Course Syllabus for ME 2510: Materials Engineering (3 hrs lectures + 1 hrs tutorial) l Instructor:...

Course Syllabus for ME 2510: Materials Engineering (3 hrs lectures + 1 hrs tutorial)

Instructor: Dr. Mahmoud M. Tash e-mail: [email protected]

» [email protected]

Homepage http://faculty.ksu.edu.sa/mtash/default.aspx http://www.eng.cu.edu.eg/users/mtash/

www.egypteducation.org TEL (+966)560558021

mailto:[email protected]

mailto:[email protected]

http://faculty.ksu.edu.sa/mtash/default.aspx

Course Specification:

Program: B. Sc. Mechanical Engineering Department: Mechanical Engineering Academic year: Level 4

A- Basic Information:

Title: Materials Engineering Code: ME 2510 Credit Hours: 3 (3,1,0) (lectures 2 + tutorial 1hr) >>Total 4 hrs/week. (Prerequisites: CHEM 101, PHYS 104)

Professional Information 1. Overall Aims of the course:

Upon completion of the course the student should be able to:

Learn what are engineering materials, their properties, processing and applications:

Know the structure and characteristics of metals, polymers and ceramics. Understand types of equilibrium-phase diagrams. Microstructures of alloys Understand the atomic imperfections and atomic movement(diffusion) Understand what is meant by mechanical properties of metals, polymers and

ceramics. Understand what is meant by heat treatment of plain-carbon steels, cast irons and

precipitation hardening.

2. Intended Learning Outcomes

a-Knowledge and understanding By the end of the course the student should be able to: a1- Differentiate between the different behaviors of engineering materials. a2- Understanding of crystal structure for materials. a3- Understanding of phase diagram for alloy systems. a4- The basic of metallic heat treatment of ferrous and non ferrous alloys. a5- What is meant by mechanical properties of materials. a6-Understanding of polymeric, ceramic and composite materials and their

applications b-Intellectual skills b1- Identify materials in engineering parts used in daily life. b2- Be familiar with the crystal structure. b3- Be familiar with the phase diagrams b4- Design the heat treatment process. b5- Identify microstructure and properties of some important alloys b6-Understand the basic of material selections

2. Intended Learning Outcomes

c- Professional and Practical Skills c1- Learning how are parts manufactured. c2- Selection of proper materials and process for specific industrial applications c3- Use of heat treatment process. c4- Use of materials testing for measuring mechanical properties

d-General and Transferable Skills d1- Material selection and evaluations. d2- Present finding of scientific research in seminars

3. Course Content (Main Topics):

Introduction to materials engineering Structure and characteristics of metals Polymers and ceramics Equilibrium-phase diagrams Microstructures of alloys Imperfections and Diffusion Mechanical properties of metals, polymers and ceramics Heat treatment of plain-carbon steels, cast irons and precipitation

hardening

4. Teaching and learning methods and Aids:

Blackboard, and Over-head.

5. Student Assessment and Grading Basis:

Attendance of lectures and tutorials is a most. Homework assignments will consist of essay questions and problem solving cases. There will be two quizzes and two midterm examination and one final test. Examinations are comprehensive, including subjects from all assigned readings, lectures, laboratory activities, and classroom demonstrations. Written exams to measure knowledge and understanding, Intellectual skills, and Professional skills. Term papers to measure Intellectual skills, Professional skills and General skills

Grading system

Mid-term exams 40% 40

Final term exam 40% 40

Section work Pop Quizzes (2)

10%10%

1010

Total 100% 100

6. References and Text Books:

8.1- Course Notes Lectures in Materials Science. 8.2- Required Books William D. Callister, Materials Sciences and Engineering- An Introduction, Jhon Wiley & Sons Inc. 1997. 8.3- Recommended Books Principles of Material Science, William Smith, 1996. Principles of Engineering Metallurgy, L. Krishna Reddy, 1996. Metallurgy for Engineers,4th edition, E.C.Rollason, 1973 8.4- Periodicals, Web Sites The Science and Engineering of Materials, 4th ed, Donald R. Askeland – Pradeep P. Phulé, © 2003

Brooks/Cole Publishing / Thomson Learning™ many internet web sites, 2002-2006

Lecture One

Introduction to Engineering Materials & Applications

Materials science is primarily concerned with the search for basic knowledge about the internal structure, properties, and processing of materials. Materials' engineering is mainly concerned with the use of fundamentals and applied knowledge of materials so that the materials can be converted into products necessary or desired by the society.

Materials in Industry: Industrial applications of materials science include materials design, cost, processing techniques (casting, rolling, welding, ion implantation, crystal growth, thin-film deposition, sintering, etc.) and analytical techniques (electron microscopy, x-ray diffraction, calorimetry, backscattering, neutron diffraction, etc.).


Materials Science & Engineering

Properties

ProcessingStructure

Performance

Materials Science

Investigating the relationship between structure and properties of materials.

Materials Engineering

Designing the structure to achieve specific properties of materials.

• Processing

• Structure

• Properties

• Performance

Processing >> Structure >> Properties >> Performance

General Categories of Engineering Materials Used Today in Manufacturing Industries

What is Materials Science and Engineering?

Materials Performance: Strength-to-weight ratio, formability, cost

Processing >>> Structure >>> Properties >>> Performance

• Composition means the chemical make-up of a material.

• Structure means a description of the arrangements of atoms or ions in a material.

• Synthesis is the process by which materials are made from naturally occurring or other chemicals.

• Processing means different ways for shaping materials into useful components or changing their properties.

What is Materials Science & Engineering?

Materials Processing

• Casting• Forging• Extrusion• Stamping• Nanotechnology• Sintering

Materials Characterization: • Diffraction with x-rays, electrons, or neutrons and various forms of spectroscopy and chemical analysis • Energy-dispersive spectroscopy (EDS),• Chromatography,• Thermogravimetric analysis,• Electron microscope analysis

Materials Properties

• Physical behavior, Response to environment

• Mechanical (e.g., stress-strain)• Thermal• Electrical• Magnetic• Optical• Corrosive

Functional Classification of Materials

• Aerospace (Composites, SiO2-Amorphous silicon, Al-alloys, Super alloys)

• Biomedical ( Titanium alloys, Stainless steels, plastics)

• Electronic Materials (Si, GaAs, BaTiO3, Conducting Polymers)

• Energy and Environmental Technology (Uo2, Ni-Cd, ZrO2, LiCoO2, Amorphous Si-H)

• Magnetic Materials (Fe, Fe-Si, NiZn and MnZn ferrites, Co-Pt-Ta-Cr)

• Optical Materials (SiO2, GaAs, Glasses, Al2O3)

• Smart Materials (NI-Ti shape memory alloys)

• Structural Materials (Steels, concrete, fiberglass, plastics, wood)


Classification of Materials-Based on Structure

1. Crystalline material is a material comprised of one or many crystals. In each crystal, atoms or ions show a long-range periodic arrangement.

2. Single crystal is a crystalline material that is made of only one crystal (there are no grain boundaries).

3. Polycrystalline material is a material comprised of many crystals (as opposed to a single-crystal material that has only one crystal). Grains are the crystals in a polycrystalline material. Grain boundaries are regions between grains of a polycrystalline material.


Properties of Materials

Mechanical properties: Elasticity and stiffness, plasticity, strength, brittleness or toughness, and fatigue.

Electrical properties: Electrical conductivity and resistivity

Magnetic properties: Paramagnetic, diamagnetic, and ferromagnetic properties.

Dielectric properties: Polarizability, capacitance, ferroelectric, piezoelectric, and pyroelectric properties.

Optical properties: Refractive index, absorption, reflection, and transmission, and birefringence (double refraction).

Corrosion, fatigue, and creep properties


© 2003 B

rooks/Cole P

ublishing / Thom

son Learning™

Strengths of various categories of materials


© 2003 B

rooks/Cole P

ublishing / Thom

son Learning™

Variation of Strengths with Temperature for various categories of materials


Materials Design and Selection

1. Density is mass per unit volume of a material, usually expressed in units of g/cm3 or lb/in.3

2. Strength-to-weight ratio is the strength of a material divided by its density; materials with a high strength-to-weight ratio are strong but lightweight.


Thank You

ATOMIC AND MOLECULAR STRUCTURE & BONDING

Atomic Bonding

Bonding: There are two types of bonds: primary and secondary. Primary bonds are the strongest bonds which hold atoms together. The three types of primary bonds are: Metallic bond, Covalent bond, and Ionic bond

Metallic Bonds

Elements in groups I and II of the periodic table, and some in group III form metallic crystals. In a metal, the outer electrons are shared among all the atoms in the solid. Each atom gives up its outer electrons and becomes slightly positively charged. The negatively charged electrons hold the metal atoms together. Since the electrons are free to move, they lead to good thermal and electrical conductivity. The metallic bonding does not have the strongly directional character of covalent bonds.


