Page # [1]

TEACHING NOTES

This is the teaching module for Elasticity, Calorimetry & Thermal Expansion . This module has to befollowed in the class. VK Bansal

ELASTICITY, CALORIMETRY & THERMAL EXPANSION

(Complete in 3 Lectures)ELASTICITY

The shape of a solid can be altered only by the application of a suitable system of forces.

ELASTICITY:If the modified shape is retained by the solid only so long as the forces act, and the original shape isregained when they cease to act, the solid is said to possess elasticity ;

PLASTICITY:if the modified shape is retained after the forces cease to act, the solid is said to be plastic.>Fluids, including liquids and gases, possess no definite shape, and therefore cannot possess elasticity.> The volume of matter, whether this matter be solid or fluid, can be altered by the application of suitableforces, and in most cases the original volume is regained when the forces cease to act ; hence, all matterpossesses volume elasticity.>Any alteration produced in the shape or volume of matter is called a strain.

Cause of elasticityWe know that in a solid, each atom or molecule is surrounded by neighboring atoms or molecules. Theseare bonded together by interatomic or intermolecular forces and stay in a stable equilibrium position.When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causinga change in the interatomic (or intermolecular) distances. When the deforming force is removed, theinteratomic forces tend to drive them back to their original positions. Thus the body regains its originalshape and size. The restoring mechanism can be visualised by taking a model of spring-ball systemshown in the Fig.. Here the balls represent atoms and springs represent interatomic forces.

Fig. Spring-ball model for the illustration of elastic behaviour of solids.If you try to displace any ball from its equilibrium position, the spring system tries to restore the ball backto its original position. Thus elastic behaviour of solids can be explained in terms of microscopic natureof the solid.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.

Page # [2]

Stress : Restoring force per unit area set up inside the body is called stress and is measured by the magnitudeof force acting on unit area of the body in equilibrium. If F is the force applied and A is the area of crosssection of the body, Magnitude of the stress = F/A. The SI unit of stress is N m2 or pascal (Pa) and itsdimensional formula is [ ML1T2 ].

Stress (Tensile/Compressive )

Tensile F F FF = AF

as F AA

Compressive FF

Strain (longitudinal)Suppose we stretch a wire by applying tensile forces of magnitude F to each end. The length of the wireincreases from L to L + L. The fractional length change is called the strain ; it is a dimensionlessmeasure of the degree of deformation.

strain = LL

Hookes law for tensile and compressive forcesSuppose we had wires of the same composition and length but different thicknesses. It would requirelarger tensile forces to stretch the thicker wire the same amount as the thinner one. We conclude that thetensile force required is proportional to the cross-sectional area of the wire (F A). Thus, the sameapplied force per unit area produces the same deformation on wires of the same length and composition.

Hookes Lawstress strain

AF

= Y LL

equation still says that the length change (L) is proportional to the magnitude of the deforming forces(F). Stress and strain account for the effects of length and cross-sectional area ; the proportionalityconstant Y depends only on the inherent stiffness of the material from which the object is composed ; itis independent of the length and cross-sectional area.

Comparing equation F = kL and AF

= Y LL

k = Y LL

A. Y is called the elastic modulus or Youngss

modulus, Y has the same units as those of stress (Pa or N/m2) since strain is dimensionless.Youngs modulus can be through of as the inherent stiffness of a material ; it measures the resistance ofthe material to elongation or compression. Material that is flexible and stretches easily (for example,rubber) has a low Youngs modulus. A stiff material (such as steel) has a high Youngs modulus ; it takesa larger stress to produce the same strain.Hookes law holds up to a maximum stress called the proportional limit. For many materials, Youngsmodulus has the same value for tension and compression. Some composite materials, such as bone andconcrete, have significantly different Youngs moduli for tension and compression. The different proper-ties of these two substances lead to different values of Youngs modulus for tension and compression.

Important Notes :1. Y is independent of size, shape of body depends only on nature and material2. Ysolid > Yliq. > Ygas

Yrigid = 3. genrally as Tempertureincreases Y

Page # [3]

Ex. A light wire of length 4m is suspended to the ceiling by one of its ends. If its crossectional area is19.6mm2, what is its extension under a load of 10kg. Youngs modulus of steel = 2 1011 Pa.

[Sol. Given ; original length L = 4m ;force F = 10 9.8 = 98 N ;

and Y = 2 1011 Nm2 ; l = ?Using the relation,

Youngs modulus (Y) = strainalIongitudinstressallongitudin

We have

Y = L/A/F

l l = YAFL

l = 611 106.19102498

= 1 104 m

= 0.1 mm ]

Ex. Two vertical rods of equal lengths, one of steel and the other of copper, are suspended from the ceiling,at a distance l apart and are connected rigidly to a rigid horizontal light bar at their lower ends.

Steel Copper

F

If AS and AC be their respective cross sectional areas, and YS and YC, their respective Youngs modulii ofelasticities, find where should a vertical force F be applied to the horizontal bar, in order that the barremains horizontal. (Fig.)

