Chapter 3 : Parallel Forces In A Plane

Example 3.3-1,Page No:64

In [1]:
import math

#Initilization of variables

W=1000 #N
Lab=1 #m
Lac=0.6 #m
theta=60 #degree #angle made by the beam with the horizontal

#Calculations

Q=(W*Lac*cos(theta*(pi/180)))/(Lab*cos(theta*(pi/180))) #N # from eq'n 2
P=W-Q #N # from eq'n 1

#Results

print"The load taken by man P is ",round(P),"N"
print"The load taken by man Q is ",round(Q),"N"

The load taken by man P is  400.0 N
The load taken by man Q is  600.0 N

Example 3.3-2,Page No:64

In [2]:
import math

#Initilization of variables

F=1000 #N
Lab=1 #m
Lbc=0.25 #m
Lac=1.25 #m

#Calculations

Rb=(F*Lac)/Lab #N # from eq'n 2
Ra=Rb-F #N # fom eq'n 1

#Results

print"The reaction (downwards)at support A is ",round(Ra),"N"
print"The reaction (upwards)at support B is ",round(Rb),"N"

The reaction (downwards)at support A is  250.0 N
The reaction (upwards)at support B is  1250.0 N

Example 3.3-3,Page No:65

In [7]:
import math

#Inilitization of variables

Lab=12 #m
Mc=40 #kN-m 
Md=10 #kN-m
Me=20 #kN-m
Fe=20 #kN #force acting at point E

#Calculations

Xa=-(Fe) #kN #take sum Fx=0
a=Me+Md-Mc #N #take moment at A
Rb=a*(Lab)**-1
Ya=-Rb #N #take sum Fy=0

#Results

print"The vertical reaction (upwards) at A is ",round(Ya,3),"kN"
print"The horizontal reaction (towards A) is ",round(Xa,2),"kN"
print"The reaction (downwards) at B is ",round(Rb,3),"kN"

The vertical reaction (upwards) at A is  0.833 kN
The horizontal reaction (towards A) is  -20.0 kN
The reaction (downwards) at B is  -0.833 kN

Example 3.3-5,Page No:66

In [11]:
import math
import numpy as np

#Initilization of variables

W=1000 #N
Lad=7.5 #m
Lae=1.5 #m
La1=3.75 #m #distance of 1st 1000N load from pt A
La2=5 #m #distance of 2nd 1000N load from pt A
La3=6 #m # distance of 3rd 1000N load from pt A

# Calculations (part1)

#using matrix to solve the given eqn's 1 & 2

A=np.array([[1 ,-2.5],[3.5 ,-5]])
B=np.array([1000,7250])
C=np.linalg.solve(A,B)

#Resuts

print"The reaction at F i.e Rf is ",round(C[0]),"N"
print"The reaction at D i.e Rd is ",round(C[1]),"N"

#Calculations (part 2)
#Consider combined F.B.D of beams AB,BC &CD. Take moment at A

Re=((W*La1)+(W*La2)+(W*La3)+(C[1]*Lad)-(C[0]*La3))/Lae #N
Ra=C[1]-Re-C[0]+(3*W) #N #Taking sum of forces in Y direction

#Results

print"The reaction at pt E i.e Re is ",round(Re),"N"
print"The reaction at pt A i.e Ra is ",round(Ra),"N" #acting vertically downwards

The reaction at F i.e Rf is  3500.0 N
The reaction at D i.e Rd is  1000.0 N
The reaction at pt E i.e Re is  833.0 N
The reaction at pt A i.e Ra is  -333.0 N

Example 3.3-7,Page No:69

In [14]:
import math

#Initilization of variables

Ws=2 #kN #weight of scooter
Wd=0.5 #kN #weight of driver
Lab=1 #m
Led=0.8 #m
Leg=0.1 #m

#Calculations

Rc=((2*Leg)+(Wd*Led))/Lab #kN #take moment at E
Ra=(2+Wd-Rc)/2 #kN # as Ra=Rb,(Ra+Rb=2*Ra)
Rb=Ra # kN

#Results

print"The reaction at wheel A is ",round(Ra,2),"kN"
print"The reaction at wheel B is ",round(Rb,2),"kN"
print"The reaction at wheel C is ",round(Rc,2),"kN"

The reaction at wheel A is  0.95 kN
The reaction at wheel B is  0.95 kN
The reaction at wheel C is  0.6 kN

Example 3.3-8,Page No:69

In [25]:
import math

#Initilization of variables

W1=15 #N #up
W2=60 #N #down
W3=10 #N #up
W4=25 #N #down
Lab=1.2 #m
Lac=0.4 #m
Lcd=0.3 #m
Ldb=0.5 #m
Lad=0.7 #m
Leb=0.417 #m #Leb=Lab-x

#Calculations

#(a) A single force

Ry=W1-W2+W3-W4 #N #take sum Fy=0
x=((-W2*Lac)+(W3*Lad)-(W4*Lab))/(Ry) #m

#(b) Single force moment at A

Ma=(Ry*x) #N-m

# Single force moment at B

Mb=W2*Leb #N-m

#Results

print"The reaction for single force (a) is ",round(Ry,2),"N"
print"The distance of Ry from A is ",round(x,3),"m"
print"The moment at A is ",round(Ma,2),"N-m"
print"The moment at B is ",round(Mb,2),"N-m"

The reaction for single force (a) is  -60.0 N
The distance of Ry from A is  0.783 m
The moment at A is  -47.0 N-m
The moment at B is  25.02 N-m

