1. The essence of the problem

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W.Brząkała: CEB007361 Lecture no.7

1 THE METHOD OF CORRESPONDING STATES OF STRESSES

for cohesive soils in the limit state of stresses

1. The essence of the problem

How to find a solution for the limit state stresses in the Coulomb-Mohr medium in cohesive soil (c > 0), if a solution to the same problem is known,

but for non-cohesive soil (c = 0)?

The term "the same problem" means that all mechanical and physical parameters are the same (ex- cept for cohesion), as well as the shape of the medium - boundary surfaces, given external loads, etc.

A non-cohesive medium that meets these requirements is called a corresponding medium; it differs from the considered cohesive medium by only one parameter: c = 0.

2D cases are considered in the plane state of displacements; so, only transversal cross-sections of a

"very long" geotechnical structures are considered, like long retaining walls, slopes, and foundation beams.

2. Governing equations and the procedure

There are 3 components of stresses σ

^z

, σ

^x

, τ in a real cohesive medium which fulfil:

1. Static equilibrium equations in 2D

+ = (1a) + = 0 (1b) 2. Algebraic limit condition for cohesive Coulomb-Mohr media

− + 4 ∙ − + ∙ sin = 2 ∙ ∙ cos ^. (2)

Fig.1. Limit state condition.

1. Static equilibrium equations in 2D

∗ + ^∗ = (1a*)

∗ + ^∗ = 0 (1b*) 2. Algebraic limit condition for non-cohesive Coulomb-Mohr media

^∗ − ^∗ + 4 ∙ ^∗ − ^∗ + ^∗ ∙ sin = 0 (2*) By making a horizontal shift of the axis τ, to the left by the value of σ

H

= c⋅ctg (ϕ) = const > 0, the stress state components in the new coordinate system σ, τ are as follows:

σ

^z

* = σ

^z

+ σ

^H

, σ

^x

* = σ

^x

+ σ

^H

, τ * = τ , or in the reverse order

σ

^z

= σ

^z

* - σ

^H

, σ

^x

= σ

^x

* - σ

^H

, τ = τ *.

In the shifted system of axes σ , τ , we deal with the corresponding non-cohesive medium in the limit state of stresses; this is seen in (1) and (2), derived from equations (1) and (2).

ϕ

τ

c

τ=σ⋅tgϕ

σ, σ ^*

0

τ ^*

σ

^H

= c ⋅ ctg ϕ

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W.Brząkała: CEB007361 Lecture no.7

2 The procedure can be summarized as follows.

a) Define the region occupied by the real cohesive medium.

Known loads τ

o

and σ

o

are applied to a part of the boundary B

o

of the cohesive medium; it can be q, σ

z

or σ

x

, or also τ , i.e. for an inclined load. On the rest of the boundary B

1

determine which load is unknown (to be found), like σ

1

, sometimes also τ

1

or some relationships between them

¹

; it can be for example the ratio τ

1

/ σ

1

= tg δ depending on the so-called frictional roughness of the wall surface

²

.

b) Formulate the identical (corresponding task in the corresponding non-cohesive medium (), modifying the acting loads applied to the boundary B*

o

, i.e. taking τ

o

* = τ

o

and σ

o

* = σ

o

+ σ

H

. Solve this problem using relevant known methods, i.e. having τ

^o

^{* and} σ

^o

^{ along B}*

^o

^,

find τ

1

* and σ

1

* along B

1

, taking c = 0.*

c) Return to the real cohesive medium using the inverse transformation:

τ

¹

⁼ τ

¹

^{* and} σ

¹

⁼ σ

¹

^{* -} σ

^H

. This way the solution is found for the cohesive medium.

3. Examples Example #1. The Coulomb solution

The method of corresponding states of stresses is exact and gives the same result as the exact Coulomb-Mohr method (discussed in Lecture 6). For the active pressure:

= ∙ ∙ + − 2 ∙ ∙ , where = !"#$ %

&"#$ % = ' &"#$ % ^{()" %} * .

For the Coulomb wall and active soil pressure, direct application of the procedure is as follows.

Point a)

B

o

= horizontal ground surface z = 0 which is loaded vertically τ

^o

= 0, σ

^o

= q = const;

B

1

= vertical smooth wall, τ

¹

= 0, σ

¹

= e

a

= ? Point b)

τ

o

* = τ

o

= 0 and σ

o

* = σ

o

+ σ

H

= q + σ

H

= q* are applied to the horizontal surface of

the corresponding (non-cohesive) medium, the parameters z, γ, ϕ and K

a

keep unchanged;

∗ = ^∗ = ₊ ∙ ∙ + ^∗ = ₊ ∙ ∙ + + ₊ ∙ _, and τ

1

* = 0.

Point c)

Reverse transformation : τ

1

= τ

1

* = 0 and σ

1

= σ

1

* - σ

^H

= e

a

* - σ

^H

= e

a

, therefore

= ₊ ∙ ∙ + + ₊ ∙ _, − _, = ₊ ∙ ∙ + − _, ∙ 1 − ₊ .

= ₊ ∙ ∙ + − ∙ ^cos _sin ∙ _1+sin ^2∙sin = ₊ ∙ ∙ + − 2 ∙ ∙ ₊ , because ^{()" %}

&"#$ % = ! "#$ %

^/

&"#$ %

^/

= !"#$ % ∙ &"#$ %

&"#$ % ∙ &"#$ % = .

Example #2. The bearing capacity coefficient N

c

for a shallow foundation beam (EC7-1) For non-cohesive soils the following formula holds: ₀ = ∙ 1 2 + ½ ⋅γ⋅B⋅ 1 3 [kPa], and the bearing capacity coefficient N

q

yields form the Prandtl solution.

