SME
Mechanics of Materials (Solid Mechanics I) · Problem type
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.”
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 (m)
r (mm)
σcr (MPa)
=
σY (MPa)
−
σY (MPa)
×
2π
2
÷
E (GPa)
Pcr (kN)
=
σcr (MPa)
×
A (mm²)
A is the column’s cross-section area, E the material’s elastic modulus (steel about 200 GPa) and σY the stress at which it yields.
λc
Transition slenderness
calculated
Formulas
λ
=
K
×
L
r
60
=
1
×
1,500
25
λc
=
√(2
×
π
2
×
E
σY
)
125.6637
=
√(2
×
π
2
×
200,000
250
)
σcr
=
σY
−
(σY ×
λ
÷ (2π))²
E
221.5034
=
250
−
(250 × 60 ÷ (2π))²
200,000
Pcr
=
σcr
×
A
443,006.8342 = 221.5034 × 2,000
Formulas are worked in MPa, mm, N. The step-by-step shows the conversions.
Assumptions
•
Johnson’s parabola is for intermediate columns, λ below λc (σcr above σY ÷ 2); from λc up use Euler’s formula.
•
K is 0.5 fixed–fixed, 0.7 fixed–pinned, 1 pinned–pinned and 2 fixed–free.
•
r is the smaller radius of gyration, √(I ÷ A); steel by default, E = 200 GPa.
•
The load is along the axis; the steps work in N, mm and MPa.
Step-by-step
WE KNOW
σY = 250 MPa, E = 200 GPa, K = 1, L = 1.5 m, r = 25 mm, A = 2,000 mm²
FIND
slenderness ratio (λ), transition slenderness (λc), critical stress (σcr), 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 = 1.5 m = 1,500 mm (1 m = 1000 mm)
Step 2 · 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
×
1,500
25
λ
=
60
Step 3 · Find transition slenderness (λc)
Use λc = √(2 × π² × E ÷ σY)
Where
Euler’s
stress
π
2
E
λ
2
falls
to
half
the
yield
strength:
the
parabola
meets
Euler’s
curve
there.
λc
=
√(2
×
π
2
×
200,000
250
)
λc
=
√(2
×
9.8696
×
200,000
250
)
λc
=
√(19.7392
×
200,000
250
)
λc
=
√(
3,947,840
250
)
λc
=
√
15
,791.36
λc = 125.6637
Step 4 · Find critical stress (σcr)
Use σcr = σY − (σY × λ ÷ (2π))² ÷ E
Johnson’s
parabola:
the
yield
strength
less
a
drop
that
grows
with
λ
squared.
σcr
=
250
−
(250 × 60 ÷ (2π))²
200,000
σcr = 221.5034171 MPa
Step 5 · Find critical load (Pcr)
Use Pcr = σcr × A
The critical stress acts over the whole section: multiply it by the area.
Pcr = 221.5034171 × 2,000
Pcr = 443,006.8342 N
Step 6 · Convert the answers
Change each answer back to the unit you chose.
Pcr = 443,006.8342 N = 443.0068 kN (1 kN = 1000 N)
Check
Put the numbers back in. Both sides should match.
60
=
1
×
1,500
25
✓
125.6637
=
√(2
×
π
2
×
200,000
250
)
✓
221.5034171
=
250
−
(250 × 60 ÷ (2π))²
200,000
✓
443,006.8 = 221.5034171 × 2,000 ✓
Topic check
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