... bearings.J 11 ¼ðp0sin2y 1 þ e cos yÞ3dy ¼p2 1 Àe2Þ3=2ð7 -13 aÞJ 12 ¼ðp0sin ycos y 1 þ e cos yÞ3dy ¼À2e 1 Àe2Þ2ð7 -13 bÞJ22¼ðp0cos2y 1 þ e cos yÞ3dy ¼p 1 þ2e2Þ2 1 Àe2Þ5=2ð7 -13 cÞThe ... yieldf ðeÞ¼4WC2mUL3¼4W 10 À3 D=2Þ2mpnDðD=4Þ3¼4  16  4900  10 À60: 018 5p  63:3ð0:075Þ2¼ 15 :15 According to the curve of f ðeÞ vs. e, for f ðeÞ 15 :15 , the eccentricity of thebearing ... ¼mnPRC2¼DL2 1 Àe2Þ2pe½p2 1 Àe2Þ 16 e2 1= 2From the right-hand side of the equation,S ¼ð2Þ2 1 À 0:82Þ2p0:8½p2 1 À0:82Þ 16  0:82 1= 2¼0:362p  0:8  3: 71 ¼ 0: 013 9b. Minimum...