Civil Engineering Reference
In-Depth Information
Figure 19.5
Work done by internal stresses on slip planes.
The work done by the internal stresses is the work dissipated by plastic straining
in the material in the thin slip surfaces that make up the compatible mechanism and,
again, undrained and drained loadingmust be considered separately. Figure 18.5 shows
short lengths of slip surfaces that have increments of displacement
w
as shown. Since
the soil is at the critical state in each case the stresses are given by Eqs. (19.1) and
(19.2) and, for drained loading, the shear and normal strains are related by Eq. (19.3).
In Fig. 19.5(b) for drained loading the water remains stationary, the work is done
by the effective stresses and hence
δ
=
τ
L
−
σ
n
L
δ
W
δ
l
δ
n
(19.10)
σ
n
and
Note that for dilation the work done by the normal stress is negative since
δ
n
are in opposite directions. From Eq. (19.10), with the volume of the slip plane
V
=
Ly
,
1
tan
ψ
c
τ
δγ
+
σ
n
δε
τ
δγ
δ
W
=
V
(
n
)
=
V
−
(19.11)
tan
φ
c
=
φ
c
and so the work dissipated by the
However, for a perfectly plastic material
ψ
c
internal stresses for drained loading is
δ
=
W
0
(19.12)
This is a very surprising result and presents difficulties which I will not explore here.
The implication is that a perfectly plastic factional material is both dissipative and
conservative, which is nonsense. The conclusion must be that the flow rule for a fric-
tional material cannot be associated. Nevertheless, the result given by Eq. (19.12) is
very convenient and it may be used to calculate upper bounds for frictional materials
like soil.
In Fig. 19.5(a) for undrained loading the increment of work done by the total stresses
τ
σ
and
is
δ
W
=
τ
L
δ
w
=
s
u
L
δ
w
(19.13)
