NotesShala Class 11 · Physics

Chapter08

Mechanical Properties of Solids

Stress, Strain and Hooke's Law · Moduli of Elasticity · Elastic Effects and Applications

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Stress, Strain and Hooke's Law

A solid has a definite shape and size. A force is needed to change (deform) its shape or size. This chapter studies how solids react to such forces. It uses the ideas of stress, strain, Hooke's law and the moduli of elasticity.

1.1Elastic Behaviour of Solids

Definition

Elasticity

Elasticity is the property of a body by which it gets back its original shape and size after the deforming force is removed. A deformation that disappears on removing the force is called an elastic deformation.

Definition

Deforming Force

A deforming force is a force that changes the shape of a body. It does not change the body's state of rest or of uniform motion.

Even a solid body changes its shape when a suitable force acts on it. You can see this when you stretch a rubber band or hang a load on a spring. A body can be deformed in four ways:

Stretchedgets longer Compressedgets shorter Bentbecomes curved Twistedturns about its axis

Interatomic (or intermolecular) forces hold the atoms of a solid together. These forces keep every atom in a stable equilibrium position. A solid reacts to a deforming force in two stages:

iWhen the Force Acts

The atoms are pushed or pulled away from their equilibrium positions. So the distances between the atoms change.

iiWhen the Force Is Removed

The interatomic forces bring the atoms back to their equilibrium positions. This is the cause of elastic behaviour.

Force applied: the gaps between atoms changeAtoms pushed out of place

Force removed: restoring forces pull the atomsAtoms go back

1.2Spring Ball Model of Solids

In the spring ball model, the atoms of a solid are treated as small balls (point masses). Springs join these balls in all three dimensions. The springs stand for the interatomic forces of attraction.

Normally, each ball rests where its potential energy is minimum. Here the interatomic force on it is zero. If a ball is moved away from this mean position, the springs attached to it pull it back. This pull is the restoring force.

Spring ball model of a solid

A ball is moved away from its mean positionSprings pull it back

Note

Microscopic View

This model explains elastic behaviour through the microscopic structure of a solid. Whenever atoms leave their equilibrium positions, the "springs" between them produce restoring forces.

1.3Stress

A deforming force shifts the atoms of a body from their normal positions. This changes its configuration (shape or size). An internal restoring force then tries to bring the body back.

Definition

Stress is the restoring force set up inside a deformed body, per unit area.

Stress

\text{Stress} = \dfrac{F}{A}

where, $F$ = restoring force (N) and $A$ = area of cross-section (m²). If the body is not permanently deformed, the restoring force is equal and opposite to the applied force. So $F$ can be taken as the external deforming force.

Units and Dimensions of Stress

SI Unit

$\text{N/m}^2 = \text{Pa}$ (newton per square metre, or pascal)

CGS Unit

$\text{dyne/cm}^2$ (dyne per square centimetre)

Dimensions

$[\text{ML}^{-1}\text{T}^{-2}]$ , the same as for pressure

Stress cannot be given one fixed directionNot a vector quantity

Breaking stress of a given materialConstant

1.4Types of Stress

Stress is of three types. The type depends on how the forces act on the body.

1. Normal (Longitudinal) Stress

Normal stress is the restoring force per unit area that acts perpendicular (normal) to the surface of a body. Normal stress is either tensile or compressive.

iTensile Stress

Equal and opposite forces pull the two ends of a rod, normal to its cross-section. The rod gets longer. A restoring force equal to $F$ develops inside. This force per unit area is the tensile stress.

\text{Tensile stress} = \dfrac{F}{A}

Pulled by

F

at both ends: length

L \to L + \Delta L

iiCompressive Stress

Equal and opposite forces push the two ends of a rod. The rod gets shorter. Again, a restoring force equal to $F$ develops inside. This force per unit area is the compressive stress.

\text{Compressive stress} = \dfrac{F}{A}

Pushed by

F

at both ends: length

L \to L - \Delta L

Note

Tensile and compressive stress together are called longitudinal stress. In both cases the net force on the body is zero. Even so, the body gets deformed.

