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- 1. transmit torque keys prevent relative motion, Rotary axial. In some construction they allow axial motion
- 2. According to shape Straight, Tapered, Rectangular, Square, Round, Dovetail.according to their intended duty as: Light duty keys Medium duty Heavy duty
- 3. b b bh Square Rectangular shallow b Tapered Two width Dove tail Woodruff Flat Saddle
- 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. Pulley shaft assembly Pressure distribution
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Thank You

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