Horizontal soil pressure

[owillis28]

A clay deposit has a unit weight of 100 lbf/ft3 and an unconfined compressive strength of 2000 lbf/ft2.

What is most nearly the depth at which the horizontal soil pressure is zero?

After some reading, I found equation 35.44, whch defines unconfined compressive strength as P/A. This would make sense if the clay deposit was in a cylinder but the problems does not provide any information. I am missing something? Is this problem really easy and I am making it seem difficult?

Maybe its just late and I need to get some sleep.

owillis

 

[Guest]

If you are thinking horizontal pressure, consider a wall with active pressure acting against it (assumed).

So this case is characterized as:

1. Clay Soil (cohesive)

2. Active pressure

What do we know?

1. Governing Equation: Horizontal Pressure = (Unit Weight)*Depth - 2*Cohesion

2. Unit Wt = 100 pcf; Unconfined Compressive Strength = 2,000 psf (Given in Problem Statement)

3. Horizontal Pressure = 0 (Given in Problem Statement)

What can we Assume?

Cohesion = 0.5*(Unconfined Compressive Strength)

[SIZE=14pt]SOLUTION[/SIZE]

So, solving first for Cohesion = 0.5*(2,000 psf) = 1,000 psf

Now that we have cohesion and we unit weight is provided in the problem statement, we can write the equation:

0 = (100 pcf)*Depth - 2*(1,000 psf)

Rearranging the equation, you can solve for depth:

Depth = 20 feet

Please let me know if you have any questions regarding the solution.

JR

 

[owillis28]

If you are thinking horizontal pressure, consider a wall with active pressure acting against it (assumed).
So this case is characterized as:

1. Clay Soil (cohesive)

2. Active pressure

What do we know?

1. Governing Equation: Horizontal Pressure = (Unit Weight)*Depth - 2*Cohesion

2. Unit Wt = 100 pcf; Unconfined Compressive Strength = 2,000 psf (Given in Problem Statement)

3. Horizontal Pressure = 0 (Given in Problem Statement)

What can we Assume?

Cohesion = 0.5*(Unconfined Compressive Strength)

[SIZE=14pt]SOLUTION[/SIZE]

So, solving first for Cohesion = 0.5*(2,000 psf) = 1,000 psf

Now that we have cohesion and we unit weight is provided in the problem statement, we can write the equation:

0 = (100 pcf)*Depth - 2*(1,000 psf)

Rearranging the equation, you can solve for depth:

Depth = 20 feet

Please let me know if you have any questions regarding the solution.

JR

Thanks, JR

I was able to come up with the answer by working backwards and using the equations that I had mentioned in my post. The only problem what that I didn't understand the problem. I was not able to find and explanation that could clear up my confusion on the problem. Thank goodness for engineerboards.com!!!!!!!!!!!

owillis

 

[squishles10]

Yeah, I had no clue on that one either. :-(

 

[owillis28]

Yeah, I had no clue on that one either. :-(

I thought I was going to be the only one that didn't understand that problem.

owillis

 

[Guest]

Yeah, I had no clue on that one either. :-(

Do you know where to go in your references to find the information I presented?

JR

 

[owillis28]

Do you know where to go in your references to find the information I presented?
JR

Just the CERM equation I referenced in the first post (equation 35.44). This equation defines unconfined compressive strength as pressure divided by area (P/A).

One of my problems with geotech is the visualization. I didn't even think to look at the horizontal pressure in terms of a wall with active pressure acting against it. Sometimes I get so tuned in to finding the correct equation to plug and chug that I don't look at the concepts behind it.

owillis

 

[squishles10]

If you are thinking horizontal pressure, consider a wall with active pressure acting against it (assumed).

Where did you get a wall from that question? You said assumed, is it always assumed that there is a wall if there is horizontal pressure?

Cohesion = 0.5*(Unconfined Compressive Strength)

Where did this come from?

 

[EL Nica PE]

Nicely done JR!

Cohesion=Unconfined Compressive Strength/2

cohesion is also referred as Undrained Shear strength (CERM Eq. 35.42 10th ed)

 

[Guest]

If you are thinking horizontal pressure, consider a wall with active pressure acting against it (assumed).

Where did you get a wall from that question? You said assumed, is it always assumed that there is a wall if there is horizontal pressure?

