A rotating shaft experiences a steady torque T = 1360LN(1,0.05) lbf · in, and at a shoulder with a 1.1-in small diameter, a fatigue stress-concentration factor Kf =1.50LN(1,0.11), Kf s= 1.28LN(1,0.11), and at that location a bending moment of M = 1260LN(1,0.05) lbf · in.

Question

A rotating shaft experiences a steady torque $T = 1360 LN\left(1,0.05\right)$ lbf · in, and at a shoulder with a 1.1-in small diameter, a fatigue stress-concentration factor $K_f =1.50LN\left(1,0.11\right) \ , \ K_{f s}= 1.28 LN\left(1,0.11\right)$ , and at that location a bending mo ment of $M = 1260 LN\left(1,0.05\right)$ lbf · in.

The material of which the shaft is machined is hot-rolled 1035 with $S_{ut}= 86.2 LN\left(1,0.045\right)$ kpsi and $S_y= 56.0 LN\left(1,0.077\right)$ kpsi.

Estimate the reliability using a stochastic Gerber failure zone.

The material of which the shaft is machined is hot-rolled 1035 with [latex]S_{ut}= 86.2 LN\left(1,0.045\right)[/latex] kpsi and [latex]S_y= 56.0 LN\left(1,0.077\right) [/latex]kpsi.

Estimate the reliability using a stochastic Gerber failure zone.

Accepted Answer

Establish the endurance strength. From Eqs. (6–70) to (6–72)

[latex]\acute{S}_e =\begin{cases} 0.506\bar{S} _{ut} LN\left(1,0,138 ight) kpsi \ or MPa & \bar{S} _{ut}\leq 212 \ kpsi \left(1460 \ MPa ight) \107 LN\left(1,0,139 ight) \ kpsi & \bar{S} _{ut}\gt 212 \ kpsi \740 LN\left(1,0,139 ight) \ Mpa & \bar{S} _{ut}\gt 1460 \ MPa \ \end{cases} [/latex]

[latex]S_e=k_ak_bk_ck_dk_f\acute{S} _e[/latex]

[latex]k_a=a\bar{S} ^b_{ut} LN \left(1,C ight) \ \ \ \left(\bar{S}_{ut} \ in \ kpsi \ or MPa ight) [/latex]

and Eq. (6–20), p. 288

[latex]k_b=\begin{cases}\left(d/0.3 ight)^{-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 ight)^{-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} [/latex]

[latex]S´ _e=0.506\left(86..2 ight) LN\left(1,0.138 ight) =43.6 LN \left(1,0.138 ight) \k_a=2.67\left(86.2 ight)^{-0.265} LN\left(1,0.058 ight) =0.820 LN\left(1,0.058 ight) \k_b=\left({1.1}/{0.30} ight)^{-0.107} =0.870\k_c=k_d=k_f=LN\left(1,0 ight) \ \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 ight) 43.6=31.1 \ kpsi\C_{se}=\left(0.058^2+0.138^2 ight) ^{{1}/{2}}=0.150\ and \ so \ S_e=31.1 LN\left(1,0.150 ight) \ kpsi[/latex]

Stress (in kpsi):

[latex]\sigma _a=\frac{32K_fM_a}{\pi d^3} =\frac{32\left(1.50 ight) LN \left(1,0.11 ight)1.26 LN \left(1,0.05 ight) }{\pi \left(1.1 ight)^3 } \\bar{\sigma }_a=\frac{32\left(1.50 ight)1.26 }{\pi \left(1.1 ight)^3 } =14.5 kpsi \C_{\sigma a}=\left(0.11^2+0.05^2 ight) ^{{1}/{2}}=0.121\ au _m=\frac{16K_{fs}T_m}{\pi d^3} =\frac{16\left(1.28 ight)LN\left(1,0.11 ight) 1.36 LN\left(1,0.05 ight) }{\pi \left(1.1 ight)^3 } \\bar{ au } _m=\frac{16\left(1.28 ight)1.36 }{\pi \left(1.1 ight)^3 } =6.66 \ kpsi\C_{ au m}=\left(0.11^2+0.05^2 ight) ^{{1}/{2}}=0.121\\acute{\bar{\sigma_a }} =\left(\bar{\sigma }^2_a+3\bar{ au }^2_a ight) ^{{1}/{2}} =\left[14.5^2+3\left(0 ight)^2 ight] ^{{1}/{2}}=14.5 \ kpsi \\acute{\bar{\sigma }} _m=\left(\bar{\sigma }^2_m+3\bar{ au }^2_m ight)^{{1}/{}2} =\left[0+3\left(6.66 ight)^2 ight] ^{{1}/{2}}=11.54 \ kpsi\ r=\frac{\acute{\bar{\sigma}_a } }{\acute{\bar{\sigma }_m } } =\frac{14.5}{11.54} =1.26[/latex]

Strength: From Eqs. (6–80)

[latex]\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} } ight)^2 } ight] [/latex]

and (6–81),

[latex]C_{S a}=\frac{\left(1+C_{S u t} ight)^{2}}{1+C_{S e}} \frac{\left\{-1+\sqrt{1+\left[\frac{2 \bar{S}_{e}\left(1+C_{S e} ight)}{r \bar{S}_{u t}\left(1+C_{S u t} ight)} ight]^{2}} ight\}}{\left[-1+\sqrt{1+\left(\frac{2 \bar{S}_{e}}{r \bar{S}_{u t}} ight)^{2}} ight]}-1[/latex]

