KEY & SPLINES DESIGN
Lec.9
Dr. Ameer Ali Kamel
Prod.Eng.& Mech. Design Dept.
Faculty of Engineering
1. TYPES OF KEYS
- Key is a piece of mild steel inserted between two
mechanical elements (usually shaft and hub) to
connect them together and transmit power from
one of them to the other.
- The power should be transmitted without any loss.
-It is inserted parallel to the axis of the shaft in
a groove or slot which called “keyway”.
Design10- shaft example of a gear animation
PRINCIPLE OF WORK
Key
Key
Keyway Key W
h
L
Hub
Hub-Key-Shaft Connection
- Keys can be classified into the following main
groups:
- Sunk keys.
- Saddle keys.
- Tangent keys.
- Round keys.
- Splines.
2. SUNK KEYS
The sunk keys are provided half in the keyway of
the shaft and the other half in the keyway of the
hub. The sunk keys have the following types:
2.1 RECTANGULAR SUNK KEY W
1:100
t
L
d 2
W= t= W
4 3
W = Width of key
t = Thickness of key.
Where: “d” is the diameter of the shaft.
If the sunk key is tapered, therefore, the
tapered top side has an inclination of 1:100.
2.2 SQUARE SUNK KEY
The main difference between rectangular and
square sunk keys is that the width (W) of the
square key is equal to its thickness (t).
i.e.; W = t
2.3 PARALLEL SUNK KEY
The parallel sunk key can be either rectangular or
square cross sectional sunk keys with a uniform
width and thickness i.e. the parallel sunk key is a
taper-less top side sunk key with a square or
rectangular cross section. The parallel sunk keys
are important for connecting the movable pulley,
gear or hubs with there carrying shafts.
2.4 GIB-HEAD KEY
It is a rectangular sunk key with a head at one end
known as gib-head. This type has the advantage
that it is more easily to removal than the other
above mentioned types.
TAPER 1:100
Shaft
1.75 t
450
Hub
Gib-head sunk key.
TAPER 1:100
Shaft
1.75 t
t
450
Hub
W
The usual proportions of the Gib-head key are:
d
W=
4
2 d
t= W=
3 6
2.5 FEATHER KEY
Movement Direction
It is a key, which is
attached to one member
of the pair and allows
Hub
the other to be movable Screw
along it. The feather
key can be screwed to
the shaft as in figure.
The proportions of the
feather key are the same
as that of the parallel
rectangular or parallel Feather key.
gib-head keys.
The following table shows the standard dimensions of
parallel, tapered and gib-head keys.
Design14- Keyway cutting in a pully
Table 1: Key dimensions according to IS 2292 and 2293-1963.
Shaft diameter up Key cross sec. Shaft diameter Key cross sec.
to (mm) W(mm) t (mm) up to (mm) W (mm) t (mm)
6 2 2 85 25 14
8 3 3 95 28 16
10 4 4 110 32 18
12 5 5 130 36 20
17 6 6 150 40 22
22 8 7 170 45 25
30 10 8 200 50 28
38 1 2 8 230 56 32
44 14 9 260 63 32
50 1 6 10 290 70 36
58 1 8 1 1 330 80 40
65 20 12 380 90 45
75 22 14 440 100 50
2.6 WOODRUFF KEY
-It is a piece from a cylindrical disc having
segmental cross section.
-It is an easily adjustable key.
Woodruff key.
Design15- Keyway cutting A Woodruff
-The woodruff key is capable of tilting in a
recess milled out in the shaft by a cutter
having the same curvature as the disc form
which the key is made (form milling cutter).
-This type is usually used in machine tool and
automobile constructions.
ADVANTAGES WOODRUFF KEY
1- Easy in assembly and disassembly
1 2 3 4
2- Its extra depth in the shaft prevents any
tendency to turn over in its keyway.
3. SADDLE KEYS
W d
t= =
3 12
Saddle key.
4. ROUND AND DOWEL PINS
- The round keys and dowel pins are circular
elements and fit into holes drilled partly in two
contact parts.
Dowel pins.
- Tapered pins are held in place by friction
between pin and reamed tapered holes.
- Round keys are usually considered to be most
appropriate for low power drives.
Dowel and Taper pins
5. SPLINES
-Spline shafts are shafts with integrated number
of keys (more than 2 keys), which fit in the
keyways, which are broached in the hub.
- Usually, the shaft has 4, 6, 10 or 16 splines.
b
D = 1.25 d D d
b = 0.25 D
-Splined shafts are stronger than the shafts
with one key. Therefore, the spline shafts are
used when the power to be transmitted is large
in proportional to the size of the shaft as in
automobile transmission and sliding gear
transmission.
-Also, axial movements of hubs with respect to
shaft can be achieved by spline shafts.
IMPORTANT NOTICES
1- Number of keys can be:
1 – 2/1800 – 4 or more.
2- Width and height of key are assumed as a
function of shaft diameter where the key length
is determined from strength equations.
D = 1.25 d
b = 0.25 D b
b
D d
d
D
6. TORQUE TRANSMISSION BY KEYS
According to stress b
analysis, keys can be F
classified into four main h/2 h
groups. These included: F
1. Rectangular fitted key Mt
in which the torque is
transmitted by means of
compressive and shear R
stresses as shown in
Figure.
