Fatigue creep
TUTORIAL 5 FATIGUE, CREEP
Fatigue Tensile stress is positive; Compressive stress is negative St e Stress ratio tio ≡ R = σmin/σmax. Mean stress ≡ σm = (σmax +σmin)/2 Æ average of the maximum and minimum stress
Stress range ≡ σr = (σmax -σmin) Æ difference between the max. and the min. stress
Stress amplitude ≡ σa = (σmax - σmin)/2 Æ half the difference between the maximum i and d the th minimum i i stress t
Alternating stress diagrams
Reversed stress
Repeated stress
[Dieter]
3. Catastrophic failure: fracture surface
2. Crack propagation: striated; each
striation indicates the position of the crack during that cycle or intermittent load. Exhibits characteristic markings called Beach or clamshell marks This is a fatigue failure
1. Crack initiation: usually at the surface near a flaw,
notch, inclusion, dislocation concentration or grain bo ndar Surface boundary. S rface looks smooth smooth.
Lab Assignment
4
Lab Assignment
3.5
Ang gle (Proportional To Strress Ampliitude)
The end results was that the paperclip exhibited no noticeable trend towards a fatigue limit. What would happen if we bent the paper clip in a different direction and did a similar sett off experiments? i t ? Would the curve look the same or different? Why?
SNC S-N Curve ffor P Paperclip li
3 2.5 2 1.5 1 05 0.5 0 1
10
Log Cycles
100 5
FATIGUE LIFE Fatigue life :
The number of cycles y p permitted at a particular stress before a material fails by fatigue. Example Steel is subjected to 90 90,000 000 psi stress, fatigue life will be 100,000 cycles. If cycle time is known, fatigue life in years can be calculated
Stress Cycles
Fatigue test
R=-1
(a) Equal stress in tension and compression, Pure sine wave ≡ Mean stress=0
(b) greater tensile stress than compressive stress, and Mean stress: σm = ( σmax + σmin ) / 2 ex:
σm = [50,000+(-10,000)]/2= 20,000 psi
Stress amplitude: σa = ( σmax – σmin ex:
)/2
σa = [50,000- (-10,000)]/2= 30,000 psi
(c) all of the stress is tensile (d) No fatigue failure if stress fluctuates in compression only
50,000 psi
-10,000 psi
Problem with Goodman Equation A tool steel is loaded in fatigue. Its ASTM fatigue S-N S N curve is given in the figure figure. Would failure be expected if the cyclic stress was between -10,000 and 60, 000 psi?
σm=0
SOLUTION
σm = (σmax+σmin)/2 Need to use Goodman equation to determine endurance limit when mean stress is not zero
⎡ ⎛ σ m σ e = σ ⎢1 − ⎜ ⎜σ f ⎢ ⎝ UTS ⎣
⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦
σe = endurance limit at mean stress
(other than zero) σf = fatigue strength for zero mean stress, (from ASTM fatigue test) f from curve; 60,000psi 60 000 i
σUTS = tensile strength (using fatigue ratio for steels : 2σfs)
Goodman relationship:
Fatigue ratio = endurance limit at zero mean stress / tensile strength σfs / σUTS = 0.5
Damage Tolerant Design steps 1. Perform NDE on all flight-critical components (if crack is found, determine initial crack size, size ai ). ) 2. Determine critical crack size to fail catastrophically, ac from fracture mechanics K IC = f σ max π a c 3 3. Calculate the expected life of the component (# of cycles) from Paris da equation = C ΔK n dN
y to failure: Number of cycles
2[(ac )( 2−n ) / 2 − (a i )( 2−n ) / 2 ] N= where n n/2 ( 2 − n)Cf n Δσ π a c = flaw size required for fracture
(Rearrange Paris equation as dN=….. and then integrate between ai and ac)
Δσ = σ max - σ min
ac obtained from
a i = initial flaw size
if σ min is compressive setσ min = 0 4. 5.
K IC = fσ max πac
Knowing the cycle time expected life can be calculated. Replace, rebuild if too close to life limit.
9
Improving Fatigue Life 1. Impose a compressive surface stress (to suppress surface cracks from growing)
S = stress amplitude Adapted from Fig. 8.24, Callister 7e.
Increasing
σm
near zero or compressive σm moderate tensile σm Larger tensile σm N = Cycles to failure
--Method 1: shot peening
--Method 2: carburizing
shot
putt surface into compression
2. Remove stress concentrators.
bad
C-rich C rich gas
better Adapted from Fig. 8.25, Callister 7e.
bad
better 10
STEADY-STATE OR MINIMUM CREEP RATE
Stage II creep rate also called ll d steady-state t d t t or minimum creep rate is an important p design g p parameter. Empirical equation relates creep rate to applied stress via a power-law l relationship l h n & ε s = K1σ
where ε&s = steady - state creep rate K 1 , and n are constants
σ = applied constant stress Qc ⎞ ⎛ K 1 = K 2 exp ⎜ − ⎟ RT ⎝ ⎠ Q C = activation energy
Universal creep equation for steady-state creep rate:
⎛ QC ⎞ n & ε s = K 2σ exp⎜ − ⎟ ⎝ RT ⎠
3. Grain boundary sliding (GBS): Grain boundaries become weak with temperature. Grains slide resulting in permanent deformation. Creep cavitation (void formation) occurs at grain boundaries as a result of grain boundary sliding.
