9826Lec.8 - Bolt Design application.pdf

‫الرمحن‬ ‫اهلل‬ ‫بسم‬
‫الرحيم‬

Bolts Design Application
Dr. Ameer Ali Kamel
Prod.Eng.& Mech. Design Dept.
Faculty of Engineering
Lec.8

Design of Cylinder Covers
 The cylinder covers may be secured by means of bolts or
studs, but studs are preferred.
 The possible arrangement of securing the cover with bolts
and studs is shown in Fig.
1. Design of bolts or studs
Let,
D = Diameter of the cylinder,
p = Pressure in the cylinder,
dc = Core diameter of the bolts
or studs,
n = Number of bolts or studs,
σtb = Permissible tensile stress
for the bolt or stud
material.

 The upward force acting on the cylinder cover ( Force
due to liquid or gas),
𝑭 =
𝝅
𝟒
𝑫𝟐
𝒑
 This force is resisted by “n” number of bolts or studs
provided on the cover.
 The resisting force offered by “n” number of bolts or
studs,
𝑭 =
𝝅
𝟒
𝒅𝒄
𝟐
𝝈𝒕𝒃 × 𝒏
Therefore,
𝝅
𝟒
𝑫𝟐𝒑 =
𝝅
𝟒
𝒅𝒄
𝟐
𝝈𝒕𝒃 × 𝒏

 From the above equation, the number of bolts or
studs may be obtained,
 If the value of “n” as obtained from the above
relation is odd or a fraction, then next higher even
number is adopted.
 The bolts or studs are screwed up tightly, along with
metal gasket or asbestos packing, in order to provide
a leak proof joint.
 Note that, due to the tightening of bolts, sufficient
tensile stress is produced in the bolts or studs. This
may break the bolts or studs, even before any load
due to internal pressure acts upon them. Therefore a
bolt or a stud less than 16 mm diameter should never
be used

The tightness of the joint also depends upon the
circumferential pitch of the bolts or studs. The
circumferential pitch should be:
𝒅 𝒄𝒊𝒓−𝒑𝒊𝒕𝒄𝒉 = ( 𝟐𝟎 ∶ 𝟑𝟎) 𝒅𝟏
where:
d1 : is the diameter of the hole in mm for bolt or
stud.
 The pitch circle diameter (dp) is usually taken as:
𝒅𝒑𝒊𝒕𝒄𝒉 = 𝑫 + 𝟐 × 𝒕 + 𝟔 × 𝒅𝟏
 The outside diameter of the cover is kept as:
𝑫𝟎 = 𝒅𝒑 + 𝟑 × 𝒅𝟏 = 𝑫 + 𝟐𝒕 + 𝟔𝒅𝟏
Where: t = Thickness of the cylinder wall

Example:
The cylinder head of a steam engine is subjected to a steam
pressure of 0.7 N/mm2. It is held in position by means of 12
bolts. A soft copper gasket is used to make the joint leak-proof.
The effective diameter of cylinder is 300 mm. Find the size of
the bolts so that the stress in the bolts is not to exceed 100
MPa.
Solution:
Given:
p = 0.7 N/mm2 ;
n = 12 ; D = 300 mm ;
σt = 100 MPa = 100 N/mm2

The external load acting on the cylinder head i.e. acting on all
the bolts (12 Bolts) is:
𝑭 =
𝝅
𝟒
𝑫𝟐
× 𝒑 =
𝝅
𝟒
𝟑𝟎𝟎 𝟐
× 𝟎. 𝟕 = 𝟒𝟗𝟒𝟗𝟎 𝑵
The external load acting on the cylinder head per bolt :
𝑷𝟐 =
𝟒𝟗𝟒𝟗𝟎
𝟏𝟐
= 𝟒𝟏𝟐𝟒 𝑵
Let
d = Nominal diameter of the bolt, and
dc = Core diameter of the bolt.
Therefore, the initial tension due to tightening of bolt,
P1 = 2840 d N ... (where d is in mm)

From the following table: K = 0.5.
i.e., for soft copper gasket with long through bolts, the
minimum value of K = 0.5.
∴ Resultant axial load on the bolt,

