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261
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261
1. Introduction.
2. Stresses in Pipes.
3. Design of Pipes.
4. Pipe Joints.
5. Standard Pipe Flanges for
Steam.
6. Hydraulic Pipe Joint for
High Pressures.
7. Design of Circular Flanged
Pipe Joint.
8. Design of Oval Flanged Pipe
Joint.
9. Design of Square Flanged
Pipe Joint.
8
C
H
A
P
T
E
R
8.18.1
8.18.1
8.1
IntroductionIntroduction
IntroductionIntroduction
Introduction
The pipes are used for transporting various fluids like
water, steam, different types of gases, oil and other chemicals
with or without pressure from one place to another. Cast iron,
wrought iron, steel and brass are the materials generally used
for pipes in engineering practice. The use of cast iron pipes
is limited to pressures of about 0.7 N/mm
2
because of its
low resistance to shocks which may be created due to the
action of water hammer. These pipes are best suited for water
and sewage systems. The wrought iron and steel pipes are
used chiefly for conveying steam, air and oil. Brass pipes, in
small sizes, finds use in pressure lubrication systems on prime
movers. These are made up and threaded to the same
standards as wrought iron and steel pipes. Brass pipe is not
liable to corrosion. The pipes used in petroleum industry are
generally seamless pipes made of heat-resistant chrome-
molybdenum alloy steel. Such type of pipes can resist
pressures more than 4 N/mm
2
and temperatures greater than
440°C.
CONTENTS
CONTENTS
CONTENTS
CONTENTS
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8.28.2
8.28.2
8.2
Stresses in PipesStresses in Pipes
Stresses in PipesStresses in Pipes
Stresses in Pipes
The stresses in pipes due to the internal fluid pressure are determined by Lame's equation as discussed
in the previous chapter (Art. 7.9). According to Lame's equation, tangential stress at any radius x,
!
t
=
22
22 2
() ()
1
() ()
io
oi
pr r
rr x
∀#
∃
%&
∋( )
(i)
and radial stress at any radius x,
!
r
=
22
22 2
() ()
1
() ()
io
oi
pr r
rr x
∀#
∋
%&
∋( )
(ii)
where p = Internal fluid pressure in the pipe,
r
i
= Inner radius of the pipe, and
r
o
= Outer radius of the pipe.
The tangential stress is maximum at the inner surface (when x = r
i
) of the pipe and minimum at
the outer surface (when x = r
o
) of the pipe.
Substituting the values of x = r
i
and x = r
o
in
equation (i), we find that the maximum tangential
stress at the inner surface of the pipe,
!
t(max)
=
22
22
[( ) ( ) ]
() ()
oi
oi
pr r
rr
∃
∋
and minimum tangential stress at the outer surface
of the pipe,
!
t(min)
=
2
22
2()
() ()
i
oi
pr
rr
∋
The radial stress is maximum at the inner
surface of the pipe and zero at the outer surface of
the pipe. Substituting the values of x = r
i
and x = r
o
in equation (ii), we find that maximum radial stress
at the inner surface,
!
r(max)
=– p (compressive)
and minimum radial stress at the outer surface of the pipe,
!
r(min)
=0
The thick cylindrical formula may be applied when
(a) the variation of stress across the thickness of the pipe is taken into account,
(b) the internal diameter of the pipe (D) is less than twenty times its wall thickness (t), i.e.
D/t < 20, and
(c) the allowable stress (!
t
) is less than six times the pressure inside the pipe ( p ) i.e.
!
t
/ p < 6.
According to thick cylindrical formula (Lame's equation), wall thickness of pipe,
t = R
1
t
t
p
p
∀#
!∃
∋
%&
!∋
%&
()
where R = Internal radius of the pipe.
The following table shows the values of allowable tensile stress (!
t
) to be used in the above
relations:
Cast iron pipes.
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263
Table 8.1. Values of allowable tensile stress for pipes of different materials.Table 8.1. Values of allowable tensile stress for pipes of different materials.
Table 8.1. Values of allowable tensile stress for pipes of different materials.Table 8.1. Values of allowable tensile stress for pipes of different materials.
Table 8.1. Values of allowable tensile stress for pipes of different materials.