2. Tangential (Shearing) Stress

A deforming force may act along a surface (tangentially). Then the shape changes, but the volume stays the same. This force per unit area is the tangential or shearing stress.

In the figure, $F$ acts along the top face of a cube. The top face slides by $\Delta x$ . The sides tilt through the angle of shear $\theta$ .

\text{Tangential stress} = \dfrac{F}{A}

Force acts along the top faceShape changes, volume same

Before and after a tangential force

F

3. Hydraulic (Bulk) Stress

Sometimes a body is pressed by a uniform force from all sides. The stress produced is called hydraulic stress or bulk stress. The volume of the body changes, but its shape does not.

The force $F$ acts normal to every point of the surface. It changes the volume of the body by $\Delta V$ .

\text{Bulk stress} = \dfrac{F}{A}

Hydraulic stress is also called volume stress. In general, stress causes a distortion: a change in the size, the shape, or both (the configuration) of the body.

Before: original volume

V

After: forces from all sides, volume falls by

\Delta V

The same force acts normally at every point of the surfaceVolume changes, shape same

1.5Strain

A deforming force changes the shape or size of a body. The body is then said to be strained or deformed.

Definition

Strain is the ratio of the change in configuration of a body to its original configuration.

Strain

\text{Strain} = \dfrac{\text{Change in configuration}}{\text{Original configuration}}

Strain is a ratio of two similar quantities. So it has no unit and no dimensions.

Three Types of Strain (Length, Volume, Shape)

1Longitudinal Strain

It is the change in length ÷ original length.

\text{Longitudinal strain} = \dfrac{\Delta L}{L}

2Volumetric Strain

It is the change in volume ÷ original volume.

\text{Volumetric strain} = \dfrac{\Delta V}{V}

3Shear Strain

A strain that changes the shape of a body is called shear strain.

It equals the relative displacement of a layer divided by its distance from the fixed layer. It is also measured by the angle of shear $\theta$ . It is the angle by which a line, first perpendicular to the fixed face, turns.

\text{Shear strain} = \tan\theta = \dfrac{\Delta L}{L}

Layers shift by

\Delta L

; edges turn through

\theta

L = 1\ \text{m}

, then shear strain

= \Delta L

Relative shift of layers 1 m apart

Note

When a load hangs from a spring, the spring gets both longitudinal and shear strain.
Strain can remain for some time even after the stress is removed. So strain lags behind stress.

1.6Hooke's Law

Robert Hooke (1635–1703) was an English physicist. Based on his experiments, he gave a law in 1679. It is now called Hooke's law.

Law

Original Statement

Within the elastic limit, the extension of a wire is directly proportional to the load hung on it.

Original Form

\text{Extension} \;\propto\; \text{Load applied}

Modified (General) Form

Later, the law was found to hold for every kind of deformation:

Stretching Compression Bending Twisting

Law

Hooke's Law (Modified Statement)

Within the elastic limit, stress is directly proportional to strain.

\text{Stress} \propto \text{Strain} \quad\Rightarrow\quad \text{Stress} = E \times \text{Strain}

Modulus of Elasticity

E = \dfrac{\text{Stress}}{\text{Strain}}

where, $E$ = a constant called the modulus of elasticity of the material

Note

The constant $E$ depends only on the nature of the material. It has three forms: Young's modulus, bulk modulus and modulus of rigidity. Part 2 explains each of them.

1.7Stress-Strain Curve

Hang a load on a wire and increase the load slowly. For small loads, the extension is proportional to the load. So stress ∝ strain.

The graph of stress against strain is called the stress-strain curve. With a growing load, the wire first behaves as an elastic body and then as a plastic body. In the end, it breaks.

A typical stress-strain curve for a metal wire

Definition

Elastic Limit

The elastic limit is the maximum deforming force (or stress) up to which a body regains its original form completely once the force is removed. Beyond this limit, the body loses its elasticity and becomes permanently deformed.

The elastic limit is a property of a body, but elasticity is a property of its material.

