He **cutting module **describes the response of a material to the application of a shear force that deforms it. Other frequently used names for shear modulus are shear modulus, shear modulus, transverse modulus of elasticity, or tangential modulus of elasticity.

When the stresses are small, the deformations are proportional to them, according to Hooke’s law, the shear modulus being the constant of proportionality. Therefore:

Shear Modulus = Shear Stress/Strain

Suppose a force is applied to the cover of a book, the other being fixed on the surface of the table. In this way, the book as a whole does not move, but is deformed when the upper cover moves with respect to the lower one by the amount *Δx*.

The book goes from having a rectangular cross section to a section in the shape of a parallelogram, as we see in the image above.

Be:

*τ = F/A*

The stress or shear stress, being *F* the magnitude of the applied force and *TO* the area on which it acts.

The deformation caused is given by the ratio:

* δ = Δx / L*

Therefore the cutting module, which we will denote as G, is:

In this way, the cutting module is in charge of measuring the resistance of the interior planes of the object that are parallel to the applied stress.

And since Δx / L is dimensionless, the units of G are the same as the units of shear stress, which is the ratio of force to area.

In the International System of Units, these units are Newton/square meter or pascal, abbreviated Pa. And in Anglo-Saxon units it is pound/square inch, abbreviated *psi*.

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**Cutting module for various materials**

Under the action of shear forces such as those described, objects offer a resistance similar to that of a book, in which the inner layers slide. This type of deformation can only occur in solid bodies, which are rigid enough to resist being deformed.

On the other hand, liquids do not offer this kind of resistance, but they can undergo volume deformations.

Below is the cutting module G in Pa for various materials frequently used in construction and in the manufacture of machinery and spare parts of all kinds:

**Experimental measurement of the shear modulus**

To find the value of the shear modulus, samples of each material must be tested and their response to the application of a shear stress examined.

The sample is a rod made of the material, with radius *R.* and length *L* known, which is fixed at one end, while the other is connected to the axis of a pulley free to rotate.

The pulley has a rope attached to the free end of which is hung a weight that exerts a force **F** over the rod through the rope. And this force in turn produces a moment ** m** on the rod, which then rotates through a small angle θ.

An assembly scheme can be seen in the following figure:

The magnitude of the moment ** m**which we denote as

*m*(not bold) is related to the rotated angle θ through the shear modulus G according to the following equation (it is derived by a simple integral):

Since the magnitude of the moment is equal to the product of the magnitude of the force F by the radius of the pulley Rp:

*M =F.Rp*

And strength is the weight that hangs **W**so:

*M = W.Rp*

Substituting into the moment magnitude equation:

We have the relationship between weight and angle:

**How to find G?**

This relationship between the variables *W* and *θ* is linear, so the different angles produced by hanging different weights are measured.

The weight and angle pairs are plotted on graph paper, the best straight line that passes through the experimental points is fitted, and the slope is calculated. *m* of said line.

From there it follows that:

**Exercises with solution**

**– Exercise 1**

A rod 2.5 meters long and with a radius of 4.5 mm is fixed at one end. The other is connected to a pulley of radius 75 cm that has a weight W of 1.3 kg hanging from it. The rotated angle is 9.5º.

With these data, it is requested to calculate the shear modulus G of the rod.

**Solution**

From the equation:

G is cleared:

And substitute the values given in the statement, taking care to express all the data in the SI International System of Units:

R = 4.5mm = 4.5 x 10 -3m

Rp = 75 cm = 0.075

To go from kilograms (which are actually kilograms – force) to newtons, multiply by 9.8:

W = 1.3 kg-force = 1.3 x 9.8 N = 12.74 N

And finally, the degrees must be in radians:

9.5 º = 9.5 x2π /360 radians = 0.1658 radians.

With all this we have:

= 2.237 x 1010 Pa

**– Exercise 2**

A cube made of gel has a side of 30 cm. One of its faces is fixed, but at the same time, a parallel force of 1 N is applied to the opposite face, which as a result moves 1 cm (see the example in the book in figure 1).

It is requested to calculate with these data:

a) The magnitude of the shear stress

b) The unit strain δ

c) The value of the shear modulus

**Solution to**

The magnitude of the shear stress is:

τ = F/A

With:

A = side2 = (30 x 10-2 cm)2 = 0.09 m2

Therefore:

τ = 1 N / 0.09 m2 = 11.1 Pa

**solution b**

The unit strain is none other than the value of δ, given by:

δ = Δx / L

The displacement of the face subjected to the force is 1 cm, then:

δ =1 / 30 = 0.0333

**solution c**

The shear modulus is the ratio between the shear stress and the unit strain:

G = Shear force/Strain

Therefore:

G = 11.1 Pa / 0.033 = 336.4 Pa

**References**

Beer, F. 2010. Mechanics of materials. McGraw Hill. 5th. Edition.

Franco García, A. Rigid Solid. Measurement of the shear modulus. Retrieved from: sc.ehu.es.

Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed Prentice Hall.

Resnick, R. (1999). Physical. Vol.1.3ra Ed. in Spanish. Continental Publishing Company SA de CV

University of Valladolid. Department of Condensed Matter Physics. Selection of problems. Retrieved from: www4.uva.es.