Mechanical properties of materials 

 

 Module 3: Physics in action

3.2 Mechanical properties of materials 

3.2 Mechanical Properties of materials

a)       Describe and explain:

I)                   Simple mechanical behavior: elastic and plastic deformation and fracture

II)                Direct evidence of the size of particles and their spacing

III)             Behavior/structure of classes of materials, limited to metals, ceramics and polymers; dislocations leading to slip in metals with brittle materials not having mobile dislocations; polymer behavior in terms of chain entanglement/unravelling

IV)             One method of measuring Young modulus and fracture stress.

b)      Make appropriate use of

I)                   The terms: stress, strain, Young modulus, tension, compression, fracture stress and yield stress, stiff, elastic, plastic, ductile, hard, brittle, tough, strong, dislocation

By sketching and interpreting:

II)                Force–extension and stress–strain graphs up to fracture

III)             Tables and diagrams comparing materials by properties

IV)             Images showing structures of materials.

c)      Make calculations and estimates involving:

I)                   Hooke’s Law, F = kx ; energy stored in an elastic material (elastic strain energy)= [math]\frac{1}{2} kx^2 [/math] energy as area under Force– extension graph for elastic materials

II)           [math]\text{Stress} = \frac{\text{Tension}}{\text{Cross-section area}} \\ \text{Strain} =\frac{\text{Extension}}{\text{Original length}} \\ \text{Toung modulus E} = \frac{\text{Stress}}{\text{Strain}}[/math]

d)     Demonstrate and apply knowledge and understanding of the following practical activities:

I)                   Plotting force–extension characteristics for arrangements of springs, rubber bands, polythene strips, etc.

II)                Determining Young modulus for a metal such as copper, steel or other wire.

 

  • I) Simple Mechanical Behavior: Elastic and Plastic Deformation and Fracture

  • ⇒ Elastic Deformation:
  • Definition:
  • – Elastic deformation occurs when a material returns to its original shape after the stress is removed.
  • Behavior:
  • – Follows Hooke’s Law: Stress is directly proportional to strain.
  • – Formula:
  • [math]σ = E.ϵ [/math]
  • Where:
  • σ: Stress (Pa).
  • E: Young’s modulus (Pa).
  • ϵ: Strain (dimensionless).
  • The material behaves elastically until it reaches the elastic limit or yield point.
  • Example: Stretching a spring or rubber band within its elastic limit.
  • ⇒ Plastic Deformation:
  • Definition:
  • – Plastic deformation occurs when a material undergoes permanent deformation and does not return to its original shape after the stress is removed.
  • Behavior:
  • – Begins after the yield point.
  • Results from the movement of dislocations in metals or permanent changes in the molecular structure of polymers.
  • Example:
  • – Bending a metal paperclip beyond its elastic limit.
  • ⇒ Fracture:
  • Definition:
  • – Fracture occurs when a material fails under stress, breaking into two or more pieces.
  • Types of Fracture:
  • 1. Brittle Fracture:
  • – Sudden failure without significant plastic deformation.
  • – Occurs in materials like ceramics or glass.
  • – Fracture surface shows no necking.
  • 2. Ductile Fracture:
  • – Accompanied by plastic deformation.
  • – Occurs in metals.
  • Fracture surface shows necking and void formation.
  • Example:
  • – Breaking of a ceramic plate (brittle) versus stretching and breaking of a copper wire (ductile).

  • II) Direct Evidence of the Size of Particles and Their Spacing

  • ⇒ Methods for Determining Particle Size and Spacing:
  • X-ray Diffraction (XRD):
  • Principle:
  • – X-rays interact with crystal lattice planes, producing diffraction patterns.
  • Use:
  • – Determines atomic spacing in crystalline solids.
  • Formula:
  • [math]nλ = 2d \, sin⁡θ [/math]
  • Where:
  • n: Order of diffraction.
  • λ: Wavelength of X-rays.
  • d: Atomic spacing.
  • θ: Angle of diffraction.
  • Electron Microscopy:
  • Principle:
  • – High-energy electrons scatter off atoms, providing high-resolution images.
  • Use:
  • – Determines particle size down to the nanometer scale.
  • Brownian Motion:
  • Observation:
  • – Random motion of particles suspended in a fluid.
  • Interpretation:
  • – Provides indirect evidence of particle size and behavior.
  • Scanning Tunneling Microscopy (STM):
  •  Measures the electron density on the surface of a material, revealing atomic-level details.