Linear Thermal Strain L/L0 = (T - T0)

as E0 (less negative)

Parabolic E vs. r shape

Larger ESmaller

Smaller ELarger

r0

r

E

asymmetry at r0 No asymmetry at r0

No affect on r(T) or V(T)

Volume Thermal Strain V/V0 = V (T - T0)

Symmetric well: No expansion possibleAtoms just vibrate back and forth!


• atoms pack in periodic, 3D arrays• typical of:

Crystalline materials...

-metals-many ceramics-some polymers

• atoms have no periodic packing• occurs for:

Noncrystalline materials...

-complex structures-rapid cooling

Si Oxygen

crystalline SiO2

noncrystalline SiO2

"Amorphous" = Noncrystalline

Atomic Packing


Summary: Bonding, Structure, Properties

Ceramics Large bond energies

Ionic and Covalent bonds large Tm, E Small

Metals Varying bond energy

Metallic bonding intermediate Tm, E,

Polymers directional properties

Covalent and Secondary secondary dominates outcome

small Tm, E large


Thank You

CRYSTAL STRUCTURE

Crystal Structure

Crystal Structure

Lattice- A collection of points that divide space into smaller equally sized segments.

Unit cell - A subdivision of the lattice that still retains the overall characteristics of the entire lattice.

Atomic radius - The apparent radius of an atom, typically calculated from the dimensions of the unit cell, using close-packed directions (depends upon coordination number).

Packing factor - The fraction of space in a unit cell occupied by atoms.

Types of Crystal Structure

Body centered cubic (BCC) Face centered cubic (FCC) Hexagonal close packed (HCP)

Crystal Structure

A number of metals are shown below with their room temperature crystal structure indicated. There are substances without crystalline structure at room temperature; for example, glass and silicone. All metals and alloys

are crystalline solids, and most metals assume one of three different lattice, or crystalline, structures as they form: body-centered cubic (BCC),

face-centered cubic (FCC), or hexagonal close-packed (HCP).

Aluminum (FCC)Chromium

(BCC)Copper (FCC) Iron (alpha)

(BCC)

Iron (gamma) (FCC) Iron (delta) (BCC)

Lead (FCC) Nickel (FCC)

Silver (FCC) Titanium (HCP)

Tungsten (BCC)

Zinc (HCP)

Number of Lattice Points in Cubic Crystal Systems

In the SC unit cell: point / unit cell = (8 corners)1/8 = 1

In BCC unit cells: point / unit cell = (8 corners)1/8 + (1 center)(1) = 2

In FCC unit cells: point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4

Crystal Structure

In SC, BCC, and FCC structures when one atom is located at each lattice point.

Relationship between Atomic Radius and Lattice Parameters

Packing Factor

In a FCC cell, there are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and the volume of the unit cell is a0 3

Crystal Structure

74.018)2/4(

)34

(4)( Factor Packing

24r/ cells,unit FCCfor Since,

)34

)(atoms/cell (4 Factor Packing

3

3

0

30

3

r

r

r

aa

Density

Density of BCC iron, which has a lattice parameter of 0.2866 nm.

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm Atomic mass = 55.847 g/mol Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cell Avogadro’s number NA = 6.02 1023 atoms/mol

32324

/882.7)1002.6)(1054.23(

)847.55)(2(

number) sadro'cell)(Avogunit of (volume

iron) of mass )(atomicatoms/cell of(number Density

cmg

Crystal Structure

Unit Cells Types

PrimitiveFace-Centered

Body-Centered End-Centered

A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition.

• Primitive (P) unit cells contain only a single lattice point.• Internal (I) unit cell contains an atom in the body center.• Face (F) unit cell contains atoms in the all faces of the planes composing the cell.• Centered (C) unit cell contains atoms centered on the sides of the unit cell.

Crystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monclinic, triclinic, trigonal) with 4 unit cell types (P, I, F, C) symmetry allows for only 14 types of 3-D lattice.

Crystal Structure

Counting Number of Atoms Per Unit Cell

Counting Atoms in 3D CellsAtoms in different positions are shared by differing numbers of unit cells.

• Vertex atom shared by 8 cells => 1/8 atom per cell.

• Edge atom shared by 4 cells => 1/4 atom per cell.

• Face atom shared by 2 cells => 1/2 atom per cell.

• Body unique to 1 cell => 1 atom per cell.

Simple Cubic

8 atoms but shared by 8 unit cells. So, 8 atoms/8 cells = 1 atom/unit cell

How many atoms/cell forBody-Centered Cubic?And, Face-Centered Cubic?

Crystal Structure

Atomic Packing Fraction for FCC

Face-Centered-CubicArrangement

APF = vol. of atomic spheres in unit cell total unit cell vol.

No. of atoms per unit cell = Volume of one atom= Volume of cubic cell = “R” related to “a” by

4/cell

4R3/3 a3

= 0.74APF =

a3

4

3( 2a/4)34

atoms

unit cell atomvolume

unit cell

volume

2a 4RUnit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

Crystal Structure

2]21[ .4

Crystallographic directions and coordinates.

Direction B1. Two points are 1, 1, 1 and 0, 0, 02. 1, 1, 1, -0, 0, 0 = 1, 1, 13. No fractions to clear or integers to reduce4. [111]

Direction A1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, -0, 0, 0 = 1, 0, 03. No fractions to clear or integers to

reduce4. [100]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 0

2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2

Crystallographic Points, Directions, and Planes.Crystal Structure

Procedure:1. Any line (or vector direction) is specified by 2 points.

• The first point is, typically, at the origin (000).

2. Determine length of vector projection in each of 3 axes in units (or fractions) of a, b, and c.• X (a), Y(b), Z(c) 1 1 0

3. Multiply or divide by a common factor to reduce the lengths to the smallest integer values, u v w.

4. Enclose in square brackets: [u v w]: [110] direction.

a b

c

DIRECTIONS will help define PLANES (Miller Indices or plane normal).

[1 1 0]5. Designate negative numbers by a bar • Pronounced “bar 1”, “bar 1”, “zero” direction.

6. “Family” of [110] directions is designated as <110>.

Crystallographic Points, Directions, and Planes.

Crystal Structure

Crystallographic Points, Directions, and Planes.

Figure 2.9 Crystallographic planes and intercepts

Plane B1. The plane never intercepts the z

axis, so x = 1, y = 2, and z = 2.1/x = 1, 1/y =1/2, 1/z = 0

3. Clear fractions:1/x = 2, 1/y = 1, 1/z = 0

4. (210)

Plane A1. x = 1, y = 1, z = 12.1/x = 1, 1/y = 1,1 /z = 13. No fractions to clear4. (111)

Plane C1. We must move the origin, since

the plane passes through 0, 0, 0. Let’s move the origin one lattice

parameter in the y-direction. Then, x = ∞ , y = -1, and z = ∞

2.1/x = 0, 1/y = 1, 1/z = 03. No fractions to clear.

4 (o1-o)

Crystal Structure

Linear Density in FCC

LD =Number of atoms centered on a direction vector

Length of the direction vector

Example: Calculate the linear density of an FCC crystal along [1 1 0].

ANSWERa. 2 atoms along [1 1 0]

in the cube.b. Length = 4R

ASKa. How many spheres along blue line? b. What is length of blue line?

LD110 2atoms

4R

12R

XZ = 1i + 1j + 0k = [110]

Crystal Structure

Planar Packing Density in FCC

Ra 22

R4

PD =Area of atoms centered on a given plane

Area of the plane

Example: Calculate the PPD on (1 1 0) plane of an FCC crystal.

• Find area filled by atoms in plane: 2R2

• Find Area of Plane: 8√2 R2

PPD 2R2

8 2R2

4 20.555

Hence,

Always independent of R!

Crystal Structure

n AVcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm3/unit cell)Avogadro's number

(6.023 x 1023 atoms/mol)

• crystal structure = FCC: 4 atoms/unit cell• atomic weight = 63.55 g/mol (1 amu = 1 g/mol)• atomic radius R = 0.128 nm

Compare to actual: Cu = 8.94 g/cm3

Result: theoretical Cu = 8.89 g/cm3

Theoretical Density,

Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3

Crystal Structure

Thank You

Imperfections in the Atomic and Ionic Arrangements

Objectives

Introduce the three basic types of imperfections: point defects, line defects (or dislocations), and surface defects.

Explore the nature and effects of different types of defects.

