# Example 2.1 Page number 21¶

In [43]:
import math
#F                force
#hd               horizontal distance
#vd               vertical  distance
#O                angle
#M                moment of force
#Taking clockwise moment as positive
#calculations
F=100.0
hd=400.0
vd=500.0
o=60.0
M=F*(math.cos(o/180.0*3.14)*vd-math.sin(o/180.0*3.14)*hd)
print '%s %.2f %s' %("\n \n  Moment is =", M ," Nmm clockwise");


Moment is = -9607.41  Nmm clockwise


# Example 2.2 Page number 21¶

In [44]:
import math
#F                force
#hd               horizontal distance
#vd               vertical  distance
#O                angle
#M                moment of force
#Taking clockwise moment as positive
#calculations
F=5000.0
o=37
M=8000.0
hd=M/(F*math.cos(o*3.14/180))
print '%s %.2f %s' %("\n \n  Distance =", hd ,"m");


Distance = 2.00 m


# Example 2.3 Page number 25¶

In [45]:
import math
#F                force
#hd               horizontal distance
#vd               vertical  distance
#O                angle
#M                moment of force
#Taking clockwise moment as positive
#calculations
F=5000.0
o=37
M=8000.0
hd=M/(F*math.cos(o*3.14/180))
print '%s %.2f %s' %("\n \n  Distance =", hd ,"m");


Distance = 2.00 m


# Example 2.4 Page number 25¶

In [46]:
import math
#R             resultant force
#Rx            resultant horizontal component
#Ry            resultant vertical component
#f1             force
#f2             force
#f3             force
#o1             angle with the line
#o2             angle with the line
#o3             angle with the line
#O              angle of resultant force with line
f1=70.0
f2=80.0
f3=50.0
o1=50.0
o2=25.0
o3=-45.0
Rx=(f1*math.cos(o1/180*3.14)+f2*math.cos(o2/180*3.14)+f3*math.cos(o3/180*3.14));
Ry=(f1*math.sin(o1/180*3.14)+f2*math.sin(o2/180*3.14)+f3*math.sin(o3/180*3.14));
R=math.sqrt(Rx**2+Ry**2)
O=math.atan(Ry/Rx)
print '%s %.2f %s' %("\n \n  Resultant Force =", R ,"N");
print '%s %.2f %s' %("\n \n  Resultant angle =", O ,"radians");


Resultant Force = 161.52 N



# Example 2.5 Page number 26¶

In [47]:
import math
#O       angle of inclined plane
#N       normal reaction
#W       weight
#F,T     forces
#Rx     resultant horizontal component
#Ry     resultant vertical component
o = 60.0
W = 1000.0
N = 500.0
F = 100.0
T = 1200.0
Rx = T-F-(W*math.sin(o/180*3.14))
Ry =  N-(W*math.cos(o/180*3.14))
print '%s %.2f %s' %("\n \n  Resultant Force  along the incline plane =", Rx ,"N");
print '%s %.2f %s' %("\n \n  Resultant Force  vertical to  the incline plane =", Ry ,"N");


Resultant Force  along the incline plane = 234.24 N

Resultant Force  vertical to  the incline plane = -0.46 N


# Example 2.6 Page number 26¶

In [48]:
import math
R=1000.0   #Resultant force
F1=500.0    #Force
F2=1000.0   #force
o=45.0*3.14/180.0              #angle resultant makes with  x axis
o1=30.0*3.14/180.0             #angle  F1 makes with  x axis
o2=60.0*3.14/180.0              #angle  F2 makes with  x axis
#F3coso3=Rcoso-F1coso1-F2sino2
#F3sino=Rsino-F1sino1-F2coso2
F3=((R*math.cos(o)-F1*math.cos(o1)-F2*math.cos(o2))**2+(R*math.sin(o)-F1*math.sin(o1)-F2*math.sin(o2))**2)**0.5
print "Force",F3,"N"
o3=180/3.14*math.atan((R*math.sin(o)-F1*math.sin(o1)-F2*math.sin(o2))/(R*math.cos(o)-F1*math.cos(o1)-F2*math.cos(o2)))
print "At an angle",o3

