Chapter 10

Problem 10.1

Given the primary normal stresses $\sigma_1 $ and $ \sigma_3 $ determine the Normal and Shear streee on a fault plane that strikes along $\sigma_2$ and dips at an angle

The direction of $\sigma_2$ is usually along the fault plane

$ \sigma_1 $ is vertical (as stated in the problem)

$ \sigma_3 $ is horizontal (as stated in the problem)

$\theta $ is the angle from the horizontal (same as the dip) for *this* problem. See figure P10.1 for help.

$\sigma_n = \dfrac{\sigma_1 + \sigma_3}{2} + \dfrac{\sigma_1 - \sigma_3}{2} \cos (2\theta)$

$\sigma_s = \dfrac{\sigma_1 - \sigma_3}{2} \sin (2\theta)$

Problem 10.2

Plot the normal and shear stress on the graph provided in P10.1

Plot the Mohr Circle with $ \sigma_1 = 100 MPa$ and $ \sigma_3 = 20 MPa$

Problem 10.3

Problem 10.5

Plot the experiments.
Draw the failure envelope. It is a straight line that touches all experiments at a single point
Meausure the slope of the line or angle of internal friction, $\phi$.
Convert $\phi$ to the columb coefficient, $\mu$ using $\mu = \tan \phi$
Finally, determine the angle of fracture plane with: $$\theta = 45^\circ + \frac{\phi}{2}$$

Problem 10.9

Oh that is a lot of text

Plot the Mohr Circle for the regional stress state $ \sigma_1 = 10 Mpa $ and $ \sigma_3 = 6 MPa $

Draw a Failure envelope with $ \mu = 0.4 $ (the slope) and a Cohesion $ C = 0 $

$ \sigma_s = \mu \sigma_1 + C $ (form of $ y= mx + b $ )

Shift the Mohr circle until it intersects with the Failure envelope. This is the same as increasing the pore fluid pressure or subtracting some pressure from the normal stresses

How far you moved the circle is the Pore fluid pressure