how do we use the beam selection tables for cantilevers 1

[tobman]

I'm having trouble interpreting what the method of analysis
is to determine which steel beam section size to use for a
cantilever problem. If I have a beam that is fixed at one
end, and has a point load at the other (let's say span of
the beam is 10 ft, and the point load is at the end of the
beam and is say 500 lbs.) The moment is then 5000 lb.ft and
the shear is 500 lbs. My question is this: Can we just
go to the beam selection tables and select a beam with this
info, and if so, what do you use for an unsupported length?
I was under the impression that the tables were just for
simply supported beam conditions with different unsupported
lengths. What would be an acceptable section to choose here
and why?

 

[Lutfi]

You need to worry about stability and untraced length. Why don't you do a simple maximum bending moment and select a beam with an acceptable section modulus?

Good luck.

 

[tobman]

I guess that's what I'm asking. I have my maximum bending
moment (5000 lb.ft) from the applied load. Can I then go
to the beam selection tables and select a section, and if so
what do you use for an unbraced length?

 

[ERV]

tobman,
A cantilevered beam as in your case(if I understand correctly), actually has multiple loading, at point of connection (A, fixed or pinned), cantilever support (B, usually pinned), and cantilever point load (c), in addition to uniform loads applied to the spans (if any). Due to multiple loading, standard tables would not apply. The unbraced length would be the center to center distance of joists attached to the beam (e.g. 5 ft o.c., etc), if decking is applied directly to the beam unbraced length would = 0.

 

[tobman]

ERV:

My beam is fixed at one end and is free at the other with
a point load on it. There is no lateral support except at
the fixed end. Again, my question isand to recap: the
length of the beam is 10 ft) can I use the beam selection
tables, and what would the section be? And if I can't use
the tables, what formula do I work with to figure out the
resisting moment of the beam in this case? Mr= Mc/Ix perhaps?

 

[SlideRuleEra]

tobman - As Lutfi has stated, the unbraced length (of the compression flange) is the primary concern. Using load case 22, Part 2, AISC Manual of Steel Construction, ASD, 9th Edition as a guide (page 2-303 in my copy) the compression flange is the bottom flange for the entire 10 foot length. Assuming that the cantilever connection provides lateral support (to the compression flange) the unbraced length is 10 feet. Please note that it does not matter if lateral support is provided at the 500 lb. loading point or not - the unbraced lenght is 10 feet in either case.
From this information the Allowable Moment in Beams Charts (page 2-175 in my copy of the AISC Manual) show that a W6x9 is the lightest section that meets the 5 kip-foot moment and 10 ft. unbraced length critera. This assumes A-36 steel and a conservative cb=1. Any beam directly above this point on the chart will work also - such as a W8x10.

Best Wishes

 

[tobman]

SlideRuleEra:

You have answered my question completely. I was under the
impression that the selection tables were only for simply
supported conditions, and that the cantilever condition did
not apply. This will make this problem a whole bunch easier, as I'm often asked to put up beams here with a bit of an over hang past the supports for
rigging purposes, and I've always been unsure of what to use for the unsupported length.

Thankyou very much.

 

[SlideRuleEra]

Your are very welcome.

 

[SlideRuleEra]

tobman - In rereading your latest post I picked up on something you may need to consider: A beam with an overhang (perhaps like load case 26 on page 2-305) is NOT the same thing as a cantilever beam (load case 22 discussed above). If your situation is like load case 26 come up with a hypothetical setup for discussion - the beam selected may be different.

 

[tobman]

SlideRuleEra:

My beam in this case is like load case 22. However, I am
asked to put up monorails quite a bit of the time (load case 26) for rigging motors and such. When I do these, I
generally use tables put up by the Monorail Manufacturer's
Association, using the max load on the monorail multiplied by an impact factor (usually 1.5). I then check for 20% of the load acting perpendicular to the monorail and 10% longitudinally down the monorail for both the monorail and the connections to the supporting beams. I then check the lower flange buckling from the charts and select my monorail. However, the charts here again do not state what to do if you have a monorail that overhangs a support with the load at the end as shown in case 26, a very common condition in the field. I have been checking the overhang by Mr=Mc/I and ensuring that I don't exceed the capacity of the monorail that way. I'm curious if the same solution as
you mentioned before using the overhang as the unsupported length and checking with the beam selection tables in the steel code to ensure that the moment and shear capacity are not exceeded would be prudent here as well.

