High Temperature Materials

Course KGP003

ByDocent. N. Menad

Dept. of Chemical Engineeringand Geosciences

Div. Of process metallurgy

Luleå University of Technology( Sweden )

Mechanical Properties of Materials

Ch. 6

Mechanical Properties

Mohs scale of Hardness

Minerals Mohs’scale

Talc 1Gypsum 2Calcite 3Fluorite 4Apatite 5Feldspar 6Quartz 7Garnet 6.5 – 7.5Beryl (Emerald) 7.5Topaz 8Corund (saphire) 9Diamant 10

Ch. 6

Mechanical Properties of Materials

IntroductionMany materials are subjected to forces or loads during the service

Ex: Al alloy ( airplane wing)The steel (in automobile axle)

The mechanical behaviour of a material reflects the relationshipbetween its response or deformation to applied load or force.

The important mechanical properties are:

Strengh, hardness, ductility and stiness

M. P Laboratory experiments ( carefully )

M. P are concern Producers and consumers of materials, research organizations, government agencies

( interpretation of their results )

Standardization of M.P are made by professional societies

In USA ASTM ( American Society for Testing and Materials )

Ch. 6

Mechanical PropertiesConcepts of stress and strain

If load is static or changes relatively slowly with time and is applied uniformly over a cross section or surface of a member, the mechanical behaviour may be performed by a simple stress-strain test.

Application of load 1. Tension 2. Compression 3. Shear

l lo

F

FAo

F

llo

F

Ao

F

F

Fθ

Ao φ

T

T

Tensile load produces an elongation and

positive linear strain

Compressive load produces contraction and

negative linear strain

Tensile load produces an Shear strain γ, γ = tan θ

Torsionaldeformation (angle

twist φ) produced by an applied torque T

Ch. 6

Mechanical PropertiesTension Tests

One of the most common mechanical stress-strain tests is performed in tension

To ascertain different mechanical properties of materials ( design )

A specimen is deformed, usually to fracture, with a gradually increasing the force applied uniaxially along the long axis. See figure 6.2

The cross section is circular or rectangular

The standard diameter 12.8 mm (0.5 in)

Length of reduced section 4 times this diameter (60 mm) 2¼”in

Ch. 6

Mechanical Properties

Standard tensile specimen with circular cross section

2”Gauge length

3/4”

3/8” Radius

diameter

Reduced section

0.505” diameter

2 ¼”

Ch. 6

Mechanical Properties

Extensometer

Load cell

Specimen

Moving crosshead

This is the equipment that can be used for the tensile stress-strain

The specimen is elongated by the moving crosshead

The load cell measures the magnitude of the applied load

The extensometer measures the elongation of specimen

The test takes several minutes

Ch. 6

Mechanical Properties

Compressive strength testsCement Compressive Strength Testing

Ch. 6

Mechanical PropertiesEngineering stress

FAo

σ =F is the force applied perpendicular

to the specimen cross section

Ao

( newtons (N) or pounds force (lbf )Is cross section area before any force applied ( m2 or in.2 )The unit of σ is MPA (SI)

1 MPA = 106 N/m2

Engineering strain

ε = li - lolo

Δl= lo

lo is the original lengthli the instantaneous length

Called just strain and is in dependent of the unit system, (sometimes in percentage)

Ch. 6

Mechanical Properties

The engineering stress-strain curve The parameters, which are used to describe the stress-strain curve of a metal, are:

Tensile strength, Yield strength or yield point, Percent elongation,Reduction of area.

The first two are strength parameters; the last two indicate ductility.

The general shape of the engineering stress-strain curve is shown in this figure . In the elastic region stress is linearly proportional to strain. When the load exceeds a value corresponding to the yield strength, the specimen undergoes gross plastic deformation.

It is permanently deformed if the load is released to zero. The stress to produce continued plastic deformation increases with increasing plastic strain, i.e., the metal strain-hardens.

