Chapter 10: Phase Diagrams and Microstructures
This presentation is the property of its rightful owner.
Sponsored Links
1 / 41

Chapter 10: Phase Diagrams and Microstructures PowerPoint PPT Presentation


  • 69 Views
  • Uploaded on
  • Presentation posted in: General

Chapter 10: Phase Diagrams and Microstructures. Temperature vs. composition behavior of the various alloy constitutes the phases. Allows design and control of heat treatment by controlling equilibrium and non-equilibrium structure.

Download Presentation

Chapter 10: Phase Diagrams and Microstructures

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Chapter 10 phase diagrams and microstructures

Chapter 10: Phase Diagrams and Microstructures

  • Temperature vs. composition behavior of the various

  • alloy constitutes the phases.

  • Allows design and control of heat treatment by controlling

  • equilibrium and non-equilibrium structure.

  • Structure(phases and microstructure) often controls properties

  • of materials, and, therefore, depends on thermal history.

  • Objectives:

  • Read and evaluate phases present at T.

  • Determine composition of phases and phase fraction.

  • Know difference between types of reactions, e.g., eutectic, eutectoid, proeutectoid.

  • Make a schematic diagram of microstructure.


Chapter 10 phase diagrams and microstructures

Phase B

Phase A

Nickel atom

Copper atom

Phase Diagrams: Issues to Address

• When we combine two elements, what equilibrium state do we get?

• In particular, if we specify...

--a composition (e.g., wt%A - wt%B), and

--a temperature (T)

then...

How many phases do we get?

What is the composition of each phase?

How much of each phase do we get?

Ordered

Phase

Solid-solution

Phase


Chapter 10 phase diagrams and microstructures

Example Phase (T-c) Diagram : Cu-Zn

• Solution – solid, liquid, or gas solutions, single phase

• Mixture – more than one phase

liquid

bcc-ss

fcc-ss

2-phase

hcp-ss

hcp

bcc

CsCl


Chapter 10 phase diagrams and microstructures

Sugar/Water Phase Diagram

10

0

Solubility

L

Limit

8

0

(liquid)

6

0

+

L

Temperature (°C)

S

(liquid solution

4

0

i.e., syrup)

(solid

20

sugar)

0

20

40

60

80

100

C

= Composition (wt% sugar)

Sugar

Water

Solubility Limit in Phase Diagram

• Solubility Limit:

Max concentration for which only a solution occurs.

• Ex: Water-Sugar

Q: What is solubility limit at 20oC?

Answer: 65wt% sugar.

If Co < 65wt% sugar:syrup

If Co > 65wt% sugar:syrup + sugar.

Adapted from Fig. 10.1,

Callister & Rethwisch 3e.

• Solubility limit increases with T:

e.g., if T = 100C, solubility limit = 80wt% sugar.


Chapter 10 phase diagrams and microstructures

Components and Phases

• Components: elements or compounds that are mixed initially (e.g., Al and Cu)

• Phases: physically and chemically distinct material regions that result (e.g., a and b).

Aluminum-Copper Alloy

Adapted from chapter-opening photograph, Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e.


Chapter 10 phase diagrams and microstructures

B(100°C,C = 70)

1 phase

D(100°C,C = 90)

2 phases

100

L

80

(liquid)

+

60

L

S

Temperature (°C)

(

liquid solution

(solid

40

i.e., syrup)

sugar)

A(20°C,C = 70)

2 phases

20

0

0

20

40

60

70

80

100

C

= Composition (wt% sugar)

Temperature (T) and Composition (C) Effects

• Changing T can change # of phases: path A to B.

• Changing Co can change # of phases: path B to D.

water- sugar

system

Adapted from Fig. 10.1,

Callister & Rethwisch 3e.


Chapter 10 phase diagrams and microstructures

Criteria for Solid Solubility: recall Hume-Rothery rules

Simple system (e.g., Ni-Cu solution)

  • Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility.

  • Ni and Cu are totally soluble in one another for all proportions.


