Engineering Materials Science Test 1

Concept:

Deformation of a material means a change in its dimensions and shape under applied load or stress. The change in shape may be linear, angular or combination of both.

The deformation may be temporary in nature i.e. recoverable or permanent i.e. irrecoverable in nature.

Further deformation may be time dependent or time independent.

• Elastic Deformation: It is small, time independent, fully recoverable and obeys Hooke’s law. It occurs in metals within their elastic limits.
• Elastomeric Deformation: It is too large, time independent, fully recoverable and does not obey Hooke’s law. It occurs in elastomers like natural rubber whose monomer is isoprene.
• Plastic Deformation/Inelastic Deformation: It is large, permanent, time independent and does not obey Hooke’s law. It occurs in metals beyond their elastic limits.
• Anelastic Deformation: It is small, fully recoverable but time dependent. It may or may not obey Hooke’s law. It occurs in rubber, plastics and metals due to thermostatic phenomena.
• Viscoelastic Deformation: It is time dependent, partially elastic and partially permanent. It obeys Hooke’s law and Newton’s law for viscous flow. It occurs in polymers.
• Creep is time dependent inelastic deformation. Creep is the slow plastic deformation of metal under constant stress or under prolonged loading.

Out of all the given phenomenon, only elastic deformation is “time-independent”. Rest all depends on time.

Concept:

\(APF = \frac{{{n_{atom}} \times {V_{atom}}}}{{{V_{unit - cell}}}}\)

Calculation:

\(0.70 = \frac{{n\; \times \;\frac{4}{3}\pi {r^3}}}{{{a^3}}}\)

\(\therefore n = \frac{{0.7\; \times \;3\; \times \;{{1.2}^3}}}{{4\pi\; \times \;{{0.4851}^3}}}\)

∴ n = 2.53

Concept:

The Gibb’s phase rule is:

C + 2 = P + F

Where, C = components, P = phases, F = degrees of freedom.

If the pressure is fixed, one independent variable is lost, and this takes the form:

C + 1 = P + F

Calculation:

For maximum number of phases, DOF = 0 (Maximum number of phases occurs at zero degree of freedom)

And for ternary component, C = 3

Thus, no of phases

C + 1 = P + F

Explanation:

1. Point defects

1. Vacancy defect
2. Frenkel defect
3. Schottky defect
4. Substitutional defect
5. Interstitial defect

Vacancy, Frenkel and Schottky defects are intrinsic defects and Substitutional, Interstitial defects are extrinsic defects.

2. Line defects

1. Edge dislocation
2. Screw dislocation

3. Surface defects

1. Grain boundary defect
2. Tilt boundary defect
3. Twin boundary defect
4. Stacking fault

4. Volume defects

1. Voids

Induction hardening: Can be split into two steps:

i) Induction heating: The first one is induction heating, in which electrically conducting metals are heated with an electromagnet.

ii) Quenching: The quenching phase follows directly after to alter the surface of material.

Characteristics:

• Heat the surface to γ-region (localised heating) using an inductor coil carrying a high frequency current in the range of 2 to 500 kHz.
• Core: Not heated, Structure remains unaltered.
• Surface converts to martensite on quenching.
• Fast heating and short hold time: Need higher austenitization temperature.
• Martensite forms in fine r (not homogenous) grains.
• Applicable only for steels especially carbon steels (0.35 – 0.7 %C)
• Little distortion and good surface finish.

Slip systems: The combination of a slip plane and its direction of slip is known as slip system. Each pattern of atomic arrangements results in different number of slip systems.

• For BCC crystal structure: 48
• For FCC crystal structure: 12
• For Hexagonal close packed structure: 3

Concept:

The correct match is,

Resilience - Capacity of a material to absorb energy elastically

Fatigue - Failure of material at loads less than that at normal conditions

Stiffness - Resistance of the material to elastic deformation

Elasticity - Ability of the material to return to the original shape when the load is removed

Resilience

• It is the total strain energy stored in a given volume of material within the elastic limit.
• On removal of a load, this energy is released. In other words, it is the area under the load-deflection curve within the elastic limit.

Fatigue

• When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue.

Stiffness

• Stiffness of a material is the resistance offered to deformation, below the elastic limit. 
• A material with a high value of Young’s modulus E is stiffer than the material with the lower value of Young’s modulus.
• Small values of E indicate flexible materials and a large value of E reflects stiffness and rigidity.

Elasticity 

• The elasticity of a metal is its power of returning to its original shape after the applied force is released.
• Properly heat-treated spring is a good example of elasticity.

Concept:

Reaction where a liquid phase decomposes to two solid phases is Eutectic. (P)

Reactions where liquid phase combines with solid phase 1 to produce another solid 2 is Peritectic (Q)

Reactions where a solid decomposed to two solid phases is Eutectoid. (R)

Reactions where two solid phases combine to produce a third solid is Peritectoid. (S)

Ease to remember:

Reactions where the term “toid” occurs deals with all phases being solid.

If ‘mono’ occurs in a reaction, then only 1 component will be the reactant and if “peri” occurs more than one reactant.

For e.g.: ‘Eu’ tec ‘toid’ – as eu occurs reactant will be one, and as toid occurs entire reaction will have solid phase.

Concept:

\({\rm{Density\;}}\left( {\rm{\rho }} \right) = \frac{{nA}}{{{N_A}{V_C}}}\)

n = effective no. of atoms in a crystal, A = atomic weight

N_A = Avogadro constant. = 6.023 × 10²³ atoms/mol, V = vol. of cubic crystal.

