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

Question

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

Answer 1

Step 1: On a plane inclined at angle $ \theta $ to the principal plane, the normal and shear stresses are $ \sigma_n = \dfrac{\sigma_1+\sigma_2}{2} + \dfrac{\sigma_1-\sigma_2}{2}\cos 2\theta $ and $ \tau = \dfrac{\sigma_1-\sigma_2}{2}\sin 2\theta $.
Step 2: These two equations describe a circle in the $ \sigma_n $-$ \tau $ plane centred at $ \left(\dfrac{\sigma_1+\sigma_2}{2}, 0\right) $, since $ \left(\sigma_n - \dfrac{\sigma_1+\sigma_2}{2}\right)^2 + \tau^2 = \left(\dfrac{\sigma_1-\sigma_2}{2}\right)^2 $.
Step 3: Comparing with the standard circle equation, the radius is the coefficient on the right-hand side:
\[ \text{Radius} = \dfrac{\sigma_1 - \sigma_2}{2} \]

Answer 2

Take a concrete pair of principal stresses, say $ \sigma_1 = 60 $ units and $ \sigma_2 = 20 $ units, for which the well-known maximum shear stress is $ \tau_{\max} = \dfrac{\sigma_1-\sigma_2}{2} = 20 $ units, and this maximum shear stress is exactly what the radius of Mohr's circle represents. Testing each option against this number:

$ \sigma_1+\sigma_2 $: Gives $ 60+20=80 $, far larger than the known maximum shear stress of 20, so this option fails the check.
$ \sigma_1-\sigma_2 $: Gives $ 60-20=40 $, exactly double the correct value of 20, confirming this is the diameter rather than the radius.
$ \dfrac{\sigma_1+\sigma_2}{2} $: Gives $ \dfrac{80}{2}=40 $, which also does not equal 20 and instead corresponds to the average (centre) stress in this example.
$ \dfrac{\sigma_1-\sigma_2}{2} $: Gives $ \dfrac{40}{2}=20 $, matching the known maximum shear stress exactly, confirming this expression correctly gives the radius.

The numeric check confirms that only $ \dfrac{\sigma_1-\sigma_2}{2} $ reproduces the correct maximum shear stress, which is the radius of Mohr's circle.

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

Answer 3

Take a concrete pair of principal stresses, say $ \sigma_1 = 60 $ units and $ \sigma_2 = 20 $ units, for which the well-known maximum shear stress is $ \tau_{\max} = \dfrac{\sigma_1-\sigma_2}{2} = 20 $ units, and this maximum shear stress is exactly what the radius of Mohr's circle represents. Testing each option against this number:

$ \sigma_1+\sigma_2 $: Gives $ 60+20=80 $, far larger than the known maximum shear stress of 20, so this option fails the check.
$ \sigma_1-\sigma_2 $: Gives $ 60-20=40 $, exactly double the correct value of 20, confirming this is the diameter rather than the radius.
$ \dfrac{\sigma_1+\sigma_2}{2} $: Gives $ \dfrac{80}{2}=40 $, which also does not equal 20 and instead corresponds to the average (centre) stress in this example.
$ \dfrac{\sigma_1-\sigma_2}{2} $: Gives $ \dfrac{40}{2}=20 $, matching the known maximum shear stress exactly, confirming this expression correctly gives the radius.

The numeric check confirms that only $ \dfrac{\sigma_1-\sigma_2}{2} $ reproduces the correct maximum shear stress, which is the radius of Mohr's circle.

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

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

Show Hint

The Correct Option is D

Approach Solution - 1

Approach Solution -2

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