The bar shown in Fig. 6–37 is machined from a cold-rolled flat having an ultimate strength of $S_{ut}= LN\left(87.6,5.74\right)$ kpsi.

The axial load shown is completely reversed.The load amplitude is $F_a= LN\left(1000,120\right)$ lbf.
(a) Estimate the reliability.
(b) Reestimate the reliability when a rotating bending endurance test shows that $S'_e=LN\left(40,2\right)$ kps.

Question

The bar shown in Fig. 6–37 is machined from a cold-rolled flat having an ultimate strength of [latex]S_{ut}= LN\left(87.6,5.74\right)[/latex] kpsi.

The axial load shown is completely reversed.The load amplitude is [latex]F_a= LN\left(1000,120\right)[/latex] lbf.
(a) Estimate the reliability.
(b) Reestimate the reliability when a rotating bending endurance test shows that [latex]S'_e=LN\left(40,2\right)[/latex] kps.

Accepted Answer

a) From Eq. (6–70),
[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) \ \left(1400 \ 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]\acute{S}_e= 0.506\bar{S} _{ut} LN\left(1,0,138 ight) = 0.506\left(87.6 ight) LN\left(1,0.138 ight)= 44.3 LN\left(1,0,138 ight) \ kpsi[/latex]
From Eq. (6–72)
[latex]k_a=a\bar{S} ^b_{ut} LN \left(1,C ight) \ \ \ \left(\bar{S}_{ut} \ in \ kpsi \ or MPa ight) [/latex]

and Table 6–10,

Table 6–10 Parameters in Marin Surface Condition Factor
[latex]k_a= aS^b_{ut} LN\left(1, C ight) [/latex]
	a			Coefficient of Variation, C
Surface Finish	kpsi	MPa	b	Coefficient of Variation, C
Ground*	1.34	1.58	−0.086	0.12
Machined or Cold-rolled	2.67	4.45	−0.265	0.058
Hot-rolled	14.5	58.1	−0.719	0.11
As-forged	39.8	271	−0.995	0.145

[latex]k_a=2.67\bar{S} ^{0.265}_{ut} LN \left(1,0.0.58 ight)= 2.67\left(87.6 ight) ^{-0.265} LN \left(1,0.0.58 ight)\=0.816 LN\left(1,0.058 ight)\ k_b= 1 [/latex] (axial loading)

From Eq. (6–73),

[latex]\left(k_c ight) _{axial}=1.23\bar{S}^{0.0778}_{ut}LN\left(1,0.125 ight) [/latex]

[latex]k_c= 1.23\bar{S} ^{-0.0778}_{ut} LN\left(1,0.125 ight) = 1.23\left(87.6 ight)^{−0.0778}LN(1,0.125)= 0.869 LN\left(1,0.125 ight)\k_d= k_f = \left(1,0 ight)[/latex]

The endurance strength, from Eq. (6–71) is

[latex]S_e=k_ak_bk_ck_d\acute{S} _e\Se=0.816 LN \left(1,0.058 ight) LN\left(1,0.125 ight)\left(1 ight)\left(1 ight)44.3 LN\left(1,0.138 ight)[/latex]

The parameters of [latex]S_e[/latex] are
[latex]\bar{S}_e=0.816\left(0.869 ight) 44.3=31.4\C_{se}=\left(0.058^2+0.125^2+0.138^2 ight)^{{1}/{2}} =0.195[/latex]

so [latex]S_e= 31.4LN(1,0.195)[/latex] kpsi.
In computing the stress, the section at the hole governs. Using the terminology
of Table A–15–1 we find [latex]{d}/{w} = 0.50[/latex], therefore [latex]K_t.= 2.18[/latex]. From Table 6–15,

Table 6–15 Heywood’s Parameter √a and coefficients ofvariation [latex]C_{Kf}[/latex] for steels
	[latex]\sqrt{a} \left(\sqrt{in} ight) [/latex]	[latex]\sqrt{a} \left(\sqrt{mm} ight) [/latex]
Notch Type	[latex]S_{ut} \ in \ kpsi[/latex]	[latex]S_{ut} \ in \ MPa[/latex]	Coefficient of Variation CKf
Transverse hole	[latex]{5}/{S_{ut}}[/latex]	[latex]{174}/{S_{ut}}[/latex]	0.10
Shoulder	[latex]{4}/{S_{ut}}[/latex]	[latex]{139}/{S_{ut}}[/latex]	0.11
Groove	[latex]{3}/{S_{ut}}[/latex]	[latex]{104}/{S_{ut}}[/latex]	0.15

,
[latex]\sqrt{a} = {5}/{Sut}= {5}/{87.6} = 0.0571 \ and C_{kf} = 0.10[/latex].
From Eqs. (6–78) and (6–79) with[latex]r=0.375[/latex] in ,

[latex]K_f=\frac{K_t}{1+\frac{2\left(K_t-1 ight) }{K_t}\frac{\sqrt{a} }{\sqrt{r} } }LN\left(1,C_{K_f} ight) =\frac{2.18}{1+\frac{2\left(2.18-1 ight) }{2.18}\frac{0.0571}{\sqrt{0.375} } } LN\left(1,0.10 ight)\=1.98 LN\left(1,0.10 ight) [/latex]

