Steels

Question 1

What are the hypereutectiod steels and why are they difficukt to machine.

Answer 1

Hypereutectiod steels are difficuklt to machine due to brittle network of pearlite.

Question 2

Describe Spherization annealing. why is it applied.

Answer 2

Spheroidization annealing: heating, soaking and cooling (very slowly). breaks up the hypereutectoid steel - cementite pearlite.

Question 3

What does the presence of Cr (chromium) or V (vanadium) do within the steel.

Answer 3

presence of Cr inhibits pearlite formatuion and therefore increases hardness and hardenability: (the ability of the material to form martensite).

Question 4

what contributes to the alloys increased shock resistance?

Answer 4

Most important property is toughness (hardness is secondary). it is therefore necessaru to have lower carbon contents than other tool steels to improve impact strength.

Question 5

what hardenability does the alloy display and what contributes to it

Answer 5

S7 has increased hardenability due to the 1.8wt% Mo and 6.5wt% Cr additions and oil quenching.

Question 6

what are the maximum temperatures for their operating temperatures?

Answer 6

200 degrees C

Question 7

List in order of heat extraction rate trandsfer air, water or oil quenching?

Answer 7

Air: 2.5-25W/m2K < oil: 20-500 W/m2K < Water: 100-15,000 W/m2K

Question 8

what is the benefit of high Cr and C cold worked tool steeks?

Answer 8

Ocidation resistance due to Cr.

Question 9

What is secondary hardening and how dies it impreove the performance of hot-worked tool steels?

Answer 9

It is the introduction of Mo and W into the grain structure to produe M-carbides which reduce the softening occuring in an alloy at high temperatures. whithou these elements diffusion of the Fe-carbides occurs readily softening the alloy at high temps.

Question 10

what is the application of HSS?

Answer 10

Mill cutters operating at high speeds

Question 11

what contributes to the materials hardness?

Answer 11

W and/or M0 provide carbide formation and “red hardness” (secodnary hardness)
-V provides increaded abrasion resistance, reduced oxidation and increased hardness.
-Co procides high temperature hardening.

Question 12

What are the three modes of failure of a tool and how do they occur?

Answer 12

Fracture failure: Cutting force becomes
excessive and/or dynamic, leading to brittle fracture. Temperature failure: Cutting temperature is too high for the tool
material. Gradual wear: Gradual wearing of the cutting tool

Question 13

Where does gradual wear occur in the tool bit and why?

Answer 13

Crater wear – occurs on top rake face
Flank wear – occurs on flank (side of tool). Abrasion - dominant cause of flank wear. Diffusion – Loss of hardening atoms
at tool-chip boundary (contributes to crater wear). Plastic deformation – contributes to flank wear

Question 14

What does the basic grain structure of cemented carbide consist of? What is the role of the individual phases within?

Answer 14

WC (α) Angular, Co (β) (Binder- Glue)

Question 15

What is the practicality of different inset shapes? (cemented carbide)

Answer 15

There is a trade off between strength, power requirements and
vibration tendency and versatility and accessibility. Increasing one decreases the other.

Question 16

What effect does increasing grain size have on the materials toughness and deformation resistance? What effect does
increasing the Co concentration have on the WC grain size? (cemented carbide)

Answer 16

The toughness increases and the deformation resistance
goes down. For an increase of 5 – 12 % the WC grain size increases from 1 – 5 um, becoming more coarse

Question 17

What is CVD & PVD? What are the influences of residual stresses in the formation of coatings by these two processes
and how do they influence the fracture resistance? (cemented carbide)

Answer 17

Chemical Vapour deposition and Physical Vapour Deposition. CVD
produces coatings with tensile stresses due to the mismatch between the thermal expansion coefficients of the coating
and substrates, lowering the fracture resistance. PVD produces compressive stresses due to the same reason, however
will show improved fracture resistance. 1

Question 18

What factors contribute to strength enhancement of the microalloyed steel?

Answer 18

Grain refinement, Sub grain formation
(means grains are very small & orientation between them is very small 0.5-1˚). Precipitation hardening