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# Six-Minute Solutions: Probs. 38 & 49

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If you have this book, I could really use help resolving some issues with their solutions. I have edition 1, printing 1 and have added in the errata from the PPI site.

In the solution to problem 38, 2nd paragraph on page 49: they state that the change in head for both layers of soil is equivalent to the piezometric head at the base of the bottom layer. They originally had that as 15', but changed it to 12'. Wouldn't the sum of the changes in head be equal to the total difference in head from the figure which appears to be 15'? I tried solving for the head loss through the upper layer using equation 205.11 in Goswami's book where H is the total head dissipated across all layers, and got h1=4.9'. I guess it comes down to how H is determined from the figure for problem 38. What's your call?

I also have a problem with their solution to question 49 on page 54. For the overburden term where the water table is between the ground surface and the footing, they use: [γd(Df) + γw(Dw-Df)]Nq.

Df is the depth to the bottom of the footing from the ground surface.

Dw is the depth to the water table from the ground surface.

According to Goswami's and Das' books, this term should be [γ(Dw) + (γsat- γw)(Df-Dw)] Nq.

They give two different answers. The second one makes sense to me, but the one from the 6-Min. solution does not. There is nothing in their errata regarding this solution, but I just don’t see how theirs could be correct.

Any help would be appreciated.

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If you have this book, I could really use help resolving some issues with their solutions. I have edition 1, printing 1 and have added in the errata from the PPI site.

In the solution to problem 38, 2nd paragraph on page 49: they state that the change in head for both layers of soil is equivalent to the piezometric head at the base of the bottom layer. They originally had that as 15', but changed it to 12'. Wouldn't the sum of the changes in head be equal to the total difference in head from the figure which appears to be 15'? I tried solving for the head loss through the upper layer using equation 205.11 in Goswami's book where H is the total head dissipated across all layers, and got h1=4.9'. I guess it comes down to how H is determined from the figure for problem 38. What's your call?

I also have a problem with their solution to question 49 on page 54. For the overburden term where the water table is between the ground surface and the footing, they use: [γd(Df) + γw(Dw-Df)]Nq.

Df is the depth to the bottom of the footing from the ground surface.

Dw is the depth to the water table from the ground surface.

According to Goswami's and Das' books, this term should be [γ(Dw) + (γsat- γw)(Df-Dw)] Nq.

They give two different answers. The second one makes sense to me, but the one from the 6-Min. solution does not. There is nothing in their errata regarding this solution, but I just don’t see how theirs could be correct.

Any help would be appreciated.

Not sure what to say. I went through the issue with #49. I tried using my DAS book for the 6 min solution and continue to get an answer that is different than the 6 min solution book. After last night I told myself that I can ony use the CERM for the shallow foundation equation. The NCEES sample exam that I have has listed or provided tables/equations for the #49 type problems. Hopefully that helps.

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Thanks for digging into it Milwaukee. So when you say that you can only use the CERM for shallow foundation equations, does that mean that you were getting the correct answer to the 6-Minute problem using that resource vs. Das? I also found that 6-minute solutions used the same equation for the overburden term (when GWT was above the footing) to solve problem 56 as well as 49. I followed my instinct, and was able to come up with the correct answer when solving for the diameter of the footing.

Bearing capacity for shallow foundation problems seem to have great potential for errors by making wrong assumptions or using the incorrect factors. q(net) vs. q(ult) in factor of safety problems has been giving me trouble, but I have learned to watch for key wording in the problems such as "ignoring correction for overburden" or "what is the factor of safety with respect to the ultimate bearing capacity". Hopefully the wording on the exam problems will be very specific, and hopefully I won't be so scattered that I miss the cues.

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Thanks for digging into it Milwaukee. So when you say that you can only use the CERM for shallow foundation equations, does that mean that you were getting the correct answer to the 6-Minute problem using that resource vs. Das? I also found that 6-minute solutions used the same equation for the overburden term (when GWT was above the footing) to solve problem 56 as well as 49. I followed my instinct, and was able to come up with the correct answer when solving for the diameter of the footing.

