What moment can a poured concrte block deliver? 4

[kingnero]

I want to install a small jib on my private property, which will be used to unload my trailor. (bags of charcoal, small things like that).

I have an existing frame with baseplate that I will be using, in combination with an electrical hoist.

I have calculated all relevant things about the jib, (moment at the baseplate, bolts, ...) however I don't know how to design correctly the concrete block to which the assembly will be bolted to.

Please look at the attached sketch,
Could someone please point me in the right direction?

What I am trying to achieve is a way to calculate the necessary dimensions, so that the block will be able to resist the moment coming from the load.

Many thanks in advance.

 

[csd72]

this is a fairly simple calculation assuming that your concrete is going to be thick and strong enough to support itself.

There are 2 things to check:

1. Resistance against overturning - take moment of resistance around the toe of the footing (i.e the edge closest to the load being lifted) and calculate this as dead load times the distance to the centre of the load. This should be at least 50% higher than the appliad moment.

2. Bearing pressure- any soil that is not obviously soft will take 50kn/square metre. This should be just a simple stress calculation of P/A + M/S but check that there is no tension otherwise you will need to use a more complex formula.

 

[a2mfk]

The trickiest part of your problem will likely be the anchor bolts. Many engineers conservatively design the back two bolts opposite of the direction of the moment to take all of the tension, with the rotation axis around the front of the plate or the front group of bolts.

If this is an existing foundation/slab, you can use expansion bolts or epoxy bolts. These are proprietary and the manufacturers have tables with load values and reductions for spacing and edge distance. If you want to use cast-in-place anchor bolts, have fun with Appendix D of ACI!

 

[kingnero]

@ csd72: If I understand your terminology correct (I'm dutch...), I should calculate the following moment: force due to the weight of the block, times the distance, which are added in red in the attached drawing?
If so, thanks very much, I will quickly have my dimensions from your information.
If I am not correct, please correct me...

Bearing pressure will be minimal, I have very small loads but I will check for it anyway.


@ a2mfk: I have calculated everything, I have chosen bolts, the mechanical side is fully covered. I only need to dig out the hole, and pour the concrete.

 

[a2mfk]

You got it right, simple as force x distance. Now the most economical way to increase your resistive moment is to increase the distance of your moment arm- by making your pad footing dimensions larger rather than thicker (though thickness adds weight). If you will only be using the jib in one direction you can increase the length of the pad just in that direction. But this also increases your concrete bending stresses, but from your description of what will be using this jib for, I don't see you having a lot of problems with just minimal reinforcement in the concrete.

 

[msquared48]

You'll get better overturning resistance leverage if you move the jib bearing plate more to the rear of the block. Also, your maximum soil bearing stress will be less. Just have to watch out that the afterset bolts are not placed too close to the edge of the concrete b lock. You may want to consider using chemical anchors, as in epoxy.

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

 

[csd72]

kingnero,

Yes, the moment you calculate in red should be more than 1.5 times the original moment you calculated.

You need to realise that at the base it is not the strength of the bolts that is critical but their anchorage into the concrete.

 

[kingnero]

I have a steel frame welded at the correct dimensions, which is drilled and tapped for the anchor bolts. With horizontal crossmembers, which will also servo to fixate the frame in the pit.

This frame will be placed in the dug out pit, at correct height, properly fixated, than concrete (reinforced with fibers) will be poured over it.

I don't worry about pull out, as this frame is massively oversized.

@ msquared: The Jib rotates at the top, so I cannot place the base plate at one side. However I have taken a moment into acocunt with a very large COS, and I will take care when using the thing. For the safety purists: if it fails (which it most likely will not), I will be the only one underneath it.

I will also make sure the concrete base is large enough, more for the leverage than for the bearing pressure (which isn't a large %)

Thanks everybody for the helpful replies, I can go on with this now...

 

[a2mfk]

Officially: We all as structural engineers offer you advice only in theory and must suggest you consult with a local professional engineer for your project.

 

[kingnero]

OK, for the record: I will do so.

 

[Bobber1]

I sent a reply but I think my company is now blocking use of this site.

 

[Bobber1]

Looks like I got in.

How close will your trailer be to the edge of your concrete footing. You may have to consider the loads from your trailer wheel on you soil pressures, too.

Also, add at least a small dynamic factor for load lifting.

Good luck.

Bob

 

[slomobile]

re: mssqaured comment about moving the anchorage away from the loaded side of the crane. If it is a true job, then it will have some sort of rotational capacity at the base, so moving it to the 'edge' would be helpful in once case, but then upon the rotation, it would become the challenge. Centered might be best depending on the angle of free rotation for the job crane itself.

Good luck.

 

[kingnero]

@ Bobber: I did take the force due to two persons on the concrete block, did not think however of the wheel of the trailor. Good point.

Also, add at least a small dynamic factor for load lifting.
The jib is rated for 1.5 tonne at the end (= three meters distance from the vertical pole), so I did all calculations with that in mind, however the electrical hoist I have can only lift 500 kg. That means, IF i should ever use the full capacity of the hoist, I will still have a COS = three.

