ELASTICITY
Inter atomic forces Elastic Modulii
Behaviour of wire under stress Elastic energy * * * * 1.1 INTRODUCTION
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.
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.
1.2 INTERATOMIC FORCES
The forces acting between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces. 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
10
10 m
.
During interaction between the two atoms, the following electrostatic forces will be active
(i)Attractive forces between the nucleus of one atom
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 F by the relations. F dU
dr
. The variation of
potential energy U(r) and interatomic force F(r) with separation r between two atoms have been shown in graphs (fig (1), fig (2))
re pu ls ion attraction F r O r0 x 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
1
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
(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.
(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 INTER MOLECULAR FORCES
The for ce between the molecules due to electrostatic interaction between the charges of the molecules are called intermolecular forces.
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 m9
The variation of intermolecular forces with distance is shown in fig.
re p u ls io n attraction F r O r0 x 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,
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.,
a 7 a 7
1 a
F orF
r r
The negative sign indicates that the force is attractive in nature.
(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. r 9 r 9
1 b
F or F
r r
1.4 : Some important defimitions
(i)Deforming Force : When an external force is applied on a body which is not free to move, the molecules of the body are forced to undergo a change in their relative 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 force is called deforming force.
(ii) Restoring force : The force developed within the body on account of r elative molecular displacement is called internal force (or) elastic force (or) restoring force.
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 rigid if the r elative positions of its constituent par ticles r emain unchanged inspite of any amount of deforming force. There is no perfect rigid body. The
nearest approach to a rigid body is diamond.
(iv) Elasticity : It is the property of material of a body by virtue of which the body regains its or iginal length, volume and shape after the deforming forces have been removed.
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 property of material of a body by virtue of which it does not regain its original shape and size (i.e it remains in the deformed state) even after the removal of deforming force 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.
1.5 Stress : 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 restoring force acting per unit area of cross–section of the deformed body is called stress.
Re
sec
storing force F
Stress
Area of cross tion 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
( ) ( ) Deforming force F Stress 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) Normal Stress : If the stress is normal to the surface, it is called normal stress. 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) Longitudinal stress : When a normal stress changes the length of a body then it is called longitudinal stress. (or)
When a force is applied normal to the cross-sectional area of the body such that its length changes then the restoring force developed per unit cross-sectional area is called longitudinal stress.
Longitudinal stress =
Deforming force (F) 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.
F F
B ody subjected to tensile force
A B C
F F
Tensile stress
A B B C
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.
F F
Body subjected to compressive force
A B C
F F
Com pressive stress
A B B C
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 stress.
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.
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)
V V V F F F F F F F F Volume stress (or) Bulk stress Force surface area F A = = = Pressure (P)
(b) Shearing stress : When the stress is tangential to the surface due to the application of forces parallel to the surface, then the stress is called tangential (or) shearing stress.
Shearing stress = Force
surface area F 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. F F fixed A A
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.
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.
Note 1.4 : q n A F e r 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’.
Longitudinal stress = Fer Fcos q
A A
To find the shearing stress, take the component of force parallel to the plane of the given area and then devide Fel by the area ‘A’.
Shearing stress = Fel Fsin q
A A
The total stress = longitudinal stress + shearing stress But not F/A.
* Problem 1.1
A steel wire of 2mm in diameter is stretched by applying a force of 72N. Find the stress in the wire. Solution : r 3 1 10 m; F=72N The stress =
2 2 3 72 1 10 F F Apr p = 6 7 2 72 2.292 10 10 Nm p . 1.6 Strain :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 atio of change pr oduced in the dimensions of a body by a system of forces or couples in equilibrium to its original dimensions is called strain.
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 is the ratio of the change in length of a body to its original 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 F F Longitudinal strain =
change in length original length If the length increases due to tensile stress, the corresponding strain is called tensile strain. If the length decreases due to compressive stress, the strain is called compressive strain.
(ii) Volume strain : It is the ratio of change in volume of body to its original 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
V- V F V F F F Volume strain =
change in volume V original volume V(iii) Shearing strain : It is defined as angle q in radians through which a plane perpendicular to the fixed surface of the cubical body gets turned under the effect of tangential force .
