Consolidated drained triaxial test

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Homework Statement


This is the problem regarding with Consolidated drained triaxial test .
I have uploaded some photos here . The first photo is from my notes . While the 2nd until the 3rd photo are from my refernce book .

Homework Equations

The Attempt at a Solution


In the first photo , I don't understand why the σ1' = σ1 ?
Since the water is drained from the soil , so i think the first red circled is incorrect ..The shear stress at step 1 should be σ1 only .

For the green circle part , why σ3 = σ3 ' ? Shouldnt it = σ'3 = σ3 + Δσ ( pressure due to drainage) ?
In 641,jpg , we can see that the effective stress at failure is σ3 + Δσ . But , i don't understand why it's σ3 = σ1 .
 
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  • Slide titled a) Consolidated-drained test (CD Test) showing total stress equals neutral pore pressure u plus effective stress, across three steps: end of consolidation, during axial stress increase, and at failure, with drainage shown and pore pressure 0.
    638.PNG
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  • Textbook page, section 12.9 Consolidated-Drained Triaxial Test, describing the CD test procedure and Skempton's pore pressure parameter B = u_c / σ3 (equation 12.18).
    639.png
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  • Two diagrams of a soil specimen in a consolidated-drained triaxial test: (a) under chamber confining pressure σ3 on all sides with u_c = 0; (b) with deviator stress Δσd applied axially, σ3 held on the radial sides, and Δu_d = 0.
    640.png
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  • Textbook page on triaxial testing of saturated soils, with two green-circled equations: total and effective confining stress equal sigma3 = sigma3', and total and effective axial stress at failure sigma3 + (delta sigma_d)f = sigma1 = sigma1'.
    641.png
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The sample starts out with the compressive stress components equal in all directions, and zero pore pressure. Then, the axial stress is increased, while the transverse stress is held constant (and water is allowed to seep out, so that the pore pressure remains zero). The only question is, why doesn't the transverse stress increase as a result of the Poisson effect. Maybe, by the nature of the test, the transverse stress is held constant.
 
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Chestermiller said:
The sample starts out with the compressive stress components equal in all directions, and zero pore pressure. Then, the axial stress is increased, while the transverse stress is held constant (and water is allowed to seep out, so that the pore pressure remains zero). The only question is, why doesn't the transverse stress increase as a result of the Poisson effect. Maybe, by the nature of the test, the transverse stress is held constant.
Do you mean in the case in the 638.PNG , the transverse stress is held constant ?
While in the 639-641 , the tranverse stress is allowed to increase ?
 
Chestermiller said:
639 doesn't say anything about the transverse stress. But in 640 and 641, yes.
Do you mean in the case in the 638.PNG , the transverse stress is held constant ?
 
Chestermiller said:
639 doesn't say anything about the transverse stress. But in 640 and 641, yes.
do you mean in 639- to 641 , all the shear stress( in all direction) is allowed to increase ? But in 638 , the tranverse stress is kept constant ?
 
fonseh said:
Do you mean in the case in the 638.PNG , the transverse stress is held constant ?
Sure.
 
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fonseh said:
do you mean in 639- to 641 , all the shear stress( in all direction) is allowed to increase ? But in 638 , the tranverse stress is kept constant ?
We are dealing with the principal stresses here, so there are no shear stresses in these directions.
 
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Chestermiller said:
We are dealing with the principal stresses here, so there are no shear stresses in these directions.
One more question , why σ1' = σ1 ? and σ1 ' = σ1 + Δσ ?