## Note on Elastic and plastic deformation

• Note
• Things to remember

### Elastic and plastic deformation

Figure

The above figure represents the variation of stress over different value of strain. In the diagram when applied stress increases linearly up to point P. the point P is known as proportion limit under the stress-strain curve. The properties of the material above the point P are called ‘Plastic behavior’ and below this point the property is known as elastic behavior. The point P separates two types of behavior.

1. #### Elastic deformation:

It is a reversible phenomena, when stress is removed the material returns to its original dimension before the application of load. The value of strain in this region is too small for metals, but for some robbers and plastics it is large.

The proportional deformation between the stress and stress is called elastic deformation. The plot diagram of stress along y-axis versus strain along x-axis results in a linear relationship. The slope of this linear segment the modulus of elasticity E corresponds to slope of this linear segment. This modulus may be thought of as stiffness, or a material’s resistance to elastic deformation. The stiffer the material, the greater the modulus, or the smaller the elastic strain that results from the application of a given stress. The modulus is an important design parameter used for computing elastic deflections.

1. #### Plastic deformation:

The limit of elastic deformation only to strains of about 0.005 for most metallic materials. When thematerials are deformed beyond this point, the stress is no longer proportional to strain permanent, or plastic deformation occurs. The transition from elastic to plastic is a gradual one for most metals. Some curvature results at the onset of plastic deformation, which increases more rapidly with rising stress.

It is an irreversible phenomenon, when stress removed the material doesn’t returns to its original dimension. There is some kind of permanent deformation in the material. From an atomic perspective, plastic deformation corresponds to breaking of bonds with original atoms neighbors and their reforming bonds with new neighbors. As large number of atoms and molecules moved relative to one another. So upon their removal of stress they do not return to original dimension.

Stress strain behavior of elastic deformation and its atomic view point:

In tensile test if the deformation is elastic the stress is directly proportional to strain produced in material which is called ‘Hook’s law’.

Within proportional point

$$\sigma\propto \epsilon$$

$$\sigma=E\epsilon\dotsm (1)$$Where E=proportionality constant=young’s modulus of elasticity or modulus of elasticity

It is measured by measuring the slope of strain-stress curve.

E=Young’s modulus of elasticity

=slope of stress-strain curve

=$$\frac{\Delta \sigma}{\Delta \epsilon}$$

Figure

The stiffness of material is measured by measuring the value of Young’s modulus of elasticity. More stiff the material higher will be the value of E (higher will be slope of S-T curve).

In some type of material there is non-linear elastic behavior i.e. the materials like polymer, ceramics the deformation is not linear but is still reversible as shown in graph.

Figure

The value of Young’s modulus of elasticity at different value of stress is obtained from the slope of S-T curve.

#### Elongation computation:

Q.A piece of copper originally 305 mm (12 in.) long is pulled in tension with a stress of 276 MPa (40,000 psi).If the deformation is entirely elastic,what will be the resultant elongation?

Solution

Because the deformation is elastic, strain is dependent on stress.The elongation $$\Delta l$$ is related to the original length$$l_\circ$$.

We have,

$$\sigma=\epsilon E=\biggl(\frac{\Delta l}{l_\circ}\biggr)E$$

$$\Delta l=\frac{\sigma l_\circ}{E}$$

The values of and l0 are given as 276 MPa and 305 mm, respectively, and the magnitude of E for copper is 110 GPa (16$$\times 10^6$$ psi).Elongation is obtained by substitution into the preceding expression as

$$\Delta l=\frac{(276 MPa)(305 mm)}{110\times 10^3} Mpa=0.7 mm$$

#### References:

Callister, W.D and D.G Rethwisch. Material Science and Engineering. 2nd. New Delhi: Wiley India, 2014.

Lindsay, S.M. Introduction of Nanoscience . New York : Oxford University Press, 2010.

Patton, W.J. Materials in industry . New Delhi : Prentice hall of India, 1975.

Poole, C.P. and F.J. Owens. Introduction To Nanotechnology. New Delhi: Wiley India , 2006.

Raghavan, V. Material Science and Engineering. 4th . New Delhi: Pretence-Hall of India, 2003.

Tiley, R.J.D. Understanding solids: The science of Materials. Engalnd : John wiley & Sons , 2004.

1. Types of deformation :

elastic deformation

plastic deformation

2.Important equations

$$\sigma\propto \epsilon$$

$$\sigma=E\epsilon$$

$$\sigma=\epsilon E=\biggl(\frac{\Delta l}{l_\circ}\biggr)E$$

$$\Delta l=\frac{\sigma l_\circ}{E}$$

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