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The given problem is an application of design of shaft based oncombined bending moment and twisting moment. Equivalent twistingmoment is calculated and based on that diameter of shaft iscalculated. Please look into the images for solution in deQ) Given data" power "P=75kW=75 xx10^(3)Wspeed in R.P.M (N)=500 r.p.mAt pulley C,Q_(1)=2Q_(2) (both vertical)At pulley D,P_(I)=2P_(2) (both horizontal)Redius of pulley C,r_(c)=220mm=0.22mRadices of pulley D,gamma_(D)=160mm=0.16mAllowable shear stress (tau_(max))=45Mpa A and B are bearing.Step -(Figure - 1)we have the-formula power P=T xx((2pi N)/(60))=>" Torque "T=(60 xx P)/(2xx pi xx N)=(60 xx75 xx10^(3))/(2xx pi xx500)=>quad T=1432.4Nm". "T.At pulley D,T=(P_(1)-P_(2))xxgamma_(D)=(2P_(2)-P_(2))xxgamma_(D)=>T=P ... See the full answer