Question

calculate t = fa = 2.4 kn·m
calculate \\( \sigma_{rs} = \frac{\sqrt{m^2 + t^2}}{w} \leq \sigma \\)
calculate \\( w = \frac{\pi d^3}{32} \\)

Explanation:

Step1: Identify the formula for \( W \)

The formula for the section modulus \( W \) of a circular cross - section under bending and torsion is given as \( W=\frac{\pi d^{3}}{32} \). We assume we know the value of \( d \) (diameter of the circular cross - section). Let's assume \( d \) is in meters (consistent with the units of \( M \) and \( T \) which are in \( kN\cdot m \)).

Step2: Calculate \( W \)

Let's assume \( d \) is a known value. For example, if \( d = 0.1m \) (100mm), then \( W=\frac{\pi\times(0.1)^{3}}{32}=\frac{\pi\times0.001}{32}\approx\frac{0.0031416}{32}\approx9.8175\times 10^{-5}m^{3} \)

Step3: Calculate the equivalent stress \( \sigma_{rs} \)

We know that \( M = 2.4kN\cdot m=2400N\cdot m \) and \( T = 2.4kN\cdot m = 2400N\cdot m \) (from \( T = f_a=2.4kN\cdot m \)). The formula for the equivalent stress in combined bending and torsion is \( \sigma_{rs}=\frac{\sqrt{M^{2}+T^{2}}}{W} \). Substituting \( M = 2400N\cdot m \), \( T = 2400N\cdot m \) and \( W=9.8175\times 10^{-5}m^{3} \) (using the example \( d = 0.1m \)):
First, calculate \( \sqrt{M^{2}+T^{2}}=\sqrt{(2400)^{2}+(2400)^{2}}=\sqrt{2\times(2400)^{2}}=2400\sqrt{2}\approx2400\times1.4142 = 3394.08N\cdot m \)
Then, \( \sigma_{rs}=\frac{3394.08}{9.8175\times 10^{-5}}\approx3.457\times 10^{7}Pa = 34.57MPa \)
(Note: If we want to check against an allowable stress \( [\sigma] \), we would compare this value with \( [\sigma] \). Also, the value of \( d \) needs to be determined based on the actual problem context, this is just an example calculation.)

Answer:

The calculation involves finding the section modulus \( W=\frac{\pi d^{3}}{32} \) first, then using \( \sigma_{rs}=\frac{\sqrt{M^{2}+T^{2}}}{W} \) to find the equivalent stress. For a given \( d \), we can compute the numerical value of \( \sigma_{rs} \) and compare it with \( [\sigma] \). (The final numerical answer depends on the value of \( d \), in the example with \( d = 0.1m \), \( \sigma_{rs}\approx34.57MPa \))