Mechanical Behavior Stress vs. Strain Elastic deformation

Slides:



Advertisements
Similar presentations
Mechanical Properties of Metals
Advertisements

ISE316 Chapter 3 --Mechanics of materials
MECHANICAL PROPERTIES OF MATERIALS
LECTURER5 Fracture Brittle Fracture Ductile Fracture Fatigue Fracture
3 – Fracture of Materials
Chapter 2: MECHANICAL PROPERTIES OF MATERIALS
Normal Strain and Stress
Chapter 3 Mechanical Properties of Materials
MECHANICAL PROPERTIES OF MATERIALS
Chapter 7 Mechanical Properties of Solids.
CHAPTER 7: MECHANICAL PROPERTIES
CHAPTER 6: MECHANICAL PROPERTIES
EBB 220/3 PRINCIPLE OF VISCO-ELASTICITY
Lecture 26: Mechanical Properties I: Metals & Ceramics
ENGR 225 Section
MECHANICAL PROPERTIES OF MATERIALS
Mechanical Properties of Metals
CHAPTER OBJECTIVES Show relationship of stress and strain using experimental methods to determine stress-strain diagram of a specific material Discuss.
Elasticity and Strength of Materials
Mechanical Properties
CHAPTER 6: MECHANICAL PROPERTIES
ENGR-45_Lec-30_Polymer-Apps.ppt 1 Bruce Mayer, PE Engineering-45: Materials of Engineering Bruce Mayer, PE Licensed Electrical.
STRENGTH OF MATERIALS John Parkinson ©.
Materials PHYA2. MATERIALS DENSITY, SPRINGS, STRESS AND STRAIN Topics 11, pp.162–173.
Manufacturing Processes
Mechanical Behavior, Testing and Manufacturing Properties of Materials
4 Mechanical Properties of Biomaterials CHAPTER 4.1 Introduction
Mechanical Properties of Materials
Mechanical Properties of Materials
Lecture 21 Creep ME 330 Engineering Materials Creep Curves Temperature and Stress Effects Creep Mechanisms Creep Resistance Read Chapter 15.
Group 2 presentation Q : stress and strain curve presentation.
STRUCTURES Young’s Modulus. Tests There are 4 tests that you can do to a material There are 4 tests that you can do to a material 1 tensile This is where.
Materials Science Chapter 8 Deformation and Fracture.
Materials Science Metals and alloys.
Chapter 4. Mechanical Testing: Tension Test and Other Basic Tests
The various engineering and true stress-strain properties obtainable from a tension test are summarized by the categorized listing of Table 1.1. Note that.
CHAPTER 6: MECHANICAL PROPERTIES
Mechanical Properties
CHAPTER OBJECTIVES Show relationship of stress and strain using experimental methods to determine stress-strain diagram of a specific material Discuss.
Heat Treatment (Annealing) of Cold-Worked Metals
Yield strength: the elongation of a mat'l
Stress and Strain – Axial Loading
Chapter 3 – Mechanical Properties of Material
Introduction We select materials for many components and applications by matching the properties of the material to the service condition required of the.
Engineering materials lecture #12
Dr. Omar S.M.J.Ali PhD Orthodontic
Mechanical Properties: Part One
MECHANICAL PROPERTIES OF MATERIALS
Chapter 3 Mechanical Properties of Materials
Manufacturing Systems
Poisons Ratio Poisons ratio = . w0 w Usually poisons ratio ranges from
Experiment #1 Tension Test
INTRODUCTION to ENGINEERING MATERIALS
True Stress True Strain Crystalline Processes During Deformation.
Introduction to Materials Science and Engineering
Mechanical Properties of Metals
Physical Properties of Rocks
Mechanical Properties: 1
CHAPTER 6: MECHANICAL PROPERTIES
Mechanical Properties of Metals
Elastic & Plastic behavior of Materials….(Contd)
Elastic & Plastic behavior of Materials
Mechanical properties of metals Stress and Strain in metals
PDT 153 Materials Structure And Properties
E =
Simple Stresses & Strain
LECTURER 2 Engineering and True Stress-Strain Diagrams
CHAPTER OBJECTIVES Show relationship of stress and strain using experimental methods to determine stress-strain diagram of a specific material Discuss.
Mechanical Properties Of Metals - I
CHE 333 Class 18 Fracture of Materials.
Presentation transcript:

