SME
Aerospace Structures
Buckling
Use this for “An aluminum tube 1.2 m long, pinned at both ends, has I = 8 × 10⁴ mm⁴. At what load does it buckle?”
Diagram
K = 1: pinned at the bottom, pinned on top.
The whole column bends in one half-wave: KL = L.
Pcr
=
π
2
EI
÷
(KL)
2
:
the
shorter
the
half-wave,
the
more
it
takes.
This is an example. Type your own values to start a new problem.
KL (m)
=
K
×
L (m)
Pcr (kN)
=
π²
×
E (GPa)
×
I (mm⁴)
(
KL (m)
)
2
A
Cross-section area
example
σcr
Critical stress
calculated
σᵧ
Yield strength
example
Formulas
KL
=
K
×
L
1,200 = 1 × 1,200
Pcr
=
π
2
×
E
×
I
÷
(
KL
)
2
38,380
=
π
2
×
70,000
×
(8
×
10
4
)
÷
(1,200)
2
σcr
=
Pcr
A
127.9
=
38,380
300
Formulas are worked in MPa, mm, N. The step-by-step shows the conversions.
Assumptions
•
The column is straight, centrally loaded, and elastic up to Pcr, so σcr must stay below σᵧ (a shorter column follows Johnson’s parabola).
•
K is 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned and 2 fixed–free.
•
I is the smaller second moment of area: the column buckles about its weaker axis.
•
Aluminum by default, E = 70 GPa and σᵧ = 270 MPa; the steps work in N, mm and MPa.
Step-by-step
WE KNOW
E = 70 GPa, I = 8 × 10⁴ mm⁴, L = 1.2 m, K = 1, A = 300 mm², σᵧ = 270 MPa
FIND
effective length (KL), critical load (Pcr), 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, mm, N).
E = 70 GPa = 70,000 MPa (1 GPa = 1000 MPa)
L = 1.2 m = 1,200 mm (1 m = 1000 mm)
Step 2 · Find effective length (KL)
Use KL = K × L
The
effective
length
is
the
half-wave
the
ends
let
the
column
bend
in:
K
times
the
length.
KL
=
1
×
1,200
KL
=
1,200
mm
Step 3 · Find critical load (Pcr)
Use Pcr = π² × E × I ÷ (KL)²
Euler’s
load:
π
2
times
the
bending
stiffness
EI,
over
the
effective
length
squared.
Pcr
=
π
2
×
70,000
×
(8
×
10
4
)
÷
(1,200)
2
Pcr
=
9.87
×
70,000
×
8 ×
10
4
1,440,000
Pcr
=
690,900
×
8 ×
10
4
1,440,000
Pcr
=
5.5272 ×
10
10
1,440,000
Pcr = 38,381.7949 N
Step 4 · Find critical stress (σcr)
Use σcr = Pcr ÷ A
The stress the column buckles at: the critical load spread over the cross-section.
σcr
=
38,380
300
σcr = 127.9 MPa
Step 5 · Convert the answers
Change each answer back to the unit you chose.
KL = 1,200 mm = 1.2 m (1 m = 1000 mm)
Pcr = 38,380 N = 38.38 kN (1 kN = 1000 N)
Check
Put the numbers back in. Both sides should match.
1,200 = 1 × 1,200 ✓
38,380
=
π
2
×
70,000
×
(8
×
10
4
)
÷
(1,200)
2
✓
127.9
=
38,380
300
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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