STRENGTHENING MECHANISM IN METALS

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1 Background Knowledge Yield Strength STRENGTHENING MECHANISM IN METALS Metals yield when dislocations start to move (slip). Yield means permanently change shape. Slip Systems Slip plane: the plane on which deformation occurs, possess the highest atomic density. Slip direction: the direction within the slip plane and is always along a line of the highest atomic density Slip systems: a crystal deforms by motion of a dislocation on a slip plane and in a certain direction slip system = slip plane + slip direction

2 [110] Example: Slip systems in FCC Slip planes: {111} plane family in FCC possesses the [011] highest atomic planar density Slip directions: <110> direction family in FCC possesses the highest atomic density [101] {111}: eight octahedral planes in a cube, only 4 of them need to be considered (the other 4 are parallel planes). <110>: total six, but, only three lie in each of the {111} slip plane. Ex: (111) slip plane contains the [011], [101], &[110] Therefore, 4 {111}planes x 3 <110>directions = 1 slip systems

3 Macroscopic slips in a single crystal Slips in a zinc single crystal Deformation of polycrystals Slip occurs in well-defined crystallographic planes within each grain, but more than one slip plane is possible and likely. In different grains, the slip planes will have different orientations because of the random nature of the crystal orientations. Microscope photograph of actual shear offsets in different grains, on surface of a copper bar.

4 Plastic Deformation in Polycrystals Before, undeformed equiaxial grains The plastic deformation has produced elongated grains

5 Slip in Single Crystals: Resolved Shear Stress Angle λ: between F & slip direction Angle φ: between F & the normal direction of slip plane The resolved force in slip direction F s F s = F cos λ The area of the slip plane Resolved shear stress τ = A s = A/cos φ Metal single crystal a number of potential slip planes exists One generally orientated most favorably largest resolved shear stress τ = σ max F A s s = ( cosλ cosφ) max Slip occurs when (crss = critical resolved shear stress) F cosλ = σ cosλ cosφ A/ cosφ τ max = τ CRSS

6 Concomitant applied normal stress σ y = (cos τ crss φ cos λ ) max Minimum stress to introduce yielding occurs when Then λ = φ = σ = τ Example 1: y crss Given: Single Crystal BCC iron Tensile stress applied along [010] direction Required: Compute the resolved shear stress along the (110) plane and [ 111] direction when a tensile stress of 5 MPa (7,500 psi) is applied. If slip occurs on (110) plane and in a [ 111] direction, and resolved shear stress is 30 MPa (4,500 psi), calculate applied tensile stress to initiate yielding. 45 o

7 Solution: φ angle between (110) plane normal and the [010] direction is 45 0 ( a / a) From triangle ABC λ= tan -1 = τ R = σ cosλ cosφ = (5MPa)(cos 45)(cos 54.7) = 1.3 MPa (3,060psi) σ 30 MPa 73.4 MPa (10,600 psi) 0 0 (cos 45 )(cos54.7 ) y = =

8 Example Problem: A FCC crystal yields under a normal stress of MPa applied in the [13] direction. The slip plane is (111) & slip direction is [101]. Determine critical resolved shear stress. Solution: cosφ = cosλ = [ 1 ( 1) [ 1 ( 1) + + 3] ( ] [ ( 1) τ c = σ y cosλ cosφ = = MPa ) + 1 1] = = = = 0.756

9 Mechanisms of Strengthening The ability of a metal to plastically deform depends on the ability of dislocations to move Reducing or inhibiting mobility of dislocations enhances mechanical strength These can be used for increasing the material strength, but ductility may be lost. When Slip is Inhibited? Dislocation motion may be inhibited by: Other dislocations (entangling) Grain boundaries Point defect (solution hardening) Other phases - Precipitates Dislocation forest

10 Strengthening of Metals There are 4 major ways to strengthen metals, and all work because they make dislocation motion more difficult. They also reduce the ductility: 1)Cold work (Strain Hardening) )Reduce grain size (Strengthening by Grain Size Reduction) 3)Add other elements in solid solution (Solid Solution Strengthening) 4)Add second phase particles (Precipitation or Age Hardening) These mechanisms may be combined. For example, the world s strongest structural material (with some ductility) is steel piano wire. It combines all four strengthening mechanisms, and can have a yield strength of 500,000 psi. One wire, 0.1 in diameter, can hold up a 4,000 lb Ford Explorer.

11 STRAIN HARDENING Ductile material becomes harder and stronger as it is plastically deformed The dislocation density expressed as total number dislocation length per unit volume mm/mm 3 increases from 10 5 to 10 6 mm - for a heat treated metal to 10 9 to mm - for a heavily deformed metal. Dislocation strain field interactions Dislocation density increases with deformation or cold working Dislocations are positioned closer together On average, dislocation-dislocation strain fields are repulsive n ~ 0.5 (FCC) n ~ 0. (BCC) n ~ 0.05 (HCP) σ n T = K ε T n = strain hardening exponent measures the ability of a metal to harden τ flow = τ 0 + k ρ disl Where ρ disl : dislocation density

12 τ crss versus density Cold Working Cold working: plastic deformation of a metal or alloy at a temperature where dislocations are created faster than they are annihilated % CW = A A 0 d 0 A 100 Where, %CW: percent of cold work A 0 : original cross-sectional area A d : area after deformation

13 Influence of Cold Working on Mechanical Properties As the yield strength and the tensile strength increases with increasing amount of cold working, the ductility of the metal decreases.

