Wind load on a single beam

[SKJ25POL]

May I ask if my wind load calculation for single W shape beam (54 ft span), 40 feet up in air laying upright is correct?

I have velocity pressure of 21.00 psf the beam is a W 24x68

I think applying a uniform wind load laterally of 54 ft x 40 psf = 2160 plf or 2.16 is the wind load on beam. May I ask if I am correct?

Thank you

 

[TehMightyEngineer]

Seems close enough to what I would expect, didn't run any numbers to see if it's correct, though as you didn't say what MPH wind zone you're in.

Just curious though, how is the wind load actually controlling the design? Is it because of the 54 ft span?

Maine EIT, Civil/Structural.

 

[PAstructural08]

Without know additional information the actual wind pressure could not be checked.

You give a wind pressure of 21 psf so where does the 40 psf come from?
Also, you shouldn't be multiplying the wind pressure by the length of the beam; it should be the design pressure x the depth of the beam. That will give you a linear load on your beam that should be applied to the length (54ft).

 

[TehMightyEngineer]

Yeah, good catch PA. It looks like you took the 21 PSF x 2 ft to get roughly the 40 PSF (which should be PLF). You then multiplied by the length to get the total load of 2160 lb or 2.16 kips. Again, this is a relatively small load for a W24x68, even in weak-axis bending. I can only assume it's being controlled by combined bi-axial bending with the long span. You should also check if there is any second-order effects on the beam.

Maine EIT, Civil/Structural.

 

[SKJ25POL]

Dear PAstructural08 (Structural), thank you for your response.
I made maistake when typing, I meant to say I have the velocity pressure of 21 psf and the length of beam is 54 feet so the liner load along the beam is 21 psf X 54 ft = 1134 plf
the 40 feet is only the elevation of beam and I stated just for FYI

Is the 1134 plf the uniform distributed load that shall be applied to beam?
Is there other method on calculating the wind on a beam at the elevation of 40 feet about ground?

The infor for the wind calcualation I used:
Wind speed 90mph
Wind Directionality factor Kd 0.85
Impoetance facror 1.15
Exposure category C
Kh 1.19
Kz 1.04
Kzt 1.0
G 0.85

Ended up qz =21.0 psf

 

[TehMightyEngineer]

Are you using the components and cladding design values?

Maine EIT, Civil/Structural.

 

[PAstructural08]

I think you're still multiplying the pressure by the length; should be 21 psf x 2ft (W24x68 depth) = 42 plf = 0.042 klf.

Also, that is a wind pressure; you should be using a design wind load. Take the velocity pressure and multiply that by a gust-effect factor and force coefficient (see section 6.5.15 ASCE 7-05). You would take that design force x 2ft = the plf load that should be applied to the beam.

 

[jayrod12]

And no. 1134 plf is not what you should be using. You should be using your uniform PSF pressure times the Height of the beam, not the length. Your uniform load along the length of the beam should be in the order of 40 PLF. This is very basic load determination.

 

[SKJ25POL]

PAstructural08 (Structural),
Agai thank you for your direction. I understand your points.
What force coefficentneeds to be used:
Fig 6-20 Solid Freestanding Walls/Solid signs
Fig 6-21 Chimmneys, Tanks, roof top,...
Fig 6-22 Open signs and Lattice Frameworkds
Fig 6-23 Trussed Towers

Which from above?

May I ask how I calculate Cf for W24x68? That's where I am stuck.

Thank you

Also wind blows toweak axis of the beam.

 

[PAstructural08]

What is this beam being used for?

A simple answer (for a single beam), I would use Fig 6-23 to calculate your force coefficient (Cf). Where your 'E' for a single beam would be unity (1) and using the formula for a square cross section Cf = 2.1.

 

[SKJ25POL]

PAstructural08 (Structural,
The beam runs paralled to each side of a belt conveyor (in air).
The story is there are two beams running parallel to a belt conveyor truss and intention is to lay the truss in these two bean and replace the bottom horizontal truss in the existing belt conveyor truss.

I don't know for wind load calculation these beams are componant/Cladding or MWFR?
Seems with the method that we are using to dtermine wind load does not matter c/c or mwfr.

I appreciate for your kind help and solid knowldge

 

[TehMightyEngineer]

It should be components and cladding.

Maine EIT, Civil/Structural.

 

[SKJ25POL]

TehMightyEngineer (Structural),
How being Componant and Clading changes the value?
I don't see anywhere in section 6.5.15 ASCE 7-05 diffrentiate between MWFR and C/C?

 

[TehMightyEngineer]

Oh, I see, you're using 6.5.15. That seems applicable and you are correct that there is no distinction between MWFRS and C/C in that section that I'm aware of.

Maine EIT, Civil/Structural.