Ch 7 — Dislocations & Strengthening Mechanisms

Slip system — FCC {111}⟨110⟩

A slip system is a slip plane combined with a slip direction inside that plane. Dislocations preferentially glide on planes of highest atomic density (smallest b / largest planar spacing) along directions of shortest b.

Why these planes?

The close-packed plane of each structure is preferred because slipping along its close-packed direction requires the smallest atomic rearrangement, lowering the Peierls stress (the lattice resistance to dislocation motion).

τ_P ∝ exp( −2π · wb )

where w is the dislocation width (proportional to planar spacing). Wider planes ⇒ wider w ⇒ lower τ_P.

Counting the slip systems — every plane listed

Slip systems = (unique slip planes) × (slip directions lying in each plane). Every plane in the family is listed below with the close-packed directions it contains. A direction only counts if it lies in the plane (its dot product with the plane normal is zero).

FCC — {111}⟨110⟩ → 4 planes × 3 directions = 12

#	Slip plane (111)	⟨110⟩ directions in that plane
1	(111)	[1̄10], [01̄1], [101̄]
2	(1̄11)	[110], [011̄], [101]
3	(11̄1)	[110], [101̄], [011]
4	(111̄)	[1̄10], [011], [101]

BCC — {110}⟨111⟩ → 6 planes × 2 directions = 12 (primary)

#	Slip plane {110}	⟨111⟩ directions in that plane
1	(110)	[1̄11], [11̄1]
2	(1̄10)	[111], [111̄]
3	(101)	[1̄11], [111̄]
4	(1̄01)	[111], [11̄1]
5	(011)	[11̄1], [111̄]
6	(01̄1)	[111], [1̄11]

SC — {100}⟨100⟩ → 3 planes × 2 directions = 6

#	Slip plane {100}	⟨100⟩ directions in that plane
1	(100)	[010], [001]
2	(010)	[100], [001]
3	(001)	[100], [010]

Cross-check: in FCC each of the 6 ⟨110⟩ directions appears in exactly 2 planes (6 × 2 = 12); in BCC each of the 4 ⟨111⟩ directions is shared by 3 planes (4 × 3 = 12). Either way the system count is the same.

Structure	Plane family	# planes	Direction	dir / plane	Slip systems
FCC	{111}	4	⟨110⟩	3	4 × 3 = 12
BCC	{110}	6	⟨111⟩	2	6 × 2 = 12
SC	{100}	3	⟨100⟩	2	3 × 2 = 6
HCP	(0001)	1	⟨112̄0⟩	3	1 × 3 = 3

Solid-solution strengthening

Foreign atoms (solutes) distort the host lattice and impose a strain field. A passing dislocation must either push through that strain field or bow around it — both cost extra stress. Substitutional solutes (different size on a lattice site) and interstitial solutes (C, N in Fe) both work; interstitials are usually 10–100× more potent per atom.

Δσ_ss ≈ G · ε^3/2 ·

\sqrt{c}

where G is shear modulus, ε is the misfit strain per solute, and c is concentration. The $\sqrt{c}$ dependence comes from random spacing between solutes scaling as 1/ $\sqrt{c}$ .

Δσ_ss contribution (Cu–Ni alloy, G=46 GPa, ε≈0.04)

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Examples

Alloy	Solute	Type	Δσ at 5 at%
Cu–Ni	Ni in Cu	substitutional	~25 MPa
Cu–Zn (brass)	Zn in Cu	substitutional	~60 MPa
Fe–C	C in α-Fe	interstitial	~300 MPa

Cold work — strain hardening

Plastic deformation at room temperature multiplies dislocations (sources like Frank–Read operate as soon as τ rises). The new dislocations interact with one another, forming forests and tangles that pin further glide. As %CW rises, the metal becomes stronger but less ductile.

%CW = A₀ − A_dA₀ × 100

τ ≈ τ₀ + α · G · b ·

\sqrt{ρ}

where ρ is the dislocation density (length / volume). Cold work raises ρ from ~10¹⁰ m⁻² (annealed) to ~10¹⁵ m⁻² (heavily worked).

Yield stress & ductility (annealed Cu baseline)

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Stress–strain trace

Pull a sample to ε₁, unload (elastic recovery), then reload: yielding now occurs at the previous flow stress, not the original σ_y. The "memory" of prior straining is stored in the dislocation network.