LESSON 2 SHAFTS, KEYS AND COUPLINGS

Unit 1 – Shafts

Overview

This chapter covers theory and application for shaft design.

Learning Objectives:

After successful completion of this module, the students should be able to:
1. Classify different types of shafts.
2. Solve shafts related problems.

Course Materials:

Transmission Shafts
The term ‘transmission shaft’ usually refers to a rotating machine element, circular in
cross section, which supports transmission elements like gears, pulleys and sprockets and
transmits power.

Some of the specific categories of transmission shafts are as follows:

• Axle is used for a shaft that supports rotating elements like wheels, hoisting drums or
rope sheaves and which is fi tted to the housing by means of bearings.
• Spindle is a short rotating shaft. The term ‘spindle’ originates from the round tapering
stick on a spinning wheel, on which the thread is twisted.
• Countershaft It is a secondary shaft, which is driven by the main shaft and from which
the power is supplied to a machine component.
• Jackshaft It is an auxiliary or intermediate shaft between two shafts that are used in
transmission of power.
• Line shaft consists of a number of shafts, which are connected in axial direction by
means of couplings.

Shaft Design on Strength Basis

Transmission shafts are subjected to axial tensile force, bending moment or torsional
moment or their combinations.

When the shaft is subjected to axial tensile force, the tensile stress is given by:

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
When the shaft is subjected to pure bending moment, the bending stresses are given by:

When the shaft is subjected to pure torsional moment, the torsional shear stress is given by:

• Maximum Principal Stress Theory Since the shaft is subjected to bending and
torsional moments without any axial force:

• Maximum Shear Stress Theory The principal shear stress is tmax.

Shaft Design on Torsional Rigidity Basis

A transmission shaft is said to be rigid on the basis of torsional rigidity, if it does not twist
too much under the action of an external torque.

In certain applications, like machine tool spindles, it is necessary to design the shaft on the
basis of torsional rigidity, i.e., on the basis of permissible angle of twist per metre length of shaft.
The angle of twist (in radians) is given by:

ASME Code for Shaft Design

One important approach of designing a transmission shaft is to use the ASME code.
According to this code, the permissible shear stress tmax. for the shaft without keyways is taken
as 30% of yield strength in tension or 18% of the ultimate tensile strength of the material,
whichever is minimum. Therefore, tmax. = 0.30 Syt or, tmax. = 0.18 Sut (whichever is
minimum).

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
Design of Hollow Shaft on Strength Basis
Hollow shafts are used to provide passage for coolants and control cables in case of
deep hole drilling and borewell drilling.

For a hollow circular cross-section:

Design of Hollow Shaft on Torsional Rigidity Basis

The design of hollow shaft on the basis of torsional rigidity is governed by the
permissible angle of twist per metre length of shaft. The angle of twist (in radians) is given by:

Examples:
1. Compute for the diameter of a line shaft to transmit 25 hp in a speed of 225 rpm design
stress is 6 ksi. (Ans. 1 7/8 in)
2. What will be the ultimate length of a 5-inch diameter steel shaft subjected to 28,000 in-lb
twisting moment to reach the maximum allowable angular deflection 0.08 degree/ft.
length, Use module of elasticity of 12 x106 psi. (Ans. 36.7 in.)
3. Compute the torsional deflection in degrees of a SAE 1040 steel shafting of 120 mm in
diameter and 1.4 m long subjected to twisting moment of 3,000 N.m. The torsional
modulus of elasticity is 80,000 MPa. (Ans. 0.148o)

Assessment:

Answer the following:

1. Determine the diameter (in) of a steel line shaft to transmit 20 hp at a speed of 300 rpm
with torsional deflection not to exceed 0.08 degree/foot length.
2. Find the torsional moment in Newton-meter developed when the shaft delivers 20 kw at
200 rpm.
3. Compute for the twisting moment in in.lb. developed when the shaft delivers 20 hp at
1200 rpm.
4. Compute for the diameter in inches of a conveyor head pulley SAE 4130 solid steel shaft
being driven by a 20 hp drive motor through a gear reducer with 120 rpm output. The
torsional deflection is 0.06 degree/foot of shaft length and the modulus of elasticity is
30x106 psi in tension.
5. Compute the diameter (in) of solid shaft transmitting 75 hp at 1800 rpm. The nature of
the load and the type of service is such that allowable Ss based on pure torsion is 6000
psi.
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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
References:

▪ V.M. Faires (1969). Design of Machine Elements. 4 th ed. EDCA Publishing and
Distributing Corporation
▪ C.G. Duaso (2013). Machine Design. Cornell Publishing House
▪ J. Tordillo (2009). Mechanical Engineering Formulas. 2nd ed. Tordillo Publishing

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
Unit 2: Keys

Overview

This chapter covers some conventional design information on keys.

Learning Objectives:

After successful completion of this module, the students should be able to:
1. Classify different types of keys.
2. Solve keys related problems.

