Modulus of Rigidity

Modulus of rigidity also known as shear modulus, is used to measure the rigidity of a given body. It is the ratio of shear stress to shear strain and is denoted by G or sometimes by S or μ. The modulus of rigidity of a material is directly proportional to its elastic modulus which depends on the material’s nature and properties.

In this article, we will talk about what Modulus of rigidity is, its formula, examples, and applications.

What is the Modulus of Rigidity (Shear Modulus)

Modulus of rigidity, also known as shear modulus, is a measure of the elastic shear stiffness of a material. It is defined as the ratio of shear stress to the shear strain. It can be used to explain why a material resists transverse deformations. A way to measure the mechanical properties of solids is to use the shear modulus of elasticity. The SI unit for the Modulus of rigidity is Pascal (Pa) or N/m². It is denoted by G, S, or μ

In simple terms, the Modulus of rigidity is a measure of how much a material resists deformation when a force is applied perpendicular to its surface.

Learn,

• Stress and Strain
• Stress strain Curve

Modulus of Rigidity Definition

Modulus of Rigidity, also known as shear modulus, is the elastic coefficient for shearing or torsion force. It gives us a measure of how rigid a body is. It is defined as the ratio of shear stress to the shear strain.

Modulus of Rigidity Formula

Formula of Modulus of Rigidity Or Shear Modulus (G) is given below:

G = τ_xy/ γ_xy = F/A/ ΔX/L = FL/AΔX

where,

• τ_xy is Shear Stress
• γ_xy is Shear Strain
• F is Force Acting on Object
• A is Area on which Force is Acting
• ΔX is Tansverse Displacement
• L is Initial Length

Learn, Poisson’s Ratio

Units and Dimension of Shear Modulus

• The SI unit for Modulus of rigidity is Pascal (Pa) or N/m²
• Modulus of rigidity can also be expressed in GigaPascal (GPa) or pounds per square inch (PSI).
• Dimensional formula for the Modulus of rigidity is M¹L^-1T^-2.

Characteristics of Modulus of Rigidity

Here are some of the important characteristics of Modulus of Rigidity:

• Modulus of rigidity is a measure of the elastic shear stiffness of a material.
• The modulus of rigidity can be experimentally determined from the slope of a stress-strain curve created during tensile tests conducted on a material sample.
• For isotropic materials, the modulus of rigidity value is determined by a torsion test.

Examples of Modulus of Rigidity

Here are some examples of modulus of rigidity for various materials:

Material	Modulus of Rigidity Value
Concrete	3 x 106 psi (21 GPa)
Wood	13 GPa
Brass	40 GPa
Rubber	0.0003 GPa
Ideal liquid	0 GPa

Modulus of Rigidity of Steel

Modulus of rigidity for steel is approximately 79 GPa (Gigapascals) or 79,000 MPa (Megapascals). This value represents the material’s resistance to shearing or torsion forces.

Modulus of Rigidity of Aluminum

Modulus of rigidity for Aluminum can vary slightly with different alloys and other factors, but typical values for the modulus of rigidity of aluminum materials used in the industry range from 24 to 28 GPa (Gigapascals), or 3.5 × 10⁶ to 4.1 × 10⁶ psi.

For example, the modulus of rigidity for aluminum 6061-T6 is approximately 24 GPa. This value represents the material’s resistance to shearing or torsion forces and is an important property for understanding the material’s behavior under shear stress.

Relation between Modulus of Elasticity and Modulus of Rigidity

The Modulus of rigidity (G) and the Modulus of elasticity (E) are related through the material’s Poisson’s ratio (v). The relationship can be expressed as:

E = 2G(1+ν)

Where,

• E is Young’s Modulus
• G is Shear Modulus
• v is Poisson’s Ratio

This equation shows that the Modulus of elasticity is related to the Modulus of rigidity and the Poisson’s ratio.

• It indicates that these elastic constants are interrelated and can be derived from each other.
• The Poisson’s ratio is a measure of the transverse strain to longitudinal strain and is typically denoted by the symbol ν.
• This relationship holds for linear, homogeneous, and isotropic materials.

