Physics of solids

Class 11 Physics, unit 7, book pages 112–126

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Solids are classified as crystalline, polycrystalline or amorphous according to how their atoms are arranged, and a crystal is described by the repeating unit cell of its lattice. When a force deforms a solid, stress and strain measure the force per unit area and the fractional change, and their ratio gives the moduli of elasticity (Young's, bulk and shear). The stress-strain curve shows the elastic and plastic behaviour of a ductile material, and the area under the force-extension graph gives the elastic potential energy stored in a stretched wire.

Learning outcomes (7)

Distinguish between the structures of crystalline, glassy, amorphous and polymeric solids.
Describe that deformation of solids in one dimension [that it is caused by a force and that in one dimension, the deformation can be tensile or compressive].
Define and use the terms stress, strain and the Young's modulus.
Describe an experiment to determine the Young modulus of a metal wire.
Describe and use the terms elastic deformation, plastic deformation and elastic limit.
Justify why and apply the fact that the area under the force-extension graph represents the work done.
Determine the elastic potential energy of a material [that is deformed within its limit of proportionality] from the area under the force-extension graph. Also state and use $E_{p} = \frac{1}{2} F x = \frac{1}{2} k x^{2}$ for a material deformed within its limit of proportionality.

7.1Classification of solids

Materials have specific uses depending on their properties, such as hardness, ductility, brittleness, malleability and their response to applied pressure. What makes steel hard and lead soft? It depends on the structure: the particular order and the bonding of the atoms in a material. This clue has made it possible to design and create materials with unique properties for modern technology.

Solid: A material that is incompressible and has a fixed shape and volume.

On the basis of the arrangement of its atoms (its structure), a solid belongs to one of three types:

crystalline solids,
polycrystalline (polymeric) solids,
amorphous (glassy) solids.

7.1.1Crystalline solids

In solids the particles are closely packed together, but the packing can follow different patterns. The arrangement of the particles can be studied by the X-ray diffraction technique.

Crystalline solids: Solids in which the constituent atoms, ions or molecules have a regular and well-defined arrangement.

The molecular and ionic structures of crystals are shown in Figure 7.1. The ionic structure of sodium chloride is a three-dimensional array in which every sodium ion ( $N a^{+}$ ) is surrounded by chloride ions ( $C l^{-}$ ) and every chloride ion by sodium ions.

Figure 7.1: The ionic crystal structure of NaCl (the book also shows the molecular structure, on the left). Large circles are chloride ions and small filled circles are sodium ions; the pattern repeats regularly in all three directions.

Examples

Salts: sodium chloride and potassium chloride.
Metals: copper, iron and zinc.
Non-metals: diamond, sulphur and mica.
Ionic compounds: NaCl and copper sulphate.
Ceramic: zirconium.
Sand and quartz.

Properties

Crystalline solids have sharp melting points.
They are anisotropic for their physical property measurement: the value of a measured property depends on the direction in which it is measured.
On cutting they give sharp cleaves (flat, clean faces).

7.1.2Polycrystalline solids

Polycrystalline (polymeric) solids: Solids whose structure lies in between order and disorder; they are partially or poorly crystalline.
Polycrystalline solid: A solid material made up of many small crystals with random orientation.

The small crystals in a polycrystalline solid are called crystallites or grains. The grains are oriented in different directions and are separated by distinct grain boundaries (Figure 7.2). Normally the grains of polycrystalline solids have sizes in the range $1 0^{2} nm$ to $1 0^{3} nm$ .

Figure 7.2: Polycrystalline structure. Inside each grain the atoms are in regular rows, but the rows point in a different direction in each grain; the grain boundaries separate the grains.

Polycrystalline materials are isotropic: they show the same properties in all directions. This is because the grains point every which way, so the direction-dependence of the single small crystals averages out over the whole piece.

Polymers

Polymers are of two types, natural and synthetic.

Natural polymers include rubber with formula $(C_{5} H_{6})_{n}$ , resin and wood.
Synthetic polymers include polythene $(C_{2} H_{4})_{n}$ , polystyrene $(C_{8} H_{8})_{n}$ and polyvinyl chloride $(C_{2} H_{3} Cl)_{n}$ .

Polymers are made by the repetition of small molecules, consisting mainly of carbon, oxygen, hydrogen and nitrogen, which join to make large chains of molecules. This process is called polymerisation. The properties of these materials include low specific gravity, a good strength-to-weight ratio, toughness, resistance to corrosion, poor conductivity (of heat and electricity) and low cost.

7.1.3Amorphous solids

The term 'amorphous' means without regular form or structure. Amorphous solids do not have a regular structure.

Amorphous solids: Solid materials whose constituent particles are arranged in a random manner.

Amorphous solids, also called glassy solids, have a structure like frozen liquids. They have a fixed volume but not a definite regular geometrical shape. Figure 7.3 in the book shows the random network of beryllium fluoride ( $Be F_{2}$ ). A latest 3-D imaging technique called 'atomic electron tomography' showed that 85% of the atoms in amorphous solids are in a disordered arrangement, but that there are pockets in which a fraction of the atoms fall into ordered 'super-clusters'. Hence these solids are called short range ordered materials.

Amorphous solids have a range of melting temperature: if a glass rod is heated, it gradually softens into a paste-like state before it becomes a viscous liquid at $800^{\circ} C$ .

Figure 7.4: The crystalline, polycrystalline and amorphous solids. Crystalline: regular rows everywhere. Polycrystalline: regular rows inside grains, with different directions in different grains. Amorphous: no regular rows.

Examples and uses

Examples of amorphous solids include glass (sometimes called a super-cooled liquid), rubber, thin film systems deposited on a substrate at low temperatures, glues and many polymers. Amorphous solids can withstand higher temperatures, without affecting their efficiency, as compared to crystalline and polycrystalline solids. If these solids are melted and then cooled slowly, they can be converted into crystalline solids.

Amorphous solids are used almost in every field of life: construction, household wares and laboratory wares, manufacturing of tyres and production of shoes. Amorphous silica is one of the best materials for converting sunlight into electricity and is used in solar panels.

Comparison of the three types

A dash means that the book does not state that property.
	Crystalline	Polycrystalline	Amorphous
Structure	Regular and well-defined arrangement	Many small crystals (grains) in random orientation	Particles arranged in a random manner
Melting	Sharp melting point	–	Range of melting temperature; softens gradually
Properties against direction	Anisotropic	Isotropic	–
Other features	Sharp cleaves on cutting	Grains of size $1 0^{2}$ to $1 0^{3} nm$	Fixed volume, no definite regular shape
Examples	NaCl, copper, iron, zinc, diamond, quartz	Most inorganic metals, many ceramics, rocks, ice	Glass, rubber, glues, many polymers

7.1.4Unit cell and crystal lattice

The crystal structure of a solid is analysed by using X-ray diffraction and Bragg's law. Solids are characterised by their structure, which is made up of some basic unit that can be defined as follows.

Unit cell: The smallest basic portion of a crystal, which repeatedly stacked together in three dimensions makes the entire crystal structure.
Basis: The atoms, molecules or ions which make up the crystal, arranged in a regular pattern.
Lattice: An imaginary geometrical structure that joins the basis; it tells us about the basic structure of the points (basis).
Crystal lattice: The structure of a crystal obtained by the repetition of the unit cell.

A unit cell has all the characteristics of the whole crystal. The crystal structure is obtained by placing the atoms on the basis. A unit cell and its corresponding crystal lattice are shown in Figure 7.5 (a).

Figure 7.5 (a): Crystal lattice and unit cell. The highlighted block is one unit cell; stacking it in all three directions builds the crystal lattice.

Types of unit cell

There are three types of unit cells (Figure 7.5 b):

Simple cubic (SC) contains eight atoms, one at each corner. Hence there is one atom per unit cell in SC.
Body centred cubic (BCC) has eight atoms at the corners and one atom at the centre of the body. Hence there are two atoms per unit cell in BCC.
Face centred cubic (FCC) has eight atoms at the corners and one atom at each face. Hence there are four atoms per unit cell in FCC.

Figure 7.5 (b): Types of unit cells: simple cubic (SC), body centred cubic (BCC) and face centred cubic (FCC).

Atoms in the three cubic unit cells
Type	Corner atoms	Other atoms	Atoms per unit cell
Simple cubic (SC)	8 at the corners	none	$8 \times \frac{1}{8} = 1$
Body centred cubic (BCC)	8 at the corners	1 at the centre of the body	$8 \times \frac{1}{8} + 1 = 2$
Face centred cubic (FCC)	8 at the corners	1 at the centre of each face (6 faces)	$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$

Edges and angles of a unit cell

A unit cell has three sides for a space lattice, $a$ , $b$ and $c$ , with three angles $α$ , $β$ and $γ$ between them (Figure 7.5 c). In terms of the different relations between the sides and the angles, a unit cell can be divided into seven distinct shapes, as shown in Figure 7.6.

Figure 7.5 (c): Unit cell with sides

a

b

c

and angles

α

β

γ

Figure 7.6: Seven shapes of unit cells.

The seven shapes of unit cells (Figure 7.6)
Shape	Sides	Angles
Simple cubic	$a = b = c$	$α = β = γ = 9 0^{\circ}$
Tetragonal	$a = b \neq = c$	$α = β = γ = 9 0^{\circ}$
Orthorhombic	$a \neq = b \neq = c$	$α = β = γ = 9 0^{\circ}$
Rhombohedral	$a = b = c$	$α = β = γ \neq = 9 0^{\circ}$
Monoclinic	$a \neq = b \neq = c$	$α = β = 9 0^{\circ}, γ \neq = 9 0^{\circ}$
Triclinic	$a \neq = b \neq = c$	$α \neq = β \neq = γ \neq = 9 0^{\circ}$
Hexagonal	$a = b \neq = c$	$α = β = 9 0^{\circ}, γ = 12 0^{\circ}$

Daily life applications of solids

Crystalline solids: diamond is the most decent example; it is used in jewellery and various industries. Quartz is extensively used in manufacturing watches and clocks. Many crystalline solids are also used as a raw material in various industries.
Amorphous solids: glass is one of the most extensively used amorphous solids, found in utensils, bottles, boxes and construction material. Rubber is another widely used amorphous solid, found in tyres, footwear and ropes, and it serves as a material in many industries.
Polycrystalline solids: most inorganic metals, many ceramics, rocks and ice are polycrystalline, which have vast applications in our daily life.

Worked example: draw a BCC unit cell and count its atoms

Draw a body centred cubic unit cell, label the two kinds of atom, and find the number of atoms that belong to one cell. How many atoms are there in a crystal block made of

2500

such cells?

Draw a cube. Put one atom at each of its eight corners and one atom at the very centre of the cube (Figure below).
A corner atom touches eight cubes, so only one eighth of it belongs to the cell being drawn. The centre atom lies wholly inside and counts as one.
$atoms per cell = 8 \times \frac{1}{8} + 1 = 1 + 1 = 2$
A block of $2500$ cells holds $2500$ times as many atoms as one cell.
$N = 2500 \times 2 = 5000$

A body centred cubic (BCC) unit cell. The eight corner atoms (dark) each belong to eight cells; the centre atom (coloured) belongs to this cell alone. Hidden edges are dashed.

2 atoms per BCC cell; 5000 atoms in 2500 cells

7.2Deformation in solids

A solid has a definite shape and volume, possesses rigidity, is not compressed easily, has closely packed particles and cannot flow. To change the shape or size of a solid, a force is required. For example, if you stretch a spring by pulling its ends, the length of the spring increases. When you release the ends of the spring, it regains its original size.

To study the mechanical properties of solids, we usually check the behaviour of solids under an applied force. Applying a force to a solid may change its dimensions, that is, it deforms the solid.

Deformation: The change in shape, length or volume of a solid when it is subjected to an external force.

For example, if we squeeze a rubber ball with our fingers, we can observe the deformation that occurs. Similarly, at the atomic level, a crystal subjected to an applied pressure may deform mainly in two forms: a slip of one cleave over the other, or twinning of the crystal. Both are shown in Figure 7.7.

Figure 7.7: Deformation in crystals. In slip, one block of atomic layers slides over the other along the slip plane. In twinning, the layers on one side of the twin plane are displaced by amounts that grow with distance from the plane. (The book also shows a rubber ball squeezed between fingers.)

The results of mechanical tests are usually expressed in terms of stress and strain, which can be defined in terms of the applied force and the resulting deformation (Section 7.2.1).

Deformation in one dimension

In one dimension the deformation is a change in length produced by a force along the length. It can be tensile (a stretching force that makes the body longer) or compressive (a squeezing force that makes the body shorter). Figure 7.9 shows both cases for a rod.

