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Participant

Join Date: Jun 2014
Posts: 3

06/25/2014 7:52 PM

Hello, we are mounting a sign to the top of a 38.5 foot pole. In the windiest conditions, 1600 lbs of force can be exerted at the top of the pole. This pole is mounted in the center (4.25 feet from each side) of a 8.5 foot wide x 8.5 foot long x 2.5 foot deep concrete base. The concrete base sits on top of the soil so the top of the pole is 41 feet from ground level. The concrete base weighs 27,094 lbs and the pole and sign weigh 5,152 lbs for a total weight of 32,246 lbs. We're trying to determine whether the soil can support the total weight of the pole, base and sign and the moment loads that are imparted in windy conditions. The load bearing capacity of the soil is 1500 lbs per square foot.

Thanks kindly, Tom

Pathfinder Tags: load moment Soil wind
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Guru

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#1

06/25/2014 10:02 PM

The soil can support 108,375 pounds over 72 square feet.

Looks OK to me.

I can't imagine wind loading being a problem, either, although I haven't looked at any numbers.

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#2

06/25/2014 11:07 PM

I'm no civil engineer but doesn't a force of 1,600 lbs at the end of a 38.5 foot long lever create an overturning moment of 61,600 lb-ft, a bit more than twice the weight of the base? That's a big mass to be suspended so high at the end of such a relatively light base, and that doesn't even take dynamics and off center movement in gusty winds into consideration.

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Join Date: Jun 2014
Posts: 3
#3

06/26/2014 2:04 AM

RAMConsult, I don't believe there's any overturning risk even in the windiest conditions. Perhaps the way to think about it is there are two moment arms working against each other. The weight of the concrete, pole, and sign form a moment arm that is equal to the total mass times the distance between the center of gravity to the fulcrum (in this instance, the edge of the concrete). So in this case that's 32,246 lbs x 4.25 ft = 137,045 lb-ft. That force is acting in the opposite direction of the wind loading and given it's over twice the torque, which you correctly calculated at 61,600, it should prevent any risk of overturning.

What I'm having trouble understanding is how to convert 61,600 lb-ft of torque to pressure on the soil. I can't quite visualize how the load is distributed across the concrete slab.

Regards, Tom

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#4

06/26/2014 5:54 PM

I think you are over-thinking this.

How can you come up with 137,045 as force?

Participant

Join Date: Jun 2014
Posts: 3
#5

06/26/2014 10:36 PM

Lyn,

To be specific, we could say gravity acting on the mass of the object is the force (just a highfalutin way of saying it weighs 32,246 lbs) and this downward force creates 137,045 lb-ft of torque. It should be remembered that the pivot point is the bottom, outside edge of the concrete pad and all torque values are derived from it. The torque that's generated by the weight of the object has to be calculated to ensure that the counter acting moment being created by the wind load won't overturn the object (in this case the sign, pole, and concrete base combined). You're point that the soil can withstand over 108,000 total lbs over 72 feet is correct but it doesn't, in itself, tell us whether that's enough to support the wind load moment.

In case you or anyone else is interested though, I did finally find an equation for calculating soil loads for just this scenario as follows:

Q = 2P/[3(B/2-e)L]

Where

Q = soil pressure in psf

P = Total weight of object (pole, base, etc)

B = Width of footing

L = Length of footing

e = eccentricity (technical term in civil engineering) It's derived by taking the torque generated by radial loads on the pole (in this instance, wind) and dividing it by the weight of the object.

So for the problem originally presented, it calculates as follows:

e = (1600*41)/32246

e = 2.034

Q = 2*32246/[3*(8.5/2 - 2.034)*8.5]

Q = 1141 psf

So the soil pressure exerted from both the weight and moment is 1141 psf which is less than our soil bearing value of 1500 psf. We appear to be in good shape!

Kind Regards, Tom

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#6

07/06/2014 9:10 AM

Interesting question! You didn't give the derivation of the eccentricity formula, so I tried working it out. Clearly no tensile stress is allowed between the slab and the soil, and I assumed the load on the soil is zero up to a line distant x from the "back" edge of the concrete slab, then increases linearly up to the front edge of the slab.

Got x = B - 3/P*(P*B/2 - Tp) = 1.853 ft and soil pressure = 2*P/(B - x)/L where Tp = torque from the flagpole.

Gratifyingly, it gives same answer, soil pressure = 1141 lb/ft2. Relation between e and x is x = 3e - B/2.

But your formula (I think) and the above are based on the torque from the flagpole square-on to the slab. But is this the worst case? Do you have a formula for maximum soil pressure when the torque is diagonal (as it obviously can be)? I tried to work it out and got maximum soil pressure 926 lb/ft2, and x = 1.602 ft but the math was pretty horrendous and I could have made a mistake, or an incorrect assumption.

Ignoring the mass of the slab and treating it as a beam in bending clearly gives higher tensile and compressive maximum stresses in the diagonal case (as the section Z-value is √2 x smaller). Also if the requirement was to ensure non-negative stress everywhere (overkill, but to help analyse the problem), by increasing the mass of the slab, the diagonal case would have to be considered. But in the diagonal case the mass of slab also gives greater moment to resist the applied torque, and my guess is this is more important, giving lower pressure and 926 lb/ft2 could be right. So I think you're safe in taking your 1141 lb/ft2 as worst case, but if you have a formula or calculation for the diagonal case I'd be interested.

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#7

07/09/2014 11:56 AM

Don't forget to include applying a Factor of Safety for both Sliding and Overturning Moment.

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#8

08/09/2014 3:38 PM

One item you appear to have neglected is wind on the pole itself which may add significantly to the calculated wind moment at the base.

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