Elasticity Study Notes

ELASTICITY PHS 111 Study Notes

CONTENT

  • Definitions
  • Some Properties of Metals
  • Hooke’s Law
  • Modulus of Elasticity, Rigidity (Shear) and Bulk Modulus
  • Energy Stored in a Strain
  • Poisson Ratio
  • Solved Problems

MATTER

  • States of Matter

    • The states of matter in bulk can be conveniently divided into solids and fluids.
    • Solids: Tend to be rigid and maintain their shape.
    • Fluids: Do not maintain their shape but flow.
    • Liquids: Flow until they occupy the lowest possible regions in their container.
    • Gases: Expand to fill their container regardless of its shape.
  • Density

    • An important property of a substance is the ratio of its mass to its volume, known as its density (ρ):
      ρ=mV\rho = \frac{m}{V}
    • The ratio of the density of a substance to that of water is called specific gravity or relative density, which is a dimensionless quantity.
    • If the volume of a substance equals the volume of water, relative density can be defined as:
      Relative Density=mass of a substancemass of an equal volume of water\text{Relative Density} = \frac{\text{mass of a substance}}{\text{mass of an equal volume of water}}

EXAMPLES

  1. Density of the Earth

    • The radius of the Earth is 6.37×106 m6.37 \times 10^6 \text{ m} and its mass is 5.98×1024 kg5.98 \times 10^{24} \text{ kg}.
    • Density calculation is required.
  2. Mass of Air in a Room

    • A room is 5 m long, 4 m wide, and 3 m high. The density of air is 1.3 kg/m31.3 \text{ kg/m}^3.
    • Volume $(V) = l \times b \times h = 5 \text{ m} \times 4 \text{ m} \times 3 \text{ m} = 60 \text{ m}^3$
    • m=Density×V=(1.3 kg/m3)(60 m3)=78 kgm = \text{Density} \times V = (1.3 \text{ kg/m}^3)(60 \text{ m}^3) = 78 \text{ kg}
  3. Density of Oil in a Flask

    • A 100 cm³ flask weighs 23.00 g when empty and 97.84 g when full of oil.
    • Mass of oil=(97.84 g)(23.00 g)=74.84 g\text{Mass of oil} = (97.84 \text{ g}) - (23.00 \text{ g}) = 74.84 \text{ g}
    • Volume of oil is 100 cm3100 \text{ cm}^3, therefore:
      Density=74.84 g100 cm3=0.784 g/cm3=748kg/m3\text{Density} = \frac{74.84 \text{ g}}{100 \text{ cm}^3} = 0.784 \text{ g/cm}^3 = 748 \text{kg/m}^3

ELASTIC PROPERTIES OF MATTER

  • When an external force is applied to a solid, it produces a change in shape, size, or volume of the solid known as deformation.

  • Elasticity: Ability of a solid (material) to regain its original shape and size after the deforming force is removed.

    • Elastic Deformation: Temporary deformation that recovers when the force is removed.
    • Plastic Deformation: Permanent deformation after the force is removed. It can be further divided into:
    • Dislocation
    • Fracture
  • Definition of Terms

    • Deformation: Change in shape and size of a material due to an external force.
    • Plasticity: Condition where material remains permanently deformed post-removal of force.
    • Common properties of materials include:
    • Hardness: Resists cutting or penetration.
    • Brittleness: Allow little bending or deformation.
    • Malleability: Can be hammered or rolled into shapes.
    • Ductility: Can be permanently bent.
    • Elasticity: Enables return to original shape.
    • Density: Mass per unit volume.
    • Fusibility: Ability to become liquid via heat.
    • Conductivity: Carries heat and electricity.

HOOKE’S LAW

  • Hooke’s Law: F=kxF = -kx
    • This law states that the restoring force (FF) from a spring is proportional to the displacement (xx) from its equilibrium position.
    • The spring constant kk characterizes the elasticity of the spring.
      k=ΔFΔxk = \frac{\Delta F}{\Delta x}

ELASTICITY: STRESS AND STRAIN

  • Stress: Refers to the force applied to deform a material per unit area, defined as: Stress=FA\text{Stress} = \frac{F}{A} where FF represents the applied force and AA is the area.
    • Units: Pascals (Pa), N/m²
    • Examples include change in length per unit length and change in volume per unit volume.
  • Strain: The relative change in dimensions (length or volume) due to applied stress.
    • Examples include change in length per unit length and change in volume per unit volume.