Outline

Point Defects Dislocations Observing Dislocations Significance of Dislocations Influence of Crystal Structure Surface Defects Importance of Defects

Point defects - Imperfections, such as vacancies, that are located typically at one (in some cases a few) sites in the crystal.

Extended defects - Defects that involve several atoms/ions and thus occur over a finite volume of the crystalline material (e.g., dislocations, stacking faults, etc.).

Vacancy - An atom or an ion missing from its regular crystallographic site.

Interstitial defect - A point defect produced when an atom is placed into the crystal at a site that is normally not a lattice point.

Substitutional defect - A point defect produced when an atom is removed from a regular lattice point and replaced with a different atom, usually of a different size.


(c) 2003 Brooks/C

ole Publishing / Thom

son Learning

Point defects: (a) vacancy, (b) interstitial atom, (c) small substitutional atom, (d) large substitutional atom, (e) Frenkel defect, (f) Schottky defect. All of these defects disrupt the perfect arrangement of the surrounding atoms.


Determine the number of vacancies needed for a BCC iron crystal to have a density of 7.87 g/cm3. The lattice parameter of the iron is 2.866 10-8 cm.

SOLUTION

The expected theoretical density of iron can be calculated from the lattice parameter and the atomic mass.

Vacancy Concentrations in Iron


SOLUTION (Continued)

Let’s calculate the number of iron atoms and vacancies that would be present in each unit cell for the required density of 7.87 g/cm3:

Or, there should be 2.00 – 1.9971 = 0.0029 vacancies per unit cell. The number of vacancies per cm3 is:


1. Dislocation - A line imperfection in a crystalline material.

2. Screw dislocation - A dislocation produced by skewing a crystal so that one atomic plane produces a spiral ramp about the dislocation.

3. Edge dislocation - A dislocation introduced into the crystal by adding an ‘‘extra half plane’’ of atoms.

4. Mixed dislocation - A dislocation that contains partly edge components and partly screw components.

5. Slip - Deformation of a metallic material by the movement of dislocations through the crystal.


The perfect crystal in (a) is cut and an extra plane of atoms is inserted (b). The bottom edge of the extra plane is an edge dislocation (c). A Burgers vector b is required to close a loop of equal atom spacings around the edge dislocation. (Adapted from J.D. Verhoeven, Fundamentals of Physical Metallurgy, Wiley, 1975.)


When a shear stress is applied to the dislocation in (a), the atoms are displaced, causing the dislocation to move one Burgers vector in the slip direction (b). Continued movement of the dislocation eventually creates a step (c), and the crystal is deformed. (Adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill, 1976.) (b) Motion of caterpillar is analogous to the motion of a dislocation.


Surface defects - Imperfections, such as grain boundaries, that form a two-dimensional plane within the crystal.

Hall-Petch equation - The relationship between yield strength and grain size in a metallic material—that is,

ASTM grain size number (n) - A measure of the size of the grains in a crystalline material obtained by counting the number of grains per square inch a magnification 100.

Small angle grain boundary - An array of dislocations causing a small misorientation of the crystal across the surface of the imperfection.

2/1K0

dy


The atoms near the boundaries of the three grains do not have an equilibrium spacing or arrangement. (b) Grains and grain boundaries in a stainless steel sample. (Courtesy Dr. A. Deardo.)


Thank You

Metallic Materials-Phase Diagrams

Engineering Alloys

Metals and alloys have many useful engineering properties and so have wide spread application in engineering designs. Iron and its alloys (principally steel) account for about 90 percent of the world's production of metals mainly because of their combination of good strength, toughness, and ductility at a relatively low cost.

Alloys based on iron are called ferrous alloys, and those based on the other metals are called nonferrous alloys. In this section we shall discuss some aspects of the processing, structure, and properties of some of the important ferrous and nonferrous alloys.

The study of metallic alloys; ferrous (steels and cast irons) or non-ferrous, requires two basic topics; phase diagram and heat treatment.

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

Phases: physically and chemically distinct material regions that result

Aluminum-Copper Alloy

(darker phase)

(lighter phase)

Phase BPhase A

Nickel atomCopper atom

Solid-solutionPhase

Ordered Phase


Cu-Ni Phase Diagram

Rule 1: If we know T and Co

• Number of phases• Composition of each phase• Amount of each phase

• Examples: Number of Phases

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L +

liquidus

solid

us

A(1100,60)B

(1250,3

5)

A(1100, 60): 1 phase:

B(1250, 35): 2 phases: L +


1st type phase Diagrams

Cu-Ni Phase Diagram

At TA: Only Liquid (L) CL = Co ( = 35wt% Ni)

At TB: Both and L CL = Cliquidus ( = 32wt% Ni here) C = Csolidus ( = 43wt% Ni here)

At TD: Only Solid () C = Co ( = 35wt% Ni)

Co = 35wt%Ni

wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)

30 40 50

TAA

DTD

TBB

tie line

433532CoCL C

Composition of phases

At TB: Both and L

At TA: Only Liquid (L) WL = 100wt%, W = 0

At TD: Only Solid () WL = 0, W = 100wt%

Co = 35wt%Ni

WL SR S

W RR S

= 27%


Amounts of phases

Lever Rule: wt. fraction of phases

• Sum of weight fractions:• Conservation of mass (Ni):

• Combine above equations:

WL W 1

Co WLCL WC

RR S

W Co CLC CL

SR S

WLC Co

C CL

• A geometric interpretation:

CoR S

WWL

CL C


Cooling in Cu-Ni Binary System

• Consider Co = 35wt%Ni.


• Upon cooling more quickly (i.e. non-equilibrium), 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.


Cu-Ni Phases and Microstructure (non-equilibrium)

Mechanical Properties of Cu-Ni

• Effect of solid solution strengthening on:

--Tensile strength (TS) --Ductility (%EL,%AR)

-Peak as a function of Co -Minimum as a function of Co

Adapted from Fig. 9.5(a), Callister 6e. Adapted from Fig. 9.5(b), Callister 6e.

Elo

ng

ati

on

(%

EL)

Composition, wt%NiCu Ni0 20 40 60 80 10020

30

40

50

60

%EL for pure Ni

%EL for pure Cu

Ten

sile

Str

en

gth

(M

Pa)

Composition, wt%NiCu Ni0 20 40 60 80 100

200

300

400

TS for pure Ni

TS for pure Cu


Binary-Eutectic Systems

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: L +


2ed type phase Diagrams

• Co < 2wt%Sn• Polycrystal of a grains.

Adapted from Fig. 9.9 and 9.10, Callister 6e.

Microstructure for Pb-Sn Eutectic Diagram

• 2wt%Sn < Co < 18.3wt%Sn• a polycrystal with fine b crystals.


160m

Micrograph of Pb-Sn eutectic microstructure

Pb-Sn system

Adapted from Fig. 9.11, Callister 6e.


Microstructure at Eutectic

Light: Sn-rich Dark: Pb-rich


Pb-Snsystem• 18.3wt%Sn < Co < 61.9wt%Sn

• Result: crystals and a eutectic microstructure


• Just above TE:

WL = (1-W) =50wt%

C = 18.3wt%Sn

CL = 61.9wt%SnS

R + SW = =50wt%

• Just below TE:C = 18.3wt%Sn

C = 97.8wt%SnS

R + SW = =73wt%

W = 27wt%

%

%

% %


Solder: Lead-Tin (Pb-Sn) microstructure

For 50 wt% Pb alloy:• Lead-rich phase (dark)• Lamellar eutectic structure of Sn-rich phase (light).

* 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.


T(°C)

(Pb-Sn System)

L + 200

Co, wt% Sn20 400

300

100

L

60

+

TE

080 100

L +

18.361.9

97.8

Cohypoeutectic

Cohypereutectic

eutectic

hypereutectic: (illustration only)

160m

eutectic: Co=61.9wt%Sn

175m

hypoeutectic: Co=50wt%Sn

eutectic micro-constituent


Fig. 10.15Fig. 10.12

Adapted from Fig. 10.15, (Illustration only)

Hypoeutectic & Hypereutectic


14

Example Problems

For a 40wt%Sn-60wt%Pb alloy at 150C, find...

