Lectures in engineering materials

Part two

Components metal:

As mention above the most Metals are compose from (Ferrous

and Nom-ferrous).

Ferrous materials contain iron, and the one elements and

Metals alloy. Ferrous alloys are extremely versatile have a

wide range of mechanical and physical properties.

Steel is an alloy of iron and carbon. Carbon steel can be

classified as:

i. Low carbon steel: (C 0.15-0.3%) , has a tensile strength of

(420 - 555 N/mm2), hardness of about (125- 140 BHN)

ii. Medium carbon steel: containing carbon from 0.35 to

0.7%

has a tensile strength of about 750 - 1230 N/mm2

iii. High carbon steel: Steel containing carbon from 0.7 to

1.5%, has a tensile strength of (580 -1400 N/mm2).

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Alloy Steel:-

Steel is considered to be alloy steel when the maximum of the

range given for the content of alloying elements exceeds one or

more of the following limits;

Mn 1.65%, Si 0.6%, Cu 0.6%, Ni= 0.7% , Cr 0.6% Mo 0.3%

Crystal structure

A crystal is a solid composed of atoms, iron or molecules

arranged in a pattern which is a repetitive in three dimensions.

The regular repetitive arrangement of atoms are described by

a three dimensional network (Space lattice). Space lattice is a

three dimensional pattern of points called lattice point.

Each point in the space lattice has identical surroundings. The

size and shape of an unit cell is described by three edge lengths

(a,b,c) and the angles (α,β,ᵧ) known as lattice parameters.

Lattice Parameter Relationships and Figures Showing Unit Cell

Geometries for the Seven Crystal Systems

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Metal crystal structure:

– The crystal system found in most elements metals are either

– BCC (Body centered cubic)

– FCC (Face centered cubic)

– HCP (Hexagonal closed pack)

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– In BCC the unit cell has one atom at each corner and one atom

at the center of the cube.

– In FCC there is one atom at each corner of the cube and one at

the center of each face.

– In HCP there are two lattice basal planes in the form of regular

hexagon with an atom at each corner of the hexagon and one

atom at the center of Basel plane another plane that provides

three additional atoms to the unit cell is situated between the top

and bottom plane.

The parameters of crystal structure:

1- The relation between Length edge a and the atomic radius

R are related through

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2- The number of atoms per unit cell, N are:

 N= 4, For the FCC crystal structure

 N= 2, For the BCC crystal structure.

 N= 6, For the HCP crystal structure.

3- The nearest-neighbor or touching atoms or coordination

number are:

for FCC = 12, for BCC = 8, HCP = 12

4- the unit cell volume VC = a3 for all crystal

5- atomic packing factor (APF)

6- density computation (𝜌):

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 theoretical density compute by this relation

NA
ρ=
VC NA

Where

n is number of atoms associated with each unit cell

A is atomic weight

Vc is volume of the unit cell

NA is Avogadro’s number (6.02 × 1023 atoms/mol)

Example: 1

Calculate the volume of an FCC unit cell in terms of the atomic

radius R.

Solution:

FCC unit cell volume VC = a3

a = 4R/√2 = 2R√2 = , VC = (2R√2)3 = 16R3√2

VC = 16R3√2

Example; 2

Compute the atomic packing factor for the FCC crystal structure
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Solution:

The APF is defined as the fraction of solid sphere volume in a

unit cell, or

volume of atoms∈a unit cell VS

volume for a sphere is = V S =¿ 43 π R ¿3

the total unit cell volume V C is 16R3√2

16 3
VS ( )πR
APF = VC
=¿ 3
=0.74
3
16 R √ 2

Example 3

Copper has an atomic radius of 0.128 nm, an FCC crystal

structure, and an atomic weight of 63.5 g/mol. Compute its

theoretical density and compare the answer with its measured

density (8.94 ).

Solution:

NA
ρ=
VC NA

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VC for FCC = 16 R3 √ 2 , N= 4

4 x 63.5
−7 3 =8.89 g/cm3
16 x √ 2 ( 0.128 x 10 ) x (6.02 ×10 )
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Home work:

Choose the correct answer

1- Low carbon steel has

a. (C 0. 5-0.3%)

b. (C 0.15-0.35%)

c. (C 0.10-0.3%)

d. All answers are false

2- Medium carbon steel has a tensile strength of about

a. 750 - 1230 N/mm2

b. 750 - 1200 N/mm2

c. 750 - 1000 N/mm2

3- In BCC the unit cell has

a. one atom at each corner and two atoms at the center of

the cube.

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 b. two atoms at each corner and one atom at the center of

the cube.

c. one atom at each corner and one atom at the center of

the cube.

4- Calculate the volume of an FCC unit cell in terms of the

atomic radius 5nm

5- Calculate the volume of an BCC unit cell in terms of the

atomic radius 4nm

6- Compute the atomic packing factor for

a. the FCC crystal structure

b. the BCC crystal structure

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