Derivation Relation Between Elastic Constants

Elastic constants are the parameters that demonstrate the relation between the stress and strain on any solid materials, to express the elastic behavior of the material. When any force is applied to a material that changes its original dimension, then the relation between the elastic constants can be used to understand the magnitude of deformation.

There are 3 elastic constants, named Young’s Modulus (Y), Bulk Modulus (B), and Shear Modulus (G). This relation between these elastic constants are expressed through a formula:

Y = 9BG/ G + 3B

Read Also: Class 11 Mechanical Properties of Solids

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Stress and Strain

When a body is subjected to a force that creates deformation in the material, then a restoring force automatically generates in the body. This restoring force is of equal strength but acts in an opposite direction to the force applied. The restoring force per unit area is called stress. If the force applied is F and the surface area in which the force is applied is A, the magnitude of stress is F/A.

The magnitude of Stress = \(\frac{F}{A}\)

There are 3 ways through which a solid can change its dimension when subjected to an external force.

Check Important Formula of Stress

Longitudinal Stress and Longitudinal Strain

• Consider a cylindrical body that is stretched by two equal forces applied normally to its cross-section area. In such a case, the restoring force developed is called tensile stress.
• When a similar cylindrical body is compressed through its cross-sectional area, then the restoring force developed is known as compressive stress.
• Both the tensile and compressive stress can be termed longitudinal stress.
• The longitudinal strain is the change in the length to the original length.

Longitudinal Strain = \(\frac{\triangle l}{l}\)

Shearing Stress and Shearing strain

• Consider a cylinder in which two equal deforming forces are applied parallel to its cross-sectional area, but in an opposite direction. Then a displacement (x) is seen between the opposite faces of the cylinder.
• The restoring force developed in this case is known as shearing stress.
• The strain produced is known as shearing strain and is defined as the ratio of relative displacement of the faces to the length of the cylinder.

Shearing Strain = \(\frac{\triangle x}{l} = tan\theta = \theta\) (Since \(\theta\) is very small of tan \(\theta\) can be considered as \(\theta\).)

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Hydraulic Stress and Volume Strain

When a solid sphere is placed in a fluid, then the pressure applied is uniform from all the sides. The fluid applies pressure in a perpendicular direction at each point of the surface. This will cause no change in the geometric shape of the solid but the volume will change. Hydraulic Stress and Volume Strain

• The restoring force developed, of magnitude equal to the hydraulic pressure is known as hydraulic stress.
• The strain produced in this case is called volume strain and is defined as the ratio of change in volume to the original volume.

Volume Strain = \(\frac{V}{V}\)

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Elastic Constants

Elasticity is the characteristics of the material and it is defined as the ratio of stress and strain. There are 3 elastic constants and all of them have been derived from the 3 ways that have been described above.

Young’s Modulus

Young’s modulus is an independent elastic constant and can be obtained experimentally. It is defined as the ratio of longitudinal stress to the longitudinal strain. 

Young’s Module \((Y) = \frac{\frac{(F)}{(A)}}{\frac{(L)}{(L)}}\)

Read Further: Longitudinal Waves

Shear Modulus

Shear Modulus is also known as modulus of rigidity and is represented by G. It is defined as the ratio of the shearing stress to the corresponding shearing strain of the material. 

Shear Modulus \((G) \)

\(= \frac{F*L}{A*\triangle x}\)

\(= \frac{\frac{F}{A}}{\theta}\)

\(= \frac{F}{A*\theta}\)

Bulk Modulus

We’ll consider the situation when the body is under hydraulic compression. It is submerged in a liquid, which leads to a decrease in the volume of the body. This is because of hydraulic stress, which also produces a strain called volume strain. So a bulk modulus can be defined as the ratio of hydraulic stress to the volume strain. It is denoted by B.

\(B = \frac{-p}{\frac{\triangle V}{V}}\)

Where, B = Bulk Modulus

p = Pressure

\(\triangle\)V = Change in volume

V = Original Volume

The negative sign indicates that the increase in pressure leads to a decrease in volume.

Also Check:

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Relation: Young’s Modulus (Y), Bulk Modulus (B), and Poisson’s ratio ()

We’ll be taking a cuboid that has a volume of l x b x t.

