# Chapter 9 : Distributed forces Moment of Inertia¶

## Example 9.4 Page No : 465¶

In :
import math

# Given Data
#Area of plate
A = 9*.75;			#in**2
y = 1./2*13.84+1/2*.75;			#in, y co-ordinate of centroid of the plate

# Calculations and Results
#All values for flange are from table from book
sumA = A+8.85;			#in**2 Total area
sumyA = y*A+0;			#in**3
Y = sumyA/sumA;			#in
#print (Y)
#Moment of inertia
#For wide flanfe
Ix1 = 291+8.85*Y**2;			#in**4
#for plate
Ix2 = 1./12*9*(3./4)**3+6.75*(7.295-3.156)**2;			#in**4
#For composite area
Ix = Ix1+Ix2;			#in**4

print "Moment of inertia Ix =  %.2e in**4 "%(Ix);

kx = math.sqrt(Ix/sumA);			#mm
print "Radius of gyration is kx =  %.1f in"%(kx);

Moment of inertia Ix =  4.86e+02 in**4
Radius of gyration is kx =  5.6 in


## Example 9.5 Page No : 466¶

In :
import math

#Given
r = 90.;			#mm, radius of half circle
b = 240.;			#mm, width
h = 120.;			#mm, height

# Calculations
#Moment of inertia of recmath.tangle
Ixr = 1./3*b*h**3;			#mm**4
#Moment of inertia of half circle
a = 4*r/(3*math.pi);			#mm
b = h-a;			#mm, Dismath.tance b from centroid c to X axis
I_AA = 1./8*math.pi*r**4;			#mm**4, Moment of inertia of half circle with respect to AA'
A = 1./2*math.pi*r**2;			#mm**2, Area of half circle
Ix1 = I_AA-A*a**2;			#mm**4, Parallel axis theorem
Ixc = Ix1+A*b**2;			#mm**4, Parallel axis theorem
#Moment of inertia of given area
Ix = Ixr-Ixc;			#mm**4

# Results
print "Moment of inertia of area about X axis is Ix =  %2.2e mm**4"%(Ix);

Moment of inertia of area about X axis is Ix =  4.59e+07 mm**4


## Example 9.7 Page No : 479¶

In :
import math

# Given Data
Ix = 10.38;			#in**4,Moment of inertia about x axis
Iy = 6.97;			#in**4,Moment of inertia about y axis

# Calculations and Results
Ixy = -3.28+0-3.28
print (Ixy)			#in in**4

#Principal axes
tan_2_theta_m = -(2*Ixy)/(Ix-Iy)
two_theta_m = math.degrees(math.atan(tan_2_theta_m))
theta_m = two_theta_m/2
print "Orientation of principle axes of section about O is Theta_m =  %.1f degree "%(theta_m);

#Principle moment of inertia, eqn 9.27
Imax = (Ix+Iy)/2+math.sqrt(((Ix-Iy)/2)**2+Ixy**2);			#mm**4
Imin = (Ix+Iy)/2-math.sqrt(((Ix-Iy)/2)**2+Ixy**2);			#mm**4

print "Principle moment of inertia of section about O are  Imax =  %.2e in**4  Imin =  %.0e  in**4"%(Imax,Imin);
#answer difference is due to roundoff

-6.56
Orientation of principle axes of section about O is Theta_m =  37.7 degree
Principle moment of inertia of section about O are  Imax =  1.55e+01 in**4  Imin =  2e+00  in**4