Unit 3 Temporary and Permanent Joints.pptx

Unit 3
Temporary and Permanent Joints

Unit Split up
1. Threaded fasteners – Bolted joints
including eccentric loading
2. Riveted joints
3. Welded joints
4. Theory of bonded joints
5. Knuckle joints
6. Cotter joints

Methods of riveting
• Cold riveting
• Hot riveting

Types of Riveted Joints
• Lap joint
• Butt joint

Screw, Fastners and Non-permanent
Joints
• Nomenclature

11/27/2022 14
Bolted joint applications

16
Bolted joints in piping connections

18
Bolted joints in piping connections

21
Bolted joints in grooved couplings

24
Bolts in wind mill
construction

25
Bolts in wind mill
construction

11/27/2022 26
Steam engine working concept

27
Bolted joint applications – steam engines

Threads
• Thread
– Helical ridge of uniform section formed on
inside or outside of cylinder or cone
• Used for several purposes:
– Fasten devices such as screws, bolts, studs,
and nuts
– Provide accurate measurement, as in
micrometer
– Transmit motion
– Increase force
32

Thread Terminology
• Screw thread
– Helical ridge of uniform section formed on
inside or outside of cylinder or cone
• External thread
– Cut on external surface or cone
• Internal thread
– Produced on inside of cylinder or cone
34

Thread Terminology
35
Major Diameter
Commonly known as the outside diameter .
On a screw thread, the major diameter is the
largest diameter of the thread on the screw or
nut.

Thread Terminology
36
Number of Threads
The number of threads per inch.

Thread Terminology
Pitch
The distance from a given point on one thread
to a corresponding point on the very next thread
37

Thread Terminology
38
Single Lead Double Lead

Thread Terminology
Crest
– Top surface joining two sides of thread
– External thread on major diameter
– Internal thread on minor diameter
39

Brown & Sharpe Worm
Thread
Used to mesh worm gears and transmit
motion between two shafts at right angles
to each other but not in same plane
40
D = .6866P
F = .335P
C = .310P

Square Thread
• Being replaced by Acme thread because
of difficulty in cutting it
• Often found on vises
and jack screws
41
D = .500P
F = .500P
C = .500P + .002

The Three-Wire Method of
Measuring 60º Threads
42

Stress on Screwed Fasteners
• Internal Stress due to screwing up forces
• Stress due to External Forces (Mostly for eccentric)
• Stress due to combination of internal and
external forces.

Internal Stress on Screw
• Bolts designed on the basis of direct tensile
forces with large FOS in order to take the
indeterminate (Unknown) stresses that are
going to occur over it.
• Internal stress/Initial tightening
stress 𝑜𝑛 𝑏𝑜𝑙𝑡, 𝑃𝑖 = 2840 𝑑
• d – Nominal dia of bolt.

External Forces on Bolt
(Same for rivets)
• Tensile Load (Max Safe axial
load)
– P = /4 dc
2 t (Stress = load/ Area)
– P = /4 (dc +dp)2 t (If PCD given)
– dc - Root dia of thread
– P = /4 dc
2 t x n (no of bolts)
• Shear Load
– Ps = /4 d2  x n
– d = Major dia of bolt

Combined tensile and shear stress
•  max = ½  t
2 + 4 2

General Design of threaded fasteners
• Core area of thread, 𝐴𝑐 = [
60 𝑃𝑏
𝜎𝑦
]
2
3
• Pb – Total load on bolt
• Pb = Pi + P . (stiffness factor K)
• Pi – Pre load/Initial load, (2840d)
• Tightening torque, T = k x Pi x d
• k – Torque coefficient factor (0.3), d – dia of
bolt

Problem on Threads
using General Formula
A steam engine of effective diameter 300mm is
subjected to a steam pressure of 1.5N/mm2.
The cylinder head is connected by 8 bolts
having yield point 330MPa and endurance limit
at 240MPa. The bolts are tightened with an
initial preload of 1.5 times of steam load.
Assume a FOS of 2, find the size of bolt
required take stiffness factor for a copper
gasket may be taken as 0.5.[May 2016 & Dec 2017]

