# Chapter no.15:Thin Cyclindrical Shell¶

## Problem no 15.1,Page no.351¶

In [13]:
import math

#Initilization of variables

D=0.8 #m #Diameter of Shell
L=3 #m #Length of shell
t=0.01 #m #thickness of metal
E=200*10**9 #Pa
p=2.5*10**6 #Pa #Internal Pressure
m=4 #Poisson's ratio

#Calculation

sigma_1=p*D*(2*t)**-1 #N/m**2 #Hoop stress
sigma_2=p*D*(4*t)**-1 #N/m**2 #Longitudinal stress

e_1=1*E**-1*(sigma_1-sigma_2*m**-1) #Hoop strain
e_2=1*E**-1*(sigma_2-sigma_1*m**-1) #Hoop strain

d=e_1*D*100 #cm #Increase in Diameter
l=e_2*L*100 #cm #Increase in Length

dell_v=2*e_1+e_2 #Volumetric strain
V=dell_v*pi*4**-1*D**2*L*10**6 #cm**3 #Increase in Volume

#Result
print"Change in Diameter is",round(d,3),"cm"
print"Change in Length is",round(l,3),"cm"
print"Change in Volume is",round(V,2),"cm**3"

Change in Diameter is 0.035 cm
Change in Length is 0.037 cm
Change in Volume is 1507.96 cm**3


## Problem no 15.2,Page no.352¶

In [21]:
import math

#Initilization of variables

D=0.8 #m #iameter of water main
w=10*10**3 #N/m**3 #Weight of Water
sigma_t=20*10**6 #MPa #Permissible stress

#Calculation

p=w*h #N/m**2 #Pressure of inside the main
t=p*D*(2*sigma_t)**-1*100 #m #Thcikness of metal

#Result
print"The Thickness of metal is",round(t,2),"cm"

The Thickness of metal is 2.0 cm


## Problem no 15.3,Page no.352¶

In [29]:
import math

#Initilization of variables

p=2*10**6 #MPa #Steam Pressure
t=0.02 #m #thickness of boiler plate
sigma_t=120*10**6 #MPa #Tensile stress
sigma_l=120*10**6 #MPa #Longitudinal stress
rho=0.90 #% #Efficiency of Longitudinal joint
rho_e=0.40 #% #Efficiency of circumferential joint

#Calculations

D_1=sigma_t*2*t*rho*p**-1 #Diameter of boiler
D_2=sigma_l*4*t*rho_e*p**-1 #Diameter of boiler

#Max diameter of boiler is equal to minimum value of diameter

#Result
print"Maximum diameter of boiler is",round(D_2,2),"m"

Maximum diameter of boiler is 1.92 m


## Problem no 15.4,Page no.352¶

In [41]:
import math

#Initilization of variables

L=0.9 #m #Length of cyclindrical shell
D=0.2 #m #Internal Diameter
t=0.008 #m #thickness of metal
E=200*10**9 #Pa
m=1*0.3**-1 #Poissoin's ratio

#Calculations

V=pi*4**-1*D**2*L #Volume of cyclinder

#Let X=2*e_1+e_2
X=dV*V**-1 #Volumetric strain   (Equation 1)

#e_1=p*D*(2*E*t)**-1*(1-1*(2*m)**-1) #Circumferential strain
#e_2=p*D*(2*E*t)**-1*(1*2**-1-1*(2*m)**-1) #Circumferential strain

#substituting above values in equation 1 we get
p=X*E*t*(D*((1-1*(2*m)**-1)+(1*4**-1-1*(2*m)**-1)))**-1*10**-3 #KN/m**2 #Pressure exerted by fluid
sigma_t=p*D*(2*t)**-1 #KN/m**2 #hoop stress

#Result
print"Pressure Exerted by Fluid on the cyclinder is",round(p,2),"KN/m**2"
print"Hoop stress is",round(sigma_t,2),"KN/m**2"

Pressure Exerted by Fluid on the cyclinder is 5956.68 KN/m**2
Hoop stress is 74458.45 KN/m**2


## Problem no 15.5,Page no.353¶

In [8]:
import math

#Initilization of variables

t=0.015 #m #Thickness of plate
sigma_t=120*10**6 #Pa #tensile stress
sigma_l=120*10**6 #Pa #Longitudinal stress
rho=0.7 #% #Efficiency of longitudinal joints
rho_l=0.3 #% #Efficiency of circumferential joints
p=2*10**6 #Pa #Internal pressure
D=1.5 #m #shell diameter

