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Chapter 3: Problem 8

An aluminum tube has inside diameter \(d_{1}=50 \mathrm{mm}\) shear modulus of elasticity \(G=27 \mathrm{GPa}, v=0.33,\) and torque \(T=4.0 \mathrm{kN} \cdot \mathrm{m} .\) The allowable shear stress in the aluminum is \(50 \mathrm{MPa}\) and the allowable normal strain is \(900 \times 10^{-6}\) (a) Determine the required outside diameter \(d_{2}\) (b) Re-compute the required outside diameter \(d_{2}\) if allowable normal stress is \(62 \mathrm{MPa}\) and allowable shear strain is \(1.7 \times 10^{-3}\)

Short Answer

Expert verified
The required outside diameter \( d_2 \) is calculated through restructuring the torque equation for both round 1 and 2 using allowable values; choose smallest \( d_2 \) found.

Step by step solution

01

Calculate Section Modulus for Torque

First, compute the polar moment of inertia for a hollow circular section. The formula is: \[ J = \frac{\pi}{32} \left( d_2^4 - d_1^4 \right) \]Then, using the relationship for allowable shear stress, we can calculate the section modulus as:\[ \tau = \frac{T \cdot c}{J} \]where \( \tau \) is the allowable shear stress and \( c = \frac{d_2}{2} \) is the distance from the center to the outer surface. Rearranging the formula gives:\[ J = \frac{T \cdot d_2}{2 \cdot \tau_{allow}} \]
02

Calculate Using Allowable Shear Stress

Insert the given values into the equation from Step 1:\[ J = \frac{4 \times 10^3 \cdot d_2}{2 \cdot 50 \times 10^6} \]This will allow you to express \( J \) in terms of \( d_2 \).
03

Shear Modulus Equation Substitution

Using the relationship between torque \( T \), polar moment of inertia \( J \), and shear modulus \( G \):\[ T = \frac{J \cdot G}{\theta} \]substituting the values provided, you determine the maximum torque and hence find the corresponding \( J \) that satisfies the allowable shear strain.
04

Check Against Normal Strain

The normal strain \( \epsilon \) is associated with \( d_2 \) using:\[ \epsilon = \frac{T \cdot d_2}{2 \cdot G \cdot J} \]Check if \( \epsilon \leq 900 \times 10^{-6} \) in terms of \( d_2 \) and derive the proper expression for \( J \).
05

Calculate for New Allowable Values

Now, re-compute \( d_2 \) using the new allowable values: \( \sigma_{allow} = 62 \mathrm{MPa} \) and \( \gamma_{allow} = 1.7 \times 10^{-3} \). Again, apply the same formulae and re-evaluate for \( d_2 \).
06

Finalize Calculation for Required \( d_2 \)

Solve the two expressions for \( J \) derived from the stress-strain limitations and choose the smallest \( d_2 \) that satisfies all constraints.

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Key Concepts

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

Shear Modulus of Elasticity
When we discuss torsion in hollow circular shafts, understanding the shear modulus of elasticity (often denoted as \( G \)) is crucial. This property tells us how rigid a material is when subjected to shear stress, a type of stress causing materials to slide along parallel internal surfaces. Here's the key takeaway about shear modulus:
  • It relates shear stress \( \tau \) to shear strain \( \gamma \).
  • Expressed as \( G = \frac{\tau}{\gamma} \).
  • In the exercise at hand, the aluminum shaft has \( G = 27 \, \text{GPa} \), meaning it is quite resistant to deformation.
This property helps us assess whether our hollow shaft can withstand the applied torque while experiencing allowable strain limits. It鈥檚 a critical component in determining how much the shaft will twist under a given load.
Shear Stress
Shear stress is vital when studying materials subjected to forces that cause twisting, like torques on circular shafts. In the exercise, the focus is on whether the shaft can endure a maximum allowable shear stress of 50 MPa without failing.For hollow shafts, shear stress \( \tau \) is calculated as:
  • \( \tau = \frac{T \cdot c}{J} \)
  • Where \( T \) is the torque, \( c \) is the radius of the shaft, and \( J \) is the polar moment of inertia.
This condition ensures the shaft is strong enough to withstand twisting forces without exceeding the material's shear strength. It's a safety measure, making sure the shaft remains safe under designed operational conditions.
Normal Strain
Normal strain refers to the deformation of the shaft鈥檚 material layers perpendicular to the applied force. For this aluminum tube, we鈥檙e keeping an eye on whether the strain stays below \( 900 \times 10^{-6} \) \( \left( \text{or } 0.0009 \right)\), because exceeding this can lead to failure.The formula linking normal strain \( \epsilon \) to the torque and geometry is:
  • \( \epsilon = \frac{T \cdot d_2}{2 \cdot G \cdot J} \)
  • Where \( d_2 \) is the outer diameter of the tube.
By substituting in different values to solve for \( d_2 \), we ensure that the designed shaft will not stretch beyond its safe limits, even under full operational load. This assessment is critical in ensuring the structural integrity of the shaft.
Polar Moment of Inertia
The polar moment of inertia \( J \) concerns how well a given shaft can resist twisting. It is part of the foundational equations that help calculate stress and strain under torsion. For a hollow circular tube, it is calculated as:
  • \( J = \frac{\pi}{32} \left( d_2^4 - d_1^4 \right) \)
  • Where \( d_2 \) and \( d_1 \) are the outer and inner diameters, respectively.
The higher the value of \( J \), the better the shaft can resist twisting. This component is crucial in determining acceptable shaft diameter \( d_2 \) to ensure goal compatibility between torsional strength and the applied load. With this knowledge, we can design shafts that are both safe and efficient in their applications.

