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College Physics for AP® Courses

5.3 Elasticity: Stress and Strain

College Physics for AP® Courses5.3 Elasticity: Stress and Strain
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  1. Preface
  2. 1 Introduction: The Nature of Science and Physics
    1. Connection for AP® Courses
    2. 1.1 Physics: An Introduction
    3. 1.2 Physical Quantities and Units
    4. 1.3 Accuracy, Precision, and Significant Figures
    5. 1.4 Approximation
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
  3. 2 Kinematics
    1. Connection for AP® Courses
    2. 2.1 Displacement
    3. 2.2 Vectors, Scalars, and Coordinate Systems
    4. 2.3 Time, Velocity, and Speed
    5. 2.4 Acceleration
    6. 2.5 Motion Equations for Constant Acceleration in One Dimension
    7. 2.6 Problem-Solving Basics for One Dimensional Kinematics
    8. 2.7 Falling Objects
    9. 2.8 Graphical Analysis of One Dimensional Motion
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  4. 3 Two-Dimensional Kinematics
    1. Connection for AP® Courses
    2. 3.1 Kinematics in Two Dimensions: An Introduction
    3. 3.2 Vector Addition and Subtraction: Graphical Methods
    4. 3.3 Vector Addition and Subtraction: Analytical Methods
    5. 3.4 Projectile Motion
    6. 3.5 Addition of Velocities
    7. Glossary
    8. Section Summary
    9. Conceptual Questions
    10. Problems & Exercises
    11. Test Prep for AP® Courses
  5. 4 Dynamics: Force and Newton's Laws of Motion
    1. Connection for AP® Courses
    2. 4.1 Development of Force Concept
    3. 4.2 Newton's First Law of Motion: Inertia
    4. 4.3 Newton's Second Law of Motion: Concept of a System
    5. 4.4 Newton's Third Law of Motion: Symmetry in Forces
    6. 4.5 Normal, Tension, and Other Examples of Force
    7. 4.6 Problem-Solving Strategies
    8. 4.7 Further Applications of Newton's Laws of Motion
    9. 4.8 Extended Topic: The Four Basic Forces—An Introduction
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  6. 5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
    1. Connection for AP® Courses
    2. 5.1 Friction
    3. 5.2 Drag Forces
    4. 5.3 Elasticity: Stress and Strain
    5. Glossary
    6. Section Summary
    7. Conceptual Questions
    8. Problems & Exercises
    9. Test Prep for AP® Courses
  7. 6 Gravitation and Uniform Circular Motion
    1. Connection for AP® Courses
    2. 6.1 Rotation Angle and Angular Velocity
    3. 6.2 Centripetal Acceleration
    4. 6.3 Centripetal Force
    5. 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
    6. 6.5 Newton's Universal Law of Gravitation
    7. 6.6 Satellites and Kepler's Laws: An Argument for Simplicity
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  8. 7 Work, Energy, and Energy Resources
    1. Connection for AP® Courses
    2. 7.1 Work: The Scientific Definition
    3. 7.2 Kinetic Energy and the Work-Energy Theorem
    4. 7.3 Gravitational Potential Energy
    5. 7.4 Conservative Forces and Potential Energy
    6. 7.5 Nonconservative Forces
    7. 7.6 Conservation of Energy
    8. 7.7 Power
    9. 7.8 Work, Energy, and Power in Humans
    10. 7.9 World Energy Use
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  9. 8 Linear Momentum and Collisions
    1. Connection for AP® courses
    2. 8.1 Linear Momentum and Force
    3. 8.2 Impulse
    4. 8.3 Conservation of Momentum
    5. 8.4 Elastic Collisions in One Dimension
    6. 8.5 Inelastic Collisions in One Dimension
    7. 8.6 Collisions of Point Masses in Two Dimensions
    8. 8.7 Introduction to Rocket Propulsion
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  10. 9 Statics and Torque
    1. Connection for AP® Courses
    2. 9.1 The First Condition for Equilibrium
    3. 9.2 The Second Condition for Equilibrium
    4. 9.3 Stability
    5. 9.4 Applications of Statics, Including Problem-Solving Strategies
    6. 9.5 Simple Machines
    7. 9.6 Forces and Torques in Muscles and Joints
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  11. 10 Rotational Motion and Angular Momentum
    1. Connection for AP® Courses
    2. 10.1 Angular Acceleration
    3. 10.2 Kinematics of Rotational Motion
    4. 10.3 Dynamics of Rotational Motion: Rotational Inertia
    5. 10.4 Rotational Kinetic Energy: Work and Energy Revisited
    6. 10.5 Angular Momentum and Its Conservation
    7. 10.6 Collisions of Extended Bodies in Two Dimensions
    8. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  12. 11 Fluid Statics
    1. Connection for AP® Courses
    2. 11.1 What Is a Fluid?
    3. 11.2 Density
    4. 11.3 Pressure
    5. 11.4 Variation of Pressure with Depth in a Fluid
    6. 11.5 Pascal’s Principle
    7. 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
    8. 11.7 Archimedes’ Principle
    9. 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
    10. 11.9 Pressures in the Body
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  13. 12 Fluid Dynamics and Its Biological and Medical Applications
    1. Connection for AP® Courses
    2. 12.1 Flow Rate and Its Relation to Velocity
    3. 12.2 Bernoulli’s Equation
    4. 12.3 The Most General Applications of Bernoulli’s Equation