Approach to General Problems on thread
• If we calculate Area, from PSG 5.42, bolt size is
calculated.
• Core area of thread, 𝐴𝑐 = [
60 𝑷𝒃
𝜎𝑦
]
2
3
• Pb – Total load on bolt (If FoS is given, x it)
• Pb = Pi + P x (stiffness factor K)
• Pi – Pre load/Initial load = 1.5.P (Given in ques)
• P = Pressure x Area (Calculate load on each
bolt)

• Approach:
• Total load on bolt = Pi + P*q
• Use Ac formula
• Compare Area with bolt size
from PSG 5.42

Problems on Bolt due to Eccentric loading
• Due to eccentric loading
the bolt Shear and fail.
• Two loads acting:
1. Primary load (Direct
Shear load) and
2. Secondary load (Shear
due to torsion)

Loads acting on Bolt
(Due to eccentric loading condition)
• Primary shear load Secondary shear load
Free body diagram

Stress on bolt
• Direct shear load (F1)

Secondary shear Load
• Ps2 (F2) = P .e. r / n. r 2
• Ps2 (F2)= P .e / n. r (If bolt is same dia)
• r = r1 = r2 = r3= r4
• If bolts are of different radius
– Ps2 = P .e r1 / r1 2 + r2 2 + r3 2 + r4 2

Loads acting on Bolt
• Primary shear load Secondary shear load

Resultant Load on each bolt
• W or FR or PR= Ps12 + Ps22 + 2 Ps1 Ps2 Cosθ
• W or FR or PR= 𝐹1
2
+𝐹2
2
+ 2 𝐹1𝐹2𝐶𝑜𝑠𝜃

Approach
• By seeing the loading condition, you need to identify
whether it will fail due to torsion and shear or shear alone
 Mandatory
• In the given loading condition, we can easily say that it will
fail only due to shear whereas the torsion is arrested due
to the eccentric design setup and the place where load is
acting.
• Here the stresses (Bending & Torsion) are combined at the
eccentric point which leads to principle stresses way of
solving the problem. i.e., Stresses are not combined at the
bolt but at the eccentric point where load is acting.
• Upper row of two bolts exert the force initially then the
bottom row bolts will exert.

4 types of Eccentric Loading
1. Load acting parallel to axis of bolt
2. Load acting perpendicular to the axis of bolt
3. Plane containing the bolts
4. Loading on bolts on circular base

Load acting parallel to axis of bolt

Load acting perpendicular to the axis
of bolt

Loading on bolts on circular base

Remember (In Eccentric loading)
• For Parallel & Perpendicular to the axis of
bolts (Type 1 & 2)
– 𝜏𝑚𝑎𝑥 =
1
2
𝜎2 + 4 𝜏2 will be used
• For Plane containing the bolt & Circular bolts
– 𝜏𝑚𝑎𝑥 =
1
2
𝜎2 + 4 𝜏2 won’t be used

Remember (Another Way)
• For this type of eccentric
loading 
• 𝜏𝑚𝑎𝑥 =
1
2
𝜎2 + 4 𝜏2 
from this find “d”
--------------------------------------
• For this of type of
eccentric loading  
•  = Load/Area * n
• W or FR or PR=
𝐹1
2
+𝐹2
2

Design the size of Bolt for the
following eccentric loading given:

Approach to eccentric type problems
• W or FR or PR= 𝑭𝟏
𝟐
+𝑭𝟐
𝟐
• 𝐹1 =
𝐿𝑜𝑎𝑑
𝑛𝑜 𝑜𝑓 𝐵𝑜𝑙𝑡𝑠
• 𝐹2 =
𝑃.𝑒.𝒓
𝜀𝑟2
• Find r from Free body diagram