[Sol. Let the force F be applied at a distance x from the steel bar, measured along the horizontal bar.Let FS and FC be the loads on steel and copper rods respectively, so

FS + FC = F ... (i)Since the rigid horizontal bar remains horizontal so, the extensions produced in the two rods and hencestrains remains same.

i.e.,SS

S

YAF

= CC

C

YAF

... (ii)

Solving (i) and (ii) FS = CCSSSS

YAYAYFA

and FC = CCSSCC

YAYAYFA

Now, taking moments about the steel bar.

FC l = Fx x = FFC l

CCSS

CC

YAYAYA

l

or x = l /

C

S

C

S

YY

AA1 ]

Page # [4]

Elastic potential energy (where F is applied in equilibrium)

dw = Fdx

xL

AYF

= LAY

x dx

F

F

l + x

F

F

w = LAY

l

0dxx

U = L2

AY 2l= 2

1

Lyl

Ll

(AL)

U = 21

(stress) (strain) (volume)

Optional Ex. A wire having a length l = 2m, and cross sectional area A = 5mm2 is suspended at one of its endsfrom a ceiling. What will be its strain energy due to its own weight, if the density and Youngs modulus ofthe material of the wire be d = 9g/cm3 and Y = 1.5 1011 Nm2?

Sol. Consider an elemental length of the wire of length dx, at a distance x from the lower end.Clearly, this length is acted upon by the external force equal to the weight of the portion of wire below it= xAdg. In equilibrium, the restoring force f = xAdg.

stress = Af

= xdg.

Now, elastic potential energy stored in the elemental length will be

dU = 21

stress strain volume

= 21

Y

)stress( 2 volume

= 21

Y

)xdg( 2 Adx = 2

1

YAgd 22

x2 dx

Total elastic potential energy U = dU

= l

0

22

YAgd

21

x2 dx xdx

xAdg

f

F

Area = A

F= 61

d2g2 Y

A 3l

Substituting the values,

U = 61

113623

105.12105)89()109(

= 3.46 107 J ]

Page # [5]

STRESS-STRAIN CURVEThe relation between the stress and the strain for a given material under tensile stress can be foundexperimentally. The applied force is gradually increased in steps and the change in length is noted. Agraph is plotted between the stress and the strain produced. The stress-strain curves vary from materialto material. These curves help us to understand how a given material deforms with increasing loads.From the graph, we can see that in the region between O to A, the curve is linear. In this region, Hookeslaw is obeyed. The body regains its original dimensions when the applied force is removed. In thisregion, the solid behaves as an elastic body.

Fig. Stress-strain curve for a metal.Beyond hookes law

In the region from A to B, stress and strain are not proportional. Nevertheless, the body still returns to itsoriginal dimension when the load is removed. The point B in the curve is known as yield point (alsoknown as elastic limit) and the corresponding stress is known as yield strength (Sy) of the material.If the tensile or compressive stress exceeds the proportional limit, the strain is no longer proportional tothe stress (figure). The solid still returns to its original length when the stress is removed as long as thestress does not exceed the elastic limit.If the stress exceeds the elastic limit, the material is permanently deformed. For still larger stresses, thesolid factures when the stress reaches the breaking point. The maximum stress that can be withstoodwithout breaking is called the ultimate strength. The ultimate strength can be different for compressionand tension ; then we refer to the compressive strength or the tensile strength of the material. A ductilematerial continues to stretch beyond its ultimate tensile strength without breaking ; the stress then de-creases from the ultimate strength (fig. (a) ). Examples of ductile solids are relatively soft metals, such asgold, silver, copper, and lead. These metals can be pulled like taffy, becoming thinner and thinner untilfinally reaching the breaking point.

Ultimatestrength

Breaking pointProportionallimit

Elastic limit

Tensile stress

Tensile strain(a)

Ductile

Ultimate strengthand breaking point

Elastic limitProportional limit

Tensilestress

(b)

Brittle

Tensile strain

{Home Work - HCV : Chapter -14 Ex 1 -11 }

Page # [6]

DIFFERENT TYPES OF STRESS

Fig. (a) Cylinder subjected to tensile stress stretches it by an amount L. (b) A cylinder subjectedto shearing (tangential) stress deforms by an angle . (c) A book subjected to a shearing stress(d) A solid sphere subjected to a uniform hydraulic stress shrinks in volume by an amount V.

There are three ways in which a solid may change its dimensions when an external force acts on it. Theseare shown in Fig. In Fig.(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. If the cylinder iscompressed under the action of applied forces, the restoring force per unit area is known as compres-sive 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. 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 LL

.

