2.  transmit torque
 keys prevent relative motion,
 Rotary
 axial.
 In some construction they allow
axial motion

3. According to shape
 Straight,
 Tapered,
 Rectangular,
 Square,
 Round,
 Dovetail.
according to their intended duty as:
 Light duty keys
 Medium duty
 Heavy duty

4. b b b
h
Square Rectangular shallow
b
Tapered Two width Dove tail
Woodruff Flat Saddle

5. Shaft Keys for light and medium duty
Taper 1/16 in per ft b
b = d/4
90
0 h= b/2
Taper 1/8 in per ft
b = d/5 to d/4
b b = d/6

6. Pulley shaft assembly
Pressure
distribution

7. Crushing strength:
 The hub is more rigid than the shaft,
The shaft will be twisted , the hub will remain
undistorted.
The pressure along the key will vary
Minimum at the free end
Maximum on the other side.
Maximum pressure : P1
The minimum pressure :P2
h
2
At L the pressure : P.  D
 At Lo the pressure equals 1P
P
to zero P2

L
L2
L0 = 2.25 D
Fig. (5.4)

8. Crushing strength Continue
 The pressure can be expressed by:
 P  P1  L tan  h
2
 Where
 D
( P1  P 2 ) P1 P1
tan    P
L2 L0 P2

L
L2
L 0 = 2 .2 5 D
F ig. (5 .4 )

9. Torque transmitted
 Considering Small length of key (dL)
h
2
dl
 D
P1
P
P2

L
L2
L 0 = 2 .2 5 D
F ig. (5 .4 )
dT  P  dL  1 D
2

10. Integrating between the limits L = 0 to L2
yields: T  1 P DL  1 DL 2 tan 
2 1 2 4 2
The pressure /unit length = b
the crushing stress the area of unit
length, (Sb 0.5h 1) then, h
P1  0 . 5 S b h
Experiments showed that length of key greater than
2.25D is not effective. The pressure at L=2.25D equals zero
and hence, P1 Sbh
tan   
L0 4 .5 D
The torque transmitted can be expressed by,
T  1 S b hDL 2  1 2
4 18
S b hL 2

11. Shear strength
The pressure on the key can be represented by a
diagram
h
2
P 1 = Ss b
where Ss is the
 D
P1
maximum shear at the P
P2
end of the key and

L
hence
L2
P Sb L 0 = 2 .2 5 D
tan   1
 s
L0 2 . 25 D F ig. (5 .4 )
The torque transmitted can be expressed by:
T  S S bDL 2  S S bL
1 1 2
2 9 2

12. Design stages:
 Based on the diameter of the shaft the standard
dimensions of a square can be determined from
table (5.1a) or table (5.1b)
 Solve for crushing strength (Obtain key length)
 It is a second order equation
 L> 2.25 D, rejected
 L< 0 one key is not enough
 L< D then take L = D.
 Check for Shear strength

13. Example:
Find suitable dimensions of a square key to fit
into 3 inch diameter shaft. The shaft transmits
7
16
95 hp at a speed of 200 rpm. The key is made of
steel SAE 1010. Take safety factor of 2.5 and stress
concentration factor k’ = 1.6

14. Solution:
 Torque = Power/ angular speed
 Torque = 63030hp
N
 Torque = 63030 x 95
200
 Torque = 29939 Lb-in
 From table (5.1b), for a shaft with a diameter of inch
7 7
 Key size : x
8 8

15. Crushing strength
 For steel SAE1010: Ss 20000 psi and Se 31000 psi
,(from table 5.2)
 Hence Sb = 2 x 31000 = 62000 psi
 Taking factor of safety = 2.5 and k’ = 1.6
 Sb = 62000/(1.6x2.5) = 15500 psi
 Crushing strength
T  1 S b hDL 
21
4 2 18 S b hL 2
29939  1 15500 x 0 . 875 x 3 . 437 L 2  18 15500 x 0 . 875 L 2
1 2
4
29939  11653.6 L 2  753.5 L 2
2

 L = 3.26 or L = 12.21
 The first root is the answer; the second one
contradicts the condition that L2 < 2.25 D.
therefore the proper length of the key is 3.5

16.  Check for shear from the equation:
T
SS 
L 2 b ( 0 . 5 D  0 . 11 L 2 )
29939
SS 
3 . 5 x 0 . 875 ( 0 . 5 x 3 . 437  0 . 113 . 5 )
 Ss = 7331 psi
 Factor of safety in shear = 20000
 2.73
7331
O.K.

17. Thank You