671017.pdf

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2 o^2 r
n
R 4

R 3

m

z

p(r,z)

h

o 1

o

t

1

Ground

Slope

Cavity expansion
source

Image source

(a)

β
φ 1

o r
22
n
R 4

R 3

m

z

p(r,z)

h

o 1

o

t
Ground

Slope

Cavity expansion
source

Image source

(b)

β

r
o 2
2

1

R n
4

R 3

m

z

p(r,z)

h
o 1

o

Ground t

Slope

Cavity expansion
source

Image source

(c)

Figure 4: The expansion of the single spherical cavity near a slope. (a) The image source is located above the ground surface. (b) The image
source is located on the ground surface. (c) The image source is located below the ground surface.


휏푟푧=

3푎^3 푞푧(푡+푟)

2푅^51

+

3푎^3 푞푧(푟−푡)

2푅 25

, (10)

휎휃=−

푎^3 푞

2푅^31


푎^3 푞

2푅^32

, (11)

푢푟=

푎^3 푞

4퐺

(

푡+푟

푅^31

+

푟−푡

푅^32

), (12)

푢푧=

푎^3 푞

4퐺

(


푅^31

+


푅^32

), (13)

where푅 1 =√(푡 + 푟)^2 +푧^2 and푅 2 =√(푡 − 푟)^2 +푧^2.
The horizontal ground and the slope are both free
surfaces,onwhichthenormalandshearstressesarezero.
Because of the symmetry, the source and its image produce
zero shear stress (휏푟푧 =0)andanonzeronormalstress
(휎푟=0̸ ) on the free surface푟=0. Thus, the presence of the
normal stress violates the free surface boundary condition


휎푟儨儨儨儨푟=0=푎^3 푞(

3푡^2

푅 05


1

푅^30

), (14)

where푅 0 =√푡^2 +푧^2.


3.2. The Expansion of a Spherical Cavity and Its Image Near
aSlope.The boundary effects of both ground surface and
slopeshouldbeinvestigatedduringtheanalysisofthecavity
expansion near the sloping ground. The problem turns to
be more difficult due to the increase of the slope boundary.
Similarly, the virtual image technique is employed to consider
the boundary effects of the slope. Taking the slope as the plane
ofsymmetry,theimagesourceissetattheimagepointas
shown inFigure 4.
The solutions of the two cavities expansion (the actual
sphericalcavityanditsvirtualimage)canbeobtained
by the principle of superposition. As a result, the stress
and displacement components in the cylindrical coordinate
system are shown to be

휎(0)푟 =

푎^3 푞

2

[

3 (푡+푟)^2

푅^53


1

푅^33

+

3 (푡+푟−푚)^2

푅^54


1

푅^34

], (15)

휎(0)푧 =

푎^3 푞

2

[

2

푅^33


3 (푡+푟)^2

푅^53

+

2

푅^34


3 (푡+푟−푚)^2

푅^54

], (16)

휏(0)푟푧=

3푎^3 푞(푧−ℎ)(푡+푟)

2푅^53

+

3푎^3 푞(푧−ℎ+푛)(푡+푟−푚)

2푅^54

,

(17)
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