  • III) Behavior and Structure of Classes of Materials

  • 1. Metals:
  • Structure:
  • – Crystalline structure with a sea of free electrons.
  • – Atoms arranged in regular lattice patterns.
  • Behavior:
  • – Ductility: Deform plastically due to dislocations (defects in the lattice).
  • – Slip: Dislocations allow atoms to slide over one another, enabling plastic deformation.
  • – High thermal and electrical conductivity due to free electrons.
  • 2. Ceramics:
  • Structure:
  • – Ionic or covalent bonds.
  • – Highly ordered but brittle lattice.
  • Behavior:
  • – Brittle: Lack mobile dislocations, making them prone to fracture.
  • – High melting points and excellent insulators.
  • – Applications: Engine components, insulators.
  • 3. Polymers:
  • Structure:
  • – Long chains of repeating molecular units.
  • – Chains can be cross-linked or entangled.
  • Behavior:
  • – Elasticity: Chains stretch and unravel under stress.
  • – Plasticity: Permanent deformation occurs when chains slide past one another.
  • – Temperature-dependent behavior:
  • Below Glass Transition Temperature ([math]T_g[/math]):
  • Above [math]T_g[/math]​: Ductile.
  • Example: Rubber band elasticity due to chain uncoiling.

  • IV) One Method of Measuring Young’s Modulus and Fracture Stress

  • 1. Measuring Young’s Modulus (E):
  • ⇒ Method: Tensile Test (Using a Wire)
  • Equipment:
  • – A long, thin wire.
  • – Meter ruler to measure length.
  • – Vernier calipers for diameter.
  • – Weights for applying force.
  • – Clamp stand to secure the wire.
  • Procedure:
  • – Measure the initial length ([math]L_o [/math]) and diameter (d) of the wire.
  • – Apply incremental weights and measure the extension ([math]∆L [/math]) for each weight.
  • – Record the force (F) applied using:
  • [math]F = mg [/math]
  • Where m is mass and g is acceleration due to gravity.
  • Plot a graph of stress [math]σ = \frac{F}{A}[/math] versus strain [math]ϵ = \frac{∆L}{L_0}[/math]
  • The slope of the linear region gives E.
  • Formula for Young’s Modulus:
  • [math] E = \frac{\sigma}{\epsilon} = \frac{F}{A} \cdot \frac{\Delta L}{L_0}[/math]
  • Where:
  • – A is the cross-sectional area of the wire:
  • [math] A = \frac{\pi d^2}{4}[/math]
  • 2. Measuring Fracture Stress:
  • ⇒ Method: Tensile Test (Continued Until Failure)
  • Procedure:
  • – Perform the tensile test as described above.
  • – Continue increasing the load until the material breaks.
  • – The maximum stress sustained before breaking is the fracture stress ().
  • Fracture Stress Formula:
  • [math]\sigma_f = \frac{F_{\text{max}}}{A}[/math]
  • Where [math]F_{max} [/math]​ is the maximum force applied before fracture.

