Material Science - short answer questions from AMIE exams

What is stress corrosion cracking (SCC)?

• Stress corrosion cracking (SCC) is the growth of crack formation in a corrosive environment. It can lead to unexpected sudden failure of normally ductile metals subjected to tensile stress, especially at elevated temperatures in the case of metals. 
• SCC is highly chemically specific in that certain alloys are likely to undergo SCC only when exposed to a small number of chemical environments. The chemical environment that causes SCC for a given alloy is often one that is only mildly corrosive to the metal otherwise. Hence, metal parts with severe SCC can appear bright and shiny, while being filled with microscopic cracks. This factor makes it common for SCC to go undetected prior to failure. 
• SCC often progresses rapidly and is more common among alloys than pure metals. The specific environment is of crucial importance, and only very small concentrations of certain highly active chemicals are needed to produce catastrophic cracking, often leading to devastating and unexpected failure.

Define anelasticity and viscoelasticity.

• Anelasticity: The time-dependent elastic behaviour is called anelasticity. In most engineering materials, there exists a time-dependent elastic component. That is elastic deformation will continue after the stress application and upon load release, some finite time is required for complete recovery. The anelasticity is due to microscopic and atomistic processes that are attendant to the deformation.
• Viscoelasticity: It is a type of deformation exhibiting the mechanical characteristics of viscous flow and elastic deformation. The phenomenon of viscoelastic behaviour is found in the polymeric material.

What is the fatigue limit of a material?

Fatigue limit: For fatigue, the maximum stress amplitude level below which a material can endure an essentially infinite number of stress cycles and not fail is called fatigue limit. in other words: The maximum stress to which a material can be subjected without fatigue, regardless of the number of cycles is called fatigue limit.

Why a polymer that is in the rubbery state has a Tg below room temperature?

• The normal state of most thermoset polymers is to be an amorphous solid at room temperature.  The arrangement of the polymer molecules is a random arrangement, meaning the polymer structure does not have a repeating arrangement of polymer chains.  
• An amorphous solid is different from a crystalline solid where the polymer molecules would be in a structured, repeating arrangement.  
• At temperatures below the Tg, the molecular chains do not have enough energy present to allow them to move around.  The polymer molecules are essentially locked into a rigid amorphous structure due to short-chain length, molecular groups branching off the chain and interlocking with each other, or due to a rigid backbone structure.   
• When heat is applied, the polymer molecules gain some energy and can start to move around.  At some point, the heat energy is enough to change the amorphous rigid structure to a flexible structure.  The polymer molecules move freely around each other.  This transition point is called the glass transition temperature.

What is corrosion fatigue?

• Corrosion fatigue is fatigue in a corrosive environment. It is the mechanical degradation of a material under the joint action of corrosion and cyclic loading. 
• Nearly all engineering structures experience some form of alternating stress and are exposed to harmful environments during their service life. The environment plays a significant role in the fatigue of high-strength structural materials like steel, aluminium alloys and titanium alloys. 
• Materials with high specific strength are being developed to meet the requirements of advancing technology. However, their usefulness depends to a large extent on the extent to which they resist corrosion fatigue. 
• The phenomenon should not be confused with stress corrosion cracking, where corrosion (such as pitting) leads to the development of brittle cracks, growth and failure. 
• The only requirement for corrosion fatigue is that the sample is under tensile stress.

What is cermet? Give examples.

• A cermet is a composite material composed of ceramic (cer) and metallic (met) materials. 
• A cermet is ideally designed to have the optimal properties of both a ceramic, such as high-temperature resistance and hardness and those of a metal, such as the ability to undergo plastic deformation. 
• The metal is used as a binder for an oxide, boride, or carbide. 
• Generally, the metallic elements used are nickel, molybdenum, and cobalt. 
• Depending on the physical structure of the material, cermets can also be metal matrix composites, but cermets are usually less than 20% metal by volume. 
• Cermets are used in the manufacture of resistors (especially potentiometers), capacitors, and other electronic components which may experience high temperatures.

What is the magnitude of maximum stress that exists at the tip of a surface crack having a radius of curvature of 0.264 nm and crack length of 1 µm when a tensile stress of 57 MPa is applied?

σ_m=2σ√(a/ρ)

=2 x 57√(1 x 10^-6/0.214 x 10^-9)

= 7016 MPa

What is thermal transformation?

Thermal transformation is heat that during a change of phase is absorbed. This occurs at constant pressure and temperature.

What is the interplanar spacing between (200), (220), (111) planes in an FCC crystal of atomic radius 1.246 Å?
d_{(h, k, l)} = a/√(h²+k²+l²)
But a = (2√2)r = 2√2 x 1.246 = 3.524 A
Hence, d₍₂₀₀₎ = 3.524/√(2²+0²+0²) = 1.762 A
Similarly, d₍₂₂₀₎ = 1.245 A
and d₍₁₁₁₎ = 2.034 A

A 45 kN force was applied on a Cu-Ni alloy tensile specimen having 12.5 mm diameter and 50 mm gauge length. Determine whether the specimen will undergo necking. Given: σ_UTS = 420 MPa and σ_y = 250 MPa.

Engineering stress = 366.89 MPa (do it yourself), mean stress = (250 + 420)/2 = 335 MPa

As engineering stress is (i) > yield stress and (ii) < UTS and (iii) more than mean stress, there will be plastic deformation and necking.

Steel has a tensile strength of 1.6 GPa. A large tensile piece of such steel has a crack of length 7 mm in the interior and the fracture at 0.6 GPa. Calculate its fracture toughness.

K=σ√(πa)= 88.95 MPa√m

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