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I want to mount two patio heaters on my terrace in front of a window wall. I cannot attach anything directly to the window wall so I am going to run an aluminium box beam between two brick columns. The beam will be 5000mm long. The beam dimensions are 80mm x 40mm with a 4mm thick wall, 13kg in weight. The heaters weigh 6.5kg each and are 1000mm long. They will be 1000mm from each end and therefore 1000 from each other. I need to drill 8 x 8mm holes for mounting brackets (2 brackets per heater, 2 holes per bracket).

I am sure the beam will deflect a little - any serious risk of collapse? Patio Heater Attachment Points Proposed Positioning

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    $\begingroup$ Welcome to Engineering! This looks like a "homework question" (notice the quotation marks). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$ – Wasabi Mar 20 '17 at 11:09
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    $\begingroup$ This is either a homework problem or something for which you should consult a structural engineer $\endgroup$ – DLS3141 Mar 20 '17 at 12:52
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    $\begingroup$ The weight of the heaters is just a rounding error, compared with what happens when somebody decides to sit on the middle of the beam - or stand on it to clean the windows! $\endgroup$ – alephzero Mar 20 '17 at 16:05
  • $\begingroup$ What is the maximum allowable bending stress of the aluminum you have? $\endgroup$ – Andyz Smith Mar 21 '17 at 13:08
  • $\begingroup$ Related: engineering.stackexchange.com/questions/10957/… $\endgroup$ – Andyz Smith Mar 21 '17 at 13:36
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Load equations

Using the 5th equation from the bottom the displacement is:

disp = Wa^3b^3/3EIL^3

We can superimpose it for each of your loads. Essentially your beam is loaded with 3.25kg (~33N) at 4 points (1,2,3,4m along).

I.e. disp(total) = disp(1st load) + disp(2nd load) ... + disp(4rth load)

For all the loads W = 33N

for the first and fourth a = 1 and b = 4 for the second and third a = 2 and b = 3

E = 69 x 10^9 (Elastic Modulus of Aluminium)

I = 7.11x10^-7 (Second moment of Area)

Ends up being:

disp = 0.001m

So at the middle x = 2.5

So displacement = 0.0025m

Sounds like it should be fine.

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  • $\begingroup$ Technically shouldn't you use distributed loads, not point loads? $\endgroup$ – JMac Mar 20 '17 at 14:04
  • $\begingroup$ @JMac If the heaters are supported by brackets, and not actually touching the beam in between the brackets, the loads are not distributed. But since the OP didn't tell us which of the 80 and 40mm dimensions is the width and the depth of the beam, that would make a lot more difference than using point loads or distributed loads. And we don't know how the beams are attached to the brick pillars either. $\endgroup$ – alephzero Mar 20 '17 at 15:56
  • $\begingroup$ As alephzero said - theyre bolted to the beam at defined points so it's definitely a point load. As for the beam orientation I just assumed he'd used the longest axis in the direction of loading and that the beam was bolted to the brick. I should probably have stated that explicitly. $\endgroup$ – m4p85r Mar 20 '17 at 23:20
  • $\begingroup$ Correct - I intent to have the longest axis be vertical. Beam will be bolted to brick with SS brackets. May be some minor twisting force as heaters will be on face not below beam. $\endgroup$ – James Cotton Mar 21 '17 at 13:23
  • $\begingroup$ I added a more precise diagram that shows the exact points of loading, but I think as @ashgetstazered pointed out, the amount of deflection should be quite tolerable. $\endgroup$ – James Cotton Mar 21 '17 at 14:01

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