Regions of the Stress-Strain Curve

1O to A: Proportional Region

The graph is a straight line. Stress is proportional to strain, so Hooke's law is fully obeyed. Point A is the proportional limit.

2O to B: Elastic Region

Remove the load anywhere before B, and the wire returns to its original length. B is the elastic limit, also called the yield point.

3B to C: Permanent Set

Let the stress go beyond B. Then remove the load, say at C. The wire does not regain its original length. A permanent strain OO′ remains. It is called the permanent set.

4C to E: Plastic Region and Fracture

Now a small increase in stress gives a large strain. This is plastic behaviour. The stress is highest at D, the ultimate tensile strength. Finally the wire breaks at E, the fracture point.

Important

Breaking Stress (Ultimate Tensile Strength)

The breaking stress of a material is the largest stress it can bear before it breaks. It is also called the ultimate tensile strength (point D on the curve).

Breaking Stress of Some Materials (N/m²)

Aluminium

2.2 \times 10^{8}

Iron

3.0 \times 10^{8}

Brass

4.7 \times 10^{8}

Phosphor bronze

5.5 \times 10^{8}

Steel

(5 \text{ to } 20) \times 10^{8}

Glass

10 \times 10^{8}

Note

Hooke's law holds only in the straight-line part of the curve. It does not hold for large strains.
The elastic limit lies very close to the proportional limit. So Hooke's law is nearly true right up to the elastic limit.
The strain in the yield region is 15 to 20 times the strain in the proportional region.

1.8Ductile and Brittle Materials

Materials can be grouped by their elastic and plastic behaviour. This topic and the next two describe these groups.

iDuctile Materials

Ductile materials have a large plastic range of extension. Their fracture point is far from the elastic limit. So they get much longer, permanently, before they break. Thin wires can be drawn from them.

e.g. Iron, copper, aluminium, silver

iiBrittle Materials

Brittle materials have a very small plastic range. They break as soon as the stress goes a little beyond the elastic limit. Their breaking point lies very close to the elastic limit.

e.g. Glass, cast iron, ceramics

Ductile: long plastic range

Brittle: breaks just after the elastic limit

1.9Elastomers

Definition

Elastomers are materials that can be stretched elastically to very large strains.

e.g. Rubber can become several times longer than it was. Yet it gets back its first length once the force is removed.

Elastomers have no clear plastic region. Their elastic region is very large, but they do not obey Hooke's law. The graph has no straight-line part.

Very large elastic strain (rubber, aorta)Does not obey Hooke's law

Stress-strain curve of an elastomer (elastic tissue of the aorta)

Note

The graph above is for the elastic tissue of the aorta, which is an elastomer. The aorta is the large blood vessel that carries blood from the heart.

1.10Malleability

A compressed solid reaches a stage beyond which it cannot regain its shape. This stage is its elastic limit for compression. Beyond it, the solid acts like a plastic body.

More compression brings a metal to its crushing point. Beyond this point, some metals can be flattened into thin sheets.

Definition

A metal that can be flattened into thin sheets by hammering or rolling is called malleable. This property is called malleability. e.g. Silver, lead.

Load-compression curve for a metal

Part 2

Moduli of Elasticity

Within the elastic limit, stress divided by strain is a constant for a material. This constant is its modulus of elasticity. This part covers its three types and Poisson's ratio.

2.1Modulus of Elasticity

Definition

The modulus of elasticity of a body is the ratio of stress to strain within the elastic limit. Another name for it is the coefficient of elasticity.

Modulus of Elasticity

E = \dfrac{\text{Stress}}{\text{Strain}}

Units and Properties

SI Unit

$\text{Nm}^{-2} = \text{Pa}$ (newton per square metre, or pascal)

Dimensions

$[\text{ML}^{-1}\text{T}^{-2}]$ , the same as for stress or pressure

Changes with the material of the bodyYes

Changes with the size of the body (length, area, volume, etc.)No

Three Types of Modulus of Elasticity

1Young's Modulus (

Y

)

For a change in length: longitudinal stress ÷ longitudinal strain.

