Hi,

You are right. I have uploaded corrected xl sheet., where 38 and 40 both used unfortunately and created trouble. As submitted journals are still under review process, hence I have deleted True Triaxial tests data provided by Professor Lade,Professor Yin and  my colleague. 

Please using Matlab plot3d.

Import  function ezimplot3 from:

http://www.mathworks.com.au/matlabcentral/fileexchange/23623-ezimplot3--implicit-3d-functions-plotter

to plot implicit function in 3D.

Otherwise, you need to express one principal stress in terms of the other principal stresses (explicit ly).

Hi,

Please see the attachment. The procedure to plot them are about similar.

Good luck.

Mohammad, there must be a mistake in the document you attached. According to the formula for M(b) (Eqn. (2)) M(b=0.2)=1.287, not 1.352. Only M(b=0)= 1.549 is correct. The remaining values are incorrect in my opinion.

Hi Krzysztof Sternik, Thank you very much for pointing errors. Please consider phi=40 rather than 38.If you find any specific discrepancies kindly let me know. 

Dear Mohammad, I followed all the calculations in your document and I found that the first value of M(b) is for friction angle 38deg when the rest is for 40deg. Nevertheless, I obtained completely different values for q, sigmas, x, y than you reported. Values of coordinates I obtained are  wrong (they give a straight line), whereas your values give correct shape of deviatoric section. But I did my calculations based on the formulas given by you.

Hi,

You are right. I have uploaded corrected xl sheet., where 38 and 40 both used unfortunately and created trouble. As submitted journals are still under review process, hence I have deleted True Triaxial tests data provided by Professor Lade,Professor Yin and  my colleague. 

Sorry for late reply for all. I have just come back after conference at Kyushyu island in Japan. Thank you so much Mr Vo Thanh Liem and great docement of Mr Mohammad Islam. 

Hi Mr Mohammad Islam, Mr Krzysztof Sternik

I also calculate based on this equation but the same case of Mr Krzysztof Sternik. The results I had received ia a straight line. What's wrong in this case?  I'm trying using this equation to solve but still not success.

After 5 value (b=0; 0.2; 0.4; 0.6:; 0.8; 1.0) I realized that you changed another value ????

If you have solution, please let me know. Thank you so much.

you can plot pi-plane using b-value or lode angle. in pi-plane complete segment is 0 to 360 degree and divided into 6 sections. For first section b=0 means compression, while for second section b=0 is the extension, again for 3rd section b=0 means compression. when you have data for three sections (0 to 60, 60 to 120 and 120 to 180) you can image them for 180 to 360. When you will consider 0 to 360 degree, you will get the complete plot. if you plot each section you may get either straight line or part of circle or section of the surface.is it clear now? if not then plz have a look the reference. don't get hesitate if you need more help

"Failure Criterion for Cross-Anisotropic Soils"

by Poul V. Lade, M.ASCE, (Prof. and Chair, Dept. of Civ. Engrg., Catholic Univ. of America, Washington, D.C. 20064. E-mail: [email protected])

Journal of Geotechnical and Geoenvironmental Engineering, Vol. 134, No. 1, January 2008, pp. 117-124, (doi: http://dx.doi.org/10.1061/(ASCE)1090-0241(2008)134:1(117))

There was a simple edtorial error reported in Mohammad's first document 'Octahedral Plane Plot.docx'. In the formula for y-coordinate there should be SQRT(6) instead of SQRT(2). Just like in Excel file.

Thank you Krzysztof Sternik for your correction.

Nguyen, for b=0..1 you get only one segment of the deviatoric cross-section. But you have all necessary data to plot the complete diagram. All you have to do is to swap sigma1 for sigma2, then for sigma3, and do it 6 times (to get 6 permutations of data columns). Each combination of columns add to the same diagram. Please look up the attached file. You'll find there each section marked with different colours.

Thanh you so much Mohammad Islam and Mr. Krzysztof Sternik. I can image this. I will try with another model. 

