Cantilever Bracket Calculations

rich866

I have a cantilever bracket, fixed at one end, free at the other.
One vertical element AB, 330mm long with a fixing at either end to a wall.
One horizontal element CD, 375mm long.
CD is fixed at 90deg to AB, 130mm from the top.
A constant, downward load W=30kg, is applied to the end of CD, l=375mm.

What is the force trying to pull the top fixing from the wall?
Does the bottom fixing work as a pivot point making this a torque calculation?

Assumptions:
The strength of the bracket will well exceed the failure rate of the top fixing, so no flex in the 40mm box sections will occur.

This bracket has been installed on a plaster board wall. The chemical fixings will take 9Kn but the plaster board will fail before this point, long before!

Before a set up a chemical anchour in the mock up of the wall to test the fixing/ board to failure I would like to know at what point I exceed my working force of the bracket/ anchour interface??

Vadar2012

Can you please give us a diagram. It's kinda hard to see what's going on. We shouldn't have to assume what it looks like.

rich866

appoligoies, left the pdf at work will attach tomorrow. Think I've solved it with two torque calculations but will attach them too just double check the math!

jehake12

Yes, I would consider the bottom fixing work as a pivot point in making your calculation. Try a quick google for "prying action" or "prying calcs". Thats a pretty common term used in civil/structural engineering for that type of action.

rich866

attached drawing of calculation. Want to know Fb, load to top fixing, when a given load is applied to the end of the cantilever arm F
Hope that makes more sense!
Given torque calculations a try, assuming that the loads I'm applying are well under the bracket components failure rating of over 1Kn, the weak link is the top fixing into the plaster board. If I know the loading on that fixing I can then move to the likely failure rate of the material
Used two torque calculations: 1. from the end of the arm to the bottom pivot point, 2. and from the torque at the pivot point back up to the top of the arm to the top fixing for pry force Fb.

Attached Files

torque calculation wall bracket 1.pdf (134.1 KB, 19 views)

nvn

rich866: Your answer for Fb is currently correct, except force is measured in units of newtons (N). Therefore, multiply your current value for Fb (and load F) by g = 9.81 m/s^2, to obtain newtons (N). Therefore, Fb = 334.4 N.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 375 mm, not 375mm. See the international standard for writing units (ISO 31-0).

rich866

Thanks nvn
Thats goood news, on the finished document I had converted the Fb load to Newtons, as the fixings tension loads are listed in KN, also added a safety factor of 1.4.

What I haven't been able to get is a pull though force for the construction of plaster board wall the bracket is to be mounted on, so I will have to construct a demo wall and physically pull the fixing out to make a statement about the maximum weight the bracket can be loaded too.

Thanks for the pointer on units, I am using this and a number of other parts of this as a project report so would be nice to be accurate. Been in the electrical industry using units for 30years and never come across that before!! learning all the time.

Thanks again

rich866

yep, just checked the standard link you attached and can't fault your point of fact. I will revise my documentation to refect this. Most useful fact of the week!

Vadar2012

I checked your calculations and get the same for Fb doing it another way. I don't understand why you did Ta that way, is there a brace there? If there wasn't Ta would equal Fb in the opposite direction.

Here's a link of something I did similar and nvn helped me with it. It might give you some ideas. I did an FEA model to determine the stresses at the wall.

http://www.physicsforums.com/showthread.php?t=625739