# Chapter 5 : Distrubuted forces centroids and centers of gravity¶

## Example 5.1 Page No : 228¶

In [2]:
import math

# Given Data
n = 4; 			# no of component
A = [120*80,120*60/2,math.pi*60*60/2,-math.pi*40*40];			#mm**2, Areas of Recmath.tangle, triangle, Semicircle, and Circle respectively
x = [60,40,60,60];			#mm, x components of centroids of Recmath.tangle, triangle, Semicircle, and Circle respectively
y = [40,-20,105.46,80];			#mm, y components of centroids of Recmath.tangle, triangle, Semicircle, and Circle respectively

sumA = 0;
sumxA = 0;
sumyA = 0;

# Calculations and Results
for i in range(n):
sumA = sumA+A[i];
sumxA = sumxA+x[i]*A[i];
sumyA = sumyA+y[i]*A[i];

# First Moment of area
Qx = sumyA;			# About X axis
print "First moments of the area are Qx =  %.0f mm**3 and Qy = %.0f mm**3 "%(Qx,Qy);

#Location of centroid
X = sumxA/sumA;			# X co-ordinate
Y = sumyA/sumA;			# Y co = ordinate
print "Co-ordinates of centroid are X =  %.1f mm and Y =  %.1f mm "%(X,Y);

First moments of the area are Qx =  506238 mm**3 and Qy = 757699 mm**3
Co-ordinates of centroid are X =  54.8 mm and Y =  36.6 mm


## Example 5.2 Page No : 229¶

In [3]:
import math

# Given Data
n = 3; 			# no of segment
L = [600,650,250];			#mm, Lengths of segment AB , BC and CA respectively
x = [300,300,0];			#mm, x components of centroids of segment AB , BC and CA respectively
y = [0,125,125];			#mm, y components of centroids of segment AB , BC and CA respectively

sumL = 0;
sumxL = 0;
sumyL = 0;

# Calculations
for i in range(n):
sumL = sumL+L[i];
sumxL = sumxL+x[i]*L[i];
sumyL = sumyL+y[i]*L[i];

#Location of centre of gravity
X = sumxL/sumL;			# X co-ordinate
Y = sumyL/sumL;			# Y co = ordinate

# Results
print "Co-ordinates of centroid are X =  %.0f mm and Y =  %.0f mm "%(X,Y);
#There is variation because of roundoff

Co-ordinates of centroid are X =  250 mm and Y =  75 mm


## Example 5.7 Page No : 242¶

In [4]:
import math

# Given Data
p = 7850.;			#kg/m**3, density of steel rim
n = 2; 			# no of component
A = [(20+60+20)*(30+20),-60*30];			#mm**2,Cross section Areas of recmath.tangle I and II

y = [375,365];			#mm, y components of centroids of Recmath.tangles I and II respectively
sumV = 0;

C = [0,0]
V = [0,0]
# Calculations
for i in range(n):
C[i] = 2*math.pi*y[i];			#mm, Dismath.tance travelled by C
V[i] = A[i]*C[i];			#mm**3, Volume of 1 component
sumV = sumV+V[i];			# mm**3, Total volume of rim

sumV = sumV*10**(-9);			#Conversion into m**3
g = 9.81;			#m/s**2, acceleration due to gravity
m = p*sumV;			#kg, mass
W = m*g;			#N, Weight

# Results
print "mass of steel is m =  %.0f kg and Wight is W =  %.0f N"%(m,W);

mass of steel is m =  60 kg and Wight is W =  589 N


## Example 5.9 Page No : 250¶

In [5]:
# Given Data
n = 2; 			# no of triangle
x = [2.,4.];			#mm, dismath.tances of centroid from point A

# Calculations and Results
sumA = 0;
sumxA = 0;
for i in range(n):
sumA = sumA+A[i];
sumxA = sumxA+x[i]*A[i];

#Location of centroid
X = sumxA/sumA;			# X co-ordinate
W = sumA;			#kN, Concentrated load
print "The equivalent concentrated mass is W =  %.0f kN and its line of action is located at a\
\n distance X =  %.1f m to the right of A "%(W,X);

# Reactions
# Applying sum(F_x) = 0
Bx = 0.;			#N
#Applying sum(M_A) = 0
By = W*X/6.;			#kN, Reaction at B in Y direction
#Applying sum(M_B) = 0
A = W*(6-X)/6;			#kN, Reaction at B in Y direction

print "The rection at A = %.1f kN, At Bx = %.1f kN and By = %.1f kN "%(A,Bx,By);

