Dislocations and mechanical properties
Terms in this set (19)
Describe why the yield strength of a material is significantly less than the "ideal" yield strength
Plastic deformation does not require concurrent breakage of every bond in the shear plane. Instead, it uses a dislocation glide.
Explain how dislocation glide causes plastic deformations in materials
when you plastically deform a material, bonds are broken and the lattice moves
all plastic deformations occur due to sheet stresses, not normal stresses
dislocation glide is the motion of dislocations that causes plastic deformation in materials
the dislocation migrates such that the crystal shape changes without mechanical fracturing or loss of crystal structure
Motion of a dislocation on its slip plane (defined by the Burger's vector and the unit tangent vector). This kind of dislocation motion does not involve mass transport.
Identify the burger's vector
Draw a clockwise loop in a perfect crystalline lattice, counting and recording the unit cell dimensions for each side of the loop.
Draw the same loop (with the same lengths for each side) around the dislocation.
The vector required to close the loop is the Burger's vector.
dislocation glide occurs in the path of least _________
crystalline planes with highest atomic packing density (atoms/area)
crystalline directions with the highest atomic packing density (atoms/length)
this = slip plane + slip direction
-- the more of these there are in a material, the easier it is to plastically deform (usually true)
To be counted as a unique slip system, slip planes must intersect and can't be colinear
plastic deformation in FCC
12 slip systems
of the 3 metal crystal systems.
slip systems intersect each other in 3D, all slip systems are close-packed planes
plastic deformation in BCC
48 slip systems
, even more flip systems than FCC and they all intersect in 3D.
because not closed packed, more resistant to dislocation motion
plastic deformation in HCP
3 slip systems
Describe different methods for strengthening a material and apply 𝜏 ∝ 𝑪𝒅𝒆𝒇𝒆𝒄𝒕𝒔 & 𝝈𝒚 = 𝟑× 𝝉𝒊
add barriers to dislocation motion:
--solution hardening: strength of material increases with square root of impurity concentration
-- precipitation hardening:
1. cutting (if precipitation is fairly small, dislocations can slice through precipitation)
2. bowing (if precipitate is large, then the dislocation encircles the precipitate and then proceeds)
-- work hardening:
as we plastically deform, we form more dislocations
strength of material increases with square root of impurity concentration
Hardening or strengthening of an alloy by the presence of a fine, uniformly distributed precipitate.
(bowing an cutting)
as we plastically deform, we form more dislocations --> need higher stress to continue plastic deformation because we work harden
total yield strength of crystalline materials
𝝉y = 𝝉lattice + 𝝉ss + 𝝉ppt + 𝝉dislocations + 𝝉GB--> so 𝝈𝒚 = 𝟑× 𝝉𝒊
Explain the characteristic shape of a stress-strain curve for a tough material
because of work hardening, as we plastically deform, we form more dislocations
therefore, we need higher stress to continue plastic deformation and yield strength increases
Explain the mechanism for fracture in materials
fracture = area under the stress strain curve
when we apply stress to a material, we plastically deform the material and propagate crack growth (which leads to fracture)
apply the equation
𝑲𝟏𝑪 = 𝝈∗ (𝝅𝒄)^1/2
K1C is a materials constant (never changes)
Knowing K1C allows one to calculate crack length (c) that will cause failure at a given stress level to fracture (𝝈) or vice-versa.
If c increases, 𝝈 decreases (and vice-versa)
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