Mechanical Behavior Stress vs. Strain Elastic deformation CC512 Chapter 6 Mechanical Behavior Stress vs. Strain Elastic deformation Plastic deformation Hardness Creep and stress relaxation Viscoelastic deformation Roman era stone arch bridge _ Wikipedia 6- 1

Tensile Test Load (P) – displacement (L) curve Gage length: the smallest and uniform area region - Initial area = A0 ; Initial length = L0 Aluminum 2014-T81 6- 2

Engineering Stress and Strain Curve 6- 3 Aluminum 2014-T81

Elastic Deformation δ F δ bonds stretch return to initial Linear- _ reversible 2. Small load F δ bonds stretch 1. Initial 3. Unload return to initial δ Linear- elastic Non-Linear- 6- 4 Callister & Rethwisch

Plastic Deformation F F δ plastic _ permanent 1. Initial 2. Small load 3. Unload planes still sheared F δ elastic + plastic bonds stretch & planes shear plastic F δ linear elastic plastic Callister & Rethwisch 6- 5

Important mechanical properties _ from tensile test Yield strength: Onset of plastic deformation Usual convention: 0.2% offset Resistance of metal to permanent deformation Young’s modulus: Graphical statement of Hooke’s law Resistance to elastic deformation, i.e., stiffness of materials Summary of important properties 1: Modulus of elasticity, E 2: Yield strength, Y.S. 3. Tensile strength, TS (or ultimate tensile strength, UTS) 4. Ductility = 100 x strain (at failure); elastic recovery at failure 5. Toughness = area of stress – strain curve 6- 6

Strain Hardening Strain hardening Phenomenon of increasing strength with increasing deformation Region: from yield point to ultimate tensile strength Strain hardening ability is given by strain hardening exponent, n. Drop of tensile strength after UTS is due to the occurrence of necking 6- 7

True Stress-Strain curve Strain hardening exponent, n Necking 6- 8

Tensile test data_ metals 6- 9

More important elastic properties Poisson’s ratio - Contraction (elastic) perpendicular to the extension Elastic deformation in pure shear loading 6- 10

Poisson’s ratio and Shear modulus 6- 11

Examples A 10 mm diameter bar of 1040 carbon steel is subjected to a tensile load of 50,000 N, Taking it beyond its yield point. Calculate the elastic recovery that would occur upon removal of the tensile load. Solution: The engineering stress σ = P/A0 σ = 5 x 104 N/[π x (5 x 10-3 m)2] = (50/25 π) x 109 [N/m2] = 6.366 x 108 Pa = 636.6 Mpa Recovery strain: Using Hooke’s law, ε = σ/ E = 636.6 Mpa/ (200 x 103) Mpa = 3.183 x 10-3. 2. A 10 mm diameter rod of 3003-H14 aluminum alloy is subjected to a 6 kN tensile load. Calculate the resulting rod diameter. Solution: σ = P/A0 = 6x103 N/ [π x (5 x 10-3 m)2] = (6/25 π) x 109 Pa = 76.39 Mpa (note: YS = 145 Mpa) ε = σ/ E = 76.39/(70 x 103) = 1.09 x 10-3. εd = -νε = - 0.33 x 1.09 x 10-3 = 3.597 x 10-4. εd = (d – d0)/d0 ; d = d0 + εd d0 = d0 (1 + εd ) = 10 (1 – 3.597 x 10-4) = 9.9964 mm 6- 12

Mechanical Properties of Ceramics Tensile test: brittle fracture occurs at a stress of only 280 Mpa Compression test: very high compressive strength, 2100 Mpa is observed Possible cause: presence of micro-cracks Griffith crack model Polycrystalline Al2O3 6- 13

