d2r(T ) = fo + s2 [ρ10 - ρ30 ν(T )]Vo .
(155)
Using eq 87, we will reduce eq 155 to
d2r(T ) = Y + s2ρ30 [ν1 - ν(T ) + (s3/η)yν1] Vo
^
(156)
where ν1 is defined as
^
ν1 (T1) = wo - ν(T1).
^
(157)
It follows from eq 156 that r is positive for T > Tm. In engineering practices the frost heave
ratio h and the water intake ratio hw are often used. These are defined as
h = r (T )/Vo
(158)
hw = fo/Vo.
(159)
According to M1, h and hw are given as
h = α o y / (d2 ηVo ) + (s2ρ30 / d2 )[ν1 - ν(T ) + (s3 / η)yν1]
^
(160)
hw = α o y / (ηVo ) - eoρ30 ν1.
^
(161)
+
SOLUTIONS IN Sp WHEN σ ≥ σc
-
+
When σ ≥ σc, the region Sp defined by eq 130 disappears, so Sp = Sp . We will consider a
line L+ (Fig. 2c) defined as
L+(α1s, αo) = [(α1, αo): α1 > α1s and αo > 0]
(162)
where (α1s, αo) ∈ Cs.
Proposition 10
Suppose that there exists a solution T1 of Problem P on L+ such that Vo > 0. Then T1 < Tσ
if fo ≥ 0, while T1 may be greater than Tσ if fo < 0.
Proof
We will write eq 98 as
1
0
(δo / Ko ) - ∫
ds fo .
T1 K1(s)T ′(s, T1)
(s,T1)
Tσ
0
K2 (s)
ds + (s2 / e1 )(Q - Y)∫
=∫
ds.
(163)
T1 K1(s)T ′(s,T1)
K1(s)
T1
Since Vo > 0, so (Q Y) > 0. It follows from eq 163 that T1 < Tσ if fo ≥ 0, while T1 may be
greater than Tσ if fo < 0. u
Since eo > 0, T1 < Tx. There are two cases: Case 1. σc ≤ σ < σx or Tσ > Tx > T1 and Case 2.
20
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