It is also the ratio of the displacement of a layer to its distance from the fixed layer. q x
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 A B G E H C D F Fixed Surface F fig (a) A x A B B F F C Fixed D q fig (b) q Shearing strain = q x * Problem 1.2
A copper wire of length 1m is stretched by 1cm. Find the strain on the wire
Solution : The strain
2 1 10 0.01 1 e L * Problem 1.3
If a platinum wire is stretched by 0.5% what is the strain on the wire?
Solution :
Fractional increase in the length e
L
= strain The strain =0.5% =0.5 0.005
100 .
1.7 Elastic limit : The maximum stress within which a body can regain its original size and shape after the removal of the deforming force 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.
1.8 Hooke’s law: Hooke’s law states that within elastic limit, the stress is directly proportional 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 : Modulus of elasticity of the material of a body is the stress in the body to produce unit strain (within the elastic limit).
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) 1pa1N / m2
C.G.S. unit is dyne / cm2
Dimensional formula is ML T1 2
1.9 Factors effecting Elasticity:
(i) ANNEALING : 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)TEMPERATURE: Normally, elasticity of the
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 stress developed and the resulting strain, 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 ratio of longitudinal str ess to the longitudinal str ain within elastic limit.
i.e., Y longitudinal stess
longitudinal strain F 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. Longitudinal stress = F A Longitudinal strain = Y = longitudinal stress longitudinal strain= F F A A
If a force is applied on a wire of radius ‘r’ by hanging a load of mass ‘M’ as shown in figure
then 2 2 Mg Mg r Y Y r p p M Problem 1.4
Show that str ess r equir ed to double the length of wire (or) to produce 100% longitudinal strain is numerically equal to Young’s modulus. Sol : 1 , 22 2 , F A Y But as 1 , Hence F Y A Problem 1.5
A load (M) suspended from a wire produces an elongation (e) in the wire then find the rise in temperature required to produce same elongation in the same wire.
sol : F A Y a
t
... 1 F AY at... 2
But from given data
a a F F t t AY AY
Note 1.5 : If two wires having lengths ; cross-1, 2 sectional areas A1, A2 and Young’s modulii Y1, Y2 are stretched by forces F1, F2 then
As 1 1 1 2 2 2 2 2 2 1 F Y F A e Y Ae Y F A e
Note 1.6 : In terms of volume V of a wire.
As 2 2 F F F F Y Y y Ae A e Ve Ve
Where ‘V’ is the volume of the wire as d m V m V d 2 F d y me
m mass of the wire
d density of material of the wire
In the above formula If V, F, Y are same for two wires then 2 2 1 1 2 2 2 e e e a
In the above formula if Y, d, F are same for two wires then 2 2 1 1 2 2 2 1 e m e m e m a Note 1.7 : As
2 2 F A F FV Y Ae A e A e 2 2 4 FV FV Y Y A e p r e as dm V m V d 2 4 p Fm Fm Y Y dA e d r eIn the above formula if F, V, Y are same for two wires then e 1
A a (or) 4 1 e r a Problem 1.6
When a body of mass ‘m’, density dB is suspended from a wire, its elongation is ‘e’ when the body is in air. I f the body is completely immersed in a non – viscous liquid of density d then its elongation is
In air In liquid F = Wa = mg 1 1 B d F W mg d as e F AY
ea F (as , A, Y are same in both cases)
1 1 mg e F e F 1 B d d mg 1 1 B d e e d Problem 1.7
Two wires of same length and radius are j oined end to end and loaded. The Young’s modulii of the materials of the two wires are Y1 and Y2. If the combination behaves as a single wire then its Young’s modulus is
W e1 e2 1 Y A 2 Y A W e 2 AY 1 2 e e e but
1 2 1 2 2 , , eq F F F e e e AY AY AY
1 2 2 eq F F F AY AY AY 1 2 2 1 1 eq Y Y Y 1 2 1 2 2 eq Y Y Y Y Y Problem 1.8Two wires of same length and radius are j oined in par allel and loaded. The Young’s modulii of the material of the wires are Y1 & Y2. If the combination is taken as a single wire then its Young’s modulus is
W 1 Y A A 2 y W 2A eq Y F = F1 + F2
2 1 2 Y A e Y Ae Y Ae 1 2 2Yeq Y Y 1 2 2 eq Y Y Y * Problem 1.9
The length of a metal wire is when the1
tension in it is T1 and is when the tension is2
T2. Then the actual length of the wire is Sol : Let its original length be ' ' .