• the phases present: a + b • the compositions of the phases:

L + L+

200

T(°C)

18.3

Co, wt% Sn 20 40 60 80 100 0

Co

300

100

L (liquid)

183°C 61.9 97.8

150

Lead-Tin (Pb-Sn) Eutectic Diagram

Solder for electronics


Answer: For a 40wt%Sn-60wt%Pb alloy at 150C, find... - phases present: a + b - compositions of phases: Ca = 11wt%Sn Cb = 99wt%Sn - relative amounts

(fractions) of each phase:


L + L+

200

T(°C)

18.3

Co, wt% Sn 20 40 60 80 100 0

Co

300

100

L (liquid)

183°C 61.9 97.8

150

11 99

R S

Lead-Tin (Pb-Sn) Eutectic Diagram

W = 59/88 = 67%

W = 29/88 = 33%


Problems Examples

Fig 1


Problems Examples

With reference to Fig. 1, answer the following questions: (Questions 1-7) The amount of α in comparison with β (amount of α: amount of β) that forms if a

90%Pb-10%Sn alloy is cooled to 0 C is

(a). 45:4 (b) 4:45 (c) 8: 98 (d) 98:8 The compositions of α and β phases at 170 C are … respectively

(a) 98.5% Pb and 11% Pb (b) 98.5% Sn and 11% Sn (c) 11%Pb and 98.5% Sn (d) None of the above

The amounts of α and β phases that form if the 30%Sn-70%Pb alloy is cooled to 0C are ……. respectively.

(a) 100 α and 0% β (b) 71.5% α and 28.5 β (c) 0 α and 100% β (d) 28.5% α and 71.5% β

The amount of primary α relative to the amount of eutectic for a 30%Sn-70%Pb alloy when it has been cooled to 0 C is

(a)25% α (b) 75% β (c) 25% β (d) 75% α For Pb-Sn alloy contains 45% Sn, which phases are in equilibrium at 100° C. What is

the relative amount of each phase? For Pb-Sn alloy contains 15% Sn, what phases, compositions and amounts are present

at 182° C, 150° C, and 50° C.


21

Result: Pearlite = alternating layers of and Fe3C phases.

120m

• 2 important points

-Eutectic (A):

-Eutectoid (B): L Fe3C

Fe3C

Fe3C

(ce

menti

te)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

austenite)

+L

+Fe3C

+Fe3C

+

L+Fe3C

(Fe) Co, wt% C0.77 4.30

727°C = Teutectoid

1148°C

T(°C)

A

B

SR

R S

Fe3C (cementite-hard)(ferrite-soft)

C

eu

tect

oid

Iron-Carbon (Fe-C) Phase Diagram

Eutectoid: from solid phase to 2-phase solid upon cooling: + Fe3C

Fig. 10.21

Fig. 10.24


From Figs. 9.21 and 9.26,Callister 6e. Fig. 10.27

Hypoeutectoid Steel


Adapted from Figs. 9.21 and 9.29,Callister 6e.

Adapted from Fig. 9.30,Callister 6e.

Hypereutectoid Steel


TE

ute

ctoi

d (

°C)

wt. % of alloying elements

Ti

Ni600

800

1000

1200

0 4 8 12

Mo SiW

Cr

Mn

wt. % of alloying elementsC

eu

tect

oid

(w

t%C

)

Ni

Ti

0 4 8 120

0.2

0.4

0.6

0.8

Cr

SiMnW

Mo

• Teutectoid changes: • Ceutectoid changes:

Adapted from Fig. 9.31,Callister 6e. Adapted from Fig. 9.32,Callister 6e.

Alloying Steel With More Elements


Problems Examples


http://www.sv.vt.edu/classes/MSE2094_NoteBook/96ClassProj/examples/no_clim.html

Problems Examples

With reference to Fig. 2, answer the following questions: Use the following iron-carbon diagram to determine the microstructure

(phases present and the percent of each phase) for the following carbon-steel alloys (1-4):

» 0.15 % C after air cooling to room temperature » 0.45% C after air cooling to room temperature » O.83 % C after air cooling to room temperature, and 8000C » 1.2 % C after air cooling to room temperature.

For an alloy containing 0.12%C, answer questions (5-6) » The alloy is designated….. … (low C - medium C - high C) steel.» This alloy contains ferrite and pearlite % at temperature below 700°C. » The alloy contains ferrite and cementite % at temperature below

700°C. Sketch the cooling curve and microstructure for alloys, Eutectoid, and

Eutectic The number of phases and relative amounts of each phase for the alloys

containing 0.2,0.4 0.8% and 1 % carbon, at 1600, 950, 700, and after equilibrium cooling to room temperature.

The structure of the alloys containing 0.2,0.4 0.8% and 1 % carbon at RT.


Self-Assessment: Phases and Composition

A Ni-50wt%Cu alloy is slowly cooled from 1400 C to 1200 C.

1. At what temperature does the first solid phase form?

2. What is the composition of the solid phase?

3. At what T does the liquid solidify?

4. What is the composition of the last remaining liquid?

start

end


Thank You

HEAT TREATMENT OF STEELS

Annealing and Normalizing Temperature Range


Effect of Annealing and Normalizing on Steel


Grain Size in Annealing and Normalizing

Annealing, coarser and less uniform

Normalizing, finer and more uniform


• Effect of wt%C

• More wt%C: TS and YS increase, %EL decreases.wt%C

0 0.5 10

50

100%EL

Impa

ct e

nerg

y (I

zod,

ft-

lb)

0

40

80

300

500

700

900

1100YS(MPa)TS(MPa)

wt%C0 0.5 1

hardness

0.77

0.77

Co>0.77wt%C Hypereutectoid

Co<0.77wt%C Hypoeutectoid

Pearlite (med)ferrite (soft)

Pearlite (med)Cementite

Hypo HyperHypo Hyper

(hard)

pearlite+ ferrite

pearlite+ Fe3C

Mechanical Behavior of Steel


• More wt%C: hardness increases, and ductility decreases.

• Fine vs coarse pearlite vs spheroidite

• Hardness: fine > coarse > spheroidite • %AR: fine < coarse < spheroidite

80

160

240

320

wt%C0 0.5 1

Bri

nell

hard

ness

fine pearlite

coarse pearlitespheroidite

0

30

60

90

wt%C0 0.5 1

Duct

ility

(%

AR

)

fine pearlite

coarse pearlite

spheroidite

Hypo Hyper Hypo Hyper

Mechanical Behavior of Steel


Hardening Temperature


Pearlite Morphology

10m

- Smaller T: colonies are larger

- Larger T: colonies are smaller

• Ttransf just below TE --Larger T: diffusion is faster --Pearlite is coarser.

• Ttransf well below TE --Smaller T: diffusion is slower --Pearlite is finer.

Adapted from Fig. 10.6 (a) and (b),Callister 6e.


Tempering of Martensite

• reduces brittleness of martensite,• reduces internal stress caused by quenching.

• decreases TS, YS but increases %AR

YS(MPa)TS(MPa)

800

1000

1200

1400

1600

1800

304050

60

200 400 600Tempering T (°C)

%AR

TS

YS

%AR

9

m

• produces extremely small Fe3C particles surrounded by


Austenite ()

Bainite ( + Fe3C plates/needles)

Pearlite ( + Fe3C layers + a proeutectoid phase)

Martensite (BCT phase diffusionless

transformation)

Tempered Martensite ( + very fine

Fe3C particles)

slow cool

moderate cool

rapid quench

reheat

Str

ength

Duct

ilit

yMartensite

T Martensite bainite

fine pearlite coarse pearlite

spheroidite

General Trends

Summary


Thank You

STEELS AND CAST IRONS

Microstructure of

Carbon steels

Pearlite >>> martensite due to quenching

STEELS

Fe-Fe3C Phase Diagram

http://www.sv.vt.edu/classes/MSE2094_NoteBook/96ClassProj/examples/no_clim.html

Cast Iron

Classification of Cast IronCast iron has higher carbon and silicon contents than steel. The carbon content of cast iron is 2.1 percent or more. Carbon exists as free graphite in all types of cast iron except in white cast iron (as intermetallic compound Fe3C called cementite).

There are four basic types of cast iron:

– 1- White cast iron ( hard, brittle, and not weldable)– 2- Grey cast iron (relatively soft, easily machined and welded. Main

applications) main applications (engine cylinder blocks, pipe, and machine tool structures)

– 3- Malleable cast iron (ductile, weldable, machinable and offers good strength and shock resistance)

– 4- Nodular or Ductile cast iron or spheroidal (ductile, malleable and weldable)

Cast Iron

White cast iron is very hard, brittle and hard machinable. Exhibits a white, crystalline fracture surface because fracture occurs along the iron carbide plates; it is the result of metastable solidification (Fe3C eutectic).• Fast cooling rate, low Si and C, carbon exist as Fe3C

Hypo-eutectic white cast iron, (x 100)

Hyper-eutectic white cast iron (x 100).

Cast Iron

Common grey iron showing ferrite (F), pearlite (P) and phosphide eutectic (PH) (x250).

