It is a property of a matter for which a material body able to regain its initial state of condition on removal of the external forces that applied on it.
2.2 Review of elastic properties of solids:
Elasticity: It is a property of a matter for which a material body able to regain its initial state of condition on removal of the external forces that applied on it.
Perfectly elastic body: The body that can regain its original state completely on the removal of the force (Figs 7 a, b).
Perfectly plastic body: The body that shows no tendency to regain its original condition on the removal of the deforming force (Fig. 7 c).
(a) (b) (c)
Fig. 7
Load: Load = External force + Own weight along the force (Fig. 8).
(a) (b) |
Fig. 8
Stress:
In the elastic limit,
i) Stress = force applied/ area = F/A (if F is applied uniformly)
When the deforming force F be inclined to the surface then: (Fig. 9 a)
ii) Tangential (shearing stress) = Fcosφ/A (Fig. 9 b).
iii) Normal stress (tensile stress) = Fsinφ/A (Fig. 9 b).
Strain: Under a stress a body undergoes a deformation in respect of length or volume or shape. The change in the dimension is described by the quantity strain.
i) Longitudinal or tensile strain = change in length/ original length = δl/L (Fig. 10 a).
ii) Volume strain = change in volume/ original volume = δv/V (Fig. 10 b).
iii) Shearing strain, θ = tanθ (Fig. 10 c).
(a) (b) (c)
Fig. 10
Hooke’s law (fundamental law of elasticity):
Within the elastic limit the stress is proportional to strain (Fig. 11). Therefore, stress/strain = constant (E). This constant is called modulus of elasticity.
Fig. 11
Stress-strain diagram of a material:
OA: The wire is perfectly elastic. AB: Stress and strain are not proportional. But OB region still exhibits elastic behavior. Beyond B: From D the wire does not come back to its original length at O but come to the position at C.
|
Fig. 12
Three types of elasticity:
(a) (b) (c)
Fig. 13
If tan θ = θ = l/L then µ =(F/A)/(l/L)=FL/AI
Poisson’s ratio: The idea is that on being stretched, a wire becomes longer but thinner.
From Fig. 14:
σ =(∆D/D)/(∆l/l)
Lateral and linear (or tangential) strains per unit stress are denoted by β and α respectively then σ = β/α.
Fig. 14
Problem 1: A wire of 2 meters long and 0.5 mm in diameter and a mass of 10 kg is suspended. It is stretched by 2.33 mm. Find out the value of the Young’s modulus of the wire.
Soln.: Let
If A is the area of cross-section and ρ is the density of the material of the wire,
then the mass of the wire = volume × density =
Breaking stress = F/A = mg/A =
or, 