7.2.1Stress and strainOften examined

Stress

Stress: The internal resistance offered by a body to resist deformation. It measures the magnitude of the force causing the deformation, and it is the applied force per unit area.

Stress is directly proportional to the applied force and inversely proportional to the area of cross section over which the force is applied. The book writes it as

Stress (δ) = \frac{Force ( F )}{Area ( A )}

The SI unit of stress is newton per square metre ( $N m^{- 2}$ ), also known as the pascal (Pa), and its dimensions are $[M L^{- 1} T^{- 2}]$ . Generally, stress can be divided into three types.

Tensile stress: the one-dimensional stretching force acting on the area of cross section of a body, which produces linear deformation in it.
Shear stress: the force which has a component acting tangentially to the area of a body and produces deformation in the shape of the body.
Volume stress: the force which acts uniformly from all directions on the area, producing a deformation in the volume of the body.

Figure 7.8: There are five types of stresses: compression, tension, shear, torsion and bending.

Strain

Due to the action of a force on a body, a deformation is produced in it.

Strain: The quantitative measure of deformation.

Strain is found by the ratio of the change to the actual value of some parameter (length, shape or volume). Hence it has no units and is a dimensionless quantity. On the basis of stress, strain can also be subdivided into three types.

Tensile strain: the fractional change in length, that is, the change in length divided by the original length. It is caused by tensile stress.
Shear strain: the tangential force acting on an area (shear stress) produces a change in the shape of a body, called shear strain.
Volume strain: the force acting at right angles to an area from all directions produces a deformation in the volume of a body, called volume strain.

Tensile strain (ε) = \frac{Change in length ( Δ L )}{Original length ( L )}

Shear strain (ε) = \frac{Displacement of sheared face}{distance from fixed face} = \frac{Δ X}{Y}

Volume strain (ε) = \frac{Change in volume ( Δ V )}{Original volume ( V )}

Stress	Force acts	Deformation	Strain
Tensile stress	along the length, stretching or squeezing	change in length	$\frac{Δ L}{L}$
Shear stress	tangentially to a face	change in shape	$\frac{Δ X}{Y}$
Volume stress	uniformly from all directions	change in volume	$\frac{Δ V}{V}$

7.2.2Modulus of elasticityOften examined

After the stress is removed, some materials regain their original shape, length or volume. This property of materials is called elasticity.

Elastic modulus (modulus of elasticity): For elastic solids, the ratio of the stress to the strain is always a constant. This constant is called the elastic modulus.

Elastic modulus = \frac{Stress}{Strain}

Since strain has no units, the modulus of elasticity has the same units as stress, that is $N m^{- 2}$ (or Pa), and the dimensions $[M L^{- 1} T^{- 2}]$ . Depending upon the type of stress, the modulus of elasticity is of three types.

Young's modulus

Young's modulus $Y$: In the case of linear deformation, the ratio of the tensile stress to the tensile strain.

Y = \frac{F / A}{Δ L / L}

(7.1)

Here $F$ is the applied force on the area $A$ of the material (say a rod) of length $L$ . The force changes the length by $Δ L$ , as shown in Figure 7.9, for both an increase and a decrease in length.

Figure 7.9: Young's modulus. A force

F

on the area

A

of a rod of length

L

changes its length by

Δ L

: an increase for a stretching force and a decrease for a squeezing force.

Bulk modulus

For three-dimensional forces the deformation occurs in all directions, as shown in Figure 7.10. In this case the ratio of the volume stress to the volume strain is called the bulk modulus.

Bulk modulus $B$: The ratio of the volume stress to the volume strain.

B = \frac{F / A}{( - Δ V / V )}

(7.2)

Here the negative sign shows the decrease in volume, but the bulk modulus is always positive.

Figure 7.10: Bulk modulus. Equal forces

F

act on every face of a body of volume

V

and reduce it to

V - Δ V

Shear modulus

Due to a tangential force, the deformation is produced in the shape of a body, as shown in Figure 7.11. In this case the ratio of the shear stress to the shear strain is called the shear modulus.

Shear modulus $S$: The ratio of the shear stress to the shear strain.

S = \frac{F / A}{( Δ X / Y )}

(7.3)

Figure 7.11: Shear modulus. A tangential force

F

on the top face (area

A

) displaces it by

Δ X

relative to the fixed bottom face at distance

Y

; the block tilts through the angle

θ

Here the strain is the ratio of the displacement of the sheared face to the height of the fixed face. From the triangle in Figure 7.11, the shear strain is expressed as

tan θ = \frac{Δ X}{Y}

For a very small angle $θ$ , $tan θ \approx θ$ . Hence $θ = Δ X / Y$ , and Eq. (7.3) takes the form

S = \frac{F}{A θ}

(7.4)

Here the angle $θ$ is called the angle of shear.

Table 7.1: Values of the modulus of elasticity of different materials
Material	Young's modulus ( $1 0^{10} N m^{- 2}$ )	Bulk modulus ( $1 0^{10} N m^{- 2}$ )	Shear modulus ( $1 0^{10} N m^{- 2}$ )
Steel	20.0	15.8	8.0
Aluminium	7.0	7.0	2.5
Copper	12.0	12.0	4.0
Iron	19.0	8.0	5.0
Brass	1.0	14.0	3.6

Example 7.1

A Masjid's minar having an area of cross-section

0.2 m^{2}

, the upper dome's weight

10, 000 N

, is made up of a material with Young's modulus

4.5 \times 1 0^{10} Pa

and mass density

2700 kg m^{- 3}

. Find the compressive stress at the cross-section located

3.0 m

below the top of the pillar, and the value of the compressive strain of the top

3.0 m

segment of the pillar.

Given

$h = 3.0 m$ height of the pillar segment
$A = 0.20 m^{2}$ area of cross-section
$ρ = 2700 kg m^{- 3}$ density
$w_{d} = 10000 N$ weight of the upper dome
$Y = 4.5 \times 1 0^{10} Pa$

To find

compressive stress $δ$
compressive strain $ε$

The load on the section is the dome plus the pillar segment above it, so first find the volume of the segment.
$V = A h = (0.20 m^{2}) (3.0 m) = 0.60 m^{3}$
Mass of the segment from the density.
$m = ρ V = (2700 kg m^{- 3}) (0.60 m^{3}) = 1.62 \times 1 0^{3} kg$
Weight of the segment.
$w_{p} = m g = (1.62 \times 1 0^{3} kg) (9.8 m s^{- 2}) = 1.59 \times 1 0^{4} N$
The normal force on the cross-section is the sum of the weight of the dome and of the segment.
$F = w_{d} + w_{p} = (1.0 \times 1 0^{4} N) + (1.59 \times 1 0^{4} N) = 2.59 \times 1 0^{4} N$
Compressive stress is force per unit area.
$δ = \frac{F}{A} = \frac{2.59 \times 1 0 ^{4} N}{0.20 m ^{2}} = 1.29 \times 1 0^{5} Pa \approx 129 kPa$
Compressive strain from the definition of Young's modulus, strain $=$ stress $/ Y$ .
$ε = \frac{δ}{Y} = \frac{1.29 \times 1 0 ^{5} Pa}{4.5 \times 1 0 ^{10} Pa} = 2.9 \times 1 0^{- 6}$

δ \approx 1.29 \times 1 0^{5} Pa \approx 129 kPa, ε \approx 2.9 \times 1 0^{- 6}

Assignment 7.1

The suspension cable to use cable cars at Murree has an unsupported span of

3.25 km

. Calculate the amount of stretch in the steel cable of diameter

6 cm

for its maximum load. The caution on weight is given by the engineers to not more than

4 \times 1 0^{6} N

Given

$L = 3.25 km = 3250 m$ length of cable taken equal to the span
$d = 6 cm = 0.06 m$ diameter
$F = 4 \times 1 0^{6} N$ maximum load
$Y = 20.0 \times 1 0^{10} N m^{- 2}$ steel, from Table 7.1

To find

stretch $Δ L$

Area of cross-section of the cable from its diameter ( $r = d /2 = 0.03 m$ ).
$A = π r^{2} = 3.14 \times (0.03 m)^{2} = 2.83 \times 1 0^{- 3} m^{2}$
Rearrange Eq. (7.1), $Y = \frac{F / A}{Δ L / L}$ , for the extension.
$Δ L = \frac{F L}{Y A}$
Put in the values.
$Δ L = \frac{( 4 \times 1 0 ^{6} ) ( 3250 )}{( 20.0 \times 1 0 ^{10} ) ( 2.83 \times 1 0 ^{- 3} )} = \frac{1.30 \times 1 0 ^{10}}{5.65 \times 1 0 ^{8}} = 23 m$

Δ L = \frac{F L}{Y A} \approx 23 m

Worked example: extension of two wires of different material

A steel wire and a copper wire are each

2.0 m

long and

1.0 mm

in diameter. Each carries a load of

50 N

. Using Table 7.1, find the extension of each wire and say which stretches more and by what factor.

Given

$L = 2.0 m$
$d = 1.0 mm$
$F = 50 N$
$Y_{s} = 20.0 \times 1 0^{10} Pa$
$Y_{c} = 12.0 \times 1 0^{10} Pa$

To find

$Δ L_{s}$ , $Δ L_{c}$ and their ratio

The area of cross section is a circle. The diameter is $1.0 mm$ , so the radius is $0.5 mm$ (not $1.0 mm$ ).
$A = π r^{2} = π (0.5 \times 1 0^{- 3})^{2} = 7.85 \times 1 0^{- 7} m^{2}$
Rearrange $Y = \frac{F / A}{Δ L / L}$ for the extension, $Δ L = \frac{F L}{Y A}$ . Steel first.
$Δ L_{s} = \frac{50 \times 2.0}{( 20.0 \times 1 0 ^{10} ) ( 7.85 \times 1 0 ^{- 7} )} = \frac{100}{1.57 \times 1 0 ^{5}} = 6.4 \times 1 0^{- 4} m$
Copper, with the same $F$ , $L$ and $A$ but a smaller $Y$ .
$Δ L_{c} = \frac{100}{( 12.0 \times 1 0 ^{10} ) ( 7.85 \times 1 0 ^{- 7} )} = \frac{100}{9.42 \times 1 0 ^{4}} = 1.06 \times 1 0^{- 3} m$
Only $Y$ differs, so the ratio of extensions is the inverse ratio of the moduli.
$\frac{Δ L _{c}}{Δ L _{s}} = \frac{Y _{s}}{Y _{c}} = \frac{20.0}{12.0} = 1.67$
The material with the larger Young's modulus is stiffer: it needs more stress for the same strain, so it stretches less.

Δ L_{s} = 0.64 mm, Δ L_{c} = 1.06 mm; copper stretches 1.67 times more

7.2.3Stress-strain curveOften examined

To study the behaviour of a material under applied stress, we use a tensile test machine, which strains the material at a fixed linear rate and records the stress. The values are then plotted on a graph by a computer attached to the machine. Strain is plotted along the horizontal axis and stress on the vertical axis. A typical stress-strain graph for a ductile material is shown in Figure 7.12.

In the initial stage of deformation the stress increases linearly with the strain, as shown by the portion OA in the graph. This portion has linear proportionality between stress and strain and obeys Hooke's law. In this region, if the applied stress is removed, the body regains its original shape.

Stress against strain for a ductile material

Figure 7.12: Stress-strain curve for a ductile material. O to A is the straight line (Hooke's law) up to the proportional limit \delta_p; O to B is the elastic region, with B the elastic limit or yield point \delta_y; beyond B is the plastic region; C is the lower yield point; D is the maximum stress, the ultimate tensile strength \delta_u; E is the fracture point with stress \delta_f.

Points on the curve

Point	Name	Meaning
A	Proportional limit, stress \delta_p	The greatest stress which a material can endure without losing the straight-line proportionality between stress and strain. After A, up to B, the straight-line proportionality ends, but if the stress is removed the body still regains its original shape.
B	Elastic limit or yield point, stress \delta_y	The maximum stress for elasticity of a material. The portion O to B, where the body can regain its original shape, is the elastic region.
C	Lower yield point	Some materials have a lower yield point.
D	Ultimate tensile strength \delta_u	The maximum stress that a material can withstand.
E	Fracture point, fracture stress \delta_f	After D the body breaks at E.

If the stress is increased beyond the yield point, the body is permanently changed and does not recover its original shape even if the stress is removed. This behaviour of materials is called plasticity: when the stress is zero, the strain is still not zero. This plastic region is the portion after point B and onward.

Elastic region: The portion O to B of the stress-strain curve, where the body has the ability to regain its original shape.
Yield point (elastic limit): The maximum limit of the stress for elasticity of a material, point B, with stress \delta_y.
Plasticity: The behaviour of a material beyond the yield point, where the body is permanently changed and the strain is not zero even when the stress is zero.
Ultimate tensile strength: The maximum stress that a material can withstand, point D, with stress \delta_u.
Fracture stress: The stress \delta_f at which the body breaks, point E.