    5. 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
    6. 12.5 The Onset of Turbulence
    7. 12.6 Motion of an Object in a Viscous Fluid
    8. 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  14. 13 Temperature, Kinetic Theory, and the Gas Laws
    1. Connection for AP® Courses
    2. 13.1 Temperature
    3. 13.2 Thermal Expansion of Solids and Liquids
    4. 13.3 The Ideal Gas Law
    5. 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
    6. 13.5 Phase Changes
    7. 13.6 Humidity, Evaporation, and Boiling
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  15. 14 Heat and Heat Transfer Methods
    1. Connection for AP® Courses
    2. 14.1 Heat
    3. 14.2 Temperature Change and Heat Capacity
    4. 14.3 Phase Change and Latent Heat
    5. 14.4 Heat Transfer Methods
    6. 14.5 Conduction
    7. 14.6 Convection
    8. 14.7 Radiation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  16. 15 Thermodynamics
    1. Connection for AP® Courses
    2. 15.1 The First Law of Thermodynamics
    3. 15.2 The First Law of Thermodynamics and Some Simple Processes
    4. 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
    5. 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
    6. 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
    7. 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
    8. 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  17. 16 Oscillatory Motion and Waves
    1. Connection for AP® Courses
    2. 16.1 Hooke’s Law: Stress and Strain Revisited
    3. 16.2 Period and Frequency in Oscillations
    4. 16.3 Simple Harmonic Motion: A Special Periodic Motion
    5. 16.4 The Simple Pendulum
    6. 16.5 Energy and the Simple Harmonic Oscillator
    7. 16.6 Uniform Circular Motion and Simple Harmonic Motion
    8. 16.7 Damped Harmonic Motion
    9. 16.8 Forced Oscillations and Resonance
    10. 16.9 Waves
    11. 16.10 Superposition and Interference
    12. 16.11 Energy in Waves: Intensity
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
    17. Test Prep for AP® Courses
  18. 17 Physics of Hearing
    1. Connection for AP® Courses
    2. 17.1 Sound
    3. 17.2 Speed of Sound, Frequency, and Wavelength
    4. 17.3 Sound Intensity and Sound Level
    5. 17.4 Doppler Effect and Sonic Booms
    6. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
    7. 17.6 Hearing
    8. 17.7 Ultrasound
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  19. 18 Electric Charge and Electric Field
    1. Connection for AP® Courses
    2. 18.1 Static Electricity and Charge: Conservation of Charge
    3. 18.2 Conductors and Insulators
    4. 18.3 Conductors and Electric Fields in Static Equilibrium
    5. 18.4 Coulomb’s Law
    6. 18.5 Electric Field: Concept of a Field Revisited
    7. 18.6 Electric Field Lines: Multiple Charges
    8. 18.7 Electric Forces in Biology
    9. 18.8 Applications of Electrostatics
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  20. 19 Electric Potential and Electric Field
    1. Connection for AP® Courses
    2. 19.1 Electric Potential Energy: Potential Difference
    3. 19.2 Electric Potential in a Uniform Electric Field
    4. 19.3 Electrical Potential Due to a Point Charge
    5. 19.4 Equipotential Lines
    6. 19.5 Capacitors and Dielectrics
    7. 19.6 Capacitors in Series and Parallel
    8. 19.7 Energy Stored in Capacitors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  21. 20 Electric Current, Resistance, and Ohm's Law
    1. Connection for AP® Courses
    2. 20.1 Current
    3. 20.2 Ohm’s Law: Resistance and Simple Circuits
    4. 20.3 Resistance and Resistivity
    5. 20.4 Electric Power and Energy
    6. 20.5 Alternating Current versus Direct Current
    7. 20.6 Electric Hazards and the Human Body
    8. 20.7 Nerve Conduction–Electrocardiograms
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  22. 21 Circuits, Bioelectricity, and DC Instruments
    1. Connection for AP® Courses
    2. 21.1 Resistors in Series and Parallel
    3. 21.2 Electromotive Force: Terminal Voltage
    4. 21.3 Kirchhoff’s Rules
    5. 21.4 DC Voltmeters and Ammeters
    6. 21.5 Null Measurements
    7. 21.6 DC Circuits Containing Resistors and Capacitors
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  23. 22 Magnetism
    1. Connection for AP® Courses
    2. 22.1 Magnets
    3. 22.2 Ferromagnets and Electromagnets
    4. 22.3 Magnetic Fields and Magnetic Field Lines
    5. 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
    6. 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
    7. 22.6 The Hall Effect
    8. 22.7 Magnetic Force on a Current-Carrying Conductor
    9. 22.8 Torque on a Current Loop: Motors and Meters
    10. 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
    11. 22.10 Magnetic Force between Two Parallel Conductors
    12. 22.11 More Applications of Magnetism
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
    17. Test Prep for AP® Courses
  24. 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
    1. Connection for AP® Courses
    2. 23.1 Induced Emf and Magnetic Flux
    3. 23.2 Faraday’s Law of Induction: Lenz’s Law
    4. 23.3 Motional Emf
    5. 23.4 Eddy Currents and Magnetic Damping
    6. 23.5 Electric Generators