Modulus of Rigidity vs Modulus of Elasticity

Comparison between Modulus of elasticity and Modulus of rigidity:

Difference between Modulus of Rigidity and Modulus of Elasticity
Property	Modulus of Elasticity	Modulus of Rigidity
Also known as	Young’s Modulus	Shear Modulus
Definition	Measure of a material’s ability to deform elastically under stress	Measure of a material’s resistance to shearing or torsion force
Symbol	E	G, S, or μ
Formula	Stress/Strain	Shear Stress / Shear Strain
SI Unit	Pascal (Pa)	Pascal (Pa)
Example	Rubber has a low modulus of elasticity, signifying more deformation under less stress	Steel has a high modulus of rigidity, signifying notable resistance to shape alterations

Relation between Bulk Modulus and Modulus of Rigidity

The Modulus of Rigidity (G) and Bulk Modulus (K) are related through the material’s Poisson’s ratio (v). The relationship can be expressed as:

G = 3K(1 – 2v)/2(1 + v)

where,

• G is the Shear Modulus
• K is the Bulk Modulus
• v is the Poisson’s Ratio

This equation shows that the Modulus of rigidity and Bulk Modulus are related through the Poisson’s ratio and some constants. This relationship holds for linear, homogeneous, and isotropic materials.

Relation Between Modulus of Rigidity and Modulus of Elasticity

The relationship between Modulus of Rigidity(G) and Modulus of Elasticity(E) is given by the formula,

E = 2G(1 + v)

where,

• G is the Shear Modulus
• E is the Modulus of Elasticity
• v is the Poisson’s Ratio

Note:

(Elastic Constant)E = 9KG/(3K+G)

Applications of Modulus of Rigidity

Some of the common applications of the Modulus of rigidity in real life include:

• Design of Structures: The shear modulus is used in designing structures such as bridges and buildings, where the material must resist forces that would cause it to deform.
• Material Selection: Material scientists and applied physicists use the concept of Modulus of rigidity to select the correct material to use for construction. For example, the smaller the force is, the easier the material will bend. It is calculated and publicly recorded for most materials.
• Calculation of Deformation and Vibrations: The shear modulus can calculate the deflection of beams and plates under transverse loads. It is also used in calculating the vibrations of plates and shells.
• Understanding Material Behavior: The shear modulus determines how elastic or bendable materials will be if they are sheared, which is essential in understanding the behavior of materials under different conditions.

Related Reads

• Shear Modulus and Bulk Modulus
• Young’s Modulus
• Elastic Behaviour of Material

Numericals on Modulus of Rigidity

Numerical 1. The area of the upper face of a rectangular block is 0.5 m x 0.5 m and the lower face is fixed. The height of the block is 1 cm. a shearing force applied to the top face produces a displacement of 0.015 mm. Find the strain, stress and the shearing force. Modulus of rigidity = η = 4.5 × 10¹⁰ N/m².

Solution:

Given,

• Area under shear = A = 0.5 m x 0.5 m = 0.25 m²
• Height of the block = h = 1 cm = 1 × 10^-2 m
• Displacement of top face = x = 0.015 mm = 0.015 × 10^-3 m = 1.5 × 10^-5 m
• Modulus of rigidity = η = 4.5 × 10¹⁰ N/m²
• Shear strain = tan θ = x/h = (1.5 × 10^-5) / (1 × 10^-2) = 1.5 × 10^-3

Modulus of rigidity = η = Shear stress / Shear strain

∴ Shear stress = η × Shear strain = 4.5 × 10¹⁰ × 1.5 × 10^-3

∴ Shear stress = 6.75 × 10⁷ N/m².

Shear stress = F/A

∴ F = Shear stress × Area

∴ F = 6.75 × 10⁷ × 0.25

∴ F = 1.69 × 10⁷ N

Ans:

• Shear strain = 1.5 × 10^-3

• Shear stress = 6.75 × 10⁷ N/m²

• Shearing force = 1.69 × 10⁷ N

Numerical 2. A metallic cube of side 5 cm, has its lower surface fixed rigidly. When a tangential force of 10⁴ kg. wt. is applied to the upper surface, it is displaced through 0.03 mm. Calculate (1) the shearing stress (2) the shearing strain and (3) the modulus of rigidity of the metal.

