Materials Engineering and Technology (MEE 1005)
Winter Semester 2021-22 MEE1005
Instructor: Ariful Rahaman
Contact Information
Instructor: Ariful Rahaman
Office: CDMM G-02
E-mail: arahaman@vit.ac.in
Tensile Test
specimen
machine
2
Types of Loading
Tensile Test
Load data is obtained from Load cell
Strain data is obtained from Extensometer
4
Important Mechanical Properties from a Tensile Test
 Young's Modulus: This is the slope of the linear portion of
the stress-strain curve, it is usually specific to each material; a
constant, known value.
 Yield Strength: This is the value of stress at the yield point,
calculated by plotting young's modulus at a specified percent of
offset (usually offset = 0.2%).
 Ultimate Tensile Strength: This is the highest value of stress
on the stress-strain curve.
 Percent Elongation: This is the change in gauge length
divided by the original gauge length.
5
Terminology
 Load - The force applied to a material during testing.
 Strain gage or Extensometer - A device used for measuring
change in length (strain).
 Engineering stress - The applied load, or force, divided by
the original cross-sectional area of the material.
 Engineering strain - The amount that a material deforms per
unit length in a tensile test.
Engineering Stress- Engineering Strain
Engineering Stress- Engineering Strain
σ = F/A0
where F is the force and A0 the original area of cross section
 The engineering stress
 The engineering strain, ε = (l-l0)/l0
where l is the length of the original gauge length under force F, and l0 is the original
gauge length.
l
l
   1 or   1 
lo
lo
LO
LO
True Stress-True Strain
9
Poissons Ratio
Problems
Q1: A steel bar is 10 mm diameter and 2 m long. It is stretched with a force of
20 kN and extends by 0.2 mm.
Calculate the true/engineering stress and true/engineering strain.
Q2:An Aluminium tensile test specimen is 5 mm diameter with a gauge length
of 50 mm. The force measure at the yield point was 982 N and the maxiumum
force was 1.6 KN. Calculate the yield stress and ultimate tensile stress. At a
point on the proportional section the extension was 0.03 mm and the force
800N. Calculate the modulus of elasticity.
Q3: A tensile test on a cold worked brass gave the following results. The
diameter of the test specimen d= 16 mm and the gauge lenth was 80 mm.
After fracture the gauge length was 85 mm and the fracture point was was 15
mm diameter. The maxiumu load was 150 KN. The load and extension at the
elastic limit were 70 KN and 0.5 mm, respectively. Calculate the modulus of
elasticity. Calculate the % of elongation and % of reduced cross sectional
area.
Q4: Derive an expression for true strain as a function of D and Do for a tensile
test specimen of round cross section, where D = the instantaneous diameter
of the specimen and Do is its original diameter.
Stress-Strain Diagram
ultimate
tensile
strength
3
 UT S
necking
Strain
Hardening
yield
strength
y
Fracture
5
2
Elastic region
slope =Young’s (elastic) modulus
yield strength
Plastic region
ultimate tensile strength
strain hardening
fracture
Plastic
Region
Elastic
Region
σ Eε
σ
E
ε
4
1
σy
E
ε 2  ε1
 ) (DL/Lo)
Strain (
Stress-Strain Diagram
• Elastic Region (Point 1 –2)
- The material will return to its original shape
after the material is unloaded( like a rubber band).
- The stress is linearly proportional to the strain in
this region.
σ Eε
σ
ε
or
σ
E
ε
: Stress(psi)
E : Elastic modulus (Young’s Modulus) (psi)
: Strain (in/in)
- Point 2 : Yield Strength : a point where permanent
deformation occurs. ( If it is passed, the material will
no longer return to its original length.)
Stress-Strain Diagram
Stress-Strain Diagram
Mild Steel
Problems
Q3: A steel bar of 25mm diameter was tested in tension and results were recorded
as, limit of proportionality = 196.32kN, load at yield = 218.13kN, ultimate load =
278.20 kN. The elongation measured over a gauge length of 100mm was
0.189mm at proportionality limit, length of the bar between gauge marks after
fracture was 112.62mm and minimum diameter was 23.64mm. Compute stress in
the specimen at various stages, Young’s modulus, % elongation and %
contraction.
Q4: Derive an expression for true strain as a function of D and Do for a tensile
test specimen of round cross section, where D = the instantaneous diameter
of the specimen and Do is its original diameter.
