# Chapter 3: Static Forces on Surfaces. Buoyancy¶

## Example 3.1, Page 65¶

In :
from __future__ import division
import math

#Initializing  the  variables
a  =  2.7;                                                                #Upper  edge
b  =  1.2  ;                                                              #Lower  edge
width  =  1.5;                                                        #Width  of  trapezoidal  plate
h  =  1.1;                                                                #Height  of  water  column  above  surface
rho  =  1000;
g  =  9.81                                                                #Acceleration  due  to  gravity
phi  =  90                                                                #Angle  between  wall  and  surface

#Calculations
A  =  0.5*(a+b)*width;                                        #Area  of  Trapezoidal  Plate
y  =  (2*(0.5*width*0.75)*0.5  +  (1.2*width)*0.75)/A;
z  =  y+h;                                                                #Depth  of  center  of  pressure
R  =  rho*g*A*z                                                      #Resultant  force

I0  =  1.2*1.5**3/12  +1.2*1.5*1.85**2  +  1.5*1.5**3/36  +  1.5*0.75*1.6**2  #Second  moment  of  area
D  =  (math.sin(math.degrees(phi)))**2*I0/(A*z);                          #depth  of  center  of  pressure
M  =  R*(1.8533-1.1);                                          #Moment  about  hinge
print "Moment about the hinge line (kN/m):",round(M/1000)

Moment about the hinge line (kN/m): 38.0


## Example 3.2, Page 67¶

In :
from __future__ import division
import math

#Initializing  the  variables
w  =  1.8;                                                                #Width  of  plate
h1  =  5;                                                                  #Height  of  plate  and  water  in  upstream
h2  =  1.5;                                                              #Height  of  water  in  downstream
rho  =  1000;
g  =  9.81  ;                                                            #Acceleration  due  to  gravity

#Calculations
def waterForce(area,meanHeight):
F  =  rho  *  g  *  area  *  meanHeight;
return F

P  =  waterForce(w*h1,h1/2)-waterForce(w*h2,h2/2);#  Resultant  force  on  gate
x  =  (waterForce(w*h1,h1/2)*(h1/3)  -  waterForce(w*h2,h2/2)*(h2/3))/P;#  point  of  action  of  p  from  bottom
R  =  P/(2*math.sin(math.radians(20)));                                          #  Total  Reaction  force
Rt  =  1.18*R/4.8;                                                #Reaction  on  Top
Rb  =  R  -  Rt  ;                                                      #Reaction  at  bottom

print "Reaction at top    (kN):",round(Rt/1000,1)
print "Reaction at bottom (kN):",round(Rb/1000,2)

Reaction at top    (kN): 72.2
Reaction at bottom (kN): 221.45


## Example 3.3, Page 70¶

In :
from __future__ import division
import math

#Initializing  the  variables
D  =  1.8;                                                                #Depth  of  tank
h  =  1.2;                                                                #Depth  of  water
l  =  3;                                                                    #Length  of  wall  of  tank
p  =  35000;                                                            #Air  pressure
rho  =  10**3;                                                          #Density  of  water
g  =  9.81;                                                              #Acceleration  due  to  gravity

#Calculations
Ra  =  p*D*l;                                                          #Force  due  to  air
Rw  =  .5*(rho*g*h)*h*l;                                    #Force  due  to  water
R  =  Ra  +  Rw;                                                        #  Resultant  force
x  =  (Ra*0.9+Rw*0.4)/R;                                    #  Height  of  center  of  pressure  from  base
print "Resultant force on the wall                     (kN) :",round(R/1000,2)
print "Height of the centre of pressure above the base (m)  :",round(x,2)

Resultant force on the wall                     (kN) : 210.19
Height of the centre of pressure above the base (m)  : 0.85


## Example 3.4, Page 72¶

In :
from __future__ import division
import math

#Initializing  the  variables
R  =  6;                                                                    #  Radius  of  arc
h  =  2*R*math.sin(math.radians(30));                                              #Depth  of  water
rho  =  10**3;                                                          #Density  of  water
g  =  9.81;                                                              #Acceleration  due  to  gravity

