Soil Bearing Capacity by SPT
Soil Bearing Capacity by SPT
(OP)
Metric Units
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?
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?





RE: Soil Bearing Capacity by SPT
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.
RE: Soil Bearing Capacity by SPT
RE: Soil Bearing Capacity by SPT
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.
RE: Soil Bearing Capacity by SPT