This document provides an introduction to machine design and the design of cotter joints. It defines key machine design terms like machine elements. It outlines the general machine design procedure and objectives. It then describes different types of stresses like tensile, compressive, shear, bending and torsional stresses. Formulas for calculating each stress type are provided. The document concludes by describing the components, applications, advantages, limitations and design procedure for cotter joints.

1. Design of Machine Elements
Introduction to Design Terms and
Design of Cotter Joint.
Gaurav Mistry
Assistant Professor
Diwaliba Polytechnic, UTU.

2. ❑ What is Machine?
▪ A machine is an assembly of various parts which transform the available energy into
desired work.
▪ It is a practical development of Mechanism consisting of number of links (elements).
❑ What do you mean by Machine Elements?
❑ Machine Design
▪ Machine elements are the links assembled with each other to form the mechanism
and transmit power/force/motion.
▪ Elements shall be designed and assembled as per the requirements.
▪ Machine Design is the creation of new and better machines and improving the
existing ones.
▪ It is the application of science and invention to the development, specifications and
constructions of machines.
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3. ❑ General Machine Design Procedure
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4. ❑ Objective of Machine Design
▪ The main objective of Machine Design is to prepare the graphical model of the
machine to satisfy the need of the user without failure and which could be produced
at an optimum cost.
▪ Depending upon the functions, the various elements are subjected to different loads.
▪ For the design of these elements, it is required to understand the types of loads and
their effect on the elements.
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5. ❑ Types of Stresses
▪ When any part or machine element is subjected to external force (load), the internal
resistance from the material of the part is produced. The resistance to external load
per unit area of cross section of the part is known as stress
𝐴 = 𝜋𝑟2
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6. ❑ Types of Stresses
Stress (σ) =
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡(𝑃)
𝐸𝑓𝑓𝑒𝑐𝑡𝑒𝑑 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓𝑡ℎ𝑒 𝑝𝑎𝑟𝑡(𝐴)
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7. ❑ Tensile stress
▪ The stress induced due to tensile load on the part is known as tensile stress.
▪ The stress induced is in opposite direction of the load and is in axial direction.
▪ It is perpendicular to the cross sectional area resisting the load (effected by the load).
𝜎𝑡 =
𝑃
𝐴
P = Tensile load (N),
A = Area of cross section (𝑚𝑚2
),
𝜎𝑡 = Tensile stress (
𝑁
𝑚𝑚2)
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8. ❑ Compression stress
▪ The stress induced due to compression load on the part is known as compressive
stress.
▪ The stress induced is in opposite direction of the load and is in axial direction.
▪ It is perpendicular to the cross sectional area resisting the load (effected by the load).
𝜎𝑐 =
𝑃
𝐴
P = Compressive load (N),
A = Area of cross section (𝑚𝑚2),
𝜎𝑐 = Compressive stress (
𝑁
𝑚𝑚2)
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9. ❑ Shear stress
▪ The stress induced due to shear effect on the part is known as shear stress.
▪ The stress induced is in perpendicular direction of the load (axial direction).
▪ It is parallel to the cross sectional area resisting the load (effected by the load).
▪ When resistance is offered by the single section, it is known as single shear and when