#Calculations (Part-1)

D_1=sigma_t*2*t*rho*p**-1    #m
D_2=sigma_l*4*t*rho_l*p**-1  #m

#Thus max diameter of shell is min of above two cases

#Calculations (Part-2)

p_1=sigma_t*2*t*rho*D**-1*10**-6    #MPa
p_2=sigma_l*4*t*rho_l*D**-1*10**-6  #MPa

#Thus Internal pressure is min of above two cases

#Result
print"Max Permissible Diameter of shell is",round(D_2,2),"m"
print"Max Permissible Internal Pressure is",round(p_2,2),"MPa"

Max Permissible Diameter of shell is 1.08 m
Max Permissible Internal Pressure is 1.44 MPa


## Problem no 15.6,Page no.354¶

In [26]:
import math

#Initilization of variables

L=3 #m #Length
D=1 #m #Internal Diameter
t=0.015 #m #thickness
p=1.5*10**6 #Pa #Internal pressure
E=200*10**9 #Pa
m=1*0.3**-1 #Poissoin's ratio

#Calculations

sigma_t=p*D*(2*t)**-1*10**-6 #MPa #Hoop stress
sigma_l=p*D*(4*t)**-1*10**-6 #MPa #Longitudinal stress

dD=(p*D**2*(2*t*E)**-1*(1-1*(2*m)**-1))*10**2 #cm #Change in Diameter
dL=p*D*L*(2*t*E)**-1*(1*2**-1-1*m**-1)*10**2 #cm #Change in Length

V=pi*4**-1*D**2*L #Volume
dV=p*D*(2*t*E)**-1*(5*2**-1-2*(m)**-1)*V*10**6 #cm #Change in Volume

#Result
print"The circumferential stresses induced is",round(sigma_t,2),"MPa"
print"The Longitudinal stresses induced is",round(sigma_l,2),"MPa"
print"The change in dimension are:D is",round(dD,3),"cm"
print"                           :L is",round(dL,4),"cm"

0.02125
The circumferential stresses induced is 50.0 MPa
The Longitudinal stresses induced is 25.0 MPa
The change in dimension are:D is 0.021 cm
:L is 0.015 cm


## Problem no 15.7,Page no.355¶

In [23]:
import math

#Initilization of variables

L=0.9 #m #Length of cyclinder
D=0.4 #m #Diameter
t=0.006 #m #thickness
p=5*10**6 #Pa #Pressure
E=100*10**9
m=3 #Poissoin's ratio
k=2.6*10**9 #Pa #Bulk modulus

#Calculations

#Let X=dV_1*V_1**-1
X=p*(0.4-2*0.006)*(2*t*E)**-1*(5*2**-1-2*m**-1) #Volumetric strain
dV_1=round(X,5)*pi*4**-1*0.388**2*L #cm**3 #Increase in volume of cyclinder
V_1=pi*4**-1*0.388**2*L #VOlume
dV_2=p*k**-1*V_1 #DEcrease in volume of oil due to increase in pressure

#Result
print"additional quantity of oil to be pumped is",round(dV,2),"cm**3"

additional quantity of oil to be pumped is 519.62 cm**3


## Problem no 15.8,Page no.356¶

In [43]:
import math

#Initilization of variables

A=1600*(3600)**-1 #Kg/sec #Amount of steam generated
v=0.24 #m**3/kg #specific volume of steam
sigma_t=4*10**6 #MPa #Tensile stress
V_1=30 #m/s #Velocity of steam
p=1*10**6 #Pa #Steam pressure

#Calculation

V=A*v #m**3/s #volume of steam
D=(V*(pi*4**-1*V_1)**-1)**0.5*100 #Diameter of pipe
t=p*D*(2*sigma_t)**-1 #Thicknes of pipe

#Result
print"Diameter of boiler is",round(D,2),"cm"
print"Thickness of steel plpe is",round(t,2),"cm"

Diameter of boiler is 6.73 cm
Thickness of steel plpe is 0.84 cm


## Problem no 15.9,Page no.359¶

In [28]:
import math

#Initilization of variables

P=14*10**3 #N #Axial pull
dL=0.0084 #cm #Elongation
L=0.25 #m #Length
p=7*10**6 #Internal pressure
dL_2=0.0034 #cm #Longation
d=0.0475 #m #Internal diameter
D=0.05 #m #External Diameter
m=0.25

