Question

example 6 the driving shaft of automobile, d = 90mm, t = 2.5mm, m_max = 1500n·m, τ=60mpa, check the strength in torsion for shaft a b m m d

Explanation:

Step1: Calculate the polar - moment of inertia for a thin - walled circular cross - section

For a thin - walled circular cross - section with outer diameter $D$ and thickness $t$, the polar - moment of inertia $I_p\approx\pi D^{3}t/4$. Given $D = 90\mathrm{mm}=0.09\mathrm{m}$ and $t = 2.5\mathrm{mm}=2.5\times10^{-3}\mathrm{m}$, then $I_p=\frac{\pi\times(0.09)^{3}\times2.5\times10^{-3}}{4}=\frac{\pi\times0.09^{3}\times2.5\times10^{-3}}{4}\mathrm{m}^{4}$.

Step2: Calculate the maximum shear stress

The formula for the maximum shear stress in torsion is $\tau_{max}=\frac{M_{max}r}{I_p}$. For a thin - walled circular cross - section, $r\approx D/2$. So $\tau_{max}=\frac{M_{max}D/2}{I_p}$. Substituting $I_p=\frac{\pi D^{3}t}{4}$ into the formula, we get $\tau_{max}=\frac{2M_{max}}{\pi D^{2}t}$. Given $M_{max}=1500\mathrm{N}\cdot\mathrm{m}$, $D = 0.09\mathrm{m}$, and $t = 2.5\times10^{-3}\mathrm{m}$, then $\tau_{max}=\frac{2\times1500}{\pi\times(0.09)^{2}\times2.5\times10^{-3}}\mathrm{Pa}$.

Step3: Evaluate the strength condition

We compare the calculated $\tau_{max}$ with the allowable shear stress $[\tau]=60\times10^{6}\mathrm{Pa}$. Calculate $\tau_{max}=\frac{2\times1500}{\pi\times(0.09)^{2}\times2.5\times10^{-3}}=\frac{3000}{\pi\times0.09^{2}\times2.5\times10^{-3}}\approx47.1\times10^{6}\mathrm{Pa}=47.1\mathrm{MPa}$.

Answer:

Since $\tau_{max}=47.1\mathrm{MPa}<[\tau]=60\mathrm{MPa}$, the shaft satisfies the strength condition in torsion.