Example 3.3-9,Page No:71

In [6]:
import math
import numpy as np

#Initilization of variables

Ra=5000 #N
Ma=10000 #Nm
alpha=60 #degree #angle made by T1 with the pole
beta=45 #degree #angle made by T2 with the pole
theta=30 #degree #angle made by T3 with the pole
Lab=6 #m
Lac=1.5 #m
Lcb=4.5 #m

#Calculations

T3=Ma/(4.5*sin(theta*(pi/180))) #N #take moment at B

# Now we use matrix to solve eqn's 1 & 2 simultaneously,

A=np.array([[-0.707, 0.866],[0.707, 0.5]])
B=np.array([2222.2,8848.8])
C=np.linalg.solve(A,B)

#Results

print"Tension in wire 1 i.e T1 is ",round(C[1],1),"N" #answer may vary due to decimal variance
print"Tension in wire 2 i.e T2 is ",round(C[0],1),"N"
print"Tension in wire 3 i.e T3 is ",round(T3,1),"N"

Tension in wire 1 i.e T1 is  8104.7 N
Tension in wire 2 i.e T2 is  6784.2 N
Tension in wire 3 i.e T3 is  4444.4 N

Example 3.3-10,Page No:76

In [32]:
import math

#Initilization of variables

w=2000 #N/m
Lab=3 #m

#Calculations

W=w*Lab/2 #N# Area under the curve
Lac=(0.6666)*Lab #m#centroid of the triangular load system
Rb=(W*Lac)/Lab #N #sum of moment at A
Ra=W-Rb #N

#Results

print"The resultant of the distibuted load lies at ",round(Lac),"m"
print"The reaction at support A is ",round(Ra),"N"
print"The reaction at support B is ",round(Rb),"N"

The resultant of the distibuted load lies at  2.0 m
The reaction at support A is  1000.0 N
The reaction at support B is  2000.0 N

Example 3.3-12,Page No:78

In [33]:
import math

# Initilization of variables

w1=1.5 #kN/m # intensity of varying load at the starting point of the beam
w2=4.5 #kN/m # intensity of varying load at the end of the beam
l=6 #m # ength of the beam

# Calculations

# The varying load distribution is divided into a rectangle and a right angled triangle

W1=w1*l #kN # where W1 is the area of the load diagram(rectangle ABED)
x1=l/2 #m # centroid of the rectangular load system
W2=(w2-w1)*l/2 #kN # where W1 is the area of the load diagram(triangle DCE)
x2=2*l/3 #m # centroid of the triangular load system
W=W1+W2 #kN # W is the resultant
x=((W1*x1)+(W2*x2))/W #m # where x is the distance where the resultant lies

#Results

print"The resultant of the distributed load system is ",round(W),"kN"
print"The line of action of the resulting load is ",round(x,1),"m"

The resultant of the distributed load system is  18.0 kN
The line of action of the resulting load is  3.5 m

Example 3.3-13,Page No:78

In [35]:
import math

# Initiization of variables

W1=10 #kN #point load acting at D
W2=20 #kN # point load acting at C at an angle of 30 degree
W3=5 #kN/m # intensity of udl acting on span EB of 4m
W4=10 #kN/m # intensity of varying load acting on span BC of 3m
M=25 #kN-m # moment acting at E
theta=30 #degree # angle made by 20 kN load with the beam
Lad=2 #m
Leb=4 #m
Laf=6 #m #distance between the resultant of W3 & point A
Lac=11 #m
Lag=9 #m #distance between the resultant of W4 and point A
Lbc=3 #m
Lab=8 #m

# Calculations

Xa=20*cos(theta*(pi/180)) #kN # sum Fx=0
Rb=((W1*Lad)+(-M)+(W3*Leb*Laf)+(W2*sin(theta*(pi/180))*Lac)+((W4*Lbc*Lag)/2))/Lab #kN # taking moment at A
Ya=W1+(W2*sin(theta*(pi/180)))+(W3*Leb)+(W4*Lbc/2)-Rb #kN # sum Fy=0
Ra=(Xa**2+Ya**2)**0.5 #kN # resultant at A

#Results

print"The horizontal reaction at A i.e Xa is ",round(Xa,2),"kN"
print"The vertical reaction at A i.e Ya is ",round(Ya),"kN"
print"The reaction at A i.e Ra is ",round(Ra),"kN"
print"The reaction at B i.e Rb is ",round(Rb),"kN"

The horizontal reaction at A i.e Xa is  17.32 kN
The vertical reaction at A i.e Ya is  10.0 kN
The reaction at A i.e Ra is  20.0 kN
The reaction at B i.e Rb is  45.0 kN

Example 3.3-14,Page No:79

In [37]:
import math

# Initilization of variables

h=4 #m #height of the dam wall
rho_w=1000 # kg/m^3 # density of water
rho_c=2400 # kg/m^3 # density of concrete
g=9.81 # m/s^2

# Calculations

P=(rho_w*g*h**2)/2 # The resultant force due to water pressure per unit length of the dam
x=(0.6666)*h #m # distance at which the resutant of the triangular load acts 
b=((2*P*h)/(3*h*rho_c*g))**0.5 # m # eq'n required to find the minimum width of the dam

# Results

print"The minimum width which is to be provided to the dam to prevent overturning about point B is ",round(b,3),"m" 

The minimum width which is to be provided to the dam to prevent overturning about point B is  1.491 m