# example 9.1 page number 286¶

In [1]:
from math import pi,atan,sqrt,cos,sin

#variable declaration

#summation of all  horizontal forces is zero & vertical forces is zero.
P1=float(10)                                     #Vertical down Load at 4m from A,KN
P2=float(15)                                     #Inclined down Load at angle 30° at 6m from A,KN
P3=float(20)                                     #Inclined down Load at angle 45° at 10m from A,KN
theta2=30
theta3=45
#horizontal,vertical component at A  is Ha,Va respectively.

Ha=P2*cos(theta2*pi/180)+P3*cos(theta3*pi/180)
Rb=(P1*4+P2*6*sin(theta2*pi/180)+P3*10*sin(theta3*pi/180))/12         #reaction at B point,KN

print "RB=",round(Rb,4),"KN"

#now vertical component
Va=P2*sin(theta2*pi/180)+P3*sin(theta3*pi/180)+P1-Rb

Ra=sqrt(pow(Ha,2)+pow(Va,2))

print "RA=",round(Ra,4),"KN"

alpha=(atan(Va/Ha))*180/pi

print "alpha=",round(alpha,2),"°"

RB= 18.8684 KN
RA= 29.989 KN
alpha= 25.21 °


# example 9.2 page number 287¶

In [2]:
from math import pi,atan,sqrt,cos,sin

#variable declaration

#summation of all  horizontal forces is zero & vertical forces is zero.
P1=float(60)                                     #inclined down to right Load at angle 60 at 1m from A,KN
P2=float(80)                                     #Inclined down to left Load at angle 75° at 3m from A,KN
P3=float(50)                                     #Inclined down to left Load at angle 60° at 5.5m from A,KN
theta1=60
theta2=75
theta3=60
thetaRb=60
#horizontal,vertical component at A is Ha,Va  respectively.

Rb=(P1*1*sin(theta1*pi/180)+P2*3*sin(theta2*pi/180)+P3*5.5*sin(theta3*pi/180))/(6*sin(thetaRb*pi/180))         #reaction at B point,KN
Ha=-P1*cos(theta1*pi/180)+P2*cos(theta2*pi/180)-P3*cos(theta3*pi/180)+Rb*cos(thetaRb*pi/180)
print "RB=",round(Rb,4),"KN"

#now vertical component
Va=P1*sin(theta1*pi/180)+P2*sin(theta2*pi/180)+P3*sin(theta3*pi/180)-Rb*sin(thetaRb*pi/180)

Ra=sqrt(pow(Ha,2)+pow(Va,2))

print "RA=",round(Ra,4),"KN"

alpha=(atan(Va/Ha))*180/pi

print "alpha=",round(alpha,2),"°"

RB= 100.4475 KN
RA= 87.0172 KN
alpha= 79.45 °


# example 9.3 page number 288¶

In [3]:
from math import pi,atan,sqrt,cos,sin

#variable declaration

#summation of all  horizontal forces is zero & vertical forces is zero.
P1=float(20)                                     #vertical down  Load  at 2m from A,KN
P2=float(30)                                     #uniform distributed load from 2m to 6m from A,KN/m(in 4m of span)
P3=float(60)                                     #Inclined down to right Load at angle 45° at 7m from A,KN

theta3=45
#horizontal,vertical component at  B is  Hb,Vb respectively.

Ra=(P1*7+P2*4*5+P3*2*sin(theta3*pi/180))/(9)     #reaction at B point,KN

print "RA=",round(Ra,4),"KN"

Hb=P3*cos(theta3*pi/180)
print "HB=",round(Hb,4),"KN"
#now vertical component
Vb=P1+P2*4+P3*sin(theta3*pi/180)-Ra
print "VB=",round(Vb,4),"KN"

Rb=sqrt(pow(Hb,2)+pow(Vb,2))

print "RB=",round(Rb,4),"KN"

alpha=(atan(Vb/Hb))*180/pi

print "alpha=",round(alpha,2),"°"

RA= 91.6503 KN
HB= 42.4264 KN
VB= 90.7761 KN
RB= 100.2013 KN
alpha= 64.95 °


# example 9.4 page number 288¶

In [4]:
#variable declaration
#Let the reactions at A be Ha, Va and Ma
#summation of all  horizontal forces is zero & vertical forces is zero.

P1=float(20)                                     #vertical down  Load  at 2m from A,KN
P2=float(12)                                     #vertical down  Load  at 3m from A,KN
P3=float(10)                                     #vertical down  Load  at 4m from A,KN
Pu=float(16)                                     #uniform distributed load from A to 2m from A,KN/m(in 2m of span)
##horizontal,vertical component at  A is  Ha,Va respectively.
print"no horizontal force ","HA=0"
Va=Pu*2+P1+P2+P3
print "VA=", round(Va,2),"KN"
Ma=Pu*2*1+P1*2+P2*3+P3*4
print "MA=", round(Ma,2),"KN-m"

no horizontal force  HA=0
VA= 74.0 KN
MA= 148.0 KN-m


# example 9.5 page number 288¶

In [5]:
#variable declaration
#Let the reactions at A be  Va and Ma
#summation of all  horizontal forces is zero & vertical forces is zero.

