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Shear stress is a type of stress that acts parallel to the cross-sectional area of a material. It arises when a force is applied tangentially and can cause the material to slide along the axis of the applied force.
To calculate shear stress (\( \tau \)), we use the formula:

• \( \tau = \frac{P}{A} \)
• where \( P \) is the applied force and \( A \) is the cross-sectional area.

In our exercise, the rod's cross section is circular. The area \( A \) is calculated as \( \frac{\pi d^2}{4} \), where \( d \) is the diameter of the rod.
Plugging this into the formula, shear stress becomes:

• \( \tau = \frac{4P}{\pi d^2} \)

This illustrates how shear stress depends on the applied force and inversely on the square of the diameter of the rod. By determining the maximum shear stress allowed, we ensure the rod does not fail when subjected to shear forces.

Calculating the minimum diameter of a rod when subjected to stresses requires consideration of both tensile and shear stresses. A circular rod must have enough diameter to withstand the maximum allowable stress, ensuring its structural integrity.
The formula used for tensile stress is:

• \( \sigma = \frac{4P}{\pi d^2} \)
• Given an allowable tensile stress (\( 118 \ \mathrm{MPa} \)), solve for \( d \) to find the rod's minimum diameter.

Let's solve it:

• Set \( \frac{4P}{\pi d^2} \leq 118 \), solve for \( d \).
• For our calculation, insert \( P = 3500 \ \mathrm{N} \) and obtain \( d \approx 6.14 \ \mathrm{mm} \).

Similarly, we solve the minimum diameter for shear stress:

• \( \tau = \frac{4P}{\pi d^2} \) with an allowable shear stress of \( 48 \ \mathrm{MPa} \).
• Insert \( P = 3500 \ \mathrm{N} \), yielding \( d \approx 9.64 \ \mathrm{mm} \).

The critical importance here is picking the larger diameter from the tensile and shear stress calculations, which in this instance is \( 9.64 \ \mathrm{mm} \). This ensures the rod satisfies both stresses.

Allowable stress refers to the maximum stress that a material can withstand under specific conditions without failing. It is crucial in ensuring designs are safe, reliable, and capable of enduring applied loads within limits.
In engineering, allowable stress is derived from the material's strength properties and is typically specified for both tensile and shear stresses:

• Tensile allowable stress: \( 118 \ \mathrm{MPa} \)
• Shear allowable stress: \( 48 \ \mathrm{MPa} \)

These values act as thresholds when designing components like rods, gears, and beams. They are safety margins based on material tests and industry standards.
For our rod scenario, the provided tensile and shear allowable stresses enable us to determine the suitable diameter to prevent material failure. By comparing calculated stresses to allowable limits, we ensure the design is safe against potential breakage due to excessive force.