# The distribution of pressure across a surface

1. Aug 7, 2014

### jeff davis

Force Distribution

Hello, I have probably a fairly simple question about how force is distributed. If i were to have some type of setup where a hydraulic cylinder with a one inch rod was pushing down (say 50 lbs) onto a plate, or something of that sort, how would that force be distributed across the plate? Would it be localized to the area of the end of the one inch rod, or would it dissipate thru out the whole plate evenly? What if the plate was attached to the one inch rod pushing down 50 lbs; would the whole plate be pushing down 50lbs at every point on it?

Thanks Guys!

2. Aug 7, 2014

### puncheex

If the plate is rigid, the force of the rod would be totally transfered to the plate; the pressure would be distributed evenly.

Yes, every part of the plate would exert 50 pounds of force. A balancing force might be concentrated at any point on the plate, or distributed over the whole plate.

3. Aug 7, 2014

### jeff davis

Thank you very much for your reply. In my mind however it is hard for me to grasp that something as small as a one inch rod pressing down onto a large plate has its force dissipated evenly across it. I am now imagining being a huge plate and having something small push down on me. You are saying that because i am rigid and large it would not be pressure in just the area of the rod but it would feel like an "x" lb of weight was over my entire surface?? Is there a good place i can read about this topic? What is this topic referred to as?

4. Aug 7, 2014

### Staff: Mentor

The force will be distributed over the area of contact of the rod with the plate only. Your assessment was correct. Ask yourself this: How could the rod possibly exert a force on the plate anywhere else?

Chet

5. Aug 8, 2014

### sophiecentaur

If both the rod and the plate are totally non-deformable, the contact could well be a single point - with infinite pressure. For real materials, there will be some deformation and the force will be shared over the whole surface of contact but there will still be local high and lower pressure areas. It's only when you have a fluid that pressure can be said to be totally uniform over a contact area.

The strain in the plate will 'spread out' within the depth of the plate material and transfer to whatever is supporting the plate - possibly on the other side or on supporting chains etc.. The details would depend on the modulus of the plate material.

6. Aug 8, 2014

### Staff: Mentor

Hi sophiecentaur.

I'm having trouble seeing how there will be point contact in the non-deformable case, if the rod has finite cross section. I agree that, if the materials are deformable, the contact stress will vary with location within the contact area.

Chet

7. Aug 8, 2014

### sophiecentaur

If you are assuming two totally flat surfaces then the pressure will be uniform. But that's ideal. A real surface is not microscopically flat. If you take a real, light, nominally flat disc and put it on a nominally flat plate, there will be just three points of contact (as with a three legged stool) where the pressure is actually, quite high. Only when you start to have a significant force (weight) between the two surfaces will there be more contact points and the areas will also increase.
(This argument applies for the laws of friction - which tell you that the fraction force is proportional to normal force over a large range of forces. It's because the increasing force causes a proportionate increase in contact area and modifying the pressure)

My point is that you can only guarantee uniform pressure when there is a fluid involved.

8. Aug 8, 2014

### Staff: Mentor

Gotcha. I think the part of your answer that the OP was looking for is that, for real materials, the force is distributed (not necessarily uniformly) over the contact area (and only over the contact area).

Chet

9. Aug 8, 2014

### 256bits

Stress analysis, and it can get quite complicated.

In many cases, initially we 'simplify' and make an assumption, that the force spreads out evenly.
At times this is all that is needed, such as if the plate is supported by some pillars, we would want to know the how much force is on each pillar and do not care about how the plate itself bends and reacts to the initial force - we assume the plate is rigid.

More detail involves more complicated calculations.

10. Aug 11, 2014

### jeff davis

Thank you guys again very much for your help. I have a new perspective on the topic now. Your answers are exactly what i was looking for. It is very good to know that it can be simplified to an extent but can also be more complicated in reality. The example of the "Three legged stool" was an excellent eye opening point for me.

Thanks again!!