Force 467.201871561 N
At an angle 61.0805191269


# Example 2.7 Page number 27¶

In [49]:
from math import cos,sin,atan,pi,asin

#variable declaration

P1=300.0
P2=500.0
thetaI=30.0*pi/180.0
thetaP2=30.0*pi/180
thetaP1=40.0*pi/180
# Let the x and y axes be If the resultant is directed along the x axis, its component in y direction is zero.
#Taking horizontal direction towards left as x axis and the vertical downward direction as y axis.
##sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

F=(P2*sin(thetaP2))/(P1)
theta=(asin((F/(cos(20*pi/180)*2)))*180/pi)-20

print"theta=",round(theta,2),"°"

theta= 6.32 °


# example2.8 page number 30¶

In [50]:
from math import cos,sin,atan,sqrt,pi

#variable declaration

P1=20.0
P2=30.0
P3=20.0
theta3=60.0*pi/180.0

#Taking horizontal direction towards left as x axis and the vertical downward direction as y axis.
##sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

Fx=20.0*cos(theta3)
Fy=P1+P2+P3*sin(theta3)

R=sqrt(pow(Fx,2)+pow(Fy,2))
print "R=",round(R,4),"KN"

alpha=atan(Fy/Fx)*180/pi
print"alpha=",round(alpha,2),"°"

#moment at A

MA=P1*1.5+P2*3.0+P3*sin(theta3)*6.0

#The distance of the resultant from point O is given by:

d=MA/R
x=d/sin(alpha*pi/180)
print"x=",round(x,3),"m"

R= 68.0592 KN
alpha= 81.55 °
x= 3.326 m


# example2.9 page number 31¶

In [51]:
from math import cos,sin,atan,sqrt

#variable declaration

thetaA=60.0*pi/180.0
thetaB=30.0*pi/180.0

#Taking horizontal direction towards left as x axis and the vertical downward direction as y axis.
##sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

Fx=PB2-PA*cos(thetaA)-PC*cos(thetaB)
Rx=-Fx

Fy=PB1+PC*sin(thetaB)-PA*sin(thetaA)
Ry=Fy

R=sqrt(pow(Rx,2)+pow(Ry,2))
print "R=",round(R,2),"KN"

alpha=atan(Fy/Fx)*180/pi
print"alpha=",round((-alpha),2),"°"

#Let x be the distance from A at which the resultant cuts AC. Then taking A as moment centre,

x=(PB1*100*sin(thetaA)+PB2*50+PC*sin(thetaB)*100)/Ry
print"x=",round(x,3),"mm"

R= 91.19 KN
alpha= 35.84 °
x= 317.023 mm


# example 2.10 page number 32¶

In [52]:
from math import sqrt,pi,atan

#variable declaration

a=1.0        #length of side,m

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Fx=HB-HD
Fy=PC-PA

R=sqrt(pow(Fx,2)+pow(Fy,2))
print "R=",round(R,2),"N"

theta=atan(Fy/Fx)*180/pi
print"theta=",round(theta),"°"

#moment at A

MA=PC*a+HD*a

#Let x be the distance from A along x axis, where resultant cuts AB.

x=MA/Fy

print"x=",round((-x),1),"m"

R= 565.69 N
theta= 45.0 °
x= 2.5 m


# example 2.11 page number 32¶

In [53]:
from math import sqrt,pi,atan,cos,sin

#variable declaration

#Let O be the centre of the encircling circle A, B, C, D, E and F. In regular hexagon each side is equal to the radius AO. Hence OAB is equilateral triangle.

angleoab=60.0*pi/180
anglecab=angleoab/2.0
theta1=anglecab
theta2=(angleoab-theta1)
theta3=theta1
theta4=theta1

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Fx=PB*cos(theta1+theta2)+PC*cos(theta2)+PD+PE*cos(theta3)+PF*cos(theta3+theta4)

Fy=-PB*sin(theta1+theta2)-PC*sin(theta2)+0+PE*sin(theta3)+PF*sin(theta3+theta4)

R=sqrt(pow(Fx,2)+pow(Fy,2))
print "R=",round(R,2),"KN"

theta=atan(Fy/Fx)*180/pi
print"theta=",round(theta),"°","i.e.",",","the resultant is in the direction x."

R= 10.0 KN
theta= 0.0 ° i.e. , the resultant is in the direction x.