Ch. 6

Mechanical Properties

Compression Tests

The tests are conducted in a manner similar to the tensile test

exception

The force is compressive and the specimen contracts along the direction of stress

FAo

σ =

ε = li - lolo

Δl= lo

Are used to compute compressive stress and strain

By convention, compressive force is taken to be negative.Produces a negative stress

lo > li From The compressive strains are negative

Ch. 6

Mechanical Properties

Shear and Torsional Tests

F

F

Fθ

AoThe shear stress τ is:

τ =F

Ao

F is the load or force

Ao is area of faces

The shear strain γ is the tangent of the strain angle θ

φ

T

T

Torsion or a variation of pure shear, wherein a structure member is twisted in the manner of the following figure

The torsional tests are performed on cylindrical solid shafts or tubes.

τ is a function of the applied torque T

γ is related to the angle of twist φ

Ch. 6

Mechanical PropertiesGeometric considerations of the stress state

θ

σ

P´

P

σ´ τ ´

It is important to underline that the stress state is a functionof the orientations of the planes See this figure

σ Applied parallel to its axis

P – P´ Plane oriented at some arbitrary angle θ

We have a complex stress state

σ´ acts to P-P´planeτ ´ acts parallel to P-P´plane

By using mechanics of materials principles , we can write the equations of these two stresses

σ´

Tensile stress is

Tensile stress or normal stress

Shear stress

= σ cos2θ σ 1 + cos2θ2

τ ´= σ sinθ cosθ = σsin2θ

2=

Ch. 6

Mechanical PropertiesElastic deformation

The behaviour of Stress-Strain

Stre

ss

Strain

Load

Slope = modulusof elasticity

unload

Stress-strain diagram showing linear elastic deformation for loading and unloading cycles

σ = E ε

The degree of strains depends on the magnitude of imposed stress

For most metals, stress and strain are proportional to each other as given by this equation

Hooke’s law

E is the modulus of elasticity or young’s modulus

45 GPa (6.5 X106 psi) and 407 GPa (59X106 psi)for W

Table 6.1, p. 118 summarises the Modulus elasticity values of some metals

Stress and strain are proportional Elastic deformation

Elastic deformation is non-permanentWhen the force applied is released, the piece returns to its original shape

Ch. 6

Mechanical Properties

Strain

σ

ε

Stre

ss

ΔσΔε =

Secant modulusBetween origin and σ 1

ΔσΔε

Tangent modulus(at σ 2)

=

At atomic levelSmall changes in the inter-atomic spacing and stretching of the inter-atomic bonds

Nonlinear elastic behaviour

Determination of the moduli

EdFdr

ro

α( inter-atomic force-separation curve)

Ch. 6

Mechanical Properties

dF

dr r0

Strongly bonded

Weaklybonded

Separation rForc

e F

0

This figure shows the force-separation curves for materials

These materials have two inter-atomic bonds

Strong and week

The for each is determined at ro

Ch. 6

Mechanical Properties

-200 0 200 400 600 800Temperature (oC)

Mod

ulus

of e

last

icity

(GPa

)

400

300

200

100

0

7060

50

40

30

20

10

0

Tungsten

Steel

Aluminum

Mod

ulus

of e

last

icity

(106

psi)

The modulus of elasticity for ceramic materials are higher than for metals. For polymers, they are lower

These differences can be explain by the different types of atomic bonding in these materials

τ = G γG is the shear modulus ( slope of the linear elastic region of the shear stress-strain curve

These values for metals are given in Table 6.1

Ch. 6

Mechanical PropertiesAnelasticity

In most engineering materials, a time-dependent elastic strain component is exist

The elastic deformation is time dependent

After application of the stress, the elastic deformation will continue, and upon load release some finite time is required for complete recovery

This time behaviour called anelasticity

For metals, the anelastic component is small (negligible)

For polymers, the amount of anelastic is very high and is called viscoelastic behaviour

Ch. 6

Mechanical Properties

Problem 6.1 page 121

lo = 305 mm (12 in.) σ = 267 MPaEcu = 110 GPa (16X106 psi),

from table 6.1

Ecu = 110 X 103 MPa

σ =Δ llo

ε E = E

Δ l =σ lo

E=

267 MPa X 305 mm

110 X 103 MPa

Δ l = 0.77 (0.03 in.)

