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Physical Geology · Problem type

Shear stress under a glacier

Use this for “Ice 300 m thick has a surface slope of 3°. What is the shear stress at its bed?”, or the thickness for 100 kPa on 2°.

Diagram

α = 3°H = 300 mτb = 141 kPaρ = 917 kg/m³g = 9.81 m/s²
Basal shear = ρgH sin α = 917 × 9.81 × 300 × sin 3° ÷ 1000 = 141.2 kPa
At or past about 100 kPa: the ice deforms and flows.
The slab is long next to its depth; the depth is not drawn to scale.
This is an example. Type your own values to start a new problem.
τb (kPa)
=
ρ (kg/m³)
×
g
×
H (m)
×
sin(
α
°)

Formulas

τb
=
ρ
×
9.81
×
H
×
sin(
α
°)
1,000
141
=
917
×
9.81
×
300
×
sin(3°)
1,000

Assumptions

•
The slope is the ice surface’s, not the bed’s; the glacier is long next to its thickness.
•
g = 9.81 m/s²; the ÷ 1,000 turns pascals into kilopascals.
•
Ice flows when the basal shear nears about 100 kPa, so thickness and slope adjust toward it.

Step-by-step

WE KNOW
H = 300 m, α = 3°, ρ = 917 kg/m³
FIND
basal shear stress (τb)
Step 1 · Find basal shear stress (τb)
Use τb = ρ × 9.81 × H × sin(α°) ÷ 1,000
The
ice
column’s
weight
per
area,
ρgH,
times
its
share
along
the
slope,
sin
α
,
in
kPa.
τb
=
917
×
9.81
×
300
×
sin(3°)
1,000
τb
=
917
×
9.81
×
300
×
0.052336
1,000
τb
=
8,995.77
×
300
×
0.052336
1,000
τb
=
2,698,731
×
0.052336
1,000
τb
=
141,240
1,000
τb = 141 kPa
Check
Put the numbers back in. Both sides should match.
141
=
917
×
9.81
×
300
×
sin(3°)
1,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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