# 7.2: Derivation of the Equation of the Elastic Curve of a Beam

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The elastic curve of a beam is the axis of a deflected beam, as indicated in Figure 7.1a.

$$Fig. 7.1$$. The elastic curve of a beam.

To derive the equation of the elastic curve of a beam, first derive the equation of bending.

Consider the portion $$cdef$$ of the beam shown in Figure 7.1a, subjected to pure moment, $$M$$, for the derivation of the equation of bending. Due to the applied moment $$M$$, the fibers above the neutral axis of the beam will elongate, while those below the neutral axis will shorten. Let $$O$$ be the center and $$R$$ be the radius of the beam’s curvature, and let $$ij$$ be the axis of the curved beam. The beam subtends an angle $$\theta$$ at $$O$$. And let $$\sigma$$ be the longitudinal stress in a filament $$gh$$ at a distance $$y$$ from the neutral axis.

From geometry, the length of the neutral axis of the beam $$ij$$ and that of the filament $$gh$$, located at a distance $$y$$ from the neutral axis of the beam, can be computed as follows:

$$i j=R \theta \text { and } g h=(R+y) \theta$$

The strain $$\varepsilon$$ in the filament can be computed as follows: $\varepsilon=\frac{g h-i j}{i j}=\frac{(R+y) \theta-R \theta}{R \theta}=\frac{y \theta}{R \theta}=\frac{y}{R}$

For a linear elastic material, in which Hooke’s law applies, equation 7.1 can be written as follows: $\frac{\sigma}{E}=\frac{y}{R}$

If an elementary area $$\delta A$$ at a distance $$y$$ from the neutral axis of the beam (see Figure 7.1c) is subjected to a bending stress $$\sigma$$, the elemental force on this area can be computed as follows: $\delta P=\sigma \delta A$

The force on the entire cross-section of the beam then becomes: $P=\int \sigma \delta A$

From static equilibrium consideration, the external moment $$M$$ in the beam is balanced by the moments about the neutral axis of the internal forces developed at a section of the beam. Thus, $M=\int(\sigma \delta A) y$

Substituting $$\sigma=\frac{E y}{R}$$ from equation 7.2 into equation 7.5 suggests the following: \begin{aligned} M &=\int\left(\frac{E}{R}\right)(y)(y)(\delta A) \\ &=\left(\frac{E}{R}\right) \int y^{2} \delta A \end{aligned}

Putting $$I=\int y^{2} \delta A$$ into equation 7.6 suggests the following: $M=\frac{E I}{R}$

where

$$I =$$ the moment of inertia or the second moment of area of the section.

Combining equations 7.2 and 7.7 suggests the following: $\frac{M}{I}=\frac{E}{R}$

The equation of the elastic curve of a beam can be found using the following methods.

From differential calculus, the curvature at any point along a curve can be expressed as follows: $\frac{1}{R}=\frac{\frac{d^{2} y}{d x^{2}}}{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}$

where

$$\frac{d y}{d x} \text { and } \frac{d^{2} y}{d x^{2}}$$ are the first and second derivative of the function representing the curve in terms of the Cartesian coordinates $$x$$ and $$y$$.

Since the beam in Figure 7.1 is assumed to be homogeneous and behaves in a linear elastic manner, its deflection under bending is small. Therefore, the quantity $$\frac{d y}{d x}$$, which represents the slope of the curve at any point of the deformed beam, will also be small. Since $$\left(\frac{d y}{d x}\right)^{2}$$ is negligibly insignificant, equation 7.9 could be simplified as follows: $\frac{1}{R}=\frac{\frac{d^{2} y}{d x^{2}}}{[1+0]^{3 / 2}}=\frac{d^{2} y}{d x^{2}}$

Combining equations 7.2 and 7.10 suggests the following: $\frac{1}{R}=\frac{M}{E I}=\frac{d^{2} y}{d x^{2}}$

Rearranging equation 7.11 yields the following: $E I \frac{d^{2} y}{d x^{2}}=M$

Equation 7.12 is referred to as the differential equation of the elastic curve of a beam.

This page titled 7.2: Derivation of the Equation of the Elastic Curve of a Beam is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by René Alderliesten (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.