Secondary shear load
• T = W x e
• T = 100 x 600
• T = 60,000 kN
Factor
• C* = T /  r 2 60,000/4*106^2 = 1.333
• PS2 = 1.333 * 106.06 = 141.05 KN

Free body diagram with Loads
25 kN
141.05 kN

Resultant Load on each bolt 1 & 2
Resultant Load on each bolt 3 & 4
• W or FR or PR= Ps12 + Ps22 + 2 Ps1 Ps2 Cosθ (θ = 45)
• W1 = W2 = 160 kN
• W or FR or PR= Ps12 + Ps22 + 2 Ps1 Ps2 Cosθ (θ = 135)
• W3 = W4 = 130 kN
• Max load, W = 160 kN
• W = /4 * d2 *  (150) * 4
• d = 56 mm (Standardize from PSG DB 5.44)
• Result: Dia of bolt is 56 mm

The structural connection shown in Fig. 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.
(APR/MAY 2018)

Approach
• Same as previous problem except
• r = 1002 + 752 (The CG is not of equidistance)
• W = 𝐹1
2
+𝐹2
2
– Since r is different, 𝐶𝑜𝑠 𝜃 =
𝐴𝑑𝑗
𝐻𝑦𝑝

Given:
• P = 10,000 N
• e = 500 mm,  = 80 MPa
• n = 4 bolts (different ‘θ’)
To Find: Design size of bolt

P = 10,000 N, e = 500 mm, n = 4 bolts
(different θ)
Approach:
1. Primary and
2. Secondary
forces
3. Resultant force
4. Find dia using
Shear force
formula

P = 10,000 N, e = 500 mm, n = 4 bolts
(different θ)
752+1002
2.5 kN
10 kN

P = 10,000 N, e = 500 mm, n = 4 bolts
(different θ),  = 80 MPa
• W or FR = /4 * d2 * 
• 12,093 = /4 * d2 * 80
• d = 13.87 mm
• d = 16 mm (Standardize from PSG DB 5.42)
• Result: M16 bolt

Same type of problem in Rivets
• A machine frame is subjected to a load of
10kN from at 500 mm eccentrically. The
distance between the each rivet is 200mm.
The max shear stress is 75MPa.Determine the
dia of rivet.

Approach
• Same as Bolt problem
• Instead or notation “r”, “l” is used.
• Finding the centroid point  different
procedure as centroid point is not given.
• Direct Load, Pd= Load / Area
• Load due to Torsion,

A wall bracket as shown in the figure is fixed to a
vertical wall by 4 bolts. Find the size of the bolt
with permissible shear stress 70MPa.

Remember (In Eccentric loading)
• If the loading is perpendicular to the axis of
bolts
1
2
𝜎2 + 4 𝜏2  from this find “d”
----------------------------------------------------------------
• If the loading is on the plane to the axis of
bolts
• W or FR or PR= 𝐹1
2
+𝐹2
2

Approach
• So, we need to find Load on 1st & 2nd set of
rows of bolts separately and find the max one
in those. (Formula when load is acting
perpendicular to axis of bolt),
• Ps1 = P .e l1 / n1. l1 2 + n2 l2
2 & Ps2 = P .e l2 / n1. l1 2
+ n2 l2
2
• Then find 𝜎 =
𝑀𝑎𝑥 𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
& 𝜏 =
𝑀𝑎𝑥 𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎 ∗𝑁𝑜 𝑜𝑓 𝑏𝑜𝑙𝑡
1
2
𝜎2 + 4 𝜏2  from this find “d”
• Lastly, from PSG 5.42, find the size of bolt.
Note: For these type of
problems, no need to find CG

A bracket is attached to the wall by 5 bolts.3 at
the top and 2 at the bottom as shown in fig. A
vertical load of 25 kN is acting at 200mm
eccentrically from the plane of bolt. Calculate
the suitable size of bolt if permissible shear
stress is 50 N/mm2.

