FE Civil review 66 – Structures - 3 Euler's formula Explained FAST!

Required:
Find the maximum permissible length L (in inches) of a square steel column such that Euler buckling stress equals the proportional limit of the material.

Given:

Proportional limit stress:
σ_pl = 50 ksi

Modulus of Elasticity:
E = 40,000 ksi

Moment of inertia of column:
I = 9 in^4

Cross-sectional area:
A = 12 in^2

End condition factor:
K = 1.0 (pinned–pinned type behavior)

We want the maximum L such that Euler critical stress = 50 ksi.

Reference Formulas to Remember (Columns – Euler Buckling)

Radius of gyration:
r = √(I / A)

Euler critical load:
P_cr = (π² E I) / (K L)²

Critical buckling stress:
σ_cr = P_cr / A
σ_cr = (π² E r²) / (K L)²

For the limiting case (maximum L):
σ_cr = σ_pl

Concept (Short Summary)

📌 YouTube Timestamps — Shear Force & Bending Moment Masterclass

00:06 Problem statement
01:06 Explanation of Euler's formula and the proportional limit
01:57 Finding Euler's formula in the FE reference manual
02:19 Breakdown of the Euler's formula terms
03:31 Listing all given values from the problem
04:02 Calculation of the radius of gyration (R)
04:10 Final calculation for the maximum permissible length (L)

Long, slender columns buckle elastically according to Euler’s formula.

Euler is valid as long as the buckling stress stays below the proportional limit.

The maximum column length is found by setting the Euler critical stress equal to the proportional limit and solving for L.

Step-by-Step Solution
1) Compute the radius of gyration r

r = √(I / A)
= √(9 / 12)
= √(0.75)
≈ 0.866 in

So:
r² = 0.75

2) Write Euler buckling stress in terms of L

Use:
σ_cr = (π² E r²) / (K L)²

Set σ_cr = σ_pl = 50 ksi and K = 1.0:

50 = (π² × 40,000 × r²) / (1 × L)²

Substitute r² = 0.75:

50 = (π² × 40,000 × 0.75) / L²

3) Solve for L

Rearrange:

L² = (π² × 40,000 × 0.75) / 50

Compute inside:

40,000 × 0.75 = 30,000

So:

L² = (π² × 30,000) / 50
L² = π² × 600

Using π² ≈ 9.8696:

L² ≈ 9.8696 × 600
L² ≈ 5,922

Now take the square root:

L ≈ √5,922
L ≈ 76.95 in

Round to a reasonable exam value:

L ≈ 77 in

Final Answer:

Maximum permissible column length:
👉 L ≈ 77 inches

Tags:

#FECivil #MechanicsOfMaterials #EulerBuckling #Columns #CivilEngineering #StructuralEngineering #EngineeringExamPrep #AverageCivilEngineerPE