Problems
Q1: Example: Mechanical Property Determinations from Stress–Strain Plot From the tensile
stress–strain behavior for the brass specimen shown in Figure below, determine the
following: (a) The modulus of elasticity (b) The yield strength at a strain offset of 0.002 (c)
The maximum load that can be sustained by a cylindrical specimen having an original
diameter of 12.8 mm (0.505 in.) (d) The change in length of a specimen originally 250 mm (10
in.) long that is subjected to a tensile stress of 345 MPa (50,000 psi)
Tensile Properties: Ductility
Ductile Materials
• A ductile material is one with a large Percentage of elongation before failure.
Material
Percentage
of
Elongation
Low-Carbon
37%
Medium-Carbon
30%
High-Carbon
25%
Ductile Materials
Properties of ductile materials:
•
•
•
•
Easily drawn into wire or hammered thin.
Easily molded or shaped.
Capable of being readily persuaded or influenced tractable.
Easily stretched without breaking in material strength.
Problem
Q3: Stress-strain data for a series of alloy steels are described in the figure
below.
(a) Which of these six steels has the highest yield strength?
(b) Which of these steels is the most ductile?
(c) For the DP 500/800 Steel, determine the:
(i) Yield stress(ii) The tensile stress
(iii) The elongation to failure
Brittle Materials
• Brittle material is one which is having very low percentage of elongation.
•
Brittle materials break suddenly under stress at a point just beyond its elastic limit.
• A Brittle material exhibits little or no yielding before failure.
• Brittle material will have a much lower elongation and area reduction than ductile
ones. The tensile strength of Brittle material is usually much less than the
compressive strength.
Problem
Q1: Three different materials, designated A, B,and C, are tested in tension using test specimens
having diameters of 0.505 in. and gage lengths of 2.0 in. (see figure).
At failure, the distances between the gage marks are found to be 2.13, 2.48, and 2.78 in.,
respectively. Also, at the failure cross sections the diameters are found to be0.484, 0.398, and
0.253 in., respectively. Determine the percent elongation and percent reduction in area of each
specimen, and then, using your own judgment, classify each material as brittle or ductile.
Problem
Q3: (a) Show, for a tensile test, that
if there is no change in specimen volume during the deformation process (i.e., Aolo = Aflf).
(b) Using the result of part (a), calculate the percent cold work (% CW) experienced by naval
brass when a strain value is 0.3.
Stress-Strain Diagram of Ceramic/Metal/Polymer
Hardness Test
 A measurement of a material’s resistance to penetration or localized plastic deformation
 Steps in hardness test:

A small indenter is forced into the surface of a material to be tested with certain load.

The depth / size of the resulting indentation is measured.

Such data are converted to a hardness number.
 The softer the material, the larger and deeper the indentation, and the lower the hardness
number.
 Hardness tests are performed more frequently than any other mechanical tests

Simple / inexpensive

nondestructive

Other mechanical properties may be estimated from hardness data.
Hardness Test
•There are many hardness tests currently in use.
•The neccessity for all these different hardness tests is due to
the need for categorizing the great range of hardness nfrom soft
rubber to hard ceramics.
Two main types: Brinell and Rockwell
Hardness Test
Current hardness tests measure either the size of indented area
or the depth of penetration
(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
The size of the indented
area is measured
The depth of penetration is
measured
Hardness Test
Brinell Hardness
Brinell-indenters
Hardness Test
Brinell Hardness
 A spherical indenter (10 mm diameter) is shot with 29
kN force at the target
 Frequently the indenter is steel, but for harder materials
it is replaced with a tungsten carbide sphere
 The diameter of the indentation is recorded
 The indentation diameter can be correlated with the
volume of the indentation.
Hardness Test
Brinell hardness (HB) is an accurate hardness
measurement for soft materials
 Indenter: a hardened steel or WC
sphere (10mm diameter)
 Load: 500-3000 kg (generally, 3000
kg)
 Restricted to softer steels or other
softer metals
Brinell Hardness
•The Brinell methods consists of indenting the metal (usually) with a 10 mm diameter steel ball subjected
to a load of 3000 kg.
•For soft materials the load is reduced to 1500 or 500 kg, as may be required to avoid too deep an
indentation
•In Brinell Test, the BHN (Brinell Hardness Number) of nearly all materials is influenced by the
–The magnitude of the indenting load,
–Diameter of the ball indenter and
–The elastic characteristics of the indenter ball.
