Critical State Soil Mechanics Phần 7 pot

128

Using the numerical values of , , , , κ λ M p0 already quoted, we get the curves of Fig. 7.13

where the points corresponding to total distortion of 1, 2, 3, 4, and 8 per cent are clearly

marked.

In comparison, we can reason that relatively larger strains will occur at each stage

of a drained test. In Fig. 7.14(a) we consider both drained and undrained compression tests

when they have reached the same stress ratio η > 0. From eq. (6.14) we have

= = ( −η)ε (ε > 0) κ & & & & M

v

v

v

v p

so that in each test there will be the same shift of swelling line for the same increment of

distortion

κ v&

ε&. The yield curves relevant to the successive shifts of swelling line are lightly

sketched in Fig. 7.14(a). We see the undrained test slanting across them from U1 to U2

while the drained test goes more directly from D1 to D2. It follows that for the same

increment of distortion ε& there will be a greater increment of η& in an undrained test than

in a drained test, which explains the different curves of q p versus ε, in Fig. 7.14(b).

Fig. 7.13 Predicted Strain Curves for Undrained Axial Compression Test of Fig. 7.12

*

For the undrained test, differentiating eq. (6.27) we get (Λη& M ) = − p& q which equals the plastic volume change

(since 0) 0 v& v v& ≡ κ and hence we have eq. (7.5). For the drained test, differentiating eq. (6.19) we get

η& = (M λ − κ ){+ v& − (λp& p)} and since q& p& = 3 we also have (3 ) (3 )( ). 2

κ η& = q& p& − qp& p = − η p& p = − η v& − v&

Eliminating p& and v& we obtain eq. (7.6).

129

We can, in fact, derive expressions for these quantities. It can readily be shown*

that for the increment U1U2 in the undrained test

( ) ( ) 0 undrained 0 M v v v

M

Λ

η = −η ε = κ & & (7.5)

whereas for D1D2 in the drained test

0 drained ( ) 3

1 1

η η ε

η

λ & M v & M

Λ = − ⎟

⎟

⎠

⎞ ⎜

⎜

⎝

⎛

− + (7.6)

(where v varies as the test progresses and is ≤v0).

Fig. 7.14 Relative Strains in Undrained and Drained Tests

Hence the ratio of distortional strains required for the same increment of η is

; (3 )

(3 )

d

0

v

M v

u

d

κ η

λ η

ε

ε

−

− + = &

& (7.7)

and we shall expect for specimens of remoulded London clay (a) at the start of a pair of

tests when

3.37

0

u

d

d 0

≅

⎪

⎭

⎪

⎬

⎫

=

=

ε

ε

η

&

&

v v

and (b) by the end when and 0.95 ,( ) 3.52. η → M vd ≅ v0 ε&

d ε&u ≅ It so happens that the minor

influences of changing values of η and v during a pair of complete tests almost cancel out;

and to all intents and purposes the ratio d u ε& ε& remains effectively constant (3.45 in this

case).

This constant ratio between the increments of shear strain d u ε& ε& will mean that the

cumulative strains should also be in the same ratio. This is illustrated in Fig. 7.15 where

results are presented from strain controlled tests by Thurairajah9

on specimens of virgin

compressed kaolin. The strains required to reach the same value of the stress ratio η in each

test are plotted against each other, giving a very flat curve which marginally increases in

slope as expected.