Chapter 11 - Keys, Couplings and Seals. 11.1 Chapter Objectives:. How attach power transmission components to shaft to prevent rotation and axial motion? Torque resistance: keys, splines, pins, weld, press fit, etc..
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11.1 Chapter Objectives:
Step 1 – Determine key size based on shaft diameter
Step 2 – Calculate required length, L, based on torque (11.4)
See Figure 11.2
For 5/16” key
Note, should also specify fillet radii and key chamfers – see Table 11-2
11.4 Design of Keys – stress analysis to determine required length:
Torque being transmitted
T = F/(D/2) or F = T/(D/2) this is the force the key must react!!!
Required Length based on Bearing Stress:
Required Length based on Shear Stress:
Typical parameters for keys:
N = 3, material 1020 CD (Sy = 21,000 psi)
Example: Specify the complete key geometry and material for an application requiring a gear (AISI OQT 1000) with a 4” hub to be mounted to a 3.6” diameter shaft (AISI 1040 CD). The torque delivered through the system is 21,000 lb-in. Assume the key material is 1020 CD (Sy = 21,000 psi) and N = 3.
Solution (note since key is weakest material, focus analysis on key!):
“Axial keys” machined into a shaft
Transmit torque from shaft to another machine element
A: Permanent Fit
B: Slide without Load
C: Slide under Load
Use this for spline design – SAE formulas based on 1,000 psi bearing stress allowable!!
Use this to get diameter. Then table 11.4 to get W, h, d
T = 1000*N*R*h
N = number of splines
R = mean radius of the splines
h = depth of the splines
Substituting R and h into torque equation:
N = 16 and d = .810D
Torque in IN-LBS/INCH of spline
Required D for given Torque
Example: A chain sprocket delivers 4076 in-lbs of torque to a shaft having a 2.50 inch diameter. The sprocket has a 3.25 inch hub length. Specify a suitable spline having a B fit.
T = kD2L
T = torque capacity in in-lbs
kD2 = torque capacity per inch (from Table 11-5)
L = length of spline in inches
Example: Specify straight spline for the previous problem (i.e. Torque = 21,000 lb-in and shaft is 3.6 in diameter.
Expensive – machining
Moderate torque capacity
Can use a key too
NO relative motion between the shafts.
Precise alignment of the shafts
Bolts in carry torque in shear. N = # of bolts.
Large shaft misalignments permissible
Key factors in selection are Torque, Angular Speed and the Operating Angle
Output not uniform wrt input
Output IS uniform wrt input