# Physics 11th Elasticity

- On 05/03/2013
- Comments (0)
- In PHYSICS

## ELASTICITY

**1. Introduction**

- We know that any solid body has definite shape and size and more or less all solid bodies can be deformed by suitable application of forces.
- Forces producing deformation in any solid body can bring about changes in length , volume or shape of the body and the body is said to be strained or deformed.
- When deforming forces are removed the body tends to recover it's original condition.
- This property of material body to regain it's original condition, on removal of deforming forces is called ELASTICITY.
- When a solid body is deformed then it's constituents i.e., atoms or molecules gets displaced from their equilibrium position causing a change in interatomic or intermolecular distances.
- Again when the deforming forces are removed interatimic forces drives atoms or molecules back to their original equilibrium position. This way body regains it's original shape and size.

DEFINITIONS

(a) Elastic forces:- The forces developed inside the body when deformed , tending to restore it's original shape are called elastic forces.

(b) Perfectly elastic bodies:- Bodies which which can recover their original condition completely on removal of deforming forces are called perfectly elastic bodies.

(c) Plastic bodies:- Bodies which does not show any tandency to recover their original condition on removal of deforming forces are called plastic bodies. - There are no perfectly elastic or plastic bodies and actual bodies lie between two extremes.
- Nearest aproach to perfectly elastic body is a qquartz fibre and perfectly plastic body is putty.

**2. Stress**

- We know that when deforming forces acts on a body, forces of internal reaction develops inside the body which tends to restore the body to it's original position.
- These internal forces developed are equal in magnitude of deforming forces and acts in direction opposite to these externally applied deforming forces.
- Stress is this restoring force applied per unit area set up inside the body and is measured by the magnitude of deforming force acting on unit area with in the elastic limits of the body.

Thus,

where F is the force applied and A is the area of crosssection of the body. - S.I. unit of stress is Nm
^{-2}or pascal (Pa). In C.G.S. system it's unit is dynes/cm^{2}. - Dimensional formula for stress is [ML
^{-1}T^{-2}]. - Stress are of two types
**(a) Normal stress:-**If elastic forces developed are perpandicular to the area of crossection of the body then the stress developed is known as normal stress. - The stress is always normal in case of change in length of wire or in case of change in volume of body shown below in figure
- The normal stress are of two types, tensile and compressive stress, accordingly as there is a increase or decrease in length or volume of body on application of force.
**(b) Tangential or shearing stress:-**Tangential or shearing developes in a body when elastic restoring forces are parallel to the cross-sectional area of the body as shown below in figure 2. - Thus when deforming force acts tangentially over an area the body gets sheared through a certain angle.

**3. Strain**

- When a body is under a system of forces or couples in equilibrium then a change is produced in the dimensions of the body.
- This fractional change or deformation produced in the body is called strain.
- Strain is a imensionless quantity.
- Strain is of three types
**(a) Longitudinal strain:-**It is defined as the ratio of the change in length to the original length. If l is the original length and Δl is the change in length then,**(b) Volume strain:-**It is defined as the ratio of change in volume to the original volume**(c) Shearing strain:-**If the deforming forces produce change in shape of the body then the strain is called shear strain. Considering Figure 2. it can also be defined as the ratio of displacement x of corner b to the transverse dimension l. Thus

or,*Shear strain = tanθ*

In practice since x is much smaller than l so, tanθ ≅ θ and the strain is simply the angle θ(measured in radians). Thus, shear strain is pure number without units as it is ratio of two lengths.

**4. Hook's Law**

- Hook's law is the fundamental law of elasticity and is stated as " for small deformations stress is proportional to strain".

Thus,*stress ∝ strain*

or,*stress/strain = constant*

This constant is known as modulus of elasticity of a given material. - Hook's law is not valid for plastic materials.
- Units and dimension of the modulus of elasticity are same as those of stress.

**5. Elastic Modulus**

- Stress required to produce a given strain in a material body depends on the nature of material under stress.
- We already know that ratio of stress to strain is known as elastic modulus of the material.
- Larger is the elastic modulus of a given material, greater would be the stress needed to produce a given strain.
- There are three different types of modulus of elasticity- Young's Modulus of elasticity, Bulk Modulus of elasticity and Modulus of Rigidity.
**(a) Young's Modulus of Elasticity** - Young's Modulus of elasticity is the ratio of longitudinal stress to longitudinal strain.
- It is denoted by Y.
- Young's Modulus of elasticity is given by

or - Let us now consider a wire of length l having area of cross-section equal to A. If the force F acting on the wire, stretches the wire by length Δ l then

and

From (1) and (2) we have Young's modulus of elasticity as

- Young's modulus of elasticity has dimensions of force/Area i.e. of pressure.
- Unit of Young's modulus is N/m
^{2}. - If area of cross-section of a wire is given by A = πr
^{2}then Young's modulus is

again if A = π r^{2}= 1cm^{2}and Δ l = l = 1cm then

Y = F

Thus, Young's modulus can also be defined as the force required to double the length of a wire of unit length and unit area of cross-section.**(b) Bulk Modulus of Elasticity** - The ratio of normal stress to volume strain within elastic limits is called Bulk Modulus of elasticity of a given material.
- It is denoted by K.
- Suppose a force F is applied normal to a surface of a body havin cross-sectional area equal to A.

If applied force bring about a change ΔV in the volume of the body and V is the original volume of the body then,

and

So, Bulk Modulus of elasticity would be,

Thus, - For gases and liquids the normal stress is caused by change in pressure i. e.,

normal stress = change in pressure ΔP.

