(More detail available from Perla, 1983; Sommerfeld, 1984 and Conway and Abrahmson, 1984)
| Perla/Sommerfeld | Conway and Abrahmson |
| Procedure | Procedure |
| Find shear layer | |
| Remove snow to height of frame on a bench | Insert frame (may need to remove new snow?) |
| Gently insert frame | Isolate column 4 sides |
| Pull Frame (repeat 21 times on each weak layer) | Pull Frame (repeat 21 times on each weak layer) |
| stress rate: 5-10 seconds to failure | stress rate: 5-10 seconds to failure |
| Measure height above shear | Measure height above shear |
|
Measure
density of snow above shear (layer weighted) |
Measure
density of snow above shear (layer weighted) |
| Use 0.65 times pull (not all authors support this) | 0.65 factor not used in this test |
| (?due to absence of gravitational component during test) | (Gravitational component is present) |
| Advantage | Advantage |
| No problem with weak new snow | Don't need to find weak layer |
| Identify the shear layer (control of experiment) | Gravitational force above shear layer is present |
| Less disturbance of weak layer during preparation | |
| Disadvantage | Disadvantage |
| Problem with crushing shear layer on insertion | Frame may not hold in new snow at top layer |
| No gravitational force above shear layer | Bending of column may influence result esp. >1.5 m column |
.
The total experimental stress to failure is the sum of the measured stress to failure and the gravity component of shear stress to failure above the shear plane = stotal during the experiment.
s total
=
σexp shear stress
= sfail
infinity + ρ' g z' Sin ( θ) + (Frame Mass
* g * 1/A) sin ( θ)
Adj.
Pulll Stress
from snow stress from mass of frame
Per
area
in frame
exptl.
per
unit area
2. Stress due to down slope gravity component (shear load)
which has not brought the snow to failure
Actual down slope shear
load
sslab shear stress
= ρ g z Sin ( θ)
Sommerfeld developed an equation to estimate additional stress
necessary to produce failure (sdiff) expressed in Newtons.
sdiff = σexp shear stress
- sslab
shear
stress = (Fs/As *
0.65
or o.56 + ρ' g z' Sin ( θ) + (Frame Mass * g *
1/A) Sin ( θ) ) - (ρ g z Sin ( θ))
sdiff = σexp shear
stress
- (ρ g z Sin ( θ))
Sommerfeld also presented an equation to estimate the new snow depth necessary to produce failure (a factor of safety should be considered)
z"= zn = sdiff / (ρn * g * Sin (θ))Conway and Abrahmson developed a shear index.
Conway and Abrahmson Shear index a = ((Fs/As* 0.65
or 0.65) + (ρ' g z' Sin (Theta)))/(ρ g z Sin (Theta)) =
S/L ratio perhaps better called strength to shear stress ratio.
Note: Although many researchers call this the strength to load
ratio, technically load is a force (newtons), so this is really
the strength to stress ratio because the units are in pascals which are
the units of stress. If you would like to calculate the strength
to stress ratio for a situation with new snow you may add ρ" g z"
Sin (Theta) to
the denominator so that new snowfall is accounted for.
F T = Tensile force required to break the snow holding
the
back of the column
AT = Area of the snow holding the back of the column after
the notches are cut out
ρ = mean snow
density in the column
g = acceleration of gravity
z = height of the column of snow above the frictionless
surface
w = width of the column of snow above the frictionless surface
b = breadth of the column of snow above the fricitonless surface
Note: z, w, b = the volume of the column of snow which when
multiplied by rg
is
a force
θ=Theta = slope
Equation for the tensile stress calculation
s T
= (F T + (ρ *
g * z * w * b * Sin ( θ)))/AT
Strength = (F + ρ g z w l sin θ)/AT . Where ρ is the
snow density,
z is the height of the connecting snow beam above the frictionless
surface,
w is the width of the beam above the frictionless surface and l
is
the length of the beam being pulled in tension above the frictionless
surface.
AT = z l, the area of the tensile surface when the break
occurs.
Again F = kg * 9.8 ms-2 which is the force in Newtons at
tensile
failure during the test. Note that the stress is a bit more
ambiguous
here because the whole pack could conceivably be pulling under
avalanche
conditions not just the column. Andy Gleason has pointed out that
researchers
he has discussed this with think that tension is unimportant because
the
release is always perpendicular to the slope so the real failure is
probably
either in compression and/or shear. This is an area for debate.
An
alternative view is that the avalanche cannot release until the tensile
strength
is exceeded so that motion on the shear plane can begin (chicken or
egg).
In any event you need to think about the difference between
compressive,
shear and tensile failure.