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# CD test formula vs Kp

## CD test formula vs Kp

(OP)
For the triaxial CD test, (σ_1^')/(σ_3^' )= (1+sin⁡〖∅'〗)/(1-sin⁡〖∅'〗 )
The term on the right side of the equaition resembles the Kp lateral earth pressure coefficient.
I want to understand why is this so. Is it because Kp relates the vertical force to horizontal force, and here at the triaxial test the vertical vs horizontal forces are related in the same way, and taht is why the formula is the same?

### RE: CD test formula vs Kp

Good observation - it's no coincidence. It looks like that because, in one of the original slope-stability methods, W.J.M. Rankine assumed essentially that same condition: that the soil was in a state of yield, with Sigma1 vertical and Sigma3 horizontal for active pressure, and the reverse of that for passive pressure. Easiest way to look at it is a Mohr circle touching the MC strength line. It was later extended to the case of a wall with friction, in which case Sigma1 and Sigma3 aren't exactly horizontal and vertical.

For the special case with a vertical frictionless wall, Coulomb's Ka and Kp work out to be the same.

Cheers!
DRG

### RE: CD test formula vs Kp

(OP)
sorry i didnt understand. are you able to expand your explanation a little further, or this is as simple as it gets? if so thanks so much anyway and I will keep checking

### RE: CD test formula vs Kp

First off, when I said "slope-stability method," that was wrong. I should have said "active and passive pressure method." Slope stability on the brain today.

Draw a Mohr circle representing failure in a Tx test. It will go from (Sigma3,0) to (Sigma1,0) and touch the strength envelope, which has slope tan phi. Now, assume that the whole mass of soil behind the wall is in yield/failure like the Tx specimen, as the fill pushes the wall away, so that the Mohr circle for any particular point is just barely touching the strength envelope, with Sigma1 being the vertical stress at that point. If the wall is frictionless and vertical, the pressure the soil exerts upon it is Sigma3 (the minor principal stress, no shear stress). You can now solve for Sigma3 as a function of Sigma1 and phi, and the ratio Sigma3/Sigma1 is Ka. For Kp, just reverse 1 and 3, so that Sigma1 will be the wall pressure and Sigma3 is the vertical.

Or, just think of the soil behind the wall as being a big triaxial test (or plane-strain compression test, more correctly). Sigma3 is the vertical stress at a particular point. Then start pushing the wall against the fill until the fill yields; the stress the wall has to apply to yield the soil is analogous to stress on the Tx load platen, Sigma1; that's the passive pressure.

For active pressure, let the fill push the wall/platen away, like a reduced triaxial extension test. The vertical stress in this case is sigma1. It remains constant while sigma3 is reduced and the diameter of the circle increases until it hits the strength envelope and the fill yields.

Better?

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