ABSTRACT
With tan/3 = —0.65 (from fig. 2.3) and el = —1 (the value of 1E11 is im -
material, as sand is considered as rate independent) we obtain E2 = 0.825, i.e.
7 —1 0 0 \ D = 0 0.825 0
0 0 0.825 I
vs. 1E l curve for El = 4% is
0(a1M2) = 1.4 A(—Ei)
OQl = 1.4x2 0(—El) or Ao-i/At = —1.4o20ci/At
6-1 = —1.40-2ei
With ei = —1 we obtain
T=
= —140, thus
—140 0 0 \ 0 0 0 0 0 0 /
kN 1112S
4. Oedometric test: given stress state: a1 = a2 = a3 = 100 kN/m2
=
T2 = T3 = —100 kN/m2. The kinematical boundary conditions of an oedometric test (D2 = D3 = 0) imply axial symmetry with T2 = T3 and T2 = T3.