  • b) Make appropriate use of:

  • I) Explanation and Use of Terms

  • 1. Stress (σ):
  • Definition:
  • – Force applied per unit area.
  • [math]σ = \frac{F}{A}[/math]
  • Where:
  • – F: Force (N).
  • – A: Cross-sectional area ([math]m^2[/math]).
  • Units: Pascals (Pa).
  • Use: Describes how much force is applied to a material.
  • 2. Strain (ϵ):
  • Definition: Measure of deformation as the fractional change in length.
  • [math]\epsilon = \frac{\Delta L}{L_0}[/math]
  • Where:
  • – ΔL: Extension (m).
  • – [math]L_0[/math]: Original length (m).
  • Units: Dimensionless (no unit).
  • Use: Indicates how much a material stretches or compresses under stress.
  • 3. Young Modulus (E):
  • Definition: Measure of stiffness, relating stress and strain in the elastic region.
  • [math]E = \frac{σ}{\epsilon}[/math]
  • Units: Pascals (Pa).
  • Use: Quantifies how stiff a material is.
  • 4. Tension and Compression:
  • Tension: Pulling force that stretches a material.
  • Compression: Pushing force that shortens a material.
  • 5. Fracture Stress ([math]σ_f[/math]​):
  • Definition:
  • – The maximum stress a material can withstand before breaking.
  • Units: Pascals (Pa).
  • Use: Indicates a material’s breaking point under stress.
  • 6. Yield Stress ([math]σ_y[/math]​):
  • Definition: Stress at which a material transitions from elastic to plastic deformation.
  • Units: Pascals (Pa).
  • Use: Determines when permanent deformation begins.
  • Material Properties:
Term Definition Example Materials
Stiff High resistance to deformation (high E) Steel Carbon fiber
Elastic Returns to original shape after deformation. Rubber, spring steel.
Plastic Undergoes permanent deformation when stressed. Copper, lead.
Ductile Can be drawn into wires (large plastic region). Gold, aluminum.
Hard Resists surface scratching or indentation. Diamond, ceramics
Brittle Breaks without significant plastic deformation. Glass, ceramics.
Tough Absorbs large amounts of energy before fracture. Kevlar, polycarbonate.
Strong Withstands large stresses before failure. Steel, titanium.

Dislocation
A defect in the crystal structure enabling plastic deformation in metals. Found in steel during slip.
  •  

    II) Sketching and Interpreting Graphs

  • 1. Force–Extension Graph
  • Axes:
  • – x-axis: Extension (m).
  • – y-axis: Force (N).
  • Elastic Region:
  • – Linear relationship ([math]F \propto \Delta L[/math]).
  • – Slope = stiffness of the material.
  • Plastic Region:
  • – Material deforms permanently.
  • Fracture Point:
  • – The material breaks.
  • Figure 1 Graph between Force (N) and extension (m)
  • 2. Stress–Strain Graph
  • Axes:
  • – x-axis: Strain (ϵ).
  • – y-axis: Stress (σ).
  • Proportional Limit: Linear section (follows Hooke’s law).
  • Elastic Limit/Yield Point: Marks the end of elastic behavior.
  • Plastic Region: Material undergoes permanent deformation.
  • Ultimate Tensile Stress (UTS): Maximum stress before necking begins.
  • Fracture Point: Material fails.
  • Figure 2 graph between stress and strain
  • 3. Comparing Materials Using Graphs
  • Brittle Material: Steep slope, fractures quickly (e.g., glass).
  • Ductile Material: Shows a long plastic region before breaking (e.g., copper).
  • Polymers: Non-linear behavior (e.g., rubber shows elastic deformation up to high strain).
  • Table diagrams comparing materials
Property Metals Ceramics Polymers
Structure Crystalline, dislocations Ionic/covalent bonds Long molecular chains
Ductility High Low Moderate (depends on type)
Elasticity Moderate Low High
Brittleness Low High Depends on type
Toughness High Low Moderate
  •  

  • IV) Images Showing Structures of Materials

  • Metal Structure:
  • – Regular crystal lattice with free electrons.
  • – Presence of dislocations.
  • Ceramic Structure:
  • – Rigid ionic/covalent bonds.
  • – No dislocations, leading to brittleness.
  • Polymer Structure:
  • – Entangled chains that can stretch, untangle, or slide depending on stress and temperature.

