**CBSE NOTES CLASS 11 CHAPTER 9**

**MECHANICAL PROPERTIES OF SOLIDS**

**Deforming Force**

A force, which when applied, produces a change in configuration of the object, is called a deforming force.

**Elasticity**

The property of a body, by virtue of which it tends to regain its original size and shape when the applied deforming force is removed, is known as **elasticity **and the deformation caused is known as **elastic **deformation.

**Restoring force**

If the system is disturbed from its equilibrium by a deforming force, the system applies a force equal in magnitude but opposite in direction to the deforming force. This is called restoring force. The **restoring force** will tend to bring the system back toward equilibrium. The **restoring force** is a function only of position of the mass or particle (i.e. extent of deformation).

**Spring ball model of elastic behavior**

We can visualize the restoring mechanism by taking a model of spring-ball system as shown in the figure. The balls represent atoms and springs represent interatomic forces. If any ball is displaced from its equilibrium position, the spring system tries to restore the ball back to its original position. This is the microscopic model of elasticity.

**Plasticity**

The bodies which have no gross tendency to regain their previous shape, and they get permanently deformed on application of deforming force, are called **plastic **and the property is called **plasticity. **

**Elastic Limit**

Elastic limit is the upper limit of deforming force upto which, if deforming force is removed, the body regains its original form completely and beyond which if the deforming force is increased the body loses its property of elasticity and gets permanently deformed.

**Perfectly Elastic Bodies**

Those bodies which regain their original configuration immediately and completely after the removal of deforming force are called perfectly elastic bodies. e.g., quartz and phosphor bronze etc.

**Perfectly Plastic Bodies**

Those bodies which do not regain their original configuration at all on the removal of deforming force are called perfectly plastic bodies, e.g., putty, paraffin wax etc.

**Stress**

The internal restoring force acting per unit area of a deformed body is called stress.

$$\mathrm{}\mathrm{Stress\; =}\mathrm{}\frac{\mathrm{}\mathrm{Restoring\; force}\mathrm{}}{\mathrm{}\mathrm{Area}\mathrm{}}$$

SI unit is N/m^{2} or Pascal.

**Types of Stress**

**Normal Stress**

If deforming force is applied normal to the area, then the stress is called normal stress.

If there is an increase in length, then stress is called **tensile stress**.

If there is a decrease in length, then stress is called **compressive stress**.

**Tangential Stress **

The restoring force per unit area developed due to the applied tangential force is known as **tangential **or **shearing stress**.

**Hydraulic Stress**

When an object is immersed in a fluid, there is a normal force acting on the surface of the object. The internal restoring force per unit area in this case is known as **hydraulic stress **and its magnitude is equal to the hydraulic pressure (applied force per unit area).

**Strain**

The fractional change in configuration is called strain.

$$\mathrm{Strain\; =}\frac{\mathrm{}\mathrm{Change\; in\; the\; configuretion}\mathrm{}}{\mathrm{}\mathrm{Original\; configuration}}$$

It has no unit and it is a dimensionless quantity.

According to the change in configuration, the strain is of three types

**Longitudinal strain**The change in the length Δ

*L*to the original length*L*of the body (cylinder in this case) is known as**longitudinal strain**.$$\mathrm{Longitudinal\; strain\; =}\frac{\mathrm{Change\; in\; length}}{\mathrm{Original\; length}}=\mathrm{}\left(\frac{\mathrm{\Delta}\mathrm{L}}{\mathrm{L}}\right)$$

**Shearing strain**It is defined as the ratio of relative displacement of the two faces, Δ

*x,*to the length of the cylinder*L*.$$\mathrm{Shearing\; strain\; =}\frac{\mathrm{\Delta}\mathrm{x}}{\mathrm{L}}=\mathrm{tan}\mathrm{\theta},\mathrm{}$$

where θ is the angular displacement of the cylinder from the vertical (original position of the cylinder).

