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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 $y'(L)$ ??

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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 $EI\frac{d^2y}{dx^2}|_{x=L} = M$.

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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.
Note : This system is hyperstatic

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  • $\begingroup$ The derivative of the deflection is not the curvature, that is the second derivative. (There is a boundary condition on the second derivative, as the bending moment is known at L.) $\endgroup$
    – ingenørd
    Apr 28, 2020 at 11:54
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    $\begingroup$ 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 $\left(\sum F_x = \sum F_y = \sum M = 0\right)$. $\endgroup$
    – Wasabi ♦
    Apr 28, 2020 at 13:20
  • $\begingroup$ You are right, thanks for pointing out my mistake $\endgroup$
    – ClariB
    Apr 29, 2020 at 6:48

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