Unit 5 threaded joint

Dr. Somnath Kolgiri
(Associate Professor)
S B Patil College of Engineering,
Indapur, Pune.

Contents
Basic types of screw fasteners, Bolts of uniform
strength, I.S.O. Metric screw threads, Bolts under
tension, eccentrically loaded bolted joint in shear,
Eccentric load perpendicular and parallel to axis of
bolt, Eccentric load on circular base, design of Turn
Buckle.

DEFINITION
Threaded joint is defined as a separable joint of
two or more machine parts that are held together by
means of a threaded fastening such as a bolt and a
nut
BASIC CONCEPTS
Screws have been used as fasteners for a long
time. Screw or thread joints are separable joints held
together by screw fastenings, such as screws, bolts,
studs and nuts, or by thread cut on the parts to be
joined.

Screws engage the
threads of nuts or of
other parts.
A nut is a threaded
fastening with internal
thread. It is screwed on
the bolt and is of a
shape designed to be
gripped by a wrench or
by hand.
Fig.5.1 Screw (bolt) and nut

TERMINOLOGY OF SCREW THREADS
Fig. 5.2 General screw-thread symbols

Where d — the largest diameter of a screw
thread;
dc — minor diameter, i.e. the smallest
diameter of a screw thread;
dp — pitch diameter;
p — pitch;
l — lead;
— Thread angle;
h — height of thread engagement;
— lead (or helix) angle.


According to their purpose, screw threads are
classified as:
(1) Fastening threads;
(2) Fastening and sealing threads;
(3) Power threads
Fig. 5.3 General screw-thread

Fig.5.4 Principal type of screw threads

SCREW FASTENINGS
Depending upon the type of screw joint
involved, screw fastenings are classed as:
(1) Screws with nuts, generally called bolts;
(2) Cap screws inserted into tapped holes in the
parts being fastened;
(3) Studs, or stud-bolts, used with nuts and
having threads on both ends.

Fig.5.5 Principal types of screw joints

With respect to the shape of their heads, screw
fastenings are divided into:
(1) Those in which the head is engaged
externally by a tool (wrench , etc.);
(2) Those in which the head is engaged
internally and from the end face;
(3) Those that prevent the screw fastening
from turing.

Setscrews are
another form of
fastener; the usual
use for them is to
prevent relative
circular motion
between two parts
such as shafts and
pulleys. They may
be used only
where the torque
requirements are
low.
Fig.5.7 Applications of setscrews

The types of points of setscrews are follows: Flat
Point , Dog Point, Cone Point, Hanger Point and
Cut Point
Fig.5.8 Setscrews with various points

The main types of nuts are as follows:
Fig.5.9 Principal types of nuts

SREWING-UP TORQUE, EFFICIENCY AND
SELF-LOCKING CONDITIONS
)( vtg
tg





v 
21 TTT 
FdT 2.0
Fig. 5.10

BOLT TIGHTENING AND INTIAL TENSION
F----initial force Fig.5.11 Torque-wrenches

PREVENTING UNINTENTIONAL
UNSCREWING OF SCREW JOINTS
Locking can be accomplished by the following
measures:
(1) By supplementary friction;
(2) by using special locking devices;
(3) by plastic deformation or welding on.

Fig.5.12 Locking devices based on the
application of supplementary friction

Fig.5.13 Locking devices using special
locking elements

• Bolts are subjected to shock and impact loads in
certain applications. The bolts of cylinder head of an
internal combustion engine
• Bolts of connecting rod are the examples of such
applications. In such cases, resilience of the bolt is
important design consideration to prevent breakage
at the threads.
• Resilience is defined as the ability of the material to
absorb energy when deformed elastically and to
release this energy when unloaded.
BOLT OF UNIFORM STRENGTH

When this bolt is subjected to tensile force, the
stress in the shank and the stress in the threaded
portion are equal. In the second method, the
diameter of the hole (d1) is obtained by equating
the cross-sectional area of the shank to that of the
threaded part. Therefore,
Fig. 5.15 Bolts of Uniform Strength

A bolted joint subjected to tensile force P is
shown in Fig. 7.12. The cross-section at the core
diameter dc is the weakest section. The maximum
tensile stress in the bolt at this cross-section is
given by,
BOLTED JOINT—SIMPLE ANALYSIS
Fig. 5.16 Bolt in Tension

