Question

The in-situ stresses are determined by the Flat-jack method by making three slots (P, Q, and R) on the wall (ABCD) of a mine gallery as shown. The in-situ stresses \( \sigma_P \), \( \sigma_Q \), and \( \sigma_R \) are determined at slot-P, slot-Q, and slot-R, respectively. If \( \sigma_P > \sigma_Q \), and \( \sigma_R = 0 \), the shear stress on the wall is

• A. \( \dfrac{\sigma_P + \sigma_Q}{2} \)
• B. \( \dfrac{\sigma_P - \sigma_Q}{2} \)
• C. \( -\left(\dfrac{\sigma_P + \sigma_Q}{2}\right) \)
• D. \( -\left(\dfrac{\sigma_P - \sigma_Q}{2}\right) \)

Hint:

Remember a flat jack reads the stress perpendicular to its own slot direction, not along it, so the 45 degree slot actually measures the 135 degree direction.

Answer 1

The Correct Option is A

Step 1: Set up what the flat jack actually measures.
A flat jack slot is cut into the rock, and the surrounding rock tries to close that slot. The pressure needed in the jack to just stop the closure equals the in-situ normal stress acting perpendicular to the slot's flat face, in the plane of the wall.
Slot-P is cut as a vertical slit, so its flat face is vertical, and it measures the normal stress acting along the horizontal direction of the wall, call this \( \sigma_x \). So \( \sigma_P = \sigma_x \).
Slot-Q is cut as a horizontal slit, so it measures the normal stress acting along the vertical direction of the wall, call this \( \sigma_y \). So \( \sigma_Q = \sigma_y \).
Slot-R is cut at 45 degrees to the horizontal. The stress it measures acts perpendicular to its own flat face, which is 90 degrees away from the slot's own line, so the direction it measures is at 135 degrees from the horizontal x-axis, not 45 degrees.

Step 2: Write the normal stress transformation formula.
For a 2D stress state with components \( \sigma_x \), \( \sigma_y \) and shear \( \tau_{xy} \), the normal stress on a plane whose outward normal makes angle \( \theta \) with the x-axis is
\[ \sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta) \]

Step 3: Apply it to slot-R's direction, \( \theta = 135^\circ \).
At \( \theta = 135^\circ \), \( 2\theta = 270^\circ \), so \( \cos(2\theta) = 0 \) and \( \sin(2\theta) = -1 \).
\[ \sigma_R = \frac{\sigma_x + \sigma_y}{2} + 0 - \tau_{xy} \]
Since \( \sigma_x = \sigma_P \) and \( \sigma_y = \sigma_Q \), this becomes
\[ \sigma_R = \frac{\sigma_P + \sigma_Q}{2} - \tau_{xy} \]

Step 4: Solve for the shear stress.
\[ \tau_{xy} = \frac{\sigma_P + \sigma_Q}{2} - \sigma_R \]
We are told \( \sigma_R = 0 \), so
\[ \tau_{xy} = \frac{\sigma_P + \sigma_Q}{2} \]
The condition \( \sigma_P > \sigma_Q \) only tells us the two normal stresses are unequal, it does not change this formula for the shear stress.

Final Answer:
The shear stress on the wall is \( (\sigma_P + \sigma_Q)/2 \), which rules out the options with a minus sign inside the bracket (those would come from mixing up \( \sigma_P \) and \( \sigma_Q \)) and the options with an overall negative sign in front (those would come from wrongly treating slot-R's measuring direction as 45 degrees instead of the correct 135 degrees). \[ \boxed{\tau = \dfrac{\sigma_P + \sigma_Q}{2}} \]