Establish the endurance strength. From Eqs. (6–70) to (6–72)
\acute{S}_e =\begin{cases} 0.506\bar{S} _{ut} LN\left(1,0,138\right) kpsi \ or MPa & \bar{S} _{ut}\leq 212 \ kpsi \left(1460 \ MPa\right) \\107 LN\left(1,0,139\right) \ kpsi & \bar{S} _{ut}\gt 212 \ kpsi \\740 LN\left(1,0,139\right) \ Mpa & \bar{S} _{ut}\gt 1460 \ MPa \ \end{cases}
S_e=k_ak_bk_ck_dk_f\acute{S} _e
k_a=a\bar{S} ^b_{ut} LN \left(1,C\right) \ \ \ \left(\bar{S}_{ut} \ in \ kpsi \ or MPa \right)
and Eq. (6–20), p. 288
k_b=\begin{cases}\left(d/0.3\right)^{-0.107}=0.879d^{-0.107} & 0.11\leq d\leq 2 \ in \\0.91d^{-0.157} & 2\leq d\leq 10 \ in \\\left(d/7.62\right)^{-0.107}=1.24d^{-0.107}& 2.79\leq d\leq 51 \ mm \\1.51d^{-0.157} & 51\leq d\leq 254 \ mm\end{cases}
S´ _e=0.506\left(86..2\right) LN\left(1,0.138\right) =43.6 LN \left(1,0.138\right) \\k_a=2.67\left(86.2\right)^{-0.265} LN\left(1,0.058\right) =0.820 LN\left(1,0.058\right) \\k_b=\left({1.1}/{0.30}\right)^{-0.107} =0.870\\k_c=k_d=k_f=LN\left(1,0\right) \\ \mathbf{S}_{e}=0.820 \mathrm{LN}(1,0.058) 0.870(43.6) \mathbf{L N}(1,0.138) \\ \bar{S} _e=0.820\left(0.870\right) 43.6=31.1 \ kpsi\\C_{se}=\left(0.058^2+0.138^2\right) ^{{1}/{2}}=0.150\\ and \ so \ S_e=31.1 LN\left(1,0.150\right) \ kpsi
Stress (in kpsi):
\sigma _a=\frac{32K_fM_a}{\pi d^3} =\frac{32\left(1.50\right) LN \left(1,0.11\right)1.26 LN \left(1,0.05\right) }{\pi \left(1.1\right)^3 } \\\bar{\sigma }_a=\frac{32\left(1.50\right)1.26 }{\pi \left(1.1\right)^3 } =14.5 kpsi \\C_{\sigma a}=\left(0.11^2+0.05^2\right) ^{{1}/{2}}=0.121\\\tau _m=\frac{16K_{fs}T_m}{\pi d^3} =\frac{16\left(1.28\right)LN\left(1,0.11\right) 1.36 LN\left(1,0.05\right) }{\pi \left(1.1\right)^3 } \\\bar{\tau } _m=\frac{16\left(1.28\right)1.36 }{\pi \left(1.1\right)^3 } =6.66 \ kpsi\\C_{\tau m}=\left(0.11^2+0.05^2\right) ^{{1}/{2}}=0.121\\\acute{\bar{\sigma_a }} =\left(\bar{\sigma }^2_a+3\bar{\tau }^2_a \right) ^{{1}/{2}} =\left[14.5^2+3\left(0\right)^2 \right] ^{{1}/{2}}=14.5 \ kpsi \\\acute{\bar{\sigma }} _m=\left(\bar{\sigma }^2_m+3\bar{\tau }^2_m \right)^{{1}/{}2} =\left[0+3\left(6.66\right)^2 \right] ^{{1}/{2}}=11.54 \ kpsi\\ r=\frac{\acute{\bar{\sigma}_a } }{\acute{\bar{\sigma }_m } } =\frac{14.5}{11.54} =1.26
Strength: From Eqs. (6–80)
\bar{S} _a=\frac{r^2\bar{S}^2_{ut} }{2\bar{S}_e } \left[-1+\sqrt{1+\left(\frac{2\bar{S}_e }{r\bar{S}_{ut} } \right)^2 } \right]
and (6–81),
C_{S a}=\frac{\left(1+C_{S u t}\right)^{2}}{1+C_{S e}} \frac{\left\{-1+\sqrt{1+\left[\frac{2 \bar{S}_{e}\left(1+C_{S e}\right)}{r \bar{S}_{u t}\left(1+C_{S u t}\right)}\right]^{2}}\right\}}{\left[-1+\sqrt{1+\left(\frac{2 \bar{S}_{e}}{r \bar{S}_{u t}}\right)^{2}}\right]}-1
\bar{S} _a=\frac{1.26^286.2^2 }{2\left(31.1\right) } \left\{-1+\sqrt{1+\left[\frac{2\left(31.1\right) }{1.26\left(86.2\right) } \right]^2 } \right\} =28.9 \ kpsi
C_{Sa}=\frac{\left(1-0.045\right)^2}{1+0.150} \frac{-1+\sqrt{1+\left[\frac{2\left(31.1\right) \left(1+0.15\right) }{1.26\left(86.2\right) \left(1+0.045\right) } \right]^2 }}{\left[-1+\sqrt{1+\left[\frac{2\left(31.1\right) }{1.26\left(86.2\right) } \right]^2 } \right] } -1=0.134
