Consolidated drained triaxial test

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The discussion revolves around the Consolidated Drained Triaxial Test, focusing on the relationships between principal stresses and effective stresses. Key points include confusion over why σ1' equals σ1 when pore pressure is zero and the implications of drainage on transverse stress. Participants clarify that during the test, the transverse stress is held constant while axial stress increases, leading to questions about the Poisson effect. The conversation highlights the distinction between effective stress and total stress, emphasizing that effective stress calculations are crucial in understanding soil behavior under these conditions. The thread seeks further clarification on these concepts, indicating a need for deeper insight into soil mechanics.
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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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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 ?
 
639 doesn't say anything about the transverse stress. But in 640 and 641, yes.
 
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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 ?
 
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fonseh said:
One more question , why σ1' = σ1 ? and σ1 ' = σ1 + Δσ ?
Because the pore pressure is zero.
 
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