I am installing a 24 ft 4x4 square quarter inch thick steel tube 4 ft into the ground with a concrete base of 28 inch diameter and just over 48 inches deep, so 20 ft of the post is above the ground. This post will be have heavy horizontal loads pulling it from the top of the post. I`m curious if the concrete base is deep enough and if reinforcement is required to keep the concrete from failing.

Thank you,

  • This link: engineersedge.com/calculators/… has been recommended on some related questions here; it requires a bit of know-how but is a fantastic DIY resource. – Fred Shope Aug 15 '19 at 19:30
  • you should be asking how deep do i need to bury a post in order to have 20ft above ground – jsotola Aug 15 '19 at 19:34
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    Is the direction of the force the same? Can the pole be down guyed? – JACK Aug 15 '19 at 19:35

Whether your concrete base will survive is irrelevant, because your beam will fail.

See: https://www.amesweb.info/StructuralAnalysisBeams/Stresses_Steel_Hollow_Structural_Sections.aspx

Input: enter image description here Output: enter image description here

Notice the highlighted field of the output. That's the stress you'll be putting on the connection with the ground, (assuming it's rigid). The yield strength of steel (where it stops bending back when unloaded) is 36,300 psi, You'd be overloading this beam 1.5 times. It will fail.

Consult an engineer to design something for your purposes.

EDIT: I've done a recalc with a cantilevered beam calculator in response to the comments below. The result is essentially identical, and any difference between the two calculations is likely rounding. Simply put, two different calculators, with two different sets of input, give the same result:

I = 8.82813 in^4, and the 2 inches is the distance from the edge to the center.

enter image description here

There are two things to note in the results:

  1. A stress of > 36000 psi, the yield strength of steel, and
  2. A deflection at the end of almost 18 inches, which throws the geometry completely out of whack, and accelerates failure.

Further EDIT: The Plastic Modulus for this section is 5.28 in^4 Multiply that by the yield strength gives a Plastic moment of 190,000 in.lbf, which is ~16000 ft.lbf, which is LESS than the moment at the support of 20000 ft.lbf.

TLDR: You're loading the entire structure into the plastic region, turning it into a plastic hinge. This WILL fail.

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  • The OP is asking for help for a “post” (cantilever) not a beam. Your references are for a 4x4 beam (material supported at both ends). – Lee Sam Aug 16 '19 at 1:36
  • Doesn't matter. Internal stresses are the same. This calculator doesn't take a length argument, nor support arguments. It's really just calculating stresses at a section, and my loads are correct. – Chris Cudmore Aug 16 '19 at 12:52
  • That is not correct. The stresses are different for cantilever beams than for simple span beams. The maximum moment for a cantilever is (P L) and for a beam it’s (P L / 4). Likewise, the maximum reaction at the base of the cantilever is (P) and for a beam it’s (P / 2). Your calculations and charts are incorrect. – Lee Sam Aug 16 '19 at 21:07
  • Show me where L comes into these calculations. – Chris Cudmore Aug 19 '19 at 13:02
  • “L” is the span. The greater the span the greater the stress (moment). – Lee Sam Aug 19 '19 at 14:55

I’d use #3 rebar because it can be bent by hand...no special bending tools or equipment is required using #3 rebar.

I’d install 4 - #3 rebar vertically and 1 - #3 rebar tie at 8” on center horizontally to keep your concrete base from cracking and failing.

Also, remember to keep all rebar a minimum of 2” from the edge of your form and 3” from the bottom. (Your steel post should also be a minimum of 3” from the bottom.) Use wire ties to keep the rebar in the correct position.

Another person has indicated that your 4x4 will fail. This is not true.

The section modulas for a 1000 lb. load at the top of a 20’ cantilever pole is:

S = M / Fb

The moment for a cantilever is: P L which is 1000# x 20’ = 20,000 #feet or 240,000 inch lbs.

S = M / Fb therefore is 240,000/36,000K or 0.0066

A 4x4 with 1/4” wall thickness is 4.0 which is more than adequate. (I didn’t check for shear, because bending will govern.)

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