 

[sframe]

Hi Tobman,
Your problem is also very common to me; and I would suggest that you use 2x cantilever length as the unsupported length. My understanding is that the compression flange is very similar to a compression member as far as stability is concerned, and the side-sway of the free-end is always a primary issue.
If the free-end is laterally braced, then you can use the cantilever length as the unsupported length. If not, then you have to double the length.
Refer to table C-C2.1, case (d), page 5-135 of AISC ASD manual, 9th edition.
Best Regards,

 

[SlideRuleEra]

tobman - Using the charts for load case 26 in addition to your other calcs is wise. Take the following hypothetical example based on load case 26:
Total Beam Length= 24 ft with
L (simple span from R(1) to R(2))= 14 ft and
a (overhang)= 10 ft
Force P= 500 lb

M(max)= 5 k-ft at R(2)
Moment along the entire 24 ft beam length is negative therefore the bottom flange is in compression for the full 24 ft length.
Assume that supports R(1) and R(2) provide lateral bracing to the bottom (compression) flange. The two unbraced lengths for this beam are 14 ft. and 10 ft. Select the (maximum) 14 ft length for use in the charts.
From the chart (page 2-175) for 14 ft unbraced lenght and 5 k-ft allowable moment the lightest beam is W6x12 with an alternate (one of many) being a W8x15.

The nice thing about the charts - they are based on decades of experience and the "funny" shape of the curves automatically takes into account much detail mathematical calculation. I recommend using these charts whenever possible.

 

[SlideRuleEra]

Just finished eating supper; so back to a variation of my above example:
Assume that supports at R(1) and R(2) DO NOT provide lateral support (say the 24 ft beam is underslung from members above by bolts thru it's top (tension) flange). Then the unbraced length is 24 feet, allowable moment is still 5 k-ft and from page 2-174 the lightest beam is W10x30 with reasonable alternates of W12x35 or W10x33.

 

[GerardJB]

Over a 30 year period of designing structures I have assiduously avoided referring to load tables, of any description, for the selection of beams.

Structural members should be designed for strength, serviceability (deflection) and durability.
Generally, my experience is that deflection/sway is the governing factor in the design of roof and floor beams.
So, whilst load tables will give an indication of the strength of a beam between lateral restraints, it is not necessarily an indicator of overall performance or a repositoty of engineering wisdom. Particularly for a cantilever.

I have found it easier and better to work "in reverse". That is initially to: assume full lateral restraint and a bending stress of 0.66Fy; establish the prescribed deflection; divide the design moment by the bending stress to give the required section modulus; derive the required moment of inertia for the load and beam conditions; from this select the beam size from the section properties (much more reliable!); recheck the allowable bending stress for the actual lateral restraint; and, design the beam connections. This is the way you control the design and deliver the optimum structure to the Client.

Good luck!

 

[tobman]

SlideRuleEra:

Just when you think you're getting the picture........
In one of your replies you said that for an "underslung" beam you would use the whole length of the beam as it's unbraced length. This doesn't make sense to me if the beam
is being used as a monorail where the load is applied to the bottom flange of the member, causing it to bend downward and thus making the bottom flange bend in tension while the top flange would bend in compression. I would thing that the compression flange would then be braced at the points of connection to the beams that are "underslinging" it. (I'm sure that's not a word, but you get the idea). Any thoughts? This happens almost everytime I put up a monorail.

 

[SlideRuleEra]

Good Morning tobman - Lets take a more detailed look. We are considering the static condition shown in Load Case 26. Load "P" (500 lb) is applied to the end of the beam. Your concerns about how and where (top or bottom flange) the load is applied are valid for the local area very near (say within a couple of inches) the 500 pound load point. However that is a different problem from the unbraced beam length. Lets assume (for now) that the beams you may consider are adequate for a 500 lb load on either the top or bottom flange.

The beam sees the following picture:
P= -500 lb
R(1)= -357 lb
R(2)= +857 lb
M(max)= -5 k-ft @ R(2)
Moment anywhere on the 24 ft long beam is negative, therefore the entire top flange is in tension and the entire bottom flange is in compression.

Now look at the details at supports R(1) & R(2). If the 24 ft beam is what I call "underslung", then both R(1) and R(2) are above the beam and are attached to only the top (tension) flange of the the 24 ft beam. Therefore the bottom (compression) flange has no lateral support anywhere along it's entire length.
This is not a problem as long as you approach the beam selection properly. A beam has the "internal" capability to resist a certain amount of lateral force. This is based on the beams material (we are assuming A36 steel) and its geometry. The equations on page 2-147 set the limits, the charts that follow just show the solutions of there equations (for one beam at a time) in graphical form. If we had the option, appropriately placed external lateral bracing allows the beam to "ignore" these equations and support greater loads. However, we are assuming that external bracing in not practical.

