- #1
saybrook1
- 101
- 4
Hi guys; I have an analytical solution for the deformation of a beam due to a couple with moments C_1 and C_2 with boundary conditions y=0 and x=±(L/2) where L≡length of the beam. The derivation from the Bernoulli-Euler equation is below:
\begin{align*}
y''=\frac{\epsilon}{2}(C_1+C_2)-\frac{\epsilon}{L}(C_1-C_2)x
\end{align*}
Where ε = 1/E*I, I=Moment of inertia, E=Young's Modulus and L= beam length.
\begin{align}
y'=\frac{\epsilon}{2}(C_1+C_2)x-\frac{\epsilon}{2L}(C_1-C_2)x^2+C_3\\
y=\frac{\epsilon}{2}(C_1+C_2)\frac{x^2}{2}-\frac{\epsilon}{2L}(C_1-C_2)\frac{x^3}{3}+C_3x+C_4
\end{align}
For y=0 at x=-L/2;
\begin{align}
0=\frac{\epsilon}{4}(C_1+C_2)(\frac{-L}{2})^2-\frac{\epsilon}{6L}(C_1-C_2)(\frac{-L}{2})^3+C_3(\frac{-L}{2})+C_4\\
C_3=-\frac{\epsilon}{4}(C_1+C_2)(\frac{-L}{2})+\frac{\epsilon}{6L}(C_1-C_2)(\frac{-L}{2})^2+\frac{2C_4}{L}
\end{align}
For y=0 at x=L/2;
\begin{align}
0=\frac{\epsilon}{4}(C_1+C_2)(\frac{L}{2})^2-\frac{\epsilon}{6L}(C_1-C_2)(\frac{L}{2})^3+C_3(\frac{L}{2})+C_4
\end{align}
Plugging in equation (4) yields:
\begin{align}
C_4=-\frac{\epsilon L^2}{16}(C_1+C_2)
\end{align}
And plugging (6) back into (4) yields:
\begin{align}
C_3=\frac{\epsilon L}{24}(C_1-C_2)
\end{align}
With both integration constants determined, we arrive at our expression for vertical displacement:
\begin{align}
y=\frac{\epsilon}{4}(C_1+C_2)(x^2-\frac{L^2}{4})+\frac{\epsilon}{6}(C_1-C_2)(\frac{Lx}{4}-\frac{x^3}{L})
\end{align}This expression let's me come up with the displacement for an undeformed beam given certain values of C_1 and C_2. I was hoping that someone may be able to help me figure out a way in which I could take the displacement data(set of y-values) of a deformed beam and apply certain values of C_1 and C_2 in order to create a new displacement profile(new y-values). Any help or a point in the right direction would be greatly appreciated. Thanks and please let me know if you need any more information.
\begin{align*}
y''=\frac{\epsilon}{2}(C_1+C_2)-\frac{\epsilon}{L}(C_1-C_2)x
\end{align*}
Where ε = 1/E*I, I=Moment of inertia, E=Young's Modulus and L= beam length.
\begin{align}
y'=\frac{\epsilon}{2}(C_1+C_2)x-\frac{\epsilon}{2L}(C_1-C_2)x^2+C_3\\
y=\frac{\epsilon}{2}(C_1+C_2)\frac{x^2}{2}-\frac{\epsilon}{2L}(C_1-C_2)\frac{x^3}{3}+C_3x+C_4
\end{align}
For y=0 at x=-L/2;
\begin{align}
0=\frac{\epsilon}{4}(C_1+C_2)(\frac{-L}{2})^2-\frac{\epsilon}{6L}(C_1-C_2)(\frac{-L}{2})^3+C_3(\frac{-L}{2})+C_4\\
C_3=-\frac{\epsilon}{4}(C_1+C_2)(\frac{-L}{2})+\frac{\epsilon}{6L}(C_1-C_2)(\frac{-L}{2})^2+\frac{2C_4}{L}
\end{align}
For y=0 at x=L/2;
\begin{align}
0=\frac{\epsilon}{4}(C_1+C_2)(\frac{L}{2})^2-\frac{\epsilon}{6L}(C_1-C_2)(\frac{L}{2})^3+C_3(\frac{L}{2})+C_4
\end{align}
Plugging in equation (4) yields:
\begin{align}
C_4=-\frac{\epsilon L^2}{16}(C_1+C_2)
\end{align}
And plugging (6) back into (4) yields:
\begin{align}
C_3=\frac{\epsilon L}{24}(C_1-C_2)
\end{align}
With both integration constants determined, we arrive at our expression for vertical displacement:
\begin{align}
y=\frac{\epsilon}{4}(C_1+C_2)(x^2-\frac{L^2}{4})+\frac{\epsilon}{6}(C_1-C_2)(\frac{Lx}{4}-\frac{x^3}{L})
\end{align}This expression let's me come up with the displacement for an undeformed beam given certain values of C_1 and C_2. I was hoping that someone may be able to help me figure out a way in which I could take the displacement data(set of y-values) of a deformed beam and apply certain values of C_1 and C_2 in order to create a new displacement profile(new y-values). Any help or a point in the right direction would be greatly appreciated. Thanks and please let me know if you need any more information.