Elastic curve equation

Every beam that carries load bends, however slightly, into a smooth curved shape called the elastic curve. A diving board dips under a person's weight, a floor joist sags a few millimetres under furniture, a bridge girder deflects under traffic — in each case the beam's straight axis becomes a curve, and that curve has a precise mathematical description tying it to the bending moment inside the beam. The elastic curve equation is the single differential equation that makes this link exact: $EI\,d^2y/dx^2 = M(x)$ . It converts the bending-moment diagram you already know how to draw into the deflected shape of the beam — the starting point for checking that a design is not just strong enough to avoid breaking, but stiff enough to avoid sagging visibly or vibrating excessively. This article builds the equation from the physics of bending and shows how to set it up for a real beam.

The core insight

Under the small-slope approximation valid for ordinary structural deflections, beam curvature equals $d^2y/dx^2$ , and since curvature also equals $M/(EI)$ from flexure theory, the two combine into $EI\,d^2y/dx^2 = M(x)$ — the differential equation whose integration gives the beam's deflected shape.

The core idea

Curvature and the small-slope approximation. The exact curvature of any plane curve $y(x)$ is

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

where: $\kappa$ is curvature (the reciprocal of the local radius of curvature $\rho$ , so $\kappa = 1/\rho$ ), and $dy/dx$ is the local slope of the beam axis. Engineering beams deflect by only a small fraction of their span, so the slope stays small everywhere, $(dy/dx)^2 \ll 1$ . Dropping that term reduces the denominator to essentially $1$ , leaving the much simpler small-slope curvature, $\kappa \approx d^2y/dx^2$ . This one approximation is what turns a messy nonlinear geometric relation into a linear differential equation — it is the single most important simplifying step in the whole topic.

Curvature and bending moment. Separately, from the same plane-sections-remain-plane assumption that gives the flexure formula $\sigma = My/I$ , elementary bending theory shows curvature is directly proportional to bending moment: $\kappa = M/(EI)$ . Here $M$ is the internal bending moment at that cross-section (read off the bending-moment diagram) and $EI$ is the flexural rigidity, the product of Young's modulus $E$ and the second moment of area $I$ about the neutral axis — a beam's inherent resistance to bending. A beam with larger $EI$ (stiffer material, or a deeper/wider cross-section) curves less for the same moment.

Combining the two — equating the two expressions for curvature — gives the governing relation of this topic, the elastic curve equation:

EI\frac{d^2y}{dx^2} = M(x)

where: $y(x)$ is the transverse deflection of the beam axis at position $x$ along the span, and $M(x)$ is the bending moment as a function of $x$ . Everything to the left of the equals sign is a property of the beam's shape (its curvature, scaled by stiffness); everything on the right is a property of the loading (the internal moment). The equation says these two must match at every point along the beam.

Sign and the double integration route. Because $EI > 0$ always, the sign of $d^2y/dx^2$ follows the sign of $M(x)$ directly. Wherever $M$ is positive (sagging, the usual convention for a simply supported beam under downward load), the curve is concave up; wherever $M$ is negative (hogging, typical near a fixed support or over a cantilever), the curve is concave down. Physically, $EI\,dy/dx$ tracks the beam's local slope, and integrating the equation once more gives $y(x)$ itself:

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

The two constants $C_1$ and $C_2$ are fixed by boundary conditions: $y=0$ at a pin or roller support, both $y=0$ and $dy/dx=0$ at a fixed support, or by symmetry (zero slope at midspan of a symmetrically loaded simply supported beam). Where $M(x)$ changes algebraic form — under a point load, at a support, or where a distributed load starts or stops — the beam must be split into separate regions, with $y$ and $dy/dx$ forced to match (continuity) at each boundary between regions. The full mechanics of carrying out these integrations, region by region, for standard loadings is the subject of the next topic, deflection of beams by double integration and superposition; here the job is to set the governing equation up correctly.

Figure

A simply supported beam sagging under a uniform load with the deflection curve exaggerated, the bending moment diagram below it as a parabola, and an inset showing that curvature equals the second derivative of deflection which equals moment divided by EI

Derivation ADeriving the elastic curve equation from curvature and flexure

The goal is to show, from first principles, that the small-slope curvature of a bent beam axis and the moment-curvature relation of flexure theory combine into a single linear differential equation $EI\,d^2y/dx^2 = M(x)$ that can be integrated twice, with two boundary conditions, to give the beam's deflected shape $y(x)$ for any given loading.

Step 1. Start from the exact geometric curvature of the deflection curve $y(x)$ : $\kappa = \dfrac{d^2y/dx^2}{\left[1+(dy/dx)^2\right]^{3/2}}$ . This holds for any smooth curve, beam or not — it is pure calculus, the standard formula for curvature of a plane curve.

Step 2. Apply the small-slope approximation. For the deflections typical of engineering beams (millimetres over metres of span), the slope $dy/dx$ stays small everywhere, so $(dy/dx)^2$ is negligible compared with $1$ . The denominator collapses to $1$ , and the curvature simplifies to $\kappa \approx d^2y/dx^2$ . This is the step that linearises the problem — without it, the governing equation would be a nonlinear ODE with no simple closed-form solution for most loadings.

Step 3. Bring in the moment-curvature relation from flexure theory. Elementary bending theory (built on the assumption that plane cross-sections remain plane after bending, the same assumption behind the flexure formula $\sigma = My/I$ ) shows that curvature is directly proportional to the internal bending moment: $\kappa = M/(EI)$ , where $M$ is the moment at that section and $EI$ is the flexural rigidity. This relation is independent of the small-slope step — it comes from the stress analysis of the cross-section, not the geometry of the whole curve.