I stated that there was a wall so you could visualize the interaction in your mind. In actuality, there could simply be an invisible plane where the horizontal forces are acting and the principles would be the same. I recommend thinking of it as a wall because then your can wrap your mind around that picture.

Cohesion = 0.5*(Unconfined Compressive Strength)
Where did this come from?

El Nica has you covered above in regards to the CERM 10th Ed. For me, it is represented in Fundamentals of Geotechnical Engineering, Das, p. 309.

Side Note: the depth at which the horizontal pressure become zero is noteworthy because it is the critical depth at which tensile cracks as the soil-wall interface will develop in a cohesive soil.

Nicely done JR!

Thanks! I get one right every now and then ..

JR

 

[squishles10]

I stated that there was a wall so you could visualize the interaction in your mind. In actuality, there could simply be an invisible plane where the horizontal forces are acting and the principles would be the same. I recommend thinking of it as a wall because then your can wrap your mind around that picture.
I still don't get this part. I got that the shear strength is 1000, but I don't know what to do with it. I don't see any equation that has horizontal pressure in it, and there's nothing in the index. If this is active pressure, what is passive pressure. Nothing is moving, that says passive to me.

El Nica has you covered above in regards to the CERM 10th Ed. For me, it is represented in Fundamentals of Geotechnical Engineering, Das, p. 309. <----- I got this part now.

JR

 

[EL Nica PE]

OK. this formula is a simply formula of tension crack with phi=0 therefore ka=1. Check Das Foundation Engineering p342 formula 6.14 fourth ed.

Let me know if you don't have this book I will scan the page for you.

 

[squishles10]

OK. this formula is a simply formula of tension crack with phi=0 therefore ka=1. Check Das Foundation Engineering p342 formula 6.14 fourth ed.Let me know if you don't have this book I will scan the page for you.

Good grief. Okay, I have Das Fundamentals of Geotech Engineering- it's on page 309, sec 9.3. When you find the right paragraph it's very clearly outlined but my goodness. Also, how do you know "phi=0 therefore ka=1"?

I've almost got it!

 

[EL Nica PE]

Good grief. Okay, I have Das Fundamentals of Geotech Engineering- it's on page 309, sec 9.3. When you find the right paragraph it's very clearly outlined but my goodness. Also, how do you know "phi=0 therefore ka=1"?
I've almost got it!

b/c the question tells it is a clay deposit therefore phi=0.

Clays=cohesive phi=0 Cu>0

sand =cohesionless phi>0 Cu=0

 

[Guest]

b/c the question tells it is a clay deposit therefore phi=0.Clays=cohesive phi=0 Cu>0

sand =cohesionless phi>0 Cu=0

Those relationships are worth having BOLDED and HIGHLIGHTED to commit to memory!! They are the key to simplifying your problems and getting parameter values that aren't specified in the problem.

JR

 

[squishles10]

Those relationships are worth having BOLDED and HIGHLIGHTED to commit to memory!! They are the key to simplifying your problems and getting parameter values that aren't specified in the problem.
JR

OHHHH- that should have been obvious! Got it- okay, took a little bit more than 6 minutes but I got it! Thanks! Just wait til I get to foundations! (And to think I started my graduate degree in geotech... *sheesh*)

 

[gte959s]

If you are thinking horizontal pressure, consider a wall with active pressure acting against it (assumed).
So this case is characterized as:

1. Clay Soil (cohesive)

2. Active pressure

What do we know?

1. Governing Equation: Horizontal Pressure = (Unit Weight)*Depth - 2*Cohesion

2. Unit Wt = 100 pcf; Unconfined Compressive Strength = 2,000 psf (Given in Problem Statement)

3. Horizontal Pressure = 0 (Given in Problem Statement)

What can we Assume?

Cohesion = 0.5*(Unconfined Compressive Strength)

[SIZE=14pt]SOLUTION[/SIZE]

So, solving first for Cohesion = 0.5*(2,000 psf) = 1,000 psf

Now that we have cohesion and we unit weight is provided in the problem statement, we can write the equation:

0 = (100 pcf)*Depth - 2*(1,000 psf)

Rearranging the equation, you can solve for depth:

Depth = 20 feet

Please let me know if you have any questions regarding the solution.

JR

Wow that was easier than I thought! This is a good thing to know for the AM section!