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

a) From Eq. (6–70),
$\acute{S}_e =\begin{cases} 0.506\bar{S} _{ut} LN\left(1,0,138\right) kpsi \ or MPa & \bar{S} _{ut}\leq 212 \ kpsi \left(1460 \ MPa\right) \ \left(1400 \ MPa\right) \\107 LN\left(1,0,139\right) \ kpsi & \bar{S} _{ut}\gt 212 \ kpsi \\740 LN\left(1,0,139\right) \ Mpa & \bar{S} _{ut}\gt 1460 \ MPa \ \end{cases}$
$\acute{S}_e= 0.506\bar{S} _{ut} LN\left(1,0,138\right) = 0.506\left(87.6\right) LN\left(1,0.138\right)= 44.3 LN\left(1,0,138\right) \ kpsi$
From Eq. (6–72)
$k_a=a\bar{S} ^b_{ut} LN \left(1,C\right) \ \ \ \left(\bar{S}_{ut} \ in \ kpsi \ or MPa \right)$

and Table 6–10,

$k_a=2.67\bar{S} ^{0.265}_{ut} LN \left(1,0.0.58\right)= 2.67\left(87.6\right) ^{-0.265} LN \left(1,0.0.58\right)\\=0.816 LN\left(1,0.058\right)\\ k_b= 1$ (axial loading)

From Eq. (6–73),

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

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

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 $K_t.= 2.18$. From Table 6–15,

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

The stress at the hole is

$\bar{\sigma } =1.98\frac{1000}{0.25\left(0.75\right) } 10^{-3}=10.56$kpsi

so stress can be expressed as $\sigma = 10.56 LN\left(1,0.156\right)$ 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

From Table A–10

the probability of failure $p_f = \Phi \left(−4.37\right) = .000 006 35$, and the reliability is

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

$\bar{S}_e = 0.816\left(0.869\right)40=28.4$ kpsi

so the endurance strength can be expressed as $S_e= 28.3LN\left(1,0.147\right)$ kpsi. From Eq. (5–43),

Using Table A–10, we see the probability of failure $p_f = \Phi \left(−4.65\right) = 0.000 001 71$,and

an increase! The reduction in the probability of failure is ${\left( 0.00000171-0.00000635\right) }/{0.00000635}=-0.73$, a reduction of 73 percent.
We are analyzing an existing design, so in part (a) the factor of safety was  $\bar{n} ={\bar{S} }/{\bar{\sigma }} ={31.4}/{10.56}=2.97$.
In part (b)$\bar{n} = {28.4}/{10.56} = 2.69$,decrease. This example gives you the opportunity toseethe role of the design factor. Given knowledge of$\bar{S},C_s,\bar{\sigma },C_{\sigma }$ and reliability (through z), the mean
factor of safety(as a design factor)separates$\bar{S} \ and \ \bar{\sigma }$ so that there liability goal is achieved.
Knowing $\bar{n}$ alone says nothing about the probability of failure.
Looking at $\bar{n} = 2.97$ and $\bar{n} = 2.69$ says nothing about the respective probabilities of failure.
The tests did not reduce $\bar{S_e}$ significantly,but reduced the variation C_ such that there liability was increased.
When a mean design factor (or mean factor of safety) defined as ${\bar{S_e} }/{\bar{\sigma }}$ 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.