Chapter 10 phase diagrams and microstructures

T(°C)

• 2 phases:

1600

(liquid)

1500

L (liquid)

(FCC solid solution)

• 3 phase fields:

1400

L

a

liquidus

+

1300

L +

a

L

solidus

a

a

1200

(FCC solid

1100

solution)

1000

wt% Ni

0

20

40

60

80

100

Cu-NiPhase Diagram: T vs. c (wt% or at%)

• Tell us about phases as function of T, Co, P.

• For this course:

--binary systems: just 2 components.

--independent variables: T and Co (P = 1atm is always used).

-- isomorphous

i.e., complete solubility of one component in another; a phase field extends from 0 to 100 wt% Ni.

Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).


Chapter 10 phase diagrams and microstructures

T(°C)

1600

1500

L (liquid)

1 phase: a

B

(1250°C, 35 wt% Ni):

(1250°C,35)

1400

liquidus

2 phases: L + a

a

+

1300

solidus

L

B

a

(FCC solidsolution)

1200

A(1100°C,60)

1100

1000

wt% Ni

0

20

40

60

80

100

Phase Diagrams: # and types of phases

• Rule 1: If we know T and Co, then we know:

--the # and types of phases present.

• Examples:

Cu-Ni

phase

diagram

A(1100°C, 60 wt% Ni):


Chapter 10 phase diagrams and microstructures

Cu-Ni system

T(°C)

A

TA

tie line

liquidus

L (liquid)

At TA

= 1320°C:

1300

a

+

L

Only Liquid (L) present

B

TB

solidus

CL = C0

( = 35 wt% Ni)

a

a

At TD

= 1190°C:

+

L

(solid)

1200

D

Only Solid (a) present

TD

C = C0

( = 35 wt% Ni)

32

35

4

3

20

30

40

50

At TB

= 1250°C:

CL

C0

C

wt% Ni

Both

and L present

CL

= C

( = 32 wt% Ni)

liquidus

C

= C

( = 43 wt% Ni)

solidus

Phase Diagrams: composition of phases

• Rule 2: If we know T and Co, then we know:

--the composition of each phase.

• Examples:

Consider C0 = 35 wt% Ni


Chapter 10 phase diagrams and microstructures

Cu-Ni system

T(°C)

A

TA

liquidus

tie line

L (liquid)

At TA

: Only Liquid (L) present

1300

a

+

L

B

WL= 1.00, Wa = 0

TB

solidus

At TD

:

Only Solid (

) present

S

R

a

a

+

WL

= 0, W

= 1.00

L

a

(solid)

1200

D

TD

At TB

:

Both

and L present

32

35

4

3

20

3

0

4

0

5

0

S

=

WL

CL

C0

Ca

wt% Ni

R

+

S

R

=

= 0.27

Wa

R

+

S

Phase Diagrams: wt. fraction of phases

• Rule 3: If we know T and C0, then can determine:

-- the weight fraction of each phase.

• Examples:

Consider C0 = 35 wt% Ni

W = wt. fraction of phase out of whole.


Chapter 10 phase diagrams and microstructures

T(°C)

tie line

liquidus

L (liquid)

1300

a

+

L

B

solidus

T

B

a

a

+

L

(solid)

1200

S

R

20

3

0

4

0

5

0

C0

CL

C

wt% Ni

The Lever Rule

• Sum of weight fractions:

• Conservation of mass (Ni):

• Combine above equations:


Chapter 10 phase diagrams and microstructures

T(°C)

tie line

liquidus

L (liquid)

1300

a

+

M

ML

L

B

solidus

T

B

a

a

+

L

(solid)

1200

R

S

S

R

20

3

0

4

0

5

0

C0

CL

C

wt% Ni

The Lever Rule: an interpretation

  • Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm

What fraction of each phase?