I am quite confident I will be on the safe side.

 

[kingnero]

One final question, as I have never used the P/A + M/S formula before:

*) I have no eccentricity (the jib will be mounted in the center of the concrete block).

*) Also no My moment (for the sake of keeping it simple, as the jib can only be turned towards one side at the time, and the footing will be square, so Mx = My)

Do I need to add the M-load (moment due to the load on the jib) and the M-block (moment due to the weight of the footing) in the formula, or not?

See attached sketch for further info.
Thanks one more time...

 

[BAretired]

P = F_load + F_concrete

M = F_load*e where e is the eccentricity of the load from center of footing. The footing has no moment about its own centroid.

A = l² since footing is square

S = l³/6 since footing is square.

This is applicable if P/A >= M/S. Otherwise, an adjustment is needed. That is because you cannot have a tension stress on the underlying soil.

In this analysis, the available pressure on the side of the footing has been ignored which is a bit conservative, but okay.

BA

 

[TXBRIDGEENG]

If you still need help, I will lay out all the steps for you. I have extensive structural and geotechnical background. Do you know what type soil do you have in your area? Clay? Sand? I won't be back until November 9, 2010. If you still need help by then, I will gladly help you in detail. I will give you a worked example of your problem for you to follow.

 

[kingnero]

@ BA: OK, I see where i made a mistake: I also took the moment of resistance of the footing into account.

However, even when I leave that one out of the equation, M/S is still larger than P/A.
(see attached sketch, again)

You say: in that case, an adjustment is necessary. I think this means I need to design a larger footing?

However, see csd72:
1. Resistance against overturning - take moment of resistance around the toe of the footing (i.e the edge closest to the load being lifted) and calculate this as dead load times the distance to the centre of the load. This should be at least 50% higher than the appliad moment.
This condition is met. Would I still need a larger footing?
M_footing=40 kN x 0.75m = 30 kNm which is larger than M_load=17.5 kNm x 150% => 26.25 kNm.

I am terribly sorry for all these questions, I can assure you I'm better in my field of experience as I can only imagine what you guys think of me now, having all those problems with a simple formula.

I have googled for this, and have found

http://www.ce-ref.com/ftg_size.htm

but there only eccentricity larger than 1/6 x footing length and smaller than 1/2 x footing length is given. I believe my problem is because my eccentricity is much larger than the footing size.

@ TXBRIDGEENG: Greatly appreciate your help, as I would like to understand this better as this really is a fascinating problem.
For your information, it is clayground, and a rather wet area as the groundwater level is not very deep. However as I am also constructing a large storage shed, I have hired a local engineer who will do the stability calcs for it (required by our local government). I will consult him for the allowable bearing pressure.


Thanks again for all your help.

 

[BAretired]

kingnero,

There are two parts to the calculation...Factor of Safety against overturning and soil pressure. In my earlier post, I was referring to soil pressure.

If M/S > P/A you do not necessarily require a larger footing. What you must do is find the effective size of your footing.

Mfooting=40 kN x 0.75m = 30 kNm which is larger than Mload=17.5 kNm x 150% => 26.25 kNm.

Working backwards;

L = 2*0.75 = 1.5m
F_load = 1.5 tonne = 1500 kg (mass) or a force of approximately 15 kN.

F_load occurs at 17.5/15 = 1.17m outside ftg.

P = 40 + 15 = 55 kN (neglecting weight of jib)
A = 55/2.25 = 24.4 kN/m^2

M = 17.5
S = 1.5^3/6 = 0.5625
M/S = 17.5/0.5625 = 31.1 kN/m^2 which is greater than P/A

Find eccentricity of load, i.e. e = M/P = 17.5/55 = 0.318 from center of ftg, or 0.75 - 0.318 = 0.432m from edge of ftg.

Effective footing is 1.5m x 3(0.432) = 1.5 x 1.29
Effective area = 1.94m^2
P/A = 55/1.94 = 28.3 kN/m^2
P/A +-M/S = 56.6 or 0 (triangular variation)

For clay soil, 56.6 kN/m^2 or about 1180 psf seems okay, so footing size is adequate.

BA

 

[BAretired]

kingnero,

Made a mistake in my last post.

Say F is load and W is footing weight.

I am assuming that F = 15 kN and W = 40 kN.

Length of jib arm = 17.5/15 + 0.75 = 1.917m (modify this if wrong)

P = W + F = 55kN (neglecting weight of jib)
c.g. of load @ 15*1.917/55 = 0.5227m from center of ftg. which is outside the middle third of the footing, so M/S > P/A.

Distance from c.g. to edge of ftg = 0.75 - 0.5227 = 0.2272m. Effective length of ftg = 3*0.2272 = 0.6816m

A(effective) = 1.5*0.6816 = 1.022m^2
P/A = 55/1.022 = 53.8 kPa

Max soil pressure = 2*53.8 = 107.5 kPa (2250 psf)
Min soil pressure = 0

The pressure may be okay if the clay is fairly stiff, but it is getting a bit on the high side.

BA