Creep Slides

EN4701 Failure Analysis
Creep and Stress Relaxation

Definitions
Creep
Stress Relaxation
Creep Testing
Temperature and Stress Effects
Parameter Methods
Larson-Miller Parameter
Stress Relaxation
Creep Resistant Materials
School of Engineering, RGU

Slide 2 of 52

Creep
material.
time-dependent strain which accompanies

application of constant stress (or load) to a
Stress Relaxation

time-dependent reduction in stress at

constant strain in a material.

Slide 3 of 52
Creep
At room temperatures most metallic materials show only very
small creep rates - usually ignored.
(apart from low-melting point metals such as lead)
With increase in temperatures creep rate increases.
Above about 0.4Tm creep becomes very significant.
Tm = melting temperature in Kelvin
In high temperature applications creep is an important failure
mechanism.
e.g. steam plant, gas turbines, nuclear and chemical processes
Other materials, e.g thermoplastic polymers, especially
sensitive to creep deformation.

Slide 4 of 52
Creep Testing
Typically consists of subjecting specimen to constant load or
stress (usually load) whilst maintaining constant temperature.
Deformation or strain measured as function of elapsed time.
Graph of strain against (log) time plotted.
For metallic material usually conducted in uniaxial tension
using specimen with same geometry as for tensile test.
Uniaxial compression tests more appropriate for brittle
materials such as ceramics.
Creep properties usually independent of loading method.
Temperature and stress level are significant.


Slide 5 of 52
Creep Testing
Typical constant load creep curve.
On load - instantaneous deformation, mostly elastic.
Graph divided into

three regions.
I primary creep
II secondary creep
III tertiary creep
II
III
in s ta n ta n e o u s
e la s t ic d e f o r m a t io n
( lo g s c a le ) t

Slide 6 of 52
Primary Creep
Primary or transient creep occurs first, typified by
continuously decreasing strain rate,
i.e. slope of curve reduces with time.
Suggests material experiencing increase in creep resistance or
strain hardening (deformation becomes more difficult as
material is strained).
II
III
e la s tic d e f o r m a t io n
( lo g s c a le ) t


Slide 7 of 52
Secondary Creep
For secondary or steady-state creep strain rate is constant
(time on log scale) and graph becomes linear.
Work-hardening process of dislocation pile-up and
entanglement balanced by recovery processes of dislocation
climb and cross-slip.
Stage II has longest duration.
II
III
( lo g s c a le ) t


Slide 8 of 52
Tertiary Creep
During tertiary creep there is acceleration of strain rate and
ultimately failure.
Failure usually called rupture.
Results from microstructural changes, e.g. grain boundary
separation, formation of internal cracks, cavities and voids.
For tensile loads neck
may form leading to
reduction in CSA and
increase in strain rate.
II
III
( lo g s c a le ) t


Slide 9 of 52
Creep Design Parameters

Most important parameter from creep test is slope of
secondary portion of creep curve.
Often called minimum or steady-state creep rate, d s
s
dt
Engineering design parameter considered for long-life
applications,
e.g. nuclear power plant components scheduled to operate for
several decades, when failure or too much strain not an option.
However, for relatively short-life creep situations it is time to
rupture, or rupture lifetime, tr that is important design
consideration,
e.g. turbine blades in aircraft and rocket motor nozzles.
Obviously, creep rupture tests take longer than those to
determine steady-state creep rate.

Slide 10 of 52
Worked Example 1
Blades of a steam turbine are 200 mm long.
Initial clearance between blade tip and housing is 0.075 mm.
b la d e

c a s in g
0 .0 7 5 m m

Slide 11 of 52
Worked Example 1
Blades elastically extend in operation by 0.02 mm.
c a s in g
0 .0 7 5 m m
b la d e
0 .0 2 m m

Slide 12 of 52
Worked Example 1

Blades elastically extend in operation by 0.02 mm.
Required minimum clearance is 0.025 mm.
c a s in g
0 .0 7 5 m m
b la d e
0 .0 2 m m
0 .0 2 5 m m
Slide 13 of 52
Worked Example 1
Calculate:
(a) max percentage creep strain allowed in blades,
(b) how long blades can operate if minimum creep strain rate is
3 x 10-6 % /h.
c a s in g
0 .0 7 5 m m
b la d e
0 .0 2 m m
0 .0 2 5 m m
Slide 14 of 52
Worked Example 1 - Solution

creep extension = initial clearance final clearance elastic extension
= 0.075 0.025 0.02

= 0.03 mm
creep extension
maximum percentage creep strain =
x 100
original length
0.03
100
200
maximum percentage creep strain = 0.015 %

