The Question is:

A uniform horizontal beam OA, of length $a$ and weight $w$ per unit length is clamped horizontally at O and freely supported at A. The transverse displacement $y$ of the beam is governed by the differential equation

$$EI \frac{d^2y}{dx^2}=\frac{1}{2}w(a-x)^2-R(a-x),$$

where $x$ is the distance along the beam measured from O, $R$ is the reaction at A, and $E$ and $I$ are physical constants.

After this they give a boundary conditions, and as one expects, to solve the equation - the maths which I have successfully completed.

My question is the description of how the beam is restricted and in which directions it can move - I don't seem to understand the mechanics of the beam.


You are given two boundary conditions here. The end O is clamped and the end A is free.

The clamped end at O means it is completely prevented from moving. This means that $y(O) = 0$ (no displacement). It also means that the beam cannot have any rotation there (it remains orthogonal to the boundary). This means $y'(O) = 0$.

The free end is free to move however the equations dictate it should move. This means there are no stresses or moments about that end. This equates to $y''(A) = 0$ and $y'''(A) = 0$.

I am not clear what the "reaction at $A$" definition is for $R$. However, if there is a point load (force or mass) located at the free end $A$ of magnitude $R$, that changes the boundary condition such that there is a moment there giving $y'''(A) = RL$.

You also have a distributed load $w$. I'll leave it to you to think about how that affects the boundary conditions, if at all.

These are typically the most commonly used boundary conditions, although other combinations are possible depending on the problem set up. For the Euler-Bernoulli beam theory, a list of boundary conditions is readily available.

But they can also be figured out by thinking about what $y$ and its derivatives are and how those relate to what is actually happening at the boundaries.

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  • $\begingroup$ Are we assuming the beam can bend into a curve? Sorry to ask this stupid question but just making sure. $\endgroup$ – Zeeshan Ahmad Feb 11 '15 at 15:26
  • $\begingroup$ @ZeeshanAhmad Yes, the governing equation covers that possibility. In fact, if it couldn't curve (meaning $y'(x) = 0$ somehow), the clamped boundary condition would imply that there is no deflection at all anywhere along the beam because that boundary enforces $y(O) = 0$$. Also note that this is for an Euler-Bernouilli beam. However, you'll find that the same ideas (can it move? can it rotate? are there forces or moments?) can be applied when you get to Timoshenko beams. $\endgroup$ – tpg2114 Feb 11 '15 at 16:02

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