Tributary are of a window with a semicircular top.

[vf3]

I could use some help resolving the tributary area of a 48" wide window that is 126" in height. If the window were rectangular I would not have a problem but the top of window is a semicircle. Any help would be appreciated as I am determining the anchor locations at the window frame to the framing.

I did a crude sketch of the window and want to know if I have the tributary area of the semicircle correct.

Thanks
Vinny

 

[Ron]

Not sure what code you're working with, but your window area as shown is 40.3 square feet. If you considered it as full rectangle, it would be 42 square feet. Not enough difference to worry about on the tributary area scale.

Your perimeter is 27.28 ft.

Your question is confusing. If you are considering fastener loading, the tributary area is for 1 fastener, so the maximum coefficients apply.

Please clarify your question and why you need to separate the tributary area of the arch from the remainder.

 

[vf3]

The loading at the sill is traianglar and the loading on a rectangular window is trapezoidal. How does the round window resolve its forces at the frame. I take the wind pressure times the area and resolve those forces to pound/inch so I know how to size the anchor clips.

 

[a2mfk]

Agree with Ron, you are over-thinking it. Any quick trib area assumption you make should not make a difference in the sizing and spacing of the fasteners. Never cut it close on window and door fasteners, these can cause the whole structure to fail if they blow in.

 

[vf3]

Both of you guys are right and I so use a 300 % safety factor but I am still trying to just understand how the pressure in the circular portion transmits to the window frame.

 

[Ron]

Assume the pressure acting on the window, in this case negatively, creates a uniform perimeter load around the window, normal to the substrate, without regard to its shape.

By the way, I typically use a safety factor of 4 for component and cladding fasteners in direct tension.

The distribution you are trying to apply is more applicable to the glass than to the frame. The load is applied to the glass by the wind, which in turn is transmitted to the frame by shear at the edge of the glass. To try to evaluate small variations in the stress distribution in the class is a waste of time.

 

[vf3]

Ron

So, you propose that I take the entire area of the window and multiply it by the factored pressure, take that load and add safety factor of 400% then take the load and divide it by the permiter to determine the loading per inch. Then size my anchors and clip assemblies appropriately?

I will use the square approach and use 42 SF as the area. The windload used in this example is 30 PSF and the safety factor is 4. Thereforer the loading from the entire window is 5,040 lbs. Using the permiter of 31.28-feet. The loading of about 161 lb/ft of the window frame.

Would that be the way that you safely compute the loads at the ourside of the window?

Thanks
Vinny

 

[Ron]

Vinny, it doesn't make any difference to the final numbers, but if your calcs will be reviewed by someone (structural engineer of record, building department, etc.), apply the factor of safety to the fastener>> (other side of the equation).

That way, it will be apparent to whomever reviews, that your are applying the factor of safety to the resistance, not to the load. If you are using manufacturer's fastener pullout tables, make sure they are given in ultimate load, not allowable load. If allowable load, you don't have to apply the FS=4, but it would be a good idea to determine if their factor of safety is 4.

 

[BAretired]

Windload = 2*30 = 60 plf on each jamb. If fastener spacing = s then load per fastener = 60s. Factored resistance required per fastener = 4*60*s (using Ron's FS).

Maintain the same spacing around the curve, i.e. s.

BA

 

[vf3]

Thanks guys!