14 Example: Given : Copper rod is cold worked such that its diameter is reduced from 15. mm to 1. mm. Determine its tensile strength and ductility CW π π = x π % From Figure TS vs CW TS = 340 MPa From Figure ductility vs CW % EL = 7 % = 35.6%

15 Effects of plastically deformed polycrystalline metal at temperature less than T m : Change in grain shape Strain hardening Increased dislocation density Stored energy When metals are plastically deformed about 5% of deformation energy is retained internally associated with dislocations. The properties of the cold worked metal (partially or totally) can be restored by: Recovery and/or Recrystallization and Grain Growth Recovery Some of the stored internal strain energy is relieved by virtue of dislocation motion as a result of enhanced atomic diffusion at elevated temperature. Effects of recovery in cold worked metals: Ductility increases Yield and tensile strength decreases slightly Hardness decreases slightly. Metal toughness increases. Electrical and thermal conductivity of the metal is recovered to their precoldworked states. There is no apparent change in the microstructure of the deformed material.

16 Recrystallization Cold worked material high dislocation density lot of stored energy very strong not very ductile process Recrystallized material low dislocation density no stored energy weak ductile After recovery grains remain at relatively high energy states Recrystallization formation of a new set of strain-free and equiaxed grains, low dislocation densities Driving force difference in internal energy between strained and unstrained material New grains form as small nuclei grow and replace parent material short range diffusion The process is a heat treating process called annealing. Annealing requires high temperature.

17 33%CW Brass 4s at 580 o C 8s at 580 o C Grain Growth after 15 min; and after 10min at 700 o C

18 Recrystallization Temperature Temperature at which recrystallization just reaches completion in one hour. 450 o C for the above example. Typically between 0.5 to 0.33 the melting point of the metal. Depends on the amount of cold work and of the impurity level of the alloy. There is a critical degree of cold work below which recrystallization can not be made to occur.

19 Notes on Recrystallization: The amount of cold work controls the initial recrystallized grain size. More cold work more stored energy easier nucleation more nucleation sites smaller grain size. The temperature and time of annealing controls the final grain size, if there is substantial growth after recrystallization. Grain growth requires diffusion, and diffusion is faster at higher temperatures. The time at temperature controls the total amount of diffusion.

20 A fine grain size has many benefits beyond strength. In general, finer grain sizes are more resistant to fatigue and fracture failures, and have more reproducible and homogeneous mechanical properties. Finally, in general, metals with fine grain size are also more easily formed in metalworking operations than metals with coarse grain sizes. Grain Growth Grains continue to grow following recrystallization at elevated temperatures Energy is reduced as grains grow in size As large grains grow small grains shrink Boundary motion short-range diffusion of atoms from one side of the boundary to the other. At a constant temperature d n d n = Kt 0 d 0 initial grain diameter at time (t) = 0 K and n are time independent constants n is generally ~

21 STRENGTHENING BY GRAIN SIZE REDUCTION Dislocations cannot penetrate grain boundaries, because the crystal planes are discontinuous at the grain boundaries. Therefore, making a smaller grain size increases strength (more obstacles and shorter mean slip distance.)

22 This can be quantified by the so-called Hall-Petch Equation: where σ y is the yield strength, d is the grain size, and σ o and k y are material constants. The increases in strength at very small grain sizes can be enormous. One are of current research is on so-called nanostructured metals, which have grain sizes from 0 to 00 nm. They can have very high strength. Influence of grain size on yield strength (brass) Strength triples as grain size goes from 100 μm to 5 μm. 100 μm 5 μm As d, σ ys and the ductility or or it is constant

23 SOLID SOLUTION STRENGTHENING Impurity atoms that go into solid solution impose lattice strains on surrounding host atoms Lattice strain field interactions between dislocations and impurity atoms result in restriction of dislocation movement This is one of the most powerful reasons to make alloys, which have higher strength than pure metals. Example: 4k gold is too soft. If we put in 16% silver and 9% copper, we get an alloy that looks just like pure gold, but is much more strong and durable. We call this 18k gold. (18/4 = 75% gold)

24 Small impurity atoms exert tensile strains (see figure below) Large impurity atoms exert compressive strains Solute atoms tend to diffuse and segregate around dislocations to reduce overall strain energy cancel some of the strain in the lattice due to the dislocations Why it works Small atoms like to live here. (they reduce lattice strain caused by the dislocation). Big atoms (or interstitials) like to live here (there is more space.) Atoms of either type diffuse to dislocations during high temperature processing, then exert forces on the dislocation later to keep them stuck.

25 Compressive Strains Imposed by Larger Substitutional Atoms Add nickel to copper, strength goes up, ductility goes down - for the same reason: dislocation mobility is decreased. Note: trade-off in properties

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