Bridge Construction Problem - Buckling Bridge A 100 ft. bridge expands 1 in. during the heat of the day. Since the ends of the bridge are embedded in rock, the bridge buckles upward and forms an arc for which the original bridge is a chord. What is the approximate distance moved by the center of the bridge?
Starting points?
 A: In my mind's eye, the bridge is across a horizontal span:

The bridge originally spanned a gap of length $L$ from $P$ to $Q$, but after growing to $L+\delta L$, it has now buckled into the arc $PSQ$ with center at $O$. From right triangle $ROQ$ with hypotenuse $OQ$ of length $r$ and $\angle ROQ=\theta$, we see that $RQ=\frac L2=r\sin\theta$ and $RO=r\cos\theta$, so the bridge has buckled from $R$ to $S$, a distance of $\delta y=r-r\cos\theta$.
The bridge now spans arc $PSQ$ of length $2r\theta=L+\delta L$. Thus
$$r=\frac{L+\delta L}{2\theta}=\frac L{2\sin\theta}$$
Solving for $\theta$,
$$\theta=\frac{L+\delta L}L\sin\theta=\sin\theta+\frac{\delta L}L\sin\theta\approx\sin\theta+\frac16\sin^3\theta$$
Using the first two terms of the series for $\arcsin$ in this last equation. Then we get
$$\sin\theta\approx\sqrt{\frac{6\delta L}L}$$ So
$$\begin{align}\delta y & =r(1-\cos\theta)=2r\sin^2\frac{\theta}2=\frac{L\sin^2\frac{\theta}2}{\sin\theta}=\frac{4L\cos^2\frac{\theta}2\sin^2\frac{\theta}2}{4\sin\theta\cos^2\frac{\theta}2} \\
 & =\frac{L\sin\theta}{4\cos^2\frac{\theta}2}\approx\frac{L\sin\theta}4\approx\frac{\sqrt6}4\sqrt{L\delta L}=\frac{\sqrt6}4\sqrt{(100)\left(\frac1{12}\right)} \\
 & \approx1.76777 ft\end{align}$$
This is my first attempt at posting an image. I hope it doesn't look too bad. You can get the first two term of the $\arcsin$ series from trigonometry and algebra alone, but it's a tough slog.
You knew in advance the answer had to look something like this from the variational principle. The length of the bridge is
$$L+\delta L\approx L+\frac{\partial L}{\partial y}\delta y+\frac12\frac{\partial^2L}{\partial y^2}(\delta y)^2$$
We know that $\frac{\partial L}{\partial y}=0$ because the shortest path between two points is a straight line, so we get $\delta y\approx k\sqrt{\delta L}$. To get consistent units, we must multiply the right hand side by the square root of a length, and the only length visible in the problem is $L$, so we have $\delta y\approx\kappa\sqrt{L\delta L}$, and we find our numerical constant $\kappa=\frac{\sqrt6}4$.
A: DRAW A SKETCH.  The  vertical line is the original, the arc is the expanded bridge.  You have some right triangles and an arc-how do the lengths compare?  You wind up comparing $\cos \theta$ to $1$ for small $\theta$.  You want the distance along the $x$ axis between the line and the arc.

A: First we have to calculate the force due to temparature effects:
$
Force = E * Δl / l
$
Reference: http://www.setareh.arch.vt.edu/safas/007_fdmtl_29_stress_due_to_temperature_change.html
Then apply the theory of buckling where nice formulas will appear:
http://www.arpnjournals.com/jeas/research_papers/rp_2007/jeas_0207_35.pdf
*
See that we are missing E (the Young’s modulus of the beam) and I (the area moment of inertia of the beam’s cross section).
*
I recommend to change the tag to "physics".
A: The two equations seems to be simple: 
$r\theta = 2r*sin{\theta /2}+1/(254*30)$
$2r*sin{\theta /2}=100$
The move of the center is:
$d=r-r*cos\theta$
Let me know if this suffice...