    7. 23.6 Back Emf
    8. 23.7 Transformers
    9. 23.8 Electrical Safety: Systems and Devices
    10. 23.9 Inductance
    11. 23.10 RL Circuits
    12. 23.11 Reactance, Inductive and Capacitive
    13. 23.12 RLC Series AC Circuits
    14. Glossary
    15. Section Summary
    16. Conceptual Questions
    17. Problems & Exercises
    18. Test Prep for AP® Courses
  25. 24 Electromagnetic Waves
    1. Connection for AP® Courses
    2. 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
    3. 24.2 Production of Electromagnetic Waves
    4. 24.3 The Electromagnetic Spectrum
    5. 24.4 Energy in Electromagnetic Waves
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
    10. Test Prep for AP® Courses
  26. 25 Geometric Optics
    1. Connection for AP® Courses
    2. 25.1 The Ray Aspect of Light
    3. 25.2 The Law of Reflection
    4. 25.3 The Law of Refraction
    5. 25.4 Total Internal Reflection
    6. 25.5 Dispersion: The Rainbow and Prisms
    7. 25.6 Image Formation by Lenses
    8. 25.7 Image Formation by Mirrors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  27. 26 Vision and Optical Instruments
    1. Connection for AP® Courses
    2. 26.1 Physics of the Eye
    3. 26.2 Vision Correction
    4. 26.3 Color and Color Vision
    5. 26.4 Microscopes
    6. 26.5 Telescopes
    7. 26.6 Aberrations
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  28. 27 Wave Optics
    1. Connection for AP® Courses
    2. 27.1 The Wave Aspect of Light: Interference
    3. 27.2 Huygens's Principle: Diffraction
    4. 27.3 Young’s Double Slit Experiment
    5. 27.4 Multiple Slit Diffraction
    6. 27.5 Single Slit Diffraction
    7. 27.6 Limits of Resolution: The Rayleigh Criterion
    8. 27.7 Thin Film Interference
    9. 27.8 Polarization
    10. 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  29. 28 Special Relativity
    1. Connection for AP® Courses
    2. 28.1 Einstein’s Postulates
    3. 28.2 Simultaneity And Time Dilation
    4. 28.3 Length Contraction
    5. 28.4 Relativistic Addition of Velocities
    6. 28.5 Relativistic Momentum
    7. 28.6 Relativistic Energy
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  30. 29 Introduction to Quantum Physics
    1. Connection for AP® Courses
    2. 29.1 Quantization of Energy
    3. 29.2 The Photoelectric Effect
    4. 29.3 Photon Energies and the Electromagnetic Spectrum
    5. 29.4 Photon Momentum
    6. 29.5 The Particle-Wave Duality
    7. 29.6 The Wave Nature of Matter
    8. 29.7 Probability: The Heisenberg Uncertainty Principle
    9. 29.8 The Particle-Wave Duality Reviewed
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  31. 30 Atomic Physics
    1. Connection for AP® Courses
    2. 30.1 Discovery of the Atom
    3. 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
    4. 30.3 Bohr’s Theory of the Hydrogen Atom
    5. 30.4 X Rays: Atomic Origins and Applications
    6. 30.5 Applications of Atomic Excitations and De-Excitations
    7. 30.6 The Wave Nature of Matter Causes Quantization
    8. 30.7 Patterns in Spectra Reveal More Quantization
    9. 30.8 Quantum Numbers and Rules
    10. 30.9 The Pauli Exclusion Principle
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  32. 31 Radioactivity and Nuclear Physics
    1. Connection for AP® Courses
    2. 31.1 Nuclear Radioactivity
    3. 31.2 Radiation Detection and Detectors
    4. 31.3 Substructure of the Nucleus
    5. 31.4 Nuclear Decay and Conservation Laws
    6. 31.5 Half-Life and Activity
    7. 31.6 Binding Energy
    8. 31.7 Tunneling
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  33. 32 Medical Applications of Nuclear Physics
    1. Connection for AP® Courses
    2. 32.1 Medical Imaging and Diagnostics
    3. 32.2 Biological Effects of Ionizing Radiation
    4. 32.3 Therapeutic Uses of Ionizing Radiation
    5. 32.4 Food Irradiation
    6. 32.5 Fusion
    7. 32.6 Fission
    8. 32.7 Nuclear Weapons
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  34. 33 Particle Physics
    1. Connection for AP® Courses
    2. 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
    3. 33.2 The Four Basic Forces
    4. 33.3 Accelerators Create Matter from Energy
    5. 33.4 Particles, Patterns, and Conservation Laws
    6. 33.5 Quarks: Is That All There Is?
    7. 33.6 GUTs: The Unification of Forces
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  35. 34 Frontiers of Physics
    1. Connection for AP® Courses
    2. 34.1 Cosmology and Particle Physics
    3. 34.2 General Relativity and Quantum Gravity
    4. 34.3 Superstrings
    5. 34.4 Dark Matter and Closure
    6. 34.5 Complexity and Chaos
    7. 34.6 High-Temperature Superconductors
    8. 34.7 Some Questions We Know to Ask
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  36. A | Atomic Masses
  37. B | Selected Radioactive Isotopes
  38. C | Useful Information
  39. D | Glossary of Key Symbols and Notation
  40. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
    22. Chapter 22
    23. Chapter 23
    24. Chapter 24
    25. Chapter 25
    26. Chapter 26
    27. Chapter 27
    28. Chapter 28
    29. Chapter 29
    30. Chapter 30
    31. Chapter 31
    32. Chapter 32
    33. Chapter 33
    34. Chapter 34
  41. Index