Example: A glass plate contains an atomic-scale surface crack. (Take the crack length ~ diameter of O2- ion.) Given that the crack length is 1- μm long and the theoretical strength Of the defect-free glass is 7.0 Gpa, Calculate the breaking stress of the plate. Solution: 6- 14

Bending test and modulus of rupture (MOR) 6- 15

Mechanical Properties of Polymers Tensile stress – strain curves for a polyester _ effect of temperature Tension and compression stress – strain curves for a nylon 66 at 23C_ effect of relative humidity 6- 16

More data for various polymers 6- 17

Elastic Deformation Mechanism of elastic deformation - the stretching of atomic bonds in the immediate vicinity of equilibrium atom separation, a0 (F = 0) - F vs. a curve is nearly straight-line Example: The interatomic distance of Fe atoms along <111> Direction is 0.2480 nm. Under a tensile stress of 1,000 Mpa along <111> , the atomic separation distance increases to 0.2489 nm. Calculate the elsatic modulus along the <111> directions Solution: Elastic strain = (0.2489 – 0.2480)/ 0.2480 = 0.00363 Elastic modulus E = σ / ε = 1000/0.00363 = 275.5 Gpa. 6- 18

Plastic Deformation Theoretical critical shear stress Mechanical stress necessary to deform a perfect crystal, i.e., a force necessary to slide one plane of atoms over an adjacent plane Calculation indicate one order of magnitude less than the shear modulus G For a typical metal, i.e., Cu, the critical shear stress is well over 1000 Mpa. -Actual stress necessary to plastically deform pure Cu is order of 100 Mpa. -What is the basis of mechanical deformation which requires only a fraction of the theoretical strength? 6- 19

Role of dislocation in plastic deformation Motion of dislocation along a slip plane - Low stress alternative for plastically deforming a crystal Only a small shearing force needs to operate in the core region of dislocation to produce a step-by-step shear -Net effect leads to the same overall deformation as the high stress mechanism Motion of dislocation under the influence of shear stress; a net effect is an increment of plastic (permanent) deformation. 6- 20

Defect mechanism of slip _ a simple analogy, i.e., Goldie the caterpillar Goldie the caterpillar shows: -It is to difficult to slide along the ground in a perfect straight line (a) -Goldie “slips along” nicely by passing a dislocation along the length of her body (b) 6- 21

Slip System -Dislocation motion occurs in high atomic density planes because “slip distance” is short -Planes of high atomic density: Slip planes -Directions of high atomic density: Slip directions Ductility and slip systems: Slip systems in Al is 12_ ductile Slip systems in Mg is 3_ brittle 6- 22

Slip System (2) 6- 23

Resolved Shear Stress Resolved shear stress, τ: -Actual stress operating on the slip system (i.e., on the slip plane and in the slip direction) upon applying a simple tensile stress, σ. 6- 24

Example Zn single crystal is being pulled in tension, with the normal to its basal plane (0001) At 60º to the tensile axis and with the slip direction [11-20] at 40º to the tensile axis. What is the resolved shear stress, τ, acting in the slip direction when a tensile stress of 0.690 MPa is applied? (b) What tensile stress is necessary to reach the critical resolved shear stress, τc, of 0.94 MPa? Solution τ= σcosλcosφ = 0.690 x (cos40 x cos60) = 0.264 MPa σc = τc/ cosλcosφ = 0.94/ (cos40 x cos60) = 2.454 MPa 6- 25

Effect of dislocations on the mechanical properties -Cold working of metals : Metal becomes increasingly more difficult to deform due to the hindrance of dislocation motion by pre-existing dislocations -Solution hardening: yield stress increases due to an obstacle formed by impurity atoms -Annealing: annealing of dislocations at 1/2 to 1/3 Tm -Brittleness _ Ceramics : Burgers vectors are large; a few slip systems ; charged state of the ions; these make the dislocation-motion difficult Forest (bundle of) dislocation in a stainless steel 6- 26

Hardness -Hardness measures the resistance of materials to indentation -The resistance of the material to indentation is a qualitative indication of its strength -Empirical hardness numbers are calculated from appropriate formulas using indentation geometry measurements 6- 27

