Mechanical Properties of Solids (PART-1)

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MECHANICAL PROPERTIES OF SOLIDS (DOC-1)

Introduction

A rigid body generally means a hard solid object having a definite shape and size. But in reality,

bodies can be stretched, compressed and bent. Even the appreciably rigid steel bar can be deformed

when a sufficiently large external force is applied on it. This means that solid bodies are not

perfectly rigid.

Elasticity and Plasticity :

A solid has definite shape and size. In order to change (or deform) the shape or size of a body, a

force is required. If you stretch a helical spring by gently pulling its ends, the length of the spring

increases slightly. When you leave the ends of the spring, it regains its original size and shape.

Elasticity

The property of a body, by virtue of which it tends to regain its original size and shape when the

applied force is removed, is known as elasticity and the deformation caused is known as elastic

deformation.

Plasticity

However, if you apply force to a lump of putty or mud, they have no gross tendency to regain their

previous shape, and they get permanently deformed. Such substances are called plastic and this

property is called plasticity. Putty and mud are close to ideal plastics. The elastic behaviour of

materials plays an important role in engineering design.

Note: For example, while designing a building, knowledge of elastic properties of materials like

steel, concrete etc. is essential.

Hooke’s Law

Robert Hooke, an English physicist (1635 - 1703 A.D) performed experiments on springs and found

that the elongation (change in the length) produced in a body is proportional to the applied force or

load.

Note:

It is very important to know the behaviour of the materials under various kinds of load from the

context of engineering design.

Stress and Strain

When forces are applied on a body in such a manner that the body is still in static equilibrium, it is

deformed to a small or large extent depending upon the nature of the material of the body and the

magnitude of the deforming force. The deformation may not be noticeable visually in many

materials but it is there.

Stress

When a body is subjected to a deforming force, a restoring force is developed in the body. This

restoring force is equal in magnitude but opposite in direction to the applied force. The restoring

force per unit area is known as stress. If F is the force applied and A is the area of cross section of

the body,

Magnitude of the stress = F/A ….(1)

The SI unit of stress is N m–2

or pascal (Pa) and its dimensional formula is [ML–1

T–2

].

Mechanical Properties of Solids

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There are three ways in which a solid may change its dimensions when an external force acts on it

i.e. three types of stress namely

1. Longitudinal stress

2. Tangential Stress

3. Hydraulic stress

These stresses are shown in the figure and are discussed below

1. Longitudinal stress

Longitudinal stress can be Further subdivided into

(i) Tensile Stress

(ii) Compressive Stress

(i) Tensile stress- In Fig.2 (a), a cylinder is stretched by two equal forces applied normal to its

cross-sectional area. The restoring force per unit area in this case is called tensile stress.

(ii) Compressive stress- If the cylinder is compressed under the action of applied forces, the

restoring force per unit area is known as compressive stress.

Tensile or compressive stress can also be termed as longitudinal stress. In both the cases, there is

a change in the length of the cylinder.

Longitudinal strain- The change in the length L to the original length L of the body (cylinder

in this case) is known as longitudinal strain.

Longitudinal strain = L

L

….(2)

2. Tangential Stress

However, if two equal and opposite deforming forces are applied parallel to the cross-sectional

area of the cylinder, as shown in Fig.2 (b), there is relative displacement between the opposite

faces of the cylinder. The restoring force per unit area developed due to the applied tangential

force is known as tangential or shearing stress.

Shearing Strain

As a result of applied tangential force, there is a relative displacement x between opposite faces

of the cylinder as shown in the Fig.2 (b).

It can also be visualised, when a book is pressed with the hand and pushed horizontally, as

shown in Fig.2 (c).

The strain so produced is known as shearing strain and it is defined as the ratio of relative

displacement of the faces x to the length of the cylinder L.

Shearing strain = x

tanL

….(3)

where the

cylinder).

Note:

Usually is very small, tan is nearly equal to angle , (if = 10o, for example, there is only

1% difference between and tan ).

Thus, shearing strain = tan ….(4)

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Fig.2: (a) A cylindrical body under tensile stress elongates by L, (b) Shearing stress on a

cylinder deforming it by an angle (c) A body subjected to shearing stress (d) A solid body

under a stress normal to the surface at every point (hydrautic stress). The volumetric strain is

V/V. but there is no change in shape.

