Question 6.34: A solid round steel bar is machined to a diameter of 1.25 in......
A solid round steel bar is machined to a diameter of 1.25 in. A groove {\frac{1}{8}} in deep with a radius of {\frac{1}{8}} in is cut into the bar. The material has a mean tensile strength of 110 kpsi. A completely reversed bending moment M = 1400 lbf · in is applied. Estimate the reliability. The size factor should be based on the gross diameter. The bar rotates.
Rotation is presumed. M and S_{u t} are given as deterministic, but notice that σ is not; therefore, a reliability estimation can be made.
From Eq. (6-70):
\mathrm{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)
\mathrm{S}_{e}^{\prime}=0.506(110){LN}(1,0.138)=55.7{LN}(1,0.138)\,kpsiTable 6-10:
{k}_{a}=2.67(110)^{-0.265}{LN}(1,0.058)=0.768{LN}(1,0.058)Based on d = 1 in, Eq. (6-20) gives
k_{b}={\left\{\begin{array}{l l}{(d/0.3)^{-0.107}=0.879d^{-0.107}}&{0.11\leq d\leq2\operatorname*{in}}\\ {0.91d^{-0.157}}&{2\lt d\leq10\ in}\\ {(d/7.62)^{-0.107}=1.24d^{-0.107}}&{2.79\leq d\leq51\ \operatorname*{mm}}\\ {1.51d^{-0.157}}&{51\lt d\leq254\operatorname*{mm}}\end{array}\right.} (6-20)
k_{b}=\left({\frac{1}{0.30}}\right)^{-0.107}=0.879Conservatism is not necessary
{S}_{e}=0.768[{LN}(1,0.058)](0.879)(55.7)[{LN}(1,0.138)]{\bar{S}}_{e}=37.6\,{\mathrm{kpsi}}
C_{S e}=(0.058^{2}+0.138^{2})^{1/2}=0.150
{S}_{e}=37.6{LN}(1,0.150)
Fig. A-15-14: D/d=1.25,\;r/d=0.125.\;\mathrm{Thus}\;\;K_{t}=1.70 and Eqs. (6-78), (6-79) and Table 6-15 give
\bar{K}_{f}=\frac{K_{t}}{1+\frac{2(K_{t}-1)}{K_{t}}\frac{\sqrt{a}}{\sqrt{r}}} (6-78)
{K}_{f}=\bar{K}_{f}{LN}\left(1,C_{K_{f}}\right) (6-79)
{K}_{f}=\frac{1.70{LN}(1,0.15)}{1+(2/\sqrt{0.125})[(1.70-1)/(1.70)](3/110)}=1.598{LN}(1,0.15)\sigma={K}_{f}\frac{32M}{\pi d^{3}}=1.598[{LN}(1-0.15)]\left[\frac{32(1400)}{\pi(1)^{3}}\right]=22.8{LN}(1,0.15)\,kpsi
From Eq. (5-43), p. 242:
z=-\frac{\mu_{\mathrm{ln}S}-\mu_{\ln\sigma}}{\left(\hat{\sigma}_{\ln S}^{2}+\hat{\sigma}_{\ln\sigma}^{2}\right)^{1/2}}=-\frac{\ln\left(\frac{\mu_{S}}{\mu_{\sigma}}\sqrt{\frac{1+C_{\sigma}^{2}}{1+C_{s}^{2}}}\right)}{\sqrt{\ln\left[\left(1+C_{s}^{2}\right)\left(1+C_{\sigma}^{2}\right)\right]}} (5-43)
z=-\frac{\ln[(37.6/22.8)\sqrt{(1+0.15^{2})/(1+0.15^{2})}]}{\sqrt{\ln[(1+0.15^{2})(1+0.15^{2})]}}=-2.37From Table A-10, p_{f}=0.008\,89
\therefore R=1-0.008\,89=0.991Note: The correlation method uses only the mean of {S}_{u t}; its variability is already included in the 0.138. When a deterministic load, in this case M, is used in a reliability estimate, engineers state, “For a Design Load of M, the reliability is 0.991.” They are in fact referring to a Deterministic Design Load.
| Table 6–10 Parameters in Marin Surface Condition Factor |
Surface Finish | \mathrm{K}_{a}=a S_{U t}^{b}\;\mathrm{LN}(\;1,\;C) | |||
| a | b | Coefficient Variation | |||
| kpsi | Mpa | ||||
| {\mathrm{Ground}}^{*} | 1.34 | 1.58 | -0.086 | 0.120 | |
| Machined or Cold-rolled | 2.67 | 4.45 | -0.265 | 0.058 | |
| Hot-rolled | 14.5 | 58.1 | -0.719 | 0.110 | |
| As-forged | 39.8 | 271 | -0.995 | 0.145 | |
| *Due to the wide scatter in ground surface data, an alternate function is {k_{a}}=0.878{{LN}}(1,0.120). Note: S_{{Ut}} in kpsi or MPa. | |||||
| Table 6–15 Heywood’s Parameter \sqrt{a} and coefficients of variation {{C}}_{K f} for steels |
Notch Type |
{\sqrt{\alpha}}({\sqrt{\mathrm{in}}}), S_{Ut} in kpsi |
{\sqrt{\alpha}}({\sqrt{\mathrm{mm}}}),
S_{Ut} in Mpa |
Coefficient of Variation C_{KF} |
| Transverse hole | 5/{{S}}_{Ut} | 174/{{S}}_{Ut} | 0.10 | |
| Shoulder | 4/{{S}}_{Ut} | 139/{{S}}_{U t} | 0.11 | |
| Groove | 3/{{S}}_{Ut} | 104/{{S}}_{Ut} | 0.15 |