Y = \dfrac{\text{Long. stress}}{\text{Long. strain}}

2Bulk Modulus (

B

K

)

For a change in volume: normal (volume) stress ÷ volumetric strain.

B = \dfrac{\text{Normal stress}}{\text{Vol. strain}}

3Modulus of Rigidity (

\eta

)

For a change in shape: tangential (shear) stress ÷ shear strain.

\eta = \dfrac{\text{Shear stress}}{\text{Shear strain}}

2.2Young's Modulus

Definition

The Young's modulus ( $Y$ ) of a material is the ratio of longitudinal stress to longitudinal strain, within the elastic limit.

Young's Modulus

Y = \dfrac{\text{Longitudinal stress}}{\text{Longitudinal strain}}

Setup — Given Information

A wire has radius $r$ and length $L$ .
A force $F$ acts along its length, that is, normal to its cross-section.
The length of the wire increases by $\Delta L$ .
Area of the circular cross-section, $A = \pi r^2$ .

Wire stretched by a force

F

Derivation

For the stretched wire:
$\text{Longitudinal stress} = \dfrac{F}{A}, \qquad \text{Longitudinal strain} = \dfrac{\Delta L}{L}$
Young's modulus is the ratio of these two:
$Y = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L} \qquad …(i)$
For a circular cross-section, $A = \pi r^2$ . Put this in (i):
$Y = \dfrac{FL}{\pi r^2\,\Delta L}$
If a load of mass $m$ stretches the wire, then $F = mg$ :
$Y = \dfrac{mgL}{\pi r^2\,\Delta L}$

Young's Modulus of a Wire

Y = \dfrac{FL}{\pi r^2\,\Delta L} = \dfrac{mgL}{\pi r^2\,\Delta L}

where, $F = mg$ = stretching force, $L$ = original length, $r$ = radius and $\Delta L$ = extension

Concept

Physical Meaning of Young's Modulus

Put $L = 1\ \text{m}$ , $A = 1\ \text{m}^2$ and $\Delta L = 1\ \text{m}$ . Then $Y = F$ . So Young's modulus equals the force needed to stretch a wire of unit length ( $1\ \text{m}$ ) and unit cross-section ( $1\ \text{m}^2$ ) by unit amount ( $1\ \text{m}$ ).

\Delta L \propto F

and

\Delta L \propto L

More force or a longer wire gives more extension

2.3Determination of Young's Modulus

The simple set-up below is used to find the Young's modulus of the material of a wire. Two long, straight wires of the same length and same radius hang side by side from a rigid support.

Reference Wire (A)

It carries the main scale M (a millimetre scale). A heavy fixed weight keeps it taut and free of kinks.

Experimental Wire (B)

It carries a vernier scale V, which slides along the main scale. Its hanger holds slotted half-kg weights.

Experimental set-up (Searle's apparatus)

Procedure

Measure the radius of the experimental wire with a screw gauge at several places. Take the average as the radius $r$ . Note the initial length $L$ .
Put a small initial load (about 1 kg) on the hanger. It keeps the wire straight and free of kinks. Note the vernier reading.
Add a half-kg weight. Let the wire stretch for about one minute. Note the new vernier reading.
The difference between the two readings gives the extension $\Delta L$ for the added weight. Repeat with more weights, one step at a time, and note each extension.

The cross-section of wire B is $A = \pi r^2$ . The stretching force is $F = mg$ .

A graph is drawn between the load and the extension. It is a straight line through the origin. This shows that the extension is directly proportional to the load.

Young's Modulus from the Experiment

Y = \dfrac{FL}{\pi r^2\,\Delta L} = \dfrac{mgL}{\pi r^2\,\Delta L}

A straight line through the origin

O

2.4Bulk Modulus

Definition

The bulk modulus ( $B$ ) of a material is the ratio of normal stress to volumetric strain, within the elastic limit. It also equals hydraulic stress ÷ hydraulic strain.

Take a body of volume $V$ and surface area $A$ . A force $F$ acts evenly over its whole surface, normal to it. The volume of the body decreases by $\Delta V$ .

