Elastic Limit and Constants MCQ Quiz - Objective Question with Answer for Elastic Limit and Constants - Download Free PDF

Explanation:

The relationship between the Modulus of Elasticity (E) and the Shear Modulus (G) for an isotropic material is given by the formula:

E=2G(1+μ)

Additional InformationModulus of Elasticity (E):

1. Definition: The Modulus of Elasticity (E), also known as Young's Modulus, is a material property that measures the stiffness of a material in response to uniaxial tensile or compressive stress. It is defined as the ratio of stress to strain in the elastic region of the material’s stress-strain curve.

2. Significance: A higher E value means the material is stiffer and deforms less under stress. It is an important factor in structural design and material selection.

3. Example: Steel has a higher E value than rubber, meaning steel is much stiffer and less deformable than rubber under the same load.

Shear Modulus (G):

1. Definition: The Shear Modulus (G), also known as Modulus of Rigidity, measures the material's response to shear stress (forces that cause deformation by sliding layers of material). It is defined as the ratio of shear stress to shear strain in the elastic region of the material’s behavior.

2. Significance: G indicates how easily a material deforms under shear forces. Materials with a high G are resistant to shape changes when subjected to shear forces.

3. Example: Steel has a higher G than rubber, meaning steel resists deformation under shear forces more effectively than rubber.

Explanation:

Ratio of Shearing Stress to Shear Strain:

• The ratio of shearing stress to shear strain is known as the modulus of rigidity (also referred to as shear modulus). It is a material property that quantifies the rigidity or stiffness of a material under shear deformation. When a material is subjected to shear stress, it undergoes angular deformation, and the modulus of rigidity determines the relationship between the applied stress and the resulting strain.
• When a material is subjected to shear stress, the forces act parallel to the surface, causing the material to deform angularly. The shear strain is the measure of the angular deformation due to the applied shear stress. The modulus of rigidity quantifies the material's resistance to this deformation. Higher values of modulus of rigidity indicate that the material is more resistant to shear deformation, whereas lower values indicate greater deformability.

Formula:

The modulus of rigidity (G) is mathematically expressed as:

G = τ / γ

Where:

• G = Modulus of rigidity (N/m² or Pa)
• τ = Shearing stress (N/m² or Pa)
• γ = Shear strain (dimensionless)

Concept:

There is a relationship between Young’s modulus \(E\), shear modulus \(G\), and Poisson’s ratio \(\nu\) given by the formula:

\(E = 2G(1 + \nu) \)

Given:

Poisson’s ratio, \(\nu = 0.5 \)

Calculation:

Substitute the given value of \(\nu\) into the formula:

\(E = 2G(1 + 0.5) = 2G \times 1.5 = 3G \)

Explanation:

Young's Modulus:

• Young's Modulus, denoted as E, is a measure of the stiffness of a material. It is a mechanical property that quantifies the ratio of tensile stress to tensile strain in the linear elastic region of a material. Mathematically, it is expressed as:

E = Stress / Strain

Where:

• Stress = Force (F) / Cross-sectional Area (A)
• Strain = Change in Length (ΔL) / Original Length (L)

Key Concept: Young's Modulus is an intrinsic property of a material, meaning it depends only on the material's nature and not on its shape, size, or dimensions (such as radius, length, or cross-sectional area). It describes the material's ability to resist deformation under tensile stress.

If the radius of a wire stretched by a load is doubled, its Young's Modulus remains unaffected. This is because Young's Modulus is an inherent property of the material and does not depend on the geometric dimensions of the object. While changing the radius alters the cross-sectional area of the wire, which in turn affects the stress and strain values individually, the ratio of stress to strain (Young's Modulus) remains constant for a given material.

Let’s break this down:

1. Stress: Stress is calculated as Force (F) divided by Cross-sectional Area (A). If the radius of the wire is doubled, the cross-sectional area increases by a factor of four (since A = πr², and r is doubled). Consequently, stress decreases by a factor of four.
2. Strain: Strain is the ratio of the change in length (ΔL) to the original length (L). The change in length is influenced by the applied force and the material's properties, but strain remains proportional to stress for elastic deformations (Hooke's Law).
3. Young's Modulus: Since Young's Modulus is the ratio of stress to strain, and both stress and strain scale proportionally with the change in cross-sectional area, their ratio (E) remains unchanged.

