Geotechnical centrifuge modelling - The Mechanics of Soils and Foundations - page 420

Figure 27.2 Scaling for the stability of a model slope.

in the model will be approximately constant with depth rather than increasing linearly
with depth as in the ground.
Another way of looking at the requirements of geotechnical modelling is through
dimensional analysis. The stability of a slope for undrained loading was described in
Sec. 21.8. For the prototype slope in Fig. 27.2(a) with height H p and slope angle i
(which is itself dimensionless), the stability depends on the undrained strength s u , the
height H p and the unit weight γ = g ρ . These can be arranged into a dimensionless

group

g ρ H

s u

N s = (27.3)

where N s is a stability number. Notice that this is exactly the same as the stability
number in Eq. (21.42). A model and a prototype are similar (i.e. they will both collapse
in the same way) if they both have the same value of N s . If the scale factor is n so that
the model height H m and the prototype height H p are related by H m = H p / n the

stability numbers can be made equal by accelerating the model in a centrifuge to ng so
that

g ρ H p

s u

ng ρ H m

s u

N s = = (27.4)

Thus the stability of the model slope illustrated in Fig. 27.2(b) will be the same as the
stability of the prototype slope in Fig. 27.2(a) and if the slopes fail they will both fail
in the same way.
The stresses, and the basic soil properties, in a prototype and in an n th scale model
will be the same if the model is accelerated in a centrifuge to ng , but time effects may
require a different scaling. There are several aspects of time in geotechnical engineering,
the most important being associated with consolidation.
Consolidation due to dissipation of excess pore pressures with constant total stresses
was discussed in Chapter 15. The rate at which excess pore pressures dissipate during
one-dimensional consolidation is given by Eq. (15.34) and for similarity the time factor
T v in the model and in the prototype should be the same. From Eq. (15.25),

c v t p
H p =

c v t m
H m

T v = (27.5)