Relation Young’s Modulus, Bulk Modulus, and Poisson's Ratio

Let \(\alpha\) = Longitudinal strain per unit stress

i.e, \(\alpha\) = Longitudinal strain / Longitudinal stress ……. (1)

Similarly, let β= Lateral strain per unit stress

i.e, β = Lateral strain / Lateral stress ………(2)

\(\alpha\) and β could be used to derive Poisson’s ratio,

I.e, Poisson’s Ratio (\(\sigma\)) = β/\(\alpha\) ……….(3)

We know that young’s modulus (Y) = Longitudinal stress / Longitudinal strain

From equation (1), it could be written as Y=1/\(\alpha\) ………..(4)

Read Also: Transverse Waves

To introduce young’s modulus, we’ll have to give longitudinal stress to the object from all directions. We’ll start by providing pressure (P) along the length from both sides. 

Longitudinal stress to the object from all directions

Because of the stress, the length of the cuboid will increase,

So, Final length = \(l + \triangle l\), where \(\triangle\)l is the change in length

We know that \(\frac{\triangle l}{l}\) = Total Strain

From equation (1), \(\frac{\triangle l}{l}\) = \(\alpha\) x stress

⇒ \(\frac{\triangle l}{l}\) =\(\alpha\) x stress (P) x l

Now Final length = l+P\(\alpha\)l …………….. (5)

Similarly Final Breadth = b-Pβb ……………..(6)

And Final Thickness = t-Pβt ……………..(7)

Check Important Notes for Viscosity

Now we’ll be providing pressure (Q) parallel to the breadth of the cuboid.

Pressure (Q) parallel to the breadth of the cuboid

Final length = l + P\(\alpha\)l – Qβl

Final Breadth = b – Pβb + Q\(\alpha\)b

Final Thickness = l – Pβt – Qβt

Read  More: Class 11 Surface Energy

Now we’ll be providing pressure (R) parallel to the thickness of the cuboid.

Pressure (R) parallel to the thickness of the cuboid

Final length = l+P\(\alpha\)l-Qβl-Rβl

Final breath = b-Pβb+Q\(\alpha\)b-Rβb

Final thickness = t-Pβt-Qβt+R\(\alpha\)t

Final Volume = lbt

(l+P\(\alpha\)l-Qβl-Rβl) (b-Pβb+Q\(\alpha\)b-Rβb)(t-Pβt-Qβt+R\(\alpha\)t )

l (1+P\(\alpha\)-Qβ-Rβ) b (1-Pβ+Q\(\alpha\)-Rβ) t (1-Pβ-Qβ+R\(\alpha\))

l [1+P\(\alpha\)- β(Q+R)] b [1+Q\(\alpha\)- β(P+R)] t [1+R\(\alpha\)- β(P+Q)]

lbt [1+\(\alpha\)P- (Q+R)+\(\alpha\)Q- (P+R)+\(\alpha\)R- (P+Q)]

lbt [1+ \(\alpha\)(P+Q+R)- β(Q+R+P+R+P+Q)]

lbt [1+ \(\alpha\)(P+Q+R)- β(2Q+2R+2P)]

lbt [\(\alpha\) (P+Q+R)-2β (P+Q+R)]

Also Read:

Change in Volume (\(\triangle\)V) = lbt [\(\alpha\) (P+Q+R)-2β (P+Q+R)]

Now we will introduce bulk’s modulus. So the pressure applied in all the direction is equal.

Thus, P=Q=R

Therefore V=lbt [3\(\alpha\)P-6βP]

= lbt (3P)(\(\alpha\)-2β)

We know that, Volumetric strain = \(\triangle\)V/V

= lbt (3P) (\(\alpha\)-2β) / lbt

= 3P (\(\alpha\)-2β)

Bulk’s modulus (B) = Hydraulic stress / Volume strain

= P/ 3P (\(\alpha\)-2β)

= 1/ 3 (\(\alpha\)-2β)

= 1 / 3 (1-2β/\(\alpha\))

= Y/ 3(1-2\(\sigma\)) [from equ. (3) and (4)]

Relation between bulk’s modulus and young’s modulus:

\(B = \frac{Y}{3(1-3\sigma)}\)

Also Read:

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Relation: Young’s Modulus (Y), Shear Modulus (G), and Poisson’s ratio (\(\sigma\))

We will consider a cube for this derivation, where length = breadth = height = l.