S.No. Pipes Allowable tensile stress (!
t
)
in MPa or N/mm
2
1. Cast iron steam or water pipes 14
2. Cast iron steam engine cylinders 12.5
3. Lap welded wrought iron tubes 60
4. Solid drawn steel tubes 140
5. Copper steam pipes 25
6. Lead pipes 1.6
Example 8.1. A cast iron pipe of internal diameter 200 mm and thickness 50 mm carries water
under a pressure of 5 N/mm
2
. Calculate the tangential and radial stresses at radius (r) = 100 mm ;
110 mm ; 120 mm ; 130 mm ; 140 mm and 150 mm. Sketch the stress distribution curves.
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 pipe,
r
o
=r
i
+ t = 100 + 50 = 150 mm
Tangential stresses at radius 100 mm, 110 mm, 120 mm, 130 mm, 140 mm and 150 mm
We know that tangential stress at any radius x,
∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗!
t
=
22 2
2
22 2 2 2 2
() () ()5 (100)
11
( ) ( ) (150) (100)
io o
oi
pr r r
rr x x
∀# ∀#
∃+ ∃
%& %&
∋∋() ()
=
2
2
2
()
41 N/mm orMPa
o
r
x
∀#
∃
%&
()
, Tangential stress at radius 100 mm (i.e. when x = 100 mm),
!
t1
=
2
2
(150)
41 43.2513MPa
(100)
∀#
∃+−+
%&
()
Ans.
Tangential stress at radius 110 mm (i.e. when x = 110 mm),
!
t2
=
2
2
(150)
41 42.8611.44MPa
(110)
∀#
∃+−+
%&
()
Ans.
Tangential stress at radius 120 mm (i.e. when x = 120 mm),
!
t3
=
2
2
(150)
41 42.5610.24MPa
(120)
∀#
∃+−+
%&
()
Ans.
Tangential stress at radius 130 mm (i.e. when x = 130 mm),
!
t4
=
2
2
(150)
41 4 2.33 9.32MPa
(130)
∀#
∃+−+
%&
()
Ans.
Tangential stress at radius 140 mm (i.e. when x = 140 mm),
!
t5
=
2
2
(150)
4 1 4 2.15 8.6 MPa
(140)
∀#
∃+−+
%&
()
Ans.
and tangential stress at radius 150 mm (i.e. when x = 150 mm),
!
t6
=
2
2
(150)
41 4 2 8MPa
(150)
∀#
∃+−+
%&
()
Ans.
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Fig. 8.1
Radial stresses at radius 100 mm, 110 mm, 120 mm, 130 mm, 140 mm and 150 mm
We know that radial stress at any radius x,
!
r
=
22 2
2
22 2 2 2 2
() () ()
5 (100)
11
( ) ( ) (150) (100)
io o
oi
pr r r
rr x x
∀# ∀#
∋+ ∋
%& %&
∋∋() ()
=
2
2
2
()
4 1 N/mm or MPa
o
r
x
∀#
∋
%&
()
, Radial stress at radius 100 mm (i.e. when x = 100 mm),
!
r1
=
2
2
(150)
41 4 1.25 5MPa
(100)
∀#
∋ + −∋ +∋
%&
()
Ans.
Radial stress at radius 110 mm (i.e., when x = 110 mm),
!
r2
=
2
2
(150)
4 1 4 0.86 3.44 MPa
(110)
∀#
∋+−∋+∋
%&
()
Ans.
Radial stress at radius 120 mm (i.e. when x = 120 mm),
!
r3
=
2
2
(150)
4 1 4 0.56 2.24 MPa
(120)
∀#
∋+−∋+∋
%&
()
Ans.
Radial stress at radius 130 mm (i.e. when x = 130 mm),
!
r4
=
2
2
(150)
4 1 4 0.33 1.32 MPa
(130)
∀#
∋+−∋+∋
%&
()
Ans.
Radial stress at radius 140 mm (i.e. when x = 140 mm),
!
r5
=
2
2
(150)
4 1 4 0.15 0.6 MPa
(140)
∀#
∋+−∋+∋
%&
()
Ans.
Radial stress at radius 150 mm (i.e. when x = 150 mm),
!
r6
=
2
2
(150)
41 0
(150)
∀#
∋+
%&
()
Ans.
The stress distribution curves for tangential and radial stresses are shown in Fig. 8.1.
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8.38.3
8.38.3
8.3
Design of PipesDesign of Pipes
Design of PipesDesign of Pipes
Design of Pipes
The design of a pipe involves the determination of inside diameter of the pipe and its wall
thickness as discussed below:
1. Inside diameter of the pipe. The inside diameter of the pipe depends upon the quantity of
fluid to be delivered.
