Calculating Forces on an L-Bracket for Supporting Large Pipework

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SUMMARY

The discussion focuses on calculating the forces acting on an L-bracket installed in a brickwork chimney to support large pipework. The user seeks guidance on resolving forces and moments, particularly around a pivot point marked as X and fixings A and B. Key insights include the importance of understanding the difference between shear and tensile forces, and the necessity of ensuring that the sum of moments (ΣM) equals zero to prevent the bracket from pulling off the wall. The safest calculation approach involves primarily considering fixing A while acknowledging the potential movement of the pivot point due to bending at fixing B.

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  • Understanding of free body diagrams
  • Knowledge of shear and tensile forces
  • Familiarity with torque calculations
  • Basic principles of structural engineering
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Engineers, construction professionals, and DIY enthusiasts involved in structural support design and installation, particularly those working with brackets and load-bearing systems.

GBE111
I am installing an L-shaped bracket into a brickwork chimney to support some large pipework. I am trying to work out the forces in each fixing to ensure that I won't be pulling the bracket off the wall once the pipes are filled with water.

I've tried drawing the free body diagram and to resolve the forces and the moments, but I'm not sure exactly how to do it. I've marked the location of where I believe the pivot point to be as X and the two fixings as A and B. Can someone please explain to me how to do this as it is a long time since I did my engineering degree!

P.S the force the fixing can withstand vertically in shear is different to the tensile force. I think this is relevant?

Many thanks,
 

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Hello GBE, :welcome:
GBE111 said:
to ensure that I won't be pulling the bracket off the wall
You have the right suspicion there: your ##\sum M = F_y L## seems only part of the story to me (this would be around an axis located on the wall; since the bottom point can't move to the left, let's take that as the preliminary potential axis of rotation)

The condition for 'no pulling offf the wall' would be ##\sum M = 0## and the counteracting torque comes from the wall pulling on plugs A (mainly) and B.

Safest calculation would ignore B (e.g. because the bracket starts to bend there, so the axis of rotation moves up to, worst case, B) and then A needs to resist pulling force $$F_y L\over y-x$$
Make sense ?
 

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