3. Hydraulic Stress

The force applied by the fluid acts in perpendicular direction at each point of the surface and the

body is said to be under hydraulic compression. The body develops internal restoring forces that

are equal and opposite to the forces applied by the fluid (the body restores its original shape and

size when taken out from the fluid). The internal restoring force per unit area in this case is

known as hydraulic stress and in magnitude is equal to the hydraulic pressure (applied force per

unit area). This leads to decrease in its volume without any change of its geometrical shape.

In Fig.2 (d), a solid sphere placed in the fluid under high pressure is compressed uniformly on all

sides.

Volume Strain

The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of

change in volume ( V ) to the original volume (V).

Volume strain V

V

….(5)

Note: Since the strain is a ratio of change in dimension to the original dimension, it has no units or

dimensional formula.

Hooke’s Law

Stress and strain take different forms in the situations depicted in the Figure above For small

deformations the stress and strain are proportional to each other. This is known as Hooke’s law.

Thus,

stress strain

stress = k × strain ….(6)

where k is the proportionality constant and is known as modulus of elasticity.

Stress Strain Curve

The relation between the stress and the strain for a given material under tensile stress can be found

experimentally. In a standard test of tensile properties, a test cylinder or a wire is stretched by an

applied force. The fractional change in length (the strain) and the applied force needed to cause the

strain are recorded. The applied force is gradually increased in steps and the change in length is

noted. A graph is plotted between the stress (which is equal in magnitude to the applied force per

unit area) and the strain produced. These curves help us to understand how a given material deforms


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with increasing loads. . The stress-strain curves vary from material to material. Analogous graphs for

compression and shear stress may also be obtained

A typical graph for a metal is shown in Fig.3.

From the graph, we can see that in the region between O to A, the curve is linear. In this region,

Hooke’s law is obeyed. The body regains its original dimensions when the applied force is

removed. In this region, the solid behaves as an elastic body.

Fig.3: A typical stress-strain curve for a metal.

In the region from A to B, stress and strain are not proportional. Nevertheless, the body still

returns to its original dimension when the load is removed. The point B in the curve is known as

yield point (also known as elastic limit) and the corresponding stress is known as yield strength

y( ) of the material.

If the load is increased further, the stress developed exceeds the yield strength and strain

increases rapidly even for a small change in the stress. The portion of the curve between B and D

shows this.

When the load is removed, say at some point C between B and D, the body does not regain its

original dimension. In this case, even when the stress is zero, the strain is not zero. The material

is said to have a permanent set. The deformation is said to be plastic deformation.

The point D on the graph is the ultimate tensile strength u( ) of the material. Beyond this point,

additional strain is produced even by a reduced applied force and fracture occurs at point E.

If the ultimate strength and fracture points D and E are close, the material is said to be brittle. If

they are far apart, the material is said to be ductile.

Fig. 4 shows stress-strain curve for the elastic tissue of aorta, present in the heart. Note that

although elastic region is very large, the material does not obey Hooke’s law over most of the

region. Secondly, there is no well defined plastic region. Substances like tissue of aorta, rubber

etc. which can be stretched to cause large strains are called elastomers

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Fig.4: Stress-strain curve for the elastic tissue of Aorta, the large tube (vessel)

carrying blood from the heart.

Elastic Moduli

The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the

material. Here we discuss the three modulii of Elasticity relating to each of the three types of stresses

1. Young’s Modulus for Longitudinal Stress

2. Shear Modulus for Shearing Stress

3. Bulk Modulus for Hydraulic Stress

1. Young’s Modulus:

The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Young’s

modulus and is denoted by the symbol Y.

σ (tensile or compressive stress) and ε (longitudinal strain) and we know

F/A

L/ L

Y

….(7)

Y = (F / A) /( L / L)

(F L) /(A L) ….(8)

Since strain is a dimensionless quantity, the unit of Young’s modulus is the same as that of stress

i.e., Nm–2

or Pascal (Pa).

2. Shear Modulus:

The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the

material and is represented by G. It is also called the modulus of rigidity.

G = shearing stress s( ) / shearing strain

G (F / A) /( x / L)

(F L) /(A x) ….(10)


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Also, x

L

Similarly, from equation

G (F / A) /

F /(A ) ….(11)

The shearing stress s can also be expressed as

s G ….(12)

The SI unit of shear modulus is Nm–2

of Pa.

