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?

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    $\begingroup$ 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°. $\endgroup$ – Air May 11 '15 at 21:25
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    $\begingroup$ As Air alluded, the real constraint is how far down the "wall" you can go. $\endgroup$ – hazzey May 11 '15 at 21:33
  • $\begingroup$ 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 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. $\endgroup$ – Kreeverp May 11 '15 at 22:00
  • $\begingroup$ 90° connected to the wall Air? $\endgroup$ – Kreeverp May 11 '15 at 22:14
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    $\begingroup$ @Air - providing a support doesn't mean that the word "cantilever" is invalid. It just means it becomes a "propped cantilever". $\endgroup$ – AndyT May 12 '15 at 12:37


  • 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


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

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    $\begingroup$ 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. $\endgroup$ – regdoug May 12 '15 at 17:53
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    $\begingroup$ 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. $\endgroup$ – Mandrill May 13 '15 at 16:27
  • $\begingroup$ @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! $\endgroup$ – Narasimham May 14 '15 at 9:35
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    $\begingroup$ @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. $\endgroup$ – Mandrill May 14 '15 at 11:42
  • $\begingroup$ 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. $\endgroup$ – NamSandStorm Apr 29 '16 at 16:15

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