Lecture 8

Question 1

if crack in material

Answer 1

stress concentration around crack is very high - very large a tip of crack

Question 2

size and shape of crack determine

Answer 2

stress concentration and whether a material can support it or not

Question 3

stress intensity factor Ki Pam^0.5 linked to

Answer 3

can be linked to the toughness of a material

Question 4

stress intensity factor equation

Answer 4

geometry factor (dimensionless) * applied stress * sqrt(pi * crack size)
Ki= Y* sigma * sqrt(pi *a)
i determines the type of crack propagation mode

Question 5

K1

Answer 5

mode 1 tensile fracture crack plane normal to direction of the largest tensile loading
mode 2 would be shear
mode 3 tearing
only look at mode 1

Question 6

K1c and typical values

Answer 6

is the critical stress intensity factor above which the material can not support brittle failure
aluminium alloy 36 MPam^0.5
steel 50MPam^0.5 can support cracks up to a larger load
low value means cannot support crack

Question 7

centre crack

Answer 7

length of crack is 2a therefore need to divide length by 2 to get a

Question 8

in an infinite plate centre crack Y is

Answer 8

1 crack length equal to twice a

l = 2a

Question 9

in an infinite plate edge crack Y is

Answer 9

1.12 crack length equal to a

l = a

Question 10

what does an infinite plate

Answer 10

a/width of plate, if value is very small much more like a infinite plate (width here is dimension parallel to crack) material around crack is well supported if a/w approaches 1 then material around crack not very well supported

Question 11

For a finite plate equation

Answer 11

can be found on formula sheet equation for both edge and centre crack this equation tends to 1 for a centre crack a/w is very small and to 1.12 for a edge crack where a/w is very small

Question 12

if calculated value of K1 is lower than K1c and if higher

Answer 12

material can support crack
if higher can not support crack material fails catastrophically brittle manner
larger crack material more likely to fail

Question 13

materials can fail below their normal strength due to

Answer 13

temperature
cracks and stress concentration
fatigue
creep

Question 14

fatigue is

Answer 14

low amplitude vibrations below the yield point of the material causes dislocation of defects to move to a similar region and therefore material fails in brittle manner

Question 15

fatigue is particularly important for

Answer 15

aircraft as goes through many cycles

Question 16

as we apply a oscillatory stress to it

Answer 16

its strength begins to fall over many cycles

Question 17

endurance limit taken at

Answer 17

10^7 cycles (10^7 oscillations) we find where material crosses this

Question 18

steels show fatigue limit of

Answer 18

40% of yield strength

Question 19

ferrous metals fatigue limit

Answer 19

drop in yield strength flattens off and end with fatigue limit below which the yield strength does not drop any further no matter how many cycles

Question 20

non ferrous metal do not

Answer 20

hit fatigue limit will just continue to decrease until material breaks - very important to suit material to your application and life span of

Question 21

creep is

Answer 21

a thermal effect - materials placed under stresses at elevated temperatures way below melting point but enough energy to cause vibrations
slowly deforms over time
load changes bonding potential and temperature helps atoms slowly move out

Question 22

creep defined as

Answer 22

time dependent and permanent deformation of materials when subjected to constant load or stress

Question 23

which is worse edge crack or centre crack of same length under same load and why

Answer 23

look at stress intensity factor equation
for edge crack a is larger as centre crack a=l/2
for edge crack geometrical factor is larger 1.12 > 1
therefore requires lower stress to achieve same stress intensity factor - material fails at lower stress