Ductile and brittle materials

Substances which undergo plastic deformation until they break are known as ductile materials. Lead, copper and wrought iron are examples.
Other materials, which break just after the elastic limit, are called brittle materials, like glass and high carbon steel.

Example 7.2

Calculate the change in length of the upper leg bone of a soccer player, having a mass of 70.0\,\mathrm{kg}, supporting 62.0\,\mathrm{kg} of his mass on this bone, assuming the bone acts as a uniform rod, 40.0\,\mathrm{cm} long and with a radius of 2.00\,\mathrm{cm}.

Given

m = 62.0\,\mathrm{kg} mass supported by the bone
Y = 9\times10^{9}\,\mathrm{Pa} Young's modulus of bone
r = 0.020\,\mathrm{m}
L = 0.400\,\mathrm{m}

To find

change in length \Delta L

The force on the bone is the weight it supports.
F=mg=(62.0\,\mathrm{kg})(9.80\,\mathrm{m\,s^{-2}})=607.6\,\mathrm{N}
Area of cross-section of the bone, a circle of radius r.
A=\pi r^{2}=(3.14)(0.020\,\mathrm{m})^{2}=1.257\times10^{-3}\,\mathrm{m^{2}}
Rearrange Eq. (7.1) for the change in length.
\Delta L=\dfrac{FL}{YA}
Put in the values.
\Delta L=\dfrac{(607.6\,\mathrm{N})(0.400\,\mathrm{m})}{(9\times10^{9}\,\mathrm{N\,m^{-2}})(1.257\times10^{-3}\,\mathrm{m^{2}})}=2.1\times10^{-5}\,\mathrm{m}\approx2\times10^{-5}\,\mathrm{m}

\Delta L=\dfrac{FL}{YA}\approx2\times10^{-5}\,\mathrm{m}

Assignment 7.2

Find the strain in your own leg's bone while lifting a 50\,\mathrm{kg} box. Relate your finding with the example given above. Does the strain in your leg turn out greater or lesser than the strain in the above case? Explain the reason.

This is a personal calculation, so the working below uses the bone of Example 7.2 (Y=9\times10^{9}\,\mathrm{Pa}, r=2.00\,\mathrm{cm}, L=40.0\,\mathrm{cm}) and a body mass supported by the bone of 62.0\,\mathrm{kg}. Replace 62.0\,\mathrm{kg} by the mass supported by your own leg bone to get your own answer.

Given

m_1 = 62.0\,\mathrm{kg} body mass on the bone
m_2 = 50\,\mathrm{kg} box
A = 1.257\times10^{-3}\,\mathrm{m^{2}} from Example 7.2
Y = 9\times10^{9}\,\mathrm{Pa}

To find

strain \varepsilon while lifting the box
compare with Example 7.2

While the box is lifted the bone supports the body and the box.
F=(m_1+m_2)g=(62.0+50)(9.8)=1.10\times10^{3}\,\mathrm{N}
Strain is stress divided by Young's modulus.
\varepsilon=\dfrac{F/A}{Y}=\dfrac{1.10\times10^{3}}{(1.257\times10^{-3})(9\times10^{9})}=9.7\times10^{-5}
Strain in Example 7.2, from the same relation.
\varepsilon_0=\dfrac{607.6}{(1.257\times10^{-3})(9\times10^{9})}=5.4\times10^{-5}
Compare the two. The area and the Young's modulus are unchanged, so the ratio of the strains is the ratio of the forces.
\dfrac{\varepsilon}{\varepsilon_0}=\dfrac{112}{62}\approx1.8

The strain while lifting the box is greater, about 1.8 times that of Example 7.2. The bone and its modulus are the same, but the force on it is larger, so the stress and therefore the strain are larger (\varepsilon=F/AY\propto F). The strain is still tiny, about 10^{-4}, so the bone remains well within its elastic region.

\varepsilon\approx9.7\times10^{-5}\ (\text{greater than }5.4\times10^{-5})

Worked example: compare ductile and brittle materials on a stress-strain graph

Sketch on one pair of axes the stress-strain curves of a ductile material and of a brittle material. State two differences you would see.

Both materials start with a straight line (Hooke's law), since stress is proportional to strain at small strain. Draw it with the same slope: the same Young's modulus.
The ductile material continues past the yield point B into a long plastic region, rises to the ultimate tensile strength at D and breaks at E, at a large strain.
The brittle material breaks soon after its elastic limit (point F), with almost no plastic region, so its curve ends early.
Differences: (i) the ductile curve has a large plastic region while the brittle curve has almost none; (ii) the ductile material stretches permanently before it breaks (large strain at fracture), the brittle material breaks with little warning at a small strain.

Ductile and brittle materials

Ductile material (coloured): straight line, yield, plastic flow, maximum stress at D, then fracture at E. Brittle material (dark): straight line and then fracture at F with almost no plastic region.

\text{Ductile: long plastic region before fracture. Brittle: breaks just after the elastic limit.}

7.3Elastic potential energyOften examined

Work done on a body stores some amount of energy in it. In the same way, the work done on an elastic body to deform it also stores energy in the body. This stored energy is known as elastic potential energy.

Elastic potential energy (strain energy): The energy stored in a material due to the strain produced in it. For a material deformed within its elastic limit, it is equal to the work done in producing the deformation.

The next three sections show how to measure the Young's modulus of a wire (7.3.1), how the work done is found from the force-extension graph (7.3.2) and how the stored energy is calculated (7.3.3).

7.3.1Modulus of metallic wireOften examined

To study the work done on a body to produce strain, consider a metallic wire which can be strained by the application of a force on it.

Figure 7.13: Young's modulus for a metallic wire. The extension of the metallic wire (accent) is read on the vernier scale against the reference wire as the variable load is increased.

Experiment: Young's modulus of a metallic wire

Take the metallic wire (the test wire) and fix it at the top. Measure its original length L and its area of cross-section A.
Connect a vernier scale to the wire.
Take another wire as a reference wire, with a scale on it, for the precise measurement of the change in length of the metallic wire when the load (weight) is added.
Add the variable load to the test wire. The force on the wire is the weight added. The wire stretches, and the vernier scale gives the extension of the wire for the increasing force.
Use Eq. (7.1) with the average force, the original length, the change in length and the area of the metallic wire to find the Young's modulus of the wire.
Y=\dfrac{F/A}{\Delta L/L}=\dfrac{FL}{A\,\Delta L}

Worked example: Young's modulus from a force-extension graph

In the experiment of Figure 7.13 a steel wire of length 2.00\,\mathrm{m} and diameter 0.80\,\mathrm{mm} gives these readings: weight 10, 20, 30, 40\,\mathrm{N} and extension 0.20, 0.40, 0.60, 0.80\,\mathrm{mm}. Plot the graph and find Young's modulus from its slope.

Given

L = 2.00\,\mathrm{m}
d = 0.80\,\mathrm{mm}

To find

Y of the wire

Plot weight on the horizontal axis and extension on the vertical axis. The points fall on a straight line through the origin (graph below). Take the slope using the end point, which is more accurate than a single reading.
\text{slope}=\dfrac{\Delta x}{\Delta F}=\dfrac{0.80\times10^{-3}\,\mathrm{m}}{40\,\mathrm{N}}=2.0\times10^{-5}\,\mathrm{m\,N^{-1}}
Since x=\dfrac{L}{YA}F, the slope equals \dfrac{L}{YA}. Find the area of cross section first.
A=\dfrac{\pi d^{2}}{4}=\dfrac{\pi(0.80\times10^{-3})^{2}}{4}=5.03\times10^{-7}\,\mathrm{m^{2}}
Solve for Y.
Y=\dfrac{L}{A\times\text{slope}}=\dfrac{2.00}{(5.03\times10^{-7})(2.0\times10^{-5})}=1.99\times10^{11}\,\mathrm{Pa}

Extension against weight

The points lie on a straight line through the origin, as Hooke's law requires. The slope \Delta x/\Delta F is used to find Y.

Y\approx2.0\times10^{11}\,\mathrm{Pa}

7.3.2Work from force-extension graphOften examined

To find the work done by the variable load on the metallic wire to produce extension, draw a graph by taking weight on the horizontal axis and extension on the vertical axis. Initially no weight is attached to the wire and it has zero extension. By continuously increasing the weight, the wire gets an extension, as shown in Figure 7.14.

Force-extension graph of a metallic wire

Figure 7.14: Force-extension graph. The weight (force) is on the horizontal axis and the extension on the vertical axis. The straight line from O to A is the proportional region; the triangle AOF (shaded) has area \tfrac12Fx.

Derivation Work done in stretching a wire

Aim. Show that the work done in stretching a wire by an extension x is W=\tfrac12Fx, which is the area under the force-extension graph.

Assumptions.

The load is increased gradually from zero.
The wire is within the proportional region (straight part OA of the graph).

In the proportional region the extension grows in proportion to the load, so the graph from O to A is a straight line through the origin. At A the load is F and the extension is x.
The work done by a variable force is the area under its graph. The area from O to A is the area of the triangle AOF, with base F and height x.
\text{Area}=\dfrac12\,F\,x\qquad(7.5\,\mathrm{a})
Check with the average force. The force rises uniformly from 0 to F, so the average force applied on the wire is half of F.
F_{avg}=\dfrac{0+F}{2}=\dfrac12F
Work done is the product of the average force and the displacement (here the extension x).
W=F_{avg}\,x\ \Rightarrow\ W=\dfrac12\,F\,x\qquad(7.5\,\mathrm{b})
Equations (7.5 a) and (7.5 b) are the same, so the work done equals the area under the graph.

ResultW=\dfrac12\,F\,x=\text{area under the force-extension graph}

The work done during the application of a load for extension in an elastic material, within the elastic limit, is equal to the area under the force-extension graph.

7.3.3Derivation of elastic potential energyOften examined

The energy stored in a material due to the strain produced in it is called the elastic potential energy. It can be calculated by using the relation for the work done during extension. For an elastic material the force is given by Hooke's law,

F=kx

The elastic potential energy in a deformed material, within the elastic limit, is equal to the work done in producing the deformation.

Derivation Elastic potential energy of a stretched material

Aim. Derive \text{P.E.}=\tfrac12kx^{2} and express it in terms of Young's modulus.

Assumptions.

The material is deformed within its elastic limit, where the force obeys Hooke's law F=kx.

The elastic potential energy equals the work done in producing the extension x, and the work done is \tfrac12Fx (Section 7.3.2).
\text{Elastic P.E.}=\text{Work done}=\dfrac12\,F\,x
By Hooke's law the force that produces the extension x is F=kx. Substitute it.
\text{P.E.}_{elastic}=\dfrac12\,(kx)\,x
Simplify.
\text{P.E.}_{elastic}=\dfrac12\,k\,x^{2}\qquad(7.6)
Now write the energy in terms of Young's modulus. From Eq. (7.1), Y=\dfrac{F/A}{x/L}, where x is the extension. Solve for the force.
F=\dfrac{YAx}{L}
Here A is the area of cross-section and L is the original length of the wire. Substitute this force in \tfrac12Fx.
\text{P.E.}=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}\qquad(7.7)

Result\text{P.E.}_{elastic}=\dfrac12\,k\,x^{2}=\dfrac12\,\dfrac{YA}{L}\,x^{2}

This elastic potential energy is also called the strain energy. Comparing F=kx with F=YAx/L shows that the force constant of the wire is k=\dfrac{YA}{L}.

Example 7.3

Calculate the strain energy stored in a 1\,\mathrm{m} long steel wire connected to the roof, holding a chandelier of weight 980\,\mathrm{N} in the main lounge of your home. The diameter of the steel wire is 0.004\,\mathrm{m} and the Young's modulus of steel is 2.0\times10^{11}\,\mathrm{N\,m^{-2}}.

