83, and 85 combine to give the force due to normal pressure for a general channel
shape, i.e.
Fpp′ =
η(x)
si η(x)
[
]
[
]
Kp g ∫ ρi (1 - p) η(x) - δ b(x, δ) dδ - ∫ ρ(1 - p) si η(x) - δ b(x, δ) dδ.
(87)
0
0
For a rectangular channel, b is constant in the vertical and equal to the top width B.
Therefore
η(x)
si η(x)
siρ[η(x) - δ] dδ - ∫ ρ[si η(x) - δ] dδ.
′ = K gB 1 - p
p (
) ∫
Fpp
(88)
0
0
Integration yields
2
η
Fpp′ = Kp gB(1 - p)siρ(1 - si )
(89)
.
2
The time integral of this net normal force in the x-direction is
(
)
t2
t2
t
Fppdt = ∫ Fpp′ - Fpp″ dt = gsi (1 - si ) ∫ (ρI3 )x - (ρI3 )x dt
2
∫
(90)
t
2
1
t
t
1
1
1
where
η2
(1 - p).
I3 = KpB
(91)
2
It should be recognized that, by setting the value of k1 = Kp in eq 85, the jam is
considered to be at its strength limit. A jam at rest could probably experience k1
values ranging from Ka, the active pressure coefficient, to Kp, the passive pressure
coefficient. For a cover in motion, k1 most likely depends on ice velocity. More
research is needed to determine the proper values of k1 to be used in different states
of ice motion.
An important further force is a shear stress τxy, produced by ice grinding against
the banks or channel sides. It relates directly to the normal stress in the x-direction
at failure, i.e.
τxy = σxk0λ = σv k0λKp
(92)
where k0 is the coefficient of lateral thrust (percentage of normal stress acting in the
x-direction that would act in the z- or cross-channel direction) and λ is the sliding
coefficient of ice on ice at the banks. Bank shear stress acts equally at either bank
(for one-dimensional formulations, at least) and may vary between section x1 and
x2. The force from the shear at the banks can be obtained from eq 84 and 92, i.e.
η2
x2
Fis = 2 ∫ k0λKpsiρg(1 - p)(1 - si )
dx.
(93)
2
x1
The time integral of this force is
40
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