Shearing stress.However, if two equal and opposite deforming forces are applied parallel to the cross-sectional area ofthe cylinder, as shown in Fig. (b), there is relative displacement between the opposite faces of thecylinder. The restoring force per unit area developed due to the applied tangential force is known astangential or shearing stress.As a result of applied tangential force, there is a relative displacement x between opposite faces of thecylinder as shown in the Fig. (b). The strain so produced is known as shearing strain and it is definedas the ratio of relative displacement of the faces x to the length of the cylinder L.

Shearing strain Lx

= tan where is the angular displacement of the cylinder from the vertical (original position of the cylinder).Usually is very small, tan is nearly equal to angle , (if = 10, for example, there is only 1%difference between and tan ).It can also be visualised, when a book is pressed with the hand and pushed horizontally, as shown in Fig.(c). Thus, shearing strain = tan

Bulk ModulusIn Fig. (d), a solid sphere placed in the fluid under high pressure is compressed uniformly on all sides.The force applied by the fluid acts in perpendicular direction at each point of the surface and the body issaid to be under hydraulic compression. This leads to decrease in its volume without any change of itsgeometrical shape. The body develops internal restoring forces that are equal and opposite to the forcesapplied by the fluid (the body restores its original shape and size when taken out from the fluid). Theinternal restoring force per unit area in this case is equal to the hydraulic pressure (applied force per unitarea). The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio ofchange in volume (V) to the original volume (V).

Volume strain vv

Page # [7]

We have seen that when a body is submerged in a fluid, it undergoes a hydraulic stress (equal in magni-tude to the hydraulic pressure). This leads to the decrease in the volume of the body thus producing astrain called volume strain.

Volume DeformationSince the fluid presses inward on all sides of the object (figure), the solid is compressed-its volume isreduced. The fluid pressure P is the force per unit surface area ; it can be through of as the volume stresson the solid object. Pressure has the same units as the other kinds of stress: N/m2 or Pa.

F=PA1

F=PA3F=PA2

F=PA1

F=PA3F=PA2

Fig. Forces on an object when submerged in a fluid

volume stress = pressure = AF

= P

The resulting deformation of the object is characterized by the volume strain, which is the fractionalchange in volume :

volume strain = volumeoriginalvolumeinchange

= VV

Unless the stress is too large, the stress and strain are proportional within a constant of proportionallycalled the bulk modulus B. A substance with a large bulk modulus is more difficult to compress than asubstance with a small bulk modulus. For solids and liquids, the volume strain dueto atmospheric pressure is, for most purposes, negligibly small (5 105 for water).

Hookes law for volume deformation

P = B VV

where V is the volume at atmospheric pressure. The negative sign. equation P = B VV

allows the

bulk modulus to be positive-an increase in the volume stress causes a decrease in volume, so V isnegative. Table lists bulk moduli for various substances.The reciprocal of the bulk modulus is called compressibility and is denoted by k. It is defined as thefractional change in volume per unit increase in pressure.

k = (1/B) = (1/p) (V/V)Bulk modulus (all three states of matter)

B = strainstressvolume

B =

VV

P

= V VP

C = B1

= V1

PV

Page # [8]

The bulk moduli of liquids are generally not much less than those of solids, since the atoms in liquids arenearly as close together as those in solids.Gases are much easier to compress than solids or liquids, so their bulk moduli are much smaller. Thebulk moduli of a few common materials are given in Table

Table : Bulk moduli (B) of some common Materials

Poissons ratio :When an elongation is produced by longitudinal stresses, a change is produced in the lateral dimensionsof the strained substance. Thus, when a wire is stretched, its diameter diminishes ; and when the longitu-dinal strain is small, the lateral strain is proportional to it. The ratio of the lateral strain to the longitudinalstrain is called Poissons ratio.

F1F1l3l2

l1

Y =

1

1

AF

ll

2

2

ll

= 3

3

ll

= 1

1

ll

Results

B = )21(3y

Page # [9]

Optional: Relation between bulk modulus & Youngs modulus of a bar is in fluid

F1l3

l1

l2

using superposition

z

yx

(i)x1

1

ll

= yP

ll1 =

zyx ll

= YP

(1 2)

(ii) ll2 = y

Pby symmetry

y1

1

ll

=

ll2 = + y

P

ll2 = y

P (1 2)

z1

1

ll

= + s yP

ll3 = y

P (1 2)

for small change in volume

(iii)

ll3 = y

PVV

=

ll1 +

ll2 +

ll3

z1

1

ll

= + yP

VV

= 3

yP

(1 2)

Ex. A uniform bar of length L and cross sectional area A is subjected to a tensile load F. If Y be the Youngsmodulus of the material of the bar and be its poissons ratio, then determine the volumertic strain.

[Sol. Longitudinal stress = AF

.

Longitudinal strain = AYF

= 1 (say) ... (i)

Now, by definition of Poissons ratio,

= strainallongitudinstrainlateral

= L/Lr/r

or r / r = L / L AYF

[From eqn. (i) ]

Since Volumetric strain = Strain in length + Twice strain in radius.

