Some Mechanical Properties of Matter

Rubber vs. Metal stress-strain behavior:

• Rubber remains elastic even when stretched to several times its original length
• No clear proportional limit (non-linear throughout)
• Lower stress for same strain compared to metals
• Can return to original length even after extreme deformation

Elastic hysteresis:

• Path followed during relaxation differs from path during stretching
• Energy absorbed during complete cycle (stretching and relaxation)
• Absorbed energy appears as heat
• Area enclosed by stress-strain loop represents energy dissipated per unit volume
• Practical application: shock absorbers using rubber padding absorb vibration energy

Amorphous/Glassy Solids:

• Lack long-range molecular ordering (unlike crystals)
• Exhibit only local ordering (4-5 molecules bonded together)
• Random arrangement of these local structural units
• Similar to liquids in lacking long-range order
• Stronger intermolecular forces than liquids, preventing fluid flow
• No well-defined melting point; gradual softening as temperature increases
• Weaker bonds break first, stronger bonds later in melting process
• Example: Glass (contains strong Si-O-Si bonds that don't extend far in space)
• May contain calcium, sodium along with silicon and oxygen in structure

In mechanical equilibrium:

1. Force balance principle:

   • External forces create internal stresses that maintain equilibrium
   • For any cross-section (ΔS), parts on either side exert equal and opposite forces on each other
   • Total resultant force and torque must be zero

2. Applications to stress types:

   • Longitudinal stress scenario:

     • When equal and opposite external forces act along the body's axis
     • Normal forces (Fₙ) develop across ΔS with Γₙ = Fₙ/ΔS
     • Each part of the body experiences equal tension/compression

   • Shearing stress scenario:

     • When equal and opposite external forces act parallel to surfaces
     • Tangential forces (Fₜ) develop across ΔS with Γₜ = Fₜ/ΔS
     • Forces must be distributed to prevent rotation
     • Results in deformation while maintaining translational and rotational equilibrium

The relationship ensures that the body deforms but does not accelerate or rotate.

Ductile vs. Brittle Materials:

Ductile Materials:

• Undergo significant plastic deformation before fracture
• Large region between elastic limit and fracture point
• Can be stretched/deformed substantially without breaking
• Absorb considerable energy before fracture (high toughness)
• Examples: Copper, aluminum, mild steel, gold

Brittle Materials:

• Fracture occurs shortly after elastic limit is exceeded
• Minimal or no plastic deformation region
• Break suddenly with little warning
• Absorb relatively little energy before fracture (low toughness)
• Examples: Cast iron, glass, ceramic, concrete

Identification Methods:

1. Stress-strain analysis: Ductile materials show extended plastic region; brittle materials show almost none
2. Percent elongation at fracture: High for ductile (>5%), low for brittle (<5%)
3. Energy absorption: Ductile materials have larger area under stress-strain curve
4. Fracture appearance: Ductile shows necking and cup-cone fracture; brittle shows flat fracture surfaces

Note: Material behavior can change with temperature, strain rate, and environmental conditions. Many ductile metals become brittle at very low temperatures.

For vulcanized rubber:

• No clear proportional limit exists in the stress-strain curve
• Non-linear relationship throughout the entire elastic range
• Stress increases at varying rates relative to strain from the beginning
• Hooke's Law (σ = Eε) cannot be applied at any point
• Young's modulus cannot be meaningfully defined as a single value
• Despite non-linearity, rubber remains elastic through very large strains (up to 8× elongation)

For metals:

• Well-defined proportional limit (point where stress-strain linearity ends)
• Clear linear region where Hooke's Law applies
• Young's modulus can be precisely calculated from the linear region
• Relatively small elastic region (typically <1% strain)
• After proportional limit, plastic deformation begins

This fundamental difference makes rubber uniquely suited for applications requiring large elastic deformations without permanent shape change.

For a wire with:

• Cross-sectional area A
• Young's modulus Y
• Natural length L
• Extension x

The tension force F is: $F = \frac{AY}{L}x$

This shows that:

• The force is directly proportional to extension (Hooke's Law)
• The force increases with greater cross-sectional area and Young's modulus
• The force decreases with greater natural length
• The proportionality constant (AY/L) is called the spring constant k

Example: A steel wire (Y = 200 GPa) with area 2 mm² and length 1 m will experience 400 N of tension when stretched by 1 mm.

The elastic potential energy U stored in a stretched wire is: $U = \frac{AY}{2L}x^2$

Where:

• A = cross-sectional area
• Y = Young's modulus
• L = natural length
• x = extension

This can be expressed in multiple equivalent forms:

• U = (1/2)(maximum force)(extension)
• U = (1/2)(stress)(strain)(volume)

Work-energy relationship:

• The work done by the external force in stretching equals the potential energy stored
• Work is calculated by integrating F·dx from 0 to x
• For a slowly stretched wire, the force increases linearly from 0 to F_max

Example: A 2 m wire with 4 mm² cross-section and Y = 2×10¹¹ N/m² stretched by 2 mm stores 0.8 J of elastic potential energy.

When the extension of an elastic wire is doubled:

• The elastic potential energy increases by a factor of 4 (quadruples)

Mathematical proof:

• For extension x, energy U = (AY/2L)x²
• For extension 2x, energy U' = (AY/2L)(2x)² = 4(AY/2L)x² = 4U

Physical explanation:

• Force increases linearly with extension: F = (AY/L)x
• Work done = ∫F·dx = area under force-extension graph
• When extension doubles:
  • Average force is doubled
  • Distance is doubled
  • Work done (and energy stored) = 2 × 2 = 4 times greater

This quadratic relationship emphasizes why small increases in deformation require disproportionately more energy, a principle crucial in structural engineering and material design.

Excess pressure inside a spherical drop: $P_2 - P_1 = \frac{2S}{R}$

Where:

• $P_2$ = pressure inside the drop
• $P_1$ = pressure outside the drop
• S = surface tension
• R = radius of the drop

Derivation principle:

• Consider forces on a hemispherical section:
  1. Force due to surface tension along periphery: $F_1 = 2\pi RS$
  2. Force due to outside pressure: $F_2 = P_1\pi R^2$
  3. Force due to inside pressure: $F_3 = P_2\pi R^2$
• At equilibrium: $F_1 + F_2 = F_3$
• Therefore: $P_2 - P_1 = \frac{2S}{R}$

Key insight: The pressure is always greater on the concave side of a curved liquid surface.

Viscosity is the property of a fluid to oppose relative motion between its layers.

Mathematical definition: $F = -\eta A \frac{dv}{dz}$

Where:

• F = viscous force between layers
• η = coefficient of viscosity
• A = area of contact between layers
• dv/dz = velocity gradient perpendicular to layers

Derivation of SI units:

• From the equation: $\eta = \frac{F}{A \cdot \frac{dv}{dz}}$
• Substituting dimensions: $[\eta] = \frac{[MLT^{-2}]}{[L^2] \cdot \frac{[LT^{-1}]}{[L]}}$
• Simplifying: $[\eta] = \frac{MLT^{-2}}{L^2 \cdot T^{-1}}$ = $ML^{-1}T^{-1}$

Therefore:

• SI unit: $\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-1}$ = Pa·s = N·s/m²
• CGS unit: dyne·s/cm² = poise (P)
• 1 poise = 0.1 Pa·s