Let the state of stress at a point in a body be the difference of two plane states of stress shown in the figure. Consider all the possible planes perpendicular to the x-y plane and passing through that point. The magnitude of the maximum compressive stress on any such plane is \( k \sigma_0 \), where \( k \) is equal to ___________ (round off to one decimal place).
Show Hint
To find the maximum stress in transformed coordinates, always use the stress transformation equations and account for both the normal and shear stresses.
We are given the difference between two states of stress. The stress states are shown as:
1. First Stress State (square with stresses \( 3\sigma_0 \)):
- In this case, the stress is uniform along both axes in the \( x \)- and \( y \)-directions.
2. Second Stress State (diamond with stresses \( 3\sigma_0 \) along \( x \) and \( y \) at 45\(^\circ\)).
We now need to calculate the resultant stress on all planes perpendicular to the \( x \)-\( y \) plane.
The stress transformation can be done using the following formula for stress on any inclined plane:
\[
\sigma_A/\sigma_B = \frac{\sigma_x + \sigma_y}{2} \pm \frac{1}{2} \sqrt{(\sigma_x - \sigma_y)^2 + 4 \tau^2}
\]
Given the stresses:
\[
\sigma_x = 1.5 \sigma_0, \quad \sigma_y = -1.5 \sigma_0, \quad \tau_{xy} = 1.5 \sigma_0
\]
Substituting the values:
\[
\sigma_A/\sigma_B = \frac{1.5 \sigma_0 + (-1.5 \sigma_0)}{2} \pm \frac{1}{2} \sqrt{(1.5 \sigma_0 - (-1.5 \sigma_0))^2 + 4(1.5 \sigma_0)^2}
\]
This simplifies to:
\[
\sigma_A/\sigma_B = \pm \frac{1}{2} \sqrt{(3 \sigma_0)^2 + 4(1.5 \sigma_0)^2}
\]
\[
= \pm \frac{1}{2} \sqrt{9 \sigma_0^2 + 9 \sigma_0^2} = \pm \frac{1}{2} \sqrt{18 \sigma_0^2} = \pm \sqrt{9 \sigma_0^2} = \pm 3 \sigma_0
\]
Thus, the maximum compressive stress is:
\[
k \sigma_0 = 2.1 \sigma_0
\]
Hence, the correct value of \( k \) is: \( \boxed{2.1} \).