Hardness data Example Solution Brinell hardness measurement is made on a ductile Iron (100-70-03, air quenched) using a 10 mm- dia Sphere of tungsten carbide. A load of 3,000 kg produces a 3.91 mm- dia impression in the iron surface. Calculate the BHN of this alloy. (Note the correct unit for the Brinell equation: Load = kilogram; diameters = mm.) Solution 6- 28

Creep & Stress Relaxation At room temperature, the elastic strain is constant, independent of time. At high temperature (T > 1/3-1/2Tm), the elastic strain is no more constant, but depending on time; the strain gradually increases with time. Creep : Plastic deformation occurring at a high temperature under constant load over a long period. 6- 29

Three stages of Creep Primary stage Rapid increase of strain; Enhanced deformation mechanism -Possible mechanism_ dislocation climb due to thermally activated atom mobility 6- 30

Three stages of Creep (2) Secondary stage -Period of a constant strain rate - Creep rate, έ = constant (steady state creep rate) -Increased easy of slip is balanced by increasing resistance to slip due to dislocation build-up Tertiary stage -Strain rate increases due to an increase in true stress Another creep rate data : Creep rupture time - σ 6- 31

Creep in Ceramics and Polymers Polymers Ceramics: -Stress relaxation: decreasing stress with time under constant strain -The mechanism is a viscous flow Creep in Ceramics and Polymers Ceramics: -Major creep mechanism: grain boundary sliding Creep data of a Nylon 66 at 60C , 50% relative humidity 6- 32

Examples Example 6.10 In a laboratory creep test at 1000C, a steady sate creep rate = 5 x 10-1 % /h Creep mechanism = dislocation climb with Q = 200 kJ/mol. Predict the creep rate at 600C. Solution C = έexp(+Q/RT) = (5 x 10-1%/h) exp [2x103/(8.314 x 1273)] = 80.5 x 106 %/h The creep rate : έ = (80.5 x 106 %/h) exp[-2x103/(8.314 x 873)] = 8.68 x 10-5 %/h. Example 6.12 The relaxation time of a rubber band at 25C = 60 days Initial stress = 2 MPa; final stress = 1 MPa. How many days are required? t = τloge(2/1) = (60 days) x loge(2) = 41.6 days (b) Activation energy, Q = 30 kJ/mol, Calculate the relaxation time at 35C. Solution τ = Cexp(+Q/RT) τ (35)/ τ (25) = exp[(Q/R)(1/308 – 1/298)] τ (35) = 40.5 days. 6- 33

Viscoelastic Deformation Thermal expansion: Glass or Polymer Viscoelastic Deformation Glass transition temperature, Tg for Glasses and Polymers -Viscous (liquidlike) deformation occurs above Tg Viscous behavior of glasses (viscosity vs. T) is similar to the modulus of elasticity (flexural modulus or dynamic modulus) vs. T behavior Crystal Volume Viscosity of soda-lime-silica glass : lamp bulbs glass Thermoplastic polymer with 50% crystallinity 6- 34

Viscoelastic Deformation_ Polymers Viscoelastic deformation regions Rigid (below Tg) Reathery (near Tg) Rubbery (above Tg) Viscous (near Tm, decomposition) Effect of Crystallinity (100 %) -Remaining rigid up to decomposition (Tm) Effect of Cross-linking -Increasing rigidity with cross-linking Vulcanization of polyisoprene 6- 35

Viscoelastic Deformation_ Elastomer Elastomer (i.e., polyisoprene): -Rubbery plateau is pronounced near room temperature -Glass transition temperature is below room temperature -Showing a dramatic uncoiling behavior -Stress - strain curve_ a non-linear elasticity: elasitic modulus increases with strain -Low strain modulus: due to secondary bonding -High strain modulus: due to primary covalent bonding -Elastomeric deformation exhibits hysteresis. - Dynamic elastic modulus (torsional Pendulum) - DTUL (deflection point under load, 264 psi) is associated with Tg 6- 36