Question

In the case of bi-axial state of normal stresses, the normal stress on 45\(^\circ\) plane is equal to ________.

• A. the sum of the normal stresses
• B. the difference of normal stresses
• C. half the sum of the normal stresses
• D. half the difference of normal stresses

Hint:

At $\theta = 45^\circ$, the normal stress is simply the average of the two normal stresses.
This average is geometrically represented by the center coordinates of Mohr's circle of stress.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to find the value of normal stress acting on an inclined plane at an angle of \(45^\circ\) under a biaxial state of stress (where only two normal stresses act on mutually perpendicular planes).

Step 2: Key Formula or Approach:
The normal stress \(\sigma_{\theta}\) on an inclined plane at an angle \(\theta\) under biaxial normal stresses \(\sigma_x\) and \(\sigma_y\) is:
\[ \sigma_{\theta} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) \]

Step 3: Detailed Explanation:

• Identify the given state of stress:
Let the normal stresses acting along the two principal axes be \(\sigma_x\) and \(\sigma_y\).
The angle of the inclined plane is \(\theta = 45^\circ\).

• Substitute \(\theta = 45^\circ\) into the normal stress transformation equation:
\[ \sigma_{45^\circ} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2 \times 45^\circ) \] \[ \sigma_{45^\circ} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(90^\circ) \]
• Since \(\cos(90^\circ) = 0\), the second term vanishes:
\[ \sigma_{45^\circ} = \frac{\sigma_x + \sigma_y}{2} \]
• This mathematical expression represents half the sum of the normal stresses.

Step 4: Final Answer:
The normal stress on the \(45^\circ\) plane is equal to half the sum of the normal stresses.