Reliability: Since S_a=28.9 LN\left(1,0.134\right) kpsi and \acute{\sigma } _a= 14.5LN\left(1,0.121\right) kpsi,Eq. (5–43), p. 250, gives
z=-\frac{\mu _{\ln S}-\mu_{\ln \sigma } }{\left(\hat{\sigma }^2_{\ln S}+ \hat{\sigma }^2_{\ln \sigma }\right)^{{1}/{2}} } =-\frac{\ln \left(\frac{\mu _S}{\mu _\sigma }\sqrt{\frac{1+C^2_\sigma }{1+C^2_S} } \right) }{\sqrt{\ln \left[\left(1+C^2_S\right)\left(1+C^2_\sigma \right) \right] } }
z=-\frac{\ln \left(\frac{\bar{S_a} }{\bar{\sigma } }\sqrt{\frac{1+C^2_\sigma }{1+C^2_S} } \right) }{\sqrt{\ln \left[\left(1+C^2_S\right)\left(1+C^2_\sigma \right) \right] } } \\=-\frac{\ln \left(\frac{28.9}{14.5}\sqrt{\frac{1+0.121^2}{1+0.134^2} } \right) }{\sqrt{\ln \left[\left(1+0.134^2\right)\left(1+0.121^2\right) \right] } } =-3.83
From Table A–10
| Table A–10 |
| Cumulative Distribution Function of Normal (Gaussian) Distribution |
| \begin{aligned} \Phi\left(z_{\alpha}\right) &=\int_{-\infty}^{z_{\alpha}} \frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{u^{2}}{2}\right) d u \\ &= \begin{cases}\alpha & z_{\alpha} \leq 0 \\ 1-\alpha & z_{\alpha}>0\end{cases} \end{aligned} |
 |
| Z_a |
0 |
0.01 |
0.02 |
0.03 |
0.04 |
0.05 |
0.06 |
0.07 |
0.08 |
0.09 |
| 0 |
0.5 |
0.496 |
0.492 |
0.488 |
0.484 |
0.4801 |
0.4761 |
0.4721 |
0.4681 |
0.4641 |
| 0.1 |
0.4602 |
0.4562 |
0.4522 |
0.4483 |
0.4443 |
0.4404 |
0.4364 |
0.4325 |
0.4286 |
0.4247 |
| 0.2 |
0.4207 |
0.4168 |
0.4129 |
0.409 |
0.4052 |
0.4013 |
0.3974 |
0.3936 |
0.3897 |
0.3859 |
| 0.3 |
0.3821 |
0.3783 |
0.3745 |
0.3707 |
0.3669 |
0.3632 |
0.3594 |
0.3557 |
0.352 |
0.3483 |
| 0.4 |
0.3446 |
0.3409 |
0.3372 |
0.3336 |
0.33 |
0.3264 |
0.3238 |
0.3192 |
0.3156 |
0.3121 |
| 0.5 |
0.3085 |
0.305 |
0.3015 |
0.2981 |
0.2946 |
0.2912 |
0.2877 |
0.2843 |
0.281 |
0.2776 |
| 0.6 |
0.2743 |
0.2709 |
0.2676 |
0.2643 |
0.2611 |
0.2578 |
0.2546 |
0.2514 |
0.2483 |
0.2451 |
| 0.7 |
0.242 |
0.2389 |
0.2358 |
0.2327 |
0.2296 |
0.2266 |
0.2236 |
0.2206 |
0.2177 |
0.2148 |
| 0.8 |
0.2119 |
0.209 |
0.2061 |
0.2033 |
0.2005 |
0.1977 |
0.1949 |
0.1922 |
0.1894 |
0.1867 |
| 0.9 |
0.1841 |
0.1814 |
0.1788 |
0.1762 |
0.1736 |
0.1711 |
0.1685 |
0.166 |
0.1635 |
0.1611 |
| 1 |
0.1587 |
0.1562 |
0.1539 |
0.1515 |
0.1492 |
0.1469 |
0.1446 |
0.1423 |
0.1401 |
0.1379 |
| 1.1 |
0.1357 |
0.1335 |
0.1314 |
0.1292 |
0.1271 |
0.1251 |
0.123 |
0.121 |
0.119 |
0.117 |
| 1.2 |
0.1151 |
0.1131 |
0.1112 |
0.1093 |
0.1075 |
0.1056 |
0.1038 |
0.102 |
0.1003 |
0.0985 |
| 1.3 |
0.0968 |
0.0951 |
0.0934 |
0.0918 |
0.0901 |
0.0885 |
0.0869 |
0.0853 |
0.0838 |
0.0823 |
| 1.4 |
0.0808 |
0.0793 |
0.0778 |
0.0764 |
0.0749 |
0.0735 |
0.0721 |
0.0708 |
0.0694 |
0.0681 |
| 1.5 |
0.0668 |
0.0655 |
0.0643 |
0.063 |
0.0618 |
0.0606 |
0.0594 |
0.0582 |
0.0571 |
0.0559 |
| 1.6 |
0.0548 |
0.0537 |
0.0526 |
0.0516 |
0.0505 |
0.0495 |
0.0485 |
0.0475 |
0.0465 |
0.0455 |
| 1.7 |
0.0446 |
0.0436 |
0.0427 |
0.0418 |
0.0409 |
0.0401 |
0.0392 |
0.0384 |
