Elastic Moduli and Properties

Elastic Moduli

As introduced with Hooke's Law, the ratio of stress to the corresponding strain within the elastic limit is a constant for a given material, called the modulus of elasticity. These moduli are specific to the type of stress and strain involved. There are three primary elastic moduli for isotropic, homogeneous materials:

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Young’s Modulus ($ Y = \frac{\text{Tensile stress}}{\text{Tensile strain}} $)

Young's modulus ($Y$) is a measure of the stiffness of a solid material under tension or compression. It is defined as the ratio of normal stress (tensile or compressive stress) to the corresponding longitudinal strain within the elastic limit.

$ Y = \frac{\text{Normal Stress}}{\text{Longitudinal Strain}} $

For a wire or rod of original length $L$ and uniform cross-sectional area $A$, subjected to a tensile force $F$ causing an extension $\Delta L$:

$ \text{Normal Stress} = \frac{F}{A} $

$ \text{Longitudinal Strain} = \frac{\Delta L}{L} $

So, Young's modulus is:

$ Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L} $

Young's modulus has the same units as stress (N/m$^2$ or Pa) since strain is dimensionless. Materials with a high Young's modulus (like steel) are stiff and resistant to stretching or compression. Materials with a low Young's modulus (like rubber) are less stiff and more easily deformed under tension or compression.

Young's modulus is primarily a property of solids. Different materials have different Young's moduli. For example, steel has a Young's modulus of around $2 \times 10^{11}$ Pa, while rubber has a Young's modulus of around $1 \times 10^6$ Pa.

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Determination Of Young’s Modulus Of The Material Of A Wire

Young's modulus of the material of a wire can be determined experimentally using a simple apparatus. The method involves measuring the extension of a wire of known original length and radius when subjected to a known stretching force (applied load).

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Experimental Setup (Searle's Apparatus)

(Image Placeholder: A diagram showing Searle's apparatus. Two wires of the same material, length, and radius are suspended from a rigid support. One is a reference wire with a fixed weight and a main scale attached at the bottom. The other is the experimental wire with a vernier scale attached at the bottom, and a weight hanger at the end. The vernier scale moves along the main scale.)

Searle's apparatus consists of two identical wires, P (reference wire) and Q (experimental wire), suspended from a rigid support. Both wires have the same original length $L$ and radius $r$. A fixed load is attached to the reference wire P to keep it taut. A main scale is attached to the bottom of wire P.

A vernier scale is attached to the bottom of the experimental wire Q, adjacent to the main scale. A weight hanger is attached to the experimental wire Q.

By using two identical wires, any change in length due to temperature variations or yielding of the support is compensated, as both wires will be affected similarly. The use of a vernier scale allows for precise measurement of the extension.

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Procedure

1. Measure the original length ($L$) of the experimental wire (from the rigid support to the point where the scale is attached).
2. Measure the diameter of the wire using a screw gauge at multiple points and find the average radius ($r$). Calculate the cross-sectional area $A = \pi r^2$.
3. Apply a small initial load to the experimental wire to make it taut and remove any kinks. Note the reading on the vernier scale against the main scale. This is the initial reading.
4. Increase the load on the experimental wire by adding weights (e.g., in steps of 0.5 kg or 1 kg). For each added load $W = Mg$ (where $M$ is the added mass), wait for the wire to extend and note the new vernier scale reading.
5. Calculate the extension $\Delta L$ for each load by subtracting the initial reading from the current reading.
6. Calculate the tensile stress ($\sigma = W/A$) and the corresponding longitudinal strain ($\epsilon = \Delta L/L$) for each load.
7. Plot a graph of stress versus strain. For elastic materials, the initial portion of the graph should be a straight line passing through the origin.
8. Calculate Young's modulus ($Y$) from the slope of the linear portion of the stress-strain graph: $ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{(W/A)}{(\Delta L/L)} $. For different loads within the elastic limit, the ratio $(W/A) / (\Delta L/L)$ should be approximately constant.
9. Repeat the process while decreasing the load to check if the wire returns to its original length upon removal of the load (elastic behaviour).