[latex]\bar{S} _a=\frac{1.26^286.2^2 }{2\left(31.1 ight) } \left\{-1+\sqrt{1+\left[\frac{2\left(31.1 ight) }{1.26\left(86.2 ight) } ight]^2 } ight\} =28.9 \ kpsi[/latex]

[latex]C_{Sa}=\frac{\left(1-0.045 ight)^2}{1+0.150} \frac{-1+\sqrt{1+\left[\frac{2\left(31.1 ight) \left(1+0.15 ight) }{1.26\left(86.2 ight) \left(1+0.045 ight) } ight]^2 }}{\left[-1+\sqrt{1+\left[\frac{2\left(31.1 ight) }{1.26\left(86.2 ight) } ight]^2 } ight] } -1=0.134[/latex]

Reliability: Since [latex]S_a=28.9 LN\left(1,0.134 ight) [/latex] kpsi and [latex]\acute{\sigma } _a= 14.5LN\left(1,0.121 ight) [/latex] kpsi,Eq. (5–43), p. 250, gives

[latex]z=-\frac{\mu _{\ln S}-\mu_{\ln \sigma } }{\left(\hat{\sigma }^2_{\ln S}+ \hat{\sigma }^2_{\ln \sigma } ight)^{{1}/{2}} } =-\frac{\ln \left(\frac{\mu _S}{\mu _\sigma }\sqrt{\frac{1+C^2_\sigma }{1+C^2_S} } ight) }{\sqrt{\ln \left[\left(1+C^2_S ight)\left(1+C^2_\sigma ight) ight] } } [/latex]

[latex]z=-\frac{\ln \left(\frac{\bar{S_a} }{\bar{\sigma } }\sqrt{\frac{1+C^2_\sigma }{1+C^2_S} } ight) }{\sqrt{\ln \left[\left(1+C^2_S ight)\left(1+C^2_\sigma ight) ight] } } \=-\frac{\ln \left(\frac{28.9}{14.5}\sqrt{\frac{1+0.121^2}{1+0.134^2} } ight) }{\sqrt{\ln \left[\left(1+0.134^2 ight)\left(1+0.121^2 ight) ight] } } =-3.83[/latex]

From Table A–10

Table A–10
Cumulative Distribution Function of Normal (Gaussian) Distribution
[latex] \begin{aligned} \Phi\left(z_{\alpha} ight) &=\int_{-\infty}^{z_{\alpha}} \frac{1}{\sqrt{2 \pi}} \exp \left(-\frac{u^{2}}{2} ight) d u \ &= \begin{cases}\alpha & z_{\alpha} \leq 0 \ 1-\alpha & z_{\alpha}>0\end{cases} \end{aligned} [/latex]
[latex]Z_a[/latex]	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 [latex]p_{f} = 0.000 065,[/latex] and the reliability is,

against fatigue,

[latex]R=1-P_f=1-0.000065=0.999935[/latex]

The chance of first-cycle yielding is estimated by interfering[latex] S_y \ with \acute{\sigma } _{max}[/latex]. The quantity [latex] \acute{\sigma } _{max}[/latex] is formed from [latex] \acute{\sigma } _{a}+ \acute{\sigma } _{m}[/latex].

The mean of [latex] \acute{\sigma } _{max}[/latex] is [latex]\bar{\acute{\sigma} } _a+\bar{\acute{\sigma} } _m = 14.5 + 11.54 = 26.04[/latex] kpsi.

The coefficient of variation of the sum is 0.121, since both COVs are 0.121, thus [latex] C_{\sigma \ max} =0.121[/latex].

We interfere [latex]S_y= 56LN\left(1,0.077 ight) [/latex] kpsi with [latex]\acute{\sigma}_{max}=26.04 LN\left(1,0.121 ight) [/latex] kpsi. The corresponding z variable is

[latex]z=-\frac{\ln \left(\frac{56}{26.4}\sqrt{\frac{1+0.121^2}{1+0.134^2} } ight) }{\sqrt{\ln \left[\left(1+0.077^2 ight)\left(1+0.121^2 ight) ight] } } =-5.39[/latex]

which represents, from Table A–10, a probability of failure of approximately 0.07358 [latex][ which represents 3.58(10^{−8})][/latex] 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 [latex]\acute{\sigma}_{a}S_a[/latex] interference and the [latex]\acute{\sigma}_{max}S_y[/latex] 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:

[latex]p_f = p\left(yield ight)+ p\left(fatigue ight) − p\left(yield \ and \ fatigue ight) \= p\left(yield ight) + p\left(fatigue ight) − p\left(yield ight) p\left(fatigue ight)\= 0.358\left(10^{-7} ight) +0.65\left(10^{-4} ight) -0.358\left(10^{-7} ight)0.65\left(10^{-4} ight) = 0.650\left(10^{-4} ight) [/latex]

[latex]R=1- 0.650\left(10^{-4} ight) =0.999935[/latex]

against either or both modes of failure.

Question 2.6.20: A rotating shaft experiences a steady torque T = 1360LN(1,0....

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