Given

L = 1\,\mathrm{m} length of the steel wire
Y = 2.0\times10^{11}\,\mathrm{Pa}
d = 0.004\,\mathrm{m} diameter
W = 980\,\mathrm{N} weight of the chandelier

To find

change in length \Delta L
strain energy U

The force on the wire is equal to the weight it supports.
F=W=980\,\mathrm{N}
Area of cross-section from the diameter (r=d/2=0.002\,\mathrm{m}).
A=\pi r^{2}=\pi\left(\dfrac d2\right)^{2}=(3.14)(0.002\,\mathrm{m})^{2}=1.257\times10^{-5}\,\mathrm{m^{2}}
Change in length from Eq. (7.1).
\Delta L=\dfrac{FL}{YA}=\dfrac{(980\,\mathrm{N})(1\,\mathrm{m})}{(2.0\times10^{11}\,\mathrm{N\,m^{-2}})(1.257\times10^{-5}\,\mathrm{m^{2}})}=3.90\times10^{-4}\,\mathrm{m}
Strain energy from Eq. (7.7), with x=\Delta L.
U=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}=\dfrac12\left[\dfrac{(2.0\times10^{11})(1.257\times10^{-5})}{1}\right](3.90\times10^{-4})^{2}
Evaluate.
U=0.191\,\mathrm{J}
Check with U=\tfrac12F\,\Delta L.
U=\dfrac12(980)(3.90\times10^{-4})=0.191\,\mathrm{J}

\Delta L=3.90\times10^{-4}\,\mathrm{m},\qquad U=0.191\,\mathrm{J}

Assignment 7.3

Calculate the work done in stretching a wire of length 5\,\mathrm{m} with cross-sectional area 1\,\mathrm{mm^{2}}, when the force applied by you on the ends of the wire produces a change in the length of the wire by 1\,\mathrm{mm}. Young's modulus of the wire is 2\times10^{11}\,\mathrm{Pa}.

Given

L = 5\,\mathrm{m}
A = 1\,\mathrm{mm^{2}}=1\times10^{-6}\,\mathrm{m^{2}}
x = 1\,\mathrm{mm}=1\times10^{-3}\,\mathrm{m} extension
Y = 2\times10^{11}\,\mathrm{Pa}

To find

work done W

The work done in stretching equals the elastic potential energy stored, Eq. (7.7).
W=\dfrac12\,\dfrac{YA}{L}\,x^{2}
Force constant of the wire.
k=\dfrac{YA}{L}=\dfrac{(2\times10^{11})(1\times10^{-6})}{5}=4\times10^{4}\,\mathrm{N\,m^{-1}}
Work done.
W=\dfrac12(4\times10^{4})(1\times10^{-3})^{2}=2\times10^{-2}\,\mathrm{J}

W=0.02\,\mathrm{J}

Worked example: strain energy when the load is doubled

A steel wire of length 3.0\,\mathrm{m} and cross-section 2.0\,\mathrm{mm^{2}} (Y=20.0\times10^{10}\,\mathrm{Pa}) is stretched by 1.5\,\mathrm{mm}. (a) Find the strain energy stored. (b) What is the energy if the load is doubled, within the elastic limit?

Given

L = 3.0\,\mathrm{m}
A = 2.0\,\mathrm{mm^{2}}=2.0\times10^{-6}\,\mathrm{m^{2}}
Y = 20.0\times10^{10}\,\mathrm{Pa}
x = 1.5\,\mathrm{mm}=1.5\times10^{-3}\,\mathrm{m}

To find

(a) U
(b) U' for double the load

(a) The wire behaves like a spring of constant k=\dfrac{YA}{L}.
k=\dfrac{(20.0\times10^{10})(2.0\times10^{-6})}{3.0}=1.33\times10^{5}\,\mathrm{N\,m^{-1}}
Strain energy is the work done, \tfrac12kx^{2}. As a check, the force is F=kx=200\,\mathrm{N} and \tfrac12Fx gives the same value.
U=\tfrac12kx^{2}=\tfrac12(1.33\times10^{5})(1.5\times10^{-3})^{2}=0.15\,\mathrm{J}
(b) Within the elastic limit the extension is proportional to the load, so doubling the load doubles x to 3.0\,\mathrm{mm}. Because U\propto x^{2}, the energy becomes four times as large, not twice.
U'=\tfrac12(1.33\times10^{5})(3.0\times10^{-3})^{2}=0.60\,\mathrm{J}=4U

Strain energy against extension

The energy grows as the square of the extension, so it is a parabola. Doubling x from 1.5 to 3.0\,\mathrm{mm} raises U from 0.15\,\mathrm{J} to 0.60\,\mathrm{J}.

U=0.15\,\mathrm{J},\qquad U'=0.60\,\mathrm{J}

Formula sheet

Stress

\delta=\dfrac{F}{A}

F: applied force; A: area of cross-section over which it acts

Units: \mathrm{N\,m^{-2}} (pascal, Pa); dimensions [\mathrm{ML^{-1}T^{-2}}]

Tensile strain

\varepsilon=\dfrac{\Delta L}{L}

\Delta L: change in length; L: original length

no unit (dimensionless)

Shear strain

\varepsilon=\dfrac{\Delta X}{Y}

\Delta X: displacement of the sheared face; Y: distance from the fixed face

no unit (dimensionless)

Volume strain

\varepsilon=\dfrac{\Delta V}{V}

\Delta V: change in volume; V: original volume

no unit (dimensionless)

Modulus of elasticity

\text{Elastic modulus}=\dfrac{\text{Stress}}{\text{Strain}}

constant for an elastic solid

Unit: \mathrm{N\,m^{-2}} (Pa)

Young's modulus

Y=\dfrac{F/A}{\Delta L/L}

F: force on area A; \Delta L: change in length of a rod of length L (Eq. 7.1)

Unit: \mathrm{N\,m^{-2}} (Pa)

Extension of a wire

\Delta L=\dfrac{FL}{YA}

Eq. (7.1) rearranged

Unit: \mathrm{m}

Bulk modulus

B=\dfrac{F/A}{(-\Delta V/V)}

the negative sign shows the decrease in volume; B is always positive (Eq. 7.2)

Unit: \mathrm{N\,m^{-2}} (Pa)

Shear modulus

S=\dfrac{F/A}{(\Delta X/Y)}

\Delta X: displacement of the sheared face; Y: height of the fixed face (Eq. 7.3)

Unit: \mathrm{N\,m^{-2}} (Pa)

Angle of shear

\tan\theta=\dfrac{\Delta X}{Y}\approx\theta

for a very small angle \theta

Unit: \mathrm{rad}

Shear modulus (small angle)

S=\dfrac{F}{A\theta}

\theta: angle of shear (Eq. 7.4)

Unit: \mathrm{N\,m^{-2}} (Pa)

Hooke's law

F=kx

k: force constant; x: extension, within the limit of proportionality

Unit: \mathrm{N}

Work done in stretching

W=\dfrac12Fx=\text{area under the force-extension graph}

F: final force; x: extension (Eq. 7.5)

Unit: \mathrm{J}

Elastic potential energy

\text{P.E.}_{elastic}=\dfrac12kx^{2}

k: force constant; x: extension (Eq. 7.6)

Unit: \mathrm{J}

Force on a wire in terms of Y

F=\dfrac{YAx}{L}

from Eq. (7.1) with extension x

Unit: \mathrm{N}

Strain energy of a wire

\text{P.E.}=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}

Y: Young's modulus; A: area of cross-section; L: original length (Eq. 7.7)

Unit: \mathrm{J}

Key terms

Solid: A material that is incompressible and has a fixed shape and volume.
Crystalline solid: A solid in which the constituent atoms, ions or molecules have a regular and well-defined arrangement; it has a sharp melting point, is anisotropic and gives sharp cleaves.
Polycrystalline solid: A solid made up of many small crystals (grains) with random orientation; its structure lies between order and disorder and it is isotropic.
Amorphous solid: A solid whose constituent particles are arranged in a random manner; it has a fixed volume but no definite regular shape and a range of melting temperature.
Anisotropic: Having physical properties whose measured value depends on the direction of measurement.
Isotropic: Having the same properties in all directions.
Crystallite (grain): One of the small crystals that make up a polycrystalline solid.
Grain boundary: The boundary between two grains oriented in different directions.
Polymerisation: The process in which small molecules join to make large chains of molecules (polymers).
Unit cell: The smallest basic portion of a crystal, which repeatedly stacked together in three dimensions makes the entire crystal structure.
Crystal lattice: The structure of a crystal obtained by the repetition of the unit cell.
Basis: The atoms, molecules or ions which make up a crystal, arranged in a regular pattern.
SC, BCC and FCC: Simple cubic, body centred cubic and face centred cubic unit cells, with 1, 2 and 4 atoms per unit cell.
Deformation: The change in shape, length or volume of a solid when it is subjected to an external force.
Slip and twinning: The two main ways a crystal deforms at the atomic level: one cleave slips over the other, or the crystal twins.
Stress: The internal resistance offered by a body to resist deformation; the applied force per unit area, \delta=F/A, measured in pascal.
Strain: The quantitative measure of deformation; the ratio of the change to the original value of a length, shape or volume. It has no unit.
Tensile, shear and volume stress: Stress that produces a change in length, in shape and in volume respectively.
Elasticity: The property of a material to regain its original shape, length or volume after the stress is removed.
Modulus of elasticity: The constant ratio of stress to strain for an elastic solid.
Young's modulus: The ratio of the tensile stress to the tensile strain, Y=\dfrac{F/A}{\Delta L/L}.
Bulk modulus: The ratio of the volume stress to the volume strain, B=\dfrac{F/A}{-\Delta V/V}.
Shear modulus: The ratio of the shear stress to the shear strain, S=\dfrac{F/A}{\Delta X/Y}.
Angle of shear: The angle \theta through which a body is sheared; \tan\theta=\Delta X/Y\approx\theta for small angles.
Hooke's law: Within the limit of proportionality the stress is proportional to the strain, or the force is proportional to the extension, F=kx.
Proportional limit: The greatest stress which a material can endure without losing straight-line proportionality between stress and strain (point A).
Elastic limit (yield point): The maximum stress up to which a material is elastic (point B).
Elastic region: The part of the stress-strain curve (O to B) where the body regains its original shape when the stress is removed.
Plasticity (plastic region): The behaviour of a material beyond the yield point: it is permanently changed and the strain is not zero even when the stress is removed.
Ultimate tensile strength: The maximum stress that a material can withstand (point D).
Fracture stress: The stress at which the body breaks (point E).
Ductile material: A material that undergoes plastic deformation until it breaks, such as lead, copper and wrought iron.
Brittle material: A material that breaks just after the elastic limit, such as glass and high carbon steel.
Elastic potential energy (strain energy): The energy stored in a material due to the strain produced in it; \tfrac12Fx=\tfrac12kx^{2} within the limit of proportionality.

Multiple choice questionsp. 125

Glue is an example of:

Acrystalline solid
Bamorphous solidAnswer
Cpoly-crystalline solid
Dductile solid

Why

Glue is an amorphous (glassy) solid. Its particles have no long-range regular arrangement, in the same way as glass and rubber in section 7.1.3.

Crystalline and poly-crystalline solids have regular arrangements (fully or partly ordered), and ductile describes how a solid deforms under stress, not its atomic structure. So B is correct.

Which of the following properties is generally exhibited by amorphous solids?

Aanisotropy
Bglass-transitionAnswer
Corderliness
Dgeometry

Why

Amorphous solids have no sharp melting point. On heating they soften gradually over a range of temperatures, which is the glass transition. So B is correct.

Anisotropy (properties that depend on direction) and orderliness belong to crystalline solids, and a regular geometry is the shape of an ordered crystal.

Silicon is found in nature in the forms of:

ASC structure
BBCC structure
CFCC structure
Dnetwork structureAnswer

Why

Silicon occurs in nature mainly as silica and silicates (sand, quartz), in which each silicon atom is bonded to four oxygen atoms and the bonds form a three-dimensional network. So D is correct.

SC, BCC and FCC are the three types of unit cell (section 7.1.4), describing how atoms are packed in a crystal. They are not forms in which silicon is found in nature.

Note. The book prints no key for the MCQs. The key D is worked out from the reasoning given here.

One end of a uniform wire of length 'L' and of weight 'W' is attached rigidly to a point in the roof and a weight 'W_1' is suspended from its lower end. If 'S' is the area of cross-section of the wire, the stress in the wire at a height '3L/4' from its lower end is:

A\dfrac{W_1}{S}
B\dfrac{W_1+(W/4)}{S}
C\dfrac{W_1+(3W/4)}{S}Answer
D\dfrac{W_1+W}{S}

Why

Cut the wire at the section that is 3L/4 above the lower end. The part of the wire below the cut has length 3L/4. The weight of a uniform wire is shared in proportion to its length, so this part weighs \dfrac{3}{4}W.

F=W_1+\dfrac{3}{4}W=W_1+\dfrac{3W}{4}

The stress at the cut is the load it carries divided by the area: \delta=\dfrac{W_1+(3W/4)}{S}, which is option C. Option B uses W/4, the weight of the upper quarter of the wire, which is above the cut and does not load it.

On suspending a weight 'mg', the length 'l' of an elastic wire with area of cross-section 'A' becomes double of its initial length. The instantaneous stress on the wire is:

A\dfrac{mg}{2A}
B\dfrac{mg}{A}Answer
C\dfrac{2mg}{A}
D\dfrac{4mg}{A}

Why

Stress is the force divided by the area. The force on the wire is the load mg and the area is A, so the stress is \dfrac{mg}{A}.

