0
$\begingroup$

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

enter image description here

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.

| improve this question | |
$\endgroup$
  • 1
    $\begingroup$ Is the pole rigid or can it bend? If it can bend, it will move regardless of the value of F1. $\endgroup$ – Wasabi♦ Sep 10 '16 at 15:47
  • $\begingroup$ @Wasabi It is rigid $\endgroup$ – univise Sep 10 '16 at 16:05
  • $\begingroup$ What variables do you want to know the relationship between? To put it another way, what values in the diagram are you prepared to have as constants? Will the type of material the ground is made of be constant? $\endgroup$ – CL22 Sep 10 '16 at 17:54
  • $\begingroup$ @Jodes I have updated the post to reflect the information requested. Shortly, yes, the pole is to be simulated in a game simulation where physics need only approximate reality with a rough degree of realism. Thus a lot of the physics can be viewed in an idealised manner, for instance we may assume the material is constant. $\endgroup$ – univise Sep 10 '16 at 21:01
2
$\begingroup$

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.

Moments with passive pressure ONLY

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ A good answer, though yet I do not understand the following: When I have the computed the pressures, how do I infer whether or how much the pole will turn? Also, could you define the meaning of the parameters h, g, and ρ? I suppose g = 9.81, ρ is the density of the soil, and h the depth in the soil we want to compute the pressure in? $\endgroup$ – univise Sep 12 '16 at 21:17
  • $\begingroup$ The answer by kamran added a picture which clarifies things. The post should begin turning when the moment about the turning point caused by the pulling force exceeds the moment from the resisting forces (from the pressures). Think of these pressures as the soil's ultimate resistance, once exceeded - the pole moves. Adding deformations might be a bit more tricky - for simplicity you could add a factor relating the displacement of the soil and the resisting force it is able to provide up to the limits given by the equations above. $\endgroup$ – ptev Sep 12 '16 at 21:26
  • $\begingroup$ So I understand that I may interpret the pressure Pp as the maximum possible magnitude for the ground's resisting force Fr(see my sketch in question)? $\endgroup$ – univise Sep 12 '16 at 21:32
  • $\begingroup$ Yes, but note the way this is distributed in the other sketch. Of course you can assume a uniform pressure distribution, but this is more realistic. Answer above edited. $\endgroup$ – ptev Sep 12 '16 at 21:49
4
$\begingroup$

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. diagram of flag pole and soil stresses.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I looked up the equation and I found out d is a distance though between what exactly? Could you please clarify what d means in this problem? Also, after some more research I understand that the passive soil bearing you mention is the passive pressure Pp mentioned in ptev's answer. Do I understand this correctly? $\endgroup$ – univise Sep 12 '16 at 21:28
  • $\begingroup$ D is embedment depth in competent soil. $\endgroup$ – kamran Sep 13 '16 at 6:27
  • $\begingroup$ Passive soil pressure must be specified by soils engineer. Normally it is in the order of 2 -3 magnitude of active soil pressure. You need a professional engineer to utilize these equation because there are restrictions that apply. $\endgroup$ – kamran Sep 13 '16 at 6:31
-1
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.