Correlations between non-standard split spoons and SPT
Correlations between non-standard split spoons and SPT
(OP)
Here in Alaska we deal with oversized granular material that makes sampling with SPT problematic. A common solution here is to use a 3" OD, 2.5" ID oversized split spoon sampler driven with a 340-lb hammer falling 30". The resulting blows "approximate" the SPT, plus we are able to collect larger samples.
The only documentation I have on this is a chart of sampler driving resistance vs. relative density that plots on log-log paper sampler-hammer ratio Rs (1X10^-5 ft^2/lb) on the y-axis and driving restistance (blows/ft) on the x-axis. I believe the chart was produced by Dames and Moore in the early 80s. I also have a xerox copy of figure 1.21 from section 1.12 of "Foundation Engineering" by an unknown author and unknown publication date that plots the same information.
Rs is defined as (Do^3-Di^3)/(144*W*H) where
Do= sampler outside diameter (inches)
Di = sampler inside diameter (inches)
W = weight of hammer (lbs)
H = hammer drop (inches)
For a given relative density, the ratio between SPT and the "Alaska setup" is ~1.1. Therefore, to get "SPT" blows, it is necessary to multiply the field blows by 1.1.
I have been in Alaska for two years and this appears to be common practice, at least among the engineers I have worked with. I would like more comfort level on this subject, so if there are any other references or similar practices you all are aware of, please let me know.
Thanks!
The only documentation I have on this is a chart of sampler driving resistance vs. relative density that plots on log-log paper sampler-hammer ratio Rs (1X10^-5 ft^2/lb) on the y-axis and driving restistance (blows/ft) on the x-axis. I believe the chart was produced by Dames and Moore in the early 80s. I also have a xerox copy of figure 1.21 from section 1.12 of "Foundation Engineering" by an unknown author and unknown publication date that plots the same information.
Rs is defined as (Do^3-Di^3)/(144*W*H) where
Do= sampler outside diameter (inches)
Di = sampler inside diameter (inches)
W = weight of hammer (lbs)
H = hammer drop (inches)
For a given relative density, the ratio between SPT and the "Alaska setup" is ~1.1. Therefore, to get "SPT" blows, it is necessary to multiply the field blows by 1.1.
I have been in Alaska for two years and this appears to be common practice, at least among the engineers I have worked with. I would like more comfort level on this subject, so if there are any other references or similar practices you all are aware of, please let me know.
Thanks!





RE: Correlations between non-standard split spoons and SPT
RE: Correlations between non-standard split spoons and SPT
This is the only chart that I have seen, too. I had put the following together forgetting that you had some of the data - so, if I may, for completeness, I am still posting this for those who don't have the book.
I have checked and Fang’s Foundation Engineering Handbook , 2nd Edition, has charts for driving of samplers through cohesionless and cohesive soils. Two standard hammers: (1) Terzaghi’s standard split spoon sampler (ID=1.375inch, OD=2inch and 140# hammer dropping 30inches) and (2) Burmister’s spoon (ID=2.93inch, OD=3.625inch and 250# hammer dropping 20inches) are plotted on the curve. Since many engineers may not have access to the book, I summarize the relationships as below.
There is a “Sampler Hammer Ratio” for cohesionless given as the ratio of <OD3 minus ID3> divided by <144xWxD>. All units in pounds and inches. For Terzaghi, the Sampler Hammer ratio is 0.895x10-5 ft2/lb; for Burmister, it is 3.12x10-5 ft2/lb. I checked the reported value for Terzaghi and got 0.893x10-5 : roundoff.
For cohesive soils, the “Sampler Hammer Ratio” is <OD2 minus ID2 > divided by <12xWxD>. Terzaghi gives 4.185x10-5 and Burmister gives 7.57x10-5.
In cohesionless: for Loose/medium dense (or, for Canadians compact) interface (Dr=40%), the Terzaghi is, familiarly, 10 blows while Burmister is 23blows). Very loose to loose interface (Dr = 20%) is 4 and 9 blows respectively. Medium Dense to Dense interface (Dr=70%) is 30 and 70 blows respectively. Dense to very dense is 50 and 110 (approx).
Similarly for clays (and I am not debating the merits of using split spoons in clays for undrained shear strengths); the interfaces at undrained shear strengths for Terzaghi and Burmister : 250psf - 2 and 2.3; 500psf – 4 and 5; 1000psf – 8 and 12; for 2000psf – 15 and 24; for 4000psf – 30 and 50; for 8000psf – 60 and 100. (Burmister is approx).
Given the above spoon and driving characteristics you can plot the Terzaghi and Burmister values given above on log-log paper: sampler ratio as the y-axis; driving resistance as the x-axis. Curves are linear. Then determine the “equivalent” blow counts at the “interfaces” for other hammer ratios by plotting.
A caveat – this is as per Fang. Unfortunately, Fang didn’t give a direct reference to these curves but they are, perhaps, from Hvorslev (1949) – my guess after scanning the chapter’s reference list. Of course, some organizations might have developed their own charts in one manner or another – by field comparisons or otherwise but this is the only reference I have run across.
RE: Correlations between non-standard split spoons and SPT
RE: Correlations between non-standard split spoons and SPT
You are using what is commonly referred to as a California sampler. Perhaps you can do an Internet search, or someone who is in the Granola State
Please see FAQ731-376 for great suggestions on how to make the best use of Eng-Tips Fora.
RE: Correlations between non-standard split spoons and SPT
RE: Correlations between non-standard split spoons and SPT
RE: Correlations between non-standard split spoons and SPT
Ashjun - remember where you live! Not all jurisdiction's practices are as restrictive. They allow a little more creativity.
RE: Correlations between non-standard split spoons and SPT
=IF(E4="SPT", IF(G$24=1.375, D4, 1.2*D4), IF(E4="U", IF(CODE(I4)=83, 0.42*D4, IF(D4>27.77, 0.54*D4, 0.00070082*D4^3 - 0.031775*D4^2 + 0.88924*D4 + 0.016121)), IF(CODE(I4)=83, 0.74*D4, IF(D4>17.24, 0.87*D4, 0.00073482*D4^3 - 0.020544*D4^2 + 1.0057*D4 + 0.0023552)) ))