Hardness Test
Typical HB values
Material
Softwood (e.g., pine)
Hardwood
Aluminum
Copper
Mild steel
18-8 (304) stainless steel annealed
Glass
Hardened tool steel
Rhenium diboride
Hardness
1.6 HBS 10/100
2.6–7.0 HBS 1.6 10/100
15 HB
35 HB
120 HB
200 HB
1550 HB
1500–1900 HB
4600 HB
Hardness Test
Disadvantages: Brinell Hardness
the test machine is very heavy.
 the area of indentation is quite large that it affects the surface quality.
This is why, sometimes it is considered as a destructive test.
the thickness of the test sample also limits its use, e.g. thin sheets will
bulge or be destroyed during the test.
for very hard materials, the test results are unreliable. The ball gets
flattened on hard surfaces.
Hardness Test
Rockwell Hardness
Rockcell Ball indenters
Basic Principle:
o The indenter moves down into position on the part surface
o A minor load is applied and a zero refernce position is established
oThe major load is applied for a specified time period
5 mm carbide ball indenter
oThe major load is released leaving the minor load applied
Major load of 100 kg (for B-scale) or 150 kg (for C-scale)
Rockcell Diamond indenters
Hardness Test
Rockwell hardness (HR) is the most widely used of all metal hardness
testing methods
 Indenter: a small hardened steel sphere or a
diamond cone
 Load: 10-150 kg
 The hardness value (based on the depth of
penetration) is read directly from either a digital
readout or from a rotary dial.
 For testing steel, the two scales most often used:
----Rockwell B scale (RB) RB 0 to RB 100: softer
low-carbon steels, Al, & other softer nonferrous
materials
---Rockwell C scale (RC) RC 20 to RC 70: hard steels
Hardness Test
Rockwell Hardness
Rockwell Hardness Scales
Scale
A
Code
HRA
Load
Indenter
Use
60 kgf
120° diamond cone
Tungsten carbide
Al, brass, and soft
steels
B
HRB
100 kgf
1/16 in diameter steel sphere
C
D
HRC
HRD
150 kgf
100 kgf
120° diamond cone
120° diamond cone
E
HRE
100 kgf
1/8 in diameter steel sphere
F
HRF
60 kgf
1/16 in diameter steel sphere
G
HRG
150 kgf
1/16 in diameter steel sphere
Harder steels
40 RHB indicates Rockwell hardness of 40 measured on B-scale
80 RHC indicates Rockwell hardness of 80 measured on C-scale
Hardness Test
Conversion of Hardness
 Conversion of hardness
 Standard hardness conversion tables
 Correlation between hardness & tensile strength
 For most steels,
UTS(MPa) = 3.45 x HB
Hardness Test
Summary of Hardness Tests
Hardness Test
Indenter
Load
Brinell hardness
(HB)
hardened steel or
WC sphere (10mm
diameter)
500-3000 kg
small hardened
Rockwell hardness
steel sphere or a
(HR)
diamond cone
10-150 kg
Problem
Q2: A cylindrical specimen of cold-worked steel has a Brinell hardness of 240.
(a) Estimate its ductility in percent elongation.
(b) If the specimen remained cylindrical during deformation and its original
radius was 10 mm, determine its radius after deformation.
Problem
Q3: (a) Show, for a tensile test, that
if there is no change in specimen volume during the deformation process (i.e., Aolo = Aflf).
(b) Using the result of part (a), calculate the percent cold work (% CW) experienced by naval
brass when a strain value is 0.3.
Mechanisms of Strengthening in Metals
 Grain size reduction
Solid solution alloying
Strain hardening
Precipitation Hardening
Strengthening by Grain Size Reduction
The grain boundary acts as a barrier to dislocation motion for two
reasons
 Difficulty for a dislocation to pass through two different grain
orientations
 The atomic disorder within a grain boundary region contributes to a
discontinuity of slip planes from one grain to another
Cont…
A fine grained material is harder and stronger than one that is coarse,
since the former has a greater total grain boundary area to impede
dislocation motion.
The relationship between the yield stress and grain size was pproposed
by Hall and Petch
σy
Yield stress
Cont…
Solid Solution Strengthening
Lattice strains produced by the introduction of solute atoms can be divided into:
Compressive lattice strain
Tensile lattice strain
Solid Solution Strengthening
Strain Hardening
Cont…
Precipitation Hardening
Precipitation hardening is accomplished by two different heat treatments.
