We have an expression:
\begin{equation} B = \frac{FL^{3}}{3EI} = \frac{N m^{3}}{3 p m^{4}} = \frac{Nm^{3}}{\frac{N}{m^{2}}m^{4}} = m \end{equation}
With constants:
- \(B\): \(m\), deflection at the point of force application
- \(F\): \(N\), force applied
- \(L\): \(m\), distance between fixed point and point of force application
- \(E\): \(p=\frac{N}{m^{2}}\), elastic modulus
- \(I\): \(m^{4}\), second moment of area
As per measured:
- \(B\): \(9.15 \cdot 10^{-4} m\)
- \(F\): \(20N\)
- \(L\): \(9.373 \cdot 10^{-2} m\)
- \(I\): \(1.37 \cdot 10^{-10} m^{4}\) = \(\frac{WH^{3}}{12}\) = \(\frac{(6.25 \cdot 10^{-3})(6.4 \cdot 10^{-3})^{3}}{12}\)
Theoretical:
- \(E\): \(7 \cdot 10^{10} P\)
As calculated:
- \(B\): \(5.74 \cdot 10^{-4} m\)