Civil Engineering Reference
In-Depth Information
strains are measured. Equation (3.32) can be differentiated to give the creep strain
rate as
c
d
ε
C
t
=
(3.33)
d
t
showing that the creep strain rate decreases with time. Many soils, particularly soft
clays and peats, show significant creep strains. These can also influence the subsequent
behaviour, as we will discuss later.
3.12 Summary
1. The basic mechanical properties of structural materials are stiffness and strength.
Stiffness relates changes of stress to changes of strain and this governs deforma-
tions and ground movements. Strength is the largest shear stress that a material
can sustain before it fails and this governs the ultimate states of collapse of
structures.
τ
c
(where
c
=
2.
Strength can be described as cohesion with
is the cohesion)
φ
is the angle of friction) or by the
Mohr-Coulomb criterion which combines cohesion and friction.
3. Purely elastic strains are recovered on unloading. In metals the elastic stress-strain
line is approximately linear so the elastic parameters
G
and
K
are approximately
constants.
4. A perfectly plastic material continues to strain with constant stresses and the vector
of plastic strain, which relates the rates of plastic shear and volumetric strains, is
normal to the current yield curve.
5. Theories for elasto-plastic straining can be obtained by adding the elastic and
plastic components of strain.
6. Most time and rate effects in soils are due to coupling of stiffness with seepage
of pore water. Creep and viscous effects are usually neglected except in peats and
other organic soils.
τ
=
σ
tan
φ
or as friction with
(where
Worked examples
Example 3.1: Stress and strain in a triaxial test
In a triaxial compression test on a
sample of soil the pore pressure is zero so total and effective stresses are equal.
The radial stress is held constant at
σ
r
=
200 kPa and the axial stress is changed
σ
a
=
from
350 kPa to 360 kPa. The strains for this increment were
δε
=
0.05% and
a
δε
0.01%.
At the start of the increment,
=−
r
q
=
σ
a
−
σ
r
=
350
−
200
=
150 kPa
p
=
1
3
(
σ
a
+
σ
r
)
1
3
(350
2
=
+
400)
=
250 kPa