Solution heat treatment is a heat treatment in which all solute atoms are
dissolved to form a single phase solid solution.
For an alloy of composition Co,
1.
heat the alloy to a temperature within a phase field—To, and wait until all the β phase is completely
dissolved.
2. rapid cool to temperature T1 to the extent that any diffusion and the formation of β phase are prevented.
3. a nonequilibrium situation of a phase solid solution supersaturated with B atoms is present.
Precipitation heat treatment
The supersaturated a solid solution is ordinarily heated to temperature T2 within the
α + β two-phase region.
2. keep temperature and the β precipitate phase begins to form as finely dispersed
particles of composition Cβ (aging).
3. after an appropriate aging time at T2, the alloy is cooled to room temperature.
How the dislocations can interact with a particle?
Strain Hardening
 Cold-Working:
A process of strain hardening at room temperature to
deform the material beyond the elastic range to obtain a
desired property.
 Increase yield strength; Decrease ductility
 Examples of cold-working: rolling, drawing, extruding,
cutting, pulling, indenting…
Q1: A cylindrical rod having an initial diameter of 6 mm is to be cold
worked (CW) such that cross-sectional area is reduced and final
diameter is 5 mm. Determine % CW. If % CW = 20 and final diameter =
5 mm, determine original diameter?
Toughness
The ability of a metal to deform plastically and to absorb energy in the
process before fracture is termed toughness.
Fracture/Failure
• The fracture of any material occurs in two steps:
Crack formation
Crack propagation
• The failure of engineering materials is classified in terms of being:
Ductile: significant plastic deformation prior to fracture
Brittle: little or no plastic deformation prior to fracture
Fracture
behavior:
%AR or %EL
Very Moderately Brittle
Ductile
Ductile
Large
• Ductile
fracture is usually
desirable!
Moderate
Ductile:
warning before
fracture
Small
Brittle:
No
warning
Example: Failure of a Pipe
• Ductile failure:
--one piece
--large deformation
• Brittle failure:
--many pieces
--small deformation
Figures from V.J. Colangelo and F.A.
Heiser, Analysis of Metallurgical Failures
(2nd ed.), Fig. 4.1(a) and (b), p. 66 John
Wiley and Sons, Inc., 1987. Used with
permission.
Ductile Fracture
 Ductile fracture of many
engineering metals results in a “cup
and cone” fracture surface.
Coalescence
Small cavity of
Initial necking formation
cavities to
form a crack.
This is created by a process
known as microvoid coalescence.
Crack propagation
Final shear fracture
Cont…
Cup-and-cone fracture in aluminum.
ductile fracture from uniaxial tensile loads
ductile fracture from shear loading
Brittle Fracture
• Brittle
fracture involves very little plastic
deformation
 The fracture surface is usually flat and
perpendicular to the applied stress.
• A brittle fracture surface often shows
Chevron markings, or a series of fan-like
ridges or “river pattern”
Chevron
Brittle fracture in a mild steel.
fan-like ridges
Cont…
• Crack propagation in brittle fracture can be either:
 Transgranular: through the grains (also called cleavage)
 Intergranular: along the grain boundaries. In both cases, the
surface usually appears shiny because the facets reflect light.
Cont…
Crack propagation through the ineterior of
grains for transgranular fracture
Ductile cast iron showing a transgranular
fracture surface
Crack propagation along grain boundaries for
Intergranular fracture
Intergranular fracture surface
IMPACT FRACTURE TESTING
 In pendulum type impact testing, the
impact load is produced by swinging
of an impact weight (W = m * g) from
initial height (h) through the arc of a
circle, thus striking and fracturing the
notched specimen. After that, the
weight reaches maximum height (h/).
Negclecting frictional losses, the
energy used to fracture the specimen
(U) is then approximately defined as:
Absorbed Energy = Initial Potential Energy – Final Potential Energy
(energy to rupture) (energy before rupture) (energy after rupture)
U = m * g * (h – h/)
 The absorbed energy (U), indicated on the scale of tester, is expressed
in joule (i.e. N*m) .This energy value is sometimes called “impact
toughness”.
62
IMPACT FRACTURE TESTING
The Impact test measures a
materials ability to absorb energy.
