SME
Aerospace Structures · Problem type
A skin panel buckling between stringers
Use this for “A 2 mm aluminum skin spans 100 mm between stringers, its edges simply supported. At what stress does it buckle?”
Diagram
3 half-waves, each about as long as the panel is wide (b).
σcr
=
kπ²
E
12(1 −
ν
2
)
×
(
t
b
)
2
:
a
thicker
or
narrower
panel
holds
more.
The wave in the section is drawn bigger than life.
This is an example. Type your own values to start a new problem.
σcr (MPa)
=
×
π²
×
E (GPa)
12
×
(1
−
ν
2
)
×
t (mm)
b (mm)
2
Formulas
σcr
=
k
×
π
2
×
E
12 × (1 −
ν
2
)
×
(
t
b
)
2
103.4
=
4
×
π
2
×
70,000
12 × (1 −
0.33
2
)
×
(
2
100
)
2
Formulas are worked in MPa. The step-by-step shows the conversions.
Assumptions
•
A flat, thin panel squeezed along its stringers, elastic up to σcr.
•
k = 4 with simply supported edges, 6.97 with clamped edges, for a panel longer than about 3b.
•
Aluminum by default, E = 70 GPa and ν = 0.33; the steps work in N, mm and MPa.
Step-by-step
WE KNOW
k = 4, E = 70 GPa, ν = 0.33, t = 2 mm, b = 100 mm
FIND
critical stress (σcr)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to MPa).
E = 70 GPa = 70,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find critical stress (σcr)
Use σcr = k × π² × E ÷ (12 × (1 − ν²)) × (t ÷ b)²
The
plate-buckling
formula:
like
Euler’s,
with
the
panel’s
width
b
in
place
of
the
length.
σcr
=
4
×
π
2
×
70,000
12 × (1 −
0.33
2
)
×
(
2
100
)
2
σcr
=
4
×
π
2
×
70,000
12 × (1 − 0.1089)
×
(
2
100
)
2
σcr
=
4
×
π
2
×
70,000
12 × 0.8911
×
(
2
100
)
2
σcr
=
4
×
π
2
×
70,000
10.6932
×
0.02
2
σcr
=
4
×
9.87
×
70,000
10.6932
×
0.0004
σcr
=
39.48
×
70,000
10.6932
×
0.0004
σcr
=
2,763,600
10.6932
×
0.0004
σcr = 258,400 × 0.0004
σcr = 103.4 MPa
Check
Put the numbers back in. Both sides should match.
103.4
=
4
×
π
2
×
70,000
12 × (1 −
0.33
2
)
×
(
2
100
)
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.