  • c) Detailed Explanation of Calculations and Estimates in Elasticity

  • ⇒ Hooke’s Law and Elastic Behavior
  • 1. Hooke’s Law:
  • Formula:
  • [math]F = kx[/math]
  • Where:
  • – F: Force applied (N).
  • – k: Spring constant or stiffness (N/m).
  • – x: Extension (m).
  • Explanation:
  • – Hooke’s Law applies within the elastic limit of a material.
  • – The spring constant k represents the stiffness of the spring or elastic material. A higher k means a stiffer material.
  • 2. Energy Stored in an Elastic Material:
  • Formula:
  • [math]U = \frac{1}{2}kx^2[/math]
  • Where:
  • – U: Elastic strain energy (J).
  • – k: Spring constant (N/m).
  • – x: Extension (m).
  • Derivation:
  • – The elastic energy is equal to the work done to stretch the material. Since force increases linearly with extension, the energy stored is the area under the force-extension graph (a triangle):
  • [math]U = \frac{1}{2}fx[/math]
  • Substituting [math]F = kx[/math]:
  • [math]U = \frac{1}{2}kx^2[/math]
  • Example Calculation:
  • – A spring with a stiffness [math]k = 200N/m[/math] is stretched by [math]x = 0.1m[/math].
  • – Energy stored:
  • [math]U = \frac{1}{2} kx^2 \\
    U = \frac{1}{2} (200) (0.1)^2 \\
    U = 1 \text{ J}[/math]
  • ⇒  Stress, Strain, and Young Modulus
  • 1. Stress (σ):
  • Formula:
  • [math]\sigma = \frac{\text{tension (Force)}}{\text{Cross-sectional area}}[/math]
  • Where:
  • – σ: Stress (Pa or [math]N/m^2[/math] ).
  • – Force: Applied force (N).
  • – Cross-sectional area: Cross-sectional area of the material ([math]m^2[/math]).
  • Example Calculation:
  • – A steel wire with a cross-sectional area [math]A = 0.0001m^2[/math] is stretched by a force [math]F = 50N[/math]
  • – Stress:
  • [math]\sigma = \frac{F}{A} \\
    \sigma = \frac{50}{0.0001} \\
    \sigma = 500,000 \text{ Pa} \text{ (or 500 kPa)}[/math]
  • 2. Strain (ϵ):
  • Formula:
  • [math]\epsilon = \frac{\text{Extension}}{\text{Original length}}[/math]
  • Where:
  • – ϵ: Strain (dimensionless, no units).
  • – Extension: Increase in length (m).
  • – Original length: Initial length of the material (m).
  • Example Calculation:
  • – A wire of original length [math]L_0 = 2m[/math] is stretched by [math]x = 0.01m[/math] .
  • – Strain:
  • [math]\epsilon = \frac{x}{L_0} \\
    \epsilon = \frac{0.01}{2} \\
    \epsilon = 0.005 \text{ (no units)}[/math]
  • 3. Young’s Modulus (E):
  • Formula:
  • [math]E = \frac{\text{Stress}}{\text{Strain}}[/math]
  • Where:
  • – E: Young’s modulus (Pa or [math]N/m^2[/math]).
  • – Stress (σ): Measure of force per unit area (Pa).
  • – Strain (ϵ): Dimensionless ratio.
  • Units:
  • – Pascals Pa or [math]N/m^2[/math].
  • Physical Meaning:
  • – Young’s modulus quantifies the stiffness of a material. Higher values indicate a stiffer material.
  • Example Calculation:
  • – A steel wire experiences a stress [math]\sigma = 500,000 \text{ Pa}, \quad \epsilon = 0.005[/math].
  • – Young’s modulus:
  • [math]E = \frac{\sigma}{\epsilon} \\
    E = \frac{500,000}{0.005} \\
    E = 100,000,000 \text{ Pa} \text{ (or 100 MPa)}[/math]
  • Energy as Area Under Force–Extension Graph
  • 1. Elastic Region:
  • For an elastic material, the area under the graph (a triangle) represents the elastic strain energy:
  • [math]U = \frac{1}{2}Fx[/math]
  • Substituting [math]F=kx[/math]:
  • [math]U = \frac{1}{2}kx^2[/math]
  • Example:
  • – A spring stretched by x=0.05 m under a force F= 20 N.
  • – Energy stored:
  • [math]U = \frac{1}{2} F x \\
    U = \frac{1}{2} (20) (0.05) \\
    U = 0.5 \text{ J}[/math]
  • 2. Beyond Elastic Region:
  • If the material undergoes plastic deformation, the total energy stored includes both elastic and plastic contributions. This is represented by the entire area under the force-extension graph.
  • ⇒ Steps for Solving Elasticity Problems
  • Identify Known Values:
  • – Determine the given quantities (e.g., force, extension, original length, etc.).
  • Calculate Stress:
  • – Use
  • [math]\sigma = \frac{F}{A}[/math]
  • Calculate Strain:
  • – Use
  • [math]\epsilon = \frac{x}{L_0}[/math]
  • Calculate Young’s Modulus:
  • – Use
  • [math]E = \frac{\sigma}{\epsilon}[/math]
  • Determine Energy Stored:
  • – Use
  • [math]U = \frac{1}{2} kx^2[/math]
  • – For elastic energy or calculate the area under the force-extension graph.
  • ⇒ Example Problem
  • Problem:
  • – A steel wire of original length [math]L_0[/math] and cross-sectional area [math]A = 0.0002m^2[/math] is stretched by [math]x = 0.002m[/math] under a force [math]F = 400N[/math] . Calculate:
  • – Stress.
  • – Strain.
  • – Young’s modulus.
  • – Energy stored in the wire.
  • Solution:
  • 1. Stress:
  • [math]\sigma = \frac{F}{A} \\
    \sigma = \frac{400}{0.0002} \\
    \sigma = 2,000,000 \text{ Pa}[/math]
  • 2. Strain:
  • [math]\epsilon = \frac{x}{L_0} \\
    \epsilon = \frac{0.002}{1.5} \\
    \epsilon = 0.00133[/math]
  • 3. Young’s Modulus:
  • [math]E = \frac{\sigma}{\epsilon} \\
    E = \frac{2,000,000}{0.00133} \\
    E = 1.5 \times 10^9 \text{ Pa} \text{ (or 1.5 GPa)}[/math]
  • 4. Energy Stored:
  • Spring constant:
  • [math]k = \frac{F}{x} \\
    k = \frac{400}{0.002} \\
    k = 200,000 \text{ N/m}[/math]
  • Energy:
  • [math]U = \frac{1}{2} kx^2 \\
    U = \frac{1}{2} (200,000) (0.002)^2 \\
    U = 0.4 \text{ J}[/math]

