# Large rotation Euler-Bernoulli beam boundary condition

Is given in Wikipedia as $$EI\frac{d^4u}{dx^4}-\frac{3}{2}EA\left(\frac{du}{dx}\right)^2\frac{d^2u}{dx^2} = q(x) ,$$ where $q(x)$ is the transverse load (assuming uniform cross-section and no axial load).

For a cantilever beam clamped at one end, assuming $q(x)$ is nonzero but smooth across the entire domain $x\in[0,L]$, the fixed boundary conditions on the left are $$u(0)=u'(0) = 0,$$ but is it true that $$u''(L)=u'''(L)=0$$ as well, i.e., zero bending moment and shear at the right boundary (free end of the beam)?

## 1 Answer

Yes you are right. To understand exactly why you are right, keep firmly in mind exactly what the "bending moment" and "shear" mean. When we say that the moment and shear are $M(x_0)$ and $\tau(x_0)$ at position $x_0$ we mean that:

1. We imagine the beam cut at the position $x_0$ and we draw free body diagrams for the two sundered sections;

2. In particular, we find that the section of the beam for $x<x_0$ exerts moment and shear $M(x_0)$ and $\tau(x_0)$ on the section of beam for $x>x_0$, and contrariwise (with a sign change). The moments come from forces normal to the cross section which are compressive on the side of the neutral axis nearer the centre of the beam's curvature, and tensile on the other side.

So, when we get to the end $x_0=L$, there is no beam for $x>L$ to exert any putative moment and shear on the section of the beam for $x<L$. Hence, your conclusion.