Let D =Inside diameter of the pipe,
v = Velocity of fluid flowing per minute, and
Q =Quantity of fluid carried per minute.
We know that the quantity of fluid flowing per minute,
Q =Area × Velocity =
2
4
Dv
.
−−
, D =
4
1.13
QQ
vv
−+
.
2. Wall thickness of the pipe. After deciding upon
the inside diameter of the pipe, the thickness of the wall
(t) in order to withstand the internal fluid pressure ( p)
may be obtained by using thin cylindrical or thick
cylindrical formula.
The thin cylindrical formula may be applied when
(a) the stress across the section of the pipe is
uniform,
(b) the internal diameter of the pipe (D) is more
than twenty times its wall thickness (t), i.e.
D/t > 20, and
(c) the allowable stress (!
t
) is more than six
times the pressure inside the pipe (p),
i.e. !
t
/p > 6.
According to thin cylindrical formula, wall thickness of pipe,
t =
or
22
!!/
ttl
pD pD
where
/
l
= Efficiency of longitudinal joint.
A little consideration will show that the thickness of wall as obtained by the above relation is
too small. Therefore for the design of pipes, a certain constant is added to the above relation. Now the
relation may be written as
t =
.
2
t
pD
C
∃
!
The value of constant ‘C’, according to Weisback, are given in the following table.
Table 8.2. Values of constant ‘Table 8.2. Values of constant ‘
Table 8.2. Values of constant ‘Table 8.2. Values of constant ‘
Table 8.2. Values of constant ‘
C’C’
C’C’
C’
.
Material Cast iron Mild steel Zinc and Lead
Copper
Constant (C) in mm 9 3 4 5
Pipe Joint
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Example 8.2. A seamless pipe carries 2400 m
3
of steam per hour at a pressure of 1.4 N/mm
2
.
The velocity of flow is 30 m/s. Assuming the tensile stress as 40 MPa, find the inside diameter of the
pipe and its wall thickness.
Solution. Given : Q = 2400 m
3
/h = 40 m
3
/min ; p = 1.4 N/mm
2
; v = 30 m/s = 1800 m/min ;
!
t
= 40 MPa = 40 N/mm
2
Inside diameter of the pipe
We know that inside diameter of the pipe,
D =
40
1.13 1.13 0.17 m 170 mm
1800
Q
v
+++
Ans.
Wall thickness of the pipe
From Table 8.2, we find that for a steel pipe, C = 3 mm. Therefore wall thickness of the pipe,
t =
. 1.4 170
36mm
2240
pD
C
−
∃+ ∃+
−
Ans.
8.48.4
8.48.4
8.4
Pipe JointsPipe Joints
Pipe JointsPipe Joints
Pipe Joints
The pipes are usually connected to vessels from which they transport the fluid. Since the length
of pipes available are limited, therefore various lengths of pipes have to be joined to suit any particular
installation. There are various forms of pipe joints used in practice, but most common of them are
discussed below.
1. Socket or a coupler joint. The most
common method of joining pipes is by means of a
socket or a coupler as shown in Fig. 8.2. A socket is
a small piece of pipe threaded inside. It is screwed
on half way on the threaded end of one pipe and the
other pipe is then screwed in the remaining half of
socket. In order to prevent leakage, jute or hemp is
wound around the threads at the end of each pipe.
This type of joint is mostly used for pipes carrying
water at low pressure and where the overall smallness
of size is most essential.
2. Nipple joint. In this type of joint, a nipple which is a small piece of pipe threaded outside is
screwed in the internally threaded end of each pipe, as shown in Fig. 8.3. The disadvantage of this
joint is that it reduces the area of flow.
Fig. 8.3. Nipple joint. Fig. 8.4. Union joint.
3. Union joint. In order to disengage pipes joined by a socket, it is necessary to unscrew pipe
from one end. This is sometimes inconvenient when pipes are long.
The union joint, as shown in Fig. 8.4, provide the facility of disengaging the pipes by simply
unscrewing a coupler nut.
Fig. 8.2. Socket or coupler joint.
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4. Spigot and socket joint. A spigot and socket joint as shown in Fig. 8.5, is chiefly used for
pipes which are buried in the earth. Some pipe lines are laid straight as far as possible. One of the
important features of this joint is its flexibility as it adopts itself
to small changes in level due to settlement of earth which takes
place due to climate and other conditions.
