinate of the centroid
y_2=d2/2 # mm # y-coordinate of the centroid
x_1=60 # mm # x-coordinate of the centroid
x_2=110 # mm # x-coordinate of the centroid
y_c=((A_1*y_1)+(A_2*y_2))/(A_1+A_2) # mm # from the bottom edge
x_c=((A_1*x_1)+(A_2*x_2))/(A_1+A_2) # mm # from the bottom edge
# (b) Moment of Inertia of the composite area about the centroidal x-axis
# Area (A_1) M.I of area A_1 about x-axis
I_x1=(b1*(d1**3))/12 # mm**4
# M.I of the area A_1 about the centroidal x-axis of the composite area (By parallel-axis theorem)
OC_1=d2+(d1/2) # mm # from the bottom edge
OC_2=d2/2 # mm # from the bottom edge
OC=y_c # mm # from the bottom edge
d_1=(d2-y_c)+(d1/2) # mm
d_2=y_c-OC_2 # mm 
I_X1=(I_x1)+(A_1*d_1**2) # mm**4
# Area(A_2) M.I of area A_2 about x-axis
I_x2=(b2*d2**3)/12 # mm**4
# M.I of the area A_2 about the centroidal x-axis of the composite area (By parallel-axis theorem)
I_X2=(I_x2)+(A_2*d_2**2) # mm**4
# COMPOSITE AREA:M.O.I of the composite area about the centroidal x-axis
I_x=(I_X1)+(I_X2) # mm**4
# (c) Moment of Inertia of the composite area about the centroidal y-axis
# Area (A_1) M.I of area A_1 about y-axis
I_y1=(d1*(b1**3))/12 # mm**4
# M.I of the area A_1 about the centroidal y-axis of the composite area (By parallel-axis theorem)
d_3=x_c-(b1/2) # mm # distance between c &c1 along x axis
I_Y1=(I_y1)+(A_1*d_3**2) # mm**4
# Area(A_2) M.I of area A_2 about y-axis
I_y2=(d2*b2**3)/12 # mm**4
# M.I of the area A_2 about the centroidal y-axis of the composite area (By parallel-axis theorem)
d_4=b1-x_c-(b2/2) # mm # distance between c &c2 along x axis
I_Y2=(I_y2)+(A_2*d_4**2) # mm**4
# COMPOSITE AREA:M.O.I of the composite area about the centroidal y-axis
I_y=(I_Y1)+(I_Y2) # mm**4
# Results
print('The M.O.I of the composite area about the centroidal x-axis is %f mm**4'%I_x)
print('The M.O.I of the composite area about the centroidal Y-axis is %f mm**4'%I_y)
# NOTE: The answer for I_x given in text book is 0.76*10**6 insted of 10.76*10**6

The M.O.I of the composite area about the centroidal x-axis is 10761666.666667 mm**4
The M.O.I of the composite area about the centroidal Y-axis is 6086666.666667 mm**4

Example 12.14 Product of Inertia

In [7]:
from __future__ import division
import math
# Initilization of variables
b=1 # cm # smaller side of the L-section
h=4 # cm # larger side of the L-section
# Calculations
# (A) RECTANGLE A_1: Using the paralel axis theorem
Ixy=0
I_xy1=(Ixy)+((h*b)*(b/2)*(h/2)) # cm**4
# (B) RECTANGLE A_2: Using the paralel axis theorem
I_xy2=(Ixy)+((b*(h-1))*(1+(3/2))*(b/2)) # cm**4
# Product of inertia of the total area
I_xy=I_xy1+I_xy2 # cm**4
# Calculations
print('The Product of inertia of the L-section is %f cm**4'%I_xy)

The Product of inertia of the L-section is 7.750000 cm**4

Example 12.15 Principal Moment of Inertia

In [9]:
from __future__ import division
import math
# Initilization of variables
I_x=1548 # cm**4 # M.O.I of the Z-section about X-axis
I_y=2668 # cm**4 # M.O.I of the Z-section about Y-axis
b=12 # cm # width of flange of the Z-section
d=3 # cm # depth of flange of the Z-section
t=2 # cm # thickness of the web of the Z-section
h=6 # cm # depth of the web of the Z-section
#Calculations
A_1=b*d # cm**2 # area of top flange
x_1=-5 # cm # distance of the centroid from X-axis for top flange
y_1=4.5 # cm # distance of the centroid from Y-axis for top flange
A_2=t*h # cm**2 # area of web
x_2=0 # cm # distance of the centroid from X-axis for the web
y_2=0 # cm # distance of the centroid from Y-axis for the web
A_3=b*d # cm**2 # area of bottom flange
x_3=5 # cm # distance of the centroid from X-axis for top flange
y_3=-4.5 # cm # distance of the centroid from Y-axis for top flange
# Product of Inertia of the total area is,
I_xy=((A_1*x_1*y_1)+(A_3*x_3*y_3)) # cm**4
# The direction of the principal axes is,
theta_m=(math.degrees(math.atan((2*I_xy)/(I_y-I_x))))/2 # degree
# Principa M.O.I
I_max=((I_x+I_y)/2)+(math.sqrt(((I_x-I_y)/2)**2+(I_xy)**2)) # cm**4
I_mini=((I_x+I_y)/2)-(math.sqrt(((I_x-I_y)/2)**2+(I_xy)**2)) # cm**4
# Results
print('The principal axes of the section about O is %f degree'%theta_m)
print('The Maximum value of principal M.O.I is %f cm**4'%I_max)
print('The Minimum value of principal M.O.I is %f cm**4'%I_mini)

The principal axes of the section about O is -35.465403 degree
The Maximum value of principal M.O.I is 3822.059509 cm**4
The Minimum value of principal M.O.I is 393.940491 cm**4