THE PRANDTL SOLUTION for a weightless wedge in limit equilibrium
1. Assumptions
Assume that soil obeys the Coulomb failure criterion =tg and:
1) is cohesionless (c=0, >0) and weightless ( = 0),
2) plane state of displacements is assumed (2D as for very long retaining walls or slopes),
3) the soil occupies a bilinear infinite wedge described by:
the angle (to the horizontal exis) the angle (to the vertical axis),
4) differential equilibrium equations are fulfilled at every point of the wedge,
5) the Coulomb failure criterion =tg 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 o to the normal, - < o < q1 = const along 0B, with the angle to the normal, - < < .
Details are presented in Fig.1.
Note that the presented situation can be rotated because there is no privileged direction like the vertical one for heavy materials ( > 0); therefore – let q > q1, with no loss of generality.
Fig.1. The Prandtl wedge.
Remark:
It is unexpected but some problems are caused here by … the angle of internal friction .
This is due to the assumed 2D state of displacements, so if a specific state of shearing happens (on a plane) – in contrast to unconfined 3D shearing with ultimate stresses 1,2,3
situated on an infinite pyramid (or cone for the Drucker-Prager ultimate model):
o
q
x + A
+
z 0
q1
B +
+
Therefore, to overcome this problem, values of the angle of internal friction should be derived experimentally from a corresponding simple direct shearing apparatus, not from cylindrical samples in 3D-cell tests.
In other words, the Prandtl model ignores the principal stresses 2 which are perpendicular to the considered plane; in a true case of the Coulomb ultimate stress pyramid (or cone) this stresses 2 do depend somehow on 1 and3.
This way, we come to the following pretty general conclusion:
-angle found from direct shearing apparatus differs from -angle obtained using cylindrical samples in 3D-cell tests. Indeed, one should remember that in practice the former are greater of several degrees (a fact useful not only in the Prandtl model context).
2. Comments
Clearly, q and q1 are not independent – in contrast to the elasticity theory, for example; 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 the triaxial apparatus where only one of two stresses is under control).
Counter-clockwise convention of signs is used, compression means positive stress .
The equilibrium equations are as follows:
σz,z+τ ,x=γ=0
σx,x+ τ ,z=0 (1) The Coulomb limit state =tg on a sliding element means the algebraic relation:
√(σz−σx)2+4⋅τ2=(σz+σx)⋅sin ϕ (2) Hence, three stress components z,x, can be potentially found from 3 equations (1),(2) and the boundary conditions along A-0-B. There is also a stress component y equal to the intermediate principal stress 2 which is – by assumption – not important in the considered model.
Formally speaking, the general situation from Fig.1 is not correctly defined: one can assume a constant loading q but it is not clear why the solution q1 would have to be constant, too.
Numerical solutions, i.e. the method of characteristics, revealed that q1 cannot be constant - the exact values are less than q1 at a close vicinity of the pole 0; therefore, a simplified assumption q1 = 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 >0.
3. Prandtl’s assumption
Instead of the Euclidean coordinates (x,y), Prandtl used the equivalent polar coordinates (,), Fig.1. This way, the stress components z,x, in the equations (1) turn into ,,
and partial derivatives /z, /x are replaced by /, /.
Prandtl assumed that all stress components do not depend on the radius (like q and q1 do along the boundaries).
If so, partial differential equations (1) turn into ordinary differential equations because all
2
/=0. The solution becomes simpler, though not simple. Now it is clear why =0 must be assumed: for > 0 stresses increase with depth like in a half-plane (special case of the wedge) where z = z.
4. Prandtl’s solution
Similarity of all Mohr circles tangent to the envelope =tg indicates that q1 is proportional to q which is greater than q1. Indeed, the Prandtl solution is in the following form:
q1=q⋅Kaq(Pr ) (3) Kaq(Pr) – the Prandtl earth-pressure coefficient, less than 1.
Define an angular parameter for the wedge:
Θ=ωαo+αo
2 +ωδ+δ
2 +ε−β
(4) where
o results from the equation: sin(o) = sin(o)/sin()
results from the equation: sin() = sin()/sin().
Usually 0 (the angle =Bw-0-Aw in Fig.2) for which:
Kaq(Pr )=cos δ−sin ϕ⋅cosωδ
cos αo+sin ϕ⋅cos ωαo⋅exp{−2⋅Θ⋅tg ϕ}
(5a) If 0 (Bw-0-Aw in Fig.2 does not exist) then:
Kaq(Pr )=cos δ−sin ϕ⋅cos ωδ
cos αo+sin ϕ⋅cos ωαo⋅cos(n )−sin ϕ⋅cos(m)
cos(n )+sin ϕ⋅cos( m) (5a) where
the angle n results from the equation: sin(n) = sin()sin(m) with m = /2 + .
5. Kinematic interpretation
Like in the triaxial apparatus, plastic yielding takes place on a local sliding area at the angle
/4 /2 to principal directions of i. In the wedge interior, the situation is not so simple because the principal directions are different from point to point or at least in some
subregions; adjacent local sliding areas continue one to another and finally create slip lines, Fig.2.
