Calculating I Beam Size

Huh? The beam is 2m long, and the load is 2m from the end of the beam? In other words, the load is at the fixed end of the beam? In that case, it is a simple shear problem. You'll need a Capt Moosie or a Codemaster to deal with such complexity.

No no the other end... no the other, other end

I can see we are going to have no end of trouble with this

Del

Thanks for that!

I think he means a cantilever.

I might have a look when I get a minute, depending on the answers to IdeaSmith's #2.

Or was it a cavalier?

Not, "from", 'at'.

munger, you said, " Did do research but time is limited"

There is never enough time to do it right the first time, but always enough time to do it over.

Sounds like homework!

Why would you have to design this and ask in an engineering forum how to do this?

What sort of safety do you want to apply?

Any dynamic loads anticipated?

Or is it a static load?

Whats the tensile strength of the steel?

Did you do a research?

Start here! https://en.wikipedia.org/wiki/I-beam

Hi IdeaSmith

1. Not homework, wish I was still in school though...

2. I don't get much exposure with design but need this for a rigging job

3. Safety factor of about 1.2 would be sufficient

4. No

5. Only static loads (oh and yes it is a cantilever)

6. Tensile strength = 480MPa

7. Did do research but time is limited to get solution at the moment and lost my notes from university

Cheers

The internet replaces your notes from university me thinks.

There was plenty of information in my links provided.

I will not do the calculation for you. Sorry but I think it will be better if you do it if you are in charge for the rigging service.

Take a higher safety factor just in case!

http://www.brighthubengineering.com/machine-design/40491-how-to-select-a-correctly-size-steel-i-beam/

Good luck!

480MPa sounds like a typical UTS for structural steel. You would need a safety factor much higher than 1.2 to get a safe working stress.

But unsurprisingly, it's limited by deflection rather than stress. The biggest I-beam I have data on to hand, 203 x 102, gives stress 130MPa, which is OK, but deflection 8.5mm, or 1/236 x beam length, a bit near the knuckle.

UB 203 x 133 x 30kg/m gives stress 106MPa, and deflection 6.8mm, or 1/295 x beam length, still a fair bit of deflection, could be acceptable, it's up to you.

But you should swat up on beam theory and make sure you're happy before going ahead!

I should have said "swot" up!

I shudder when I hear, "don't know how, it is just a rigging exercise" - to me this means, "someone could get killed".

It looks like you do not understand what you are doing if you think a safety factor of 1.2 is likely to be enough. I was a design engineer in years gone by and don't recall ever using a factor that low.

Just one question to alert you to a potential risk, and one that even the odd professional has got caught on - "is there a possibility of a torsional mode/stability failure, and or how is this to be avoided" ?

I see this calculation:

Length of beam x load x span x safety factor x torsional load : then a miracle happens = answer.

A well-considered response. Always nice to learn something from another's thought processes..thanks!

Safety factor of about 1.2 would be sufficientI always go for a minimum of 2. Even then its iffy. As load is applied repetitively the contact area is work hardned and yield/tensile values change. We have had this problem with crane hooks.

I think, you must pass the licensure exam first, even if design of prototype involve only you riding on it. Dangerous stuff when life is at risk (your dog's, your neighbor's, your wife's, your children's life)

I suggest you must reflect on this first.

http://www.engineeringtoolbox.com/cantilever-beams-d_1848.html

Yes, I keep forgetting all the forumulae are available online these days. Call me old fashioned, but I would refer to my CISC or AISC handbook of steel construction. Have to make sure you know if it is guided or unguided.

The above is correct in simple terms, but there is a mode of failure not provided for in the formula, and a trap for the naive - that of the section twisting (and typically moving laterally) to bend around the minor axis - torsional stability failure. Regrettably, for that reason, the answer should not be rated GA.

Exactly correct. The failure mode is directly linked to high stress in the compression flange, and the laterally unsupported length of the section between lateral bracing mechanisms.

....additionally...

http://nptel.ac.in/courses/105106112/6_beams/3_behaviour_of_steel_beams.pdf

Nope, I'm not touching this one. Pay me to figure it out for you.