Wind load on a single beam
Wind load on a single beam
(OP)
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
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






RE: Wind load on a single beam
Just curious though, how is the wind load actually controlling the design? Is it because of the 54 ft span?
Maine EIT, Civil/Structural.
RE: Wind load on a single beam
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).
RE: Wind load on a single beam
Maine EIT, Civil/Structural.
RE: Wind load on a single beam
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
RE: Wind load on a single beam
Maine EIT, Civil/Structural.
RE: Wind load on a single beam
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.
RE: Wind load on a single beam
RE: Wind load on a single beam
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.
RE: Wind load on a single beam
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.
RE: Wind load on a single beam
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
RE: Wind load on a single beam
Maine EIT, Civil/Structural.
RE: Wind load on a single beam
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?
RE: Wind load on a single beam
Maine EIT, Civil/Structural.