Bearers on Brick Piers

[ausdesign]

If the brick columns are internal (free standing & not engaged piers to an outer skin) then as silent said an M10 bolt is the nominal minimum requirement.
If it is a truss roof then there is no uplift requirements to take into consideration.
If it a pitched roof then specific (stronger) tie down of the bearer to the pier may be required depending upon the calculated wind uplift.

[brynk]

They taught us how to design trusses and calculate all the compression and tension values at each connection.
They taught us how to calculate all the tension stresses (and shear I think) in a suspended and cantilevered concrete slab.
They taught us how to calculate all the shear and tension stresses in a steel beam with point loads at random places along the beam, and they taught us how to design it by using the values that we got, against tables for the properties of steel sections. They also went into steel columns as well. I've still got the book with all the tables of the properties of steel sections somewhere.
And they taught us how to design a steel structure supporting an overhead set of traffic lights, ie a column that's bolted to a footing on the footpath, extending up, with a beam bolted on top that goes out over the roadway supporting the traffic lights. We had to calculate the torsional forces imposed on the column by wind forces acting on the traffic lights.

all of that as well as bending moments, allowable deflections, etcet - they teach first principles to aspiring builders now - rather than tables we used all the equations in the codes (steel framing, concrete, formwork, timber structures) with the various constants and multipliers. pain in the posterior - i just over-engineer it all anyway (sorry, build a safety factor into it ) cause most times the slightly additional cost will cover you against failure - what do they say? never enough time to do it right but always enough time to do it over!

As I've said, my memory's a bit cloudy, and it seemed a bit pointless to me wasting the time learning all that stuff since we weren't allowed to use it anyway.

hehehe its so you dont stuff around; and go and get an engineer to design it for you

[pawnhead]

rather than tables

Oh they weren't simply span tables if that's what you think. In fact I don't think there were any spans in them, but I can't recall. They taught us the same way that an engineer would design them back in the day. We had to do a lot of calculations and match the properties that we required to the properties given in the book.
I can't remember what the book was called but it cost me a bit.

If you're an engineer then you must have a book that you look up the properties of say a 360UB57 in?

What if I said how much load could you put on a 264 UB 97 before it fails?
There's no such thing so it's not in any books. Could you work it out if I told you that it has 50mm thick flanges, and a 5mm thick web, and it spans ten metres?
How many apples (Newtons) can you stack in the centre before it fails?
That's all the information you'll need to make one and test it if you can't work it out, but I suppose that you probably could work it out.
They didn't teach us that though.

[thebuildingsurv]

When i did the building diploma at tafe I had to do basic calculations, and when I did the degree it was very complex. I failed first time. I still have never used any of it and would know how to today. Isnt an bending moment when an male architect and male engineer agree on something and want to show how much they like each other mardi gras style.

[brynk]

no the ones i thought you were speaking of are those 'logarithmic suitability tables' that you can find in the old aussie institute of steel handbook, if you are lucky enough to have one (for christ's sake don't lend it to anyone!!), where you work off span and load on the x & y axes then find the beam that suits. haven't seen any for ages my old course coordinator who is a civil engineer and commercial builder for many years gave me a copy of one once & showed me how to use it. this probably explains why the course material calls for us to use the codes as an engineer would.

to give you an idea of how i would tackle that problem;

264UB97, 50mm flanges and a 5mm web. hmmm, sounds expensive -
my gut feeling for such a beam is that the web is undersized - the flanges seem awfully large in comparison & i would be raising an eyebrow or two if the engineer handed me such an instruction then i would be talking to them about why it has to be, and suitable equivalents

now since saying that i have looked up my handy 'section properties guide for universal beams' available from onesteel's website in the technical section (http://www.onesteel.com/productsupersearch.asp - enable javascript & give yourself a good hour to go through all the amazing stuff they have!). i'm guessing there is a limitation on the height of the space available for such a specific beam-depth. so the nearest 'off-the-shelf' equivalent is:
250UB 37.3, overall depth 256mm, flange 10.9mm, web 6.4mm; there is a whole buch of other stuff like the section modulus and a few other columns with I's and Z's and S's that i can't remember what stands for - lost in the booze-addled fog - that you plug into the equations. if at this point i was feeling particularly masochistic and felt like punishing myself because i had been a bad boy, then i would proceed to the standard for steel structures - as4100 - and painstakingly work my way through the equations, pausing oft to exclaim aloud that combination of three words that starts with 'what the ' and ends with 'k?' and maybe i would reach a point where i could give you an answer that might be in the field of play if i'd held my nose right, but more than likely gone over the dead-ball line due to that amazing defense mechanism - glazed eyballs

if however, i was having a good day, then i would proceed to ask myself... why are the flanges so bloody thick? maybe because you are using it as a beam to support some sort of insanely huge ub rollers carrying a massive load? maybe you are building your own a380-eater - i dunno. in anycase - the beam alone will be insufficient - that is my gut feeling, it needs more to support its own weight over a 10m span without deflecting beyond suitability (disregarding a pre-camber ofcourse - but then if there is a height limitation then how to get it into place), let alone considering the weight that it is going to support. the structural engineer can tell you exactly how many apples (yeah i like that one - mind if i use it?) it can carry but the builder will be asking why it can't be done differently in the first place.