the resistance is offered by the two sections, it is known as double shear.
𝜏 =
𝑃
𝐴
(Single shear)
𝜏 = Shear stress (
𝑁
𝑚𝑚2)
𝜏 =
𝑃
2𝐴
(Double shear)
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10. ❑ Bending stress
▪ When the external load is acting in the lateral direction perpendicular to the axis of
the part (beam), it creates bending and the stresses induced across the section of the
part is known as bending stress.
▪ When a beam experiences load like that shown in figure one the top fibres of the
beam undergo a normal compressive stress. The stress at the horizontal plane of the
neutral is zero. The bottom fibres of the beam undergo a normal tensile stress. It can
be concluded therefore that the value of the bending stress will vary linearly with
distance from the neutral axis.
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11. ❑ Bending stress
▪ When an external load causes only bending, the element is said to be subjected to
pure bending. The pure bending equation is given as follow:
𝑀 𝑏
𝐼
=
𝜎 𝑏
𝑦
=
𝐸
𝑅
or
Where,
▪ 𝑀 𝑏 = Bending moment,
▪ 𝐼 = Moment of Inertia,
▪ 𝜎 𝑏 = Bending Stress,
▪ 𝑦 = The distance between the
extreme fibre of the member from it’s
neutral axis,
▪ 𝐸 = Modulus of elasticity,
▪ 𝑅 = Radius of curvature,
▪ Z = section modulus.
▪ Due to bending the fibres below the neutral axis
are subjected to tensile stress and the fibres
above the neutral axis are subjected to
compressive stress.
▪ At neutral axis the stress will be zero.
𝜎𝑏 =
𝑀 𝑏
𝐼
𝑦
But
𝐼
𝑦
= Z , Therefore, 𝜎𝑏 =
𝑀 𝑏
𝑍
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12. ❑ Torsional shear stress
▪ When the machine member is subjected to torque or twisting moment, the material
fibres are twisted and the resistance offered by the member is known as torsional
shear stress
𝑀𝑡
𝐽
=
𝐺𝜃
𝐿
=
𝜏
𝑟
or
Where,
▪ 𝑀𝑡 = Torque or twisting moment on
the shaft,
▪ 𝐽 = Polar Moment of Inertia,
▪ 𝜏 = Maximum Shear Stress,
▪ 𝑟 =radius of the shaft,
▪ 𝐺 = Modulus of rigidity,
▪ 𝜃 = Angle of twist,
▪ L = Length of shaft under twisting
▪ 𝑍 𝑝 =polar section modulus.
𝜏=
𝑀𝑡
𝐽
𝑟
But
𝐽
𝑟
= 𝑍 𝑝 , Therefore, 𝜏 =
𝑀𝑡
𝑍 𝑝
𝑀𝑡
𝑀𝑡
𝑀𝑡
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13. ❑ Factor of Safety
▪ In order to prevent the permanent deformation or failure of the part, the magnitude
of the induced stress should be well below the yield point stress
▪ It is defined, in general, as the ratio of the maximum stress to the working stress.
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆𝑎𝑓𝑒𝑡𝑦 =
𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑆𝑡𝑟𝑒𝑠𝑠
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑟 𝑆𝑎𝑓𝑒 𝑆𝑡𝑟𝑒𝑠𝑠 𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑆𝑡𝑟𝑒𝑠𝑠
𝑓𝑠 =
𝜎 𝑚𝑎𝑥
[𝜎]
Mathematically,
𝐹𝑜𝑟 𝐷𝑢𝑐𝑡𝑖𝑙𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, 𝑓𝑠 =
𝜎 𝑦
[𝜎]
𝐹𝑜𝑟 𝐵𝑟𝑖𝑡𝑡𝑙𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, 𝑓𝑠 =
𝜎 𝑢
[𝜎]
𝐹𝑜𝑟 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔, 𝑓𝑠 =
𝜎−1
[𝜎]
𝜎 𝑚𝑎𝑥 = 𝜎 𝑦 (Yield Point Stress)
𝜎 𝑚𝑎𝑥 = 𝜎 𝑢 (Ultimate Stress)
𝜎 𝑚𝑎𝑥 = 𝜎−1 (Endurance Stress)
𝑓𝑠 =
𝜏 𝑢
[𝜏]
, 𝜏 𝑢 = ultimate shear stress and [𝜏] = allowable shear stress 𝜏 𝑢 = 0.577 𝜎 𝑢
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14. ▪ Numerical related to simple machine elements design are to be solved in class. (refer notes)
❑ Stress on bolt under tensile load
D d
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c
cA
A
A
After substituting the area
equation in
we can find
the core diameter 𝑑 𝑐.
Later the nominal diameter
d can be obtained by,
𝑑 𝑐 = 0.84 d
𝜎𝑡 =
𝑃
𝐴
If we say M16 bolt, then here 16 stands for 16 mm nominal diameter (d) of the bolt.

