Introduction to Engineering Materials ENGR2000 Chapter 7: Dislocations and Strengthening Mechanisms Dr. Coates
An edge dislocation moves in response to an applied shear stress dislocation motion
7.1 Introduction What is plastic deformation on a microscopic scale? Net movement of atoms in response to an applied stress Inter-atomic bonds are broken & reformed Motion of dislocations
7.2 Basic Concepts Slip the process by which plastic deformation is produced by dislocation motion. Slip plane The crystallographic plane along which the dislocation line moves.
An edge dislocation moves in response to an applied shear stress Applied stress is perpendicular to dislocation line Dislocation motion is parallel to applied stress Macroscopic plastic deformation results
An edge dislocation moves in response to an applied shear stress
A screw dislocation moves in response to an applied shear stress Dislocation motion is perpendicular to applied stress Macroscopic plastic deformation results
For a mixed dislocation, would the direction of motion be parallel or perpendicular if subjected to an applied shear stress?
Dislocation Density dislocatio n density Metals crystals totaldislocation length per unit volume 10 3 mm -2 Ceramic crystals 10 2 mm -2-10 4 mm -2 Silicon single crystals 0.1-1 mm -2 All metals and alloys contain some dislocations. What metal fabrication processes might cause this? Solidification, rapid cooling, plastic deformation
7.3 Characteristics of Dislocations When metals are plastically deformed ~5 % of deformation energy is retained as internal energy (stored strain energy associated with dislocations) Remainder is dissipated as heat energy
Lattice strains - due to an edge dislocation
Two edge dislocations of the same sign and lying on the same slip plane Pushed out! compression Pulled in! tension
Two edge dislocations of opposite sign and lying on the same slip plane
Dislocation density and Plastic Deformation During plastic deformation, dislocation density increases significantly ~ 10 10 mm -2 Existing dislocations multiply Grain boundaries, internal defects, surface irregularities (scratches, etc.) act as stress concentrations serve as dislocation formation sites
Slip plane 7.4 Slip Systems Preferred plane on which dislocation motion occurs. Plane with highest planar density. Slip direction Preferred directions on the slip plane along which dislocation motion occurs. Direction with highest linear density. Slip system = slip plane + slip direction
slipdirections: 110 Slip system in a FCC crystal planes: close packed family of 111
Slip system in a BCC crystal close packed family of slip directions : 111 # of slip systems 12 planes : 110
What are the planar densities along the slip planes for the FCC crystal structure?
PD 111 FCC # of atoms centered on a plane area of plane 1 1 3 3 2 6 1 4R 2R 3 2 4R 2 2 3 0.29 R 2
Compare the planar densities of a (111) plane with the (110) plane in an FCC crystal structure PD 110 # of atoms centered on a FCC area of plane 2 0.177 2 2 8R 2 R plane
Metals with FCC or BCC structures are ductile while those with HCP structures are normally brittle why?
7.5 Slip in Single Crystals Resolved shear stress: - shear stressdue to applied stress - shear stressalong the slip direction on the slip plane R cos cos where is the applied stress. In general, 90 0.
One slip system is generally oriented most favorably for slip (largest resolved shear stress) R cos cos max max Slip in a single crystal commences most favorably when the resolved shear stress reaches some critical value R crss max crss represents the minimum shear stress required to initiate slip and is a material property that determines when yielding occurs y crss cos cos max The minimum applied stress to introduce yielding occurs when j 45 0 for a single crystal stressed in tension
Resolved shear stress and Critical resolved shear stress Resolved shear stress is the shear component of an applied tensile or compressive stress resolved along a slip plane that is other than perpendicular or parallel to the applied stress axis. The critical resolved shear stress is the value of resolved shear stress at which yielding begins; it is a material property.
Slip in a crystal along a number of equivalent slip systems Each step results from the movement of a large number of dislocations along the same plane!
Polished single crystal: Slip lines
Example 7.1 Consider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction. (a) Compute the resolved shear stress along a (110) plane and in a 111 direction when a tensile stress of 52 MPa is applied.
Example Given : direction of slip plane: R cos cos R slip direction : 52 MPa? Solution : direction of 7.1(a) 110 111 110. 010 110 010 111. 010 111 010 cos cos applied stress: normal 1 2 1 3 010 21.3MPa toslip plane: 110
Example 7.1(b) b) If slip occurs on a (110) plane and in a 111 direction, and the critical resolved shear stress is 30 MPa, calculate the magnitude of the applied tensile stress necessary to initiate yielding.
