Elasticity 1
December 13, 2016 | Author: Sesha Sai Kumar | Category: N/A
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Elasticity theory part for the intermediate, eamcet, iit, jee mains students and for various other competitive exams by ...
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PHYSICS - I C
ELASTICITY
ELASTICITY *
I nter atomic for ces
* Elastic M odulii
1
* Behaviour of wir e under str ess * Elastic ener gy
1.1 I NTRODUCTI ON Elasticity deals with property of a material, its strength and ability to withstand against external forces which are acting on it. While selecting a suitable material for a project, an engineer is always interested to know its strength. The strength of a material may be defined as an ability to resist its failure under the action of external forces. As a matter of fact the properties of a material under the action of external forces are very essential, for an engineer, to enalbe him, in designing him all types of structures and machines.The properties of matter like elasticity, surface tension, viscocity, can be studied well with the help of interatomic and intermolecular forces.
F
The for ces acting between the atoms due to electr ostatic inter action between the char ges of the atoms ar e called inter atomic for ces. Thus interatomic forces are electrical in nature. The interatomic forces are active if the distance between the two atoms is of the order of atomic size 1010 m . During interaction between the two atoms, the following electrostatic forces will be active (i)Attractive forces between the nucleus of one atom AKASH MULTIMEDIA
lsion
1.2 I NTERATOM I C FORCES
repu
The forces acting among charged particles in an atom are responsible for structure of atom. The electromagnetic forces acting among atoms are responsible for the structure of molecules. The electromagnetic forces acting among the molecules are responsible for the structure of matter and their elastic behaviour.
and electrons of the other atom. These attractive forces tend to decrease the potential energy of the pair of atoms. (ii) The repuslive forces between the nucleus of one atom with the nucleus of another atom and electrons of one atom with the electrons of the other atom. These repulsive forces tend to increase the potential energy of pair of atoms. The potential energy U is related with the force dU F by the relations. F . The variation of dr potential energy U(r) and interatomic force F(r) with separation r between two atoms have been shown in graphs (fig (1), fig (2))
x O
r0
r
attraction fig.(2) From graphs the following points are observed. (i)At large distances, the potential energy is negative and becomes more negative as r decrease. It implies that interatomic force in this region is 5
PHYSICS - I C
ELASTICITY
attractive. For a particular value of r denoted by x in fig.(2), the attractive interatomic force becomes maximum. After this distance x, the attractive force starts decreasing rapidly with the further decrease in the value of r
the force of attraction becomes maximum. After this distance, the force of attraction decreases and becomes zero at a distance r0. It is found that foce of attraction between the molecules varies inversely as the seventh power of intermolecular distance r, i.e.,
(ii) At a distance r0 the potential energy attains minimum value (maximum negative value). At this stage, the two atoms will be in a state of equilibrium. The distance r0 is called as normal or equilibrium distance. At this distance, the attractive force between two atoms will become zero.
1 a orFa 7 The negative sign indicates that 7 r r the force is attractive in nature.
(iii) As the distance is further decreased below r0, the potential energy starts increasing, becomes zero for a particular value of r and after this becomes positive. In this region, the interatomic force is repulsive, the repulsive force increases very rapidly as the distance between the two atoms decreases. So, the two atoms cannot be fused together easily. therefore, atoms are regarded as hard elastic spheres. 1.3 I NTER M OL ECUL AR FORCES The for ce between the molecules due to electr ostatic inter action between the char ges of the molecules ar e called inter molecular for ces. Thus intermolecular forces are also electrical in origin. These forces are active if the separation between two molecules is of the order of molecular size 10 9 m The variation of intermolecular forces with distance is shown in fig. repu
F
lsion
x O
r0
r attraction fig.(2)
(i) For large distnace r, the intermolecular force is neglisibly small. As the distance decreases, the force of attraction increases. At a particular distance x, AKASH MULTIMEDIA
Fa
(ii) When the distance between the molecules becomes less than r0, the force becomes repulsive in nature. The repulsive force increases very rapidly with decrease in intermolecular distance. It is found that repuslive force varies inversely as the ninth power of r, ie. Fr
1 b or Fr 9 9 r r
1.4 : Some impor tant defimitions (i)Defor ming For ce : When an exter nal for ce is applied on a body which is not fr ee to move, the molecules of the body ar e for ced to under go a change in their r elative positions. Due to this change, the body may suffer a change in length (or ) volume (or ) shape. Such a body is said to be deformed. The applied for ce is called defor ming for ce. (ii) Restor ing for ce : The for ce developed within t he body on account of r elat ive molecular displacement is called inter nal for ce (or ) elastic for ce (or ) r estor ing for ce. At equilibrium the restoring force developed in a body is equal and opposite to the deforming force applied on the body. (iii) Rigid body : A body is said to be r igid if the r elative positions of its constituent par ticles r emain unchanged inspite of any amount of defor ming force. There is no perfect rigid body. The nearest approach to a rigid body is diamond. (iv) Elasticity : I t is the pr oper ty of mater ial of a body by vir tue of which the body r egains its or iginal length, volume and shape after the defor ming for ces have been r emoved. 6
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If a body regains its original length, volume and shape completely when the deforming forces are removed, then the body is said to be a perfectly elastic body. There is no perfectly elastic body in nature. The nearest approach to a perfectly elastic body is “Quartz fiber”. (v) Reason for elasticity : In a solid, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to neighbouring molecules. These forces are known as intermolecular forces. The two molecules in their equilibrium positions are at certain separation (r = r0) called inter – molecular seperation. At this separation the potential energy is minimum. On applying the deforming forces, the molecules either come closer or go far apart from each other. In both the cases potential energy of molecules is greater than the minimum. Since every system tends to remain in the state of minimum potential energy, the molecules has a tendency to come back to its original position. Tendency of the body to recover its original configuration can be interpreted as due to the presence of some forces known as restoring forces acting in a direction opposite to that of deforming forces. This gives rise to the property of elasticity. When the deforming forces are removed, these restoring forces bring the molecules of solid to their respective equilibrium positions (r = r0) and hence the body regains its original form. (vi) Plasticity : The pr operty of material of a body by vir tue of which it does not r egain its or iginal shape and size (i.e it r emains in the defor med state) even after the r emoval of defor ming for ce is called plasticity. If a body does not have any tendency to recover its original configuration on the removal of deforming force, then the body is said to be a perfectly plastic body. There is no perfectly plastic body in nature, the nearest approach to a perfect plastic body is putty. Note – 1.1 : Most of the bodies are neither perfectly elastic nor perfectly plastic. They are partially elastic. AKASH MULTIMEDIA
ELASTICITY
1.5 Str ess : When a deforming force is applied on a body, there will be relative displacement of the particles. Due to property of elasticity an internal restoring force is developed which tends to restore the body to its original state. Definition :The internal r estoring force acting per unit ar ea of cr oss–section of the defor med body is called str ess. Stress
Re storing force F Area of cross section A
At equilibrium, as the restoring force is equal in magnitude and opposite to external deforming force, stress can also be equal to external deforming force per unit area on a body. If ‘F’ is the external deforming force applied on the Area ‘A’ of the body then Stress
Deforming force ( F ) Area ( A)
* Stress is a tensor quantity * SI unit of stress is Pascal (N/m2) * C.G.S. unit of stress is dyne cm–2. * Dimensional formula of stress is (ML–1T–2) * The units and dimensions of stress are same as that of pressure. Stress developed in a body depends upon how the external forces are applied over it. On this basis there are two types of stresses. They are (a) Normal stress (b) Tangential ( or) shearing stress. (a) Nor mal Str ess : I f the str ess is nor mal to the sur face, it is called nor mal str ess. The stress is always normal in the case of change in length of a wire (or) volume of a body. The normal stress can further be compressive (or) tensile depending upon whether it produces a decrease (or) increase in length or volume. (i) L ongitudinal str ess : When a nor mal str ess changes the length of a body then it is called longitudinal str ess. (or ) 7
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When a force is applied nor mal to the cr osssectional ar ea of the body such that its length changes then the r estor ing for ce developed per unit cr oss-sectional ar ea is called longitudinal stress. Deforming force (F)
Longitudinal stress =
Area of cross - section (A)
It can be further divided into two types, they are tensile stress and compressive stress. 1) If a rod is stretched by two equal forces applied normal to its cross–sectional area,the restoring force developed per unit area in this case is called tensile stress. A
As shown in figure, a small solid sphere is placed in a fluid such that it 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 is said to be under hydraulic compression. This leads to decrease in its volume without any change of its geometrical shape. The internal restoring force per unit area in this case is known as bulk stress and is equal to the hydraulic pressure in magnitude (applied force per unit area) F
F
C
B
V
F
F
V V
F
F
F
F B ody subjected to tensile force A
B
B
Volume stress F Force = = (or) surface area A Bulk stress
C
F
F Tensile stress
Ex : A string fixed at one end and stretched at the other end experience tensile stress. 2) If a rod is compressed under the action of applied forces, the restoring force per unit area is called compressive stress. A
C
B
F
F Body subjected to compressive force A
B
B
F
C
F Com pressive stress
Ex: The pillars of a building experience compressive stress. (ii) Volume (or) Bulk stress : When a normal stress changes the volume of a body then it is called volume str ess. When forces of equal magnitude act on a body normally from all directions, the volume of the body changes. The body develops internal restoring forces that are equal and opposite to the forces applied. The internal restoring force per unit area in this case is known as volume stress. AKASH MULTIMEDIA
F F
= Pressure (P) (b) Shear ing stress : When the str ess is tangential to the sur face due to the application of for ces par allel to the sur face, then the str ess is called tangential (or ) shear ing stress. F Force = Shearing stress = surface area A If two equal and opposite deforming forces are applied parallel to the two surfaces of the cube as shown in the figure, there is relative displacement between the opposite faces of the cube. The restoring force per unit area developed due to the applied tangential force is known as tangential (or) shearing stress. A
F
F
A fixed
Note 1.2 : If deforming force is applied on a body such that normal stress is developed in a body, then the length or volume of the body may change. 8
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Note 1.3 : If deforming force is applied tangential to the surface, such that tangential stress is developed in a body, then the shape of the body may change. F Note 1.4 : n
er
A
q
F
F
When a force ‘F’ acts at an angle ' q ' with outward normal n to the area A as shown in figure. In this case, the stress will have the normal and tangential components. To find the linear (or) longitudinal stress, take the component of the force perpendicular to the plane of a given area A, then divide this component ( Fer ) by the area ‘A’. F F cos q Longitudinal stress = A A To find the shearing stress, take the component of force parallel to the plane of the given area and then devide F by the area ‘A’. F F sin q Shearing stress = A A The total stress = longitudinal stress + shearing stress
Being the ratio of two similar quantities, strain is a dimensionless quantity and has no unit. Like stress, strain is a tensor. Strain is classified into three types depending upon the change produced in a body, they are (i) Longitudinal strain (ii) Volume strain (iii) Shearing strain (i) Longitudinal strain : It isthe ratio of the change in length of a body to its or iginal length. Consider a wire of length ' ' and is suspended from a rigid support. Let a stretching force ‘F’ be applied normally to its face. Let the wire suffer a change in its length. F
er
el
F
F
el
But not F/A. * Pr oblem 1.1 A steel wir e of 2mm in diameter is str etched by applying a for ce of 72N. Find the str ess in the wir e. Solution : r 1103 m; F=72N F F 72 The stress = 2 2 A pr p 1103 72 2.292107 Nm2 . p 106 1.6 Str ain : When the forces (or) a torque acting upon a body causes relative displacements of its particles, a change in length (or) volume (or) shape is produced. The body is then said to be strained. Def : The r at io of change pr oduced in t he dimensions of a body by a system of for ces or couples in equilibr ium to its or iginal dimensions is called str ain.