This quality is often referred to as
the Toughness of the material.
Charpy
Izod
Two standardized test: Izod and Charpy
Tested samples
Ductile to Brittle Transition Temperature (DBTT)
 Deformation should be continuous across grain boundary in polycrystals for
their ductile behaviour
 5 independent slip systems required (absent in HCP and ionic materials)
FCC crystals remain ductile upto 0 K
 Common BCC metals become brittle at low temperatures or at v.high
strain rates

f , y →
y
f
Brittle
Ductile
T →
DBTT
Ductile – brittle transition temperature (DBTT)
 Ductile  y < f  yields before fracture
 Brittle  y > f  fractures before yielding
f , y →
Ductile to Brittle Transition Temperature (DBTT)
f
y (BCC)
y (FCC)
T →
DBTT
No DBTT
 Ductile  y < f  yields before fracture
 Brittle  y > f  fractures before yielding
Ductile to Brittle Transition Temperature (DBTT)
• Increasing temperature...
--increases %EL and Kc
• Ductile-to-Brittle Transition Temperature (DBTT)...
Impact Energy
FCC metals (e.g., Cu, Ni)
BCC metals (e.g., iron at T < 914°C)
polymers
Brittle
More Ductile
High strength materials (  y > E/150)
Temperature
Ductile-to-brittle
transition temperature
Design Strategy:
Stay Above The DBTT!
• Pre-WWII: The Titanic
Reprinted w/ permission from R.W. Hertzberg,
"Deformation and Fracture Mechanics of Engineering
Materials", (4th ed.) Fig. 7.1(a), p. 262, John Wiley and
Sons, Inc., 1996. (Orig. source: Dr. Robert D. Ballard,
The Discovery of the Titanic.)
• Problem:
• WWII: Liberty ships
Reprinted w/ permission from R.W. Hertzberg,
"Deformation and Fracture Mechanics of Engineering
Materials", (4th ed.) Fig. 7.1(b), p. 262, John Wiley and
Sons, Inc., 1996. (Orig. source: Earl R. Parker,
"Behavior of Engineering Structures", Nat. Acad. Sci.,
Nat. Res. Council, John Wiley and Sons, Inc., NY,
1957.)
Used a type of steel with a DBTT ~ Room temp.
Fatigue
 Fatigue
is a form failure that occurs in structures subjected to
dynamic and fluctuating stresses.
 Estimated to comprise approximately 90% of all metallic failures.
Fatigue failure is brittle like in nature even in normally ductile metals.
The fatigue occurs by the initiation and propagation of cracks.
Fatigue under cyclic/repeated loading
 Cracks generally grow under repeated loading
 Trucks passing over bridges,
 Bicycle pedals
 May result failure or fracture: fatigue fracture
 Periodic inspections required for fatigue critical systems
 Thermal fatigue: repeated heating and cooling can cause a cyclic
stress due to differential thermal expansion and contraction
Fatigue
Fatigue failures are often easy to identify.
The fracture surface near the origin is usually smooth (Beach mark-crack
initiation point). The surface becomes rougher as the crack increases in
size.
Concentric line patterns: the slow cyclic build up of crack growth from a
surface intrusion. Line patterns are on a much finer scale and show the
position of the crack tip after each cycle.
Granular portion of the fracture surface: rapid crack propagation at the
time of catastrophic failure
Crack Initiation and Propagation
Three distinct steps of fatigue failure
 crack initiation, wherein a small crack forms at some point of high stress concentration
Crack propagation, during which this crack advances incrementally
with each stress cycle
 Final failure, which occurs very rapidly once the advancing crack has reached a
critical size
The total number of cycles to failure is the sum of cycles at the first
and the second stages
Number of cycles for crack initiation
Number of cycles to failure
Nf = Ni + Np
Number of cycles for crack propagation
Fatigue: Cyclic Stresses
Revised stress cycle; in which the stress alternates from
A maximum tensile stress (+) to a maximum compressive
Stress(-) of equal magnitude.
Repeated stress cycle; in which maximum and minimum
stresses are asymmetrical relative to the zero stress level.
 Random stress cycle
Concepts
 Constant amplitude stressing
 Mean stress
 Stress amplitude (half of the
range) variation about the
mean
 Stress ratio R, Amplitude ratio
Fatigue Test
Fatigue testing apparatus for making rotating bending tests
 Result is plotted as S (stress) vs N(number of cycles to failure
The most important fatigue data for engineering designs are the S-N curves, which is the Stress-Number
of Cycles curves.
