Materials 2

Question 1

Line Tension

Answer 1

T = Gγ²πb²
= πGb² / 8 (for γ = 0.5)

Question 2

Creep stress relaxation

Question 3

Creep Diffusion

Question 4

Taylor Factor

Answer 4

σ_y = 3 τ_y
Gross tensile yield stress from gross shear yield stress for polycrystalline materials with multiple grains.

Question 5

Basquin’s Law

Question 6

LCF Equations

Question 7

Stress Triaxiality Ratio

Answer 7

T = σm / σeq

High stress triaxiality ratios, therefore high equivalent stresses, promote brittle fracture over yielding.

Question 8

Inglis Solution
Stress concentration factor around a hole

Answer 8

K_A = σ_max / σ = ( 1 + 2 a/b )

Where b is radius in direction of stress
and a is radius in direction perpendicular to stress
K_A = 3 for a circle

Question 9

Rupture Time

Question 10

Fatigue/creep damage

Question 11

Variables critically resolved shear stress:
τr = Fcos(λ) / A/cos(Φ) = σ cos(λ) cos(Φ)

Answer 11

τr = critically resolved shear stress, MPa
F = Force on a single crystal, N
A = effective area of F, m^2
σ = Tensile Stress (F/A)
λ = Angle between slip direction and applied force, °
Φ = Angle between normal of slip plane and applied force, °

(Same can be done for yield stress, just use σy instead of σ)

Question 12

Variables in Solid Solution Hardening strength increase:
τ_y = fi/b + fss/b

Answer 12

τ_y = Critical shear stress to initiate dislocation slip (MPa)
fi = intrinsic lattice resistance per unit length (N/m)
fss = solute solution resistance per unit length N/m
b = Burger’s Vector (m)

Question 13

Precipitation Hardening

Answer 13

Small precipitates of another compound (usually partly made of the parent metal and with an added element) exist within the main metal lattice.
Smaller precipitates better dispersed tend to make for stronger, harder lattices.
This can be shown by an increase in f (friction) with a decrease in L (gap length)

Question 14

Line tension to be pushed through a gap in precipitates:
τbL = 2T

Answer 14

τ = Applied shear stress, MPa
T = Line Tension (J/m)
L = Gap between obstacles, m
b = Burgers vector, m

Question 15

Work Hardening

Answer 15

Caused by interpenetration of dislocations between slip systems
Also interaction of dislocation stress fields (where there is tension/compression around a dislocation), as same signs can repel.

Question 16

Shear Stress due to dislocation density (for work hardening)
τ_y = τ_i + α G b ρ^1/2

Answer 16

τ_y = Critical shear stress, MPa
τ_i = Initial stress due to other strngthening methods
α = A material constant
G = Shear modulus, GPa
b = Burger’s Vector, m
ρ = Dislocation density, m^-2

Question 17

Hall-Petch for shear stress

Answer 17

Can just substitute shear stress for tensile stress in standard hall-petch relationship

Question 18

Heat treatment - Rcovery

Answer 18

High temperature
Atomic diffusion happens
Causes relief in internal strain energy because…
Dislocations move and annihilate
Reduction in dislocation density
Material softens

Question 19

Heat treatment - Recrystallisation

Answer 19

Formation of a new set of equiaxial grains (equal dimensions in all directions)
Strain free grains!
Very low dislocation density

First, small grains form as very small nuclei
Then the small grains grow to consume the larger parent grains
Due to a difference in dislocation energy throughout the deformed/undeformed material

Question 20

Recrystallisation Temperature

Answer 20

The temperature at which recrystallisation can completely occur in 1 hour
(Typically 1/3 to 1/2 of melting temp)

Question 21

Recrystallisation grain size equation

Answer 21

dⁿ - d₀ⁿ = K t
K and n are fixed constants, t in minutes

Question 22

Effective Modulus at Plane Stress

Answer 22

E’ = E

Question 23

Effective Modulus at Plane Strain

Answer 23

E’ = E/(1-v²)

Question 24

Yielding or brittle fracture likelihood ratio

Answer 24

σ1/σ_eq

Question 25

Tresca equivalent stress

Answer 25

Max( σ1 ; σ2 ; σ1-σ2 )

Question 26

Von-Mises equivalent stress

Answer 26

sqrt( σ1² + σ2² - σ1 σ2)

Question 27

Mean / deviatoric stress explanation

Answer 27

Each stress is comprised of the global mean stress plus the deviatoric stress for that direction

Question 28

Uniaxial conditions (for stress triaxiality)

Answer 28

σ_m = σ1 / 3 σ_eq = σ1 T = 1/3

Question 29

Hydrostatic tension conditions (for stress triaxiality)

Answer 29

σ_m = 3 σ1 / 3 = σ1 σ_eq = 0 T = ∞

Question 30

Variables in stress concentration factor equation K = σ Y sqrt(a)

Answer 30

K is stress concentration factor σ is stress Y is some geometric function (provided) a is crack length At critical K, you can have it due to a critical length at fixed load, or critical load at fixed length