We know e T
1
T1 .... (1)
2
T ... (2)2 From (1) & (2) 1 1 2 1 2 1 2 1 2 2 T T T T T T
2 1 1 2 2 1 T T T T 2 1 1 2 2 1 T T T T Problem 1.10The length of a rubber cord is metres1
when the tension in it is 4N and metres when2
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
From (1) & (2) 1 1 2 2 4 5 5 4 4 5 5142 ... 4
From (1) & (3)
1 1 2 1 3 3 1 2 5 4 4 4 9 5 4 9 1 1 2 3 1 2 5 4 4 5 4 9 2 1 3 1 2 4 4 4 5 4 9 2 1 3 1 2 36 36 4 20 16 2 1 3 3 2 1 20 16 4 5 4 * Problem 1.11A steel wire of 1mm diameter and of length 1m is stretched by applying force of 10N. If the increase in length is 0.064mm, find (i) the stress, (ii) the strain and (iii) the Young’s modulus of the wire. Solution : r = 0.5 x 10–3m; L=1m F = 10N ; e=0.064 x 10–3m i) Stress = 2 F F Apr =
2 10 p = 7 1.273 10 N m-2 (ii) Strain = 3 0.064 10 1 e L = 0.064 x 10–3 (iii) Y= 7 3 1.273 10 0.064 10 Stress Strain = 1.989x1011Nm–2 * Problem 1.12A steel wire of diameter 1 mm and length 2m is stretched by applying a force of 2kg wt. Calculate (i) the increase in length of the wire, (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) Sol: 1 3 r 10 m; L 2m; 2 F = 2kg wt = 2 x 9.8N; Y = 2 x 1011 Nm–2 i) 2 F L Y r e 2 2 3 11 FL 2 9.8 2 e r Y 1 10 2 10 2 = 2.495x10–4m
ii) The strain=
4 e 2.495 10 L 2 = 4 1.248 10
iii) The stress = Y x strain
= 4 11 2.495 10 2 10 2 = 7 2 2.495 10 Nm
*Problem 1.13
What mass must suspended from the free end of a steel wire of length 2m and diameter 1mm to stretch it by 1mm? (Y=2x1011Nm–2) Sol: r 1 10 m; L3 2m; 2 2 Mg L Y r e
2 Y r e M gL 2 11 1 3 3 2 10 10 1 10 2 9.8 2 2 10 100 8.015kg 9.8 4 39.2 *Problem 1.14A brass wire of length 5m and cross section 1mm2 is hung from a rigid support, with a brass
weight of volume 1000 cm3 hanging from the
other end. Find the decrease in the length of the wir e, when the br ass weight is completely immersed in water.
(Ybrass=1011 Nm–2; g=9.8ms–2; 3 water 1gcm
)
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 = ? Young's modulus, Y F L A e MgL e AY .
