Question 6.39: A 1/4 by 1 1/2 -in steel bar has a 3/4 -in drilled hole loca......
A {\frac{1}{4}} by 1{\frac{1}{2}} -in steel bar has a {\frac{3}{4}} -in drilled hole located in the center, much as is shown in Table A–15–1. The bar is subjected to a completely reversed axial load with a deterministic load of 1200 lbf. The material has a mean ultimate tensile strength of {\bar{S}}_{u t}=80\,\mathrm{kpsi}.
(a) Estimate the reliability.
(b) Conduct a computer simulation to confirm your answer to part a.
S_{u t}=80\,\mathrm{kpsi}
(a) Strength
{k}_{a}=2.67(80)^{-0.265}{LN}(1,0.058)=0.836{LN}(1,0.058)k_{b}=1
{k}_{c}=1.23(80)^{-0.078}{LN}(1,0.125)=0.874{LN}(1,0.125)
{S}_{a}^{\prime}=0.506(80){L}{N}(1,0.138)=40.5{LN}(1,0.138)\,kpsi
{S}_{e}=0.836[{LN}(1,0.058)](1)[(0.874{L}{N}(1,0.125)][40.5{L}{N}(1,0.138)]
\bar{S}_{e}=0.836(1)(0.874)(40.5)=29.6\,\mathrm{kpsi}
C_{S e}=(0.058^{2}+0.125^{2}+0.138^{2})^{1/2}=0.195
Stress: Fig. A-15-1; d/w=0.75/1.5=0.5,\,\,K_{t}=2.17. From Eqs. (6-78), (6-79) and Table 6-15
\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{2.17{LN}(1,0.10)}{1+(2/{\sqrt{0.375}})[(2.17-1)/2.17](5/80)}}=1.95{LN}(1,0.10)\sigma_{a}=\frac{{K}_{f}F_{a}}{(w-d)t},\quad C_{\sigma}=0.10
\bar{\sigma}_{a}=\frac{K_{f}F_{a}}{(\omega-d)t}=\frac{1.95(1.2)}{(\ 1.5-0.75)(0.25)}=12.48\;\mathrm{kpsi}\;
S_{a}=S_{e}=29.6\,\mathrm{kpsi}
z=-\frac{\ln\left(\bar{S}_{a}/\bar{\sigma}_{a}\right)\sqrt{\left(1+C_{\sigma}^{2}\right)/\left(1+C_{S}^{2}\right)}}{\sqrt{\ln\left(1+C_{\sigma}^{2}\right)\left(1+C_{S}^{2}\right)}} =-{\frac{\ln[(29.6/12.48)\sqrt{(1+0.10^{2})/(1+0.195^{2})}]}{\sqrt{\ln(1+0.10^{2})(1+0.195^{2})}}}=-3.9
From Table A-20
p_{f}=4.481(10^{-5})R=1-4.481(10^{-5})=0.999\,955
(b) All computer programs will differ in detail.
| 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 |