The height of the nut h can be determined by
equating the strength of the bolt in tension with the
strength in shear. The procedure is based on the
following assumptions:
(i) Each turn of the thread in contact with the nut
supports an equal amount of load.
(ii) There is no stress concentration in the
threads.
(iii) The yield strength in shear is equal to half of
the yield strength in tension (Ssy = 0.5Syt).
(iv) Failure occurs in the threads of the bolt and
not in the threads of the nut. Substituting,

• The strength of the bolt in tension is given by,
The threads of the bolt in contact with the nut are sheared
at the core diameter dc. The shear area is equal to (p dc
h), where h is the height of the nut.
The strength of the bolt in shear is given by,
Therefore, for standard coarse
threads, the threads are equally
strong in failure by shear and
failure by tension, if the height of
the nut is approximately 0.4 times of
the nominal diameter of the bolt. The
height of the standard hexagonal nut is (0.8d). Hence, the
threads of the bolt in the standard nut will not fail by shear.

Rewriting the height of the standard nut, The
The design of the bolt consists of determination of
correct size of the bolt. The size of the bolt is given by
the nominal diameter d and pitch p. In design
calculations, many times the core diameter dc is
determined. Therefore, it is necessary to convert the
core diameter dc into the nominal diameter d.
The correct relationship for ISO metric screw
threads is as follows
Since there are two unknowns on the right hand side, it
is not possible to find out the value of d by knowing the
value of dc. Therefore, the following approximate
relationship can be used,

Example 1. An electric motor weighing 10 kN is lifted
by means of an eye bolt as shown in Fig. 5.17 The eye
bolt is screwed into the frame of the motor. The eye bolt
has coarse threads. It is made of plain carbon steel 30C8
(Syt = 400 N/mm2) and the factor of safety is 6.
Determine the size of the bolt.
Fig. 5.17 General screw-thread

Example 2. Two plates are fastened by means of two
bolts as shown in Fig. 5.18. The bolts are made of
plain carbon steel 30C8 (Syt = 400 N/mm2) and the
factor of safety is 5. Determine the size of the bolts if,
Fig. 5.18

In structural connections, a group of bolts is frequently
employed, as shown in Fig. 5.19. Let A1, A2, ..., A5
be the cross-sectional areas of the bolts. The co-
ordinates (x1, y1), (x2, y2), ..., (x5, y5) indicate the
position of bolt-centres with respect to the origin. G is
the centre of gravity of the group of bolts. The co-
ordinates (x, y) indicate the location of the centre of
gravity.
ECCENTRICALLY LOADED BOLTED JOINTS
IN SHEAR

The primary and secondary
shear forces are added by
vector addition method to get
the resultant shear forces P1,
P2, P3, and P4.
In this analysis, it is assumed that the components
connected by the bolts are rigid and the bolts have the
same crosssectional area.
Fig. 5.21

Example 1. The structural connection shown in Fig. 7.16 is subjected to an eccentric
force P of 10 kN with an eccentricity of 500 mm from the CG of the bolts. The
centre distance between bolts 1 and 2 is 200 mm, and the centre distance between
bolts 1 and 3 is 150 mm. All the bolts are identical. The bolts are made from plain
carbon steel 30C8 (Syt = 400 N/mm2) and the factor of safety is 2.5. Determine the
size of the bolts.

Example 2. A steel plate subjected
to a force of 5 kN and fixed to a
channel by means of three identical
bolts is shown in Fig. 7.19(a). The
bolts are made from plain carbon
steel 45C8 (Syt = 380 N/mm2) and
the factor of safety is 3. Specify the
size of bolts

A bracket, fixed to the steel structure by means of four bolts, is shown in
Fig. 7.20(a). It is subjected to eccentric force P at a distance e from the
structure. The force P is perpendicular to the axis of each bolt. The lower
two bolts are denoted by 2, while the upper two bolts by 1. In this
analysis, the following assumptions are made:
(i) The bracket and the steel structure are rigid.
(ii) The bolts are fitted in reamed and ground holes.
(iii) The bolts are not preloaded and there are no tensile stresses due to
initial tightening.
ECCENTRIC LOAD PERPENDICULAR
TO AXIS OF BOLT

The force P results in direct shear force on the bolts. Since the bolts are
identical, the shear force on each bolt is given by,

The bolts denoted by 1 are
subjected to maximum force. In
general, a bolt, which is located at
the farthest distance from the tilting
edge C, is subjected to maximum
force. Equations (7.9) and (7.10)
give shear and tensile forces that
act on the bolt due to eccentric load
perpendicular to the axis of the
bolts. The direct shear stress in the
bolt is given by,

Example 2. A wall bracket is
attached to the wall by means of
four identical bolts, two at A and
two at B, as shown in Fig. 7.21.
Assuming that the bracket is held
against the wall and prevented
from tipping about the point C by
all four bolts and using an
allowable tensile stress in the bolts
as 35 N/mm2, determine the size
of the bolts on the basis of
maximum principal stress theory.

