Lecture_12

Lecture_12 - NE 125 Lecture 12 NE Mechanical Behaviour...

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Unformatted text preview: NE 125: Lecture 12 NE Mechanical Behaviour: Elastic and Plastic Deformation (cont’d) Instructor: William K. O’Keefe, P.Eng. [email protected] “Office Hours”: Mondays 1:30 to 2:20 RCH 106 Teaching Assistant (TA): Hua (Leanne) Wei [email protected] Reminder: Mid term exam Tuesday February 12, 7 – 9 PM D. Aasen to W. Lee MC 4059 S. Legge to N. Zaver MC 4061 Week 5 Reading: Shackelford 6th Ed, Chapter 6; sections 6.4 - 6.6 Assignment 5 problem set due: 11 February 2008 NE 125: Introduction to Materials Science and Engineering Introduction Review : Stress vs Strain Engineering Stress (σ ) σ= P A0 Engineering Strain l −l 0 ε= l0 Illustrations from Shackelford “Introduction to Materials Science for Engineers 6th Ed., Used by permission NE 125: Introduction to Materials Science and Engineering Introduction Review : Stress vs Strain Tensile test yields several useful material properties: 1. 1. 1. 1. 1. Modulus of Elasticity (E) Yield Strength (Y.S.) Tensile Strength (T.S.) (Max. Stress) Ductility (% elongation at failure) Toughness = ∫ σdε 1. Resilience Resilience U r= ∫ σdε 0 εy Ur = modulus of resilience Illustrations from Shackelford “Introduction to Materials Sciende for Engineers 6th Ed., Used by permission NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem from Lecture 11 (Elastic Recovery) Tensile Properties: (From Mid-Term Exam 2007) Tensile (From The mechanical properties of a certain metal were investigated. For the metal tested (data in The table), calculate the elastic recovery for the specimen upon removal of the load of 59000 N. elastic ( 4890) N ∆y m= = = 479412 N mm ∆x ( 50.8102 − 50.8000 ) mm Therefore the slope when removing the load will be – 479412 Determine final elongation Estimate slope in initial region of data using first two data points ∆y ∆y 59000 N m= ⇒ ∆x = =− = −0.1231 mm ∆x m 479412 N mm The magnitude of the elastic strain that is The recovered is the elastic (strain) recovery recovered ∆l .1231 ε= = = 0.00242 l0 50.8 NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 1 Mechanical Properties of Materials Which of the following materials will: (a) Experience the greatest percent reduction in area and why? (b) Which is the strongest and why? (c) Which is the stiffest and why? Material Yield Strength [Mpa] Tensile Strength [Mpa] Strain at Fracture Fracture Strength [Mpa] Elastic Modulus [Mpa] A B C D E 310 340 100 120 415 550 700 850 Fractures before yielding 0.23 0.40 0.15 0.14 265 105 500 720 650 210 150 310 210 350 NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 1 Mechanical Properties of Materials (a) (b) (c) Material B is most ductile; highest strain at fracture Material D is strongest; highest yield strength and tensile strength Material E is stiffest; has highest modulus of elasticity (E) Yield Strength [Mpa] Tensile Strength [Mpa] Strain at Fracture Fracture Strength [Mpa] Elastic Modulus [Mpa] Material A B C D E 310 340 100 120 415 550 700 850 Fractures before yielding 0.23 0.40 0.15 0.14 265 105 500 720 650 210 150 310 210 350 For engineering design purposes, if a strong material is needed, which is more For important, a high yield strength or a high tensile strength? important, NE 125: Introduction to Materials Science and Engineering Introduction Stress vs Strain Design stress and Safe Working Stress For engineering design purposes, the yield strength limits the utility of the material. the Elastic deformation is non-permanent. The capacity of a material to deform elastically when exposed to loads (eg. metals) is a useful property as a structural material. metals) Once plastic deformation has occurred, (i.e. yielding has Once occurred) the component often loses its utility (may not function properly as designed) function The design stress (σ d) is related to maximum expected The is stress (σ max) by a design factor F. A material selected for stress by design application must have a yield stress in excess of the design stress design σ d = Nσ max Alternatively, a safe working stress (σ w) is sometimes w) used expressed in terms of the yield strength and a factor of safety (F) factor σ YS σ w= F Illustrations from Shackelford “Introduction to Materials Sciende for Engineers 6th Ed., Used by permission NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 2 Safe working stress A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 13,300 N. Determine the minimum required wire diameter assuming a factor of safety of 2. The steel has a yield strength of 860 MPa. yield NE 125: Introduction to Materials Science and Engineering Introduction Mechanism of Elastic Deformation F attraction Elastic strain is due to stretching of interatomic bonds interatomic The modulus of elasticity is a measure of the resistance of adjacent atoms to being pulled apart apart FN 0 r E dF (r =r 0 ) dr repulsion σ= F = Eε A ⇒ F ∝ε ⇒ FPx F = −kx (Hooke’s Law) NE 125: Introduction to Materials Science and Engineering Introduction Mechanism of Elastic Deformation F attraction E