CHAPTER four
Torque Transmitting Joints:
Keys ,Spline Joints; Pin Joints
;Interference Fit
4.1 introduction
• are used as joining machine members and
transmitting torque between the machine members
• are used on shafts to secure rotating elements, such
as gears, pulley , wheels, coupling etc.
4.2 Key Joints
• A key is a fastening
inserted into the keyway
of two mating parts,
it is used to transmit
torque from a shaft to
a hub or vice versa.
Types of Keys
1. Sunk keys
2. Saddle keys
3. Tangent keys,
4. Round keys
5. Splines.
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.
1. Rectangular sunk key. A rectangular sunk key
is shown
Width of key, w = d / 4 ; and thickness of key, t =
2w / 3 = d / 6
Sunk Keys
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.
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
Sunk Keys
• The usual proportions of the gib head key are: Width, w
= d/4; and thickness at large end, t = 2w/3 = d/6
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.
Sunk Keys
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
• 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.
• 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.
Sunk Keys
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 can not 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.
A flat saddle key is a taper key which fits in a
keyway in the hub and is flat on the shaft
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.
Tangent Keys
• The tangent keys are fitted in pair at right angles
as shown
• 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, 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.
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.
2. Forces (F) due to the torque transmitted by the shaft.
These forces produce shearing and compressive (or
crushing) stresses in the key.
Forces acting on a Sunk 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
• 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. 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
• Due to the power transmitted by the shaft, the key may fail
due to shearing or crushing.
Strength of a Sunk Key
• 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
T = F × d/2 = l × w × τ × d/2
• Considering crushing of the key, the tangential crushing force
acting at the circumference of the shaft:
F = Area resisting crushing × Crushing stress = l × t/2 × σc
∴ Torque transmitted by the shaft
T = F × d/2 = l × t/2 × σc× d/2
Strength of a Sunk Key
• The key is equally strong in shearing and crushing, if
l × w × τ × d/2 = l × t/2 × σc× d/2
w/t = σc/2τ
• The permissible crushing stress for the usual key
material is at least twice the permissible shearing
stress
• If we have w = t, i.e. 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
Strength of a Sunk Key
The shearing strength of key:
T = l × w × τ × d/2
and torsional shear strength of the shaft:
T = π/16 × τ1 × d3
Equating both, we get:
l × w × τ × d/2 = π/16 × τ1 × d3
When the key material is same as that of the shaft, then τ = τ1.
l = 1.571 d
Effect of Keyways
•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.
• 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 :
4.3 Pin Joint
• A pin joint is a solid cylinder-shaped device,
similar to a bolt,
• used to connect objects at the joint area.
• allows each object to rotate at the point of
joint connection.
Pin Joints
Pin for joining
Cylindrical pin Taper pin
safety pin
4.4 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
• These shafts usually have four, six, ten or
sixteen splines
• 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
4.5 INTERFERENCE FIT
• An interference fit results in an interference between
two mating parts under all tolerance conditions.
• In an interference fit, the two parts will always
interfere with each other, requiring a force or press fit
Interference Fit
• In an interference fit, the two parts
will always interfere with each
other, requiring a force or press fit
• As pairs, calculate the
• Hole Tolerance.________
• Shaft Tolerance.________
Allowance.____________
Clearance._____________
• for the fit shown to the right.
………………
26
Interference Fit
• In an interference fit, the
two parts will always
interfere with each other,
requiring a force or press
fit
• As pairs, calculate the
• Hole Tolerance. .013
• Shaft Tolerance. .016 Allowance.
-.037 Clearance. -.008
27
Example 1
• Design the rectangular key for a shaft of 50mm diameter. The shearing
and crushing stresses for the key material are 42MPa and 70MPa
Given:
d = 50mm; τ = 42MPa = 42 N/mm2; σc = 70MPa = 70N/mm2
Solution:
From Table , we find that for a shaft of 50mm diameter:
Width of key, w = 16mm
and thickness of key, t = 10mm
• Now the length of key is obtained by considering the key in shearing
and crushing:
Example 1
Considering shearing of the key. The shearing strength (or torque
transmitted) of the key:
T = l × w × τ × d/2 = l × 16 × 42 × 50/2
= 16 800l N-mm
The torsional shearing strength (or torque transmitted) of the shaft:
T = π/16 × τ1 × d3 = π/16 × 42 (50)3
= 1030000 N-mm
We get:
16 800l = 1030000
l = 61.31 mm
• Now considering crushing of the key.
Example 1
The crushing strength (or torque transmitted) of the key:
T = l × t/2 × τ × d/2 = l × 10/2 × 70 × 50/2
= 8750l N-mm
Equating:
8750l = 1030000
l = 117.7mm
Taking larger of the two values, we have length of key:
l = 117.7 say 120mm
d
T l . w . . (i )
2
2T
l
d .w.
Considering crushing of the key
d
T l . w . . (i )
2
2T
l
d .w.
Considering the maximum length, the length of the length of the
key……….
