The purpose of this post is to present a quick derivation of the static Euler-Bernoulli beam equation. This is something I’ve reviewed many times before and kept forgetting, so I decided to write it up once and for all to ensure that I can remember at least one cornerstone of civil engineering.
Geometrically, it is intuitively clear that when a beam is bent by external forces and torques, some sections will become longer (hence experiencing internal tension) while other sections will become shorter (hence experiencing internal compression). By the intermediate value theorem, there thus exists a section which becomes neither longer nor shorter, instead possessing a bend-invariant length; this is called the neutral axis of the beam (strictly speaking, the neutral axis is not a 1D line, but a 2D surface cutting through the center of the beam so perhaps a better name would have been neutral surface). The neutral axis of the beam therefore experiences neither internal tension nor internal compression, and so is defined by experiencing zero internal stress \(\sigma_{\text{int}}=0\). The idea is to then understand how the internal stress field \(\sigma_{\text{int}}(y)\) varies as a function of the distance \(y\) above (or below) the neutral axis (where we’ve just established it’s zero). Intuitively, this should depend not only on \(y\), but also on the stiffness of the beam material (this is why it’s harder to bend stiffer beams!). Since the internal tensions and internal compressions are essentially uniaxial stresses along the beam, one should invoke Hooke’s law:
$$\sigma_{\text{int}}(y)=E\varepsilon(y)$$
where the engineering strain \(\varepsilon(y)\) at a distance \(y\) from the neutral axis may be approximated as \(\varepsilon(y)=\kappa y\) where \(\kappa\) is the local curvature of the beam (and the key reason for introducing it is that it is independent of \(y\), being only a function of position \(x\) along the neutral axis). Thus, the key finding here is that the internal stress field varies linearly about the neutral axis:
$$\sigma_{\text{int}}(y)=\kappa Ey$$
Throughout this, the analysis has been for some fixed \(x\), and looking at the variation in \(y\). To globalize this from \(y\to x\), one could stick with \(\sigma_{\text{int}}(x,y)=\kappa(x)Ey\) but this would still be mucking up \(x\) with \(y\). So in order to get rid of \(y\) and be able to just worry about \(x\), it is necessary to find some property that characterizes each cross section of the beam. Staring at the picture of the internal stress field, it is clear that one useful metric is provided by the internal torque \(\bar{\tau}_{\text{int}}(x)\) about the neutral axis (also called the internal bending moment by engineers), which by definition is:
$$\bar{\tau}_{\text{int}}(x)=\iint_{\text{cross section at }x}y\sigma_{\text{int}}(x,y)dA$$
Introducing the second moment of area \(\bar I(x):=\iint_{\text{cross section at }x}y^2dA\) about the neutral axis, this leads to the result:
$$\bar{\tau}_{\text{int}}(x)=E\bar I(x)\kappa(x)$$
which washes out all \(y\)-dependence as desired. The product \(E\bar I\) is called the flexural rigidity of the beam. Letting \(\delta(x)\) denote the profile of the bent beam, one has the approximation \(\kappa(x)\approx d^2\delta/dx^2\), so (suppressing the \(x\)-dependence):
$$\frac{d^2\delta}{dx^2}=\frac{\bar{\tau}_{\text{int}}}{E\bar{I}}$$
Finally, although this form is generally sufficient (thanks to the “system nitpick trick”), it is also possible to express the internal torque \(\bar{\tau}_{\text{int}}\) about the neutral axis directly in terms of the (perhaps more intuitive) linear external transverse force density \(f_{\text{ext}}^{\perp}(x)\) along the beam. Specifically, because the beam is static, translational equilibrium in the vertical direction gives \(f_{\text{ext}}^{\perp}(x)=\frac{dF_{\gamma}}{dx}\) where \(F_{\gamma}(x)\) is the vertical shear force at position \(x\) and rotational equilibrium gives \(F_{\gamma}(x)=\frac{d\bar{\tau}_{\text{int}}}{dx}\). Altogether, this yields a 4th-order ODE which is traditionally called the static Euler-Bernoulli beam equation:
$$\frac{d^4\delta}{dx^4}=\frac{f_{\text{ext}}^{\perp}}{E\bar I}$$
Evidently, a more descriptive name for the static Euler-Bernoulli beam equation would be the static small deflection beam equation since the most important assumption underlying it is that the deflection \(\delta(x)\) of the beam is small everywhere.
Example: Within the Euler-Bernoulli small deflection regime, what is the shape of a homogeneous cantilever beam of density \(\rho\), Young’s modulus \(E\), and length \(L\) with an \(a\times a\) square cross section in the presence of Earth’s surface gravitational field \(g\)? One option is to insert \(f_{\text{ext}}^{\perp}=\rho ga^2\) and integrate the static Euler-Bernoulli beam equation from \(0\) to \(L\) with respect to \(x\). Because the internal shear force \(F_{\gamma}(0)=\rho ga^2L\) at the fixed end must balance the weight of the cantilever, this gives a boundary condition to fix the first integration constant. Integrating again, and now using the boundary condition \(\bar{\tau}_{\text{int}}(0)=\rho ga^2L^2/2\), one gets a second boundary condition. Then integrating twice more and using the cantilever boundary conditions \(\delta(0)=d\delta/dx(0)=0\) yields the deflected beam profile as a quartic polynomial in \(x\):
$$\delta(x)=\frac{\rho g}{2a^2E}x^2(6L^2-4Lx+x^2)$$
with maximum deflection \(\delta(L)=\frac{3\rho gL^4}{2a^2E}\), so if one wished to minimize this, then a suitable merit index for materials selection would be \(E/\rho\).