The doubling of the length tells us about the strain (the wire has stretched a lot), not about the stress. Options C and D double or quadruple the force, and option A halves it, so none of them is right.

The diameter of a brass rod is 4\,\mathrm{mm} and Young's modulus of brass is 10\times10^{10}\,\mathrm{N\,m^{-2}}. The force required to stretch by 0.1\,\% of its length is:

A36\pi\,\mathrm{N}
B40\pi\,\mathrm{N}
C360\pi\,\mathrm{N}
D400\pi\,\mathrm{N}Answer

Why

From Y=\dfrac{F/A}{\Delta L/L} the force is F=YA\dfrac{\Delta L}{L}. The area is A=\pi\left(\dfrac{d}{2}\right)^{2}=\pi(2\times10^{-3})^{2}=4\pi\times10^{-6}\,\mathrm{m^{2}} and the strain is \dfrac{\Delta L}{L}=0.1\,\%=1\times10^{-3}.

F=(10\times10^{10})(4\pi\times10^{-6})(1\times10^{-3})=400\pi\,\mathrm{N}\approx1.26\times10^{3}\,\mathrm{N}

So D is correct. Options A and B are a tenth of the correct value, which is a power-of-ten slip in the strain or the modulus, and C is only about 10\,\% too small.

Note. The brass modulus here is 10\times10^{10}\,\mathrm{N\,m^{-2}}, the real value. Table 7.1 and numerical 7.3 print 1.0\times10^{10}\,\mathrm{N\,m^{-2}}, which is a misprint for 10\times10^{10}\,\mathrm{N\,m^{-2}}.

A cantilever beam has a load at the free end. The strain energy in the beam is due to:

AbendingAnswer
Bshearing
Cstretching
Delongation

Why

The load at the free end bends the beam. The fibres on the upper side are stretched and those on the lower side are compressed, and the energy stored in this bending is the strain energy. So A is correct.

Shearing (B) is a sliding of layers and is not the main deformation of a beam under an end load. Stretching and elongation (C and D) describe a change in length of a wire, not a bending beam.

When a pressure of 100 atmospheres is applied on a spherical ball of rubber and its volume reduces to 0.01\,\%, the bulk modulus of the material of rubber in \mathrm{dyne\,cm^{-2}} is:

A1\times10^{12}Answer
B10\times10^{12}
C20\times10^{12}
D100\times10^{12}

Why

The volume is reduced by 0.01\,\%, so \dfrac{\Delta V}{V}=1\times10^{-4}. The applied pressure is the volume stress. One atmosphere is 1.013\times10^{6}\,\mathrm{dyne\,cm^{-2}}, so p=100\,\mathrm{atm}=1.013\times10^{8}\,\mathrm{dyne\,cm^{-2}}.

B=\dfrac{p}{\Delta V/V}=\dfrac{1.013\times10^{8}}{1\times10^{-4}}\approx1\times10^{12}\,\mathrm{dyne\,cm^{-2}}

So A is correct. The bulk modulus is positive, as the negative sign in its formula only shows the decrease in volume.

Note. 'Volume reduces to 0.01\,\%' is read as a reduction by 0.01\,\% (that is, \Delta V/V=10^{-4}). Reading it as the volume becoming 0.01\,\% of its original value would give a strain close to 1, which does not suit an elastic solid.

The property of a material to store strain energy is called:

Aductility
Bhardening
CresilienceAnswer
Dstiffness

Why

Resilience is the ability of a material to store strain energy within its elastic limit and to give it back when the load is removed. So C is correct.

Ductility (A) is the ability to be drawn into a wire without breaking, hardening (B) is the increase in strength by deformation, and stiffness (D) is the resistance to deformation (the modulus), not the energy it stores.

Gradient of the force-extension graph has:

Aincreasing value
Bdecreasing value
Cvariable value
DconstantAnswer

Why

Within the elastic limit the graph is a straight line through the origin, F=kx, so its gradient \dfrac{\Delta F}{\Delta x}=k=\dfrac{YA}{L} is constant. This gradient is the stiffness of the wire. So D is correct.

The gradient would change only if the graph curved, which happens beyond the limit of proportionality.

Short questionsp. 125–126

13 marks

Why do most of the solids prefer to be in the crystalline state? What is glass transition in amorphous solids?

Solution

Most solids prefer the crystalline state because it is the most stable (lowest energy) arrangement. In a crystal each atom, ion or molecule sits where the attraction from its neighbours balances the repulsion, so the potential energy of the solid is at a minimum. The regular, close packing is what a solid settles into when it cools slowly.

The glass transition is the behaviour of an amorphous solid on heating. It has no sharp melting point. It softens gradually over a range of temperatures, first into a soft, paste-like state and then into a viscous liquid (a glass rod softens in this way before it becomes a viscous liquid at about 800\,^\circ\mathrm{C}).

23 marks

Why the window glasses of old buildings show milky appearance over time?

Solution

Window glass is an amorphous solid, with its particles arranged at random. Over a long time the random structure slowly changes: small crystalline regions form inside the glass, which is the conversion of an amorphous solid into a crystalline one by slow change.

These small crystals have different optical properties from the glass around them, so they scatter light. The scattered light makes the glass look milky or cloudy, which is why old window glass looks milky.

33 marks

In the elastic body, which one is more fundamental: the stress or the strain? Explain.

Solution

Stress is more fundamental. Stress is the cause: the force applied to the body, per unit area. Strain is the effect: the fractional change in shape, length or volume that the stress produces.

For an elastic body the strain is fixed by the stress through the modulus of elasticity, \text{strain}=\dfrac{\text{stress}}{\text{modulus}}. The stress is what we apply, and the strain follows from it. Strain has no units, while stress is a physical quantity with the units \mathrm{N\,m^{-2}}.

43 marks

Can tensile stress produce volumetric strain? Explain.

Solution

Yes. A tensile stress changes the length of the body, and it also changes its cross-sectional area (a stretched body becomes thinner). Since the volume is V=AL, both changes alter the volume.

\dfrac{\Delta V}{V}\approx\dfrac{\Delta L}{L}+\dfrac{\Delta A}{A}

The volumetric strain is therefore not zero in general. For ordinary solids the change in volume is small, and the sum of the two terms is usually a small increase in volume.

53 marks

What does the slope of the force-extension graph represents? Explain.

Solution

The slope of the force-extension graph is \dfrac{F}{x}, the force needed per unit extension. Its unit is \mathrm{N\,m^{-1}}. Within the elastic limit this slope is constant and is called the stiffness k of the wire.

\text{slope}=\dfrac{F}{x}=k=\dfrac{YA}{L}

This follows from F=\dfrac{YA}{L}x (equation 7.1 rearranged). So the slope depends on the material (Y), the area A and the original length L of the wire, and not on the load itself.

63 marks

Is strain energy always positive or can it be negative? Justify your answer.

Solution

Strain energy is always positive. It is the energy stored in a body by its deformation, and it is equal to the area under the force-extension graph:

U=\dfrac12Fx=\dfrac12kx^{2}

Here k>0, so \dfrac12kx^{2}\geq0 for any extension x, whether it is positive or negative. In compression both F and x are negative, and their product Fx is still positive. The energy is given to the body by the work done on it, and this work is positive, so the stored energy cannot be negative.

73 marks

On which types of loads the strain energy depends? Explain.

Solution

Strain energy depends on the type of load, that is on the kind of stress applied, and on the size of the load. For each kind of stress the stored energy is the work done in the matching deformation:

Tensile or compressive load (change in length): U=\tfrac12F\,\Delta L, or \tfrac12\times\text{stress}\times\text{strain} per unit volume.
Shearing load (change in shape): \tfrac12 F\,\Delta X, or \tfrac12\times\text{shear stress}\times\text{shear strain} per unit volume.
Volume load (change in volume): \tfrac12\,p\,\Delta V, where p is the applied pressure.

Within the elastic limit the energy also depends on the magnitude of the load, because U=\tfrac12 \dfrac{YA}{L}x^{2} grows as the square of the extension.

82025 Annual I, Set D, Q.2 (x)3 marks

Why are suspension bridges assign a period of use? Is it dangerous to use the bridge after that period?

Solution

The cables of a suspension bridge carry the huge, repeated loads of traffic, wind and their own weight. Each load cycle stresses the metal, and over many years tiny cracks start and grow. This fatigue reduces the strength and the elastic limit of the wires, so they can be damaged by loads they used to carry safely. A period of use is therefore assigned, and the cables are inspected and replaced when it ends.

Yes, it is dangerous to keep using the bridge after that period, because the wires may have been permanently strained or weakened and may break at a load well below the design load.

93 marks

If we want to break a wire, why we use repeated bending? How does it affect the wire?

Solution

Bending a wire stresses it beyond its elastic limit at the bend, so the bend is deformed plastically (permanently). Bending it back and forth reverses this stress many times.

Each cycle makes the metal harder and more brittle at the bend, and cracks start and grow. The wire then breaks at a stress much smaller than its ultimate tensile strength, which is called fatigue failure. This is why repeated bending breaks a wire much more easily than a single steady pull.

103 marks

What is the significance of modulus of elasticity?

Solution

The modulus of elasticity, \text{stress}/\text{strain}, measures the stiffness of a material, that is how strongly it resists deformation. A material with a large modulus needs a large stress to produce a given strain.

Steel has a Young's modulus of 20\times10^{10}\,\mathrm{Pa}, about three times that of aluminium (7\times10^{10}\,\mathrm{Pa}). This makes steel the better choice for cables, beams and bridges, where the deformation must be small. The modulus therefore helps engineers select the best material for a job.

112026 Annual I, Set D, Q.2 (xi)3 marks

What is meant by crystal lattice.

Solution

A crystal lattice is the imaginary, regular three-dimensional structure formed by joining the points of the basis (the atoms, ions or molecules). The unit cell is the smallest part of a crystal that, repeated in three directions, makes the whole crystal, so the crystal lattice is what results when the unit cell is repeated in all directions.

Comprehensive questionsp. 126

Explain the three types of solids: crystalline, poly-crystalline and amorphous solids.

Solution

Crystalline solids

In a crystalline solid the atoms, ions or molecules are arranged in a regular and well-defined pattern that repeats throughout the whole solid. Such an arrangement is studied by X-ray diffraction. Examples are salts (sodium chloride, potassium chloride), metals (copper, iron, zinc), diamond, sulphur, mica, quartz and sand.

Crystalline solids have a sharp melting point, they are anisotropic (their properties depend on direction), and on cutting they give sharp cleaves.

Polycrystalline solids

A polycrystalline solid is made of many small crystals, called crystallites or grains, with random orientations. Neighbouring grains meet at grain boundaries. Its structure lies between order and disorder, so it is partly or poorly crystalline. Grains are typically 10^{2} to 10^{3}\,\mathrm{nm} in size. Polycrystalline solids are isotropic, showing the same properties in all directions. Most inorganic metals, many ceramics, rocks and ice are polycrystalline.

Amorphous solids

An amorphous solid (also called a glassy solid) has its particles arranged in a random manner. It has no regular geometrical shape, but it has a fixed volume. It behaves like a frozen liquid: it has short-range order only, and it has no sharp melting point, softening gradually instead. Glass, rubber, glues, thin films deposited at low temperature and many polymers are amorphous. Amorphous solids are isotropic.

Figure 7.4: The crystalline, polycrystalline and amorphous solids.

Comparison of the three types of solids.
Property	Crystalline	Polycrystalline	Amorphous
Arrangement	Regular, repeating over the whole solid	Regular within each grain, grains oriented at random	Random, short-range order only
Melting	Sharp melting point	Sharp melting point	Softens over a range of temperature
Properties	Anisotropic	Isotropic	Isotropic
Examples	Salt, diamond, quartz, mica	Most metals, ceramics, rocks, ice	Glass, rubber, glues, polymers

Note. The book calls polycrystalline solids 'polymeric' in section 7.1.2. Polymers are a different class of materials (long chains of molecules), so the word is probably a misprint for 'poly-crystalline'.

22026 Annual I, Set D, Q.2 (xi)

What is unit cell? Explain its types.

Solution

Unit cell

A unit cell is the smallest basic portion of a crystal which, repeated in three dimensions, builds the entire crystal structure. It has all the characteristics of the whole crystal. The atoms, molecules or ions that make up the crystal are called the basis. The regular arrangement of points that joins the basis is the lattice, and placing the atoms on the lattice gives the crystal. Repeating the unit cell gives the crystal lattice (Figure 7.5(a)).

Types of unit cell

There are three types of cubic unit cell, shown in Figure 7.5(b).