For small θ,

Shearing strain = tan θ ≈ θ

**Volumetric strain**The strain produced by a hydraulic stress is called

**volume strain**and is defined as the ratio of change in volume (Δ*V*) to the original volume (*V*).$$\mathrm{Volume\; strain\; =}\frac{\mathrm{Change\; in\; volume}}{\mathrm{Originl\; volume}}=\frac{\mathrm{\Delta}\mathrm{V}}{\mathrm{V}}$$

**Hooke’s Law**

Within the elastic limits, the stress is proportional to the strain.

Stress ∝ Strain

or Stress = E × Strain

where, E is the **modulus of elasticity **of the material of the body.

Hooke’s law is an empirical law and is found to be valid for most materials.

**TYPES OF MODULUS OF ELASTICITY**

**Young’s Modulus**

The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as **Young’s modulus **and is denoted by the symbol *Y*.

$$\mathrm{Y}\mathrm{}=\frac{\mathrm{Normal\; stress}}{\mathrm{Longitudinal\; strain}}$$

$$\mathrm{Y}\mathrm{}=\frac{\mathrm{\sigma}}{\mathrm{\epsilon}}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{\Delta}\mathrm{L}/\mathrm{L}}=\mathrm{}\frac{\mathrm{F}\mathrm{L}}{\mathrm{A}\mathrm{\Delta}\mathrm{L}}$$

Its unit is N/m^{2} or Pascal

**Shear Modulus**

The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by G. It is also called the modulus of rigidity.

$$\mathrm{G}\mathrm{}=\frac{\mathrm{Shearing\; stress}\left({\mathrm{\sigma}}_{\mathrm{s}}\right)}{\mathrm{Shearing\; strain}}$$

$$\mathrm{G}\mathrm{}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{\Delta}\mathrm{x}/\mathrm{L}}=\frac{\mathrm{F}\mathrm{}\times \mathrm{}\mathrm{L}}{\mathrm{A}\mathrm{}\times \mathrm{}\mathrm{\Delta}\mathrm{x}}$$

$$\mathrm{Or\; G}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{\theta}}=\frac{\mathrm{F}}{\mathrm{A}\mathrm{}\times \mathrm{}\mathrm{\theta}}$$

The shearing stress σ_{s} can also be expressed as σ_{s} = *G *× θ

SI unit of shear modulus is N m^{–2} or Pa.

For most materials *G *≈ *Y*/3.

**Bulk Modulus of Elasticity**

The ratio of hydraulic stress to the corresponding volume strain is called *bulk modulus*. It is denoted by symbol *B*.

$$\mathrm{B}\mathrm{}=\mathrm{}\u2013\frac{\mathrm{\Delta}\mathrm{p}}{\mathrm{\Delta}\mathrm{V}/\mathrm{V}}$$

The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs. That is, if *Δp *is positive, *ΔV *is negative. Thus for a system in equilibrium, the value of bulk modulus *B *is always positive.

SI unit of bulk modulus is the same as that of pressure *i.e.*, N m^{–2} or Pa.

**Compressibility**

Compressibility of a material is the reciprocal of its bulk modulus of elasticity.

Compressibility is given by,

$$\mathrm{}\mathrm{k}\mathrm{}=\mathrm{}\left(\frac{1}{\mathrm{B}}\right)=\mathrm{}\u2013\mathrm{}\left(\frac{1}{\mathrm{\Delta}\mathrm{p}}\right)\times \mathrm{}\left(\frac{\mathrm{\Delta}\mathrm{V}}{\mathrm{V}}\right)$$

Its SI unit is N^{-1}m^{2} and CGS unit is dyne^{-1} cm^{2}.

- Steel is more elastic than rubber. Solids are more elastic and gases are least elastic.
- For liquids. modulus of rigidity is zero.
- Young’s modulus (Y) and modulus of rigidity (η) are possessed by solid materials only.

**Stress-Strain Curve**

- A graph is plotted between the stress (which is equal in magnitude to the applied force per unit area) and the strain produced. A typical graph for a metal looks like given in the figure.

