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.