Volumetric strain = LL

+ rr2

= AYF

2

AY

F= AY

F (1 2). ]

{Home Work - HCV : Chapter -14 Ex 12 - 15,Question & Short Answer 1 6Obj. - I 1 10Obj. - II 1 }

Page # [10]

CALORIMETRYA system is said to be isolated if no exchange or transfer of heat occurs between the system and itssurroundings. When different parts of an isolated system are at different temperature, a quantity of heattransfers from the part at higher temperature to the part at lower temperature. The heat lost by the partat higher temperature is equal to the heat gained by the part at lower temperature. Calorimetry meansmeasurement of heat. When a body at higher temperature is brought in contact with another body atlower temperature, the heat lost by the hot body is equal to the heat gained by the colder body, providedno heat is allowed to escape to the surroundings.A device in which heat measurement can be made is called a calorimeter. It consists a metallic vesseland stirrer of the same material like copper or aluminium. The vessel is kept inside a wooden jacketwhich contains heat insulating materials like glass wool etc. The outer jacket acts as a heat shield andreduces the heat loss from the inner vessel. There is an opening in the outer jacket through which amercury thermometer can be inserted into the calorimeter. The following example provides a method bywhich the specific heat capacity of a given solid can be determinated by using the principle, heat gainedis equal to the heat lost. Water equivalent of a calorimeter means amount of heat required for the equiva-lent amount of water to raise temperature by 1C.

Ex. A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so thatthe sphere is at 100 C. It is then immediately transferred to 0.14 kg copper calorimeter containing0.25 kg of water at 20 C. The temperature of water rises and attains a steady state at 23 C. Calculatethe specific heat capacity of aluminium.

Ans. In solving this example we shall use the fact that at a steady state, heat given by an aluminium sphere illbe equal to the heat absorbed by the water and calorimeter. Mass of aluminium sphere (m1) = 0.047 kgInitial temp. of aluminium sphere = 100 CFinal temp. = 23 CChange in temp (T ) = (100 C 23 C) = 77 CLet specific heat capacity of aluminium be sAl.The amount of heat lost by the aluminiumsphere = m1sAlT = 0.047 kg sAl 77CMass of water (m2) 0.25 kgMass of calorimeter (m3) = 0.14 kgInitial temp. of water and calorimeter = 20CFind temp. of the mixture = 23CChange in temp. (T2) = 23C 20C = 3CSpecific heat capacity of water (sw) = 4.18 10

3 J kg1 K1Specific heat capacity of copper calorimeter = 0.386 103 J kg1 K1The amount of heat gained by water and calorimeter = m2 sw T2 + m3scu T2

= (m2sw + m3scu) (T2)= (0.25 kg 4.18 103 J kg1 K1 + 0.14 kg 0.386 103 J kg1 K1 ) (23C 20C)

In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained bycalorimeter.So, 0.047 kg sN 77C

= (0.25 kg 4.18 103 J kg1 K1 + 0.14 kg 0.386 103 J kg1 K1) (3C)sAl = 0.911 kJ kg

1 K1

Ex. A 5g piece of ice at 20C is put into 10g of water at 30C. Assuming that heat is exchanged onlybetween the ice and the water, find the final the final temperature of the mixture. Specific heat capacity ofice = 2100 J/kg-C specific heat capacity of water = 4200 J/kg-C and latent heat of fusion ofice = 3.36 105 J/kg.

Sol. The heat given by the water when it cools down from 30C to 0C is(0.01 kg) (4200 J/k-C) (30C) = 1260 J.

The heat required to bring the ice to 0C is (0.005 kg) (2100 J/kg-C) (20C) = 210 J.The heat required to melt 5g of ice is (0.005 kg) (3.36 105 J/kg) = 1680.We see that whole of the ice cannot be melted as the required amount of heat is not proved by the water.Also, the heat is enough to bring the ice to 0C. Thus the final temperature of the mixture is 0C withsome of the ice melted.

Page # [11]

THERMAL EXPANSIONThermal Expansion

When matter is heated without change in state, it usually expands. According to atomic theory of matter,asymmetry in potential energy curve is responsible for thermal expansion as with rise in temperature sayfrom T1 to T2 the amplitude of vibration and hence energy of atoms increases from E1 to E2 and hence theaverage distance between the from r1 to r2.

r0r1r2 T2

T1T0

0+

E2E1E0

r

E

Due to this increase in distance between atoms, the matter as a whole expands. Had the potential energycurve been symmetrical, no thermal expansion would have taken place in spite of heating.

Normal SolidsTo varying extents, most materials expand when heated and contract when cooled. The increase in anyone dimension of a solid is called linear expansion, linear in the sense that the expansion occurs along aline. A rod whose length is L0 when the temperature is T0 when the temperature increases to T0 + T, thelength becomes L0 + L, where T and L are the magnitudes of the changes in temperature andlength, respectively.Conversely, when the temperature decreases to T0 T, the length decreases to L0 L.For small temperature changes, experiments show that the change in length is directly proportional to thechange in temperature (L T). In addition, the change in length is proportional to the initial length ofthe rod,

L0L

Equation L = L0T expresses the fact that L is proportional to both L0 and T(L L0T) byusing a proportionality constant , which is called the coefficient of linear expansion.