- In the region between O to A, the curve is linear. In this region, Hooke’s law is obeyed.
- In the region from A to B, stress and strain are not proportional, but the body returns to its original dimension when the load is removed. The point B in the curve is known as
**yield point**(also known as**elastic limit**) and the corresponding stress is known as**yield strength**(*S*_{y}) of the material. - If the load is increased further (curve between B and D), the stress developed exceeds the yield strength and strain increases rapidly even for a small change in the stress. When the load is removed, say at some point C between B and D, the body does not regain its original dimension. In this case, even when the stress is zero, the strain is not zero. The material is said to have a
**permanent set**. The deformation is said to be**plastic deformation**. The point D on the graph is the ultimate**tensile strength**(*S*_{u}) of the material. - Beyond this point, additional strain is produced even by a reduced applied force and fracture occurs at point E. If the ultimate strength and fracture points D and E are close, the material is said to be brittle. If they are far apart, the material is said to be ductile.
- The stress-strain curves vary from material to material.
- These curves help us to understand how a given material deforms with increasing loads.

**Limit of Elasticity**

The maximum value of deforming force for which elasticity is present in the body is called its limit of elasticity.

**Breaking Stress**

The minimum value of stress required to break a wire, is called breaking stress.

Breaking stress is fixed for a material but breaking force varies with area of cross-section of the wire.

$$\mathrm{Safety\; factor\; =}\frac{\mathrm{Breaking\; stress}}{\mathrm{Working\; stress}}$$

**Elastic Fatigue**

The property of an elastic body by virtue of which its behavior becomes less elastic under the action of repeated alternating deforming force is called elastic fatigue.

**Ductile Materials**

The materials which show large plastic range beyond elastic limit are called ductile materials, e.g., copper, silver, iron, aluminum, etc.

Ductile materials are used for making springs and sheets.

**Brittle Materials**

The materials which show very small plastic range beyond elastic limit are called brittle materials, e.g., glass, cast iron, etc.

**Elastomers**

The materials for which strain produced is much larger than the stress applied, with in the limit of elasticity are called elastomers, e.g., rubber, the elastic tissue of aorta, the large vessel carrying blood from heart.

Elastomers have no plastic range.

**Elastic Potential Energy in a Stretched Wire**

The work done in stretching a wire is stored in form of potential energy of the wire.

Potential energy U = Average force × Increase in length

= $\frac{1}{2}$ FΔL

= $\frac{1}{2}$ Stress × Strain × Volume of the wire

Elastic potential energy per unit volume

u = $\frac{1}{2}$ × Stress × Strain

= $\frac{1}{2}$(Young’s modulus) × (Strain)^{2}

Elastic potential energy of a stretched spring = $\frac{1}{2}$ *k x*^{2},

where, k = Force constant of spring and *x* = Change in length.

**Poisson’s Ratio**

When a deforming force is applied at the free end of a suspended wire of length 1 and radius R, then its length increases by dL but its radius decreases by dR. Now two types of strains are produced by a single force.

- Longitudinal strain = $\frac{\mathrm{\Delta}\mathrm{L}}{\mathrm{L}}$
- Lateral strain = – $\frac{\mathrm{\Delta}\mathrm{R}}{\mathrm{R}}$

$$\mathrm{Poisson\text{'}s\; ratio\; (\sigma )\; =}\frac{\mathrm{Lateral\; strain}}{\mathrm{Longotudinal\; strain}}\mathrm{}=\mathrm{}\u2013\mathrm{}\frac{\mathrm{\Delta}\mathrm{R}/\mathrm{R}}{\mathrm{\Delta}\mathrm{L}/\mathrm{L}}$$

The theoretical value of Poisson’s ratio lies between – 1 and 0.5. Its practical value lies between 0 and 0.5.