# example 2.12 page number 33¶

In [54]:
from math import sqrt,pi,atan,cos,sin

#variable declaration

a=10.0       #side length,mm

# If theta1, theta2 and theta3 are the slopes of the forces 2 kN, 5 kN and 1.5 kN forces with respect to x axis, then

theta1=atan(a/a)
theta2=atan((3*a)/(4*a))
theta3=atan((a)/(2*a))

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Fx=P1*cos(theta1)+P3*cos(theta2)-P2*cos(theta3)

Fy=P1*sin(theta1)-P3*sin(theta2)-P2*sin(theta3)

R=sqrt(pow(Fx,2)+pow(Fy,2))
print "R=",round(R,2),"N"

alpha=atan(Fy/Fx)*180/pi
print"alpha=",round((-alpha),2),"°"

#Distance d of the resultant from O is given by
#Rd=sum of moment at A

d=((a*3)*P1*cos(theta1)+(5*a)*P3*sin(theta2)+P2*(a)*sin(theta3))/(4.66)
print"d=",round(d,2),"mm"

#Note: To find moment of forces about O, 2 kN force is resolved at it’s intersection with y axis and 5 kN and 1.5 kN forces are resolved at their intersection with x axis, and then Varignon theorem is used

R= 4.66 N
alpha= 28.99 °
d= 42.73 mm


# example 2.13 page number 34¶

In [55]:
from math import sqrt,pi,atan,cos,sin

#variable declaration

AB=1.0
BC=2.0
CD=1.0
#length are in m

# Let x and y axes be selected
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Rx=0
Ry=PA+PB+PC+PD

R=sqrt(pow(Rx,2)+pow(Ry,2))
print "R=",round(R,2),"KN"

#Taking clockwise moment as positive,
#sum of moment at A

MA=(0)*PA+(AB)*PB+PC*(AB+BC)+PD*(AB+BC+CD)
print"MA=",round(MA,2),"KN-m"

# The distance of resultant from A is,

x=MA/R
print "x=",round(x,1),"m"

R= 150.0 KN
MA= 270.0 KN-m
x= 1.8 m


# example 2.14 page number 35¶

In [56]:
from math import sqrt,pi,atan,cos,sin

#variable declaration

AB=2.0
BC=2.0
CD=4.0
DE=2.0
#length are in m

# Let x and y axes be selected
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Rx=0
Ry=PA-PB+PC-PD+PE

R=sqrt(pow(Rx,2)+pow(Ry,2))
print "R=",round(R,2),"KN","in y-direction"

#Taking clockwise moment as positive,
#sum of moment at A

MA=(0)*PA-(AB)*PB+PC*(AB+BC)-PD*(AB+BC+CD)+PE*(AB+BC+CD+DE)

print"MA=",round(MA,2),"KN-m"

# The distance of resultant from A is,

x=MA/R
print "x=",round(x),"m"

R= 100.0 KN in y-direction
MA= 300.0 KN-m
x= 3.0 m


# example 2.15 page number 35¶

In [57]:
from math import pi,atan,sin ,cos,sqrt

#variable declaration
a=150.0
theta=60.0*pi/180
#assume Resulat R at distance x from O,
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Rx=P1*cos(theta)-H
Ry=-P3-P2-P1*sin(theta)

R=sqrt(pow(Rx,2)+pow(Ry,2))
print "R=",round(R,2),"KN","in y-direction"

alpha=atan(Ry/Rx)*180/pi
print"alpha",round(alpha,2),"°"

#Let the point of application of the resultant be at a distance x from the point O along the horizontal arm. Then,

x=(P1*sin(theta)*(2*a)+P2*a-P3*a*cos(theta)+H*a*2*sin(theta))/(-Ry)
print"x=",round(x,3),"mm"

R= 2671.19 KN in y-direction
alpha 80.3 °
x= 141.195 mm


# example 2.16 page number 36¶

In [58]:
from math import pi,atan,sin ,cos,sqrt

#variable declaration
ah=4.0
a1=2.0
a2=4.0
a3=5.0
a=7.0
#assume Resulat R at distance x from O,
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Rx=H
Ry=P1-P2+P3

print "Ry=",round(Ry,2),"KN","downward"

#Let x be the distance from O where the resultant cuts the base.
#moment at O
x=(H*ah+P1*a1-P2*a2+P3*a3)/(Ry)

print"x=",round(x,3),"m"

print "The resultant passes through the middle third of the base i.e., between 7/3m, and 2*7/3 m.Hence, the dam is safe."

Ry= 1420.0 KN downward
x= 4.127 m
The resultant passes through the middle third of the base i.e., between 7/3m, and 2*7/3 m.Hence, the dam is safe.


