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Euler's formula both ends fixed, with limitations

Short answer

For both ends fixed, effective length L_e = L/2, so P_cr = 4π²EI/L². Euler's formula is limited to long columns undergoing elastic buckling; it ignores material nonlinearity, imperfections, shear, and axial shortening, and is unsafe for short and intermediate columns.

Derive Euler's formula of critical load for a steel column with both ends fixed. Also explain the limitations to the use of this formula.

Euler's crippling load is the smallest axial load at which an initially straight, slender column buckles. For both ends fixed, the ends are held in position and direction.

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

When the column buckles, the fixed ends develop an end moment M0M_0 (fixing moment) while the load PP stays axial. Take the origin at the lower fixed end A, with yy the lateral deflection at distance xx.

   P
   |  B  (fixed, M0)
   |  /\
   | (  y
   |  \/
   A  fixed, M0
   P

The moment at the section is M=M0−PyM = M_0 - Py, so

EId2ydx2=M0−Py  ⇒  d2ydx2+k2y=M0EI,k2=PEIEI\frac{d^2y}{dx^2} = M_0 - Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{M_0}{EI},\qquad k^2 = \frac{P}{EI}

The solution is

y=C1cos⁡kx+C2sin⁡kx+M0Py = C_1\cos kx + C_2\sin kx + \frac{M_0}{P}

Boundary conditions:

  • At x=0x=0: y=0⇒C1=−M0Py = 0 \Rightarrow C_1 = -\dfrac{M_0}{P}.
  • At x=0x=0: dydx=0⇒C2k=0⇒C2=0\dfrac{dy}{dx} = 0 \Rightarrow C_2 k = 0 \Rightarrow C_2 = 0.

So

y=M0P (1−cos⁡kx)y = \frac{M_0}{P}\,(1 - \cos kx)
  • At x=Lx = L: y=0y = 0 (the top end is at the same lateral position) gives cos⁡kL=1\cos kL = 1.

The smallest non-zero solution is kL=2πkL = 2\pi:

PL2EI=4π2\frac{PL^2}{EI} = 4\pi^2 Pcr=4π2EIL2=π2EI(L/2)2P_{cr} = \frac{4\pi^2EI}{L^2} = \frac{\pi^2EI}{(L/2)^2}

The effective length is Le=L/2L_e = L/2, so a fixed-fixed column carries 4 times the load of a hinged-hinged column of the same length. The buckled shape has inflection points at L/4L/4 from each end (at x=L/4x = L/4 and x=3L/4x = 3L/4).

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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