CHAPTER 4 1/1/2016. Mechanical Properties of Metals - I. Processing of Metals - Casting. Hot Rolling of Steel. Casting (Cont..)

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1 Processing of Metals - Casting CHAPTER 4 Mechanical Properties of Metals - I Most metals are first melted in a furnace. Alloying is done if required. Large ingots are then cast. Sheets and plates are then produced from ingots by rolling Wrought alloy products. Channels and other shapes are produced by extrusion. Some small parts can be cast as final product. Example :- Automobile Piston. 1 2 Casting (Cont..) Hot Rolling of Steel Casting Process Casting mold Cast parts Hot rolling Greater reduction of thick nee in a single pass. Rolling carried out at above recrystallization temperature. Ingots preheated to about 12 C. Ingots reheated between passes if required. Usually, series of 4 high rolling mills are used. 3 4 Cold Rolling of Metal Sheet Cold Rolling (Cont..) 5 Cold rolling is rolling performed below recrystallization temperature. This results in strain hardening. Hot rolled slabs have to be annealed before cold rolling. Series of 4 high rolling mills are usually used. Less reduction of thickness. Needs high power. 6 Initial metal thickness inal metal thickness % Cold work = Initial metal thickness x 1 1

2 Extrusion orging 7 Metal under high pressure is forced through opening in a die. Common products are cylindrical bar, hollow tubes from copper, aluminum etc. Normally done at high temperature. Indirect extrusion needs less power however has limit on load applied. Die Direct Extrusion Indirect Extrusion Metal Container Container Metal 8 Metal, usually hot, is hammered or pressed into desired shape. Types:- Open die: Dies are flat and simple in shape * Example products: Steel shafts Closed die: Dies have upper and lower impression * Example products: Automobile engine connection rod. orging increases structural properties, removes porosity and increases homogeneity. Direct orging Metal Indirect orging Die Drawing Stress and Strain in Metals Wire drawing :- Starting rod or wire is drawn through several drawing dies to reduce diameter. % cold work = Change in cross-sectional area X 1 Original area Deep drawing:- Used to shape cup like articles from flats and sheets of metals. Wire or rod Carbide nib Metals undergo deformation under uniaxial tensile force. Elastic deformation: Metal returns to its original dimension after tensile force is removed. Plastic deformation: Metal is deformed to such an extent such that it cannot return to its original dimension. 9 1 Engineering Stress and Strain Poisson s Ratio A Engineering stress σ = A Δl (Average uniaxial tensile force) A (Original cross-sectional area) Units of Stress are PSI or N/m 2 (Pascals) 1 PSI = 6.89 x 1 3 Pa Engineering strain = ε = Change in length Original length Units of strain are in/in or m/m.. w Poisons ratio = ( lateral ) ( longitudinal) w w w Usually poisons ratio ranges from.25 to.4. Example: Stainless steel.28 Copper.33 y z

3 Shear Stress and Shear Strain S (Shear force) τ = Shear stress = A (Area of shear force application) Tensile test Strength of materials can be tested by pulling the metal to failure. Load cell 13 Amount of shear displacement Shear strain γ = Distance h over which shear acts. Elastic Modulus G = τ / γ 14 Extensometer orce data is obtained from Load cell Strain data is obtained from Extensometer Specimen Tensile Test (Cont) Mechanical Properties Modulus of elasticity (E) : Stress and strain are linearly related in elastic region. (Hooke s law) Commonly used test specimen Typical stress-strain curve E = σ (Stress) ε (Strain) Stress Higher the bonding strength, higher is the modulus of elasticity. Strain Linear portion of the stress strain curve Examples: Modulus of elasticity of steel is 27 GPa Modulus of elasticity of aluminum is 76 GPa Δε Δσ E = Δσ Δε Yield Strength Ultimate tensile strength Yield strength is strength at which metal or alloy show significant amount of plastic deformation. Ultimate tensile strength (UTS) is the maximum strength reached by the engineering stress strain curve. Necking starts after UTS is reached. Al 224-Tempered 17.2% offset yield strength is that strength at which.2% plastic deformation takes place. Construction line, starting at.2% strain and parallel to elastic region is drawn to.2% offset yield strength. 18 More ductile the metal is, more is the necking before failure. Stress increases till failure. Drop in stress strain curve is due to stress calculation based on original area. S t r e s s (MPa) Necking Point Al 224-Annealed Strain Stress strain curves of Al 224 with two different heat treatments. Ductile annealed sample necks more 3

