chapter 11: kinetics, microstructure, heat...

MEEN 3344, Introduction to Materials Science

Chapter 11: Kinetics, Microstructure, Heat Treating

1. Take Roll (pass out attendance sheet) 2. Show & tell? 3. Questions from Chapter 10 homework problems? How do you harden Steel? How do you soften steel? Kinetics, Microstructure, Heat Treating

• Kinetics – rate at which a process occurs • Non-equilibrium states – caused by rapid cooling, and/or addition of pressure. May be aided by addition of small amounts of certain elements. • The most common non-equilibrium process is that of heat-treating (quenching). • Micro-structure – the shape and composition of a material at the level of an individual grain.

Phase Transformation: What happens if you do not follow equilibrium cooling? Can we get other phases?


The Fe – C phase diagram revisited By cooling quickly, we can keep the austenite structure, or obtain a new, harder phase/structure. …. Will not spend time on, or require you to go over all of the equations in the first part of chapter 11 on nucleation.


• Grains are formed at nucleation sites. By adding a small amount of a grain-refiner we can get smaller and more uniform grains. • Adding a nucleation catalyst reduced the energy it takes to cause grain growth, resulting in more uniform growth


Isothermal Transformation Diagrams

Transformation time as a function of Temperature…..


Phase transformation or Heat Treatment of Steel


Prob 11.5 For this problem, we are given, for the austenite-to-pearlite transformation, two values of y and two values of the corresponding times, and are asked to determine the time required for 95% of the austenite to transform to pearlite. The first thing necessary is to set up two expressions of the form of Equation 11.17, and then to solve simultaneously for the values of n and k. In order to expedite this process, we will rearrange and do some algebraic manipulation of Equation 11.17. First of all, we rearrange as follows:

1 − y = exp − ktn( ) Now taking natural logarithms ln (1 − y) = − ktn

Or

− ln (1 − y) = ktn which may also be expressed as ln 1

1 − y

⎛

⎝ ⎜

⎞

⎠ ⎟ = kt n

Now taking natural logarithms again, leads to ln ln 1

1 − y

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ = ln k + n ln t

which is the form of the equation that we will now use. Using values cited in the problem statement, the two equations are thus

ln ln 1

1 − 0.2

⎡

⎣ ⎢

⎤

⎦ ⎥

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= ln k + n ln(280 s) ln ln 1

1 − 0.6

⎡

⎣ ⎢

⎤

⎦ ⎥

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= ln k + n ln(425 s)

Solving these two expressions simultaneously for n and k yields n = 3.385 and k = 1.162 × 10-9. Now it becomes necessary to solve for the value of t at which y = 0.95. One of the above equations—viz


− ln (1 − y) = ktn may be rewritten as

t n = −ln (1 − y)

k

And solving for t leads to t = −

ln (1 − y)k

⎡ ⎣ ⎢

⎤ ⎦ ⎥ 1/n

Now incorporating into this expression values for n and k determined above, the time required for 95% austenite transformation is equal to

t = −

ln (1 − 0.95)1.162 × 10−9

⎡

⎣ ⎢

⎤

⎦ ⎥ 1/3.385

= 603 s


The whole IT / TTT Diagram Pearlite is formed at temperatures greater than at the nose of the TTT (IT) diagram, Bainite is formed below the nose, Martensite is formed if the material is quenched quick enough, below a certain point.


Summary of Terms, learn these terms!!

Martensite is the new term here…. It is a BCT (body-centered tetragonal) structure, that forms upon quenching from austenite. (very hard and strong)


Pearlite formed at various temperatures 2-phase layered material. formed by diffusion Bainite is formed below the nose. It consists of needles or plates, and is relatively hard and brittle.


Spheroidite - from pearlite or bainite that is soaked at a temperature below the eutectic for many hours (18 – 24 hrs)


Martensite: The Ms line is where the martensitic phase change starts, M50 is the 50% temperature, and M90 is the 90% phase change temperature, and Mf is the temperature where the martensitic transformation is finished.

Martensitic transformations are also found in ceramics, other metal alloys an in rocks in nature.


Progress and structure of the martensitic transformation Note the needle-like structure

• The FCC/BCT transformation occurs without diffusion, so it can be very fast. • Volume always increases when austenite transforms to martensite…… what problems could this cause?

Tempered Martensite • martensite can be tempered by raising the temperature of the steel part until cementite begins to form.…. and spheroidite can be formed. • The tempering reduces some internal stresses, and makes the steel less brittle. Note what happens as the temperature and time goes up.


Use of an IT (TTT) diagram (Example 11.2)


11.10 Do some on chalkboard Example At the right is shown the

isothermal transformation diagram for a eutectoid iron-carbon alloy, with time-temperature paths that will yield (a) 100% coarse pearlite; (b) 50% martensite and 50% austenite; and (c) 50% coarse pearlite, 25% bainite, and 25% martensite.

11.11 We are asked to determine

which microconstituents are present in a 1.13 wt% C iron-carbon alloy that has been subjected to various isothermal heat treatments.

(b) Spheroidite (c) Bainite and martensite


Review, Questions & Comments When does one get pearlite? Bainite? Martensite?

• Changing the amount of carbon in a steel, or the amount of other alloy components can move the nose of the TTT diagram to the right or left, and change the temperature at which pearlite, bainite, and martensite start to form. • Hardenability is the capability of the steel to easily form martensite. If the nose of the TTT diagram is moved to the right the material can be hardened easier, and at a slower rate. • Since martensite forms due to a rapid temperature change, the thermal conductivity of the steel is important. Some steels will only form martensite when in thin strips. Others will harden through thick sections.


