**Relation among elastic constants** : ( Y, σ, k, n, x)

Young’s modulus, bulk modulus and Rigidity modulus of an elastic solid are all together are known as elastic constants. When a deforming force is acting on a solid continuously, it results in the change in its original dimension. In such cases, we can use the relation between elastic constants to understand the magnitude of deformation.

**1. Relation among Y, σ, k –**

**2. Relation among Y, σ, n –**

**3. Relation among Y, σ, x –**

**4. Relation among k, n, σ –**

**5. Relation among Y, k, n –**

**6. Relation among x, n, σ –**

**7. Relation among n, k, x –**

***** IF NECESSARY, I WILL UPLOAD THE PROOF OF THE ABOVE RELATIONS.**

## Limiting value of Poisson’s ratio (σ) –

The Poisson’s ratio of a stable, isotropic and linear elastic material is greater than −1 or less than 0.5 for the requirement for Young’s modulus, the shear modulus and bulk modulus to have positive values. Most materials have Poisson’s ratio have the values between 0 and 0.5.

We know, Y = 3k (1 – 2σ)

Since Y and k are +ve, hence

1 – 2σ ≥ 0

– 2σ ≥ -1

σ ≤ 1/2

Again, Y = 2n(1+σ)

Since Y and n are +ve, hence

1 + σ ≥ 0

σ ≥ -1

Considering both cases, we may write

-1 ≤ σ ≤ 1/2

Note :

“σ is negative” which means a longitudinal extension is associated with a lateral extension, which is practically impossible.

Practical or acceptable limit of σ is

0 ≤ σ ≤ 1/2

**A solid cylinder is extended longitudinally such that its volume remains constant. Find its Poisson’s ratio.**

Sol-

Let us consider a solid cylinder of length l and radius r is extended longitudinally by an amount dl and its radius is decreases by dr, then

Poisson’s ratio

Volume of the cylinder,

Differentiating,

**Thermal Stress**

In mechanics and thermodynamics, **thermal stress** is mechanical stress created by any change in temperature of a material. These stresses can lead to fracturing or plastic deformation depending on the other variables of heating, which include material types and constraints.^{} Temperature gradients, thermal expansion or contraction and thermal shocks are things that can lead to thermal stress. This type of stress is highly dependent on the thermal expansion coefficient which varies from material to material. In general, the greater the temperature change, the higher the level of stress that can occur. Thermal shock can result from a rapid change in temperature, resulting in cracking or shattering.

If a rod is fixed at its one end and is subjected to temperature change across it, then a stress will be developed

within it, which is called thermal stress.

Calculation-

Let us consider a rod of length L and its temperature is increased by t, then its length is increased by l.

Hence longitudinal strain = l/L

If α be the linear coefficient of thermal expansion of the material of the rod, then

**Thermal stress = Yαt**

**A steel rod of length 5 meters fixed rigidly between two supports. The coefficient of linear expansion of steel is 12 x 10^-6 / °C . Calculate the stress is the rod for an increase of temperature of 40°C Given Y = 2 x 10^11 N/m².**

Sol-

Given,

Thermal stress

## Rupturing **tensile stress**

Stress rupture is the sudden and complete failure of a material under stress. During testing, the sample is held at a specific load level and temperature for a previously determined amount of time. In stress rupture testing, loads can be applied by tensile bending, flexural, bi-axial, or hydro-static methods.

If a thin rod is rotated uniformly with an angular velocity, then a stress is developed within the rod. Then the stress is known as tensile stress.

Calculation-

Let a thin uniform rod of length l and mass m rotates uniformly with an average velocity w.

Tension at A, at a distance x from rotating axis

= centripetal force due to the portion AB

So, Tensile stress at a distance x

**Torsion of a cylinder**

Torsion is actually the twisting of an object due to an externally applied torque. In sections perpendicular to the torque axis, the resultant shear stress in this section is perpendicular to it’s radius.

Let us consider (solid) of radius a and length l· Due to an external torque cylinder is twisted by an angle Φ.

Let us consider an inner cylinder of radius r (r<a). Due to twisting Φ, let a point A on the inner cylinder is displaced to A’ which produces a shearing strain θ.

From figure,

So, shearing angle, θ = rΦ/l ———– 1

Due to elastic property, a restoring torque is produced which tends to prevent the twisting of the cylinder.

To calculate resisting tongue let us take a coaxial cylindrical shell of radius r and thickness dr.

If F be the tangential force for elementary portion – dx of the shell, then

tangential stress = F/dr.dx ————– 2

Hence rigidity modules (n)

Moment of force on that elementary portion

Restoring torque over the whole perimeter of the elementary shell,

Summing up all such elementary shells, torque over cylinder ;

Note :

where

called **torsional rigidity.**

c = τ/Φ i.e. torsional couple per unit twist is called torsional rigidity.

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