# example 2.17 page number 37¶

In [59]:
from math import pi,atan,sin ,cos,sqrt

#variable declaration
a=1.0

theta=45.0*pi/180.0
#The roof is inclined at 45° to horizontal and loads are at 90° to the roof. Hence, the loads are also inclined at 45° to vertical/horizontal.

#assume Resulat R at distance d from A,
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
Rx=(P1+P2+P3+P4+P5+P6+P7+P8)*cos(theta)
Ry=-(P1+P2+P3+P4)*sin(theta)+(P5+P6+P7+P8)*sin(theta)

print "R=",round(Rx,3),"KN"
#and its direction is horizontal
#Let R be at a distance d from the ridge A
#moment at A
d=((P1*3*cos(theta)*a+P2*cos(theta)*2*a+P3*cos(theta)*a)*2)/(Rx)

print"d=",round(d,1),"m"," Resultant is a horizontal force of magnitude",round(Rx,3)," at",round(d,1)," m below A."

R= 42.426 KN
d= 1.5 m  Resultant is a horizontal force of magnitude 42.426  at 1.5  m below A.


# example 2.18 page number 37¶

In [60]:
from math import sqrt,pi,atan,cos,sin

#variable declaration
#The two 40 kN forces acting on the smooth pulley may be replaced by a pair of 40 kN forces acting at centre of pulley C and parallel to the given forces, since the sum of moments of the two given forces about C is zero

PD=50.0        #inclined at 30.0 at distance 2m form A,KN
PE=20.0        #inclined at alpha at distance xm form A,KN

thetaA=45.0*pi/180.0
thetaB=60.0*pi/180.0
thetaC1=30.0*pi/180.0
thetaC2=20.0*pi/180.0
AC=3.0
AB=6.0

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

R=sqrt(pow(Fx,2)+pow(Fy,2))
print "R=",round(R,2),"KN"

alpha=atan(Fy/Fx)*180/pi
print"alpha=",round(alpha,2),"°"

#Let the resultant intersect AB at  a distance x from A. Then,

print"x=",round(X,2),"m"

print"The equilibriant is equal and opposite to the resultant in which E = 116.515 kN, alpha= 76.82° and ","x=",round(X,2),"m"

R= 116.52 KN
alpha= 76.82 °
x= 1.48 m
The equilibriant is equal and opposite to the resultant in which E = 116.515 kN, alpha= 76.82° and  x= 1.48 m


# example 2.19 page number 42¶

In [61]:
from math import sin,cos,pi

#Free body diagram of the sphere  shows all the forces moving away from the centre of the ball. Applying Lami’s theorem to the system of forces.

#variable declaration
W=100.0                #weight of sphere,N
theta=15.0*pi/180      #angle of inclination of string with wall

T=(W*sin((pi/2)))/sin((pi/2)+theta)
R=(W*sin((pi-theta)))/sin((pi/2)+theta)
print"T=",round(T,2),"N"
print"R=",round(R,2),"N"

#The above problem may be solved using equations of equilibrium also. Taking horizontal direction as x axis and vertical direction as y axis,

#Notes:
#1. The string can have only tension in it (it can pull a body), but cannot have compression in it (cannot push a body).
#2. The wall reaction is a push, but cannot be a pull on the body.
#3. If the magnitude of reaction comes out to be negative, then assumed direction of reaction is wrong. It is acting exactly in the opposite to the assumed direction. However, the magnitude will be the same. Hence no further analysis is required. This advantage is not there in using Lami's equation. Hence, it is advisable for beginners to use equations of equilibrium, instead of Lami's theorem even if the body is in equilibrium under the action of only three forces.

T= 103.53 N
R= 26.79 N


# example 2.20 page number 43¶

In [62]:
from math import sin,cos,pi

#The body is in equilibrium under the action of applied force P, self-weight 1500 N and normal reaction R from the plane. Since R, which is normal to the plane, makes 30° with the vertical (or 60° with the horizontal),

#variable declaration
W=1500.0                #weight of block,N
theta=30.0*pi/180      #angle of inclination

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

R=W/cos(theta)
print"R=",round(R,2),"N"

P=R*sin(theta)
print"P=",round(P,2),"N"

#Note: Since the body is in equilibrium under the action of only three forces the above problem can be solved using Lami’s theorem

R= 1732.05 N
P= 866.03 N


# example 2.21 page number 42¶

In [63]:
from math import sin,cos,pi

#A bar can develop a tensile force or a compressive force. Let the force developed be a compressive force S (push on the cylinder).