Example Given : direction of slip plane: crss ys cos ys slip direction : Solution : direction of cos? 110. 010 110 010 111. 010 111 010 crss 7.1(b) / applied stress: 110 30 MPa 111 normal 1 2 1 3 010 cos cos 73.4 MPa toslip plane: 110
How does polycrystalline material compare single crystal in terms of strength? Why? Slip lines on surface of a polycrystalline specimen of copper that was polished and then deformed
Equiaxed Elongated Alteration of the grain structure of polycrystalline material as a result of plastic deformation
Deformation by Twinning Displacement magnitude distance from twin plane Will occur on a definite plane and in a specific direction that depend on crystal structure
Single crystal subjected to a shear stress (a) deformation by slip (b) deformation by twinning (a) Crystallographic orientation same above and below plane before/after deformation, distinct atomic spaces (b) Crystallographic reorientation across twin plane, < interatomic separation Mechanical twinning occurs in BCC and HCP at low temperatures and at high rates of loading, twinning may introduce new slip systems!
Mechanisms of Strengthening in Metals The ability of a metal to plastically deform depends on the ability of dislocations to move. Restricting dislocation motion renders a material harder and stronger.
What do you think happens to a dislocation when it encounters a grain boundary? Direction change slip plane discontinuity Would the phenomena be different at a high angle grain vs low angle grain boundary?
Would a fine grained material be stronger or weaker than a course grained material? Why? Hall - Petch equation : where d isthe average grain diameter, ys 0 and 0 k k y y d 1/ 2 are constants (material). Not valid for coarse grain material or extremely fine grain polycrystalline material From graph, what direction is grain size increasing? Is grain size linearly changing with yield strength?
Impurity If there are impurity atoms that go into either substitutional or interstitial solid solutions, do you think this would strengthen or weaken the solid solution? Why?
Impurity atoms impose lattice strains on surrounding host atoms -restricts dislocation movement (a)tensile strains (left), compressive strains(right) (b) partial cancellation of dislocation strains->restricts dislocation movement
7.10 Strain Hardening A ductile material becomes harder and stronger as it is plastically deformed. Work hardening Cold working Temperature at which deformation occurs is cold relative to the melting temperature of the metal The effects of strain hardening may be removed by an annealing heat treatment.
Percent cold work Percent cold work : % CW where A and A d 0 A 0 A A 0 d 100 is the original area of cross -section is the area after deformation.
Strain hardening - on a stress-strain curve
From the onset of tothepoint of K n t t where n is and K is Strain-hardening exponent - measure of the ability of a metal to strain harden plasticdeformation necking : the strain - hardening the strength coefficien t. exponent
Example 7.2 Use graphs below to determine the tensile strength and ductility (%EL) of a cylindrical copper rod if it is cold worked such that the diameter is reduced from 15.2 mm to 12.2 mm.
Example 7.2 Percent cold work : % CW 2 0 From figure 7.17 : uts d 340 MPa % EL 7% d d 2 0 2 d 100 35.6%
Recovery, Recrystallization, and Grain Growth Plastic deformation of a material produces Change in grain shape Strain hardening Increase in dislocation density Change in material properties
Recovery, Recrystallization, and Grain Growth These may be reverted back by appropriate heat treatment. Restoration processes Recovery Recrystallization Grain growth
7.11 Recovery High temperature process Enhanced atomic diffusion Dislocation motion Stored internal energy is relieved Reduction in number of dislocations Physical properties are recovered to their precoldworked states
7.12 Recrystallization High temperature process Formation of a new set of strain-free grains Low dislocation density
Heat treatment time = 1 h
Recrystallization Temperature - temperature at which recrystallization reaches completion in 1 h.
Recovery and Recrystallization During recovery, there is relief of internal strain energy by dislocation motion; there are virtually no changes in the grain structure. During recrystallization, a new set of strain-free grains forms and the material becomes softer and more ductile.
7.13 Grain Growth By the migration of grain boundaries Via diffusion of atoms from one side of the boundary to the other Directions of boundary movement and atomic motion are opposite to each other Larger grains grow at the expense of smaller ones
Grain growth proceeds more rapidly as temperature increases why?
Design Example 7.1 A cylindrical rod of non cold-worked brass having an initial diameter of 6.4 mm is to be cold worked by drawing such that the cross-sectional area is reduced. It is required to have a cold-worked yield strength of at least 345 MPa and a ductility in excess of 20 % EL; in addition, a final diameter of 5.1 mm is necessary. Describe a manner in which this procedure may be carried out.
Design Example 7.1 Given : - brass rod d d 0 f 6.4 mm 5.1mm % EL 20% ys 345 MPa
Design Example 7.1 Percent cold work : % CW From figure : ys % EL A 410 MPa 8% - not satisfactory 0 A 0 A f 100 36.5%
Percent cold work required for ys 345MPa: % CW ~ 20% Percent cold work required for %EL 20% : % CW ~ 23% Using a %CW 21.5% d ' 0 A ' 0 A A 5.8mm f f 21.5% : 100 - initial diameter of the second drawing - final diameter of the first drawing