Lecture13 - ME 382 Lecture 13 EFFECTS OF ALLOYING ON YIELD...

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Unformatted text preview: ME 382 Lecture 13 EFFECTS OF ALLOYING ON YIELD STRENGTH 1. Solid-solution hardening • • Occurs with interstitial solid solutions or with substitutional solid solutions Local compressive stress field around interstitial solute atoms • • • Local stress field when substitutional solute atom is different size from solvent atoms Example: Bronze consists of Sn dissolved in Cu Sn atoms larger than Cu • Stress field interacts with stress field of dislocation • Dislocation line distorts to minimize energy (at cost of extra length) 6/ii/06 1 ME 382 Lecture 13 • This is minimum energy position ∴ • Extra work is needed to move the line from this configuration Shear strength is given by: shear strength k { !{ # " 3/ 2 m misfit strain concentration c { • Increase concentration above equilibrium (supersaturated solution) for enhanced σY 2. Precipitation hardening Coherent or semi-coherent precipitates • If small precipitates develop they will generally have a mismatch strain with matrix (Coherent or semi-coherent pttes. have some lattice alignment with host) • • Strengthening mechanism similar to solid-solution hardening As precipitate size ↑ ⇒ mismatch strain energy ↑ ∴ More efficient strengthening ⇒ yield strength ↑ Incoherent precipitates • As precipitate size ↑, mismatch energy becomes so large that it is more favorable to form incoherent precipitate • Energy cost: Increased surface energy between precipitate and matrix Energy gain: No misfit strain energy ∴ Bigger pttes with more strain energy will make this transition • • If no misfit strain ⇒ no interaction with dislocation? Two ways for dislocation to pass through large precipitate ( i) ( ii) Shear through precipitate Bow around precipitate 6/ii/06 2 ME 382 Lecture 13 • Shear strength dictated by lowest stress required to shear or bow Cutting through precipitates or dispersions • Assume particles of diameter do with a shear resistance of kp (N.m-2) • • • • Force (per unit width) resisting shear of the particle: fp = kp b (N m-1) Assume particles separated by a distance L in matrix with shear resistance km (N.m-2) Force (per unit width) resisting shear of the matrix: Consider length L of dislocation line fm = km b (N m-1) • Force required to move dislocation of length through the alloy of matrix & particle: fc L= fp x do + fm x (L -do) • • But, force required to move dislocation is related to shear yield strength by fc = τc b Therefore, shear yield strength of alloy associated with cutting of particles given by: ∴ " c = k c # k p ( do / L) + k m = σYc/2 (if L >> do) (1) • • ∴ Soft particles provide little additional strengthening ! If coarsening (growth) of particles occurs: do ↑, but L ↑ so do /L = constant Contribution to yield strength from cutting not changed by precipitate growth But yield strength proportional to amount of second phase. 6/ii/06 3 ME 382 Lecture 13 Bowing around precipitates or dispersions • Draw a free-body diagram for a length L of dislocation line • By equilibrium: ∴ ∴ 2T sin ! = f " L Gb sin! = " bL 2 != Gb sin" L (θ = 90°) • Maximum value of τ for equilibrium is Gb/L Dislocation can then bow out without limit Portions of the loop cancel, leaving loops around particles and straight dislocation Dislocation line moves without impediment to next particles • • Shear strength from bowing: "b = Gb + k m = σYb/2 L (2) If coarsening (growth) of particles occurs: do ↑ ⇒ L ↑ ∴ τb ↓ i!precipitate growth occurs f 6/ii/06 4 ME 382 Lecture 13 • Increase in volume fraction of precipitates ⇒ L↓ ∴ • • τb increases with volume fraction of precipitates (Eqns. 1 & 2) Yield strength determined by lowest value of τb or τc Bowing occurs if τb < τc I.e., if do > Gb/ k p (when particles are very big or very hard) • Cutting occurs if τYc < τYb I.e., if do < Gb/ k p (when particles are very small or very soft) Summary of hardening by alloying • • Generally, yield strength increases with amount of alloying Interstitial or substitutional solid - solution Dislocations hindered by local stresses • Small precipitates - more efficient hardening by same mechanism (while precipitates are coherent or semi-coherent) • Large precipitates & dispersions have no mismatch stresses Dislocations can cut through small soft particles Dislocations can bow past large hard particles Yield strength goes down when coarsening of incoherent precipitates occurs. 6/ii/06 5 ...
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This note was uploaded on 01/26/2011 for the course ME 382 taught by Professor Garikipati during the Winter '08 term at University of Michigan.

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