for a weightless wedge in the limit state of stresses

THE PRANDTL SOLUTION

for a weightless wedge in the limit state of stresses

1. Assumptions

Assume that the subsoil obeys the Coulomb failure criterion and:

1) is cohesionless (c = 0, ϕ > 0) and weightless ( γ = 0),

2) plane state of displacements is considered (2D as for very long retaining walls or slopes),

3) the subsoil fills a bilinear infinite wedge defined by:

the angle ε (to the horizontal axis), the angle β (to the vertical axis),

4) differential equilibrium equations are fulfilled at every point of the wedge,

5) the Coulomb failure criterion τ = σ⋅tgϕ is fulfilled at every point of the wedge (plastic yielding),

6) the external loadings applied to the boundaries 0B and 0A are uniform:

q = const along 0A, with the angle α

to the normal, - ϕ < α

< ϕ , q

= const along 0B, with the angle δ to the normal, - ϕ < δ < ϕ . Details are shown in Fig.1.

Note that the presented situation can be rotated because there is no privileged direction like the vertical one for heavy materials (γ > 0); therefore, let q > q

, where q is known and q

should be found, with no loss of generality.

Fig.1. The Prandtl wedge (here there is: β

α

< 0, ε , δ > 0).

.

Comment:

unexpectedly, this is the angle of internal friction ϕ which makes some problems here.

In this 2D formulation, it is the angle of internal friction in the assumed plane state of deformation, i.e. for a rather specific imposed sliding surface; this angle can be (or actually should be) determined simply in a direct shear apparatus (box apparatus) with blocked displacements in one direction, which is a relevant analogy to the 2D situation.

In other words, in the Prandtl problem the principal stress σ

in the direction perpendicular to the considered plane is omitted, however it occurs in the general limit state condition in 3D cases (points representing the stress in the limit state lie on the Coulomb-Mohr infinite pyramid, i.e. σ

is involved similarly as are σ

and σ

) - see Fig.2:

β

ε

q

q

+

+ ∞

+ ∞

+∞

0 δ

α

z

x

A

B

ρ

ω

Fig.2. Coulomb limit state surface in 3D.

2. Comments

Clearly, q and q

are not independent – in contrast to the elasticity theory; one of them is assumed as a given loading, the other one must be found in such a way that the plastic yielding takes place (the same happens in triaxial tests on cylindrical samples, where only one of two stresses σ

or σ

is under control).

Counter-clockwise convention of signs is used, compression means positive stress σ.

The equilibrium equations are as follows:

0 , ,

0 ,

,

z x x

x z z

= τ + σ

= γ

= τ +

σ (1)

The Coulomb limit state means the algebraic relation:

( ^{σ + σ} ) ^{⋅ ϕ}

= τ

⋅ + σ

−

σ ) 4 sin

(

(2) Hence, 3 stress components σ

, σ

, τ can be potentially found from 3 equations (1),(2) and the boundary conditions along A-0-B. There is also a stress component σ

equal to the

intermediate principal stress σ

which is – by assumption – not important in the considered model. There are only several simple situations that can be solved analytically, the so-called Coulomb wall in this number; and still less for γ > 0.

The model is statically determinate, it does not analyze displacements.

Formally speaking, the general situation from Fig.1 is not correctly defined: one can apply a constant loading q along 0A but it is not clear why the solution q

along 0B would have to be also constant. Numerical solutions, i.e. the method of characteristics, reveal that for the assumed q = const, the solution q

cannot be constant - the exact values are less than q

at a close vicinity of the pole 0; therefore, it is said that the simplified assumption q

= const is on the safe side.

Effective solving of the equations (1),(2) with predefined loading on a part of the boundary needs, generally, computer algorithms – especially for γ > 0.

3. The Prandtl assumption

Instead of the Euclidean coordinates (x,y), Prandtl used the equivalent polar coordinates ( ρ , ω ), Fig.1. This way, the stress components σ

, σ

τ in the equations (1) turn into σ

, σ

τ

and two partial derivatives ∂ / ∂ z, ∂ / ∂ x are replaced by ∂ / ∂ρ , ∂ / ∂ω . Prandtl assumed that:

all stress components σ

, σ

τ

in the wedge do not depend on the radius ρ This means that the angle ϕ

determined in the triaxial test may be systematically different than the ϕ

determined in the box apparatus; and indeed:

ϕ

≅ ϕ

+ 2

÷ 4

in loose soils, ϕ

≅ ϕ

+ 4

÷9

in dense soils.

It is worth remembering, not only

in the context of the Prandtl

solution.

(like q and q

do along the boundaries, which are also the radii). If so, the partial differential equations (1) turn into ordinary differential equations because everywhere ∂/∂ρ = 0.

The solution becomes simpler, though not so simple. Now it is clear why γ = 0 must be assumed: for γ > 0 stresses increase with depth like in a half-plane (special case of the wedge) where σ

= γ⋅z.