3. Bulk Modulus:

The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus. It is

denoted by symbol B.

B = – p/(ΔV/V) ….(13)

The negative sign indicates the fact that with an increase in pressure, a decrease in volume

occurs. That is, if p is positive, ΔV is negative. Thus for a system in equilibrium, the value of

bulk modulus B is always positive.

SI unit of bulk modulus is the same as that of pressure i.e., Nm–2

or Pa.

Compressibility

The reciprocal of the bulk modulus is called compressibility and is denoted by k. It is defined as the

fractional change in volume per unit increase in pressure.

k (1/ B) (1/ p) ( V / V) ….(14)

Compressibility of solids

Solids are least compressible whereas gases are most compressible.

The incompressibility of the solids is primarily due to the tight coupling between the neighbouring

atoms. The molecules in liquids are also bound with their neighbours but not as strong as in solids.

Molecules in gases are very poorly coupled to their neighbours.

Applications of Elastic Behaviour of Materials

The study of the elastic behaviour of materials plays an important role in everyday life. For example

while designing a building, the structural design of the columns, beams and supports require

knowledge of strength of materials used. Some of the uses are discussed below:

Rope of Crane

Cranes used for lifting and moving heavy loads from one place to another have a thick metal rope to

which the load is attached. The rope is pulled up using pulleys and motors. Suppose we want to

make a crane, which has a lifting capacity of 10 tonnes or metric tons (1 metric ton = 1000 kg). How

thick should the steel rope be? We obviously want that the load does not deform the rope

permanently. Therefore, the extension should not exceed the elastic limit.

Loading of beams

A bridge has to be designed such that it can withstand the load of the flowing traffic, the force of

winds and its own weight. Similarly, in the design of buildings use of beams and columns is very

common. In both the cases, the overcoming of the problem of bending of beam under a load is of

prime importance. The beam should not bend too much or break.

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Buckling

But on increasing the depth, unless the load is exactly at the right place (difficult to arrange in a

bridge with moving traffic), the deep bar may bend as shown in Fig.7 (b). This is called buckling. To

avoid this, a common compromise is the cross-sectional shape shown in Fig.7(c). This section

provides a large load bearing surface and enough depth to prevent bending.

Fig.7: Different cross-sectional shapes of a beam. (a) Rectangular section of a bar;(b) A thin

bar and how it can buckle; (c) Commonly used section for a load bearing bar.

Use of Pillars

Use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends

as shown in Fig.8 (a) supports less load than that with a distributed shape at the ends [Fig.8 (b)]. The

precise design of a bridge or a building has to take into account the conditions under which it will

function, the cost and long period, reliability of usable materials etc.

Fig.8: Pillars or columns: (a) a pillar with rounded ends, (b) Pillar with distributed ends.


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SOLVED EXAMPLES

PRACTICE QUESTIONS

QUESTION 1: A structural steel rod has a radius of 10mm and a length of 1.0m A 100 kN force

stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s

modulus, of structural steel is 2.0 × 1011

N.m–2

.

Solution: We assume that the rod is held by a clamp at one end, and the force F is applied at the

other end, parallel to the length of the rod. Then the stress on the rod is given by

2

F FStress

A r

3

22

100 10 N

3.14 10 m

= 3.18 × 108 N m

–2

The elongation,

(F / A)L

LY

8 2

11 2

3.18 10 N m (1m)

2 10 N m

= 1.59 × 10–3

m

= 1.59 mm

The strain is given by

Strain = L/ L

= (1.59 × 10–3

m)/(1m)

= 1.59 × 10–3

= 0.16%

QUESTION 2: A copper wire of length 2.2 m and a steel wire of length 1.6m, both of diameter 3.0

mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm.

Obtain the load applied.

Solution: The copper and steel wires are under a tensile stress because they have the same tension

(equal to the load W) and same area of cross-section A. we have stress = strain × Young’s modulus.

Therefore

C C C S S SW / A Y ( L / L ) Y ( L / L )

where the subscripts c and s refer to copper and stainless steel respectively. Or,

C S S C C SL / L (Y / Y ) (L / L )

Given C SL 2.2m, L 1.6m,

From Table 9.1m, YC = 1.1 × 1011

N.m–2

, and

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YS = 2.0 × 1011

N.m–2

. 11 11

C SL / L (2.0 10 /1.1 10 ) (2.2/1.6) 2.5.