dF (r = r 0 ) dr Elastic strain is due to stretching of interatomic bonds Elastic The modulus of elasticity is a measure of the resistance of adjacent atoms to being pulled apart E is higher for metals with primary bonding compared to polymers (Van der Waals interactions) polymers FN 0 r Ti Cu HDPE LDPE repulsion E 107 GPa 110 GPa 830 MPa 170 MPa How will E be affected by How temperature? temperature? Metallic bonding Metallic (primary bonding) Intermolecular bonding in polyethylene1 (Secondary bonding) [1] From Shackelford “Introduction to Materials Science for Engineers” 6th Ed.; Used by permission NE 125: Introduction to Materials Science and Engineering Introduction Mechanism of Elastic Deformation E dF (r = r 0 ) dr Elastic strain is due to stretching of interatomic bonds Elastic The modulus of elasticity is a measure of the resistance of adjacent atoms to being pulled apart Interatomic bonds become weaker as the internal energy (kT) of the solid increases; (eg. lattice vibrations) vibrations) Towards the melting point of the material, the modulus of Elasticity diminishes abruptly modulus E T General trend of E(T) for metals; Tmp note we will deal with thermal behaviour in greater detail in Chapter 7 and polymer properties in Ch. 13 NE 125: Introduction to Materials Science and Engineering Introduction Review: Edge and Screw Dislocations Linear Defects: Edge Dislocation Screw Dislocation Screw Red connects lattice in planes above dislocation line, blue connects lattice below dislocation line; when viewed down dislocation line, atoms take helical path hence the name “screw” dislocation Dislocations: Dislocations: Lattice distortions are centered about a linear defect. The magnitude of the lattice strain diminishes with lattice distance from the dislocation line distance [1] Figure from Shackelford Introduction to Materials Science for Engineers 6th Ed.; Used by permission NE 125: Introduction to Materials Science and Engineering Introduction Plastic Deformation and Dislocations Edge Dislocation Plastic deformation is permanent Plastic deformation caused by the movement of dislocations (slip) from applied shear stress stress The number of plastic dislocations The increases significantly during plastic deformation deformation Dislocation Density – the total dislocation length per unit volume (1/mm2) length th NE 125: Introduction to Materials Science and Engineering Introduction Plastic Deformation and Dislocations Edge Dislocation Plastic deformation results from the movement of planes of atoms as a result of applied Plastic stress on the material. Interatomic bonds are broken and re-formed as atoms change position position Plastic deformation is permanent deformation caused by the the movement of dislocations (slip) due to an applied shear stress stress [1] From Shackelford “Introduction to Materials Science for Engineers” 6th Ed.; Used by permission NE 125: Introduction to Materials Science and Engineering Introduction Plastic Deformation and Dislocations In an edge dislocation, the direction of motion of the dislocation is parallel to the shear stress In whereas in a screw dislocation, the direction of motion is perpendicular to the shear stress whereas Direction of motion Direction of motion Screw dislocation Edge dislocation NE 125: Introduction to Materials Science and Engineering Introduction Plastic Deformation and Dislocations Lattice Strain Most of the energy associated with the shear work inducing the dislocation is dissipated as heat. However, Most some energy is stored in the lattice as strain energy some Atoms in compression relative to Atoms ideal crystal lattice (compressive strain field) strain All metals and alloys have some Inherent dislocations and strain fields due to thermal stresses from cooling during casting processes Strain fields are important in the strengthening of materials Slip plane Atoms experience tension relative Atoms to ideal crystal (tensile strain field) field) [1] graphic obtained from Shackelford “Introduction to Materials Science for Engineers”, 6th Edition; used by permission NE 125: Introduction to Materials Science and Engineering Introduction Plastic Deformation and Dislocations Slip, Slip Planes and Slip Systems Slip – plastic deformation arising from the motion of dislocations due to an applied external Slip shear stress shear The energy of dislocation depends on the crystallographic direction The most probable crystallographic direction of a dislocation, is the direction The of highest atomic density (i.e. the slip direction) (i.e. P2 Eb= b Similarly, the energy of dislocation depends on the crystallographic plane. Specifically, the plane with the greatest planar density is favoured for slip. The plane resulting in the lowest dislocation energy (most probable) is known as the slip plane slip The slip system is the combination of the slip plane and the slip direction The slip NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 3 Sample Slip Systems Slip Which of the following is the slip system for the simple cubic structure? Explain {100} 110 {110} 110 {100} 010 {110} 111 NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 3 Sample Slip Systems Slip Which of the following is the slip system for the simple cubic structure? Explain {100} 110 {110} 110 {100} 010 {110} 111 [010] [100] NE 125: Introduction to Materials Science and Engineering Introduction Burgers Vectors for Slip Process Burgers Slip Systems Slip We can define a Burgers vector for the slip processes for the various We crystal structures, in terms of the lattice parameters crystal a b ( fcc) = 110 2 a b (bcc) = 111 2 a b (hcp) = 11 20 3 a Eg. fcc Eg. b d (111) c f a bc def (111) plane Note that other slip systems are possible at elevated temperature for these crystal structures Note NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 4 Sample Slip systems Shackelford 6-37 Shackelford Identify the 12 individual slip systems for the alternate slip system for bcc metals { 211} 11 1 NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 4 Sample Slip systems Shackelford 6-37 Shackelford Identify the 12 individual slip systems for the alternate slip system for bcc metals { 211} 11 1 ( 211) (121) (112) ( 211) (121) (112 ) First identify the 12 non-parallel planes: ( 2 1 1) ( 1 21) ( 1 12) ( 21 1 ) (12 1 ) (1 1 2) The slip directions which pair with these planes to complete the slip systems are normal to the slip planes and will give zero dot products with the normal vectors: ( 211) [ 1 11] (121) [1 1 1] (112) [11 1 ] ( 211)[111] (121)[111] (112 )[111] ( 2 1 1)[11 1 ] ( 1 21)[11 1 ] ( 1 12)[1 1 1] ( 21 1 )[1 1 1] (12 1 )[ 1 11] (1 1 2)[ 1 11] NE 125: Introduction to Materials Science and Engineering Introduction Slip in Single Crystals Slip Resolved Shear Stress Resolved Components of shear stress exist in all directions Components except for λ or φ = 90° Magnitude of resolved shear stress depends on applied tensile stress and orientation of slip plane and slip direction and Slip occurs when the resolved stress exceeds a Slip critical value (the critical resolved shear stress ) critical λ Normal to plane φ Slip direction Schmid’s Law: τ R= σ cos φ cos λ (resolved shear stress) NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 5 Sample Slip systems (From Assignment #4) Slip Shackelford 6-27 A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction. If the applied stress is 0.5 MPa, what will be the resolved shear stress along the [101] direction within the 11 1 plane? ( ) τ R= σ cos φ cos λ (resolved shear stress) NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 5 Sample Slip systems (From Assignment #4) Slip Shackelford 6-27 A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction. If the applied stress is 0.5 MPa, what will be the resolved shear stress along the [101] direction within the 11 1 plane? ( ) τ R= σ cos φ cos λ The direction of the applied load is [111] The slip plane is The (resolved shear stress) φ is the angle between the normal to the slip plane and the direction of the applied load is λ Is the angle between the slip direction and the direction of the applied load Is (11 1 ) [11 1 ] [ ] The vector normal to the slip plane is The The slip direction is [101] (we are asked to determine component of resolved shear stress in this direction) The Therefore, φ is the angle between the vectors [111] and 11 1 λ is the angle between the vectors [101] and [111] is NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 5 Sample Slip systems (From Assignment #4) Slip Shackelford 6-27 A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction. If the applied stress is 0.5 MPa, what will be the resolved shear stress along the [101] direction within the 11 1 plane? τ R= σ cos φ cos λ ( ) (resolved shear stress) From Appendix 1, Al has FCC structure Since Al has a cubic structure, we can use rules of Euclidean Geometry (i.e. Vector dot Product) Determine φ The dot product of the vectors gives [111] • The dot product also = |a| |b| cos φ [11 1 ] = 12 +12 -12 = 1 a = 12 + 12 + 12 = 3 ∴ cos φ = 1 1 = 33 3 b = 12 + 12 + ( − 1) = 3 2 NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 5 Sample Slip systems (From Assignment #4) Slip Shackelford 6-27 A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction. If the applied stress is 0.5 MPa, what will be the resolved shear stress along the [101] direction within the 11 1 plane? τ R= σ cos φ cos λ ( ) (resolved shear stress) From Appendix 1, Al has FCC structure Since Al has a cubic structure, we can use rules of Euclidean Geometry (i.e. Vector dot Product) Determine λ The dot product of the vectors gives [111] • [101] The dot product also = |a| |b| cos λ = 12 +02 +12 = 2 a = 12 + 12 + 12 = 3 ∴ cos λ = 2 32 b = 12 + 0 2 + 12 = 2 NE 125: Introduction to Materials Science and Engineering Introduction Sample Problem 5 Sample Slip systems (From Assignment #4) Slip Shackelford 6-27 A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction. If the applied stress is 0.5 MPa, what will be the resolved shear stress along the [101] direction within the 11 1 plane? ( ) τ R= σ cos φ cos λ (resolved shear stress) 1 2 = ( 0.5MPa ) = 0.136 3 2 3 (MPa) ...
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