#Calculation

t=(D-d)*2**-1 #thickness  od tube
A=pi*4**-1*(D**2-d**2) #Area of tube
sigma=P*A**-1 #stress
e=dL*(L)**-1 #strain
E=sigma*e**-1 #Modulus of Elasticity
sigma_1=p*d*(2*t)**-1 #Hoop stress
sigma_2=p*d*(4*t)**-1 #Longitudinal stress

m=-(sigma_1*(dL_2*L**-1*E-sigma_2)**-1) #POissoin's ratio\

#Let X=1*m**-1
X=1*m**-1 #Poissoin's ratio

#Result
print"The value of Poissoin's ratio is",round(X,3)


The value of Poissoin's ratio is 0.277


## Problem no 15.10,Page no.357¶

In [37]:
import math

#Initilization of variables

D=0.8 #m #Diameter
t=0.01 #m #Thickness
p=5*10**6 #Pa #Pressure
m=1*0.25**-1
E=200*10**9 #Pa

#Calculations

sigma_1=5*10**6*0.8*(4*0.01)**-1 #stress
sigma_2=sigma_1
e_1=sigma_1*E**-1-sigma_2*(m*E)**-1 #strain
e_v=3*e_1
V=4*3**-1*pi*(D*2**-1)**3 #m**3  tress
dell_v=e_v*V*10**6 #cm**3

#Result

Volume of additional Fluid 301.593 cm**3


## Problem no 15.11,Page no.358¶

In [29]:
d=0.3 #m #Diameter
D=0.003 #m #Diameter of steel wire
t=0.006 #m #thickness
sigma_w=8*10**6 #Pa #Stress
p=1*10**6 #Pa #Internal pressure
E_s=200*10**9 #Pa #Modulus of Elasticity for steel
E_c=100*10**9 #Pa #Modulus of Elasticity for cast iron
m=1*0.3**-1

#Calculations

sigma_p=(sigma_w*pi*2**-1*d)*(2*t)**-1 #compressive hoop stress
sigma_l=p*d*(4*t)**-1 #Longitudinal stress

#when internal presure is apllied Let sigma_w_1=Tensile in wire and sigma_p_1=tensile hoop in wire
#sigma_p_1*2*t+sigma_w_1*2*d**-1*pi*4**-1*d**2=p*D

#After substituting values and further simplifying we get
#1.2*sigma_p_1+0.471*sigma_w_1=3000    Equation 1

#1*E_c**-1(sigma_p_1-sigma_1*m**-1+sigma_p)=1*E_s**-1(sigma_w_1-sigma_w)

#After substituting values and further simplifying we get
#sigma_p_1-0.5*sigma_w_1=1.36*10**6
#sigma_p_1=0.5*sigma_w_1-3.39*10**6   Equation 2

#From Equation 2 substituting value of sigma_p_1 in Equation 1

sigma_w_1=(40.68*10**3+0.3*10**6)*(10.71238*10**-3)**-1
sigma_p_1=0.5*sigma_w_1-3.39*10**6

#Let X=sigma_p_1 and Y=sigma_w_1
X=sigma_p_1*10**-6 #MPa #Stresses in pipe
Y=sigma_w_1*10**-6 #MPa #Stresses in wire

#Result
print"Stress in the pipe is",round(X,2),"MN/m**2"
print"Stress in the wire is",round(Y,2),"MN/m**2"

Stress in the pipe is 12.51 MN/m**2
Stress in the wire is 31.8 MN/m**2


## Problem no 15.12,Page no.359¶

In [30]:
import math

#Initilization of variables

D=0.038 #m #External Diameter
d=0.035 #m #Internal Diameter
d_1=0.0008 #m #Steel wire diameter
p=2*10**6 #pa #Pa #Internal Pressure
sigma_t_1=7*10**6 #Pa #Circumferential stress
#E_s=1.6*E_s
m=0.3

#Calculation

t=(D-d)*2**-1 #m Thickness

#sigma_t*2*t=pi*d*2**-1*sigma_w
#From Above equation we get

#sigma_t=0.419*sigma_w   (Equation 1)

sigma_w_1=(p*d-sigma_t_1*2*t)*(2*d_1**-1*pi*4**-1*d_1**2)**-1 #stress in wire
sigma_l=p*d*(4*t)**-1 #Longitudinal stress in tube

#Now Equating equations of strain in tube and wire we get
sigma_w=-(1.6*(sigma_t_1-sigma_l*m)-sigma_w_1)*1.67**-1*10**-6

#Result
print"The Tension at which wire must have been wound is",round(sigma_w,2),"MPa"

The Tension at which wire must have been wound is 20.0 MPa