w2 (T1) = s1(y - e5 )a4 = [s1η / (s2s3 ν1)]e6 a4
^
˙
(195)
s1 = (k1 / ko )(1 - s2 ) > 0
(196)
e5 (T1) = -ηeoo / (s2s3 ) > 0
(197)
e6 (T1) =
(-eo ν1)
^
(198)
T1
1
0
a3 (T1) = ∫
ds
(199)
T1 K1(s)(1 + e4)2
0
a4 (T1) = ∫
ds.
(200)
T1 K1(s)(1 + e4)2
The sign of w2 is the same as e6 and e6 is a property of a given soil. First we will consider
the case where the following condition holds true:
e6 (T1) ≥ 0
for T1 < Ts.
(201)
-
Suppose that T1 is a solution of Proposition 15. Then W1(T1) > 0 and g2 (T1) < 0 on Lc in this
˙
-
case. Therefore, this solution is unique. From eq 120 we find that T1 < 0 on Lc .
Next we will study the case in which eq 201 does not hold true. We will examine the
-
behavior of W1(T1) on Lc . When α1 approaches α1s, T1 approaches Ts, α1s is given by eq 168.
Using eq 88, we reduce eq 168 to
α1s = (e1s/s1)αo
(202)
where
e1s = e1(Ts).
(203)
We will write W1 as
W1(T1) = (a3 - eo a4 )yαo + (s1α1 - e1α o )(y - e5 )a4
˙
˙
(204)
when α1 approaches α1s, the second term of the right-hand side of eq 204 vanishes and
W1(T1) approaches W1(Ts) given as
W1(Ts ) = (a3 - eo a4 )yα o > 0.
˙
(205)
-
It follows from eq 205 that W1(T1) > 0 in a neighborhood of α1 = α1s on Lc . It is easy to find
that W1 vanishes when α1 becomes infinite.
-
Since the second term of the right-hand side of eq 204 is negative on Lc , it is possible
-
that W1(T1) may become negative at some point on Lc . If such is the case, then there ex-
-
ists at least one point (α1g, αo) on Lc where W1 vanishes because W1 is a continuous func-
tion of α1. From eq 193 we obtain
α1g = -(w1 / w2 )α o .
(206)
From eq 202 and 206 we obtain
(α1g - α1s ) / α o = a3 ηe7 [1 - (a4 / a3 )eos ] / (-s2 s3 ν1w2 )
(207)
^
25
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