The plan view of a link is the same as in Prob. 6–30; however, the forces F are completely reversed, the reliability goal is 0.998, and the material properties are Sut = 64LN(1, 0.045) kpsi and Sy = 54LN(1, 0.077) kpsi. Treat Fa as deterministic, and specify the thickness h.

Question

The plan view of a link is the same as in Prob. 6–30; however, the forces F are completely reversed, the reliability goal is 0.998, and the material properties are [latex]S _{u t}[/latex] = 64LN(1, 0.045) kpsi and [latex]S _{y}[/latex] = 54LN(1, 0.077) kpsi. Treat Fa as deterministic, and specify the thickness h.

Accepted Answer

This is a very important task for the student to attempt before starting Part 3. It illustrates the drawback of the deterministic factor of safety method. It also identifies the a priori decisions and their consequences.
The range of force fluctuation in Prob. 6-30 is - 16 to + 5 kip, or 21 kip. Let the repeatedly-applied [latex]F_{a}[/latex] be 10.5 kip. The stochastic properties of this heat of AISI 1018 CD are given in the problem statement.

Function	Consequences
Axial	[latex]F_{a}=10.5 kip[/latex]
Fatigue load	[latex]\begin{aligned}&C_{F a}=0 \&C_{k c}=0.125\end{aligned}[/latex]
Overall reliability R ≥ 0.998;	[latex]z=-3.09[/latex]
with twin fillets [latex]R \geq \sqrt{0.998}=0.999[/latex]	[latex]C_{K f}=0.11[/latex]
Cold rolled or machined surfaces	[latex]C_{k a}=0.058[/latex]
Ambient temperature	[latex]C_{k d}=0[/latex]
Use correlation method	[latex]C_{\phi}=0.138[/latex]
Stress amplitude	[latex]\begin{aligned}&C_{K f}=0.11 \&C_{\sigma a}=0.11\end{aligned}[/latex]
Significant strength [latex]S_{e}[/latex]	[latex]C_{S e}=\left(0.058^{2}+0.125^{2}+0.138^{2} ight)^{1 / 2}=0.195[/latex]

Choose the mean design factor which will meet the reliability goal. From Eq. (6-88)

[latex]\begin{aligned}C_{n} &=\sqrt{\frac{0.195^{2}+0.11^{2}}{1+0.11^{2}}}=0.223 \\bar{n} &=\exp \left[-(-3.09) \sqrt{\ln \left(1+0.223^{2} ight)}+\ln \sqrt{1+0.223^{2}} ight] \\bar{n} &=2.02\end{aligned}[/latex]

In Prob. 6-30, it was found that the hole was the significant location that controlled the analysis. Thus,

[latex]\begin{aligned}& \sigma _{a}=\frac{ S _{e}}{ n } \&\bar{\sigma}_{a}=\frac{\bar{S}_{e}}{\bar{n}} \quad \Rightarrow \quad \bar{K}_{f} \frac{F_{a}}{h\left(w_{1}-d ight)}=\frac{\bar{S}_{e}}{\bar{n}}\end{aligned}[/latex]

We need to determine [latex]\bar{S}_{e}[/latex]

[latex]\begin{aligned}\bar{k}_{a} &=2.67 \bar{S}_{u t}^{-0.265}=2.67(64)^{-0.265}=0.887 \k_{b} &=1 \\bar{k}_{c} &=1.23 \bar{S}_{u t}^{-0.0778}=1.23(64)^{-0.0778}=0.890 \\bar{k}_{d} &=\bar{k}_{e}=1 \\bar{S}_{e} &=0.887(1)(0.890)(1)(1)(0.506)(64)=25.6 kpsi\end{aligned}[/latex]

From the solution to Prob. 6-30, the stress concentration factor at the hole is [latex]K_{t}=2.68[/latex].
From Eq. (6-78) and Table 6-15

[latex]\begin{aligned}&\bar{K}_{f}=\frac{2.68}{1+\frac{2(2.68-1)}{2.68} \frac{5 / 64}{\sqrt{0.2}}}=2.20 \&h=\frac{\bar{K}_{f} \bar{n} F_{a}}{\left(w_{1}-d ight) \bar{S}_{e}}=\frac{2.20(2.02)(10.5)}{(3.5-0.4)(25.6)}=0.588\end{aligned}[/latex]

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Eq. (6-78) : [latex]\bar{K}_{f}=\frac{K_{t}}{1+\frac{2\left(K_{t}-1 ight)}{K_{t}} \frac{\sqrt{a}}{\sqrt{r}}}[/latex]

Eq. (6-88) : [latex]\bar{n}=\exp \left[-z \sqrt{\ln \left(1+C_{n}^{2} ight)}+\ln \sqrt{1+C_{n}^{2}} ight] \doteq \exp \left[C_{n}\left(-z+C_{n} / 2 ight) ight][/latex]

Table 6–15 Heywood’s Parameter √a and coefficients of variation [latex]C_{K f}[/latex] for steels	Notch Type	[latex]\sqrt{ a }(\sqrt{ ext { in }}) ,S_{ ext {ut }} ext{in kpsi}[/latex]	[latex]\sqrt{ a }(\sqrt{ m m }) , S_{ ext {ut }} ext{in MPa}[/latex]	Coefficient of Variation [latex]C_{K f}[/latex]
	Transverse hole	[latex]5 / S_{u t}[/latex]	[latex]174 / S_{u t}[/latex]	0.1
	Shoulder	[latex]4 / S_{u t}[/latex]	[latex]139 / S_{u t}[/latex]	0.11
	Groove	[latex]3 / S_{u t}[/latex]	[latex]104 / S_{u t}[/latex]	0.15

Function	Consequences
Axial	$F_{a}=10.5 kip$
Fatigue load	$\begin{aligned}&C_{F a}=0 \\&C_{k c}=0.125\end{aligned}$
Overall reliability R ≥ 0.998;	$z=-3.09$
with twin fillets $R \geq \sqrt{0.998}=0.999$	$C_{K f}=0.11$
Cold rolled or machined surfaces	$C_{k a}=0.058$
Ambient temperature	$C_{k d}=0$
Use correlation method	$C_{\phi}=0.138$
Stress amplitude	$\begin{aligned}&C_{K f}=0.11 \\&C_{\sigma a}=0.11\end{aligned}$
Significant strength $S_{e}$	$C_{S e}=\left(0.058^{2}+0.125^{2}+0.138^{2}\right)^{1 / 2}=0.195$

Table 6–15 Heywood’s Parameter √a and coefficients of variation $C_{K f}$ for steels	Notch Type	$\sqrt{ a }(\sqrt{\text { in }}) ,S_{\text {ut }} \text{in kpsi}$	$\sqrt{ a }(\sqrt{ m m }) , S_{\text {ut }} \text{in MPa}$	Coefficient of Variation $C_{K f}$
	Transverse hole	$5 / S_{u t}$	$174 / S_{u t}$	0.1
	Shoulder	$4 / S_{u t}$	$139 / S_{u t}$	0.11
	Groove	$3 / S_{u t}$	$104 / S_{u t}$	0.15

Question 6.70: The plan view of a link is the same as in Prob. 6–30; howeve...

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