In this type of joint, the spigot end of one pipe fits into the
socket end of the other pipe. The remaining space between the
two is filled with a jute rope and a ring of lead. When the lead
solidifies, it is caulked-in tightly.
5. Expansion joint. The pipes carrying steam at high
pressures are usually joined by means of expansion joint. This
joint is used in steam pipes to take up expansion and contraction
of pipe line due to change of temperature.
In order to allow for change in length, steam pipes are not rigidly clamped but supported on
rollers. The rollers may be arranged on wall bracket, hangers or floor stands. The expansion bends, as
shown in Fig. 8.6 (a) and (b), are useful in a long pipe line. These pipe bends will spring in either
direction and readily accommodate themselves to small movements of the actual pipe ends to which
they are attached.
Fig. 8.6. Expansion bends.
Fig. 8.7. Expansion joints.
The copper corrugated expansion joint, as shown in Fig. 8.7 (a), is used on short lines and is
satisfactory for limited service. An expansion joint as shown in Fig. 8.7 (b) (also known as gland and
stuffing box arrangement), is the most satisfactory when the pipes are well supported and cannot sag.
Fig. 8.5. Spigot and socket joint.
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6. Flanged joint. It is one of the most widely used pipe joint. A flanged joint may be made with
flanges cast integral with the pipes or loose flanges welded or screwed. Fig. 8.8 shows two cast iron
pipes with integral flanges at their ends. The flanges are connected by means of bolts. The flanges
have seen standardised for pressures upto
2 N/mm
2
. The flange faces are machined
to ensure correct alignment of the pipes.
The joint may be made leakproof by
placing a gasket of soft material, rubber
or convass between the flanges. The
flanges are made thicker than the pipe
walls, for strength. The pipes may be
strengthened for high pressure duty by
increasing the thickness of pipe for a short
length from the flange, as shown in Fig. 8.9.
For even high pressure and for large
diameters, the flanges are further strengthened by ribs or stiffners as shown in Fig. 8.10 (a). The ribs
are placed between the bolt holes.
Fig. 8.10
For larger size pipes, separate loose flanges screwed on the pipes as shown in Fig. 8.10 (b) are
used instead of integral flanges.
Fig. 8.8. Flanged joint.
Fig. 8.9. Flanged joint.
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7. Hydraulic pipe joint. This type of joint has oval flanges and are fastened by means of two bolts,
as shown in Fig. 8.11. The oval flanges are usually used for small pipes, upto 175 mm diameter. The
flanges are generally cast integral with the pipe ends. Such joints are used to carry fluid pressure varying
from 5 to 14 N/mm
2
. Such a high pressure is found in hydraulic applications like riveting, pressing, lifts
etc. The hydraulic machines used in these installations are pumps, accumulators, intensifiers etc.
Fig. 8.11. Hydraulic pipe joint.
8.58.5
8.58.5
8.5
Standard Pipe Flanges for SteamStandard Pipe Flanges for Steam
Standard Pipe Flanges for SteamStandard Pipe Flanges for Steam
Standard Pipe Flanges for Steam
The Indian boiler regulations (I.B.R.) 1950 (revised 1961) have standardised all dimensions of
pipe and flanges based upon steam pressure. They have been divided into five classes as follows:
Class I : For steam pressures up to 0.35 N/mm
2
and water pressures up to 1.4 N/mm
2
. This is
not suitable for feed pipes and shocks.
Class II : For steam pressures over 0.35 N/mm
2
but not exceeding 0.7 N/mm
2
.
Class III : For steam pressures over 0.7 N/mm
2
but not exceeding 1.05 N/mm
2
.
Class IV : For steam pressures over
1.05 N/mm
2
but not exceeding 1.75 N/mm
2
.
Class V : For steam pressures from
1.75 N/mm
2
to 2.45 N/mm
2
.
According to I.B.R., it is desirable that
for classes II, III, IV and V, the diameter of
flanges, diameter of bolt circles and number
of bolts should be identical and that
difference should consist in variations of the
thickness of flanges and diameter of bolts
only. The I.B.R. also recommends that all
nuts should be chamfered on the side bearing
on the flange and that the bearing surfaces
of the flanges, heads and nuts should be true.