4 Percent Elongation Percent elongation is a measure of ductility of a material. It is the elongation of the metal before fracture expressed as percentage of original length. Percent Reduction in Area Percent reduction area is also a measure of ductility. The diameter of fractured end of specimen is measured using caliper. % Elongation = inal length* initial Length* Initial Length % Reduction Initial area inal area = Area inal area Measured using a caliper fitting the fractured metal together. Example:- Percent elongation of pure aluminum is 35% or 776-T6 aluminum alloy it is 11% Percent reduction in area in metals decreases in case of presence of porosity Stress-strain curves of different metals True Stress True Strain Tensile Test 118 Steel (Low Carbon) True stress and true strain are based upon instantaneous cross-sectional area and length. True stress = σ t = A i (instantaneous area) 21 True strain = ε t = i d li A Ln Ln l A True stress is always greater than engineering stress. i 22 Tensile strength = 44 MPa Modulus of elasticity = 25 GPa Reduction in area = 4% Elongation = 15% Tensile Test 145 Steel (High Carbon) Hardness and Hardness Testing Hardness is a measure of the resistance of a metal to permanent (plastic) deformation. General procedure: Press the indenter that is harder than the metal into metal surface. Withdraw the indenter 23 Tensile strength = 696 MPa Modulus of elasticity = 27 GPa Reduction in area = 4% Elongation = 1% 24 Measure hardness by measuring depth or width of indentation. Rockwell hardness tester 4

5 Hardness Tests Hardness: Virtual Lab Module Click below to view the virtual lab module related to Rockwell hardness tester and scales Hardness: Virtual Lab Module Hardness: Virtual Lab Module Click below to view the virtual lab module related to Rockwell hardness testing procedure: Inserting indenter and adding weights. Click below to view the virtual lab module related to Rockwell hardness testing procedure: Applying loads Hardness: Virtual Lab Module Plastic Deformation in Single Crystals Click below to view the virtual lab module related to Rockwell hardness testing: Interactive simulation. Plastic deformation of single crystal results in step markings on surface slip bands. Atoms on specific crystallographic planes (slip planes) slip to cause lip bands. Slip bands

6 Slip Bands and Slip Planes Slip bands in ductile metals are uniform (occurs in many slip planes). Slip occurs in many slip planes within slip bands. Slip planes are about 2A thick and are offset by about 2A Slip Mechanism During shear, atoms do not slide over each other. The slip occurs due to movement of dislocations. Wall of high dislocation density igure 5.32 Dislocation cell structure in lightly deformed aluminum Slip in Crystals Slip Systems Slip occurs in densely or close packed planes. Lower shear stress is required for slip to occur in densely packed planes. If slip is restricted in close planes, then less dense planes become operative. Less energy is required to move atoms along denser planes. Close packed plane Slip systems are combination of slip planes and slip direction. Each crystal has a number of characteristic slip systems. In CC crystal, slip takes place in [111] octahedral planes and (11) directions. 33 Non-close-packed plane 34 4 [111] type planes and 3 (11) type directions. 4 x 3 = 12 slip systems. Slip Systems in BCC Crystal Critical Resolved Shear Stress 35 Slip planes and directions for CC BCC crystals are not close packed. The slip predominantly occurs in [11] planes that has highest atomic density. If HCP crystals have high c/a ratio, slip occurs along basal planes [1]. or crystals with low c/a ratio, slip also occurs in [11] and [111] planes. 36 Critical resolved shear stress is the stress required to cause slip in pure metal single crystal. Depends upon: Crystal Structure Atomic bonding characteristics Temperature Orientation of slip planes relative to shear stress Slip begins when shear stress in slip plane in slip direction reaches critical resolved shear stress. This is equivalent to yield stress. Example :- Zn HCP % pure.18 MPa Ti HCP 99.99% pure 13.7 MPa Ti HCP 99.9% pure 9.1 MPa 6

