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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, Timoshenko 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. 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.

$$\delta(\frac{L}{2})=\frac{F(\frac{L}{2})^3}{24EI}-(\frac{FK}{AGL}+\frac{FL}{24EI})(\frac{L}{2})^2+\frac{KF(\frac{L}{2})}{GA}$$

$$\delta(\frac{L}{2})=\frac{FL^3}{192EI}-(\frac{FK}{AGL}+\frac{FL}{24EI})(\frac{L^2}{4})+\frac{KFL}{2GA}$$

I then run a simple simulation to check the validity of the equation. $L=30mm,H=8mm,T=2mm,$6061 $(E=69GPa)$,$F=50N,K=\frac{5}{6},v=0.33$

I get an anticipated displacement of $-0.815\mu m$, which seems inherently wrong as normally a positive number indicates a displacement in the direction of the force vector. In addition when I ran the simulation I got a displacement of $1.98\mu m$ in the direction of the force vector.

So I am not confident in my approach on this. I believe there are a few locations I might have made mistakes but I am unsure:

  1. I am not confident in the $C_1$ value.
  2. I am not confident in that I took a whole indefinite integral of the whole "equation", I am unsure if the operation needs to be more like: $\int (\frac{Fx}{4Ei}-\frac{M'}{EI})+\int \frac{K}{GA}\frac{dv}{dx}$ which would result in 2 constants per integration.
  3. I am not confident that I have applied the Timoshenko beam analysis properly on this problem.

Any insight would be much appreciated!

Fixed Fixed Beam With Point Load At L/2 enter image description here

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1 Answer 1

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I was able to get a hold of Roark's Formulas for Stress and Strain (9th Edition), and there is a section for (8.10) Beams of Relatively Great Depth.

For this problem that is end supported, center load (I have changed variables to match my problem variables):

$\delta_{shear} = \frac{FLK}{4AG}$

So the total deflection is

$\delta_{max}=\delta_{bending}+\delta_{shear}=\frac{FL^3}{192EI}+\frac{FLK}{4AG}$

I ran a handful of simple simulations and I was getting less then 2-3% difference between the hand calculations vs simulations. I played around a bit trying to understand why $\delta_{shear}$ has a $\frac{1}{4}$ in it. I believe it is because the max moment of the shear is at the middle and is $\frac{F}{2}*\frac{L}{2}$. However I have not dug into that deeply. I hope this answer is able to help some one!

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  • $\begingroup$ It has 1/4 in it, because compared to one end fixed with the other one free with the force, your case is like 2 of these (twice the stiffness), but with halved lengths (again twice the stiffness). $\endgroup$ Sep 20, 2022 at 16:21
  • $\begingroup$ That makes sense, but I wasn't quite able to follow the steps from equation 8.10-1 to the end equation. This is what I was thinking but not sure: From the text book: Ys = F * Integral (V* v)dx/(AG) (F is form factor, V is vertical shear due to actual loads, v is vertical shear due to unit load, A area, G is modulus of rigidity). As you integrate and and get FVx/AG + C. y(0)=0 C=0. Max deflection is at x = L/2 y(x) = F* V* (L/2)/AG However to highlight "V is the vertical shear due to actual loads", so I belive Vshear = Vload/2 as Vshear max is Vload/2. y(x) = F*(Vload/2)*(L/2)/AG $\endgroup$ Sep 21, 2022 at 3:05

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