=
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change in length
Longitudinal strain =
original length
I f the length incr eases due to tensile str ess, the cor r esponding str ain is called tensile str ain. I f the length decr eases due to compressive str ess, the strain is called compr essive str ain. (ii) Volume str ain : I t is the r atio of change in volume of body to its or iginal volume. Let ‘v’ be the volume of a given body. Under the action of a normal stress, let the change in volume of the body be V .Then F
F
V- V
V
F
F
change in volume V Volume strain = original volume V 9
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(iii) Shear ing str ain : I t is defined as angle q in r adians thr ough which a plane per pendicular to the fixed sur face of the cubical body gets tur ned under the effect of tangential for ce . I t is also the r atio of the displacement of a x layer to its distance from the fixed layer. q
Consider a cube of material fixed at its lower face and acted upon by a tangential force ‘F’ at its upper surface as shown in figure (a). The upper surface is displaced relative to lower surface by ‘x’ as show in the figure (b). The perpendicular distance between upper and lower fixed layer is ' ' . Then F
G F
A
Fixed Surface fig (a)
C
x
A
F
H
E D
A
B
F
q
q C
B
Fixed fig (b)
Shearing strain = q
D
x
* Pr oblem 1.2 A copper wir e of length 1m is str etched by 1cm. Find the str ain on the wir e Solution : The strain
e 1102 0.01 L 1
* Pr oblem 1.3 If a platinum wir e is stretched by 0.5% what is the str ain on the wir e? Solution : e Fractional increase in the length = strain L 0.5 0.005 . The strain =0.5% = 100 AKASH MULTIMEDIA
1.8 Hooke’s law: Hooke’s law states that within elastic limit, the str ess is dir ectly pr opor tional to the strain stress strain (for small deformations) stress = E x (strain) stress E strain
Where E is proportionality constant and it is also called modulus of elasticity. Def : M odulus of elasticity of the mater ial of a body is the str ess in the body to pr oduce unit str ain (within the elastic limit).
B
F
1.7 Elastic limit : The maximum str ess within which a body can r egain its or iginal size and shape after the r emoval of the deforming for ce is called elastic limit. If the stress developed in a body excedes this limit, then it will not get the initial size and shape completely, even after the removal of deforming force.
E depends on the nature of the material, temperature and impurities. It is independent of dimentions of the body. SI unit of ‘E’ is Pascal (Pa) 1pa 1N / m 2 C.G.S. unit is dyne / cm2 Dimensional formula is ML1T 2 1.9 Factor s effecting Elasticity: (i) ANNEAL I NG : The processes responsible for making uniform structures from a given sample reduce the elasticity of a material while those responsible for generating smaller regular units inside the sample increase the elasticity. Hence annealing decreases elasticity while hammering and rolling increases it. (ii)I M PURI TI ES: The impurity having higher elasticity than the sample to which it is added increases the elasticity while the impurity with smaller elasticity decreases the elasticity of the sample. (iii)TEM PERATURE: Normally, elasticity of the material gets decreased with rise in temperature. 10
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However, INVAR STEEL is a material whose elastic behaviour is not affected by rise in temperature.
1.10 Types of modulii Depending on the type of str ess developed and the r esulting str ain, we have the following three modulii of elasticity. (i) Young’s modulus (ii) Bulk modulus (iii) Modulus of Rigidity 1.11 Young’s modulus (Y) : I t is the r atio of longitudinal str ess to the longitudinal str ain within elastic limit. longitudinal stess i.e., Y longitudinal strain F
Pr oblem 1.5 A load (M ) suspended fr om a wir e pr oduces an elongation (e) in the wir e then find the r ise in temperature required to produce same elongation in the same wir e.
F
Consider a wire of length ' ' and cross sectional area ‘A’. One end of the wire fixed to rigid support and a stretching force ‘F’ is applied normally to its face as shown in the figure. Due to the stretching force, the length of the wire changes by . Then at equilibrium. F Longitudinal stress = A Longitudinal strain = F longitudinal stress A F Y = longitudinal strain = A If a force is applied on a wire of radius ‘r’ by hanging a load of mass ‘M’ as shown in figure
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F Y 1 , Hence A
But as
F A Y sol :
Mg 2 pr Mg Y 2 then Y p r
Pr oblem 1.4 Show that str ess r equir ed to double the length of wire (or) to pr oduce 100% longitudinal str ain is numer ically equal to Young’s modulus. Sol : 1 , 2 2 F A 2 , Y
M
F .......1 AY
a
t
at.......2
But from given data
F F at t AY AYa Note 1.5 : If two wires having lengths 1 , 2 ; crosssectional areas A1, A2 and Young’s modulii Y1, Y2 are stretched by forces F1, F2 then F Y1 F1 1 A2 e2 As Y Ae Y F A e 2 2 2 2 1
Note 1.6 : In terms of volume V of a wire. As
Y
F F F 2 F2 Y y Ae Ve A e Ve
Where ‘V’ is the volume of the wire m m as d V V d 2 F d y me 11
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m mass of the wire d density of material of the wire In the above formula If V, F, Y are same for two e1 12 2 e a wires then e2 22 In the above formula if Y, d, F are same for two wires 2 2 e1 1 m2 then e a m e2 2 m1 Note 1.7 : As Y
Pr oblem 1.7 Two wires of same length and r adius are j oined end to end and loaded. The Young’s modulii of the mater ials of the two wires are Y 1 and Y 2. I f the combination behaves as a single wire then its Young’s modulus is
e1
F F A FV 2 2 Ae Ae Ae
FV FV Y 2 4 2 Ae pr e m m as d V V d
Y1 A
e Y2 A
e2
2 AY
Y
Y
W
Fm Fm Y dA e d p 2 r 4e
In the above formula if F, V, Y are same for two wires then ea
1 1 ea 4 (or) A r
Pr oblem 1.6 When a body of mass ‘m’, density dB is suspended from a wir e, its elongation is ‘e’when the body is in air. I f the body is completely immer sed in a non – viscous liquid of density d then its elongation is
In liquid
In air F = Wa = mg as e
F AY
d F 1 W mg 1 d B
e a F (as , A, Y are same in both cases)
d 1 mg d 1 1 1 dB 1 e F e e dB e F mg
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W
e e1 e2 F 2 F F e , e1 , e2 but AYeq AY1 AY2 F 2 F F AYeq AY1 AY2 2 1 1 Yeq Y1 Y2 Yeq
2Y1Y2 Y1 Y2
Pr oblem 1.8 Two wires of same length and r adius are j oined in par allel and loaded. The Young’s modulii of the mater ial of the wires areY 1 & Y 2. I f the combination is taken as a single wir e then itsYoung’s modulus is Y1
y2 A
A W
Yeq
2A W
F = F1 + F2 Y 2 A e Y1 Ae Y2 Ae 2Yeq Y1 Y2 Yeq
Y1 Y2 2 12
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* Pr oblem 1.9 The length of a metal wir e is 1 when the tension in it is T 1 and is 2 when the tension is T 2. Then the actual length of the wire is Sol : Let its original length be ' ' .