When weight hung in a liquid,
Weight of the body in the liquid = Mg - Vg where V is the volume of the body
This is the force, F acting on the wire i.e., F = Mg – Vg
Increase in length of the wire,
/ Mg v g L e 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
Mg V g L
MgL AY AY V gL AY Here, V = 1000cm3 = 1000 x 10-6m3 =1gcm–3=1x103kgm–3;g=9.8ms–2;L=5m 2 6 2 11 2 A1mm 1 10 m ;Y 1 10 Nm The decrease in length= 6 3 6 11 1000 10 1 10 9.8 5 1 10 1 10 = 49 x 10-5 m = 0.49 mm *Problem 1.15
A copper wire and a steel wire of radii in the ratio 1:2, lengths in the ratio 2:1 are stretched by the same force. If the Young's modulus of copper = 1.1 x 1011Nm–2 find the ratio of their extensions (youngsmodulus of steel = 2 x 1011 N/m2). Sol: we know 2 FL e r Y 2 1 1 2 2 2 2 1 1 e L r Y F e L r Y F Here r1:r2= 1:2, L1:L2 = 2:1 Y1=1.1x1011 Nm–2; Y2 = 2.0x1011Nm–2 2 11 1 11 2 e 2 2 2.0 10 16 160 e 1 1 1.1 10 1.1 11 e1:e2 = 160:11 *Problem 1.16
An aluminium wire and a steel wire of the same length and cross-section are j oined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in the length of the composite wire is 2.7mm, find the increase in the length of each wire.
(YA1=2x1011Nm–2, Y
steel=7x1011Nm–2) Sol: Total increase in length, e = e1 + e2.
we know e FL AY
As F, A, L are same for both the wires. So, e 1 Y 1 2 2 1 e Y e Y = 11 11 2 10 20 7 10 7 , 1 2 20 e e 7 substituting in e1 + e2 = 2.7 mm 1 2 20 e e 2.7mm 7 2 27e 2.7mm 7 e2 = 0.7 mm 1 2 20 20 e e 0.7 2.0mm 7 7 * Problem 1.17
A block of mass 1 kg is fastended to one end of a wire of cross - sectional area 2 mm2 and is rotated in a vertical circle of radius 20 cm. The speed of the block at the bottom of the circle is 3.5 m s–1. Find the elongation of the wire when the block is at the bottom.
Sol: i) Tension at the bottom of the circle,
2 mv T mg r
2 1 3.5 1 9.8 0.2 = 61.25 + 9.8 = 71.05 NThis tension in the string is equal to the force, F i.e. F = 71.05N, L =r = 0.2m.
The increase in length ,
6 11 71.05 0.2 2 10 2 10 FL e AY = 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
6 11 12.25 0.2 2 10 2 10 FL e AY = 5 0.6125 10 m * Problem 1.18
A steel wire of length 2 m and cross sectional area 2 mm2 is fixed at one end and stretched by suspending a block of mass 2 kg on the surface of the moon. If the Ypung’s modulus of steel is 2 x 1011 N m–2 find the increase in the length of the steel wire.
(g on the moon 1
6
of the g on the earth)
Sol: L = 2 m ; A = 2 mm2 = 2 x 10–6 m2; M = 2 kg ; g on the moon = 9.8 2 6 6 g ms 11 2 2 10 N ; ? Y e m
The increase in length ,
6 11 2 9.8 2 6 2 10 2 10 FL MgL e AY AY = 1.633 x 10–5m Problem 1.19
One end of a uniform wire of length ‘L’ and mass ‘M’ is attached rigidly to a point in the roof and a load of mass ‘m’ is suspended from its lower end. If A is the area of cross - section of the wire then find the stress in the wire at height ‘x’ from its lower end (x < L)
L
(L-x)
x
m
P
Tension in the string at point ‘P’ is T = wt of load + wt of wire of length ‘x’
M T mg xg L Stress at P = T/A = mg Mxg A AL Problem 1.20
A metal ring of radius ‘r’ and cross - sectional area ‘A’ is to be fitted on to a wooden circular disc of radius R (R > r). If the Young’s modulus of the material of the ring is Y the force with which the metal ring expands is
(R > r) r
Initial length of wire 2 rp
for it to be fitted onto a wodden disc, its final length must be 1 2 Rp
1 2 e p Rr , Y F Ae 2p YA YAe F
2p R r
YA R r r r
YA R r F r Problem 1.21If two wires are arranged as shown in the figure. What are the elongations of upper and lower wires
Sol : for lower wire F = m2g
2 2 2 2 2 F m g e e Ay Ay m1 m2 1 2 2 y 1 y A A For upper wire F = (m1 +m2)g
1 2
1 1 1 1 m m g F e e Ay Ay Problem 1.22As shown in adjucent figure if a load of mass (m) is attached at lower end of lower wire. Then find the displacements of the points B, C, D are
Sol : As shown in figure elongation of first wire
1 1 1 mg e Ay 1 2 2 y 1 y A D A m A B C 3 3 y elongation of 2nd wire
2 2 2 mg e Ay elongation of 3rd wire
