Question

example 1 a transmission shaft is rotating at n =300r/min. knowing its input power is p1=500kw, and its output are p2=150kw, p3=150kw, p4=200kw, try to plot the internal torque diagram. solution: ①calculate the external torque

Explanation:

Step1: Convert rotation speed to rad/s

The rotation speed $n = 300r/min=\frac{300\times2\pi}{60}rad/s = 10\pi rad/s$.

Step2: Use the formula $m=\frac{P}{\omega}$ to calculate external - torques

For the input power $P_1 = 500kW = 500\times10^{3}W$, the external torque $m_1=\frac{P_1}{\omega}=\frac{500\times10^{3}}{10\pi}N\cdot m=\frac{50000}{\pi}N\cdot m$.
For $P_2 = 150kW = 150\times10^{3}W$, $m_2=\frac{P_2}{\omega}=\frac{150\times10^{3}}{10\pi}N\cdot m=\frac{15000}{\pi}N\cdot m$.
For $P_3 = 150kW = 150\times10^{3}W$, $m_3=\frac{P_3}{\omega}=\frac{150\times10^{3}}{10\pi}N\cdot m=\frac{15000}{\pi}N\cdot m$.
For $P_4 = 200kW = 200\times10^{3}W$, $m_4=\frac{P_4}{\omega}=\frac{200\times10^{3}}{10\pi}N\cdot m=\frac{20000}{\pi}N\cdot m$.

Step3: Analyze the internal - torque in different sections

In section $CD$: The internal torque $T_{CD}=m_4=\frac{20000}{\pi}N\cdot m$.
In section $BC$: $T_{BC}=m_4 + m_3=\frac{20000 + 15000}{\pi}N\cdot m=\frac{35000}{\pi}N\cdot m$.
In section $AB$: $T_{AB}=m_4 + m_3+m_2=\frac{20000 + 15000+15000}{\pi}N\cdot m=\frac{50000}{\pi}N\cdot m$.
To plot the internal - torque diagram, we take the length of the shaft on the x - axis and the internal torque on the y - axis. Mark the values of internal torque at the boundaries of each section ($A$, $B$, $C$, $D$) and connect them with horizontal lines (since in a uniform - shaft under these loading conditions, the internal torque is constant within each section).

Answer:

The internal torque in section $CD$ is $\frac{20000}{\pi}N\cdot m$, in section $BC$ is $\frac{35000}{\pi}N\cdot m$, and in section $AB$ is $\frac{50000}{\pi}N\cdot m$. Plot these values on a graph with the shaft - length on the x - axis and internal torque on the y - axis.