Solution : r = 0.5 x 10–3m; L=1m F = 10N ; e=0.064 x 10–3m 10 F F i) Stress = 2 = p 2 = A pr
We know e T 1 T1 .... (1)
2 T2 ...... (2) From (1) & (2) 1 T 1 T2 1 T2 T1 2 T1 2 T2 T T T2 1 T1 2 T2 T1 2 1 1 2 T2 T1 Pr oblem 1.10 The length of a r ubber cor d is 1 metr es when the tension in it is 4N and 2 metr es when the tension is 5N. Then the length in meters when the tension is 9 N is sol : Let ' ' be the original length and 3 be the length of the wire when the tension is 9N. We know e F
1 4.......1 , 2 5....... 2
3 9.......3 1 4 From (1) & (2) 5 51 5 4 2 4 2 5 1 4 2 ....... 4
-2 1.27310 7 N m
e 0.064 103 (ii) Strain = L 1 = 0.064 x 10–3
Stress 1.27310 7 (iii) Y= Strain 0.064103 = 1.989x1011Nm–2 * Pr oblem 1.12 A steel wir e of diameter 1 mm and length 2m is str etched by applying a for ce of 2kg wt. Calculate (i) the incr ease in length of the wir e, (i i ) t he st r ai n and (i i i ) t he st r ess. (g = 9.8 ms–2, Y = 2 x 1011 N m–2) 1 3 Sol: r 10 m; L 2m; 2 F = 2kg wt = 2 x 9.8N; Y = 2 x 1011 Nm–2
i)
1 5 1 4 2 4 1 4 From (1) & (3) 9 5 4 9 1 3 3 2 51 4 2 4 4 2 4 1 4 1 3 51 4 2 9 3 51 4 2 9
36 2 361 4 3 201 16 2 20 2 16 1 4 3 3 5 2 4 1
* Pr oblem 1.11 A steel wir e of 1mm diameter and of length 1m is str etched by applying for ce of 10N. I f the incr ease in length is 0.064mm, find (i) the str ess, (ii) the strain and (iii) theYoung’s modulus of the wir e. AKASH MULTIMEDIA
Y e
F L r 2 e
FL 2 9.8 2 2 2 r Y 1 3 10 2 1011 2
= 2.495x10–4m ii)
The strain=
e 2.495104 L 2
= 1.248104 iii) The stress = Y x strain 4 11 2.49510 2 10 = 2
= 2.49510 7 Nm 2 13
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* Pr oblem 1.13 What mass must suspended fr om the fr ee end of a steel wir e of length 2m and diameter 1mm to str etch it by 1mm? (Y=2x1011Nm–2) 1 Mg L 3 Sol: r 10 m; L 2m; Y 2 2 r e
M
Y r 2 e gL
1 2 3 210 10 1103 2 9.82 11
102 100 8.015kg 9.8 4 39.2 * Pr oblem 1.14 A br ass wir e of length 5m and cr oss section 1mm2 is hung fr om a r igid suppor t, with a br ass weight of volume 1000 cm3 hanging fr om the other end. Find the decr ease in the length of the wir e, when t he br ass weight is complet ely immer sed in water. (Ybrass=1011 Nm–2; g=9.8ms–2; water 1gcm 3 ) Sol: When a weight is hung in air from the other end of a wire, F = Mg. The increase in length of the wire, e = ? FL Young's modulus, Y Ae MgL e . AY When weight hung in a liquid, Weight of the body in the liquid = Mg - V g where V is the volume of the body This is the force, F acting on the wire i.e., F = Mg – V g Increase in length of the wire, e/
Mg vg L
AY which is less than the increase in length of the wire when the weight is in air. Decrease in length = e in air - e| in liquid
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MgL Mg Vg L AY AY
VgL AY Here, V = 1000cm3 = 1000 x 10-6m3 =1gcm–3=1x103kgm–3;g=9.8ms–2;L=5m
A 1mm2 1106 m2 ;Y 11011 Nm2 The decrease in length 1000106 110 3 9.8 5 = 1106 11011 = 49 x 10-5 m = 0.49 mm * Pr oblem 1.15 A copper wir e and a steel wir e of r adii in the r atio 1:2, lengths in the r atio 2:1 ar e stretched by the same force. I f the Young's modulus of copper = 1.1 x 1011Nm–2 find the r atio of their extensions (youngsmodulus of steel = 2 x 1011 N/m2). FL Sol: we know e 2 r Y e1 L1 r 2 r1 e 2 L 2
2
Y2 F Y F 1
Here r1:r2= 1:2, L1:L2 = 2:1 Y1=1.1x1011 Nm–2; Y2 = 2.0x1011Nm–2 2 e1 2 2 2.01011 16 160 e 2 1 1 1.11011 1.1 11
e1:e2 = 160:11 * Pr oblem 1.16 An aluminium wir e and a steel wir e of the same length and cr oss-section ar e j oined end to end. The composite wir e is hung fr om a r igid suppor t and a load is suspended fr om the fr ee end. I f the incr ease in the length of the composite wir e is 2.7mm, find the incr ease in the length of each wir e. (Y A1=2x1011Nm–2, Y steel =7x1011Nm–2) Sol: Total increase in length, e = e1 + e2. e1 + e2 = 2.7 mm 14
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FL AY 1 As F, A, L are same for both the wires. So, e Y e1 Y2 21011 20 20 e1 e 2 = , 11 e 2 Y1 710 7 7 substituting in e1 + e2 = 2.7 mm
we know e
20 e1 e 2 2.7mm 7 e2 = 0.7 mm e1
27e 2 2.7mm 7
* Pr oblem 1.18 A steel wire of length 2 m and cross sectional ar ea 2 mm2 is fixed at one end and str etched by suspending a block of mass 2 kg on the sur face of the moon. I f theYpung’s modulus of steel is 2 x 1011 N m–2 find the increase in the length of the steel wir e. 1 (g on the moon of the g on the earth) 6 Sol: L = 2 m ; A = 2 mm2 = 2 x 10–6 m2;
20 20 e 2 0.7 2.0mm 7 7
M = 2 kg ; g on the moon =
* Pr oblem 1.17 A block of mass 1 kg is fastended to one end of a wir e of cr oss - sectional ar ea 2 mm2 and is r otated in a ver tical cir cle of r adius 20 cm. The speed of the block at the bottom of the cir cle is 3.5 m s–1. Find the elongation of the wir e when the block is at the bottom. Sol: i) Tension at the bottom of the circle,
mv 13.5 mg 19.8 r 0.2 2
T
2
= 61.25 + 9.8 = 71.05 N This tension in the string is equal to the force, F i.e. F = 71.05N, L =r = 0.2m. The increase in length , FL 71.050.2 e AY 2106 21011
= 3.553 x 10–5m ii)
Tension at the top of the circle, T = Tension at the bottom – 6 mg = 71.05 – 6 x1x9.8 = 71.05 – 58.8 = 12.25 N. F = 12.25 N; L = 0.2 m. The increase in length e
FL 12.25 0.2 AY 2106 21011
g 9.8 2 ms 6 6
N ;e ? m2 The increase in length , Y 21011
e
FL MgL 29.82 AY AY 62106 21011
= 1.633 x 10–5m Pr oblem 1.19 One end of a unifor m wire of length ‘L’and mass ‘M ’is attached r igidly to a point in the r oof and a load of mass ‘m’is suspended from its lower end. I f A is the area of cross - section of the wire then find the str ess in the wir e at height ‘x’fr om its lower end (x < L ) Tension in the string at point ‘P’ is (L-x)
T = wt of load + wt of wire of length ‘x’ M x T mg xg L mg Mxg m Stress at P = T/A = A AL Pr oblem 1.20 A metal ring of radius ‘r’and cross- sectional area ‘A’ is to be fitted on to a wooden circular disc of r adius R (R > r ). I f the Young’s modulus of the mater ial of the r ing is Y the force with which the metal r ing expands is P L
x
R
r
(R > r)
= 0.6125105 m AKASH MULTIMEDIA
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Initial length of wire 2p r for it to be fitted onto a wodden disc, its final length must be 1 2p R F Ae YA R r
e 1 2p R r , Y F
F
YAe YA2 p R r 2pr YA R r
r
r
Pr oblem 1.21 I f two wir es ar e ar r anged as shown in the figur e. What ar e the elongations of upper and lower wir es Sol : for lower wire F = m2g
Pr oblem 1.23 A copper wire of negligible mass, length , cr oss - sectional ar ea (A) is kept on a smooth horizontal table with one end fixed, a ball of mass ‘m’ is attached at other end. The wir e and the ball are rotated with angular velocity ‘w ’. If wire elongates by then find Young’s modulus of wir e. I f on incr easing the angular velocity fr om w to w 1 the wir e br eakdown, obtain br eaking str ess Sol : a) r F = T = mrw2
m w 2 F as in small , F mw 2 y Ae
F2 m g e e 2 2 Ay2 Ay2
A y1 1 m1 A 2 For upper wire F = (m1 +m2)g y2 m2 e
m m2 g 1 F 1 e 1 Ay1 Ay1
C
elongation of 2nd wire 3 A y3 mg 2 e2 D Ay2 m elongation of 3rd wire mg 3 e3 Ay3 displacement of B is e1 displacement of C is e1 + e2 displacement of D is e1 + e2 + e3
A
b) We know Breaking stress Breaking force = Area of cross sec tion m w 1 = A
Pr oblem 1.22 As shown in adjucent figur e if a load of mass (m) is attached at lower end of lower wir e. Then find the displacements of the points B, C, D ar e Sol : As shown in figure elongation of first wire A y1 1 mg 1 B e1 2 Ay1 A y2
AKASH MULTIMEDIA
y
mw 2
2
Pr oblem 1.24 A stone of mass (m) is attached to one end of a small wir e of length and cr oss - sectional area (A) suspended ver tically. The stone is now r otated in hor izontal plane such that the wir e mak es an angle ' q ' with ver t ical. Find the increase in length of wire if itsYoung’s modulus is Y. mg Sol : From fig. T cos q mg T cos q S T sin q mRw 2
q L
e
e
F T AY AY mg AY cos q
T T cos q q
R
T sin q
mg
16
PHYSICS - I C
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Note : If ' q ' was not given but R, w and m was given then in such case T 2 cos 2 q T 2 sin 2 q mg mRw 2
T mg mRw 2
and use e
2
2 2
ii) equal str ains in both wir es Young’s modulus of br ass = 1 x 1011 N/m2 Young’s modulus of steel = 2 x 1011 N/m2
Brass wire
Steel wire
A
T2 B
x
As the whole system is in equilibrium, so t = 0. Taking moment of all the forces acting on
T1 x T2 2 x 0
Pr oblem 1.25 A light r od of length 2 m is suspended fr om the ceiling hor izontally by means of two ver tical wir es of equal length tied to its ends. One of the wir es is made of steel and is of cross–section 10–3 m2 and the other is of br ass of cr oss – section 2 x 10–3m2. Find out the position along the r od at which a weight may be hung to pr oduce; i) equal str ess in both wir es
C (2 x)
Sol : Suppose a1 and a2 are the cross - sectional areas, and Y1 and Y2 are the Young’s moduli of steel and brass wire respectively. Let T 1 and T 2 are tensions in the steel and brass wires respectively. Let x is distance of the position of the hanging weight from the steel wire. i) First case : For equal stress in both wires, we have T1 T2 a1 a2 T T2 (or) 13 10 2103
......... (ii)
Solving equations (i) and (ii), we get 4 x m 3 ii) Second case : For equal strains in both the wires e1 = e2 T1 T 2 a1Y1 a 2 Y2 T1 T2 11 3 10 210 210 1011 (or) T1 = T2 ........... (iii)
(or)
3
From equations (ii) and (iii) , we get x=1m Pr oblem 1.26 A steel wire of area of cross-section A and length 2Lis clamped fir mly between two points seper ated by a distance ' 2L' . A body is hung from the middle point of the wir e such that the middle point sags by a distance x. Calculate the mass of the body and the angle made by the str ing with the horizontal
W
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......... (i)
the rod about C, we have
2
T Ay
T1
(or) T2 2T1
L
L
x T
T Mg
Since ' ' is small sin tan
x L
y
F L . A e
F
YAe YA 2 2 1/ 2 L x L L L
17
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ELASTICITY
So tension in the wire (due to elasticity)
1pa 1N / m 2 YA YA x 2 x2 F L 1 L L L 2 L 2 L L 2L
F
YAx 2 2 L2 for small angles sin
2F mg 2.
