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Like
Mach
Attack, this question provides an opportunity to derive an
interesting engineering formula for yourself. You might see it
quite quickly, but don't be too concerned if you don't.
Understanding the formula $\psi =
\int \kappa ds$
Carrying on from the
hint,
$\int_{s=0}^{\phi r} \kappa ds = \int_{0}^{\phi r} \frac{1}{r} ds =
\phi$, the angle swept out at the centre of the circle. If we then
draw a straight line from $s=0$ to $s=\phi r$, we find that $\phi =
\psi$, where $\psi$ is the angle the beam has deviated from a
straight line. Thus the formula is proved for circles. For a more
general curve, as we zoom in around a point, we can approximate the
curve better and better by a circle, until in the limit of zooming
in to a single point it matches exactly.
Find $v$ in termsof $\kappa$ and
$x$
Deriving the formulae in this problem is all to do with looking at
small limits, i.e. considering a distance $\delta s$ along the
curve which is very small. This is a common trick in engineering,
and at the end we will integrate over all of these small sections
to get the full result.
When $\delta s$ is small, we can approximate the situation by a
triangle:
[Note about the minus sign in the above diagram: this is simply
convention that vertical displacement is positive up, but curvature
and angle are positive when the beam is bent down, such that the
top of the beam is in tension.]
Thus $- \delta v = \delta s \sin \psi (x)$.
For small deflections, such as we are considering, $\psi$ is small
so $\sin\psi \approx \psi$ and $\delta x = \delta s \cos \psi
\approx \delta s$. Thus $ - \delta v \approx \delta x \psi
(x)$.
To get total deflection we need to sum all these small deflections
along the length of the beam, which is where integrating comes in.
We find $v = - \int{\psi}dx$
And since $\psi = \int{\kappa}dx$, we find the formula $v = -
\int{\int{\kappa}}dx dx$. You could alternatively write this as
$\kappa = -\frac{d^2v}{dx^2}$.
You can apply this formula to a real problem in
Euler's Buckling
Formula.