ROHINI COLLEGE OF ENGINEERING & TECHNOLOGY

UNIT II
SHAFTS AND COUPLINGS
CHAPTER 2

Introduction on Key
A key is a piece of mild steel inserted between the shaft and hub or boss of the
pulley to connect these together in order to prevent relative motion between them. It is
always inserted parallel to the axis of the shaft. Keys are used as temporary fastenings
and are subjected to considerable crushing and shearing stresses. A keyway is a slot or
recess in a shaft and hub of the pulley to accommodate a key.

Types of Keys
The following types of keys are important from the subject point of view:
1. Sunk keys
2. Saddle keys
3. Tangent keys
4. Round keys and
5. Splines.
We shall now discuss the above types of keys, in detail, in the following pages.

Sunk Keys
The sunk keys are provided half in the keyway of the shaft and half in the keyway
of the hub or boss of the pulley. The sunk keys are of the following types:
1. Rectangular sunk key. A rectangular sunk key is shown in Fig. 2.1. The usual
proportions of this key are
Width of key, w = d / 4 and thickness of key, t = 2w / 3 = d / 6
where d = Diameter of the shaft or diameter of the hole in the hub.
The key has taper 1 in 100 on the top side only.

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Fig 2.1 Rectangle Sunk Key.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 471]

2.Square sunk key. The only difference between a rectangular sunk key and a
square sunk key is that its width and thickness are equal, i.e.

w=t=d/4

3. Parallel sunk key. The parallel sunk keys may be of rectangular or square
section uniform in width and thickness throughout. It may be noted that a parallel
key is a taperless and is used where the pulley, gear or other mating piece is
required to slide along the shaft.

4. Gib-head key. It is a rectangular sunk key with a head at one end known as gib
head. It is usually provided to facilitate the removal of key. A gib head key is
shown in Fig. 2.2 (a) and its use in shown in Fig. 2.2 (b).

Fig 2.2 Gib-head key.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 471]

The usual proportions of the gib head key are

Width, w = d / 4

and thickness at large end, t = 2w / 3 = d / 6

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5. Feather key. A key attached to one member of a pair and which permits
relative axial movement is known as feather key. It is a special type of parallel
key which transmits a turning moment and also permits axial movement. It is
fastened either to the shaft or hub, the key being a sliding fit in the key way of the
moving piece.

Fig 2.3 Feather key.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 472]

The feather key may be screwed to the shaft as shown in Fig. 2.3 (a) or it may have
double gib heads as shown in Fig. 2.3 (b). The various proportions of a feather key
are same as that of rectangular sunk key and gib head key.

6. Woodruff key. The woodruff key is an easily adjustable key. It is a piece from
a cylindrical disc having segmental cross-section in front view as shown in Fig.
2.4. A woodruff key is capable of tilting in a recess milled out in the shaft by a
cutter having the same curvature as the disc from which the key is made. This key
is largely used in machine tool and automobile construction.

Fig 2.4 Woodruff key.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 473]

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The main advantages of a woodruff key are as follows:

1. It accommodates itself to any taper in the hub or boss of the mating piece.

2. It is useful on tapering shaft ends. Its extra depth in the shaft *prevents
any tendency to turn over in its keyway.

The disadvantages are

1. The depth of the keyway weakens the shaft.

2. It cannot be used as a feather.

Saddle keys

The saddle keys are of the following two types

1. Flat saddle key, and

2. Hollow saddle key.

1. Flat saddle key: A flat saddle key is a taper key which fits in a keyway in the
hub and is flat on the shaft as shown in Fig. 2.5. It is likely to slip round the shaft
under load. Therefore, it is used for comparatively light loads.

Fig 2.5 Saddle key. Fig 2.6 Tangent key.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 473]

2. Hollow saddle key: A hollow saddle key is a taper key which fits in a keyway
in the hub and the bottom of the key is shaped to fit the curved surface of the shaft.
Since hollow saddle keys hold on by friction, therefore these are suitable for light
loads. It is usually used as a temporary fastening in fixing and setting eccentrics,
cams etc.

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Tangent Keys
The tangent keys are fitted in pair at right angles as shown in Fig. 2.6. Each key is
to withstand torsion in one direction only. These are used in large heavy duty shafts.

Round Keys

The round keys, as shown in Fig. 2.7(a), are circular in section and fit into holes
drilled partly in the shaft and partly in the hub. They have the advantage that their
keyways may be drilled and reamed after the mating parts have been assembled. Round
keys are usually considered to be most appropriate for low power drives.

Fig 2.7 Round Keys

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 474]

Sometimes the tapered pin, as shown in Fig. 2.7 (b), is held in place by the friction
between the pin and the reamed tapered holes.

Splines

Sometimes, keys are made integral with the shaft which fits in the keyways broached in
the hub. Such shafts are known as splined shafts as shown in Fig. 2.8. These shafts usually
have four, six, ten or sixteen splines. The splined shafts are relatively stronger than shafts
having a single keyway. The splined shafts are used when the force to be transmitted is
large in proportion to the size of the shaft as in automobile transmission and sliding gear
transmissions. By using splined shafts, we obtain axial movement as well as positive
drive is obtained.

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Fig 2.8 Splines.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 474]

Forces acting on a Sunk Key

When a key is used in transmitting torque from a shaft to a rotor or hub, the following
two types of forces act on the key:

1. Forces (F1) due to fit of the key in its keyway, as in a tight fitting straight key
or in a tapered key driven in place. These forces produce compressive stresses in
the key which are difficult to determine in magnitude.

2. Forces (F) due to the torque transmitted by the shaft. These forces produce
shearing and compressive (or crushing) stresses in the key.

