# MATERIAL SCIENCE@ YEE PIN by black90star

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```									ASSIGNMENT 1        BMT 101 MATERIAL SCIENCE   NOVEMBER SEMESTER 2009

SCHOOL OF ENGINEERING AND
TECHNOLOGY INFRASTURCTURE

BMT 101
MATERIALS SCIENCE

ASSIGNMENT 1

NOVEMBER SEMESTER 2009

NAME: TAN YEE PIN
MATRIC NO: KI 091393
LECTURER’S NAME: MS. AZMA FATINI ABDUL AZIZ
ASSIGNMENT 1                      BMT 101 MATERIAL SCIENCE             NOVEMBER SEMESTER 2009

Question 1:
Below, atomic radius, crystal structure, electro negativity and the most common valence are
tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.

Elements        Atomic Radius      Crystal Structure     Electro negativity        Valence
Ni              0.1246                FCC                    1.8                  +2
C                0.071
H                0.046
O                0.060
Ag              0.1445                FCC                     1.9                  +2
Al              0.1431                FCC                     1.5                  +3
Co              0.1253                HCP                     1.8                  +2
Pt              0.1387                FCC                     2.2                  +2
Table 1
Which of these elements would you expect to form the following with nickel:
a) A substitutional solid solution having complete solubility
b) An interstitial solid solution

Solution:
For complete substitutional solubility the following criteria must be met:
1) the difference in atomic radii between Ni and other element (ΔR%) must be less than ±15%
2) the crystal structures for both atom types must be the same
3) the difference in the electronegativity of both elements must be less
4) the valences should be the same, or nearly the same which a metal have more tendency to
dissolve another metal of higher valency than one of a lower valency.

Below are tabulated, for the various elements, these criteria.
Elements      Difference in Atomic          Crystal         Electronegativity        Valence
Ni                                        FCC                   1.8                  2+
C                  -43%                   HEX                   2.5                  4+
H                  -63%                     -                   2.1                  1+
O                  -52%                     -                   3.5                  2-
Ag                  16%                   FCC                  +0.1                  2+
Al                  15%                   FCC                  -0.3                  3+
Co                 0.6%                   HCP                    0                   2+
Pt                  11%                   FCC                  +0.4                  2+

(ΔR %) = (Atomic Radii of Others Elements - Atomic Radii of Ni) x 100 %

(a) Pt is the only element that meets all of the criteria and thus forms a substitutional solid
solution having complete solubility
(b) C, H, and O form interstitial solid solutions. These elements have atomic radii that are
significantly smaller than the atomic radius of Ni.
ASSIGNMENT 1                          BMT 101 MATERIAL SCIENCE              NOVEMBER SEMESTER 2009

Question 2:
Below are shown three different crystallographic planes for a unit cell of some hypothetical
metal. The circles represent atoms:

(0.39nm)                (0.32nm)                  (0.36nm)

(0.20nm)                  (0.30nm)               (0.25nm)

(110)                    (101)                  (011)

a) To what crystal system does the unit cell belong?
b) What would this crystal structure be called?

Solution:

(110)                                (101)                          (011)

a) The crystal system of unit cell is orthorhombic which a ≠ b ≠ c.
b) The crystal structure is face-centered cubic crystal structure (FCC).
ASSIGNMENT 1                     BMT 101 MATERIAL SCIENCE           NOVEMBER SEMESTER 2009

Question 3:
A specimen of magnesium having a rectangular cross section of dimension 3.2mm x 19.1mm
is deformed in tension. Using the load-elongation data tabulated as follows, complete parts (i)
through (v).

0                                             63.50
1380                                            63.53
2780                                            63.56
5630                                            63.62
7430                                            63.70
8140                                            63.75
9870                                            64.14
12850                                            65.41
14100                                            66.68
14340                                            67.95
13830                                            69.22
12500                                            70.49
Fracture
Table 2
(i) Plot the data as engineering stress versus engineering strain.
(ii) Compute the modulus of elasticity.
(iii) Determine the yield strength at a strain offset of 0.002.
(iv) Determine the tensile strength of this alloy.
(v) Determine the elongation of the specimen when a tensile stress of 100MPa is applied.

