Failure Plane and Active Earth Pressure 1

[geomane]

For the rankine active earth pressure case, the wall moves a sufficient distance away from the soil mass causing the soil to fail on a plane of 45 + phi/2 with the horizontal. This is true for a granular soil with no cohesion. If you completely removed the wall away from the soil, the soil should stabilize at the phi angle (stable slope angle = phi angle).

We had a question come up in the office today of why wouldn't the active earth pressure failure plane equal the phi angle, when the wall only slightly moved away from the soil mass (enough movement for the active case). The way I look at it is, if the the wall is moved only enough to cause the soil to fail, you still have confining pressure, which keeps your mohr circle to the right (higher stresses) than zero. So at the point where the mohr circle touches the mohr-circle failure envelope, that point is at angle of 45 + phi/2 measured from the horizontal.

I'm having trouble explaining the case with the wall completely removed in terms of mohrs circles and failure plains, other than when you remove the wall completely, you remove the confining pressures.

Can someone please provide a better explanation on this?

 

[fattdad]

. . . but the active wedge is not a failure plane?

f-d

ípapß gordo ainÆt no madre flaca!

 

[geomane]

fattdad,

Maybe I was using the wrong terminology. The slips lines or theoretical failure lines may be what you are used to seeing?

The active wedge slip/failure line for the rankine state is at an angle of 45 + phi/2 degrees from the horizontal. The rankine active and passive states can be approximated as an element of soil being failed in a triaxial test. The specimen would be brought to failure by decreasing the confining pressure while the axial stress remains constant for the active state, and for the passive state the confining pressure would increase while holding the axial stress constant. So, I guess this is why I sometimes use the term failure plane, because the soil is assumed to theoretically fail along the same plane when comparing a retained soil mass in the active state and a triaxial specimen undergoing lateral extension. Soil Mechanics by Lambe and Whitman has a good section on this.

I'm still not sure if I can compare the slip/failure lines of the rankine active state to the slope angle of a mound of granular soil.

But correct me if I am wrong.

 

[DaveAtkins]

I am not sure about the theory behind it, but the active wedge has an angle of 45 + phi/2 from the horizontal, as you stated. However, a mound of soil will reach equilibrium at an angle to the horizontal equal to the angle of repose, which is approximately equal to phi.

DaveAtkins

 

[fattdad]

Again, there is no slip plain on the active or the passive wedge. Yes, the 45(+/-) informs the active and passive wedge, but. . . it's not like some angle of repose! (I don't even like that term!).

I do understand triaxial extension and compression tests. I understand stress paths. I just don't try to correlate these to active and passive wedge geometry.

Maybe I need to learn something new?

f-d

ípapß gordo ainÆt no madre flaca!

 

[geomane]

Maybe my question can't be answered because I'm comparing apples to oranges.

 

[aeoliantexan]

I haven't drawn Mohr circles or made calculations, but here is the way I understand it: We are talking about cohesionless soil (clean sand) with no free water present. The wall holds the soil in place with a pressure greater than active. If the wall deflects sufficiently, the shear stress equals the soil strength first along a plane at the 45 + phi/2 angle. With all due respect, FD, I think slip does occur there, and the active wedge slips down slightly. In fact, there may be multiple parallel shear planes throughout the active wedge. The soil now exerts active pressure on the wall, and the wall responds with an equal but opposite pressure, which keeps the wedge in place. The wedge exerts pressure on the soil behind it, which keeps that soil in place without shearing. If the wall is removed, the active wedge collapses and no longer exerts pressure on the soil behind it. That soil can slip at flatter angles until the angle of repose is reached.

 

[fattdad]

I take no issue with aeoliantexan.

f-d

ípapß gordo ainÆt no madre flaca!

 

[ebaftj01]

Geomane, aeoliantexan is correct. They are 2 separate but distinct concepts. Think of the slip/failure plane as the beginning of the end (oh sh** moment), and the angle of repose as the end condition (news anchors Monday night quarterbacking).

 

[GeotechNerd86]

Cohesionless soils cannot stand vertically without confinement (Wall?) unless you have some capillary cohesion holding it together. If the stress state (circle) is below the failure line a soil wedge shearing onthe angle of rupture is being held back. The stress state will be violated if you excavate vertically and σ3 = 0 and therefore you cannot explain it with Mohr Circles. The best way to look at this is with the infinite slope equation (Limit Equilibrium, tanɸ/tanβ without cohesion) for a FoS= 1 where you will see the soils will fail at the friction angle/angle of repose. If the soils are completely saturated it will fail at 0.5xtanɸ/tanβ and fail at an angle of almost halve the angle of repose (For ɸ=30) .