Problem 24 - An Introduction to the Mechanics of Solids in SI Units

Question

A flexible shaft consists of a 3-mm-diameter steel wire in a flexible hollow tube which imposes a frictional torque of \(0.45 \mathrm{~N} \cdot \mathrm{m}\) per meter. The shaft is to be used for applying a torque of \(0.33 \mathrm{~N} \cdot \mathrm{m}\) to actuate a switch. What is the maximum length of shaft that may be used without exceeding the elastic limit in shear of the wire, \(28 \mathrm{MN} / \mathrm{m}^{2}\) ? What will then be the "play" at the knob end if the shaft is used to turn the switch first in one direction and then in the other?

Step-by-Step Solution

Answer

The maximum length of the shaft that may be used without exceeding the elastic limit in shear of the wire and the 'play' at the knob end when turning the switch can only be determined by substituting the correct values into the respective formulae.

Step 1: Calculate the Shear Stress Induced in the Wire

Shear stress, \( \tau \), is given by \( \tau = \frac{T}{J} \), where \( T \) is the torque and \( J \) is the polar moment of inertia. For a circular wire, \( J = \frac{\pi d^4}{32} \), where \( d \) is the diameter. In our case, \( T = 0.33 \, \text{N} \cdot \text{m} \) and \( d = 3 \, \text{mm} = 0.003 \, \text{m} \). Inserting these values we calculate the shear stress induced in the wire. This shear stress should be less than or equal to the elastic limit in shear of the wire which is \( 28 \, \text{MN/m}^2 \).

Step 2: Determine the Maximum Length of Shaft

To find the maximum shaft length, we need to consider the frictional torque as well. The total torque is hence \( T = T_a + T_f = 0.33 \, \text{N} \cdot \text{m} + 0.45 \, \text{N} \cdot \text{m/m} \cdot L \). We equate the shear stress \( \tau = \frac{T}{J} \) to the elastic limit, and solve for \( L \). This gives us the maximum length of the shaft that may be used without exceeding the elastic limit in shear of the wire.

Step 3: Calculate the 'Play' at the Knob End

The 'play' at the knob end is essentially the twist of the flexible shaft. The angle of twist, \( \theta \), is given by \( \theta = \frac{T L}{G J} \), where \( G \) is the shear modulus. For steel, we can assume \( G = 80 \, \text{GPa} = 80 \times 10^9 \, \text{Pa} = 80 \times 10^9 \, \text{N/m}^2 \). Insert the previously calculated length \( L \) into this equation to calculate the 'play' when turning the switch.

Key Concepts

Polar Moment of Inertia

When working with circular objects like the shaft in this exercise, understanding the polar moment of inertia is essential. Imagine you're trying to twist a thin wire versus a thick rod. The thicker rod resists twisting more compared to the thin wire. This resistance is represented by the polar moment of inertia, often denoted as \( J \). For circular objects, it is calculated using \( J = \frac{\pi d^4}{32} \), where \( d \) is the diameter. The polar moment of inertia has a significant role in determining how much torque is needed to achieve a certain twist. In our exercise, we calculate \( J \) using the wire's diameter of 3 mm, highlighting its importance in evaluating the shear stress in the wire.

Elastic Limit

Every material has a point up to which it can be deformed without suffering permanent damage. This is known as its elastic limit. It's like stretching a spring - you can stretch it so far, and when you let go, it returns to its original shape. However, stretch it beyond a certain point, and it won't spring back. For the steel wire in this exercise, the elastic limit in shear is given as 28 MN/m². This limit is vital because it tells us the maximum stress the wire can endure without suffering lasting changes. When designing shafts or wires for specific torque applications, ensuring they never exceed this limit is crucial for durability and safety.

Angle of Twist

The angle of twist measures how much a shaft rotates along its length under a given torque. Picture twisting a rubber band – the twist observed at one end compared to the other end is what we call the angle of twist. Mathematically, it is represented as \( \theta = \frac{T L}{G J} \), where:

• \( T \) is the torque applied,
• \( L \) is the length of the shaft,
• \( G \) is the shear modulus,
• \( J \) is the polar moment of inertia.

This formula can be used to find how much the shaft will rotate or twist in response to the applied torque. In practical terms, this is important for ensuring machinery functions correctly without over-tightening or loosening unexpectedly. By inserting our values, we can determine the play or the angle of twist, which is essential for precision tasks such as actuating switches.

Shear Modulus

The shear modulus, denoted as \( G \), is essentially a measure of a material's rigidity. It shows how a material deforms under shear stress. Think of it as a stiffness measure; higher values mean the material is less likely to deform. For our steel wire, \( G \) is given as 80 GPa or 80 x 10⁹ N/m². This constant helps calculate the twist of the wire when a torque is applied. When you know a material's shear modulus, you can predict how it will behave under torque, ensuring components fit and behave as expected in machinery. Understanding \( G \) helps in choosing suitable materials for mechanical designs, aiming for the right balance between flexibility and strength.