2 YAx2 x . mg , 2L2 L
x Mg L YA
Mg Tan YA
1/ 3
So at lowest point
, Tan
So 8pv 2 / 5.12 1.99.4p 8p 9.8 x L
Mg Tan1 YA
1/ 3
Pr oblem 1.27 A spher e of r adius 0.1 m and mass 8p kg is attached to the lower end of a steel wir e of length 5.0 m and diamet er 10–3m. The wire is suspended fr om 5.22 m high ceiling of a r oom. When the spher e is made to swing as a simple pendulum, it j ust gr azes the floor at its lowest point. Calculate the velocity of the spher e at the lowest position. Y for steel = 1.994 x 1011 N/m2.
T
5.22m
0.2m
mv2 CFF r
mg
Sol : As the length of the wire is 5m and diameter 2 x 0.1 = 0.2 m and at lowest point it grazes the floor which is at a distance 5.22 m from the roof, the increase in length of the wire at lowest point L 5.22 5 0.2 = 0.02 m AKASH MULTIMEDIA
[as q 0 ]
But here r = 5 + 0.02 + 0.1 = 5.12 m
YAX 3 M 3 Lg 1/ 3
,
and as equation of circular motion of a mass ‘m’ tied to a string in a vertical plane is
mv 2 / r T mg
YAx 2 mg 2 L2
x 3 Mg L3 YA
2
mv 2 / r T mg cos q
2T sin mg 2T mg
1.9941011 p 5104 0.02 YA T L 199.4p N L 5
i.e., v 2 1215.12 / 8 77.44 , so v = 8.8 m/s. 1.12 Elongation of wir e due to its own weigt: Consider a wire of length ' ' and cross sectional area ‘A’. If density of its material is ‘d’ then weight of the wire W A dg
dx
x Let the wire is hanging from the rigid support. The wire extends due to self weight Let us consider an element of thickness ‘dx’ at a distance ‘x’ from the free end. The weight of the wire of length ‘x’ is w1 = (Ax)dg The extension of the element due to this weight is
w1 dx xAdg dx dg de x dx AY AY Y
dg The total extension e de Y x dx 0 0
dg W dg2 x dx = Y , e also e 2 AY 2Y 0 18
PHYSICS - I C
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* Note 1.8 : The above formula can also derived by considering total weight at center of mass and using effective length / 2 .
Y
F / 2 Ae
cm
/2 wt
e
F mg 2 AY 2 AY
A dg 2 AY
2 e dg , 2Y
Note 1.9 : In the above case, if a force ‘F’ is applied at the lower end in addition to its weight then the total elongation is
2dg F e 2Y AY
1.13 Ther mal Str ess:
a) Find the for ce with which the r ods acts on each other at heigher temper atur e. b) Find the lengths of the r ods at the higher temperature. Assume that there is no change in the crosssectional ar ea of the r ods and the r ods do not bend. Ther e is no defor mation of walls. 1
a Y aY 2
A
2 2
1 1
B
Sol : a) Due to heating the increases in length of the composite rod will be
I 1a1t 2a2 t 1a1 2a2 t ... (1) due to compressive force ‘F’ from the walls, due to elasticity, the decrease in length will be F F F D 1 2 1 2 .... (2) AY1 AY2 A Y1 Y2 As the length of the composite rod remains unchanged the increase in length due to heating must be equal to decrease in length due to compression.
When a rod whose ends are rigidly fixed such that it is prevented from expansion or contraction undergoes a change in temperature, due to thermal expansion or contraction a compressive or tensile stress is developed init. Due to this thermal stress the rod will exert a large force on the supports. If the change in temperature of a rod of length ' ' is q o C . Then aq as a Thermal strain = q Thermal stress = Y (thermal strain) Thermal stress =
Y aq
Force F YAaq Pr oblem 1.28 Two rodsof different metals, having the same area of cross - section A, are placed end to end between two massive walls as shown in fig. If the temper atur e of both the r ods ar e now r aised by t 0 C then AKASH MULTIMEDIA
F 1a1 2a2 t 1 2 A Y1
F
b)
Y2
A 1a1 2 a2 t 1 2 Y1 Y2
As initially the length of one rod in L1 and due
to heating its length increases by 1 H a11t , while due to compression its length decreases by
1 C
F 1 AY1
so its final length
11 1 1 H 1 C F 1 = 1 1a1t AY 1
19
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1.14 Analogy of rod as a spr ing :
for the other rod
2 2 H 2 C 1 2
12 2 2a2t
We know Y
F2 AY2
Note 1.10: In the above problem length of composite rod remains unchanged, i.e. 11 12 1 2 . but that of individual rods changes i.e. 11 1 and 12 2 . Note 1.11 : In the above problem if the displacement of junction point was asked, This displacement is equal to change in length of any one of the rod F 1 1 1a1t AY1 where F 1a1 2a2 t 1 2 A Y Y 1 2
a1Y1 a2Y2 If 1 2 then Y Y t 1 2 * Pr oblem 1.29 A st eel wir e, 2mm in diameter, is j ust st r et ched bet ween t wo f i xed poi nt s at a temperatur e of 300C. Deter mineits tension when the temper atur e falls to 200C. (Coefficient of linear expansion of steel = 0.000011/0C; Young's modulus for steel = 2.0 x 1011 Nm–2) F Sol: Thermal stress = Yt A Tension in the wire YA t 2 t1 . Here, Y =2.0x1011 Nm–2 ;
11106 / 0 C; t2 = 300C; t1 = 200C; radius=1mm=1x10–3m; A r 110 10 m The tension in the wire 2
3 2
6
= 69.14N AKASH MULTIMEDIA
K
AY (or) F
=
F
AY constant, depends on type of material and geomentry of rod. F k (or) (F = kx) AY Where k is t he equivalent spring constant.
Pr oblem 1.30 A mass ‘m’is attached with r od as shown in figur e. This mass is slightly stretched and released then find the time per iod. (Y isYoung’s modulus of rod, A is cross sectional area of rod, ' ' is its length).
