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# Shaft keys

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### Shaft keys

1. 1.  transmit torque keys prevent relative motion,  Rotary  axial.  In some construction they allow axial motion
2. 2. According to shape Straight, Tapered, Rectangular, Square, Round, Dovetail.according to their intended duty as: Light duty keys Medium duty Heavy duty
3. 3. b b bh Square Rectangular shallow b Tapered Two width Dove tail Woodruff Flat Saddle
4. 4. 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
5. 5. Pulley shaft assembly Pressure distribution
6. 6. 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)
7. 7. 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 )
8. 8. 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
9. 9. Integrating between the limits L = 0 to L2yields: 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 than2.25D is not effective. The pressure at L=2.25D equals zeroand 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
10. 10. Shear strengthThe pressure on the key can be represented by adiagram h 2P 1 = Ss bwhere Ss is the  D P1maximum shear at the P P2end of the key and  Lhence 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
11. 11. 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
12. 12. 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
13. 13. 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
14. 14. 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
15. 15.  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.
16. 16. Thank You