Fig.2. Slip lines in the wedge B-0-A interior.
q
0 A
q1
Aw
B
Bw
4
There are three families (or two) of smooth slip lines:
“passive” zone A-0-AW with straight slip lines,
„active” zone B-0-BW with straight slip lines,
transition zone Bw-0-Aw, called the Prandtl fan consisting of radii coming from the pole 0 and logarithmic spirals – this family does not appear for 0 when Bw- 0-Aw does not exist.
/4+/2
q
6. Applications
Example #1: the Coulomb results for active earth pressure.
Assume for the wedge: = 0, = 0, o = 0, = 0.
The situation means a quarter-plane x >0, z >0 vertically loaded by q with a horizontal soil pressure q1 on the wall surface.
In details, o = 0, =0 and = 0 in the equation (4); the transition zone does not exist or is reduced to one line, there are only straight slip lines.
The loading q is given, the loading q1 is to be found whereas q1 < q so it is the active state of soil pressure on a smooth vertical wall (“The Coulomb wall”), q1 = ea.
Simple substitution to (5) results in:
q1=q⋅Kaq(Pr )=q⋅1−sin ϕ 1+sin ϕ
which coincides with the Coulomb solution:
ea = Ka(z + q) = qKa as far as = 0.
The same formula holds also for >0 – see details in the Example #3.
Example #2: the passive Prandtl pressure.
The default option was as follows:
q is applied to the soil surface, q1 appears on the wall surface as the ultimate (minimal) value; q1 < q corresponds to the outward movement of the wall.
In the opposite interpretation, the situation can be reversed: q1 is given and q > q1 is to be found for the ultimate case of stresses, i.e. q is as great as possible (for the given q1).
Therefore, the equation (3) results in:
q= q1
K(Pr )aq (3*) or just simply
q=q1⋅K(pqPr ) where Kpq
(Pr )
= 1 K(aqPr ) .
Conclusion:
like in the Coulomb method, there is: Kp Ka = 1, which is not true in the Poncelet (Műller- Breslau, PN-83/B-03010); in the Prandtl method the active and the passive soil pressure mean in fact the same limit equilibrium, only the axes are rotated:
z = 1 and x = 3 … means the active state,
x = 1 and z = 3 … means the passive state.
0
Fig.3. The Coulomb slip lines.
6
dq dq 0 A
B
dq
h L
Example #3: Difficulties with heavy soils
The case of >0 is beyond the scope of the Prandtl method. Note that the assumption 0 could be acceptable only for (relatively) great loading q, i.e. for q >> L.
For retaining walls the situation is usually just opposite.
Sometimes the following “trick” with the superposition principle can be helpful, like the one for the Poncelet method (layered soils).
For =0 in Fig.4 assume that dq= dz and use (3) in the form dq1 =dq Kaq(Pr) for different levels 0<z<h.
The following integration q1≅∫Kaq
(Pr )dq=K(Pr )aq ∙ γ ∙∫
0 h
dz =K(Pr )aq ∙ γ ∙h=Kaq(Pr)∙ γ ∙ L∙ cos β
suggests that Kaγ(Pr )≈Kaq(Pr )⋅cosβ .
This is an approximate approach because dq would be reduced near the wall surface 0B and the superposition principle is not always justified here; for rather smooth wall surface and for
close to 0o the approximation error is not very big.
For the Coulomb wall this method is exact Kaγ(Pr )=Kaq(Pr)=Ka .
Comments on the superposition principle:
in limit equilibrium (and plasticity) the superposition principle is generally not true.
If a massive block resting on horizontal soil has a friction coefficient then:
the shearing force T1 = N causes sliding (normal force N equals the weight of the block),
alternatively, the force T2 = -N also causes sliding but the sum T1 +T2 =0 is neutral.
Similarly for stress components – with one important exception: the superposition principle holds if two limit states are coaxial, i.e. with the same principal directions of the stress tensor (draw Mohr’s circles to see this).
Fig.4. Subdivision into elementary strips dz.
Example 4: Bearing capacity of a shallow foundation beam
A very long foundation beam of the width B (plane displacements) has a certain bearing capacity R [kN/m] or in terms of stresses r = R/B [kPa] where
r=c ∙ Nc+q ∙ Nq+1
2∙ γB∙ B ∙ Nγ
1. Assume that c = 0, so Nc does not matter. If c > 0, the coefficient Nc can be derived from the equivalent states principle, Nc = ctg(Nq – 1) like in the Eurocode EC7-1.
2. Let B = 0. If B > 0, the exact analytical solution does not exist, the coefficient N needs approximate calculation (typical for the Prandtl method) which can lead to different results.
3. Consider the uniform vertical loading r [kPa] along the half-line 0-A, not only under the foundation area B. The overburden loading on 0-B equals q [kPa]. The equation (3*) can be adopted having in mind that r >> q. Here, = 0, = -/2, o = 0, = 0; the Prandtl wedge is reduced to the half-plane, Fig.5. Moreover, o = 0, =0 and = 0 – (-/2) =
/2 > 0 in the formula (4).
Fig.5. Bearing capacity calculation.
Due to (3*):
r=q ∙ K(Pr )pq
Therefore, using (5a):
Nq=K(Pr )pq = 1
Kaq(Pr)=1+sin φ
1−sin φ∙exp {π ∙ tgφ } like in the Eurocode EC7-1.
width B r r q
0
B A
Bw
Aw
8
