SME
Mechanics of Materials (Solid Mechanics I) · Problem type
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?”
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.
r (mm)
=
I (mm⁴)
÷
A (mm²)
λ
=
×
L (m)
r (mm)
σcr (MPa)
=
π²
×
E (GPa)
(
)
2
A is the column’s cross-section area, I its second moment of area about the weaker axis, and E the material’s elastic modulus (steel about 200 GPa).
σY
Yield strength
example
Formulas
r
=
√(
I
A
)
31.6228
=
√(
2 ×
10
6
2,000
)
λ
=
K
×
L
r
158.1139
=
1
×
5,000
31.6228
σcr
=
π
2
×
E
λ
2
78.9568
=
π
2
×
200,000
158.1139
2
Formulas are worked in MPa, mm. The step-by-step shows the conversions.
Assumptions
•
Euler’s formula needs σcr below the proportional limit; design uses it only while σcr ≤ σY ÷ 2 (λ ≥ λc = √(2π²E ÷ σY)), the Johnson formula below λc.
•
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, so r = √(I ÷ A) is about the weaker axis.
•
Steel by default, E = 200 GPa; the steps work in N, mm and MPa.
Step-by-step
WE KNOW
E = 200 GPa, A = 2,000 mm², I = 2 × 10⁶ mm⁴, K = 1, L = 5 m, σY = 250 MPa
FIND
radius of gyration (r), slenderness ratio (λ), Euler 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).
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
L = 5 m = 5,000 mm (1 m = 1000 mm)
Step 2 · Find radius of gyration (r)
Use r = √(I ÷ A)
The
radius
of
gyration
is
where
the
whole
area
would
sit
to
give
the
same
I
.
r
=
√(
2 ×
10
6
2,000
)
r
=
√
1
,000
r
=
31.62278
mm
Step 3 · Find slenderness ratio (λ)
Use λ = K × L ÷ r
The slenderness ratio is the effective length over the radius of gyration; it has no unit.
λ
=
1
×
5,000
31.62278
λ
=
158.1139
Step 4 · Find Euler critical stress (σcr)
Use σcr = π² × E ÷ λ²
Euler’s
load
Pcr
=
π
2
EI
÷
(
KL
)
2
over
the
area,
with
I
=
Ar²:
π
2
E
λ
2
.
σcr
=
π
2
×
200,000
158.1139
2
σcr
=
9.87
×
200,000
25,000
σcr
=
1,974,000
25,000
σcr = 78.9568 MPa
Check
Put the numbers back in. Both sides should match.
31.62278
=
√(
2 ×
10
6
2,000
)
✓
158.1139
=
1
×
5,000
31.62278
✓
78.9568
=
π
2
×
200,000
158.1139
2
✓
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Intermediate columns: the Johnson formula
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Stress transformation, combined loading and pressure vessels
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