3 3 3 mg e Ay displacement of B is e1 displacement of C is e1 + e2 displacement of D is e1 + e2 + e3 Problem 1.23A copper wire of negligible mass, length
, cross - sectional area (A) is kept on a smooth horizontal table with one end fixed, a ball of mass ‘m’ is attached at other end. The wire and the ball are rotated with angular velocity ‘w’. If wireelongates by then find Young’s modulus of
wire. If on increasing the angular velocity from
w to w1 the wire breakdown, obtain breaking stress
Sol : a) r F = T = mrw2
w2 m as in small , w2 F m F y Ae
w2
m y Ab) We know Breaking stress
= sec
Breaking force Area of cross tion
= 2 1 w m A Problem 1.24
A stone of mass (m) is attached to one end of a small wire of length
and cross - sectional area (A) suspended vertically. The stone is now rotated in horizontal plane such that the wire makes an angle ' 'q with ver tical. Find the increase in length of wire if its Young’s modulus is Y.Sol : From fig. cos
cos q q mg T mg T 2 sin q w T mR F T e AY AY mg cos T q sin T q R L T S q q cos mg e AY q
Note : If ' 'q was not given but R, w and m was given then in such case
2
2 2 2 2 2 2 cos sin T qT q mg mRw
2
2
2 T mg mRw and use e T Ay Problem 1.25A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross–section 10–3 m2 and the other is of brass of cross – section 2 x 10–3m2. Find out the position along the rod at which a weight may be hung to produce; i) equal stress in both wires
ii) equal strains in both wires
Young’s modulus of brass = 1 x 1011 N/m2 Young’s modulus of steel = 2 x 1011 N/m2
(2 x) x W A B T1 T2 Brass wire Steel wire C
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 T1 and T2 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 1 2 1 2 T T a a (or) 13 2 3 10 2 10 T T (or) T22T1 ... (i)
As the whole system is in equilibrium, so
t
= 0. Taking moment of all the forces acting on the rod about C, we have
1 2 2 0
T xT x ... (ii) Solving equations (i) and (ii), we get
4 3 x m
ii) Second case : For equal strains in both the wires
e1 = e2 1 2 1 1 2 2 T T a Y a Y (or) 3 1 11 23 11 10 2 10 2 10 10 T T (or) T1 = T2 ... (iii)
From equations (ii) and (iii) , we get x = 1 m
Problem 1.26
A steel wire of area of cross-section A and length 2Lis clamped firmly between two points seperated by a distance ' L'2 . A body is hung from the middle point of the wire such that the middle point sags by a distance x. Calculate the mass of the body and the angle made by the string with the horizontal L L T T Mg x Since ' ' is small x sin tan L F L y . A e
2 2
1 2 / YAe YA F L x L L L2 1pa 1N / m 2 2 2 1 2 2 YA x YA x F L L L L L L L L 2 2 2 YAx F L 2 2 2 2 2 2 2 T sin mg
T mg for small angles sin F mg YAx . mg L 2 2 2 2 YAx x . mg L L , 3 3 YAX M L g 3 3 x Mg YA L , 1 3 / x Mg L YA , x Tan L 1 3 1 3 1 / / Mg Mg Tan Tan YA YA Problem 1.27
A sphere of radius 0.1 m and mass 8p kg is attached to the lower end of a steel wire of length
5.0 m and diamet er 10–3m. The wire is suspended
from 5.22 m high ceiling of a room. When the sphere is made to swing as a simple pendulum, it j ust grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. Y for steel = 1.994 x 1011 N/m2. 5.22m 0.2m T 2 mv 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
5.22 5 0.2
L
= 0.02 m
So tension in the wire (due to elasticity)
2 11 4 1.994 10 5 10 0.02 199.4 5 YA T L N L p p and as equation of circular motion of a mass ‘m’ tied to a string in a vertical plane is
2
/ cos mv r T mg q So at lowest point
2
/ mv r T mg [as q ]0 But here r = 5 + 0.02 + 0.1 = 5.12 m So
8pv2/ 5.12
1.99.4p8p9.8
i.e., v 2
121 5.12 / 8
77.44, so v = 8.8 m/s.1.12 Elongation of wire 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 dgdx
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
1 w dx xAdg dx dg de x dx AY AY Y The total extension
0 0 dg e de x dx Y
= 0 dg x dx Y
, 2 2 dg e Y also 2 W e AY* Note 1.8 : The above formula can also derived by
considering total weight at center of mass and using effective length / 2 .