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Most popular questions from this chapter

A hollow aluminum shaft (see figure) has outside diameter \(d_{2}=100 \mathrm{mm}\) and inside diameter \(d_{1}=50 \mathrm{mm}\) When twisted by torques \(T,\) the shaft has an angle of twist per unit distance equal to \(2^{\circ} / \mathrm{m}\). The shear modulus of elasticity of the aluminum is \(G=27.5 \mathrm{GPa}\) (a) Determine the maximum tensile stress \(\sigma_{\max }\) in the shaft. (b) Determine the magnitude of the applied torques \(T\)

A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions \(b=150 \mathrm{mm}\) and \(h=100 \mathrm{mm} .\) The wall thickness \(t\) is constant and equal to \(6.0 \mathrm{mm}\) (a) Determine the shear stress in the tube due to a torque \(T=1650 \mathrm{N} \cdot \mathrm{m}\) (b) Determine the angle of twist (in degrees) if the length \(L\) of the tube is \(1.2 \mathrm{m}\) and the shear modulus \(G\) is 75 GPa.

The drive shaft for a truck (outer diameter \(60 \mathrm{mm}\) and inner diameter \(40 \mathrm{mm}\) ) is running at \(2500 \mathrm{rpm}\) (see figure) (a) If the shaft transmits \(150 \mathrm{kW}\), what is the maximum shear stress in the shaft? (b) If the allowable shear stress is \(30 \mathrm{MPa}\), what is the maximum power that can be transmitted?

\(3.5-11\) Two circular aluminum pipes of equal length \(L=610 \mathrm{mm}\) are loaded by torsional moments \(T\) (see figure). Pipe 1 has outside and inside diameters \(d_{2}=76 \mathrm{mm}\) and \(d_{1}=64 \mathrm{mm},\) respectively. Pipe 2 has a constant outer diameter of \(d_{2}\) along its entire length \(L\) and an inner diameter of \(d_{1}\) but has an increased inner diameter of \(d_{3}=67 \mathrm{mm}\) over the middle third. Assume that \(E=72 \mathrm{GPa}, v=0.33,\) and allowable shear stress \(\tau_{\mathrm{a}}=45 \mathrm{MPa}\) (a) Find the maximum acceptable torques that can be applied to Pipe \(1 ;\) repeat for Pipe 2 (b) If the maximum twist \(\phi\) of Pipe 2 cannot exceed \(5 / 4\) of that of Pipe \(1,\) what is the maximum acceptable length of the middle segment? Assume both pipes have total length \(L\) and the same applied torque \(T\) (c) Find the new value of inner diameter \(d_{3}\) of Pipe 2 if the maximum torque carried by Pipe 2 is to be \(7 / 8\) of that for Pipe 1 (d) If the maximum normal strain in each pipe is known to be \(\varepsilon_{\max }=811 \times 10^{-6},\) what is the applied torque on each pipe? Also, what is the maximum twist of each pipe? Use original properties and dimensions.

A hollow aluminum tube used in a roof structure has an outside diameter \(d_{2}=104 \mathrm{mm}\) and an inside diameter \(d_{1}=82 \mathrm{mm}\) (see figure). The tube is \(2.75 \mathrm{m}\) long, and the aluminum has shear modulus \(G=28\) GPa. (a) If the tube is twisted in pure torsion by torques acting at the ends, what is the angle of twist (in degrees) when the maximum shear stress is 48 MPa? (b) What diameter \(d\) is required for a solid shaft (see figure) to resist the same torque with the same maximum stress? (c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft?
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