SME
Mechanics of Materials (Solid Mechanics I)
Column buckling
Use this for “A steel column 3 m long, pinned at both ends, has I = 2 × 10⁶ 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.
Pcr (kN)
=
π²
×
E (GPa)
×
I (mm⁴)
(
×
L (m)
)
2
KL
Effective length
calculated
Formulas
KL
=
K
×
L
3,000 = 1 × 3,000
Pcr
=
π
2
×
E
×
I
÷
(
KL
)
2
438,649.0845
=
π
2
×
200,000
×
(2
×
10
6
)
÷
(3,000)
2
Formulas are worked in MPa, mm, N. The step-by-step shows the conversions.
Assumptions
•
Straight, loaded along its axis and slender: λ = KL ÷ r at least λc = √(2π²E ÷ σY), so σcr ≤ σY ÷ 2 (stockier: the Johnson formula).
•
K is 0.5 fixed–fixed, 0.7 fixed–pinned, 1 pinned–pinned and 2 fixed–free.
•
I is the smaller second moment of area: the column buckles about its weaker axis.
•
Steel by default, E = 200 GPa; the steps work in N, mm and MPa.
Step-by-step
WE KNOW
E = 200 GPa, I = 2 × 10⁶ mm⁴, L = 3 m, K = 1
FIND
effective length (KL), critical load (Pcr)
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 = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
L = 3 m = 3,000 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
×
3,000
KL
=
3,000
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
×
200,000
×
(2
×
10
6
)
÷
(3,000)
2
Pcr
=
9.87
×
200,000
×
2 ×
10
6
9,000,000
Pcr
=
1,974,000
×
2 ×
10
6
9,000,000
Pcr
=
3.948 ×
10
12
9,000,000
Pcr = 438,649.0845 N
Step 4 · Convert the answers
Change each answer back to the unit you chose.
KL = 3,000 mm = 3 m (1 m = 1000 mm)
Pcr = 438,649.0845 N = 438.6491 kN (1 kN = 1000 N)
Check
Put the numbers back in. Both sides should match.
3,000 = 1 × 3,000 ✓
438,649.1
=
π
2
×
200,000
×
(2
×
10
6
)
÷
(3,000)
2
✓
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Intermediate columns: the Johnson formula
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More problem types
Intermediate columns: the Johnson formula
Use this for “A pinned steel column 1.5 m long has r = 25 mm, A = 2,000 mm² and σY = 250 MPa. Find its critical stress and load.”
Slenderness and Euler’s critical stress
Use this for “A pinned steel column 5 m long has A = 2000 mm² and I = 2 × 10⁶ mm⁴. Find its slenderness ratio and critical stress. Does Euler’s formula apply if σY = 250 MPa?”