Learning Objectives

By the end of this section, you will be able to:

  • State Hooke's law.
  • Explain Hooke's law using graphical representation between deformation and applied force.
  • Discuss the three types of deformations such as changes in length, sideways shear, and changes in volume.
  • Describe with examples the Young's modulus, shear modulus, and bulk modulus.
  • Determine the change in length given mass, length, and radius.

We now move from consideration of forces that affect the motion of an object (such as friction and drag) to those that affect an object's shape. If a bulldozer pushes a car into a wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important characteristics are observed. First, the object returns to its original shape when the force is removed—that is, the deformation is elastic for small deformations. Second, the size of the deformation is proportional to the force—that is, for small deformations, Hooke's law is obeyed. In equation form, Hooke's law is given by

F=kΔL,F=kΔL, size 12{F=kΔL} {}
5.26

where ΔLΔL size 12{ΔL} {} is the amount of deformation (the change in length, for example) produced by the force FF size 12{F} {}, and kk size 12{k} {} is a proportionality constant that depends on the shape and composition of the object and the direction of the force. Note that this force is a function of the deformation ΔLΔL size 12{ΔL} {} —it is not constant as a kinetic friction force is. Rearranging this to

Δ L = F k Δ L = F k size 12{ΔL= { {F} over {k} } } {}
5.27

makes it clear that the deformation is proportional to the applied force. Figure 5.12 shows the Hooke's law relationship between the extension ΔLΔL size 12{ΔL} {} of a spring or of a human bone. For metals or springs, the straight line region in which Hooke's law pertains is much larger. Bones are brittle and the elastic region is small and the fracture abrupt. Eventually a large enough stress to the material will cause it to break or fracture.

Hooke's Law

F = kΔL , F = kΔL , size 12{F=kΔL} {}
5.28

where ΔLΔL size 12{ΔL} {} is the amount of deformation (the change in length, for example) produced by the force FF size 12{F} {}, and kk size 12{k} {} is a proportionality constant that depends on the shape and composition of the object and the direction of the force.

Δ L = F k Δ L = F k size 12{ΔL= { {F} over {k} } } {}
5.29
Line graph of change in length versus applied force. The line has a constant positive slope from the origin in the region where Hooke's law is obeyed. The slope then decreases, with a lower, still positive slope until the end of the elastic region. The slope then increases dramatically in the region of permanent deformation until fracturing occurs.
Figure 5.12 A graph of deformation ΔLΔL size 12{ΔL} {} versus applied force FF size 12{F} {}. The straight segment is the linear region where Hooke's law is obeyed. The slope of the straight region is 1k1k size 12{ { {1} over {k} } } {}. For larger forces, the graph is curved but the deformation is still elastic— ΔLΔL size 12{ΔL} {} will return to zero if the force is removed. Still greater forces permanently deform the object until it finally fractures. The shape of the curve near fracture depends on several factors, including how the force FF size 12{F} {} is applied. Note that in this graph the slope increases just before fracture, indicating that a small increase in FF size 12{F} {} is producing a large increase in LL size 12{L} {} near the fracture.

The proportionality constant kk size 12{k} {} depends upon a number of factors for the material. For example, a guitar string made of nylon stretches when it is tightened, and the elongation ΔLΔL size 12{ΔL} {} is proportional to the force applied (at least for small deformations). Thicker nylon strings and ones made of steel stretch less for the same applied force, implying they have a larger kk size 12{k} {} (see Figure 5.13). Finally, all three strings return to their normal lengths when the force is removed, provided the deformation is small. Most materials will behave in this manner if the deformation is less than about 0.1% or about 1 part in 103103 size 12{"10" rSup { size 8{3} } } {}.

Diagram of weight w attached to each of three guitar strings of initial length L zero hanging vertically from a ceiling. The weight pulls down on the strings with force w. The ceiling pulls up on the strings with force w. The first string of thin nylon has a deformation of delta L due to the force of the weight pulling down. The middle string of thicker nylon has a smaller deformation. The third string of thin steel has the smallest deformation.
Figure 5.13 The same force, in this case a weight (ww size 12{w} {}), applied to three different guitar strings of identical length produces the three different deformations shown as shaded segments. The string on the left is thin nylon, the one in the middle is thicker nylon, and the one on the right is steel.

Stretch Yourself a Little

How would you go about measuring the proportionality constant kk size 12{k} {} of a rubber band? If a rubber band stretched 3 cm when a 100-g mass was attached to it, then how much would it stretch if two similar rubber bands were attached to the same mass—even if put together in parallel or alternatively if tied together in series?

We now consider three specific types of deformations: changes in length (tension and compression), sideways shear (stress), and changes in volume. All deformations are assumed to be small unless otherwise stated.

Changes in Length—Tension and Compression: Elastic Modulus

A change in length ΔLΔL size 12{ΔL} {} is produced when a force is applied to a wire or rod parallel to its length L0L0 size 12{L rSub { size 8{0} } } {}, either stretching it (a tension) or compressing it. (See Figure 5.14.)