Common unit for the coefficient of linear expansion : C

1 = (C)1

Ex. A brass scale correctly calibrated at 15C is employed to measure a length at a temperature of 35C. Ifthe scale gives a reading of 75cm, find the true length. (Linear expansively of brass = 2.0 105C)1)

[Sol. Let the distance between two fixed divisions on the scale at 15C be L1 and that at 35C be L2.Clearly, (L2 L1) = L1 (35 15)or L2 = L1 (1 + 20 2.0 10

5)= L1 (1.0004)

i.e., at 35C, an actual length of L2 will be read as L1 (note), due to the increased separation of thedivisions of the scale. In other words, the observed length will be less than the actual length.Given : L1 = 75 cm L2 = 75 (1.0004) cm

= 75.03 cm ]

Page # [12]

Ex. Estimate the time lost or gained by a pendulum clock at the end of a week when the atmospherictemperature rises to 40C. The clock is known to give correct time at 15C and the pendulum is of steel.(Linear expansively of steel is 12 106 / C).

[Sol. Rise in temperature = 40 15 = 25C

Time lost per second = 21

= 21

(12 106 / C) (25C)

= 150 106 s/sTherefore, time lost per week (i.e., 7 86400 s)

= 150 106 s/s 7 86400 s= 90.72 s ]

Ex. A glass rod when measured with a zinc scale, both being at 30C, appears to be of length 100cm. If thescale shows correct reading at 0C, determine the true length of the glass rod at (a) 30C and (b) 0C.( for glass = 8 106 /C and for zinc 26 106 / C)

Sol. At 30C, although the reading shown by the zinc scale corresponding to the length of the glass rod is100cm, but the actual length would be more than 100cm, the reason being the increased separationbetween the markings, owing to a rise in temperature (from 0C to 30C).Now, an actual (original at 0C) length of 100cm on the zinc scale (or more precisely, two markings ordivisions on the scale, separated by a distance of 100cm) would, at a temperature of 30C, correspondto a length given by

l = 100 (1 + 26 106 30) cm= 100.078 cm

The true length of the glass rod at 30C is 100.078 cm.Now, at 0C, the length of glass rod would be lesser than that at 30C,

Using lt = l0 (1 + t), l0 = t1t

l

The length of the rod at 0C, will be

l0 = )301081(cm078.100

6 = 100.054 cm ]

The Bimetallic StripA bimetallic strip is made from two thin strips of metal that have different coefficients of linear expansion,as figure shows.

SteelBrass

(a) (b) Heated (c) Cooled

(a) A bimetallic strip and how it behaves when (b) heated and (c) cooled

Page # [13]

Often brass [ = 19 106 (C)1] and steel [ = 12 106 (C)1] are selected. The two pieces arewelded or riveted together. When the bimetallic strip is heated, the brass, having the larger value of ,expands more than the steel. Since the two metals are bonded together, the bimetallic strip bends into anarc as in part b, with the longer brass piece having a larger radius than the steel piece. When the strip iscooled, the bimetallic strip bends in the opposite direction, as in part c.

Mathematical Solution

2dR = L0 (1 + 1)

l

d

d

2dR = L0 (1 + 2)

2dR2dR

=

2

1

11

R

(R+d/2) (Rd/2)

Find R

{Home Work - HCV : Chapter - 25 }

The expansion in length is called linear expansion. The expansion in area is called area expan-sion. The expansion in volume is called volume expansion

= = =

Ex. Show that the coefficient of area expansion, (A/A)/T. of a rectangular sheet of the solid is twice itslinear expansively, 1.

Ans.

Consider a rectangular sheet of the solid material of length a and breadth b (Fig.). When the temperatureincreases by T, a increases by a = a 1T and b increases by b = b T. From figure the increasein area

Page # [14]

A = A1 + A2 + A3A = a b + b a + (a) (b)

= a 1b T + b 1 aT + (1)2 ab (T)2

= 1ab T (2 + 1 T) = 1 A T (2 + 1 T)Since 1 ; 10

5 K1, the product 1 T for fractional temperature is small in comparison with 2 and maybe neglected. Hence,

AA

T1

= 21

Volume Thermal ExpansionThe volume of a normal material increases as the temperature increases. Most solids and liquids behavein this fashion. By analogy with linear thermal expansion, the change in volume V is proportional to thechange in temperature T and to the initial volume V0, provided the change in temperature is not toolarge. These two proportionalities can be converted into equation V = V0T with the aid of aproportionality constant , known as the coefficient of volume expansion. The algebraic form of thisequation is similar to that for linear expansion, L = L0T.

Volume thermal expansionThe volume V0 of an object change by an amount V when its temperature changes by an amount T.

