# Boundary conditions of an elastic bar

I was following some online lecture relating to an elastic bar with length $L$ that obey the differential equation $\displaystyle \frac{d^{2}u}{dx^{2}} = f(x)$, where $f(x)$ is its own weight or some load. The professor during that online lecture states that $u(0)=0$, $u(L)= 0$ as a boundary condition if the bar is fixed on both ends (this part I got it). But if one end is free then boundary conditions will change to $u''(x) = f(x)$ and $\displaystyle \frac{du(0)}{dx} = 0$ (?). Anyone could tell me why is that the case? How come boundary conditions changes from $u(0)=u(L)=0$ fixed end to $u’(0) = 0 = u(L)$ free end? Thank you!

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At a free end there can be no force imparted. You also need to distinguish between a supported end and a fixed end. A supported end forces the end to stay in position, so $u=0$, but $u'$ can be non-zero. A fixed end forces the end to stay in position and the bar does not bend, so $u=0, u'=0$. As a free end may move, you may have $u \neq 0$, but as there is no force you must have $u''=0$. $u'$ at a free end is set by the other conditions and can be non-zero (think of a bar with one end fixed and horizontal and the other end free-it will sag, so $u'$ is non-zero at the free end). The two ends of the bar may have different conditions, so apply the correct one at $0$ and $L$.
The primes are $\frac {d}{dx}$, not $\frac {d}{dt}$. $u'$ is the slope. A bar simply supported at the ends will sag in the middle, so $u'(0) \lt 0, u'(\frac{L}{2})=2, u'(L) \gt 0$. A bar fixed at the zero end and free at $L$ will have $u'(0)=0$ because of the boundary condition, but $u' \lt 0$ for all other points. $u''$ is the bend in the bar. – Ross Millikan May 24 '11 at 17:27
@Heber: good question. It is a typo an you are right $u'(L/2)=0$ in that case. – Ross Millikan May 24 '11 at 22:26