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Scaffolding Query

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tempworks78

Structural
Jan 29, 2010
1
i'm involved in temporary works and have been asked to check the calcs of a small facade retention scaffold. my query is, how large a percentage of wall self weight should the transverse bracing be required to take? the calcs ive been given show 10%, i may be wrong but this figure seems rather small
 
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A classic masonry façade not slashed by the works, nor hammered by flying weights, is quite able to sustain (vertically) itself. So plan for sustentation of void openings, and then you have to add bracing to prevent failures in buckling ways plus any lateral forces that can happen during the works. Of the vertical forces, you may need to sustain even the 100% with a safety factor if you plan a very destructive demolition or set of works. The lateral forces are determined in ordinary way following wind and earthquake; lateral stability itself for the sheer weight is usually meager compared to the lateral forces to consider, and 10% of the vertical forces for vertical stability is maybe 10 to 20 times what required, not considering the lateral forces. But the lateral forces you need to count on, and cover at temporary works safety factor.

Do not forget also to consider some amount of differential settlement; shored façades are usually besides excavations and some coordinated evaluation of the likely settlements should be included in the study; for the kind of excavation works that work well, it is likely the differential settlement center to end of any part or the whole will be less than 1/400 or otherwise cracks would start to be quite apparent.
 
If this is masonry, how many 4 foot high lifts are there above the arch that the scaffold has to support?

The reason I ask is that if there is more than one, after the first lift cures and the second is applied, there may be enough strength in the first to allow arching action to occur, limiting the load seen to the scaffold. This could also be extrapolated to the lateral bracing to reduce that load.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
 
tempworks78,

I used to have to calculate this type of load from scretch for concrete panels in fire.

You can reasonably think of the brick wall as a similar situation overall.

Now taking the wall as a solid straight rectangle allow for a reasonable out of plumb of say 5% or 1 in 20. Ignoring any base fixity the mass of the wall will then act at Height/20 from the centreline of the equivalent vertical wall.

You then get an overturning moment of Wt.x Ht./20

Taking the reaction at half the height you get
Rn.xHt/2 = Wt.x Ht./20
or Rn = Wt./10 or 10% of weight

I would only use this if the wall had serious out of straightness issues otherwise I would usually use half of this for an old wall and possibly none for a perfectly plumb wall.
 
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