- The property by virtue of which material bodies regain their original dimensions (size, shape or both) after removal of deforming force is called elasticity.
- e.g. Rubber, Steel, Aluminum, Sponge etc.
- When a body is acted upon by deforming forces the shape and/or size of the body changes. But, if the deforming forces are removed, the body retains its new shape and size. Such body is called as plastic body and the property is called as plasticity.
- e.g. Plaster of Paris, Clay, Mud, Plastic, etc. shows plasticity.
- When a body is acted upon by deforming forces, the shape and size of the body do not get altered, whatever may be the magnitude of deforming forces. Such body is called as a rigid body and the property is known as rigidity.
Elasticity in Detail:
- The net internal elastic force (restoring force) acting per unit area of the surface which is subjected to deformation is called stress.
Stress = F / A
- S.I. Unit of stress is N m-2 or Pa (pascal) and its dimensions are [L-1M1T-2].
- The change in dimension per unit original dimension of a body subjected to deforming forces is called as Strain.
Strain = Change in dimension / Original dimension
- it is unitless, dimensionless quantity.
Hooke’s Law of Elasticity:
- Statement: Within the elastic limit, the stress developed in the body is directly proportional to the strain produced in the body.
- Explanation :
By Hooke’s Law, Stress ∝ Strain
∴ Stress = Constant x Strain
∴ Stress / Strain = Constant
- This constant of proportionality is called as the modulus of elasticity or coefficient of elasticity.
- It is the upper limit of deforming force up to which if the deforming force is removed, the body regains its original shape and size completely.
- When the deforming forces are such that there is a change in the length of the body, then the stress produced in the body is called longitudinal stress. Longitudinal stress is further classified into two types. Tensile stress and compressive stress.
Longitudinal stress = F / A
- Its S.I. Unit of stress is N m-2 or Pa (pascal) and its dimensions are [L-1M1T-2].
- The change in length per unit original length of a body subjected to deforming forces is called as Strain.
- Mathematically the longitudinal strain is given by
Longitudinal strain = Change in length(l) / Original length (L)
- It is unitless dimensionless quantity
Young’s Modulus of Elasticity:
- Within elastic limit, the ratio of the longitudinal stress to the corresponding longitudinal strain in the body is always constant, which is called as Young’s modulus of elasticity.
- It is denoted by letter “Y” or “E”. Its S.I. Unit of stress is N m-2 or Pa (pascal) and its dimensions are [L-1M1T-2].
Youn’s modulus of elasticity = Longitudinal stress / Longitudinal strain
- Young’s modulus of elasticity is not defined for liquids and gases.
Expression for Young’s Modulus of Elasticity of the Material of a Wire:
- Consider a wire of length ‘L’ and radius of cross-section ‘r’ is fixed at one end and stretched by suspending a load of ‘mg’ from the other end. Let ‘‘ be the extension produced in the wire when it is fully stretched.
- Now, by the definition of Yong’s modulus of elasticity we have This is an expression for Young’s modulus of elasticity of a material of a wire.
- This is an expression for Young’s modulus of elasticity of a material of a wire.
- The ratio of transverse strain to the corresponding longitudinal strain is called Poisson’s ratio.
