5
$\begingroup$

I was solving a problem and came across some confusion regarding the point of application of Normal Force. In Classical Mechanics 101, we had always treated the Normal Force as acting on a point (which can be called as the "centre" of the Normal Force), and it was calculated by applying Newton's Laws (Balancing forces and torques).

But how to I determine the exact distribution of the Normal Force on the surface of a body if the contact is not at points?

Because there can be a lot of distributions that may result in the same centre for the Normal force. So, is this model for Normal force incomplete?

Edit

The problem I was solving was this. Imagine a book kept on a piece of wood such that the book covers only half of the piece. I can figure out the centre of Normal Reaction but not how it's distributed over the surface.

$\endgroup$
1
$\begingroup$

You have to consider the elastic properties of the contact to determine the distribution of forces. Actually, it is the distribution of pressure you are interested, since you sum up the pressures to get force, and sum of the moment of pressure to get torque. Imagine a contact area with the center is on the contact point, the z-axis normal to the contact, and the xy plane along the contact.

$$ \begin{aligned} F_z & = \int P(x,y) \, {\rm d} A \\ M_x & = \int (-y) P(x,y) \,{\rm d} A \\ M_y & = \int (+x) P(x,y)\, {\rm d} A \end{aligned} $$

So given a pressure distribution $P(x,y)$ you can find the forces, but how to get the pressure distribution?

The most basic example is an object pushing on a flat plane. Consider the case where there is a known overlap amount $a(x,y)$ as a function of location on the contact

overlap

As a result, the surface of the object (red curve) must deform by $\delta(x,y)$ in order to conform to the overlap. Surface pressure (yellow curve) must be applied to the object to do this

pressure

The pressure deflection relationship was developed by Boussinesq and it looks like this

$$ \delta(x,y) = \frac{1-\nu^2}{\pi E} \iint \frac{P(u,v)}{r} {\rm d} A $$

where $r=\sqrt{(x-u)^2+(y-v)^2}$ and $\nu$ is the Poisson's ratio of the material and $E$ is the Elastic modulus.

A special case for the above is the Hertzian contact which deals with a sphere on a plane, sphere on a sphere, cylinder on a cylinder (2D contact) and general football shape (two radii) contact.

The general case can only be treated numerically with the contact are divided into a grid and a giant system of equations is developed of the form $\delta_i = C_{i j} P_j$ (Hartnett solution).

The steps are as follows:

  1. Assume a certain amount of overlap $a(x,y)$ derived from the maximum penetration amount.
  2. Set the deflections equal to the overlap, and solve for the pressures $P(x,y)$.
  3. Calculate the total load and moments, and adjust the overlap accordingly. If the total load is less than the applied load then increase the overlap, and if the total moment is less than the applied torque then tilt the object
  4. If the forces and moments are still imbalanced, then go to step 2 and repeat until convergence has been achieved.

See this detailed paper for one example on numerical contact analysis. And here is a paper from NIST with some theory on the general contact problem.

$\endgroup$
  • $\begingroup$ So a body which is flat would have a $delta(x,y)$ a constant which would be equal to it's maximum penetration amount? $\endgroup$ – netflix_and_physics Aug 18 at 9:32
  • $\begingroup$ so, if $\delta$ is a costant, is not the system of equations underdetermined? $\endgroup$ – Wolphram jonny Aug 18 at 17:01
  • $\begingroup$ $\delta(x,y)$ is a known function. It is the overlap map and it depends on the approach (overlap at origin center) and the two tilt angles. These three degrees of freedom correspond to the three known forces and moments applied. So the system is solvable, as in each iteration the error is reduced. $\endgroup$ – ja72 Aug 18 at 19:47
0
$\begingroup$

The distribution of a force over the surface of a body (as opposed to a "point") is called stress. If the force is normal (perpendicular) to the body, it is called normal stress and is given as

$$σ=\frac{F}{A}$$

When you deal with forces on deformable bodies (and all bodies are deformable) the force is never applied to a "point". A point has no area. From the equation you can see that if $A$ is zero, you would have an infinite applied stress.

The deformation δ for uniaxial loading due to the applied stress is given by:

$$δ=\frac{PL}{AE}$$

where $P$ is the load (normal force), $L$ is the length of the member, $A$ is the cross sectional area and $E$ is the modulus of elasticity of the material which is a measure of its stiffness and $\frac{P}{A}$ is the normal stress.

So if the load were applied to a point (zero Area) the body would need to be infinitely stiff (an idea rigid body) in order to have no deformation. There is no such thing as an ideal rigid body, at least at the macroscopic level.

Hope this helps.

$\endgroup$

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.