When a body is subjected to bi-axial stress, i.e. direct stress (P1) and (P2) in two mutually perpendicular planes accompanied by a simple shear stress (q), then maximum normal stress is 

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BPSC AE Paper IV General Engineering 19 Dec 2024 Official Paper
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  1. \(\rm \frac{P_1-P_2}{2}+\frac{1}{2} \sqrt{\left(P_1+P_2\right)^2+4 q^2} \)
  2. \( \rm \frac{P_1-P_2}{2}-\frac{1}{2} \sqrt{\left(P_1+P_2\right)^2+4 q^2}\)
  3. \( \rm \frac{P_1+P_2}{2}-\frac{1}{2} \sqrt{\left(P_1-P_2\right)^2+4 q^2} \)
  4. \(\rm \frac{P_1+P_2}{2}+\frac{1}{2} \sqrt{\left(P_1-P_2\right)^2+4 q^2}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{P_1+P_2}{2}+\frac{1}{2} \sqrt{\left(P_1-P_2\right)^2+4 q^2}\)
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Detailed Solution

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Concept:

When a body is subjected to bi-axial stresses (direct stresses P1 and P2 ) along with a shear stress q , the maximum and minimum normal stresses can be determined using stress transformation equations. The maximum normal stress is particularly important for failure analysis.

Given:

  • Direct stress in the first direction, \( P_1 \)
  • Direct stress in the second direction, \( P_2 \)
  • Shear stress, \( q \)

Step 1: Understand the Stress State

The given stress state consists of two normal stresses P1 and P2 acting in mutually perpendicular directions, along with a shear stress q .

Step 2: Use the Formula for Principal Stresses

The maximum and minimum normal stresses (principal stresses) are given by:

\[ \sigma_{\text{max}}, \sigma_{\text{min}} = \frac{P_1 + P_2}{2} \pm \frac{1}{2} \sqrt{(P_1 - P_2)^2 + 4q^2} \]

Step 3: Identify the Maximum Normal Stress

The maximum normal stress corresponds to the positive root of the expression:

\[ \sigma_{\text{max}} = \frac{P_1 + P_2}{2} + \frac{1}{2} \sqrt{(P_1 - P_2)^2 + 4q^2} \]

 

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