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Two shafts are made of the same material. The cross section of shaft \(A\) is a square of side \(b\) and that of shaft \(B\) is a circle of diameter b. Knowing that the shafts are subjected to the same torque, determine the ratio \(\tau_{A} / \tau_{B}\) of maximum shearing stresses occurring in the shafts.

Short Answer

Expert verified
The ratio \( \tau_A / \tau_B \) of maximum shearing stresses is \( \frac{3\pi}{16} \), approximately 0.589.

Step by step solution

01

Determine the formula for the maximum shearing stress for a square cross-section

For a shaft with a square cross-section of side length \( b \), the maximum shearing stress \( \tau_A \) in the shaft due to applied torque \( T \) is calculated using the polar moment of inertia \( J \). The formula for a square is \( J = \frac{b^4}{6} \), giving \( \tau_A = \frac{T \cdot b}{2J} = \frac{3T}{b^3} \).
02

Determine the formula for the maximum shearing stress for a circular cross-section

For a shaft with a circular cross-section of diameter \( b \), the maximum shearing stress \( \tau_B \) is determined using \( J = \frac{\pi b^4}{32} \). Hence, \( \tau_B = \frac{T}{J} \cdot \frac{b}{2} = \frac{16T}{\pi b^3} \).
03

Calculate the ratio of maximum shearing stresses \( \tau_A / \tau_B \)

Substitute the expressions for \( \tau_A \) and \( \tau_B \) from Steps 1 and 2 into the ratio: \( \frac{\tau_A}{\tau_B} = \frac{\frac{3T}{b^3}}{\frac{16T}{\pi b^3}} \). This simplifies to \( \frac{\tau_A}{\tau_B} = \frac{3\pi}{16} \).
04

Simplify the expression for the ratio

Simplifying the equation further, we compute \( \frac{3\pi}{16} \), ensuring you understand \( \pi \approx 3.14159 \), which makes the final approximate ratio \( \frac{3 \cdot 3.14159}{16} \approx 0.589 \).

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Key Concepts

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

Torque
Torque is a fundamental concept in mechanical engineering and physics, describing a force that causes an object to rotate about an axis. It is essentially the rotational equivalent of linear force and is crucial when analyzing rotational motion.

Torque is calculated by multiplying the force applied to an object by the distance from the point of rotation to where the force is applied. The formula is given by:
  • \( \tau = F \times r \)
Where \( \tau \) is the torque, \( F \) is the force applied, and \( r \) is the lever arm or the distance from the axis of rotation.

Understanding torque is important because it helps us analyze how different forces affect an object's motion and stability. In the context of the exercise, torque is the same for both shafts because they are subjected to the same rotational force. Consequently, comparing their reactions to the same torque helps us understand their performance and efficiency.
Polar Moment of Inertia
The Polar Moment of Inertia is a measure used to predict an object's resistance to torsional loading or twisting. It essentially describes how the area of a cross-section is distributed around an axis. For engineers, it is vital because it directly influences the shaft's ability to resist stress and deformation when torque is applied.

Depending on the cross-section shape, the Polar Moment of Inertia can have varying formulas. For example:
  • Square shaft \( J = \frac{b^4}{6} \)
  • Circular shaft \( J = \frac{\pi b^4}{32} \)
In general, a larger Polar Moment of Inertia indicates that the material is better equipped to resist twisting.

In the exercise, the different calculations of \( J \) between the square and circular shafts are crucial to understanding why their stress responses differ, even though the same torque is applied. This concept helps explain why engineers opt for different cross-sectional shapes depending on the mechanical requirements.
Cross-Sectional Area
Cross-sectional area is a key characteristic that affects a material's mechanical properties and its behavior under load. It refers to the surface area of the shape obtained when an object is sliced through perpendicular to its axis.

The cross-sectional area determines how stress is distributed in a material when forces such as tension, compression, or shear are applied. In cases involving torque, the cross-sectional shape and area play a significant role in determining how well a shaft can handle the applied loads.

- For a square cross-section, the area is \( b \times b = b^2 \).
- For a circular cross-section, the area is \( \pi \left( \frac{b}{2} \right)^2 \).

In the exercise, the different areas for the square and circular shafts influence how each one handles shearing stress due to torque. When analyzing the maximum shearing stress, understanding the cross-sectional area is critical, as it helps illustrate why certain shapes are more effective in specific torque scenarios. By recognizing how the shape alters the stress distribution, students gain insight into designing efficient mechanical components.

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Most popular questions from this chapter

A 2.5 -m-long steel shaft of 30 -mm diameter rotates at a frequency of 30 Hz. Determine the maximum power that the shaft can transmit, knowing that \(G=77.2 \mathrm{GPa}\), that the allowable shearing stress is \(50 \mathrm{MPa}\), and that the angle of twist must not exceed \(7.5^{\circ}\)

A 1.5 -m-long tubular steel shaft \((G=77.2 \mathrm{GPa})\) of 38 -mm outer diameter \(d_{1}\) and \(30-\mathrm{mm}\) inner diameter \(d_{2}\) is to transmit \(100 \mathrm{kW}\) between a turbine and a generator. Knowing that the allowable shearing stress is \(60 \mathrm{MPa}\) and that the angle of twist must not exceed \(3^{\circ},\) determine the minimum frequency at which the shaft can rotate.

Under normal operating conditions a motor exerts a torque of magnitude \(T_{F}=1200\) lb -in. at \(F\). Knowing that \(r_{D}=8\) in., \(r_{G}=3\) in., and the allowable shearing stress is \(10.5 \mathrm{ksi}\) in each shaft, determine the required diameter of \((a)\) shaft \(C D E,(b)\) shaft \(F G H\)

Shafts \(A\) and \(B\) are made of the same material and have the same length and cross-sectional area, but \(A\) has a circular cross section and \(B\) has a square cross section. Determine the ratio of the maximum values of the angles \(\phi_{A}\) and \(\phi_{B}\) when the two shafts are subjected to the same maximum shearing stress \(\left(\tau_{A}=\tau_{B}\right)\) Assume both deformations to be elastic.

The solid rod \(A B\) has a diameter \(d_{A B}=60 \mathrm{mm}\) and is made of a steel for which the allowable shearing stress is 85 MPa. The pipe \(C D,\) which has an outer diameter of \(90 \mathrm{mm}\) and a wall thickness of \(6 \mathrm{mm},\) is made of an aluminum for which the allowable shearing stress is 54 MPa. Determine the largest torque \(\mathbf{T}\) that can be applied at \(A\)

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