EGR250 HomeWork 4 PDF

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Quang Nguyen Quang Nguyen Section 10 Collaboration with: None

1.

EGR 250-Homework No.3

September 20,2016 Instructor: Dr. P. N. Anyalebechi

The photomicrograph of the microstructure of an annealed low carbon steel sample taken at 150x magnification is shown in Figure 1. (a) Use the mean linear intercept method to determine the average grain diameter. (b) Using the ASTM Comparative Chart method, determine the ASTM grain size number. (c) Using the ASTM Comparative Chart method, determine the grain size index.

150x Figure 1: Photomicrograph showing the microstructure of an annealed low carbon steel. Solution: We are asked to determine the average grain diameter using the mean linear intercept method, to determine the ASTM grain size number using the ASTM Comparative Chart method, and to determine the grain size index using the ASTM Comparative Chart method, given the following Magnification

=

150x 1

Quang Nguyen

In Figure 1, there are 46 full grains and 32 half grains

The formula to calculate the number of grain in given photomicrograph is: number of grain=number of full grain+

number of half grain 2

So, the number of grain in photomicrograph in Figure 1 is: number of grain=46 grains+

32 half grains =62 grains 2

According to measurement, the photomicrograph has dimension of or

77 mm ×101 mm ,

3.03 ∈× 3.976 ∈¿ .

The formula to calculate the area of rectangular shape is: A=a ×b where A = area of the rectangular shape, a and b are the length of the sides of the rectangular shape. So the area of the photomicrograph (in

2 mm ) is:

A=77 mm× 101mm =7777 mm2 So the area of the photomicrograph (in

2 ¿ ) is: 2

A=3.03∈× 3.976∈¿12.05 ¿

(a) To determine the average grain diameter using the mean linear intercept, 10 lines which include 5 equally horizontal lines and 5 equally vertical lines are drawn across the given photomicrograph.

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Quang Nguyen

Figure 1a:Photomicrograph with grain number and lines

The length of each horizontal line is 100 mm The length of each vertical line is 75 mm

Number of grains in line 1 is 7 grains Number of grains in line 2 is 7 grains Number of grains in line 3 is 8 grains Number of grains in line 4 is 9 grains Number of grains in line 5 is 8 grains Number of grains in line 6 is 7 grains Number of grains in line 7 is 8 grains Number of grains in line 8 is 6 grains Number of grains in line 9 is 8 grains Number of grains in line 10 is 6 grains

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Quang Nguyen

The average grain diameter is indicated by the equation: ´d= L MN where

´d = the average grain diameter, L = the length of each line segment, M =

magnification of the photomicrograph, and N = number of grains or grain boundaries.

So the average grain diameter in line 1 is: ´d = 100 mm =0.09524 mm 1 150 ∙ 7 So the average grain diameter in line 2 is: ´d = 100 mm =0.09524 mm 2 150 ∙ 7 So the average grain diameter in line 3 is: ´d 3= 100 mm=0.08333 mm 150∙ 8 So the average grain diameter in line 4 is: ´d 4 = 100 mm=0.07407 mm 150 ∙ 9 So the average grain diameter in line 5 is: ´d = 100 mm=0.08333 mm 5 150∙ 8 So the average grain diameter in line 6 is: ´d 6= 75 mm =0.07143 mm 150∙ 7 So the average grain diameter in line 7 is: ´d 7= 75 mm =0.06250 mm 150∙ 8

So the average grain diameter in line 8 is: ´d 8= 75 mm =0.08333 mm 150∙ 6

So the average grain diameter in line 9 is:

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Quang Nguyen

´d 9= 75 mm =0.06250 mm 150∙ 8

So the average grain diameter in line 10 is: ´d = 75 mm=0.08333 mm 10 150 ∙ 6

The formula to determine the average value is:

∑¿ number of values average=¿ So the average grain diameter in given photomicrograph is: ´d= 0.09524 mm ∙ 2+ 0.08333 mm ∙ 4 + 0.07407 mm+0.07143 mm+0.06250 mm ∙2 10 ´ 0.07943mm ∴ d=¿

The average grain diameter in given photomicrograph is 0.07943mm

(b) The number of grains per area is determined by the equation: g=

number of grains area

where g = number of grains per area

So the number of grain per area (in ¿2 ) is: gi=

62 grains =5.145 grains/¿2 2 12.05¿

So the number of grain per area (in mm2 ) is: gm=

62 grains =7.97 ×10−3 grains / mm 2 2 7777 mm

The formula calculate the grain size number is: N=2

n −1

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Quang Nguyen

where N = the number of grain observed in an area of 1¿ 2 ( 645 mm2 ) on a photomicrograph taken at a magnification of 100 times, and n = an integer referred to as the ASTM grain-size number. At a given magnification: M 100 ¿ ¿ N=¿ where M = magnification of the photomicrograph and

gi = number of grains per

area (in ¿2 ) So the number of grain observe in an area of 1 ¿2 150 100 ¿ ¿ N=¿

is:

So the ASTM grain size number is: 11.6 =2n−1 ∴n−1=log 2 ( 11.6 ) =3.54 ∴n=¿ 4.54

So the ASTM grain size number is 4.54

(c) The formula calculate the grain size number is: m=2G +3 e

where m = the number of grain per mm2 at magnification of 1x, and

G e = grain

size index At a given magnification: M 1 ¿ ¿ m=¿ where M = magnification of the photomicrograph and area (in mm2 ) 6

gm = number of grains per

Quang Nguyen

So the number of grain observe in an area of 1 mm2

is:

150 1 ¿ ¿ N=¿ So the grain size index is: 179.3=2

G E +3

∴G E +3=log 2 ( 179.3 )=7.49 ∴G E=¿ 4.49

So the grain size index is 4.49

2. The grain size of engineering materials, especially metals, affects their mechanical, chemical, and physical properties. In fact, according to the Hall-Petch empirical relationship, yield strength is inversely proportional to the square root of grain diameter. Briefly describe the methods used in industry to control grain size in metal products. Solution: In industry, grain size are control by three main methods. The first method is controlling the rate of solidification from the liquid phase. This method will have to decrease the grain size, thus, the yield strength is improved. The second method is addition of grain refiner to molten metal prior to casting. Adding grain refiner helps increase the yield strength, and the ductility of material by decreasing grain size. Finally, the last method is using appropriate heat treatment in plastic deformation. This method uses large amount of deformation to minimize opportunity for grain grow, so it helps decrease grain size. Moreover, increase the time in annealing process makes the grain size smaller. 3.

In a certain crystalline material, the vacancy concentration at 25oC is one fourth that at 80oC. At what temperature would the vacancy concentration be three times that at 80oC?

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Quang Nguyen

Solution: We are asked to determine the temperature at which the vacancy concentration of a certain crystalline material be three times that at 80 oC, given the following: Vacancy concentration at 25oC

1 4

=

Vacancy concentration at

80oC The formula to convert Celsius degree to Kelvin degree is: Temperature degree ∈Kelvin= temperature degree ∈ Celsius +273 So, 25oC

=

298oK

80oC

=

353oK

The vacancy concentration is determined by formula: C v =e

(

−Q v ) RT

where C v = vacancy concentration, a dimension quantity,

Q v = energy required to

produce a vacancy, in J/mol, R = gas constant, 8.314 J/mol.K; and T = absolute temperature in K.

So, The vacancy concentration at 25oC is: (

C v at 25 C =e

−Qv ) o R ∙298 K

o

The vacancy concentration at 80oC is: (

C v at 80 C=e

−Q v ) o R ∙353 K

o

Because the vacancy concentration at 25oC is one fourth that at 80oC: ∴C v at 25 C o

(

e

−Qv ) o R ∙298 K

1 C 4 vat 80 C

=

o

1 ( R ∙ 353 e 4

−Q v

=

8

o

K

)

Quang Nguyen

( ln [ e ( ln [ e

−Qv R ∙ 298o K −Qv R ∙ 298o K

)]

=

)]

=

1 ln [ e(R ∙ 353 K )] 4 1 4 e −Q v ¿+ ln[ ¿ ¿ ] R ∙ 353 o K ln ¿ −Q v 1 ln + 4 R ∙ 353 o K 1 ln 4 1 ln 4 1 ln 4 1 ln 4 −1 1 ) ( + o o K 353 298 K o 2651.452 K −Qv

o

(

−Q v

( ( ( (

=

o

−Q v o

R ∙ 298 K −Q v

−

R ∙ 298 K −Q v

=

o

R ∙ 353 K Qv + o o R ∙ 298 K R ∙ 353 K Qv 1 −1 + ) ∙( o R 298 K 353o K Qv R

Qv R

= =

)

) ) ) )

()

=

=

Qv ¿ 2651.452o K R

So,

Let T is the temperature would the vacancy concentration be three times that at 80oC. The vacancy concentration at ToC is: C v at T C =e

(

−Qv ) o R∙T K

o

Because the vacancy concentration at ToC is three times that at 80oC: ∴C v at T

=

3 ∙C v at 80 C

−Q v ) R ∙T

=

( 3∙e

=

( ln [3 ∙ e

(

e

( )] ln [e −Qv R ∙T

o

9

−Qv o R ∙ 353 K

)

−Q v o R ∙ 353 K

)]

Quang Nguyen

( R ∙T )] ln [e −Qv

3 e

=

¿+ ln[ ¿ ¿ −Q v R∙T −Q v

ln (3 ) +

−Q v o

R ∙ 353 K ln ¿ −Q v

)

]

o

R∙ 353 K

=

ln (3 )

=

ln (3 )

=

ln (3 )

−1 1 ) + T 353 o K 1 −1 + T 353o K −1 T 1 T T

=

ln (3 )

=

ln (3 ) 2651.452o K ln (3 )

T

=

−Q v − R ∙ T R ∙ 353o K −Q v Qv + R ∙ T R ∙ 353 o K Q v −1 1 ∙( ) + R T 353o K

Because

=

(

Qv ¿ 2651.452o K , so: R

2651.452o K ∙(

=

1 o 2651.452 K 353 K −ln ( 3) 1 + o 2651.452 K 353 o K 1 −ln ( 3 ) 1 + o 2651.452 K 353o K 413.5 o K o

