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Euler's formula both ends hinged, with limitations

Short answer

Euler's critical load for both ends hinged is P_cr = π²EI/L², derived from EI y'' = -Py. Limitations: applies only to slender columns deforming elastically; neglects initial crookedness, load eccentricity, and shear; it overestimates the load for short or intermediate columns.

Derive the expression for the Euler's formula for crippling load on a column with both ends hinged condition. Explain the limitation of Euler's Formula also.

Euler's critical (crippling) load is the smallest axial load at which an initially straight, slender column becomes unstable and buckles sideways.

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column AB of length LL be hinged at both ends, with the load PP producing a small deflection yy at distance xx from A.

   P
   |  B
   |  /\
   |  \ y
   |  /
   A /
   P
 (A origin, x along column)

Bending moment at the section: M=−PyM = -Py (taking sagging moment as positive, the moment due to PP tends to reduce the curvature, so it is negative). Using the differential equation of the elastic curve:

EId2ydx2=−Py  ⇒  d2ydx2+k2y=0,k2=PEIEI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2 y = 0,\qquad k^2 = \frac{P}{EI}

The solution is

y=C1cos⁡kx+C2sin⁡kxy = C_1\cos kx + C_2\sin kx

Boundary conditions:

  • At x=0x = 0, y=0⇒C1=0y = 0 \Rightarrow C_1 = 0.
  • At x=Lx = L, y=0⇒C2sin⁡kL=0y = 0 \Rightarrow C_2\sin kL = 0.

For a buckled shape C2≠0C_2 \ne 0, so sin⁡kL=0\sin kL = 0 and kL=nπkL = n\pi. The smallest non-zero load is for n=1n = 1:

kL=π  ⇒  PL2EI=π2k L = \pi \;\Rightarrow\; \frac{PL^2}{EI} = \pi^2 Pcr=π2EIL2P_{cr} = \frac{\pi^2 EI}{L^2}

Here II is the least moment of inertia, since the column buckles about the axis of least resistance. The shape is a half sine wave, y=C2sin⁡(πx/L)y = C_2\sin(\pi x/L), with maximum deflection at mid-span.

Limitations of Euler's formula

  1. It applies only to long (slender) columns. In terms of stress, σcr=π2Eλ2\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2} must not exceed the proportional limit, so λ≥πE/σp\lambda \ge \pi\sqrt{E/\sigma_p} (about 80 to 100 for mild steel). For short and intermediate columns it over-estimates the load.
  2. It assumes a perfectly straight column, a perfectly axial load and homogeneous material. Real columns have initial crookedness and eccentricity, so they fail at a lower load.
  3. It is valid only within the elastic range (Hooke's law), so σcr\sigma_{cr} cannot exceed the proportional limit.
  4. It ignores the effect of direct compression, shear deformation and imperfect end conditions.
  5. For very slender columns it is accurate, and for intermediate columns empirical formulas (Rankine, Johnson, Secant, straight-line) are used instead.

✓ Answer checked by Claude Opus 5.5 (high effort)

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