Bearing capacity for shallow foundation problems seem to have great potential for errors by making wrong assumptions or using the incorrect factors. q(net) vs. q(ult) in factor of safety problems has been giving me trouble, but I have learned to watch for key wording in the problems such as "ignoring correction for overburden" or "what is the factor of safety with respect to the ultimate bearing capacity". Hopefully the wording on the exam problems will be very specific, and hopefully I won't be so scattered that I miss the cues.

From what I gather, the Sample NCEES exams usually give you tables, charts or the N-factors. I imagine that they will also state to use the Meyerhof, Terzaghi or general bearing capacity formula on the exam. It appears that the 6-min solution since it is produced by PPI generally only uses the CERM manual equations but I have gone through the NCEES sample exams using DAS and Knowles books and got the correct answer.

I also have seen that the overburden and FS are important to watch out for, as I have tripped over the same issues. Just my two cents, without having taken the real exam yet.

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Thanks for digging into it Milwaukee. So when you say that you can only use the CERM for shallow foundation equations, does that mean that you were getting the correct answer to the 6-Minute problem using that resource vs. Das? I also found that 6-minute solutions used the same equation for the overburden term (when GWT was above the footing) to solve problem 56 as well as 49. I followed my instinct, and was able to come up with the correct answer when solving for the diameter of the footing.

Bearing capacity for shallow foundation problems seem to have great potential for errors by making wrong assumptions or using the incorrect factors. q(net) vs. q(ult) in factor of safety problems has been giving me trouble, but I have learned to watch for key wording in the problems such as "ignoring correction for overburden" or "what is the factor of safety with respect to the ultimate bearing capacity". Hopefully the wording on the exam problems will be very specific, and hopefully I won't be so scattered that I miss the cues.

From what I gather, the Sample NCEES exams usually give you tables, charts or the N-factors. I imagine that they will also state to use the Meyerhof, Terzaghi or general bearing capacity formula on the exam. It appears that the 6-min solution since it is produced by PPI generally only uses the CERM manual equations but I have gone through the NCEES sample exams using DAS and Knowles books and got the correct answer.

I also have seen that the overburden and FS are important to watch out for, as I have tripped over the same issues. Just my two cents, without having taken the real exam yet.

Speaking of problems with overburden, there are a couple of problems in 6-min solutions geotech depth that have me scratching my head:

Problem 58: For a raft foundation at a depth of 2', why do they subtract the overburden γDf from qult to get qnet before applying the factor of safety to get qall? Wouldn't the area above the raft foundation be excavated, hence no soil above the raft to diminish the ultimate bearing capacity?

Problems 63 & 72: 63 is an individual deep pile capacity problem using the beta method, problem 72 is a pile group settlement problem. In both cases, the ground surface is below some depth of water (bottom of a harbor or reservoir). In problem 63, to calculate the effective stress at the midpoint of the pile length, they add overburden for the depth of water above the harbor bed. The pore pressure that is subtracted from the total stress is then calculated from the water surface, and the additional pore pressure then cancels out the overburden from the water above the ground surface.

In problem 72, we have to calculate P'o at the middle of the clay layer. There is 5' of water above the ground surface. The ΔP' is based on the distributed pressure of the pile group plus the water level in the reservoir is increased to 10'. In this case, when calculating P'o, they do not add overburden for the 5' of water above the ground surface, but they do calculate the pore pressure from the water surface which gives a value lower that the effective stress (based on the way it was calculated in problem 63). When calculating ΔP', they increase the pore pressure due to the effect of the additional 5' of water depth, but once again do not add overburden due to the water above the ground surface. So, the way I calculated it (based on the method used in 63) I got P'o=1358psf & ΔP'=645psf. The solution had P'o=1046psf & ΔP'=312psf. In either case, the difference between the two is 713psf. However, when these values are plugged into the log portion of the settlement equation, they produce a different outcome by about 2".

There seems to be an inconsistency between the approach used in problem 63 and the one used in 72.

Can anyone help clear this up?