Figure a is a cylindrical rod standing on its end with a height of L sub zero. Two vectors labeled F extend away from each end. A dotted outline indicates that the rod is stretched by a length of delta L. Figure b is a similar rod of identical height L sub zero, but two vectors labeled F exert a force toward the ends of the rod. A dotted line indicates that the rod is compressed by a length of delta L.
Figure 5.14 (a) Tension. The rod is stretched a length ΔLΔL size 12{ΔL} {} when a force is applied parallel to its length. (b) Compression. The same rod is compressed by forces with the same magnitude in the opposite direction. For very small deformations and uniform materials, ΔLΔL size 12{ΔL} {} is approximately the same for the same magnitude of tension or compression. For larger deformations, the cross-sectional area changes as the rod is compressed or stretched.

Experiments have shown that the change in length (ΔLΔL size 12{ΔL} {}) depends on only a few variables. As already noted, ΔLΔL size 12{ΔL} {} is proportional to the force FF size 12{F} {} and depends on the substance from which the object is made. Additionally, the change in length is proportional to the original length L0L0 size 12{L rSub { size 8{0} } } {} and inversely proportional to the cross-sectional area of the wire or rod. For example, a long guitar string will stretch more than a short one, and a thick string will stretch less than a thin one. We can combine all these factors into one equation for ΔLΔL size 12{ΔL} {}:

ΔL=1YFAL0,ΔL=1YFAL0, size 12{ΔL= { {1} over {Y} } { {F} over {A} } L rSub { size 8{0} } } {}
5.30

where ΔL ΔL size 12{ΔL} {} is the change in length, FF size 12{F} {} the applied force, YY size 12{Y} {} is a factor, called the elastic modulus or Young's modulus, that depends on the substance, AA size 12{A} {} is the cross-sectional area, and L0L0 size 12{L rSub { size 8{0} } } {} is the original length. Table 5.3 lists values of YY size 12{A} {} for several materials—those with a large YY size 12{A} {} are said to have a large tensile strength because they deform less for a given tension or compression.

Material Young's modulus (tension–compression)Y ( 10 9 N/m 2 ) ( 10 9 N/m 2 ) Shear modulus S ( 10 9 N/m 2 ) ( 10 9 N/m 2 ) Bulk modulus B ( 10 9 N/m 2 ) ( 10 9 N/m 2 )
Aluminum 70 25 75
Bone – tension 16 80 8
Bone – compression 9
Brass 90 35 75
Brick 15
Concrete 20
Glass 70 20 30
Granite 45 20 45
Hair (human) 10
Hardwood 15 10
Iron, cast 100 40 90
Lead 16 5 50
Marble 60 20 70
Nylon 5
Polystyrene 3
Silk 6
Spider thread 3
Steel 210 80 130
Tendon 1
Acetone 0.7
Ethanol 0.9
Glycerin 4.5
Mercury 25
Water 2.2
Table 5.3 Elastic Moduli1

Young's moduli are not listed for liquids and gases in Table 5.3 because they cannot be stretched or compressed in only one direction. Note that there is an assumption that the object does not accelerate, so that there are actually two applied forces of magnitude FF size 12{F} {} acting in opposite directions. For example, the strings in Figure 5.14 are being pulled down by a force of magnitude ww size 12{w} {} and held up by the ceiling, which also exerts a force of magnitude ww size 12{w} {}.

Example 5.3 The Stretch of a Long Cable

Suspension cables are used to carry gondolas at ski resorts. (See Figure 5.15) Consider a suspension cable that includes an unsupported span of 3020 m. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is 3.0×106 N3.0×106 N size 12{3 "." 0 times "10" rSup { size 8{6} } " N"} {}.

Ski gondolas travel along suspension cables. A vast forest and snowy mountain peaks can be seen in the background.
Figure 5.15 Gondolas travel along suspension cables at the Gala Yuzawa ski resort in Japan. (credit: Rudy Herman, Flickr)

Strategy

The force is equal to the maximum tension, or F=3.0×106 NF=3.0×106 N size 12{F=3 "." 0 times "10" rSup { size 8{6} } " N"} {}. The cross-sectional area is πr2=2.46×103 m2πr2=2.46×103 m2 size 12{πr rSup { size 8{2} } =2 "." "46" times "10" rSup { size 8{ - 3} } " m" rSup { size 8{2} } } {}. The equation ΔL=1YFAL0ΔL=1YFAL0 size 12{ΔL= { {1} over {Y} } { {F} over {A} } L rSub { size 8{0} } } {} can be used to find the change in length.

Solution

All quantities are known. Thus,

Δ L = 1 210 × 10 9 N/m 2 3 . 0 × 10 6 N 2.46 × 10 –3 m 2 3020 m = 18 m . Δ L = 1 210 × 10 9 N/m 2 3 . 0 × 10 6 N 2.46 × 10 –3 m 2 3020 m = 18 m .
5.31

Discussion

This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these environments.

Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in the bone shearing or snapping. The behavior of bones under tension and compression is important because it determines the load the bones can carry. Bones are classified as weight-bearing structures such as columns in buildings and trees. Weight-bearing structures have special features; columns in building have steel-reinforcing rods while trees and bones are fibrous. The bones in different parts of the body serve different structural functions and are prone to different stresses. Thus the bone in the top of the femur is arranged in thin sheets separated by marrow while in other places the bones can be cylindrical and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained compressions in bone joints and tendons.