A Y
k
m
T 2p
AY
m
m m T 2p k AY
1.15 Bulk modulus (K ) : I t is defined as the r atio of the volume str ess (nor mal str ess) to the volume str ain within the elastic limit volume stress K volume strain F
2
2.0 1011 11106 106 30 20
stress F Y strain A
V
F F
F
V V
F
F F
F
20
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ELASTICITY
when a solid (or) fluid is subjected to a change in pressure, its volume changes but the shape remains unchanged. The force per unit area, applied normally and uniformaly to the surface of the body i.e pressure gives the stress and the change in volume per unit volume gives the volume strain. Thus if the volume 'V' of the body decreases by an amount V when the pressure on its surface is increased uniformaly by p , then in equilibrium. volume stress = p volume strain
Bulk modulus
P V V
The bulk modulus of a gas in adiabatic condition is defined as adiabatic Bulk modulus of elasticity. For adiabatic process PV = constant differentiating both sides P gV g 1dV V g dp 0
P gV g1dV V g dp
P gV gV 1dV V g dp
V V
K
1.17 Adiabatic Bulk modulus of elasticity Ef :
1 dV dp V dp dp g p gp dV dV E g p f V V gp
K V
P V
The negative sign shows that with increase in pressure, the volume decreses. Note 1.12 : All the states of matter possess bulk modulus K solids K liquids K gases
Hence, adiabatic Bulk modulus of elasticity is equal to ‘ g ’ times pressure Note 1.14 : Ratio of adiabatic to isothermal Bulk modulus of elasticity
Note 1.13 : Gases have two bulk modulli, they are
Ef
1.16 I sother mal Bulk modulus of elasticity Eq :
Eq
The bulk modulus of a gas in isothermal condition is defined as isother mal Bulk modulus of eleasticity. We know for isother mal process PV = constant Differentiating both sides pdV + Vdp = 0 PdV = – V dp
P
P
dp dV V dp Eq P dV V
Hence isot hermal elasticity is equal to pressure. AKASH MULTIMEDIA
gP g P
g > 1 E f Eq
adiabatic bulk modulus of elasticity is g times to the isothermal bulk modulus of elasticity. 1.18 Compr essibility : The reciprocal of bulk modulus iscalled compressibility. i.e 1 Compressibility = Bulk mod ulus 1 V V P O.F = M–1LT2 S.I unit of compressibility is N–1 m2 Note 1.15 : A rigid body and an ideal liquid are incompressible i.e., compressibility is zero implies bulk modulus is infinite * 1.19 Density of compr essed liquid : If a liquid of density ' r ', volume V and bulk modulus 'K' is compressed, then its density increases m density r V 21
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1 r V -------- (1) V r V But by definition of bulk modulus r
K
V P V P ----------------(2) V V K r
P
from (1) and (2) r K
r1 r P P r1 r r r K K P r1 r 1 K 1 Also r r 1 CP where ‘C’ is the com-
pressibility.
Sol : V 103 m3 ; V 106 m3 ; P 10 atm = 1010 5 1106 Nm2 P V K = V
1106 103 110 9 Nm2 V= 6 10 * Pr oblem 1.32 Deter mine the pr essur e r equir ed to r educe the given volume of water by 1 % . Bulk modulus of water is 2 10 9 N m-2 V 1 , K= 2 10 9 N m-2 P = ? V 100
P V K V P K V , V 9 = 210
4 V R V p R3 , 3 3 V R Now by definition of bulk modulus B V
* Pr oblem 1.31 A volume of 10–3 m 3 is subj ected to a pressure of 10 atmospher e. The changein volume is 10-6 m3. Find the bulk modulus of water. (Atmospheric pressure= 1105 N m–2 )
Sol :
Pr oblem 1.33 A solid sphere of r adius 'R' made of a mater ial of bulk modulus B is sur r ounded by a liquid in a cylindr ical container. A massless piston of ar ea 'A' floats on the sur face of the liquid. Find the fr actional change in the r adius of the spher e dR , when a mass M is placed on the piston to R compr ess the liquid. Sol : As for a spherical body
1 2 107 N m -2 100
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P V P Mg i.e V V B AB
Mg as P A
dR 1 V dR Mg R 3 V R 3 AB
Pr oblem 1.34 A uniform pressure 'P' is exer ted on all sides of a solid cube at temper atur e t 0C . By what amount should the temper atur e of the cube be r aised in or der to br ing its volume back to the volume it had befor e the pr essur e was applied, if the bulk modulus and coefficient of volume expansion of the mater ial ar e B and g r espectively. Sol : As by definition of bulk modulus P B V V , with increase in pressure decrease in volume of the cube will be given by V
VP , (as P P ) B
Now with rise in temperature due to thermal expansion, volume increases so if ' q ' is the rise in V temperature then V V gq as g V q As the volume of the cube remains constant
VP P V gq q B gB 22
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ELASTICITY
Pr oblem 1.35 When a r ubber ball of volumeV, bulk modulus 'K ' is taken to a depth 'h' in water, then decr ease in its volume is P m K Pa V v V
V
phrg
PV
h m
K
hr g V K
v1
Pa p
hr g V So decrease in volume of ball K 1.20 M odulus of r igidity h (or ) shear modulus Within elastic limit, the r atio of shear ing str ess to the shear ing str ain is called modulus of r igidity of the mater ial of the body. shearing stress h shearing strain A
F
cube to be slightly displaced or sheared relative to one another, each line such as AB or CD in the cube is rotated through an angle ' q ' by this shear.. The shearing strain is the angle q in radians through which a line normal to a fixed surface has turned. For small values of angle
AA1 x AB L shear stress F / A F h shear strain q Aq shearing strain q
* In this case shape of a body changes but its volume remains unchanged. * Only solids can exhibit a shearing as these have definite shape. * Pr oblem 1.36 A 5.0 cm cube of substance hasits upper face displaced by 0.65 cm, by a tangential force of 0.25 N. Calculate the modulus of r igidity of t he substance. FL FL F Sol : h = , A = L2, h 2 A L L Here, L 5.0 102 m 0.65102 m ; F=0.25 N.
A
F
0.25 0.2510 4 = 5.0102 0.65102 3.25 –2 = 769.2 N m
fixed
A1
A
h
C
x L q
F
B
C1
F
q Fixed
D
consider a cube of material fixed at its lower face and acted upon by a tangential force 'F' at its upper surface having area A as shown in the figure. F// el F A A As shown in above figure, the shearing force 'F' causes the consecutive horizontal layers of the Shearing stress
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* Pr oblem 1.37 A tangential for ce of 2100 N is applied on a sur face of ar ea 3106 m 2 which is 0.1 m fr om a fixed face. The for ce pr oduces a shift of 7mm of upper sur f ace wi t h r espect t o bot t om. Calculate the modulus of rigidity of the material. Sol: F = 2100 N ; A = 3106 m2 ; L =0.1m; 7103 m. h
FL 2100 0.1 =1x1010 Nm–2 A 3106 7103 23
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* Pr oblem 1.38 A steel plate of face ar ea 2 cm2 and the thickness 1.0cm is fixed r igidly at the lower face. A tangential force of 10 N is applied on the upper sur face . Find the later al displacement of the upper sur face with r espect to the lower sur face. Rigidity modulus of steel = 8.41010 Nm -2 Sol : A 2cm2 2 104 m2 ; L=1.0 cm= 1102 m; F=10 N. h 8.41010 N m-2 ; ? FL h A The lateral displacement of the upper face with respect to the lower face is
FL Ah
101102 2104 8.41010 1 107 m = 5.952109 m. = 16.8 Pr oblem 1.39 Calculate the for ce ‘F’ needed to punch a 1.46 cm diameter hole in a steel plate1.27 cm thick (as shown in fig). The ultimate shear str ength of steel is 345 M N/m2 =
Type of stress
Stress
Sol : As in punching, shear elasticity is involved, the hole will be punched it F11 ultimate shear stress A
F
F11 > (shear stress) X Area F11 min = (3.45 X 108) 2prL A 2prL
8 2 2 = 3.4510 23.140.7310 1.2710 200KN
1.21 Some impor tant points on modulus of elasticity 1) Young's modulus (Y) and rigidity modulus ( h ) exist only for solids but not for liquids. This is because liquids and gases cannot be deformed along one dimension only and also cannot sustain ( shear strain). Bulk modulus (K) exists for all states of mater (solids, liquids and gases) 2) Gases being most compressible are least elastic while solids are most elastic. Esolid Eliquid E gas
Strain
Elongation or Two equal and compression opposite forces Tensile parallel to force perpendicular or direction to opposite compressive faces L / L (longitudinal strain) s F / A Two equal and opposite forces parallel to opposite surfaces [forces Shearing in each case such Pure shear, q that total force and total torque on the body vanishes] Bulk
Forces perpendicular everywhere to the Volume change surface, force per (compression or elongation) unit area (pressure) same everywhere V / V
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Change in Elastic Name of State of shape volume modulus modulus Matter
No
Yes
Y
Yes
No
G
No
Yes
F L AL
Young's modulus
Solid
F q
Shear modulus
Solid
A
p B V / V
Solid, Bulk liquid modulus and gas
24
PHYSICS - I C
3) For a perfectly rigid body as L, V or f 0 , So Y, K or h will be ' a ' i.e elasticity of a rigid body is infinite. 4) Greater the value of modulli of elasticity, more elastic is the material. 1 1 1 , Ka But as Y a and h a for a f L V constant stress, smaller change of shape or size for a given stress corresponds to greater elasticity. ex :(1) For same load, more elongation is produced in rubber wire than in steel wire of same cross-section hence steel is more elastic than r ubber. ex : (2) Water is more elastic than air as volume change in water is less for same applied pressure 5) The value of moduli of elasticity is independent of the magnitude of the stress and strain. It depends on the nature of the material of the body. 6) For a given material there can be different moduli of elasticity depending on the type of stress applied and the strain produced. 7) In a suspension br idge as there is a stretch in the ropes by the load of the bridge, the elasticity involved is linear or tensile. 8) In an automobile tyr e as air is compressed the elasticity involved is volume, i.e., bulk. 9) In transmitting power an automobile shaft is sheared as it rotates, so the elasticity involved is shear, i.e., rigidity. 10) When a coiled spr ing is stretched, the deformation of the wire of the spring is in the form of a twisting strain so the elasticity involved is shear, i.e., rigidity. 11) In a water lift pump as the water is compressed, the elasticity involved is volume, i.e., bulk 12) The shape of r ubber heels changes under stress, the elasticity involved is shear, or rigidity. 1.22 Poisson's r atio : When a wire is stretched by a force along its length, then its length increases and the radius (or) diameter decreases as shown in the figure.
