Question

A solid round bar with a diameter of 2.32 in has a groove cut to a diameter of 2.09 in, with a radius of 0.117 in. The bar is not rotating. The bar is loaded with a repeated bending load that causes the bending moment at the groove to fluctuate between 0 and 25,000 lbf·in. The bar is hot-rolled AISI 1095, but the groove has been machined. Determine the factor of safety for fatigue based on infinite life using the modified Goodman criterion and the factor of safety for yielding.

Answer 1

Answer:

nf=1.11 (Goodman criterion)

ny=2.41 (factor of safety for fatigue)

Explanation:

From the table A-20 Deterministic ASTM minimum tensile and yield strengths for HR and CD steels, we have approximately:

Sut=120 kpsi

Sy=66 kpsi

Due Sut<1400 Mpa, the endurance limit is:

$Se=0.5Sut=0.5*120=60 kpsi$

The surface condition modification factor is:

$ka=a(Sut)^{b}=2.7(120)^{-0.265} =0.759$

The effective diameter is:

$de=0.37d=0.37*2.09=0.7733 in$

The size factor is:

$kb=0.879de^{-0.107} =0.879(0.7733)^{-0.107} =0.9$

The endurance limit at critical location is:

$See=ka*kb*kc*kd*kd*ke*kf*Se=0.759*0.9*1*60=40.986 kpsi$

$\frac{D}{d}=\frac{2.32}{2.09} =1.11\\\frac{r}{d}=\frac{0.117}{2.09} =0.056$

From Figure A-15-15 chart, the Kf = 2.1

The notch sensitivity is:

$\sqrt{a}=0.246-3.08(10^{-3})Sut+1.51(10^{-5})Sut^{2}-2.67(10^{-8})Sut^{3}$

$\sqrt{a}=0.246-3.08(10^{-3})(120)+1.51(10^{-5})(120^{2})-2.67(10^{-8})(120^{3})=0.048$

$q=\frac{1}{1+\frac{\sqrt{a} }{\sqrt{r} } }=\frac{1}{1+\frac{0.048}{\sqrt{0.117} } }=0.877$

The fatigue stress is:

$Kf=1+q(Kt-1)=1+0.877(2.1-1)=1.96$

The moment of inertia is:

$I=\frac{\pi }{64} d^{4}=\frac{\pi }{64} (2.09^{4})=0.936 in^{4}$

The maximum stress is:

$omax=\frac{M*c}{I}=\frac{25000*\frac{2.09}{2} }{0.936} =27911.32 psi=27.911 kpsi$

The mean stress is:

$om=Kf\frac{omax+omin}{2} =1.96\frac{27.911+0}{2}=27.35 kpsi$

The alternate stress is:

$oa=Kf|\frac{omax-omin}{2}|=27.35 kpsi$

The fatigue factor using Goodman is:

$\frac{1}{nf}=\frac{oa}{See}+\frac{om}{Sut}\\\frac{1}{nf}=\frac{27.35}{40.986}+\frac{27.35}{120}$

nf=1.11

the factor of safety is:

$ny=\frac{Sy}{omax}=\frac{66}{27.35} =2.41$