B = \dfrac{\text{Normal stress}}{\text{Volumetric strain}} = \dfrac{F/A}{\Delta V/V}

Volume

V

becomes

V - \Delta V

Bulk Modulus

B = -\dfrac{pV}{\Delta V}

where, $p = \dfrac{F}{A}$ = normal pressure. The minus sign shows that the volume decreases when the stress increases.

Note

The negative sign shows that the volume falls as the stress rises. So, for a system in equilibrium, $B$ is always positive.
A perfectly rigid body and an ideal liquid both have an infinite bulk modulus.

Compressibility

Definition

The compressibility ( $k$ ) of a material is the reciprocal of its bulk modulus.

Compressibility

k = \dfrac{1}{B} = -\dfrac{\Delta V}{pV}

SI Unit

$\text{N}^{-1}\text{m}^{2}$

CGS Unit

$\text{dyne}^{-1}\text{cm}^{2}$

Dimensions

$[\text{M}^{-1}\text{LT}^{2}]$

2.5Modulus of Rigidity

Definition

The modulus of rigidity (or shear modulus) of a material is the ratio of tangential (shear) stress to shear strain, within the elastic limit. It is written as $G$ or $\eta$ .

Take a cube with its lower face fixed. A tangential force $F$ acts on its upper face of area $A$ .

The upper face shifts by $AA' = \Delta L$ . The vertical sides turn through an angle $\theta$ . This angle is the shear strain.

Lower face

BC

fixed; top shifts by

\Delta L

Modulus of Rigidity

\eta \;(\text{or } G) = \dfrac{\text{Tangential (shear) stress}}{\text{Shear strain}}

Derivation

The tangential (shear) stress on the upper face is:
$\text{Shear stress} = \dfrac{F}{A}$
In the figure, $AB = DC = L$ and $\angle ABA' = \theta$ . For a small angle, the shear strain is:
$\theta \approx \tan\theta = \dfrac{AA'}{AB} = \dfrac{\Delta L}{L}$
So the modulus of rigidity is:
$\eta = \dfrac{F/A}{\theta} = \dfrac{F}{A\theta} = \dfrac{F}{A}\cdot\dfrac{L}{\Delta L}$

Shear Modulus

\eta = \dfrac{FL}{A\,\Delta L}

2.6Comparison of the Elastic Moduli

iSolids

A solid has all three moduli of elasticity: Young's modulus $Y$ , bulk modulus $B$ and modulus of rigidity $\eta$ .

iiFluids (Liquids and Gases)

A fluid can change only its volume, not its shape. So a fluid has only the bulk modulus $B$ . For an ideal liquid, the modulus of rigidity is zero.

Summary: Stress, Strain and the Elastic Moduli

Type of stress	Nature of force	Strain	Change in shape	Change in volume	Elastic modulus	State of matter
Tensile / Compressive	Two equal and opposite forces ⊥ to opposite faces	Longitudinal, $\dfrac{\Delta L}{L}$	Yes	No	Young's modulus $Y = \dfrac{FL}{A\,\Delta L}$	Solid
Shearing	Two equal and opposite forces ∥ to opposite faces (net force and torque zero)	Pure shear, $\theta$	Yes	No	Shear modulus $\eta = \dfrac{F}{A\theta}$	Solid
Hydraulic (Bulk)	Forces ⊥ everywhere on the surface; pressure the same all over	Volumetric, $\dfrac{\Delta V}{V}$	No	Yes	Bulk modulus $B = -\dfrac{pV}{\Delta V}$	Solid, liquid and gas

2.7Poisson's Ratio

A loaded wire becomes longer and thinner. Its length increases, and its diameter decreases. The strain along the force is the longitudinal strain. The strain at right angles to the force is the lateral strain.

Definition

The ratio of lateral strain to longitudinal strain is called Poisson's ratio ( $\sigma$ ).

iLongitudinal Strain

The length increases from $L$ to $L + \Delta L$ .

