13.1 Elastic curve equation and boundary conditions

Elastic Curve Equation and Boundary Conditions

The elastic curve equation describes how a beam deflects under load. It connects the beam's deformed shape to the applied loading, material stiffness, and geometry, making it the starting point for nearly every deflection analysis you'll do in this course. Boundary conditions then let you pin down the specific solution for a given support configuration.

Elastic Curve Equation for Beams

Derivation and Assumptions

The elastic curve is the deformed shape of the beam's neutral axis under load. To derive the governing equation, we rely on Euler-Bernoulli beam theory, which makes two key assumptions:

• Plane sections remain plane and perpendicular to the neutral axis after deformation (no warping of cross-sections).
• The material is linearly elastic, homogeneous, and isotropic (Hooke's law applies everywhere).

These assumptions work well for slender beams with small deflections relative to their length.

Starting from equilibrium and the moment-curvature relationship (covered below), you can write the elastic curve as a fourth-order differential equation:

$\frac{d^4y}{dx^4} = \frac{q(x)}{EI}$

where:

• $y$ = transverse deflection of the beam
• $x$ = position along the beam's length
• $q(x)$ = distributed load intensity (force per unit length)
• $E$ = elastic (Young's) modulus of the material
• $I$ = second moment of area (moment of inertia) of the cross-section

Simplification and Loading Conditions

If the flexural rigidity $EI$ is constant along the beam and you already know the bending moment function $M(x)$, you can work with the simpler second-order form:

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

This is the version you'll use most often. You write $M(x)$ from your shear and moment diagrams, then integrate twice to get $y(x)$.

The equation applies to any loading type. You just need the correct expression for $M(x)$:

• Concentrated (point) loads produce piecewise-linear moment diagrams
• Distributed loads (uniform, linearly varying, etc.) produce curved moment diagrams
• Concentrated moments cause jumps in the moment diagram

Boundary Conditions for Beam Deflection

Types of Boundary Conditions

Boundary conditions capture how the beam is supported and constrained at specific points. Without them, you can't solve for the integration constants that appear when you integrate the elastic curve equation.

The four most common support types and their associated conditions:

"Free" in the table means that quantity is not constrained to a known value at that point.

Application of Boundary Conditions

The number of boundary conditions you need matches the number of integration constants. If you start from the second-order form $\frac{d^2y}{dx^2} = \frac{M(x)}{EI}$, two integrations produce two constants ($C_1$ and $C_2$), so you need two boundary conditions. If you start from the fourth-order form, you'll have four constants and need four boundary conditions.

Here's how this plays out for two classic cases:

Cantilever beam (fixed at $x = 0$, free at $x = L$):

• $y(0) = 0$ (no deflection at the wall)
• $\frac{dy}{dx}(0) = 0$ (no rotation at the wall)
• If using the fourth-order form, add: $M(L) = 0$ and $V(L) = 0$ at the free end

Simply supported beam (pins at both ends):

• $y(0) = 0$ (no deflection at left support)
• $y(L) = 0$ (no deflection at right support)
• If using the fourth-order form, add: $M(0) = 0$ and $M(L) = 0$

Curvature, Moment, and Rigidity Relationship

Beam Curvature

Curvature ($\kappa$) measures how sharply the beam bends at a given point. It equals the reciprocal of the radius of curvature $\rho$:

$\kappa = \frac{1}{\rho}$

The exact curvature expression involves both $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$, but under the small-deflection assumption (slopes much less than 1), it simplifies to:

$\kappa \approx \frac{d^2y}{dx^2}$

This approximation is what makes the elastic curve equation linear and solvable by direct integration.

Moment-Curvature Relationship

The link between internal bending moment and curvature is:

$M = EI\,\kappa$

or equivalently:

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

This is the equation that ties everything together. Flexural rigidity $EI$ quantifies a beam's resistance to bending. It depends on both the material (through $E$) and the cross-section shape (through $I$).

Common moment of inertia formulas you'll need:

• Rectangular cross-section (width $b$, height $h$): $I = \frac{bh^3}{12}$
• Solid circular cross-section (radius $r$): $I = \frac{\pi r^4}{4}$

A steel beam with a deep I-section has a much larger $EI$ than a wooden beam of similar weight, which is why steel beams deflect less under the same load.

Integration Constants for Beam Deflection

Solving the Elastic Curve Equation

Each integration of the elastic curve equation introduces one unknown constant. These constants carry physical meaning:

• Integrating $\frac{d^2y}{dx^2} = \frac{M(x)}{EI}$ once gives the slope equation $\frac{dy}{dx}$, with constant $C_1$
• Integrating again gives the deflection equation $y(x)$, with constant $C_2$

$C_1$ and $C_2$ encode the beam's slope and deflection at whatever reference point you choose. You find their values by plugging in your boundary conditions.

Determining Integration Constants: Step-by-Step

1. Write $M(x)$ from your free-body diagram and equilibrium equations.
2. Substitute into $EI\frac{d^2y}{dx^2} = M(x)$ and integrate once to get $EI\frac{dy}{dx} = \int M(x)\,dx + C_1$.
3. Integrate again to get $EI\,y = \iint M(x)\,dx\,dx + C_1 x + C_2$.
4. Apply boundary conditions by substituting known values of $y$ and $\frac{dy}{dx}$ at specific $x$-locations.
5. Solve the resulting system of equations for $C_1$ and $C_2$.
6. Substitute back into the general solution to get the specific deflection equation $y(x)$.

Example: Cantilever with a point load $P$ at the free end

The moment function is $M(x) = -P(L - x)$ (taking the fixed end as $x = 0$). After two integrations:

$EI\,y(x) = -P\left(\frac{Lx^2}{2} - \frac{x^3}{6}\right) + C_1 x + C_2$

Applying $y(0) = 0$ gives $C_2 = 0$. Applying $\frac{dy}{dx}(0) = 0$ gives $C_1 = 0$. The specific solution is:

$y(x) = \frac{P}{6EI}\left(x^3 - 3Lx^2\right)$

The maximum deflection occurs at the free end ($x = L$):

$y_{\max} = -\frac{PL^3}{3EI}$

The negative sign indicates downward deflection (in the direction of the load). This is one of the most commonly referenced deflection formulas, so it's worth committing to memory.