671017.pdf

0305 10 15 20 25

n=50kPa

n= 100kPa

n=150kPa

n= 200kPa

n=250kPa

n=300kPa

n= 350kPa

Interface shear displacement (mm)

n=50kPa

n= 100kPa

n=150kPa

n= 200kPa

n=250kPa

n=300kPa

n= 350kPa

0305 10 15 20 25

Interface shear displacement (mm)

0

0. 5

1

1. 5

2

2. 5

3

4

3. 5

Shear stress (kPa)

0

20

40

60

80

100

120

Vertical

deformation (mm)

Figure 5: Test results for the interface between clay and #2 plate (initial normal stress of 400 kPa).

0305 10 15 20 25

Interface shear displacement (mm)

0305 10 15 20 25

0

0. 5

1

1. 5

2

2. 5

3

3. 5

Vertical

displacement (mm)

Shear displacement (mm)

Shear stress (kPa)

0

5

10

15

20

25

30

35

ni=400kPa

ni=300kPa

ni= 200kPa

ni= 100kPa

− 0. 5

ni=100kPa

ni=200kPa

ni= 300kPa

ni= 400kPa

Figure 6: Test results for the interface between plate #0 and clay (applied normal stress of 100 kPa).

in which푑휎푛and푑휏are the incremental normal and shear
stresses, respectively, and푑푢푛and푑푢푠are the incremental
normal and tangential displacements, respectively. All the
parameters can be obtained directly from the test. While the
interface is not purely smooth, the researchers [ 7 , 12 , 17 , 18 ]
found that a shear band exists near the interface during
shear. The thickness푡wasobservedtoequalfivetimesthe
diameters of the sand at the interface between the sand and
the structure. For the interface with clay, determining the
value of푡has not been studied. In the direct shear test, the
thickness of the shear band cannot be determined because
theshearislimitedalongtheplane.Asaresult,thevalueof

푡is set to a constant value in the sections below. An internal

strain was assumed, even in the shear band. Consider

푑휀푛=

푑푢푛

푡

,

푑휀푠=

푑푢푠

푡

,

(2)

where푑휀푛and푑휀푠are the incremental normal and shear

strain, respectively. Thus, the matrix can be expressed as

[퐾]=푡[퐷푒푝]. (3)