Saturday, February 11, 2017

Bouncing balls a little beyond the elastic range

This is an edit to wikipedia

==Predicting the coefficient from material properties==
When colliding objects do not have a center of gravity that is inline with their direction of motion and point of impact, energy that would have been available for the post-collision velocity difference will be lost to rotation and friction. This section will consider only spherical objects colliding directly either with other spherical objects or with a flat surface to avoid rotation and friction losses.

When soft objects strike hard objects, most of the energy available for for the post-collision velocity different will be stored in the soft object and the value of the COR will depend on how efficient the soft object is at storing the compression energy without losing it to heat and plastic deformation. A rubber ball will bounce a lot better off concrete than glass, but the COR of glass-on-glass is a lot better than rubber-on-rubber. So when wanting to know the COR of an object, it's good to impact it with an object that is much harder. For this reason the [[Leeb rebound hardness test]] impacts test samples with a tip of tungsten carbide, one of the hardest substances available. There is no perfectly hard material, and the COR depends on both objects, so for ideal testing and theory, determining the COR of a material depends on both objects being that same material.

Many materials are assumed to be perfectly elastic when their yield strength is not approached during impact. The impact energy is theoretically stored only in the spring-effect of elastic compression and results in e=1. In practice a various stainless steels have a large variation well below e=1. Amorphous metals can achieve e = 0.95 or higher. The elastic range can be exceeded at low velocities because all the kinetic energy is concentrated at the point of impact. If the velocity is above 1 m/s, the yield strength of metals is usually exceeded in part of the contact area, losing energy to "plastic deformation" by not remaining in the elastic region. To account for this, the following method estimates the percent of the initial impact energy that did not get lost. Approximately, it divides how easy a volume of the material can store energy in compression (1/{\text{elastic modulus}}) by how well it can stay in the elastic range (1/{\text{yield strength}}):

% \text{impact energy available for restitution} \propto \frac{\text{yield strength}}{\text{elastic modulus}}

For a given material density and velocity this results in:

\text{coefficient of restitution} \propto \sqrt{\frac{\text{yield strength}}{\text{elastic modulus}} }

To be more precise, these and two more quantities can be shown to be important when predicting the COR at moderate velocities. A high ''yield strength'' allows the material to stay in the elastic region at higher energies. A lower ''elastic modulus'' allows a larger surface area of contact during impact so the energy is distributed to a larger volume at the contact point which helps prevent the yield strength from being exceeded. A ''lower velocity'' increases the coefficient by needing less energy to be absorbed. A ''lower density'' also means less initial energy needs to be absorbed. The density instead of mass is used because the volume of the sphere cancels out with the volume of the affected volume at the contact area.

Combining these four variables, a theoretical estimation of the coefficient of restitution can be made when a ball is dropped onto a surface of the same material.http://itzhak.green.gatech.edu/rotordynamics/Predicting%20the%20coefficient%20of%20restitution%20of%20impacting%20spheres.pdf

* e = coefficient of restitution
* Sy = dynamic yield strength (dynamic "elastic limit")
* E' = effective elastic modulus
* ρ = density
* v = velocity at impact
* μ = Poisson's ratio

e = 3.1 \left(\frac{S_{y}}{1}\right)^\frac{5}{8} \left(\frac{1}{E'}\right)^\frac{1}{2} \left(\frac{1}{v}\right)^\frac{1}{4} \left(\frac{1}{\rho}\right)^\frac{1}{8}

E' = \frac{E}{1-\mu^2}

This applies for a direct impact and when:

0.001 < \frac{\rho v^2}{S_y} < 0.1

Although the accuracy of this equation is not good, it is easy to calculate and accurately predicts the relative coefficient for many materials, even at velocities above and below its intended range.

Theoretical coefficient of restitution solid spheres dropped 1 meter (v= 4.5 m/s). Values > 1.0 indicates the equation indicates the equation has errors.http://www-mdp.eng.cam.ac.uk/web/library/enginfo/cueddatabooks/materials.pdf Yield strength instead of dynamic yield strength was used.

{|
|'''Metals and Ceramics:'''
|'''Predicted COR, e'''
|-
|silica glass
|1.36 to 1.71
|-
|Alumina
|0.45 to 1.63
|-
|silicon nitride
|0.38 to 1.63
|-
|silicon carbide
|0.47 to 1.31
|-
|highest amorphous metal
|1.27
|-
|tungsten carbide
|0.73 to 1.13
|-
|magnesium alloys
|0.5 to 0.89
|-
|titanium alloy grade 5
|0.84
|-
|aluminum alloy 7075-T6
|0.75
|-
|glass (soda-lime)
|0.69
|-
|glass (borosilicate)
|0.66
|-
|nickel alloys
|0.15 to 0.70
|-
|stainless steel alloys
|0.23 to 0.62
|-
|zinc alloys
|0.21 to 0.62
|-
|cast iron
|0.3 to 0.6
|-
|copper alloys
|0.15 to 0.55
|-
|titanium grade 2
|0.46
|-
|tungsten
|0.37
|-
|aluminum alloys 3003 6061, 7075-0
|0.35
|-
|zinc
|0.21
|-
|nickel
|0.15
|-
|copper
|0.15
|-
|aluminum
|0.1
|-
|lead
|0.08
|-
|
|-
|
|}

Plastics and rubbers will give higher values than their actual values because they are not as ideally elastic as metals, glasses, and ceramics because of heating during compression. So the following is only a guide to ranking of polymers.

'''Polymers''' (overestimated compared to metals and ceramics):

* polybutadiene (golf balls shell) 11.8
* butyl rubber 6.24
* EVA 4.85
* silicone elastomers 2.80
* polycarbonate 1.46
* nylon 1.28
* polyethylene 1.24
* Teflon 1.21
* polypropylene 1.14
* ABS 1.12
* acrylic 1.06
* PET 0.95
* polystyrene 0.87
* PVC 0.86

For metals the range of speeds to which this theory can apply is about 5 to 100 m/s which is a drop of 1 to 500 meters, provided the sphere is small enough for Hertzian contact theory to apply (see page 366http://www.ewp.rpi.edu/hartford/~ernesto/S2015/FWLM/Books_Links/Books/Johnson-CONTACTMECHANICS.pdf) But the above rankings that it provides remain accurate.

Dropping hard spherical objects onto a softer surface (lower elastic modulus) which also has a lower coefficient of restitution will reduce the apparent coefficient of restitution of the dropped object. For example, most rubber and plastic balls have a lower coefficient than glass and some metal alloys, but when dropped on wood or cement, the softer material will bounce higher. This is because the harder objects distribute the impact energy over a much smaller contact area, losing energy to heat by exceeding the elastic range of the floor.

For metals, the theoretically perfect elastic range (the coefficient theoretically equals 1.0 and the above equation does not apply) is when the velocity is less than

v = \left(26 \frac{S_y}{\rho} \left(\frac{S_{y}}{E'}\right)^4 \right)^{0.5}

which is less than 0.1 m/s.

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