Course Materials:

Keys
A key can be defined as a machine element which is used to connect the transmission shaft to
rotating machine elements like pulleys, gears, sprockets or flywheels.

• Saddle Keys A saddle key is a key which fits in the keyway of the hub only.
• Sunk Keys A sunk key is a key in which half the thickness of the key fits into the keyway
on the shaft and the remaining half in the keyway on the hub.
• Feather Key A feather key is a parallel key which is fixed either to the shaft or to the hub
and which permits relative axial movement between them.
• Woodruff Key A Woodruff key is a sunk key in the form of an almost semicircular disk of
uniform thickness.

Design of Square and Flat Keys

Although there are many types of keys, only square and flat keys are extensively used in
practice.

The exact location of the force P on the surface AC is unknown. In order to simplify the
analysis, it is assumed that the force P is tangential to the shaft diameter. Therefore,

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
 The design of square or flat key is based on two criteria; failure due to shear stress and
failure due to compressive stress. The shear stress t in the plane is given by:

The compressive stress in the key is given by,

Design of Kennedy Key

The Kennedy key consists of two square keys. The hub is bored off the centre and the
two keys force the hub and the shaft to a concentric position. Kennedy key is used for heavy
duty applications.

Splines
Splines are keys which are made integral with the shaft. They are used when there is a
relative axial motion between the shaft and the hub. The gear shifting mechanism in automobile
gearboxes requires such type of construction.

The torque transmitting capacity of splines is given by,

Examples:
A key is to be designed for a 12.7 cm shaft, which will transmit power of 150 kW at 360
rpm. If the allowable shear stress for the key is 920 kg/cm 2, the allowable compressive stress is
1200 kg/cm2, the width of key is 3.175 cm and height is 2.22 cm.
1. Determine the torque in kN.m. (Ans. 3.98 kN.m)
2. Determine the force in kN. (Ans. 62.566 kN)
3. What is the length of key based on the allowable shearing stress? (Ans. 21.87 mm)
4. What is the length of key based on the allowable compressive stress? (Ans. 47.90 mm)

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
 5. What axial force to remove the hub from the shaft if the coefficient of friction is 0.45?
(Ans. 56.4 kN)

Activities:

Assignment Answer the following:

A pulley is keyed to a 2 ½ inches diameter shaft by a 7/16 inch by 5/8 inch by 3 inches
flat key. The shaft rotates at 50 rpm. The allowable shearing stress for the key is 22 ksi. The
allowable compressive stress for the key, hub and shaft are 66 ksi, 59 ksi and 72 ksi,
respectively.
1. Determine the torque (in.lbs) based on shearing stress for the key.
2. Determine the torque (in.lbs) based on compressive stress for the key.
3. What torque (in.lbs) transmitted by the shaft?
4. Find the torque (in.lbs) that can be carried by the hub.
5. What is the safe torque (in,lbs)?

Assessment:

A rectangular key used in a pulley connected to aline shaft with a power of 7.46 kW at a
speed of 1200 rpm. If the shearing stress of the shaft and key are 30 MPa and 240 MPa.
1. Find the torque in N.mm.
2. What is the diameter (mm) of the shaft?
3. What force in Newtons?
4. Determine the width (mm) of key.
5. Determine the length (mm) of key.

References:

▪ V.M. Faires (1969). Design of Machine Elements. 4 th ed. EDCA Publishing and
Distributing Corporation
▪ C.G. Duaso (2013). Machine Design. Cornell Publishing House
▪ J. Tordillo (2009). Mechanical Engineering Formulas. 2nd ed. Tordillo Publishing

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
Unit 3: Coupling

Overview

This chapter will describe some typical couplings that are used to connect shafts.

Learning Objectives:

After successful completion of this module, the students should be able to:
1. Classify different types of coupling.
2. Solve coupling related problems.

Course Materials:

Couplings
A coupling can be defined as a mechanical device that permanently joins two rotating
shafts to each other. The most common application of coupling is joining of shafts of two
separately built or purchased units so that a new machine can be formed.

Muff Coupling
Muff coupling is also called sleeve coupling or box coupling.

Muff coupling has following advantages:

• It is the simplest form of coupling with only two parts, viz., sleeve and key. It is simple to
design and manufacture.
• It has no projecting parts except the keyhead. The external surface of the sleeve is
smooth. This is an advantage from the standpoint of safety to the operator.
• It has compact construction with small radial dimensions.
• It is cheaper than other types of coupling.

Muff coupling has following disadvantages:

• Muff coupling is difficult to assemble or dismantle. The sleeve has to be either shifted
over the shaft by half of its length or the ends of the shafts have to be drawn together or
apart by half length of the sleeve.
• It is a rigid type of coupling and requires accurate alignment of shafts. It cannot tolerate
misalignment between the axes of two shafts. The misalignment of shafts, caused by
inaccurate assembly, induces forces, which tend to bend the shafts.
• Since there is no flexible element in the coupling, it cannot absorb shocks and vibrations.
It can be used only where the motion is free from vibrations.
• It requires more axial space compared with flange couplings.

