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(cos
2
sin
2
a =
,t
cos
sin
=
,t
(cos
)/(2cos
))
=
,t
/(2cos
)
(9.35a)
(cos
2
sin
2
b =
,t
sin
cos
=
,t
(sin
)/(2sin
))
=
,t
/(2sin
)
(9.35b)
Hence, the strains in the wedge can be expressed by
xx,t
= u
xx,t
= a/
sin
=
,t
/
sin(2
)
(9.36a)
yy,t
= u
yy,t
= b/
cos
=
,t
/
sin(2
)
(9.36b)
Easily, one may verify that for distortion
,t
=
0, and for the volume constraint
yy,t
=
0 holds
.
Next, the plastic work rate is elaborated. Inside the wedge and
along the interface OB' a constant cohesion
k
is assumed (note, 0
<
xx,t
<
/2)
yy,t
)
2
,t
2
]
dV
W
,t
= k
OAB
[(
k
OB
|
s| ds =
xx,t
,t
sin
-1
(2
= 2k|
)|
OAB
dV
k|
,t
cotg(2
)|
OB
sds
,t
sin
-1
(2
)|
h
2
tan
)|
h
2
/2cos
2
= k|
k|
,t
cotg(2
,t
|h
2
{tan
)|/2cos
2
= k|
/
sin(2
))
|cotg(2
}
,t
|h
2
{1
)|
} (2cos
2
)
-1
= k|
|cotg(2
(9.37)
It can be shown that this expression is minimum for
=
/
4. Then,
W
,t
=
,t
|h
2
. If along the axis OA a rigid border is present, the work along it has to be
accounted for. Assuming there a cohesion
k'
this contribution becomes
k|
W
,t
= k'
OA
|bs|ds = k'|
,t
|
/
(2sin
)
OA
s dy
= k'|
,t
|
/
(2sin
)
OA
y/
(cos
)
dy
,t
|h
2
/(2sin(2
,t
|h
2
= k'|
)) = ½
k'|
(9.38)
which attains a minimum at
=
/
4. Finally, the minimum total work rate
becomes
,t
|h
2
W
,t
W
,t
=
(
k
½
k'
)
|
(9.39)
Rankine wedge
An important element of a visco-plastic stationary motion with an analytical
solution is a Rankine wedge as shown in Fig 9.8. After a small rotation
,t
dt
the
wedge OAB is supposed to undergo a plastic deformation by one-directional
frictional sliding and it moves into shape OA'B, while the outside remains rigid.
Along the interface OA slip takes place against a rigid border, compatible with
volume constraint conditions. For convenience, the angle
is restricted to 0
<
<
/2, without loss of practicality. Geometric considerations yield
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