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window fastener load calculations 1

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wangbungal

Mechanical
Jun 24, 2020
5
Hello all, i copied this thread from the fasteners thread, just wondering if you could help me start my window fasteners load calculation journey, so I have a window residential, i did the ASCE 7 wind load
calculations manually already, i have a max wind load of 62 psf and tributary area of 52 inches height by 37 inches wide tributary area, and an Ultracon fastener 1/4 inch diameter
with Fy tensile yield strength of 155 KSI and Fu ultimate tensile of 177 KSI, my case is how do i calculate if i alter the parameters such as increasing the shim between the window and concrete structure
or proving that at a certain shim gap my current 1/4 inch dia fastener is still capable of sustaining the load? can you show me a direction, let me know please. thank you
 
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IMO, properly placed shim will ensure the fasteners subject to shear force only. If leave the gap unfilled, the fasteners will experience bending on top of shear, a lousy situation to be in.
 
Your tributary area for a fastener is too large. For windows, the tributary area is quite small, resulting in higher factors.

As retired13 noted, the shim spacing at the perimeter is key to resolving fastener issues unless you are using flanged windows. A shim spacing for fasteners in shear is typically limited to 1/4 inch. More than that induces bending in the fastener so then you have to consider combined shear and bending loading, thus significantly reducing the fastener capacity.

To answer your question directly, take the load on the window and transpose that to perimeter loading. Then divide the load by the number of fasteners anticipated. That will give you the load per fastener. Given that load, you can calculate the shear on the screw and calculate the bending moment if the shim gap exceeds 1/4 inch. Then apply a unity equation for shear and bending to see if the fastener complies.

 
Ron said:
To answer your question directly, take the load on the window and transpose that to perimeter loading

Ron,

At what aspect ratio (L/W), do you need to consider the fasteners along the longer window side dimension will take more load due to the greater stiffness in the shorter dimension?
 
Because i was just considering for shear I use the formula sigma = Force load P / Cross sectl area of fastener if you increase shim gap on a window that is flush and no flanged, for bending how do you calculate?
 
It is same as a cantilever beam, the moment equal to force per fastener times lever arm (the gap).
 
The stress due to bending is sigma = My/I. If you're a mechanical engineer hopefully you remember that from your first year at university/college?
 
Agent666 i am aware of that formula, my view is the bending with regards to the eccentricity factor of it, i have been reading books AISC and state codes, but could not fathom how to incorporate that effective ‘e’ factor, and what table or book to get it from, is that something like a rule of thumb or there is a specific formula to get it?
 
sigma = My/W people (MOMENT / MODULUS) and not moment / inertia, for God's sake!!
 
Well I'm not quite sure what you mean by effective eccentricity factor, e is usually used for the dimension or measurement of the eccentricity. Like e = 20mm, therefore M = shear x e. Can you post an example of what you are referring to?
 
Johnie134, no one said that. SIGMA = My/I period!

I/y = elastic modulus, where Y = max Y from centroidal axes.
 
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If you see this example, he use e = 0.75 in and then effective ‘e’ =0.375 in
 
I think what they are saying is that they are assuming the fixing is going into double curvature, fixed at both ends with the same moment developed. This would entirely depend on how things were fixed and the type of fixings.

I tend to think while some double bending may in fact occur, that most designers would ignore the effect as it is very hard to quantify. Screwing into wood, the maximum moment in the fixing likely occurs some distance into the wood in reality.

Fixings are cheap, so if you need a few more, just go for it.
 
What is the thickness of window frame? The e will than be bframe+1/4" (gap).
 
Thank you all, now it makes more sense to me now, thickness of the frame is 1-1/2 in.
 
You need to check the document (BCCO) for the definition of "e", and the assumption in using e/2 for moment calculation.
 
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