Thus, bulk Modulus is

here negative sign indicates that the volume decreases if pressure increases and vice-versa. - For extremely small changes in pressure and volume, the Bulk Modulus is given by

- Reciprocal of Bulk Modulus is called compressibility of substance. Thus,
**(c) Modulus of Rigidity** - When a body is sheared, the ratio of tangential stress to the shearing strain within elastic limits is called the Modulus of Rigidity.
- If lower face of the rectangular block shown below in the figure, is fixed and tangential force is applied at the upper face of area A, then shape of rectangular block changes.

So,

shearing strain = θ ≅ tanθ

or,

Thus,

**6. Poisson's Ratio**

- When two equal and opposite forces are applied to a body in a certain direction , the body extends along that direction and at the same time it cintracts along the perpandicular direction.
- The fractional change in length of the body in the direction of the applied forces is longitudinal strain and fractional change in the perpandicular direction of the force applied is called lateral strain.
- The ratio of lateral strain to the longitudinal strain is called poisson's ratio which is constant for material of that body.
- So when a body is subjected to strain , say an elongation, it also suffers contraction in parpandicular direction.
- Within elestic limits, lateral strain β is is proportional to longitudinal strain α.hence

σ = β α

longitudinal strain = α = Δl l

lateral strain = β = ΔD D - hence poisson's ratio is

σ = lΔD DΔl

**7. Stress-Strain Digram**

- In case of solids if we go on increasing stress continually then a point is reached at which strain increases more and more rapidly and Hook's law is no longer obeyed.
- Thus, the stress at which linear relationship between stress and strain ceases to hold is referred as elastic limit of material for the stress applied.
- If the elastic limit of material is exceeded it will fail to recover its original shape or size on removal of stress and would acquire a permanent set.
- Any type of stress can be plotted against appropriate strain and the shape of resulting stress- strain digrams would have shapes, depending on the kind of material.
- Simple stress- strain digram for a bar or wire is shown below in the figure.

(i). Portion OA is the straight line which clearly shows that stress produced is directly proportional to strain i.e., Hook's law is perfectly obeyed upto A and on removal of stress wire or bar will recover its original condition. Point A is called Proportionality limit

(ii). As soon as proportionality limit is crossed beyond point A, the strain increases more rapidly than stress and curve AB in graph shows that extension of wire in this limit is partly elastic and partly plastic and point B is the elastic limit of the material. Thus if we start decreasing load from point B the graph does not come to O via path BAO instead it traces straight line BG. So that there remains a residual strain. This is called permanent set.

(iii). If we continue to increases the stress beyond point B then for little or no increase in stress the strain increases rapidly upto point C.

(iv). Further increase of stress beyond point C produces a large increase in strain untill a point E is reached at which fracture takes place and from B to D material is said to undergo plastic flow which is irreversible.**Conclusion :** - The wire exhibits elasticity from O to b and plasticity from b to d. If the distance between b and d is more, then the metal is ductile. If the distance between b and d is small, then metal is brittle.
- The substances which break as soon as the stress is increased beyond elastic limit are called brittle substances eg: glass, cast iron, high carbon steel.
- The substances which have a large plastic range are called ductile substances. Eg: copper, lead, gold, silver, iron, aluminium. Ductile materials can be drawn into wires. Malleable materials can be hammered into thin sheets. Eg: gold, silver, lead.

Solved Examples

**Question 1**. A block of gelatin is 60 mm by 60 mm by 20 mm when unstressed. A force of .245 N is applied tangentially to the upper surface causing a 5mm displacement relative to the lower surface.The block is placed such that 60X60 comes on the lower and upper surface. Find the shearing stress,shearing strain and shear modulus

a) (68.1 N/m

^{2},.25,272.4 N/m

^{2})

b) (68 N/m

^{2},.25,272 N/m

^{2})

c) (67 N/m

^{2},.26,270.4 N/m

^{2})

d) (68.5 N/m

^{2},.27,272.4 N/m

^{2})

Solution:

Shear stress=F/A=.245/36*10

^{-4}==68.1 N/m

^{2}

Shear strain= tan?= d/h=5/20=.25

Shear modulus (S) =shear stress/shear strain=272.4 N/m

^{2}

**Question 2**.A steel wire of diameter 4mm has a breaking strenght of 4X10

^{5}N. The breaking strenght of similar steel wire of diameter 2 mm is a.1X10

^{5}N.

b.4X10

^{5}N.

c.16X10

^{5}N.

d. none of the these

**Solution 2**.

Breaking strenght is proportional to square of diameter,Since diameter becomes half,Breaking strenght reduced by 1/4/ Hence A is correct.

**Question 3**.What is the SI unit of modulus of elasticity of a substance?BR> a. Nm

^{-1}BR> b. Nm

^{-2}

c. Jm

^{-1}BR> d. Unitless quantity

**Solution 3**

Answer is b

**Question 4**A thick uniform rubber rope of density 1.5 gcm

^{-3}and Young Modulus 5X1010

^{6}Nm

^{-2}has a length 8 m. when hung from the celing of the room,the increase in length due to its own weight would be ? a. .86m

b. .2m

c. .1m

d. .096m

**Solution 4**The weight of the rope can be assumed to act at its mid point. Now the extension x is proportional to the original lenghth L. if the weight of the rope acts at its mid point,the extension will be that produced by the half of the rope.So replacing L by L/2 in the expression for Young 's Modulus ,we have Y=FL/2Al or l=FL/2AY Since F=mg=ρV=ρAL Therefore l=gL

^{2}ρ/2Y Substituing the values,we get l=.096m