Example 3. A bracket is fastened to the steel structure by means of six
identical bolts as shown in Fig. 7.22(a). Assume the following data:
l1 = 300 mm l2 = 200 mm l3 = 100 mm l = 250 mm P = 50 kN
Neglecting shear stress, determine the size of the bolts, if the maximum
permissible tensile stress in any bolt is limited to 100 N/mm2.

Example 3. A crane-runway bracket
is fastened to the roof truss by
means of two identical bolts as
shown in Fig. 7.23. Determine the
size of the bolts, if the permissible
tensile stress in the bolts is limited
to 75 N/mm2.

Example.5 Cast iron bracket, supporting the transmission shaft and the
belt pulley, is fi xed to the steel structure by means of four bolts as
shown in Fig. 7.25(a). There are two bolts at A and two bolts at B.
The tensions in slack and tight sides of the belt are 5 kN and 10 kN
respectively. The belt tensions act in a vertically downward
direction. The distances are as follows, l1 = 50 mm l2 = 150 mm l =
200 mm The maximum permissible tensile stress in any bolt is 60
N/mm2. Determine the size of the bolts.

Many times, a machine component is made with a circular base, which is
fastened to the structure by means of bolts located on the circumference
of a circle. Flanged bearings of the machine tools and the A round flange
bearing fastened by means of four bolts is shown in Fig. 7.26(b). It is
subjected to an external force P at a distance l from the support. The
following assumptions are made: (i) All bolts are identical.
(ii) The bearing and the structure are rigid.
(iii) The bolts are not preloaded and there is no tensile stress due to initial
tightening.
(iv) The stress concentration in the threads is neglected.
(v) The bolts are relieved of shear stresses by using dowel pins.
ECCENTRIC LOAD ON CIRCULAR
BASE

As shown in Fig. 7.26(a), when the load tends to tilt the bearing
about the point C, each bolt is stretched by an amount (d), which is
proportional to its vertical distance from the point C. Or,
Therefore, it can be concluded that the resisting force acting on any
bolt due to the tendency of the bearing to tilt, is proportional to its
distance from the tilting edge. If P1 P2 … are the resisting forces
induced in the bolts,

where C is the constant of proportionality. Equating the moment
due to the external force P about C with the moments due to
resisting forces,

Example 1. A round flange bearing,
as shown in Fig. 7.26(b), is fastened
to the machine frame by means of
four cap screws spaced equally on a
300 mm pitch circle diameter. The diameter of the flange is 400 mm. The
external force P is 25 kN, which is located at a distance of 150 mm from
the machine frame. There are two dowel pins to take shear load. The cap
screws are relieved of all shear force. Determine the size of the cap
screws, if the maximum permissible tensile stress in the cap screw is
limited to 50 N/mm2.

Example 2. A pillar crane, shown in Fig. 7.27, is
fastened to the foundation by means of 16 identical bolts
spaced equally on 2 m pitch circle diameter. The
diameter of the pillar flange is 2.25m. Determine the size
of the bolts if a load of 50 kN acts at a radius of 7.5 m
from the axis of the crane. The maximum permissible
tensile stress in the boltis limited to 75 N/mm2.

Example 3. Figure 7.28 shows the
bracket used in a jib crane to connect
the tie rod. The maximum force in
the tie rod is 5 kN, which is inclined
at an angle of 30° with the
horizontal. The bracket is fastened by
means of four identical bolts, two at
A and two at B. The bolts are made
of plain carbon steel 30C8 (Syt = 400
N/mm2) and the factor of safety is 5.
Assume maximum shear stress
theory and determine the size of the
bolts.

