## Problem no 9.1,Page no.232¶

In [11]:
import math

#Initilization of variables

D=15 #cm #External Diameter
t=2 #cm #Thickness
L=6 #m #Length of cyclinder
E=80*10**9 #Pa
alpha=1*1600**-1
sigma_c=550*10**6 #Pa #compressive stress

#Calculations

d=D-2*t #m #Internal Diameter
A=pi*4**-1*(D**2-d**2)*10**-4 #m**2 #Areaof Tube
I=pi*64**-1*(D**4-d**4)*10**-4 #m**4 #M.I of tube

P_R=sigma_c*A*(1+alpha*(L*k**-1)**2)**-1 #According to Rankine's Formula

#The Answer in Textbook is incorrect for P_R

#Now again from Rankine's Formula
#As K=I*A**-1,so substituting in below equation
#Thus Stress calculated from Euler's Formula cannot exceed the yield stress of 550 MPa
L=(pi**2*E*k**2*(550*10**6)**-1)**0.5*10**-2 #m #Length of cyclinder

#Result
print"The Length of strut is",round(L,2),"cm"

The Length of strut is 1.76 cm


## Problem no 9.2,Page no.233¶

In [9]:
import math

#Initilization of variables

L=1.5 #m #Length of steelbar
b=2 #cm #bredth of steelbar
d=0.5 #cm #depth of steelbar
sigma=320 #MPa #Yield point
E=210 #GPa #modulus of Elasticity of steelbar

#Calculations

I_min=b*d**3*12**-1*10**-8 #m**4 #Moment of Inertia

#Let dell=Central Deflection

#M=P*dell #Max Bending moment
#After substituting value in above equation we get
#M=191.9*dell

A=b*d*10**-4 #m**2 #Area of steel bar
sigma_1=P*A**-1*10**-6 #Mpa #Direct stress

Z=b*d**3*10**-6  #Section modulus
#sigma_2=M*Z**-1 #N/m**2 #Bending stress
#After substituting value in above equation we get
#sigma_2=dell*2302.8*10**6 #N/m**2

#sigma=sigma_1+sigma_2
#Now substituting value of Bending stress and direct stress in above equation we get

#320*10**6=1.919*10**6+2302.8*10**6*dell
dell=((320*10**6-1.919*10**6)*(2302.8*10**6)**-1)*10**2 #cm #Central Deflection

#Result
print"Maximum Central Deflection is",round(dell,2),"cm"

Maximum Central Deflection is 13.81 cm


## Problem no 9.3,Page no.233¶

In [32]:
import math

#Initilization of variables

dell=1 #cm #Deflection
FOS=4 #Factor of safety
E=210 #GPa #Modulus of Elasticity of steel bar

#Flange Dimensions
b=30 #cm #width of flange
d=5 #cm #depth of flange

#Web Dimensions
d_1=100 #cm #Depth of web
t_1=2 #cm #Thcikness of web

#Calculations

I_xx=(0.3*1.1**3-0.28*1**3)*12**-1 #m**4 #M.I about x-x axis
I_yy=2*0.05*0.3**3*12**-1+1*0.02**3*12**-1 #m**4 #M.I about y-y axis

#From the equation of deflection we get
L=(dell*10**-2*384*E*10**9*I_xx*(5*40*10**3)**-1)**0.25 #m #Length of beam

#Result

The Safe Load is 2336127.78 N


## Problem no 9.5,Page no.235¶

In [33]:
import math

#Initilization of variables

L=4 #m #Length of column
FOS=5 #Factor of safety
#d=0.8*D #Internal diameter is 0.8 times Extarnal Diameter
sigma_c=550 #MPa #Compressive stress
alpha=1*1600**-1 #constant

#Calculations

#A=pi*4**-1(D**2-d**2) #m**2 #Area of hollow cyclinder
#After substituting value of d we get

#A=pi*0.09*D**2

#I=pi*64**-1*(D**4-d**4) #m**4 #Mo,ent Of Inertia
#After substituting value of d we get d we get

#I=0.009225*pi*D**4

#After substituting value of I and A and further simplifying we get
#K=0.32*D

#Now using the Relation we get
#P=sigma_c*A*(1+alpha*(l_e*k)**2)**-1 #Rankines Formula
#Now Substituting values in above equation we get
#125*10**4=550*10**6*pi*0.09*D**2*(1+1*1600**-1*((2*0.32)**2)**-1)**-1

#Further simplifying and rearranging we get
#D**4-0.008038*D**2-0.0001962397=0

a=1
b=-0.008038
c=-0.0001962397

X=b**2-4*a*c

D_1=((-b+X**0.5)*(2*a)**-1)**0.5*10**2
D_2=((-b-X**0.5)*(2*a)**-1)**0.5

#Thus Diameter cannot be negative, discard value of D_2
d=0.8*D_1

#Result
print"The Minimum Diameter is",round(d,2),"cm"

