 # Elasticity

Welcome to class!

In today’s class, we will be talking about elasticity. Enjoy the class!

### Elasticity #### Definition of elastic properties of a material

When a material is stretched by a force, its size or shape may change. The larger the force applied, the greater the deformation. Some materials can regain their original shape or size when the straitening force is removed. Such materials are called Elastic Material.

Elasticity is the ability of a material to return to its original shape and dimension upon removal of the applied force.

Elastic properties of materials are very useful in many applications. The application of this property of materials includes spring balance of a clock, spiral spring used in the shock absorber of cars, materials for bridges, etc are shooting of stones and in a catapult.

• ##### Elasticity:

The ability of a material to return to its original shape and dimension upon removal of the applied force. This ability varies from one material to another.

• ##### Elastic limit (E):

The load above which the material no longer recovers its original shape and dimension upon removal of the load.

• ##### Yield point (Y):

The load at which the material begins to show a substantial deviation from Hooke’s law, i.e. the beginning of plastic deformation.

The maximum load which the material can support,

• ##### Fracture point (F):

The load at which the material breaks into two, i.e. it fractures.

#### Hooke’s law

This state that, the extension of a wire is directly proportional to the load applied provided the elastic limit is not exceeded.

Mathematically,

f = ke

k is called the stiffness or force constant of the material. The unit is N.

##### Stiffness or force constant

Stiffness or Force constant of an elastic material is the force required to give a unit extension.

k =f/e

The unit is Nm-1

###### Worked examples

(1) A force of 0.8N is stretching an elastic spring by 2cm. find the elastic constant of the spring.

Solution

From Hooke’s law

F = ke

F = 0.8N,

e = 2cm = 0.02m

0.8 = 0.02k

k =0.8 / 0.02

k = 40Nm-1

(2) A spiral spring extends by 5m under a load of 60N. When the load is replaced by a steel block, the new extension is 7m, the weight of the steel block is.

Solution

From Hooke’s law

F = ke

F = 60N

e = 5m

60 = 5k

k = 60 / 5

k = 12Nm-1

therefore;

F = ke

k = 12N/m, e = 7m

Weight of the block = ke

Weight of the block = 12 x 7 = 84N.

#### Stress and strain

Stress is defined as the force acting on a unit cross-sectional area of the material. If a tensile force F is applied to a material of cross-sectional area A.

Tensile stress = Force / Area

The unit is N/m2

Strain is the ratio of the extension (x) to the original length (L), i.e.

Tensile strain = extension / length

It has no unit

##### Young’s modulus

The young’s modulus of elasticity is the ratio of tensile stress to tensile strain within the elastic region.

Young’s modulus (E) =tensile stress / tensile strain

Young’s modulus (E) = (force * length) / (area * extension)

The unit is N/m2

##### Energy stored in a stretched body

Whenever an elastic material is stretched, work is done on the material. This work is stored in the material as potential energy. For instance, when you stretch a catapult it has potential energy, which can throw a stone. The stored potential energy is thereby converted to the kinetic energy of the stone.

##### Work done

The work done when a wire or bar undergoes an extension x under an applied force (F) is

Work done = ½ Fx = ½ kx2

##### Elastic potential energy

The elastic potential energy or energy stored in an elastic bar or spring is equal to the work done in stretching the bar by a distance x. i.e.

Elastic potential energy = ½ kx2

Therefore, when Hooke’s law is obeyed, the work done is the area under the force vs extension plot.

###### Worked examples

(1) A spring is stretched 40mm by a force of 15N. What is the work done by the force?.

Solution

W = ½ Fe

(e) = 40 mm = 0.04m

F= 15N

W = ½ x 15 x 0.04

W = 0.3J

(2) If a force of 8N extends a spring by 0.4m. Calculate the force constant of the spring and the energy stored in the spring.

Solution

F = 8N, e = 0.4m

F = ke

k = F/e = 8/0.4

k = 20N/m

Energy stored = ½ ke2

= ½ x 20 x (0.4)2

= 1.6J

(3) A rubber band of length 20cm has a load of 10N tied to one end while suspended freely at the other end. If the length of the band now increases to 25cm, calculate the stiffness of the rubber band. What will be the new length when a load of 15N is suspended from its end?

Solution

Extension = new length – original length

= 25cm – 20cm = 5cm = 0.05cm

Force = 10N

F = ke

10N = k (5cm)

k = 2Ncm-1

(ii) if 15 – 2e

Then e = 7.5cm or 0.75m

The new length will 20 cm + 7.5cm = 27.5cm

(4) A force of 2N stretches an elastic material by 30mm. what additional force will stretch the material to 35mm? Assume that the elastic limit is not exceeded.

Solution

From Hooke’s law

F = 2N, e = 30mm

F = ke

k = F/e = 2/0.03 = 66.67(N/m)

let the force stretching the material 35mm be:

F = ke

F = 2.33N

Therefore additional force = 2.33N – 2.00N

=0.33N

We hope you enjoyed the class.

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