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From equations (i) and (ii), we have
!
t2
× ∋(d.t =
2
()
4
pd
∋
∀
)!
t2
=
4
pd
t
∀
or
2
4
t
pd
t
∀
#
!
If %
c
is the efficiency of the circumferential joint, then
t =
2
4
tc
pd
∀
!∀%
From above we see that the longitudinal stress is half of the circumferential or hoop stress.
Therefore, the design of a pressure vessel must be based on the maximum stress i.e. hoop stress.
Example 7.1. A thin cylindrical pressure vessel of 1.2 m diameter generates steam at a
pressure of 1.75 N/mm
2
. Find the minimum wall thickness, if (a) the longitudinal stress does not
exceed 28 MPa; and (b) the circumferential stress does not exceed 42 MPa.
Solution. Given : d = 1.2 m = 1200 mm ; p = 1.75 N/mm
2
; !
t2
= 28 MPa = 28 N/mm
2
;
!
t1
= 42 MPa = 42 N/mm
2
(a) When longitudinal stress (
!!
!!
!
t2
) does not exceed 28 MPa
We know that minimum wall thickness,
t =
2
. 1.75 1200
4428
t
pd
∀
#
!∀
= 18.75 say 20 mm
Ans.
(b) When circumferential stress (
!!
!!
!
t1
) does not exceed 42 MPa
We know that minimum wall thickness,
t =
1
. 1.75 1200
25 mm
2242
t
pd
∀
##
!∀
Ans.
Example 7.2.
A thin cylindrical pressure vessel of 500 mm diameter is subjected to an internal
pressure of 2 N/mm
2
. If the thickness of the vessel is 20 mm, find the hoop stress, longitudinal stress
and the maximum shear stress.
Solution. Given : d = 500 mm ; p = 2 N/mm
2
; t = 20 mm
Cylinders and tanks are used to store fluids under pressure.
Pressure Vessels
n
229
Hoop stress
We know that hoop stress,
!
t1
=
.2500
2220
pd
t
∀
#
∀
= 25 N/mm
2
= 25 MPa
Ans.
Longitudinal stress
We know that longitudinal stress,
!
t2
=
.2500
4420
pd
t
∀
#
∀
= 12.5 N/mm
2
= 12.5 MPa
Ans.
Maximum shear stress
We know that according to maximum shear stress theory, the maximum shear stress is one-half
the algebraic difference of the maximum and minimum principal stress. Since the maximum principal
stress is the hoop stress (!
t1
) and minimum principal stress is the longitudinal stress (!
t2
), therefore
maximum shear stress,
&
max
=
12
–
25 – 12.5
22
tt
!!
# = 6.25 N/mm
2
= 6.25 MPa
Ans.
Example 7.3.
An hydraulic control for a straight line motion, as shown in Fig. 7.4, utilises a
spherical pressure tank ‘A’ connected to a working cylinder B. The pump maintains a pressure of
3 N/mm
2
in the tank.
1. If the diameter of pressure tank is 800 mm, determine its thickness for 100% efficiency of
the joint. Assume the allowable tensile stress as 50 MPa.
Fig. 7.4
2. Determine the diameter of a cast iron cylinder and its thickness to produce an operating
force F = 25 kN. Assume (i) an allowance of 10 per cent of operating force F for friction in the
cylinder and packing, and (ii) a pressure drop of 0.2 N/mm
2
between the tank and cylinder. Take safe
stress for cast iron as 30 MPa.
3. Determine the power output of the cylinder, if the stroke of the piston is 450 mm and the time
required for the working stroke is 5 seconds.
4. Find the power of the motor, if the working cycle repeats after every 30 seconds and the
efficiency of the hydraulic control is 80 percent and that of pump 60 percent.
