T
K2 (s)
1
0
σ
go (σ) = ∫
ds - (η / s3 )∫
ds
(128)
K1(s)
Tx K1(s)
Tx
where Tσ = σ/γ as defined by eq 37. We will define σx as
σx = γTx .
(129)
It follows from eq 128 that Tx < Tσ or σx > σ when eq 123 holds true.
When Tx < Tσ , go (σ) is a decreasing and continuous function of σ. We will define σc as
0 K2 (s) - (η / s3 )
Tc
K2 (s)
1
0
go (σc ) = ∫
ds - (η / s3 )∫
ds = 0, or σc = ∫
ds
(130)
K1(s)
Tx K1(s)
K1(s)
Tx
Tx
where Tc is defined as
Tc = σc/γ > Tx.
(131)
It is noted that Tc is uniquely determined by the properties of a given soil and that Tc > Tx or
σc < σx. Using go, when σ < σc, we will write αoc of Proposition 5 as
α oc = (s3Ko / δo )(σc - σ).
(132)
-
+
When σ < σc, the Lc (α1s, αoc) divides Sp into two regions Sp (eo > 0) and Sp (eo < 0) (Fig.
-
+
-
2b). The region Sp disappears if σ ≥ σc. In terms of αo we may define Sp and Sp as
{
}
+
Sp = (α1 , α o ) ∈ Sp : α o > α oc , eo > 0
(133)
{
}
-
Sp = (α1 , α o ) ∈ Sp : α o < α oc , eo < 0 .
(134)
+
SOLUTIONS IN Sp WHEN σ < σc
+
We will consider a line Lc defined as
L+ (α1s , α o ) = { (α1 , α o ) : α1 > α1s and α o > α oc > 0}.
(135)
c
+
+
It is clear that Lc belongs to Sp where T1 < Tx.
Proposition 6
+
In Sp T1 < Tx < Tσ if σ < σc .
Proof
+
Suppose that there exists a point in Sp such that Tx ≥ Tσ. We will write eq 128 as
1
0
σc = σ - γ (Tx - Tσ ) - (η / s3 )∫
ds.
(136)
K1(s)
Tx
It follows from eq 136 that σ > σc. This contradicts the assumption. u
Proposition 7
+
Suppose that a solution of Problem P exists on Lc , then Vo > 0 and T1 > 0 on L+ .
~
c
17
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