The Delft Sand, Clay & Rock Cutting Model, 2019a

The Delft Sand, Clay & Rock Cutting Model.
d 1 (0.64 0.56h b/ h i) (0.0164 0.0085h b/ h i) SPT10
(α = 30 degrees) (6-93)
d 1 (0.83 0.45h b/ h i) (0.0268 0.0085h b/ h i) SPT10
(α = 45 degrees) (6-94)
d 1 (0.99 0.39h b/ h i) (0.0503 0.0099h b/ h i) SPT10
(α = 60 degrees) (6-95)
With: SPT10 = the SPT value normalized to 10 m water depth.
Lambe & Whitman (1979), page 78) and Miedema (1995) give the relation between the SPT value, the relative
density RD (0-1) and the hydrostatic pressure in two graphs, see Figure 6-29. With some curve-fitting these graphs
can be summarized with the following equation:
2.52
SPT 0.243 82.5 g z 10 RD
(6-96)
And:
l
4.12
SPT
RD
82.5 l
g z 10
0.397
(6-97)
Lambe & Whitman (1979), (page 148) and Miedema (1995) give the relation between the SPT value and the angle
of internal friction, also in a graph, see Figure 6-28. This graph is valid up to 12 m in dry soil. With respect to the
internal friction, the relation given in the graph has an accuracy of 3 degrees. A load of 12 m dry soil with a density
of 1.67 ton/m 3 equals a hydrostatic pressure of 20 m.w.c. An absolute hydrostatic pressure of 20 m.w.c. equals 10
m of water depth if cavitation is considered. Measured SPT values at any depth will have to be reduced to the
value that would occur at 10 m water depth. This can be accomplished with the following equation (see Figure
6-30):
SPT
282.5
10 82.5 z
l g z 10
SPT
(6-98)
With the aim of curve-fitting, the relation between the SPT value reduced to 10 m water depth and the angle of
internal friction can be summarized to:
0.01753 SPT
51.5 25.9 e 10
(+/- 3 degrees) (6-99)
For water depths of 0, 5, 10, 15, 20, 25 and 30 m and an available power of 100 kW the production is shown
graphically for SPT values in the range of 0 to 100 SPT. Figure 6-31 shows the specific energy and Figure 6-32
the production for a 45 degree blade angle.
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