The total elongation is given to be

4

C SL L 7.0 10 m

Solving the above equations, 4 4

C SL 5.0 10 m, and L 2.0 10 m.

Therefore

W = C C C(A Y L ) / L

3 2 4 11(1.5 10 ) [(5.0 10 1.1 10 ) / 2.2]

= 1.8 × 102N

QUESTION 3: A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force

(on its narrow face) of 9.0 × 104 N. The lower edge is riveted to the floor. How much will the upper

edge be displaced?

Solution: The lead slab is fixed and the force is applied parallel to the narrow face as shown in Fig.

The area of the face parallel to which this force is applied is

A = 50cm × 10cm

= 0.5m × 1.0m

= 0.05m2

Therefore, the stress applied is

= (9.4 × 104N/0.05m

2)

= 1.80 × 106 N.m

–2

We know that shearing strain = ( x / L) = Stress /G.

Therefore the displacement x (Stress × L)/G

= (1.8 × 106 N m

–2 × 0.5m)/(5.6 × 10

9N m

2

= 1.6 × 10–4

m = 0.16mm

QUESTION 4: The average depth of Indian Ocean is about 3000m. Calculate the fractional

compression, V / V, of water at the bottom of the ocean, given that the bulk modulus of water is

2.2 × 109 N m

–2. (Take g = 10 m s

–2)

Solution: The pressure exerted by a 3000 m column of water on the bottom layer

p = hp g = 3000 m × 1000 kg m–1

s–2

= 3 × 107 kg m

–1 s

–2

= 3 × 107 N m

–2

Fractional compression V / V,is 7 2 9 2V / V stress / B (3 10 N m ) /(2.2 10 Nm )

21.36 10 or 1.36%

Additional Solved Examples

Example 1: How much will a 3.0m long copper wire elongate if a weight of 10 kg is suspended

from one end and the other end is fixed? The diameter of the wire is 0.4mm. Given: Y for copper =

1011

Nm–2

and g = 9.8 m s–2

.

Solution: M = 10 kg, F = Mg = 10 × 9.8 N = 98 M

r = 0.2 nm = 0.2 × 10–3

m, Y = 1011

Nm–2

l = 3, l = ?


10

Now, F l

YA l

or 2

F ll

r Y

3 2 11

98 3 7l m

22 (0.2 10 ) 10

3

98 21m 0.02339m 2.339cm

88 10

Example 2: A copper wire of length 2.2m and a steel wire of length 1.6m, both of diameter 3.0 mm,

are connected end to end. When stretched by a load, the net elongation is found to be 0.7mm. Obtain

the load applied. Given: Young’s modulus of copper = 1.1 × 1011

Nm–2

and Young’s modulus of

steel = 2.0 × 1011

Nm–2

.

Solution: The copper and steel wires are under a tensile stress because they have the same tension

(equal to the load W) and the same area of cross-section A.

Young’s modules = stress

strain

or stress = Young’s Modulus × strain

c sc s

c s

L LWY Y

A L L

[The subscripts c and s refer to copper and stainless steel respetively.]

Now, c s c

s c s

L Y L

L Y L

= 11

11

2.0 10 2.22.5

1.1 10 1.6

Net elongation = Lc + Ls = 0.7 mm = 7 × 10–4

m

Now, 2.5 Ls + Ls = 7 × 10–4

m

or 4

4

s

7 10L m 2 10 m

3.5

Again, Lc = 2.5 Ls = 2.5 × 2 × 10–4

m = 5 × 10–4

m

Now, W = c c

c

Y L

L

3 2 11 422(1.5 10 ) (1.1 10 )(5 10 )

7 N2.2

= 1.8 × 102 N

Example 3: A sphere contracts in volume by 0.01% when taken to the bottom of sea 1km deep. Find

the bulk modulus of the material of the sphere. Given : density of sea water is 1g cm–3

, g = 980 cm

s–2

.

Solution: 3 5V 0.01

, h 1km 10 m 10 cmV 100

11

3 5 21g cm ; P 10 1 980dynecm ,K ?

5

2P P V 10 980 100K dyne cm

V / V V 0.01

= 98 × 1010

dyne cm–2

Example 4: An Indian rubber cube of side 7cm has one side fixed while a tangential force equal to

the weight of 200kg is applied to the opposite face. Find the shearing strain produced and te distance

through which the strained side moves. Given: modulus of rigidity for rubber is 2 × 107

dyne cm–2

.