15. ❑ Basic Series and Progression Ratio
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16. ❑ Preferred Numbers
Φ = 𝑅(
1
𝑁−1
)
Where,
Φ = Progression Ration,
𝑅 = Range Ratio,
R =
𝐿𝑎𝑠𝑡 𝑇𝑒𝑟𝑚
𝐹𝑖𝑟𝑠𝑡 𝑇𝑒𝑟𝑚
,
N = Number of Terms in Geometric Series
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17. ❑ Cotter Joint
▪ A cotter joint is a temporary fastening and is used to connect rigidly two co-axial rods or bars
which are subjected to axial tensile or compressive forces.
▪ A cotter is a flat wedge shaped piece of rectangular cross-section and its width is tapered
(either on one side or both sides) from one end to another for an easy adjustment.
▪ As shown in the figure, the cotter joint mainly consist of three parts: (1) Spigot end (Male
Part), (2) Cotter and (3) Socket end (Female Part).
Design of Machine Elements
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Click on Image for Video

18. ❑ Cotter Joint
▪ Socket is a hollow cylindrical end of the
joint.
▪ Spigot is a solid cylindrical shape
assembled with the hole of the socket.
▪ On the perpendicular direction of the axis
of the joint, a rectangular shape hole is
made in both the ends which receives the
wedge shape element (cotter) to fit both
the ends.
➢ Applications of the cotter joint:
1. To connect two co – axial rods having
straight line motion.
2. To connect piston rod and cross head
of steam engine.
3. For cotter foundation bolts.
4. To connect two parts of the split
flywheel.
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19. ➢ Advantages of the cotter joint:
1. The joint can be opened/disassembled when needed.
2. Joint can be made robust by using forged spigot and socket ends using wedge
shape cotter to carry heavy loads.
3. Using sleeve, cotter joint can be used to connect two pipes used in structural
works.
❑ Cotter Joint
➢ Limitations of the cotter joint:
1. It is expensive to form the spigot and socket and hence it is relatively costly.
2. Requires more time in manufacturing the joints.
3. It is not advisable to use when the connected rods are subjected to rotary
motion.
➢ Material for the cotter joint:
▪ Generally, all the three parts of the joint – spigot, socket and cotter are made
from the same material i.e. mild steel (M.S.) or wrought iron.
▪ In case of heavy loads and to limit the overall size of the joint, it may be made
from alloy steel.
▪ Compared to spigot and socket end, the cotter is kept relatively weaker so that it
can fail earlier than spigot and socket and can be replaced easily and
economically.
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20. ❑ Cotter Joint
➢ Design Procedure:
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, and
σc = Permissible crushing stress for the cotter material.
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It is assumed that the load is uniformly
distributed over the various cross-sections of
the joint. And so the effect of bending in
cotter is neglected.
Click on Image for Video

21. ❑ Cotter Joint
➢ Under the action of tensile/compressive load, considering failure of
spigot the various dimensions are obtained.
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22. ❑ Cotter Joint
➢ Under the action of tensile/compressive load, considering failure of
socket the various dimensions are obtained.
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23. ❑ Cotter Joint
➢ Under the action of tensile/compressive load, considering failure of
cotter the various dimensions are obtained.
➢ Neglecting the failure of cotter due to bending load
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24. ❑ Cotter Joint Design
1. Failure of the rods in tension:
The area resisting the tension in the rods is
Tensile strength of the rod is:
Equating this to load (P), we have:
From this equation, diameter of the rods (d)
may be determined.
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25. 2. Failure of spigot end in tension across the weakest section (at slot):
Since the weakest section of the spigot is that section which has a slot in it for the cotter,
therefore
Area resisting tension of the spigot across the slot is
and tensile strength of the spigot across the slot is
Equating this to load (P), we have
From this equation, the diameter of spigot or inside diameter of socket
(d2) may be determined.
Note : In actual practice, the thickness of cotter is usually taken as
d2 / 4.
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❑ Cotter Joint Design

26. 3. Failure of spigot rod at slot (or cotter) in crushing under tension load:
We know that the area that resists crushing of a spigot rod
(or cotter)
∴ Crushing strength
Equating this to load (P), we have
From this equation, the induced crushing stress may be checked.
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❑ Cotter Joint Design

27. 4. Failure of spigot rod at slot at the area parallel to the tension load under
shear:
Since the rod end is in double shear, therefore the area
resisting shear of the rod end
and shear strength of the rod end
Equating this to load (P), we have
From this equation, the distance from the end of the slot to the end of the spigot (a) may
be
obtained.
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❑ Cotter Joint Design

