It's been a while since I worked with Mohr's circle. Hope I still remember enough to help you.
If σ1 and σ3 are 40 and 110 respectively, with τ being 20, you can draw the Mohr's circle, with centre at (40+110)/2=75 and radius = √((110-75)²+20²)=40.3 kpa
The principal stresses are therefore
=75 ± 40.3
=34.7/115.3
Remembering that pore pressure shifts the Mohr's circle to the left, what would be the magnitude of the pore pressure to reduce the minor principal stress to zero?
total normal stresses of 110 and 40 kpa are applied to two planes at right angles in an element of soil. the shear stress acting on those palnes is 20 kpa. what must be the value of the pore pressure in the element if the minor principle stress in the elemnt is to be zero?
3 answers
a large soil sample obtained from a borrow pit has a wet mass of 26.50kg. the in place volume occupied by the sample is 0.013 m3. a small portion of the sample is used to determine the water content; the wet mass is 135g, and after drying in an oven, the mass is 117g.
a) determine the soil's water content.
b) determine the soil wet and dry density for conditions at the borrow pit.
a) determine the soil's water content.
b) determine the soil wet and dry density for conditions at the borrow pit.
a. 15.4%
b.2038.5 kg/m^3
c.1766.5kg/m^3
b.2038.5 kg/m^3
c.1766.5kg/m^3