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Soil Bearing Capacity by SPT

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gotlboys

Civil/Environmental
May 31, 2015
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Meyerholf's formula when B<=1.22m ----> qa = 12NFd where Fd=1+0.33D/B <=1.33 and B>=1.22m ---> qa=8N[(B+0.305)/B]^2(Fd).

Assume N=5 blows,D=2m and B>1.22m. The only variable we can play with is B. It could be larger or smaller but not less than 1.22m for second eq applicability.
Trial 1. Let B=1.5m, Fd=1+0.33(2/1.5)=1.44, use 1.33. Thus qa=8*5[(1.5+0.305)/1.5]^2(1.44)=77.03 kPa

Trial 2. Let B=2m, Fd=1.825, use 1.33. Thus qa=70.66 kPa

Can anyone explain what is the implication of the difference between the trials given that B=2m is larger in width but gives lesser allowable bearing capacity?
 
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I had trouble following your method because of your wording, however, take your qa equation and break it apart.

qa=8N[(B+0.305)/B]^2(Fd) can also be expressed as qa=8N[((B/B)+(0.305/B))]^2(Fd)

You are saying that when you increase b your qa goes down well look at what happens as your B goes to infinity. B/B will remain 1 however 0.305/B will get closer to zero as B increases which means as B approaches infinity the section i brought up goes lower until it theoretically equals one. The same thing happens when you are equating Fd. If B gets larger than Fd will theoretically get smaller. I think you have run the equation incorrectly.

In reality if Depth stays the same and B gets larger than your bearing capacity should go up. Id check your equation.
 
If the Meyerhof equations are such that it limiting settlements (I have not gone into this for a long time) or at least taking this into account to a degree, the bigger (wider) the footing, the more soil depth beneath is stressed and the bearing pressure (note, I say pressure for serviceabity purposes rather than capacity which is a strength concept) would have to decrease. Just look at the standard charts of N vs bearing pressure for 1 inch (25 mm) settlement as you find in Bowles or many other texts. By the way - if you have N = 5 for a granular soil, you must think about punching conditions as well.
 
Appreciate the replies.

Blackbear09, I am glad that you are going to check the equation and I do agree that if depth remains unchanged for an increasing size in B the capacity has to be larger but look at the curves BigH suggests, I suppose these are what he meant. In the sample problem, the curve shows that as B increases qa consequently decreases.
Extracted from Bowles text book.
 
You might be interested in a paper by Zekkos, Bray and Kiureghian (University of California, Berkeley) entitled: "Reliability of shallow foundation design using the standard penetration test" - Google the title and you should find it. It explains using N for bearing capacity (shear), bearing pressure (serviceability - i.e., settlement) and then goes on to reliability index, etc.
 
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