max stress due to point load on concrete? 1

[kingnero]

Is there an easy way to check for the maximum allowable point load force on concrete?

Situation is as follows: a heavy workbench will be resting on the floor, supported on four bolts (bolted head down in the feet of the bench).
The bolts will enable to level the workbench. So all the weight of the bench rests on the four bolt heads.

Each bolt will see a max. load of 12.5 kN (static load), x2 (safety) = 25 kN. (situation in which the bench rests on three points is foreseen in this).
Area of the bolt head is around 520 mm², so a local 50 Mpa.


Concrete will be poured 20 cm (8") thick, of the (european) C25/30 quality, and will be reinforced with fibres.

If necesssary, I can still adjust the bolt sizes, I doubt I can change the prescribed concrete...

Could someone point me into the right direction for this?
If anymore information in needed, please do ask...
I'd like to see the method instead of a ready answer, so I can refer to this if later ever necessary.

Putting metal plates under the bolt heads to distribute the load is not an option as the bench will also have four casters, and will be rolled around regularly. The bolts will always be used whenever (and wherever) the bench remains stationary.

 

[Lion06]

I'm assuming that tha anchors at each leg are embedded? If so, and if they are headed in the concrete, you can just use the bearing equation - 0.85f'c*Ab. You're Ab will be the area of the bolt head. It will have fairly substantial capacity. If your loads are higher, use the pullout equation from ACI App. D (this isn't pulling out, but the failure mode would be the same just in the opposite direction) or bump your bolt size.

 

[kingnero]

there are no anchors, as I believe embedded means poured into the concrete?

The bench is moveable on four castors, however when stationary the four bolts will be unscrewed so that the bench rests at the bolt heads.
So disregard the castors, and suppose that the weight on the worktable rests at the four bolt heads.

see attached image ...

 

[desertfox]

Hi kingnero

I would just take the area of the bolt head and divide it into the force in each bolt, thats the pressure on the concrete, then compare that value with the concrete's safe stress.

desertfox

 

[Grahammachin]

To BS 5950 clause 4.13.1 (it is the steelwork code but stay with me on this), the concrete foundation bearing strength under a baseplate may be taken as 0.6 fcu (in the case of C25/30 concrete fcu is 30 N/mm2).

This is for column bases and so in theory is relevant for your application.

Therefore your allowable bearing stress is 0.6*30 = 18N/mm2

The only thing you would need to look at then is your punching shear, which will depend on your reinforcement, cover and whether the slab is suspended or ground bearing and if ground bearing what the sub-strata is.

Though I may be over thinking the last bit.

 

[kieran1]

If the stress under the bolts is excessive, you could provide a bearing plate under each support condition

Kieran

 

[kingnero]

Thanks for your replies.

As I said before, bearing plates are not an option as there is no way to make sure the operators will actually use them.

Grahammachine, If I should stay below 18 Mpa, and I've got a local 50 Mpa, that means I should enlarge the bolt area by about 3.

I am already using an M30 (typo in my first post), I'll have to see if I can accomodate an even larger bolt (d > 42 mm).

 

[dik]

Further to Kieran... having a bearing plate under the bolt head will also prevent the bolt from damaging the concrete surface when it is adjusted.

Dik

 

[Grahammachin]

An M30 bolt has a cross sectional area of 706mm2, the bolt head has an area of 1832mm2 (sizes taken from the Steel Detailers manual of 46mm across the flats and 53mm from corner to corner) giving a bearing pressure of 35.4 & 13.6N/mm2 respectively.

An M24 gives a head area of 1122mm2 (36mm across the flats and 42mm from corner to corner) giving a bearing pressure of 22.3N/mm2

The sizes can be found here

http://www.roymech.co.uk/Useful_Tables/Screws/Hex_Screws.htm

So you may want to check the values.

Graham

 

[trainguy]

If this is a fibre-reinforced slab on grade, there's a software package I used literally 20 yrs ago called Airport (by the Canadian Portland Cement Association) which performs exactly the checks you need for the type of point loading you're referring to.

There must also be many newer code equations that I'm not familiar with.

tg

 

[msquared48]

You could weld the bearing plates to the bolt heads if needed, then there is no problem with moving the assembly.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[kingnero]

True, the M30 already has a head with d>42.
key width is 46, so circular area of 1662 mm².
So the bearing pressure is below 18 Mpa.

Should I worry about shear/punch through? I really doubt that, as it's 8" thick on stabilised soil (not dug out).

@ msquared48: If I welded the bearing plates to the bolt heads, I would also have trouble adjusting the bolts... the feet are threaded for the bolts, it isn't a plate with a hole, and two nuts (so the bolt would remain in the same position).


Thanks everybody for your replies...

 

[BAretired]

A Factor of Safety of 1.5 should be sufficient, so 12.5*1.5 = 18.75kN is the factored load. The factored bearing resistance can be multiplied by (A2/A1)^0.5 but not more than 2 when the supporting surface is wider on all sides than the loaded area. So the factored bearing stress is 30.6MPa and the factored load is 15.9 if A1 = 520 mm^2.

Wouldn't hurt to increase the bearing area to 620 mm^2.

BA

 

[trainguy]

What is the c/c spacing of the 4 legs?

Could you get the 4 legs loaded at once, each with 25kN factored load?

Punching may not occur, but isn't it your job to prove, by calculation, that it doesn't?

tg

 

[Ron]

Mike is exactly right...but those items are available off the shelf. They are just load levelers. You can also get them with rubber "feet".

 

[kingnero]

@ BA: I'm using a FOS = 2 because there is dynamic loading (when working on the car, ...)
What is represented by A1 and A2?

@ Trainguy:
cc distance is about a 3 by 1 meter rectangle. perhaps give or take an inch.
And No that is not possible, 25 kN is if the table should balance on two diagonal opposite feet.

Yes and no, I do not have to prove it but would still like to check. This isn't exactly for work, I'm doing a friend a favor after hours but I still want to make sure everything is done as correctly as possible. I do know there are risks involved hence the checking.