#variable declaration
W=10.0                #weight of Roller,KN

thetaS=30.0*pi/180.0

#Since there are more than three forces in the system, Lami’s equations cannot be applied. Consider the components in horizontal and vertical directions.
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

S=(-H+IL*cos(theta))/cos(thetaS)
print"S=",round(S,3),"N"

print"Since the value of S is negative the force exerted by the bar is not a push, but it is pull (tensile force in bar) of magnitude",round(-S,3) ,"kN."

R=W+IL*sin(theta)-S*sin(thetaS)
print"R=",round(R,3),"kN"

S= -0.058 N
Since the value of S is negative the force exerted by the bar is not a push, but it is pull (tensile force in bar) of magnitude 0.058 kN.
R= 14.979 kN


# example2.22 page number 44¶

In [64]:
from math import cos,sin,pi,asin

#The pulley C is in equilibrium under the action of tensile forces in CA and CB and vertical downward load 200 N. The tensile forces in segment CA and CB are the same since the pulley is frictionless. Now consider the equilibrium of pulley C
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

#variable declaration
AB=3.0
BI=1.0
ACB=5.0  #Length of cord,m
DE=3.0
BE=4.0
theta=asin(4.0/5.0)
#assume T is tension in string making angle theta1 & theta2,solving horizontal we find theta1=theta2,lets called them theta ,as triangleCFD=triangle=CFA.so, CD=AC

HI=BI*DE/BE
AH=DE-HI
x=AH/2
print"x=",round(x,3),"m"

T=L/(2*sin(theta))
print"T=",round(T),"N"

x= 1.125 m
T= 125.0 N


# example2.23 page number 45¶

In [65]:
from math import sin ,acos, pi,cos

#When the roller is about to turn over the curb, the contact with the floor is lost and hence there is no reaction from the floor. The reaction R from the curb must pass through the intersection of P and the line of action of self weight, since the body is in equilibrium under the action of only three forces (all the three forces must be concurrent).

#variable declaration
W=2000.0               #weight of roller,N
h=150.0                # height of curb,mm
OC=r-h
AO=r

alpha=acos(OC/AO)

#angleOAB=angleOBA,Since OA=OB,
angleOBA=(alpha)/2

#the reaction makes 30° with the vertical
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

R=W/cos(angleOBA)
P=R*sin(angleOBA)

print"P=",round(P,2),"N"

#Least force through the centre of wheel: Now the reaction from the curb must pass through the centre of the wheel since the other two forces pass through that point. Its inclination to vertical is theta = 60°. If the triangle of forces ABC  representing selfweight by AB, reaction R by BC and pull P by AC, it may be observed that AC to be least, it should be perpendicular to BC. In other words, P makes 90° with the line of action of R.
#From triangle of forces ABC, we get
P=W*sin(alpha)
print "P=",round(P,2),"N"

P= 1154.7 N
P= 1732.05 N


# example 2.24 page number 47¶

In [66]:
from math import sin,cos,pi

#variable declaration
thetabc=30.0*pi/180.0
thetabd=60.0*pi/180.0
thetaed=45.0*pi/180.0
#Free body diagrams of points B and D . Let the forces in the members be as shown in the figure. Applying Lami’s theorem to the system of forces at point D,

T1=PD*sin(pi-thetabd)/sin(thetaed+(pi/2)-thetabd)
T2=PD*sin(pi-thetaed)/sin(thetaed+(pi/2)-thetabd)

print "T1=",round(T1,2),"N"
print "T2=",round(T2,2),"N"

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

T3=(PB+T2*cos(thetabd))/cos(thetabc)
print "T3=",round(T3,2),"N"

T4=(T2*sin(thetabd))+T3*sin(thetabc)
print "T4=",round(T4,2),"N"

T1= 224.14 N
T2= 183.01 N
T3= 336.6 N
T4= 326.79 N


# example 2.25 page number 47¶

In [67]:
from math import sin,cos,pi,acos,sqrt

#variable declaration

CD=2.0
AC=1.5
BD=1.0
AB=4.0

x=((pow(AC,2)-pow(BD,2))/4)+1
y=sqrt(pow(AC,2)-pow(x,2))

alpha=acos(x/AC)
beta=acos((CD-x)/BD)

