# GEOTECH HELP PLEASE



## boo (Oct 27, 2010)

can you please somebody tell me how can find the UNIT WEIGHT OF SOIL WITH HAVING THE

WEIGHT OF SOIL= 3.2 AND SG=2.66 water content =12.8%

THANKS


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## PEin2010 (Oct 27, 2010)

unit weight of soil = specific gravity * density of water = 2.66 * 62.4 = 165.98 lb/ ft^3


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## civilized_naah (Oct 27, 2010)

PEin2010 said:


> unit weight of soil = specific gravity * density of water = 2.66 * 62.4 = 165.98 lb/ ft^3


No, that's the unit weight of a solid block of soil - with no voids

The data (WEIGHT OF SOIL= 3.2 AND SG=2.66 water content =12.8%) is not sufficient. You need volume info also.


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## PEin2010 (Oct 27, 2010)

agree ---that is the unit weight of solids in the soil composition.

@ civilized_naah, can't you calculate it this way?

so if you draw the mass-volume diagram

you can assume for 1 cu.ft of soil.

mass of soilds = 165.98 lb

given the water content of 12.8%, mass of water = mass of soilds * water content = 21.25 lb

mass of voids = 0

and so total mass = 165.98 + 21.25 = 187.23 lbs for 1 cu.ft. or in other words, wet unit weight of soil = 187.23 lb/ cft?

Thanks


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## Badger (Oct 27, 2010)

I agrree with civilized_nah. You need more information either S,e, or V.

Something is missing from your question, you have weight of soil = 3.2; what is the size of the sample?


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## phillyphan (Oct 27, 2010)

Badger said:


> I agrree with civilized_nah. You need more information either S,e, or V.
> Something is missing from your question, you have weight of soil = 3.2; what is the size of the sample?


Is this for a Proctor Test?

If so, the volume is 1/30 cu. ft.


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## PEin2010 (Oct 28, 2010)

:mail-296:

can somebody please comment on my answer below? Did I do it right?

Thank you!



PEin2010 said:


> agree ---that is the unit weight of solids in the soil composition.
> @ civilized_naah, can't you calculate it this way?
> 
> so if you draw the mass-volume diagram
> ...


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## RJs (Oct 28, 2010)

Time to relax and get ready for tomorrows exam. One of my senior studied SO hard a day before the exam and he had headache on the exam day and had to go to exam room taking tylenol.


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## Badger (Oct 28, 2010)

PEin2010 said:


> :mail-296:
> 
> can somebody please comment on my answer below? Did I do it right?
> 
> ...


Well if you are making the assuption Vv=Vw, or totally saturated, where S=1, you can use e=w(SG)/S = .128(2.66)/1 so e=0.3405

Now you can get the dry unit wgt Yd = (SG)Yw/(1+e) Yd = (2.66)62.4/(1.3405) Yd = 123.82 pcf

Ysat = (1=w)Yd =139.7pcf

You might note the dry unit weight of sand is 100pcf, wet ~125 pcf, gravel dry 95, wet 125, wet clay 125 pcf. And solid basalt is about 188 pcf and solid granite ~168 pcf.

So when you consider that and get wet weight of 187 pcf and a dry weight of 167 pcf, you might have missed something.

I had to work a lot of these problems, because I did the same thing you did.

I hope this helps.


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## littlewheels4 (Nov 5, 2010)

Badger said:


> PEin2010 said:
> 
> 
> > :mail-296:
> ...



I believe this is correct, but the original post is a little vague. I am not sure if he is asking for wet or dry unit weight. I am guessing that he is looking for dry unit weight based on given water content.


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