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Beam / structure Calculations

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RustyH

Mechanical
Oct 7, 2013
58
Good Afternoon,

Over the years since qualifying as a mechanical engineer, Ive wanted to learn more about structural mechanics. We learnt a fair bit about about beam analysis when I done my degree 15 years ago, and the fact I went down a thermal engineering route means Ive had to dig out all the old structural mechanics books.

I've decided to set myself a challenge of designing a mezzanine floor, nothing more than a learning exercise and not to actually be assembled and used!

Ignoring all the requirements for seismic assessment, floor loading assessment etc. I just wanted to check if I was going about things the right way. Do you start from the tops, working down. So knowing your dead weights (structure weight) and live loads, then first carry out SFD and BMD on the joists, then on the main beams, then the pillars?

Thanks in advance for any help

 
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What are the acronyms SFD and BMD?

A confused student is a good student.
 
Steel or wood?

A confused student is a good student.
 
Hi,

Sorry, SFD = shear force diagram, BMD = bending moment diagram

I'm was thinking of calculating a steel main structure with timber joists, but I guess it doesn't matter as its just a learning exercise at the moment.

A program would definitely be useful, but I want to try and learn and understand the theory / fundamentals first

One questions, I notice there a certain regs that give KN/m2 values for minimum floor loads, but how do you apply this as a UDL when UDL's are KN/m
 
RustyH,
Your process is logical.
Multiply the specified loads/area by the spacing of your structural load carrying members to get uniform loads per unit length for your design. Google "tributary width" & "tributary area".
 
Thanks mate,

So say you had 400mm joist widths, would you base calculations for a joist on the 400mm area that could act up on it
 
Tributary width / area - just looked it up, I got ya, so I was on the right path with what I just said.

So as toy said, just just multiply out the KN/m2 value by the Tributary area acting on the beam.

Definitely something new I've learnt here. Thank you
 
Using inches feet & pounds (sorry), if your joists are spaced at 16" on center, and the specified load is 40 pounds per square foot live load and 20 pounds per square foot dead load, then your joists must support (40+20) x 16/12 = 80 pounds per lineal foot.
 
Got you, that makes sense

I am right in thinking dead loads are loads permanently in place, such as weight of beams, weight of walls or structures built on top of the beams, etc

And live loads are every day usage loads, such as people, storage, etc
 
You are correct in that assessment as well. Think of it as your dead loads being the stuff you can accurately measure and be reasonably confident in the magnitude. The live loads are the uncertain loads, might be there or might not, could be in excess of your design load or you could never have your actual live load even come close to your design.

It's more a question of confidence when choosing what is a dead load and what is a live load. But your thinking is the general consensus.
 
Thanks for all your help so far everyone.

As already discussed, Im starting with a Joist calculation, and made it with a UDL and single point load, with two supports (one offset from end)

I've had my head in the books and think Ive figured out how to do the BM and SF diagrams and the equations to go with it. Please see the attached for any comments

I am a little stuck now mind,. Where do I take it from here to work out if a specific beam profile and material can with stand this loading with out failure.
Do I work out the aera of moment of inetia and calculate the maximum Tensile and Shear stresses?
Do I need to work out the beam deflection as well, or providing the beam is not in a faulure mode then all is ok and deflection is not needed?

Scan_20150210.jpg
 
Right, I've tried to take a step further and do the stress calculations (see below)

I've based this on formulas from Roark's

Am I in the right ball park here??

Scan_20150211.jpg
 
You are in the right ballpark (though I'll caveat that by saying I've only glanced at your calculations) for the old fashioned working stress design.

Modern design for the vast majority of the world is now limit states. For a fundamental understanding of what is going on, you should jiay keep going with working stress as you have done so far.
 
Thanks for the input

Can you elaborate on what Limit State disign is?

Also, from the above stress, is it possible to determine if a beam would fail or not, do you just compare the calculated stresses against the ultimate tensile stress of the material
 
You're asking me to lay out two years of Uni... Pretty tough. Limit States and split factor LRFD (not the same thing) have been discussed here at length. Not to be flippant, but a search will give you plenty to read...
 
by the way, am I right in assuming that the stresses calculated above are enough to determine if a beam will fail or not by comparing with the Ultimate Tensile Strength of the material?
 
That's what CEL was alluding to... Limit states design factors your applied loads up, and your material resistance down, and then you check the calculated stresses versus the reduced resistance.

The old days, you took the applied loading and analyzed for that directly, then compared to an Allowable (or working) stress that was a reduced value from the ultimate stress.

Your analysis appears correct (at a brief glance) however actually determining whether it passes the design codes requires a bit more digging into the codes. For wood in the USA you'll want to be looking at the NDS, in Canada it's CSA O86. Other materials have their own governing codes.
 
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