Solution. Given : p = 3 N/mm
2
; d = 800 mm ; % = 100% = 1 ; !
t1
= 50 MPa = 50 N/mm
2
;
F = 25 kN = 25 × 10
3
N ; !
tc
= 30 MPa = 30 N/mm
2
: %
H
= 80% = 0.8 ; %
P
= 60% = 0.6
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A Textbook of Machine Design
1. Thickness of pressure tank
We know that thickness of pressure tank,
t =
1
. 3 800
24 mm
2. 2501
t
pd
∀
##
!% ∀ ∀
Ans.
2. Diameter and thickness of cylinder
Let D = Diameter of cylinder, and
t
1
= Thickness of cylinder.
Since an allowance of 10 per cent of operating force F is provided for friction in the cylinder
and packing, therefore total force to be produced by friction,
F
1
= F +
10
100
F = 1.1 F = 1.1 × 25 × 10
3
= 27 500 N
We know that there is a pressure drop of 0.2 N/mm
2
between the tank and cylinder, therefore
pressure in the cylinder,
p
1
= Pressure in tank – Pressure drop = 3 – 0.2 = 2.8 N/mm
2
and total force produced by friction (F
1
),
27 500 =
4
∋
× D
2
× p
1
= 0.7854 × D
2
× 2.8 = 2.2 D
2
) D
2
= 27 500 / 2.2 = 12 500 or D = 112 mm
Ans.
We know that thickness of cylinder,
t
1
=
1
.2.8112
5.2 mm
2230
tc
pD
∀
##
!∀
Ans.
3. Power output of the cylinder
We know that stroke of the piston
= 450 mm = 0.45 m (Given)
and time required for working stroke
=5 s (Given)
)Distance moved by the piston per second
=
0.45
0.09 m
5
#
Jacketed pressure vessel.
Pressure Vessels
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231
We know that work done per second
= Force × Distance moved per second
= 27 500 × 0.09 = 2475 N-m
) Power output of the cylinder
= 2475 W = 2.475 kW Ans. (∵ 1 N-m/s = 1 W)
4. Power of the motor
Since the working cycle repeats after every 30 seconds, therefore the power which is to be
produced by the cylinder in 5 seconds is to be provided by the motor in 30 seconds.
) Power of the motor
=
HP
Power of the cylinder 5 2.475 5
0.86 kW
30 0.8 0.6 30
∀# ∀#
%∀% ∀
Ans.
7.67.6
7.67.6
7.6
Change in Dimensions of a Change in Dimensions of a
Change in Dimensions of a Change in Dimensions of a
Change in Dimensions of a
Thin CylindrThin Cylindr
Thin CylindrThin Cylindr
Thin Cylindr
ical Shell due to an Interical Shell due to an Inter
ical Shell due to an Interical Shell due to an Inter
ical Shell due to an Inter
nalnal
nalnal
nal
PrPr
PrPr
Pr
essuressur
essuressur
essur
ee
ee
e
When a thin cylindrical shell is subjected to an internal pressure, there will be an increase in the
diameter as well as the length of the shell.
Let l = Length of the cylindrical shell,
d = Diameter of the cylindrical shell,
t = Thickness of the cylindrical shell,
p = Intensity of internal pressure,
E = Young’s modulus for the material of the cylindrical shell, and
∗ = Poisson’s ratio.
The increase in diameter of the shell due to an internal pressure is given by,
+d =
2
.
1–
2. 2
pd
tE
∗
,−
./
01
The increase in length of the shell due to an internal pressure is given by,
+l =
1
–
2. 2
pdl
tE
,−
∗
./
01
It may be noted that the increase in diameter and length of the shell will also increase its volume.
The increase in volume of the shell due to an internal pressure is given by
+V = Final volume – Original volume =
4
∋
(d + +d)
2
(l + +l) –
4
∋
× d
2
.l
=
4
∋
(d
2
.+l + 2 d.l.+d ) (Neglecting small quantities)
Example 7.4. Find the thickness for a tube of internal diameter 100 mm subjected to an internal
pressure which is 5/8 of the value of the maximum permissible circumferential stress. Also find the
increase in internal diameter of such a tube when the internal pressure is 90 N/mm
2
.