Approach
• Methodology same as previous problem.
• Only n1 & n2 differs.

AU Problem on Bolts & Rivets
Figure shows a solid forged bracket
to carry a vertical load of 13.5 kN
applied through the centre of hole.
The square flange is secured to the
flat side of a vertical stanchion
through four bolts. Estimate the
tensile load on each top bolt and
the maximum shearing force on
each bolt. Find the bolt size, if the
permissible stress is 65 MPa in
shear. All dimensions in mm
(NOV/DEC 2016)

Approach
1
2
𝜎2 + 4 𝜏2
• 𝜏 = Resultant load/Area*n
• 𝜎 = Resultant load/Area
• FR= 𝐹1
2
+𝐹2
2
• F1 = P/n
• F2 = P.e/n.r

Approach
1
2
𝜎2 + 4 𝜏2
• 𝜏 = Shear load/Area
• 𝜎 = Direct load/Area
• Shear load = P/n
• Direct load =

• A cast iron bracket, as shown in figure, supports a
load of 10 kN. It is fixed to the horizontal channel
by means of four identical bolts, two at A and two
at B. The bolts are made of steel 30C8 whose
yield strength is 400 MPa and the factor of safety
is 6. Determine the major diameter of the bolts if
dc = 0.8d. (NOV/DEC 2020 AND April/May 2021)

• The structural connection shown in Fig. 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. (APR/MAY 2018)

• A steam engine of effective diameter 300 mm is
subjected to a steam pressure of 1.5 N/mm2. The
cylinder head is connected by 8 bolts having yield
point 330 MPa and Endurance limit at 240 MPa.
The bolts are tightened with an initial preload of
1.5 times the steam load. A soft copper gasket is
used to make the joint leak proof. Assuming a
factor of safety 2, find the size of bolt required.
The stiffness factor for copper gasket may be
taken as 0.5. (NOV/DEC 2015) (APR/MAY 2016)

AU Problem on Bolts
• 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. 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. (NOV/DEC 2010)

AU Problem on Rivets
Find the efficiency of a double riveted lap joint
of 6 mm plates with 20 mm diameter rivets
having a pitch of 55 mm. Assume Permissible
tensile stress in plate is 120 MPa, Permissible
shearing stress is 90 MPa and Permissible
crushing stress is 180 MPa. (APR/MAY 2019)

Formulas for Rivet (Efficiency Prob)
• Thickness of Cover plate, t1= 0.625 x t
– t – thickness of plate
• Diameter of rivet, d = 6.05√𝑡
• Pitch of rivet, P = 3 x d
• Margin of rivet, e = 1.5 x d
• Distance between adjacent rows of rivets, Pb = 3 x d
• Efficiency of rivet, η = Least of Ft, Fs, Fc / P. t. σ
• Shearing force, Fs = i . n. π/4 d^2. τ
– i = 2 for double rivet, n = 1 for lap joint
• Tearing force, Ft = (P – d) x t x σ
• Crushing force, Fc = i x d x t x σ

AU Problem on Rivets
Design a double riveted butt joint with
two cover plates for the longitudinal
steam of a boiler shell 1.5 m in diameter
subjected to a steam pressure of 0.95
N/mm2. Assume joint efficiency as 75%,
allowable tensile stress in the plate 90 MPa;
Compressive stress 140 MPa and Shear stress in
the rivet 56 MPa. (NOV/DEC 2016)

A wall crane with a pin joint tie rod
is as shown in Fig. The crane hook
is to take a maximum load 35 kN,
when the load is at a distance of 2
m from the wall. The tie rod and
pin are made of steel FeG 250 (Syt
= 250N/mm) and the factor of
safety is 5. Calculate the diameter
of the tie rod and the pin.
(APR/MAY 2017)