V = V0Twhere is the coefficient of volume expansion.Common Unit for the coefficient of volume Expansion : (C)1

The unit for , like that for , is (C)1. Values for depend on the nature of the material. The values of for liquids are substantially larger than those for solids, because liquids typically expand more thansolids, given the same initial volumes and temperature expansion is three times greater than thecoefficient of linear expansion : = 3.If a cavity exists within a solid object, the volume of the cavity increases when the object expands, justas if the cavity were filled with the surrounding material. The expansion of the cavity is analogous to theexpansion of a hole in a sheet of material. Accordingly, the change in volume of a cavity can be foundusing the relation V = V0T, where is the coefficient of volume expansion of the material thatsurrounds the cavity.

The expansion of holesAn interesting example of linear expansion occurs when there is a hole in a piece of solid material. Weknow that the material itself expands when heated. But what about the hole? Does it expand, contract,or remain the same? Following example provides some insight into the answer of this question

ExampleDo holes expand or contract when the temperature increases?Figure (a) shows eight square tiles that are arranged to form a square pattern with a hole in the centre. Ifthe tiles are heated, what happens to the size of the hole?

Hole

(a) unheated

Expanded hole

(b) Heated (c)

9th tile(heated)

Page # [15]

Reasoning and solution We can analyze this problem by disassembling the pattern into separate tiles,heating, it is evident from figure (b) that the heated pattern expands and so does the hole in the centre. Infact, if we had a ninth tile that was identical to and also heated like the others, it would fit exactly into thecentre hole, as figure (c) indicates. Thus, not only does the hole in the pattern expand, but it expandsexactly as much as one of the tiles. Since the ninth tile is made of the same material as the others, we seethat the hole expands just as if it were made of the material of the surrounding tiles.The thermal expansion of the hole and the surrounding material is analogous to a photographicenlargement ; in both situations everything is enlarged, including holes.Thus, it follows that a hole in a piece of solid material expands when heated and contracts when cooled,just as if it were filled with the material that surrounds it. If the hole is circular, the equation L = L0Tcan be used to find the change in any linear dimension of the hole, such as its radius or diameter. Exampleillustrates this type of linear expansion.

Asking questionFigure shows a cross-sectional view of three cylinders, A, B and C. Each is made from a differentmaterial ; one is lead, one is brass, and one is steel. All three have the same temperature, and barely fitinside each other. As the cylinders are heated to the same, but higher, temperature, cylinder C falls off,While cylinder A becomes tightly wedged to cylinder B. Given, lead has the greatest coefficient of linearexpansion, followed by brass, and then by steel.Which cylinder is made from which material?

Brass

Steel

Lead

A

BC

(a)

Brass

Steel

Lead A

BC

(b)

Reasoning and solution We need to consider how the outer and inner diameters of each cylinderchange as the temperature is raised. With respect to the inner diameter, we will be guided by the fact thata hole expands as if it were filled with the surrounding material. These data indicate that the outer andinner diameters of the lead cylinder change the most, while those of the steel cylinder change the least.Since the steel cylinder expands the least, it cannot be the outer one, for if it were, the greater expansionof the middle cylinder would prevent the steel cylinder from falling off. The steel cylinder also cannot bethe inner one, because then the greater expansion of the middle cylinder would allow the steel cylinder tofall out, contrary to what is observed. The only place left for the steel cylinder is in the middle, whichleads to the two possibilities in figure. In part a, lead is on the outside and will fall off as the temperatureis raised, since lead expands more than steel. On the other hand, the inner brass cylinder expands morethan the steel that surrounds it and becomes tightly wedged, as observed. Thus, one possibility is A =brass, B = steel, and C = lead.In part b of the drawing, brass is on the outside. As the temperature is raised, brass expands more thansteel, so the outer cylinder will again fall off. The inner lead cylinder has the greatest expansion and willbe wedged against the middle steel cylinder. A second possible answer, then is A = lead, B = steel, andC = brass.

Page # [16]

Thermal StressEx. A brass rod of length 1m is fixed to a vertical wall, at one end, with the other end free to expand. When

the temperature of the rod is increased by 120C, the length increases by 2cm. What is the strain?Sol. After the rod expands, to the new length there are no elastic forces developed internally in it.

So, strain = 0.