P1=float(15)                                     #vertical down  Load  at 3m from A,KN
P2=float(10)                                     #vertical down  Load  at 5m from A,KN
M=float(30)                                      #CW moment at 4m distance from A, KN-m
Pu=float(20)                                     #uniform distributed load from A to 2m from A,KN/m(in 2m of span)
##horizontal,vertical component at  A is  Ha,Va respectively.
print"no horizontal force ","HA=0"
Va=Pu*2+P1+P2
print "VA=", round(Va,2),"KN"
Ma=Pu*2*1+P1*3+P2*5+M
print "MA=", round(Ma,2),"KN-m"

no horizontal force  HA=0
VA= 65.0 KN
MA= 165.0 KN-m


# example 9.6 page number 289¶

In [6]:
#variable declaration

#As supports A and B are simple supports and loading is only in vertical direction, the reactions RA and RB are in vertical directions only.

#summation of all  horizontal forces is zero & vertical forces is zero.

P1=float(30)                                     #vertical down  Load  at 1m from A,KN
P2=float(40)                                     #vertical down  Load  at 6.5m from A,KN
Pu=float(20)                                     #uniform distributed load from 2m to 5m from A,KN/m(in 3m of span).

Rb=(Pu*3*3.5+P1*1+P2*6.5)/5
print "RB=", round(Rb,2),"KN"
Ra=Pu*3+P1+P2-Rb
print "RA=", round(Ra,2),"KN"

RB= 100.0 KN
RA= 30.0 KN


# example 9.7 page number 289¶

In [7]:
#variable declaration
#Let the reactions at A be  Va and Ma.
#summation of all  horizontal forces is zero & vertical forces is zero.

P1=float(60)                                     #vertical down  Load  at 4m from A to right,KN
P2=float(20)                                     #vertical down  Load  at 11m from A to right,KN
M=float(30)                                      #CW moment at 7m distance from A to right, KN-m
Pu=float(20)                                     #uniform distributed load from A to 2m from A to left ,KN/m(in 2m of span)
##horizontal,vertical component at  A is  Ha,Va respectively.
print"no horizontal force ","HA=0"
Vb=(-Pu*2*1+P1*4+P2*11+M)/9
print "VB=", round(Vb,2),"KN"
Va=Pu*2+P1+P2-Vb
print "VA=", round(Va,2),"KN"

no horizontal force  HA=0
VB= 50.0 KN
VA= 70.0 KN


# example 9.8 page number 290¶

In [8]:
from math import pi,atan,sqrt,cos,sin

#variable declaration

#summation of all  horizontal forces is zero & vertical forces is zero.

P1=float(30)                                     #Inclined down Load at angle 45° to left at 5m from A,KN
Pu=float(20)                                     #uniformly distributed load from 6m to 8m from A ,KN,(2m of span)
theta1=45
M=40                                             #ACW moment at 3m from A, KN-m
#horizontal,vertical component at A  is Ha,Va respectively.

Rb=(M+P1*5*sin(theta1*pi/180)+Pu*2*7)/6         #reaction at B point,KN

print "RB=",round(Rb,4),"KN"

Ha=P1*cos(theta1*pi/180)

#now vertical component
Va=P1*sin(theta1*pi/180)-Rb+Pu*2

Ra=sqrt(pow(Ha,2)+pow(Va,2))

print "(Negative sign show that the assumed direction of VA is wrong. In other words, VA is acting vertically downwards). "

Va1=-1*Va
print "RA=",round(Ra,4),"KN"

alpha=(atan(Va1/Ha))*180/pi

print "alpha=",round(alpha,2),"°"

RB= 71.011 KN
(Negative sign show that the assumed direction of VA is wrong. In other words, VA is acting vertically downwards).
RA= 23.3666 KN
alpha= 24.79 °


# example 9.9 page number 290¶

In [9]:
#variable declaration

#summation of all  horizontal forces is zero & vertical forces is zero.

#Let the left support C be at a distance x metres from A.

P1=float(30)                        #vertical down load at A,KN
Pu=float(6)                         #uniform distributed load over whole span,KN/m,(20m of span)
P2=float(50)                        #vertical down load at B, KN

#Rc=Rd(given) reaction at C & D is equal.

Rc=(P1+P2+Pu*20)/2
Rd=Rc

#taking moment at A
x=(((Pu*20*10+P2*20)/100)-12)/2

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

X= 5.0 m