AU Problem on Bolts & Rivets (Part C)
A steel plate is subjected to a force of 5 kN and fixed to a channel
by means of 3 identical bolts as shown in figure. The bolts are
made from plain carbon steel for which yield stress in tension is
380 N/mm2 and factor of safety is 3. Determine the size of the
bolts. (15) (NOV/DEC 2018)

• A bracket is bolted to a column by six bolts as
shown in fig. It carries a load of 50 KN at a
distance of 150 mm, from the center of
column. If the maximum stress in the bolts is
to be limited to 150 MPa, determine the
diameter of the bolt. (NOV 2021)

Fig shows a bracket fixed on a steel column by
means of 3 bolts of same size. If the permissible
tensile and shear stress are limited to 75 N/mm2
and 55 N/mm2 respectively. Find the size of
bolts. (NOV/DEC 2017)

Welding
• A welded joint is a permanent joint which is obtained
by the fusion of the edges of the parts to be joined
together, with or without the application of pressure
and a filler material.
• The heat required for the fusion of the material is
obtained by burning gas(gas welding) or by an
electric arc(electric arc welding).
• Welding is extensively used in fabrication as an
alternative to riveted and bolted joints.
• Welding also used as a repair medium

Types of welding processes
• Heat – Fusion welding
• Thermit welding, Gas welding and Electric arc
welding (Shielded arc and Unshielded arc).
• Heat and Pressure
• Forge welding
• Diffusion welding

Types of Welded Joints
1. Lap Joint
• The lap joint or the fillet joint is obtained by overlapping the plates
and then welding the edges of the plates. The cross‐section of the
fillet is approximately triangular. The fillet joints may be
1. Single transverse fillet,
2. Double transverse fillet, and
3. Parallel fillet joints.

Butt Joint
• The butt joint is obtained by placing the plates edge to edge. In butt
welds,
the plate edges do not require beveling if the thickness of plate is less
than 5 mm. On the other hand, if the plate thickness is 5 mm to 12.5 mm,
the edges should be beveled to V or U‐groove on both sides. The butt
joints may be
• 1. Square butt joint,
• 2. Single V‐butt joint
• 3. Single U‐butt joint,
• 4. Double V‐butt joint, and
• 5. Double U‐butt joint

Other Joints
• The other type of welded joints are corner joint, edge joint and
T‐joint as
shown in Fig. below.
The main considerations involved in the selection of
weld type are:
1. The shape of the welded component required,
2. The thickness of the plates to be welded, and
3. The direction of the forces applied.

Refer PSG 11.1 – Weld Symbols

Strength of Transverse Fillet
Welded Joints
• The transverse fillet welds are designed for
tensile strength.
• Let us consider a single and double transverse
fillet welds as shown in Fig. (a) and (b)
respectively.

In order to determine the strength of the fillet joint, it is assumed that
the section of fillet is a right angled triangle ABC with hypotenuse
AC making equal angles with other two sides AB and BC.
• The enlarged view of the fillet is shown in Fig. below The length of
each side is known as leg or size of the weld and the perpendicular
distance of the hypotenuse from the intersection of legs (i.e. BD) is
known as throat
thickness. The
minimum area
of the weld is
obtained at the
throat BD, which
is given by the
product of the
throat thickness and length of weld.
t = Throat thickness (BD),s = Leg or size of weld,= Thickness of
plate, and l = Length of weld,

From Fig. below, we find that the throat thickness,
Cos 45 = t/s  t = 0.707 s (In few books, t = 0.707 h)
Minimum area of the weld or throat area,
A = Throat thickness × Length of weld = t × l = 0.707 s × l
If σt is the allowable tensile stress for the weld metal, then
the tensile strength of the joint for single fillet weld,
F = Throat area × Allowable tensile stress = 0.707 s × l × σt
and tensile strength of the joint for double fillet weld,
F = 2 × 0.707 s × l × σt = 1.414 s × l × σt
D
45
h or s
h or s
t
h or s
h or s