The stress at the hole is

[latex]\sigma =K_f\frac{F}{A} =1.98 LN\left(1,0.10 ight) \frac{1000 LN\left(1,0.12 ight) }{0.25\left(0.75 ight) } [/latex]

[latex]\bar{\sigma } =1.98\frac{1000}{0.25\left(0.75 ight) } 10^{-3}=10.56[/latex]kpsi

[latex]C_\sigma =\left(0.10^2+0.12^2 ight) ^{{1}/{2}}=0.156[/latex]

so stress can be expressed as [latex]\sigma = 10.56 LN\left(1,0.156 ight)[/latex] kpsi.
The endurance limit is considerably greater than the load-induced stress, indicating that finite life is not a problem. For interfering lognormal-lognormal distributions,
Eq. (5–43), p. 250, gives

[latex]z=-\frac{\ln \left(\frac{\bar{S_e} }{\bar{\sigma } } \sqrt{\frac{1+C^2_\sigma }{1+C^2_{S_e}} } ight) }{\sqrt{\ln \left[\left(1+C^2_{S_e} ight)\left(1+C^2_\sigma ight) ight] } } =-\frac{\ln \left(\frac{31.4 }{10.56} \sqrt{\frac{1+0.156^2 }{1+0.195^2} } ight) }{\sqrt{\ln \left[\left(1+0.195^2 ight)\left(1+0.156^2 ight) ight] } }=4.37[/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 \&=\left\{\begin{array}{ll} \alpha & z_{\alpha} \leq 0 \1-\alpha & z_{\alpha}>0 \end{array} ight.\end{aligned}[/latex]
[latex]Z_a[/latex]	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5000	0.4960	0.4920	0.4880	0.4840	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	0.4090	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	0.3520	0.3483
0.4	0.3446	0.3409	0.3372	0.3336	0.3300	0.3264	0.3238	0.3192	0.3156	0.3121
0.5	0.3085	0.3050	0.3015	0.2981	0.2946	0.2912	0.2877	0.2843	0.2810	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	0.2420	0.2389	0.2358	0.2327	0.2296	0.2266	0.2236	0.2206	0.2177	0.2148
0.8	0.2119	0.2090	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	0.1660	0.1635	0.1611
1.0	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	0.1230	0.1210	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	0.0250	0.0244	0.0239	0.0233

the probability of failure [latex]p_f = \Phi \left(−4.37 ight) = .000 006 35[/latex], and the reliability is

[latex]R=1-0.0000035=0.99999365[/latex]

(b) The rotary endurance tests are described by [latex]\acute{S}_e=40LN\left(1,0.05 ight)[/latex] kpsi whose mean
is less than the predicted mean in part a. The mean endurance strength [latex]\bar{S}_e[/latex] is

[latex]\bar{S}_e = 0.816\left(0.869 ight)40=28.4[/latex] kpsi

[latex]C_{se}=\left(0.058^2+0.125^2+0.05^2 ight)^{{1}/{2}}=0.147[/latex]

so the endurance strength can be expressed as [latex]S_e= 28.3LN\left(1,0.147 ight)[/latex] kpsi. From Eq. (5–43),

[latex]z=-\frac{\ln \left(\frac{28.4}{10.56}\sqrt{\frac{1+0.156^2}{1+0.147^2} } ight) }{\sqrt{\ln \left[\left(1+0.147^2 ight)\left(1+0.156^2 ight) ight] } } =-4.65[/latex]

Using Table A–10, we see the probability of failure [latex]p_f = \Phi \left(−4.65 ight) = 0.000 001 71[/latex],and

[latex]R=1-0.00000171=0.99999829[/latex]

an increase! The reduction in the probability of failure is [latex]{\left( 0.00000171-0.00000635 ight) }/{0.00000635}=-0.73[/latex], a reduction of 73 percent.
We are analyzing an existing design, so in part (a) the factor of safety was [latex]\bar{n} ={\bar{S} }/{\bar{\sigma }} ={31.4}/{10.56}=2.97[/latex].
In part (b)[latex]\bar{n} = {28.4}/{10.56} = 2.69[/latex],decrease. This example gives you the opportunity toseethe role of the design factor. Given knowledge of[latex]\bar{S},C_s,\bar{\sigma },C_{\sigma }[/latex] and reliability (through z), the mean
factor of safety(as a design factor)separates[latex]\bar{S} \ and \ \bar{\sigma }[/latex] so that there liability goal is achieved.
Knowing [latex]\bar{n} [/latex] alone says nothing about the probability of failure.
Looking at [latex]\bar{n} = 2.97 [/latex] and [latex]\bar{n} = 2.69 [/latex] says nothing about the respective probabilities of failure.
The tests did not reduce [latex]\bar{S_e}[/latex] significantly,but reduced the variation [latex]C_[/latex] such that there liability was increased.
When a mean design factor (or mean factor of safety) defined as [latex]{\bar{S_e} }/{\bar{\sigma }}[/latex] is said to be silent on matters of frequency of failures, it means that a scalar factor of safety
by itself does not offer any information about probability of failure. Nevertheless,some engineers let the factor of safety speak up, and they can be wrong in their conclusions.

Question 2.6.19: The bar shown in Fig. 6–37 is machined from a cold-rolled fl...

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