TYPES OF STRESS

  1. Tensile Stress: Arises when forces are directed away from each other (lengthening).
  2. Compressive Stress: Arises when forces are directed toward each other (shortening).

ELASTIC LIMIT

  • The elastic limit is the maximum stress without permanent deformation.
  • Ultimate Strength: Greatest stress experienced without breaking.
  • Shear Modulus: Not applicable to fluids (gases and liquids) as they do not possess shear modulus like solids.

MODULUS OF ELASTICITY

  • Elastic deformation (strain) is directly proportional to the applied force (stress) when the elastic limit is not exceeded:
    Modulus of Elasticity=StressStrain\text{Modulus of Elasticity} = \frac{\text{Stress}}{\text{Strain}}
  • Commonly measured modulus types include:
    • Young’s Modulus (Y): Measures longitudinal elasticity.
    • Bulk Modulus (K): Measures volumetric elasticity.
    • Shear Modulus (G): Measures shear elasticity.

EXAMPLES OF ELASTICITY MEASUREMENT

Example 1: A steel wire 10 m long and 2 mm in diameter is stressed by a weight of 200 N, causing a strain of 3.08 mm.

  • Calculate the stress:
    • A=π(0.002m)24    A=3.14×106m2A = \frac{\pi(0.002 m)^2}{4} \implies A = 3.14 \times 10^{-6} m^2
    • Stress=FA=200N3.14×106m2=6.37×107Pa\text{Stress} = \frac{F}{A} = \frac{200 N}{3.14 \times 10^{-6} m^2} = 6.37 \times 10^7 Pa

Example 2: Calculate Young's Modulus. If Stress is 6.37×107Pa6.37 \times 10^7 Pa and Strain is 3.08×1043.08 \times 10^{-4}:

  • E=StressStrain=6.37×107Pa3.08×104=207×109PaE = \frac{\text{Stress}}{\text{Strain}} = \frac{6.37 \times 10^7 Pa}{3.08 \times 10^{-4}} = 207 \times 10^9 Pa.

Example 3: A block of unknown material with dimensions 60 mm x 60 mm x 20 mm is under a shearing force of 0.245 N with displacement of 5 mm.

  • Calculate the shear modulus:
    • Shear stress=0.245NArea=0.245N60 mm×20 mm=2450N/m2\text{Shear stress} = \frac{0.245 N}{\text{Area}} = \frac{0.245 N}{60\text{ mm} \times 20\text{ mm}} = 2450 N/m^2.

ENERGY STORED IN A STRAIN

  • When materials are stretched or compressed, they store energy.
  • Work done on the elastic material is stored as potential energy.
  • The energy stored can be described as:
    • Energy per unit volume for tensile strain:
      E=12FxV=12kx2E = \frac{1}{2} \frac{F x}{V} = \frac{1}{2} kx^2
  • For shear strain:
    E=13FΔxAE = \frac{1}{3} \frac{F \Delta x}{A}

POISSON'S RATIO

  • Poisson's ratio relates to elastic moduli (B, S, Y).
  • Defined as:
    σ=3B2S6B+2S\sigma = \frac{3B - 2S}{6B + 2S}
    Y=2S(1+σ)Y = 2S(1 + \sigma)
    Y=3B(12σ)Y = 3B(1 - 2\sigma)

ADDITIONAL SOLVED PROBLEMS

  • Example: To find the pressure required to reduce the volume of a 1 kg water from 100 to 99 L where bulk modulus for water is 2x1011N/m22 x 10^{11} N/m^2. The change in volume can be calculated.

  • Example: A solid brass sphere: Calculate changes in volume subjected to different pressures, utilizing the Bulk modulus of seawater.

  • Example: A slingshot potential energy: Calculate energy stored when stretched and use it to find speed at which stone leaves the slingshot.

  • Example: The force needed to punch a hole in steel, given shear stress.

Conclusion

  • The concepts of elasticity, stress and strain, and energy storage are fundamental in understanding material properties and their applications in various fields including engineering, construction, and materials science.