/ 2
F Y Ae
2 2 mg F e AY AY / 2 wt cm A
2 dg A Y , 2 2 dg e Y 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 2 2 dg F e Y AY 1.13 Thermal Stress:
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 o
C q . Then Thermal strain = a q as a q
Thermal stress = Y (thermal strain) Thermal stress =
Y
a q
Force
F
YA
a q
Problem 1.28
Two rods of 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 temperature of both the rods are now raised by
0
t C
then
a) Find the force with which the rods acts on each other at heigher temperature.
b) Find the lengths of the rods at the higher temperature.
Assume that there is no change in the cross-sectional area of the rods and the rods do not bend. There is no deformation of walls.
1
2
1 1
aY a Y2 2
A B
Sol : a) Due to heating the increases in length of the composite rod will be
I 1 1a t 2a2 t
1 1a 2a2
t ... (1)due to compressive force ‘F’ from the walls, due to elasticity, the decrease in length will be
1 2 1 2 1 2 1 2 D F F F AY AY A Y Y .... (2)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.
1 2 1 1 2 2 1 2 a a t F A Y Y
1 1 2 2
1 2 1 2 a a A t F Y Yb) As initially the length of one rod in L1 and due to heating its length increases by
1
H a1 1t, while due to compression its length decreases by
11
1
C F AY
so its final length
1 1 1 1 H 1 C = 1 1 1 1 1 a t F AYfor the other rod
1 2 2 2 H 2 C 1 2 2 2 2 2 2 a t F AYNote 1.10: In the above problem length of composite
rod remains unchanged, i.e. 1 1
1 2 1 2
. but that of individual rods changes i.e. 1
1 1
and 1
2 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
1 1 1 1 1 F t AY a where
1 1 2 2
1 2 1 2 t F A Y Y a a If 1 2 then 1 1 2 2 1 2 Y Y t Y Y a a * Problem 1.29A steel wire, 2mm in diameter, is j ust st r et ched bet ween t wo f i xed poi nt s at a temperature of 300C. Determine its tension when the temperatur e falls to 200C. (Coefficient of linear expansion of steel = 0.000011/0C; Young's modulus for steel = 2.0 x 1011 Nm–2)
Sol: Thermal stress = F Y t
A
Tension in the wire Y A t
2t1
. Here, Y =2.0x1011Nm–2; 6 0 11 10 / C; t2 = 300C; t1 = 200C; radius=1mm=1x10–3m;
2 2 3 6 2 A r 1 10 10 mThe tension in the wire
11 6 6
2.0 10 11 10 10 30 20
= 69.14N
1.14 Analogy of rod as a spring :
We know Y stress Y F strain A (or) AY F
=
F K AYconstant, depends on type of material and geomentry of rod.
F k (or) (F = kx) Where AY k is t he equivalent spring constant. Problem 1.30
A mass ‘m’ is attached with rod as shown in figure. This mass is slightly stretched and released then find the time period. (Y is Young’s modulus of rod, A is cross sectional area of rod, ' ' is its length). A m Y m AY k 2 2 m m T T k AY p p 1.15 Bulk modulus (K) :
It is defined as the ratio of the volume stress (normal stress) to the volume strain within the elastic limit volume stress K volume strain V VV F F F F F F F F
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 pres-sure 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 V V Bulk modulus P P K K V V V V
The negative sign shows that with increase in pressure, the volume decreses.