\dfrac{\Delta L}{L}

iiLateral Strain

The diameter decreases from $D$ to $D - \Delta D$ .

-\dfrac{\Delta D}{D}

Poisson's Ratio

\sigma = \dfrac{\text{Lateral strain}}{\text{Longitudinal strain}} = \dfrac{-\Delta D/D}{\Delta L/L} = -\dfrac{L\,\Delta D}{D\,\Delta L}

The minus sign shows that the two strains are in opposite senses. One is an increase, and the other is a decrease. $\sigma$ is a ratio of two strains, so it has no unit and no dimensions. It is a constant for a material, as long as the material stays elastic, homogeneous and isotropic.

For theoretical purposes

-1 < \sigma < 0.5

For practical purposes

0 < \sigma < 0.5

Relations between the Elastic Constants

\text{(i)}\;\; Y = 3B(1 - 2\sigma) \qquad\qquad \text{(ii)}\;\; Y = 2\eta(1 + \sigma)

\text{(iii)}\;\; \sigma = \dfrac{3B - 2\eta}{2\eta + 6B} \qquad\qquad \text{(iv)}\;\; \dfrac{9}{Y} = \dfrac{1}{B} + \dfrac{3}{\eta}

where, $Y$ = Young's modulus, $B$ = bulk modulus, $\eta$ = modulus of rigidity and $\sigma$ = Poisson's ratio

Part 3

Elastic Effects and Applications

What changes elasticity, the energy stored in a wire, and everyday uses of elasticity.

3.1Factors Affecting Elasticity

iHammering and RollingElasticity ↑

Both processes break the crystal grains into smaller pieces. This raises the elasticity.

iiAnnealingElasticity ↓

Annealing makes the crystal grains bigger. This lowers the elasticity.

iiiPresence of Impurities↑ or ↓

An impurity can increase or decrease the elasticity. It depends on the nature of the impurity.

ivTemperatureElasticity ↓

For most materials, elasticity decreases as the temperature rises. Exception: special nickel–steel alloys such as elinvar (a kind of invar steel). Their elasticity stays almost the same when the temperature changes.

3.2Elastic After Effect

Definition

The delay of a body in getting back its original state, after the deforming force is removed, is called the elastic after effect.

When the deforming force is removed, some bodies return to their original state at once. Others take a long time.

iQuartz FibreNegligible effect

It returns to its normal state immediately when the twisting torque on it stops.

iiGlass FibreLarge effect

It needs hours to recover its original state once the force is removed.

3.3Elastic Fatigue

Definition

Elastic fatigue is the loss of strength of a material when it goes through repeated alternating strains.

iMetal Wire

Bend a metal wire again and again in opposite directions. It slowly loses strength due to elastic fatigue. Finally it breaks.

iiRailway Bridges

Railway bridges are declared unsafe after a long period of use. This prevents accidents due to elastic fatigue.

3.4Energy Stored in a Deformed Body

Concept

Elastic Potential Energy

Stretching a wire brings interatomic forces into play. They oppose the change. So work must be done against these restoring forces. This work is stored in the wire as elastic potential energy.

Setup — Given Information

A force $F$ on a wire of length $L$ increases its length by $\Delta L$ .
At the start, the wire has no internal restoring force.
As the length grows by $\Delta L$ , the internal force rises from $0$ to $F$ .

Derivation

The average internal force during the stretching is:
$F_{avg} = \dfrac{0 + F}{2} = \dfrac{F}{2}$
Work done = average force × increase in length. It is stored as elastic potential energy $U$ :
$U = \dfrac{F}{2}\times \Delta L = \dfrac{1}{2}F\,\Delta L$
Let $A$ be the area of cross-section. Multiply and divide by $AL$ :
$U = \dfrac{1}{2}\cdot\dfrac{F}{A}\cdot\dfrac{\Delta L}{L}\times AL$
Here $\dfrac{F}{A} = \text{Stress}$ , $\dfrac{\Delta L}{L} = \text{Strain}$ and $AL = \text{Volume}$ . So:
$U = \dfrac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume}$

Elastic Potential Energy Stored

U = \dfrac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume}

The energy stored in unit volume of the wire is called its elastic energy density ( $\mu$ ).

