Question 6.32: Solve Prob. 6–1 if the ultimate strength of production piece......
Solve Prob. 6–1 if the ultimate strength of production pieces is found to be {S}_{u t}=245{LN} (1, 0.0508) kpsi.
Given {S}_{u t}=245{LN}(1,0.0508)\,\mathrm{kpsi}
From Table 7-13: a = 1.34, b = −0.086, C = 0.12
{ k}_{a}=1.34{\bar{S}}_{u t}^{-0.086}{{L N}}(1,0.120) =1.34(245)^{-0.086}{LN}(1,0.12) =0.835{{L N}}(1,0.12)
k_{b}=1.02 (as in Prob. 6-1)
Eq. (6-70) {S}_{e}=0.835(1.02){L}{N}(1,0.12)[107{L}{N}(1,0.139)]
{S}_{e}^{\prime}=\left\{\begin{array}{l l}{{0.506\bar{S}_{u t}{L}{N}(1,0.138)~\mathrm{kpsi~or~MPa}}}&{{\bar{S}_{u t}\leq212~\mathrm{kpsi~(1460~MPa)}}}\\ {{107{LN}{(1,0.139)~kpsi}}}&{{\bar{S}_{u t}\gt 212~\mathrm{kpsi}}}\\ {{{7}40{LN}{(1,0.139)~MPa}}}&{{\bar{S}_{u t}\gt 1460~\mathrm{MPa}}}\end{array}\right. (6-70)
\bar{S}_{e}=0.835(1.02)(107)=91.1\,{\mathrm{kpsi}}Now
C_{S e}\doteq(0.12^{2}+0.139^{2})^{1/2}=0.184{S}_{e}=91.1{L}{N}(1,0.184)\,\mathrm{kpsi}
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