JEE Physics Mechanical Properties Of Solids Notes Set 02

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Elastic and Plastic Behaviour

When a deforming force acts on a body, its atoms are pushed away from their equilibrium separations and internal restoring forces appear. A body that returns completely to its original shape once the force is removed is perfectly elastic (quartz fibre is the nearest example). A body that keeps the deformation is perfectly plastic (putty). Real materials lie in between, and every material has an elastic limit: the largest load up to which it still recovers fully.

Stress and Strain

Stress is the restoring force per unit area, \( \sigma=F/A \), measured in N/m². Tensile and compressive stresses act normal to the surface, while shear stress acts tangentially. Strain is the fractional deformation and has no unit.

Type of strainDefinition
Longitudinal (linear)\( \Delta L/L \)
Volumetric\( \Delta V/V \)
Shear\( \theta=\Delta x/h \) (angle of shear)

Stress differs from pressure: pressure is always normal and compressive, a scalar; stress can be normal or tangential, tensile or compressive, and is a tensor quantity.

Hooke's Law and Moduli of Elasticity

Within the proportional limit, stress is proportional to strain. The constant ratio is the modulus of elasticity of the material.

ModulusFormulaApplies to
Young's modulus \( Y \)\( \dfrac{F/A}{\Delta L/L}=\dfrac{FL}{A\,\Delta L} \)change in length
Bulk modulus \( B \)\( -\dfrac{\Delta P}{\Delta V/V} \)change in volume; compressibility \( =1/B \)
Shear modulus \( \eta \)\( \dfrac{F/A}{\theta} \)change in shape

Solids have all three moduli, liquids and gases have no shear modulus, and a perfectly rigid body has infinite moduli. Steel is more elastic than rubber because its Young's modulus is much larger.

Stretching of a Wire

A wire of length \( L \) and area \( A \) behaves like a spring of force constant \( k=YA/L \). The extension under load \( F \) is \( \Delta L=\dfrac{FL}{AY} \). For the same load, the extension is proportional to the length and inversely proportional to the square of the diameter. The breaking force is the breaking stress times the area, so it depends on area only.

Poisson's ratio \( \sigma \) is lateral strain divided by longitudinal strain (in magnitude). Under a longitudinal strain \( \varepsilon \) the volume strain is \( \varepsilon(1-2\sigma) \), and \( \sigma \) lies between 0 and 0.5.

Elastic Potential Energy

The work done in stretching a wire slowly is stored as elastic energy:

\( U=\tfrac12F\,\Delta L=\tfrac12\dfrac{YA(\Delta L)^2}{L} \)

Energy per unit volume: \( u=\tfrac12\,\text{stress}\times\text{strain}=\tfrac12Y\varepsilon^2=\dfrac{\sigma^2}{2Y} \). If a hanging mass \( M \) is lowered slowly until it stops, it loses \( Mg\,\Delta L \) of potential energy, half of which is stored in the wire and half dissipated as heat.

Thermal Stress

If a rod is clamped rigidly and its temperature changes by \( \Delta T \), the prevented thermal strain \( \alpha\Delta T \) creates a stress \( \sigma=Y\alpha\Delta T \) and a force \( F=YA\alpha\Delta T \). Neither depends on the length of the rod. For two rods held end to end with the junction fixed, equal forces need \( Y_1A_1\alpha_1=Y_2A_2\alpha_2 \). To keep a body's volume constant while heating, a pressure \( P=3\alpha\Delta T\,B \) is needed. A ring of strip tension \( T \) exerts a force \( 2T \) across a diameter.

Stress-Strain Graph

In the initial straight part (proportional limit) Hooke's law holds. Beyond it the curve bends but the wire still recovers up to the elastic limit. Past the yield point, plastic flow begins and a permanent set remains on unloading, until the wire reaches the ultimate stress and fractures at the breaking point. Materials with a long plastic region are ductile; those that break soon after the elastic limit are brittle.

Useful Results

1. Wire in series with a spring: \( \frac{1}{k_{eff}}=\frac1k+\frac{L}{YA} \).
2. Speed of longitudinal waves in a solid: \( v=\sqrt{Y/\rho} \).
3. Wave speed on a stretched wire with strain \( \varepsilon \): \( v=\sqrt{Y\varepsilon/\rho} \).
4. A pendulum whose wire stretches under an extra load obeys \( (T'/T)^2=1+\Delta L/L \).
5. Weight scales as \( L^3 \), leg area as \( L^2 \), so stress grows as \( L \) when all dimensions are scaled by \( L \).

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