Determine contact area

[abaqususer1981]

Imagine you have two block with square cross sections with side length a. One is fixed and cannot deform. The other is free to move and rotate.
Both blocks have initially two surfaces in full contact and no sliding is allowed between these two surfaces.

A bending moment in each principal inertia axis are applied to the block free to rotate, M1 and M2 bending moments, as well as a axial tensile force, Nt, so that this block deforms over the fixed block.

My question is: How can I determine the contact area? To make things easy assume that only one bending moment, M, is applied plus the axial tensile force.

I'm struggling to find how the axial tensile force changes the contact area by an analytical equation. Anyone can help?

Thanks

Joao

 

[corus]

To see the contact area try plotting CPRESS or COPEN and adjust the contours to start from zero or 1e-6. The actual contact area may be difficult to calculate as it may pass across only parts of some elements.

 

[abaqususer1981]

Hi corus! Thanks for your reply. The thing is I'm trying to develop a joint finite element (spring like element) that simulates the contact between two blocks (simulated by beam elements). So I need to know when the elements are separating apart and to know that I need to calculate the contact area. I'm reading the book "Analytical Method in Reinforced Concrete" as it seems the author gives formulas for this problem.

I continue to be open to suggestions.

 

[abaqususer1981]

In particular, the stresses can be determine in elastic regime by:

σ = Nt/At - M1*y/I - M2*x/I

Know At is the contact area, but I don't think I is the inertia of the entire cross section of the block since only part of it is being deformed (opposes to deformation).

 

[corus]

If you're using a spring like element then just set the stiffness to be sero when the force is negative (ie. in tension) and so only acts in compression when in contact.

 

[BAretired]

In the absence of either moment or axial tension force, the contact area is a*a and the bearing stress is W/a² where W is the weight of the block.

With an applied moment about one axis, the block will overturn if the applied moment reaches Wa/2 after which the contact area = 0.

If the applied moment is less than Wa/6, the contact area will remain a*a but stress will be W/a² ± My/I

If the applied moment is greater than Wa/6 but less than Wa/2, the contact area will be b*a where b is less than a. Stress will vary linearly from 0 to 2W/ab. Dimension b may be calculated from the relationship that W(a/2-b/3) = M.

If tension Nt exceeds W, the movable block will lift off and contact area = 0.

If tension Nt is less than W, the effective weight of the movable block is W-Nt and the above relationships will apply with effective weight used as W.

If moments are applied about both axes, the contact area will not be rectangular, making stress determination a bit more difficult.

BA

 

[paddingtongreen]

Another way of looking at it:

This used to arise as a foundation problem and the thing to remember was the middle third rule. For a single moment, if you divide the moment by the axial force, you get an eccentric location for the application of the force. If this is within the middle third of the original contact area, it will stay in contact. If it is outside the middle third of the original contact area, a reduced contact area is formed with the force applied at the edge of the middle third. If there is moment about both axes, the middle third reduces to a kern. The kern is the area marked 1 from here

The main thing to remember is that centroid of the pressure diagram must be directly under the eccentric axial force point.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[abaqususer1981]

I appreciate all the answers.

Corus: that's just a very rude simplification to make in my opinion. If I wanted to do that there was no point of developing a spring-like element.

Regarding other comments, please do not forget that the axial load is a tension load, not compression, so for an eccentricity less than B/6 the section is in fact completely in tension and not otherwise in compression. The thing that is difficult is to determine the elastic stresses due to the bending moments considering the effective contact area. I'm still not convinced that the inertia of the section remains the same...

 

[BAretired]

abaquser,

Your last message makes no sense at all. We know the axial force is tensile. The section cannot be completely in tension or it would lift off. It is not difficult to determine bending stress for moment about a single axis.

Of course the inertia doesn't remain the same if the contact area decreases. Why would anyone try to convince you otherwise?

