Machine Design I (MCE-C 203)

Mechatronics Dept., Faculty of Engineering, Fayoum University

Dr. Ahmed Salah Abou Taleb Lecturer, Mechanical Engineering Dept., Faculty of Engineering, Fayoum University

1 Course Outlines Design of detachable joints: ( threaded joints , keys and splines). 

 Keys  Pins.  Splines.

2 Introduction How attach power transmission components to shaft to prevent rotation and axial motion? Torque resistance: keys, splines, pins, weld, press fit, etc..

Axial positioning: retaining rings, locking collars, shoulders machined into shaft, etc….

3 Keys • A key is the piece inserted in an axial direction between a shaft and hub of the mounted machine element such as pulley or gear etc., to prevent relative rotation…. • Keys are temporary fastening and are always made of mild steel because they are subjected to shearing and compressive stresses caused by the torque they transmit. • a keyway is the groove cut in the shaft or hub to accommodate a

key. 4 Keys

5 Keys Design

• keys are sunk in the shaft and the hub.

Let D = diameter of the shaft - width of the key W = D/4

 Rectangular cross-section nominal thickness H = (2/3)W = (1/6)D  Square cross-section:

nominal thickness H = W = D/4 6 Keys Design Step 1 – Determine key size based on shaft diameter. Step 2 – Calculate required length, L, based on torque.

Bearing stress

T = F*(D/2) or F = T/(D/2) this is the force the key Shear stress must react!!!

Required Length based on Required Length based on Shear Stress: Bearing Stress: 2T 4T L  where  d 0.5Sy / N L  where  d  Sy / N  d DW  d DH N = 3 7 Keys Design

8 Keys Design

9 Splines “Axial keys” machined into a shaft  Transmit torque from shaft to another machine element Advantages: •Can carry higher torque for given diameter (vs keys). •Lower stress on attachment (gear) •Better fit, less vibration (spline integral to shaft so no vibrating) •May allow axial motion while reacting torque Disadvantage: •Cost

•Impractical to use as fuse 10 Splines Design Torque capacity is based bearing stress on the sides of the splines T = 1000*N*R*h N = number of splines 1  D  d  D  d R = mean radius of the splines R     1 2  2  4 h = depth of the splines h  (D  d) 2

 D  d  D  d   D2  d2    T  1000N    1000N   4  2   8 

11 Splines Design

A: Permanent Fit B: Slide without Load C: Slide under Load

12 Splines Design T = kD2L T = torque capacity kD2 = torque capacity per unit length (from Table 11-5) L = length of spline in inches

13 Splines Design

14 Splines Design