Diagram indicating the structures of iron resulting from variation of silicon and carbon contents

Cast Iron

Microstructure of cast iron

Martensitic iron (Ni-hard).(x 200)

different forms of graphite

Coarse graphite flakes. (x 60)

Temper carbon in a malleable iron;(x 100)

Cast Iron

γ transformation due to cooling

F.C A.C W.C

Thank You

Nonferrous Alloys

Nonferrous Alloys

Aluminum Alloys Magnesium and Beryllium Alloys Copper Alloys Nickel and Cobalt Alloys Titanium Alloys Refractory and Precious Metals

Aluminum Alloys

Magnesium alloys are used in aerospace applications, high-speed machinery, and transportation and materials handling equipment.

Instrument grade beryllium is used in inertial guidance systems where the elastic deformation must be minimal; structural grades are used in aerospace applications; and nuclear applications take advantage of the transparency of beryllium to electromagnetic radiation. Beryllium is expensive, brittle, reactive, and toxic.

Magnesium and Beryllium Alloys

Magnesium Alloys

Blister copper - An impure form of copper obtained during the copper refining process.

Applications for copper-based alloys include electrical components (such as wire), pumps, valves, and plumbing parts, where these properties are used to advantage.

Brass - A group of copper-based alloys, normally containing zinc as the major alloying element.

Bronze - Generally, copper alloys containing tin, can contain other elements.

Copper Alloys

Copper Alloys

Nickel and cobalt alloys are used for corrosion protection and for high-temperature resistance, taking advantage of their high melting points and high strengths.

Superalloys - A group of nickel, iron-nickel, and cobalt-based alloys that have exceptional heat resistance, creep resistance, and corrosion resistance.

Nickel and Cobalt Alloys

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The effect of temperature on the tensile strength of several nickel-based alloys.


Titanium’s excellent corrosion resistance provides applications in chemical processing equipment, marine components, and biomedical implants such as hip prostheses.

Titanium is an important aerospace material, finding applications as airframe and jet engine components.

Titanium alloys are considered biocompatible (i.e., they are not rejected by the body). By developing porous coatings of bone-like ceramic compositions known as hydroxyapatite, it may be possible to make titanium implants bioactive (i.e., the natural bone can grow into the hydroxyapatite coating).

Titanium Alloys

Titanium Alloys

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The effect of temperature on the yield strength of selected titanium alloys.

Titanium Alloys

Refractory metals – These include tungsten, molybdenum, tantalum, and niobium (or columbium), have exceptionally high-melting temperatures (above 1925oC) and, consequently, have the potential for high-temperature service.

Applications of Refractory metals include filaments for light bulbs, rocket nozzles, nuclear power generators, tantalum- and niobium-based electronic capacitors, and chemical processing equipment.

Precious Metals - These include gold, silver, palladium, platinum, and rhodium.From an engineering viewpoint, these materials resist corrosion and make very good conductors of electricity.

Refractory and Precious Metals

Refractory and Precious Metals

Cobalt-Chromium Alloys

Dental Metals - Amalgam 65% silver (minimum)

29% tin, 6% copper, 2% zinc, 3% mercury (maximum) 45 - 55 % mercury - 45 % silver 15 % tin

Dental Metals - Gold Nickel-Titanium Alloys Tantalum, Platinum and other noble metals

Dental Metals

Thank You

Mechanical Behavior of Materials

Know the concepts of mechanical properties of materials.

Understand the factors affecting the mechanical properties.

Be aware of the basic testing procedures that engineers use to evaluate many of these properties.

Objective

Part One Outline

Mechanical Properties of Materials

Stress-Strain Diagram & Properties

Bend Test of Materials

Hardness Test of Materials


Tension Testing Machine Tensile Specimens

Adapted from Callister 7e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.)


Engineering Stress Strain Diagram For A High-Strength Aluminum Alloy.

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A unidirectional force is applied to a specimen in the tensile test by means of the moveable crosshead. The cross-head movement can be performed using screws or a hydraulic mechanism


Mechanical property data obtained from the tensile test are of engineering importance for structural design

Mechanical property data obtained from the tensile test and the engineering stress-strain diagram are:

1. modulus of elasticity2. yield strength at 0.2 percent offset3. ultimate tensile strength4. percent elongation at fracture5. percent reduction in area at fracture

- Stress () = Force or load per unit area of cross-section.- Strain () = Elongation change in dimension per unit length- Young’s modulus (E)= The slope of the linear part of the stress-strain curve in the elastic region

(stress) = E * (strain)or E = (stress)/ (strain) (units of psi or pa)


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Comparison of the elastic behavior of steel and aluminum. For a given stress, aluminum deforms elastically three times as much as does steel


Engineering stress-strain. Elastic range in stress-strain.


Engineering stress-strain curve, showing various featuresEngineering stress-strain curve, showing various features

Yield stress (Y), Ultimate tensile strength (UTS), and Fracture.Yield stress (Y), Ultimate tensile strength (UTS), and Fracture.

1. Elastic and Plastic, 2. Uniform elongation and Necking. 1. Elastic and Plastic, 2. Uniform elongation and Necking.


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Tensile stress-strain curves for different materials. Note that these are qualitative


Alloying a metal with other metals or nonmetals and heat treatment can greatly affect the tensile strength and ductility of metals. For metals with a thick cross section such as plate, a 0.50-in-diameter round specimen is commonly used. For metal with thinner cross sections such as sheet, a flat specimen is used. A 2-in gage length within the specimen is the most commonly used gage length for tensile tests.

During the tensile test, after necking of the sample occurs, the engineering stress decreases as the strain increases, leading to a maximum engineering stress in the engineering stress-strain curve. Thus, once necking begins during the tensile test, the true stress is higher than the engineering stress.

• Engineering stress σ = P/A0 and • Engineering strain ε =(l-l0)/l0 • True stress σT = F/Ai = σ (1+ ε) and

• True strain εT =ln (li/l0) = ln (1+ ε)



Engineering stress-strain curves for some metals and alloys

Chapter 4, mechanical properties of metals



Comparison between engineering and tue stress-strain curve

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Determining the 0.2% offset yield strength in gray cast ion, and (b) upper and lower yield point behavior in a low-carbon steel


Typical yield strength values for different engineered materials. (Source: Reprinted from Engineering Materials I, Second Edition, M.F. Ashby and D.R.H. Jones, 1996, Fig. 8-12, p. 85. Copyright © Butterworth-Heinemann. Reprinted with permission from Elsevier Science.)


Resilience, Ur

Ability of a material to store energy – Energy stored best in elastic region

If we assume a linear stress-strain curve this simplifies to


yyr2

1U

y dUr 0


Modulus of Resilience (MModulus of Resilience (MRR):): is defined as the area under the elastic region and is defined as the area under the elastic region and

is the elastic specific energy (in.lb./in.is the elastic specific energy (in.lb./in.33), it is a measure of the amount of elastic ), it is a measure of the amount of elastic energy that can be stored in each cubic inch of the specimen.energy that can be stored in each cubic inch of the specimen.

E

Y

EeeEdeeEde e

ee

22E

EYM

:)Y/e(E modulus sYoung' Using

2|

2M

2

2

2

R

0

20

0

2

00

R0

00

For spring steel, MFor spring steel, MRR = 385 in.lb./in. = 385 in.lb./in.33 or 1355 in.lb./lb. For rubber, M or 1355 in.lb./lb. For rubber, MRR = 1680 = 1680

385 in.lb./in.385 in.lb./in.33 or 48,000 in.lb./lb.. Rubber can store much more energy per or 48,000 in.lb./lb.. Rubber can store much more energy per unit volume or weight than can steel. unit volume or weight than can steel.


Ductility of metals is most commonly expressed as percent elongation and percent reduction in area. The percent elongation and percent reduction in area at fracture is of engineering importance not only as a measure of ductility but also as an index of the quality of the metal.

Percent elongation is the amount of elongation that a tensile specimen under goes during testing provides a value for the ductility of a metal.

Percent reduction in area is usually obtained from a tensile test using a specimen 0.50 in (12.7 mm) in diameter.

x 100L

LLEL%o

of

100xA

AARA%o

fo -=



Localized deformation of a ductile material during a tensile test produces a necked region. The micrograph shows necked region in a fractured sample



The stress-strain behavior of brittle materials compared with that of more ductile materials


Example Problem



Figure 6.10 The stress-strain curve for an aluminum alloy from Table 6-1


Example Problem


Example Problem

Young’s Modulus of Aluminum Alloy

From the data in Example 6.1, calculate the modulus of elasticity of the aluminum alloy. Use the modulus to determine the length after deformation of a bar of initial length of 50 in. Assume that a level of stress of 30,000 psi is applied.