𝑷 = 𝑷𝟏 + 𝑲 × 𝑷𝟐
= 𝟐𝟖𝟒𝟎𝒅 + 𝟎. 𝟓 × 𝟒𝟏𝟐𝟒
= 𝟐𝟖𝟒𝟎 𝒅 + 𝟐𝟎𝟔𝟐 𝑵
The load on the bolt (P):
𝟐𝟖𝟒𝟎 𝒅 + 𝟐𝟎𝟔𝟐 =
𝝅
𝟒
𝒅𝒄
𝟐
𝝈𝒕
=
𝝉
𝟒
𝟎. 𝟖𝟒 𝒅 𝟐 × 𝟏𝟎𝟎
= 𝟓𝟓. 𝟒𝒅𝟐 (taking dc=0.84 d)
∴ 𝟓𝟓. 𝟒 𝒅𝟐
− 𝟐𝟖𝟒𝟎 𝒅 − 𝟐𝟎𝟔𝟐 = 𝟎
𝐨𝐫 𝒅𝟐
− 𝟓𝟏. 𝟑𝒅 − 𝟑𝟕. 𝟐 = 𝟎
𝒅 =
𝟓𝟏. 𝟑 ± 𝟓𝟏. 𝟑 𝟐 + 𝟒 × 𝟑𝟕. 𝟐
𝟐
=
𝟓𝟏. 𝟑 ± 𝟓𝟐. 𝟕
𝟐
= 𝟓𝟐 𝒎𝒎
The bolt size is M52

2. Design of cylinder cover plate
 The thickness of the cylinder cover plate (t1) and the
thickness of the cylinder flange (t2) may be
determined as discussed below:
 Let us consider the semi-cover plate as shown in Fig.
 The internal pressure in the cylinder tries to lift the
cylinder cover while the bolts or studs try to retain it
in its position.

 But the centers of pressure of these two loads do not
coincide. Hence, the cover plate is subjected to
bending stress.
 The point X is the center of pressure for bolt load and
the point Y is the center of internal pressure.
We know that the bending moment at A-A,

We know that the bending moment at A-A,
𝑀 =
𝑇𝑜𝑡𝑎𝑙 𝐵𝑜𝑙𝑡 𝐿𝑜𝑎𝑑
2

Lthread = 2d + 6 L≤125
= 2d + 12 125 ≤ L ≤ 200
= 2d+ 25 L ≥ 200

Tension Joints—The External Load
 Let us now consider what
happens when a bolt subject to
an external tensile load “P”.
 It is to be assumed, of course, that the clamping Force
(the preload “Fi”), has been correctly applied by
tightening the nut before P is applied.

Assume that:
Fi = Preload.
P = External tensile load.
Pb = Portion of “P” taken by
bolt.
Pm = portion of „P” taken by members.
Fb = Pb + Fi = Resultant bolt load.
Fm = Pm − Fi = Resultant load on members.
C = Fraction of external load “P” carried by bolt.
1 − C = Fraction of external load “P” carried by
members.

 The load “P” is tension, and it causes the connection to
stretch, or elongate, through some distance “δ”.
 This elongation can be related to the stiffness by
recalling that k is the force divided by the deflection.
Thus:
𝛅 =
𝐏𝐛
𝐤𝐛
and 𝛅 =
𝐏𝐦
𝐤𝐦
𝐏𝐦 = 𝐏𝐛
𝐤𝐦
𝐤𝐛
Since P = Pb + Pm
And,
Pm = P – Pb = (1-C)P

Relating Bolt Torque to Bolt Tension
 High preload is very desirable in important bolted
connections.
 Next, it is important to consider the means of ensuring
that the preload is actually developed when the parts
are assembled.
 If the overall length of the bolt can actually be
measured with a micrometer when it is assembled.
 The bolt elongation due to the preload “Fi“ can
be computed using the formula:
δ = Fi l/(AE)
 Then the nut is simply tightened until the bolt
elongates through the distance δ.

 This ensures that the desired preload has been
attained.
 The elongation of a screw cannot usually be measured,
because the threaded end is often in a blind hole.
[[[It is also impractical in many cases to measure
bolt elongation.]]]
 In such cases the wrench torque, required to develop
the specified preload, must be estimated.
 Then torque wrenching, pneumatic-impact wrenching,
or the turn-of-the-nut method may be used.

9826Lec.8 - Bolt Design application.pdf

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