Determine contact area
Determine contact area
(OP)
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
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






RE: Determine contact area
RE: Determine contact area
I continue to be open to suggestions.
RE: Determine contact area
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).
RE: Determine contact area
RE: Determine contact area
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
RE: Determine contact area
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
RE: Determine contact area
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...
RE: Determine contact area
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
RE: Determine contact area
Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
RE: Determine contact area
"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....
RE: Determine contact area
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
RE: Determine contact area
RE: Determine contact area
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
RE: Determine contact area
RE: Determine contact area
BA
RE: Determine contact area
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.
RE: Determine contact area
RE: Determine contact area
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.
RE: Determine contact area
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.
RE: Determine contact area
I think I know what you are trying to do. I think you are calculating the plus/minus pressures from overturning and thinking that if the plus is bigger than the average uplift "pressure" it will maintain contact over that small area. THIS IS NOT TRUE.
Foe equilibrium, ΣF=0 and ΣM=0. If your W-Nt <0 you don't have equilibrium.
Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin
RE: Determine contact area
RE: Determine contact area
BA
RE: Determine contact area
1) Why is the first block a square and not a planar surface?
2) What prevents sliding?
3) The first block is infinitely rigid, but no information is provided about the properties of the second block.
4) There is no mention of where the moments are applied. Do you mean couples?
5) There is no mention of weight or other force to offset the tensile load.
6) There is no mention of any means to react the moment load (couple?)
7) There is no mention of where the Tensile load acts. Does it change position as the moment (couple) is applied?
Since this is a fundamental problem, the inability to find it already solved would indicate there is no general purpose equation.
You have access to FEM/FEA. Create test cases for your example and run contact analyses, then use the output contact area and input loads to develop an equation that predicts contact area based on the loads and all the other assumptions you are making in posting the problem. This is the way experimentally derived formulae are created and should work for this case.
**************************
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