<PERSON> Fixed Fixed Total Beam Deflection
I am trying to solve a problem where there is a fixed-fixed (clamped-clamped) beam with a point load in the middle (at L/2). The L/H is less then the 10 required for normal beam bending analysis and then results in, what I believe requires, <PERSON> beam analysis.
I went a head and attempted to derive the total displacement in a similar fashion to what would be done in a normal beam bending problem. I start at the moment equation (2nd derivative of displacement, which I will be using the symbol of $\delta$ for.):
$$M(x) = \frac{d^2\delta}{dx^2}=\frac{d^2\delta_{bending}}{dx^2}+\frac{d^2\delta_{shear}}{dx^2}=\frac{Fx}{4EI}-\frac{M'}{EI}+\frac{K}{GA}\frac{dv}{dx}$$ We integrate: $$\theta(x)=\frac{d\delta}{dx}=\frac{d\delta_{bending}}{dx}+\frac{d\delta_{shear}}{dx}=\frac{Fx^2}{8EI}-\frac{M'x}{EI}+\frac{KV}{GA}+C_1$$ We know that the shear $V$ is $=\frac{F}{2}$
We integrate:
$$\delta(x)=\delta_{bending}+\delta_{shear}=\frac{Fx^3}{24EI}-\frac{M'x^2}{2EI}+\frac{KFx}{2GA}+C_1x+C2$$
We can apply some boundary conditions:
$$\delta(0)=0=C_2$$
We apply another boundary condition: $$\theta(0)=0=\frac{KF}{2GA}+C_1 $$
This is the first spot where I am uncertain. 955
I believe that $C_1 = \frac{KF}{2GA}$, not $C_1 = \frac{-KF}{2GA}$ as if it is the 2nd case all of the shear displacement goes to 0. So I continued assuming $C_1 = \frac{KF}{2GA}$
$$\delta(L)=0=\frac{FL^3}{24EI}-\frac{M'L^2}{2EI}+\frac{KFL}{2GA}+C_1L$$ $$\delta(L)=0=\frac{FL^3}{24EI}-\frac{M'L^2}{2EI}+\frac{KFL}{2GA}+\frac{KFL}{2GA}$$ $$\delta(L)=0=\frac{FL^3}{24EI}-\frac{M'L^2}{2EI}+\frac{KFL}{GA}$$
Now we solve for M'
$$M'=\frac{2EI}{L^2}(\frac{FL^3}{24EI}+\frac{KFL}{GA})$$ $$M'=\frac{FL}{12}+\frac{2KFEI}{GAL}$$
We plug this back into the $\delta(x)$
$$\delta(x)=\frac{Fx^3}{24EI}-\frac{\frac{FL}{12}+\frac{2KFEI}{GAL}}{2EI}x^2+\frac{KFx}{2GA}+C_1x$$ $$\delta(x)=\frac{Fx^3}{24EI}-\frac{\frac{FL}{12}+\frac{2KFEI}{GAL}}{2EI}x^2+\frac{KFx}{2GA}+\frac{KFx}{2GA}$$ $$\delta(x)=\frac{Fx^3}{24EI}-(\frac{FK}{AGL}+\frac{FL}{24EI})x^2+\frac{KFx}{GA}$$
$\delta$ max should be at $\frac{L}{2}$. So we plug it all back in. 374