Bracket Connections

[davetrotter]

hi, ive devised a system whereby i intend to hang peripherals weighing approx 400kg from a steel bracket(see sketch). My concern is the possibility of the screw fixings pulling out of the wall. Can someone advise what loads are generated by this arrangement? my colleague suggested only downward loads are created and so pullout from the wall isnt a problem, but im not so sure.

thanks

 

[davetrotter]

ERROR: the load is 40kg not 400kg as stated

thanks

 

[handleman]

Your colleague should never be trusted again for any advice regarding anything mechanical. There will certainly be pullout forces on the screws.

If the load is 40kg, you should not mount this to drywall. It must be anchored in studs.

If there is a possibility of anyone standing under this 40kg, get someone qualified involved in the design.



-handleman, CSWP (The new, easy test)

 

[vpl]

Speaking from personal experience, not only must it be mounted into studs, the screws need to be long enough to ensure that it won't pull loose.

Patricia Lougheed

******

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[davetrotter]

can anyone tell me what loads are generated in each screw by this arrangement then? i require this info so the manufacturer can advise on a sufficient type of screw

thanks

 

[handleman]

I don't think there is enough information here to properly size a screw.

Practically, if this is not a safety-critical application and you are going to screw it to the wall (rather than putting screws into the wall and trying to hang it from them) then drywall type screws that are long enough to penetrate into the (assumed to be wooden) studs by at least 1.5" should be adequate.

If you try to put the screws in the wall first and then hang the thing from them, you have to size every screw so that it can hold up the entire load by itself. The reason is that load sharing cannot be even due to variations in screw location/depth in the wall.

-handleman, CSWP (The new, easy test)

 

[desertfox]

Hi davetrotter

I have uploaded a file which will give you the max load on the screw, the screw is located at the very top of the vertical leg, if you require the load in any other screw just multiply [μ] in the file I have uploaded by the distance from the pivot point.
I have assumed there is only one screw acting at each dimension you have indicated.
Now I have assumed that the bracket is rigid and all the load is taken by the screws, also you need to find the shear load on the screws which is simply the 40kg * 9.81/5.
In practice of course the bracket itself will deflect but I don't know how long your bracket is in the other plane, nor do I know how the 40kg is applied is a point load or is it actually a distributed load over some length however I think you need to worry more about the bracket deflecting,then the load in the screws.

desertfox

 

[EspElement]

What desertfox calculated is the 40kg (88 lb) coming down on the bracket is now creating a torque point at the bottom corner.. then the load is calculated by a rotational force out of the wall at the differant screw positions.. the higher (farther away from the corner) you position the bolts.. that less load on them due to mechanical advantage

4mm thick doesnt seem to be very much for the load u are putting on it.. as fox said, it really depends on how wide your bracket is on the other plane.. u should gather the inertia of the flat of the bracket that is taking the load and determine how much deflection will be generated with a simple beam calc

 

[Ltwine]

In order to call it a bracket, you will need a diagonal connecting from near the load point to the vertical member.
As pointed out by desertfox, the screws have less problem than the horizontal arm, which will end up in the vertical position upon loading.

 

[desertfox]

Hi

EspElement, Actually the fixing furthest away from the pivot point in this case takes the most load.
Rather than I ramble on and explain it have a look at this example:-

http://school.mech.uwa.edu.au/~dwright/DANotes/threads/fatigue/exampleB.html#top

regards

desertfox

 

[davetrotter]

thanks for the replies, and desertfox for the uploaded calculations, most helpful

 

[davetrotter]

one further query

my screws have a capacity of 9N and so the desertfox calcs prove the top screw fails(12.55N). am i right in thinking though this screw cannot fail unless the next screw does and the next and so on...what im trying to say is, i have 5 screws @ 9N = 45N total....the total force in all the screws adds up to 32.1N, therefore the screws cannot fail. im probs way off the mark but was trying to think of it logically

thanks

 

[flash3780]

Dave,
You'll be better to add screws. Desertfox's calculations look correct, so it suggests that the top screw will fail.

If you eliminate that screw (since it fails) and perform the calculation again, you'll notice that the load on the rest of the screws goes up, and more screws fail. The whole thing will let go.

Also, have you considered the bending on your bracket (s=My/I or s=6M/bh^2 for a rectangular x-section). You'll want to make sure that you don't fail your bracket in that way as well.

-Chris

 

[desertfox]

Hi davetrotter

You need either bigger screws or more of them.
The screws should be sized on the one carrying the greatest load, which is the one at the top furthest from the pivot point.
Think of it this way, the top srew fails and therefore can carry no load, so now you have 4 screws and they have to carry the additional load, further the value of ? would need to be recalculated for carrying the full moment on 4 screws,so its a domino effect working against you.
What size screws are you using to only be rated at 9N each?
The reason I ask is you have a 40kg=400N approx component to support.
More importantly I think you have a even bigger problem with the bracket,we could do with some more details.

desertfox

 

[davetrotter]

the wall material is extremely poor so im sacking the idea off and getting a desk. thanks for your help, much appreciated.
regards
dave

 

[desertfox]

hi davidtrotter

Okay your welcome, sounds like we have saved you from a headache anyway.

desertfox

 

[EspElement]

Desertfox, Thank you for ur posts i was misled by this application because of the advantages of torque and not seeing the true uses.. I am going to break down that document you posted later.. cause im not grasping all of the concepts on it.. I am going to school for engineering and work as a mechanical designer currently.. been learning the ropes thru the good ole machinaries handbook! This is obviously a much more indepth approach

So if you dont mind i would like to ask a few questions im sure u can explain pretty easy.. i understand the farther away from that point the more stress is on that bolt due to the angle that is applied verses if being close.. but lets say i put a bunch of bolts far away from the rotational point there.. would the factor of shearing come into play more since its not being allowed to rotated do to the mechanical advantage there?

 

[desertfox]

Hi EspElement

If you put a horizontal line of bolts at the furthest point on the bracket, then each of those bolts will equally share the tensile load due to the moment.
The shearing forces however act at right angles to the bolt heads and whatever the total vertical force down ie:- 40kg * 9.81 N in this case here, then you just divide the total number of bolts into the 40kg*9.81N, that would be the case even if the bolts are set at different vertical positions, in this analysis its only the tensile force in the bolt due to rotating of the bracket that changes, due to the bolts vertical position relative to the pivot point.

regards

desertfox

 

[flash3780]

You may wish to read up on Newton's Laws of Motion (second law in particular) as well as D'Alembert's principle.

When you talk about shearing, you're referring to the force on the components that would make them fail in the same way as a shear pin on your lawn mower blade. In the case of your bracket, the shear force on the bolts is simply the total load divided by the number of bolts.

What you're seeing in desertfox's calculations is the moments summed about a point (a moment or torque is simply a force that is applied at a distance away from your summation point).

You'll pick this stuff up when you take your statics/dynamics classes.

Two things that you always need to remember when designing something:
You can't push a rope, and F always equals m x a.

 

[EspElement]

I guess because ur example was an ecetric load.. it created alot of strange vectors on the bolts on the bracket? I cant wait to learn that stuff.. i can learn on my own.. and i should so im prepared for classes! lol