Think of tie line as a lever


Chapter 10 phase diagrams and microstructures

T(°C)

L: 35wt%Ni

L (liquid)

a

130

0

A

+

L

L: 35 wt% Ni

B

a: 46 wt% Ni

35

46

C

32

43

D

L: 32 wt% Ni

24

36

a

a: 43 wt% Ni

+

120

0

E

L

L: 24 wt% Ni

a: 36 wt% Ni

a

(solid)

110

0

35

20

3

0

4

0

5

0

wt% Ni

C0

Phase and Microstructure (equilibrium)

Example: Cooling in Cu-Ni Binary

• Consider microstuctural changes that accompany the cooling of a

C0 = 35 wt% Ni alloy

  • From liquid, solid phase nucleates.

  • From solid, other phases can nucleate.

  • Like ice, many grains of solid form.

  • wt% of SOLUTE given by line dropped from boundaries

  • Fraction of PHASES present given by the “lever rule”.

  • Microstructure different depending on cool slowly or quench.


Chapter 10 phase diagrams and microstructures

Phases and Microstructure (non-equilibrium)

  • Upon cooling quickly (i.e. nonequilibrium), microstructure has range of composition depending on when it was formed.

  • Inside nucleus of solid phase higher composition (in Cu-Ni case) due to its creation at higher T.

  • Outside part of growing solid phase nucleus has lower composition due to its forming at lower T.

Core-like development


Chapter 10 phase diagrams and microstructures

Cored versus Equilibrium Phases

• Ca changes as we solidify.

• Cu-Ni case:

First a to solidify has Ca = 46 wt% Ni.

Last a to solidify has Ca = 35 wt% Ni.

• Fast rate of cooling:

Cored structure

• Slow rate of cooling:

Equilibrium structure


Chapter 10 phase diagrams and microstructures

60

%EL

for pure Cu

400

%EL

for

50

pure Ni

TS for

Elongation (%EL)

40

pure Ni

Tensile Strength (MPa)

300

30

TS for pure Cu

200

20

0

20

40

60

80

100

0

20

40

60

80

100

Cu

Ni

Cu

Ni

Composition, wt% Ni

Composition, wt% Ni

Recall: Mechanical Properties of Cu-Ni

• Effect of solid solution strengthening on:

--Ductility (%EL,%AR)

--Tensile strength (TS)

-Peak as a function of Co

-Minimum as a function of Co


Chapter 10 phase diagrams and microstructures

Cu-Ag

system

T(°C)

1200

L (liquid)

1000

a

L

+

a

b

L

+

779°C

b

800

TE

8.0

91.2

71.9

600

a

+

b

400

200

80

100

0

20

40

60

CE

C

,

wt% Ag

cooling

heating

Binary-Eutectic Systems

has a special composition

with a minimum melting T.

  • Ex: Cu-Ag

  • 3 single-phase regions

  • (L, , )

  • Limited solubility

  • : mostly Cu

  • : mostly Ag

  • TE: no liquid below TE.

  • cE: composition for min. melting T (Eutectic).

Eutectic: direct from liquid to 2-phase solid upon cooling: L   + 


Chapter 10 phase diagrams and microstructures

T(°C)

300

L (liquid)

a

L

+

a

b

b

L

+

200

183°C

18.3

61.9

97.8

C- C0

150

S

R

S

=

W

=

R+S

C- C

100

a

+

b

99 - 40

59

=

=

= 0.67

99 - 11

88

100

0

11

20

60

80

99

40

C0- C

R

W

C

C

C0

=

=

C, wt% Sn

C - C

R+S

40 - 11

29

=

= 0.33

=

99 - 11

88

Solder for electronics

Example 1: Lead-Tin (Pb-Sn) Eutectic Diagram

• For a 40wt%Sn-60wt%Pb alloy at 150oC, determine...

--phases present: a + b

--compositions of phases:

Ca = 11 wt% Sn

Cb = 99 wt% Sn

-- relative amount of each phase:

Use the “Lever Rule”

Adapted from Fig. 10.8, Callister & Rethwisch 3e.