Point a)

The considered region is just the half-plane B0A, being a special case of the Prandtl wedge; it has the horizontal boundary and vertical loading, Fig.2.

1 σ

1

does not mean - exceptionally - principal stress, but generally just a normal stress along B0.

2 The angle δ is usually a fraction of the angle ϕ being constant in this method, therefore the angle δ also does

not change.

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W.Brząkała: CEB007361 Lecture no.7

3 B

o

= 0B, τ

o

= 0, σ

o

= q (known),

B

1

= 0A, τ

¹

= 0, σ

¹

= q

f

= ? Point b)

τ

o

* = τ

o

= 0 and σ

o

* = σ

o

+ σ

H

= q + σ

H

= q* , τ

1

* = 0.

Therefore for c* = 0 there is ^∗ = ₀ ^∗ = ^∗ ∙ 1 2 + ½⋅γ⋅ B ⋅1 3 = ∙ 1 2 + 4 ∙ 1 2 + ½⋅γ⋅ B ⋅1 3 . Point c)

Reverse transformation: = 0 = ₀ ^∗ − 4 = ∙ 1 2 + ∙ 1 5 + ½⋅γ⋅ B ⋅1 3 ,

where the following coefficient is introduced 1 5 = 67 ∙ 81 2 − 19, the same as in EC7-1.

The other bearing capacity coefficients N

q

, N

γ

are not changed during this procedure, because they depend only on the angle ϕ which is constant.

Fig.2. Bearing capacity problem in the corresponding medium (c= 0).*

Example #3. Pressure of a weightless cohesive soil on a retaining wall

This situation recalls the general case of the weightless Prandtl wedge and shows significant effects of the vector nature of the pressure

³

.

Symbols and sign convention are taken as previously by Coulomb-Prandtl.

Fig.3. The Prandtl wedge.

For c > 0, the active soil pressure q

1

< q is to be found in its vector form (both value and angle).

Point a)

The weightless wedge A0B loaded by a constant q along A0 is considered.

Given: ϕ , c

,

q, α

^{o ,}

β , ε , δ = η⋅ϕ , where η ≤ 1 is a certain constant.

3 The previous issues were actually scalar ones, because there was τ = 0 along the boundary, that is, only one component of the soil pressure vector was of interest – the normal component σ

¹

.

q ^*

q f * q f *

B A

B

w

A

w

0 β

ε

q

1

+

+ ∞

+ ∞ δ 0

α ^o

A

B

c > 0, γ = 0

δ

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W.Brząkała: CEB007361 Lecture no.7

4 B

o

= 0A: τ

o

= q⋅sin(α

o

), σ

o

= q⋅cos(α

o

).

B

1

= 0B: q

1

= ? Point b)

σ

^H

= c ⋅ ctg( ϕ ) increases the normal component of the boundary load σ

^o

^* ⁼ σ

^o

+ σ

^H

and – keeping unchanged τ

^o

^* ⁼ τ

^o

– generates a new vectorial boundary load along 0A

∗ = : ∗ + : ∗

Comment:

for the corresponding medium along 0A, the angle of q to the normal is reduced; instead of the real* value α

^{o =}

arctg( τ

^o

^/ σ

^o

) for q, there is a virtual value α

^o

* = arctg( τ

^o

^*/ σ

^o

^*) ≤ α

^o

for q (in absolute values of* both angles α , of course).

The problem is being solved in the corresponding medium, so the formulae (4),(5) from Lecture 7 (Coulomb-Prandtl) can be used.

Taking ϕ , c = 0

,

q,* α

^o

*

,

β , ε , δ = η⋅ϕ one gets the solution q

1

* = const along 0B, which has two following components σ

1

* = q

1

*⋅cos(δ) and τ

1

*= q

1

*⋅sin(δ).

Point c)

The reverse transformation goes as usually: σ

1

= σ

1

* - σ

H

, τ

1

= τ

1

. This solution defines a vectorial loading q*

1

= const along 0B

= + = ;<=6

which has an angle δ

^c

to the normal; this angle is greater than the frictional roughness δ , because tg( δ

^c

) = τ

1

/ σ

1

= τ

1

*/( σ

1

* - σ

^H

) > τ

1

*/ σ

1

* = tg( δ ).

As the effect:

cohesion c > 0 causes a reduction of the length of the vector q

1

and increases its angle to the normal;

but this is potentially possible that δ = ϕ, therefore, this inclination can be greater in cohesive soils than the internal friction angle

⁴

.

Example 4. Pressure of a heavy cohesive soil on a retaining wall

The procedure is similar to the previous example, but the matter is complicated by the lack of an exact solution for the non-cohesive soil - necessary step in Point b). In general, it cannot be even the standard Poncelet solution from Lecture 6, because it concerns only the vertical load q applied to the ground surface ( α

^o

= -ε ); for the angle |α

^o

* | < |α

^o

| it will no longer be a vertical load on the

corresponding medium (unless ε = 0).

However, it is easy to predict the features of such a solution for soil pressure on the wall: along the wall 0B soil pressure will increase, both τ

¹

* = τ

¹

and σ

¹

*. Thus σ

¹

= σ

¹

* - σ

^H

will also increase, but the influence of this σ

^H

correction will decrease relatively, because σ

^H

is a constant value.

So tg ( δ

^c

) > tg ( δ ) but tg ( δ

^c

) → tg ( δ ).

Of course, there is nothing special about it, because the same yields from the Coulomb condition τ = σ⋅ tg ϕ + c; the cohesion c becomes of little importance if σ is "very large".

4

Similarly to the angle of the resultant force in GEO stability against sliding on a cohesive soil layer.

1. The essence of the problem

W.Brząkała: CEB007361 Lecture no.7

1 THE METHOD OF CORRESPONDING STATES OF STRESSES

for cohesive soils in the limit state of stresses