  • d) Demonstration and Application of Practical Activities

  • I. Plotting Force–Extension Characteristics for Different Materials

  • ⇒ Objective:
  • To investigate the force–extension characteristics of different materials such as springs, rubber bands, and polythene strips.
  • ⇒ Materials Required:
  • – Spring(s), rubber bands, polythene strips.
  • – Clamp stand with a ruler.
  • – Weights or force meter.
  • – Hook for attaching materials.
  • – Graph paper or plotting software.
  • ⇒ Methodology:
  • Setup:
  • – Attach the material to a fixed support using a clamp stand.
  • – Suspend a force meter or weights at the free end of the material.
  • – Place a ruler beside the material to measure extension.
  • Procedure:
  • – Apply an increasing force in small increments (e.g., by adding weights).
  • – Measure the extension of the material for each applied force.
  • – Record the force (F) and corresponding extension (x) in a table.
  • For Different Materials:
  • – Spring:
  • Measure within its elastic limit to verify Hooke’s Law (F = kx).
  • – Rubber Bands:
  • Observe non-linear force–extension behavior due to elastic hysteresis.
  • – Polythene Strips:
  • Examine their non-elastic or plastic deformation behavior.
  • – Plotting Results:
  • Plot force (F) on the y-axis and extension (x) on the x-axis.
  • – The graph for:
  • Springs: A straight line through the origin (linear region) showing Hooke’s Law.
  • Rubber bands: A non-linear curve due to elastic hysteresis.
  • Polythene: A curve showing limited elasticity followed by permanent deformation.
  • ⇒ Interpretation:
  • Determine the spring constant (k) for springs from the slope of the linear region.
  • Identify elastic and plastic behavior for other materials based on the shape of the curve.

