Page 127 - Tribology in Machine Design
P. 127
Friction, lubrication and wear in lower kinematic pairs 113
Solution
For uniform pressure, p=0.07MPa; the total axial force is
Effective radius
Number of pairs of active surfaces n a = 2, then
friction couple =fRn ar m= 0.25 x 1270 x 2 x 0.084 = 53.34 Mm.
Assuming uniform acceleration during the time required to reach full speed
from rest
It should be noted that energy is dissipated due to clutch slip during the
acceleration period. This can be shown as follows:
the angle turned through by the constant speed driving shaft during the
period of clutch slip is
the angle turned through by the machine shaft during the same
2
2
period = iat =£ 11.6 x 2.71 =42.6 radn, thus
thus
total energy supplied during the period of clutch slip
= energy dissipated + kinetic energy
= 2267 + 2267=4534Nm.
