A solid 1.50-in.-diameter shaft is used in an aircraft engine to transmit 160 hp at 2,800 rpm to a propeller that develops a thrust of 1,800 lb. Determine the magnitudes of the principal stresses and the maximum shear stress at any point on the outside surface of the shaft.

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Accepted Answer

Section properties:

[latex]A=\frac{\pi}{4}(1.5 in .)^{2}=1.767146 in .^{2} \quad J=\frac{\pi}{32}(1.5 in .)^{4}=0.497010 in .^{4}[/latex]

Normal and shear stress magnitudes:

[latex]\sigma=\frac{P}{A}=\frac{1,800 lb }{1.767146 in .{ }^{2}}=1,018.592 psi ( T )[/latex]

The torque in the propeller shaft is:

[latex]T=\frac{P}{\omega}=\frac{(160 hp )\left(\frac{550 lb - ft / s }{1 hp } ight)}{\left(\frac{2,800 rev }{ min } ight)\left(\frac{2 \pi rad }{1 rev } ight)\left(\frac{1 min }{60 s } ight)}=300.121 lb - ft[/latex]

[latex] au=\frac{T c}{J}=\frac{(300.121 lb - ft )(1.5 in . / 2)(12 in . / ft )}{0.497010 in .^{4}}=5,434.676 psi[/latex]

(sense of shear stress cannot be established definitively from the information given)

Principal stress calculations:
The principal stress magnitudes can be computed from Eq. (12.12):

[latex]\begin{aligned}\sigma_{p 1, p 2} &=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2} ight)^{2}+ au_{x y}^{2}} \&=\frac{(1,018.592)+(0)}{2} \pm \sqrt{\left(\frac{(1,018.592)-(0)}{2} ight)^{2}+(5,434.676)^{2}} \&=509.296 \pm 5.458 .487\end{aligned}[/latex]

[latex]\begin{aligned}&\sigma_{p 1}=5,970 psi ext { and } \sigma_{p 2}=-4,950 psi \& au_{\max }=5,460 psi \quad( ext { maximum in-plane shear stress }) \&\sigma_{ ext {avg }}=509 psi ( T ) \quad ext { (normal stress on planes of maximum in-plane shear stress) }\end{aligned}[/latex]

Question 15.5: A solid 1.50-in.-diameter shaft is used in an aircraft engin...

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