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The beam is fixed on both ends and there is a link in the middle. There is a force F on the link.

My approach to solve this problem:

A free body diagram is cut at the link and made into 2 systems.

Horizontal forces are ignored.

Sum of force in vertical direction for system 1 on the left:

$A+L(link)-F = 0$ (1)

The sum of moments about A:

$ Ma+L*l-F*l=0 $

For the Right side I do the same as above.

The problem is I keep getting nowhere and running in circles. I calculated the displacement for the left side and also the right side.

I equaled them to tr and solve for unknown variables, as the displacement at the link (length l) has to be same for both the sides, but still can't cancel out the variables to solve for the reaction of the forces. My goal is to solve in terms of F for the reaction forces.

enter image description here

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    $\begingroup$ What did you do in that first equation? What's $L(link)$? Is it just the length from A to C? Why is it (a length) in a sum of forces equation? $\endgroup$ – Wasabi May 7 '17 at 20:23
  • $\begingroup$ Link L is a joint movable so the force F makes it go down! So its reaction force is up in a free body diagram. $\endgroup$ – gorobo May 7 '17 at 20:29
  • $\begingroup$ I cut the above Freebody diagram at the Joint C so i use the reaction forces equal and opposite on both the sides of the Beam. $\endgroup$ – gorobo May 7 '17 at 20:31
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    $\begingroup$ The stiffness of a cantilever is proportional to $1/L^3$, so it should be obvious the that the load $F$ splits into $8F/9$ and $F/9$, otherwise the displacements at C won't be equal. $\endgroup$ – alephzero May 7 '17 at 22:22
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The problem with your a approach is, that this is a statically indeterminate system. Basically, you could for example turn one fixed bearing into a simple support and the structure would still be statically stable. Remove another reaction force and the system turns into a mechanism. Thus the degree of static indeterminacy is $n=1$.

As a consequence, the reaction forces depend on the materials elastic properties and cannot be determined simply by solving the forces and moment equations. My approach was using the force method, assuming the elastic modulus $E$ and the area moment of inertia $I$ to be constant throughout the entire beam.

For this more general case of beam-lengths $a$ and $b$ I got the following solutions:

$$ A=F\frac{b^3}{a^3+b^3} \qquad B=F\frac{a^3}{a^3+b^3} $$

Thus for the case $a=L$ and $b=2L$:

$$ A=\frac{8}{9}F \qquad B=\frac{1}{9}F $$

PS: It may look like the solution is not depending on the elastic properties of the beam, but this is due to the assumption of a homogeneous beam, allowing me to cancel out $EI$ during my calculations.

force-method: (see attachment)

The principle behind the force method is, that you remove e.g. the moment restraint at point $B$ (replace it with a simple support, so it becomes determinate) and then calculate the moment distribution of this determinate system. Then, by implementing a unitary moment $X_1=1$, you calculate the required moment $X_1=M_B$ to satisfy the conditions of a fixed restraint: no rotation at $B$ $\quad \to\delta_1=\delta_{10}+X_1\cdot\delta_{11}=0 $

note: $\delta_1$ is the rotation at $B$, which must be zero, $\delta_{10}$ is the rotation caused by $F$ in the reduced system (marked as GS), $X_1\cdot\delta_{11}$ is the rotation caused by the moment $X_1$ (marked as ÜS), thus when you solve for $X_1$ you get $M_B$

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  • $\begingroup$ Did you solve it each part separatly? Moment at hinge should be zero no matter is it determinate or undeterminate beam. $\endgroup$ – Katarina Jan 11 '18 at 20:41
  • $\begingroup$ I added a more detailed description and a file, which shows the method I used. Your statement about the moment being zero at the hinge is correct. $\endgroup$ – Andrew Jan 12 '18 at 12:53

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