Concept:

The relation between Elastic constant - E, K, G(or C) and μ:

\(E = 2G (1 + μ)\)

\(E = 3K (1 - 2μ)\)

\({\rm{E}} = \frac{{9{\rm{KC}}}}{{3{\rm{K}} + {\rm{C}}}}\)

Where,

E = Young's Modulus of Rigidity = Stress/strain

C = Shear Modulus or Modulus of rigidity = Shear stress / Shear strain

μ = Poisson’s ratio = - lateral strain / longitudinal strain

K = Bulk Modulus of elasticity = Volumetric stress / Volumetric strain​​

Poisson’s ratio:

\(μ = {- {lateral strain \over longitudinal strain}}\)

Calculation:

Given data

d = 25 mm and L = 1.2 m

E = 2 x 105 N/mm2.

longitudinal strain = 4 times the lateral strain

Poisson’s ratio:

\(μ = {- {lateral strain \over longitudinal strain}} = {1 \over 4}\) = 0.25

Now using the relationship of elastic constant

\(E = 2G (1 + μ)\)

 \(G = \frac{E}{{2\left( {1 + \mu } \right)}} = \frac{{2 \times {{10}^5}}}{{2\left( {1 + 0.25} \right)}}\)

G = 0.8 × 105 N/mm2

Explanation:

Poisson's Ratio:

Poisson's ratio is the ratio of transverse strain to longitudinal strain.

\(μ = - \frac{{{\epsilon_{(lateral)}}}}{{{\epsilon_{(longitudinal)}}}}\)

The value of Poisson's ratio for different materials is as follows:

Explanation:

Isotropic material means a material having identical values of a property in all directions. Glass and metals are examples of isotropic materials.

Homogeneous material is a material of uniform composition throughout that cannot be mechanically separated into different materials.

Anisotropic material's properties such as Young's Modulus, change with the direction along the object. Common examples of anisotropic materials are wood and composites.

Viscoelastic material have both properties of elasticity and viscosity.

Concept:

Relationship between modulus of elasticity (E), modulus of rigidity (G), and Poisson's ratio (μ) is described by:

E = 2G (1 + μ)

Calculation:

Given:

μ = 0.4

E = 2G (1 + μ)

\(\frac{{\rm{G}}}{{\rm{E}}} = \frac{1}{{2\left( {1 + {\rm{μ }}} \right)}} = \frac{1}{{2\left( {1 + 0.4} \right)}} = 0.357\)

Important Points

Other relationships between various elastic constants are:

• \({\bf{E}} = \frac{{9{\bf{KG}}}}{{3{\bf{K}}\; + \,{\bf{G}}\;}}\)
• E = 3K(1 - 2μ)

​where K is the bulk modulus of elasticity.

Explanation:

Isotropic Material: 

• Isotropic materials have identical material properties in all directions at every given point.
• When a specific load is applied at any point of isotropic materials, it will exhibit the same strength, stress, strain, young’s modulus, and hardness in the x, y, or z-axis direction.

Examples: metals, glasses.

• Almost every metal exhibits isotropic behaviour at the micro-level because all single-crystal systems are isotropic with respect to mechanical properties.
• There are a few exceptional cases where metals exhibit anisotropic behaviour at the micro-level.
• Random localized distribution of anisotropic behaviour at the micro-level, nullify each other and the net effect at the macro level is isotropic behaviour.

For an isotropic, homogeneous, and elastic material obeying Hook's law, the number of independent elastic constants is 2.

It means to fully define the elastic behaviour of isotropic material only 2 elastic constants are good enough.

E = 2G (1 + μ) = 3K(1 - 2μ)

As seen in the above equations there are a total of 4 elastic constants.

If we know two of them, then we can find other constants also.

∴ there are 2 independent elastic constants required to define the stress-strain relationship for isotropic material.