The number of bolts in all cases should be a
multiple of four. The I.B.R. recommends that for 12.5 mm and 15 mm bolts, the bolt holes should be
1.5 mm larger and for higher sizes of bolts, the bolt holes should be 3 mm larger. All dimensions for
pipe flanges having internal diameters 1.25 mm to 600 mm are standardised for the above mentioned
classes (I to V). The flanged tees, bends are also standardised.
The Trans-Alaska Pipeline was built to carry oil across the
frozen sub-Arctic landscape of North America.
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Note: As soon as the size of pipe is determined, the rest of the dimensions for the flanges, bolts, bolt holes,
thickness of pipe may be fixed from standard tables. In practice, dimensions are not calculated on a rational
basis. The standards are evolved on the basis of long practical experience, suitability and interchangeability. The
calculated dimensions as discussed in the previous articles do not agree with the standards. It is of academic
interest only that the students should know how to use fundamental principles in determining various dimen-
sions e.g. wall thickness of pipe, size and number of bolts, flange thickness. The rest of the dimensions may be
obtained from standard tables or by empirical relations.
8.68.6
8.68.6
8.6
Hydraulic Pipe Joint for High PressuresHydraulic Pipe Joint for High Pressures
Hydraulic Pipe Joint for High PressuresHydraulic Pipe Joint for High Pressures
Hydraulic Pipe Joint for High Pressures
The pipes and pipe joints for high fluid pressure are classified as follows:
1. For hydraulic pressures up to 8.4 N/mm
2
and pipe bore from 50 mm to 175 mm, the flanges
and pipes are cast integrally from remelted cast
iron. The flanges are made elliptical and secured
by two bolts. The proportions of these pipe joints
have been standardised from 50 mm to 175 mm,
the bore increasing by 25 mm. This category is
further split up into two classes:
(a) Class A: For fluid pressures from
5 to 6.3 N/mm
2
, and
(b) Class B:
For fluid pressures from
6.3 to 8.4 N/mm
2
.
The flanges in each of the above classes
may be of two types. Type I is suitable for pipes
of 50 to 100 mm bore in class A, and for 50 to
175 mm bore in class B. The flanges of type II
are stronger than those of Type I and are usually
set well back on the pipe.
2. For pressures above 8.4 N/mm
2
with
bores of 50 mm or below, the piping is of wrought
steel, solid drawn, seamless or rolled. The flanges
may be of cast iron, steel mixture or forged steel. These are screwed or welded on to the pipe and are
square in elevation secured by four bolts. These joints are made for pipe bores 12.5 mm to 50 mm
rising in increment of 3 mm from 12.5 to 17.5 mm and by 6 mm from 17.5 to 50 mm. The flanges and
pipes in this category are strong enough for service under pressures ranging up to 47.5 N/mm
2
.
In all the above classes, the joint is of the spigot and socket type made with a jointing ring of
gutta-percha.
Notes: The hydraulic pipe joints for high pressures differ from those used for low or medium pressure in the
following ways:
1. The flanges used for high pressure hydraulic pipe joints are heavy oval or square in form, They use two or
four bolts which is a great advantage while assembling and disassembling the joint especially in narrow space.
2. The bolt holes are made square with sufficient clearance to accomodate square bolt heads and to allow
for small movements due to setting of the joint.
3. The surfaces forming the joint make contact only through a gutta-percha ring on the small area provided
by the spigot and recess. The tightening up of the bolts squeezes the ring into a triangular shape and makes a
perfectly tight joint capable of withstanding pressure up to 47.5 N/mm
2
.
4. In case of oval and square flanged pipe joints, the condition of bending is very clearly defined due to
the flanges being set back on the pipe and thickness of the flange may be accurately determined to withstand the
bending action due to tightening of bolts.
Hydraulic pipe joints use two or four bolts which is
a great advantage while assembling the joint
especially in narrow space.
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271
8.78.7
8.78.7
8.7
Design of Circular Flanged Pipe JointDesign of Circular Flanged Pipe Joint
Design of Circular Flanged Pipe JointDesign of Circular Flanged Pipe Joint
Design of Circular Flanged Pipe Joint
Consider a circular flanged pipe joint as shown in Fig. 8.8. In designing such joints, it is assumed
that the fluid pressure acts in between the flanges and tends to separate them with a pressure existing at
the point of leaking. The bolts are required to take up tensile stress in order to keep the flanges together.
The effective diameter on which the fluid pressure acts, just at the point of leaking, is the
diameter of a circle touching the bolt holes. Let this diameter be D
1
. If d
1
is the diameter of bolt hole
and D
p
is the pitch circle diameter, then
D
1
= D
p
– d
1
, Force trying to separate the two flanges,
F =
2
1
()
4
Dp
.