Example 6.3 SOLUTION


Ductility of an Aluminum Alloy

The aluminum alloy in Example 6.1 has a final length after failure of 2.195 in. and a final diameter of 0.398 in. at the fractured surface. Calculate the ductility of this alloy.




The effect of temperance (a) on the stress-strain curve and (b) on the tensile properties of an aluminum alloy


True Stress and True Strain Calculation

Compare engineering stress and strain with true stress and strain for the aluminum alloy in Example 6.1 at (a) the maximum load and (b) fracture. The diameter at maximum load is 0.497 in. and at fracture is 0.398 in.



SOLUTION (Continued)


The following example illustrates the differences between engineering stress and strain The following example illustrates the differences between engineering stress and strain ((ee, e), e) and true stress and strain data and true stress and strain data ((TT, , )) when plotting the results of a tensile test when plotting the results of a tensile test

AA00 = 0.056 in. = 0.056 in.22, A, Af f = 0.016 in.= 0.016 in.22, L, L00 = 2.00 in., e = 49%, = 2.00 in., e = 49%, ee = F/ A = F/ A0 0 ,, TT = F/ A = F/ A

= ln(L= ln(L00 + + L)/ LL)/ L00)), At fracture )), At fracture ff = ln(A = ln(A00/A/Aff), By constant volume A = A), By constant volume A = A00LL00/L /L

F L e Alb in. in./in. psi in.2 in./in. psi

1600 0.00 0.00 28571 0.0560 0.000 285712500 0.02 0.01 44642 0.0554 0.010 451263000 0.08 0.04 53571 0.0538 0.039 557623600 0.20 0.10 64286 0.0509 0.095 707274200 0.40 0.20 75000 0.0467 0.182 899374500 0.60 0.30 80357 0.0431 0.262 1044084600 0.86 0.42 82143 0.0392 0.358 1173473300 0.98 0.49 58928 0.0160 1.253 206250

ee TT

(Note: Columns 1 and 2 are measured data and the rest are calculated)(Note: Columns 1 and 2 are measured data and the rest are calculated)

0

50000

100000

150000

200000

250000

0.000 0.200 0.400 0.600 0.800 1.000 1.200 1.400

True Strain

Tru

e S

tre

ss

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

0.00 0.10 0.20 0.30 0.40 0.50 0.60

Engineering Strain

En

gin

ee

rin

g S

tre

ss

.• Factor of safety, N

Ny

working

Often N isbetween1.2 and 4

• Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5.

Design or Safety Factors

4

0002202 /d

N,

5

Ny

working

1045 plain

carbon steel: y = 310 MPa

TS = 565 MPa

F = 220,000N

d

Lo

d = 0.067 m = 6.7 cm


Bend Test for Materials

Bend Test for Brittle Materials

1. Bend test - Application of a force to the center of a bar that is supported on each end to determine the resistance of the material to a static or slowly applied load.

2. Flexural strength or modulus of rupture -The stress required to fracture a specimen in a bend test.

3. Flexural modulus - The modulus of elasticity calculated from the results of a bend test, giving the slope of the stress-deflection curve.



The bend test often used for measuring the strength of brittle materials, and (b) the deflection δ obtained by bending



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Stress-deflection curve for Mg0 obtained from a bend test



Compression:Compression: Many manufacturing processes such as forging, rolling, Many manufacturing processes such as forging, rolling, extrusion, are performed with the workpiece subjected to compressive forces. extrusion, are performed with the workpiece subjected to compressive forces. Compression test, in which the specimen is subjected to compressive load, Compression test, in which the specimen is subjected to compressive load, gives information useful for these processes. gives information useful for these processes. When the results of compression When the results of compression tests and tension tests on ductile metals are compared, the true stress-true tests and tension tests on ductile metals are compared, the true stress-true strain curves for the two tests coincide. This comparability does not hold true strain curves for the two tests coincide. This comparability does not hold true for brittle materials, which are generally stronger and more ductile in for brittle materials, which are generally stronger and more ductile in compression than in tensioncompression than in tension


Bending (Flexure):Bending (Flexure): Bend or flexure test is commonly used for brittle Bend or flexure test is commonly used for brittle materials. It usually involves a specimen that has a rectangular cross-section materials. It usually involves a specimen that has a rectangular cross-section and is supported at its ends. The load is applied vertically, at either one point and is supported at its ends. The load is applied vertically, at either one point or two: as a result, these tests are referred to as three-point and four point or two: as a result, these tests are referred to as three-point and four point bending, respectively. The longitudinal stresses in these specimens are tensile bending, respectively. The longitudinal stresses in these specimens are tensile at their lower surfaces and compressive at their upper surfaces. at their lower surfaces and compressive at their upper surfaces.

The stress at fracture in bending is known as the The stress at fracture in bending is known as the modulus of rupturemodulus of rupture, or , or transverse rupture strength.transverse rupture strength.



Hardness of Materials

Hardness:Hardness: Hardness is defined as Hardness is defined as resistance to permanent indentationresistance to permanent indentation. It gives . It gives a general indication of the strength of the material and of its resistance to a general indication of the strength of the material and of its resistance to scratching and to wear. Example, steel is harder than aluminum, and scratching and to wear. Example, steel is harder than aluminum, and aluminum is harder than lead. aluminum is harder than lead. Several methods have been developed to Several methods have been developed to measure the hardness of materials.measure the hardness of materials.



Hardness and StrengthHardness and Strength: Studies have shown that (in the same units) the : Studies have shown that (in the same units) the hardness of a cold-worked metal is about three times its yield stress, Y: for hardness of a cold-worked metal is about three times its yield stress, Y: for annealed metals, it is about five times Y. annealed metals, it is about five times Y. A relationship has been established A relationship has been established between the ultimate tensile strength (UTS) and the Brinell hardness (HB) for between the ultimate tensile strength (UTS) and the Brinell hardness (HB) for steels. In SI units, steels. In SI units, UTS = 3.5*(HB), where UTS is in Mpa. Or UTS = 500*(HB), where UTS is in UTS = 3.5*(HB), where UTS is in Mpa. Or UTS = 500*(HB), where UTS is in psi and HB is in kg/mm2, as measured for a load of 3000 kg.psi and HB is in kg/mm2, as measured for a load of 3000 kg.



Hardness-Testing ProceduresHardness-Testing Procedures: The following considerations must be taken for : The following considerations must be taken for hardness test to be meaningful and reliable:hardness test to be meaningful and reliable:

1.1. The zone of deformation under the indenter must be allowed to develop The zone of deformation under the indenter must be allowed to develop freely.freely.

2.2. Indentation should be sufficiently large to give a representative hardness Indentation should be sufficiently large to give a representative hardness value for the bulk material.value for the bulk material.

3.3. Surface preparation is necessary, if conducting Rockwell test and other Surface preparation is necessary, if conducting Rockwell test and other tests, except Brinell test.tests, except Brinell test.




Temperature EffectsTemperature Effects: Increasing the temperature generally has the following effects : Increasing the temperature generally has the following effects on stress-strain curves:on stress-strain curves:

a.a. It raises ductility and toughnessIt raises ductility and toughness

b.b. It lowers the yield stress and the modulus of elasticityIt lowers the yield stress and the modulus of elasticity

c.c. It lowers the strain-hardening exponent of most metalsIt lowers the strain-hardening exponent of most metals

Rate-of-Deformation EffectsRate-of-Deformation Effects: Deformation rate is defined as the speed at which a : Deformation rate is defined as the speed at which a tension test is being carried out, in units of, say, m/s or ft/min.tension test is being carried out, in units of, say, m/s or ft/min.

Strain rateStrain rate is defined as the true strain that the material undergoes per unit time. is defined as the true strain that the material undergoes per unit time.

The strain rate is a function of the specimen length. A short specimen elongates The strain rate is a function of the specimen length. A short specimen elongates proportionately more during the same time period than does a long specimen.proportionately more during the same time period than does a long specimen.

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When a ductile material is pulled in a tensile test, necking begins and voids form – starting near the center of the bar – by nucleation at grain boundaries or inclusions. As deformation continues a 45° shear lip may form, producing a final cup and cone fracture


Thank You

Impact Testing of Materials

Impact test - Measures the ability of a material to absorb the sudden application of a load without breaking.

Impact energy - The energy required to fracture a standard specimen when the load is applied suddenly.

Impact toughness - Energy absorbed by a material, usually notched, during fracture, under the conditions of impact test.

Fracture toughness - The resistance of a material to failure in the presence of a flaw.



