# Boundary conditions for this beam deflection problem

consider the diagram above. I know that deflection is described by the equation $$EI\frac{d^4y}{dx^4}=p(x)$$ where $$L$$ is the length of the beam, $$E$$ is the Young's module, $$I$$ is the moment of inertia, $$p$$ is the distribution of load and $$M$$ is the torque.

I have the boundary conditions

$$y(0)=y'(0)=0$$ which means that the left end of the beam is immovable.

$$y(L)=0$$ which mean that the right end of the beam is supported by something.

Then, what would be the condition for $$y'(L)$$ ??

Here, $$y'(L)$$, i.e., the slope at $$x=L$$ cannot be determined beforehand. However, moment is prescribed at $$x=L$$. So, the boundary condition will be $$EI\frac{d^2y}{dx^2}|_{x=L} = M$$.
You have no imposed boundary condition on $$y'$$ on the right side of the beam. The derivative of a beam deflection (i.e the curvature) is not constrained by a simple support.
• Also, the system isn't hyperstatic because vertical displacement is blocked twice. If it were, any simply-supported beam would be hyperstatic. The system is hyperstatic because there are 4 constraints $\left(\delta_x(0) = \delta_y(0) = \theta(0) = \delta_y(L) = 0\right)$ and only three equilibrium equations $\left(\sum F_x = \sum F_y = \sum M = 0\right)$. – Wasabi Apr 28 '20 at 13:20