FL

D

D D

F

L

top view

Note : A change in shape/size i.e., dimensions need not necessarily imply a strain. For example, if a bodyis heated to expand, its volume change, as it acquires a new size, due to expansion. However, thestrain remains zero. Unless and until, internal elastic forces operate, to bring the body to theoriginal state, no strain exists. When a body is heated, the total energy of molecule increase,owing to an increase in the kinetic energy of the molecules. This results in a shift (increment) ofthe equilibrium distance of molecules and the body acquires a new shape and size, in theexpanded form, whereby the molecules are in zero force state. Hence, there is no strain. How-ever, if the body is resistricted to expand, during the process of heating, then the molecules be-come strained, and even if there is no apparent change in dimensions of the body, there isstrain. In such cases, strain is measured as the ratio. In dimension that would have occured, andthe change in dimension that would have occured, had the body been free to expand or contract,to the original dimension.When a metal rod is heated or cooled it tends to expand or contract. If it is left free to expand orcontract, no temperature stresses will be induced. However, if the rod be restricted to change its length,then temperature stresses are generated within it. Stress induced due to temperature change can beunderstood as follows:

L lA B

F F

Consider a uniform rod AB fixed rigidly between two supports. (fig.) If L be its length, the coefficientof linear expansion, then a change in temperature of , would tend to bring a change in its length byl = L. Had the rod been free (say one of its ends) its length would have changed by l. Now, let aforce be gradually applied so as to restore the natural length. Since the rod, tends to remain in the newstate, due to a change in temperature, so when a force F is applied, thermal stress is induced. Inequilibrium.

AF

= )L( ll YY [ stress = strain Y]

Neglecting l in comparison to L,

F = LAl

Y = AY

Now, if the two ends remain fixed, then this external force is provided from the support.

Clearly strain = L

=

Page # [17]

Ex. The stress on a steel beamA steel beam is used in the roadbed of a bridge. The beam is mounted between two concrete supportswhen the temperature is 22C, with no room provided for thermal expansion.

Beam

Concretesupport

Concretesupport

What compressional stress must the concrete supports apply to each end of the beam, if they are tokeep the beam from expanding when the temperature rises to 42C?

Sol. Reasoning Recall from that the stress( force per unit cross-sectional area or F/A) required to changethe length L0 of an object by an amount L is

Stress = Y 0LL

where Y is Youngs modulus. If the steel beam were free to expand because of the change intemperature, the length would change by L = L0T. Because the concrete supports do not permitany expansion, they must supply a stress to compress the beam by an amount L. Thus,

Stress = Y 0LL

= Y 0

0

LTL

= YYT

Youngs modulus and the coefficient of linear expansion for steel are Y = 2.0 1011 N/m2 and = 12 106 (C)1, respectively. The change in temperature from 22 to 42C is T = 20C. Thethermal stress isStress = YT = (2.0 1011 N/m2) [12 106(C)1] (20C) = 4.8 107 N/m2]

Ex. A rod of length 2m is at a temperature of 20C. Find the free expansion of the rod, if the temperature isincreased to 50C. Find the temperature stresses produced when the rod is (i) fully prevented to ex-pand, (ii) permitted to expand by 0.4 mm. Y = 2 1011 Nm2 ; = 15 106 per C.

Sol. Free expansion of the rod = = (15 106 /C) (2m) (50 20)C= 9 104 m = 0.9 mm

(i) If the expansion is fully prevented

then strain = 2109 4

4.5 104

temperature stress = strain Y= 4.5 104 2 1011 = 9 107 Nm2

(ii) If 0.4 mm expansion is allowed, then length restricted to expand = 0.9 0.4 = 0.5 mm

Strain = 2105 4

= 2.5 104

Temperature stress = strain Y= 2.5 104 2 1011 = 5 107 Nm2

Ex. Two rods made of different materials are placed between massive walls. The cross section of the rodsare A1 and A2, their lengths L1 and L2 coefficients of linear expansion 1 and 2 and the modulii ofelasticity of their materials Y1 and Y2 respectively. If the rods are heated by tC, find the force F withwhich the rods act on each other.

Page # [18]

Sol. Let the first rod expand slightly (say by length l ) and the second rod get compressed by the sameamount (since net elongation / compression of the rods is zero.) Natural increase in length of the first rod (after being heated) when free to expand would havebeen 1L1t. The expansion allowed is just l (where l < 1L1t). Amount of elongation restricted = 1L1t l

Strain = lengthoriginalrestrictedelongation

= )t1(LtL

11

11

l

Since 1t

Page # [19]

However, after sometime, the liquid level eventually rises to Z. The entire phenomenon can be under-stood as follows: Upon being heated, the container gets heated first and hence expands. As a result, thecapacity of the flask increases and hence the liquid level falls.After sometime, the heat gets conducted from the vessel to the liquid and hence liquid also expandsthereby rising its level eventually to Z. Since, the volume expansivity of liquids, in general, are far morethan that of solids, so the level Z will be above the level X.

Effect of temperature on densityWhen a solid or liquid is heated, it expands, with mass remaining constant. Density being the ratio ofmass to volume, it decreases. Thus, if V0 and Vt be the respective volumes of a substance at 0C andtC and if the corresponding values of densities be 0 and t, then the mass m of the substance is givenby

m = V00 = VttBut Vt = V0 (1 + t), so t = 0 (1 + t)

1

Ex. The volume of a glass vessel filled with mercury is 500 cc, at 25C. What volume of mercury willoverflow at 45C?the coefficients of volume expansion of mercury and glass are 1.8 104 / C and 9.0 106 / Crespectively.