The equivalent concentrated mass is W =  18 kN and its line of action is located at a
distance X =  3.5 m to the right of A
The rection at A = 7.5 kN, At Bx = 0.0 kN and By = 10.5 kN


## Example 5.10 Page No : 251¶

In [7]:
# Given Data
t = 0.3;			#m thickness of dam
g = 9.81;			# m/s**2, acceleration due to gravity
p1 = 2400.;			#kg/m**3, density of concrete
p2 = 1000.;			#kg/m**3, density of water
W1 = 0.5*2.7*6.6*t*p1*g/1000;			#kN, Weight of concrete component 1
W2 = 1.5*6.6*t*p1*g/1000;			#kN, Weight of concrete component 2
W3 = 1./3*3*5.4*t*p1*g/1000;			#kN, Weight of concrete component 3
W4 = 2./3*3*5.4*t*p2*g/1000;			#kN, Weight of water
P = 0.5*2.7*6.6*t*p1*g/1000;			#kN, pressure force exerted by water

# Calculations and Results
# Applying sum(F_x) = 0
H = 42.9;			#kN, Horizontal reation at A
#Applying sum(Fy) = 0
V = W1+W2+W3+W4;			#kN, Vertical Reaction at A
print "The horizontal reaction is H = %.1f kN \
\nVertical rection at A V = %.1f kN "%(H,V);
#Applying sum(M_A) = 0
M = W1*1.8+W2*3.45+W3*5.1+W4*6-P*1.8;			#kN.m, Moment at A

# We can replace force couple system by math.single force acting at dismath.tance right to A
d = M/V;			# m Dismath.tance of resultant force from A

print "The moment about A is M = %.1f kN.m anticlockwise and  \
\nif we replace it by force couple system resultant, s distance from A is d =  %0.2f m  "%(M,d);
#Difference is because of round off

The horizontal reaction is H = 42.9 kN
Vertical rection at A V = 202.8 kN
The moment about A is M = 626.5 kN.m anticlockwise and
if we replace it by force couple system resultant, s distance from A is d =  3.09 m


## Example 5.11 Page No : 263¶

In [9]:
import math

# Given Data
n = 3; 			# no of component
l = 100.;			#mm length of cylinder
V = [0.5*4./3*math.pi*(r)**3,math.pi*r*r*l,-math.pi/3*r*r*l];			#mm**3, Volumes of Hemisphere, cylinder and cone respectively
x = [-3./8*r,l/2.,3./4*l];			#mm, x components of centroids of Hemisphere, cylinder and cone respectively

sumV = 0;
sumxV = 0;

# Calculations
for i in range(n):
sumV = sumV+V[i];
sumxV = sumxV+x[i]*V[i];

#Location of centre of gravity
X = sumxV/sumV;			# X co-ordinate

# Results
print "Co-ordinates of centroid are X =  %.0f mm "%(X);

Co-ordinates of centroid are X =  15 mm


## Example 5.12 Page No : 264¶

In [10]:
import math

# Given Data
l = 4.5; 			# in in
b = 2.;			#in
h = .5;			#in
a_I = l*b*h
a_II = ((1./4)*math.pi*b**2*h)
a_III = -math.pi*(h**2)*h
a_IV = -math.pi*(h**2)*h
V = [a_I, a_II, a_III, a_IV]
#print (V)

x = [.25,1.3488,.25,.25];			#in, x components of centroids of part I,II , III and IV respectively
y = [-1,-0.8488,-1,-1];			#in, y components of centroids of part I,II , III and IV respectively
z = [2.25,0.25,3.5,1.5];			#in, z components of centroids of part I,II , III and IV respectively

# Calculations and Results
y = [0,0,0,0]
for i in range(4):
temp = 0
sum_xV = 0
sum_xV = V[i]*x[i]
y[i] = sum_xV

x = sum(y)
print "The sum of x*V = %f  in**4 "%(x)

for i in range(4):
temp = 0
sum_zV = 0
sum_zV = V[i]*z[i]
y[i] = sum_zV

z = sum(y)
print "The sum of z*V = %f  in**4 "%(z)

for i in range(4):
temp = 0
sum_yV = 0
sum_yV = V[i]*y[i]
y[i] = sum_yV

s = sum(y)
print "The sum of y*V = %f  in**4 "%(s)

The sum of x*V = 3.047341  in**4
The sum of z*V = 8.554204  in**4
The sum of y*V = 46.950413  in**4