Figure 7.5: Simple cubic (SC), body-centred cubic (BCC) and face-centred cubic (FCC) unit cells.

Simple cubic (SC): eight atoms, one at each corner. Each corner atom is shared by eight cells, so the cell holds 8\times\tfrac18=1 atom.
Body-centred cubic (BCC): eight corner atoms and one atom at the centre of the body. The corners give 8\times\tfrac18=1 atom and the centre gives 1, so there are 2 atoms per cell.
Face-centred cubic (FCC): eight corner atoms and one atom at each face. The corners give 1 atom and each face atom is shared by two cells, so the six faces give 6\times\tfrac12=3. There are 4 atoms per cell.

Atoms per unit cell

Count the atoms in SC: 8 corners, each shared by 8 cells.
N_{SC}=8\times\tfrac18=1
Count the atoms in BCC: corners, plus 1 at the centre which belongs to this cell alone.
N_{BCC}=8\times\tfrac18+1=2
Count the atoms in FCC: corners, plus 6 faces each shared by 2 cells.
N_{FCC}=8\times\tfrac18+6\times\tfrac12=4

The book also describes the unit cell by its three sides a, b, c and three angles \alpha, \beta, \gamma. These give seven distinct shapes of unit cell (simple cubic, tetragonal, orthorhombic, rhombohedral, monoclinic, triclinic and hexagonal, Figure 7.6). The cubic cells above are the three types that are needed in the atom counting.

32025 Annual I, Set D, Q.2 (ix) OR

Explain the terms: stress, strain and modulus of elasticity.

Solution

Stress

The stress \delta is the force applied per unit area of a body, so the internal resistance of the body balances the applied force:

\delta=\dfrac{F}{A}

Its SI unit is \mathrm{N\,m^{-2}}, also called the pascal (Pa), and its dimensions are [ML^{-1}T^{-2}]. There are three types of stress:

Tensile stress: a force acting at right angles to a cross-section that stretches the body and changes its length.
Shear stress: a force with a component acting tangentially to the area, which changes the shape of the body.
Volume stress: a force acting on all faces of the body, which changes its volume.

Strain

The strain is the quantitative measure of the deformation: the ratio of the change in a parameter (length, shape or volume) to its original value. It has no units. The three types are:

\varepsilon_{\text{tensile}}=\dfrac{\Delta L}{L},\qquad \varepsilon_{\text{shear}}=\dfrac{\Delta X}{Y},\qquad \varepsilon_{\text{volume}}=\dfrac{\Delta V}{V}

Modulus of elasticity

For an elastic solid the ratio of stress to strain is constant. This constant is the modulus of elasticity:

\text{modulus of elasticity}=\dfrac{\text{stress}}{\text{strain}}

Since strain has no units, the modulus has the units of stress, \mathrm{N\,m^{-2}}. Depending on the type of stress there are three moduli:

Young's modulus Y=\dfrac{F/A}{\Delta L/L} for linear deformation (equation 7.1).
Bulk modulus B=\dfrac{F/A}{-\Delta V/V} for volume deformation (equation 7.2). The minus sign shows the decrease in volume, so B is always positive.
Shear modulus S=\dfrac{F/A}{\Delta X/Y} (equation 7.3). For a small angle of shear \theta, \tan\theta\approx\theta=\dfrac{\Delta X}{Y}, so S=\dfrac{F}{A\theta} (equation 7.4).

Table 7.1: Modulus of elasticity of different materials (book values).
Material	Young's modulus (10^{10}\,\mathrm{N\,m^{-2}})	Bulk modulus (10^{10}\,\mathrm{N\,m^{-2}})	Shear modulus (10^{10}\,\mathrm{N\,m^{-2}})
Steel	20.0	15.8	8.0
Aluminium	7.0	7.0	2.5
Copper	12.0	12.0	4.0
Iron	19.0	8.0	5.0
Brass	1.0	14.0	3.6

Note. The book\'s Table 7.1 prints brass Young\'s modulus as 1.0\times10^{10}\,\mathrm{N\,m^{-2}}. This is a misprint for 10\times10^{10}\,\mathrm{N\,m^{-2}}, the real value of brass (the value used in MCQ 6 and in the book\'s answer to numerical 7.3).

Draw and discuss the stress-strain curve for a ductile material.

Solution

A tensile test machine stretches a specimen at a fixed linear rate and records the stress. A computer plots the strain on the horizontal axis and the stress on the vertical axis. The graph below is the typical curve for a ductile material such as copper, lead or wrought iron.

Figure 7.12: Stress-strain curve for a ductile material. A proportional limit, B elastic limit (yield point), C lower yield point, D ultimate tensile strength, E fracture point.

Discussion of the curve

OA (proportional limit): stress is proportional to strain, so the curve is a straight line and Hooke's law holds. The slope of this line is the Young's modulus. If the stress is removed, the body regains its original shape.
OB (elastic region): the body returns to its original shape when the stress is removed. The point B is the elastic limit or yield point, the maximum stress for which the material is elastic.
BCD (plastic region): beyond B, part of the deformation is permanent. Some materials show a lower yield point at C. The stress then rises to D, the ultimate tensile strength, the largest stress the material can withstand.
Fracture at E: after D the body breaks at the fracture point E.

Materials that undergo plastic deformation before they break are ductile, for example lead, copper and wrought iron. Materials that break soon after the elastic limit, such as glass and high carbon steel, are brittle. In the elastic region the strain is small: the permanent set at the end of the elastic region is less than 1\,\%.

52026 Annual I, Set B, Q.2 (ii) OR

Explain plastic and elastic deformation by drawing stress-strain curve.

Solution

Elastic deformation

A body is elastically deformed if it regains its original shape, length or volume when the stress is removed. In the region OA of the stress-strain curve the stress is proportional to the strain (Hooke's law), and up to the elastic limit B the body returns completely to its original size.

Plastic deformation

If the stress is increased beyond the elastic limit B, the body does not regain its original shape when the stress is removed. Some of the deformation is permanent. This is plastic deformation (plasticity), and it occurs in the region after B. The strain does not return to zero, even though the stress is zero.

Showing both on the curve

Load the specimen up to a point P beyond B, then unload it. The unloading path is a straight line parallel to the initial straight line OA, because the modulus of elasticity is the same. The unloading line meets the strain axis at R, so the strain OR remains after the stress has been removed. This is the permanent (plastic) strain. The strain OP-OR is recovered elastically.

Loading to P beyond the elastic limit, then unloading. The unloading line is parallel to OA, and OR is the permanent strain.

So the elastic part of the deformation disappears when the stress is removed, while the plastic part remains.

Justify that 'the area under the force-extension curve is equal to the work'.

Solution

Statement

The work done in stretching a wire is the area under its force-extension graph. We justify this for a wire loaded slowly, within its elastic limit.

Derivation Work done as the area under the force-extension graph

Aim. Show that the work done in extending a wire by x is equal to the area under its force-extension graph.

Assumptions.

The force increases steadily from zero to F as the extension grows from zero to x, so no kinetic energy is produced.
The wire is within its limit of proportionality, so the graph is a straight line through the origin.

Force-extension graph of a wire within its elastic limit. The shaded triangle is the work done in extending it by x.

Work is force times displacement. When the force changes, the work done in a small extension \Delta x at force F is F\,\Delta x. This is the area of a thin strip under the graph.
\Delta W=F\,\Delta x
Adding up all the strips from zero to x gives the total area under the graph. Because the graph is a straight line through the origin, this area is a triangle with base x and height F.
\text{Area}=\dfrac12\times\text{base}\times\text{height}=\dfrac12\,x\,F
The same answer follows from the average force. The force rises uniformly from 0 to F, so its average value is \dfrac{0+F}{2}, and the work is the average force times the displacement.
W=F_{\text{avg}}\,x=\dfrac{0+F}{2}\,x=\dfrac12Fx

ResultW=\dfrac12Fx=\text{area of the triangle under the force-extension graph}

The area under the graph equals the work done, and this work is stored in the wire as elastic potential energy (section 7.3.3).

72026 Annual I, Set B, Q.2 (vi)2026 Annual I, Set D, Q.4 OR2025 Annual I, Set B, Q.5 OR

Derive a relation for the elastic potential energy of a material.

Solution

Derivation

Derivation Elastic potential energy of a stretched wire

Aim. Derive U=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2} for the energy stored in a wire of length L and area A stretched by x.

Assumptions.

The wire is stretched slowly, so no kinetic energy is produced and the stored energy equals the work done.
Hooke's law holds within the elastic limit, F=kx.
The wire is uniform, with original length L and area of cross-section A.

Force-extension graph of a wire within its elastic limit. The shaded triangle is the work done in extending it by x.

The energy stored equals the work done in extending the wire. This work is the area under the force-extension graph (Q.6), which is a triangle.
U=W=\tfrac12Fx\qquad(7.5)
Hooke's law gives the force at extension x as F=kx. Substituting for F gives the energy in terms of k.
U=\tfrac12(kx)x=\tfrac12kx^{2}\qquad(7.6)
Express the force in terms of the material. Young's modulus (equation 7.1) gives \dfrac{F/A}{x/L}=Y, so F=\dfrac{YA}{L}x and the stiffness is k=\dfrac{YA}{L}.
F=\dfrac{YAx}{L}
Substitute this force into U=\tfrac12Fx.
U=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}\qquad(7.7)

ResultU=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2}

The energy depends on the square of the extension, so doubling the extension gives four times the energy. This energy is also called the strain energy, and it is represented by the area under the stress-strain curve.

Describe an experiment to determine the Young's Modulus of metallic wire.

Solution

Aim

To determine the Young's modulus Y of a metallic wire from the force, the length, the area of cross-section and the extension of the wire.

Apparatus

A long test wire of the metal, hung from a fixed support, with a vernier scale attached to it.
A reference wire hung beside it, with a scale on it. The vernier reading on the test wire is compared with this scale for a precise extension.
Slotted weights for the load, a metre rule to measure the length, and a micrometer screw gauge to measure the diameter.

Figure 7.13: Young's modulus for a metallic wire.

Procedure

Measure the original length L of the test wire with the metre rule, from the fixed support to the point where the load is attached.
Measure the diameter d at several places along the wire with the micrometer screw gauge. Take the average and calculate the area A=\dfrac{\pi d^{2}}{4}.
Set the vernier reading to zero with no load on the test wire.
Add a load of mass m. The force is F=mg. Record the extension \Delta L shown by the vernier.
Increase the load in equal steps, recording F and \Delta L for each. Keep the load within the elastic limit, and check that the wire returns to zero extension when the load is removed.
For each pair of readings calculate Y=\dfrac{FL}{A\,\Delta L}, and take the average.

Result

Each reading gives a value of Young's modulus from equation 7.1:

Y=\dfrac{F}{A}\cdot\dfrac{L}{\Delta L}=\dfrac{FL}{A\,\Delta L}

If \Delta L is plotted against F, the points lie on a straight line through the origin whose slope is \dfrac{L}{YA}. Then Y=\dfrac{L}{A\times\text{slope}}, which gives the same value as the average of the readings.

Precautions

The wire must hang vertically and be free of kinks.
Read the vernier and the scale at eye level, to avoid a parallax error.
Use loads only within the elastic limit, so that the extension is elastic and the readings are reliable.

Numerical problemsp. 126

Calculate the mass of picture frame hanging from a steel nail (Y_S=20\times10^{10}\,\mathrm{Pa}), which is sufficient to produce a force providing shear strain, as shown in figure here. Given that the nail bends only 1.80\,\mu\mathrm{m}. Also discuss the result.

Solution

Given

Y_S = 20\times10^{10}\,\mathrm{Pa}
d = 1.50\,\mathrm{mm}\ \Rightarrow\ r=0.75\,\mathrm{mm}
\Delta x = 1.80\,\mu\mathrm{m}
Y = 5.00\,\mathrm{mm}\ \text{(distance from the wall to the frame)}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

m, the mass of the frame

The shear strain is the displacement of the sheared face divided by the distance from the fixed face.
\theta=\dfrac{\Delta x}{Y}=\dfrac{1.80\times10^{-6}}{5.00\times10^{-3}}=3.6\times10^{-4}
The shear stress is the force divided by the area of the nail's cross-section. Use S=\dfrac{F/A}{\theta} and make F the subject.
F=S\,A\,\theta
Area of the cross-section of the nail, with radius r=0.75\,\mathrm{mm}.
A=\pi r^{2}=\pi(0.75\times10^{-3})^{2}=1.767\times10^{-6}\,\mathrm{m^{2}}
Substitute. The force is the weight of the frame, so this is the force on the nail.
F=(20\times10^{10})(1.767\times10^{-6})(3.6\times10^{-4})=127\,\mathrm{N}
The weight of the frame is mg=F.
m=\dfrac{F}{g}=\dfrac{127}{9.8}

m\approx13\,\mathrm{kg}

Book answer. 5.2\,\mathrm{kg}

Note. The book's answer of 5.2\,\mathrm{kg} does not follow from its own data. With the printed diameter 1.50\,\mathrm{mm} the force is 127\,\mathrm{N} and the mass is about 13\,\mathrm{kg}. A mass of 5.2\,\mathrm{kg} needs a force of about 51\,\mathrm{N}, and taking 1.50\,\mathrm{mm} as the radius gives about 52\,\mathrm{kg}. The book's answer is probably a misprint. The discussion in the book (a fairly massive picture, a bend of 1.80\,\mu\mathrm{m} shows the strength of steel) still holds.