(i)
Let n = Number of bolts,
d
c
= Core diameter of the bolts, and
!
t
= Permissible stress for the material of the bolts.
, Resistance to tearing of bolts
=
2
()
4
ct
dn
.
!−
(ii)
Assuming the value of d
c
, the value of n may be obtained from equations (i) and (ii). The
number of bolts should be even because of the symmetry of the section.
The circumferential pitch of the bolts is given by
p
c
=
p
D
n
.
In order to make the joint leakproof, the value of p
c
should be between 20
1
d
to 30
1
d
,
where d
1
is the diameter of the bolt hole. Also a bolt of less than 16 mm diameter should never be used
to make the joint leakproof.
The thickness of the flange is obtained by considering a segment of the flange as shown in Fig. 8.8 (b).
In this it is assumed that each of the bolt supports one segment. The effect of joining of these
segments on the stresses induced is neglected. The bending moment is taken about the section X-X,
which is tangential to the outside of the pipe. Let the width of this segment is x and the distance of this
section from the centre of the bolt is y.
, Bending moment on each bolt due to the force F
=
F
y
n
−
(iii)
and resisting moment on the flange
= !
b
× Z
(iv)
where !
b
= Bending or tensile stress for the flange material, and
Z = Section modulus of the cross-section of the flange =
2
1
()
6
f
xt
−
Equating equations
(iii) and (iv), the value of t
f
may be obtained.
The dimensions of the flange may be fixed as follows:
Nominal diameter of bolts, d = 0.75 t + 10 mm
Number of bolts, n = 0.0275 D + 1.6 (D is in mm)
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Thickness of flange, t
f
= 1.5 t + 3 mm
Width of flange, B = 2.3 d
Outside diameter of flange,
D
o
= D + 2t + 2B
Pitch circle diameter of bolts,
D
p
= D + 2t + 2d + 12 mm
The pipes may be strengthened by providing greater thickness near the flanges
equal to
2
f
tt
∃
01
23
45
as shown in Fig. 8.9. The flanges may be strengthened by providing ribs equal to thickness of
2
f
tt
∃
,
as shown in Fig. 8.10 (a).
Example 8.3. Find out the dimensions of a flanged joint for a cast iron pipe 250 mm diameter
to carry a pressure of 0.7 N/mm
2
.
Solution. Given: D = 250 mm ; p = 0.7 N/mm
2
From Table 8.1, we find that for cast iron, allowable tensile stress, !
t
= 14 N/mm
2
and from
Table 8.2, C = 9 mm. Therefore thickness of the pipe,
t =
. 0.7 250
9 15.3 say 16 mm
2214
t
pD
C
−
∃+ ∃+
!−
Ans.
Other dimensions of a flanged joint for a cast iron pipe may be fixed as follows:
Nominal diameter of the bolts,
d = 0.75 t + 10 mm = 0.75 × 16 + 10 = 22 mm Ans.
Number of bolts, n = 0.0275 D + 1.6 = 0.0275 × 250 + 1.6 = 8.475 say 10 Ans.
Thickness of the flanges, t
f
= 1.5 t + 3 mm = 1.5 × 16 + 3 = 27 mm Ans.
Width of the flange, B = 2.3 d = 2.3 × 22 = 50.6 say 52 mm Ans.
Outside diameter of the flange,
D
o
= D + 2t + 2B = 250 + 2 × 16 + 2 × 52 = 386 mm
Ans.
Pitch circle diameter of the bolts,
D
p
= D + 2t + 2d + 12 mm = 250 + 2 × 16 + 2 × 22 + 12 mm
= 338 mm
Ans.
Circumferential pitch of the bolts,
p
c
=
338
106.2 mm
10
p
D
n
.−
.−
++
Ans.
In order to make the joint leak proof, the value of p
c
should be between 20
1
d
to 30
1
d
where d
1
is the diameter of bolt hole.
Let us take d
1
= d + 3 mm = 22 + 3 = 25 mm
,
1
20 20 25 100 mm
d
++
and
1
30 30 25 150 mm
d
++
Since the circumferential pitch as obtained above (i.e. 106.2 mm) is within
11
20 to 30
dd
,
therefore the design is satisfactory.
