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The A-36 steel tubular shaft is 2 \(\mathrm{m}\) long and has an outer diameter of \(50 \mathrm{mm}\). When it is rotating at \(40 \mathrm{rad} / \mathrm{s}\), it transmits \(25 \mathrm{kW}\) of power from the motor \(M\) to the pump \(P\) Determine the smallest thickness of the tube if the allowable shear stress is \(\tau_{\text {allow }}=80 \mathrm{MPa}\)

Short Answer

Expert verified
The smallest thickness of the tube should be about \(10.9 \mathrm{mm}\).

Step by step solution

01

Calculate the Torque

The power transmitted by the shaft, given by \(P=25 \mathrm{kW}\), is related to the angular velocity, \( \omega=40 \mathrm{rad} / \mathrm{s} \), and torque, \( T \), by the equation \( P=T\omega\). Therefore, we can calculate the torque transmitted using the equation \( T = \frac{P}{\omega} \).\nSo, \( T = \frac{25,000}{40} = 625 \mathrm{Nm}\).
02

Set up the shear stress equation

The stress produced in the tube is computed by the equation \( \tau = \frac{T}{J} \). Here, \( T \) is the torque, calculated in the previous step, and \( J \) is the polar second moment of area, determined by \( J=\frac{\pi}{32}(d_o^4 -d_i^4)\), where \(d_o\) and \(d_i\) are the outer and inner diameters respectively.
03

Rewriting the equation for tube thickness

We seek to find the thickness \(t = \frac{d_o-d_i}{2}\). It can be rewritten from the diameter equation as \(d_i = d_o - 2t\).\nSo, substituting \(d_i\) into the equation of \(J\) in Step 2, we can rewrite it as \(J = \frac{\pi}{32}\left(d_o^4 - (d_o - 2t)^4 \right)\).
04

Calculate the inner diameter

Given that the maximum allowable shear stress is \( \tau_{\text {allowable}} = 80 \mathrm{MPa}\), equating this with the stress equation, it becomes 80,000,000 \(= \frac{625}{\frac{\pi}{32}(d_o^4 - (d_o-2t)^4)}\). Solving this equation for \(d_i = d_o - 2t\), where \( d_o = 50 \mathrm{mm} \), we get, \(d_i = 28.2\, \mathrm{mm} \).
05

Calculate the tube thickness

After calculating the inner diameter in the previous step, we can compute the tube thickness using the equation \( t = \frac{d_o-d_i}{2} \).\nSo, \( t = \frac{50 - 28.2}{2} = 10.9 \mathrm{mm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Shear Stress
Shear stress is a fundamental concept in mechanical engineering. It represents the internal force per unit area in materials, when they are under tangential forces. In this exercise, shear stress occurs in the tubular shaft when torque is applied, due to the rotational motion transmitted through the shaft.
To calculate shear stress, use the formula:
  • \( \tau = \frac{T}{J} \)
Where:
  • \( \tau \): Shear stress
  • \( T \): Torque applied
  • \( J \): Polar moment of inertia
In our step-by-step solution, we equate allowable shear stress, \( \tau_{\text{allow}} = 80 \text{ MPa} \), to ensure safety and performance. Calculating shear stress helps determine if a material can resist the applied torque without failure.
Torque Calculation
Torque is essential in mechanics, especially concerning rotating shafts. It refers to the rotational force applied to an object, causing it to turn around an axis. Understanding torque calculations help in designing components to safely transmit power.
  • The exercise provides power, \( P = 25 \text{ kW} \), and angular velocity, \( \omega = 40 \text{ rad/s} \).
  • Torque \( T \) is computed using \( T = \frac{P}{\omega} \).
Substituting values, we find:
  • \( T = \frac{25,000}{40} = 625 \text{ Nm} \).
This torque value is vital for further calculations like shear stress and ensures the component's ability to perform tasks efficiently.
Polar Moment of Inertia
Polar moment of inertia, commonly denoted as \( J \), is a key property in torsional dynamics. It quantifies an object's resistance to twisting and is especially crucial for cylindrical objects like shafts.
  • The formula for \( J \) in a tube is \( J = \frac{\pi}{32}(d_o^4 - d_i^4) \).
Where:
  • \( d_o \) is the outer diameter.
  • \( d_i \) is the inner diameter.
In practical situations, the polar moment of inertia helps engineers understand how a shaft will behave under load, ensuring structural integrity. In the exercise, computing \( J \) allows us to determine the minimum wall thickness required to sustain safe operation while under torsional stress. This knowledge ultimately contributes to designing effective mechanical systems.

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Most popular questions from this chapter

The A-36 steel bolt is tightened within a hole so that the reactive torque on the shank \(A B\) can be expressed by the equation \(t=\left(k x^{2 / 3}\right) \mathrm{N} \cdot \mathrm{m} / \mathrm{m},\) where \(x\) is in meters. If a torque of \(T=50 \mathrm{N} \cdot \mathrm{m}\) is applied to the bolt head, determine the constant \(k\) and the amount of twist in the \(50-m m\) length of the shank. Assume the shank has a constant radius of \(4 \mathrm{mm}\)

\(\quad\) A steel tube having an outer diameter of 2.5 in. is used to transmit 9 hp when turning at 27 rev / min. Determine the inner diameter \(d\) of the tube to the nearest \(\frac{1}{8}\) in. if the allowable shear stress is \(\tau_{\text {allow }}=10 \mathrm{ksi}\)

A motor delivers 500 hp to the shaft, which is tubular and has an outer diameter of 2 in. If it is rotating at \(200 \mathrm{rad} / \mathrm{s}\) determine its largest inner diameter to the nearest \(\frac{1}{8}\) in. if the allowable shear stress for the material is \(\tau_{\text {allow }}=25\) ksi.

The propellers of a ship are connected to an A-36 steel shaft that is \(60 \mathrm{m}\) long and has an outer diameter of \(340 \mathrm{mm}\) and inner diameter of \(260 \mathrm{mm}\). If the power output is \(4.5 \mathrm{MW}\) when the shaft rotates at \(20 \mathrm{rad} / \mathrm{s},\) determine the maximum torsional stress in the shaft and its angle of twist.

The shaft has a radius \(c\) and is subjected to a torque per unit length of \(t_{0},\) which is distributed uniformly over the shaft's entire length \(L\). If it is fixed at its far end \(A\), determine the angle of twist \(\phi\) of end \(B\). The shear modulus is \(G\)

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