Bed Load Pit Analysis
Assumptions
Fully submerged spherical particle on a flat bed prevented from lateral motion by a step
of height h.
Neglect lift forces.
Assume flow is undisturbed.
Free body diagram
Particle has contact forces at A and B.
Particle has body forces of weight and buoyancy.
Drag force caused by liquid flow past particle.
Neglect friction at point B (this will make sense shortly).
Contact point A has a normal and tangential force; the tangential force is a frictional
force resisting sliding (hence A is an instantaneous rotation center).
Diagram of geometry at A
Geometry at A relates the radius and height to the contact angle; the angle is defined by r
and h.
r  h  r cos 
l  r sin 
b  cos 1 (
rh
)
r
Force balance of system while the contact force at B is non-zero (but approaching zero).
 F  F  A sin   A cos 
 F  (W  B)  B  A cos   A sin 
 M  (W  B)l  B l  F (r  h)
x
d
y
y
A
y
d
The term (W-B) is the submerged weight, Ws
Now assume that By vanishes (particle is just out of contact with bottom of pit), then the
system is equivalent to:
The force balance now becomes
 F  F  A sin   A cos 
 F  W  A cos   A sin 
 M  W l  F (r  h)
x
d
y
s
A
s
d
The moment equation provides the necessary information to determine Fd as the particle
begins to rotate about the moment center. Set the moment to zero (equilibrium) and
analyze.
Ws
l
 Fd
(r  h)
Substitute in geometry and the definition of submerged weight
4 3
r (  s   l ) g tan   Fd
3
Assume the drag force is proportional to momentum.
1
 lV 2 Ap
2
Cd is the constant of proportionality, the the velocity is some kind of average near the
particle. Cd will probably range between 0.1 and 2.0.
Fd  Cd
Set the two expressions equal to each other and solve for velocity, this velocity will be a
critical velocity, if flow is above this critical velocity the particle should begin to rotate, if
below the particle should be “stable”.
V 
8
  l 1
rg tan  ( s
)( )
3
l
Cd
Except for Cd the right hand side is determined by the particle dimension r and the step
height h.
The plot below is a family of curves of critical velocity for particles of different
dimensions (1 centimeter to 10 meters), versus a dimensionless step height. The plot was
produced using the Cd of 0.5. (Sphere at Reynolds numbers ranging from 1000-10000).
critical velocity m/s
100
10
1
r=0.01 meters
r=0.1 meters
r=1.0 meters
r=10.0 meters
0.1
0.0
0.2
0.4
0.6
0.8
1.0
dimensionless height (h/r)
Assuming this analysis is at all close, a 1cm particle in a pit that is ¼ particle diameter
deep should not move at flow velocities less than 1 meter/second. As the pit depth
decreases, the mobilization velocity decreases in a log-linear fashion.
This analysis is testable in the physical model (we will still have to guess Cd).
The case of shielded flow is essentially the same; a sketch without explanation is below.
What changes?
1. Projected area is reduced:
Ap  r 2  r 2  r 2 sin  cos 
2. Line of action of drag force changes
2r (sin  cos2   sin  )
r 
3(    sin  cos  )
3. Moment arm of drag force is increased
Fd [( r  h)  r ]
4. Resulting equation is
8 3
r (  s   l ) gr sin 
3
V 
2r (sin  cos2   sin  )
Cd  l {r 2  r 2  r 2 sin  cos  }[r cos  
]
3(    sin  cos  )