Calculation:

Given:

a = 0.3 nm = 3 × 10^-10 m = 3 × 10^-8 cm

A = 11.90 g/cc

\(n = \frac{{p\;{N_A}{V_c}}}{A} = \frac{{11.90\; \times \;6.023\; \times\; {{10}^{23}}\; \times \;{3^3}\; \times \;{{10}^{ - 24}}}}{{49}}\)

∴ n = 3.94 ∼ 4

FCC has 4 effective atoms.

∴ Structure is FCC [Face-centred cubic]

Concept:

Percentage empty space = 1 - Atomic pacing factor × 100

Calculation:

Given:

R = 0.208 nm, Structure = BCC

Now,

The relation for BCC crystal is 4R = √3a

Where a = side of the crystal and R = atomic radius

Now,

4R = √3a

\(R\; = \frac{{\sqrt 3 }}{4}a\)

\(\therefore a = \frac{{0.208 \times 4}}{{\sqrt 3 }}\)

∴ a = 0.48

Now,

\({\rm{APF}} = \frac{{{\rm{Total\;volume\;of\;atoms}}}}{{{\rm{Volume\;of\;unit\;cell}}}}\)

\({\rm{APF\;}} = \frac{{{\rm{n}}\frac{4}{3}{\rm{\pi }}{{\rm{R}}^3}}}{{{{\rm{a}}^3}}}{\rm{\;}}\left( {{\rm{n}} = {\rm{effective\;number\;of\;atoms\;inside\;unit\;cell}}} \right){\rm{\;For\;BCC\;n}} = 2\)

\(\therefore \;APF\; = \frac{{2 \times \frac{4}{3}\pi {R^3}}}{{{a^3}}}\)

\(\therefore \;APF = \frac{{2 \times \frac{4}{3}\pi {{\left( {0.208} \right)}^3}}}{{{{0.48}^3}}}\;\)

∴ APF = 0.6816

Now,

Percentage empty space = (1 - Atomic pacing factor) × 100

∴ Percentage empty space = (1 – 0.6816) × 100

∴ Percentage empty space = 31.83

Explanation:

Density \(\left( \rho \right) = \frac{{\eta \cdot {A_w}}}{{N \cdot {V_{uc}}}}\)

∴ No. of atoms \({\rm{\;}} = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4\) (Refer geometry)

(V_uc volume of unit cell)

V_uc = a³

Now,

∴ \(\rho = \frac{{4A}}{{N{a^3}}}\)

For FCC

\(r = \frac{{a\sqrt 2 }}{4} \Rightarrow a = \frac{{4r}}{{\sqrt 2 }}\)

∴ \(\rho = \frac{{4{A_w}}}{{N \cdot {{\left( {\frac{{4r}}{{\sqrt 2 }}} \right)}^3}}}\)

Explanation:

For a face-centred cubic (FCC) structure, there is a fixed relationship between atomic radius and the lattice parameter.

 An FCC unit cell has atoms of each of six cubic faces.

From geometry

 \(r = \frac{{a\sqrt 2 }}{4}\)

∴ \(a = \frac{{4r}}{{\sqrt 2 }}\)

∴ \( a = \frac{{4\: \times \:0.143}}{{\sqrt 2 }}\)

∴ a = 0.404 nm

Explanation:

Given data spacing (1, 1, 1)

Inter planer distance

\(\lambda = \frac{a}{{\sqrt {{h^2} + {k^2} + {\lambda ^2}} }}\)

\(\lambda = \frac{a}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \frac{a}{{\sqrt 3 }}\)

\(\therefore \lambda = \frac{a}{{\sqrt 3 }}\)

For FCC

\(a = \frac{{4R}}{{\sqrt 2 }} = \frac{{4\; \times\; 0.128}}{{\sqrt 2 }}\)

a = 0.362 nm            (1 A° = 0.1 nm)

\(\therefore \lambda = \frac{{0.362}}{{\sqrt 3 }}\)

∴ λ = 0.20 g nm ≈ 2 A° 

Concept:

For tetragonal crystal structure, lattice parameters are a = b ≠ c & α = β = γ = 90°

If c/a < √0.67, then edge along z-axis will be close packed direction.

⇒ 2 r = c

If c/a lies between √0.67 and √2, then body diagonal will be close packed direction.

⇒ 4 r = (2a² + c²)^1/2

If c/a > √2, then edge along base will be close packed direction.

⇒ 2 r = a

Now,

For body centered tetragonal,

One atom lies at the center of the body

At all the eight corners, eight atoms will be there and each atom will be shared by 8 unit cells

No. of atoms per unit cell = 1 + (8 × (1/8)) = 2 (Option 1)

Calculation:

Given:

c = 1.2 a ⇒ c/a = 1.2

Here c/a lies between √0.67 and √2, then body diagonal will be close packed direction.

⇒ 4 r = (2a² + c²)^1/2 = (2a² + (1.2a)²)^1/2 = 1.85 a

⇒ r = 0.463 a (Option 3)

\(APF = \frac{{{N_{atoms}}{V_{atoms}}}}{{{V_{unit\;cell}}}} = \frac{{2 \times \frac{4}{3} \times \pi \times {{\left( {0.463a} \right)}^3}}}{{a \times a \times 1.2a}} = 0.693\)   (Option 2)

Mistake point:

When body diagonal is mentioned close packed direction, we can’t take 4r = √3a which holds good for BCC but not BCT