Chapter 10 phase diagrams and microstructures

T(°C)

300

L (liquid)

a

L

+

220

a

b

b

R

L

+

S

200

CL - C0

46 - 40

183°C

=

W

=

= 0.21

a

CL - C

46 - 17

100

a

+

b

100

17

46

0

20

40

60

80

C

CL

C0

C, wt% Sn

C0 - C

23

=

WL

=

= 0.79

CL - C

29

Example 2: Pb-Sn Eutectic System

• For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine:

-- phases present:

a + L

-- phase compositions

Ca = 17 wt% Sn

CL = 46 wt% Sn

-- relative amt of phases


Chapter 10 phase diagrams and microstructures

T(°C)

L: C0wt% Sn

400

L

a

L

300

L

a

+

a

200

(Pb-Sn

a: C0wt% Sn

TE

System)

100

b

+

a

0

10

20

30

C

,

wt% Sn

C0

2

(room T solubility limit)

Microstructure for Eutectic Diagrams: Pb-Sn

• For alloys with C0 < 2 wt% Sn

• Result: at room temperature -- polycrystalline with grains of

a phase having C0

• Note: liquid-to-solid creates grain boundaries.


Chapter 10 phase diagrams and microstructures

L: C0 wt% Sn

T(°C)

400

L

L

300

a

L

+

a

a: C0wt% Sn

a

200

TE

a

b

100

b

+

a

Pb-Sn

system

0

10

20

30

C

,

wt% Sn

C0

2

(sol. limit at T

)

18.3

room

(sol. limit at TE)

Microstructure for Eutectic Diagrams: Pb-Sn

• For alloys with

2 wt% Sn < C0< 18.3 wt% Sn

• Result:

at T’s in a + b range

-- polycrystalline with agrains

and small b-phase particles

• Note: Solubility depends on T.


Chapter 10 phase diagrams and microstructures

Micrograph of Pb-Sn

T(°C)

eutectic

L: C0 wt% Sn

microstructure

300

L

Pb-Sn

system

a

L

+

a

b

L

200

183°C

TE

100

160m

a

: 97.8 wt% Sn

Adapted from Fig. 10.14, Callister & Rethwisch 3e.

: 18.3 wt%Sn

0

20

40

60

80

100

97.8

18.3

CE

C, wt% Sn

61.9

Microstructure at Eutectic

• For alloy of composition C0 = CE

• Result: Eutectic microstructure (lamellar structure)

-- alternating layers (lamellae) of a and b phases.

Light: Sn-rich

Dark: Pb-rich

Adapted from Fig. 10.13, Callister & Rethwisch 3e.


Chapter 10 phase diagrams and microstructures

Lamellar Eutectic Structure

Diffusion local

Adapted from Figs. 10.14 & 10.15, Callister & Rethwisch 3e.


Chapter 10 phase diagrams and microstructures

• Just above TE :

L

T(°C)

L: C0

wt% Sn

Ca

= 18.3 wt% Sn

L



CL

= 61.9 wt% Sn

300

L

Pb-Sn

system

S

Wa

L

+

= 0.50

=

R

+

S

a

b

WL

= (1-

W

)

= 0.50

L

+

a

R

S

200

TE

S

R

• Just below TE :

C

= 18.3 wt% Sn

100

+

a

primary

C

= 97.8 wt% Sn

a

eutectic

S

b

eutectic

W

= 0.73

=

R

+

S

0

20

40

60

80

100

W

= 0.27

18.3

61.9

97.8

Adapted from Fig. 10.16, Callister & Rethwisch 3e.

C, wt% Sn

Microstructure “below” Eutectic (hypoeutectic)

• For alloys for 18.3wt%Sn < Co < 61.9wt%Sn

• Result: a crystals and a eutectic microstructure


Chapter 10 phase diagrams and microstructures

Solder: Lead-Tin (Pb-Sn) microstructure

  • For 50 wt% Pb alloy:

  • Lead-rich  phase (dark)

  • Lamellar eutectic structure

  • of Sn-rich  phase (light).