Solution: l = 7 cm, F = 200 kgf

= 200 × 1000 × 981 dyne,

= 2 × 107 dyne cm

–2

Area of the face of the cube, A = l2 = 7 cm × 7 cm = 49 cm

2

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F F 200 1000 981or rad 0.2 rad

A A 49 2 10

shearing strain, l

l

or l = l = 7 × 0.2 cm = 1.4 cm

Example 5: A uniform steel wire of length 2.5m and of density 8g cm–3

weights 50g. When

stretched by a force of 10kgf, the length increases by 2 mm. Calculate Young’s modulus for steel.

Solution: 3 6 3m 50 50

V cm 10 m8 8

6 2V 50 1

a 10 ml 8 2.5

2

6 3

10 9.8 2.5Y Nm

2.5 10 2 10

10 24.9 10 N m

Example 6: A rubber rope of length 8m is hung from the ceiling of a room. What is the increase in

length of the rope due to its own weight? (Given: Young’s modulus of elasticity of rubber

= 5 × 106N m

–2 and density of rubber = 1.5 × 10

3 kg m

–3. Takes g = 10 m s

–2.)

Solution: Mg L / 2

YA L

(Length is taken as L

2 because weight acts at C.G.)

Now, M = AL

[For the purpose of calculation of mass, the whole of geometrical length L is to be considered].

2AL gL gL

Y or L2A L 2Y

3

6

1.5 10 10 8 8m

2 5 10

= 9.6 × 10–2

m

= 9.6 × 10–2

× 103 mm = 96 mm


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Example 7: A composite wire of uniform diameter 3.0 mm consisting of a copper wire of length

2.2m and a steel wire of length 1.6m stretches under a load by 0.7mm. Calculate the load, given that

the Young’s modulus for copper is 1.1 × 1011

Pa and for steel is 2.0 × 1011

Pa.

Solution: 3

c sl l 0.7 10 m

c s c

s c s

l Y lAlso,

l Y l

1

11

2 10 2.22.5

1.1 10 1.6

or lc = 2.5 ls ; 3.5 ls = 0.7 × 10–3

or ls = 3

40.7 10m 2 10 m

3.5

Now, ss 2

s s

FlY

r l

or 11 3 3 42 10 3.14 1.5 10 1.5 10 2 10

N1.6

= 1.77 × 102 N

Example 8: Determine the fractional change in volume as the pressure of the atmosphere (1×105 Pa)

around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the

metal is 1.25 ×1011

N m–2

.

Solution: P = 105 N m

-2,

K = 1.25 × 1011

N m–2

K = P

V

V

or 5

11

V P 10

V K 1.25 10

= 8 × 10–7

Example 9: A spherical ball contracts in volume by 0.0098% when it is subjected to a pressure of

100 atmosphere. Calculate its bulk modulus?

Solution: V 0.0098

, P 100atm, K ?V 100

P

KV / V

or K = 6100 100

atm 1.02 10 atm0.0098

Example 10: By applying a tangential force of 5kg to the upper surface of a cube of side 5cm with

respect to the bottom. Calculate the shearing stress, strain and shear modulus.

Solution: F = 5 kg wt = 5 × 9.8 N,

a = 25 × 10–4

m2,

l = 0.5 cm, l = 5 cm

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Shear modulus

2

4

5 9.8 5N m

25 10 0.5

= 1.96 × 105 N m

–2

Example 11: A box-shaped piece of gelatin dessert has a top area of 15 cm2 and a height of 3 cm.

When a shearing force of 0.50N is applied to the upper surface, the upper surface displaces 4mm

relative to the bottom surface. What are the shearing stress, shearing strain and the shear modulus for

the gelatin?

Solution: A = 15 × 10–4

m2, l = 3 × 10

–2 m,

F = 0.50 N, l = 4 × 10–3

m

Stress = 2

4

0.50N m

15 10

= 333.3 N m–2

Strain = 3

2

l 4 100.133

l 3 10

Shear modulus 2

2

4 3

stress 0.50 3 10N m

strain 15 10 4 10

5 2 21.510 N m 2500 N m

60

Example 12: A load of 4.0 kg is suspended from a ceiling through a steel wire of radius 2.0 mm.

