KEY & SPLINES DESIGN

Prof. Dr. Mohamed Omar Mousa

El-Minia University
.Prod. & Mech. Design Dept

Prof. Dr. Mohamed Omar Mousa February 2006

 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
Prof. Dr. Mohamed Omar Mousa February 2006
 PRINCIPLE OF WORK

Key
Key
Keyway Key W

h
L

Hub

Hub-Key-Shaft Connection
Design12- Keyway Cutting A Straight Type Design11- Keyway Cutting
Prof. Dr. Mohamed Omar Mousa February 2006
- 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:

Design13- Keyway Broaching a keyway

Prof. Dr. Mohamed Omar Mousa February 2006

 2.1 RECTANGULAR SUNK KEY W
1:100
t
L

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.
Design02- shaft example Pumps Animation
Prof. Dr. Mohamed Omar Mousa February 2006
 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.
Prof. Dr. Mohamed Omar Mousa February 2006
 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
t
5
1.
.Gib-head sunk key
Prof. Dr. Mohamed Omar Mousa February 2006
 TAPER 1:100
Shaft

1.75 t

t
450
Hub
W
t
5
1.

The usual proportions of the Gib-head key are:

Prof. Dr. Mohamed Omar Mousa February 2006

 2.5 FEATHER KEY Movement Direction
It is a key, which is
attached to one
member of the pair and
Hub
allows the other to be Screw
movable along it. The
feather key can be
screwed to the shaft
as in proportions
The figure. 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.
Prof. Dr. Mohamed Omar Mousa Design14- Keyway cutting in a pully February 2006
 Table 1: Key dimensions according to IS 2292 and 2293-1963.
.Key cross sec Shaft diameter .Key cross sec Shaft diameter
t (mm) W (mm) up to (mm) t (mm) W(mm) up to (mm)

14 25 85 2 2 6
16 28 95 3 3 8
18 32 110 4 4 10
20 36 130 5 5 12
22 40 150 6 6 17
25 45 170 7 8 22
28 50 200 8 10 30
32 56 230 8 2 1 38
32 63 260 9 14 44
36 70 290 10 6 1 50
40 80 330 1 1 8 1 58
45 90 380 12 20 65
50 100 440 14 22 75
Prof. Dr. Mohamed Omar Mousa February 2006
 2.6 WOODRUFF KEY

-It is a piece from a cylindrical disc having

segmental cross section.

- It is an easily adjustable key.

.Woodruff key

Prof. Dr. Mohamed Omar MousaDesign15- Keyway cutting A Woodruff February 2006
 - 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.

Prof. Dr. Mohamed Omar Mousa February 2006

 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.

Prof. Dr. Mohamed Omar Mousa February 2006

 3. SADDLE KEYS

.Saddle key

Prof. Dr. Mohamed Omar Mousa February 2006

 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

Prof. Dr. Mohamed Omar Mousa February 2006

 - 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

Prof. Dr. Mohamed Omar Mousa February 2006

 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

Prof. Dr. Mohamed Omar Mousa February 2006

 - 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.

Prof. Dr. Mohamed Omar Mousa February 2006

 IMPORTANT NOTICES
1- Number of keys can be:
1 – 2/180 0 – 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.

3- Strength of key material should be less

than that of shaft and hub materials.

Prof. Dr. Mohamed Omar Mousa February 2006

 D = 1.25 d
b = 0.25 D b
b

D d
d

Prof. Dr. Mohamed Omar Mousa February 2006

 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

Prof. Dr. Mohamed Omar Mousa February 2006
 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
Prof. Dr. Mohamed Omar Mousa February 2006
 3. Tapered keys, in which
the torque is transmitted
R
by means of friction
induced by compressive
stress as in Figure.
Transmission of torque F
due to frictional forces
Mt
generated by taper sides p
keys.

Transmission of torque due to

frictional forces generated
by taper sides keys.

Prof. Dr. Mohamed Omar Mousa February 2006

 4. Tapered keys fitted on R’
the sides and round keys, R
F
in which torque is
transmitted by the F’
simultaneous action of
compressive and shear Mt
stresses and friction as
shown in Figure. R’
R
F

F’
Mt

Prof. Dr. Mohamed Omar Mousa February 2006

 7. FORCES ACTING ON SUNK KEYS

L W

’F
W
F’ << F F
t
F
F’ neglected

’F

Prof. Dr. Mohamed Omar Mousa February 2006

 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.

Prof. Dr. Mohamed Omar Mousa February 2006

 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.

Prof. Dr. Mohamed Omar Mousa February 2006

 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.
τ : Shear stress for the material of the key.
σ : Compressive stress of the material of the key.

Prof. Dr. Mohamed Omar Mousa February 2006

 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) τ
i.e.; Torque transmitted = T
’F
W
F
t
F

’F

Prof. Dr. Mohamed Omar Mousa February 2006

 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

’F
W
F
t
F

’F

Prof. Dr. Mohamed Omar Mousa February 2006

 Then, the key can be equally strong in both of shear
and crushing, if :
Crushing stress = Shear stress

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., σ = 2τ
Therefore,
W=t
Prof. Dr. Mohamed Omar Mousa February 2006
 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:

----------- (Equ. I)

And, the torsion shear strength of the shaft is:

----------- (Equ. II)

Prof. Dr. Mohamed Omar Mousa February 2006

 where,
τ : The shear strength of the key.
τ’ : The shear strength of the shaft material.

Equ. I = Equ. II
, Therefore

--------- Equ. III

Take: W = d/4, then;

--------- Equ. IV

Prof. Dr. Mohamed Omar Mousa February 2006

 Equation (IV) can be used to determine the key
length.
For special cases when the material of the shaft
and key is similar,

And, if w = d/4 then,

Prof. Dr. Mohamed Omar Mousa February 2006

 SOLVED .9
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.
Solutio
n
:Given
.P = 20 h . p. N = 960 r.p.m
D = 40 mm L = 75 mm
τall = 560 kp/cm 2 σ all = 1120 kp/cm2

Prof. Dr. Mohamed Omar Mousa February 2006

 Design of keyway.1

W × 560 × 4/2 × 7.5 = 1492

W = 0.17 cm = 1.7 mm

Prof. Dr. Mohamed Omar Mousa February 2006

 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

Prof. Dr. Mohamed Omar Mousa February 2006

 Notice: The value of “σ” is twice “τ”.

Prof. Dr. Mohamed Omar Mousa February 2006

 EFFECT OF .10
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;

Prof. Dr. Mohamed Omar Mousa February 2006

 “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.

Prof. Dr. Mohamed Omar Mousa February 2006

 END
Prof. Dr. Mohamed Omar Mousa February 2006