4. The Prandtl solution

Similarity of all Mohr circles tangent to the envelope τ = ±σ⋅tgϕ indicates that q

is proportional to q. Indeed, the Prandtl solution is in the following form:

(Pr) aq

1

q K

q = ⋅ (3a) K

– the Prandtl earth-pressure coefficient, active earth-pressure since q

< q.

Define an angular parameter for the Prandtl wedge:

= + + − (4) where

ω

results from the equation: sin( ω

) = sin( α

)/sin( ϕ ) ω

results from the equation: sin( ω

) = sin( δ )/sin( ϕ ).

Usually Θ ≥ 0 (the open angle Θ = ∠B

0A

in Fig.2) for which:

{ − ⋅ Θ ⋅ ϕ }

ω ⋅

⋅ ϕ + α

ω

⋅ ϕ

−

= δ

α

δ

exp 2 tg

cos sin cos

cos sin K cos

o o

(Pr)

aq

(5a)

If Θ ≤ 0 ( ∠ B

0A

in Fig.2 does not exist) then:

) m cos(

sin ) n cos(

) m cos(

sin ) n cos(

cos sin cos

cos sin K cos

o o

(Pr)

aq

+ ϕ ⋅

⋅ ϕ

⋅ − ω

⋅ ϕ + α

ω

⋅ ϕ

−

= δ

α

δ

(5b)

where the angle n results from the equation:

sin(n) = sin( ϕ ) ⋅ sin(m) with m = π /2 + Θ .

5. Kinematic interpretation

Like in the triaxial tests, plastic yielding takes place on local sliding areas at the angle π/4 ± ϕ/2 to principal directions of σ

. In the wedge interior, the situation is not so simple because the principal directions are generally different from point to point or at least in some sub- regions; adjacent local sliding lines continue one to another and finally create continuous slip lines, Fig.2.

There are three families (or sometimes two) of smooth slip lines:

• “passive” zone A-0-A

with straight slip lines,

• „active” zone B-0-B

with straight slip lines,

• transition zone B

-0-A

, called the Prandtl transition zone consisting of radii coming

from the pole 0 and logarithmic spirals; this family does not appear for Θ ≤ 0

when ∠ B

0A

does not exist.

Fig.3. Slip lines in the wedge B-0-A interior.

Because local shearing happens everywhere in the wedge interior, so the slip lines drawn above occur de facto very densely.

6. Applications

Example #1: the Coulomb results for active earth pressure.

Assume for the wedge: ε = 0, β = 0, α

= 0, δ = 0.

The situation means a quarter-plane x > 0, z > 0 vertically loaded by q with a horizontal soil pressure q

on the wall surface.

In details, ω

= 0, ω

= 0 and Θ = 0 in the equation (4); the transition zone Bw-0-Aw does not exist or is reduced to one line; there are only straight slip lines, as assumed by Poncelet.

The loading q is given, the loading q

is to be found whereas q

< q, so it is the active state of soil pressure on a smooth vertical wall (called as “the Coulomb wall”), q

= e

.

Simple substitution to (5) results in:

ϕ +

ϕ

⋅ −

=

⋅

= 1 sin

sin q 1

K q q

which coincides with the Coulomb solution:

e

= K

⋅ (γ⋅z + q) = q⋅ K

as far as γ = 0.

The same formula can be also derived from the Prandtl solution for γ > 0 – see details in the Example

#3 – which is, however, a unique situation.

The "standard" Coulomb active pressure wedge consists here of a "pressure" wedge at the wall (green lines) and a "resistance" wedge at the surface (red ones, as in Fig.3), but in total they give one triangle and straight slip lines going smoothly from the red to the green.

The Prandtl transition zone does not occur here, it is reduced to one line of zero thickness, 0Aw = 0Bw.

In this case - quite exceptionally - one can derive the exact formula for the active earth- pressure of the heavy soil by making use of the Prandtl solution, because the principle of superposition in the stress limit state holds true; more comments about the superposition law

B

0 A

B

A

π /4+ ϕ/2 q

B 0 A

q

q

Fig.4. The Coulomb slip lines.

Example #2: the passive Prandtl solution comes directly from the active one.

The default option was as follows:

q is applied to the soil surface, q

appears on the wall surface as the ultimate (minimal) value; q

< q corresponds to the outward movement of a rigid wall 0B

. In the opposite interpretation, the situation can be reversed: q

is given and q > q

is to be found for the ultimate case of stresses, i.e. q is as great as possible (for the given q

).

Therefore, the equation (3a) results directly in:

(Pr) aq

1

K

q = q (3b)

or just simply

(Pr) pq

1 K

q

q= ⋅

where

(Pr) aq (Pr)

pq

K

K = 1 .

Conclusion:

like in the Coulomb method, there is always: K

⋅ K

= 1,

which is not true in the Poncelet approach (Műller-Breslau, PN-83/B-03010); in the Prandtl method the active and the passive soil pressures mean in fact the same limit equilibrium, only the axes are rotated

:

• σ

= σ

and σ

= σ

? means the active soil pressure on a vertical wall,

• σ

= σ

and σ

= σ

? means the passive soil pressure on a vertical wall.