**Relation between Y, B, G and σ**

$$\left(\mathrm{i}\right)\mathrm{}\mathrm{Y}\mathrm{}=\mathrm{}3\mathrm{B}\left(1\mathrm{}\u2013\mathrm{}2\mathrm{\sigma}\right)$$

$$\left(\mathrm{i}\mathrm{i}\right)\mathrm{}\mathrm{Y}\mathrm{}=\mathrm{}2\mathrm{}\mathrm{G}\mathrm{}\left(1\mathrm{}+\mathrm{}\mathrm{\sigma}\right)$$

$$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\mathrm{}\mathrm{\sigma}\mathrm{}=\frac{3\mathrm{B}\mathrm{}\u2013\mathrm{}2\mathrm{G}}{2\mathrm{G}\mathrm{}+\mathrm{}6\mathrm{B}}$$

$$\left(\mathrm{i}\mathrm{v}\right)\mathrm{}\mathrm{}\frac{9}{\mathrm{Y}}=\frac{1}{\mathrm{B}}+\frac{3}{\mathrm{G}}\mathrm{}\Rightarrow \mathrm{}\mathrm{}\mathrm{Y}\mathrm{}=\mathrm{}\frac{9\mathrm{}\mathrm{B}\mathrm{}\mathrm{G}}{\mathrm{G}\mathrm{}+\mathrm{}3\mathrm{B}}$$

**Thermal Stress**

- When temperature of a rod fixed rigidly at its both ends is changed, then the produced stress is called thermal stress.
Thermal stress = F/A = Y α Δθ

where,

α = coefficient of linear expansion of the material of the rod

Δθ = change in temperature

Y = Youngs modulus

- When temperature of a gas enclosed in a vessel is changed, then the thermal stress produced is equal to change in pressure (Δp) of the gas.
Thermal stress = Δ p = B γ Δ θ

where,

B = bulk modulus of elasticity and

γ = coefficient of cubical expansion of the gas.

**Interatomic force constant**K = Y r

_{o},where, r

_{o}= interatomic distance.

**Applications of Elastic Behaviour of Materials**

- Beams used in construction of bridges, as supports have a cross-section of the type I. This cross-sectional shape provides a large load bearing surface and enough depth to prevent bending. This shape reduces the weight of the beam without sacrificing the strength and hence reduces the cost.
- A heap of sand or a hill have a pyramidal shape
**How thick should the steel rope of the crane be?**The load should not deform the rope permanently. Therefore, the extension should not exceed the elastic limit.

Mild steel has yield strength (

*S*_{y}) of about 300 × 10^{6}N m^{–2}. Thus, the area of cross-section (*A*) of the rope should at least be$$\mathrm{Elastic\; limit,}{\mathrm{S}}_{\mathrm{y}}\mathrm{}=\frac{\mathrm{Load\; to\; be\; lifted}}{\mathrm{Area\; of\; cross-section}}$$

$$=\frac{\mathrm{M}\mathrm{g}}{\mathrm{\pi}{\mathrm{r}}^{2}}$$

A should be ≥ $\frac{\mathrm{W}}{{\mathrm{S}}_{\mathrm{y}}}$ = $\frac{\mathrm{M}\mathrm{g}}{{\mathrm{S}}_{\mathrm{y}}}$

$$=\frac{{10}^{4}\mathrm{}\mathrm{k}\mathrm{g}\mathrm{}\times \mathrm{}10\mathrm{}\mathrm{m}\mathrm{}{\mathrm{s}}^{2}}{300\mathrm{}\times \mathrm{}{10}^{6}\mathrm{}\mathrm{N}\mathrm{}{\mathrm{m}}^{-2}}$$

= 3.3 × 10

^{-4}m^{2 }This corresponds to a radius of about 1 cm for a rope of circular cross-section.

Generally a large margin of safety (of about a factor of ten in the load) is provided.

Thus a thicker rope of radius about 3 cm is recommended.

A single wire of this radius would practically be a rigid rod.

So the ropes are always made of a number of thin wires braided together, like in pigtails, for ease in manufacture, flexibility and strength.

- A bridge has to be designed such that it can withstand the load of the flowing traffic, the force of winds and its own weight. Similarly, in the design of buildings use of beams and columns is very common. In both the cases, the overcoming of the problem of bending of beam under a load is of prime importance.
The beam should not bend too much or break.

A beam loaded at the centre and supported near its ends.

