--- title: "Euler's formula: both ends fixed" subject: "Strength of Materials" chapter: 8 key: "semester_3/strength_of_materials:euler-both-ends-fixed" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-both-ends-fixed site: BCT Engineering --- # Euler's formula: both ends fixed Strength of Materials · Chapter 8: Column Theory - Asked in: 2081 Baisakh, 6 marks; 2075 Asoj, 8 marks (asked 2 times) - Video: [Understanding Buckling](https://www.youtube.com/watch?v=21G7LA2DcGQ) — The Efficient Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-both-ends-fixed - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8 ## Question Derive the expression for the Euler's formula for crippling load on a column of length L with both ends fixed. ## Short answer With both ends fixed, the end moments lead to P_cr = 4π²EI/L² = π²EI/(L/2)². Effective length is L/2, and inflection points appear at quarter points. ## Answer **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 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 $M_0$ (fixing moment) while the load $P$ stays axial. Take the origin at the lower fixed end A, with $y$ the lateral deflection at distance $x$. ```text P | B (fixed, M0) | /\ | ( y | \/ A fixed, M0 P ``` The moment at the section is $M = M_0 - Py$, so $$ EI\frac{d^2y}{dx^2} = M_0 - Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{M_0}{EI},\qquad k^2 = \frac{P}{EI} $$ The solution is $$ y = C_1\cos kx + C_2\sin kx + \frac{M_0}{P} $$ Boundary conditions: - At $x=0$: $y = 0 \Rightarrow C_1 = -\dfrac{M_0}{P}$. - At $x=0$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2 k = 0 \Rightarrow C_2 = 0$. So $$ y = \frac{M_0}{P}\,(1 - \cos kx) $$ - At $x = L$: $y = 0$ (the top end is at the same lateral position) gives $\cos kL = 1$. The smallest non-zero solution is $kL = 2\pi$: $$ \frac{PL^2}{EI} = 4\pi^2 $$ $$ P_{cr} = \frac{4\pi^2EI}{L^2} = \frac{\pi^2EI}{(L/2)^2} $$ The effective length is $L_e = L/2$, so a fixed-fixed column carries 4 times the load of the same hinged column. The buckled shape has points of inflection at $L/4$ from each end (at $x = L/4$ and $x = 3L/4$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the equation have M0? A. Fixed ends develop a fixing moment, so M = M0 - Py B. Fixed ends have no moment C. Moment M = -Py as for hinged D. M = P L Answer: A. Held-in-direction ends provide a restraining moment M0. (Key point: End moment) 2. Which conditions are used at x = 0? A. y = 0 and y'' = 0 B. y = 0 and dy/dx = 0, giving y = (M0/P)(1 - cos kx) C. Slope only D. y = M0/P Answer: B. Fixed end has no deflection and no slope. (Key point: Boundary conditions) 3. What condition at x = L gives the smallest root? A. sin kL = 0 so kL = pi B. cos kL = 0 so kL = pi/2 C. cos kL = 1 so kL = 2 pi D. tan kL = kL so kL = 4.493 Answer: C. cos kL = 1 first non-zero at 2 pi. (Key point: Smallest root) 4. What are P_cr and effective length? A. pi^2EI/L^2, Le = L B. pi^2EI/4L^2, Le = 2L C. 2 pi^2EI/L^2, Le = 0.7L D. 4 pi^2 EI/L^2, Le = L/2, four times the hinged-hinged load Answer: D. Substituting kL = 2 pi gives P_cr = 4 pi^2EI/L^2. 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