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TABLE 2 USE FACTORS FOR THE TORQUE RATING OF UNIVERSAL JOINTS

Intermittent Running Conditions
Suggested Use Factor

SPEED
R.P.M.
ANGLE OF OPERATION-DEGREE
0 3 5 7 10 15 20 25 30
1800 9 20 34 45 - - - - -
1500 8 16 28 39 - - - - -
1200 7 13 22 32 40 - - - -
900 6 11 16 23 34 - - - -
600 5 8 11 15 22 34 40 - -
300 4 5 7 8 11 16 22 28 34
100 3 4 4 5 6 8 9 11 12

Continuous Running Conditions
Suggested Use Factor

SPEED
R.P.M.
ANGLE OF OPERATION-DEGREE
0 3 5 7 10 15 20 25 30
1800 18 40 68 90 - - - - -
1500 16 32 55 78 - - - - -
1200 14 26 44 64 80 - - - -
900 12 21 32 46 68 - - - -
600 10 15 22 30 44 68 80 - -
300 8 10 14 16 22 32 44 55 68
100 6 7 8 10 12 15 18 22 24

5.1 Example 2 Universal Joint Selection for Continuous Operation
A single universal joint is to transmit a continuously acting torque of 20 in-lbs., while operating  at an angle of 15° and at a speed of 600 RPM. Select a suitable joint.
      From Table 2 for continuous operation, the use factor Is given as 68. Note that there are blank spaces in the table, If the combination of operating angle and speed results in a blank entry in the table, this combination should be avoided.
The required torque is (68) (20) = 1360 in-lbs. There is no shock load and the dynamic factor of 2 does not
apply in this case.
      From the catalogue, it is seen that there are two joints meeting this specification: 508-0500 and 508—D516, both with a torque capacity of 1700 in-lbs. The first has a solid-shaft construction and the second a bored construction. The choice depends on the application.

5.2 Example 3: Universal Joint Selection for Intermittent Operation With Shock Loading
A single universal joint is to transmit 1/4 horsepower at 300 RPM at an operating angle of 15° Select a suitable joint for intermittent operation with shock loading.

            Here we make use of the equation:

Torque =  Horsepower x 63,025 in-lbs.
                            RPM
Hence, operating torque = ( 0.25) (63,025 ) = 52.5 n-lbs. From Table 2, for intermittent
                                               (300)

loads (300 RPM, 15°), the use factor is 16. Due to shock loading there should be an additional dynamic factor of 2.
     Hence, the rated torque = (52.5) (16) (2) = 1680 in-lbs. Thus the same joints found in the previous example are usable in this case.

5.3 Example 4: Determining the Maximum Speed of an Input Shaft
A universal joint is rated at 250 in-lbs., and operates at an angle of 12°, driving a rotating mass, which can be represented (together with the inertia of the driven shaft) by a steel, circular disc, radius r = 6”, thickness t = 1/2",
attached to the driven shaft, How fast can the input shaft turn if the inertia torque is not to exceed 50% of rated torque?

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