# Hooke's Law and plastic materials

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1. Jan 9, 2016

### MBBphys

Hi,
All materials exhibit elastic deformation up to a certain limit, beyond which they exhibit plastic deformation.
Some materials, such as plasticine, have extremely tiny elastic regions, so we call them 'plastic materials'.
Some, like rubber, have large elastic regions, so we call them 'elastic materials'.
For some materials, within their elastic region, the force is directly proportional to the extension, and we say that these materials obey Hooke's Law.

Is that right? Or is it that plastic materials have no elastic region?
Would much appreciate your help :)

2. Jan 9, 2016

### SteamKing

Staff Emeritus
Consulting a dictionary would go a long way to clearing up your confusion here.

A material is said to be elastic if, when it is loaded, it spontaneously springs back to its original shape after the load is removed. If a material is loaded and takes up a new shape that is retained after the load is removed, then the material is said to exhibit plastic behavior.

Hooke's Law states that the amount of strain or deformation a material undergoes is proportional to the force applied, much like a spring is stretched by a given amount if a unit load is applied, it will stretch twice that amount if twice the load is applied.

If this loading occurs in the elastic region of the material, when the load is removed, the material will resume its original shape.
If a material is loaded beyond the elastic region, the amount of deformation or strain is no longer directly proportional to the amount of force applied, and when the force is removed, the material will retain what is called a permanent set, since the material is no longer in the elastic region.

The stress-strain diagram above is typical of the behavior of mild steel under load. If the material is loaded beyond the yield stress, you would find the strain produced by running a horizontal line from the stress axis over to the curve to find the intersection. When the load producing this stress is removed, you would find the permanent set by drawing a line from the intersection point down to the strain axis, but instead of a vertical line, this line would have a slope equal to the Young's modulus of the material.

3. Jan 10, 2016

### Staff: Mentor

In my judgment, your assessment is correct.

4. Jan 10, 2016

### MBBphys

Thanks for your reply, but I am confused about that image that you provided regarding the stress-strain graph for a ductile material.
So, you keep increasing force on the wire and measuring the extension; you get to the elastic limit. Then you keep increasing the force, but why do we have the second yield point (the 'hump' in the graph)? How come at the same stress level we have one lower strain value and one higher strain value?
Then, why is it that the breaking point (fracture) is at a lower stress level than the ultimate tensile strength level? How is it possible that it does not break at a higher force but then breaks at a lower stress level?
I am really confused about that diagram; the stress-strain graph for brittle and polymeric materials is quite easy, but this ductile material one is so difficult to grasp :(
I would really be grateful if you could explain that graph to me. Thank you so much for your help

5. Jan 10, 2016

### SteamKing

Staff Emeritus
That particular graph of stress-strain behavior was for a typical mild steel, which of course, is an alloy of iron, carbon, and other trace elements. The shape of this curve is not typical of other common metals, such as aluminum, which does not have all of the jogs that mild steel does. Aluminum also does not exhibit the typical elastic region behavior that mild steel does. The Young's modulus for aluminum is determined using the value for the stress which corresponds to an arbitrary value of strain (typically 0.2%)

This behavior is due to the nature of steel. Since it's an alloy chiefly of iron and carbon, after the material reaches yield, there is a dislocation in the structure of the metal, as the carbon atoms mixed in with the iron atoms are knocked out of position in the steel crystal.

https://en.wikipedia.org/wiki/Stress–strain_curve

There are two different ways to calculate the stress used in the typical stress-strain diagram for mild steel, and curves for both methods are plotted below:

Curve A is plotted using the apparent stress in the test sample, which is based on its original dimensions. When a steel test piece is loaded beyond its elastic limit, it begins to deform in a plastic manner, which is not confined solely to changes in the length of the sample. The cross section of the sample also changes dimension, which phenomenon is called 'necking'. A circular cross section, for example, becomes smaller in diameter, thus reducing its area from the original value, which also increases the actual calculated stress for a given tensile load.

If the computed stress during the test uses the actual area of the sample as it changes diameter, then the plot of stress-strain becomes Curve B, where the stress continues to increase after yield until the sample finally breaks.

6. Jan 15, 2016

### MBBphys

Thank you for your comprehensive reply! The bit about the cross sectional area was what I didn't get at first; thank you :)