Solution:

Given,

• Area under shear = A = 5 cm x 5 cm = 25 cm² = 25 × 10^-4 m²
• Height of the block = h = 5 cm = 5 × 10^-2 m
• Displacement of top face = x = 0.03 mm = 0.03 × 10^-3 m = 3 × 10^-5 m
• Shearing Force = 10⁴ kg-wt = 10⁴ × 9.8 N

Shear stress = F/A

∴ Shear stress = (10⁴ × 9.8)/( 25 × 10^-4)

∴ Shear stress = 3.92 × 10⁷ N

Shear strain = tanθ = x/h = (3 × 10^-5 ) / (5 × 10^-2 ) = 6 × 10^-4

Modulus of rigidity = η = Shear stress / Shear strain

η = (3.92 × 10⁷) / (6 × 10^-4) = 6.53 × 10¹⁰ N/m²

Ans:

• Shear stress = 3.92 × 10⁷ N
• Shear strain = 6 × 10^-4
• Modulus of rigidity = 6.53 × 10¹⁰ N/m²

Numerical 3. A metal plate has an area of face 1m x 1m and thickness of 1 cm. One face of a larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced thereby is 0.005 cm. Find the shearing stress, strain and magnitude of the tangential force applied. Modulus of rigidity of metal is ϒ = 8.4 × 10¹⁰ N/m²

Solution:

Given,

• Area under shear = A = 1 m x 1 cm = 1 m²
• Thickness of plate = h = 1 cm = 1 × 10^-2 m
• Displacement of top face = x = 0.005 cm = 0.005 × 10^-2 m = 5 × 10^-5 m
• Modulus of rigidity = η = 8.4 × 10¹⁰ N/m²

Shear strain = tanθ = x/h = (5 × 10^-5) / (1 × 10^-2) = 5 × 10^-3

Modulus of rigidity = η = Shear stress / Shear strain

∴ Shear stress = η × Shear strain = 8.4 × 10¹⁰ × 5 × 10^-3

∴ Shear stress = 4.2 × 10⁸ N/m².

Shear stress = F/A

∴ F = Shear stress × Area

∴ F = 4.2 × 10⁸ ×1

∴ F = 4.2 × 10⁸ N

Ans:

• Shear Strain = 5 × 10^-3
• Shear Stress = 4.2 × 10⁸ N/m²
• Shearing Force = 4.2 × 10⁸ N

IIT JEE Questions on Modulus of Elasticity

Q1. Modulus of Rigidity of Diamond is

1. Too less
2. Greater than all matters
3. Less than all matters
4. Zero

Ans: 2

Q2. Which statement is true for a metal

1. Y < n
2. Y = n
3. Y > n
4. Y < 1/n

Ans: 3

As Y = 2n (1 + σ)

Q3. The Young’s Modulus of the material of a wire is 6 x 10¹² N/m² and there is no transverse strain in it, then it’s modulus of Rigidity will be

1. 3 x 10¹² N/m²
2. 2 x 10¹² N/m²
3. 10¹² N/m²
4. None of the above

Ans: 1

Y = 2n (1 + σ) for no transverse strains ( s = 0 )

Y = 2n ⇒ n = Y/2 = 3 x 10¹² N/m²

Q4. A cube of aluminum of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

1. 0.02
2. 0.1
3. 0.005
4. 0.002

Ans: 4

Shearing Strain φ = x/l = 0.02cm/10cm = 0.002 cm

Modulus of Rigidity-FAQs

1. What is Shear Modulus?

Shear Modulus also called Modulus of Rigidity measures a material’s resistance to deformation under shear stress.

2. What is Difference between Modulus of Rigidity and Modulus of Elasticity?

Modulus of rigidity focuses on shear deformation, while modulus of elasticity deals with linear deformation or stretching.

3. What is Bulk Modulus and Rigidity Modulus?

Bulk modulus measures a material’s resistance to volume change, while rigidity modulus (shear modulus) assesses resistance to shear deformation.

4. What is Modulus of Rigidity Formula?

Modulus of rigidity formula is, G = τ / γ

5. What is Unit of Modulus of Rigidity?

Modulus of rigidity unit is pascals (Pa) or gigapascals (GPa).

6. What is the Modulus of Rigidity of Ideal Liquid?

The modulus of rigidity for an ideal liquid is zero. Liquids don’t resist shear deformation like solids, so they have no shear modulus.

7. Modulus of Rigidity is defined as the Ratio of?

Modulus of rigidity is defined as the ratio of shear stress to the corresponding shear strain in a material. It measures how a material deforms under shear forces.