if you convinced me that it had to be done that way then i'd give you a price on having some steel made up & figure out the way to put it in for you so you can go ahead and hoist your jet-engines into place...

r's brynk

[Dirty Doogie]

I'm about halfway through doing the "engineering design" component of my builders/building designers liscence. I showed one of my texts to a real engineer and he said he'd forgotten all that stuff LOL.

[Dirty Doogie]

LOL Brynk!

One of the questions on the assesment test is " design an 18 metre lintel & support seating for the front of a townhouse complex. The lintel is on the ground floor with another 2 stories of block construction residental space above. refer to attached plans and specifications" etc etc

Ok so you start working through the loads and double checking - its not long before you realise that such a lintel is going to cost about $160000 plus transport and placing costs.

then the numbers start making your head spin so you go out and get some lunch and you notice as you look around that there aren"t any 18 metre lintels anywhere. You go driving around further and no 18 mtre lintels can seen.

There are 18 mtr glass fronts hanging in a curtain wall, 18 mtre openings under canterlevers but no 18 mtre load bearing lintels. Why are they getting students to design 18 mtr lintels if no one uses them? An email goes off to the tutor....... am i misreading the question?

A few days later the tutors reply comes through ..." the objective of the question is to demonstrate an understanding of the engineering principals that must be bought to bear when asked to design a building element. The correct response to this question also involves looking at material supply options beyond those generally available ie steel sizes/grades custom made.

In practice I agree that an 18 metre lintel in the situation described would be improbable due to costs."

Ah OK ! My calculations were indicating I'd need 2 UB each 950 mm x 400mm 35 mm web 55mm top flange 70 mm bottom flange with 20 mm compression webbing at 400 centres - weighing --- that cant be right! 23 tonn!!s How silly is that?? Can any one make that up??

I sent an email ofF to One steel with my engineering specs - one week later they send back an email saying their engineers looked at the problem and have attached a solution as well as seating details. In essense they have solved the problem and answered the question for me ..... I just had to laugh! they even gave me a price delivered to Brisbane - $198,000.

So for assesment I'm going to hand in One steels solution plus my own initial calculations - becuase lets face it - in the real world a builder or designer would do exactly the same thing LOL

[DvdHntr]

What if I said how much load could you put on a 264 UB 97 before it fails?
There's no such thing so it's not in any books. Could you work it out if I told you that it has 50mm thick flanges, and a 5mm thick web, and it spans ten metres?
How many apples (Newtons) can you stack in the centre before it fails?
That's all the information you'll need to make one and test it if you can't work it out, but I suppose that you probably could work it out.
They didn't teach us that though.

Yes, of course I could work it out. That is part of what first principles means. So what you did was use the effective modulus from the AISC or BHP handbook. I don't think they would have taught you how to calculate the elastic and plastic section modulus. But I would also need to know how it is restrained and the centres of the internal restraints. Also, what the grade of the steel is.

[DvdHntr]

I showed one of my texts to a real engineer and he said he'd forgotten all that stuff LOL.

Does he work in structural design? I can't believe that if he does he could forget it. Although some do rely on computer programs alot.

[DvdHntr]

In the framing manual, tie down is specified for all internal piers. The bearers are tied to the piers, the joists are tied to the bearers, the walls are tied to the joists and the trusses are tied to the walls. If you miss out any of those, it seems pointless to have tie down at all.

Any decent bricky should have done thousands of them. From what I've seen, the rod through the pier is the most common method. You want something that is going to be solid and I don't think dynabolting a bracket to a single brick at the top of the pier is going to satisfy that.

Maybe you should consider using steel piers? The Duragal or Uni pier systems satisfy the tie down requirements. You wont need a bricky at all then.

The piers are being supported off an existing slab. It is exposed so there will be little to no uplift. I only want to restrain the sliding. My other thought was to just skew some masonry nails through the bearer and into the top of the pier.

I have other brickwork so I need the bricky anyway.

