Moduli of Elasticity, Physics tutorial

Young's Modulus:

When the force (F) is applied to end of the wire of cross-sectional area A and having the initial length L, we state tensile stress on wire as force per unit area.

Tensile stress = Force/Area

= F/a

Where F is estimated in Newton's (N) and Area is in meter-square (m2). Thus unit of tensile stress is Nm-2.

Cross-sectional area A of wire is stated as,

A = (πD2/4)πr2

Where, r is a radius of the cross-section and D is a diameter of cross-section.

If on the application of the force F, the wire is extended by (e) then we would state the tensile strain as

Tensile strain = Extension/initial length

= e/l

Force in the Bar Because of Contraction or Expansion:

When the iron bar is heated and is prohibited from contracting as it cools, experiments have illustrated that significant force is applied at ends of the bar. It is possible to get force exerted with knowledge of Young Modules E, cross-sectional area A, the coefficient of linear expansion α (linear expansivity) and change in temperature because of contraction or expansion, ΔΘ.

Force F is given by

E = F/A ÷ e/L

Therefore E =F/A X L/e

Therefore F = EAe/L.................Eq.1

Also coefficient of linear expansion is provided as

α therefore ΔL/LΔΘ = e/LΔΘ

Therefore e = αLΔΘ.................Eq.2

Substituting Eq.2 in Eq.1, we get

Therefore F = EA/L X αLΔΘ

F = EAαΔΘ.................Eq.3

Energy Stored in a Wire:

Work done is stated as product of force and distance moved by the force that is:

Work done = Force x distance

For a given wire, work in stretching it is provided by

Work done = Average force x extension

= (F/2)x e

The force increases from zero to the given value, if elastic limit is not exceeded. Thus amount of work done in stretching the wire, or amount of energy stored in wire is provided by

W = Fe/2.................Eq.4

Energy stored is gain in molecular Potential Energy of molecules because of displacement from their mean positions. Energy store could be stated in terms of Young Modulus.

Now, substitute value of F from Eq.1 in Eq.4, we get

W = 1/2 x (EAe/L)x e

W = 1/2 x (EAe2/L) .................Eq.5

Bulk Modulus of Elasticity:

While Young Modulus of elasticity is related with solid objects, bulk modulus manages liquids and gases. When the liquid or gas is subjected to the increase pressure the substance contracts. The change in bulk thus happens. Bulk strain is stated as change in volume (ΔV) of the gas (or liquid) to original volume V.

Bulk strain = Change in volume of gas or liquids/Original volume of gas or liquid

As change in volume is -ΔV

And original volume is V,

Therefore the, Bulk Strain = -ΔV/V .................Eq.6

Negative sign indicates that volume decreases. Bulk stress is stated as increase in force per unit area.

Therefore Bulk stress = Increase in force/Area = ΔF/A.................Eq.7

Then Bulk Modulus of elasticity K is stated as

K = Bulk Stress/Bulk strain

= ΔP ÷ -ΔV/Δ

K = -ΔVP/ΔV.................Eq.8

Limit of ΔP/ΔV as ΔP and ΔV tend to zero is dp/dv

Limit ΔP/ΔV = dp/dV

ΔP, ΔV→0

Therefore k = -Vdp/dv.................Eq.9

Reciprocal of bulk modulus of elasticity K is dp/dv that is called as compressibility of material.


1/K = (1/V)(dV/dP) .................Eq.10

Bulk Modulus of a Gas - Isothermal Bulk Modulus:

When pressure and volume of gas changes at constant temperature, the expression is as follows:

PV = Constant

It is also called as Boyle's Law. Additionally, it is also related with an isothermal condition. On differentiating the expression with respect to volume V. We get

PdV/dV + VdP/dV = 0.................Eq.11

Therefore P + Vdp/dV = 0

Therefore P = -Vdp/dV = K.................Eq.12

Bulk modulus of elasticity is provided as

K = -Vdp/dV

Therefore P = -VdP/dV = k.................Eq.13

Therefore isothermal bulk modulus of elasticity is equivalent to pressure P.

Bulk Modulus of the Gas - Adiabatic Modulus:

By adiabatic, we signify the system whereby heat is not permitted to escape in it or get out of it. Pressure, and volume of a gas can be associated in such a way that

PV3/4 = Constant.................Eq.14

Where γ = Cp/Cv

Cp/Cv is the ratio of molar heat capacities of the gas at constant pressure CP and constant volume CV. The state is explained as adiabatic when no heat is permitted to leave system or enter system of a gas. External work done is entirely at expense of the internal energy of gas. Result is that gas cools down starting from PVγ = Constant

On differentiating the Eq.14 with respect to V, we get

Therefore P d(Vγ)/dV + Vγ d(P)/dV = 0.................Eq.15

Therefore P x γVγ-1 + Vγ d(P)/dV = 0

Therefore P γ = -Vγ /Vγ-1dP/dV = 0

On arranging the terms, we get

= -VγV/VγdP/dV

Note that Vγ-1 = VγV-1 = Vγ/V

Therefore Pγ = -VdP/dV

But K = -VdP/dV

Therefore Pγ = K.................Eq.16

Hence the adiabatic Bulk Modulus is

K = Pγ

The values of K at isothermal and adiabatic conditions are given below:

For air at normal pressure,

K = 1.0 x 105Nm-2 for isothermal bulk modulus, while

K = 1.4 x 105Nm-2 for adiabatic bulk modulus.

These values are of order of 105 times smaller than liquids as gases are much more compressible.

Shear Modulus of Elasticity or Modulus of Rigidity:

Shear modulus, numerical constant which explains elastic properties of the solid under application of transverse internal forces like arise, for instance, in torsion, as in twisting a metal pipe about lengthwise axis. Within such a material any tiny cubic volume is somewhat distorted in such a way that two of the faces slide parallel to each other a small distance and two other faces change from squares to diamond shapes. Shear modulus is the measure of ability of the material to resist transverse deformations and is valid index of elastic behavior only for small deformations, after which material is able to return to original configuration. Large shearing forces lead to flow and permanent deformation or fracture. Shear modulus is also called as rigidity.

Mathematically shear modulus is equivalent to quotient of shear stress divided by shear strain. Shear stress, in turn, is equivalent to shearing force F divided by area A parallel to and in which it is acted, or F/A. Shear strain or relative deformation is the measure of change in geometry and in this situation is stated by trigonometric function, tangent (tan) of angle Θ (theta), that denotes amount of change in the 90°, or right, angles of minute representative cubic volume of unstrained material. Mathematically, shear strain is stated as tan Θ or equivalent, by definition, x/y. Shear modulus itself may be stated mathematically as

Shear modulus = (shear stress)/(shear strain) = (F/A)/(x/y).

Shear modulus may be stated in units of pounds per square inch (generally abbreviated to psi); common SI units are Newton per square metre (N/m2). Value of shear modulus for aluminum is approx 3.5 × 106 psi, or 2.4 × 1010 N/m2. By comparison, steel under shear stress is more than three times as rigid as aluminum.

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