In-class problem For maximum and minimum stresses of 600 and 200 mega-pascals (MPa) oriented as a vertical vector and a horizontal, E-W striking vector (respectively), determine the normal and shear stresses on a plane oriented North-South, 45 degrees East. It helps to first draw a block diagram. So max stress is oriented vertically and equal to 600 MPA Min stress is horizontal, oriented east-west and = 200 MPa

For max and min stresses of 600, 200 (MPa) oriented as a vertical vector and a horizontal, E-W striking vector (respectively), determine the normal and shear stresses on a plane oriented North-South, 45E. It helps to first draw a block diagram. So max stress is oriented vertically and equal to 600 MPA Min stress is horiz., east-west = 200 Mpa Key: Normal stress = mean stress: 400 MPa; Shear stress 200 MPa

sn = (600 + 200) - (600 - 200) cos 90 2 = 400 ss = (600 - 200) sin 90 For the minimum and maximum principle stresses of 600 and 200 megapascals (MPa) oriented as a vertical vector and a horizontal, E-W striking vector (respectively), determine the normal and shear stresses on a plane oriented North-South, 45 degrees East sn = (600 + 200) - (600 - 200) cos 90 2 = 400 ss = (600 - 200) sin 90 2 = 200

Mean Stress Differential Stress For the stress state in the previous problem, determine the differential stress and mean stress. Start by plotting the solution for normal and shear stresses on the Mohr Stress Diagram. Mean Stress Differential Stress

Determine whether decreasing the dip of the fault will decrease or increase the shear stress acting on it.

Determine whether decreasing the dip of the fault will decrease or increase the shear stress acting on it. It will decrease the shear stress since decreasing the dip of the fault would decrease theta, the angle between the largest principal stress and the normal to the plane the stress is acting on, (the plane is rotating closer to 90 deg. or normal to the maximum stress)

In-class problem 2 Discuss how a change in differential stress might affect whether a rock might be more or less likely to break. It may help by arbitrarily varying the stresses and looking at how they plot on the circle, or by imagining stress on a cube. If the differential stress is decreased (making the stresses closer in value, so the mohr circle shrinks), it would make it less likely to intersect a failure envelope and break. If the differential stress was increased, there would be a bigger difference between the two stresses, the Mohr circle would increase in diameter and more likely to intersect the failure envelope making it more likely to break.

Failure envelope

In-class problem 2 Now discuss whether increasing the mean stress would cause a rock to break more readily. Would this be more or less likely with increasing depth in the crust? Increasing the mean stress would move the circle to the right (increasing normal) stress, resulting in the rock being less likely to break. Given the shape of the envelope, increasing the mean stress will never bring the rock closer to failure (assuming differential stress is not increased as well). Mean stress increases with depth in the crust (and we are assuming differential stress does not increase as well). Intuitively, the difference in stress would have to increase for a rock to break under a higher mean stress.

Draw the stress state where the minimum and maximum stresses are both equal to 600 MPa. The principle stresses are thus equal to the mean stress and differential stress is zero.