Penetration of a stake as it goes deeper into the soil

[byarble]

I was quite curious when I noticed that a stake gets progressively harder to drive the deeper it goes into the soil. When I tried looking up ballistics formulas they seemed to disagree and state that there should be a force that should work to penetrate all depths. Does the force increase because a stake increases in friction since more of it comes into contact with the soil? Thus in something like steel where the forces of breaking it would far exceed those of friction, would it just be as the ballistic formula states? What are the forces involved in putting a stake in the ground?

 

[Lnewqban]

Welcome, @byarble !

Friction is the main thing to overcome in that case, unless your stake hits solid rock.

Power actuated pins into steel beams are also hold in place by friction.
The normal force comes from the elasticity of the deformed metal into which the pin is forced.

 

[Frabjous]

Soils also get compacted which increases their resistance. Also at the surface, there is room for material to flow which makes things easier.

 

[Rive]

I noticed that a stake gets progressively harder to drive the deeper it goes into the soil.

As an addition for all things already mentioned above, the top layers of soil are usually more loose than below.

 

[Baluncore]

When I tried looking up ballistics formulas they seemed to disagree and state that there should be a force that should work to penetrate all depths.

That depends on the analysis, and the meaning or context of "penetrate all depths".

Since the projectile has a fixed length, once it has fully entered the solid material, friction is constant along the path.
The limit to projectile penetration is loss of energy as it opens and moves along a path through the target material. At some point, the projectile will melt, fragment, or it will stop when momentum reaches zero.

A ground stake is subjected to friction, proportional to the depth of penetration, plus the point force, needed to open the ground. The increasing force needed to penetrate, is transferred along the stake, so the stake must have a sufficient section, to avoid buckling or bending during insertion.

 

[byarble]

Welcome, @byarble !

Friction is the main thing to overcome in that case, unless your stake hits solid rock.

Power actuated pins into steel beams are also hold in place by friction.
The normal force comes from the elasticity of the deformed metal into which the pin is forced.

View attachment 323507

I think I get it now so I have 1 more question, in the case of projectiles, friction would increase with projectile length yes?

 

[Baluncore]

I think I get it now so I have 1 more question, in the case of projectiles friction would increase with projectile length yes?

Yes, but that assumes the target is thicker than the length of the projectile.
For sheet targets, thinner than the length of the projectile, the frictional force will be proportional to the thickness of the sheet.

 

[Lnewqban]

I think I get it now so I have 1 more question, in the case of projectiles, friction would increase with projectile length yes?

I believe it would not, as friction does not depend on area.
I may be wrong, because I don't know anything about how different types of projectiles perform.

 

[Rive]

in the case of projectiles, friction would increase with projectile length yes?

The impact is a science of its own. Sometimes you get wider hole than the projectile (common for high speed impacts), sometimes the bullet can even stuck... Too many variables for a general question.

Ps.: for any stake-like (slow) event: yes.

 

[Baluncore]

I believe it would not, as friction does not depend on area.

That is true, for a constant force such as weight. But for a slow projectile, the crowd pressure on the projectile is a function of target elasticity, which will close the hole on the projectile or Hilti DX fastner, as it penetrates and passes. A longer projectile has a greater area subjected to the confining pressure from the target, so the force is greater, in proportion to contact length.

 

[byarble]

That depends on the analysis, and the meaning or context of "penetrate all depths".

Since the projectile has a fixed length, once it has fully entered the solid material, friction is constant along the path.
The limit to projectile penetration is loss of energy as it opens and moves along a path through the target material. At some point, the projectile will melt, fragment, or it will stop when momentum reaches zero.

A ground stake is subjected to friction, proportional to the depth of penetration, plus the point force, needed to open the ground. The increasing force needed to penetrate, is transferred along the stake, so the stake must have a sufficient section, to avoid buckling or bending during insertion.

so I got another question, would friction be proportional purely to surface area of contact or the volume of displacement, I feel like a blade which has a lot of surface area would perform better in penetration than a cylinder with comparatively less surface area but more volume, wouldn't this also mean that there is a material property with the SI units of kg m^-2 s^-2

 

[Baluncore]

... , I feel like a blade which has a lot of surface area would perform better in penetration than a cylinder with comparatively less surface area but more volume, ...

"Perform better" is meaningless, unless you specify an application and criteria for assessment.

The path must be deformed beyond yield by initial penetration, so sectional area is important. But at low speeds, the material will elastically spring back, against the entire surface area, making surface area also important.

At high speeds, the surface will not have time to pinch the sides and tail of a projectile, so the analysis must then be different.

Points without an edge penetrate, but do not cut. Flat blades cut roots, ligaments, pipes, cables and arteries.

If there is a fibrous structure in the material, that is turned along the path of the penetration, then the material may later pinch and so lock, preventing extraction. That happens with nails in wood, and bayonets in bone. It also makes wood screws harder to initially reverse.

There is no single SI unit that I know of with dimensions of; Length^-2⋅Mass⁺¹⋅Time^-2

 

[jrmichler]

I was quite curious when I noticed that a stake gets progressively harder to drive the deeper it goes into the soil. When I tried looking up ballistics formulas they seemed to disagree and state that there should be a force that should work to penetrate all depths. Does the force increase because a stake increases in friction since more of it comes into contact with the soil? Thus in something like steel where the forces of breaking it would far exceed those of friction, would it just be as the ballistic formula states? What are the forces involved in putting a stake in the ground?

Since you are interested in stakes driven into the ground, ballistics calculations do not apply. Unless, of course, that stake is fired into the ground at high velocity. The forces to push a stake into the ground can be calculated using the methods of soil mechanics. This image, from Introductory Soil Mechanics and Foundations 3rd Edition by Sowers and Sowers is for piles, which are very large stakes.

The rest of the chapter goes into the details of calculating the total load the pile can support. That load is the same as the force to drive it deeper. The procedure is to separately calculate the force from skin friction and end bearing, then add the two forces. Since the force due to skin friction is proportional to the side area in contact with the soil, the force to drive the stake deeper must increase as the stake is driven deeper.

I see that the 4th edition of that book is available from Amazon. It's a good buy if you want to get deeper into the subject.