7.1 Phasing
In order to obtain a constant angular-velocity ratio (1:1) between input and Output
shafts, proper phasing of the joints is required. This phasing can be described as
follows: two cardan joints in series will transmit a constant angular velocity ratio (1:1)
between two intersecting or nonintersecting shafts (see Figure 2), provided that the angle
between the connected shafts and the intermediate shaft are equal ( b = b’ ) and that when yoke 1 lies in the plane of the Input and
intermediate shafts, yoke 2 lies in the plane of the intermediate shaft and the output
shaft .
If shafts 1 and 3 intersect, yokes 1 and 2 are coplanar.
When the above phasing has been realized, torsional and
inertial excitation is reduced to a minimum. However, inertia excitation will Inevitably
remain in the intermediate shaft 2. because this shaft has the angular acceleration of the
output shaft of a single universal joint the first of the two joints in series). It is for
this reason that guidelines exist limiting the max. angular accleratlons of the
intermediate shaft. Depending on the application values between 300 rad/sec˛ and values
In excess of 1000 rad/sec˛ have been advocated. In light, Industrial drives the allowable
speed may be higher. For an accurate determination of allowable speed, a stress
determination is necessary
7.2 Example 5: DetermIning the Maximum Speed of an Input Shaft In a Series
In a drive consisting of two universal joints in series, phased so as to produce a
constant (1:1) angular velocity ratio between input and output shafts, the angle between
the intermediate shaft and input (and output) shaft is 20°. If the max. angular
acceleration of the intermediate shaft is not to exceed 1000 rad/sec2. what is the upper
limit of the speed of the input shaft?
From Table 1, with b = 20°, we find amax/w˛ = 0.1250.
Since amax = 1000 rad/sec˛,
w˛ = (amax)(O.125O)
= (1000)/(0.1250) = 8000 rad/sec˛.
Hence, w = (8000)˝ = 89.4 rad/sec. = (89A)(60) = 854
RPM.
(2) (p)
Hence, the speed of the input shaft should not exceed
854 RPM. When the joint angle is less than or equal to 10°,
Table 3 can be used as an alternative.
7.3 Example 6
Same as problem 5, except operating angle is 10°. Here we can use Table 3. The
intersection of b = 10° and the 1000 sec˛ curve yields N @ 1800
RPM Hence, the speed of the input shaft should not exceed 1800 RPM. A more exact
calculation, as in Example 5. yields N = 1726 RPM. For practical purposes however, the
value obtained from Table 3 is entirely satisfactory.