The 6061-T6 aluminum alloy solid shaft is fixed at one end but free at the other end. If the shaft has a diameter of 100 mm, determine its maximum allowable length L if it is subjected to the eccentric force P=80 kN.

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Section Properties.

[latex]A=\pi\left(0.05^{2} ight)=2.5\left(10^{-3} ight) \pi \mathrm{m}^{2} \I=\frac{\pi}{4}\left(0.05^{4} ight)=1.5625\left(10^{-6} ight) \pi \mathrm{m}^{4} \r=\sqrt{\frac{I}{A}}=\sqrt{\frac{1.5625\left(10^{-6} ight) \pi}{2.5\left(10^{-3} ight) \pi}}=0.025 \mathrm{~m} \e=0.1 \mathrm{~m} \quad c=0.05 \mathrm{~m}[/latex]

For a column that is fixed at one end and free at the other, K=2. Thus,

K L=2 L

Buckling. The critical buckling load is [latex]P_{\mathrm{cr}}=80 \mathrm{kN}[/latex]. Applying Euler's equation,

[latex]P_{\mathrm{cr}}=\frac{\pi^{2} E I}{(K L)^{2}} \80\left(10^{3} ight)=\frac{\pi^{2}\left[68.9\left(10^{9} ight) ight]\left[1.5625\left(10^{-6} ight) \pi ight]}{(2 L)^{2}} \L=3.230 \mathrm{~m}[/latex]

Euler's equation is valid only if [latex]\sigma_{\mathrm{cr}}<\sigma_{Y}[/latex]

[latex]\sigma_{\mathrm{cr}}=\frac{P_{\mathrm{cr}}}{A}=\frac{80\left(10^{3} ight)}{2.5\left(10^{-3} ight) \pi}=10.19 \mathrm{MPa}<\sigma_{Y}=255 \mathrm{MPa}[/latex]

Yielding. Applying the secant formula,
[latex]\sigma_{\max }=\frac{P}{A}\left[1+\frac{e c}{r^{2}} \sec \left[\frac{K L}{2 r} \sqrt{\frac{P}{E A}} ight] ight]\255\left(10^{6} ight)=\frac{80\left(10^{3} ight)}{2.5\left(10^{-3} ight) \pi}\left[1+\frac{0.1(0.05)}{0.025^{2}} \sec \left[\frac{2 L}{2(0.025)} \sqrt{\frac{80\left(10^{3} ight)}{68.9\left(10^{9} ight)\left[2.5\left(10^{-3} ight) \pi ight]}} ight] ight]\\sec 0.4864 L=3.0043\L=2.532 \mathrm{~m}=2.53 \mathrm{~m}\quad (controls)[/latex]

Question 13.57: The 6061-T6 aluminum alloy solid shaft is fixed at one end ...

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