A 5\,\mathrm{m} long aluminum wire (Y_A=7\times10^{10}\,\mathrm{Pa}) of diameter 3\,\mathrm{mm}, supports a 40\,\mathrm{kg} mass. To achieve same elongation in a copper wire (Y_C=12\times10^{10}\,\mathrm{Pa}) of the same length under the same weight, find the diameter of copper wire.

Solution

Given

L = 5\,\mathrm{m}\ \text{(both wires)}
Y_A = 7\times10^{10}\,\mathrm{Pa}
d_A = 3\,\mathrm{mm}
m = 40\,\mathrm{kg}\ \text{(same for both)}
Y_C = 12\times10^{10}\,\mathrm{Pa}

To find

The two wires carry the same force F, have the same length L and must stretch by the same \Delta L. Use \Delta L=\dfrac{FL}{YA} for each wire.
\dfrac{FL}{Y_AA_A}=\dfrac{FL}{Y_CA_C}
F and L cancel, so the product YA must be the same. Solve for the area of the copper wire.
A_C=A_A\,\dfrac{Y_A}{Y_C}
Since A=\dfrac{\pi d^{2}}{4}, the diameters are related by a square root.
d_C=d_A\sqrt{\dfrac{Y_A}{Y_C}}=3\sqrt{\dfrac{7}{12}}
Evaluate.
d_C=3\times0.764=2.29\,\mathrm{mm}

d_C\approx2.29\,\mathrm{mm}

Book answer. 2.29\,\mathrm{mm}

Note. The mass and the length cancel, so the answer does not depend on them. The copper wire is thinner than the aluminium wire because copper is stiffer.

A 4\,\mathrm{kg} mass is hanging with the help of a 1.5\,\mathrm{m} long steel wire (Y_S=20\times10^{10}\,\mathrm{Pa}) another mass of 6\,\mathrm{kg} is hanging from the 4\,\mathrm{kg} mass using 1\,\mathrm{m} long brass wire (Y_B=1.0\times10^{10}\,\mathrm{Pa}). Both wires have diameter 0.25\,\mathrm{cm}. Calculate the total elongation in the whole system of wires, also calculate the total strain energy.

Solution

Given

L_S = 1.5\,\mathrm{m}
Y_S = 20\times10^{10}\,\mathrm{Pa}
L_B = 1\,\mathrm{m}
Y_B = 10\times10^{10}\,\mathrm{Pa}
d = 0.25\,\mathrm{cm}=2.5\,\mathrm{mm}
m_1 = 4\,\mathrm{kg}\ \text{(hung from the steel wire)}
m_2 = 6\,\mathrm{kg}\ \text{(hung from the 4 kg mass)}
g = 9.8\,\mathrm{m\,s^{-2}}

To find

total extension \Delta L
total strain energy U

Area of each wire, with radius 1.25\,\mathrm{mm}.
A=\pi\left(\dfrac{d}{2}\right)^{2}=\pi(1.25\times10^{-3})^{2}=4.91\times10^{-6}\,\mathrm{m^{2}}
The steel wire holds both masses, 4+6=10\,\mathrm{kg}. The brass wire holds only the 6\,\mathrm{kg} below it.
F_S=(10)(9.8)=98\,\mathrm{N},\qquad F_B=(6)(9.8)=58.8\,\mathrm{N}
Extension of the steel wire.
\Delta L_S=\dfrac{F_SL_S}{Y_SA}=\dfrac{98\times1.5}{20\times10^{10}\times4.91\times10^{-6}}=1.50\times10^{-4}\,\mathrm{m}
Extension of the brass wire.
\Delta L_B=\dfrac{F_BL_B}{Y_BA}=\dfrac{58.8\times1}{10\times10^{10}\times4.91\times10^{-6}}=1.20\times10^{-4}\,\mathrm{m}
Total extension of the system.
\Delta L=\Delta L_S+\Delta L_B=1.50\times10^{-4}+1.20\times10^{-4}=2.70\times10^{-4}\,\mathrm{m}
Strain energy of each wire is \tfrac12F\,\Delta L, and the total is the sum.
U=\tfrac12(98)(1.50\times10^{-4})+\tfrac12(58.8)(1.20\times10^{-4})=7.3\times10^{-3}+3.5\times10^{-3}
Evaluate.
U\approx1.09\times10^{-2}\,\mathrm{J}\approx10.9\,\mathrm{mJ}

\Delta L\approx2.70\times10^{-4}\,\mathrm{m},\qquad U\approx10.9\,\mathrm{mJ}

Book answer. 2.79\times10^{-4}\,\mathrm{m}, 11.02\,\mathrm{mJ}

Note. The printed brass modulus 1.0\times10^{10}\,\mathrm{Pa} is a misprint for 10\times10^{10}\,\mathrm{Pa}, the real value of brass and the value used in MCQ 6. With this value the answer above matches the book, so the book's answer follows from the correct value. With g=10 the extension is 2.75\times10^{-4}\,\mathrm{m}. The steel wire carries 10 kg and the brass wire 6 kg.

A rod with a cross-sectional area of 90\,\mathrm{mm^{2}} and a length of 3\,\mathrm{m}. If a stress of 300\,\mathrm{MPa} is applied to stretch the rod, then find the strain energy if Young's modulus of the rod is 200\,\mathrm{GPa}.

Solution

Given

A = 90\,\mathrm{mm^{2}}=9\times10^{-5}\,\mathrm{m^{2}}
L = 3\,\mathrm{m}
\delta = 300\,\mathrm{MPa}=3\times10^{8}\,\mathrm{Pa}
Y = 200\,\mathrm{GPa}=2\times10^{11}\,\mathrm{Pa}

To find

strain energy U

Strain from Y=\dfrac{\delta}{\varepsilon}.
\varepsilon=\dfrac{\delta}{Y}=\dfrac{3\times10^{8}}{2\times10^{11}}=1.5\times10^{-3}
Extension of the rod.
\Delta L=\varepsilon L=1.5\times10^{-3}\times3=4.5\times10^{-3}\,\mathrm{m}
Force on the rod.
F=\delta A=3\times10^{8}\times9\times10^{-5}=2.7\times10^{4}\,\mathrm{N}
Strain energy is the area under the force-extension graph, U=\tfrac12F\,\Delta L.
U=\tfrac12(2.7\times10^{4})(4.5\times10^{-3})=60.75\,\mathrm{J}

U\approx61\,\mathrm{J}\ (60.75\,\mathrm{J})

Book answer. 12.15\,\mathrm{J}

Note. The book's answer, 12.15\,\mathrm{J}, is one fifth of the correct value. Check with the volume form: U=\tfrac12\delta\varepsilon\,V=\tfrac12(3\times10^{8})(1.5\times10^{-3})(2.7\times10^{-4})=60.75\,\mathrm{J}, with V=AL=2.7\times10^{-4}\,\mathrm{m^{3}}.

A force of 500\,\mathrm{N} is applied to one end of a cylindrical steel rod of diameter 50\,\mathrm{cm}. What is the tensile stress?

Solution

Given

F = 500\,\mathrm{N}
d = 50\,\mathrm{cm}=0.50\,\mathrm{m}

To find

tensile stress \delta

Radius of the rod.
r=\dfrac{d}{2}=0.25\,\mathrm{m}
Area of the cross-section.
A=\pi r^{2}=\pi(0.25)^{2}=0.196\,\mathrm{m^{2}}
Tensile stress is the force divided by the area.
\delta=\dfrac{F}{A}=\dfrac{500}{0.196}=2.55\times10^{3}\,\mathrm{N\,m^{-2}}

\delta\approx2.5\times10^{3}\,\mathrm{N\,m^{-2}}

Book answer. 2.5\times10^{5}\,\mathrm{N\,m^{-2}}

Note. The book's answer is 100 times larger. It is what a diameter of 5\,\mathrm{cm} gives: A=1.96\times10^{-3}\,\mathrm{m^{2}} and \delta=2.5\times10^{5}\,\mathrm{N\,m^{-2}}. The printed '50 cm' is therefore probably a misprint for 5\,\mathrm{cm}. The answer above uses the diameter as printed.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 51 mark

Which of the following is a correct way to represent Young's Modulus?

A\dfrac{F/A}{\Delta L/L}Answer
B\dfrac{F.A}{\Delta L}
C\dfrac{F}{A.\Delta L}
D\dfrac{F.\Delta L}{L.A}

Why

Young's modulus is the ratio of the tensile stress to the tensile strain, Y=\dfrac{F/A}{\Delta L/L} (equation 7.1), so option A is correct.

Option D, \dfrac{F\,\Delta L}{L\,A}=\dfrac{F}{A}\times\dfrac{\Delta L}{L}, is the stress multiplied by the strain, not divided by it. Options B and C do not have the form stress over strain.

2025 Annual I, Set B, MCQ 81 mark

Young's Modulus, Y=

A\dfrac{\Delta L/L}{F/A}
B\dfrac{FA}{\Delta L\cdot L}
C\dfrac{\Delta L\cdot L}{FA}
D\dfrac{F/A}{\Delta L/L}Answer

Why

Young's modulus is the ratio of the tensile stress to the tensile strain: Y=\dfrac{F/A}{\Delta L/L}, which is option D.

Option A, \dfrac{\Delta L/L}{F/A}, is the reciprocal: strain over stress. Options B and C are products of F and A or of \Delta L and L, not a ratio of stress to strain.

2026 Annual I, Set B, Q.2 (ii) OR3 marks

Distinguish between the terms elastic deformation and plastic deformation.

Solution

Both terms describe how a solid responds to a force. The difference is whether the body recovers its shape when the force is removed.

Point	Elastic deformation	Plastic deformation
Meaning	The body regains its original shape, length or volume when the stress is removed.	The body is permanently changed and does not regain its original shape when the stress is removed.
Region of the stress-strain curve	From O to the elastic limit (yield point) B; the part OA also obeys Hooke's law.	Beyond the yield point B, up to the fracture point E.
Strain when the stress is removed	The strain returns to zero.	The strain is not zero even though the stress is zero (permanent set).

2025 Annual I, Set B, Q.2 (iv)3 marks

Differentiate between crystalline solids and amorphous solids. (Give at least three differences.)

Solution

Solids are crystalline or amorphous according to how orderly their particles are arranged.

Any three of these differences earn the 3 marks.
Point	Crystalline solids	Amorphous solids
Arrangement of particles	Regular and well-defined arrangement of atoms, ions or molecules.	Particles arranged in a random manner (a frozen liquid).
Melting point	Sharp melting point.	No sharp melting point; they soften gradually over a range of temperature.
Cutting or breaking	Give sharp cleaves (flat, clean faces).	Give odd, often twisted surfaces.
Properties against direction	Anisotropic: a measured property depends on the direction.	No regular structure to give a marked direction-dependence.
Examples	Sodium chloride, copper, iron, diamond, quartz.	Glass, rubber, glues, many polymers.

2026 Annual I, Set B, Q.2 (vi)3 marks

Derive the relationship P.E = \dfrac{1}{2}\left[\dfrac{YA}{L}\right]x^2

Solution

The elastic potential energy of a stretched wire equals the work done in stretching it. The work is found from the force-extension graph (the area of the triangle), and Young's modulus is then used to write the force in terms of the material.

Derivation Elastic potential energy of a stretched wire

Aim. Derive U=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2} for the energy stored in a wire of length L and area A stretched by x.

Assumptions.

The wire is stretched slowly, so no kinetic energy is produced and the stored energy equals the work done.
Hooke's law holds within the elastic limit, F=kx.
The wire is uniform, with original length L and area of cross-section A.

Force-extension graph of a wire within its elastic limit. The shaded triangle is the work done in extending it by x.

