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530.352 Materials Selection
Lecture #26 High Temperature Creep - II
Tuesday
November 15th, 2005
Sherby-Dorn Equation
ss = C n exp (- Qdiffusion / RT)
.
Temperature dependence
Stress dependence
Constants
Use this equation to calculate creep rate at
any given or new stress or temperature !!
Example of creep based design:
Ni-base superalloys that are used for jet turbine
applications exhibit Qcreep = 320 kJ/mol and n=5.
• What is the creep rate at 925 oC and 350 MPa if C=1.7x10-7
and R=8.314 J/mol-oC ?
• What would the creep rate be if the stress were increased by
25 MPa ?
• What would the creep rate be if the temperature were increased
by 25 oC ?
• If your boss wanted to increase the operating temperature by
50 oC, how much would you have to decrease the stress to
maintain the same creep rate ?
Creep rate; T=925 oC and =350 MPa
ss = C n exp (- Q / RT)
.
ss = 1.7x10-7 3505 exp( -320,000/8.314 x 1198 K )
= 1.7x10-7 x 5.25x1012 x 11.1x10-15
= 10-8 sec-1
.
Is 10-8 sec-1 fast ?
year
per
or
year
year
days
day
hours
hour
ss
ss
%
31
31
.
0
25
.
365
24
sec
600
,
3
sec
10
1
8








Is short for a service life but long for a graduate student
-- must extrapolate from short tests to long times !!
Increasing by 25 MPa :
ss-1 = C 1
n exp (- Q / RT1)
.
ss-2 = C 2
n exp (- Q / RT2)
.
1 / 2 = (1/2 )n = (350/375)5 = 0.708
2 = 1.4 x 10-8 sec
. .
.
Increasing by 25 oC:
ss-1 = C 1
n exp (- Q / RT1)
.
ss-2 = C 2
n exp (- Q / RT2)
.
518
.
0
1223
1
1198
1
314
.
8
000
,
320
exp
1
1
exp
2
1
2
1
2
1












 

























T
T
R
Q
2 = 1.92 x 10-8 sec
Changing both T and  :
ss-1 = C 1
n exp (- Q / RT1)
.
ss-2 = C 2
n exp (- Q / RT2)
.
MPa
T
T
R
Q
n
307
1223
1
1198
1
314
.
8
000
,
320
exp
350
1
1
1
exp
2
5
2
2
1
2
1
2
1












 





































Creep Mechanisms (metals and ceramics)
Diffusion creep
Dislocation creep (power-law creep)
Stress Relaxation
Creep Fracture
Diffusion creep

bulk crystal
diffusion
d1
grain boundary
diffusion
d


Dislocation creep
glide
glide
climb
Diffusion assisted climb important:
climb
1. Annihilation:
poof !
2. By passing obstacles:
Dislocation Climb:
Stress Relaxation
total = el + pl = 0 so el = - pl
el =  / E and pl = c = A n (@ cont. T)
-n d = - AE dt
1-n | - AE t |0
. .
. .
. .

time

time

p
el
o
t
Tertiary creep :
c
time
tertiary
creep
Creep damage starts
Design against creep (metals)
Minimize T / Tmelting
to slow diffusion, climb, and creep.
Arrange for large grain sizes to slow
diffusion.
Use precipitates (oxide particles) and
solid solutions to slow dislocations.
Creep in ceramics :
Very little dislocation motion - mostly
diffusion creep or something else.
Glassy phases (oxides) that form at
grain boundaries soften and high T
and lead to grain boundary sliding.
Design against creep (ceramics)
Similar to metals, reduce diffusion and
dislocation motion, but must also ...
Reduce/control grain boundary phases.
Creep Mechanisms (polymers)
Tg replaces Tm at the critical T
and Tg is often close to RT !!!
Viscous flow is like creep:
 = C 1 exp (-Qv / RT)
Qviscous not QDiffusion,
Qvicsous is Q slide lumpy molecules past one another
n = 1 for Newtonian viscous flow
.
Design Against Creep (polymers)
Increased degree of cross-linking ->
increased Tg and less creep.
High molecular weight -> high viscosity
-> low creep rate.
Crystalline polymers better than glassy.
Add fibers or particles to make
composites !!

