Question

Which planes carry maximum normal stress and what is shear stress on these planes

• A. Maximum shear planes, Maximum
• B. Principal planes, zero
• C. Oblique planes, Minimum
• D. Oblique planes, Maximum

Hint:

On Mohr's Circle, the principal stresses are located where the circle crosses the horizontal axis. Since it sits directly on the horizontal axis, the vertical coordinate (which represents shear stress) must equal zero!

Answer 1

The Correct Option is B

Concept: In stress analysis and continuum mechanics, when a material body is subjected to complex multi-axial loading conditions, the internal stress state varies depending on the orientation plane inside the material.

• Principal Planes: Defined as the specific orientation planes where the normal stress vector reaches its maximum or minimum mathematical values.

• Principal Stresses: The maximum and minimum normal stresses acting on these principal planes, denoted by \( \sigma_1 \) and \( \sigma_2 \).

• A fundamental property of these principal planes is that the shear stress (\( \tau \)) is identically zero.

Step 1: Using Mohr's Circle transformation equations to analyze stress behavior.
Let us analyze this using the standard 2D stress transformation equations for a plane oriented at an angle \( \theta \): \[ \sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta) \] \[ \tau_\theta = -\frac{\sigma_x - \sigma_y}{2}\sin(2\theta) + \tau_{xy}\cos(2\theta) \]

Step 2: Finding the maximum condition for normal stress.
To find the orientation angle \( \theta \) that maximizes the normal stress \( \sigma_\theta \), we take its derivative with respect to \( \theta \) and set it equal to zero: \[ \frac{d\sigma_\theta}{d\theta} = 0 \] Differentiating the normal stress expression: \[ \frac{d\sigma_\theta}{d\theta} = -2 \cdot \frac{\sigma_x - \sigma_y}{2}\sin(2\theta) + 2\tau_{xy}\cos(2\theta) = 0 \] Simplifying this derivative equation: \[ -(\sigma_x - \sigma_y)\sin(2\theta) + 2\tau_{xy}\cos(2\theta) = 0 \quad \cdots (1) \]

Step 3: Comparing the derivative to the shear stress equation.
Now let us compare equation (1) directly with our expression for the transformed shear stress \( \tau_\theta \): \[ \tau_\theta = -\frac{\sigma_x - \sigma_y}{2}\sin(2\theta) + \tau_{xy}\cos(2\theta) \] Notice that we can rewrite equation (1) by factoring out a 2: \[ 2 \cdot \left[ -\frac{\sigma_x - \sigma_y}{2}\sin(2\theta) + \tau_{xy}\cos(2\theta) \right] = 0 \] The expression inside the brackets is exactly equal to \( \tau_\theta \). Therefore, this simplifies to: \[ 2\tau_\theta = 0 \quad \Rightarrow \quad \tau_\theta = 0 \]

Step 4: Evaluating the options.
This proof shows that on the specific planes where normal stress is maximized (defined as Principal Planes), the shear stress values must equal zero. This matches the statement in Option (B).