Pressure done by a bar/beam inserted in a continuous medium (wall) Related to a DIY project, I'm facing the following question: 

Assume a metal bar (beam) of square section,  $a \times a$, is inserted partially in a continuous medium (lets say, a wall) in a way that $h$ of the beam is inside the medium and $l$ is outside of the wall, and a force $F=200 N$ is done in the other extreme of the bar, perpendicular to it. The question is: how the pressure ( the force per unit area, $N/mm^2$ ) is distributed in the contact surfaces between the cantilever beam and the wall ?
Note: we could say that the beam has no weight, no friction forces and the system is stable, no displacements. An infinite wall in width, height and thickness. If easier, we can consider a beam of circular section instead of square.
Inside the wall the beam has 5 surfaces: right, left, top, bottom and back. In the ideal case of no friction, I think forces in right and left surfaces will be null. No idea if forces in back and top are also null. Force in bottom must allow to keep fixed all system, with an unknown distribution of pressure on it ( uniform? maximum at $h$ deep? ). 
I've no knowledge to solve it as a continuous problem. A strong simplification that came to my mind is consider the system as equivalent to a lever with lengths $h$ and $l$, being the support axe the red line in the draw. That means to the force in the other extreme of the beam (green line in the draw) will be $F' = F \frac {l}{h}$. By example if h=5cm and l=50 cm, then $F'=200*50/5=2000N$. 
Background: if we stand a TV of 20 kg using a $l=50 cm$ beam, how to estimate the forces over the wall ?
 A: I'm not sure what you mean by distribution of the "pressure", but this is a static problem involving a cantilever beam. The beam inside the wall is ignored. First from a static equilibrium perspective, there has to be an upward reaction force exerted by the wall of 200 N for vertical equilibrium to counter the downward 200 N force. There also has to be a counter-clock wise moment reaction at the wall of 200N x 50 cm = 10000 N.cm to counteract the clockwise moment (the "rotational effect") of 10000 N.cm due to the 200 N load.
The beam is subjected to an internal bending moment as well as a vertical shear force throughout the exposed portion.  The vertical shear force divided by the beam cross section is the shear stress. Stress has units of force per unit area, just like pressure. So maybe that is what you are thinking of. The bending moment creates tensile and compressive stresses at the extremes (top and bottom bottom of the beam, respectively).
Bending moment is a maximum at the wall and zero at the free end. Shear force is constant throughout and equals 200 N. If you look at the shear and moment diagrams for the cantilever with a force at the end, you will see the shear is constant and the moment linearly increasing progressing from the free end to the wall.   
This is the best I can give you without more detail, and without being sure what you mean by pressure. You really need to take a statics and mechanics of materials course to progress.
Hope this helps.
A: It is a good pratice in structural engineering to check that both the beam material and  the material supporting the beam can withstand the loads. The admissible pressure for the wall of course  depends on the type of material and could vary from say few tens of bars for bricks to over 100 bars for concrete. Solving exactly your problem might not be so easy.  A possible simple approach is to model the wall material as a distribution of very stiff springs reacting at the bottom and on the top of the beam to the moment and shear stress imposed by the load. I would suggest to use the reacting stress on the beam as shown in  diagram a). The resulting reactions on the beam are shown in b). The 2 equations of equilibrium (moment and resultant) make it possible to solve for $R_{top}$ and $R_{bot}$. Since $R_{bot}=R_{top}+ P$, the max stress will be at the bottom and worth $\sigma_{max}=2R_{bot}/ha$, $ha$ being the surface of the beam resting on the wall material.
A: This shows a combination of a steel rod and a concrete wall in 3D. The rod is embedded in the wall 1/6 of the length. Section of a rod 1x1. We take the effective force divided by the cross-sectional area as 1. The middle picture is the distribution of the vertical component of the deformations in the wall; in the right-hand picture, this is the distribution of the $\sigma_{zz}$ stress component. It can be seen that in such a situation, a stress increase of 30 times can be obtained. The calculations are performed using FEM and Mathematica 12.