0 .0 7 5 m m
c a s in g
0.03 mm
b la d e
0 .0 2 m m
0 .0 2 5 m m
Slide 15 of 52

0.015
3 10 6
strain
life
strain rate
life = 5000 hours
0 .0 7 5 m m
c a s in g
0.03 mm
b la d e
0 .0 2 m m
0 .0 2 5 m m
Slide 16 of 52
Stress and Temperature Effects

Stress and temperature play important part in creep behaviour.
Shape of curve will depend on temperature and stress.
main factors controlling work-hardening and recovery processes.
As temperature and/or stress increase creep behaviour changes

in a similar way
instantaneous strain increases (stress only)
secondary creep rate increases

secondary stage shortens
tertiary creep begins earlier
rupture occurs earlier
in c r e a s in g t e m p e r a t u r e
a n d /o r s tre s s
( lo g s c a le ) t

Slide 17 of 52
Alternative Graphing
Results of creep rupture tests often presented on graphs of
stress versus rupture lifetime
both quantities on logarithmic axes.
Stress (MPa)
Linear relationship seen to occur at each temperature.

Not always case over relatively large stress ranges.
Rupture Lifetime (hours)


Slide 18 of 52
Alternative Graphing
Results of creep rupture tests can also be presented on graphs
of stress vs. steady-state creep rate.
both quantities on logarithmic axes.
Stress (MPa)
Linear relationship seen to occur at each temperature.

Not always case over relatively large stress ranges.
Steady-State Creep Rate (% / 1000 h)


Slide 19 of 52
Tutorial Questions
You can now attempt Q1 Q4 on the tutorial sheet.
Q1 & Q2 use figure 8.38
Q3 & Q4 - use figure 8.39


Slide 20 of 52
Effect of Temperature
With increase in temperature, creep rate increases.
softening processes can take place more easily.
Creep rate closely linked to Arrhenius equation:
s Ae H RT
s = secondary creep rate
H = activation energy for creep for material
R = universal gas constant = 8.31 J/mol K
T = absolute temperature
A = constant
Taking logs of equation gives:

ln s ln A
H
RT

Slide 21 of 52
Effect of Stress
Secondary creep rate also increases with increasing stress.
Relationship usually expressed using the power law equation:
s B n
B and n are constants (n usually between 3 and 8).
ln s ln B n ln
Plot of ln s versus ln will

result in straight line of slope n.
.
lo g 1 0 ( s)
lo g 1 0 B
lo g
10

Slide 22 of 52
Effect of Stress and Temperature

Arrhenius equation and power law equation can be combined
into one equation
Effects of strain rate, stress and temperature are combined:
s Dn e H RT
D = constant.
ln s
H
ln D n ln
RT
All 3 equations above are not universally applicable.

Constants A, B, D and H are not true constants as values
depend on stress, temperature range and metallurgical
variables.

Slide 23 of 52
Worked Example 2
Alloy steel bar 1500 mm long and 2500 mm2 CSA.
Subjected to axial tensile load of 8.9 kN at temp. of 600C.
Determine value of creep elongation in 10 years using power
n
B
law relationship s
B = 26 x 10-12 and n = 6.0 at 600C.
B determined using hours for time and MPa for stress.
Solution
applied stress,

P
8900
3.56 MPa
A 2500

Slide 24 of 52

Using power law equation: strain rate, s Bn
6 .0
s 26 10 12 3.56 26 10 12 2036
s 52.94 10 9 / h
Duration of test, t = 10 x 365 x 24 = 87600 hours

Strain, = 52.94 x 10-9 x 87600
= 4.637 x 10-3
Original length,
L = 1500 mm
Total elongation, = L = 1500 x 4.637 x 10-3
= 6.96 7 mm

Slide 25 of 52
Tutorial Questions


Slide 26 of 52
Parameter Methods
Engineers often have to confirm that a component will
withstand use at elevated temperatures for given lifetime.
For some applications - furnace equipment or steam lines may be considerable number of years.
Impractical to test a component for long periods of time
(~20+ years) and form of data extrapolation needed.
Simplest method is to:
test at proposed temperature,

calculate secondary (minimum) creep rate,
assume this will continue for proposed lifetime,
calculate whether or not resulting creep strain is acceptable.


Slide 27 of 52
Parameter Methods
One obvious disadvantage ignores tertiary creep and
rupture.
Number of workers have proposed methods of accelerated
creep testing to overcome this.
Tests carried out at higher temperatures than used in practice.
Results used to predict creep-life or creep-strain over longer
period at lower temperatures.