Another biological example of Hooke's law occurs in tendons. Functionally, the tendon (the tissue connecting muscle to bone) must stretch easily at first when a force is applied, but offer a much greater restoring force for a greater strain. Figure 5.16 shows a stress-strain relationship for a human tendon. Some tendons have a high collagen content so there is relatively little strain, or length change; others, like support tendons (as in the leg) can change length up to 10%. Note that this stress-strain curve is nonlinear, since the slope of the line changes in different regions. In the first part of the stretch called the toe region, the fibers in the tendon begin to align in the direction of the stress—this is called uncrimping. In the linear region, the fibrils will be stretched, and in the failure region individual fibers begin to break. A simple model of this relationship can be illustrated by springs in parallel: different springs are activated at different lengths of stretch. Examples of this are given in the problems at end of this chapter. Ligaments (tissue connecting bone to bone) behave in a similar way.

The strain on mammalian tendon is shown by a graph, with strain along the x axis and tensile stress along the y axis. The stress strain curve obtained has three regions, namely, toe region at the bottom, linear region between, and failure region at the top.
Figure 5.16 Typical stress-strain curve for mammalian tendon. Three regions are shown: (1) toe region (2) linear region, and (3) failure region.

Unlike bones and tendons, which need to be strong as well as elastic, the arteries and lungs need to be very stretchable. The elastic properties of the arteries are essential for blood flow. The pressure in the arteries increases and arterial walls stretch when the blood is pumped out of the heart. When the aortic valve shuts, the pressure in the arteries drops and the arterial walls relax to maintain the blood flow. When you feel your pulse, you are feeling exactly this—the elastic behavior of the arteries as the blood gushes through with each pump of the heart. If the arteries were rigid, you would not feel a pulse. The heart is also an organ with special elastic properties. The lungs expand with muscular effort when we breathe in but relax freely and elastically when we breathe out. Our skins are particularly elastic, especially for the young. A young person can go from 100 kg to 60 kg with no visible sag in their skins. The elasticity of all organs reduces with age. Gradual physiological aging through reduction in elasticity starts in the early 20s.

Example 5.4 Calculating Deformation: How Much Does Your Leg Shorten When You Stand on It?

Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius.

Strategy

The force is equal to the weight supported, or

F=mg=62.0 kg9.80 m/s2=607.6 N,F=mg=62.0 kg9.80 m/s2=607.6 N, size 12{F= ital "mg"= left ("62" "." 0`"kg" right ) left (9 "." "80"`"m/s" rSup { size 8{2} } right )="607" "." 6``N} {}
5.32

and the cross-sectional area is πr2=1.257×103m2πr2=1.257×103m2 size 12{πr rSup { size 8{2} } =1 "." "257"` times "10" rSup { size 8{ - 3} } m rSup { size 8{2} } } {}. The equation ΔL=1YFAL0ΔL=1YFAL0 size 12{ΔL= { {1} over {Y} } { {F} over {A} } L rSub { size 8{0} } } {} can be used to find the change in length.

Solution

All quantities except ΔLΔL size 12{ΔL} {} are known. Note that the compression value for Young's modulus for bone must be used here. Thus,

Δ L = 1 9 × 10 9 N/m 2 607 . 6 N 1. 257 × 10 3 m 2 (0.400 m) = 2×10−5 m. Δ L = 1 9 × 10 9 N/m 2 607 . 6 N 1. 257 × 10 3 m 2 (0.400 m) = 2×10−5 m. alignl { stack { size 12{ΔL= { {1} over {9 times "10" rSup { size 8{9} } " N/m" rSup { size 8{2} } } } times { {"607" "." "6 N"} over {1 "." "257" times "10" rSup { size 8{ - 3} } " m" rSup { size 8{2} } } } times 0 "." "400 m"} {} # =0 "." "002" times "10" rSup { size 8{ - 3} } " m" {} } } {}
5.33

Discussion

This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table 5.3 have larger values of Young's modulus YY size 12{Y} {}. In other words, they are more rigid and have greater tensile strength.

The equation for change in length is traditionally rearranged and written in the following form:

FA=YΔLL0.FA=YΔLL0. size 12{ { {F} over {A} } =Y { {ΔL} over {L rSub { size 8{0} } } } } {}
5.34

The ratio of force to area, FAFA size 12{ { {F} over {A} } } {}, is defined as stress (measured in N/m 2 N/m 2 ), and the ratio of the change in length to length, ΔLL0ΔLL0 size 12{ { {ΔL} over {L rSub { size 8{0} } } } } {}, is defined as strain (a unitless quantity). In other words,

stress=Y×strain.stress=Y×strain. size 12{"stress"=Y times "strain"} {}
5.35

In this form, the equation is analogous to Hooke's law, with stress analogous to force and strain analogous to deformation. If we again rearrange this equation to the form

F=YAΔLL0,F=YAΔLL0, size 12{F= ital "YA" { {ΔL} over {L rSub { size 8{0} } } } } {}
5.36

we see that it is the same as Hooke's law with a proportionality constant

k=YAL0.k=YAL0. size 12{k= { { ital "YA"} over {L rSub { size 8{0} } } } } {}
5.37

This general idea—that force and the deformation it causes are proportional for small deformations—applies to changes in length, sideways bending, and changes in volume.

Stress

The ratio of force to area, FAFA size 12{ { {F} over {A} } } {}, is defined as stress measured in N/m2.