#Calculations
Rh  =  (rho*g*h**2)/2;                                          #  Resultant  horizontal  force  per  unit  length
R  =  (Rh**2+Rv**2)**0.5;                                        #  Resultant  force  on  gate
theta  =  180/math.pi*math.atan(Rv/Rh);                                      #Angle  between  resultant  force  and  horizontal

print "Magnitute of resultant force                  (kN/m)   :",round(R/1000,2)
print "Direction of resultant force to the horizontal(Degrees):",round(theta,2)

Magnitute of resultant force                  (kN/m)   : 179.45
Direction of resultant force to the horizontal(Degrees): 10.27


## Example 3.5, Page 75¶

In :
from __future__ import division
import math

#Initializing  the  variables
B  =  6;                                                                    #  Width  of  pontoon
L  =  12;                                                                  #Length  of  pontoon
D  =  1.5;                                                                #Draught  of  pontoon
Dmax  =  2;                                                              #Maximum  permissible  draught
rhoW  =  1000;                                                        #Density  of  fresh  water
rhoS  =  1025;                                                        #Density  of  sea  water
g  =  9.81;                                                              #Acceleration  due  to  gravity

#Calculations
def Weight(D):
W  =  rhoW*g*B*L*D;
return W

W  =  Weight(D);                                                    #  Weight  of  pontoon  in  fresh  water  =  weight  of  water  displaced
Ds  =  W/(rhoS*g*B*L);                                        #Draught  in  sea  water
L  =  Weight(Dmax)  -  Weight(D);                      #  maximum  load  that  can  be  supported

print "Weight of pontoon (kN) :",round(W/1000,1)
print "Draught in sea    (m)  :",round(Ds,2)

Weight of pontoon (kN) : 1059.5
Draught in sea    (m)  : 1.46


## Example 3.6, Page 80¶

In :
from __future__ import division
import math

#Initializing  the  variables
D  =  1.8;                                                                #  Diameter  of  buoy
H  =  1.2;                                                                #Height  of  buoy
W  =  10*10**3;                                                        #Weight  of  buoy
G  =  0.45;                                                              #  Center  of  gravity
rho  =  1025;                                                          #Density  of  sea  water
g  =  9.81;                                                              #Acceleration  due  to  gravity

#Calculations
Z  =  4*(W+L)/(rho*g*math.pi*D**2);                        #  Depth  of  Immersion
BG  =  (math.pi*D**4/64)/(math.pi*D**2*Z/4);
Z  =  0.5*Z  +BG;                                                #  Position  of  combined  center  of  gravity
Z1  =  ((W+L)*Z-0.45*W)/L;                              #Maximum  height  of  load  above  bottom

print "Maximum height of center of gravity above bottom (m) :",round(Z1,3)

Maximum height of center of gravity above bottom (m) : 1.748


## Example 3.7, Page 83¶

In :
from __future__ import division
import math

#Initializing  the  variables
l  =  20;                                                                  #  Length  of  barage
b  =  6;                                                                    #Width  of  barage
r  =  3;                                                                    #Radius  of  circular  top  of  barage
W  =  200*10**3;                                                      #Weight  of  empty  barage
d1  =  0.8;                                                              #  Depth  of  water  in  1st  half
d2  =  1;                                                                  #  Depth  of  water  in  2nd  half
rho  =  1000;                                                          #Density  of  water
R  =  0.8;                                                                #Relative  density  of  liquid
g  =  9.81;                                                              #Acceleration  due  to  gravity
ZG  =  0.45;                                                            #  Center  of  gravity  of  barage

#Calculations
I00  =  l*b**3/12  +math.pi*b**4/128;
ICC  =  l*(.5*b)**3/12;
L  =  d1*rho*g*l*b/2*(d1+d2);                          #  Weight  of  liquid  load
W  =  L  +  W;                                                          #Total  weight
A  =  l*b  +math.pi*r**2/2;                                          #  Area  of  plane  of  waterline
V  =  W/(rho*g);                                                  #  Volume  of  vessel  submerged
D  =  V/A  ;                                                              #Depth  submerged
ZB  =  .5*D;                                                            #Height  of  center  of  buoyancy
NM  =  ZB-ZG  +(1/V)*(I00-R*2*ICC);                #  Effective  metacentric  height
P  =  R*rho*g*l*b/2*(d2-d1);                            #overturning  moment
theta  =  math.atan(P*1.5/(W*NM))*180/math.pi;                      #Angle  of  roll
# converting into degrees and minutes

Angle of roll is 2.0 degrees 0.37 minutes