Question 6.19: The bar shown in Fig. 6–37 is machined from a cold-rolled fl...

(a) From Eq. (6–70), S′_{e} = 0.506 \overline {S}_{ut}LN(1, 0.138) = 0.506(87.6)LN(1, 0.138)

S′{e}= \begin{cases} 0.506 \overline {S}_{ut} LN(1, 0.138) kpsi or MPa & \overline {S}_{ut} ≤ 212 kpsi (1460 MPa) \\ 107LN(1, 0.139) kpsi & \overline {S}_{ut} > 212 kpsi \\ LN(1, 0.139) MPa & \overline {S}_{ut} > 1460 MPa \end{cases}             (6-70)

= 44.3LN(1, 0.138)  kpsi

From Eq. (6–72) and Table 6–10,

k_{a} = a \overline{S}^{b}_{ut} LN(1,C) ( \overline{S}_{ut} in kpsi or MPa)                       (6–72)

Table 6–10   Parameters in Marin Surface Condition Factor

*Due to the wide scatter in ground surface data, an alternate function is k_{a}0.878LN(1, 0.120). Note: S_{ut} in kpsi or MPa.

k_{a} = a \overline{S}^{-0.265}_{ut} LN(1,0.058)=2.67(87.6)^{−0.265}LN(1, 0.058)

= 0.816LN(1, 0.058)
k_{b} = 1 (axial loading)

From Eq. (6–73),

(k_{c})_{axial} = 1.23 \overline{S}^{−0.0778}_{ut} LN(1, 0.125)                       (6–73)

k_{c}= 1.23 \overline{S}^{−0.0778}_{ut} LN(1, 0.125)=1.23(87.6)^{−0.0778}LN(1, 0.125)

= 0.869LN(1, 0.125)
k_{d} = k_{f} = (1, 0)

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

S_{e} = k_{a}k_{b}k_{c}k_{d}k_{f} S′_{e}                 (6–71)

S_{e} = 0.816LN(1, 0.058)(1)0.869LN(1, 0.125)(1)(1)44.3LN(1, 0.138)

The parameters of S_{e} are

\overline{S}_{e} = 0.816(0.869)44.3 = 31.4  kpsi
CS_{e} = (0.058^{2} + 0.125^{2} + 0.138^{2})^{1/2} = 0.195

so S_{e} = 31.4LN(1, 0.195) kpsi.

In computing the stress, the section at the hole governs. Using the terminology of Table A–15–1 we find d/w = 0.50, therefore K_{t}\dot{=}2.18. From Table 6–15, \sqrt{a}=5/S_{ut} =5/87.6 =0.0571 and C_{kf}=0.10. From Eqs. (6–78) and (6–79) with r = 0.375 in,

Table A–15    Charts of Theoretical Stress-Concentration Factors K*_{t}

​*Factors from R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

Table 6–15  Heywood’s Parameter \sqrt {a} and coefficients of variation C_{Kf} for steels

\overline{K}_{f} =\frac {K_{t}}{1 + \frac {2(K_{t} − 1)}{K_{t}} \frac {\sqrt{a}}{\sqrt{r}}}              (6–78)

K_{f} =\overline{K}_{f} LN (1,C_{K_{f}})                (6–79)

K_{f}=\frac {K_{t}}{1 + \frac {2(K_{t} − 1)}{K_{t}} \frac {\sqrt{a}}{\sqrt{r}}} LN (1,C_{K_{f}})=\frac {2.18}{1 +\frac {2(2.18 − 1)}{2.18} \frac {0.0571}{\sqrt{0.375}}}LN(1, 0.10)

= 1.98LN(1, 0.10)

The stress at the hole is

σ= K_{f} \frac{F}{A} = 1.98LN(1, 0.10) \frac{1000LN(1, 0.12)}{0.25(0.75)}

\overline{σ} = 1.98 \frac {1000}{0.25(0.75)}10^{−3} = 10.56   kpsi

C_{σ} = (0.10^{2} + 0.12^{2})^{1/2} = 0.156

so stress can be expressed as σ= 10.56LN(1, 0.156) kpsi.^{34}

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. 242, gives

z = −\frac {μ ln S − μ ln σ }{(\hat{σ}^{2}ln S + \hat{σ}^{2}ln σ)^{1/2}} = −\frac {ln\left(\frac {μ_{S}}{μ_{σ}} \sqrt {\frac {1 + C^{2}_{σ}}{1 +C^{2}_{S}}}\right)}{\sqrt{ln[(1 +C^{2}_{S})(1 + C^{2}_{σ})]}}               (5–43)

z =−\frac {ln\left(\frac {\overline{S}_{e}}{\overline{σ}} \sqrt {\frac {1 + C^{2}_{σ}}{1 +C^{2}_{S_{e}}}}\right)}{\sqrt{ln[(1 +C^{2}_{S_{e}})(1 + C^{2}_{σ})]}}=−\frac {ln\left(\frac {31.4}{10.56} \sqrt {\frac {1 + 0.156^{2}}{1 +0.195^{2}}}\right)}{\sqrt{ln[(1 +0.195^{2})(1 + 0.156^{2})]}}= −4.37

From Table A–10 the probability of failure p_{f} =Φ (−4.37) = .000 006 35, and the reliability is

R = 1 − 0.000 006 35 = 0.999 993 65

Table A–10
Cumulative Distribution Function of Normal (Gaussian) Distribution

(b) The rotary endurance tests are described by S′_{e} = 40LN(1, 0.05) kpsi whose mean is less than the predicted mean in part a. The mean endurance strength \overline{S}_{e} is

\overline{S}_{e}= 0.816(0.869)40 = 28.4  kpsi
C_{S_{e}} = (0.058^{2} + 0.125^{2} + 0.05^{2})^{1/2} =0.147

so the endurance strength can be expressed as S_{e} = 28.3LN(1, 0.147) kpsi. From Eq. (5–43),

z=−\frac {ln\left(\frac {28.4}{10.56} \sqrt {\frac {1 + 0.156^{2}}{1 +0.147^{2}}}\right)}{\sqrt{ln[(1 +0.147^{2})(1 + 0.156^{2})]}}= −4.65

Using Table A–10, we see the probability of failure p_{f} =Φ (−4.65) = 0.000 001 71, and

R = 1 − 0.000 001 71 = 0.999 998 29

an increase! The reduction in the probability of failure is (0.000 001 71 − 0.000006 35)/0.000 006 35 = −0.73, a reduction of 73 percent.We are analyzing an existing design, so in part (a) the factor of safety was \overline {n} = \overline {S}/\overline{σ} = 31.4/10.56 = 2.97. In part (b) \overline {n}= 28.4/ 10.56 = 2.69, a decrease.This example gives you the opportunity to see the role of the design factor. Given knowledge of \overline {S}, C_{S}, \overline{σ}, C_{σ}, and reliability (through z), the mean factor of safety (as a design factor) separates \overline {S}  and \overline{σ} so that the reliability goal is achieved. Knowing \overline {n} alone says nothing about the probability of failure. Looking at \overline {n}= 2.97 and \overline {n} = 2.69 says nothing about the respective probabilities of failure. The tests did not reduce \overline {S}_{e} significantly, but reduced the variation C_{S} such that the reliability was increased.
When a mean design factor (or mean factor of safety) defined as \overline {S}_{e}/ \overline{σ} 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.

34 Note that there is a simplification here. The area is not a deterministic quantity. It will have a statistical
distribution also. However no information was given here, and so it was treated as being deterministic.