A 1/4 by 1 1/2 -in steel bar has a 3/4 -in drilled hole located in the center, much as is shown in Table A–15–1. The bar is subjected to a completely reversed axial load with a deterministic load of 1200 lbf. The material has a mean ultimate tensile strength of Sut = 80 kpsi. (a) Estimate the

Question

A [latex]\frac{1}{4}[/latex] by 1 [latex]\frac{1}{2}[/latex] -in steel bar has a [latex]\frac{3}{4}[/latex] -in drilled hole located in the center, much as is shown in Table A–15–1. The bar is subjected to a completely reversed axial load with a deterministic load of 1200 lbf. The material has a mean ultimate tensile strength of [latex]\bar{S}_{u t}=80 kpsi[/latex].
(a) Estimate the reliability.
(b) Conduct a computer simulation to confirm your answer to part a.

Accepted Answer

[latex]\begin{aligned}&F_{a}=1200 lbf \&S_{u t}=80 kpsi\end{aligned}[/latex]

(a) Strength

[latex]\begin{aligned}& k _{a}=2.67(80)^{-0265} L N (1,0.058)=0.836 L N (1,0.058) \&k_{b}=1 \& k _{c}=1.23(80)^{-0.0778} L N (1,0.125)=0.875 L N (1,0.125)\end{aligned}[/latex]

[latex]\begin{aligned}S _{e}^{\prime} &=0.506(80) L N (1,0.138)=40.5 L N (1,0.138) kpsi \S _{e} &=[0.836 L N (1,0.058)](1)[0.875 L N (1,0.125)][40.5 L N (1,0.138)] \\bar{S}_{e} &=0.836(1)(0.875)(40.5)=29.6 kpsi \C_{S e}&=\left(0.058^{2}+0.125^{2}+0.138^{2} ight)^{1 / 2}=0.195\end{aligned}[/latex]

Stress: Fig. A-15-1; [latex]d / w=0.75 / 1.5=0.5, K_{t}=2.18[/latex]. From Eqs. (6-78), (6-79) and Table 6-15

[latex]K _{f}=\frac{2.18 L N (1,0.10)}{1+\frac{2(2.18-1)}{2.18} \frac{5 / 80}{\sqrt{0.375}}}=1.96 L N (1,0.10)[/latex]

[latex]\begin{aligned}& \sigma _{a}= K _{f} \frac{F_{a}}{(w-d) t}, \quad C_{\sigma}=0.10 \&\bar{\sigma}_{a}=\frac{\bar{K}_{f} F_{a}}{(w-d) t}=\frac{1.96(1.2)}{(1.5-0.75)(0.25)}=12.54 kpsi \&\bar{S}_{a}=\bar{S}_{e}=29.6 kpsi\end{aligned}[/latex]

[latex]\begin{aligned}z &=-\frac{\ln \left[\left(\bar{S}_{a} / \bar{\sigma}_{a} ight) \sqrt{\left(1+C_{\sigma}^{2} ight) /\left(1+C_{S}^{2} ight)} ight]}{\ln \left[\left(1+C_{\sigma}^{2} ight)\left(1+C_{S}^{2} ight) ight]} \&=-\frac{\ln \left[(29.6 / 12.48) \sqrt{\left(1+0.10^{2} ight) /\left(1+0.195^{2} ight)} ight]}{\sqrt{\ln \left[\left(1+0.10^{2} ight)\left(1+0.195^{2} ight) ight]}}=-3.9\end{aligned}[/latex]

From Table A-20, [latex]p_{f}=4.81\left(10^{-5} ight) \Rightarrow R=1-4.81\left(10^{-5} ight)=0.999955[/latex]

(b) All computer programs will differ in detail.

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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. (6-79) : [latex]K _{f}=\bar{K}_{f} L N \left(1, C_{K_{f}} 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

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 6.71: A 1/4 by 1 1/2 -in steel bar has a 3/4 -in drilled hole loca...

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