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Elastic Modulus>Young’s Modulus

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Topics

  • Definition
  • Formula Derivation
  • Table  
  • Example 1
  • Example 2
Maharashtra State Board: Class 11

Introduction

There are three types of moduli of elasticity corresponding to stress and strain related to changes in length, volume, and shape. Young's modulus is specifically the modulus of elasticity related to the change in length of an object (like a metal wire or rod) under a deforming force. Due to this relationship, it is also known as the elasticity of length.

Maharashtra State Board: Class 11

History/Origin

  • Name Origin: It is named after the British physicist Thomas Young.
  • Timeline: Thomas Young lived from 1773 to 1829.
Maharashtra State Board: Class 11

Definition: Young's Modulus

"Young’s modulus is the ratio of longitudinal stress to longitudinal strain."

Maharashtra State Board: Class 11

Formula: Young's modulus

The mathematical expression for Young's modulus (Y) is:

Y = \[\frac{MgL}{\pi r^2l}\]

Where:

  • Y = Young’s Modulus
  • M = Mass of the load attached
  • g = Acceleration due to gravity
  • L = Original length of the wire
  • r = Radius of the wire cross-section
  • l = Extension or elongation produced in the wire
Maharashtra State Board: Class 11

Characteristics

  • Elasticity of Length: Specifically relates to changes in length (longitudinal).
  • Resistance Indicator: Indicates the resistance of an elastic solid to elongation or compression.
  • Material Property: Useful for characterizing objects subjected to tension or compression.
  • Scope: It is a property of solids only.
  • SI Unit: N/m2 (Newton per square meter) or Pascal (Pa).
  • Dimensions: [L−1M1T−2].
Maharashtra State Board: Class 11

Explanation and Derivation

Consider a metal wire suspended from a rigid support. When a load is attached to the free end, it creates a deforming force that stretches the wire.

  1. Force Applied: A load of mass M exerts a downward force F = Mg.
  2. Longitudinal Stress: This force acts on the cross-sectional area A = πr2.
    Stress = \[\frac{\text{Applied Force}}{\mathrm{Area}}=\frac{Mg}{\pi r^2}\]
  3. Longitudinal Strain: The wire extends by length l from its original length L.
    Strain = \[\frac{\text{Change in length}}{\text{Original length}}=\frac{l}{L}\]
  4. Final Ratio: Young's modulus is calculated by dividing the stress by the strain.

Maharashtra State Board: Class 11

Example 1

Problem: Compare Young's modulus of a brass wire (Length 4.5m, Area 3 × 10−5m2) and a copper wire (Length 5.0m, Area 4 × 10−5m2) stretched by the same load and producing the same elongation.

Solution:

  1. Identify constants: Force (F) and elongation (l) are the same for both.
  2. Formula for Brass (YB): YB = \[\frac{F\times4.5}{3\times10^{-5}\times l}\]
  3. Formula for Copper (YC): YC = \[\frac{F\times5.0}{4\times10^{-5}\times l}\]
  4. Take the ratio (\[\frac {Y_B}{Y_C}\]): Divide the Brass equation by the Copper equation.
  5. Result: The ratio is 1.2.
Maharashtra State Board: Class 11

Example 2

Problem: A 20m wire with area 1.25 × 10−4m2 stretches by 1 × 10−4munder a load of 2.5 kg.

Solution:

  1. Given Data:
    L = 20 m
    A = 1.25 × 10−4 m2
    F = m × g = 2.5 × 9.8 N
    l = 10−4 m
  2. Calculation:
    Substitute values into Y = \[\frac{FL}{Al}\]
    Y = \[\frac{(2.5\times9.8)\times20}{(1.25\times10^{-4})\times(10^{-4})}\]
  3. Result: Y = 3.92 × 1010 N/m2

Test Yourself

Related QuestionsVIEW ALL [31]

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by `(Ypir^4)/(4R) . Y` is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

The stress-strain graphs for materials A and B are shown in Figure

The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young’s modulus?

(b) Which of the two is the stronger material?

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by `(Ypir^4)/(4R) . Y` is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Read the following statements below carefully and state, with reasons, if it is true or false

The Young’s modulus of rubber is greater than that of steel;

A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 × 1011 Nm–2.)

A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m–3 (Young’s modules Y = 2 × 1011 Nm–2).

A uniform metal rod of 2 mm2 cross section is heated from 0°C to 20°C. The coefficient of linear expansion of the rod is 12 × 10-6/°C, it's Young's modulus is 1011 N/m2. The energy stored per unit volume of the rod is ______.

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