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Solving the confusion on Bowles equation for swelling pressure

Solving the confusion on Bowles equation for swelling pressure

Solving the confusion on Bowles equation for swelling pressure

Hello everybody. I am writing this post related to the thread158-157942: Joseph E.Bowles foundation analysis and design-swelling clay help. in which there is a confusion about the sum of two quantities on the equation for correlating LL, w and density to the swelling pressure of the soil, which appears on Bowles Foundation analysis and design 5th. edition.

A few months ago I was trying to calculate the swelling pressure of a clay, found the thread and it gave me a lot of headaches. So here it goes my explanation:

Clearly, 2.132 + 2.096 is not equal to 0.226. But 2.132 + 2.096 = 4.228, so: logPs = 4.228 and: Ps = 16,904. AND THEN, IF YOU DIVIDE BY 10,000 you get 1.69 kg/cm^2, which matches approximately the result on the book: 1.683 kg/cm^2.

So, I suspect that the original equation, as it is written on Bowles, gives the result in kg/m^2. That's a possible explanation.

But I confirmed this assumption using other sources. Searching the internet (a lot...) I found the equation, supposedly by Komornik & David (1969 or 1980) on at least two technical papers, one of them from Turkey, I think, but with -1.868 instead of +2.132 as the constant.

That's: logPs = 0.0208LL + 0.665Rhod - 0.0269wn - 1.868, which is the same equation of Bowles but with the result divided by 10,000. See below for a demonstration.

Demonstration: log(Ps x 10,000) = [2.132 + 0.0208LL + 0.665Rhod - 0.0269wn] (kg/cm^2)

Then, by the theorem of logarithms log(MN) = logM + log N,

logPs + log(10,000) = [2.132 + 0.0208LL + 0.665Rhod - 0.0269wn] (kg/cm^2)

logPs + 4 = [2.132 + 0.0208LL + 0.665Rhod - 0.0269wn] (kg/cm^2)

finally, clearing:

logPs = 0.0208LL + 0.665Rhod - 0.0269wn - 1.868 (kg/cm^2)

So, in brief, you either use the equation on Bowles 5th edition, but dividing by 10,000 the result in order to obtain kg/cm^2, or you use the formula with the -1.868 (which is what they do in Bowles; they just don't explain it). :)

RE: Solving the confusion on Bowles equation for swelling pressure

Just saw that the NAVFAC manuals have examples for calculating pressures due to expansive soils...

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