Elastic Energy Density

\mu = \dfrac{U}{\text{Volume}} = \dfrac{1}{2}\,\text{Stress}\times\text{Strain} = \dfrac{1}{2}\,Y\times(\text{Strain})^2

The last form uses Stress $= Y \times$ Strain.

3.5Applications of Elastic Behaviour

Elastic behaviour matters in engineering and in daily life. A structure must carry its load without permanent deformation or breaking. So its design depends on the elastic limit of the material used. Four important uses are given below.

(i) Parts of Machinery

A metal part of a machine is never loaded beyond the elastic limit of its material.

Stress kept within the elastic limitPart works properly ✓

Stress taken beyond it: the machine does not work properlyPermanently deformed ✗

(ii) Metal Ropes in Cranes

Cranes lift and move heavy loads using metal ropes. The thickness of such a rope is chosen from its elastic limit and a suitable factor of safety.

Note

Factor of Safety

The factor of safety is also called the safety factor (SF). It tells how much extra load a structure can carry beyond its expected (actual) load.

(iii) Design of Bridges and Beams

A bridge or a beam must carry moving traffic, resist the wind and also support its own weight. So its bending under a load is very important. It must not bend too much, and it must not break.

Take a beam of length $l$ with a rectangular cross-section of breadth $b$ and depth $d$ . It rests on two supports. A load $w$ hangs from its middle. The beam sags (is depressed) by an amount $\delta$ .

Beam loaded at the middle sags by

\delta

Depression of a Centrally Loaded Beam

\delta = \dfrac{w\,l^{3}}{4\,Y\,b\,d^{3}}

where, $l$ = length, $b$ = breadth, $d$ = depth, $Y$ = Young's modulus and $w$ = load at the centre. The sag $\delta$ varies inversely as $b$ and inversely as $d^{3}$ .

Sag and breadth:

\delta \propto \dfrac{1}{b}

Weak dependence

Sag and depth:

\delta \propto \dfrac{1}{d^{3}}

Strong (cube) dependence

Remember

Key Design Point

Since $\delta \propto 1/d^{3}$ but only $\delta \propto 1/b$ , increasing the depth $d$ reduces the sag much more than increasing the breadth $b$ .

Bending Resistance and Elastic Modulus

A beam carries a load by resisting bending. The greater this resistance, the more load the beam can carry. This resistance depends on three things:

Elastic modulusof material Cross-section shapeof beam Ultimate stressof material

(iv) Maximum Height of a Mountain

The elastic behaviour of the Earth sets the greatest possible height of a mountain on Earth. It is about 10 km. The base of a mountain is not under uniform compression. So the rocks at the base feel a shearing stress. Under this stress, the rocks can flow.

Setup — Given Information

Height of the mountain $= h$ . Density of its material, $\rho \approx 3\times10^{3}\ \text{kg/m}^3$ .
Elastic limit of a typical rock $= 30\times10^{7}\ \text{N/m}^2$ . Take $g = 9.8\ \text{m/s}^2$ .

Derivation

The weight of the mountain presses on its base. The stress there is:
$\text{Stress at base} = h\rho g$
At the greatest height, this stress just equals the elastic limit of the rock:
$h\rho g = 30\times10^{7}$
Put $\rho = 3\times10^{3}\ \text{kg/m}^3$ and $g = 9.8\ \text{m/s}^2$ :
$h \times 3\times10^{3} \times 9.8 = 30\times10^{7}$
Solve for $h$ :
$h = \dfrac{30\times10^{7}}{2.94\times10^{4}} \approx 1.0\times10^{4}\ \text{m} \approx 10\ \text{km}$

Maximum Height of a Mountain

h_{\max} \approx 10\ \text{km}

Above this height, the stress at the base would cross the elastic limit of the rock. The rock would then flow. So mountains on Earth cannot be much taller.

End of Chapter 8 · notesshala.in