BA

 

[paddingtongreen]

I went back and read the OP again. If the one block is just sitting on the other and you apply a net axial uplift force, you have zero contact area, your block is flying through the air. The moments make no difference unless you have a net axial downward force i.e. there must be pressure for the moments to add to and subtract from.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[abaqususer1981]

Well, then please rethink your statement:

"If tension Nt exceeds W, the movable block will lift off and contact area = 0. " because this is what doesn't make sense and most importantly is wrong.
If the load is tensile but the eccentricity of the bending moment is such that the section is partially in compression then equilibrium can be met, just do a simple FEA model if you unsure of it.
Besides "W(a/2-b/3) = M" is only true if the stresses can remain elastic which must be verified....

 

[BAretired]

Well, then please rethink your statement:

"If tension Nt exceeds W, the movable block will lift off and contact area = 0. " because this is what doesn't make sense and most importantly is wrong.
If the load is tensile but the eccentricity of the bending moment is such that the section is partially in compression then equilibrium can be met, just do a simple FEA model if you unsure of it.
Besides "W(a/2-b/3) = M" is only true if the stresses can remain elastic which must be verified....

Read what paddingtongreen said. If the tension exceeds the weight of the block, the block will be flying through the air. What I said earlier makes complete sense and is not wrong.

I do not need to do a simple FEA model to resolve a simple issue and I am not unsure of it.

It is implicit in your original question that we are talking about elastic materials. How do you propose to verify it?

BA

 

[abaqususer1981]

If you have a lateral load applied to the top body (or this body suffers a rotation) and then it is stretched, the applied stress due to tension load is Nt/Ac, where Ac is the compressed area, which will be added to the stresses due to the bending moment. If the tensile stresses from the applied tension load are such that are larger than the compression ones from the bending moment then the section lifts, otherwise it will move but equilibrium can be met.

 

[BAretired]

If you have a lateral load applied to the top body (or this body suffers a rotation) and then it is stretched, the applied stress due to tension load is Nt/Ac, where Ac is the compressed area, which will be added to the stresses due to the bending moment. If the tensile stresses from the applied tension load are such that are larger than the compression ones from the bending moment then the section lifts, otherwise it will move but equilibrium can be met.

That is incorrect. If moment is applied first the compressed area will be Ac. When tension is then added, Ac will be reduced, so that is not the way to solve the problem.

You have to start with the tension. If the tension exceeds the weight of the block, forget about it...there is no solution because the block is no longer in contact with the base.

If the tension is less than the weight of block, then the effective weight of the block is W-Nt. If you now add moment, you can calculate Ac based on the effective weight.

BA

 

[abaqususer1981]

Yes, Ac will be reduced so you have to iterate, new tension stresses, new Ac..... and it will converge to a non zero Ac or to a zero Ac.

 

[BAretired]

You do not have to iterate. You start with the effective weight and calculate the stress based on the magnitude of the moment. It is a one shot deal.

BA

 

[abaqususer1981]

Let's say you apply a vertical compressive axial load to the top of a column together with a horizontal load (with a smaller value). At the bottom the column has a baseplate (square in plan).
If you apply these loads and let the column deform a lot you will start to see tensile forces at the column base plus very large bending moments due to large second order effects. By your reasoning the baseplate should lift off but it still manages to rotate until the contact between the baseplate and the ground is zero.

Although it does not completely make my point it shows that if you use the forces at a given section and assume that for Nt > W there is no equilibrium you may obtain bad results.

 

[abaqususer1981]

Of course, contact between surfaces must not be frictionless.

 

[CELinOttawa]

Sure, but you problem statement does not involve compression... While others believed it did, they tried to help. Now that you seem hopelessly confused, they are giving up. I'm surprised BA put up with this so long.

Please either draw a sketch, clarify that you're also welding the two blocks together, or go away and play with your fake physics elsewhere. Some of us prefer experience and discussions based on the real thing. Computers are only a useful tool if you have the experience to recognise garbage in when you see sh1t come out.

 

[abaqususer1981]

Sorry but the only person who seems desesperate is you, chill and take your depression pills. Look, I don't want to waste my time with you but as a benefit to all the community let me say that my original question is quite clear: tension axial load + bending moment.

The feedback from other users is not sufficient for the problem I have in hand but it is something for me to think about.

Feel free to write toilet paper phrases as much as you like. I just ask you to stay a way from dynamic contact problems with large displacements and deformations because you seem to be tickheaded.