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