Material Science - short answer questions from AMIE exams
Suggest one suitable material for each of the following purposes with justifications:
- File cabinet
- Water tap
- Manhole cover
- Garden chair
- Glass cutter
A suitable material is
- Steel, board
- cast iron
- Wrought iron, concrete, bamboo, timber
- Low alloy steel
Atomic radii of two metal atoms are 0.128 nm and 0.133 nm respectively. Find out whether they form a solid solution, and if they form, state what type of solid solution it is.
They will form substitutional solid solutions because the sizes of atoms are nearly the same.
Atomic radii of two metal atoms are 0.1278 nm (Cu) and 0.1431 nm (Al) respectively. Find out whether they form a solid solution, and if they form, state what type of solid solution it is.
Only 20% of copper atoms can be replaced by aluminium because there is a substantial difference in size.
Write down the slip plane and slip direction (one plane and one direction) of Nickel (only Miller indices). How many slip systems are there in Nickel?
Why has ferrite very low solubility of carbon, while austenite has high solubility of carbon?
Austenite is an FCC Crystal structure whereas Ferrite possesses a BCC crystal structure. Carbon is an interstitial impurity with respect to Iron i.e., it settles in the interstitial voids of Iron lattice owing to the large difference in the atomic radii of Carbon and Iron. Since the size of the available voids is more in FCC than in BCC, much Carbon can settle in them. Hence the solubility of Carbon is more in Austenite (2.14%) than in Ferrite (0.0225%).
Given activation energy, Q of 142 kJ/mol, for the dilution of carbon in FCC iron and an initial temperature of 1000 K, find the temperature that will increase the diffusion coefficient by a factor of 10. [R = 8.314 J/(mol.K)]
Hint: D = D0e-Q/RT
where D is diffusion coefficient, D0 is proportionality constant, Q is the activation energy, T is temperature.
A tension test recorded an engineering strain of 0.0046 against the engineering stress of 345 MPa of material within its elastic range. Find out the elastic modulus of the material and the type of metallic alloy (like iron base etc.).
Elestic modulous = stress/strain = 345/0.0046 = 75000 MPa
It is a Silica base of Metallic alloy.
The final thickness of a hard copper sheet is 1.0 mm. It was produced by cold working with 25% deformation. What was the starting thickness of the metal before cold working?
Hint: %CW = [(A0-Af)/A0] x 100
where A0 and Af are original and final areas respectively.
The fracture toughness equation of a material is given by Kfc = σf√(π.a). If the material has a strength of 300 MPa and a fracture toughness of 3.6 MPa√m, find out the largest internal crack in microns the material will support without cracking. σf = strength (MPa), a = crack size, m
K = σf√(π.a)
⇒a = K2/(πσf2)
= 3.62 = (π x 300)
= 4.58 x 10-5 m
Explain why ceramics are hard.
Positive metallic ions (atoms that have lost electrons) and negative nonmetallic ions (atoms that have gained electrons) develop strong attractions for each other. Each cation (positive) surrounds itself with anions (negative). Considerable force is usually required to separate the two. Ceramic materials tend to be hard.
Ceramic materials are usually ionic or covalent bonded materials and can be crystalline or amorphous. A material held together by either type of bond will tend to fracture before any plastic deformation takes place, which results in poor toughness in these materials. Additionally, because these materials tend to be porous, the pores and other microscopic imperfections act as stress concentrators, decreasing the toughness further, and reducing the tensile strength. These combine to give catastrophic failures, as opposed to the normally much more gentle failure modes of metals.
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