Read Further Class 11 Mechanical Properties of Fluids

Since ABCD is a square, so ∠DCB = 90° and ∠DCA = 45°

Now to introduce shear modulus, we’ll apply two opposite forces parallel to the cross-sectional area of the cube. Which will cause a displacement between the opposite faces of the cube.

Two opposite forces parallel to the cross-sectional area of the cube

Stress = F/A

Let l = displacement between the opposite faces

Angle formed after displacement = \(\theta\)

We know that when shearing stress is applied then the shearing strain is Tan \(\theta\) = l / L

Since is very small. Therefore Tan \(\theta\) = \(\theta\) = l / L

Also Check:

The change is very small, so we’ll consider that ∠D’C’A = 45° (D’ and C’ is the sides after displacement)

Now we’ll draw a perpendicular line from C to AC’

From CRC’

Sin 45° = RC’/l

⇒ l=RC' / sin 45°

⇒ l=RC'2

Since AC ≈ AR

RC’ is the elongation along with diagonal AC

Let \(\alpha\) = Longitudinal strain per unit stress

i.e, \(\alpha\) = Longitudinal strain / Longitudinal stress ……. (1)

Similarly, let β= Lateral strain per unit stress

i.e, β = Lateral strain / Lateral stress ………(2)

Read Important Difference for Transverse and Longitudinal Waves

\(\alpha\) and β could be used to derive Poisson’s ratio,

I.e, Poisson’s Ratio \((\alpha) = \frac{\beta}{\alpha}\) ……….(3)

We know that young’s modulus (Y) = Longitudinal stress / Longitudinal strain

From equation (1), it could be written as Y=1/\(\alpha\) ………..(4)

RC' = \(\alpha\) x T x (AC) + \(\beta\) x T x (AC)

⇒ RC'=T(AC)(\(\alpha\)+\(\beta\))

⇒ \(l/\sqrt2\) =T (\(l/\sqrt2\)) (\(\alpha\)+\(\beta\))

⇒ l/L= T x 2 x (\(\alpha\)+\(\beta\))

⇒ \(\theta\) = 2T (\(\alpha\)+\(\beta\))

⇒ T/\(\theta\) = 1/ 2(\(\alpha\)+\(\beta\))

⇒ G = 1/ 2\(\alpha\)(1+\(\beta\)/\(\alpha\))

⇒ G = Y/ 2(1+\(\sigma\))

Relation between Shear modulus and Young’s modulus is:

\(G = \frac{Y}{2(1+\sigma)}\)

Also Read:

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Relation Derivation of Elastic Constants

We’ll be considering the two formulas derived in the above two relations:

B = Y/ 3(1-2\(\sigma\))

⇒ B = 1/ 3(\(\alpha\)-2\(\beta\)) …….. (1)

The value of \(\alpha\) and \(\beta\) is the same as in the above two derivations.

\(\alpha\) = Longitudinal strain per unit stress

β = Lateral strain per unit stress

Y = 1/ \(\alpha\)………. (2)

G = Y/ 2(1+\(\sigma\))

⇒ G = Y/ 2(\(\alpha\)+\(\beta\)) ………..(3)

Read More: Elastic and Inelastic Collision

From equation (1)

B = 1/ 3(\(\alpha\)-2\(\beta\))

⇒ \(\alpha\)-2\(\beta\) = 1/3K ………… (4)

From equation (3)

G = Y/ 2(\(\alpha\)+\(\beta\))

⇒ 2\(\alpha\) + 2\(\beta\) = 1/G …………. (5)

Now adding equ. (4) and equ. (5)

(\(\alpha\)-2\(\beta\)) + (2\(\alpha\)+2\(\beta\)) = 1/3B + 1/G

⇒ 3\(\alpha\) = 1/3B + 1/G

⇒ 3/Y = 1/3B + 1/G

⇒ 3/Y = G + 3B/ 3BG

⇒ Y = 9BG/ G + 3B

Thus, the relation between elastic constants is Y = 9BG/ G + 3B

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