SME
Aerospace Structures · Problem type
Short columns: the Johnson parabola
Use this for “An aluminum strut has σᵧ = 350 MPa, E = 70 GPa and KL ÷ r = 40. At what stress does it fail?”
Diagram
K = 1: pinned at the bottom, pinned on top.
The whole column bends in one half-wave: KL = L.
This is an example. Type your own values to start a new problem.
λ
=
×
L (mm)
r (mm)
σcr (MPa)
=
σᵧ (MPa)
−
(
σᵧ (MPa)
)
2
4
×
π²
×
E (GPa)
×
λ
2
λₜ
Transition slenderness
calculated
Formulas
λ
=
K
×
L
r
40
=
1
×
500
12.5
λₜ
=
√(2
×
π
2
×
E
σᵧ
)
62.83
=
√(2
×
π
2
×
70,000
350
)
σcr
=
σᵧ
−
(
σᵧ
2
4 ×
π
2
×
E
)
×
λ
2
279.1
=
350
−
(
350
2
4 ×
π
2
× 70,000
)
×
40
2
Formulas are worked in MPa. The step-by-step shows the conversions.
Assumptions
•
λ = KL ÷ r, the effective length over the smaller radius of gyration.
•
K is 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned and 2 fixed–free.
•
Johnson’s parabola holds up to λₜ; a more slender column buckles elastically, by Euler’s formula.
•
The load is along the axis; the steps work in MPa.
Step-by-step
WE KNOW
σᵧ = 350 MPa, E = 70 GPa, K = 1, L = 500 mm, r = 12.5 mm
FIND
slenderness ratio (λ), transition slenderness (λₜ), 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 slenderness ratio (λ)
Use λ = K × L ÷ r
The effective length KL over the radius of gyration.
λ
=
1
×
500
12.5
λ
=
40
Step 3 · Find transition slenderness (λₜ)
Use λₜ = √(2 × π² × E ÷ σᵧ)
Where
Euler’s
stress
π
2
E
λ
2
falls
to
half
the
yield
strength:
the
parabola
meets
Euler’s
curve
there.
λₜ
=
√(2
×
π
2
×
70,000
350
)
λₜ
=
√(2
×
9.87
×
70,000
350
)
λₜ
=
√(19.74
×
70,000
350
)
λₜ
=
√(
1,381,800
350
)
λₜ
=
√
3
,948
λₜ = 62.83
Step 4 · Find critical stress (σcr)
Use σcr = σᵧ − (σᵧ² ÷ (4 × π² × E)) × λ²
Johnson’s
parabola:
the
yield
strength
less
a
drop
that
grows
with
λ
squared.
σcr
=
350
−
(
350
2
4 ×
π
2
× 70,000
)
×
40
2
σcr
=
350
−
350
2
4 × 9.87 × 70,000
×
40
2
σcr
=
350
−
350
2
39.48 × 70,000
×
40
2
σcr
=
350
−
350
2
2,763,600
×
40
2
σcr
=
350
−
122,500
2,763,600
×
1,600
σcr = 350 − 0.04433 × 1,600
σcr = 350 − 70.928
σcr = 279.1 MPa
Check
Put the numbers back in. Both sides should match.
40
=
1
×
500
12.5
✓
62.83
=
√(2
×
π
2
×
70,000
350
)
✓
279.1
=
350
−
(
350
2
4 ×
π
2
× 70,000
)
×
40
2
✓
Topic check
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