AKASH MULTIMEDIA
ELASTICITY
The r atio of change in r adius (or ) diameter to the or iginal r adius (or ) diameter is called later al strain L
r (D)
r r D D
F
D Lateral strain D (or)
r r
The ratio of change in length to the original length is called longitudinal strain Longitudinal strain Lateral strain is directly proportional to the longitudinal strain lateral strain a (longitudinal strain) Lateral strain = s (longitudinal strain) Where ' s ' is poisson's ratio. It depends on the nature of the material. Poisson's ratio ( s ) is defined as the ratio of lateral strain to longitudinal strain. Lateral strain poisson ' s ratio s Longitudinal strain D D s negative sign indicates that the radius or diameter of the wire decreases when it is stretched. Poissons's ratio has no units and dimensions as it is ratio of two strains. The theroetical limits of poisson's ratio are from – 1 to + 0.5. But its practical limits are from 0 to 0.5 and generally between 0.2 and 0.4. 1.23 Relation among volume strain, Lateral strain and poisson's r atio : Consider a wire of length ' ' and radius 'r', then its volume V p r 2 (1) V r 2 V r 25
PHYSICS - I C
ELASTICITY
r r s But we know r s r V 2s V
V 12s V
V 0, Note 1.16 : If a material has s 0.5 then V Þ V 0 , there in no change in the volume of the body and the material is said to be incompressible.
The fractional increase in volume, V 0.01(1–2x0.30)=0.01x0.40=0.004. V 1.24 BEHAVI OUR OF A M ETAL WI RE UNDER I NCREASI NG L OAD Consider a metal wire having its upper end fixed to a rigid support and loaded at the lower end by attaching a weight hanger. Let the load be increased gradually. To study the behaviour of the metal wire under increasing load, a graph is plotted between the stress on the Y – axis and the strain on the X – axis. In general, the curve shown in figure is obtained for ductile materials that can be drawn into wires.
* Pr oblem 1.40 A 3 cm long copper wir e is str etched to incr ease its length by 0.3 cm. find the later al strain in the wire, if the Poisson’s r atio for copper is 0.26. Sol : L = 3cm ; L 0.3cm; s 0.26. Longitudinal strain
s
L 0.3 0.1 L 3
LateralStrain LongitudinalStrain
The Lateral Strain = s LongitudinalStrain = 0.260.1 = 0.026. * Pr oblem 1.41 Find the fr actional incr ease in volume of a wir e of cir cular cr oss section if its longitudinal str ain is 1% . s 0.30 Sol: we know that dV dL 1 2s V L
Here,
dL 1 1% 0.01; s 0.30. L 100
AKASH MULTIMEDIA
1. From O to P the graph is a straight line showing that stress is proportional to strain i.e., the wire obeys Hooke's law upto the point P. So, P is called the proportionality limit of the wire. 2. From P to E as the graph is slightly curved, the stress is not proportional to strain. If the load is removed at any point between O and E it will regain its natural length. The point E is called the elastic limit. In case of some materials, the wire may obey Hooke's law upto E coinciding E with P i.e., P will be the elastic limit for such materials. 3. On increasing the load beyond elastic limit, the graph is more curved upto the point Y called yield point. From E to Y, the wire does not obey Hooke's law indicating that for a small increase in load there is greater increase in length. If the load applied on the wire is removed between E and Y, the wire does not regain its natural length completely. It will have a permanent increase in length. This behaviour of 26
PHYSICS - I C
ELASTICITY
the wire is shown by the dashed line which is a straight line that cuts the x–axis not at O but at O1. OO1 is the permanent set. | Strain OO | Permanent increase in length OO 4. When the wire crosses the point Y strain increases rapidly without any increase in the load. So, yield point is defined as the point beyond which strain increases rapidly without any increase in load. Beyond the point Y as the wire becomes thin and the stress for the same load becomes larger and larger increasing the strain further and further. If the load is not removed the strain increases continuously till the wire reaches a point T. The stress corresponding to T is called the tensile strength of the given material. The tensile strength is ratio of maximum load to which the wire may be subjected by slowly increasing the load to the original area of crosssection of the wire. 5. Beyond the point T, the thinning of the wire is no longer uniform and the wire shows necks. Immediately, as this occurs, the stress decreases automatically and the part TB is obtained. At B the wire ultimately breaks. B is called breaking point. 6. If large deformation occurs between the elastic limit and the breaking point, the material is ductile. Ex : copper, silver, gold etc., 7. If the deformation between the elastic limit and the breaking point is very small or if the wire breaks immediately after crossing the elastic limit, the material is brittle. Ex : glass, ceramic etc. 1.25 BREAK I NG STRESS : i) The breaking stress of a wire is the maximum stress at which the wire breaks. ii)
Breaking stress =
Breaking Force initial area of cross section
iii) Breaking force = Breaking stress x area of cross section. iv) Breaking stress a) depends only on the nature of material of the wire b) is independent of the length and area of cross-section of the wire. v) Breaking force a) is independent of length of the wire b) depends on the area of cross-section and nature of material of the wire. AKASH MULTIMEDIA
vi) Breaking force is proportional to area of crosssection. vii) If we cut a cable that can support a maximum load of W into two equal parts, then each part can support a maximum load of W. viii) A very long wire suspended vertically may break due to its own weight The maximum length of a wire that can hung without breaking under its own weight is mg Ag g breaking stress = A A
Breaking Stress rg
Note 1.17 : A metal rope of density ‘ b ’ has breaking stress (B:S). This rope is used to measure the depth of the sea. Then the depth of the sea that can be measured without breaking is r r mg1 Arb g1 rb rb B. S. = grb r A A
B.stress grb r
(Whoe PL is the density of sea water) * Pr oblem 1.42 Find the greatest length of the wir e made of mater ial of br eaking str ess 8x108Nm –2 and density 8x103kgm–3 that can be suspended from a rigid support without breaking. (g = 10 ms–2) Sol: Greatest length of the wire without breaking.
L
Braking Stress rg
Here, breaking stress = 8 x 108 Nm–2; r = 8 x 103 kg m–3; g = 10 ms–2.
L
8108 1104 m 10km 3 810 10 27
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* Pr oblem 1.43 A block of mass 1 kg is fastended to one end of a copper wir e of cr oss- sectional ar ea 1 mm2 and is r otated in a ver tical cir cle of r adius 20 cm. I f the br eaking str ess of copper is 5 x 108 Nm–2, find the maximum number of r evolutions the block make in the minute without the str ing breaking. Sol : Maximum tension on the string = Breaking stress x Area of cross - section = 5x108 x1x10 –6= 500 N. When a body revolves in a horizontal circle, Tension on the string = Centripetal force 500 = mr w 2 Where m = 1 kg, r = 0.20m, w = ? 500 = 1 x 0.2 x w 2 500 w2 2500 0.2 Maximum angular speed, w 50 rad s1 . t= 60 s,n= ? 2p n 2p n w 50 , t 60 The maximum number of revolut ions , n
50 60 1500 477.4rpm. 2p p
1.26 Elastic Fatigue : When a body is subjected to a repeated stress, even within the elastic limit, it becomes weak since it loses its elastic property to some extent temporarily. If greater stress is applied on the body without knowing present state, cracks develop within the body and it breaks. This occurs even for a stress lesser than that of breaking stress. Ex : When a metal wire is bent once, it may not break. But, it breaks when it is bent repeatedly at the same point. This weakness or the state of temporary loss of elastic nature of the body when subjected to repeated stress is called “elastic fatigue”. If the material is given some rest, i.e., kept in unstrained state for some time, it regains its original elastic nature. AKASH MULTIMEDIA
1.27 EL ASTI C AFTER EFFECT: When stress is removed the strain does not reduce to zero at once. It takes some time for the strain to become zero after the removal of stress, the delay in recovering back to the original condition on removal of deforming force is called elastic aftereffect. This effect is very much dominant in glass while it is totally absent in quartz, phosphor bronze, silver and gold. 1.28 Str ain energy : When a wire has natural length, the potential energy corresponding to the atomic and molecular forces is minimum. When the wire is deformed, internal forces called restoring forces are set up and work is to be done against these forces to produce the deformation. This work done is stored in the wire as potential energy which is called strain energy. Str ain Ener gy is the ener gy stor ed in a body due to its defor mation : 1.29 Expr ession for str ain ener gy or wor k done in str etching a wir e : Consider a metal wire of length L and cross–sectional area A fixed at one end and is stretched by an external force applied at the other end. The force is so adjusted that the wire is only slowly stretched. This ensures that at any time during the extension the external force is equal to the tension in the wire. When the extension is e, the wire is under a longitudinal stress F/A, where F is the tension at that instant of time. The strain is x/L. Let the force acting on a wire suspended from a rigid support be F. The work done in increasing its length by ‘de’ is dW = Fde The total work done in increasing the length of wire by ‘e’ is obtained by integrating the above expression between the limits 0 and ‘e’. e
W dW Fde e
= 0
0
YAe de L
Y FL Ae
YA e 2 YAe e 1 Fe L 2 L 2 2 28
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This is stored as strain energy in the wire. 1 Strain energy in the wire = Fe 2 Strain energy per unit volume of the wire
1 Fe 1 F e . 2 AL 2 A
1 Stress Strain 2 2 1 Stress strain Stress = Y 2 Y 1 2 = strain Y 2 When the external force is withdrawn, the stress disappears and the strain energy appears as heat. The above relation holds good for longitudinal, volume and shearing strains. Note 1.18: i) Work done in stretching a wire,
w=
1 2
ii) w = 12
x stretching force x extension. 1 F 2 1 F2 1 YAe 2 Fe = = 2 AY 2 pr 2 y 2 l
iii) w = 12 x stress x strain x volume of the wire iv) Area under F-e graph gives the work done or the strain energy sto red in t he wire. Area =
1 2
F.e = W
* Pr oblem 1.44 I f Young’s modulus of the mater ial of a wire is 1.2 1011 N m–2, calculate the wor k done in str etching the wir e of length 3 m and cr oss- sectional ar ea 4 mm2 when it is suspended ver tically and a load of 8 kg is attached to its lower end. Sol : Y = 1.21011 Nm -2 : L =3 m; A = 4mm2 = 4106 m2 ; M= 8 kg. 1 work done = stretching force increase in 2 length. FL But increase in length, e AY AKASH MULTIMEDIA
ELASTICITY
1 F 2 L 1 Mg L The work done = 2 AY 2 AY 2
8 9.8 3 1 2 4106 1.21011 = 0.0192 J 2
Pr oblem 1.45 A metal wire of length and cross-sectional area A has mass m. I t is stretched by an amount ‘e’by a load of mass M . I f the wir e br eaks at the point of suspension due to the load then find the r ise in temper atur e of the wir e ? 1 1 Strain energy = F.e Mg e 2 2 As the wire breaks strain energy stored in the wire appears in the form of heat. 1 Mg e mst [s = specific heat of wire] 2
t
Mg e 2 ms
Pr oblem 1.46 A stone of mass ‘m’ is pr oj ected fr om a r ubber catpult of length ‘ ’and cr oss-sectional area A stretched by an amount ‘e’. I f Y be the young’s modulus of r ubber then find the velocity of pr oj ection of stone ? Solution : 1 Strain energy = F.e 2 1 YAe 1 YAe2 e = 2 2 As the stone is released, the strain energy in the catpult appears in the form of kinetic energy of the stone. 1 YAe 2 1 2 mv 2 2 2 YAe YAe2 v2 , v m m 29
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1.30 Exper imental deter mination of Young’s modulus (Y) – Sear le’s appar atus Young’s modulus of the material of a wire can be experimentally determined by Searle’s method.
