Chapter 6: Plastic Deformation in Metals Flashcards
Shear stress for slip in pure, close-packed metals
1MPa
How is plastic deformation produced?
Relative motion of one part of material against another
Why is yield stress so high in metals?
Formation and motion of dislocations due to shear stress.
Schmid’s law for slip (resolved shear stress)
𝜏R = σcosλcosΦ
σ = applied tensile stress
λ = angle between the slip direction and the stress axis Φ = angle between the normal to the slip plane and stress axis
Critical resolved shear stress
Shear stress required to cause a dislocation to move in a pure single crystal of a metal. Maximum at 45°
Relationship of plane roughness to motion
As the planar density decreases, resistance to motion in shear increases
How many slip systems for good ductility?
5
Number of slip systems in BCC
12
Number of slip systems in FCC
12
Number of slip systems in HCP
3
What does strengthening do?
Reduce plastic deformation
Grain boundary strengthening
Strengthening materials by decreasing their average crystal (grain) size. Described by the Hall-Petch equation.
Hall-Petch equation
σys = σ0 + kd^(-1/2)
σ0 = 3𝜏
k = Hall-Petch coefficient (MNm^(-3/2))
d = grain size (m)
Solid solution strengthening
Strengthening a metal by adding solutes of different size and them on the lattice (substitutional solid solution) or between the lattice points (interstitial solid solution) to achieve strengthening through interaction with defects (such as dislocations).
Solid solution strengthening equation
𝜏ys = 𝜏0 + βGsqrt(c)
β = property of the solvent/solute combination
c = concentration of solute