[brynk]

then the numbers start making your head spin so you go out and get some lunch and you notice as you look around that there aren"t any 18 metre lintels anywhere. You go driving around further and no 18 mtre lintels can seen.

...

Ah OK ! My calculations were indicating I'd need 2 UB each 950 mm x 400mm 35 mm web 55mm top flange 70 mm bottom flange with 20 mm compression webbing at 400 centres - weighing --- that cant be right! 23 tonn!!s How silly is that?? Can any one make that up??

...

keep driving all the way to canberra then go to the national gallery's underground carpark & you will see one similar to these in action

...
I sent an email ofF to One steel with my engineering specs - one week later they send back an email saying their engineers looked at the problem and have attached a solution as well as seating details. In essense they have solved the problem and answered the question for me ..... I just had to laugh! they even gave me a price delivered to Brisbane - $198,000.

>>>
hahah damn why didnt i think of that - you will probably get top marks for that answer

>>>

So for assesment I'm going to hand in One steels solution plus my own initial calculations - becuase lets face it - in the real world a builder or designer would do exactly the same thing LOL

dvdhntr,

how high off the ground are your piers?

do the limit state design equations bring suitability of a member in under the elastic modulus in the equation? or the plastic? i'm guessing there is a fair margin of safety built in - what do you think the safety factor would be if you went to 100% of capacity according to the equations?

haha julia gillard just announced they have a special 500-million dinar fund to alleviate infrastructure costs in new developments being passed on to first-home-buyers... gazes into his crystal ball and sees civil engineers' and contractors' fees mystically going up as we speak...

r's brynk

[DvdHntr]

The highest they get are 230. I will have special termite barriers installed to meet the BCA requirement. The walls on the flooring will be non load bearing, I am not touching the roof structure and not tying into it. I cannot see any load lifting the bearers off the piers, so the only concern is sliding of the flooring. The actual weight of the flooring and walls should hold it in place but I wanted to know if there was something that could stop the sliding. Silent C's suggestion would but I think it is overkill.

The design by elastic method uses the effective section modulus. In order to calculate it you need to know both the elastic and plastic modulus. Elastic is the moment of inertia divided by the centroid and the plastic is calculated by the parallel axis theorem. Then once you calculate the nominal sectional capacity you then assess the restraints to find the member capacity. That is the basic procedure for elastic analysis. You can then calculate plastic moment if you have an indeterminate structure and there is also second order effects that come into play when you go plastic.

If you are talking factors of safety, it depends on the nominated duration of the load.

Permanent actions (dead loads) have a 1.2 factor applied but that is coupled with a material phi factor (0.9 for steel). Imposed actions (live loads) have a factor of 1.5. And they are combined so the factor of safety could be around 1.3-1.8. But that is for a statically determinate structure on one member. More indeterminate and complex the structures would have safety factors in excess of 2.

Well as I am not practising civil engineering that doesn't really concern me.

[silentC]

Silent C's suggestion would but I think it is overkill.

It's not my suggestion, those are the nominal fixings for bearers on brick piers as specified in the framing manual. It doesn't qualify it by saying that "if this" or "if that", it's quite clear that bearers are to be tied down to all internal piers using the nominal fixing method as a minimum, regardless of any other factors.

If you want to do something different, then you're on your own. I don't think a masonry nail into a brick is going to hold anything much. As you no doubt know, bricks are very strong under compression but weak under side loads or under tension. If you nail a bearer to a brick at the top of the pier and the bearer wants to move sideways or back and forth, the brick wont stop it.

[ausdesign]

I hope thats 230 centimeters.

[pawnhead]

- mind if i use it?)

You'd have to ask my old teacher. He's the one that always used to tell us that one Newton = one apple, now how many apples blah blah blah,,,,

but the builder will be asking why it can't be done differently in the first place.

Yes, I was attempting to make it look a bit ridiculous. It's purely a theoretical exercise, and if it can't carry it's own weight, then how many helium balloons (antigravity apples) are required to be attached to the centre to prevent failure.
I'm not really that interested in the answer BTW so you don't actually have to go to the trouble

edit:

Yes, of course I could work it out. That is part of what first principles means. So what you did was use the effective modulus from the AISC or BHP handbook. I don't think they would have taught you how to calculate the elastic and plastic section modulus. But I would also need to know how it is restrained and the centres of the internal restraints. Also, what the grade of the steel is.

Aha!
I thought you guys were pretty cluey.
Engineering only covered an hour a week for two years, so I'm sure that there was a lot more to learn than what they taught us. It was only a small percentage of the entire BS that we were taught. e.g. we spent six months (an hour per week) studying building history back to the Roman days.