Chapter 5 thin cylinders 6th semester

STRENGTH OF MATERIAL ARSHAD ALI (arshadalibuitems@yahoo.com)
CHAPTER5THINCYLINDERS
Chapter 5
Thin Cylinders or
Thin Shells
1

1. Stress Analysis Mar 2012
2. Stress Transformation Apr 2012
3. Mid Term Exam Apr 2012
4. Columns May 2012
5. Theory of Yielding / Failure May 2012
6. Thin cylinder June 2012
7. Thick Cylinder Jun 2012
8. Curved Beam Jun 2012
9. Torsion Jun 2012
10. Fatigue Jul 2012
COURSE PLANS 2

– Cylindrical pressure vessels
– Hydraulic cylinders
– Gun barrels
– High Pressure
fluid pipes
Thin Cylinder under internal pressure 3

Thin Cylinder under internal pressure
– A pressurized cylinder is considered a thin-walled vessel if
the wall thickness is less than one-twentieth of its diameter
– When a thin-walled cylinder is subjected to internal pressure, three
components of stresses will act in on the internal cylinder, namely the
1. Circumferential or hoop stress
2. Radial stress 3. Longitudinal stress
4
1
20
1
20
thickness
inside diameter
t
d


t
d

Stresses in Cylinders 5
H
L
H = Hoop Stress or tangential stress
L = Longitudinal Stress
R = Radial Stress
Act parallel to the longitudinal
axis of the shell
Act tangential to the
circumference of the shell
Act radially and are too small
and can be neglected

Hoop Stress
6

Hoop Stress 7
Total force due to internal pressure
The total force due to hoop stress
to counter balance the above force
pressure projected area
p d L
p pressure
d diameter
L length
 
  



H HL t L t      
1
2
Projected
area
Hoop stress
H
Force = Stress  Area
t
t

Hoop Stress 8
2
2
H
H
p d L L t
p d
t
where
p pressure
d inside diameter
t thickness


    






For maintaining equilibrium the equation 1 must equal to equation 2
H
H
p
d
Hoop Stress

Longitudinal Stress
9

Longitudinal Stress 10
Consider the free body diagram in the transverse section of the tank:
The total force acting at the rear of the tank must equal to the total
longitudinal stress on the wall .
Total force on the end of the cylinder due to internal pressure
2
4
i
i
Force pressure projected area
d
F p
where
p pressure
d internal diameter

 
 


1

2 2
4 4
o i
L
Resisting Force Stress Area
d d
Resisting Force
 

 
 
   
 
The total force due to longitudinal stress
2
 
 
 
2 2 2
2 2 2
2 2 2
2
2 2
4 4 4
4 4
2
i o i
L
i L o i
i L o i
i L i i
d d d
p
p d d d
pd d d
pd d t d
  

 



 
    
 
   
 
   
   di
do

 
 
 
2
2 2
2 2 2 2
2 2
2
2
2
2 2 4
4 4
4
4
4
4
i L i i
i L i i i
i L i
i L i
i
L
i
i
L
L
pd d t d
pd d d t t d
pd d t t
pd d t
pd
d t
pd
t
or
p d
t







   
  
       
 






Longitudinal Stress
2
H
p d
t




Hoop Stress
compare

Problems
13

CIVILENGINEERINGBUITEMSQUETTA
14
Calculate the bursting pressure for a cold drawn seamless steel tubing of 60 mm inside
diameter with 2mm wall thickness. The ultimate strength of steel is 380 MN/m2
2
60
2
?
380H
d mm
t mm
p
MN
m





Problem 1
Data
2
H
p d
t




2
2
380 2 2
2
25.33
60
H
MN
mm
t MNmp
d mm m
  
 
  
Formula
Answer

CIVILENGINEERINGBUITEMSQUETTA
15
Calculate the thickness of the metal required if a cast iron main 800 mm in diameter for
water at a pressure head of 100 m if the maximum permissible tensile stress 20 MN/m2 and
weight of water is 10 kN/m3
3
2
800
10
100
20
?
2
H
H
d mm
kN
w
m
h m
MN
m
t
p d
t










Problem 2
Data
3
2
3
3
6
2
2
2
10 100 800
2 20
10
10 100 0.8
10
2 20
20
H
H
p d
t
w h d
t
kN
m mm
mt
MN
m
N
m m
mt
N
m
t mm





 


 


 



Formula
Answer
Solution

Stress in Sphere 16
2
4
Force pressure projected area
d
F p
where
p pressure
d internal diameter

 
 


Resisting Force Stress Area
dt 
 
 
2
4
4
d
p dt
p d
t

 

  




Problem 3 17
A large spherical tank (see figure) contains gas at a pressure of 450 psi. The tank is
42 ft in diameter and is constructed of high-strength steel having a yield stress in
tension of 80 ksi. Determine the required thickness (to the nearest 1/4 inch) of the
wall of the tank if a factor of safety of 3.5 with respect to yielding is required
y
d = Diameter of tank = 42ft
p = Internal Pressure= 450psi
= yield stress= 80 ksi
n = Factor of Safety= 3.5
MinimumWall Thickness= t =?

Data
4
p d
t




Formula
2
2
4
450 42
80
4
450 42
4 80
450 42 12
2.5
4 80 1000
p d
t
psi ft
ksi
t
psi ft
t
ksi
lb
in
int in
lb
in










 
 
 

18
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Chapter 5 thin cylinders 6th semester

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