The energy stored equals the work done in extending the wire. This work is the area under the force-extension graph (section 7.3.2), which is a triangle.
U=W=\tfrac12Fx\qquad(7.5)
Hooke's law gives the force at extension x as F=kx. Substituting for F gives the energy in terms of k.
U=\tfrac12(kx)x=\tfrac12kx^{2}\qquad(7.6)
Express the force in terms of the material. Young's modulus (equation 7.1) gives \dfrac{F/A}{x/L}=Y, so F=\dfrac{YA}{L}x and the stiffness is k=\dfrac{YA}{L}.
F=\dfrac{YAx}{L}
Substitute this force into U=\tfrac12Fx.
U=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}\qquad(7.7)

ResultU=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2}

2025 Annual I, Set B, Q.5 OR1+6 = 7 marks

What is meant by elastic potential energy? Find a relation for the elastic potential energy of a material.

Solution

Elastic potential energy (1 mark)

The elastic potential energy (strain energy) is the energy stored in a material due to the strain produced in it. For a material deformed within its elastic limit it is equal to the work done in producing the deformation.

Relation for the elastic potential energy (6 marks)

The work done in stretching a wire by an extension x is the area under its force-extension graph. Hooke's law and Young's modulus then give the energy in terms of the material.

Derivation Elastic potential energy of a stretched wire

Aim. Derive U=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2} for the energy stored in a wire of length L and area A stretched by x.

Assumptions.

The wire is stretched slowly, so no kinetic energy is produced and the stored energy equals the work done.
Hooke's law holds within the elastic limit, F=kx.
The wire is uniform, with original length L and area of cross-section A.

Force-extension graph of a wire within its elastic limit. The shaded triangle is the work done in extending it by x.

The energy stored equals the work done in extending the wire. This work is the area under the force-extension graph (section 7.3.2), which is a triangle.
U=W=\tfrac12Fx\qquad(7.5)
Hooke's law gives the force at extension x as F=kx. Substituting for F gives the energy in terms of k.
U=\tfrac12(kx)x=\tfrac12kx^{2}\qquad(7.6)
Express the force in terms of the material. Young's modulus (equation 7.1) gives \dfrac{F/A}{x/L}=Y, so F=\dfrac{YA}{L}x and the stiffness is k=\dfrac{YA}{L}.
F=\dfrac{YAx}{L}
Substitute this force into U=\tfrac12Fx.
U=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}\qquad(7.7)

ResultU=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2}

2026 Annual I, Set D, MCQ 61 mark

Which of the following is a correct way to represent Young's Modulus?

A\dfrac{F.\Delta L}{L.A}
B\dfrac{F/A}{\Delta L/L}Answer
C\dfrac{F.A}{\Delta L}
D\dfrac{F}{A.\Delta L}

Why

Young's modulus is Y=\dfrac{\text{tensile stress}}{\text{tensile strain}}=\dfrac{F/A}{\Delta L/L}, which is option B.

Option A, \dfrac{F\,\Delta L}{L\,A}, equals stress times strain instead of stress divided by strain. Options C and D do not have the form stress over strain.

2025 Annual I, Set D, MCQ 81 mark

Glass is an example of ______ solid.

ACrystalline
BAmorphousAnswer
CPoly-crystalline
DDuctile

Why

Glass is an amorphous solid. Its particles have no long-range regular arrangement and it softens over a range of temperature instead of melting sharply (the book calls glass a super-cooled liquid).

Crystalline (A) and poly-crystalline (C) solids have fully or partly ordered structures. Ductile (D) describes how a solid deforms under stress, not its structure.

2025 Annual I, Set D, MCQ 91 mark

Young's Modulus, Y=

A\dfrac{\Delta L/L}{F/A}
B\dfrac{FA}{\Delta L\cdot L}
C\dfrac{\Delta L\cdot L}{FA}
D\dfrac{F/A}{\Delta L/L}Answer

Why

Young's modulus is the ratio of tensile stress to tensile strain: Y=\dfrac{F/A}{\Delta L/L}, option D.

Option A is the reciprocal of this (strain over stress). Options B and C are not ratios of stress to strain.

2026 Annual I, Set D, Q.2 (ii)1x3 = 3 marks

Give Comparison of amorphous and polymeric solids. (Three points)

Solution

The book calls polymeric solids polycrystalline solids: their structure lies between order and disorder, while an amorphous solid has no regular structure at all.

Point	Amorphous solids	Polymeric (polycrystalline) solids
Structure	The particles are arranged in a random manner, like a frozen liquid; only short-range order is present.	Made of many small crystals (grains) in random orientation, separated by grain boundaries; partly ordered. Polymers are long chains made by the repetition of small molecules (polymerisation).
Behaviour	No sharp melting point: a glass rod softens gradually over a range of temperature. No definite regular geometric shape, and broken surfaces are irregular.	Isotropic, because the grains point in all directions. They have low specific gravity, a good strength-to-weight ratio, toughness and resistance to corrosion.
Examples	Glass, rubber, glues.	Natural polymers (rubber, resin, wood) and synthetic polymers (polythene, polystyrene, polyvinyl chloride).

2026 Annual I, Set D, Q.2 (vi)1.5+1.5 = 3 marks

Briefly explain brittle and ductile materials. (In context of elastic and plastic deformation)

Solution

Both words describe what happens to a solid as the stress is raised beyond its elastic limit. They are compared on the stress-strain curve.

Ductile materials (1.5 marks). A ductile material undergoes plastic deformation until it breaks. After the elastic limit B it has a long plastic region (B to D to E), so it can be drawn into thin wires without breaking and it gives a warning before fracture. Examples: lead, copper and wrought iron.

Figure 7.12: Stress-strain curve for a ductile material. A proportional limit, B elastic limit (yield point), C lower yield point, D ultimate tensile strength, E fracture point.

Brittle materials (1.5 marks). A brittle material breaks just after the elastic limit, with almost no plastic region between the yield point and the fracture point. Its fracture occurs suddenly, with little permanent deformation. Examples: glass and high carbon steel.

2025 Annual I, Set D, Q.2 (ix) OR3 marks

Briefly describe the terms, stress, strain and Young's modulus.

Solution

Stress (1 mark). Stress is the force applied per unit area of cross-section, \delta=\dfrac{F}{A}, the internal resistance a body offers to deformation. Its SI unit is the pascal (\mathrm{N\,m^{-2}}). It is tensile, shear or volume stress.

Strain (1 mark). Strain is the quantitative measure of deformation: the ratio of the change in a parameter to its original value, for example tensile strain \varepsilon=\dfrac{\Delta L}{L}. It has no units.

Young's modulus (1 mark). For linear deformation it is the ratio of the tensile stress to the tensile strain; it is constant within the elastic limit:

Y=\dfrac{F/A}{\Delta L/L}=\dfrac{FL}{A\,\Delta L}

Its unit is the pascal, the same as that of stress, since strain has no unit.

2025 Annual I, Set D, Q.2 (x)3 marks

Why are the suspension bridges written with their period of use? Why after that period it is dangerous to use the bridge? (Elastic potential)

Solution

Why a period of use is written (about 1.5 marks). The wires of a suspension bridge carry heavy loads of traffic and wind again and again. Within the elastic limit a stretched wire stores strain energy U=\tfrac12kx^{2} and returns to its length when the load goes. After years of repeated loading the metal suffers fatigue: tiny cracks grow and the wire is permanently strained, so its elastic limit and strength fall. A period of use is therefore assigned, after which the cables are inspected and replaced.

Why it is dangerous afterwards (about 1.5 marks). Beyond that period the wires may already be in the plastic region, with a permanent set, and they can store less strain energy. A load that was safe before may now stretch them past the elastic limit, or even break them, so the bridge may collapse. It is therefore dangerous to use the bridge after that period.

2026 Annual I, Set D, Q.2 (xi)2+1 = 3 marks

Briefly describe: a. Unit cell and its types b. Crystal lattice

Solution

(a) Unit cell and its types (2 marks). A unit cell is the smallest basic portion of a crystal which, repeatedly stacked together in three dimensions, makes the entire crystal structure. It has all the characteristics of the whole crystal. There are three types of cubic unit cell (Figure 7.5 (b)):

Figure 7.5: Simple cubic (SC), body-centred cubic (BCC) and face-centred cubic (FCC) unit cells.

Simple cubic (SC): eight atoms, one at each corner. Each corner atom is shared by eight cells, so the cell holds 8\times\tfrac18=1 atom.
Body-centred cubic (BCC): eight corner atoms and one atom at the centre of the body: 8\times\tfrac18+1=2 atoms.
Face-centred cubic (FCC): eight corner atoms and one atom at the centre of each face. Each face atom is shared by two cells: 8\times\tfrac18+6\times\tfrac12=4 atoms.

(b) Crystal lattice (1 mark). The crystal lattice is the structure of a crystal obtained by the repetition of the unit cell in all three directions. It is the regular geometrical arrangement of the points (basis: the atoms, molecules or ions) of the crystal.

2026 Annual I, Set D, Q.4 OR2+4 = 6 marks

Briefly explain Elastic Potential Energy. Derive its mathematical formula.

Solution

Elastic potential energy (2 marks)

Derivation of the formula (4 marks)

The energy stored equals the work done in stretching the wire, and this work is the area under the force-extension graph.

Derivation Elastic potential energy of a stretched wire

Aim. Derive U=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2} for the energy stored in a wire of length L and area A stretched by x.

Assumptions.

The wire is stretched slowly, so no kinetic energy is produced and the stored energy equals the work done.
Hooke's law holds within the elastic limit, F=kx.
The wire is uniform, with original length L and area of cross-section A.

Force-extension graph of a wire within its elastic limit. The shaded triangle is the work done in extending it by x.

The energy stored equals the work done in extending the wire. This work is the area under the force-extension graph (section 7.3.2), which is a triangle.
U=W=\tfrac12Fx\qquad(7.5)
Hooke's law gives the force at extension x as F=kx. Substituting for F gives the energy in terms of k.
U=\tfrac12(kx)x=\tfrac12kx^{2}\qquad(7.6)
Express the force in terms of the material. Young's modulus (equation 7.1) gives \dfrac{F/A}{x/L}=Y, so F=\dfrac{YA}{L}x and the stiffness is k=\dfrac{YA}{L}.
F=\dfrac{YAx}{L}
Substitute this force into U=\tfrac12Fx.
U=\dfrac12\left[\dfrac{YA}{L}\right]x^{2}\qquad(7.7)

ResultU=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2}

Exam focus

Derive the elastic potential energy of a stretched wire, U=\tfrac12Fx=\tfrac12kx^{2}=\tfrac12\dfrac{YA}{L}x^{2}, using the force-extension graph. (long)
The most repeated question of the unit: 2026 Set B Q.2 (vi) (3 marks), 2026 Set D Q.4 OR (6 marks) and 2025 Set B Q.5 OR (7 marks). Same as comprehensive question 7.
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Young's modulus as a ratio: Y=\dfrac{F/A}{\Delta L/L} (stress over strain). (mcq)
Asked as an MCQ in all four new-syllabus papers: 2026 Set B MCQ 5, 2026 Set D MCQ 6, 2025 Set B MCQ 8 and 2025 Set D MCQ 9. The wrong options are the reciprocal or stress times strain.
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Crystalline, polymeric (polycrystalline) and amorphous solids: differences with examples. (short)
2026 Set D Q.2 (ii) (amorphous and polymeric), 2025 Set B Q.2 (iv) (crystalline and amorphous, three differences) and 2025 Set D MCQ 8 (glass). Learn the comparison table.
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Elastic and plastic deformation, ductile and brittle materials, with the stress-strain curve. (long)
2026 Set B Q.2 (ii) OR (elastic and plastic deformation) and 2026 Set D Q.2 (vi) (brittle and ductile). Comprehensive questions 4 and 5 give the full curve with points O, A, B, C, D, E.
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Define stress, strain and Young's modulus. (short)
2025 Set D Q.2 (ix) OR (3 marks); same as comprehensive question 3. Give formula, unit and dimensions of each.
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Unit cell and its types (SC, BCC, FCC) and the crystal lattice. (short)
2026 Set D Q.2 (xi) (2+1 marks); comprehensive question 2 and short question 11. Remember 1, 2 and 4 atoms per cell.
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Why are suspension bridges assigned a period of use? Is it dangerous to use the bridge after it? (short)
2025 Set D Q.2 (x), the same as short question 8. Link it to fatigue and the loss of the elastic limit.
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Justify that the area under the force-extension graph is the work done, W=\tfrac12Fx. (long)
Not asked as a separate question in the 2025 or 2026 papers, but it is the first step of every elastic potential energy derivation that was asked, and comprehensive question 6.
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Numericals on Young's modulus and strain energy: \Delta L=\dfrac{FL}{YA} and U=\tfrac12F\,\Delta L. (numerical)
No numerical from this unit was in the 2025 or 2026 papers, but the book's numericals and Example 7.3 use these relations and any of them can be set.
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