Solution:
(i) Area = (3.2 x 10¯³)(19.1 x 10¯³)
= 61.12 x 10 m²

Load (N)          Length (mm)         Stress = Force (load)      Length (m)       Strain = l - l
Area                                          l
0                 63.50                     0                63.50 x 10 ³            0
1380               63.53                 22.58 x 10           63.53 x 10 ³       0.47 x 10 ³
2780               63.56                 45.48 x 10           63.56 x 10 ³       0.94 x 10 ³
5630               63.62                 92.11 x 10           63.62 x 10 ³       1.89 x 10 ³
7430               63.70                121.56 x 10           63.70 x 10 ³       3.15 x 10 ³
8140               63.75                133.18 x 10           63.75 x 10 ³       3.94 x 10 ³
9870               64.14                161.49 x 10           64.14 x 10 ³      10.08 x 10 ³
12850               65.41                210.24 x 10           65.41 x 10 ³      30.08 x 10 ³
14100               66.68                230.69 x 10           66.68 x 10 ³      50.08 x 10 ³
14340               67.95                234.62 x 10           67.95 x 10 ³      70.08 x 10 ³
13830               69.22                226.28 x 10           69.22 x 10 ³      90.08 x 10 ³
12500               70.49                204.52 x 10           70.49 x 10 ³     110.08 x 10 ³
Fracture
The data is plotted in graph.
ASSIGNMENT 1                      BMT 101 MATERIAL SCIENCE            NOVEMBER SEMESTER 2009

(ii) E = Slope of the Graph
= Δ σ/ Δε
= (σ2 - σ1 ) / (ε 2 - ε 1)
= (80.00 – 0 ) MPa / ( 1.70 x 10 ³ - 0)
= 47.05 x 10 Pa
E = 47.05 GPa

Hence, the modulus of elasticity is 47.05 GPa.

(iii) The 0.002 strain offset line is constructed as show on graph, its intersection with the
stress-strain curve is at approximately 148MPa.

(iv) The tensile strength is the stress at the maximum on the engineering stress-strain curve
which is 234.62MPa.

(v) The corresponding strain is accomplished by locating the stress point on the stress-strain
curve point A which is approximately 2.20 x 10 ³.

Δl = εlo
= (2.20 x10 ³) ( 63.50 x 10 ³)
= 139.7 x 10 m
ASSIGNMENT 1                     BMT 101 MATERIAL SCIENCE           NOVEMBER SEMESTER 2009

Question 4:
A cylindrical rod 500mm long, having a diameter of 12.7mm is to be subjected to a tensile
load. If the rod is to experience neither plastic deformation nor an elongation of more than
1.3mm when the applied load is 29,000N, which of the four metals or alloys listed below are
possible candidate(s)?

Material        Modulus of Elasticity (GPa)     Yield Strength (MPa)     Tensile Strength (MPa)
Aluminum Alloy                 70                           255                       420
Brass Alloy                 100                           345                       420
Copper                    110                           210                       275
Steel Alloy                 207                           450                       550

Solution:
σ=F/A
= 29000N / { ����(12.7 x 10 ³/ 2) ²}
= 29000N / { ����(6.35 x 10 ³) ²}
= 228.93 x 10 Pa
= 228.93 MPa

All the metal alloys listed, only aluminum, brass and steel have yield strengths greater than
engineering stress, 228.93 MPa. Hence, the copper is not a candidate. By using the equation σ
= E ε and Δl = εlo, the elongation produced in aluminum, brass, and steel is determine
whether or not this elongation is less than 1.3 mm. The new equation is formed:

σ=Eε
σ = E Δl
lo
Δl = σ lo
E
Alloy                  Formula: Δl = σ lo                            Conclusion
E

Aluminum         (228.93 MPa)(500 x 10 ³ mm)                  Thus, aluminum is not a candidate.
70 GPa
=1.64 x 10 ³ m
=1.64 mm
Brass            (228.93 MPa)(500 x 10 ³ mm)                  Thus, brass is a candidate.
100 GPa
=1.15 x 10 ³ m
=1.15 mm
Steel            (228.93 MPa)(500 x 10 ³ mm)                  Thus, steel is a candidate.
207 GPa
=0.55 x 10 ³ m
=0.55mm
Therefore, all of these four alloys, only brass and steel which is the possible candidates
satisfy the stipulated criteria.
ASSIGNMENT 1                   BMT 101 MATERIAL SCIENCE          NOVEMBER SEMESTER 2009

Question 5:
Construct the hypothetical phase diagram for metals A and B between room temperature
(20°C) and 700°C given the following information:

   The melting temperature of a metal A is 480°C
   The maximum solubility of B in A is 18 wt% B, which occurs at 420°C
   The solubility of B in A at room temperature is 0 wt% B
   A eutectic occurs at 420°C and 46 wt% B – 54 wt% A
   The melting temperature of metal B is 600°C
   The maximum solubility of A in B is 22 wt% A, which occur at 420°C
   The solubility of A in B at room temperature is 3 wt% A

Referring to the same diagram, for a 90 wt% B – 10 wt% A alloy at 550°C
(a) What is/are the phase/phases present?
(b) What is the composition of the phase?
(c) Sketch the microstructure that would be observed for condition of very slow cooling
from the liquid phase. Label all phases present.

Solution:
(a) The phase that present is liquid phase.
(b) The composition of the phase is 54 wt% A.
(c) The microstructure that would be observed for condition of very slow cooling from
the liquid phase is sketch in graph paper. The phases are labeled.

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