POISSON'S RATIO: COEFFICIENT, FORMULAS, VALUES, EXAMPLES - PHYSICAL - 2025

The Poisson 's ratio is a dimensionless quantity, characteristic of every material. It is an indication of the deformation of a piece of material before the application of certain forces.

When a piece of material that is subjected to tension, or compression, undergoes a deformation, the quotient between the transverse deformation and the longitudinal deformation is precisely the Poisson's ratio.

Figure 1. Poisson's ratio measures the relationship between longitudinal stretch and transverse narrowing. (Prepared by Ricardo Pérez)

For example, a rubber cylinder that is subjected to tension at its ends stretches in the longitudinal direction, but narrows transversely. Figure 1 shows a bar whose original dimensions are: length L and diameter D.

The bar is subjected to a tension T at its ends, and as a consequence of this tension it undergoes a stretching, so that the new length is L '> L. But when it is stretched, its diameter also narrows to the new value: D '<D.

The quotient between the stretch (positive) and the narrowing (negative) multiplied by (-1), is a positive number between 0 and 0.5. This number is the so-called Poisson's ratio ν (Greek letter nu).

Poisson's ratio formula

To calculate the Poisson's ratio, it is necessary to determine the longitudinal and transverse strain.

The longitudinal strain ε _L is the stretch divided by the original length:

ε _L = (L '- L) / L

Similarly, the transverse strain ε _T is the radial narrowing divided by the original diameter:

ε _T = (D '- D) / D

Therefore, the Poisson's ratio is calculated by the following formula:

ν = - ε _T / ε _L

Relationship with modulus of elasticity and modulus of rigidity

The Poisson's ratio ν is related to the modulus E of elasticity (or Young's modulus) and the modulus of rigidity G, by the following formula:

Poisson's ratio value for materials

Figure 2. Stainless steel has a Poisson's ratio between 0.30 and 0.31. Source: Pixabay.

Calculation examples

Example 1

A bar of a certain plastic material has a length of 150 mm and a circular section of 20 mm in diameter. When subjected to a compression force F of 612.25 kg-f, a shortening of 14 mm is observed and simultaneously an increase of 0.85 mm in the diameter of the bar.

Calculate:

a) Longitudinal strain.

b) The transverse strain.

c) The Poisson's ratio of that material.

d) Young's modulus of elasticity corresponding to the material.

e) The modulus of rigidity for that plastic.

Solution to

Recall that the longitudinal strain εL is the stretch divided by the original length:

εL = (L '- L) / L

εL = (-14 mm) / 150 mm = -0.0933

Note that the longitudinal strain is dimensionless, and in this case it was negative because there was a decrease in its longitudinal dimension.

Solution b

Similarly, the transverse strain εT is the radial taper, divided by the original diameter:

εT = (D '- D) / D

εT = (+0.85 mm) / 20 mm = 0.0425

The transverse strain has been positive because there has been an increase in the diameter of the bar.

Solution c

For the calculation of Poisson's ratio we must remember that it is defined as the negative of the quotient between the transverse deformation and the longitudinal deformation:

ν = - εT / εL

ν = - 0.0425 / (-0.0933) = 0.4554

It should be remembered that Poisson's ratio is a positive dimensionless number and for most materials it is between 0 and 0.5.

Solution d

Young's modulus of elasticity, denoted by the letter E, is the constant of proportionality in Hooke's law. By E, the normal stress σL is related to the strain εL, as follows:

σL = E εL

The normal stress is defined as the quotient between the normal force (in this case parallel to the axis of the bar) and the cross-sectional area:

σL = F / A = F / (π / 4 * D ^ 2)

In this exercise, the force F is 612.25 kg-f, which must be converted to newtons, which is the SI unit of force:

F = 612.25 kg-f = 612.25 * 9.8 N = 6000 N = 6 kN

For its part, the cross section of area A is:

A = (π / 4 * D ^ 2) = (3.1416 / 4) * (20 * 10 ^ -3 m) ^ 2 = 3.1416 * 10 ^ -4 m ^ 2

Finally the normal stress applied to the bar is:

σL = F / A = 6000 N / 3.1416 * 10 ^ -4 m ^ 2 = 19.098.593 Pa = 19.098 MPa

To calculate Young's modulus of elasticity, we solve for E from Hooke's law σL = E εL:

E = σL / εL = 19,098,593 Pa / 0.0933 = 204.7 MPa

Solution e

The modulus of rigidity G is related to Young's modulus E and Poisson's ratio ν by this formula:

E / (2 G) = 1 + ν

From there we can solve for G:

G = E / (2 (1 + ν)) = 204.7 MPa / (2 (1 + 0.4554)) = 70.33 MPa

Example 2

There is a copper cable with a diameter of 4 mm and 1 m long. Knowing that the Young's modulus of copper is 110,000 MPa and that its Poisson's ratio is 0.34, estimate the stretching and narrowing in diameter that the wire undergoes when a weight of 100 kg-f is hung on it.

Solution

First, it is necessary to calculate the normal tensile stress that the weight exerts on the wire, following this formula:

σL = F / A = F / (π / 4 * D ^ 2)

The force F is 980 N and the cross sectional area is:

A = (π / 4 * D ^ 2) = (3.1416 / 4) * (4 * 10 ^ -3 m) ^ 2 = 1.2566 * 10 ^ -5 m ^ 2

Then the tensile stress is:

σL = 980 N / 1.2566 * 10 ^ -5 m ^ 2 = 77,986,000 Pa

Calculation of Strain of Wire

Young's modulus of elasticity, denoted by the letter E, is the constant of proportionality in Hooke's law that relates the normal stress σL to the strain εL:

σL = E εL

From there the longitudinal strain of the copper wire can be solved:

εL = σL / E = 77.986 MPa / 110000 MPa = 7.09 * 10 ^ -4

Calculation of transverse strain

On the other hand, to know the transverse strain, the Poisson's ratio is applied:

ν = - εT / εL

Finally, the transverse strain is:

εT = –ν εL = - 0.34 * 7.09 * 10 ^ -4 = -2.41 * 10 ^ -4

Calculation of absolute cable stretch

Finally, to know the absolute stretch of the cable, the following relationship must be applied:

ΔL = εL * L = 7.09 * 10 ^ -4 * 1 m = 7.09 * 10 ^ -4 m = 0.709 mm

That is to say, with that weight the cable barely stretched 0.709 millimeters.

Calculation of decrease in diameter

To obtain the absolute shrinkage in diameter we use the following formula:

ΔD = εT * D = -2.41 * 10 ^ -4 * 4 mm = -9.64 * 10 ^ -4 mm = -0.000964 millimeters.

This narrowing in diameter is so small that it is difficult to see with the naked eye, even its measurement requires a high precision instrument.

References

1. Beer F.. Mechanics of materials. 5th. Edition. 2010. Mc Graw Hill. 1-130.
2. Hibbeler R. Mechanics of materials. Eighth edition. Prentice Hall. 2011. 3-60.
3. Gere J. Mechanics of materials. Eighth edition. Cengage Learning. 4-220.
4. Giancoli, D. 2006. Physics: Principles with Applications. 6th Ed. Prentice Hall. 238-242.
5. Valera Negrete, J. 2005. Notes on General Physics. UNAM. 87-98.