Take E = 205 kN/mm
2
and ∗ = 0.29. Neglect longitudinal strain.
Solution. Given : p = 5/8 × !
t1
= 0.625 !
t1
; d = 100 mm ; p
1
= 90 N/mm
2
; E = 205 kN/mm
2
= 205 × 10
3
N/mm
2
; ∗ = 0.29
Thickness of a tube
We know that thickness of a tube,
t =
1
11
0.625 100
.
31.25 mm
22
t
tt
pd
!∀
##
!!
Ans.
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A Textbook of Machine Design
Increase in diameter of a tube
We know that increase in diameter of a tube,
+d =
2
2
1
3
90 (100) 0.29
1– 1– mm
2. 2 2
2 31.25 205 10
pd
tE
∗
,− 2 3
#
./
45
01 6 7
∀∀∀
= 0.07 (1 – 0.145) = 0.06 mm
Ans.
7.77.7
7.77.7
7.7
Thin SpherThin Spher
Thin SpherThin Spher
Thin Spher
ical Shells Subjected to an Interical Shells Subjected to an Inter
ical Shells Subjected to an Interical Shells Subjected to an Inter
ical Shells Subjected to an Inter
nal Prnal Pr
nal Prnal Pr
nal Pr
essuressur
essuressur
essur
ee
ee
e
Consider a thin spherical shell subjected to an internal pressure as shown in Fig. 7.5.
Let V = Storage capacity of the shell,
p = Intensity of internal pressure,
d = Diameter of the shell,
t = Thickness of the shell,
!
t
= Permissible tensile stress for the
shell material.
In designing thin spherical shells, we have to determine
1. Diameter of the shell, and 2. Thickness of the shell.
1. Diameter of the shell
We know that the storage capacity of the shell,
V =
4
3
× ∋ r
3
=
6
∋
× d
3
or
1/3
6
V
d
,−
#
./
∋
01
2. Thickness of the shell
As a result of the internal pressure, the shell is likely to rupture along the centre of the sphere.
Therefore force tending to rupture the shell along the centre of the sphere or bursting force,
= Pressure × Area = p ×
4
∋
× d
2
(i)
and resisting force of the shell
= Stress × Resisting area = !
t
× ∋ d.t
(ii)
Equating equations (i) and (ii), we have
p ×
4
∋
× d
2
= !
t
× ∋ d.t
or
.
4
t
pd
t
#
!
If % is the efficiency of the circumferential
joints of the spherical shell, then
t =
.
4.
t
pd
t
#
!%
Example 7.5. A spherical vessel 3 metre
diameter is subjected to an internal pressure of
1.5 N/mm
2
. Find the thickness of the vessel required
if the maximum stress is not to exceed 90 MPa. Take
efficiency of the joint as 75%.
Solution. Given: d = 3 m = 3000 mm ;
p = 1.5 N/mm
2
; !
t
= 90 MPa = 90 N/mm
2
; %(= 75% = 0.75
Fig. 7.5. Thin spherical shell.
The Trans-Alaska Pipeline carries crude oil 1, 284
kilometres through Alaska. The pipeline is 1.2
metres in diameter and can transport 318 million
litres of crude oil a day.
Pressure Vessels
n
233
We know that thickness of the vessel,
t =
. 1.5 3000
4. 4900.75
t
pd
∀
#
!% ∀ ∀
= 16.7 say 18 mm
Ans.
7.87.8
7.87.8
7.8
Change in Dimensions of a Change in Dimensions of a
Change in Dimensions of a Change in Dimensions of a
Change in Dimensions of a
Thin SpherThin Spher
Thin SpherThin Spher
Thin Spher
ical Shell due to an Interical Shell due to an Inter
ical Shell due to an Interical Shell due to an Inter
ical Shell due to an Inter
nalnal
nalnal
nal
PrPr
PrPr
Pr
essuressur
essuressur
essur
ee
ee
e
Consider a thin spherical shell subjected to an internal pressure as shown in Fig. 7.5.
