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Q: Steel Beam Sizing ( Answered,   1 Comment )
Question  
Subject: Steel Beam Sizing
Category: Science > Physics
Asked by: josephc68-ga
List Price: $10.00
Posted: 28 Oct 2006 12:57 PDT
Expires: 27 Nov 2006 11:57 PST
Question ID: 777843
I am doing some reinforcement in my house to compensate for some poor
building practices.  I need to support a 2nd floor and attic load
without exceeding the depth of my joists, 7 1/4".  I used an
engineered lumber program to compute my loads but do not know how to
convert this load information to size a steel beam.  I am pretty sure
that a W6x25 would carry the load but would like this confirmed and to
see the math behind this.

Deflection Criteria: Specified(LL:L/360,TL:L/240). 

Load Group: Primary Load Group
                                   _________________
                                    ^  14' 11.50"  ^ 
Max. Vertical Reaction Total (lbs) 6473          6473
Max. Vertical Reaction Live (lbs)  4550          4550
Required Bearing Length in       2.47(W)         2.47(S)
Max. Unbraced Length (in)                35         

    Loading on all spans, LDF = 0.90 , 1.0 Dead 
Shear at Support (lbs)                1569   -1569  
Max Shear at Support (lbs)            1896   -1896  
Member Reaction (lbs)              1896          1896
Support Reaction (lbs)             1923          1923
Moment (Ft-Lbs)                          7091       

Loading on all spans, LDF = 1.00 , 1.0 Dead + 1.0 Floor 
Shear at Support (lbs)                5281   -5281  
Max Shear at Support (lbs)            6384   -6384  
Member Reaction (lbs)              6384          6384
Support Reaction (lbs)             6473          6473
Moment (Ft-Lbs)                          23872      
Live Deflection (in)                     0.486      
Total Deflection (in)                    0.691

Request for Question Clarification by redhoss-ga on 28 Oct 2006 16:03 PDT
Just out of curiosity, what were the loads that you used to get these
results. I can answer your question.
Answer  
Subject: Re: Steel Beam Sizing
Answered By: redhoss-ga on 30 Oct 2006 08:14 PST
 
Okay josephc68, I guess you don't want to tell me how you got your
numbers. We will just use your numbers and go ahead.

The maximum bending moment is:

M = Rx x L/2 = 6473 x 89.75 = 580,952 in-lbs

The required section modulus is:

S = M/allowable stress = 580,952/(36,000 x .55) = 29.34 in^3

The allowable deflection is L/360 = 179.5/360 = 0.5 in

Using this deflection to calculate the required moment of inertia:

Where P = 2 x Rx = 12,946 lb
      E is the modulus of elasticity of steel = 30,000,000 psi
      
I = PL^3/48ED = 12,946 (179.5^3)/48 (30,000,000)(0.5) = 104 in^4

The properties for a W6x25 are:
S = 16.7 in^3
I = 53.3 in^4

So, you see that the W6x25 is not adequate. It looks like your best
choice would be a W8x40 with S = 35.5 and I = 146.

Look this over and if you have something else you would like to try
please ask for a clarification.

Redhoss

Request for Answer Clarification by josephc68-ga on 30 Oct 2006 11:10 PST
Sorry Redhoss, I didn't get my email notification that there was an
answer waiting for me.  For simplicity my loads are a 10' tributary
with 30 psf live load and 12 psf dead load.  Do I have any other
options in terms of a steel beam that would be less than a 2x8 (7.25")
in height?

In what sense is the W6x25 inadequate, deflection?  The reason I ask
is that I found a table online that shows the W6x25 can support 652
lb/ft over a 16' span.

Thanks.

Clarification of Answer by redhoss-ga on 30 Oct 2006 14:03 PST
I wish that you had started with the info that you state in your
clarification request. If you take your new numbers here is what we
have:

w (load in lbs. per foot) = 42 psf x 10 = 420 lb/ft

So, the end reactions would be:

Rx = 420 x 14.958 / 2 = 3,141 lbs.

This is less that half the 6,473 number that your "engineered lumber
program" gave you. I don't know what went wrong, but it yielded some
bogus numbers. Here is a re-calculation using the correct numbers:

The maximum bending moment is:

M = Rx x L/2 = 3,141  x 89.75 = 281,905 in-lbs

The required section modulus is:

S = M/allowable stress = 281,905/(36,000 x .55) = 14.24 in^3

The allowable deflection is L/360 = 179.5/360 = 0.5 in

Using this deflection to calculate the required moment of inertia:

Where P = 2 x Rx = 6,282 lb
      E is the modulus of elasticity of steel = 30,000,000 psi
      
I = PL^3/48ED = 6,282 (179.5^3)/48 (30,000,000)(0.5) = 50.46 in^4

Now the W6x25 looks just fine. Good choice.
Comments  
Subject: Re: Steel Beam Sizing - josephc68-ga
From: poodleheadmikey-ga on 18 Nov 2006 19:37 PST
 
Joe,

I just wanted to gently point out that if it ever comes around again
that you are restricted by height and so cannot get the load rating
you need - just use two beams side-by-side.  Even without joining
plates they are fairly close to being double the strength of one beam.

PHM
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