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]

kingnero,

The problem with P/A +M/s is that you will get tension in the soil if the M/s is greater than P/a and unlike steel, soil cannot take tension.

The result is that instead of having 2 stress triangles of opposite mmagnitude (one tension and one compression) you end up with one compression triangle and a section of footing beyond this that has zero soil pressure. The centre of the triangle will coincide with the centre of the reaction. The rest is just basic mathematics.

 

[kingnero]

@ BA: thanks very much of walking me through this with your calcs.

I now understand that, if the case of P_total falls out ouf the middle third of the footing, the formula differs from the original:
It becomes: Q_max = 2x P/A_eff ; so the +-M/S part falls away.

It's not that hard, however it helps if someone walks you through that scenario.

Thanks again for your time and explications.

 

[csd72]

Kingnero,

Thats P/Aeff where Aeff is the width of the loaded area I mentioned above.

 

[BAretired]

kingnero,

I wouldn't say +-M/S falls away. It becomes 2*P/A_eff because M/S = P/A. Thus, P/A_eff + or - M/S_eff becomes 2*P_eff/A or 0 (triangular variation of soil pressure across effective length)

BA

 

[kingnero]

I really don't want to stretch this discussion out any further, as I already have all the answers I need, however when I do the maths, I get: P/A = 53.8 kN/m² (exactly like you said),
but:
S = width x length² /6 ; with length = 3 times the distance edge to c.g. of load (S= 1.5 x (3 x .2272)² / 6 = .11622 m³)
so
M/S = 17.5 kNm / .11622 m³ which is 150 kN/m².

So I still must be interpreting something wrongly.

@csd72: I understood everything up until Aeff (incl.), however I am having more problems with the second half of the formula, especially when I saw BA took the P/A part twice... didn't see that one coming!

 

[BAretired]

Don't worry about stretching the discussion. It is always best to fully understand something even if it takes a bit longer.

The moment of 17.5kNm is the moment of F(load) about the edge of the footing.

The c.g. of load occurs 0.2272m inside the edge of the footing. That means the eccentricity from the center of the effective footing is 0.2272/2 = 0.1136m. The moment you want is P*e = 55*0.1136 = 6.248 kNm and M/S = 6.248/0.11622 = 53.8 kPa which is equal to P/A.

BA

 

[BAretired]

In more general terms, say you have a footing L x B with the jib normal to B.

The total load of footing, crane and load is P which occurs at the third point of footing, i.e. at B/3 from the edge.

A = L*B
S = L*B₂/6
e = B/2 - B/3 = B/6 where e is the eccentricity of total load from the center of the effective footing

P/A = P/L*B
M/S = P*B/6 * 6/L*B₂2 = P/L*B

So P/A = M/S Q.E.D.

BA

 

[BAretired]

The jib is normal to L, not B.

BA

 

[BAretired]

Trying to type too fast...sorry. The equations should read:

A = L*B
S = L*B²/6
e = B/2 - B/3 = B/6 where e is the eccentricity of total load from the center of the effective footing

P/A = P/L*B
M/S = P*B/6 * 6/L*B² = P/L*B

So P/A = M/S Q.E.D.

BA

 

[csd72]

Wow, I am almost confused from those last posts and I know at least 3 different ways to solve this problem!

kingnero,

The reason why you double it is because it is a trangular distribution and you maximum stress is twice the average.

Just remember that the centroid of the triangle is one third from the largest end and that this needs to coincide with the location of the applied reaction.

 

[kingnero]

BA, many thanks for showing me exactly where I went wrong (I took the wrong moment, again, in my last post).

I've tried this with varying loads and surfaces, just to practice this, and it went just as it should. I feel confident now to show my calculations to (obligatory by the government) hired engineer, to have him comment on the allowable soil bearing pressure.

I hope he won't make a problem out of this (as he is only paid to give his findings/calculations about the shed foundations to the government, not to me).

Thanks again for writing this up, I think this post qualifies excellently for a how-to or faq, for future reference for people asking similar questions. Maybe a thought for the moderators?