In a fatigue test, a specimen is subjected to a cyclic stress of a certain form and amplitude and the
number of cycles to failure is determined.
The number of cycles, N, to failure is a function of the stress amplitude, S.
A plot of S versus N is called the S-N curve.
Fatigue: S-N Curve
Fatigue Design Parameters
• Fatigue limit, Sfat:
S = stress amplitude
--no fatigue if S < Sfat
unsafe
case for
steel (typ.)
Sfat
safe
103
• Sometimes, the
fatigue limit is zero!
105
107
109
N = Cycles to failure
S = stress amplitude
unsafe
safe
103
105
107
109
N = Cycles to failure
Adapted from Fig.
8.19(a), Callister 7e.
case for
Al (typ.)
Adapted from Fig.
8.19(b), Callister 7e.
Fatigue: S-N Curve
Fatigue Limit:
•For some materials such as BCC steels, the S-N curves become horizontal when the stress amplitude is decreased to a
certain level.
•This stress level is called the Fatigue Limit, or Endurance Limit.
Fatigue Strength:
For materials, which do not show a fatigue limit such as Al, Cu, and Mg (non-ferrous alloys), fatigue strength is specified as
the stress level at which failure will occur for a specified number of cycles, where 107 cycles is often used.
• Fatigue life: indicates how long (no. of cycles) a component survives a particular stress.
Fatigue strength: is applicable to a component with No endurance limit. It is the maximum stress for which
fatigue will not occur at a particular number of cycles, in general, 108 cycles for metals.
Endurance ratio: the endurance limit is approximately ¼ to ½ the tensile strength.
Factors that affect fatigue life
Magnitude of stress (mean, amplitude...)
Quality of the surface (scratches, sharp transitions and edges).
--Method 1: shot peening
--Method 2: carburizing
C-rich gas
shot
put
surface
into
compression
Solutions:
Polishing (removes machining flaws etc.)
Introducing compressive stresses (compensate for applied tensile stresses) into thin
surface layer by “Shot Peening”- firing small shot into surface to be treated. High-tech
solution - ion implantation, laser peening.
Case Hardening - create C- or N- rich outer layer in steels by atomic diffusion from the
surface. Makes harder outer layer and also introduces compressive stresses
Optimizing geometry - avoid internal corners, notches etc.
2. Remove stress
concentrators.
bad
better
Adapted from
Fig. 8.25, Callister 7e.
bad
better
Factors that affect fatigue life
Thermal Fatigue: is induced at elevated temperatures by fluctuating thermal stresses. The
Origin of these thernmal stresses is the restraint to the dimensional expansion and cotraction
Improving Fatigue Life
Eliminate restraint by design
 use materials with low thermal expansions.
Corrosion Fatigue: failure that occurs by the simultaneous action of a cyclic stress and
Chemical attack.
Improving Fatigue Life
 add protective surface coating
 decrease corrosiveness of medium
Add residual compressive stresses
Creep
 Time -dependent deformation which occurs when materials are loaded above 0.4 Tmelt

Sample deformation at a constant stress () vs. time
,
0
t
Primary Creep: slope (creep rate)
decreases with time.
Secondary Creep: steady-state
i.e., constant slope.
Tertiary Creep: slope (creep rate)
increases with time, i.e. acceleration of rate.
Creep: Parameters
Most of component life spent here.
Strain rate is constant at a given T, σ
Steady state creep rate
On the other hand, for many short life creep situations, time to rupture
or the rupture lifetime tr, is the dominant design
Creep: Stress and Temperature Effects
With increasing stress and temperature
The instantaneous strain at the time of stress
application increases
Steady state creep rate is increased
The rupture lifetime is diminished
stress exponent (material parameter)
activation energy for creep
(material parameter)
strain rate
Applied stress
material const.
Alloys For High Temperature Use
Gas turbine, high temperature steam boilers, heat treating furnaces,
aircraft, missiles etc
Creep is generally minimized in materials with:
 high melting temperatures
 high elastic modulus
 large grain sizes
 Following materials are resilient to creep:
 Refractory metals, like Nb, Mo, W etc.
 Stainless Steel
 Superalloys , Co-Ni based