Strain

The ratio of the change in length to length, ΔLL0ΔLL0 size 12{ { {ΔL} over {L rSub { size 8{0} } } } } {}, is defined as strain (a unitless quantity). In other words,

stress=Y×strain.stress=Y×strain. size 12{"stress"=Y times "strain"} {}
5.38

Sideways Stress: Shear Modulus

Figure 5.17 illustrates what is meant by a sideways stress or a shearing force. Here the deformation is called ΔxΔx size 12{Δx} {} and it is perpendicular to L0L0 size 12{L rSub { size 8{0} } } {}, rather than parallel as with tension and compression. Shear deformation behaves similarly to tension and compression and can be described with similar equations. The expression for shear deformation is

Δx=1SFAL0,Δx=1SFAL0, size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {}
5.39

where SS size 12{F} {} is the shear modulus (see Table 5.3) and FF size 12{F} {} is the force applied perpendicular to L0L0 size 12{L rSub { size 8{0} } } {} and parallel to the cross-sectional area AA size 12{A} {}. Again, to keep the object from accelerating, there are actually two equal and opposite forces FF size 12{F} {} applied across opposite faces, as illustrated in Figure 5.17. The equation is logical—for example, it is easier to bend a long thin pencil (small AA size 12{A} {}) than a short thick one, and both are more easily bent than similar steel rods (large SS size 12{S} {}).

Shear Deformation

Δx=1SFAL0,Δx=1SFAL0, size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {}
5.40

where SS size 12{S} {} is the shear modulus and FF size 12{F} {} is the force applied perpendicular to L0L0 size 12{L rSub { size 8{0} } } {} and parallel to the cross-sectional area AA size 12{A} {}.

Bookcase sheared by a force applied at the bottom right toward the bottom left, and at the top left toward the top right.
Figure 5.17 Shearing forces are applied perpendicular to the length L 0 L 0 and parallel to the area A A , producing a deformation Δx Δx . Vertical forces are not shown, but it should be kept in mind that in addition to the two shearing forces, FF size 12{F} {}, there must be supporting forces to keep the object from rotating. The distorting effects of these supporting forces are ignored in this treatment. The weight of the object also is not shown, since it is usually negligible compared with forces large enough to cause significant deformations.

Examination of the shear moduli in Table 5.3 reveals some telling patterns. For example, shear moduli are less than Young's moduli for most materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young's modulus, but it is as large as that of steel. This is one reason that bones can be long and relatively thin. Bones can support loads comparable to that of concrete and steel. Most bone fractures are not caused by compression but by excessive twisting and bending.

The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper part of the body. The spinal column has normal curvature for stability, but this curvature can be increased, leading to increased shearing forces on the lower vertebrae. Discs are better at withstanding compressional forces than shear forces. Because the spine is not vertical, the weight of the upper body exerts some of both. Pregnant women and people that are overweight (with large abdomens) need to move their shoulders back to maintain balance, thereby increasing the curvature in their spine and so increasing the shear component of the stress. An increased angle due to more curvature increases the shear forces along the plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge shaped disc below the last vertebrae) is particularly at risk because of its location.

The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can withstand compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors or during earthquakes. Modern structures were made possible by the use of steel and steel-reinforced concrete. Almost by definition, liquids and gases have shear moduli near zero, because they flow in response to shearing forces.

Example 5.5 Calculating Force Required to Deform: That Nail Does Not Bend Much Under a Load

Find the mass of the picture hanging from a steel nail as shown in Figure 5.18, given that the nail bends only 1.80 µm 1.80 µm size 12{ left (1 "." "80" times "10" rSup { size 8{ - 6} } m right )} {}. (Assume the shear modulus is known to two significant figures.)

Diagram showing the side view a nail in a wall, deformed by the weight of a picture hanging from it. The weight w of the picture is downward. There is an equal force w upward on the nail from the wall. The nail is 1 point five zero millimeters thick. The length of the nail that is outside the wall is five point zero zero millimeters. The deformation delta x of the nail as a result of the picture is 1 point eight zero micrometers.
Figure 5.18 Side view of a nail with a picture hung from it. The nail flexes very slightly (shown much larger than actual) because of the shearing effect of the supported weight. Also shown is the upward force of the wall on the nail, illustrating that there are equal and opposite forces applied across opposite cross sections of the nail. See Example 5.5 for a calculation of the mass of the picture.

Strategy

The force FF size 12{F} {} on the nail (neglecting the nail's own weight) is the weight of the picture ww size 12{w} {}. If we can find ww size 12{w} {}, then the mass of the picture is just wgwg size 12{ { {w} over {g} } } {} . The equation Δx=1SFAL0Δx=1SFAL0 size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {} can be solved for FF size 12{F} {}.

