L=0.6m,Fa=2kN,n=1.5,N=104cycles,Sut=770MPa,Sy=420MPa
First evaluate the fatigue strength.
Se′=0.5(770)=385MPa
Se′=0.5(770)=385MPa
ince the size is not yet known, assume a typical value of kb=0.85 and check later. All other modifiers are equal to one.
Se=
=0.488(0.85)(385)=160MPa
Sut=770/6.89=112kpsi
f=0.83
a=Se(fSut)2=160[0.83(770)]2=2553MPa
b=−31log(SefSut)=−31log(1600.83(770))=−0.2005
Sf=aNb=2553(104)−0.2005=403MPa
Now evaluate the stress.
Mmax=(2000N)(0.6m)=1200N⋅m
σa=σmax=IMc=B36M=b37200Pa
σmax=0.03037200×10−6=267Mpa
ny=σmaxSy=267420=1.57
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