Slide 28 of 52
Most well known method is that of Larson and Miller and
based on Arrhenius equation:
Ae H RT
Assuming strain rate, s, proportional to reciprocal of time to
rupture (as strain rate increases time to rupture decreases):
1
tr
1
H RT
De
Combining this with Arrhenius equation:
tr
D = constant


Slide 29 of 52
1
De H RT
tr
Taking logs (to base 10): log10 t r log10 D 0.4343
1
Rearranging: log10 t r C m
T
log D)
H1
RT
C = constant (=
10
m is function of stress (H assumed stress related)

T log10 t r C m
Multiplying both sides by T:
T log10 tr C
known as Larson-Miller parameter

(or Larson-Miller rupture parameter) and is function of stress.

Slide 30 of 52
T log10 tr C
For ferrous metals C usually between 15 and 30 - with time to
rupture, tr, in hours.
C can be found from intercept of graph of log10 tr vs 1/T for
tests at same stress level.
C often assumed = 20.
Results from range of tests carried out at different
temperatures and stresses shown on a Larson-Miller plot.
Sometimes called a master curve.


Slide 31 of 52
Stress (MPa)
Stress (MPa)
Larson-Miller Master Curves
103 T(20 + log tr) (K-h)

S-590 Iron
103 T(20 + log tr) (K-h)

18-8 Mo stainless steel
Slide 32 of 52
Worked Example 3
Lives of Nimonic 90 turbine blades tested under varying
conditions of stress and temperature shown in table.
stress (MPa)
180
180
300
350
temperature (C)
750
800
700
650
life (h)
3000
500
5235
23820
Use info. to produce master curve based on Larson-Miller

parameter.
Calculate expected life of blade subjected to stress of 250 MPa
at temperature of 750C.


Slide 33 of 52

Larson-Miller parameter is T log10 tr C
First step is to determine constant C.
Plot graph of log10 tr against 1/T
T = temperature in Kelvin for a series of tests at same stress.
C = intercept on graph.
Only two results at same stress (180 MPa) so we can calculate
C or draw a graph.
stress (MPa)
180
180
300
350
temperature (C)
750
800
700
650
life (h)
3000
500
5235
23820
Slide 34 of 52

Calculation Method
T1(log10 t r1 C) T2 (log10 t r 2 C)
1023 (log10 3000 C) 1073 (log10 500 C)
1023(3.477 + C) = 1073(2.699 + C)
3557 + 1023C = 2896 + 1073C
50C = 661
C = 13.22


Slide 35 of 52

Graphical Method
Extracting data for stress of 180 MPa from table gives:
Plot graph of log10 tr vs 1/T for tests at same stress.
stress (MPa)
180
180
temperature, T (C)
750
800
life, tr (h)
3000
500
1/T (/K)
977.5 x 10-6
932.0 x 10-6
log10 tr
3.477
2.699
3.5
C = 13.222
3.4
3.3
y = 0.0171x - 13.222
3.2
log10(tr)
Plotting graph in Excel

and including equation of
line gives intercept as
-13.222
3.1
3
2.9
2.8
2.7
2.6
2.5
930
940
950
960
970
980
1/T (1/K x 10^6)


Slide 36 of 52

Now plot graph of log10 vs T(log10 tr +C) which is Larson-Miller
master curve.
In tabular form:
stress (MPa)
180
180
280
350
temp. (C)
750
800
700
650
life (h)
3000
500
5235
23820
T log10 t r C x10
17.083
17.083
16.484
16.244
As expected Larson-Miller parameter is same for same stress.

Only 3 points to plot.


Slide 37 of 52

1000
stress (MPa)
Stress = 250 MPa

LM parameter 16.7 x 103
250
100
16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9 17
17.1 17.2
Larson-Miller parameter T(log t + C) / 10^3


Slide 38 of 52

Stress = 250 MPa
LM parameter 16.7 x 103
For a temperature of 750 C (1023 K) calculate expected life:

16.7 x 103 = 1023(log10tr + 13.22)
log10tr = 16.32 13.22 = 3.10
tr = 1272 hours


Slide 39 of 52
Tutorial Questions


Slide 40 of 52
Stress Relaxation
Stress relaxation is time-dependent reduction in stress at
constant strain in a material.
It is important in situations involving, e.g.
cylinder head bolts or rivets in pressure vessels at high
temperatures.
Under these conditions

assumed strain remains
constant but over time
stress reduces
Maybe to level such that
integrity of structure is
compromised.
s tre s s
m in
tim e


Slide 41 of 52
Stress Relaxation
Consider 2 plates held together by
bolt deformed by stress 0.
Stress produces initial strain, 0,
which is all elastic.
0
E
At elevated temperatures and under

steady-state creep, bolt will
,
elongate at a strain rate,
dictated by power law.
d c
B n
dt


Slide 42 of 52
Stress Relaxation
If thickness of plates constant,
creep strain, c, will reduce elastic
initial strain to give time-dependent
elastic strain, e:
e = 0 - c
Since creep strain decreases elastic
component of initial strain, e, a
corresponding decrease in stress
must also result.