The value of Young's modulus obtained gives a quantitative measure of the material's stiffness under longitudinal stress.

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Shear Modulus

Shear modulus ($G$ or $\eta$), also known as the modulus of rigidity, measures the resistance of a material to shearing deformation. It is defined as the ratio of tangential stress (shear stress) to the corresponding shear strain within the elastic limit.

$ G = \frac{\text{Tangential Stress}}{\text{Shear Strain}} $

Consider a rectangular block of height $h$ and base area $A$. If a tangential force $F_{parallel}$ is applied to the top surface, while the bottom surface is fixed, the top surface shifts by a distance $\Delta x$ relative to the bottom surface. The shear strain is the angle $\phi$ (in radians) such that $\tan\phi \approx \phi = \Delta x / h$ for small angles.

$ \text{Tangential Stress} = \frac{F_{parallel}}{A} $

$ \text{Shear Strain} = \phi = \frac{\Delta x}{h} $ (for small $\Delta x$)

So, the shear modulus is:

$ G = \frac{F_{parallel}/A}{\Delta x/h} = \frac{F_{parallel} \cdot h}{A \cdot \Delta x} $

The units of shear modulus are also N/m$^2$ or Pa. Shear modulus is relevant only for solids, as fluids cannot resist static shear stress (their shear modulus is zero).

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Bulk Modulus ($B$ or $K$)

Bulk modulus ($B$ or $K$) measures the resistance of a material to uniform compression or volume change when subjected to uniform pressure from all sides (hydrostatic stress). It is defined as the ratio of the bulk stress (change in pressure) to the corresponding volumetric strain within the elastic limit.

$ B = \frac{\text{Bulk Stress}}{\text{Volumetric Strain}} $

If a body is subjected to a uniform increase in pressure $\Delta P$, its volume decreases by $\Delta V$ from the original volume $V$. The bulk stress is $\Delta P$. The volumetric strain is $\Delta V/V$. The negative sign is introduced to make the bulk modulus positive, as an increase in pressure ($\Delta P > 0$) causes a decrease in volume ($\Delta V < 0$).

$ B = \frac{\Delta P}{-\Delta V/V} = -V \frac{\Delta P}{\Delta V} $

The units of bulk modulus are also N/m$^2$ or Pa. Bulk modulus is relevant for solids, liquids, and gases.

The reciprocal of the bulk modulus is called compressibility ($k = 1/B$). Compressibility measures how easily a substance can be compressed. Gases are much more compressible than liquids and solids, so they have much lower bulk moduli (higher compressibility).

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Poisson’s Ratio ($\nu$ or $\sigma$)

When a material is subjected to tensile stress in one direction, it not only extends in that direction (longitudinal strain) but also contracts in the perpendicular directions (lateral strain). Similarly, under compression, it shortens in the direction of compression and expands in the perpendicular directions.

Poisson's ratio ($\nu$, often denoted by $\sigma$ in some texts, but $\nu$ is more common to avoid confusion with stress) is the ratio of lateral strain to the corresponding longitudinal strain within the elastic limit.

$ \nu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} $

The negative sign is included because tensile longitudinal strain (positive $\Delta L/L$) causes a compressive lateral strain (negative change in width/diameter), and vice versa. This makes Poisson's ratio a positive value for most common materials.

If a wire of original length $L$ and original diameter $D$ undergoes a longitudinal strain $\epsilon_L = \Delta L/L$ due to tensile stress, its diameter changes by $\Delta D$. The lateral strain is $\epsilon_{lat} = \Delta D/D$. The change in diameter $\Delta D$ will be negative (decrease) for positive $\Delta L$ (increase).

