Metals- Formulae Flashcards
Stress and strain relations in normal and shear
σ=Eε
τ=Gγ
Stored energy from graph and formulae
Area under linear region of stress strain graph
1/2 Eε^2
1/2 Gγ^2
Form of a Burgers vector
b=a/2 [uvw]
Where a is lattice constant
Magnitude of Burgers vector
The magnitude of the vector form of the Burgers vector.
Remember the a/2 on outside multiplies it
Force acting on dislocation and derivation
F=τb
Edge dislocation length L experiences a force per unit length F.
Total force acting over dislocation is FL.
Work done is force x distance d the dislocation moves so FLd.
Must be balanced against work done by shear stress to shear the crystal by a distance b which is τb. This acts over entire slip plane area so τLdb.
Equate works so FLd=τLdb
Get F=τb
Elastic strain energy associated with dislocation
Λ=aGb^2
About 1/2 Gb^2
Slip systems of fcc, bcc, hcp, diamond
FCC: {111}<1-10>
BCC: {110}<1-11>
HCP: {001}<100>
Diamond: {111}<1-10>
Schmid’s law and derivation
τcrss=σyCosφCosλ
Have a cylinder with force acting along axis and top CSA=A.
Slip plane normal at angle φ to force and slip direction along slip plane at angle λ to force.
τR=Force/A. Force parallel to slip direction Fcosλ.
Active area of slip plane=A/cosφ
τR=Fcosλ/(A/cosφ)=F/A CosφCosλ=σCosφCosλ
Value of τR when slip occurs is τcrss where σ=σy
Schmid factor
CosφCosλ
Frank’s rule
b3^2 less than b1^2+b2^2
b1^2 greater than b2^2+b3^2
Dislocation density
ρ=1/L^2
Where L is average space between two dislocations so only a single dislocation present in simple cubic array with area L^2
Line tension of a dislocation
T=aGb^2
About 1/2 Gb^2
Strength of alloy relative to dislocation density
σY proportional to ρ^1/2
Hall-Petch equation
σY=σ0+k/d^x
Contribution of solid solution strengthening to yield strength
σY proportional to C^1/2
Where C is concentration of the element in the alloy
