Question #314425

A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of ±2 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a factor of safety of 1.5 at 100°C, what dimension should the square cross section have? Neglect any stress concentrations at the support end.




1
Expert's answer
2022-03-22T23:36:03-0400

L=0.6m,Fa=2kN,n=1.5,N=104cycles,Sut=770MPa,Sy=420MPaL=0.6 m,F a ​ =2 kN,n=1.5,N=10 4 cycles, S ut ​ =770 MPa,S y ​ =420 MPa

First evaluate the fatigue strength.

Se′​=0.5(770)=385MPaS e ′ ​ =0.5(770)=385 MPa

Se′​=0.5(770)=385MPaS e ′ ​ =0.5(770)=385 MPa

ince the size is not yet known, assume a typical value of kb=0.85k_{b}​ = 0.85 and check later. All other modifiers are equal to one.


Se=S e ​ =

=0.488(0.85)(385)=160MPa=0.488(0.85)(385)=160 MPa

Sut=770/6.89=112kpsiS ut ​ =770/6.89=112 kpsi


f=0.83f = 0.83

a=(fSut)2Se=[0.83(770)]2160=2553MPaa=\frac{\left(f S_{u t}\right)^{2}}{S_{e}}=\frac{[0.83(770)]^{2}}{160}=2553 MPa

b=13log(fSutSe)=13log(0.83(770)160)=0.2005b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right)=-\frac{1}{3} \log \left(\frac{0.83(770)}{160}\right)=-0.2005

Sf=aNb=2553(104)0.2005=403MPaS_{f}=a N^{b}=2553\left(10^{4}\right)^{-0.2005}=403 MPa

Now evaluate the stress.

Mmax=(2000N)(0.6m)=1200NmM max ​ =(2000 N)(0.6 m)=1200 N⋅m


σa=σmax=McI=6MB3=7200b3Paσ_a =σ _{max}=\frac{Mc}{I} = \frac{6M}{B^3}= \frac{7200}{b^3} Pa


σmax=72000.0303×106=267Mpa\sigma_{max}= \frac{7200}{0.030^3}\times 10^-6= 267 Mpa


ny=Syσmax=420267=1.57n_y= \frac{S_y}{\sigma_{max}}=\frac{420}{267}=1.57


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