Additional Information

Orthotropic material

• A material is said to be orthotropic if it has three different properties in three mutually perpendicular directions.
• There are 9 independent elastic constants required to define the stress-strain relationship for orthotropic material.

Anisotropic material

• A material is said to be anisotropic if it has different properties in each direction.
• There are 21 independent elastic constants required to define the stress-strain relationship for an anisotropic material.

Concept:

The relation between E, K, and μ is given by:

E = 2N (1 + μ)

E = 3K (1 - 2μ)

\({\rm{E}} = \frac{{9{\rm{KC}}}}{{3{\rm{K}} + {\rm{C}}}}\)

Where,

E = Young's Modulus of Rigidity, i.e. E = Stress / strain

N = Shear Modulus or Modulus of rigidity  i.e. N = Shear stress / Shear strain

μ = Poisson’s ratio, μ = -lateral strain / longitudinal strain

K = Bulk Modulus of elasticity,  i.e. K = Volumetric stress / Volumetric strain

Calculation:

Here, the modulus of rigidity is given as N

So,

E = 2N (1 + μ)

Put E = 2N, we get μ = 0

Now take, E = 3K (1 - 2μ)

E = 3 K (1 - 0)

∴ K = E / 3

Concept:

The relationship between elastic modulus, bulk modulus and Poisson's ratio is given by

\(E=3K(1-2μ)\)

The relationship between elastic modulus, modulus of rigidity and Poisson's ratio is given by

\(E=2G(1+μ)\)

The relationship between elastic modulus, modulus of rigidity and bulk modulus is given by

\(E={9KG\over{G+3K}}\)

Hence, combining all these equations and solving it together we can get,

\(\begin{array}{l} μ = \frac{{3K - 2G}}{{6K + 2G}} \end{array}\)

Calculation:

Given data;

E= 240 GPa

μ = 0.2

The relationship between elastic modulus, modulus of rigidity and Poisson's ratio is given by

\(E=2G(1+μ)\)

\(G={E\over2(1+μ)}\)

\(G={240\over2(1+0.2)}\)

G= 100 GPa

Explanation:

Poisson's ratio is the ratio of transverse strain to longitudinal strain.

\(\mu = - \frac{{{\epsilon_{(lateral)}}}}{{{\epsilon_{(longitudinal)}}}}\)

For perfectly isotropic elastic material, Poisson’s ratio is 0.25 but for most of the materials.

The value of Poisson's ratio lies in the range of 0 to 0.5.

Poisson's ratio for various materials are:

• Cork: 0.0
• Aluminium: 0.31
• Cast iron: 0.21 – 0.26
• Steel: 0.27 – 0.30
• Stainless steel: 0.30 – 0.31
• Copper: 0.33
• Rubber: 0.5

Explanation:

For an isotropic, homogeneous and elastic material obeying Hook's law, the number of independent elastic constants is 2 (Young’s Modulus and Poisson’s Ratio).

Concept:

E = 2G(1 + v)

where E = Modulus of elasticity, G = Modulus of rigidity, ν = Poisons ratio.

Calculation:

Given:

Poisson's ratio, v = 0.25

the ratio of modulus of rigidity to modulus of elasticity = \(\frac{G}{E}\)

\(\frac{G}{E} = \frac{1}{{2\left( {1 + v} \right)}}\)

\(\frac{G}{E} = \frac{1}{{2\left( {1 + 0.25} \right)}}\)

\(\frac{G}{E} = 0.4\)

Hence the required value of the ratio of modulus of rigidity to modulus of elasticity is 0.4.

Explanation

Given: -

Shearing stress, τ = 100 MPa

Modulus of elasticity, E = 200 GPa

Poisson’s ratio, μ = 0.25

We know,

Modulus of rigidity (G) is defined as the ratio of shearing stress to the shearing strain.

Also,

\( \Rightarrow G = \frac{{200}}{{2\left( {1 + 0.25} \right)}} = 80\;GPa\)

Then,
\(G = \frac{\tau }{\gamma }\)

\( \Rightarrow \gamma = \frac{{100}}{{80 \times {{10}^3}}} = 0.00125\;rad\)