Lecture 6 plastic region

Question 1

True stress and strain

Answer 1

not related to initial length and area, use force/current cross sectional area, use change in length as length from current to previous
assume volume stays constant

Question 2

issues with engineering stress strain

Answer 2

okay with small stress and strain but not very accurate after

Question 3

true strain = (epsilon little t)

Answer 3

ln (1+engineering strain)

ln (1+ epsilon)

Question 4

true stress = (sigma little t)

Answer 4

(1+engineering strain) engineering stress

Question 5

dont use true stress strain as

Answer 5

it would require to measure the length and area all the time which would be difficult to do can then change to true stress and strain

Question 6

true stress strain diagram

Answer 6

stress always rising which makes more sense

UTS is no longer easy definable on true stress strain curve

Question 7

two types of strength

Answer 7

ultimate tensile strength max engineering stress a material can support
yielding strength elastic limit is reached onset of plastic deformation
measured in MPa steels -1000s bone cast iron 100s - aluminium 10s

Question 8

why is yield strength hard to measure

Answer 8

should mark region where plastic region begins - bonds begin to break material begins to permanently deform
hard to pinpoint due to as plastic region get unstable region
get upper lower yield strength but dependent on method of doing test

Question 9

offset yield strength or proof strength

Answer 9

0.002 or 2% strain follow maintain YMs then when hit stress strain curve that point is offset yield strength, sometimes unstable region is much higher therefore use larger offset (this will be defined)

Question 10

yield strength different to most material properties as

Answer 10

not constant can alter on the material using workhardening

Question 11

what is work hardending

Answer 11

heat to make material ductile
stress beyond the yield strength below UTS
becomes permanently deformed 
yield strength permanently increased 
take back to elastic region and repeat
can take all the way upto the UTS

Question 12

after work hardening

Answer 12

much larger elastic region yield strength higher but area under curve is less therefore material more brittle

Question 13

how to take fit plastic part of curve up to UTS

Answer 13

take log curve of true stress and strain
get straight line
log of true stress = log K (intercept) + n log true strain
where n is the gradient of line constant for different materials

Question 14

plastic region curve fit what happens at UTS

Answer 14

true strain = n (gradient of line) 
can use this to find UTS
true stress=  K*true strain ^n = Kn^n
UTS stress (sigma u) = true strain / exp (n)
K (MPa) is phenomenological

Question 15

Definition of phenomenological

Answer 15

need to do experiment on material and then fit with equation - no way of telling based on materials other properties

Question 16

Theoretical strength

Answer 16

strength a material should be at

Question 17

How to calculate theoretical strength

Answer 17

all fracture req breaking of atomic bond
hookes law obeyed to failure therefore we go all the way up elastic region to stress which is just high enough to break material
area under stress strain curve = elastic energy per unit volume
energy (U) = 0.5 stress*strain * volume = 0.5 stress^2/ Yms
volume of material being fractured = cross sectional area * atomic distance
U/area= 0.5 stress^2 * lattice constant/ Yms
this energy per unit area at theoretical strength = energy per unit area to create new surface (2gamma)
this can be used to work the theoretical strength

Question 18

theoretical strength =

Answer 18

2* sqroot(YM * energy to create a surface (gamma)/ lattice constant (ao)) but this is an overestimate due to force energy curve not being linear so we generally use
sqroot(YM * energy to create a surface (gamma)/ atomic distance (ao))

Question 19

roughly the theoretical strength =

Answer 19

YMS/3

Question 20

where do most materials fail in comparison the theoretical strength

Answer 20

most material fail before theoretical strength

due to issues with crystal structure

Question 21

issues within crystal structure x4

Answer 21

most crystal structures are not perfect have;
vacancy
interstitial atom
substitutional impurity atom
interstitial impurity atom
which lead to 
local weaknesses

Question 22

vacancy

Answer 22

atom missing in crystal structure - atoms around missing bond local change in structure
local weakness

Question 23

interstitial atom

Answer 23

atom no sitting on a real lattice site in the middle of somewhere causes bonds around it to be stretched
local weakness

Question 24

substitutional impurity atom

interstitial atom

Answer 24

when forming materials what species of atoms that you designed to be in there but when forming impurities get in ie carbon
changes local bonding structure local weakness
also could sit in middle

Question 25

power law

Answer 25

true stress = K true strain^n

Question 26

power law hardening parameters and how to find them

Answer 26

K and n
log K = the y intercept of the log graph of the true stress strain plastic region curve curve remember to convert!!
n = is equal to the gradient of the log graph of the stress stress strain plastic region curve

Question 27

true uniform strain

Answer 27

true strain at the UTS

Question 28

theoretical strength typical values

Answer 28

gamma (energy to create a surface) = 1J*m^-2
Youngs modulus = 10^11 Pa
ao (lattice space) = 10^-10