Solution

Solving the equation Δx=1SFAL0Δx=1SFAL0 size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {} for F F , we see that all other quantities can be found:

F=SAL0Δx.F=SAL0Δx. size 12{F= { { ital "SA"} over {L rSub { size 8{0} } } } Δx} {}
5.41

S is found in Table 5.3 and is S=80×109 N/m2S=80×109 N/m2 size 12{S="80" times "10" rSup { size 8{9} } " N/m" rSup { size 8{2} } } {}. The radius rr size 12{r} {} is 0.750 mm (as seen in the figure), so the cross-sectional area is

A=πr2=1.77×106m2.A=πr2=1.77×106m2. size 12{A=πr rSup { size 8{2} } =1 "." "77" times "10" rSup { size 8{ - 6} } m rSup { size 8{2} } } {}
5.42

The value for L0L0 size 12{L rSub { size 8{0} } } {} is also shown in the figure. Thus,

F = ( 80 × 10 9 N/m 2 ) ( 1 . 77 × 10 6 m 2 ) ( 5 . 00 × 10 3 m ) ( 1 . 80 × 10 6 m ) = 51 N. F = ( 80 × 10 9 N/m 2 ) ( 1 . 77 × 10 6 m 2 ) ( 5 . 00 × 10 3 m ) ( 1 . 80 × 10 6 m ) = 51 N. size 12{F= { { \( "80" times "10" rSup { size 8{9} } " N/m" rSup { size 8{2} } \) \( 1 "." "77" times "10" rSup { size 8{ - 6} } " m" rSup { size 8{2} } \) } over { \( 5 "." "00" times "10" rSup { size 8{ - 3} } " m" \) } } times \( 1 "." "80" times "10" rSup { size 8{ - 6} } " m" \) ="51"N} {}
5.43

This 51 N force is the weight w w of the picture, so the picture's mass is

m=wg=Fg=5.2 kg.m=wg=Fg=5.2 kg. size 12{m= { {w} over {g} } = { {F} over {g} } =5 "." 2" kg"} {}
5.44

Discussion

This is a fairly massive picture, and it is impressive that the nail flexes only 1.80 µm 1.80 µm —an amount undetectable to the unaided eye.

Changes in Volume: Bulk Modulus

An object will be compressed in all directions if inward forces are applied evenly on all its surfaces as in Figure 5.19. It is relatively easy to compress gases and extremely difficult to compress liquids and solids. For example, air in a wine bottle is compressed when it is corked. But if you try corking a brim-full bottle, you cannot compress the wine—some must be removed if the cork is to be inserted. The reason for these different compressibilities is that atoms and molecules are separated by large empty spaces in gases but packed close together in liquids and solids. To compress a gas, you must force its atoms and molecules closer together. To compress liquids and solids, you must actually compress their atoms and molecules, and very strong electromagnetic forces in them oppose this compression.

A cube with area of cross section A and volume V zero is compressed by an inward force F acting on all surfaces. The compression causes a change in volume delta V, which is proportional to the force per unit area and its original volume. This change in volume is related to the compressibility of the substance.
Figure 5.19 An inward force on all surfaces compresses this cube. Its change in volume is proportional to the force per unit area and its original volume, and is related to the compressibility of the substance.

We can describe the compression or volume deformation of an object with an equation. First, we note that a force “applied evenly” is defined to have the same stress, or ratio of force to area FAFA size 12{ left ( { {F} over {A} } right )} {} on all surfaces. The deformation produced is a change in volume ΔVΔV size 12{ΔV} {}, which is found to behave very similarly to the shear, tension, and compression previously discussed. (This is not surprising, since a compression of the entire object is equivalent to compressing each of its three dimensions.) The relationship of the change in volume to other physical quantities is given by

ΔV=1BFAV0,ΔV=1BFAV0, size 12{ΔV= { {1} over {B} } { {F} over {A} } V rSub { size 8{0} } } {}
5.45

where B B is the bulk modulus (see Table 5.3), V0V0 size 12{V rSub { size 8{0} } } {} is the original volume, and FAFA size 12{ { {F} over {A} } } {} is the force per unit area applied uniformly inward on all surfaces. Note that no bulk moduli are given for gases.

What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-grade diamonds by compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline structure into the more tightly packed pattern of diamonds. In nature, a similar process occurs deep underground, where extremely large forces result from the weight of overlying material. Another natural source of large compressive forces is the pressure created by the weight of water, especially in deep parts of the oceans. Water exerts an inward force on all surfaces of a submerged object, and even on the water itself. At great depths, water is measurably compressed, as the following example illustrates.

Example 5.6 Calculating Change in Volume with Deformation: How Much Is Water Compressed at Great Ocean Depths?

Calculate the fractional decrease in volume (ΔVV0ΔVV0 size 12{ { {ΔV} over {V rSub { size 8{0} } } } } {}) for seawater at 5.00 km depth, where the force per unit area is 5 . 00 × 10 7 N / m 2 5 . 00 × 10 7 N / m 2 size 12{5 "." "00" times "10" rSup { size 8{7} } N/m rSup { size 8{2} } } {} .

Strategy

Equation ΔV=1BFAV0ΔV=1BFAV0 is the correct physical relationship. All quantities in the equation except ΔVV0ΔVV0 are known.

Solution

Solving for the unknown ΔVV0ΔVV0 gives

ΔVV0=1BFA.ΔVV0=1BFA. size 12{ { {ΔV} over {V rSub { size 8{0} } } } = { {1} over {B} } { {F} over {A} } } {}
5.46

Substituting known values with the value for the bulk modulus B B from Table 5.3,

Δ V V0 = 5.00 × 10 7 N/m 2 2 . 2 × 10 9 N/m 2 = 0.023=2.3%. Δ V V0 = 5.00 × 10 7 N/m 2 2 . 2 × 10 9 N/m 2 = 0.023=2.3%.
5.47

Discussion

Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress.

Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way.

Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here.

Footnotes

  • 1 Approximate and average values. Young's moduli Y Y size 12{Y} {} for tension and compression sometimes differ but are averaged here. Bone has significantly different Young's moduli for tension and compression.
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