#Applying Lami’s theorem to the system of forces acting at point C

T1=PC*sin(pi/2)/sin(pi-alpha)
T2=PC*sin((pi/2)+alpha)/sin(pi-alpha)
T3=T2*sin(pi/2)/sin((pi/2)+beta)
W=T2*sin(pi-beta)/sin((pi/2)+beta)

print "W=",round(W,2),"N"

W= 2863.64 N


# example 2.26 page number 49¶

In [68]:
from math import sin,cos,pi,atan

#variable declaration

PB=20.0                       #vertical loadng at point B,KN
PC=30.0                       #vertical loadng at point C,KN

thetaab=30.0 *pi/180.0
thetabc=50.0*pi/180.0

#applying lami's thereom

T1=PB*sin(thetabc)/sin(pi-thetabc+thetaab)
T2=PB*sin(pi-thetaab)/sin(pi-thetabc+thetaab)
theta=atan((T2*sin(thetabc))/(PC-T2*cos(thetabc)))*180/pi

print "T1=",round(T1,2),"KN"

print "T2=",round(T2,2),"KN"

#Writing equations of equilibrium for the system of forces at C

print "theta=",round(theta,2),"°"

T3=(PC-T2*cos(thetabc))/cos(theta*pi/180)
print "T3=",round(T3,2),"KN"
#mistake in book

T1= 44.8 KN
T2= 29.24 KN
theta= 63.42 °
T3= 25.04 KN


# example 2.27 page number 49¶

In [69]:
from math import sin,cos,pi,atan

#variable declaration

PB=20.0                   #vertical loadng at point B,KN

PC=25.0                    #vertical loadng at point C,KN

thetaab=30.0*pi/180.0

#Writing equations of equilibrium for the system of forces at joints B and C

print "T3=",round(T3,2),"KN"

print "T1=",round(T1,2),"KN"

print"theta=",round(theta,2),"°"

print "T2=",round(T2,2),"KN"

T3= 22.5 KN
T1= 38.97 KN
theta= 54.79 °
T2= 23.85 KN


# example 2.28 page number 50¶

In [70]:
from math import sin,cos,atan,pi

#variable declaration
W=600.0                      #weight of cyclinder,N
a=600.0                      #mm
b=300.0                      #mm

#Free body diagram of sphere and frame

##sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

RB=600.0
#As the frame is in equilibrium under the action of three forces only, they must be concurrent forces. In other words, reaction at D has line of action alone OD. Hence, its inclination to horizontal is given by:
print"RB=",round(RB,2),"N"
alpha=atan((a-r)/r)
print"alpha=",round(alpha,4),"°"

RD=W/sin(alpha)
print"RD=",round(RD,3),"N"

RC=RD*cos(alpha)
RA=RC
print"RC=",round(RC),"N"
print"RA=",round(RA),"N"

RB= 600.0 N
alpha= 1.249 °
RD= 632.456 N
RC= 200.0 N
RA= 200.0 N


# example 2.29 page number 51¶

In [71]:
from math import sin,cos,pi,acos,asin

# Let O1 and O2 be the centres of the first and second spheres. Drop perpendicular O1P to the horizontal line through O2. show free body diagram of the sphere 1 and 2, respectively. Since the surface of contact are smooth, reaction of B is in the radial direction, i.e., in the direction O1O2. Let it make angle a with the horizontal. Then,

#Variable declaration

W=100.0                     #weight of spheres,N

d=360.0                     # horizontal channel having vertical walls, the distance b/w,mm

O1A=100.0
O2D=100.0
O1B=100.0
BO2=100.0

O2P=360.0-O1A-O2D
O1O2=O1B+BO2

alpha=acos(O2P/O1O2)

###sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
RB=W/sin(alpha)
RA=RB*cos(alpha)
print"RB=",round(RB,2),"N"
print"RA=",round(RA,2),"N"

RC=100+RB*sin(alpha)

RD=RB*cos(alpha)

print"RC=",round(RC),"N"

print"RD=",round(RD,2),"N"

RB= 166.67 N
RA= 133.33 N
RC= 200.0 N
RD= 133.33 N


# example2.30 page number 52¶

In [72]:
from math import  sin,cos,pi

# Two cylinders, A of weight 4000 N and B of weight 2000 N rest on smooth inclines. They are connected by a bar of negligible weight hinged to each cylinder at its geometric centre by smooth pins

