1. Introduction
  2. \to \textbf{Elasticity} is the tendency of restoring / recovering the original shape and size of deformed body after the deforming force is removed.\\\to \textbf{Perfectly elastic body} regains its original shape and size completely after removal of deforming force. Eg: Quartz fiber, Phosphor bronze etc.\\\to \textbf{Perfectly plastic body} does not regain its original shape and size at all. Eg: mud, wax, plastics etc.\\

  3. \textbf{Stress} \\\toRestoring force applied in a unit area\\Stress(S) = \frac{ Force(F)}{Area(A)}=FA\\\toStress is a tensor quantity with dimension [M L^{-1} T^2]and unit N/m^2.\\

  4. \textbf{ Types of stress are:} \\

    \toNormal stress: Normal force per unit cross-section area.\\\toTangential or Shearing stress:Tangential force applied parallel to surface per unit area of that surface producing shearing (change in shape).\\\toVolume stress: Force per unit total surface area producing change in volume.\\

  5. \textbf {Strain} \\The change in dimension of body per unit original dimension is strain.\\Strain = \frac{change -in -configuration}{ original- configuration} \\Strain is unit less and dimension less physical quantity.

  6. \textbf{Types of strain are:}\\\to \textbf{Longitudinal strain :} Change in length per unit initial length. Strain (long.) =\frac{ \Delta L}{L} \\\to \textbf{Volumetric Strain :} Change in volume per unit original volume and is given by \frac{\Delta V}{V} \\\to \textbf{Shearing Strain :} Change in position per unit original position. It is generally represented by angle θ through which a body initially fixed is twisted under the action of tangential deforming force. Shearing force = \theta = \frac{x}{L}

Hooke's law

  1. It states that, "Within elastic limit, extension produced is directly proportional to force applied."\\i.e. e \propto F \to F = -k.e ; where k is force constant.

  2. It also states, within elastic limit, stress is directly proportional to the strain\\i.e. Stress \propto Strain so, \frac{stress}{strain}= E (modulus of elasticity)

  3. \textbf {Types of modulus of elasticity:} \\\to \textbf {Young's modulus of elasticity (Y):} \\It is the ratio of normal stress to the longitudinal strain, within the elastic limit.\\Y = \frac{normal -stress}{ longitudinal -strain} =\frac{\frac{F}{A}} {\frac{\Delta L}{L}}= \frac{FL}{A \Delta L}

    \textbf{NOTE :} \\\toIt is property of solid body only.\\\toYoung's modulus of elasticity increases on mixing the impurity in the solid and decreases on increasing temperature of solid body.\\

    \to \textbf {Bulk modulus of elasticity (B):} \\Ratio of normal stress to volumetric strain within elastic limit.\\B = \frac{normal -stress}{ volumetric -strain} =\frac{\frac{F}{A}} {\frac{-\Delta V}{V}}= \frac{\Delta P}{\frac{-\Delta V}{V}} \\( -ve sign indicates decreasing of volume on increasing of applied force)\\

    \textbf{NOTE :} \\\toBulk modulus is property of solid, liquid and gases.\\\toGases have isothermal and adiabatic bulk modulus of elasticity.\\\toK is minimum for gases (maximum volume change) and maximum for solid (minimum volume change)\\\toReciprocal of bulk modulus of elasticity is Compressibility.\\

  4. \textbf{Modulus of rigidity (η)} \\\toRatio of tangential stress to shearing strain within elastic limit.\\\to η = \frac{tangential- stress}{shearing -strain}= \frac{\frac{F}{A}}{\theta} =\frac{F}{A \theta }
    \toIt is property of solid materials only. And its value is zero for liquid.\\

    \textbf{Poisson's ratio (σ) :} \\\to The ratio of lateral strain to longitudinal strain\\\to σ = \frac{lateral- strain}{ longitudinal- strain} =\frac{\frac{-\Delta r}{r}} {\frac{\Delta L}{L}}= \frac{-\Delta r l}{\Delta l r}\\
    (-ve sign indicates decrease of radius in expense of increasing length for constant volume)\\\to Theoretical value of σ lies in between -1 to + \frac{1}{2} but practical lies between 0 to + \frac{1}{2} . \\\to Fractional change in volume : \frac{\Delta V}{V} =(1-2σ)x \frac{\Delta L} {L} ( so, change in volume is zero when the value of σ 0.5 )

Energy stored during elastic deformation

  1. A step towards visualization

  2. Think of a triangle with extension (x) as its base and the force applied on body to produce that extension be represented by height of triangle. Where, force (F) = kx And k is the force constant of that elastic body. Then, area of that triangle represents the work done in stretching the body by x.\\

    So, elastic potential energy stored ( E_{pe} ) = \frac{1}{2} k x^2