Poisson’s Ratio = Lateral strain / Longitudinal strain
Problems Based on Stress, Strain and Young’s Modulus
- A wire 2 m long and 1 mm in diameter, when stretched by weight of 8 kg has its length increased by 0.24 mm. Find the stress, strain and Young’s modulus of the material of the wire. g = 9.8 m/s² . Ans: (2.5× 107 N/m², 1.2 × 10-4 , 2.08 × 1011 N/m²)
- A wire of length 2 m and cross-sectional area 10-4 m² is stretched by a load 102 kg. The wire is stretched by 0.1 cm. Calculate longitudinal stress, longitudinal strain and Youn’s modulus of the material of wire. Ans : (1 × 107 N/m², 5 × 10-4 , 2 × 1010 N/m²)
- A mild steel wire of radius 0.5 mm and length 3 m is stretched by a force of 40 N. calculate a) longitudinal stress, b) longitudinal strain c) elongation produced in the body if Y for steel is 2.1 × 1011 N/m². Ans: (6.238 × 107 N/m², 2.970 × 10-4, 0.891 mm)
- A metal wire 1 m long and of 2 mm diameter is stretched by a load of 40 kg. If Y = 7 × 1010 N/m² for the metal, find the (1) stress (2) strain and (3) force constant of the material of the wire. Ans: (1.25 × 108 N/m², 1.78 × 10-3, 2.2 × 105 N/m)
- What must be the elongation of a wire 5m long so that the strain is 1% of 0.1? If the wire has cross-selection of 1mm² and is stretched by 10 kg-wt, what is the stress? Ans: (5 mm, 9.8 × 107 N/m²)
- A brass wire of length 2 m has its one end, fixed to a rigid support and from the other end a 4 kg wt is suspended. If the radius of the wire is 0.35 mm, find the extension produced in the wire. g = 9.81m/s² ,Y = 11 × 1010 N/m² Ans: (0.185 m)
- A wire of length 1.5 m and of radius 0.4 mm is stretched by 1.2 mm on loading. If the Young’s modulus of its material is 12.5 × 1010 N/m². , find the stretching force. Ans: (50.27 N)
- What force is required to stretch a steel wire 1 cm2 in cross-section to double its length? Y = 2× 1011 N/m². Assume Hooke’s law. Ans: (2× 107 N)
- Find the maximum load which may be placed on a tungsten wire of diameter 2 mm so that the permitted strain does not exceed 1/1000. Young’s modulus for tungsten = Y = 35 × 1010 N/m². Ans: (1100 N)
- A mass of 2kg is hung from a steel wire of radius 0.5 mm and length 3m. Compute the extension produced. What should be the minimum radius of wire so that elastic limit is not exceeded? Elastic limit for steel is 2.4 × 108 N/m², Y for steel = Y = 20 × 1010 N/m² Ans: (0.3743 mm, 0.1612 mm)
- A wire is stretched by the application of a force of 50 kg wt/sq. cm. What is the percentage increase in the length of the wire? Y = 7 × 1010 N/m², g = 9.8 m/s². Ans: (0.007 percent)
- A compressive force of 4 × 104 N is exerted at the end of a bone of length 30 cm and 4 cm² square cross-sectional area. What will happen to the bone? Calculate the change in length of a bone. The compressive strength of bone is 7.7 × 108 N/m² and Young’s modulus of bone is 1.5 × 1010 N/m². Ans: (2 mm)
- The radius of a copper bar is 4 mm. What force is required to stretch the rod by 20% of its length assuming that the elastic limit is not exceeded? Y = 12 × 1010 N/m². Ans: (1.207× 106 N)
- If the strain to be 1% of 0.1, find the change in length of a wire 5m long and 1 mm² in cross-section when the stretching force is 10 kg-wt. Y = 4.9 × 1011 N/m². and g=9.8 m/s². Ans: (1 mm)
- Elastic limit is exceeded when the strain in a wire (Y=14 × 1011 N/m² ) exceeds 1/2000. If the area of the cross-section of the wire is 0.02 cm², find the maximum load that can be used for stretching the wire without causing a permanent set. Ans: (1400 N)
- Elastic limit of steel is exceeded when the stress on given steel wire exceeds 8.26 × 108 N/m². Can a steel wire ( Y = 2 × 1011 N/m² ) 2m long be stretched by 10 mm without exceeding the elastic limit? Ans: (N0)
- Young’s modulus of the material of a wire is 9.68 × 1010 N/m². A wire of this material of diameter 0.95 mm is stretched by applying a certain force. What should be the limit of this force if the strain is not to exceed 1 in 1000? Ans: (68.62 N)
- The elastic limit of copper is 1.5 × 108 N/m². A copper wire is to be stretched by a load of 10 kg. find the minimum diameter the wire must have if the elastic limit is not to be exceeded. Ans: (0.912 mm)
- What would be the greatest length of a steel wire which when fixed at one end can hang freely without breaking? Density of steel = 7800 kg/m³. Breaking stress for steel = 7.8 × 108 N/m². Ans: (1.021 × 104 m)
Problems on Poisson’s Ratio