#variable declaration

WA=4000.0              #weight of cylinder A,N
WB=2000.0              #weight of cylinder B,N

thetaWA=60.0*pi/180.0  #inclination of wall with cylinderA,°
thetaWB=45.0*pi/180.0  #inclination of wall with cylinderB,°
thetaAb=15.0*pi/180.0  #angle inclination bar with cylinder A ,N
thetaBb=15.0*pi/180.0  #angle inclination bar with cylinder B ,N

#he free body diagram of the two cylinders. Applying Lami’s theorem to the system of forces on cylinder A, we get

C=WA*sin(thetaWA)/sin(thetaWA+(pi/2)-thetaAb)

#Consider cylinder B. Summation of the forces parallel to the inclined plane
P=(-WB*cos(thetaWB)+C*cos(thetaWA))/cos(thetaBb)
print"P=",round(P,1),"N"

P= 1071.8 N


# example 2.31 page number 55¶

In [73]:
from math import pi,atan,sin ,cos,sqrt

# The 12 m boom AB weighs 1 kN, the distance of the centre of gravity G being 6 m from A. For the position shown, determine the tension T in the cable and the reaction at B

#variable declaration

theta=15.0*pi/180
AG=6.0             #Length of boom AB is 12m
GB=6.0
thetaAB=30.0*pi/180.0
thetaABC=15.0*pi/180.0
#sum of moment at A

T=(PB*(AG+GB)*cos(thetaAB)+WAB*AG*cos(thetaAB))/(sin(thetaABC)*12)
print"T=",round(T,4),"KN"

#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up
HA=T*cos(thetaABC)
VA=WAB+PB+T*sin(thetaABC)

RA=sqrt(pow(RA,2)+pow(RA,2))
print "RA=",round(RA,2),"KN"

alpha=atan(VA/HA)*180/pi
print"alpha",round(alpha,2),"°"

T= 10.0382 KN
RA= 188.56 KN
alpha 32.17 °


# example2.32 page number 56¶

In [74]:
from math import sin,cos,pi

#variable declaration

#A cable car used for carrying materials in a hydroelectric project is at rest on a track formed at an angle of 30° with the vertical. The gross weight of the car and its load is 60 kN and its centroid is at a point 800 mm from the track half way between the axles. The car is held by a cable .  The axles of the car are at a distance 1.2 m. Find the tension in the cables and reaction at each of the axles neglecting friction of the track.

W=60.0       #gross weight  of car,KN
theta=60.0*pi/180.0

T=W*sin(theta)
print"T=",round(T,4),"KN"

#Taking moment equilibrium condition about upper axle point on track, we get

R1=(-T*600.0+W*sin(theta)*800.0+W*cos(theta)*600.0)/1200.0
print"R1=",round(R1,4),"KN"

#Sum of forces normal to the plane = 0, gives
R2=W*cos(theta)-R1
print"R2=",round(R2,4),"KN"

T= 51.9615 KN
R1= 23.6603 KN
R2= 6.3397 KN


# example 2.33 page numnber 56¶

In [75]:
from math import sin,cos,acos,pi

# A hollow right circular cylinder of radius 800 mm is open at both ends and rests on a smooth horizontal plane. Inside the cylinder there are two spheres having weights 1 kN and 3 kN and radii 400 mm and 600 mm, respectively. The lower sphere also rests on the horizontal plane.
# Join the centres of spheres, O1 and O2 and drop O1D perpendicular to horizontal through O2.

#variable declaration
R=800.0
W1=1.0
r1=400.0
W2=3.0
r2=600.0
O1O2=1000              #mm
O2D=600                #mm

#If alpha is the inclination of O2O1 to horizontal
alpha=acos(O2D/O1O2)

#Free body diagrams of cylinder and spheres are shown.  Considering the equilibrium of the spheres.
#Sum of Moment at O2

R1=W1*O2D/(O1O2*sin(alpha))
#sum of vertical Fy & sum of horizontal forces Fx is zero
#Assume direction of Fx is right
#Assume direction of Fy is up

R2=R1
R3=W1+W2
#Now consider the equilibrium of cylinder. When it is about to tip over A, there is no reaction from ground at B. The reaction will be only at A.

#Sum of Moment at A

W=R1*O1O2*sin(alpha)/R

print"W=",round(W,2),"KN"

W= 0.75 KN