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Urgent Question Soil Mechanics -- Please Help
A saturated over-consolidated silty clay is estimated to have the effective strength parameters
c'=5 kPa
∅'=25 kPa
A specimen of the silty is mounted into a triaxial cell and full consolidated to 400 kPa with a cell pressure of 500 kPa and a back pressure of 100 kPa. With the drainage then turned off, the specimen is slowly loaded axially failure. At failure, the porewater pressure was measured to be 160 kPa and the diameter of the specimen was calculated to be 103.47 mm.
Estimate
1. the effective deviator stress at failure (σ1'-σ3')
2. the total deviator stress at failure (σ1-σ3), and
3. the load which would be needed to be applied to the loading piston to cause failure.
σ1=σ'1+u
deviator stress =deviator load/area
cell pressure =500 kPa, back pressure=100 kPa, σ3= 500-100=400 kPa
u(failure)=160 kPa
Area at failure = (∏ x diameter^2)/4=(∏x103.47^2)/4=8408.50 mm^2
1. Deviator stress= deviator load/area= (400 x 10^3)/8408.50=47.57 kPa
σ'3 at failure = 400-160=240 kPa
σ'1 at failure= σ'3 +deviator stress = 240 +47.57=287.57 kPa
2. σ1=σ1'+u=287.57+160=447.57 kPa σ3=σ3'+u =240+160=400 kPa
3. Not sure how to do this. Was thinking of drawing the mohr circles using the previous calculations. However I feel like there is something wrong with the above as when I sketched the mohr circles I got a strange circle.
I would greatly appreciate any help on this question as I am getting prepared for a final exam in my engineering degree. My exam is in 2 days and I would really like to know if I am on the right track or not. Thanks in advance.
Homework Statement
A saturated over-consolidated silty clay is estimated to have the effective strength parameters
c'=5 kPa
∅'=25 kPa
A specimen of the silty is mounted into a triaxial cell and full consolidated to 400 kPa with a cell pressure of 500 kPa and a back pressure of 100 kPa. With the drainage then turned off, the specimen is slowly loaded axially failure. At failure, the porewater pressure was measured to be 160 kPa and the diameter of the specimen was calculated to be 103.47 mm.
Estimate
1. the effective deviator stress at failure (σ1'-σ3')
2. the total deviator stress at failure (σ1-σ3), and
3. the load which would be needed to be applied to the loading piston to cause failure.
Homework Equations
σ1=σ'1+u
deviator stress =deviator load/area
The Attempt at a Solution
cell pressure =500 kPa, back pressure=100 kPa, σ3= 500-100=400 kPa
u(failure)=160 kPa
Area at failure = (∏ x diameter^2)/4=(∏x103.47^2)/4=8408.50 mm^2
1. Deviator stress= deviator load/area= (400 x 10^3)/8408.50=47.57 kPa
σ'3 at failure = 400-160=240 kPa
σ'1 at failure= σ'3 +deviator stress = 240 +47.57=287.57 kPa
2. σ1=σ1'+u=287.57+160=447.57 kPa σ3=σ3'+u =240+160=400 kPa
3. Not sure how to do this. Was thinking of drawing the mohr circles using the previous calculations. However I feel like there is something wrong with the above as when I sketched the mohr circles I got a strange circle.
I would greatly appreciate any help on this question as I am getting prepared for a final exam in my engineering degree. My exam is in 2 days and I would really like to know if I am on the right track or not. Thanks in advance.