• The parallel fillet welded joints are designed for shear strength. Consider a
double parallel fillet welded joint as shown in Fig. (a). The minimum area of
weld or the throat area, A = 0.707 s × l
• If τ is the allowable shear stress for the weld metal, then the shear strength
of the joint for single parallel fillet weld,
F = Throat area × Allowable shear stress = 0.707 s × l × τ
and shear strength of the joint for double parallel fillet weld,
F = 2 × 0.707 × s × l × τ = 1.414 s × l × τ
F = 0.707 s × l × τ F = 2 x 0.707 s × l × τ
F = 1.414 s × l × τ
F = 2 x 0.707 s × l × τ
F = 1.414 s × l × τ

Here t is the thickness of the plate. The torque P . e will cause
secondary shearing stress τ at the weld end, A, which will be
greater than shearing stress at any other point in the weld,
A is at a distance of R from CG.
It can be shown that G = P e r2 /J
Here, J is the polar moment of inertia of fillet weld about G. AS
shown in Figure ,
τ1 and τ2 at A act at an angle θ.
Since nature of both these stresses is same, they can be
added vectorially. The resultant stress τA is
τ max = [τ1
2 + τ2
2 + 2 τ1 τ2 cosθ]½, cos θ = r1/r2

PSG DB 11.3 Stress in Weld joints

PSG DB 11.5 & 11.6
Properties of Weld treated as Line

Approach
• Length of transverse Weld, l1 = b- t
• σt = P1 / A1, P1 – Load by single transverse
weld
• P = P1 + P2
• τ = P2 /A2, P2 – Load by double parallel weld
• A2 = 2 x 0.707 x h x l2

Welding Joints under Eccentric Loading
1. Weld Joints subjected to moment acting on the plane
(Bending & Shear Stress)
2. Weld Joints subjected to moment acting normal to the plane
3. Weld Joints subjected to direct shear, bending and torsional

General formula for Welding
1
2
𝜎𝑏
2
+ 4 𝜏2
• 𝜎𝑏 =
𝑀𝑏
𝑍
,
– Mb – Bending Moment (PxL) = Load x distance
– Z – Sectional modulus – PSG 11.5 & 11.6
(In that page, “t” has to be multiplied)
• 𝜏 =
𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
, Area should be multiplied by “t”
• Refer PSG 11.3 to 11.6 pages for other formula

Case 1 - Weld Joints subjected to
moment acting on the plane

Approach
1
2
𝜎𝑏
2
+ 4 𝜏2
• 𝜏 =
𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
• A = t(2b+2d) = 0.707 h (2b+2d)
• 𝜎𝑏 =
𝑀𝑏
𝑍
,
– Mb = P x L
– For Z – PSG 11.6 (multiply with “t”)
– Ans will be in terms of “h”

direct shear, bending and torsional

Approach
1
2
𝜎𝑏
2
+ 4 𝜏2
• 𝜏 =
𝐿𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
• A = π𝐷 x t
• 𝜎𝑏 =
𝑀𝑏
𝑍
• Mb = P x L
– For Z – PSG 11.6 (multiply with “t”)
– Ans will be in terms of “h”

moment acting normal to the plane

Eccentric load in Welding
Direct Shear Stress
• Direct shear load (F1)

Remember (In Eccentric loading Case 2)
• Direct Shear, 𝜏1 =
𝑃
𝐴
• Shear due to Torsion, 𝜏2 =
𝑀𝑡 ∗ 𝒓
𝐽
– 𝑀𝑡 = Load x eccentric distance (e)
– J – PSG 11.5
• Resultant Shear Stress,
𝜏𝑟= 𝜏1
2
+𝜏2
2
+ 2 𝜏1𝜏2𝐶𝑜𝑠𝜽

Types of Loads on adhesive joints

Knuckle joint
Two or more rods subjected to tensile
and compressive forces are fastened
together
Their axes are not in alignments
but meet in a point
The joint allows a small
angular moment of one
rod relative to another
It can be easily
connected and
disconnected
Applications: Elevator chains, valve rods, etc

