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Q: Roof support ( Answered 5 out of 5 stars,   0 Comments )
Subject: Roof support
Category: Reference, Education and News > Teaching and Research
Asked by: hankconti-ga
List Price: $50.00
Posted: 13 Sep 2005 07:43 PDT
Expires: 13 Oct 2005 07:43 PDT
Question ID: 567547
I'm bulding a barn 50' x 120' the 120 section have an opening of 30'
for an 30' x 30' adition. Roof trusses sits on 2' on center. Reaction
on the trusses is 3300 lbs. Can you help me with sizing an I-beam for
this opening?

Request for Question Clarification by redhoss-ga on 14 Sep 2005 16:36 PDT
I can help you, but I need to know something first. The 3,300 lb.
number sounds a little high if it is per end. What is your live load
and dead load. Could the 3,300 lbs. be a total number, say 1,650 lbs.
per end. Are you in a heavy snow area. You would have to be in a 60
PSF area to get 3,300 lbs. per side. Once we get past this, the rest
is easy.

Clarification of Question by hankconti-ga on 15 Sep 2005 04:32 PDT
Loading (psf)
TCLL 30.0
TCDL 15.0
BCLL  0.0
BCDL 10.0
Subject: Re: Roof support
Answered By: redhoss-ga on 16 Sep 2005 06:51 PDT
Rated:5 out of 5 stars
Hello hankconti, sorry it took so long to get back with you. We will
use your truss reaction of 3,300 lbs. First we will calculate the
maximum bending moment:

Using the formula for a simple beam-uniformly distributed load from
the AISC (American Institute of Steel Construction):

M = (w x L^2)/8 
Where "w" in your case is the total load on the beam divided by the
length or w = 3,300 lb x 14 (trusses on beam)/30 ft = 1540 lb/ft

M = (1540 x 30^2)/8 = 173,250 ft-lb

Section modulus = M / allowable stress
For a standard structural beam (yield stress = 36 ksi) the allowable
stress is usually taken as 0.6 x 36 ksi of 22,000 psi

Section modulus = 173,250/22,000 =  7.85 in.^3

This was an interesting exercise, but in your case the beam design is
controlled by deflection. The amount of deflection should be limited
D = L/360 or (30 ft x 12 in/ft)/360 = 1 in.

The formula for maximum deflection is:

D = 5wL^4/384 EI 
Where E (Modulus of Elasticity) is a constant for steel = 30,000,000 psi
and I is the "Moment of Inertia" required.

Solving the above equation for I we get:

I = 5wL^4/384 ED

I = ((5 x 1540 x 30^4)/(384 x 30,000,000 x 1)) x 1728 in^3/Ft^3 
Note: the 1728 factor is required to get the units correct

I = 935.55 in^4

In a wide flange beam a good choice would be a W21x49 with I = 971 in^4.
This is a beam that is 21 inches deep and weighs 49 lb. per ft.

I hope you have been able to follow this and I will be glad to answer
any questions you might have. Please ask for a clarification if you
need additional help.

Good luck with your project, Redhoss
hankconti-ga rated this answer:5 out of 5 stars

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