# Euler-Bernoulli Beam

### Summary

\begin{align} w(x) &= \frac{dV}{dx} \\ V(x) &= \frac{dM}{dx} \\ M(x) &= \kappa EI \\ \kappa (x) & = -\frac{d \varphi}{dx} \\ \varphi (x) &= \frac{dv}{dx} \end{align}

### Assumptions

Euler-Bernoulli beam theory assumes:

• small deflections and
• insignificant shear deflections

Shear deflections become significant as the span:depth ratio increases. If large shear deflections may be an issue, consider using the Timoshenko beam formulation.

If deflections are large, consider using a geometrically non-linear analysis.

### Derivation Figure 1

Consider an infinitecimally short section of a beam with force per unit length, $$w$$ as Figure 1.

Resolve vertical forces

\begin{align} V + w \delta x & = V + \frac{dv}{dx} \delta x \\ w &= \frac{dv}{dx} \label{A}\tag{Eq.1}\\ V &= \int w\,dx \end{align}

Resolve moments about left hand side of segment

\begin{align} \frac{w}{2} \delta x^{2} + M + \frac{dM}{dx}\delta x & = (V + \frac{dv}{dx} \delta x) \delta x + M \\ \frac{w}{2} \delta x^{2} + \frac{dM}{dx}\delta x & = (V + \frac{dv}{dx} \delta x) \delta x \\ \frac{w}{2} \delta x + \frac{dM}{dx} & = V + \frac{dv}{dx} \delta x \end{align}

Subsitute $$w = \frac{dv}{dx}$$ as $$\ref{A}$$.

\begin{align} \frac{1}{2} \frac{dV}{dx} \delta x + \frac{dM}{dx} & = V + \frac{dV}{dx} \delta x \\ \frac{dM}{dx} & = V + \frac{1}{2} \frac{dV}{dx} \delta x\ \\ \end{align}

$$\delta x$$ terms can be ignored as infinitecimal.

\begin{align} \frac{dM}{dx} & = V \\ M & = \int V \, dx \end{align}

For derivation of displacement relationships, see #small-deflection-approximation.