I have a simple wall mount I want to analyze using hand calculations (pure beam bending if possible) that looks like this:

enter image description here

I want to be able to model it as if it was in a state of pure bending with a point load at the very end of the single bar section. The two bar section is fixed to the wall. However, I'm a bit confused with how to treat the "pin" (for lack of a better word) that connects the two sections together. Obviously, it's purpose is to transfer load to the two bars mounted to the wall, but I'm not sure if it would be a poor assumption to assume 50% of the load is transferred to each bar, and that the only force transferred through the pin is a straight downwards force and a moment.

I know it's probably a pretty basic/simple problem, but I was hoping someone could provide some insight on how I could deal with the pin and the two bars using hand calculation. Thanks!


2 Answers 2


Let's say your tv is 50lbs and the middle lever is 24"long and 6" high and the distance the hinge or shaft extends beyond on top and bottom before they reach the support members on the wall is also 6".

The moment is $ M= PL= 50*2' =100lbsft \$

and the shear is 50 which is most likely only supported by the lower arm of the support.

This moment will also cause pulling, tensile stress on top which wants to bend or cut the shaft, and compression on the bottom which again wants to push the shaft in.

if we call the tensin T and compression C $ \quad T=C=100/(6/12)=200lbs $

Thes forces need to be supported by the two brackets on the wall. a shear of 50 lbs and pulling on the top and pushing on the bottom of 100lbs each, assuming they are also 6" wide.

  • $\begingroup$ Hi, thanks for your input. I was wondering why you suggest that the shearing is only supported by the bottom arm? $\endgroup$
    – Nick6373
    Commented Mar 29, 2020 at 23:38
  • $\begingroup$ Unless you have special bearings, like lateral weight bearing on top one won't block sliding down. $\endgroup$
    – kamran
    Commented Mar 29, 2020 at 23:49

This answer addresses design concerns as well as structural reactions.

As the pictures are shown below, $T, C, R$ are the global reactions required for the anchorage design, and $V, M$ are internal reactions for checking the connection rod/pin and the tube steel. Note the local reactions $H$ & $m$ may not be significant comparing to $V$ and $M$ if $d$ is small, but do check them anyway.

Do not forget to add adequate safety factors.

enter image description hereenter image description here


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