I'm struggling with this very basic equation, hopefully someone with a fresh mind can help me out:
So Eulers beam theory for a simply supported load with 2 equal symmetric point forces is given by:
$\delta=\frac{Pa(3L^2-4a^2)}{24EI}$ in relation to this diagram (http://www.structx.com/Beam_Formulas_009.html)
Using a given a=0.005 m, L=0.017 m, P = 207N, E of 3.8E9 Pa, and I of 1.257E-11 $m^4$ , I can calculate $\delta=1.39E-03$, where P is calculated using a known yield stress.
Now, if i want to work backwards to derive strain from a given displacement, i can do the following:
$\sigma = E\varepsilon = \frac{P}{A}$, $I=\frac{\pi r^4}{4}$, $A=\pi r^2$
Rearranging Beam theory above, I get:
$P=\frac{24EI\delta}{a(3L^2-4a^2)}$
Subbing in the above 3 into the rearranged beam theory, I should get:
$\varepsilon=\frac{6r^2\delta}{a(3L^2-4a^2)}$
Now using the same yield stress used above to calculated the force, I get $\varepsilon = 4.342E-03$.
Using Hookes law: $E=\frac{\sigma}{\varepsilon}$, I get a calculated E value of 3.8E10, a result 10 times greater than the E i started off with.
I CANNOT workout why this is happening, if anyone could shed some light on this, my headache will really appreciate it.
Thankyou
You used a wrong formula to calculate the stress in the beam.
In your calculate, you used $\sigma =\frac{P}{A}$. This is assuming that the stress is uniform in the cross section of the beam.
However, you are dealing with a beam deflection problem. The stress is linear varied through the cross section. The max formula is:
$\sigma = \frac{My}{I}$
where $M$ is the moment at the cross section. $y$ is the coordinate in the vertical direction of the cross section.