The Minimum Diameter is 10.91 cm


## Problem no 9.6,Page no.236¶

In [54]:
import math

#Initilization of variables

d=0.04 #m #Internal Diameter of tube
D=0.05 #m #External Diameter of tube
L=2 #m #Length of tube
l=3 #m #Length of strut

#Calculations

A=pi*4**-1*(D**2-d**2) #m**2 #Areaof Tube
I=pi*64**-1*(D**4-d**4) #m**4 #M.I of tube
sigma_c=P*A**-1 #Pa #Compressive stress

l_e=L*2**-1 #m #According to given condition i.e Both ends fixed

#Now from crippling Load Equation we get
alpha=((sigma_c*A*P_2**-1-1)*((l_e*k**-1)**2)**-1)*10**4

#Now Crippling Load when L=3 m Is used as strut
l_e_2=l*(2**0.5)**-1
P_3=sigma_c*A*(1+alpha*10**-4*(l_e_2*k**-1)**2)**-1

print"The Value of constant value alpha is",round(alpha,2)
print"The Crippling Load of Tube is",round(P_3,2),"N"

The Value of constant value alpha is 1.33
The Crippling Load of Tube is 71954.46 N


## Problem no 9.8,Page no.239¶

In [130]:
import math

#Initilization of variables

D=0.038 #m #External diameter
d=0.035 #m #Internal diameter
E=210*10**9 #Pa
e=0.002 #m #Eccentricity
L=1.5 #m #Lenght of tube

#Calculations

A=pi*4**-1*(D**2-d**2) #m**2 column
I=pi*64**-1*(D**4-d**4) #m**4 #M.I of column
m=(P*(E*I)**-1)**0.5

#Let X=secmL*2**-1
X=(1*(cos(m*L*2**-1))**-1)
M=P*e*X #N-m #MAx Bending Moment
sigma_1=P*A**-1*10**-6 #Pa #Direct stress
sigma_2=M*0.019*I**-1*10**-6 #Pa #Bending stress

sigma_c_max=(sigma_1+sigma_2) #MPa #Max compressive stress

#Result
print"The Max stress developed is",round(sigma_c_max,2),"MPa"

The Max stress developed is 246.79 MPa


## Problem no 9.9,Page no.239¶

In [125]:
import math

#Initilization of variables

L=5 #m #Length of column
D=0.2 #m #External Diameter
d=0.14 #m #Internal diameter
e=0.015 #m #Eccentricity
E=95 *10**9 #Pa

#Calculations

L_2=L*2**-1 #m #half length of column
A=pi*4**-1*(D**2-d**2) #m**2 column
I=pi*64**-1*(D**4-d**4) #m**4 #M.I of column
m=(P*(E*I)**-1)**0.5

#Let X=secmL*2**-1
X=(1*(cos(m*L_2*2**-1))**-1)
M=P*e*X #N-m #MAx Bending Moment
sigma_1=P*A**-1*10**-6 #Pa #Direct stress
sigma_2=M*0.1*I**-1*10**-6 #Pa #Bending stress

sigma_c_max=(sigma_1+sigma_2) #MPa #Max compressive stress

print"The Max stress developed is",round(sigma_c_max,2),"MPa"

The Max stress developed is 17.65 MPa


## Problem no 9.10,Page no.240¶

In [98]:
import math

#Initilization of variables

L=3 #m #Length of strut
b=0.04 #m #Width of rectangle
d=0.10 #m #Depth if rectangle
P=100*10**3 #N #Axial thrust
E=210*10**9 #Pa

#Calculations

A=b*d #m**2 #Area of strut
I=b*d**3*12**-1 #m**4 #M.I
m=(P*(E*I)**-1)**0.5

#Let X=secmL*2**-1
X=(1*(cos(m*L*2**-1))**-1)

M=w*E*I*P**-1*(X-1)*3**-1 #N*m #Max Bending Moment
sigma_1=P*A**-1 #Pa #Direct stress
sigma_2=M*0.05*I**-1 #Pa #Bending stress

sigma_c_max=sigma_1+sigma_2 #Max compressive stress

#If the Eccentricity of thrust is neglected
M_2=w*L**2*(3*8)**-1 #Max Bending moment
sigma_2_2=M_2*0.05*I**-1 #Pa #Bending stress

sigma_c_max_2=(sigma_1+sigma_2_2)*10**-6 #Pa

#Let Y=Percentage error
Y=((sigma_c_max-sigma_c_max_2*10**6)*sigma_c_max**-1)*100

#Result
print"Max stress induced is",round(sigma_c_max_2,2),"MPa"
print"The Percentage Error is",round(Y,3),"%"

Max stress induced is 81.25 MPa
The Percentage Error is 9.638 %