How much does a steel spring wire bend due to it's own weight? I'm working on an art project and I need to know how much a steel spring wire bends dependent on it's diameter. It's point of support is in the middle and it's lengths is 1.20m. How much lower are the end points in relation to the center?
1. A diameter of 1.20mm
2. A diameter of 1.50mm

Edit:
I've found the following values for spring steel: $p = 7800 kg/m^3$, $E = 2.1×10^{11} N/m^2$. But I've trouble reproducing your results, I'm not sure if I assembled you equation right.
Edit
Corrected formula in W|A
 A: The approach to this question would be with Euler-Bernoulli beam theory, which in your case simply says that the beam's vertical deflection $z(x)$ satisfies the equation 
$$EI \frac{d^4 z}{dx^4}= \lambda g$$
where $\lambda$ is the linear density of the wire, $E$ is the wire's Young Modulus, and $I$ is the second moment of area. For a circular wire of radius $R$, you have $I=\frac{\pi}{2}R^4$ and $\lambda = \pi R^2 \rho$ where $\rho$ is the volume density. Plugging that to the equation yields
$$\frac{E R^2}{2 \rho g}\frac{d^4 z}{dx^4}= 1 $$
Solving this equation with the right boundary conditions gives
$$z(x)=\frac{\rho g}{12 E R^2}x^2\left(x^2-6 L^2 \right) $$
and therefore the deflection at $x=L$ is $$z(x=\pm L)=-\frac{5 L^4 \rho g}{12 E R^2}$$
I don't know what do you mean by tensile strength, but I suspect that it is not enough in order to determine $E$. Instead, if you take $E=200\times 10^9 N/m^2$ from wikipedia and $\rho=8000 kg/m^3$ (also from wikipedia) and plug that all in, you'll have ~1m for the 1.20mm thick wire and 60cm for the 1.50mm thick wire.
Edit: I forgot to carry $g$ in my answer, and therefore my previous answer was wrong by a factor of $g$. I now updated the values, which are large compared to the length of the wire. This makes the beam-equation inaccurate, and one should solve the problem without assuming small deflections. It is possible, and my (new) answer gives a fair estimation of the result.
