specific gravity from consolidation test 1

[danyul]

is it possible to get the specific gravity of the soil from the consolidation test? and if so what is the procedure?

 

[BigH]

As a very quick first thought - why do you want the specific gravity? Is it to obtain an "e" (void ratio) value? If so, plot your test results as logp' vs strain and the resulting straight lines become Cc/(1+eo) and Cr/(1+eo) - the initial void ratio is already "built in."

 

[dgillette]

Nineteen times out of twenty, you can assume the specific gravity is 2.66 to 2.7. The twentieth time, when you have unusual mineralogy, that assumption will mess you up.

 

[danyul]

thanks for the replies...
yes BigH i was thinking of how to get the initial void ratio. now correct me if im wrong but dont you need the eo to draw the curve? to get the Cc and Cr...

since i can draw the strain versus p curve then 0 strain is the "top" of the curve?

 

[BigH]

danyul, see Bowles 5th edition, section 2-10.3, fig 2-16a and fig 2-16b. This will explain it to you.

 

[fattdad]

There is no "need" for initial or final void ratio to properly interpret a consolidation curve (or use the data from one for a settlement calculation). The sample is 1-in thick so dial readings (in inches) reflect strain rate. If you plot strain v. log pressure you then get a relationship that is directly parallel to the plot of void ratio v. log pressure. As already pointed out the slope on a strain curve and the slope on the void ratio curve are related by 1/(1+e).

If you have the slope of the strain curve (what I call the consolidation ratio v. the consolidation index) you can then do a settlement calculation using the ratio value (without any consideration of void ratio).

Hope this helps (or adds to the other's contributions).

f-d

¡papá gordo ain’t no madre flaca!

 

[danyul]

ok i understand how to use the strain vs logp to calculate settlement... i was more curious about the specific gravity... the program i use asks for the specific gravity which then calculates the eo.

so previously i had been using the data with some assumptions to calc the Gs:

from the saturated sample... wet density - dry density = water

water/62.4= weight

((take the last reading after the rebound + 0.5") - weight)*dry density = density of solids

density of solid/62.4= Gs
after the consol is it safe to assume 100% saturation?
i usually get "high" Gs values of 2.8-3.2... this is why im asking... thanks

 

[fattdad]

From an assumed specific gravity and a known dry or saturated density you can calculate void ratio. While the odometer sample is maintained under water, I'm not certain full saturation is a safe bet. There are lots of clay samples below the phreatic surface that are unsaturated. Hence the need for backpressure saturation in triaxial testing.

f-d

¡papá gordo ain’t no madre flaca!

 

[BigH]

Quirks of relying "solely" on computer programs.

 

[escrowe]

I had a nice little Excel spread sheet set up to calculate Gs based on the consol sample data. You do have to assume total saturation at the end of the test cycle. So I would essentially work backwards to determine Gs and e before calculating the consolidation test results.

 

[emmgjld]

I have used many consol test data to 'estimate' specific gravity. I will put stress on 'estimate'.
My first assumption, backed up with several flex wall perm tests, is the final saturation of my normal soils will be between 87% to 96%. In a few cases, 100% saturation has been assumed, usually when the specific gravity seemed too low.

 

[danyul]

so i guess thats a no. not without assuming 100% saturation which isnt the case...

 

[emmgjld]

As an aside, 10-12 years ago, my brother put together and Autocadd drawing for plotting Proctor Curves. For each Zero Air Voids Curve at a particular Specific Gravity, he had a layer graphing an Air Voids Curve at 86% & 96%. When the proctor curves were plotted, the 'saturated' down sides of the curves usually plotted within this 86% & 96% zone, actually, usually below 94% saturation.

 

[mudman54]

anything you can test rather than guess is better. why not just do the specific gravity test? it doesn't vary by much unless you got pure feldspar sand. ralph peck once said "beware of the oddball" which you can never find by guessing. once you have the s.g. you can calculate your saturation from the test and begin to evaluate if it is reasonable compared to your in-situ test data.

 

[BigH]

mudman - not necessarily true - i.e., doesn't vary by much. Our saprolite clay here has Gs values of 2.8 to 3.1 - a lot depends on the make-up minerals. Again, I question if you really need the Gs anyway. You plot e-logp to get Cc - then in your settlement computation you have to divide Cc by 1+eo. Or, you plot strain-log p and you get CR which is [ Cc / (1+eo) ] directly. So what need Gs??

 

[danyul]

ok maybe im missing something simple here... i understand how to use the strain vs logp curve to get my Cc' and Cr' values. also i understand that these are equal to Cc/(1+eo) and Cr/(1+eo).

now BigH and fattdad say to use the Cc and Cr from the e vs logp curve then i can find the eo. ok simple...

now am i mistaken or dont you need the eo to plot the e vs logp curve in the first place?

 

[BigH]

danyul - what fattdad and I are saying - you don't need to plot the e-logp'. Just use the strain-logp'.
I see your dilemma - in that you have a computer program that requires the eo rather than letting you plug in CR and RR values. If you assume eo say, of 2.65, use it to plot the e-logp' curve. Obtain Cc. To see the senstivity of Cc on eo, assume 2.75 and plot - get Cc(2.75) and assume 2.7. Expect Cc will not be "much different" given the accuracy of plots, etc. Then you can get eo by eo = (Cc/CR)-1 . To make your computer .

 

[danyul]

BigH-thanks for the gentle reply... ok i see your theory but somethings not working

example:
an actual consol test, from the strain vs logp i got a Cc'=4.49 and Cr'=0.735=CR

now my program asks for Gs actually then calcs the eo. so i plugged in different Gs and the program reports eo,Cc and Cr for the e vs logp
Gs eo Cr Cc
2.6 1.895 2 13
2.7 2.006 2 13
2.8 2.117 2 14
2.9 2.229 2 14
3.0 2.340 2 15

now i assume these number are rounded... so this is what you expected but how do you know which Gs is the right one? if you use the Cr=2 then you get a low Gs of 2.4ish which isnt right.

now if you use the eo with the strain Cc' and Cr' you can calc the exact Cc and Cr. curiously the Cc/Cr for each Gs are the same.

thanks for your time...

 

[BigH]

CR is the compression ratio derived from the strain-logp' curve. (not Csubr).

 

[danyul]

bigH im not sure what you mean... i use Cr' and Cc' from the strain vs logp and Cr and Cc from the e vs logp

im assuming you use CR as my Cr' not Cr