Design of Knuckle joints
(Failures)
1. Rod
1. Tensile load
2. Single eye end
1. Tensile load
2. Shearing load
3. Crushing load
3. Double eye end
1. Tensile load
2. Shearing load
3. Crushing load
4. Knuckle Pin
1. Double Shear load

Step 1 – Design of Rod

Step 2.1 – Design of Single eye end

• Shearing load

• Crushing Load

Step 3.1 – Design of Double eye end

• Shear load

• Crushing load

Step 4 – Design of Knuckle Pin

Cotter joints
• A cotter is a flat wedge-shaped piece of
steel
• This is used to connect rigidly two rods
which transmit motion in the axial
direction, without rotation.
• These joints may be subjected to tensile
or compressive forces along the axes of
the rods
• It is uniform in thickness but tapering in
width , generally on one side only. Usually
the taper is 1 in 30.

Applications:
 Piston rod to the crosshead of a steam engine,
 Piston rod and its extension as a tail or pump rod
 Valve rod and its stem
 Strap end of connecting rod
 Foundation bolts to fasten heavy machinery to
foundations …..etc.,
Cotter joints

TYPES
• Socket and Spigot cotter joint
• Sleeve and cotter joint
• Gib and cotter joint

Sleeve and cotter joint
The enlarged ends of the rods butt against each other with a common sleeve over them
•The rod ends are enlarged to take care of the weakening effect caused by slots
For circular rods

Socket and Spigot Cotter Joint
Slots are wider than the cotter.
Cotter pulls the rod and socket tightly together
Clearance: must be provided for adjustment.(2 to 3 mm)
Proportions
cotter thickness = (1/3)diameter of rod
cotter width = rod diameter

Gib and cotter joint for rectangular rods
One bar end is made in the form of a strap
A Gib is used along with the cotter.
The thickness of the gib and cotter
are same

Gib and cotter joint for rectangular rods
When the cotter alone (i.e. without gib) is driven, the friction between its ends and
the inside of the slots in the strap tends to cause the sides of the strap to spring open
(or spread) outwards as shown dotted in Fig.

Design of Socket and Spigot Cotter Joint
.

Design of Socket and Spigot Cotter
Joint
• Design of Shaft
– Tensile force
• Design of Spigot
– Tensile force
– Crushing force
• Design of Cotter
– Crushing
– Shear
• Design of Socket
– Tensile
– Shear

Let P = Load carried by the rods,
• d = Diameter of the rods,
• d1 = Outside diameter of socket,
• d2 = Diameter of spigot or inside diameter of socket,
• d3 = Outside diameter of spigot collar,
• t1 = Thickness of spigot collar,
• d4 = Diameter of socket collar,
• c = Thickness of socket collar,
• b = Mean width of cotter,
• t = Thickness of cotter,
• l = Length of cotter
• a = Distance from the end of the slot to the end of rod,
• σt = Permissible tensile stress for the rods material
• τ = Permissible shear stress for the cotter material
• σc = Permissible crushing stress for the cotter material
Design of Socket and Spigot joint

1.Failure of the rods in tension
P
P

2. Failure of spigot in tension across the
weakest section (or slot)

3. Failure of the rod or cotter in crushing

4.Failure of the socket in tension across the slot

5. Failure of cotter in shear

6. Failure of the socket collar in crushing

7. Failure of socket end in shearing

8. Failure of rod end in shear

9.Failure of spigot collar in crushing

10. Failure of the spigot collar in shearing

11. Failure of cotter in bending

Unit 3
Temporary and Permanent
Joints
Bolted joints Rivted joints Welded joints Knuckle joints
Cotter
joints

Unit 3 Temporary and Permanent Joints.pptx

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Unit 3 Temporary and Permanent Joints.pptx