Subject: Re: " I " beam size needed

From: [email-masked] (Curlyindy)
Date: 26 Mar 1996 12:44:51 -0500

Let's get down to it:
Assuming 'I' beam means a steel beam and the loading condition is truely
1000 lbs on a 30 ft span, Use a W10x12 .  The solution is:

R=V=P/2=(1000 lb)/2= 500 lbs reaction at each end of beam
M=PL/4=(1000 lb)(360 in)/4= 90,000 lb in. of maximum bending moment
F=0.66xFy=0.66x(36,000 lbs/sq. in)= 23760 lbs/sq.in allowable bending
stress
S=M/Fy=(90,000 lb. in.) / (23,760 lbs./sq.in) = 3.78 in^3 (cubic in.)
Section Modulus
d=L/360=360 in / 360= 1 in. deflection at point of load in center (worst
case)
I=PL^3/48Ed=((1000 lb)(360 in)^3) / 48 (29x10^6 lb/sq.in)(1in)=33.5 in^4 =
   Moment of Inertia.

Checking the American Institute of Steel Construction (A.I.S.C.) Manual of
Steel Construction  (the "green book") one and the 'properties for
designing' section,
The following sizes of steel beam would work:
   W8x13  ....  I=39.6 in^4 ...  S=9.91 in^3
   W10x12 ...  I=53.8 in^4 ...  S=10.9 in^3
   W12x14 ...  I=88.6 in^4 ...  S=14.9 in^3

Use a W10x12 steel beam. Depth=10 in.  Width= 4 in. Weight = 12x30=360
lbs.    Cost. $180 (assuming $.50/lb)  Call your local steel fabricator or
supplier for availability.  Grab your freind who has a trailer and go pick
the sucker up.

Notes:

1.  (2) 2x4's of Construction Grade lumber would work to hold up the ends
of the beam although one should not necessarily bear steel on wood (shrink
and swell).  Brace the 2x4's at least at midpoint for 8' long studs.

2.  If the weight is distibuted over the length of the beam rather than at
the center (i.e. floor joists on a beam rather than a hoist beam on which
to wrap a chain to yank an engine), don't worry.  A uniformly loaded beam
would have less stress. If the beam needs to carry more load, don't worry.
 As a friend of mine is fond of saying,"What's it going to do .... BEND". 
Just don't overdo it too much.

3.  Please call me at 317-570-0035 if any of the above conditions or
assumptions are not true and we will re-figur' the problem.

4.  Dean Mellinger - Architect  (a.k.a. CurlyIndy)

5.  The beam will work if the above assumptions are correct but I can't
take liability since I don't know the particulars of the actual
conditions.  You should work with a local engineer or architect (one with
salt).

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From: [email-masked] (RonR908326)
Date: 26 Mar 1996 17:31:31 -0500

"CurlyIndy"   Good try but you goofed.  For a W10x12 the maximum Lc (The
length at which you can use 0.66Fy) is only 3.9 feet and the maximum Lu
(the length at which you can use 0.6Fy is only 4.3 feet).  Since it's
unlikely to be braced laterally you have to design for an unsupported
length of 30 feet, for which the lightest satisfactory section is W5x19.

Give the guy the phone number for a good structural engineer next time.
Ronald M. Roberts, P.E.
Structural Engineer
17503 NE 137th Street
Redmond, WA  98052-2182
Email: [email-masked]

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From: [email-masked] (Paul R. Culley)
Date: Thu, 28 Mar 1996 14:08:46 GMT

In article <[email-masked]> [email-masked] (Curlyindy) writes:
>From: [email-masked] (Curlyindy)
>Subject: Re: " I " beam size needed
>Date: 26 Mar 1996 12:44:51 -0500

>1000 lbs on a 30 ft span, Use a W10x12 .

Can someone provide a "decoder" for the above designation i.e. W10x12?
My engineer put some beams on my house plan at W12x22 and W12x30 and I am 
curious as to the actual size.

[email-masked]

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From: [email-masked]
Date: 29 Mar 1996 14:32:11 GMT

>curious as to the actual size.

W12 is a rough designation of the depth of the beam. (12") The W means that
it is a Wide Flange member (as opposed to the older, S or standard I beam)
the x22 and x30 refer to the weight per foot of the beams i.e. 22 pounds and 
30 pounds.

The rolls used to make these beams are adjusted to make the flanges thicker
or thinner depending on the strength (weight) needed. To keep things simple,
the web connecting the two flanges is a constant depth. Thus, there is some 
variation in overall depth of these members (the extreme case is a w14x730
which is 22.5" deep)

>[email-masked]

----------------------------------------------------------------------------
W. E. (Ted) Blockley AIA, Architect           Email: [email-masked]
Bakersfield CA                                Phone: (805) 322-9373
----------------------------------------------------------------------------

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From: [email-masked] (Curlyindy)
Date: 28 Mar 1996 14:49:06 -0500

For a W12x22 and W12x30 the decoder is:
WAAxBB where:
W=wide flange (you can have C for channels, MC for misc channels, L for
angle etc.
AA=depth in inches (nominal, actual depths vary by fractions of an inch)
BB=weight in lbs per foot

So a W12x22 is a Wide Flange steel beam, 12" deep, and weighs 22 lbs/ft.  
     a W12x30 is the same but weighs 30 lbs per foot.

roughly, steel costs about $0.50 / lb so a 10 ft long W12x22 would cost
$110