1) If we you equation as discussed about we can change sigma axias order. When we use lode angle we can draw one segment of the deviatoric cross-section (with lode angle changr from -30deg to 30deg). So how we can draw draw with 5 others segment (Because lode angle only change from -30deg to 30deg we can't use lode angle from 0 to 360 as Mohammad Islam suggested)

2) With the eq. (1) (word file - principle stress devided 2 components - mean stress p and deviatoric stress q). It means each model which has the different deviatoric stress q ??? (because other component wouldn't change)

Hello Nguyen

Have you obtained a satisfactory tool or formulae for plotting the L-D and M-N criteria in the deviatoric plane?  If you have not succeeded or are unsatisfied with the one you have because of limitation of lode angle say so with a post and I will supply you with the information.

I"m still not succeeded now. Please show me some advice about this. Thank you so much Mr. Amaechi J. Anyaegbunam

Dear Nguyen

Attached is the required information. The equations will plot the whole curves without the need for any modification. The equations have been checked out by plotting with Matlab. I believe there are no typographical errors. Keep p (mean stress) at a constant value to trace the curve on the deviator plane. Vary p also to get a 3D surface of the yield criteria.  Good luck to you.

Please check the attached file.

thank you so much for your kind support Amaechi J. Anyaegbunam. I will try it.

Dear Nguyen.

The principle behind these equations are quite easy but the implementation demands a lot of complex algebraic manipulation. Let the deviator plane be defined by x, y, s axis where the s-axis represents the hydrostatic axis, and let the principal stresses be the usual cartesian axes. s and p are related by s = p* sqrt(3). Then these two sets of axes (I will provide a pictorial representation later) are related by

σ1 =  [2x + f]/ sqrt(6)

σ2 = [ -x + y*sqrt(3) + f ]/sqrt(6)

σ3 = [ -x - y*sqrt(3) + f ]//sqrt(6)

where f = s*sqrt(2) = p*sqrt(6). It is immediately obvious that the 1st stress invariant is

I1 = σ1 + σ2 + σ3 = 3p as expected

Substitute these expressions for the principal stresses into the other invariants namely I2 and I3.  I2 = σ1σ2 + σ2σ3 + σ3σ1; I3 = σ1σ2σ3

Finally substitute the resulting expressions into the yield criterion.

A cumbersome formula will be obtained that , fortunately, can be solved for x. The real problem lies in determining yb and ya the coordinates of the bottom and apex of the curve. In the case of L-D criterion I had to solve a cubic equation in closed-form.  Note that All yield criteria are made to pass through the apex of a Mohr-Coulomb (M-C) deviatoric trace corresponding to triaxial compression. Consequently, ya is same for all yield criteria and equals ya = 2f*sin φtc/(3-sin φtc), where f = p√6.   It is also known that the M-N criterion passes through the vertices of the M-C hexagon hence for the M-N: yb = - 2f*sinφtc/(3+sinφtc). Where φtc = friction angle from triaxial compression test. Thus, the cumbersome formulas I gave for yb and ya in the case of M-N in my pdf file could be disregarded for they agree with the M-C values. However, the yb I gave for L-D stands, and I don't know of a simple method for obtaining it. Attached a jpeg file of the plots of the different criteria I obtained with my formulas. 

     Wishing you all the best.

Hi,

 if you know the range for -30 to +30 degree you will get 0 to 360 degree as it is also possible adopting b-value. Add 30 degree with -30 you will get 0 degree which corresponds to b=0 for first segment and +60 degree corresponds to b=1.Please have a look the book that I have referred in personal communication page number 16 and 17, Chapter 2. I hope you will find your answer in equation 2.27 to 2.37.If you need further help let me know. Actually I am submitting PhD thesis by this month, therefore I haven't seen still you are in problem.

For derivation of equation :

p=f(s1,s2,s3)

q=f(s1,s3)

b=f(s1,s2,s3)

Three equation, three unknown. It is easy to solve. For more details read following two books as I have used and in my opinion they are good book if you interested to dig out insight view of M-N

The SMP Concept-Based 3D Constitutive Models for Geomaterials Hardcover – March 6, 2006

by Hajime Matsuoka (Author), De'an Sun (Author)

Be the first to review this item

ISBN-13: 978-1563273742 ISBN-10: 0415395046 

Constitutive Modeling of Geomaterials: Principles and Applications Hardcover – August 6, 2012

by Teruo Nakai (Author)

ِDrawing with the components of σ in the D, E and

F directions by the inner products of σ.

Davis, R.O., and Selvadurai, A.P.S. (2002). Plasticity and geomechanics. Cambridge University Press.