Garbuzova KR1 Option 08
📂 Tests
👤 Михаил_Перович
Product Description
Option 8
1. What is the essence of the process of modification? Give an example of use
Modifiers for improving the properties of foundry aluminum alloys.
2. What is the difference between cold and hot plastic deformation? Describe
Features of both types of deformation.
3. Draw a diagram of the iron-iron carbide state, indicate the structural
Components in all areas of the diagram, describe the transformations and construct a curve
Cooling (using a phase rule) for an alloy containing 5.0% C. What is the structure of this
Alloy at room temperature and how such an alloy is called?
4. Carbon steel 35 and U8 after quenching and tempering have a martensite structure
Tempering and hardness: the first 45 HRC, the second - 60 HRC. Using the iron-
Iron carbide and taking into account the transformations occurring during tempering, indicate the temperature
Quenching and tempering temperature for each steel. Describe the transformations taking place in these
Steels in the quenching and tempering process, and explain why steel U8 has a greater hardness than
Steel 35.
5. Steel 40 was quenched from temperatures of 760 and 840? C. Using the chart
State of iron-cementite indicate which structures are formed in each case. Explain
Reasons for the formation of different structures and recommend the optimal mode of heating for quenching
Of this steel.
1. What is the essence of the process of modification? Give an example of use
Modifiers for improving the properties of foundry aluminum alloys.
2. What is the difference between cold and hot plastic deformation? Describe
Features of both types of deformation.
3. Draw a diagram of the iron-iron carbide state, indicate the structural
Components in all areas of the diagram, describe the transformations and construct a curve
Cooling (using a phase rule) for an alloy containing 5.0% C. What is the structure of this
Alloy at room temperature and how such an alloy is called?
4. Carbon steel 35 and U8 after quenching and tempering have a martensite structure
Tempering and hardness: the first 45 HRC, the second - 60 HRC. Using the iron-
Iron carbide and taking into account the transformations occurring during tempering, indicate the temperature
Quenching and tempering temperature for each steel. Describe the transformations taking place in these
Steels in the quenching and tempering process, and explain why steel U8 has a greater hardness than
Steel 35.
5. Steel 40 was quenched from temperatures of 760 and 840? C. Using the chart
State of iron-cementite indicate which structures are formed in each case. Explain
Reasons for the formation of different structures and recommend the optimal mode of heating for quenching
Of this steel.
No Reviews Yet
Be the first to leave a review for this product!
Related Products
Targ S.M. 1989 K2 variant 57
Seller: Timur_ed
Targ S.M. 1989 K2 variant 52
Seller: Timur_ed
Answers to IDZ 9.1 option 9 Ryabushko part 2
Seller: plati-goods
Answers to IDZ 8.4 option 9 Ryabushko part 2
Seller: plati-goods
Answers to IDZ 8.3 option 9 Ryabushko part 2
Seller: plati-goods
Targ S.M. 1989 K4 variant 99
Seller: Timur_ed
Targ S.M. 1989 C2 option 99
Seller: Timur_ed
Targ S.M. 1988 K3 option 65
Seller: Timur_ed
More from this Seller
Solution of task 4.2.13 from the collection of Kepe OE
For Education Students
Solution of task 4.3.14 from the collection of Kepe OE
For Education Students
Solution of task 3.2.11 from the collection of Kepe OE
For Education Students
Solution of task 4.3.15 from the collection of Kepe OE
For Education Students
Solution of task 4.3.13 from the collection of Kepe OE
For Education Students
Solution of task 4.2.18 from the collection of Kepe OE
For Education Students