  • II. Determining the Young Modulus for a Metal (e.g., Copper, Steel, or Other Wire)

  • ⇒ Objective:
  • To calculate the Young modulus (E) of a metal wire using experimental measurements of stress and strain.
  • ⇒ Materials Required:
  • – Metal wire (e.g., copper, steel) of known diameter.
  • – Fixed support with a pulley system.
  • – Weights or a force meter.
  • – Meter ruler or vernier calipers for length and diameter measurements.
  • – Micrometer screw gauge for precise diameter measurement.
  • ⇒ Methodology:
  • Setup:
  • – Attach one end of the metal wire to a fixed support.
  • – Pass the wire over a pulley with weights or a force meter attached at the free end.
  • – Mark an initial reference point along the wire for measuring elongation.
  • Initial Measurements:
  • – Measure the original length ([math]L_0[/math]​) of the wire using a ruler.
  • – Measure the diameter (d) of the wire using a micrometer screw gauge.
  • – Calculate the cross-sectional area (A) using:
  • [math]A = \frac{\pi d^2}{4}[/math]
  • Procedure:
  • – Apply increasing forces by adding weights or adjusting the force meter.
  • – For each applied force (F), measure the corresponding extension (x) using the ruler.
  • – Record the data in a table of force (F), extension (x), and stress-strain values.
  • Calculations:
  • – Stress (σ):
  • [math]\sigma = \frac{F}{A}[/math]
  • where F is the applied force and A is the cross-sectional area.
  • – Strain (ϵ):
  • [math]\epsilon = \frac{x}{L_0}[/math]
  • where x is the extension and ​ is the original length.
  • – Young’s Modulus (E):
  • [math]E = \frac{\sigma}{\epsilon}[/math]
  • – Plotting Results:
  • Plot stress (σ) on the y-axis and strain (ϵ) on the x-axis.
  • The gradient of the linear region gives the Young modulus (E).
  • Example Calculation:
  • Given:
  • – Original length,[math]L_0 = 1.5 \text{ m}[/math]
  • – Diameter,[math]d = 0.001 \text{ m}[/math]
  • – Force applied, [math]F = 50 \text{ N}[/math]
  • – Extension,[math]x = 0.002 \text{ m}[/math]
  • Calculate:
  • – Cross-sectional area:
  • [math]A = \frac{\pi d^2}{4} \\ A = \frac{(3.14) (0.001)^2}{4} \\ A = 7.85 \times 10^{-7} \text{ m}^2[/math]
  • – Stress:
  • [math]\sigma = \frac{F}{A} \\ \sigma = \frac{50}{7.85 \times 10^{-7}} \\ \sigma = 6.37 \times 10^7 \text{ Pa}[/math]
  • – Strain:
  • [math]\epsilon = \frac{x}{L_0} \\ \epsilon = \frac{0.002}{1.5} \\ \epsilon = 0.00133[/math]
  • – Young’s Modulus:
  • [math]E = \frac{\sigma}{\epsilon} \\ E = \frac{6.37 \times 10^7}{0.00133} \\ E = 4.79 \times 10^{10} \text{ Pa}[/math]
  • ⇒ Precautions:
  • Ensure the wire is taut before starting measurements.
  • Avoid overloading the wire to prevent permanent deformation.
  • Use precise instruments (micrometer screw gauge, force meter) for accurate measurements.
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