28. 5. Failure of spigot collar cross section in crushing under compressive load:
We know the area that resists crushing of the collar
and crushing strength of the collar
Equating this to load (P), we have
From this equation, the diameter of the spigot
collar (d3) may be obtained.
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❑ Cotter Joint Design

29. 6. Failure of the spigot collar at the circumferential area parallel to the
compressive load under shear:
We know the area that resists shearing of the collar
and shearing strength of the collar,
Equating this to load (P) we have
From this equation, the thickness of spigot collar (t1) may be obtained
Design of Machine Elements
❑ Cotter Joint Design

30. 7. Failure of the socket in tension across the slot:
We know that the resisting area of the socket across the
slot,
Tensile strength of the socket across the slot
Equating this to load (P), we have
From this equation, outside diameter of socket (d1)
may be determined.
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❑ Cotter Joint Design

31. 8. Failure of the socket collar near slot in crushing under
tension load:
We know that area that resists crushing of
socket collar
and crushing strength
Equating this to load (P), we have
From this equation, the diameter of socket
collar (d4) may be obtained.
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❑ Cotter Joint Design

32. 9. Failure of the socket collar at slot at the area parallel to the tension load
under shear:
Since the socket end is in double shear,
therefore area that resists shearing of socket
collar
and shearing strength of socket collar
Equating this to load (P), we have
From this equation, the thickness of socket
collar (c) may be obtained.
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❑ Cotter Joint Design

33. 10. Failure of cotter in shear under tensile load:
Since the cotter is in double shear, therefore shearing
area of the cotter
and shearing strength of the cotter
Equating this to load (P), we have
From this equation, width of cotter (b) is
determined.
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❑ Cotter Joint Design

34. 11.The length of cotter (l) in taken as 4 d.
12. The taper in cotter should not exceed 1 in 24. In case the greater taper is required, then a
locking device must be provided.
13.The draw of cotter is generally taken as 2 to 3 mm.
Notes:
1. When all the parts of the joint are made of steel, the following proportions in terms of
diameter of the rod (d) are generally adopted :
d1 = 1.75 d , d2 = 1.21 d , d3 = 1.5 d , d4 = 2.4 d , a = c = 0.75 d , b = 1.3 d, l = 4 d , t = 0.31 d ,
t1 = 0.45 d , e = 1.2 d.
Taper of cotter = 1 in 25, and draw of cotter = 2 to 3 mm.
2. If the rod and cotter are made of steel or wrought iron, then τ = 0.8 σt and σc = 2 σt may be
taken.
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❑ Cotter Joint Design

35. Gaurav Mistry 35
❑ Cotter Joint Numerical
❖ Design and draw a cotter joint to support a load varying from 30 kN in compression to
30 kN in tension. The material used is carbon steel for which the following allowable
stresses may be used. The load is applied statically. Tensile stress = compressive stress =
50 MPa ; shear stress = 35 MPa and crushing stress = 90 MPa.
Solution: Given : P = 30 kN = 30 × 103 N ; σt = 50 MPa = 50 N / mm2 ; τ = 35 MPa = 35 N /
mm2 ; σc = 90 MPa = 90 N/mm2
Design of Machine Elements

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❑ Cotter Joint Numerical
Design of Machine Elements
3

37. Gaurav Mistry 37
❑ Cotter Joint Numerical
Design of Machine Elements
4
5
Spigot
Spigot
Spigot

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❑ Cotter Joint Numerical
Design of Machine Elements
6
7

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❑ Cotter Joint Numerical
Design of Machine Elements
8
9
10

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❑ Cotter Joint Numerical
Design of Machine Elements
11
12
In all the above relations, it is assumed that the load is uniformly distributed over
the various cross-sections of the joint. But in actual practice, this does not happen
and the cotter is subjected to bending. Which is neglected in the above case.
In order to find out the bending stress induced, it is assumed that the load on the
cotter in the rod end is uniformly distributed while in the socket end it varies from
zero at the outer diameter (d4) and maximum at the inner diameter (d2), as shown
in Fig.

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Design of Machine Elements
Or cotter

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Design of Machine Elements
REFERENCES:
1. A textbook of Machine design, R. S. Khurmi, S. Chand.
2. Design of Machine Elements, S. B. Soni, Atul prakashan.
3. www.google.com