Hi everybody
The image below is a cross section through a glass door, and the door can be up to 3M wide x 3M tall which would then weigh 360KG.
If the door weighed 360KG and the plate at the base was 3MM Stainless Steel Flat x 3M in Length how would I calculate the deflection of the plate at the pivot point?
And, what if the plate was 3MM aluminium or 5MM Aluminium or the door was 200KG, etc…
Any help would be much appreciated.
Why would you place a pivot point under the weakest part of the assembly?
Let me understand: Is that 50mm of insulated glass with a spacer that’s 6mm of glass one side and 10mm the other? You need to know the weight of the 6mm and the 10mm plates separately.
If you make each of the objects a solid in SketchUp, you could get the volumes and, with the density of the materials you could get their masses and the combined mass along with centers of gravity using TIG’s Center of Gravity plugin. Maybe that would get you started.
This is a sliding door and the pivot point represents the wheels, and generally there are 2 each at either end.
The weight for each sheet is calculated as Glass Thickness in MM X Surface Area in Square Metres x 2.5.
So in the 3M x 3M example, the 6MM sheet weighs 135KG & the 10MM weighs 225KG.
I disregard the weight of the spacer bar.
Hi Dave
I can calculate the masses fairly easily, i.e. surface area of glass in metres squared, X glass thickness in mm, X 2.5
Its the plate at the bottom I am interested in.
Understood. You would need to do some finite element analysis which SketchUp can’t. Knowing where the CofG of the assembly will help.
Thanks very much for that - I will study it with interest.
But aside from calculating the loads, what I really want to know is what effect those loads have on the plate, and what thickness does the plate have to be to carry those loads, and whether I can get away with aluminium or does it have to be steel? Obviously a strip 10mm thick would support the door easily but I want to know the minimum thickness I can get away with safely. Ideally I would like to use 3mm.
Hopefully that makes sense.
Looking at this quickly, the maximum deflection won’t be at the centre if that’s the pivot point—it will be at the free end. You should treat it as a cantilever, with the pivot point fixed and the free end supported by the 3 mm beam carrying the glass.
If the calculated deflection under the expected load exceeds allowable limits, the beam will need to be strengthened. From experience, 3 mm stainless steel over a 3 m span at this loading seems far too flexible.
Assumptions made:
For a quick check, you can model half the door (1.5 m) with half the weight (180 kg) acting as a point load at the free end. This will give an idea of deflection. My gut feeling is that a 3 mm stainless section at 1.5 m with a 180 kg point load at the end will be too flexible.
Additional factors to consider include how the plate is fixed, as load transfer and the stiffness of the glass panels themselves will influence overall performance.
An angle section of stainless would be much better to resisting bending.
Edit: Seconds thoughts on this
The glass wouldn’t act as a point load so you can look at it as a UDL load along a cantilever which would give a better representation on how the glass load will act on the beam.
Thank you very much for your response.
I am going to draw an axo view of the proposed setup this evening, to provide a clearer explanation of the problem.
I don’t think I have explained it clearly.
Oh, wait, is this a pivot door, i.e. hinged in the middle so the whole door weight comes to one point there?
No - in hindsight that was a poor description by me. The door is supported on 2 double wheel carriages either end, and the carriages themselves are fixed to the plate. I will draw it up again this evening to make the setup clearer
I’m wondering if the rigidity of the glass itself has to be factored in. That is, if the glass doesn’t sag the support won’t bend.
This looks like a question for a qualified engineer.
Ah. Good question…
Yes pretty much, i think what is suggesting now is there are two roller wheel each end and wants to know if it will sag between them. The glass is large so wont bend when standing on its side. Effectively the plate at the bottom will be a way of fixing the wheels and hold the glass in place. The load will be the wheels.
The way to look at it is would the glass take its own weight if the 3mm plate wasn’t there..? probably yes so the plate isn’t doing much its just a way to connect it all.
Best way to start is to draw a free body diagram, see where there the loads are and where the reaction forces are.
Do you have a 3D model? If this are sliding glass doors they should have railing if they’re not sliding doors the pivot should be on a different place and if the doors are heavy they should have also some railing or wheels mechanism.
Kevin? why don’t you build it as a 3d model in sketchup… even if it is rough, post it here and most of us will be able to more effectively help.. I mean, that is the advantage of SketchUp.. 3d modelling !
PS wouldn’t the most severe loading be on the other plane, mid-point between the roller wheels.
(and the solution add more roller wheels ?)
glass slider.skp (69.0 KB)
PS can you use 6mm glass in a frame 3 x 3 metres in size? I used to work in my Dad’s glazing business… not sure I would want to handle that !
