How much force can a pole in ground take before falling?

Question

Consider I have a pole stuck in the ground as illustrated below:

When a force F1 is exerted on the pole, its bottom end will push against the ground it's stuck in; the ground then exerts a resistance force. Intuitively, if F1 is very strong then the pole's upper end will move in F1's direction and if F1 is very weak, the ground resistance Fr keeps the pole in place.

Given some measures of the pole, specifically, how deep it is (L2) and what radius r it has (it is a cylinder), I would like to be able to determine the magnitude required by F1 to move the pole's end.

I have asked a similar question, with more physical details on another stack exchange site, too. Here is the link.

Update:

Assume that the ground is made of only one material, with uniform density, disregarding any "dirty" effects, such as moisture, or temperature. The length of the pole, its radius (it is a cylinder), and its depth L2 are all constant.

Let's view the pole as a lever with its pivot where air, ground, and pole meet. Then we have F1 * L1 = F2 * L2 when the lever is in balance. Now, I would like to find out at what force F1 the pole starts moving; the pole is moving when F1 * L1 > Fr* L2.

The problem would be solved if we find a means to compute Fr. My idea was that as the pole (in the ground)pushes towards the right at every height 0 > L >= L2 (L = 0 on dotted area where air and ground meet), there is a resistance force Fr(L) at each height (view illustration here). For big L this force is small while for small L the force grows. I assume Fr(L) must grow in a linear fashion as it does with levers (F1 * L1 = F2 * L2). Assuming that, Fr could be computed by F2 = 0∫L2 Fr(L) dL.

The solution need not be that exact, the force need only be computed approximately: the force will be used by game simulation the physics need only appear real. All that is required is a means to compute what resistance force Fr the pulling force F1 must overcome to move the pole.

Answer 1

Depending on the kind of engineer you ask you will get many different options to approach this - your question is pretty open-ended. As a Geotechnical Engineer I'd say your pole can be looked at as a retaining wall. Now, if the ground is modelled as real "dirt" it will have some give. What would happen is that the pole will rotate about some point in the ground (again - you can go into this in more depth) and also bend (or not if rigid).

Now this rotation causes a the rod to displace the soil, which exerts "passive" or "active" lateral pressure, depending on whether it is stressed further of allowed to relax (i.e. rod moving into soil region or moving away from it). These pressures are given by the coefficients of passive and active earth pressure ($K_{p}$ and $K_{a}$) and the vertical effective stress ($\sigma_{v}' = hg\rho$). The coefficients are found from the angle of friction of the soil ($\phi'$) as:

$$ K_{a} = \dfrac{1-\sin \phi'}{1+\sin \phi'} $$ and $$ K_{p} = \dfrac{1+\sin \phi'}{1-\sin \phi'} $$ So passive pressure would be $P_{p} = K_{p}\sigma'$ and so on. A rigid rod will then rotate around the point where all the forces on it are balanced. You can assume some value of $\phi'$ and be on your way. These pressure will grow linearly with depth, so the resulting forces from the different active/passive wedges will act at 1/3 from the bottom of each wedge. Everything can now be expressed in terms of the distance between the pivot and the ground level. Equilibrium around the pivot will give you its location and then the sizes of the forces can be found.

The picture above is how I would approach this (neglecting the active soil wedges, if included they mirror the passive ones with a limiting pressure as given above). From equilibrium about the orange point you can find its depth and then the values of the pressures (with appropriate Fr and soil parameter $\phi'$ assumed). Or you can assume a value of x and find the Fr that causes movement. Many options possible.

Answer 2

this is a typical engineering problem and has an equation approved by IBC numbered 18-1.
$$d = 0.5 A{1 + [1 + (4.36h/A)]1/2}$$ The calculation process is I first calculate A =2.3* post load/(allowed passive soil bearing value*diameter of caisson)
Then you plug A into EQ 18-1.

The post resists overturning momentum mainly by rotating around a pivot under the soil surface, usually near 1/3 total embedment depth.

So if you have a lateral force acting at height H trying to turn the pole clockwise the top part of the pole embedded in dirt will compress the soil to right with a triangular stress area and the rest of embedded pole will press the dirt to left.
This way of analysis has been tested in many different types of soil and shown to be a good assumption.

However there are limitations and conditions that needs to be observed. I recommend you google the IBC code 18-1. Just bear in mind this code is not concerned with vertical loads or torsion.

Answer 3

The first scenario is that the pole is very firmly anchored into the ground in a hard and reliant surface to a good depth for example if it is embedded in a large mass of concrete. Here the pole will bend over its whole length until the elastic limit in reached an then fail at some point colse to the ground surface and form a '[plastic hinge1' and then continue to bend at this point untill it fails completely.

The second scenario it that the pole itself can tolerate substantially more force than the ground in which it is embedded. This is a lot more complex to model as 'dirt' can have very variable physical properties depending on its composition and condition.

For example a dry clay soil with lots of rubble in it can behave in a similar way to concrete and it will resist any force well up to a certain point at which it will suddenly crack and fail.

For the purposes of game simulation if you want a simple method it would probably make sense to consider that the pole pivots somewhere near ground level and just pick a maximum value for Fr and compare moments F1*L1 and Fr*L2.

For more detail you could say that you set a value A for F1 which is able to start shifting the pole and a higher value B which knocks it over completely at which it is no longer fixed to the ground (ie becomes a detached object lying on the ground) and any vale in between causes a certain angle of lean. ie if A is 100 force units and B is 200 force units it leans over at 45 degrees.

This approach allows you an easy way to adjust values without having to run a complex soil mechanics simulator in your game.

Another simple and physically reasonable approach would be to assume that the pole has a permanently attached plug of material around it below ground level and just treat that as a dead weight, as if it was placed in a bucket of concrete siting on the ground.