A Continuous Cooling Curve shows hardening rate Cooling Rate Notes:

• Carbon and other elements shift the nose to the left • If less than .25% carbon, can’t cool steel fast enough • Typically only a “Pearlite nose” is shown • Addition of other elements may produce a “bainite nose”


11.28 This problem asks that we briefly describe the simplest continuous cooling heat treatment procedure that would be used in converting a 4340 steel from one microstructure to another. Solutions to this problem require the use of Figure 11.29.

(a) In order to convert from (martensite + ferrite + bainite) to (martensite + ferrite + pearlite + bainite) it is necessary to heat above about 720°C, allow complete austenitization, then cool to room temperature at a rate between 0.02 and 0.006°C/s.

(b) To convert from (martensite + ferrite + bainite) to spheroidite the alloy must be heated to about 700°C for several hours.

(c) In order to convert from (martensite + bainite + ferrite) to tempered martensite it is necessary to heat to above about 720°C, allow complete austenitization, then cool to room temperature at a rate greater than 8.3°C/s, and finally isothermally heat treat the alloy at a temperature between about 400 and 550°C (Figure 11.35) for about one hour.


Mechanical behavior of constituents as a function of carbon


Notes • Martensite – hardest and strongest, most brittle, no ductility • Bainite – has a finer structure and is stronger and harder than pearlite, has a good combination of strength, ductility. • Cementite – much harder and more brittle than pearlite, but reduces impact strength and toughness • Layer thickness of each material also affects behavior. • Coarse pearlite is more ductile than fine pearlite • Spheroidite – spherical nodules, lower in strength and stiffness than pearlite, softest and weakest of steel micro-structures, very ductile and tough.


Two hardening and tempering processes Note that the second is a “mar-tempering process” and is used to relieve thermal gradients and minimize surface cracking.


Precipitation Hardening again….


A material at temperature T1 is quenched to T2. The α phase is “frozen” in place. At T2 the extra component B is allowed to age or precipitate out to a desired amount. If T2 is above room temp, it is called artificial aging. If T2 is at room temp, it is called natural aging. The component B can fill either interstitial or regular vacancies. When B precipitates out, it prevents the formation of the β phase or a softer, weaker phase. The filled vacancies tend to stretch the crystal structure, adding energy to the system.


Example of Precipitation Hardening

• There are a number of PH aluminum alloys. • Typically the presence of fine, randomly distributed precipitates yield higher

strength.

Some Metals that PH-harden: • Some Stainless Steels, aluminum-copper, copper-berillium, copper-tin,

and magnesium-aluminum alloys.


Polymers and Phase Transformation


Example Questions 11.6 This problem asks us to consider the percent recrystallized versus logarithm of

time curves for copper shown in Figure 10.2. (a) The rates at the different temperatures are determined using Equation (10.2),

which rates are tabulated below: Temperature (°C ) Rate (min)-1

135 0.105 119 4.4 x 10-2 113 2.9 x 10-2 102 1.25 x 10-2 88 4.2 x 10-3 43

3.8 x 10-5 (b) These data are plotted below.


The activation energy, Q, is related to the slope of the line drawn through the data points as

Q = − Slope (R) where R is the gas constant. The slope of this line is -1.126 x 104 K, and thus

Q = − −1.126 x 104 K( )(8.31 J/mol - K)

= 93,600 J/mol (c) At room temperature (20°C), 1/T = 3.41 x 10-3 K-1. Extrapolation of the data

in the plot to this 1/T value gives

ln (rate) ≅ − 12.8 which leads to

rate ≅ exp (−12.8) = 2.76 x 10-6 (min)-1 But since

rate =1

t0.5

then t0.5 =

1rate

=1

2.76 x 10−6 (min)−1 daysx 250min1062.3 5 ==


This problem inquires as to the possibility of producing an iron-carbon alloy of eutectoid composition that has a minimum hardness of 200 HB and a minimum ductility of 25%RA. If the alloy is possible, then the continuous cooling heat treatment is to be stipulated. According to Figures 11.31(a) and (b), the following is a tabulation of Brinell hardnesses and percents reduction of area for fine and coarse pearlites and spheroidite for a 0.76 wt% C alloy. Microstructure HB %RA Fine pearlite 270 22 Coarse pearlite 205 29 Spheroidite 180 68 Therefore, coarse pearlite meets both of these criteria. The continuous cooling heat treatment which will produce coarse pearlite for an alloy of eutectoid composition is indicated in Figure 11.28. The cooling rate would need to be considerably less than 35°C/s, probably on the order of 0.1°C/s.


This problem inquires as to the possibility of producing a precipitation-hardened 2014 aluminum alloy having a minimum tensile strength of 425 MPa (61,625 psi) and a ductility of at least 12%EL. In order to solve this problem it is necessary to consult Figures 11.43(a) and (b). Below are tabulated the times required at the various temperatures to achieve the stipulated tensile strength.

Temperature (°C)Time Range (h) 260 <0.5 204 <15 149 1-1000 121 >35-? With regard to temperatures and times to give the desired ductility: Temperature (°C)Time Range (h) 260 <0.02, >10 204 <0.4, >350 149 <20 121 <1000

chapter 11: kinetics, microstructure, heat...

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