$ \nu = -\frac{\Delta D/D}{\Delta L/L} $

Poisson's ratio is a dimensionless quantity, as it is a ratio of two strains. Its value typically ranges from 0 to 0.5 for isotropic materials. For most metals, $\nu$ is around 0.25 to 0.35. For an ideal incompressible material (where volume does not change under tension or compression), Poisson's ratio is 0.5 (e.g., ideal rubber). Cork has a Poisson's ratio close to zero.

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Elastic Potential Energy In A Stretched Wire

When a wire is stretched or compressed within its elastic limit by an applied force, work is done by the force. This work done is stored in the wire as elastic potential energy. This stored energy is recovered when the deforming force is removed and the wire returns to its original state.

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Calculation of Stored Energy

Consider a wire of original length $L$ and cross-sectional area $A$. Let it be stretched by an amount $\Delta L$ by an applied force. If the stretching is done slowly, the applied force is equal in magnitude to the restoring force in the wire at any point during the stretching process.

According to Hooke's Law, the force $F$ required to produce an extension $x$ in the wire is proportional to the extension within the elastic limit. From $Y = \frac{F/A}{x/L}$, the force $F(x)$ at extension $x$ is $F(x) = \frac{YA}{L} x$. This is similar to Hooke's Law for a spring, $F=kx$, where the effective spring constant for the wire is $k = \frac{YA}{L}$.

The work done ($dW$) in increasing the extension by an infinitesimal amount $dx$ is $dW = F(x) dx = \frac{YA}{L} x \, dx$.

The total work done in stretching the wire from $x=0$ to $x=\Delta L$ is:

$ W = \int_{0}^{\Delta L} dW = \int_{0}^{\Delta L} \frac{YA}{L} x \, dx $

$ W = \frac{YA}{L} \int_{0}^{\Delta L} x \, dx = \frac{YA}{L} \left[ \frac{x^2}{2} \right]_{0}^{\Delta L} $

$ W = \frac{YA}{L} \frac{(\Delta L)^2}{2} = \frac{1}{2} \frac{YA}{L} (\Delta L)^2 $

This work done is stored as elastic potential energy ($U$) in the stretched wire:

$ U = \frac{1}{2} \frac{YA}{L} (\Delta L)^2 $

We can express this energy in terms of stress and strain:

From $Y = \frac{\sigma}{\epsilon}$, we have $\sigma = Y\epsilon$. Also, $\Delta L = \epsilon L$. Substituting these into the energy formula:

$ U = \frac{1}{2} Y A L \left(\frac{\Delta L}{L}\right)^2 = \frac{1}{2} Y (AL) \epsilon^2 $

$ U = \frac{1}{2} Y \epsilon^2 \times (\text{Volume}) $

Also, from $Y = \frac{\sigma}{\epsilon}$, $\epsilon = \frac{\sigma}{Y}$. Substituting this:

$ U = \frac{1}{2} Y \left(\frac{\sigma}{Y}\right)^2 (\text{Volume}) = \frac{1}{2} Y \frac{\sigma^2}{Y^2} (\text{Volume}) = \frac{1}{2} \frac{\sigma^2}{Y} (\text{Volume}) $

And, substituting $\epsilon = \sigma/Y$ and $\sigma = Y\epsilon$ into $U = \frac{1}{2} Y \epsilon^2 (\text{Volume})$, or directly from $W = \frac{1}{2} F \Delta L$ (since $F$ increases linearly from 0 to $F_{final}$):

$ U = \frac{1}{2} F_{final} \Delta L $

$ U = \frac{1}{2} (\sigma A) (\epsilon L) = \frac{1}{2} \sigma \epsilon (AL) = \frac{1}{2} \sigma \epsilon (\text{Volume}) $

This last expression, $U = \frac{1}{2} \sigma \epsilon \times \text{Volume}$, or the energy density (energy per unit volume), $\frac{U}{\text{Volume}} = \frac{1}{2}\sigma\epsilon$, is a very useful form. It shows that the elastic potential energy stored per unit volume is equal to half the product of stress and strain.

The elastic potential energy is stored in the deformed bonds between the atoms or molecules of the material. This energy is released when the material returns to its original shape.

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