The peak
displacement lead (or lag), peak angular-velocity ratios (max. and mm.) and peak
angular-acceleration ratios are shown as a function of operating angle in Table 1, which
is reproduced from “The Analytical Design of Universal Joints” by S.J. Baranyl,
Design News, Sept. 1, 1969.
The table should always be
consulted for exact numerical values. As a qualitative guideline, It may be kept in mind
that for small operating angles (say up to 10), the angular.displacement error (max. lead
or lag), the deviation of the max. and mm. angular velocity ratios from unity, and the
maximum angular acceleration ratio are very nearly proportional to the square of the
operating angle.
The static torque
transmitted by the output shaft Is equal to the product of the input torque and the
angular.velocity ratio.
The angular acceleration gives rise to an
inertia torque, as well as to vibrations. The inertia torque typically would be equal to
the sum of the product of the angular acceleration of the output shaft (rad/sec2) and the
polar mass moment of inertia of the output shaft (in.lb.sec2) and the output torque (with
the units ir.dicated the torque would be given In in.lbs).
The inertia loading often determines the
ultimate limit on the speed of operation of the joint. Recommended speed limits vary
depending on operating angle, transmitted power and nature .f the application. Recommended
peak angular accelerations of the driven shaft vary from 300 radlsec2 to over 2000
rad/sec2 in power drives. In light instrument drives, the allowable accele. rations may be
higher. For an accurate determination of allowable speed, a stress determination is
necessary.
4.1 Example 1: Determining the
Maximum Inertia Torque
A universal joint operates at 250 RPM with an operating angle of 10. Find the max. angular
displacement lead (or lag), max. and mm. angular-velocity of output shaft and max.
angular acceleration of output shaft.
If
the system drives an inertial load so that the total inertial load seen by the output
shaft (including its own inertia) can be represented by a steel, circular disc attached to
the output shaft (radius r = 3", thickness t = 1/4"), find the max. inertia
torque of the drive.
From Table 1 with b = 10, the max. displacement lead (or
lag) = 0.439º = 26.3’. The max. and mm. angular-velocity ratios are given as 1.0154
and 0.9848, respectively. Hence, the corresponding output-shaft speeds are:
Qmax = (250) (1.0154) = 254 RPM
Qmin = (250) (0.9848) = 246 RPM.