Creep cavities formed at grain boundaries in an austentic stainless steel
Usually it is difficult to observe creep damage except cavitation due to GBS It is creepp damage
Data Extrapolation-Larson-Miller parameter •Creep tests are of long duration and it is impractical to collect data over long times times, e.g several years. solution: perform creep rupture tests at higher temperatures under same stress for shorter times (accelerated tests) extrapolate for service conditions through Larson-Miller analysis Larson-Miller parameter:
T(C + log l tr) - C is a constant (~20), p ((K)) - T is temperature - tr is the creep rupture life (hours) plot log σ versus L-M parameter Used to find rupture time at any temperature for a given stress.
LarsonMiller parameter for ductile cast iron
Appendix
14
14
2. Fail-Safe (Damage Tolerant) Fatigue Design Design for adequate service life after fatigue damage has occurred. 1 1.
Determination of inspection schedule based on Non-Destructive Testing – Initial defect size. – Fatigue crack growth rate. Paris Equation – Critical defect size for failure failure.
NDE: Utrasonic inspection, X-Ray fluorescence, thermography : 2.
Re-assessment of the residual life after inspection allows the total service life to be extended
Crack Grow C wth Rate da a/dN
Fracture Toughness
Fatigue Growth rate
n~3 for steel, n~4 for aluminum aluminum, C = material constant C = 1.62 × 1012 for steel.
Paris equation
da = C ΔK n dN ΔKth
n ~ 2 to 4
15 Stress Intensity Range ΔK
15
Creep Sample deformation at a constant stress (σ) vs. time σ,ε
σ
0
t
Primary Creep: slope (creep rate) decreases with time time. Secondary Creep: steady-state i.e., constant slope. Tertiary Creep: slope (creep rate) 16rate. increases with time, i.e. acceleration of
Adapted from Fig. 8.28, Callister 7e. 16
Creep • Occurs at elevated temperature, T > 0.4 Tm
tertiary primary
secondary
elastic
Adapted from Figs. 8.29, Callister 7e.
17
17
Fatigue Tensile stress is positive; Compressive stress is negative St e Stress ratio tio ≡ R = σmin/σmax. Mean stress ≡ σm = (σmax +σmin)/2 Æ average of the maximum and minimum stress
Stress range ≡ σr = (σmax -σmin) Æ difference between the max. and the min. stress
Stress amplitude ≡ σa = (σmax - σmin)/2 Æ half the difference between the maximum i and d the th minimum i i stress t
Alternating stress diagrams
Reversed stress
Repeated stress
[Dieter]
3. Catastrophic failure: fracture surface
2. Crack propagation: striated; each
striation indicates the position of the crack during that cycle or intermittent load. Exhibits characteristic markings called Beach or clamshell marks This is a fatigue failure
1. Crack initiation: usually at the surface near a flaw,
notch, inclusion, dislocation concentration or grain bo ndar Surface boundary. S rface looks smooth smooth.
Lab Assignment
4
Lab Assignment
3.5
Ang gle (Proportional To Strress Ampliitude)
The end results was that the paperclip exhibited no noticeable trend towards a fatigue limit. What would happen if we bent the paper clip in a different direction and did a similar sett off experiments? i t ? Would the curve look the same or different? Why?
SNC S-N Curve ffor P Paperclip li
3 2.5 2 1.5 1 05 0.5 0 1
10
Log Cycles
100 5
FATIGUE LIFE Fatigue life :
The number of cycles y p permitted at a particular stress before a material fails by fatigue. Example Steel is subjected to 90 90,000 000 psi stress, fatigue life will be 100,000 cycles. If cycle time is known, fatigue life in years can be calculated
Stress Cycles
Fatigue test
R=-1
(a) Equal stress in tension and compression, Pure sine wave ≡ Mean stress=0
(b) greater tensile stress than compressive stress, and Mean stress: σm = ( σmax + σmin ) / 2 ex:
σm = [50,000+(-10,000)]/2= 20,000 psi
Stress amplitude: σa = ( σmax – σmin ex:
)/2
σa = [50,000- (-10,000)]/2= 30,000 psi
(c) all of the stress is tensile (d) No fatigue failure if stress fluctuates in compression only
50,000 psi
-10,000 psi
Problem with Goodman Equation A tool steel is loaded in fatigue. Its ASTM fatigue S-N S N curve is given in the figure figure. Would failure be expected if the cyclic stress was between -10,000 and 60, 000 psi?
