A pair of straight teeth spur gears, 200 involute full depth teeth is to transmit 10kW at 200rpm of the pinion. The velocity ratio is 3. The module value is 6mm and the face width is equal to 10 times module value. The number of teeth of pinion is 30 and the material combination factor K is 1.32 N/mm2. The service factor is 0.8 and the deformation factor C =80 N/mm. a) Determine the dynamic load for the pinion ( enter the value in kN) Answer kN b) Determine the wear load for the pinion load ( enter the value in kN) Answer kN c) Is the design satisfactory?. i) Yes, the wear load for the pinion is greater than the dynamic load. ii) Yes, the wear load for the pinion is less than the dynamic load. iii) Yes, the wear load for the pinion is equal to dynamic load.

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Solution - Given data arePower transmitted P=10kW=10000WSpeed of pinion in apm N_(p)=2008pmVelocity ratio VR=3module m=6mmFace width b=10 m=10 xx6=60mmNumber of teeth of pirion T_(p)=30Material combination factor K=1.32N//mm^(2)service factor C_(S)=0.8Deformation factor C=80N//mmcacculation for the dynamic load for the pinjonTangential load W_(T)=(P xxC_(S))/(v)- earation (i)here V=(piD_(p)Np)/(60)=(pi mpi_(p)N_(p))/(60)V=(pi xx6xx30 xx200)/(60)v=1884mm//sv=(1884)/(100)m//sv=1.884m//sso Tangential load W_(T)=(10000 xx0.8)/(1.884){:[W_(T)=(80.0)/(1.884)],[W_(T)=4246.28N]:}Dynamic load for the pinion is given by,{:[W_(D)=W_(T)+(21 V(bc+W_(T)))/(21 V+sqrt(bc+W_(T)))],[W_(D)=4246.28+(21 xx1.884 xx(60 xx80+4246^(@)))/(21 xx1.884+sqrt(60 xx80)+4246.28)]:}{:[W_(D)=4246.28+(39.564 xx(4800+4246.28))/(39.564+sqrt(4800+4246.28))],[W_(D)=4246.28+(39.564 xx9046.28)/(39.564+sqrt9046.28)],[W_(D)= ... See the full answer