W Flange I Beam

[dynotime]

Hello to all. I have Beam check but am running into a Question that I cannot seem to
Find in the Program. I have a w14 Beam with 6 3/4 breath and 13 7/8 Depth
Can someone tell me how much Weight in a 20 ft Span with the beam being suported at 1.33 ft. I would like to know what amout of weight at .010 Deflection
.050 Deflection .100 Deflection Thanks Greg I can also be reache at
dynotime@micro-net.com

 

[dik]

Need a little more information...

What is the thickness of the flange and web? Need this to determine what beam it is... any idea of the material grade?
Can you provide a little more information on the supports and the span distance? Is it OK to assume the loading is uniform? any point loads on the span? Is the beam laterally supported to keep it from twisting or toppling?

 

[dynotime]

Hello dik: The Flange is 1/4 thick and the Web is 3'16 Thick, I do not know the grade of the Material. Please use the weakest. The loading will be 15 Point loads and the span will be 20 Feet. The beam will also be supported at the 15 point loads to keep the beam from toppling. Greg

 

[boo1]

Supported laterally?

 

[dik]

If the flange thickness were 0.385 (3/8"+) then the section could be a W14x30. This has an Ix = 291 in^4 and an Sx = 42.1 in^3.

For ungraded steel, in Canada, the max allowable stress is 210MPa or about 30ksi.

The maximum limit states moment capacity of the beam is Mr = 30x42.1x0.9/12 = 96.1 ft-k.

For a 20' span, the maximum factored uniformly distributed load (UDL), qf = 8x96.1/20^2 = 1.923 klf. This would be equivalent to a service loading of approximately 1.37 klf.

delta for UDL = .00624*M*L^2/I where M in ft-k, L in ft. and I in in^3.

for delta = 0.1" we have: .00624*Ms*20^2/291 = 0.1 and service moment, Ms = 11.66 ft-k and the service UDL, qs = 8x11.66/20^2 = 0.233 klf... For a reasonably stiff beam, design deflections could be in the order of 5/8" and not 1/10". The service loading for this would be in the order of 1.2 klf.

The deflections of 0.05 and 0.01 are likely outside irregularities of the beam itself... also the beam may not be a W14x30, if the dimension of 0.25" is correct. The above will give you an idea of how to calculate this stuff...

Since I'm confused by big and little numbers, before implementing any of the above, please confirm it with a local registered structural engineer; the life you save may be your own!

 

[SlideRuleEra]

dynotime - One way to approach this problem is to go back to basics. Compute the the beams properties based on the measurements that you have (the accuracy of the dimensions will determine the outcome however). With the defections that you are talking about (max 0.1 inches over a span of 240 inches) the beams materal and unbraced length will be of minor importance. The most important factor will be the Modulus of Elasticity which AISC assumes to be a constant for all the common steels.

I'll use AISC 9th Edition, ASD as a reference:
1. On page 6-19 compute the moment of inertia for flanges: "Equal Rectangles - Axis through the CG". I get 157 in^4 (but please check my math in all cases).
2. On page 6-17 compute moment of inertia for the web: "Rectangle - Axis through center" Get 37 in^4.
3. Total moment of inertia: 194 in^4.
4. Compute crossectional area of flanges + web. Assume 1 foot length and using 490 lb/ft^3 for steel, page 6-8, compute weight per ft. Get 20 lb/ft.
5. Use beam diagram for "Simple Beam - Uniformly Distributed Load", page 2-296 (15 point loads over 20 feet will be remarkably close to a uniform load - besides you had to make assumptions like this for a solution, in reasonable time on a slide rule).
6. Compute deflection for the self weight of the beam. Get 0.013 in. Exceeds the 0.01 in. requirement without any applied load.
7. By ratioing the 0.013 deflection with the required 0.05 in. come up with about 58 lb/ft live load + the beam weight. Transforms to 77 lb. for each of the 15 point loads.
8. By the same logic get 135 lb/ft (180 lb/point load) for the 0.1 in. deflection.

Based on computed section modulus the bending stress will be well under 4 ksi.

Please check my math, but even with these simplifying assumptions the results should be surprisingly accurate (I'll bet within 2% of an exact solution).

 

[boo1]

I see a difference in Moment of Inertia in the two post. Its too late for me to try it, but maybe tomorrow...

 

[dik]

boo...

I couldn't find a 14" dp section with a flange thickness of 1/4"... My old steel book gives I for W14x30 as 291 in^4, but the flange thickness is 0.385... The method posted should be generally OK... The other moment of inertia calculated based on the rectangular sections should be approximately 291 - 0.125x6.75x50x2 or about 200. I'd guess his 194 is likely in the ballpark...

 

[dynotime]

Thanks to all I went with a w12 19 Thanks to all Greg