The impact test: (a) The Charpy and Izod tests, and (b) dimensions of typical specimens


Ductile to brittle transition temperature (DBTT) - The temperature below which a material behaves in a brittle manner in an impact test.

Notch sensitivity - Measures the effect of a notch, scratch, or other imperfection on a material’s properties, such as toughness or fatigue life.


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Results from a series of Izod impact tests for a super-tough nylon thermoplastic polymer


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The Charpy V-notch properties for a BCC carbon steel and a FCC stainless steel.


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The area contained within the true stress-true strain curve is related to the tensile toughness. Although material B has a lower yield strength, it absorbs a greater energy than material A.


Fracture mechanics - The study of a material’s ability to withstand stress in the presence of a flaw.

Fracture toughness - The resistance of a material to failure in the presence of a flaw.


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Schematic drawing of fracture toughness specimens with (a) edge and (b) internal flaws


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The fracture toughness Kc of a 3000,000psi yield strength steel decreases with increasing thickness, eventually leveling off at the plane strain fracture toughness Klc


Fatigue of Materials

• S-N-Curve, the most commonly used stress ratio is R = (S min/S max). If the stresses are fully reversed, then R = -1. If the stresses are partially reversed, R = a negative number less than 1. If the stress is cycled between a maximum stress and no load, R = zero. If the stress is cycled between two tensile stresses, R = a positive number less than 1.

• The basic method of presenting engineering fatigue data is by means of the S-N curve, a plot of stress S against the number of cycles to failure N. The value of stress that is plotted can be σa, σmax, or σmin. The stress values are usually nominal stresses, i.e., there is no adjustment for stress concentration.

Fatigue Test Setup



The S-N fatigue curve for an acetal polymer


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Examples of stress cycles. (a) Equal stress in tension and compression, (b) greater tensile stress than compressive stress, and (c) all of the stress is tensile


The highest stress at which a run out (non-failure) is obtained is taken as

the fatigue limit. For materials without a fatigue limit the test is usually terminated for practical

considerations at a low stress where the life is about 108 or 5x108 cycles.

The S-N curve is usually determined with about 8 to 12 specimens.

Fatigue limit (endurance limit) occurs for some materials (some Fe and Ti allows). In this case, the S-N

curve becomes horizontal at large N. The fatigue limit is maximum stress amplitude below which the material never fails, no matter how large the

number of cycle is.



The S-N curves for a tool steel and an aluminum alloy


Creep of Materials

Creep Behavior

Creep is a time-dependent and permanent deformation of materials when subjected to a constant load at a high temperature (>0.4Tm). Examples: turbine blades, stream generators.

Stages of CreepCreep Testing



The effect of temperature or applied stress on the creep curve

Creep Behavior



A typical creep curve

Secondary/steady-state creep is of longest duration and is the most important parameter of the creep behavior in long-life applications

έ=Δε/Δt

Stages of Creep

• Primary/transient creep. • Secondary/steady-state creep.• Tertiary creep.


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Results from a series of creep tests. (a) Stress-rupture curves for an iron-chromium-nickel alloy and (b) the Larson-Miller parameter for ductile cast iron

Creep Behavior


Thank You

CERAMICS MATERIALS

Ceramics properties High hardness, (high strength, stiffness, wear resistance) Brittle, Low ductility or malleability i.e. low plasticity, Electrical and thermal insulating, Chemical stability, and high melting temperatures Some ceramics are translucent, window glass (based on silica). Lower density than most metals, Low resistance to fracture, highly resistant to compressive loads. Corrosion resistance Ceramics are hard, brittle, totally elastic and are heat resistant.At extremely low temperature, exhibit superconductivity. Due to high resistance to heat, application in furnace linings. Ceramics are often used as protective coatings to other materials.

CERAMICS MATERIALS

Electrical Ceramics, insulators, electrical devices, Superconductors. Coatings, Biocompatible coatings (fusion to bone), Self-lubricating bearings Abrasives Piezoelectric materials are lead zirconate titanate and barium titanate. Design of high-frequency loudspeakers, Transducers for sonar, and Actuators for atomic force and scanning tunneling microscopes. Semiconducting ceramics are also employed as gas sensors. Corrosion resistant applications, Windows, Television screens, Magnetic materials (audio/video tapes, disks, etc.), Magnets. Ceramic fibers, graphite and aluminum oxide, fiber-reinforced composites Pottery, clay, glasses, vitreous enamels, and Cutting tools. Chemically Bonded Ceramics (e.g. cement and concrete) Structural Ceramics, Whitewares (e.g. porcelains), Engineering ceramics

1- Oxides (SiO2, Al2O3, Fe 2O3, MgO, SrTiO3, MgAl2O4, YBa2Cu3O7-x)2- Carbides (SiC, WC, TiC), Borides, Nitrides (Si3N4, TiN, AlN, GaN, BN),3- Composites: Particulate reinforced

CERAMICS MATERIALS

Applications of Ceramics

http://www.mse.cornell.edu/courses/engri111/fiber.htm

http://www.mse.cornell.edu/courses/engri111/compo.htm

Traditional Ceramics Primary products are fired clay (pottery, tableware, brick,

and tile), cement, and natural abrasives such as alumina Glass is also a silicate ceramic material and is sometimes

included among traditional ceramics

Raw Materials for Traditional Ceramics Mineral silicates, such as clays of various compositions, and

silica, such as quartz, are among the most abundant substances in nature and constitute the principal raw materials for traditional ceramics

Another important raw material for traditional ceramics is alumina

CERAMICS MATERIALS

Clay and Silica as a Ceramic Raw Material Clays consist of fine particles of hydrous aluminum silicate,

Most common clays are based on the mineral kaolinite, (Al2Si2O5(OH)4)

When mixed with water, clay becomes a plastic substance that is formable and moldable. When heated to a sufficiently elevated temperature (firing ), clay fuses into a dense, strong material. Thus, clay can be shaped while wet and soft, and then fired to obtain the final hard product

Silica is available naturally in various forms, most important is quartz, the main source of quartz is sandstone, Low in cost; also hard and chemically stable

Principal component in glass, and an important ingredient in other ceramic products including whiteware, refractories, and abrasives

CERAMICS MATERIALS

Alumina as a Ceramic Raw Material Bauxite - most alumina is processed from this mineral, which is

an impure mixture of hydrous aluminum oxide and aluminum hydroxide plus similar compounds of iron or manganese. Bauxite is also the principal source of metallic aluminum

Corundum - a more pure but less common form of Al2O3, which contains alumina in massive amounts

Alumina ceramic is used as an abrasive in grinding wheels and as a refractory brick in furnaces

Traditional Ceramic Products Pottery and Tableware Brick and tile Refractories Abrasives

CERAMICS MATERIALS

New Ceramics

Ceramic materials developed synthetically over the last several decades

The term also refers to improvements in processing techniques that provide greater control over structures and properties of ceramic materials

In general, new ceramics are based on compounds other than variations of aluminum silicate, which form most of the traditional ceramic materials

New ceramics are usually simpler chemically than traditional ceramics; for example, oxides, carbides, nitrides, and borides

Thin films of many complex and multi-component ceramics are produced using different techniques such as sputtering, sol-gel, and chemical-vapor deposition (CVD).

Fibers are produced from ceramic materials for several uses: as a reinforcement in composite materials, for weaving into fabrics, or for use in fiber-optic systems.

CERAMICS MATERIALS

Oxides Ex. Properties

Aluminum oxide Al2O3 high strength and hardness, high stiffness, high thermal stability

magnesium oxide MgO high thermal stability

Mullite Al6Si2O13 Low coefficient of thermal expansion, high thermal stability

silicon dioxide SiO2 Low density, transparency

Zirconium dioxide

ZrO2 high toughness when transformation toughened

Carbides Ex. Properties

Diamond C high strength, stiffness, low coefficient of thermal expansion,

Graphite C high strength, stiffness, low coefficient of thermal expansion

silicon carbide SiC high strength and hardness, high stiffness

tungsten carbide WC high strength and hardness

Nitrides Ex. Properties

Boron nitride BN very high strength and hardness, very high stiffness

silicon nitride Si3N4 high strength, hardness, stiffness and high thermal stability

CERAMICS MATERIALS

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When silica crystallizes on cooling, an abrupt change in the density is observed. For glassy silica, however, the change in slope at the glass temperature indicates the formation of a glass from the undercooled liquid. Glass does not have a fixed Tm or Tg. Crystalline materials have a fixed Tm and they do not have a Tg.

CERAMICS MATERIALS

Adapted from Fig. 13.6, Callister, 7e.