0.0375 |
0.0367 |
| 1.8 |
0.0359 |
0.0351 |
0.0344 |
0.0336 |
0.0329 |
0.0322 |
0.0314 |
0.0307 |
0.0301 |
0.0294 |
| 1.9 |
0.0287 |
0.0281 |
0.0274 |
0.0268 |
0.0262 |
0.0256 |
0.025 |
0.0244 |
0.0239 |
0.0233 |
| 2 |
0.0228 |
0.0222 |
0.0217 |
0.0212 |
0.0207 |
0.0202 |
0.0197 |
0.0192 |
0.0188 |
0.0183 |
| 2.1 |
0.0179 |
0.0174 |
0.017 |
0.0166 |
0.0162 |
0.0158 |
0.0154 |
0.015 |
0.0146 |
0.0143 |
| 2.2 |
0.0139 |
0.0136 |
0.0132 |
0.0129 |
0.0125 |
0.0122 |
0.0119 |
0.0116 |
0.0113 |
0.011 |
| 2.3 |
0.0107 |
0.0104 |
0.0102 |
0.0099 |
0.00964 |
0.00939 |
0.00914 |
0.00889 |
0.00866 |
0.00842 |
| 2.4 |
0.0082 |
0.00798 |
0.00776 |
0.00755 |
0.00734 |
0.00714 |
0.00695 |
0.00676 |
0.00657 |
0.00639 |
| 2.5 |
0.00621 |
0.00604 |
0.00587 |
0.0057 |
0.00554 |
0.00539 |
0.00523 |
0.00508 |
0.00494 |
0.0048 |
| 2.6 |
0.00466 |
0.00453 |
0.0044 |
0.00427 |
0.00415 |
0.00402 |
0.00391 |
0.00379 |
0.00368 |
0.00357 |
| 2.7 |
0.00347 |
0.00336 |
0.00326 |
0.00317 |
0.00307 |
0.00298 |
0.00289 |
0.0028 |
0.00272 |
0.00264 |
the probability of failure is p_{f} = 0.000 065, and the reliability is,
against fatigue,
R=1-P_f=1-0.000065=0.999935
The chance of first-cycle yielding is estimated by interfering S_y \ with \acute{\sigma } _{max}. The quantity \acute{\sigma } _{max} is formed from \acute{\sigma } _{a}+ \acute{\sigma } _{m}.
The mean of \acute{\sigma } _{max} is \bar{\acute{\sigma} } _a+\bar{\acute{\sigma} } _m = 14.5 + 11.54 = 26.04 kpsi.
The coefficient of variation of the sum is 0.121, since both COVs are 0.121, thus C_{\sigma \ max} =0.121.
We interfere S_y= 56LN\left(1,0.077\right) kpsi with \acute{\sigma}_{max}=26.04 LN\left(1,0.121\right) kpsi. The corresponding z variable is
z=-\frac{\ln \left(\frac{56}{26.4}\sqrt{\frac{1+0.121^2}{1+0.134^2} } \right) }{\sqrt{\ln \left[\left(1+0.077^2\right)\left(1+0.121^2\right) \right] } } =-5.39
which represents, from Table A–10, a probability of failure of approximately 0.07358 [ which represents 3.58(10^{−8})] of first-cycle yield in the fillet.
The probability of observing a fatigue failure exceeds the probability of a yield failure, something a deterministic analysis does not foresee and in fact could lead one to expect a yield failure should a failure occur.
Look at the \acute{\sigma}_{a}S_a interference and the \acute{\sigma}_{max}S_y interference and examine the z expressions.
These control the relative probabilities. A deterministic analysis is oblivious to this and can mislead. Check your statistics text for events that are not mutually exclusive, but are independent, to quantifyn the probability of failure:
p_f = p\left(yield\right)+ p\left(fatigue\right) − p\left(yield \ and \ fatigue\right) \\= p\left(yield\right) + p\left(fatigue\right) − p\left(yield\right) p\left(fatigue\right)\\= 0.358\left(10^{-7}\right) +0.65\left(10^{-4}\right) -0.358\left(10^{-7}\right)0.65\left(10^{-4}\right) = 0.650\left(10^{-4}\right)
R=1- 0.650\left(10^{-4}\right) =0.999935
against either or both modes of failure.