Let d = Diameter of the spherical shell,
t = Thickness of the spherical shell,
p = Intensity of internal pressure,
E = Young’s modulus for the material of the spherical shell, and
∗ = Poisson’s ratio.
Increase in diameter of the spherical shell due to an internal pressure is given by,
+d =
2
.
4.
pd
tE
(1 – ∗)
(i)
and increase in volume of the spherical shell due to an internal pressure is given by,
+V = Final volume – Original volume =
6
∋
(d + +d)
3
–
6
∋
× d
3
=
6
∋
(3d
2
× +d) (Neglecting higher terms)
Substituting the value of +d from equation
(i), we have
22 4
3.
(1 – ) (1 – )
64. 8.
dpd pd
V
tE tE
23
∋∋
+# ∗# ∗
45
67
Example 7.6. A seamless spherical shell, 900 mm in diameter and 10 mm thick is being filled
with a fluid under pressure until its volume increases by 150 × 10
3
mm
3
. Calculate the pressure
exerted by the fluid on the shell, taking modulus of elasticity for the material of the shell as
200 kN/mm
2
and Poisson’s ratio as 0.3.
Solution. Given : d = 900 mm ; t = 10 mm ; +V = 150 × 10
3
mm
3
; E = 200 kN/mm
2
= 200 × 10
3
N/mm
2
; ∗ = 0.3
Let p = Pressure exerted by the fluid on the shell.
We know that the increase in volume of the spherical shell (+V),
150 × 10
3
=
4
8
pd
tE
∋
(1 – ∗) =
4
3
(900)
81020010
∋
∀∀ ∀
p
(1 – 0.3) = 90 190 p
) p = 150 × 10
3
/90 190 = 1.66 N/mm
2
Ans.
7.97.9
7.97.9
7.9
ThicThic
ThicThic
Thic
k Cylindrk Cylindr
k Cylindrk Cylindr
k Cylindr
ical Shells Subjected to an Interical Shells Subjected to an Inter
ical Shells Subjected to an Interical Shells Subjected to an Inter
ical Shells Subjected to an Inter
nal Prnal Pr
nal Prnal Pr
nal Pr
essuressur
essuressur
essur
ee
ee
e
When a cylindrical shell of a pressure vessel, hydraulic cylinder, gunbarrel and a pipe is subjected
to a very high internal fluid pressure, then the walls of the cylinder must be made extremely heavy or
thick.
In thin cylindrical shells, we have assumed that the tensile stresses are uniformly distributed
over the section of the walls. But in the case of thick wall cylinders as shown in Fig. 7.6 (a), the stress
over the section of the walls cannot be assumed to be uniformly distributed. They develop both
tangential and radial stresses with values which are dependent upon the radius of the element under
consideration. The distribution of stress in a thick cylindrical shell is shown in Fig. 7.6 (b) and (c). We
see that the tangential stress is maximum at the inner surface and minimum at the outer surface of the
shell. The radial stress is maximum at the inner surface and zero at the outer surface of the shell.
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In the design of thick cylindrical shells, the following equations are mostly used:
1. Lame’s equation; 2. Birnie’s equation; 3. Clavarino’s equation; and 4. Barlow’s equation.
The use of these equations depends upon the type of material used and the end construction.
Fig. 7.6. Stress distribution in thick cylindrical shells subjected to internal pressure.
Let r
o
= Outer radius of cylindrical shell,
r
i
= Inner radius of cylindrical shell,
t = Thickness of cylindrical shell = r
o
– r
i
,
p = Intensity of internal pressure,
∗ = Poisson’s ratio,
!
t
= Tangential stress, and
!
r
= Radial stress.
