Geomechanics- direct shear test

In summary: The values in the lab manual are for cohesive soils.thanksIn summary, the sand sheared when the compressive stress was increased from 0.75 to 1 and 2 kg/cm2. The shear stress was higher at 0 than at any other compressive stress.
  • #1
Dell
590
0
i am taking a geomechanics course, as a part of the course we are required to do labs, one of which is to test the shearing stress of a sand using the direct shear method.

we repeated the experiment 3 times changing the compressive stress each time, 0.75, 1 and 2 kg/cm2

from each compressive stress we found the appropriate shearing stress and then plotted the shear- compressive stress graph,

from the 3 points i did not get a straight line, and after using MATLAB to calculate a linear approximation i found that at a compressive stress of 0, the shearing stress was larger than 0, (i got 0.4) from what i understand this means that the sand has cohesion of 0.4 which as far as i understand it is not meant to have. C is meant to be 0

what could i be doing wrong?
 
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  • #2
Theoretically, c should be zero, but if the sand is dense, the graph is slightly curved, and if this is an undrained test, then the curvature is even more noticeable, with the origin (s,p) at (0,0). You might want to try it using a low value for the compressive stress. The value of c = .4 seems higher than you can account for due to experimental errors.
 
  • #3
thanks jay, unfortunately i don't have another opportunity to use the lab before i have to hand my report in, so i won't be able to use a lower compressive value,
as far as the density goes, the sand i used was 1.55 [g/cm2]
the 3 results were as follows

compression=0.75 ; shear= 0.6263
compression=1 ; shear=0.9648
compression=2 ; shear=1.2368

i don't really have anything to compare these results to either(they may all be totally wrong) so any further advice would be appreciated
thanks again
 
  • #4
Your sand appears more loose than dense, where the angle of internal friction is somewhere around 25 to 30 degrees.,. The tangent of the internal friction angle should approximate the slope of the (assumed) straight line curve from the test. Your slope gives a higher angle for the internal friction angle, especially in consideration of point 2. I guess you can explain it in your error analysis.
 
  • #5
thanks, do these results seem at all probable or have i maybe processed the results incorrectly? i don't see where i could have made a mistake, yet i don't see any reason for there to be such errors in measurements, the entire experiment was done in a lab and all the readings were done electronically and sent automatically to ms excel. from there i graphed the horizontal and veritcal displacement and shear force divided by changing area. from the graphs i took the maximum shear stress (from each experiment) and made the shear/ compressive graph.
 
  • #6
What value do you get for the slope of the graph when you fit the points to a straight line?
 
  • #7
it gives around 40 degrees
 
  • #8
Seems a bit high. I think since you were using a cohesionless soil like sand that you could start your graph at (0,0) and draw the interpolated straight line using the other values. It's been many a year when I last did such a test, so I can't recall. Where did you get the values to use for the confining compressive stresses?
 

FAQ: Geomechanics- direct shear test

What is a direct shear test in geomechanics?

A direct shear test is a laboratory experiment used to determine the shear strength and deformation behavior of soil or rock materials. It involves applying a shear force to a specimen and measuring the resulting displacement and stress.

How is a direct shear test performed?

In a direct shear test, a rectangular specimen of the soil or rock material is placed between two parallel plates called shear boxes. A normal load is applied to the top plate, while a shear force is applied to the bottom plate until failure occurs. The resulting displacement and stress data are recorded and analyzed.

What is the purpose of a direct shear test?

The purpose of a direct shear test is to determine the shear strength and deformation characteristics of soil or rock materials. This information is important in engineering applications, such as designing foundations, retaining walls, and slopes, as well as in geotechnical hazard assessments.

What factors can affect the results of a direct shear test?

The results of a direct shear test can be affected by several factors, including the type and properties of the soil or rock material, the sample preparation method, the normal load and shear force applied, and the testing conditions such as temperature and moisture content.

What are the advantages of a direct shear test?

Direct shear tests have several advantages, including being relatively simple and quick to perform, requiring a small amount of sample material, and providing direct measurements of shear strength parameters. They can also be used to evaluate the effect of different factors on shear strength, such as the influence of soil structure or the effectiveness of soil improvement techniques.

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