Shallow Drilled Piers: Uplift, Lateral, and Overturning

[Delchi]

Scoreboards for school sports fields often have 2 ft diameter x 6 ft deep concrete "drilled piers" as foundation elements. The holes are drilled with a Bobcat-driven auger. I have questions about how to determine the capacity of these small, shallow piers in the following two applications:

Case 1: Wide flange columns are embedded in the concrete and flag-pole up to support the sign. In this case, the pier would be subjected to lateral, overturning, and some gravity. How can these capacities of the pier be determined?

Case 2: The piers are used in pairs to eliminate the overturning forces on a single pier. In this case, the leeward pier is primarily subject to gravit + wind forces while windward pier is subject to uplift. On these shallow piers, how can you determine the pier's resistance to uplift due to skin friction?

Most of the soils in our area are clayey sand or sandy clay.

Thanks!

 

[JAE]

This might help - take a look - a nomograph for laterally loaded piers in the ground.

 

[aeoliantexan]

Delchi, for Case 2, I would calculate the ultimate uplift resistance two ways and use the lowest result:

A. The unit skin friction at any depth z is W*z*Ka*tan(Alpha), where W=unit weight of the soil, Ka is the active pressure coefficient (roughly 0.3 for sand), and alpha is the friction angle between the sand and the concrete, say roughly 2/3 Phi. Phi is the angle of internal friction for the soil, perhaps 30 to 35 degrees for medium dense sand. You multiply the unit skin friction times the perimeter of the pier and integrate it over the full depth of embadment. This comes out to be 0.5*Pi*D*W*d^2*Ka*tan(Alpha), where D=Diameter of pier and d=embedment depth. I get abuy 1700 pounds for your 2foot by six foot pier. You can add the dead weight of the concrete. Remember, if any of the soil is below the water table, its unit weight is the bouyant unit weight, which is roughly half the total unit weight. If the water table is at the ground surface, the above value is cut in half.

B. Calculate the weight of the inverted conical volume of soil surrounding the pier with the sides sloping about 1 foot horizontally for every two feet vertically. I get about 7065 pounds. Again, you can add the weight of the concrete. Again, the number goes down if the water table is above the bottom of the pier.

Then I would apply a safety factor of at least 3 to get an allowable uplift load.

Check the Navy Manual, DM-7.1 and DM-7.2. It is availably on line.

 

[aeoliantexan]

Delchi,

Sorry, I posted the message before checking it. The weight of the cone is more like 14,100 pounds. Friction still controls.
Please run your own calcs; mine were intended as an example.

 

[kslee1000]

Delchi:

If subjects to high wind, the score board should be set on truss type structure, or as a minimum, each leg should consist of vertical post and slant kicker (like y). For the latter, ideally the pier should be aligned with the kicker to increase its effectiveness. Both piers (front & back) would then subject to simple compression (soil bearing), and/or tension (shear friction), depends on the wind direction.

 

[Delijosi]

The first method suggested by Aeoliantexan is the right one. Reason is that friction between soil and pile material in this case is lower than friction within the soil.

So use the first method.

 

[Delchi]

I appreciate the comments!

I will, of course, do all my own calculations, though I appreciate the examples. It appears that many of these scoreboard foundations are nowhere near the size they should be for the winds we have in the Front Range area.