Beam / structure Calculations

[RustyH]

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

 

[RustyH]

Thanks Gents,

I will do some more digging, but glad Im on the right path

 

[ajc2014]

Just for the sake of safety..... No you don't compare the stresses caused by applied loads to the ultimate tensile strength. Instead it's compared to the allowable tensile stress, which is the yield strength * Factor of safety. As I'm sure you have or will read. This is the older method, Allowable stress design or ASD. My terms may not be 100% because I myself don't use this method, but that's the idea behind it.

 

[RustyH]

Thanks ajc,

I guess what Ive done so far has not been wasted, but will do some reading on the more modern approaches

 

[CELinOttawa]

Also, depending on how it is set up, the allowable stress is the Fy DIVIDED by a F.O.S..... IE: 350MPa/4, with four being the "Omega", or Safety Factor (aka Factor of Safety)

 

[ukbridge]

Just wanted to add my support in this thread, think you've set yourself a really cool goal here rooting for ya buddy. In a (extemely small) nut shell yes, work your way down from the top. First work out what the structure needs to be designed for, get your loads first after doing the conceptual design. A knowledge of structural behaviour will come, eventually!

I recommend grabbing a good designers manual for the relevant codes and working your way through that, you'll learn how calcs are done in practice and all the various checks that are done. It won't cover everything but will get you off to a good start. Best of luck! Pete

 

[RustyH]

Thanks for the positive words everyone, I'm certainly trying, typical I was taught old school solid mechanics!! Haha. Themis new system is mind boggling!!

Can I ask,
What is meant by Characteristic Load Capacity?
What is meant by the term 'design'. I.E: 'design shear stress' or 'design bending stress'. Is it just referring to the design in question, or does it refer to predetermined design constraints??

 

[CELinOttawa]

The "design" in most terminology indicates that something has been factored. "Ultimate" is typically the corollary term.

Characteristic stresses and strengths are normally a material property point or somewhat arbitrary issue. Be warned this does tend to change between traditions and eras (read: countries and code generations), and should be read co textually. The most common characteristic stresses or strengths are found in offset materials (such as those which are brittle or have no distinct yield plateau) and large sampling groups on existing structures, where the 5% value is taken to be characteristic. The same is done with all wood/timber grades as well.

 

[RustyH]

Apologies......whats the 5% value mean (I did try a google search but not alot came up)

 

[Ingenuity]

whats the 5% value mean

I think CELinOttawa is referring to the statistical approach, like 95% confidence level.

 

[CELinOttawa]

Yup... Effectively in that CONTEXT (damn autocorrect!) The "characteristic" should be seen as indicating the value used in design. It is a practical thing when the material is brittle or highly variable, such as aluminium and timber, respectively...

 

[hokie66]

RustyH,
I have a different take as to 'characteristic' and 'design'. I am older than some of these folks, but I was taught that characteristic loads are unfactored. When you factor the loads, they become design loads. You use characteristic loads in serviceability assessment, and design loads for strength.

 

[RustyH]

So, am I right in thinking that the characteristic value is the value from the material datasheet / properties, and the design value is the end value after applying safety factors, etc

If so, if the design value calculated merely by the characteristic value * Safety Factor

 

[RustyH]

I found this about "design" strength values (which Im reading over and over to try and understand!)

Design Strength values
The characteristic strengths, Xk, are converted to design values, Xd, by dividing by a partial factor, γM and multiplying by a factor kmod.

Xd = Kmod(Xk / yM)

Values for these factors are included in the tables below.

Note: γM is not simply a partial factor for materials but also takes account of modelling and geometric uncertainties.

kmod = modification factor to strength values, allowing for load duration and moisture content

The eurocode , like BS 5268, allows the design strength determined using equation this be multiplied by a number of other factors as appropriate such as kcrit , kv , kc,90 and the loading sharing factor, ksys,where several equally spaced similar members are able to resist a common load. Typical members which fall into this category may include joists in flat roofs or floors with a maximum span of 6m and wall studs with a maximum height of 4m

The design values for the stiffness are obtained as follows

Ed = Emean / γM
Gd = Gmean / γM



Table for partial factor γM

Design situation Partial factor γM
Fundamental combinations ...
Solid timber 1.3
Glued laminated timber 1.25
Laminated veneer lumber (LVL),plywood, OSB 1.2
Particle board 1.3
Fibreboard hard 1.3
Fibreboard medium 1.3
Fibreboard, MDF 1.3
Fibreboard , soft 1.3
Connections 1.3
Punched metal plate fasteners 1.25
Accidental combinations 1.0
Serviceability limit states 1.0

 

[RustyH]

Good Evening All,

I have another quick question.

I'm starting to understand alot more the design limits concept of analysis, and found some great examples on gooooooogle.

In addition, I've downloaded a decent beam analysis spreadsheet that is based on this design limit approach. The spreadsheet is specifically for timber as thats what I will be doing an analysis on first and based on EN-1995.

The thing Im really struggling with is as follows

The spreadsheet give you an option to analyse based on a timber beam or a timber joist. If I set it to a timber beam, I have no issue in understanding how it works out the deflection (which is by the standard form in Roark). BUT, if I select it as a joist, the deflection is completely different (its less that half of as a beam). I cant for the life of me work out or understand why that would be (and if it should)

 

[RustyH]

Scrap that last post, Ive just figured it out.

For a Joist, it was taking the UDL as a per square value and then multiplying it by the joist spacing to get the direct UDL acting on 1 joist.

I'd already accounted for this and was entering the my calculated per unit UDL (which was instead being treated as a per square force)

 

[Triangled]

Hmmm.... Of course don't know what software you downloaded, so a wild guess.,,, some of the engineered wood spreadsheets took deflection credit for the joist by assuming some composite action with a wood diaphragm. If that's your case, you should be able to toggle that feature off and get a deflection linearly related to your moment if inertia, and also relatable to your weed beam calc.

 

[CELinOttawa]

I'm impressed Rusty... Keep it up!

 

[Triangled]

ahhhhhhhh......, seeing your scrap that last post.... Then also scrap my last post!! Good luck!

 

[RustyH]

I know I said 1 question, but just want to swing back round to Deflection again

I can calculate deflection by using roark's tables. (Which I assume is classed at instantaneous deflection)

But Ive noticed some examples work out the deflections individually (Dead Load Deflection, Live Load Deflection, Dead Load Shear only Deflection, Live Load Shear only Deflection) then sum them up at the end to give a final deflection.

Is there a need / advantage for this?


Also, I notice in the Standards, they mention deflection by creep. With it being timber in the first instance, I would imagine creep is quite important, but I can’t seem to find any reference to working this out properly.

 

[CELinOttawa]

Okay, so knowing specific components of the total defection will allow us to address them. A good example is cambering of steel beams, as well as some glulam "cambers", which are created during manufacturing.

As for the calculation, in many materials this is as straightforward as the elastic formulae you find in Roarke's, but not in timber and a few others (eventually you should look into th e plethora of methods in reinforced concrete!). Timber has now a number of factors applied to the elastic defection to increase this to a more realistic estimate.

Keep reading; You're doing great!