Geoscience Reference
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- The strain in layer 4 is larger than 10% and (5.11) should be used which gives a
settlement contribution of
H
(
1
bottom
)/2 = 3(0.253+0.241)/2 = 0.741 m.
- With the large strain concept the total settlement
S
becomes 1.241 m. The linear
strain concept overestimates the settlement when large strain occurs, in this case
with 9.5%. In fact, one should use (5.11) instead of (5.12) and the best result is
obtained automatically.
1
top
application 5.2
A fill, which imposes a load of 50 kN/m
2
, compresses a clay layer of 5 m thick.
The wet unit weight of the clay is
= 15 kN/m
3
and the compressibility index of
the clay is
= 0.2. Give an estimate of the final settlement by considering only
stresses at the mid-height of the layer. Next, refine this estimate by considering two
equally thick sublayers, each 2.5 m thick.
application 5.3
Does an equal numerical value of natural strain respectively linear strain
correspond to greater, equal or less compression for the natural strain case?
application 5.4
To how much (linear) compression does a natural strain of 120% correspond?
application 5.5
An oil tank is built at a site containing a soft sublayer, which was preloaded
before construction (Fig 5.3). The preload equals the full tank load. How much will
the tank settle on its first filling to full tank load? Assume stresses at mid-height of
the soft material as representative for the whole layer. Sand fill: Thick 2m,
= 20
kN/m
3
,
= 16 kN/m
3
,
C
c
' /(1+
e
0
) =
0.01,
C
c
/(1+
e
0
) = 0.1. Full tank load: 100 kPa. Phreatic level: 1m below surface
d
= 16 kN/m
3
. Soft sublayer: Thickness 5m,
phreatic level
sandfill
soft material
Figure 5.3
application 5.6
A soft layer with compressibility coefficient
C
c
/(1+
e
0
) and unit weight
'
1
dips
slightly through stiffer soil (unit weight
'
2
) as shown (Fig 5.4). Give an expression
for the compression due to this layer as function of the horizontal position
x
.
Quantify the compression at left, centre and right, and explain the trend. Use
H
=
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