- When a brass rod of diameter 6 mm is subjected to a tension of 5 × 103 N, the diameter changes by 3.6 × 10-4 cm. Calculate the longitudinal strain and Poisson’s ratio for brass given that Y for the brass is 9 × 1010 N/m². Ans: (1.96 × 10-3 , 0.31)
- A metal wire of length 1.5 m is loaded and an elongation of 2 mm is produced. If the diameter of the wire is 1 mm, find the change in the diameter of the wire when elongated. σ = 0.24. Ans : (3.2 × 10-7 m)
- A metallic wire ( Y = 20 × 1010 N/m². and σ = 0.26) of length 3 m and diameter 0.1 cm is stretched by a load of 10 kg. Calculate the decrease in diameter of the wire. Ans: (1.62 × 10-7 m)
- A copper wire 3m long and 1 mm² in cross-section is fixed at one end and a weight of 10 kg is attached at the free end. If Y for copper is 12.5 × 1010 N/m² and σ = 0.25 find the extension, lateral strain and the lateral compression produced in the wire. Ans : (2.352 mm, 1.96 × 10-4, 2.21 × 10-7 m)
- A wire of diameter 2 mm and length 5 m is stretched by a load of 10 kg. Find the extension produced in the wire if Y = 12 × 1010 N/m². If σ = 0.35 for the material of the wire, find the lateral contraction. Ans : (1.3 mm, 1.82 × 10-7 m)
- Find the longitudinal stress to be studied to a wire to decrease its diameter uniformly by 1%. Poisson’s ratio = 0.25, Young’s modulus = 2 × 1011N/m² . Ans : ( 8 × 109 N/m²)
- A copper wire 3 m long is stretched to increase its length by 0.3 cm. Find the lateral strain produced in the wire. If Poisson’s ratio for copper is 0.26. Ans: (3.6 × 10-4)
- A steel wire having cross-sectional area 1 mm² is stretched by 10 N. Find the lateral strain produced in the wire. Young’s modulus for steel is 2 × 1011 N/m² and Poisson’s ratio is 0.291. Ans: (0.291 × 5 × 10-5 , 1.455 × 10-5)
- A load 1 kg produces a certain extension in the wire of length 3 m and radius 5 × 10-4 m. How much will be the lateral strain produced in the wire? Given Y = 7.48 × 1010 N/m², σ = 0.291. Ans: (4.86 × 10-5)
Problems Based on Composite Wires
- For two wires of the same material, both the radii and lengths are in the ratio 1:2. What should be the ratio of stretching forces on the wires if equal extensions are to be produced in the two? Ans: (1:2)
- Two wires of different material, but of the same cross-section and length are stretched by the same force. Find the ratio of Young’s moduli are of their material if the elongations produced in them are in the ratio 3:1. Ans: (1:3)
- Two wires are of the same material. Wire 1 if four times longer than the wire 2 but wire 1 has a diameter double that of wire 2. Compare stresses and elongations produced in the wires when under the same load. Ans: (equal)
- Two wires of the same material have lengths in the ratio 2:1 and diameters are in the ratio 2:1 Find the ratio of extensions produced in the wires when the stretching forces acting on them are in the ratio 2:1. Ans: (equal)
- A brass wire (Y = 11 × 1010 N/m²) and a steel wire (22 × 1010 N/m²) of the same length and cross-section are joined end to end. The composite wire is hung from a rigid support and a weight is suspended from the free end. Find the extension in each wire if the increase in the length of the composite wire is 0.279 cm. Ans: (0.186 cm , 0.o93 cm)
- A uniform brass wire (Y = 10 × 1010 N/m²) and a uniform steel wire (Y = 20 × 1010 N/m²) each of length 3.14 m and diameter 2 × 10-3 m are joined end to end to form a composite wire is hung from a rigid support and a load suspended from the free end. If the increase in the length of the composite wire is 6 × 10-3 m, find the increase in the length of each wire. Ans: (4 mm, 2 mm)
- A light rod 1 m long is suspended horizontally by two wires of the same length and of the same cross-section but of different materials. The Young’s modulus of the material of one wire is 30 × 1010 N/m² and that of the other is 20 × 1010 N/m². At what point should a weight W be hung from the rod so that it still remains horizontal? Ans : (0.4 m from first wire)
- A weightless rod 105 cm long is suspended horizontally by two wires P and Q of equal length. The cross-section of P is 1 mm² and that of Q is 2 mm². From what point on the rod should a weight be suspended in order to produce equal strains in P and Q? Yp= 2× 1011 N/m²; YQ = 1011 N/m². Ans : (Midpoint)
- A brass wire of length 5 m and of sectional area 1 mm2 is hung from a rigid support with a brass weight of volume 1000 cc hanging from the other end. Find the decrease in length of the wire when the brass weight is completely immersed in water. Y = 1011 N/m². Take g = 10 m/s². Density of brass = 8400 kg/m3 and of water = 1000 kg/m3. Ans : (0.5 mm)