Euler’s buckling (crippling) load forms: Identify the correct Euler critical load expression P_cr for a prismatic column (modulus E, second moment I, actual length L) under different end conditions, using the effective length concept.

Difficulty: Easy

Correct Answer: All of the above (each matches its end condition)

Explanation:


Introduction / Context:
Euler’s column theory gives the elastic buckling (critical) load for slender, perfectly straight columns with ideal end conditions. The formula uses an effective length L_e that depends on end restraints. Greater end fixity reduces L_e and increases the buckling load.


Given Data / Assumptions:

  • P_cr = π² E I / L_e².
  • Idealized end conditions; column remains elastic up to buckling.
  • No initial imperfections or eccentricity (theoretical model).


Concept / Approach:

Effective length L_e embodies end restraint: hinged-hinged L_e = L; fixed-fixed L_e = L/2; fixed-free (cantilever) L_e = 2 L; fixed-hinged L_e = L/√2. Substituting each L_e into P_cr = π² E I / L_e² yields the respective expressions.


Step-by-Step Solution:

Both ends fixed: L_e = L/2 ⇒ P_cr = π² E I / (L/2)².Both ends hinged: L_e = L ⇒ P_cr = π² E I / L².One end fixed, other free: L_e = 2 L ⇒ P_cr = π² E I / (2 L)².One end fixed, other hinged: L_e = L/√2 ⇒ P_cr = π² E I / (L/√2)².


Verification / Alternative check:

Ordering by restraint: fixed-fixed has 4× the hinged-hinged capacity; fixed-hinged has 2×; fixed-free has 1/4×, consistent with classical tables.


Why Other Options Are Wrong:

Each individual option is correct for its stated end condition; only the combined choice confirms all correctly.


Common Pitfalls:

Mixing L with L_e; forgetting that the cantilever (fixed-free) is the weakest case (largest effective length).


Final Answer:

All of the above (each matches its end condition)

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