Note 1.12 : All the states of matter possess bulk
modulus
solids liquids gases
K K K
Note 1.13 : Gases have two bulk modulli, they are 1.16 Isothermal Bulk modulus of elasticity
Eq :The bulk modulus of a gas in isothermal con-dition is defined as isothermal Bulk modulus of eleasticity. We know for isothermal process
PV = constant
Differentiating both sides pdV + Vdp = 0 PdV = – V dp
dp P dV V dp P E P dV V q Hence isot hermal elasticity is equal to pressure.
1.17 Adiabatic Bulk modulus of elasticity
Ef :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
1 0 PgVgdV V dpg 1 P Vg gdV V dpg 1
P V V dV
g
g
V dp
g 1 p dV dp V g
dp dp p p dV dV V V g g EfgpHence, adiabatic Bulk modulus of elasticity is equal to ‘g’ times pressure
Note 1.14 : Ratio of adiabatic to isothermal Bulk
modulus of elasticity E P E P f q g g g > 1 EfEq
adiabatic bulk modulus of elasticity is g times to the isothermal bulk modulus of elasticity.
1.18 Compressibility :
The reciprocal of bulk modulus is called com-pressibility. i.e Compressibility = 1 mod Bulk 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
in-compressible i.e., compressibility is zero implies bulk modulus is infinite
*1.19 Density of compressed liquid :
If a liquid of density 'r', volume V and bulk modulus 'K' is compressed, then its density increases
density m
V r
1 V V V r r r --- (1) But by definition of bulk modulus K V P V P V V K ---(2) from (1) and (2) P K r r
1 1 P P K K r r r r r r 1 1 r r P KAlso r1r
1 C P
where ‘C’ is the com-pressibility.* Problem 1.31
A volume of 10–3 m3 is subj ected to a pressure of 10 atmosphere. The change in volume is 10-6 m3. Find the bulk modulus of water.
(Atmospheric pressure= 5 1 10 N m–2 ) Sol : 3 3 6 3 10 ; 10 ; 10 V m V m P atm = 5 6 2 10 10 1 10 Nm K = P V V V= 6 3 9 2 6 1 10 10 1 10 10 Nm * Problem 1.32
Determine the pressure required to reduce the given volume of water by 1 %. Bulk
modu-lus of water is 9 2 10 N m-2 Sol : 1 , 100 V V K= 9 2 10 N m-2P= ? K P V V , V P K V = 2 109 1 2 107 100 N m -2 Problem 1.33
A solid sphere of radius 'R' made of a mate-rial of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area 'A' floats on the surface of the liquid. Find the fractional change in the radius of the sphere
dR R
, when a mass M is placed on the piston to
compress the liquid.
Sol : As for a spherical body
3 4 3 V pR , V 3 R V R
Now by definition of bulk modulus
. P V P Mg Mg B V i e as P V V B AB A 1 3 3 dR V dR Mg R V R AB Problem 1.34
A uniform pressure 'P' is exerted on all sides of a solid cube at temper ature t0C . By what amount should the temperature of the cube be raised in order to bring 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 material are B and g respec-tively.
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
B
, (as P P)
Now with rise in temperature due to thermal expansion, volume increases so if 'q' is the rise in temperature then V V as V V g q g q
As the volume of the cube remains constant
VP P
V
Problem 1.35
When a rubber ball of volume V, bulk modu-lus 'K' is taken to a depth 'h' in water, then de-crease in its volume is
P K V V
P V
V K h m v Pa p h gr m v1 Pa p h g V K r So decrease in volume of ball h g Vr K
1.20 Modulus of rigidity
h (or) shear modulusWithin elastic limit, the ratio of shearing stress to the shearing strain is called modulus of rigidity of the material of the body.
shearing stress shearing strain h F F fixed A A L A A1 C C1 F B F D Fixed x q q
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.
