Question

If \( \sigma_1 \) and \( \sigma_2 \) are two principal stresses, then the radius of Mohr’s circle is

• A. \( \sigma_1 + \sigma_2 \)
• B. \( \sigma_1 - \sigma_2 \)
• C. \( \dfrac{\sigma_1 + \sigma_2}{2} \)
• D. \( \dfrac{\sigma_1 - \sigma_2}{2} \)

Hint:

In Mohr’s circle, diameter represents principal stresses, and radius gives the maximum shear stress.

Answer 1

The Correct Option is D

Step 1: Understanding Mohr’s circle.
Mohr’s circle is a graphical representation of the state of stress at a point. The principal stresses \( \sigma_1 \) and \( \sigma_2 \) lie at the extreme ends of the diameter of Mohr’s circle.
Step 2: Determining the center of Mohr’s circle.
The center of Mohr’s circle lies at the average of the principal stresses: \[ \text{Center} = \frac{\sigma_1 + \sigma_2}{2} \]
Step 3: Calculating the radius.
The radius of Mohr’s circle is half the difference of the principal stresses: \[ \text{Radius} = \frac{\sigma_1 - \sigma_2}{2} \]
Step 4: Conclusion.
Hence, the radius of Mohr’s circle is \( \dfrac{\sigma_1 - \sigma_2}{2} \).

Answer 2

Mohr's circle plots normal stress on the horizontal axis and shear stress on the vertical axis, with the two principal stresses \( \sigma_1 \) and \( \sigma_2 \) marking the two ends of its horizontal diameter. Checking each option against this construction:

1. \( \sigma_1 + \sigma_2 \): This equals twice the coordinate of the circle's centre on the stress axis, not its radius. If \( \sigma_1 = \sigma_2 \), this sum stays nonzero even though the circle should shrink to a point (radius zero), so it cannot be the radius.
2. \( \sigma_1 - \sigma_2 \): Since \( \sigma_1 \) and \( \sigma_2 \) sit at opposite ends of the circle, this difference is the full diameter, not the radius. Using it directly would double the true maximum shear stress the circle represents.
3. \( \dfrac{\sigma_1 + \sigma_2}{2} \): This is the average normal stress, which locates the centre of the circle on the horizontal axis. It does not vanish when \( \sigma_1 = \sigma_2 \), so it cannot describe the size (radius) of the circle.
4. \( \dfrac{\sigma_1 - \sigma_2}{2} \): Half the distance between the two diametrically opposite points equals the distance from the centre to either point, which is exactly the radius. It correctly reduces to zero when \( \sigma_1 = \sigma_2 \), matching the physical case of equal biaxial stress where no shear acts on any plane.

Only \( \dfrac{\sigma_1 - \sigma_2}{2} \) satisfies both the geometric definition of a radius (half of a diameter) and the physical check of vanishing when the two principal stresses coincide.

Therefore, the correct answer is \( \dfrac{\sigma_1 - \sigma_2}{2} \).