L

L + 

 + 

* Why is Liquid-phase ~62.9wt%Sn and -phase ~16.3wt%Sn at 180 C?

* For fraction of total  phase (both eutectic and primary), use the Lever Rule.


Hypo eutectic hyper eutectic

Hypoeutectic & Hypereutectic

Adapted from Fig. 10.8,

Callister & Rethwisch 3e.

(Figs. 10.14 and 10.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)


Chapter 10 phase diagrams and microstructures

Mg2Pb

Intermetallic Compounds

Adapted from

Fig. 10.20, Callister & Rethwisch 3e.

Intermetallic compounds exists as a line on the phase diagram - not an area - because of stoichiometry (i.e. c of a compound is a fixed value).


Chapter 10 phase diagrams and microstructures

cool

cool

cool

heat

heat

heat

  • Eutectoid – one solid phase transforms to two other solid phases

    • S2S1+S3

    •  + Fe3C (For Fe-C, 727C, 0.76 wt% C)

intermetallic compound - cementite

  • Peritectic - liquid and one solid phase transform to a second solid phase

    • S1 + LS2

    •  + L (For Fe-C, 1493°C, 0.16 wt% C)

Eutectic, Eutectoid, & Peritectic

  • Eutectic - liquid transforms to two solid phases

    • L +  (For Pb-Sn, 183C, 61.9 wt% Sn)


Chapter 10 phase diagrams and microstructures

Peritectic transformation  + L

Eutectoid transformation  + 

Eutectoid, & Peritectic

Cu-Zn Phase diagram

Adapted from Fig. 10.21, Callister & Rethwisch 3e.


Chapter 10 phase diagrams and microstructures

Ceramic Phase Diagrams

MgO-Al2O3 diagram:

Adapted from Fig. 10.24, Callister & Rethwisch 3e.


Chapter 10 phase diagrams and microstructures

Iron-Carbon (Fe-C) Phase Diagram

T(°C)

1600

Adapted from Fig. 10.28,Callister & Rethwisch 3e.

d

L

1400

g

+L

A

- Eutectic (A):

g

L+Fe3C

1200

1148°C

(austenite)

Þ

g

+

L

Fe3C

A

- Eutectoid (B):

1000

g

+Fe3C

Fe3C (cementite)

a

g

Þ

a

+

Fe3C

g

g

+

800

g

a

g

g

B

727°C = T

eutectoid

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

C, wt% C

(Fe)

Result: Pearlite is

4.30

0.76

alternating layers of and Fe3C phases

120 mm

a

Fe3C (cementite-hard)

a

(ferrite-soft)

• 2 points


Chapter 10 phase diagrams and microstructures

T(°C)

1600

d

L

1400

(Fe-C

g

+L

g

g

g

System)

1200

L+Fe3C

1148°C

(austenite)

g

g

g

1000

g

g

+Fe3C

g

g

Fe3C (cementite)

a

g

g

800

727°C

a

a

a

g

g

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

a

C, wt% C

(Fe)

C0

0.76

pearlite

Hypoeutectoid

100 mm

steel

pearlite

proeutectoid ferrite

Adapted from Fig. 10.34, Callister & Rethwisch 3e.

Hypoeutectoid Steel

Adapted from Figs. 10.28 and 10.33


Chapter 10 phase diagrams and microstructures

T(°C)

1600

d

L

1400

(Fe-C

g

+L

g

System)

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Wa= s/(r+s)

Wa’= S/(R+S)

Fe3C (cementite)

a

r

s

g

g

800

Wg=(1 - Wa)

W=(1 – Wa’)

727°C

a

a

a

g

g

R

S

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

a

C, wt% C

(Fe)

C0

0.76

Wpearlite=Wg

pearlite

Hypoeutectoid

100 mm

steel

Fe3C

pearlite

proeutectoid ferrite

Adapted from Fig. 10.34, Callister & Rethwisch 3e.