T

Specific volume

Supercooled Liquid

solid

T m

Liquid(disordered)

Crystalline (i.e., ordered)

T g

Glass (amorphous solid)

Specific volume (1/r) vs Temperature (T)

Sheet forming – continuous draw– originally sheet glass was made by “floating” glass on a pool of

mercury


CERAMICS MATERIALS

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Different techniques for processing of advanced ceramics.

CERAMICS MATERIALS

Processing of Advanced Ceramics

©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

Processes for shaping crystalline ceramics: (a) pressing, (b) isostatic pressing, (c) extrusion, (d) jiggering, and (e) slip casting.

CERAMICS MATERIALS

Thank You

POLYMERIC MATERIALS

POLYMER composed primarily of C and H, they have low melting temperature, poor thermal and electrical conductors. Some are crystals, many are not, high plasticity, a few have good elasticity. Some are transparent, some are opaque. Low density structures.

Main applications

• Films, foams, paints, fibers, and structural materials.

• Microelectronics industry, fabrication of semiconductor devices.

• Plastics, Liquid crystals, Adhesives and glues

• Containers and Water-resistant coatings (latex)

• Moldable products (computer casings, telephone handsets)

• Clothing (vinyl , polyesters, nylon), Biomaterials (organic/inorganic)

• Low-friction materials (teflon), Synthetic oils and greases

• Gaskets and O-rings (rubber), Soaps and surfactants

POLYMERIC MATERIALS

Industrially Important Polymers

Polymer Example Applications

Polyethylene (PE) Electrical wire insulation, flexible tubing, bottles

Polypropylene (PP) carpet fibers, liquid containers (cups, buckets, tanks), pipes

Polystyrene (PS) packaging foams, egg cartons, lighting panels, electrical components

Polyvinyl chloride (PVC) bottles, pipes, valves, electrical wire insulation, toys, raincoats, automobile roofs

POLYMERIC MATERIALS

Mechanisms of PolymerizationPolymerization is the formation of chemical linkages between relatively small molecules or monomers to form very large molecules or polymers. These linkages are formed by either one or two of the following two types: addition or condensation.

Addition polymerization process is characterized by the simple combination of molecules without the generation of any by-products as a result of the combination. The original molecules do not decompose to form reaction debris. When units of single monomers are hooked together, the resulting product is a homopolymer, such as polyethylene, that is made from the ethylene monomer. When two or more polymers are used in the process, the product is a co-polymer.

Condensation polymerization involves the chemical reaction of two or more chemicals to form a new molecule. The chemical union of two molecules can be only achieved by the formation of a by-product molecule with atoms from the two molecules to create the link for the polymerization to continue. This chemical reaction produces a condensate or non-polymerizable byproduct, usually water. A catalyst is often required to start and maintain the reaction. It can also be used to control the reaction rate.

Structure–Property Relationships in Thermoplastics

Branched polymer - Any polymer consisting of chains that consist of a main chain and secondary chains that branch off from the main chain.

Crystallinity is important in polymers since it affects mechanical and optical properties.

Tacticity - Describes the location in the polymer chain of atoms or atom groups in nonsymmetrical monomers.

Liquid-crystalline polymers - Exceptionally stiff polymer chains that act as rigid rods, even above their melting point.

POLYMERIC MATERIALS

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The relationship between the density and the temperature of the polymer shows the melting and glass temperatures. Note that Tg and Tm are not fixed; rather, they are ranges of temperatures.

POLYMERIC MATERIALS

Both Tm and Tg increase with increasing chain stiffness.

Regularity – effects Tm only


Melting vs. Glass Transition Temp.

POLYMERIC MATERIALS

Table 4 Melting and glass temperature ranges (0C) for selected

thermoplastics and elastomers

• Thermoplastics: -- little crosslinking -- ductile -- soften w/heating -- polyethylene polypropylene polycarbonate polystyrene

• Thermosets: -- large crosslinking (10 to 50% of mers) -- hard and brittle -- do NOT soften w/heating -- vulcanized rubber, epoxies, polyester resin, phenolic resin

Adapted from Fig. 15.19, Callister 7e. (Fig. 15.19 is from F.W. Billmeyer, Jr., Textbook of Polymer Science, 3rd ed., John Wiley and Sons, Inc., 1984.)

Thermoplastics vs. Thermosets

Callister, Fig. 16.9

T

Molecular weight

Tg

Tmmobile liquid

viscous liquid

rubber

tough plastic

partially crystalline solid

crystalline solid

Polymer Additives

Improve mechanical properties, processability, durability, etc.

Fillers– Added to improve tensile strength & abrasion resistance,

toughness & decrease cost– ex: carbon black, silica gel, wood flour, glass, limestone, talc, etc.

• Plasticizers – Added to reduce the glass transition

temperature Tg

– commonly added to PVC - otherwise it is brittle

POLYMERIC MATERIALS

Polymer Additives

Stabilizers– Antioxidants, Antistatic Agent– UV protectants, Catalysts

• Lubricants – Added to allow easier processing – “slides” through dies easier – ex: Na stearate

• Colorants – Dyes or pigments

• Flame Retardants– Cl/F & B

POLYMERIC MATERIALS

• Reinforcements

POLYMERIC MATERIALS

Industrially Important Glasses

Glass Ex. Properties

Silica glassSiO2 Used for optical fibers when it is very pure

Soda-lime glass SiO2-Na2O-

CaO

standard glass used for bottles and windows due to its low cost and easy manufacturing

Borosilicate glass

SiO2-B2O3 thermal shock resistance (glassware) and low coefficient of thermal expansion

Lead glass SiO2-PbO high index of refraction

Polymer Crystallinity

Polymers rarely 100% crystalline Too difficult to get all those chains

aligned

• % Crystallinity: % of material that is crystalline. -- TS and E often increase with % crystallinity. -- Annealing causes crystalline regions to grow. % crystallinity increases.

Adapted from Fig. 14.11, Callister 6e.(Fig. 14.11 is from H.W. Hayden, W.G. Moffatt,and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, John Wiley and Sons, Inc., 1965.)

crystalline region

amorphousregion

POLYMERIC MATERIALS

Polymer Processing and Recycling

Forming Processes for Thermoplastics:• Extrusion• Blow Molding• Injection Molding• Thermoforming

Forming Processes for Thermosetting polymers:• Calendaring• Spinning• Compression Molding• Transfer Molding

POLYMERIC MATERIALS

POLYMERIC MATERIALSExtrusion: The polymer is heated to the liquid state and forced through a die under pressure resulting in an endless product of constant cross section. Examples include: tubing, pipes, window frames, sheet, and insulated wire.

Film Blowing: Using the same method as extrusion the material coming out of the die is blown into a film. An example is plastic wrap.

Injection molding: Similar to extrusion, the polymer is heated to the liquid state, but it is prepared in metered amounts, and the melt is forced into a mold to create the part. It is not a continuous process. Many toys are made by injection molding.

Blow molding: The melted polymer is put into a mold, and then compressed air is used to spread the polymer into the mold. It is used to make many containers such as plastic soda containers and milk jugs.

Compression molding: Solid polymer is placed in a mold; the mold is heated and puts pressure on the polymer to form the part.

Reaction injection molding: Liquid monomers are placed in the mold avoiding the need to use temperature to melt the polymer or pressure to inject it. The monomers polymerize in the mold forming the part.

Processing Plastics - Molding

Injection molding– thermoplastic & some thermosets

Adapted from Fig. 15.24, Callister 7e. (Fig. 15.24 is from F.W. Billmeyer, Jr., Textbook of Polymer Science, 2nd edition, John Wiley & Sons, 1971. )

POLYMERIC MATERIALS

Processing Plastics – Extrusion

Adapted from Fig. 15.25, Callister 7e. (Fig. 15.25 is from Encyclopædia Britannica, 1997.)

POLYMERIC MATERIALS

Blown-Film Extrusion

Adapted from Fig. 15.26, Callister 7e. (Fig. 15.26 is from Encyclopædia Britannica, 1997.)

POLYMERIC MATERIALS

Mechanical Propertiesbrittle polymer

plasticelastomer

FS of polymer

10% that of metals

Strains – deformations > 1000% possible (for metals, maximum strain ca. 10% or less)

elastic modulus – less than metal


POLYMERIC MATERIALS

• Decreasing T... -- increases E -- increases TS -- decreases %EL

• Increasing strain rate... -- same effects as decreasing T.

©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

The stress-strain curve for 6,6-nylon, a typical thermoplastic polymer.

POLYMERIC MATERIALS

©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under

license.

Necks are not stable in amorphous polymers, because local alignment strengthens the necked region and reduces its rate of deformation.

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Course Syllabus for ME 2510: Materials Engineering (3 hrs lectures + 1 hrs tutorial) l Instructor:...

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