The distribution of the forces along the length of the key is not uniform because the forces
are concentrated near the torque-input end. The non-uniformity of distribution is caused
by the twisting of the shaft within the hub.

Fig 2.9 Forces acting on a sunk key.

[Source: “A Textbook of Machine Design by R.S. Khurmi J.K. Gupta, Page: 475]

The forces acting on a key for a clockwise torque being transmitted from a shaft to

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a hub are shown in Fig. 2.9. In designing a key, forces due to fit of the key are neglected
and it is assumed that the distribution of forces along the length of key is uniform.

Strength of a Sunk Key

A key connecting the shaft and hub is shown in Fig. 2.10.

Let T = Torque transmitted by the shaft,

F = Tangential force acting at the circumference of the shaft,

d = Diameter of shaft,

l = Length of key,

w = Width of key.

t = Thickness of key, and

τ and σc = Shear and crushing stresses for the material of key.

A little consideration will show that due to the power transmitted by the shaft, the key
may fail due to shearing or crushing.

Considering shearing of the key, the tangential shearing force acting at the circumference
of the shaft,

F = Area resisting shearing × Shear stress = l × w × τ

∴ Torque transmitted by the shaft,

d
T=F×
2

d
T=l×w×τ× ...(i)
2

Considering crushing of the key, the tangential crushing force acting at the circumference
of the shaft,
t
F = Area resisting crushing × Crushing stress = l × × σc
2

∴ Torque transmitted by the shaft,

d
T=F×
2

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t d
T = l × × σc × ...(ii)
2 2

The key is equally strong in shearing and crushing, if

d t d
l × w × τ × = l × × σc × ...[Equating equations (i) and (ii)]
2 2 2
w σc
= ...(iii)
t 2τ

The permissible crushing stress for the usual key material is at least twice the permissible

shearing stress. Therefore, from equation (iii), we have w = t. In other words, a square
key is equally strong in shearing and crushing.

In order to find the length of the key to transmit full power of the shaft, the shearing
strength of the key is equal to the torsional shear strength of the shaft.

We know that the shearing strength of key,

d
T=l×w×τ× ...(iv)
2

and torsional shear strength of the shaft,

π
T= × τ1 × d 3
16

... (Taking τ1 = Shear stress for the shaft material)

From equations (iv) and (v), we have

d π
l×w×τ× = × τ1 × d3
2 16

π τ1 d2
l= ×
8 w×τ

πd τ1
l= ×
2 τ
τ1
l = 1.571d × ... (Taking w = d/4) ...(vi)
τ

When the key material is same as that of the shaft, then

τ = τ1.

∴ l = 1.571 d ... [From equation (vi)]

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Effect of Keyways

A little consideration will show that the keyway cut into the shaft reduces the load
carrying capacity of the shaft. This is due to the stress concentration near the corners of
the keyway and reduction in the cross-sectional area of the shaft. It other words, the
torsional strength of the shaft is reduced. The following relation for the weakening effect
of the keyway is based on the experimental results by H.F. Moore.
w h
e = 1- 0.2 ( ) – 1.1( )
d d

where e = Shaft strength factor.

It is the ratio of the strength of the shaft with keyway to the strength of the same shaft
without keyway,

w = Width of keyway,

d = Diameter of shaft, and

Thickness of key (t)
h = Depth of keyway =
2

It is usually assumed that the strength of the keyed shaft is 75% of the solid shaft,
which is somewhat higher than the value obtained by the above relation. In case the
keyway is too long and the key is of sliding type, then the angle of twist is increased in
the ratio kθ as given by the following relation:
w h
kθ = 1+ 0.4 ( ) + 0.7( )
d d

where kθ = Reduction factor for angular twist.

Problem 2.1
A 15 kW, 960 r.p.m. motor has a mild steel shaft of 40 mm diameter and the extension
being 75 mm. The permissible shear and crushing stresses for the mild steel key are 56
MPa and 112 MPa. Design the keyway in the motor shaft extension. Check the shear
strength of the key against the normal strength of the shaft.
Given Data:
P = 15 kW = 15 × 103 W
N = 960 r.p.m.

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d = 40 mm
l = 75 mm
τ = 56 MPa = 56 N/mm2
σc = 112 MPa = 112 N/mm2
We know that the torque transmitted by the motor,
P×60 15×103 ×60
T= = = 149 N-m
2πN 2π×960

T = 149 × 103 N-mm

Let w = Width of keyway or key.
Considering the key in shearing. We know that the torque transmitted (T),
d
149 × 103 = l × w × τ ×
2
40
= 75 × w × 56 ×
2

= 84 × 103 w
∴ w = 149 × 103 / 84 × 103
∴ w = 1.8 mm
This width of keyway is too small. The width of keyway should be at least d / 4.
d 40
w= =
4 4

w = 10 mm.
Since σc = 2τ, therefore a square key of w = 10 mm and t = 10 mm is adopted.
According to H.F. Moore, the shaft strength factor,
w h
e = 1- 0.2 ( ) – 1.1( )
d d

w t
e = 1- 0.2 ( ) – 1.1( ) ... (h = t/2)
d 2d

10 10
e = 1- 0.2 ( ) – 1.1( )
20 2×40

e = 0.8125

∴ Strength of the shaft with keyway,

π
= × τ × d3 × e
16
π
= × 56 × 403 × 0.8125
16

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= 571844 N

and shear strength of the key

d
=l×w×τ×
2
40
= 75 × 10 × 56 ×
2

= 840000 N
Shear strength of the key 840000
∴ =
Normal strength of the shaft 571844
Shear strength of the key
∴ = 1.47
Normal strength of the shaft

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