Find the tensile stress developed in the wire when equilibrium is achieved. Take

2g 3.1 ms

Solution: Tension in the wire is

F = 4.0 × 3.1 N

The area of cross section is

A = r2 = × (2.0 × 10

–3 m)

2

= 4.0 × 10–6

m2

Thus, the tensile stress developed

= 2

6

F 4.0 3.1N m

A 4.0 10

= 3.1 × 106 N m

–2

Example 13: A steel wire of length 2.0m is stretched through 2.0mm. The cross-sectional area of the

wire is 4.0 mm2. Calculate the elastic potential energy stored in the wire in the stretched condition.

Young modulus of steel = 2.0 × 1011

N m–2

.

Solution: The strain in the wire 3l 2.0 mm

10l 2.0 m

The stress in the wire = Y × strain

= 2.0 × 1011

N m–2

× 10–3

= 2.0 × 108 N m

–2

The volume of the wire = (4 × 10–6

m2) × (2.0 m)

= 8.0 × 10–6

m3


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The elastic potential energy stored

= 1

stress strain volume2

= 8 2 3 6 312.0 10 N m 10 8.0 10 m

2

= 0.8 J

Example 14: Two wires of equal cross section but one made of steel and the other of copper, are

joined end to end. When the combination is kept under tension, the elongations in the two wires are

found to be equal. Find the ratio of the lengths of the two wires. Young modulus of steel = 2.0 × 1011

N m–2

and that of copper = 1.1 × 1011

N m–2

.

Solution: As the cross sections of the wires are equal and same tension exists in both, the stresses

developed are equal. Let the original lengths of the steel wire and the copper wire be Ls and Lc

respectively and the elongation in each wire be l.

11 2

s

l stress

L 2.0 10 N m

….(i)

and 11 2

c

l stress

L 1.1 10 N m

….(ii)

Dividing (ii) by (i)

Ls/Lc = 2.0/1.1 = 20 : 11

Example 15: One end of a steel rod of radius R = 9.5 mm and length L = 81 cm is held in a vise. A

force of magnitude F = 62 kN is then applied perpendicularly to the end face (uniformly across the

area) at the other end. What are the stress on the rod and he elongation L and strain of the rod?

Solution: The stress is the ratio of the magnitude F of the perpendicular force to the area A. The

ratio is the left side. The elongation L is related to the stress and Young’s modulus E by Eq. (F/A

= E L/L). Strain si the ratio of the elongation to the initial length L.

To find the stress, we write

Stress = 4

2 3 2

F F 6.2 10 N

A R ( ) (9.5 10 m)

= 2.2 × 108 N/m

2

The yield strength for structural steel is 2.5 × 108 N/m

2, so this rod is dangerously close to its yield

strength.

We find the value of Young’s modulus for steel. Then from Eq. we find the elongation:

8 2

11 2

(F / A)L (2.2 10 N / m )(0.81m)L

E 2.0 10 N / m

= 8.9 × 10–4

m = 0.89 mm

For the strain, we have

4L 8.9 10 m

L 0.81m

= 1.1 × 10–3

= 0.11%

You can see that at just 0.1% strain, the rod is at yield point. This is the reason we cannot see any

change in the dimensions of the metal objects under normal situations.

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Example 16: A solid sphere is initially kept in open air, and the pressure exerted on it by air is 1.0 ×

105 N/m

2 (atmospheric pressure). The sphere is lowered into the ocean to a depth where the pressure

is 200 times the atmospheric pressure. The volume of the sphere in air is 0.5 m3. What is the change

in the volume once the sphere is submerged? Given that bulk modulus is 6.1×1010

N/m2.

Solution: What happens when you squeeze a tennis ball in your hand, its volume reduces as it

shrinks. We can expect same thing to happen to the sphere when it is subjected to high pressure from

the ocean water. We need to perform a simple calculation using Eq.

iV PV

B

Substituting the numerical values:

3 7 2 5 2

10 2

(0.50m )(2.0 10 N / m 1.0 10 N / m )V

6.1 10 N / m

= –1.6 × 10–4

m–3

The negative sign indicates that the volume of the sphere decreases when submerged.

Mechanical Properties of Solids (PART-1)

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Transcript of Mechanical Properties of Solids (PART-1)