Extension
Descr iption : Two wires of the same material, length and area of cross section, suspended from a rigid support carry at their lower ends, two rectangular metal frames as shown in figure. One of the wires is called experimental wire and the other wire is called reference wire. The frame attached to the reference wire carries a constant weight to keep the wire stretched without any kinks. The frame attached to the experimental wire carriers a hanger, over which slotted weights can be slipped as required. A spirit level is hinged with one end to the frame attached to the reference wire and rests horizontally on the tip of a micrometer screw which can be worked in the frame attached to the experimental wire along a vertical scale marked in millimeter.
relative to the other frame and the air bubble shifts to one side. The micrometer screw is now adjusted to take back the air bubble to the centre and the micrometer screw reading is noted. 4) The experiment is repeated at least five times every time increasing the load by half a kilogram weight. Readings of the micrometer screw are noted while increasing and decreasing the load and mean reading is found. 5) The difference between the first and second readings gives the increase in length or extension produced in the experimental wire when the load is increased by half a kilogram weight. The difference between the first and third readings gives the extension for a load of one kilogram weight. 1 Similarly, the extensions for 1 , 2, .... kg wt are 2 found. 6) A graph is plotted between the load and extension. The graph is a straight line and gives the elongation ‘e’ for a load Mg.
Load
7) The radius, r, of the experimental wire is found by using a screw gauge and measuring the diameter at 6 or 7 places of the wire. The length, L, of the experimental wire is measured with the help of a meter scale. 8) Substituting the values of r, L and
Mg in e
the formula of Young’ modulus. Wor king : 1) A suitable load is kept on the hanger so that the experimental wire is straight without kinks. 2) The micrometer screw is adjusted so that the air bubble in the spirit level comes in the centre. The reading of the micrometer screw is noted. 3) Half a kilogram weight is then slipped into the hanger. This elongates the experimental wire. The frame attached to the experimental wire moves down AKASH MULTIMEDIA
Y
F L Mg L = ( F = Mg and A = pr 2 ) Ae pr e
the Young’s modulus of the material is calculated. Sour ces of er r or and their minimisation : There are two sources of error in the experiment. i) The support may yield when the load is attached at the lower end of the experimental wire and the 30
PHYSICS - I C
measured value of increase in length may not be correct. ii) While the experiment is carried out, temperature may change which causes some increase in length. The measured value of increase in length becomes incorrect. Both the errors are minimized by using the reference wire. The yield of support or the change of temperature affects both the experimental and reference wires. The relative increase of the experimental wire with respect to the reference wire will give correct increase in length. L ong Answer Questions 1.
Shor t Answer Questions
2. 3.
Ver y Shor t Answer Questions 1. 2. 3. 4. 5. 6. 7.
Define Hooke's law of elasticity. Descr ibe Sear le’s method to deter mine the Young's modulus of the mater ial of a wir e.
1.
ELASTICITY
8. 9. 10.
Def i ne H ook e' s l aw of el ast i ci t y, pr opor tionality limit, per manent set and Br eaking str ess. Define modulus of elasticity, stress, strain the Poission's ratio.
11.
Define defor mation and defor ming for ce. M ention the differ ence between elastic and plastic bodies.
14.
4.
Descr ibe the behaviour of a wir e under gr adually incr easing load.
5.
DefineYoung's modulus, Bulk modulus and Rigidity modulus.
6.
Define str ess and explain the types of str ess.
7.
Define str ain and explain the types of strain.
8.
Define strain energy and derive the equation for the same.
9.
“Steel exhibits mor e elastic natur e than r ubber ” Explain.
10. I f a wire is bent continuously at a par ticular point in opposite dir ections, it br eaks. Explain. AKASH MULTIMEDIA
12. 13.
15.
Define Hooke's law of elasticity. State the units and dimensions of str ess. State the units and dimensions of modulus of elasticity. State the units and dimensions of Young's modulus. State the units and dimensions of Rigidity modulus. State t he units and dimensions of Bulk modulus. State the examples of near ly per fectly elastic and plastic bodies . State the theor etical limits of Poisson's ratio. State the pr actical limits of Poisson's r atio. What is str ain ener gy? State its expr ession in ter ms of the applied for ce and extension. Expr ess str ain ener gy per unit volume in terms of stress and strain, stress and Young's modulus. What is elastic fatigue? What ar e the sour ces of er r or in Sear le's exper iment? How ar e the er r or s eliminated in Sear le's exper iment? State the examples of ductile and br ittle materials. Assess Your self
1.
Why is a spr ing made of steel but not of copper ? Ans. As Young’s modulus of steel is greater than that of copper, the strain produced is small for a given stress in case of steel than copper. Hence steel is prefered 2. A cable is cut to half its or iginal length. Can each par t suppor t the same maximum load as the or iginal cable ? Ans. Yes. Since the breaking stress is constant for a given material and the breaking load = breaking stress x area of cross section; the maximum load remains the same as the area of cross section does not change. 31
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3.
When a spr ing is str etched, what type of str ain is pr oduced ?
Ans. When a spring is stretched there is neither change in the length of the wire forming the coil nor change in its volume. The change takes place in the shape of coil producing shear strain. 4.
Can a liquid offer r esistance to shear ing str ess?
Ans. No. It cannot offer permanent resistance forces to change its shape. 5.
Can a liquid offer r esistance to bulk str ess?
Ans. Yes. It can offer very great resistance forces tending to decrease its volume. 6.
Can a gas offer resistance to shearing str ess?
Ans. No. 7. Can a gasoffer great resistanceto bulk stress? Ans. No. It can offer small resistance forces tending to decrease its volume. 8. What will be the modulus of elasticity of a r igid body? Ans. Infinity. 9. Can a shear str ain be expr essed in ter ms of tensile str ain and compr essive str ain ? Ans.Yes. If shear strain is equal to q , tensile and q compressive strains are each equal to . 2 10. I f a body is per fectly incompr essible what will be its value of Poisson’s r atio ? Ans. For a body to be perfectly incompressible, V V 0 .We know that 1 2s V V
. Hence 1 2s 0 s 0.5 . 11. Br idges ar e declar ed unsafe after long use. Explain. Ans.After long use, the bridge loses its elastic strength. It develops large strains corresponding to the same usual values of stress and the bridge may collapse. AKASH MULTIMEDIA
12. I s ther e any incr ease in temper ature when a wir e br eaks ? Ans.Yes. Since the strain energy is converted into heat energy. Additional topic's for AI EEE ** (i) Elastic hysteresis : As a result of elastic after effect strain in a material lags behind the stress to which it is subjected, this phenomenon of lagging behind of strain with respect to stress is called elastic hysteresis. (ii) Hyster esis Cur ve : When a ductile material is loaded and then unloaded, stress-strain graph of the material is as shown in Figure. It is seen that at the time of unloading strain is larger than that at the time of loading for the same stress. This lagging is known as hysteresis. The area enclosed by hysteresis curve represents the hysteresis loss during the process. This energy is lost as heat. The material is selected depending upon the type of use. For example when rubber is used as shock absorber then we want that large quantity of energy of mechanical vibrations which are impressed upon it is dissipated as heat. In this case, rubber having the stress-strain curve as shown in Fig . is choosen. For air craft tyres where wear and tear matters, rubber whose hysteresis curve has low area, is choosen (fig.)
