The tubular shaft has an inner diameter of 15 mm . Determine to the nearest millimeter its minimum outer diameter if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft.

Question

The tubular shaft has an inner diameter of 15 mm . Determine to the nearest millimeter its minimum outer diameter if it is subjected to the gear loading. The bearings at A and B exert force components only in the y and z directions on the shaft. Use an allowable shear stress of [latex]T_{ ext {allow }}=70 \mathrm{MPa}[/latex], and base the design on the maximumshear-stress theory of failure.

Accepted Answer

[latex]I=\frac{\pi}{4}\left(c^{4}-0.0075^{4} ight) and J=\frac{\pi}{2}\left(c^{4}-0.0075^{4} ight)[/latex]

[latex]T_{ ext {allow }}=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2} ight)^{2}+ au_{x y}^{2}}[/latex]

[latex] au_{ ext {allow }}=\sqrt{\left(\frac{M c}{2 l} ight)^{2}+\left(\frac{T c}{J} ight)^{2}}[/latex]

[latex] au_{ ext {allow }}^{2}=\frac{M^{2} c^{2}}{4 I^{2}}+\frac{T^{2} c^{2}}{J^{2}}[/latex]

[latex] au_{ ext {allow }}^{2}\left(\frac{c^{4}-0.0075^{4}}{c} ight)^{2}=\frac{4 M^{2}}{\pi^{2}}+\frac{4 T^{2}}{\pi^{2}}[/latex]

[latex]\frac{c^{4}-0.0075^{4}}{c}=\frac{2}{\pi au_{ ext {allow }}} \sqrt{M^{2}+T^{2}}[/latex]

[latex]\frac{c^{4}-0.0075^{4}}{c}=\frac{2}{\pi(70)\left(10^{6} ight)} \sqrt{75^{2}+50^{2}}[/latex]

[latex]c^{4}-0.0075^{4}=0.8198\left(10^{-6} ight) c[/latex]

Solving, c = 0.0103976 m

d = 2 c=0.0207952 m = 20.8 mm

Use d = 21 mm

Question 11.39: The tubular shaft has an inner diameter of 15 mm . Determine...

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