Example 4. A bracket, subjected
to a force of 5 kN inclined at an
angle of 60° with the vertical, is
shown in Fig. 7.29. The bracket is fastened by means of four identical
bolts to the structure. The bolts are made of plain carbon steel 30C8 (Syt
= 400 N/mm2) and the factor of safety is 5 based on maximum shear
stress. Assume maximum shear stress theory and determine the size of
the bolts.

Example 5. A rigid bracket subjected to a vertical force of 10 kN is
shown in Fig. 7.30(a). It is fastened to a vertical stanchion by means of
four identical bolts. Determine the size of the bolts by maximum shear
stress theory. The maximum permissible shear stress in any bolt is
limited to 50 N/mm2.

Step I Problem Specification: It is required to design a turnbuckle for
connecting the tie rods in the roof truss. The maximum pull in the tie
rods is 50 kN. of a circle.
Step II Construction : The construction of the turnbuckle is shown in
Fig. 7.31. It consists of a central portion called coupler and two rods.
One rod has right-hand threads while the other rod has left-hand threads.
The threaded portions of the rods are screwed to the coupler at the two
ends. As the central coupler rotates, the rods are either pulled together or
pushed apart depending upon the direction of the rotation of the coupler.
The outer portion of the coupler is given hexagonal shape so that it can
be rotated by means of a spanner Sometimes a hole is provided in the
coupler as indicated by a dotted circle in the figure. Instead of using a
spanner, a tommy bar is inserted in this hole to rotate the coupler. The
turnbuckle is used for connecting two rods which are in tension and
which require slight adjustment in length during the assembly. Some of
its applications are as follows:
DESIGN OF TURNBUCKLE

(i) to tighten the members of the roof truss;
(ii) to tighten the cables or the stay ropes of electric distribution poles ;
and
(iii) to connect the tie rod to the jib in case of jib-cranes.

Step III Selection of Materials The coupler has a relatively complex
shape and the economic method to make the coupler is casting. Casting
reduces the number of machining operations. Grey cast iron of grade FG
200 (Sut = 200 N/mm2) is selected as the material for the coupler. The
rods are subjected to tensile force and torsional moment. From strength
considerations, plain carbon steel of grade 30C8 (Syt = 400 N/mm2)
is selected for the rods.
Step IV General Considerations Many times, turnbuckle is subjected to
rough handling in use. Sometimes, workers use a pipe to increase the
length of the spanner and tighten the rods. Some workers even use a
hammer to apply force on the spanner. To account for this ‘missuse,’ a
higher factor of safety of 5 is used in the present design. The coupler and
the rods are provided with ISO metric coarse threads. Coarse threads are
preferred because of the following advantages:
1. Coarse threads are easier to cut than fine threads;
2. They are less likely to seize during tightening;
3. There is more even stress distribution.

The rods are tightened by applying force on the wrench handle and
rotating the hexagonal coupler. The expression for torque required to
tighten the rod with specific tension P can be derived by suitable
modification of the equation derived for the trapezoidal threads. The
torque required to overcome thread friction in case of trapezoidal threads
is given by Eq. (6.13) of Chapter 6.
where d is the nominal diameter of
the threads. The coefficient of
friction varies from 0.12 to 0.20 depending upon the surface fi nish and
accuracy of the thread profi le and lubrication. Assuming, u = 0.15
and substituting above values in
Eq. (a), The above expression is
used to find out torsional moment
at each end of the coupler.
Step V Design of Rods :
The free body diagram of forces
acting on the rods and the
coupler is shown in Fig. 7.32.

Each rod is subjected to a tensile force P and torsional moment Mt. In the
initial stages, it is not possible to find out torsional moment. Considering
only tensile force, where A is the tensile stress area of the
threaded portion of the rod and st is the permissible tensile stress. The
rod is made of steel 30C8 (Syt = 400 N/mm2) and factor of safety is 5.
Therefore,

The factor of safety is satisfactory.
Therefore, the nominal diameter and
the pitch of the threaded portion of
the rod should be 42 mm and 4.5
mm respectively. In Fig. 7.31, the
length of the threaded portion of the
rod in contact with coupler threads
is denoted by l. It is determined by
shearing of the
threads at the minor diameter dc.
Equating shear resistance of the
threads to the tension in the rod,

References
• Shigley, Joseph; Mechanical Engineering
Design, Seventh Edition, 2003,McGraw
Hil, pg 396
• VB Bandari; Design of Machine Elements,
1994, Tata McGraw Hill, pg 175
66

Unit 5 threaded joint

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