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I want to affix a cantilever to wall. I will support the other end of the cantilever with a strut made of wood, that attaches to some point on the wall below the cantilever, as shown in this sketch (click for full resolution):

At what angle will the strut provide the greatest vertical strength/support for the free end of the cantilever?

Kreeverp
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    Adding a member as shown in the illustration means there is no longer any "free end" or cantilever beam. Once you add that member, the structure becomes a frame. Terminology aside, is the cross-section of the added member fixed or variable? What material are you using? Did you try any calculations? Without more details the answer is trivial: 90°. – Air May 11 '15 at 21:25
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    As Air alluded, the real constraint is how far down the "wall" you can go. – hazzey May 11 '15 at 21:33
  • Very interesting. If you are adding to the other end, it is no longer a cantilever. The material is wood - but I am interested in what exactly is the strongest way to support the loads on a cantilever? This tower https://www.google.ca/search?tbm=isch&q=niagara+falls+observation+tower has material added to the end, but I think it is still considered a cantilever. I do see that it has support coming out the other end - I am not interested in that. What I want is to affix to the wall - have no legs - and want to understand what is the best way to support the load. Thank you for helping me. – Kreeverp May 11 '15 at 22:00
  • 90° connected to the wall Air? – Kreeverp May 11 '15 at 22:14
  • I edited some of your clarification into the question. You can also edit, using the grey link underneath the tags on your question, which goes to [this page](http://engineering.stackexchange.com/posts/2780/edit). I was giving the angle between the tabletop and the support, so 90° would be a vertical column (struts are compression members) but that was not a serious answer since I suspected you wanted "no legs." – Air May 11 '15 at 23:16
  • As for "cantilever" it's no big deal - yes, the Niagara tower is a cantilever truss, and the more horizontal your added support is, the more your structure resembles a simple cantilever beam. – Air May 11 '15 at 23:34
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    @Air - providing a support doesn't mean that the word "cantilever" is invalid. It just means it becomes a "propped cantilever". – AndyT May 12 '15 at 12:37
  • @AndyT Yep, I was very lazy with that comment and I'll eat crow. There's also still a free end, technically. Practically speaking, a DIY like this should be overbuilt so it's idiot-proof, in which case these distinctions don't really matter, but I should have been more careful anyway. – Air May 12 '15 at 18:32
  • Yes... don't bother providing distinction to us idiots :D Just came here to learn about cantilevers and supports from experts. – Kreeverp May 12 '15 at 20:05

1 Answers1

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Assumptions

  • The angle between the wall and the strut is $\theta$
  • $a$ is the depth of the table top
  • $P$ is the weight on the table top, applied at the edge furthest from the wall
  • The strut will fail when it buckles, which implies $F_{\text{max}}=\frac{\pi^2EI}{L^2}$ where $L$, $E$ and $I$ are the length, the elastic modulus, and the moment of area, respectively, of the strut

Analysis

The axial force on the strut will be $F=\frac{P}{\cos\theta}$. The length of the strut will be $L=\frac{a}{\sin\theta}$. Combining both equations with the equation for buckling we have: $(EI)_{\text{required}}=\frac{Pa^2}{\pi^2\sin^2\theta \cos\theta}$.

$EI$ is the stiffness of the strut. The most efficient strut will be one for which $(EI)_{\text{required}}$ is minimized. The lowest $(EI)_{\text{required}}$ occurs when $\sin^2\theta \cos\theta$ is maximized and that is when $\theta=\sin^{-1}\sqrt{\frac{2}{3}}$ so the most efficient angle is $\theta\approx54.7^{\circ}$ strut

wythagoras
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Mandrill
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    The equations here are also useful even if 54.7° is infeasible. You can determine the required strut cross-section from $I = \frac{Pa^2}{E\pi^2 \sin^2\theta\cos\theta}$. For a square members $I = w^4/12$, where $w$ is the side length. – regdoug May 12 '15 at 17:53
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    I think it is very important to say that will be a $P tan\theta$ force trying to move the table top away from the wall and this force is higher than the weight itself when $\theta>45º$ (the angle increases as the strut is made shorter) so for $P=100$N and $\theta=54.7º$ this force will be $141$N. So be careful. – Mandrill May 13 '15 at 16:27
  • @Mandrill I am not seeing the point here. Firstly is $EI$ changing and depending on inclination $\theta$, and that is why you differentiated it? If it is constant, why did you differentiate it? Secondly, at 54.7 deg it is *weakest* but not strongest! – Narasimham May 14 '15 at 09:35
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    @Narasimham I should have said $EI_{required}$. 54.7º is the angle the requires the lowest EI value, all other angles requires a stronger strut. If the angle requires less EI it means a thinner strut can do the job. – Mandrill May 14 '15 at 11:42
  • Should one also consider the stress an anchor detail is going to experience? As you change the location of the lower reaction, the magnitude of the two reactions will change. – SlydeRule Apr 29 '16 at 16:15