The shear modulus describes the response of a material to the application of a shear stress that deforms it. Other frequently used designations for shear modulus are shear, shear, transverse elasticity or tangential elastic modulus.
When the stresses are small, the strains are proportional to them, according to Hooke’s law, with the shear modulus being the constant of proportionality. Therefore:
Shear modulus = shear / warp tension
Suppose one force is applied to the cover of a book, the other being fixed to the table surface. In this way, the book as a whole does not move, but deforms when the top cover moves relative to the bottom by the amount Δx .
The book goes from a rectangular cross section to a parallelogram section, as we can see in the image above.
Be:
τ = F / A
The stress or shear stress, where F is the magnitude of the applied force and A is the area in which it acts.
The deformation caused is given by the quotient:
δ = Δx / L
Therefore, the cut module, which we will designate as G, is:
And since Δx / L has no dimensions, the units of G are the same as the 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’s pound/square inch, abbreviated psi .
Cutting module for various materials
Under the action of shear forces such as those described, objects offer a book-like resistance in which the inner layers slide. This type of deformation can only occur in solid bodies, which have enough rigidity to resist deformation.
On the other hand, liquids do not offer this type of resistance, but can suffer volume deformations.
Below is the G in Pa cutting module for various materials frequently used in the construction and manufacture of machines and spare parts of all types:
Experimental measurement of the shear modulus
To determine the shear modulus value, samples of each material must be tested and their response to the application of a shear stress examined.
The specimen is a rod made of the material, with a known radius R and length L , which is fixed at one end while the other is connected to the shaft of a freely rotating pulley.
The pulley has a cable attached to the free end, the weight of which is suspended, which exerts a force F on the rod through the cable. And this force, in turn, produces a moment M in the rod, which then rotates a small angle θ.
A diagram of the assembly can be seen in the following figure:
The magnitude of the moment M , which we call M (without bold), is related to the rotated angle θ through the shear modulus G, according to the following equation (deduced by a simple integral):
Since the magnitude of the moment is equal to the product of the modulus of force F times the radius of the pulley R p :
M = FR p
And force is the weight that locks W , so:
M = WR p
Substituting in the momentum 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 that the different angles produced by the suspension of different weights are measured.
The weight and angle pairs are plotted on graph paper, the best line passing through the experimental points is adjusted and the slope m of that line is calculated.
Exercises with solution
– Exercise 1
A rod 2.5 meters long and 4.5 mm in radius is attached to one end. The other is connected to a 75 cm radius pulley that has a pendant weight W of 1.3 kg. The rotated angle is 9.5º.
With these data, it is requested to calculate the modulus of cut G of the shank.
Solution
From the equation:
G is clean:
And the values provided in the declaration are replaced, taking care to express all data in the SI International System of Units:
R = 4.5 mm = 4.5 x 10 -3 m
R p = 75 cm = 0.075
To go from kilograms (which are actually kilograms – force) to newton, 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 ° = 9.5 x2π / 360 radians = 0.1658 radians.
With all this you have:
= 2,237 x 10 10 Pa
– Exercise 2
A cube of gel measures 30 cm on a side. One of its faces is fixed, but at the same time, a parallel force of 1 N is applied to the opposite face, which displaces it by 1 cm (see the book example in figure 1).
You are asked to calculate with this data:
a) The magnitude of the shear stress
b) The unit deformation δ
c) The value of the shear modulus
Solution for
The magnitude of the shear stress is:
τ = F / A
With:
A = side 2 = (30 x 10 -2 cm) 2 = 0.09 m 2
Therefore:
τ = 1 N / 0.09 m 2 = 11.1 Pa
Solution b
The unit deformation 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, so:
δ = 1/30 = 0.0333
Solution c
The shear modulus and the ratio between shear stress and unit strain:
G = shear stress / strain
Therefore:
G = 11.1 Pa / 0.033 = 336.4 Pa