[Sol. The volume of mercury overflowing will be the expansion of mercury relative to the glass vessel (i.e.,apparent expansion).Now, since (V)a = (V)r (V)cApparent expansion (V)a will be

(V)a = V11T VCCT

= 500 cc (180 9) C

10 6

(45 25)C

= 1.71 ccThus, 1.71 cc of mercury overflows. ]

Dulong and Petits MethodIt is a special method of determining the cubical expansivity of a liquid without the intervention of thecontainer. fig. shows the experimental arrangement consisting of a glass U tube filled with the experimen-tal liquid to the same height in the two limbs. Both the limbs are enclosed in two different jackets,maintained at steam point (100C) and ice point (0C) by means of steam and ice cold water circulatedin the two jackets respectively.

Page # [20]

The horizontal portion of the U tube is covered with a blotting paper continuously soaked in cold water,to avoid heat flow from the liquid in the limb at a higher temperature of that at a lower temperature. Thetwo jackets contain accurate thermometers to yield steady readings, the levels of the liquid in two limbsare recorded (say h1 and h2). Since, the base of the liquid columns in the two limbs are at the samehorizontal level, pressures, at those points with be the same. If 0 and 100 be the densities of the liquid at0C and 100C and P0 the atmospheric pressure, then

P0 + h10g = h2100g + 0or h10 = h2100

100 = 100.10

So, h2 = h1 (1 + 100 )

or = 1

12

h100hh

The experiment can be carried, even with temperatures other than the ice and the steam point. If T1Cand T2C be the temperatures of the liquid columns in the two limbs, corresponding to heights h1 and h2respectively, then

P0 + h11g = P0 + h22gor h11 = h22

or1

01

T1h

= 2

02

T1h

or1

2

hh

= 1

2

T1T1

or = )ThTh(hh

1221

12

]

Ex. An automobile radiatorA small plastic container, called the coolant reservoir, catches the radiator fluid that overflows when anautomobile engine becomes hot (see figure).

Radiator

An automobile radiator and a coolant reservoir for catching the overflow from the radiator

The radiator is made of copper, and coolant has a coefficient of volume expansion of = 4.10 104 (C)1. If the radiator is filled to its 15-quart capacity when the engine is cold (6.0 C),how much overflow from the radiator will spill into the reservoir when the coolant reaches its operatingtemperature of 92C?Reasoning When the temperature increases, both the coolant and radiator expand. If they were toexpand by the same amount, there would be no overflow. However, the liquid coolant expands morethan the radiator, and the amount of overflow is the amount of coolant expansion minus the expansion ofthe radiator cavity.

Page # [21]

Sol. When the temperature increases by 86C, the coolant expands by an amountV = V0T = [4.10 10

4 (C)] (15 quarts) (86 C) = 0.53 quartsThe volume of the radiator cavity expands as if it were filled with copper [ = 51 106 (C)1.The expansion of the radiator cavity is

V = V0T = [51 106 (C)1] (15 quarts) (86 C) = 0.066 quarts

The amount of coolant overflow is 0.53 quarts 0.066 quarts = 0.46 quartsAnomolous expansion of waterWhile most substances expand when heated, a few do not. For instant, if water at 0C is heated, itsvolume decreases until the temperature reaches 4C. Above 4 C water behaves normally, and itsvolume increases as the temperature increases.Because a given mass of water has a minimum volume at 4C, the density (mass per unit volume) ofwater is greatest at 4 C, as figure shows.

999.6999.7999.8999.9

1000.0

0 2 4 6 8 10

Maximum density at 4C

Temperature, C

Den

sity

, kg/

m3

The density of water in the temperature range from 0 to 10C. Water has a maximum density of999.973kg/m3 at 4C. (This value for the density is equivalent to the often quoted density of 1.000grams per milliliter)When the air temperature drops, the surface layer of water is chilled. As the temperature of the surfacelayer drops toward 4C, this layer becomes more dense than the warmer water below. The denserwater sinks and pushes up the deeper and warmer water, which in turn is chilled at the surface. Thisprocess continues until the temperature of the entire lake reaches 4C. Further cooling of the surfacewater below 4C makes it less dense than the deeper layers ; consequently, the surface layer does notsink but stays on top. Continued cooling of the top layer to 0C leads to the formation of ice that floatson the water, because ice has a smaller density than water at any temperature. Below the ice, however,the water temperature remains above 0C. The sheet of ice acts as an insulator that reduces the loss ofheat from the lake, especially if the ice is covered with a blanket of snow, which is also an insulator. Asa result, lakes usually do not freeze solid, even during prolonged cold spells, so fish and other aquatic lifecan survive.

{Home Work - HCV : Chapter -23 Ex 10 - 34,Question & Short Answer Leave 4, 5Obj. - I Leave 1 to 4Obj. - II Leave 4 }

Note : For any suggestion or correction please contact Amit Gupta or give it in computer room.