Skip to main content

Euler's formula: fixed-hinged strut with limitation

Short answer

Derivation gives P_cr ≈ 20.19EI/L², effective length ~0.7L. Limitations: valid only for slenderness ratio above ~80-100, for perfectly straight elastic columns with axial load; it overestimates for short columns and ignores imperfections.

Derive the Euler's formula for critical load for a strut with one end fixed another hinged. Also mention the limitation for using this formula.

Euler's critical load for a column fixed at one end and hinged 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 have its top end B hinged and its bottom end A fixed, length LL. When it buckles, the fixed end develops a moment, and a horizontal reaction HH appears at the hinge to keep B in position. Take the origin at the hinged end B, with xx measured downward and yy the lateral deflection.

   P
   |  B  hinged (H)
   |  )
   |  ) y
   |  )
   A  fixed
   P

Moment at the section: M=−Py+HxM = -Py + Hx, so

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

The solution is

y=C1cos⁡kx+C2sin⁡kx+HPxy = C_1\cos kx + C_2\sin kx + \frac{H}{P}x

Boundary conditions:

  • At x=0x=0: y=0⇒C1=0y = 0 \Rightarrow C_1 = 0.
  • At x=Lx=L (fixed end): y=0⇒C2sin⁡kL+HLP=0y = 0 \Rightarrow C_2\sin kL + \dfrac{HL}{P} = 0
  • At x=Lx=L: dydx=0⇒C2kcos⁡kL+HP=0\dfrac{dy}{dx} = 0 \Rightarrow C_2k\cos kL + \dfrac{H}{P} = 0

Eliminate C2C_2 and HH between the last two equations:

tan⁡kL=kL\tan kL = kL

This transcendental equation is solved by trial. The smallest non-zero root is

kL=4.493  ⇒  PL2EI=4.4932=20.19kL = 4.493 \;\Rightarrow\; \frac{PL^2}{EI} = 4.493^2 = 20.19 Pcr=20.19 EIL2≈2π2EIL2=π2EI(0.7L)2P_{cr} = \frac{20.19\,EI}{L^2} \approx \frac{2\pi^2EI}{L^2} = \frac{\pi^2EI}{(0.7L)^2}

The effective length is Le≈0.7LL_e \approx 0.7L. The bending moment is zero where sin⁡kx=0\sin kx = 0, so the inflection point lies at x=π/k≈0.7Lx = \pi/k \approx 0.7L from the hinged end, that is, about 0.3L0.3L from the fixed end.

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)

All Strength of Materials chapter 8 questions →