d(x)
Fp1′ = g ∫ ρ[D(x) - δ] b(x, δ) dδ .
(41)
0
For a rectangular channel, b is constant in the vertical and equal to the top width B.
Therefore
d(x)
Fp1′ = g ∫ ρ[D(x) - δ] Bdδ .
(42)
0
[
]
Fp1′ = ρgB d2 2 + dsi η .
(43)
Thus, the time integral for the net pressure Fp1 is
(
)
t2
t2
t
′
″
∫ Fp1dt = ∫ Fp1 - Fp1 dt = g ∫ (ρI1)x - (ρI1)x dt
2
(44)
2
1
t
t
t
1
1
1
where
[
]
I1 = B d2 2 + dsi η .
(45)
Two gravity forces act vertically on the water control volume. The first acts on
the bottom surface of the control volume (the bed). It is attributable to the combined
weight of water, ice, and pore water above. The second acts on the upper surface of
the control volume (the bottom of the jam). It is attributable to the weight of the ice
and pore water above. The horizontal component of the first gravity force is
x2
x2
(
)
Fg1 = ∫ ρgA + ρi gAi (1 p) + ρgAi si p Sodx = ∫ ρg(A + Ai si )Sodx
(46)
x1
x1
where So is bed slope
yb
So = -
.
(47)
x
For the period t1 to t2
t2
t2 x2
∫ Fg1dt = ∫ ∫ ρg(A + Ai si )Sodxdt .
(48)
t1
t1 x1
The gravity force attributable to the weight of ice and pore water acting on the
upper surface of the control volume (in the x-direction) is
x2
x2
[
]
Fg2 = ∫ ρi gAi (1 - p)Sib + ρgAi si pSib dx = ∫ [ρgAi siSib ] dx
(49)
x1
x1
where Sib is the slope of the jam underside
(yb + d) = -
d
yb
d
Sib = -
x + x = So - x .
(50)
x
Substituting eq 50 into 49 and integrating over the period t1 to t2 gives
t2
t2 x2
d
∫ Fg2dt = ∫ ∫ ρgAi si So - dxdt .
(51)
x
t1
t1 x1
33
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