σm=0
SOLUTION
σm = (σmax+σmin)/2 Need to use Goodman equation to determine endurance limit when mean stress is not zero
⎡ ⎛ σ m σ e = σ ⎢1 − ⎜ ⎜σ f ⎢ ⎝ UTS ⎣
⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦
σe = endurance limit at mean stress
(other than zero) σf = fatigue strength for zero mean stress, (from ASTM fatigue test) f from curve; 60,000psi 60 000 i
σUTS = tensile strength (using fatigue ratio for steels : 2σfs)
Goodman relationship:
Fatigue ratio = endurance limit at zero mean stress / tensile strength σfs / σUTS = 0.5
Damage Tolerant Design steps 1. Perform NDE on all flight-critical components (if crack is found, determine initial crack size, size ai ). ) 2. Determine critical crack size to fail catastrophically, ac from fracture mechanics K IC = f σ max π a c 3 3. Calculate the expected life of the component (# of cycles) from Paris da equation = C ΔK n dN
y to failure: Number of cycles
2[(ac )( 2−n ) / 2 − (a i )( 2−n ) / 2 ] N= where n n/2 ( 2 − n)Cf n Δσ π a c = flaw size required for fracture
(Rearrange Paris equation as dN=….. and then integrate between ai and ac)
Δσ = σ max - σ min
ac obtained from
a i = initial flaw size
if σ min is compressive setσ min = 0 4. 5.
K IC = fσ max πac
Knowing the cycle time expected life can be calculated. Replace, rebuild if too close to life limit.
9
Improving Fatigue Life 1. Impose a compressive surface stress (to suppress surface cracks from growing)
S = stress amplitude Adapted from Fig. 8.24, Callister 7e.
Increasing
σm
near zero or compressive σm moderate tensile σm Larger tensile σm N = Cycles to failure
--Method 1: shot peening
--Method 2: carburizing
shot
putt surface into compression
2. Remove stress concentrators.
bad
C-rich C rich gas
better Adapted from Fig. 8.25, Callister 7e.
bad
better 10
STEADY-STATE OR MINIMUM CREEP RATE
Stage II creep rate also called ll d steady-state t d t t or minimum creep rate is an important p design g p parameter. Empirical equation relates creep rate to applied stress via a power-law l relationship l h n & ε s = K1σ
where ε&s = steady - state creep rate K 1 , and n are constants
σ = applied constant stress Qc ⎞ ⎛ K 1 = K 2 exp ⎜ − ⎟ RT ⎝ ⎠ Q C = activation energy
Universal creep equation for steady-state creep rate:
⎛ QC ⎞ n & ε s = K 2σ exp⎜ − ⎟ ⎝ RT ⎠
3. Grain boundary sliding (GBS): Grain boundaries become weak with temperature. Grains slide resulting in permanent deformation. Creep cavitation (void formation) occurs at grain boundaries as a result of grain boundary sliding.
Creep cavities formed at grain boundaries in an austentic stainless steel
Usually it is difficult to observe creep damage except cavitation due to GBS It is creepp damage
Data Extrapolation-Larson-Miller parameter •Creep tests are of long duration and it is impractical to collect data over long times times, e.g several years. solution: perform creep rupture tests at higher temperatures under same stress for shorter times (accelerated tests) extrapolate for service conditions through Larson-Miller analysis Larson-Miller parameter:
T(C + log l tr) - C is a constant (~20), p ((K)) - T is temperature - tr is the creep rupture life (hours) plot log σ versus L-M parameter Used to find rupture time at any temperature for a given stress.
LarsonMiller parameter for ductile cast iron
Appendix
14
14
2. Fail-Safe (Damage Tolerant) Fatigue Design Design for adequate service life after fatigue damage has occurred. 1 1.
Determination of inspection schedule based on Non-Destructive Testing – Initial defect size. – Fatigue crack growth rate. Paris Equation – Critical defect size for failure failure.
NDE: Utrasonic inspection, X-Ray fluorescence, thermography : 2.
Re-assessment of the residual life after inspection allows the total service life to be extended
Crack Grow C wth Rate da a/dN
Fracture Toughness
Fatigue Growth rate
n~3 for steel, n~4 for aluminum aluminum, C = material constant C = 1.62 × 1012 for steel.
Paris equation
da = C ΔK n dN ΔKth
n ~ 2 to 4
15 Stress Intensity Range ΔK
15
Creep Sample deformation at a constant stress (σ) vs. time σ,ε
σ
0
t
Primary Creep: slope (creep rate) decreases with time time. Secondary Creep: steady-state i.e., constant slope. Tertiary Creep: slope (creep rate) 16rate. increases with time, i.e. acceleration of
Adapted from Fig. 8.28, Callister 7e. 16
Creep • Occurs at elevated temperature, T > 0.4 Tm
tertiary primary
secondary
elastic
Adapted from Figs. 8.29, Callister 7e.
17
17