Slide 43 of 52
Stress Relaxation
Initial strain, 0, is constant, so if equation is differentiated wrt
time:
d e
d c
dt
dt
e = /E where is time-dependent, instantaneous stress:
d
d E
c
dt
dt
E assumed constant.
d
1 d
c B n
Incorporating power law: E dt
dt


Slide 44 of 52
Stress Relaxation
Therefore:
Rearranging:
d
EB n
dt
1
d EBdt
n
Integrate both sides of equation:
1
d EB dt
n
1
EBt K
n1
n 1
K = constant of integration
To find K consider initial condition.


Slide 45 of 52
Stress Relaxation
1
EBt K
n1
n 1
When time, t = 0, stress, = 0
1
EB 0 K
n1
n 1 0
K
1
n 1 n01
1
1
EBt
n 1
n 1
n 1 n01
Substituting back and rearranging gives:
1
1
EB n 1 t
n 1
n 1
0

Slide 46 of 52
Worked Example 4
Steel bolt clamping two rigid plates together
Held at temperature of 1000C.
Use power law equation
At 28 MPa, n = 3.0 and
= 0.7 x 10-9 /h
Calculate stress remaining in bolt after 9000 hours if initially

tightened to stress of 70 MPa.
Assume E = 200 GPa.


Slide 47 of 52

Bn
Power law equation is:
0.7 10
B 28 10
6 3
0.7 10 9
32
B
3
.
189
10
21952 1018
For stress relaxation:
n 1
1
1
2
70 10 6
n0 1
EB n 1 t
200 10 9 3.189 10 32 2 9000
1
16
3
.
188
10
2

= 56 MPa

Slide 48 of 52

Several factors affect creep characteristics of metals, e.g.
- melting temperature
- elastic modulus
- grain size
In general:
higher melting temperature (Tm)

greater elastic modulus (E)
larger grain size
the more a material is resistant to creep.


Slide 49 of 52

Materials especially resistant to creep and used in high
temperature applications include:
Stainless steels
Refractory metals (niobium, molybdenum, tungsten, tantalum,
osmium)
Superalloys (alloys of cobalt, nickel or iron)
However, there are practical difficulties with some:

Tungsten is difficult to machine
Molybdenum forms volatile oxides
Osmium is very expensive.
Therefore, nickel and cobalt more extensively used.


Slide 50 of 52

Advanced processing techniques can increase creep
resistance, e.g. turbine blade
directional solidification producing elongated grains
possible to create single-crystal blade


Slide 51 of 52
Tutorial Questions
You can now attempt Q11 on the tutorial sheet.


Slide 52 of 52

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EN4701 Failure Analysis

Creep and Stress Relaxation

Creep and Stress Relaxation

School of Engineering, RGU

Creep and Stress Relaxation

time-dependent strain which accompanies

Creep and Stress Relaxation

time-dependent reduction in stress at

School of Engineering, RGU

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

Graph divided into

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

Creep Design Parameters

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

School of Engineering, RGU

Blades of a steam turbine are 200 mm long.

Worked Example 1 - Solution

= 0.075 0.025 0.02

maximum percentage creep strain = 0.015 %

Worked Example 1 - Solution

life = 5000 hours

Stress and Temperature Effects

As temperature and/or stress increase creep behaviour changes

secondary creep rate increases

School of Engineering, RGU

Linear relationship seen to occur at each temperature.

Rupture Lifetime (hours)

School of Engineering, RGU

Linear relationship seen to occur at each temperature.

Steady-State Creep Rate (% / 1000 h)

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

Creep rate closely linked to Arrhenius equation:

Creep and Stress Relaxation

School of Engineering, RGU

Plot of ln s versus ln will

Taking logs of equation gives:

School of Engineering, RGU

Effect of Stress and Temperature

All 3 equations above are not universally applicable.

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU

Worked Example 2 - Solution

Duration of test, t = 10 x 365 x 24 = 87600 hours

School of Engineering, RGU

Creep and Stress Relaxation

School of Engineering, RGU