# What angle - for a strut - provides the greatest vertical strength/support for a cantilever?

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?

• 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
• 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 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. May 11 '15 at 22:00
• 90° connected to the wall Air? May 11 '15 at 22:14
• @Air - providing a support doesn't mean that the word "cantilever" is invalid. It just means it becomes a "propped cantilever". May 12 '15 at 12:37

## 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 $\theta\approx54.7^{\circ}$ • 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. May 12 '15 at 17:53
• 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. 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! May 14 '15 at 9:35
• @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. 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. Apr 29 '16 at 16:15