Example #3: Difficulties (almost not to overcome) with heavy soils The case of γ > 0 is beyond the scope of the Prandtl method.

1) Note that in practice the assumption γ ∼ 0 could be acceptable only for (relatively) great loading q, say for q >> γ⋅ L.

2) For retaining walls the situation is usually just opposite.

Let ε denote the surface angle, the length L along the wall 0B is taken as the leading parameter for the study.

If the constant vertical load (stress) is taken as dq = γ⋅dz⋅cos(ε) = γ⋅dL⋅cos(β)⋅cos(ε)

and it is applied at a certain depth in the wedge A-0-B interior, along a line parallel to the boundary 0A, then below the point P an active earth-pressure occurs dq

= K

⋅dq = K

⋅ γ⋅ cos(β)⋅cos(ε)⋅dL due to (3).

The sum or simply integral of such effects over 0÷ L can approximate the active soil- pressure along the wall surface 0B.

Define = ∙ cos β ⋅cos ε . Finally:

Rys.5. Model przybliżony dla γ > 0.

= !

=

∙ % ∙ cos β ⋅cos ε dL = ∙ % ∙

1one can rotate this wedge freely, because if the medium is weightless, then there is no direction distinguished by the vertical force of gravity; it is also seen in (4) that the solution does not depend

dq A 0

B H

P

dq

z β

L

dz dL

ε

dz

The approximate nature of the solution results from the fact that the balance of vertical forces omits some parts of the vertical loads transmitted to the wall which has an inclined or rough surface; soil pressure is therefore slightly overestimated.

In Coulomb conditions, of course, this is the exact method, because the smooth vertical wall does not transfer any vertical force and constant vaues of dq = γ⋅dz = γ⋅dL are correct.

Comments on the superposition principle:

in limit stress equilibrium (and plasticity) the superposition principle is generally not true, it requires caution – in contrast to linear elasticity.

If a block of the weight N

lies on a horizontal rough surface with a kinematic friction coef- ficient µ, and a horizontal force T

= µ⋅N

moves it steadily to the right, it is a limit state (slip).

Of course, the force T

= −µ⋅ N

will also move the block with the weight N

to the left (slip).

If one sticks such two blocks together (2N

weight) and apply both horizontal forces, this will not cause a limit state, because T = T

+ T

= 0. The sum of the limit states is not a limit state here. But in the case when both forces operate in the same direction, the sum of the limit states is a limit state, because: T = µ⋅ N, if T = T

+ T

, N = 2N

,

where here T

= µ⋅ N

, T

= µ⋅ N

.

More generally: the same is true for stresses, if the principal directions and signs are identical, both positive σ

are parallel and both positive σ

are parallel to each other

: (

= ∙ (

, (

= ∙ (

and (

= (

+ (

, (

= (

+ (

cause that also (

= ∙ (

.

Example 4: Bearing capacity of a shallow foundation beam

This is not immediately seen, but the commonly used standard formula in EC7-1 for the bearing capacity R

[kN/m] it is just the Prandtl solution for a wedge A0B, so the problem is closely related to ... retaining walls.

A very long foundation beam of the width B (plane displacements) has a certain bearing capacity R

[kN/m] or in terms of stresses q

= R

/ B [kPa] where

!

= + ∙ ,

+ !

∙ , + ½⋅γ⋅ B ⋅, [kPa] (6) 1. Assume that c = 0, so N

does not matter. If c > 0, the coefficient N

can be derived as in

Lecture 7 from the equivalent states principle, N

= ctg ϕ⋅ (N

– 1) like in the Eurocode EC7-1.

2. Let γ

= 0. If γ

> 0, the exact analytical solution does not exist, the coefficient N

needs one of approximate calculations (it is typical for the Prandtl method), which can lead to different values of N

.

3. It must be shown that !

= !

∙ , , so N

has to be found.

The load q under the beam foundation is extended to the right, from B to infinity, because the bearing capacity of the beam is governed by a small loading q

; the subsoil move- ment will occur towards the left side and the surface heave will occur there.

The minimal depth of the founding level is known and it equals h

, therefore q

= γ⋅ h

. Strength of the subsoil situated next to the foundation is ignored, the backfill is only a ballast. In this case, q

, ε = 0, β = - π / 2, α

= 0, δ = 0 are known, thus such are ω

= 0, ω

= 0 and finally Θ = 0 - (- π /2) = π / 2 > 0 in formula (4). The Prandtl wedge is horizontal now and it creates a horizontal half-plane A-0-B, Fig.6.

Of course, we're looking for !

= ! =

∙ !

.

Using (5a):

, =

= 1

= 1 + sin 6

1 − sin 6 ∙ 789: ∙ tan 6=

like in the Eurocode EC7-1.

Fig.6. Bearing capacity calculation (2D).

q

q

q

B A

B

A

0