7 Schmid s Law Twinning 37 The relationship between uniaxial stress action on a single cylinder of pure metal single crystal and resulting resolved shear stress produced on a slip system is given by τ r = Shear orce Shear Area r. Cos Cos. Cos A A / Cos A 1. Cos. Cos Normal to Slip plane A 1 =Area of Slip plane A Slip direction r r A 1 38 In twinning, a part of atomic lattice is deformed and forms mirror image of lattice next to it. Distance moved by atoms is proportional to their distance from twinning plane. Deformation from twinning is small. Twinning reorient the slip system. Twining if important was of deformation in HCP crystals due to lesser slip planes. Effects of Grain Boundaries on Strength Hall-Petch Equation Grain boundaries stop dislocation movement and hence strengthen the metals. ine grain size is desirable, and hence metals are produced with finer grains. iner the grains, superior are the mechanical properties (at room temperature). More isotropic properties Less resistant to corrosion and creep Hall-Petch equation - Empirical y = o + k / (d) 1/2 ; y = yield strength d = average grain diameter Stress-strain curve of single and polycrystalline copper 39 Slip bands in polycrystalline Aluminum grains Dislocations piled up against grain boundaries in stainless steel 4 o and k are constants for a metal. o 7 MPa and k =.74 MPa.m 1/2 for mild steel. Effects of Plastic Deformation Plastic deformation results in shearing of grains relative to each other. The grains elongate in rolling direction. Dislocations get rearranged. Grain structure at different regions of cartridge brass rolled into a wedge Effect of Cold Work on Tensile Strength Number of dislocations are increased by cold work. Dislocation movements are hindered by both grain boundaries and other dislocations Strain Hardening 118-Cold Rolled 118-Annealed Stress-Strain curves of 118 steel

8 Solid Solution Strengthening Addition of one or more metals can increase the strength of metals. Solute atoms, on case of substitutional solid solution, create stress fields around themselves and hinder the dislocation movement. Distortion of lattice and clustering of like atoms also impede dislocation movement. Example: Solid solution of 7 wt.% Cu & 3 wt.% Zn (cartridge brass) has tensile strength of 5 MPa. Tensile strength of unalloyed copper is 33 MPa. Recovery and Recrystallization Cold worked metals become brittle. Reheating, which increases ductility results in recovery, recrystallization and grain growth. This is called annealing and changes material properties (Adapted from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 2d ed., Wiley, 1976, p.228.) Structure of Cold Worked Metals Recrystallization Strain energy of cold work is stored as dislocations. Heating to recovery temperature relieves internal stresses (Recovery stage). Polygonization (formation of sub-grain structure) takes place. Dislocations are moved into lower energy configuration. Structure of 85% Cold worked metal TEM of 85% Cold worked metal Polyganization Dislocations Slip bands Grain Boundaries If metal is held at recrystallization temperature long enough, cold worked structure is completely replaced with recrystallized grain structure. Two mechanisms of recrystallization Expansion of nucleus Migration of grains Expansion More deformed region Migration Structure and TEM of Recrystallized metal 45 Structure of stress relived metal TEM of stress relived metal 46 igure 6.5 Nucleus of recrystallized grain 47 Effects on Mechanical Properties Annealing decreases tensile strength, increases ductility. Example: 5% cold rolled Tensile strength 52MPa(75 ksi) Ductility 3% 85% Cu & 15% Zn Annealed 1 4 C Tensile strength 31MPa(45 ksi) Ductility 38 % actors affecting recrystalization: Amount of prior deformation Temperature and time Initial grain size Composition of metal igure acts About Recrystallization A minimum amount of deformation is needed. Smaller the deformation, higher the recrystallization temperature. Higher the temperature, lesser is the time required. Greater the degree of deformation, smaller are the recrystallized grains. Larger the original grain size, greater amount of deformation is required to produce equivalent temperature. Recrystallization temperature Continuous annealing increases with purity of metals. 8

9 Superplasticity in Metals Mechanism of Superplasticity 49 At elevated temperature and slow loading, some alloys deform 2%. Annealed Ti alloy Elongates 12% at room temperature Elongates up to 117% at 87 o C and 1.3x1-4 /s loading rate. Conditions: very fine grain size (5-1 microns) * Highly strain sensitive * Temperature above.5 Tm * Slow strain rate 5 Very limited dislocation activity Deformation mechanism: Grain boundary sliding Grain boundary diffusion Sliding and rotation of individual grains. Applications: Metal forming operations. Blow forming to produce automobile hoods. Grains before and after deformation Nanocrystalline Metals Average grain diameter < 1 nm Results in high strength and hardness, and Superplasticity. If grain diameter reduces from 1 microns to 1 nm, yield strength of copper increases 31 times. Very difficult to produce nanocrystalline metals. If d < 5 nm, elastic modulus drops as more atoms are in grain boundary Hall-Petch equation is invalid in lower nanocrystalline range. Negative Hall-Petch effect might take place 51 9

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