---
title: "Differential equation of the elastic curve"
subject: "Strength of Materials"
chapter: 7
key: "semester_4/strength_of_materials_mechanical:beam-deflection-differential-equation"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/beam-deflection-differential-equation
site: BCT Engineering
---

# Differential equation of the elastic curve

Strength of Materials · Chapter 7: Deflection of Beams by Integration Method

- Asked in: year not given, 6 marks
- Video: [Introduction to beam deflection and the elastic curve equation (double integration method)](https://www.youtube.com/watch?v=EWhL-mixfaI) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/beam-deflection-differential-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-7

## Question

Derive the differential equation of the elastic curve EI d2y/dx2 = M for a beam. State the sign convention and the boundary conditions for (i) a cantilever, (ii) a simply supported beam, and (iii) a propped cantilever. How are continuity conditions used when the bending moment expression changes along the span?

## Short answer

From flexure formula M/I = E/R and curvature approximation, EI d²y/dx² = M. Sign: x right, y upward, sagging M positive. Cantilever: y=0, dy/dx=0 at fixed end. Simply supported: y=0 at supports. Propped cantilever: y=0, dy/dx=0 at fixed end, y=0 at prop. Continuity: slope and deflection equal at junctions; Macaulay's brackets unify constants.

## Answer

#### Derivation
When a beam bends, the neutral surface takes the shape of a curve, the elastic curve. From the flexure formula

$$
\frac{M}{I} = \frac{E}{R} \;\Rightarrow\; \frac1R = \frac{M}{EI}
$$

From calculus the curvature of the curve $y(x)$ is

$$
\frac1R = \frac{d^2y/dx^2}{\left[1+(dy/dx)^2\right]^{3/2}}
$$

The slope $dy/dx = \theta$ is very small for stiff beams, so $(dy/dx)^2 \ll 1$:

$$
\boxed{EI\frac{d^2y}{dx^2} = M}
$$

Integrating once gives the slope and twice gives the deflection:

$$
EI\frac{dy}{dx} = \int M\,dx + C_1, \qquad EI\,y = \iint M\,dx\,dx + C_1x + C_2
$$

#### Sign convention
$x$ to the right, $y$ positive **upward**, sagging bending moment positive. With this convention the downward deflection comes out negative. (Some books take $y$ downward and use $EI\,y'' = -M$.) The magnitude is the same.

#### Boundary conditions
| Support | Condition |
|---|---|
| (i) Cantilever, fixed at $x=0$ | $y = 0$, $dy/dx = 0$ at the wall |
| (ii) Simply supported at $x=0$ and $x=L$ | $y = 0$ at both supports (slope is not zero) |
| (iii) Propped cantilever: fixed at $A$, prop at $B$ | $y_A=0$, $\theta_A = 0$, $y_B = 0$ |

#### Continuity (Macaulay's method)
If the loading changes along the span, $M$ has a different expression in each segment, giving extra constants. At the junction between two segments both the slope and the deflection must be equal on either side. Macaulay's method writes $M$ for the whole beam in one expression using brackets $\langle x-a\rangle^n$ that are zero for $x<a$. The bracket is integrated as a whole, so there are only two constants $C_1,C_2$ for the entire beam. A uniformly distributed load that stops before the end must be extended to the end and balanced by an equal and opposite load.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What approximation turns the curvature expression into EI d2y/dx2 = M?
   A. The radius R is taken as infinite
   B. The slope dy/dx is small, so (dy/dx)^2 is negligible against 1
   C. The modulus E is taken as constant so curvature vanishes
   D. The bending moment is taken as uniform
   Answer: B. Curvature 1/R = y''/[1+(y')²]^1.5 reduces to y'' when y' is small, and 1/R = M/EI. (Key point: Deriving EI y'' = M)

2. Which boundary conditions apply to a simply supported beam?
   A. y = 0 and slope = 0 at both supports
   B. Slope = 0 at both supports only
   C. y = 0 at both supports; the slope is not zero
   D. y = 0 at one support and slope = 0 at the other
   Answer: C. Simple supports prevent deflection but allow rotation. (Key point: Boundary conditions)

3. Which conditions apply to a propped cantilever fixed at A with a prop at B?
   A. yA = 0, yB = 0 only
   B. Slope at A = 0 and slope at B = 0
   C. yA = 0, yB = 0, slope at B = 0
   D. yA = 0, slope at A = 0, yB = 0
   Answer: D. The fixed end gives zero deflection and slope; the prop gives zero deflection at B. (Key point: Propped cantilever)

4. What is true at the junction of two segments and how does Macaulay's method help?
   A. Slope and deflection are equal on both sides; one expression with bracket terms leaves only two constants
   B. Slope is equal but deflection may jump; each segment keeps its own constants
   C. Bending moment is zero at the junction; four constants are needed
   D. Deflection is equal but slope changes; two constants per segment
   Answer: A. Continuity of slope and deflection is built into integrating the brackets as a whole, so only C1 and C2 remain. (Key point: Continuity in Macaulay's method)

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## Related in other subjects

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