Ka = (1 - sin ϕ]) / (1 + sin ϕ) Kp = (1+ sin ϕ) / (1 - sin ϕ)

Thank you very much. Sorry for the mixed languages used. You may refer to the following link - a more clear explanation and voice. And maybe you can understand what I'm trying to explain. kzbin.info/www/bejne/mqfHfHujjrKMobs Thank you..

Thank you for your question. In my understanding, back pressure is used to allow water to flow into the sample at the beginning of the CD test. In the CD triaxial test, the drainage valve will be kept open during the consolidation and shearing stages. So, porewater is permitted to drain out until porewater pressure, u =0. Thus, there is no (u) were recorded at the end of the CD test. And we will have the effective strength parameter from the CD test. σ' = σ - u, u = 0 --> σ' = σ You may refer to this video kzbin.info/www/bejne/iGGTf2agl76ZsJo - I have explained in this video. Unlike the CU Test, during the shearing stage, the drainage valve is closed (that refer to U for Undrained), so the porewater pressure (u) will be recorded at the end of the test (negative porewater pressure or positive porewater pressure). Hope this answer your query. Thank you.

Glad it was helpful!

Sorry for the very late reply, you may refer to this video for the CD test data example. Thank you.. kzbin.info/www/bejne/jHjXfYdqe56UqtU

Thank you. Sorry for the late reply ..yes, the dimension for the thickness of the clay layer is given in the question as shown in the figure. That figure is a part of the question, as you can read from line 2-3 (The figure shows the arrangement of the pumping well and observation wells, together with relevant dimensions.). Hope it answered your question. Thank you again :)