mee1005 lecture4_devaprakasam

DEVAPRAKASAM DEIVASAGAYAMProfessor of Mechanical Engineering

Room:11, LW, 2nd FloorSchool of Mechanical and Building Sciences

Email: [email protected], [email protected]

MEE1005: Materials Engineering and Technology (2:0:0:2)

Devaprakasam D, Email: [email protected], Ph: +91 9786553933

MEE1005 MATERIALS ENGINEERING AND TECHNOLOGY

Purpose of Engineering /Education/Research

There are two important purposes and driving force behind the Engineering/Education/Research:

1. Minimum consumption of Energy and Materials without sacrificing the efficiency and functionality.

2. Maximum conversion of Energy from one form to the other, to identify or design highly efficient process or system.

Purpose of Engineering /Education/Research

UNIT-I : Structure of Materials

Introduction to engineering materials – significance of structure propertycorrelations in all classes of engineering materials, Unit Cells, MetallicCrystal Structures, Density Computations, Crystal Systems,Crystallographic Points, Crystallographic Directions, CrystallographicPlanes, Linear and Planar Densities, Close-Packed Crystal Structures,Crystalline and Non-crystalline Materials, Single Crystals, PolycrystallineMaterials, Imperfection in solids – Point, Line, Surface and Volumedefects - Polymorphism and Allotropy

Structure Property Correlations

= mass/volume= (atoms in the unit cell, n ) x

(mass of an atom, M) /(the volume of the cell, Vc)

Density Computations

Atoms in the unit cell, n = 4 (FCC)

Mass of an atom, M = A/NAA = Atomic weight (amu or g/mol)Avogadro number NA = 6.023 1023 atoms/mol

The volume of the cell, Vc = a3 (FCC)a = 2R2 (FCC)R = atomic radius

AcNVnA

Density of a crystalline material, = density of the unit cell

= (atoms in the unit cell, n ) (mass of an atom, M) / (the volume ofthe cell, Vc)

Density Computations Summarized

Atoms in unit cell, n = 2 (BCC); 4 (FCC); 6 (HCP)

Mass of atom, M = Atomic weight, A, in amu (or g/mol). Translatemass from amu to grams divide atomic weight in amu by Avogadronumber

NA = 6.023 1023 atoms/mol

Volume of the cell, Vc = a3 (FCC and BCC)a = 2R2 (FCC); a = 4R/3 (BCC)where R is the atomic radius AcNV