APPENDIX A:Geomatric Properties of an Area

Example A.1:Page no. 786

In [7]:
#Given
#From fig. A-4(a)   The given dimensions are
l1=8       #inch
l2=3       #inch
l3=10      #inch
l4=5       #inch
l5=11.5    #inch
l6=2       #inch

#calculation
ymean1=((l4*l3*l6)+(l5*l2*l1))/((l3*l6)+(l2*l1))
#From fig. A-4(b)
l1_=-8       #inch
l2_=3       #inch
l3_=10      #inch
l4_=-1.5    #
l5_=2       #inch
ymean2=((l4_*l2_*-l1_)+(l1_*l3_*l5_))/((l2_*-l1_)+(l3_*l5_))
d=ymean1-ymean2     #Depth of beam
#From fig. A-4(c)
la=8       #inch
lb=6.5       #inch
lc=10       #inch
ld=13        #
le=5        #inch
lf=3        #inch
ymean3=((lb*ld*la)-2*(le*lc*lf))/((ld*la-2*(lc*lf)))
print"Location of centroid in fig (a)is",ymean1,"inch" 
print"Location of centroid in fig (b)is",ymean2,"inch"
print"Location of centroid in fig (c)is",ymean3,"inch"

Location of centroid in fig (a)is 8.54545454545 inch
Location of centroid in fig (b)is -4.45454545455 inch
Location of centroid in fig (c)is 8.54545454545 inch

Example A.2:Page no 789

In [17]:
#Given
#Dimension in the fig.A-7 a
#The given dimensions are
l1=8       #inch
l2=8.55    #inch
l3=10      #inch
l4=5       #inch
l5=1.5     #inch
l6=2       #inch
l7=4.45    #inch

#Calculation
Ix1=(1/12.0*l6*l3**3)
A1=l6*l3
dy1=(l2-l4)
Ix2=(1/12.0*l1*(l5+l5)**3)
A2=l1*(l5+l5)
dy2=(l7-l5)
I1=(Ix1+A1*dy1**2)+(Ix2+A2*dy2**2)
print I1

#Dimension in the fig.A-7 b
l1_= 13      #inch
l2_= 3      #inch
l3_=10      #inch
l4_=5       #inch
l5_= 2   #inch
l6_= 6.5      #inch
l7_=4.45    #inch
l8_=8.55    #inch
l9_=6.5    #inch

Ix1_=(1/12.0*l1_*(l2_+l5+l2_)**3)
A1_=l1_*(l2_+l5+l2_)
dy1_=(l8_-l9_)
Ix2_=(1/12.0*l2_*(l3_)**3)
A2_=l2_*(l3_)
dy2_=(l7_-l4_)
I2=(Ix1+A1*dy1**2)+(Ix2+A2*dy2**2)

#Result
print"Moment of inertia for fig a is",round(I1,0),"inch**4"
print"Moment of inertia for fig a is",round(I2,0),"inch**4"

645.576666667
645.576666667
Moment of inertia for fig a is 646.0 inch**4
Moment of inertia for fig a is 646.0 inch**4

Example A.3 Page no: 790

In [2]:
#Given
#From fig A-8(a)
#The given dimensions are
l1=100    #mm
l2=400    #mm
l3=600    #mm
dx=250    #mm
dy=200     #mm

#Calculation
#Rectangle A:
Ix1=(1/12.0*l1*(l2-l1)**3)
Ady=(l1*(l2-l1)*dy**2)
Ix=(Ix1+Ady)
Iy1=(1/12.0*(l2-l1)*l1**3)
Adx=(l1*(l2-l1)*dx**2)
Iy=(Iy1+Adx)

#Rectangle B:
Ix_=(1/12.0*l3*l1**3)
Iy_=(1/12.0*l1*l3**3)

#Rectangle C
Ix3=(1/12.0*l1*(l2-l1)**3)
Ady_=(l1*(l2-l1)*200**2)
Ix3_=(Ix3+Ady_)
Iy3=(1/12.0*(l2-l1)*l1**3)
Adx_=(l1*(l2-l1)*dx**2)
Iy3_=(Iy3+Adx)

#Total Moment of inertia
Itx=(Ix+Ix_+Ix3_)
Ity=(Iy+Iy_+Iy3_)

#Result
print"Moment of inertia across x is ",Itx,"mm**4"
print"Moment of inertia across y is ",Ity,"mm**4"

Moment of inertia across x is  2900000000.0 mm**4
Moment of inertia across y is  5600000000.0 mm**4

Example A.4 :page no. 793

In [3]:
#Given
#From fig A-12 a
#The given length of sides are
l1=100      #mm
l2=300      #mm
dy=200      #mm
dx=250      #mm

#Calculation
#Rectangle A
Ixy1=0
A1=l1*l2
Ixy1=Ixy1+A1*(-dx)*dy

#Rectangle B
Ixy2=0
A2=0
Ixy2=Ixy2+A2*dx*dy

#Rectangle D
Ixy3=0
A3=l1*l2
Ixy3=Ixy3+A3*(dx)*(-dy)
Ixy=Ixy1+Ixy2+Ixy3

#Result
print"The moment of inertia is",Ixy,"mm**4"

The moment of inertia is -3000000000 mm**4

Example A.5 :page no. 796

In [22]:
#Given
#From fig A-15 and From Example A.3 and A.4
Ix=2.9*10**9           #moment of inertia along x
Iy=5.6*10**9           #moment of inertia along y
Ixy=-3*10**9           #moment of inertia along xy

#Calculation
import math
#Using eq. A11
import math
thetaP1=1/2.0*math.atan(-Ixy*2/(Ix-Iy))*100
#As shown in fig. A-15
thetaP2=-32.9      #degree
Imax=(Ix+Iy)/2.0+math.sqrt((((Ix-Iy)/2.0)**2)+Ixy**2)
Imin=(Ix+Iy)/2.0-math.sqrt((((Ix-Iy)/2.0)**2)+Ixy**2)

#Result
print"Maximum moment of inertia is",Imax,"mm**4"
print"Minimum moment of inertia is",Imin,"mm**4"

Maximum moment of inertia is 7539756829.92 mm**4
Minimum moment of inertia is 960243170.081 mm**4

Example A.6 :page no. 799

In [33]:
#given
#From fig. A-17 a and Example  A.3 and A.4
Ix=2.9*10**9           #mm**4, moment of inertia
Iy=5.6*10**9
Ixy=-3*10**9

#Calculation
import math
d=(Ix+Iy)/2.0   #distance of centre of circle       
#from fig A-17 b
BC=1.35
AB=3
CA=math.sqrt(BC**2+AB**2)

#the circle intersect the I axis at point (7.54,0) and  (0.960,0) hence
Imax=7.54*(10**9)  #mm**4
Imin=0.960*(10**9)  #mm**4
thetap1=1/2.0*(180-(math.atan(AB/BC))*180/math.pi)

#Result
print"The maximum moment of inertia is",Imax,"mm**4"
print"The minimum moment of inertia is",Imin,"mm**4"
print"The angle is ",round(thetap1,1)

The maximum moment of inertia is 7540000000.0 mm**4
The minimum moment of inertia is 960000000.0 mm**4
The angle is  57.1