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

A flat jack slot always measures the normal stress acting perpendicular to the plane of the slot, never the stress along the slot's own length. Slot-P is a vertical cut, so it reads the horizontal normal stress \( \sigma_x \). Slot-Q is a horizontal cut, so it reads the vertical normal stress \( \sigma_y \). Slot-R is cut along the 45 degree diagonal, so the direction it actually reads is perpendicular to that diagonal, which sits at 135 degrees from the horizontal axis.

On Mohr's circle, a plane at 135 degrees in the real wall plots at twice that angle, 270 degrees, measured around from the point that represents \( \sigma_x \). Going 270 degrees around the circle from \( \sigma_x \) lands on the point directly below the circle's centre, where the shear stress reaches its full negative value while the normal stress sits exactly at the centre value.

1. The centre of Mohr's circle is the average normal stress, \( (\sigma_x + \sigma_y)/2 = (\sigma_P + \sigma_Q)/2 \).
2. At the point 270 degrees around, the reading is that centre value minus the shear, so \( \sigma_R = (\sigma_P + \sigma_Q)/2 - \tau \).
3. We are told \( \sigma_R = 0 \), so \( \tau = (\sigma_P + \sigma_Q)/2 \).

Notice this result does not depend on which of \( \sigma_P \) or \( \sigma_Q \) is bigger. The condition \( \sigma_P > \sigma_Q \) in the question only tells us the wall is not under equal biaxial stress in both directions, it is not an extra condition that changes the formula.

Let's summarize:

• Slot-P and slot-Q, cut at 0 and 90 degrees, hand us \( \sigma_x \) and \( \sigma_y \) directly.
• Slot-R, cut along the diagonal, actually reads the stress at 135 degrees, not 45 degrees, because the flat jack measures perpendicular to its own slot.
• Setting \( \sigma_R = 0 \) in the transformation equation isolates the shear stress as \( (\sigma_P + \sigma_Q)/2 \).

So the shear stress on the wall is \( (\sigma_P + \sigma_Q)/2 \).