Direct Shear, consolodation problem. 1

[wiggythecat]

Hi,
I previously worked in the UK using BS1377 for small shear box testing. I always hand calced my consolidation using T100 method.
I am now working in Australia and the consolidations are worked out using the T50 as per AS1289.
I am wondering would basing your shearing times on 50% consolidation allow for enough pore water pressure to dissapate to not have any effect on the sample?
The difference between the the shearing times for T50's and T100's is vast. I ran all my shear boxes on clay in the UK over night if I had the option. And am surpised to see a shear on a clay done in five hours.
I wonder is anyone can enlighten me as to why this is and why the AS standards vary so much for the BS?

 

[Erdbau]

I don't know either standard. What are the equations? I am assuming that the t50 is multiplied by a higher factor than the t100. ASTM D3080, which I use, suggests time of failure = 50*t50. I always run clays overnight regardless of what I get from the consolidation (unless it tells me that overnight is too fast, then even longer). You really want to be sure that you are getting drained behavior.

 

[wiggythecat]

AS1289 t50*50 (same as AST), BS1377 the consolidation is plotted on height v square root time and 100% consolidation is obtained from graph in mins. (I can post the equation for you if you wish to see it).

I recently had a clay sample and worked out the consolidation time to be appx 48 hours, using the T100 method. The same sample was sent to another lab and they sheared the sample over 6 hours.

I am relieved to hear that you opt to shear clays as slow as possible.

I am still somewhat confused with the t50 calcs, surely the sample is not fully consolidated at the point when the rate of strain is obtained? In reality the sample is 100% consolidated when it is shearing not 50%.

 

[Erdbau]

When you say "the consolidation time to be appx 48 hours" do you mean t100? How big is the shear box? I've never seen a t100 of 48 hours.

The fact that the shearing rate is based off of t50 does not mean that the sample is not fully consolidated before you start shearing.

In ASTM (and I assume AS 1289), you obtain t50 using the plot of displacement versus log time by finding t0 and t100 and their corresponding displacements. d50 = (d0 + d100)/2 and t50 is the time on the curve corresponding to d50. Does this sound like what you are doing?

There is no magic behind 50*t50. It has been determined to be "slow enough" to dissipate pore pressures. They could have just as easily based it of of any other percentage of consolidation t100, t90, etc. Consolidation theory (explained in soil mechanics text) gives the relationships between percent consolidation. In other words, if you know what t50 is, you can calculate what t90 is.

First step is to check how you are determining t50.

 

[wiggythecat]

Hi, shear box is 100mm². In the UK our clay samples could run up to 3 days (very rare tho), mostly overnight tho 2 days not uncommon. And yes I did the mean the T100, excuse my phrasing, new to site and in my own admittance my maths is not the greatest.

Yes, it sounds like the same method for obtaining the T50.

The shearing loads are not peaking (I am assuming that this is because they are run to fast). Using T100 method, the sample would 99% of the time peak and then fall, so you were sure you had the definate peak value. Using the T50 method the samples are still increasing in load when full travel is reached.

The difference between the shearing rates for the two methods is vast. In my little experience of using the T50 method, it seems way to fast.
I am not doubting what your saying and am grateful of any help, tho who actually determined that T50 method was slow enought to dissapate pore pressure?

My T50 calcs are actually worked out by a computer programme and have had several engineers check them so I am confident that they are correct.

T100 constant is 12.7 and yes can see what your saying about the consolidation, just got a nagging feeling, something isnt right.

 

[Erdbau]

Your UK shear rates seem reasonable, and are generally what I do.

For that size shear box, t100 = 48 hours does not seem reasonable. Time to shear the sample = 48 hours, maybe.

The discrepancy between using t50 and t100 should not be that great, so I'm not sure what is going on. According to consolidation theory, t90 should be about 4.28*t50. If you compare the multiplication factors, 50/12.7 = 3.9 which is close to 4.28, so the two methods should be predicting approximately the same shearing time.

At this point, you should ask the lab to provide the consolidation versus time plot and check their method versus your calculation.