Load Conversion to UDL 2

[palk7 EIT]

Hi,

For converting a snowdrift load (Let's say 94 PSF at the peak point) into UDL pressure to simplify the calculation, subtracted the peak with the balanced snow load (lets say 35 PSF)and multiply it by (2/3) or (1/2) and add it to the balanced snow load: which approach do you think is somewhat reasonable and justifiable by simple mechanics approach?

1). (94-35)*0.67 + 35

or

2). (94-35)*0.5 + 35

My thought was that a triangular load to convert it into a uniformly distributed pressure is to multiply it by 0.5 to attain a UDL of that load, but I saw previously where (2/3) was used in the multiplication is there a reason for it?

Thank you!

 

[rb1957]

2/3rds of a triangular dist'n will be reacted at one end (due to the centroid of a triangle).

another day in paradise, or is paradise one day closer ?

 

[jayrod12]

In our office, for preliminary sizing of stuff, e.g. open webbed steel joists, we take the triangle build-up and spread it evenly over the entire joist length.

In your case that would be (94-35)*L_drift*0.5/L_joist

Add that to the 35 PSF base load and multiply by the joist spacing. We would then use that number for sizing the depth of the joists.

Now when it comes to designing the beams supporting these joists, we do a proper load distribution taking into account the drift. That would be where the 2/3 (or 1/3) would come in; lever arm of a triangle.

 

[BAretired]

w_max = 94psf; w_min = 35psf; dw = 94-35 = 59psf
span = L; length of snowdrift = a

For a joist running parallel to the fall line of the drift, consider w=35psf over full span plus line load of 59*a/2 plf @ a/3 from w_max.

For joists running normal to the fall line of drift, calculate the actual snow load at the member location.

BA

 

[palk7 EIT]

Its basically for the beam, so in that case, the drift is closer to the beam side (i.e)2/3 [centroid], so would it be [35+(94-35)*0.67]*Tributary width of the beam? conservatively

Thank you!

 

[JLNJ]

Determining the actual moments under triangular loading is not all that difficult. Why simplify and run the risk of errors?

 

[jayrod12]

so would it be [35+(94-35)*0.67]*Tributary width of the beam?

No, that would not be correct.

It is dependent on the span of the joists and the length of drift. Really, without a framing plan it's tough to give you a set answer.

If the drift is exactly the same length as the joist, then yes 2/3 precisely goes to one side, and 1/3 to the other. However, since the drift is likely less than the joist length it's possible a larger proportion goes to one support. Regardless, using your method would not be the correct method.

For example, say your drift length is 20 feet, but your joists framing into this beam span 60 feet. Then the UDL on the beam at the drift side of the joist is 30*35+59*(20*0.5)*(60-20/3)/60 = 1575 PLF. If you do the 2/3 "rule" you are trying to use that would be 2236 PLF. so yes, conservative but to me, way too conservative to the point of being incorrect.

These are basic statics questions. Are you sure you should be performing these calculations?

 

[palk7 EIT]

Yep got that, I won't do something that I am not aware of it or doesn't understand it, until then I try to learn it to the end of it and make sense out of it. Yeah its basic stuff but, thank you very much for the help it cleared out

 

[rb1957]

"I won't do something that I am not aware of it or doesn't understand it" ? really ?? how do you know what you don't know ?

How do you miss something you're not aware of ? You miss it, 'cause you're unaware of it, and then someone makes you aware of it.
It's called "learning".

another day in paradise, or is paradise one day closer ?

 

[palk7 EIT]

Yep, that is learning exactly, I really appreciate your help!

 

[BAretired]

A single concentrated load plus a uniform load is about as simple as it gets.

BA