A bar of length

*l*, breadth*b*, and depth*d*when loaded at the centre by a load*W*will have depression$$\mathrm{\delta}\mathrm{}=\mathrm{}\frac{\mathrm{W}{\mathrm{L}}^{3}}{4\mathrm{}\mathrm{Y}\mathrm{}\mathrm{b}{\mathrm{d}}^{3}}$$

For δ to be small, L should be small, Y should be large, b and d should be large

**Cantilever**A beam clamped at one end and loaded at free end is called a cantilever.

Depression at the free end of a cantilever is given by

$$\mathrm{\delta}\mathrm{}=\frac{\mathrm{w}{\mathrm{L}}^{3}}{3\mathrm{Y}{\mathrm{I}}_{\mathrm{G}}}$$

where, w = load,

L = length of the cantilever,

Y = Young’s modulus of elasticity, and

I

_{G}= geometrical moment of inertia.For a beam of rectangular cross-section having breadth b and thickness d,

$${\mathrm{I}}_{\mathrm{G}}\mathrm{}=\frac{\mathrm{b}{\mathrm{d}}^{3}}{12}$$

For a beam of circular cross-section area having radius r,

$${\mathrm{I}}_{\mathrm{G}}\mathrm{}=\frac{\mathrm{\pi}\mathrm{}{\mathrm{r}}^{4}}{4}$$

**Why a mountain cannot be higher than 10 km?**At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is

*hρg*where*ρ*is the density of the material of the mountain and gThe material at the bottom experiences this force in the vertical direction, and the sides of the mountain are free.

Now the elastic limit for a typical rock is 30 × 10

^{7}N m^{-2}.Equating this to

*h ρ g*, with ρ = 3 × 10^{3}kg m^{-3}gives*h ρ g*= 30 × 10^{7}N m^{-2}Or

*h =*$\frac{30\mathrm{}\times \mathrm{}{10}^{7}\mathrm{}\mathrm{N}\mathrm{}{\mathrm{m}}^{-2}}{3\mathrm{}\times \mathrm{}{10}^{3}\mathrm{}\mathrm{k}\mathrm{g}\mathrm{}{\mathrm{m}}^{-3}\mathrm{}\times \mathrm{}10\mathrm{}\mathrm{m}\mathrm{}{\mathrm{s}}^{-2}}$= 10 km

**Hollow shaft vs Solid Shaft**

Torque τ required, to produce, a unit twist in a solid shaft of radius r, length L and of a material of modulus of rigidity G is given by,

$$\mathrm{\tau}=\mathrm{}\frac{\mathrm{\pi}\mathrm{}\mathrm{G}\mathrm{}{\mathrm{r}}^{4}}{2\mathrm{L}}$$

For a hollow cylindrical shaft

$$\mathrm{\tau \u2019}\mathrm{}=\frac{\mathrm{\pi}\mathrm{}\mathrm{G}\mathrm{}\left({{\mathrm{r}}_{2}}^{4}-\mathrm{}{{\mathrm{r}}_{1}}^{4}\right)}{2\mathrm{L}}$$

If the two shafts are made of same and equal amount of material, then there volumes will be same,

πr^{2}L = π(r_{2}^{2 }- r_{1}^{2})L

Or, r^{2} = r_{2}^{2} –r_{1}^{2 }

$$\frac{\mathrm{\tau \u2019}}{\mathrm{\tau}}\mathrm{}=\frac{\mathrm{\pi}\mathrm{}\mathrm{G}\mathrm{}\left({{\mathrm{r}}_{2}}^{4}-\mathrm{}{{\mathrm{r}}_{1}}^{4}\right)}{2\mathrm{L}}/\frac{\mathrm{\pi}\mathrm{}\mathrm{G}\mathrm{}{\mathrm{r}}^{4}}{2\mathrm{L}}$$

$$=\frac{{{\mathrm{r}}_{2}}^{2}+\mathrm{}{{\mathrm{r}}_{1}}^{2}}{{\mathrm{r}}^{2}}\mathrm{}1$$