Step 4. Equate the two expressions for $\kappa$ . Since both Step 2 and Step 3 give curvature at the same point, $d^2y/dx^2 \approx M/(EI)$ , and rearranging gives the elastic curve equation $EI\,d^2y/dx^2 = M(x)$ . Because $M(x)$ is known once the bending-moment diagram is drawn, this is a second-order ordinary differential equation in $y(x)$ with a known right-hand side, solvable by direct integration. Verified: setting $dy/dx \to 0$ in the exact curvature formula recovers $\kappa \to d^2y/dx^2$ exactly, confirming the small-slope limit is consistent with the general definition, and dimensional check $[EI][y][x^{-2}] = [\text{Pa}][\text{m}^4][\text{m}][\text{m}^{-2}] = \text{N}\,\cdot\,\text{m}$ matches the units of $M(x)$ .

Worked example

A simply supported beam of length $L$ carries a uniformly distributed load $w$ per unit length, giving the standard parabolic bending-moment distribution $M(x) = \dfrac{wLx}{2} - \dfrac{wx^2}{2}$ , with $x$ measured from the left support. Set up the elastic curve equation, find the slope expression $EI\,dy/dx$ , and state what further step is needed to get the full deflection curve $y(x)$ .

Step 1. Write the governing equation by substituting $M(x)$ directly into $EI\,d^2y/dx^2 = M(x)$ :

EI\frac{d^2y}{dx^2} = \frac{wLx}{2} - \frac{wx^2}{2}

Step 2. Integrate once with respect to $x$ to get an expression for the slope, introducing one constant of integration $C_1$ :

EI\frac{dy}{dx} = \frac{wLx^2}{4} - \frac{wx^3}{6} + C_1

Step 3. Fix $C_1$ using a known condition. This load and support arrangement is symmetric about midspan, so the slope must vanish at $x = L/2$ . Substituting $x = L/2$ and setting the expression to zero:

0 = \frac{wL(L/2)^2}{4} - \frac{w(L/2)^3}{6} + C_1 = \frac{wL^3}{16} - \frac{wL^3}{48} + C_1 \;\;\Rightarrow\;\; C_1 = -\frac{wL^3}{24}

Step 4. Write the completed slope equation and state the final step. Substituting $C_1$ back in:

EI\frac{dy}{dx} = \frac{wLx^2}{4} - \frac{wx^3}{6} - \frac{wL^3}{24}

This is the slope of the elastic curve at any point $x$ . To get the deflection $y(x)$ itself, this expression must be integrated once more with respect to $x$ , introducing a second constant $C_2$ , which is fixed using the boundary condition $y = 0$ at $x = 0$ (the left support carries no deflection). That second integration and the resulting deflection formula belong to the deflection-of-beams topic that follows this one; the job here — setting up and taking the first integral of the elastic curve equation — is complete.

Common traps

Forgetting that the small-slope approximation is what makes the equation linear — it silently assumes deflections and slopes stay small, which is true for ordinary structural beams but would fail for something like a highly flexible cable or a beam bent into a near-circle.
Mixing up a sign convention for $M(x)$ between the bending-moment diagram and the elastic curve equation — a sign slip flips the predicted concavity and can turn a sagging prediction into a hogging one.
Applying a zero-slope condition at a pin or roller support, which only fixes $y=0$ ; a zero-slope condition is only valid at a fixed (built-in) support, or at a point of symmetry such as beam midspan under symmetric loading.
Integrating one $M(x)$ expression across a beam whose moment actually changes form partway along — under a point load or at the edge of a partial distributed load, the beam must be split into separate regions, each integrated separately, with slope and deflection matched at the boundary between regions.

Ask the tutor: If you are unsure how to split a beam into regions and match slope and deflection at the boundaries when the moment equation changes form, ask the AI tutor to walk through a point-loaded beam example with you.

Why does the small-slope approximation matter so much for this equation?

Without it, curvature is $\kappa = \dfrac{d^2y/dx^2}{[1+(dy/dx)^2]^{3/2}}$ , a nonlinear expression that generally has no simple closed-form solution. Dropping $(dy/dx)^2$ because it is tiny for ordinary beam deflections turns this into the linear equation $EI\,d^2y/dx^2 = M(x)$ , which can be integrated directly. This is why the method works well for stiff structural beams but would need the full nonlinear curvature for something like a very slender cantilever deflecting through a large angle.

How is the elastic curve equation different from the flexure formula I already know?

The flexure formula $\sigma = My/I$ gives the bending stress at a point within a cross-section, for a single known value of $M$ . The elastic curve equation instead treats $M$ as a function of position, $M(x)$ , and asks how the whole beam axis deflects along its length. They share the same underlying moment-curvature relation $\kappa = M/(EI)$ , but flexure theory answers a stress question at one section while the elastic curve equation answers a shape question over the whole span.

Why do I need two integrations and two boundary conditions instead of one?

The elastic curve equation is second-order in $y$ , so solving it by direct integration produces two constants of integration, one from each integration step. Each constant needs one piece of independent information to pin it down — a known slope or a known deflection at some point. That is exactly what boundary conditions (at supports) or symmetry conditions (at midspan) supply, one for each constant, so two conditions are always needed to fully determine $y(x)$ .

Elastic curve equation

The core idea

Worked example

Common traps

Before this

Where this leads