1. How many iron atoms are there per carbon atom in 4130 steel? Show your work. (1 points) Iron = 97.03% to 98.22, % iron =97.625%. Carbon = 0.280 to 0.330, % Carbon = 0.305%.
How many iron atoms are there per carbon atom in 4130 steel?
The Big Twist Final Project- Phase 1
1. How many iron atoms are there per carbon atom in 4130 steel? Show your work. (1 points)
Iron = 97.03% to 98.22, % iron =97.625%
Carbon = 0.280 to 0.330, % Carbon = 0.305%
Wt of fe = 97.625g, wt of C = 0.305g. Molecular wt of fe= 55.85g/mol, molecular wt of C= 12g/mol.
No. of Fe atoms = 97.625g/55.85g * 6.022×10^23 atoms = 10.522×10^23 atoms
No. of C atoms = 0.305g/12g*6.022×10^23 atoms = 0.153×10^23
Finally, No. of iron atoms per carbon atom = 10.522×10^23/0.153×10^23 = 68.77
2. For 1018 steel, how many iron atoms are there per carbon atom? Show your work. (1 points)
1018 steel is a plain carbon steel with0.18 wt%C.
We can assume that the remainder is iron (99.82 wt%Fe).
(0.18g C/99.82g Fe)*((mol C/12g C)/ (mol Fe/56 g Fe)) = (0.015mol C/1.7825 mole Fe) = 0.0084C/Fe This means there are ~ 8 C atoms for every 1000 Fe atoms. Or, said differently, ~125 Fe atoms for every C atom.
3. How can steel be processed to produce a martensite microstructure? Explain at least one reason why this structure has high strength. (2 points)
This structure has high strength because it is heavily saturated with carbon. In order to produce a martensite structure, steel has to be heated to really high temperatures to form the high temperature phase (austenite) and then martensite will form when it is cooled very quickly.
4. What happens to the microstructure of steel during the tempering process? (1 point)
Tempering is a process of applying heat to steel while keeping it below its eutectoid temperature to make the product tougher. During the tempering process martensite structure changes to ferrite and cementite which reduces the brittleness thus making the metal tougher.
5. What is the effect of tempering on strength and hardness? (1 point)
Cold: increased strength, decreased ductility, same elastic modulus and increased dislocation density. With the decrease of tempering temperature, the hardness of the steel increases and the and strength increases.
Hot: no strengthening, anisotropic behavior, poor surface finish and reduced dimensional accuracy. With the increase of tempering temperature, the hardness of the steel decreases and the toughness increases.
6. In one paragraph or less, explain what a tensile test is and why it was used by your team to test the samples of 4130 steel (Table 1). (1 point) A tensile test is a form of tension testing and is a materials science test whereby controlled tension is applied to a sample until it fully fails. It was used by our team so we would be able to record a yield and tensile strength, while also finding out the hardness and type of fracture the steel incurred. This will allow us to safely test the steel’s limitations.
7. Define yield strength and ultimate tensile strength. (1 point)
Yield strength: A stress value obtained graphically that describes no more than a specified amount of deformation (usually 0.002). Also known as the offset yield strength.
Ultimate tensile strength: The stress obtained at the highest applied force is the tensile strength, which is the maximum stress on the engineering stress strain curve.
8. Using only a few sentences, explain what a hardness test is and why it was used by your team (Table 1). (1 point)
A hardness test measures resistance to surface penetration of materials by a hard object. There are two types of hardness test (Brinell Hardness Test) and (Rockwell Hardness Test). We used a Rockwell Hardness Test (HRB) to determine how the materials properties such as strength, ductility, and wear resistance, changed throughout the thermal processing experiment.
9. Identify two possible sources of error for the results presented in Table 1. (1 point)