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Euler's formula: fixed-free column with limitations

Short answer

Derivation gives P_cr = π²EI/(4L²) = π²EI/(2L)². Limitations: valid only for long, slender columns with stress below the proportional limit; assumes perfect straightness, axial load, homogeneous elastic material; ignores imperfections, shear, and direct compression.

Derive Euler critical buckling load formula for a column having one end fixed and the other end free. Discuss the limitations of Euler buckling formula.

Euler's critical load for a column fixed at one end and free at the other is derived below.

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, length LL, have its lower end A fixed and its upper end B free. Under the load PP, the free end deflects by δ\delta. Take the origin at the fixed end A, with yy measured from the original axis.

      P
      v
   B  )  <- free end, deflection delta
      )
      ) y
      )
   A  fixed

At a section distance xx from A, the lever arm of PP is (δ−y)(\delta - y), so M=P(δ−y)M = P(\delta - y):

EId2ydx2=P(δ−y)  ⇒  d2ydx2+k2y=k2δ,k2=PEIEI\frac{d^2y}{dx^2} = P(\delta - y) \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = k^2\delta,\qquad k^2 = \frac{P}{EI}

The solution is

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

Boundary conditions:

  • At x=0x=0: y=0⇒C1=−δy=0 \Rightarrow C_1 = -\delta.
  • At x=0x=0: dydx=0⇒C2=0\dfrac{dy}{dx} = 0 \Rightarrow C_2 = 0.

So y=δ (1−cos⁡kx)y = \delta\,(1 - \cos kx).

  • At x=Lx=L: y=δ⇒δ=δ(1−cos⁡kL)⇒cos⁡kL=0y = \delta \Rightarrow \delta = \delta(1 - \cos kL) \Rightarrow \cos kL = 0

The smallest solution is kL=π2kL = \dfrac{\pi}{2}:

PL2EI=π24\frac{PL^2}{EI} = \frac{\pi^2}{4} Pcr=π2EI4L2=π2EI(2L)2P_{cr} = \frac{\pi^2EI}{4L^2} = \frac{\pi^2EI}{(2L)^2}

The effective length is Le=2LL_e = 2L, so this column carries one quarter of the load of a hinged-hinged column of the same length.

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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