Question

The block diagrams below show states of stress on soil elements. For each state of stress, perform the following tasks:

1. Sketch a Mohr circle of stress

2. Identify the major and minor principal stresses, σ1 and σ3, and the maximum shear stress, τmax

3. Compute the mean stress and shear stress invariants, p and q. Note that p = 1⁄2(σ1+ σ3) and q = σ1-σ3 for two-dimensional stress conditions where we ignore the intermediate principal stress, σ2.

4. Identify the pole for each diagram.

5. For the block diagrams in parts (a) through (d), compute the normal and shear stresses on a block diagram rotated 30° from horizontal, as shown in part (e) below. Compute these stresses using the coordinate transformation equations, and using a graphical solution based on the Mohr circle.

40 kPa 50 KPa > 10 kPa 20 kPa 30 kPa 40 kPa 100 kPa 60 kPa 0 kPa 30 kPa 7? 20 kPa 45

Answer 1

Solution:

2. Major and minor principal stresses:

a. major = 50 , minor =30 , no shear stress

b. $\sigma$ x = 40, $\sigma$ y = 20, shear =10, max shear stress = ( major - minor) / 2

c. $\sigma$ x = 40 , $\sigma$ y=20 , shear stress is always positive so 10 kPa

for both b and c ,

$\sigma$ 1 / $\sigma$ 2 = ( $\sigma$ x + $\sigma$ y )/2 +/- sqrt ( ( $\sigma$ x - $\sigma$ y)²/4 +   $\tau$ ² }

$\sigma$ 1 = 44.13 kPa , $\sigma$ 2 = 15.858 kPa , max shear stress = 14.136kPa

d. when a stress block is rotated by an angle 45 degrees then then it lies on principal plane,

here, it is a case of pure shear

$\sigma$ 1 = shear stress = 30 kPa

$\sigma$ 2 = -shear stress = -30kPa

3. p and q :

a. p = ( $\sigma$ 1 + $\sigma$ 3) /2 , q = $\sigma$ 1 - $\sigma$ 3

$\sigma$ 3 = $\sigma$ 2 in 2 -D

p = 40kPa , q= 20 kPa

for b and c, p = 29.994kPa , q = 28.272kPa

d. p = 0 , q= 60kPa

5. diagram in a is rotated 30 degrees

normal stress $\sigma$ x' = ( $\sigma$ x + $\sigma$ y)/2 + ( $\sigma$ x - $\sigma$ y)/2 * cos2 $\theta$ + $\tau$ *sin2 $\theta$

as there is no shear stress originally in the block the shear stress component $\tau$ = 0

so putting the values , $\sigma$ x = 30 , $\sigma$ y = 50

$\sigma$ x' = 40 + 5 = 45kPa

$\sigma$ y' = ( $\sigma$ x + $\sigma$ y)/2 - ( $\sigma$ x - $\sigma$ y)/2 * cos2 $\theta$

= 40 - 5 = 35kPa

shear stress = -sin2 $\theta$ * ( $\sigma$ x - $\sigma$ y) /2

= -8.66 kPa