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Snow Load on Piping

Snow Load on Piping

Snow Load on Piping

(OP)
I am having trouble understanding the ASCE 7-16 Figure 7.13-2a/b figures for determining snow loads on a Pipe. I cannot get the units to work out using the ASCE figure.

I did my own math to calculate the area of the snow above the pipe given a 70° repose and multiplied by the ASCE density from equation 7.7-1. That produces reasonable numbers but does not match with the 7.13-2a/b numbers.

Could somebody walk me through this? (e.g., Pf = 50psf, D = 10in, angle = 70°)

RE: Snow Load on Piping

If Pf = 50psf, D = 10in, angle = 70°,


Ground Snow Loads, pg, should be around 70 psf and γ =0.13pg + 14 = 23 lb∕ft3

Diameter or Width = 10 in. = 10/12=0.83 ft < 0.73pf ∕γ =0.73*50/23=1.58 ft

⇒ 7.13-2a is applicable and Qs=1.37Dγ =1.37*0.83*23=26 lb/ft

Does this post answer to your problem?

PS. Always remember to use consistent units...

RE: Snow Load on Piping

(OP)
HTURKAK - thank you for the response. It is helpful.

What would the equation be for Qs if D > .73pf/gamma ???

RE: Snow Load on Piping

Quote (Lee Hutz

...........What would the equation be for Qs if D > .73pf/gamma ???)


This is basic knowledge of highschool geometry.. and wish should not be asked.

If D > .73pf/gamma The attachment at you first post 7.13-2b is applicable.

Assume Diameter or Width = 24 in. = 24/12=2.0 ft > 0.73pf ∕γ =0.73*50/23=1.58 ft and If Pf = 50psf, angle = 70°,

Snow area = h*( w1+w2)/2 =(50/23)*( 2+2-2* (50/23 )TAN 20 )*0.5=2.62 sq-ft

and Qs=2.62*23=60 lb/ft

RE: Snow Load on Piping

Not familiar with this particular section of ASCE 7-16 so this interests me, but the way I read it, especially the last sentence of section 7.7.1, the drift load (lbf/ft2) pd = hd*γ, where hd is the height of the drift (ft) and γ is the snow density (lbf/ft3).

The "critical" diameter would be;
pf = 50 lbf/ft2
γ = 23 lbf/ft3
Dc = 0.73*pf/γ = 1.586 ft (19 in)

In the first example with the triangular load distribution, it states "load = 1.37*D*gammma", or equivalently, as height = 1.37*D; pd = hd*γ. Using the numbers from above:
D = 0.83 ft
γ = 23 lbf/ft3
pd = 1.37*D*γ = 1.37*(0.83 ft)*(23 lbf/ft3)= 26.1 lbf/ft2
Distributed load along length of pipe = pd*D = (26.1 lbf/ft2)*(0.83 ft) = 21.7 lbf/ft

In the second instance with the trapezoidal load distribution, it appears to me the load is simply the "pf" as previously calculated:
D = 2 ft
pf = 50 lbf/ft2
Distributed load along length of pipe = pf*D = (50 lbf/ft2)*(2 ft) = 100 lbf/ft

As an aside, if you use the formulas both at the critical diameter, you can see there is no discontinuous jump:
D = 1.586 ft
pd1 = 1.37*D*γ = 1.37*(1.586 ft)*(23 lb/ft3) = 49.97 lbf/ft2
pd2 = pf = 50 lbf/ft2
e.g. pd1 = pd2 at the critical diameter

Anyway, again, not familiar with this section of ASCE 7-16 as I don't typically deal with snow loads on piping, but this discussion interested me; so I thought I would look at it.

----------------------------------
Not making a decision is a decision in itself

RE: Snow Load on Piping

(OP)
CSK62
That was my confusion as well; whether to use a trapezoidal area like HTURKAK did (where I was confused on setting the "h"-value), or just to use pf. It wasn't clear to me by the Figure. The continuity at the "critical diameter" you pointed out does seem to back up your argument.

In the spreadsheet I made, the loads don't vary much either way, so just trying to do it "per code". The biggest difference is subtracting the segment of the pipe area from the triangular/trapezoidal(rectangular?) area, which is not permitted by the code.

Thank you both for your input.

RE: Snow Load on Piping

Quote (csk62


In the first example with the triangular load distribution, it states "load = 1.37*D*gammma", or equivalently, as height = 1.37*D; pd = hd*γ. Using the numbers from above:
D = 0.83 ft
γ = 23 lbf/ft3
pd = 1.37*D*γ = 1.37*(0.83 ft)*(23 lbf/ft3)= 26.1 lbf/ft2
Distributed load along length of pipe = pd*D = (26.1 lbf/ft2)*(0.83 ft) = 21.7 lbf/ft
In the second instance with the trapezoidal load distribution, it appears to me the load is simply the "pf" as previously calculated:
D = 2 ft
pf = 50 lbf/ft2
Distributed load along length of pipe = pf*D = (50 lbf/ft2)*(2 ft) = 100 lbf/ft

)



Distributed load along length of pipe for D= 0.83 ft = 1.37Dγ *D /2=(26.1 lb/ft)*(0.83 ft /2)*(1 ft)=10.83 lb/ft

Distributed load along length of pipe for D= 2.0 ft = Snow cross section area* density* 1 ft length = 2.62 *23*1 = 60 lb/ft

RE: Snow Load on Piping

Looking at this again, it is clear to me that I should have just ignored the picture in ASCE and taken it from engineering principles.

Triangular distribution
h = (D/2)*sin(70)/sin(20) = 1.373*D
A = 1/2*b*h
A = (1/2)*D*(1.373*D) = 0.687*D^2
For D = 0.83 ft --> A = 0.473 ft2
For γ = 23 lbf/ft3
Q = A*γ = 10.9 lbf/ft (distributed load along piping length) - agrees with HTURKAK answer

Trapezoidal Distribution
I think where the confusion may lie stems from the picture. Whereas the triangular picture shows "h" as a height (ft) and a clear geometric area (ft2) can be calculated, the trapezoidal picture shows the height as "pf" which is a distributed load (lbf/ft2).


----------------------------------
Not making a decision is a decision in itself

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