Repeat Prob. 6–68, with the bar subject to a completely reversed torsional moment of 2000 lbf · in.

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Accepted Answer

From Prob. 6-68:

[latex]\begin{aligned}S _{e}^{\prime} &=23.2 L N (1,0.138) kpsi \k _{a} &=0.644 L N (1,0.11) \k_{b} &=0.936\end{aligned}[/latex]

Eq. (6-74): [latex]k _{c}=0.328(76)^{0.125} L N (1,0.125)=0.564 L N (1,0.125)[/latex]

Eq. (6-71): [latex]S _{e}=[0.644 L N (1,0.11)](0.936)[0.564 L N (1,0.125)][23.2 L N (1,0.138)][/latex]

[latex]\begin{aligned}&\bar{S}_{e}=0.644(0.936)(0.564)(23.2)=7.89 kpsi \&C_{S e}=\left(0.11^{2}+0.125^{2}+0.138^{3} ight)^{1 / 2}=0.216\end{aligned}[/latex]

Table A-16: [latex]d / D=0, a / D=(3 / 16) / 1.5=0.125, A=0.89, K _{t s}=1.64[/latex]

From Eqs. (6-78) and(7-79), and Table 6-15

[latex]K _{f s}=\frac{1.64 L N (1,0.10)}{1+\frac{2(1.64-1)}{1.64} \frac{5 / 76}{\sqrt{3 / 32}}}=1.40 L N (1,0.10)[/latex]

Table A-16:

[latex]\begin{aligned}J_{ ext {net }} &=\frac{\pi A D^{4}}{32}=\frac{\pi(0.89)\left(1.5^{4} ight)}{32}=0.4423 in ^{4} \ au_{a} &= K _{f s} \frac{T_{a} D}{2 J_{ ext {net }}}=1.40[ L N (1,0.10)] \frac{2(1.5)}{2(0.4423)}=4.75 L N (1,0.10) kpsi\end{aligned}[/latex]

From Eq. (6-57):

[latex]z=-\frac{\ln (7.89 / 4.75) \sqrt{\left(1+0.10^{2} ight) /\left(1+0.216^{2} ight)}}{\sqrt{\ln \left[\left(1+0.10^{2} ight)\left(1+0.216^{2} ight) ight]}}=-2.08[/latex]

Table A-10, [latex]p_{f}=0.0188, \quad R=1-p_{f}=1-0.0188=0.981[/latex]

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Eq. (6-57): [latex]\sum \frac{n_{i}}{N_{i}}=c[/latex]

Eq. (6-71): [latex]S _{e}= k _{a} k_{b} k _{c} k _{d} k _{f} S _{e}^{\prime}[/latex]

Eq. (6-74): [latex]\left( k _{c} ight)_{ ext {torsion }}=0.328 \bar{S}_{u t}^{0.125} L N (1,0.125)[/latex]

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 6.69: Repeat Prob. 6–68, with the bar subject to a completely reve...

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