Ch. 6

Mechanical Properties

x

y

z

Δlz2

loz

σz

σz

Elastic Properties of Materials

loxΔlx2

The compressive strain εx and εyare determinedIf the applied stress is unixaial (in the z direction) and the material is isotropic. εx = εy

υεx

εz=εyεz

=

Poisson’s ratio

should be 1/4For isotropic materials υυ max

( no net volume change)= 0.50 For metals and alloys υ ( 0.25 – 0.35)=

See table 6.1 page118

E = 2G(1+υ)For isotropic materials shear and elastic moduli are related each other

Ch. 6

Mechanical Properties Plastic deformationTensile Properties

Yielding and Yielding strength

Strain

Stre

ss

Lower yieldpoint

Upper yield point

σyP

Strain

0.002

Stre

ss

Elastic Plastic

σy

The elastic deformation in most structures will result by the application of stress.

It is important to know the behaviour of this stress at which plastic deformation begins, or where the phenomena of yielding occurs

Ch. 6

Mechanical PropertiesTensile Strength

Stre

ss

Stráin

MTS

Typical engineering stress-strain behaviour to fracture. Point F.

TS is tensile strength at point M.

TS may vary from 50MPa (700 psi) for Al to 300 MPa (450 000 psi) for steels

Ex. Pr. 6.3 page 126

Ch. 6

Mechanical PropertiesDuctility

It is an important parameter for mechanical property

Stre

ssStrain

Brittle

Ductile

C

B´

C´

B

A

Tensile stress-strain behaviour and ductile

materials loaded to fracture

Material with no plastic deformation upon fracture, called brittle

See this figure

Ductility of material may be expressed quantitatively (% elongation or % reduction in area)

lf - lo

lo% EL = X 100

lf is fracture length and lo is the original gauge length

% RA =Ao - Af

AoX 100

Ch. 6

Mechanical Properties

Determination of ductility of materials is important because:

1. It indicates to a designer the degree to which a structure will deform plastically before fracture

2. It specifies the degree of allowable deformation during fabrication operations

Brittle materials have a fracture strain < 5%

Table 6.2 gives typical mechanical properties of several metals and alloys in annealed state

Ch. 6

Mechanical Properties

Resilience

Materialdeformation

Material deformed elastically, upon unloading

Energy

Absorption

It is the modulus of resilience, strain energy per unit volume required to stress a material from an unload

state up to point of yielding

Ch. 6

Mechanical Properties

εy StrainSt

ress

0.002

σ y

Ur = σ εd

εy

0

The modulus of resilience for a specimen. From this figurer

Ur = εyσ y1/2

If the elastic region is linear

σ = E ε σyεE

= Ur = σ y1/2σyE =

σ 2y

2E

Ch. 6

Toughness

True stress and strain

Ex. Pr. 6.4, 6.5 p. 133

Ch. 6

Fatigue Crack Initiation

Ch. 6

Strengthening/Hardening Mechanisms

Strain Hardening Control of Grain Size

The ability of a crystalline material to plastically deform largely depends on the ability for dislocation to move within a material. Therefore, impeding the movement of dislocations will result in the strengthening of the material. There are a number of ways to impede dislocation movement, which include:

1. controlling the grain size (reducing continuity of atomic planes) 2. strain hardening (creating and tangling dislocations) 3. alloying (introducing point defects and more grains to pin dislocation)

Ch. 6

Strengthening/Hardening Mechanisms

Effects of Elevated Temperature on Strain Hardened Materials

When strain hardened materials are exposed to elevated temperatures, the strengthening that resulted from the plastic deformation can be lost. This can be a bad thing if the strengthening is needed to support a load. However, strengthening due to strain hardening is not always desirable, especially if the material is being heavily formed since ductility will be lowered. Heat treatment can be used to remove the effects of strain hardening. Three things can occur during heat treatment

Recovery . Re-crystallization . Grain growth

Ch. 6