So we have the following conditions:
Unbraced length of compression flange = 24 ft
M(max) = -5000 k-ft

From the chart on page 2-174 the lightest beam is W10x30 (note that no beam on page 2-175 meets the critera).
Note that a check of the bending stess in the W10x30 is only about 1.85 ksi instead of the normally allowable 24 ksi (with adequate lateral bracing). This is because of a dramatic "downrating" dictated by the very long (for this beam) unbraced length.

At this time it is appropriate to get back to the details of the 500 lb loading (top or bottom flange, etc.)

I hope this helps, if not let me know, I'll be happy to keep working at it with you.

 

[tobman]

SlideRuleEra:

The math all checks out for your example. I even sketched out the deflected shape and I was surprised to see that the bottom flange for a monorail would not be braced for it's entire length. The only problem I'm having now is that it just doesn't seem right. By that I mean that if you had a monorail that was underslung over say three or four spans of beams (which I don't think is too uncommon) your unbraced length which you are saying is the whole beam is going to be quite long, and would lead to some pretty hefty beams. What would you use for an unsupported length for say a beam underslung from three supporting beams(10 ft span each) with an overhang of 5 ft?

 

[SlideRuleEra]

tobman - Good question - So now we are looking at a 25 ft beam that is continuous (and underslung) across supports at 0 ft, 10 ft and 20 ft. the last five feet are cantilevered. Load is -500 lb at the 25 ft point.
Load case 26 is for a simple beam so does not apply for the above continuous example. I ran it through a software program (I like BeamBoy Freeware for this type problem).

The beam sees the following:
P= -500 lb @ 25 ft point
0 ft to 12 ft - moment is positive (top flange in compression, bottom flange in tension)
12 ft to 25 ft - moment is negative (top flange in tension, bottom flange in compression)
M(max)= 2.5 k-ft @ the 3rd support (20 ft location)
Supports at 0 ft and 10 ft provide lateral bracing to the compression (top) flange.
Support at 20 ft DOES NOT provide lateral bracing to the compression (bottom) flange.

Therefore criteria to use in the tables is:
M(max)= 2.5 k-ft
Unbraced length of compression flange (from 12 ft to 25 ft) = 13 ft

From chart on page 2-175 the lightest beam is W6x9 (barely). Looks like a W6x12 would be a better choice.

Of course we have been ignoring the weight of the beam. In "real life" you must include it (the charts do not), usually as a uniform distibuted load for the entire beam length. With software this is no problem - you just make an educated guess as to the beam weight when you start and then check it(and adjust your calcutaions based on the accuracy of your intial guess).

 

[atvargas]

Tobman: recently I needed to do a jib crane then I calculate it as I found in Marks Manual (Mechanical Engineer)Page 5-30 Table 5.2.4. & Table 5.2.5 (Novena Edición en Español)
these are the formulas:
I found a practical method as follows:
You need to calculate beam first like a member supported at both ends at 8 times the capacity. (500#)*8=4000#
then using this formula:

+++++++++++++++++++++++++++++++++++1795AD
Maximum Safety Load (in Pounds)MSL:------=
++++++++++++++++++++++++++++++++++++++L

Where:MSL: Maximum Safety Load
A: Cross Secction Area
D: Height of beam
L: Distance between supports

then, suppose a size beam of 6":
..4000*10 40000
A=-------= ------=3.714 in2
..1795*6 10770

Area of :IS beam 6&quot;x12.5, A=3.61 in2<3.714, then use
IS 6&quot;x17.25 A=5.02 in2>3.714 in2

The load capacity of a member in cantiliver is 1/8 of capacity when supported at both ends

use this: (1795AD/L)/8

Jib crane is working now in a good condition after selection of beam with this method.

I think this can help you

 

[tobman]

SlideRuleEra:

I haven't seen the size of your neck, but I would guess that it is quite thick to support such a large brain. My thanks again for clearing up the conditions of &quot;unbraced lengths&quot;. I feel very confident in using the checks that I have been using for monorails now, and the use of the beam selection tables with the unbraced lengths as you have indicated. A great and thorough job of answering my questions!!!