Shear stresS =S6HPQ=tau(6a) Axial Normal stres.s = 70 HPa Bending stress =+-146HPaVonmises stress{:[" Alternating stress "=sqrt(6_(a)^(2)+3tau^(2))=140HPa],[" Meanstress "=sqrt(6m_(2)^(2)+3tau_(m)^(2))],[quadsigma_(m)=sqrt((140)^(2)+3(56)^(2))=119.62HPa.]:}Endurance limit modytying fectorSurtate factor a(" sut ")^(b)=4.51(5.25)^(-0.265)=0.858" size tactor "=(0.858-0.000837(63.5))=0.806Cocid fuctor 1{:[S_(e)= ... See the full answer