SME
Aerospace Structures · Problem type
Skin stress in a pressurized fuselage
Use this for “A fuselage 2 m in radius with 1.6 mm skin is pressurized to 60 kPa above the outside air. Find the hoop and axial stress.”
Diagram
Hoop
stress
σₕ
=
pr
t
round
the
wall
is
twice
the
axial
σₐ
=
pr
2t
.
Wall
drawn
thicker
than
to
scale
(
r
t
=
1,250).
This is an example. Type your own values to start a new problem.
σₕ (MPa)
=
Δp (kPa)
×
r (m)
t (mm)
σₐ (MPa)
=
Δp (kPa)
×
r (m)
2
×
t (mm)
Formulas
σₕ
=
Δp
×
r
t
75
=
60
×
2
1.6
σₐ
=
Δp
×
r
2 ×
t
37.5
=
60
×
2
2 × 1.6
Assumptions
•
A thin cylinder: the skin is far thinner than the radius, so the stress is even through it.
•
Δp is the cabin pressure minus the outside pressure; the skin alone carries it.
•
The units cancel in Δp × r ÷ t: kPa × m ÷ mm is MPa, as is MPa × mm ÷ mm.
Step-by-step
WE KNOW
Δp = 60 kPa, r = 2 m, t = 1.6 mm
FIND
hoop stress (σₕ), axial stress (σₐ)
Step 1 · Find hoop stress (σₕ)
Use σₕ = Δp × r ÷ t
Cut
the
skin
along
its
length:
Δp
×
2
r
per
unit
length
is
held
by
two
walls
t
thick.
σₕ
=
60
×
2
1.6
σₕ
=
120
1.6
σₕ = 75 MPa
Step 2 · Find axial stress (σₐ)
Use σₐ = Δp × r ÷ (2 × t)
Cut across the fuselage: Δp on the end, πr², is held by the ring of skin 2πrt.
σₐ
=
60
×
2
2 × 1.6
σₐ
=
60
×
2
3.2
σₐ
=
120
3.2
σₐ = 37.5 MPa
Check
Put the numbers back in. Both sides should match.
75
=
60
×
2
1.6
✓
37.5
=
60
×
2
2 × 1.6
✓
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