//
Fel F Shearing stress
A A
As shown in above figure, the shearing force 'F' causes the consecutive horizontal layers of the
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
1 AA x shearing strain AB L q / shear stress F A F shear strain A h q q
* In this case shape of a body changes but its vol-ume remains unchanged.
* Only solids can exhibit a shearing as these have definite shape.
* Problem 1.36
A 5.0 cm cube of substance has its upper face displaced by 0.65 cm, by a tangential force of 0.25 N. Calculate the modulus of r igidity of the substance. Sol : h=FL A , A = L2, 2 FL F L L h Here,L5.0 10 2m 2 0.65 10 m ; F=0.25 N. 2 2 0.25 5.0 10 0.65 10 h = 4 0.25 10 3.25 = 769.2 N m–2 * Problem 1.37
A tangential force of 2100 N is applied on a
surface of area 6
3 10 m 2 which is 0.1 m from
a fixed face. The force produces a shift of 7mm of upper sur face wi th r espect to bot t om. Calculate the modulus of rigidity of the material. Sol: F = 2100 N ; A = 6 3 10 m2 ; L =0.1m; 3 7 10 m. 6 3 2100 0.1 3 10 7 10 FL A h =1x1010 Nm–2
* Problem 1.38
A steel plate of face ar ea 2 cm2 and the thickness 1.0cm is fixed rigidly at the lower face. A tangential force of 10 N is applied on the upper surface . Find the lateral displacement of the upper surface with respect to the lower surface.
Rigidity modulus of steel = 10
8.4 10 Nm -2 Sol : 2 4 2 2 2 10 A cm m ; L=1.0 cm= 2 1 10 m; F=10 N. 10 8.4 10 h N m-2 ; ? FL A h
The lateral displacement of the upper face with respect to the lower face is FLAh
= 2 4 10 10 1 10 2 10 8.4 10 = 1 10 7 16.8 m = 9 5.952 10 m. Problem 1.39
Calculate the force ‘F’ needed to punch a 1.46 cm diameter hole in a steel plate 1.27 cm thick (as shown in fig). The ultimate shear strength of steel is 345 M N/m2
Sol : As in punching, shear elasticity is involved, the
hole will be punched it 11 F A
ultimate shear stress
F
F11 > (shear stress) X Area
F11 min = (3.45 X 108)
2 rLp
A2prL
=
3.45 10 2 3.14 0.73 10 8
21.27 10 2200KN1.21 Some important 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 elas-tic while solids are most elaselas-tic.
solid liquid gas
E E E
Type of
stress Stress Strain
Change in
shape volume modulusElastic
Name of modulus State of Matter Tensile or compressive
Two equal and opposite forces perpendicular to opposite faces Elongation or compression parallel to force direction (longitudinal strain) Young's modulus Solid
s F A/
L L/ F L Y A L ShearingTwo equal and opposite forces parallel to opposite
surfaces [forces in each case such that total force and
total torque on the body vanishes]
Pure shear,q G F modulusShear Solid
A q Yes No Bulk Forces perpendicular everywhere to the surface, force per unit area (pressure)
same everywhere Volume change (compression or elongation) Bulk modulus Solid, liquid and gas No Yes / p B V V / V V Yes No
3) For a perfectly rigid body as L, V or
0
f , 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.
But as a 1 , a 1 Y K L V and 1 h a f for a
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-sec-tion hence steel is more elastic than rubber. 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 indepen-dent of the magnitude of the stress and strain. It de-pends 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 bridge 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 tyre 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 spring is stretched, the de-formation 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 com-pressed, the elasticity involved is volume, i.e., bulk
12) The shape of rubber heels changes under stress, the elasticity involved is shear, or rigidity.
1.22 Poisson's ratio :
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.
The ratio of change in radius (or) diameter to the original radius (or) diameter is called lat-eral strain L (D) r rr 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 lon-gitudinal 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.
' s Lateral strain poisson s ratio Longitudinal strain D D s negative sign indicates that the radius or diam-eter 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 ratio :
Consider a wire of length '' and radius 'r', then its volume 2 (1) V pr 2 V r V r