Load
. Extension
(iii) Bending of Beam : Beam is the structural member which can carry transverse load. A simply supported beam is supported at its ends. A cantilever beam is fixed at one end.
32
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(iv) Deflection of beam : Deflection of beam at its centre due to load placed as shown in Figure.
(vi) A hollow shaft is stronger than a solid shaft made of same mater ial and of equal volume.
W 3 for simply supported beam and 48YI
Consider a solid bar or shaft of radius r, length l and made of material of modulus of rigidity . It can be proved that the torque required to produce
d
W 3 for cantilever beam where I is called 3YI geometric moment of area. d
3
i) For rectangular cross - section I
bd 12
pr 4 I ii) For circular cross - section 4
d
unit twist in the bar is given by t
If the shaft is hollow with internal and external radii r1 and r2 respectively, then torque required to produce unit twist is given by
b
r
(v) Twisting of a shaft : Let us consider a shaft of length and radius r,, whose one end is rigidly clamped and torque t is applied at the free end. Because of this the free end is twisted by and angle q .
phr 4 .....(1) 2
t
ph r24 r14
2
............(2)
Length of the hollow shaft is the same from eqs (1)and (2) we have 2 2 2 2 ' r24 r14 r2 r1 r2 r1 ........(3) r4 r4
As the two shafts are made from equal material, of same volume. Hence pr 2 p r22 r12 or r 2 r22 r12 .....(4)
Substituting the value of r 2 in eq. (3) we get
From the diagram, arc s r q f where q angle of twist and f angle of shear
t ph r 4 q 2 where h modulus of rigidity.. Note 1.19: One end of the rod is fixed. The other free end is twisted through an angle ' q ' by applying a torque ' t ' then the work done on the rod (or) energy stored in the rod is
Torsional rigidity of shaft
1 W 2
where is in radians. AKASH MULTIMEDIA
2 2 2 2 ' r2 r1 r2 r1 r22 r12 1 r2 r22 r12 r 2
r
2
r22 r12 and r22 r12 r 2
' Therefore, torque required to twist a hollow cylinder is more than required to twist a solid cylinder. So, a hollow shaft is more stronger than a solid shaft. Due to this reason, electric poles are made hollow. (vii) REL ATI ON BETWEEN ‘Y’, n, AND K : (OPTI ONAL ) Consider a unit cube with sides parallel to the axes OX,OY and OZ.Let the forces P,Q and R are acting along X,Y and Z axes respectively. Since the area of each face is unit, the force acting on each face is equal to stress. Each force produces 33
PHYSICS - I C
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elongation in its direction and contraction in the other two directions. We know that,
P changein length Y 1 2 P 1 2 original length 1 Y
But, volume strain = 3 Longitudinal strain
Bulk modulus K
Young’s modulus
Longitudinal stress LongitudinalStrain
Longitudinal strain
Longitudinalstress Young's modulus
Extension along X -axis = P/Y
Lateral strain = longitudinal strain.
Lateral strain Poisson ' s ratio Longitudinal strain P Y P Compression along Y and Z axis Y P Elongation along each of Y and Z axes Y Similarly, the elongations in other directions are tabulated as follows. STRAIN PRODUCED ALONG Stress P along X-axis
X-axis P Y
Q along Y-axis
s
Q Y
R along Z-axis
s
R Y
Y-axis
Z-axis
P s Y Q Y
s
P Y
s
Q Y
s
R Y
when P, Q and R P s Q s act simultaneously Y Y Q R Y Y P R
(i)
If P = Q = R, then Elongation produced in each side P s P 2 P 1 2s Y Y Y
Longitudinal strain AKASH MULTIMEDIA
R Y R s P Q Y Y
3P 1 2 Y
Normal stress volumestrain
P Y 3P 3 1 2s 1 2 s Y
Y 3 K 1 2 1 ii) Two forces, one elongative and the other compressive force constitute shear, i.e., If Q = – P and R = 0, then shear will be produced in the cube. Now the linear strain P P P 0 1 Y Y Y Shearing strain 2 linear strain
2P 1 Y Tangential stress Rigidity modulus Shearing strain P n 2P 1 Y Y 1 2 2n Fom equation (1) Y 1 2 3 3K and from equation 2 Y 2 2 4 n On adding equations 3 and 4
Y Y 3 1 1 3 3K n Y n 3K 1 1 1 5 Y 3n 9K 9 nK Y 6 3K n
34
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ELASTICITY
(viii) L imiting values of Poisson’s r atio : Y 1 2 From equation (3) 3K Y and K are positive, 1 2 must be positive 1 or 2 1 or 2 Y 1 From equation (2) 2n Y and n are positive, 1 must be positive 1 0 1 or 1 so 1 . But can never be 2 negative. 1 1 or 0 Poisson’s ratio lies between 0 and 2 2 Note : From (1) and (2) 3K 1 2 2 n 1
3K 6Ks 2n 2ns 3K 2n 6 K s 2ns
BC
and CD
70103 1 1210
11
50103 2 121011
= 4.5 x 10–7 + 3.5 x 10–7 + 5.0 x 10–7 = 13 x 10–7 m Pr oblem 1.48 A uniform elastic plank moves over a smooth hor izontal plane due to a constant for ce F 0 distr ibuted unifor mly over the end face. The sur face area of the end face is equal to A and Young’s modulus of the mater ial is Y. Find the compr essive str ain of the plank in the dir ection of acting for ce. F0
M
x
3K 2n 7 . 2 3 K n Pr oblem 1.47 A steel r od of cr oss-sectional ar ea 1m2 is acted upon by forces shown in the fig. Deter mine the total elongation of the bar. Take Y = 2.0 x 1011 N/m2. 60 kN
C
10 kN
1m
50 kN
2m
Sol: The action of forces on each part of rod is shown in fig. 10 kN
20 kN
50 kN
60 kN 60 kN
50 kN
50 kN
60 kN 60 kN
50 kN
70 kN 70 kN
We know that the extension due to external force F is given by F AB AY 60103 1.5 AB 4.5107 m 11 1210 AKASH MULTIMEDIA
M L
F Fx x 0 0 M L
Fx
Fx dx
Elongation of the element F0 x dx F dx d x L AY AY Total elongation L
60 kN 50 kN
Sol : The force at any section is due to the inertia behind the section. The stress therefore increases from zero to maximum at the end where force is applied. Consider a small element of length dx at a distance x from the free end. The force Fx = ma =
D 20 kN
1.5m
m dx
A
5.0107 m
The total extension AB BC CD
3K 2 n 2 3K n
B
3.5107 m
0
F0 x dx F0 L2 (or) F0 L ALY 2 AY 2 ALY
Pr oblem 1.49 A slightly conical wir e of length and r adius r 1 and r 2 is str etched by two for ces applied par allel to length in opposite dir ections and nor mal to end faces. I f Y denotes the Young’s modulus, then find the elongation of the wir e. 35
PHYSICS - I C
ELASTICITY L
Sol : F
r2
r
r1
r
F
F
F
x dx
dx
Consider an element of length dx at distance x as shown in fig. The radius of the sect ion r r rx r1 2 1 x
The extension of the element F dx Fdx 2 d p rx Y AxY Total extension
0
Fdx r r p r1 2 1 x Y 2
Pr oblem 1.51 The tension in the r od will not be constant but will var y from point to point. At the free end, i.e., r = L, it will be min = 0 while at the other end r = 0, it wil be max = 2p106 N . (b) Now if dy is the elongation in the element of length dr at position r where tension is T, by definition of Young’s moulus, dy T stress as strain dr AY Y Which in the light of Eqn. (1) gives
F p r1r2Y
Pr oblem 1.50 A thin unifor m metallic r od of length 0.5 m and r adius 0.1 m r otates with an angular velocity 400 r ad/s in a hor izontal plane about a ver tical axis passing thr ough one of its ends. Calculate tension in the r od and the elongation of the r od. The density of mater ial of the r od is 104 kg/m3 and theYoung’s modulus is 2 x 1011 N/m|2. w Sol :(a) Consider an element of length dr at a distance ‘r’ r dr from the axis of rotation as L shown in fig. The centripetal force acting on this element will be dT dmrw 2 r Adr rw 2 As this force is provided by tension in the rod (due to elasticity), so the tension in the rod at a distance r from the axis of roataion will be due to the centripetal force due to all elemetns between x = r to x = L. centripetal force due to all elemetns between x = r to x = L. AKASH MULTIMEDIA
1 2 2 2 2 i.e., T r Aw rdr 2 r Aw L r .......(1) r 2 1 2 1 4 2 r 2 T 10 p 10 400 So here 2 2 1 8p 106 r 2 N 4
1 rw 2 2 L r 2 dr 2 Y so the elongation of the whole rod dy
rw 2 L 2Y
L
1 rw 2 L3 Y
2 2 L r dr 3 0
4 1 10 400 0.53 1 103 m Here L 11 3 2 10 3 2
SYNOPSIS 1.
Ri gi d body : A body whose shape and size cannot be changed, however large the applied force may be, is called rigid body. There is no perfectly rigid body in nature.
2.
Defor mation force : A force which changes the size or shape or both of a body without moving it as a whole is called deformation force.
3.
Restor ing for ce : The force which restores the size and shape of the body when deformation forces are removed is called restoring force. Deformation force and restoring force are not action reaction pair. 36
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