Steel beam design

[btomcik]

Hi. I've been away from steel design for a few years and have recently begun reviewing it again, as I'm going to need it for my new job. I am stuck on what appears to be a simple beam problem in my steel textbook and was wondering if you could help.

Select the most economical W section with A36 steel. Beam is a cantilever, fixed at one end. Length is 12'. At the tip of the cantilever is a 16k load. Beam is braced on its upper flange at the fixed end and at the tip of the cantilever end only.

Answer is a W18x60 but I get stuck with the whole Lc<Lb<Lu, or Lb>Lu thing. Can someone walk me through the process of getting to a W18x60 please? Thank you.

 

[dgkhan]

btomcik.
If you are planning to do steel design yourself, try to understand these paarmeters yourself. They are very basics. Come up with Q's after that, more precise ones.

 

[btomcik]

I have been reading and trying to understand the parameters. I know they are basic fundamentals, but I'm stuck on this one problem and maybe if someone could just enlighten me a little I'd understand things much better.

Anybody else?

 

[JrStructuralEng]

btomcik, what is the total length, and the cantilevered length. Is there a distributed load on the beam or just a point load?

 

[btomcik]

Total length is 12'. Cantilever is 12'. Only load is a point load of 16k at the end of the cantilever.

Thanks for your help!

 

[dgkhan]

JrStr...
1.Your questions were already answered by btomcik.
2. Let me know also with AISC ref, if UDL or point load has
anything to do with unbraced length moment capacities.
3.This forum has some prerequisite b4 posting q's
4. If too many these question show up than we will loose JAE, slide rule era, USFE and others who really know.

 

[JrStructuralEng]

From the beam tables a W460x89(W18x60) has at an unsupported length of 3.66m (12ft) has a Mr= ~ 545KN*m

Is that all you are asking

 

[btomcik]

Well, sort of. I redid the problem on my own this morning and here is the process I followed. I don't know if this is the right way to get to the W18x60. If not, please advise where I went wrong and how to correct my approach.

1) Moment is 16k*12' = 192 k-ft
2) S = M/Fb, assume for trial that Fb = 0.66Fy = 0.66*36 = 23.76ksi
3) Sx = 192*12/23.76 = 97 in3
4) page 2-11, W18x55 has Sx = 98.3 in3 > 97 in3, and MR = 195 k-ft > 192 k-ft
5) However, Lc = 7.9' and Lu = 12.1'. Actual unbraced length is 12', so Lc<Lb<Lu so initial assumption of Fb = 0.66Fy is wrong.
6) Section F1-3, use equation F1-5 -- Fb =0.6Fy = 21.6ksi
7) Sx = 192*12/21.6 = 106.67 in3 -- page 2-11 choose next largest W18, which is W18x60.

Thank you.

 

[dgkhan]

looks like you are right. Check for deflection also.

 

[PEinc]

Don't forget to add the load from the weight of the beam also.

 

[JrStructuralEng]

dgkhan, I don't use AISC, and also use limit states design...but anyway...I believe unsupported length is exactly what this problem deals with. What do you think it deals with??? and give me references??? Look at the details. He is not providing support to the compression flange of the cantilevered beam. Since 12ft exceeds Lu the moment capacity is reduced for an unsupported length of 12ft.

A W18x60 has a factored moment capacity of 545kN*m @ 12ft (approx 363kN*m allowable or 260k*ft).

Cheers

 

[Lion06]

How in the world are you providing a cantilevered beam in which only the top flange is braced at the support? What is bracing the tip of the cantilever?
When checking deflection of cantilevered beams with no backspan (like yours), I also check the rotation at the joint and add that rigid body rotation into the deflection at the tip (columns do rotate).

 

[SlideRuleEra]

btomcik - From the page numbers that you cite, it looks like you are using AISC 9th Edition. Since this practice example is for your new job, on the job you may have to use AISC 13th Edition which has replaced the 9th Ed.

Be that as it may, and working your example using the 9 Edition, there is a nice (graphical) shortcut that takes care of all unbraced length and other concerns (reducing the chance for math errors). See page 2-168. Knowing just the moment (192 k-ft) and the unbraced length (12 ft), the graph points you to the W18x60 as the most economical section.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[JrStructuralEng]

I agree Sliderule, the charts are the easiest way to size beams for varying unbraced lengths.

 

[DonPhillips]

Not all jurisdictions have adopted the 13th edition.


Don Phillips

http://worthingtonengineering.com

 

[btomcik]

Thank you all for your help and your insight. I appreciate it.

The work I will be doing will be as part of my job designing temporary structures for construction. I'm working for a bridge contractor. With that being said, I don't think it is mandatory for me to use the 13th edition, unlike if I were working for an engineering firm doing designs for permanent structures. What is everybody's take on that?

Thanks again.

 

[SlideRuleEra]

When working for a Contractor, I would consider using the old 9th Edition to be a BIG advantage. As time goes by, temporary members will have holes burned in them, get "dings" from being handled often, welded splices, etc.

To compensate for this "deterioration" it is easy to lower the allowable stress. For example, as a bridge contractor in the 1970's, I never loaded "reused" A36 beams to more than 19 ksi. Of course it's a judgment call.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[RonRoberts]

Might also want to review:

http://www.aisc.org/MSCTemplate.cfm...tomSource/Faq/SteelInterchange.cfm&FaqID=2519

and

http://www.ce-ref.com/s-beam-asd.htm

 

[asixth]

Design the beam for deflections first, and then design for strength. In my experience, steel design is governed by deflections the majority of the time (not always)

 

[dgkhan]

btomcik,

beam top flange is barced at both ends. Can you explain how it is braced at free end/