F = Mt / R Rectangular sides keys.
2. Tangential keys, in
which the torque is
transmitted by means of
compressive stress alone
as shown in Figure. h
R
F
Mt
F = Mt / R
Tangential keys.
3. Tapered keys, in which
the torque is transmitted
R
by means of friction
induced by compressive
stress as in Figure.
F
Transmission of torque
due to frictional forces
Mt
generated by taper sides p
keys.
Transmission of torque due to
frictional forces generated by
taper sides keys.
4. Tapered keys fitted on R’
the sides and round keys, R
F
in which torque is
transmitted by the F’
simultaneous action of
Mt
compressive and shear
stresses and friction as
shown in Figure. R’
R
F
F’
Mt
7. FORCES ACTING ON SUNK KEYS
L W
F’
W
F’ << F F
t
F
F’ neglected
F’
Therefore, when key transmitted torque between
shaft and hub, the following forces appear:
A. Force “ F’ ” due to the fit of the key in its keyway
(Compressive - difficult to determine in magnitude
- small). F’
B. Forces “F” that F
generate due to the F
transmitted torque.
F = Torque /radius
F’
These forces produced shearing and compressive
(crushing) stresses in the key.
8. STRENGTH OF SUNK KEY
During the design of sunk key, the following assumptions
should be taken into consideration:
1. The forces due to fit (F’) are small and negligible.
2. The forces are uniform distributed along the
length of the key.
Let us consider the following;
T : Torque transmitted by the system.
F : Tangential force acting on the key at the
circumference of the shaft.
D :Diameter of the shaft.
L : Length of the key.
W : Width of the key.
t : Thickness of the key.
: Shear stress for the material of the key.
: Compressive stress of the material of the key.
Now, considering shear of the key:
The tangential shearing force acting on the
circumference of the shaft can be computed as
follows:
F = Area resisting shearing x shear stress
F = (L.w)
i.e.; Torque transmitted = T
F’
d F W
T =F F
t
2
d
T = LW F’
2
Considering, crushing of key, the tangential crushing
force acting on the circumference of the shaft can
be determined as follows:
F = Crushing area x crushing stress
t
F=L F’
2 W
F
t
F
d t d
T=F = L
2 2 2
F’
Then, the key can be equally strong in both of shear
and crushing, if :
Crushing stress = Shear stress
d t d
LW = L
2 2 2
It is important to notice that the permissible crushing
stress for the usual key material is at least twice the
permissible shear stress.
i.e., = 2
Therefore,
W=t
i.e. the square key is equally strong in shearing
and crushing.
To calculate the length of the key to transmit full
power of the shaft, the shearing strength of the
shaft is equal to the torsion shear strength of the
shaft.
The shear strength of the key is:
d
T =LW ----------- (Equ. I)
2
And, the torsion shear strength of the shaft is:
T= ' d3 ----------- (Equ. II)
16
where,
: The shear strength of the key.
’ : The shear strength of the shaft material.
Equ. I = Equ. II
, Therefore
2
' d
L= --------- Equ. III
8 W
Take: W = d/4, then;
' --------- Equ. IV
L = 1.57d
Equation (IV) can be used to determine the key
length.
For special cases when the material of the shaft
and key is similar,
= '
d2
& L=
8W
And, if w = d/4 then,
L= d = 1.571 d
2
9. SOLVED PROBLEM
A 20 h.p., 960 revolution per min. motor has a mild steel
shaft of 40 mm diameter and extension being 75mm.
The permissible shear and crushing stresses for the
mild steel key are 560 kp/cm and 1120 kp/cm . Design
the keyway in the motor shaft extension. Check the
shear strength of the key against the normal strength
of the shaft.
Solution
Given:
P = 20 h . p. N = 960 r.p.m.
D = 40 mm L = 75 mm
all = 560 kp/cm2 all = 1120 kp/cm2
20x 4500
T= = 14.92 kp.m
2x 960
= 1492 kp.cm
1.Design of keyway
d
T =LW
2
As the width of the keyway is too small, then,
“W” should be 0.25 d
i.e.
W = d/4 = 4/4 = 1 cm = 10 mm
Checking the shear strength of the key against
the normal strength of the shaft
Checking the shear strength of the key against the
normal strength of the shaft
d
Shear strength of the key LW
= 2
Normal strength of the shaft
' d3
16
8LW
=
d3
8 x7.5x1
=
x 43
= 1.2
Notice: The value of “” is twice “”.
10. EFFECT OF KEYWAYS
Cutting of keyways in the shafts tends to reduce the
load carrying capacity of the shaft due to the
occurrence of the stress concentration. Therefore, a
shaft strength factor is determined from
experimental results which can be expressed as
follows;
W h
e = 1 − 0.2 − 1.1 1
d d
“e” is the ratio of the strength of the shaft with
keyway to the strength of the same shaft without
keyway.
where:
e : Shaft strength factor.
W : Width of keyway.
d : Shaft diameter.
h : Depth of keyway.
However, it is usually assumed that the strength of the
keyed shaft is 75% that of the solid shaft without
keyway, which is higher than the value obtained by the
above relation.
END
