Steel beam design
Steel beam design
(OP)
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.
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.






RE: Steel beam design
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.
RE: Steel beam design
Anybody else?
RE: Steel beam design
RE: Steel beam design
Thanks for your help!
RE: Steel beam design
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.
RE: Steel beam design
Is that all you are asking
RE: Steel beam design
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.
RE: Steel beam design
RE: Steel beam design
RE: Steel beam design
A W18x60 has a factored moment capacity of 545kN*m @ 12ft (approx 363kN*m allowable or 260k*ft).
Cheers
RE: Steel beam design
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).
RE: Steel beam design
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.
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RE: Steel beam design
RE: Steel beam design
Don Phillips
http://worthingtonengineering.com
RE: Steel beam design
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.
RE: Steel beam design
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.
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RE: Steel beam design
http://ww
and
http://www.ce-ref.com/s-beam-asd.htm
RE: Steel beam design
RE: Steel beam design
beam top flange is barced at both ends. Can you explain how it is braced at free end/