All the above mentioned equations are now discussed, in detail, as below:
1. Lame’s equation. Assuming that the longitudinal fibres of the cylindrical shell are equally
strained, Lame has shown that the tangential stress at any radius x is,
2222
22 2 22
()– () ()() –
()–() ()–()
ii oo i o i o
t
oi oi
pr pr r r p p
rr x rr
23
!# ∃
45
67
While designing a tanker, the pressure added by movement of the vehicle also should be
considered.
Pressure Vessels
n
235
and radial stress at any radius x,
2222
22 2 22
()– ( ) ()() –
–
()–() ()–()
ii oo i o i o
r
oi oi
pr pr r r p p
rr x rr
23
!#
45
67
Since we are concerned with the internal pressure ( p
i
= p) only, therefore substituting the value
of external pressure, p
o
= 0.
) Tangential stress at any radius x,
22
22 2
() ()
1
()–()
io
t
oi
pr r
rr x
23
!# ∃
45
67
(i)
and radial stress at any radius x,
22
22 2
() ( )
1
()–()
23
!# 8
45
67
io
r
oi
pr r
rr x
(ii)
We see that the tangential stress is always a tensile stress whereas the radial stress is a compressive
stress. We know that the tangential stress is maximum at the inner surface of the shell (i.e. when
x = r
i
) and it is minimum at the outer surface of the shell (i.e. when x = r
o
). Substituting the value of
x = r
i
and x = r
o
in equation (i), we find that the *maximum tangential stress at the inner surface of the
shell,
22
()
22
[( ) ( ) ]
()–()
oi
tmax
oi
pr r
rr
∃
!#
and minimum tangential stress at the outer surface of the shell,
2
()
22
2()
()–()
i
tmin
oi
pr
rr
!#
We also know that the radial stress is maximum at the inner surface of the shell and zero at the
outer surface of the shell. Substituting the value of x = r
i
and x = r
o
in equation (ii), we find that
maximum radial stress at the inner surface of the shell,
((((!
r(max)
= – p (compressive)
and minimum radial stress at the outer surface of the shell,
!
r(min)
=0
In designing a thick cylindrical shell of brittle material (e.g. cast iron, hard steel and cast
aluminium) with closed or open ends and in accordance with the maximum normal stress theory
failure, the tangential stress induced in the cylinder wall,
!
t
= !
t(max)
=
22
22
[( ) ( ) ]
()–()
oi
oi
pr r
rr
∃
Since r
o
= r
i
+ t, therefore substituting this value of r
o
in the above expression, we get
22
22
[( ) ( ) ]
()–()
ii
t
ii
pr t r
rt r
∃∃
!#
∃
!
t
(r
i
+ t)
2
– !
t
(r
i
)
2
= p (r
i
+ t)
2
+ p (r
i
)
2
(r
i
+ t)
2
(!
t
– p)=(r
i
)
2
(!
t
+ p)
2
2
()
–
()
it
t
i
rt p
p
r
∃!∃
#
!
* The maximum tangential stress is always greater than the internal pressure acting on the shell.
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A Textbook of Machine Design
–
it
it
rt p
rp
∃!∃
#
!
or
1
–
t
it
p
t
rp
!∃
∃#
!
)
–1
–
t
it
p
t
rp
!∃
#
!
or
–1
–
t
i
t
p
tr
p
23
!∃
#
45
!
45
67
(iii)
The value of !
t
for brittle materials may be taken as 0.125 times the ultimate tensile
strength (!
u
).
We have discussed above the design of a thick cylindrical shell of brittle materials. In case of
cylinders made of ductile material, Lame’s equation is modified according to maximum shear stress
theory.