Hypoeutectoid Steel


Chapter 10 phase diagrams and microstructures

T(°C)

1600

d

L

1400

g

+L

g

g

g

1200

L+Fe3C

1148°C

g

g

(austenite)

g

1000

g

g

+Fe3C

g

g

Fe3C (cementite)

Fe3C

800

g

g

a

g

g

600

a

+Fe3C

400

C0

0

1

2

3

4

5

6

6.7

C, wt%C

0.76

(Fe)

pearlite

Hypereutectoid

60mm

steel

pearlite

proeutectoid Fe3C

Adapted from Fig. 10.37, Callister & Rethwisch 3e.

Hypereutectoid Steel


Chapter 10 phase diagrams and microstructures

T(°C)

1600

d

L

1400

g

+L

g

Fe3C

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Wg=x/(v + x)

Fe3C (cementite)

x

v

800

g

W=(1-Wg)

g

Fe3C

a

g

g

V

X

600

a

+Fe3C

400

C0

0

1

2

3

4

5

6

6.7

C, wt%C

0.76

(Fe)

pearlite

Wpearlite=Wg

Wa= X/(V+X)

Hypereutectoid

60mm

W=(1 - Wa)

steel

Fe3C’

pearlite

proeutectoid Fe3C

Adapted from Fig. 10.37, Callister & Rethwisch 3e.

Hypereutectoid Steel


Chapter 10 phase diagrams and microstructures

Example Problem Steel

  • For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following:

  • The compositions of Fe3C and ferrite ().

  • The amount of cementite (in grams) that forms in 100 g of steel.

  • The amounts of pearlite and proeutectoid ferrite () in the 100 g.


Chapter 10 phase diagrams and microstructures

1600

d

L

1400

T(°C)

g

+L

g

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Fe3C (cementite)

800

Amount of Fe3C in 100 g

= (100 g)WFe3C

= (100 g)(0.057) = 5.7 g

727°C

R

S

600

a

+Fe3C

400

0

1

2

3

4

5

6

6.7

C0

C

C, wt% C

CFe C

3

Solution to Problem

a) Use RS tie line just below Eutectoid

Ca = 0.022 wt% CCFe3C = 6.70 wt% C

  • Use lever rule with the tie line shown


Chapter 10 phase diagrams and microstructures

1600

d

L

1400

T(°C)

g

+L

g

1200

L+Fe3C

1148°C

(austenite)

1000

g

+Fe3C

Fe3C (cementite)

Amount of pearlite in 100 g

= (100 g)Wpearlite

= (100 g)(0.512) = 51.2 g

800

727°C

V

X

600

a

+Fe3C

400

C0

0

1

2

3

4

5

6

6.7

C

C

C, wt% C

Solution to Problem

c) Using the VX tie line just above the eutectoid and realizing that

C0 = 0.40 wt% CCa = 0.022 wt% CCpearlite = C = 0.76 wt% C


Alloying steel with more elements

Alloying Steel With More Elements

• Teutectoid changes:

• Ceutectoid changes:

(wt% C)

Ti

Si

Mo

(°C)

Ni

W

Cr

Cr

eutectoid

Eutectoid

Si

Mn

W

Mn

Ti

Mo

C

T

Ni

wt. % of alloying elements

wt. % of alloying elements

Adapted from Fig. 10.38,Callister & Rethwisch 3e. (Fig. 10.38 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

Adapted from Fig. 10.39,Callister & Rethwisch 3e. (Fig. 10.39 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)


Chapter 10 phase diagrams and microstructures

Summary

Phase Diagrams and Microstructures

• Phase (T vs c) diagrams are useful to determine:

- the number and types of phases,

- the wt% of each phase,

- and the composition of each phase

for a given T and composition of the system.

• Alloying to produce a solid solution usually

- increases the tensile strength (TS)

- decreases the ductility.

• Binary eutectics and binary eutectoids allow for

a range of microstructures.

- Alloy composition and annealing temperature and time

determine possible microstructure(s). Slow vs. Fast.

- In 2-phase regions, use “lever rule” to get wt% of phases.

- Varying microstructure affects mechanical properties.


  • Login