Pipes and Pipe Joints
n
273
Example 8.4. A flanged pipe with internal diameter as 200 mm is subjected to a fluid pressure
of 0.35 N/mm
2
. The elevation of the flange is shown in Fig. 8.12. The flange is connected by means
of eight M 16 bolts. The pitch circle diameter of the bolts is 290 mm. If the thickness of the flange is
20 mm, find the working stress in the flange.
Solution. Given : D = 200 mm ; p = 0.35 N/mm
2
; n = 8 ;* d = 16 mm ; D
p
= 290 mm ; t
f
= 20 mm
First of all, let us find the thickness of the pipe. Assuming the pipe to be of cast iron, we find from
Table 8.1 that the allowable tensile stress for cast iron, !
t
= 14 N/mm
2
and from Table 8.2, C = 9 mm.
, Thickness of the pipe,
t =
. 0.35 200
911.5say12mm
2214
t
pD
C
−
∃+ ∃+
!−
Since the diameter of the bolt holes (d
1
) is taken larger than the nominal diameter of the bolts
(d), therefore let us take diameter of the bolt holes,
d
1
= d + 2 mm = 16 + 2 = 18 mm
Fig. 8.12
and diameter of the circle on the inside of the bolt holes,
D
1
= D
p
– d
1
= 290 – 18 = 272 mm
, Force trying to separate the flanges i.e. force on 8 bolts,
F =
22
1
( ) (272) 0.35 20 340 N
44
Dp
++
Now let us find the bending moment about the section X-X which is tangential to the outside of
the pipe. The width of the segment is obtained by measuring the distance from the drawing. On
measuring, we get
x = 90 mm
and distance of the section X-X from the centre of the bolt,
y =
290 200
12 33 mm
22 2 2
p
D
D
t
01 0 1
∋∃+ ∋ ∃+
23 2 3
45 4 5
Let !
b
= Working stress in the flange.
We know that bending moment on each bolt due to force F
=
20340
33 83 900 N-mm
8
F
y
n
−+ − +
(i)
* M16 bolt means that the nominal diameter of the bolt (d) is 16 mm.
274
n
A Textbook of Machine Design
and resisting moment on the flange
=
2
1
()
6
bb f
Zxt
!− +!− −
=
2
1
90 (20) 6000 N-mm
6
bb
!− − + !
(ii)
From equations (i) and (ii), we have
!
b
= 83 900 / 6000
= 13.98 N/mm
2
= 13.98 MPa Ans.
8.88.8
8.88.8
8.8
Design of Oval Flanged Pipe JointDesign of Oval Flanged Pipe Joint
Design of Oval Flanged Pipe JointDesign of Oval Flanged Pipe Joint
Design of Oval Flanged Pipe Joint
Consider an oval flanged pipe joint as shown
in Fig. 8.11. A spigot and socket is provided for
locating the pipe bore in a straight line. A packing of
trapezoidal section is used to make the joint leak
proof. The thickness of the pipe is obtained as
discussed previously.
The force trying to separate the two flanges has
to be resisted by the stress produced in the bolts. If a
length of pipe, having its ends closed somewhere
along its length, be considered, then the force
separating the two flanges due to fluid pressure is
given by
F
1
=
2
4
Dp
.
−−
where D = Internal diameter of the pipe.
The packing has also to be compressed to make the joint leakproof. The intensity of pressure
should be greater than the pressure of the fluid inside the pipe. For the purposes of calculations, it is
assumed that the packing material is compressed to the same pressure as that of inside the pipe.
Therefore the force tending to separate the flanges due to pressure in the packing is given by
F
2
=
22
1
() ()
4
DDp
.
∀#
−∋
()
where D
1
= Outside diameter of the packing.
, Total force trying to separate the two flanges,
F = F
1
+ F
2
2222
11
() () ()
44 4
Dp D Dp Dp
.
∀#
+− −∃ ∋ +
()
Since an oval flange is fastened by means of two bolts, therefore load taken up by each bolt is
F
b
= F/2. If d
c
is the core diameter of the bolts, then
F
b
=
2
()
4
ctb
d
.
!
where !
tb
is the allowable tensile stress for the bolt material. The value of !
tb
is usually kept low to
allow for initial tightening stress in the bolts. After the core diameter is obtained, then the nominal
diameter of the bolts is chosen from
*tables. It may be noted that bolts of less than 12 mm diameter
* In the absence of tables, nominal diameter =
Core diameter
0.84
Oval flanged pipe joint.
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