According to this theory, the maximum shear stress at any point in a strained body is equal to
one-half the algebraic difference of the maximum and minimum principal stresses at that point. We
know that for a thick cylindrical shell,
Maximum principal stress at the inner surface,
!
t
(max)
=
22
22
[( ) ( ) ]
()–()
oi
oi
pr r
rr
∃
and minimum principal stress at the outer surface,
!
t(min)
=– p
) Maximum shear stress,
22
22
() ()
[( ) ( ) ]
–(– )
–
()–()
22
oi
tmax tmin
oi
max
pr r
p
rr
∃
!!
&#& # #
22 22 2
22 22
[() ()] [()–()] 2 ()
2[( ) – ( ) ] 2[( ) – ( ) ]
oi oi o
oi oi
pr r pr r pr
rr rr
∃∃
##
2
22
()
()–()
i
ii
pr t
rt r
∃
#
∃
(∵ r
o
= r
i
+ t)
or &(r
i
+ t)
2
– &(r
i
)
2
= p(r
i
+ t)
2
(r
i
+ t)
2
(& – p)=&(r
i
)
2
2
2
()
–
()
i
i
rt
p
r
∃
&
#
&
–
i
i
rt
rp
∃
&
#
&
or
1
–
i
t
rp
&
∃#
&
)
–1
–
i
t
rp
&
#
&
or
–1
–
i
tr
p
23
&
#
45
&
67
(iv)
The value of shear stress (&) is usually taken as one-half the tensile stress (!
t
). Therefore the
above expression may be written as
–1
–2
t
i
t
tr
p
23
!
#
45
!
45
67
(v)
From the above expression, we see that if the internal pressure ( p) is equal to or greater than
the allowable working stress (!
t
or &), then no thickness of the cylinder wall will prevent failure.
Thus, it is impossible to design a cylinder to withstand fluid pressure greater than the allowable
working stress for a given material. This difficulty is overcome by using compound cylinders (See
Art. 7.10).
Pressure Vessels
n
237
2. Birnie’s equation. In
case of open-end cylinders (such
as pump cylinders, rams, gun
barrels etc.) made of ductile
material (i.e. low carbon steel,
brass, bronze, and aluminium
alloys), the allowable stresses
cannot be determined by means of
maximum-stress theory of failure.
In such cases, the maximum-strain
theory is used. According to this
theory, the failure occurs when the
strain reaches a limiting value and
Birnie’s equation for the wall
thickness of a cylinder is
(1 – )
–1
–(1 )
t
i
t
p
tr
p
23
!∃ ∗
#
45
!∃∗
45
67
The value of !
t
may be taken
as 0.8 times the yield point stress
(!
y
).
3. Clavarino’s equation.
This equation is also based on the
maximum-strain theory of failure,
but it is applied to closed-end cyl-
inders (or cylinders fitted with
heads) made of ductile material.
According to this equation, the
thickness of a cylinder,
(1 – 2 )
–1
–(1 )
t
i
t
p
tr
p
23
!∃ ∗
#
45
!∃∗
45
67
In this case also, the value of !
t
may be taken as 0.8 !
y
.
4. Barlow’s equation. This equation is generally used for high pressure oil and gas pipes.
According to this equation, the thickness of a cylinder,
t = p.r
o
/(!
t
For ductile materials, !
t
= 0.8 !
y
and for brittle materials !
t
= 0.125 !
u
, where !
u
is the ultimate
stress.
Example 7.7. A cast iron cylinder of internal diameter 200 mm and thickness 50 mm is
subjected to a pressure of 5 N/mm
2
. Calculate the tangential and radial stresses at the inner, middle
(radius = 125 mm) and outer surfaces.
Solution. Given : d
i
= 200 mm or r
i
= 100 mm ; t = 50 mm ; p = 5 N/mm
2
We know that outer radius of the cylinder,
r
o
= r
i
+ t = 100 + 50 = 150 mm
Oil is frequently transported by ships called tankers. The larger tank-
ers, such as this Acrco Alaska oil transporter, are known as super-
tankers. They can be hundreds of metres long.
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