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
For the sleeve of muff coupling, the standard proportions used in practice are as follows:

Design Procedure for Muff Coupling

The basic procedure for finding out the dimensions of the muff coupling consists of the
following steps:
• Calculate the diameter of each shaft by the following equations:

• Calculate the dimensions of the sleeve by the following empirical equations,

o Also, check the torsional shear stress induced in the sleeve by the following
equations:

• Determine the standard cross-section of flat sunk key. The length of the key in each
shaft is one-half of the length of the sleeve. Therefore:

Clamp Coupling
The clamp coupling is also called compression coupling or split muff coupling.

The clamp coupling has the following advantages:

• It is easy to assemble and dismantle.
• It can be easily removed without shifting the shaft in axial direction, unlike solid muff
coupling.
• As compared with flange coupling, clamp coupling has small diametral dimensions.

The disadvantages of clamp coupling are as follows:

• There is difficulty in dynamic balancing of the coupling. Therefore, it is not possible to
use the clamp coupling for high-speed applications.
• Clamp coupling is unsuitable for shock loads.
• It is necessary to provide a guard for the coupling to comply with the factory regulation
act.

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
Clamp coupling is usually designed on the basis of standard proportions for sleeve halves and
clamping bolts.
• For sleeve halves:

• For clamping bolts:

o The clamping force of each bolt is given by:

Design Procedure for Clamp Coupling

The basic procedure for finding out the dimensions of clamp coupling consists of the
following steps:
• Calculate the diameter of each shaft by the following equations:

• Calculate the main dimensions of the sleeve halves by using the following empirical
equations:

• Determine the standard cross-section of the flat key. The length of the key in each shaft
is one-half of the length of sleeve. Therefore:

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
 With these dimensions of the key, check the shear and compressive stresses in
the key.

• Calculate the diameter of clamping bolts.

Rigid Flange Couplings

A flange coupling consists of two flanges—one keyed to the driving shaft and the other
to the driven shaft.

The rigid flange couplings have the following advantages:

• Rigid coupling has high torque transmitting capacity.
• Rigid coupling is easy to assemble and dismantle.
• Rigid coupling has simple construction. It is easy to design and manufacture.

The rigid flange couplings have the following disadvantages:

• It is a rigid type of coupling. It cannot tolerate misalignment between the axes of two
shafts.
• It can be used only where the motion is free from shocks and vibrations.
• It requires more radial space.

Design Procedure for Rigid Flange Coupling

The basic procedure for finding out the dimensions of the rigid flange coupling consists of
the following steps:
• Shaft Diameter Calculate the shaft diameter by using the following two equations:

• Dimensions of Flanges Calculate the dimensions of the flanges by the following

empirical equations:

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
 o The inner and outer diameters of the hub are d and dh respectively. The torsional
shear stress in the hub is given by,

o The flange at the junction of the hub is under shear while transmitting the
torsional moment Mt.

• Diameter of Bolts Decide the number of bolts using the following guidelines:

• Dimensions of Keys Determine the standard cross-section of flat key. The length of the
key in each shaft is lh. Therefore,

o With these dimensions of the key, check the shear and compressive stresses in
the key

Examples:
A flange coupling connects two 2-inch diameter shaft. The flanges are fitted with 6 bolts
of SAE 1040 steel on a 7 inches bolt circle. The shaft runs at 300 rpm and transmits 45 hp.
Assume a factor of safety of 5, ultimate tension of 70,000 psi, and ultimate shear of 55,000 psi.
1. What is torque transmitted (in-lbs.) (Ans. 9,453.8 in.lbs)
2. Determine the total force transmitted in pounds. (Ans. 2,701 lbs)
3. Determine the force (lbs) per bolt. (Ans. 450 lbs)
4. Find the diameter (in.) of bolts required. (Ans. ¼ in)
5. How thick should the flange be? (Ans. 0.14 in.)

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME
Assessment:

A flanged coupling has an outside diameter of 200 mm and connects two 40 mm shafts.
The four 16 mm bolts on a 140 mm bolt circle. The radial flange thickness is 20 mm. If the
torsional stress in the shaft is not to exceeds 26 MPa.
1. What torque (N/m) transmitted by the shaft?
2. Determine the power (kw) that can be transmitted at 600 rpm.
3. Determine the shearing stress (MPa) in the bolts if uniformly distributed.
4. Determine the bearing pressure (MPa) in the bolts.

References:

▪ V.M. Faires (1969). Design of Machine Elements. 4 th ed. EDCA Publishing and
Distributing Corporation
▪ C.G. Duaso (2013). Machine Design. Cornell Publishing House
▪ J. Tordillo (2009). Mechanical Engineering Formulas. 2 nd ed. Tordillo Publishing

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SUBJECT: MEEN 30223 – MACHINE ELEMENTS 2
PREPARED BY: JHAY AHR C. FLORES, ME