--- title: "Euler's formula both ends fixed, with limitations" subject: "Strength of Materials" chapter: 8 key: "semester_3/strength_of_materials:euler-both-ends-fixed-limitation" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-both-ends-fixed-limitation site: BCT Engineering --- # Euler's formula both ends fixed, with limitations Strength of Materials · Chapter 8: Column Theory - Asked in: 2074 Asoj, 6+2 marks - Video: [Euler's Formula For Columns (Assumptions) || What is Euler's Formula](https://www.youtube.com/watch?v=Bt8I-AYdy-M) — Gear Institute Mechanical Engineering Videos - 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-limitation - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8 ## Question 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. ## 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. ## 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 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 a hinged-hinged column of the same length. The buckled shape has inflection points at $L/4$ from each end (at $x = L/4$ and $x = 3L/4$). #### Limitations of Euler's formula 1. It applies only to **long (slender) columns**. In terms of stress, $\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2}$ must not exceed the proportional limit, so $\lambda \ge \pi\sqrt{E/\sigma_p}$ (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 $\sigma_{cr}$ 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is M = M0 - Py? A. No end moments B. Fixed ends provide a fixing moment M0 C. Free ends D. M0 is axial Answer: B. Fixed ends resist rotation. (Key point: End moment) 2. What is the smallest root? A. pi B. pi/2 C. kL = 2 pi D. 4.493 Answer: C. cos kL = 1. (Key point: Root) 3. What is P_cr? A. pi^2EI/L^2 B. 2 times hinged C. pi^2EI/4L^2 D. 4pi^2EI/L^2, four times hinged Answer: D. Le = L/2. (Key point: Result) 4. What is the main limitation of Euler's formula? A. It holds only for long, slender columns where critical stress is below the proportional limit B. It holds for short columns C. It overestimates nothing for short columns D. It holds beyond the elastic range Answer: A. For short and intermediate columns it overestimates the load, so empirical formulas are used. (Key point: Limitation) ## Related in Strength of Materials - [Euler's formula: both ends fixed](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-fixed) — Strength of Materials, chapter 8, asked 2× - [Euler's formula both ends hinged, with limitations](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-hinged-limitation) — Strength of Materials, chapter 8 - [Euler's formula: fixed-free column with limitations](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-fixed-free-limitations) — Strength of Materials, chapter 8 - [Euler's formula: both ends hinged](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-hinged) — Strength of Materials, chapter 8, asked 4× - [Euler's formula: fixed-hinged strut with limitation](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-one-fixed-one-hinged-limitation) — Strength of Materials, chapter 8, asked 2× - [Bar fixed at both ends: load shares, displacements](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acircular-bar-fixed-both-ends-50kn) — Strength of Materials, chapter 3 ## Related in other subjects - [Column with battens, 1500 kN, 8 m, fixed ends](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Acolumn-rolled-i-1500kn-8m-fixed) — Design of Steel and Timber Structure, chapter 6 - [Euler's buckling load for columns](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Aeuler-crippling-load-pinned-column) — Strength of Materials, chapter 10 - [Tubular column load and eccentric loading](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Atube-column-and-eccentric-short-column) — Strength of Materials, chapter 10 - [Fixed-end column 4 m, 1000 kN](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Afixed-end-column-4-m-1000) — Design of RCC Structure, chapter 10