Differential equation of the elastic curve

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

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.