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Creep II.ppt

  • 1. 530.352 Materials Selection Lecture #26 High Temperature Creep - II Tuesday November 15th, 2005
  • 2. Sherby-Dorn Equation ss = C n exp (- Qdiffusion / RT) . Temperature dependence Stress dependence Constants Use this equation to calculate creep rate at any given or new stress or temperature !!
  • 3. Example of creep based design: Ni-base superalloys that are used for jet turbine applications exhibit Qcreep = 320 kJ/mol and n=5. • What is the creep rate at 925 oC and 350 MPa if C=1.7x10-7 and R=8.314 J/mol-oC ? • What would the creep rate be if the stress were increased by 25 MPa ? • What would the creep rate be if the temperature were increased by 25 oC ? • If your boss wanted to increase the operating temperature by 50 oC, how much would you have to decrease the stress to maintain the same creep rate ?
  • 4. Creep rate; T=925 oC and =350 MPa ss = C n exp (- Q / RT) . ss = 1.7x10-7 3505 exp( -320,000/8.314 x 1198 K ) = 1.7x10-7 x 5.25x1012 x 11.1x10-15 = 10-8 sec-1 .
  • 5. Is 10-8 sec-1 fast ? year per or year year days day hours hour ss ss % 31 31 . 0 25 . 365 24 sec 600 , 3 sec 10 1 8         Is short for a service life but long for a graduate student -- must extrapolate from short tests to long times !!
  • 6. Increasing by 25 MPa : ss-1 = C 1 n exp (- Q / RT1) . ss-2 = C 2 n exp (- Q / RT2) . 1 / 2 = (1/2 )n = (350/375)5 = 0.708 2 = 1.4 x 10-8 sec . . .
  • 7. Increasing by 25 oC: ss-1 = C 1 n exp (- Q / RT1) . ss-2 = C 2 n exp (- Q / RT2) . 518 . 0 1223 1 1198 1 314 . 8 000 , 320 exp 1 1 exp 2 1 2 1 2 1                                        T T R Q 2 = 1.92 x 10-8 sec
  • 8. Changing both T and  : ss-1 = C 1 n exp (- Q / RT1) . ss-2 = C 2 n exp (- Q / RT2) . MPa T T R Q n 307 1223 1 1198 1 314 . 8 000 , 320 exp 350 1 1 1 exp 2 5 2 2 1 2 1 2 1                                                   
  • 9. Creep Mechanisms (metals and ceramics) Diffusion creep Dislocation creep (power-law creep) Stress Relaxation Creep Fracture
  • 10. Diffusion creep  bulk crystal diffusion d1 grain boundary diffusion d  
  • 11. Dislocation creep glide glide climb Diffusion assisted climb important: climb 1. Annihilation: poof ! 2. By passing obstacles:
  • 13. Stress Relaxation total = el + pl = 0 so el = - pl el =  / E and pl = c = A n (@ cont. T) -n d = - AE dt 1-n | - AE t |0 . . . . . .  time  time  p el o t
  • 15. Design against creep (metals) Minimize T / Tmelting to slow diffusion, climb, and creep. Arrange for large grain sizes to slow diffusion. Use precipitates (oxide particles) and solid solutions to slow dislocations.
  • 16. Creep in ceramics : Very little dislocation motion - mostly diffusion creep or something else. Glassy phases (oxides) that form at grain boundaries soften and high T and lead to grain boundary sliding.
  • 17. Design against creep (ceramics) Similar to metals, reduce diffusion and dislocation motion, but must also ... Reduce/control grain boundary phases.
  • 18. Creep Mechanisms (polymers) Tg replaces Tm at the critical T and Tg is often close to RT !!! Viscous flow is like creep:  = C 1 exp (-Qv / RT) Qviscous not QDiffusion, Qvicsous is Q slide lumpy molecules past one another n = 1 for Newtonian viscous flow .
  • 19. Design Against Creep (polymers) Increased degree of cross-linking -> increased Tg and less creep. High molecular weight -> high viscosity -> low creep rate. Crystalline polymers better than glassy. Add fibers or particles to make composites !!