= =

−

The formula to convert Celsius degree to Kelvin degree is: Temperature degree ∈Kelvin= temperature degree ∈ Celsius + 273

So, 413.5oK

140.5oC

=

So the temperature at which the vacancy concentration of a given crystalline material be three times that at 80oC is 140.5o C

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Quang Nguyen

4. (a) Determine the interplanar spacing and the length of the Burgers vector for slip on the expected slip systems in aluminum. The lattice parameter for aluminum is 0.404958 nm. (b) Repeat the calculation assuming that the slip system is a (110)/[ 11¯ 1 ]. (c) What is the ratio between the shear stresses required for the two slip systems. Assume k in the Peieerls-Nabarro stress equation is 2.

Solution: We are asked to determine the interplanar spacing and the length of the Burgers vector for slip on the expected slip systems in aluminum and on the slip system (110)/[ 11¯ 1 ], and to determine the ratio between the shear stresses required for the two slip systems, given the following: Lattice parameter of Aluminum

=

0.404958 nm

k in the Peieerls-Nabarro stress equation

=

2

Aluminum atomic radius is 0.143 nm Assume aluminum has Face-centered cubic (FCC) crystal structure. The relationship between lattice parameter and atomic radius in FCC crystal structure is determined by the formula: a o=

4r √2

∴r = where

√ 2 ∙ ao 4 a o = lattice parameter, and r = atomic radius

So the atomic radius of aluminum in the case aluminum has FCC crystal structure is: r= √

2∙ 0.404958 nm =0.143 nm 4

So the aluminum found in this case matched the actual atomic radius of aluminum. So aluminum has FCC crystal structure.

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Quang Nguyen

(a) The expected slip system in aluminum is (111)/[110] The formula to determine the interplanar spacing in cubic crystal system is d hkl=

ao

√ h +k 2 +l2 2

where d = interplanar spacing between adjacent slip planes,

a o = lattice parameter, and h,

k, l represent the Miller indices of the adjacent planes being consider.

So the interplanar spacing on expected slip system in aluminum is: d hkl=

0.404958 nm =¿ 0.2338nm √ 12+12 +12

In direction [110] in FCC crystal system, the formula to calculate the magnitude of Burgers vector is: b=

ao √2 2

where b = magnitude of Burgers vector and

a o = lattice parameter

So the magnitude of Burgers vector on expected slip system in aluminum is: b=

0.404958 nm √ 2 =0.2863nm 2

So interplanar spacing on expected slip system in aluminum is 0.2338nm, the magnitude of Burgers vector on expected slip system in aluminum is 0.2863nm

(b) The formula to determine the interplanar spacing in cubic crystal system is d hkl=

ao

√ h +k 2 +l2 2

where d = interplanar spacing between adjacent slip planes,

a o = lattice parameter, and h,

k, l represent the Miller indices of the adjacent planes being consider. So the interplanar spacing on the slip system (110)/[ 1 ¯1 1 ] in aluminum is:

12

Quang Nguyen

d hkl=

0.404958 nm =¿ 0.2863nm √ 12 + 12 + 02

¯ ] in FCC crystal system, the formula to calculate the magnitude of In direction [ 111 Burgers vector is: b=ao √ 3 where b = magnitude of Burgers vector and

a o = lattice parameter

So the magnitude of Burgers vector on the slip system (110)/[ 11¯ 1 ] in aluminum is: b=0.404958 nm √ 3 =0.7014nm So interplanar spacing on the slip system (110)/[ 11¯ 1 ] in aluminum is 0.2863nm, the magnitude of Burgers vector on the slip system (110)/[ 11¯ 1 ] in aluminum is 0.7014nm

(c) The Peierls-Nabarro stress require to move the dislocation from one equilibrium location to another is given by the equation: τ =c e where

−kd b

τ

= shear stress required to move the dislocation, d = interplanar spacing

between adjacent slip planes, b = magnitude of the Burgers vector, and c and k are constant for the material.

So the shear stress required for the expected slip system in aluminum is: τ =c e

−2 ∙ 0.2338nm 0.2863 nm

=0.1953 ∙ c

So the shear stress required for slip system (110)/[ 1 ¯1 1 ] in aluminum is: τ =c e

−2 ∙ 0.2863nm 0.7014 nm

=0.4420 ∙ c

The ratio between the shear stresses required for the two slip systems is determined by the formula: 13

Quang Nguyen

ratio= where

τ1 τ2 τ 1 = shear stress of expected slip system in aluminum, and

τ 2 = shear stress

¯ ] in aluminum of the slip system (110)/[ 111 So the ratio between the shear stresses required for the two slip systems, which are expected slip system in aluminum and the slip system (110)/[ 11¯ 1 ] in aluminum is 0.1953 ∙ c ratio= =¿ 0.442 0.4420 ∙ c So the ratio between the shear stresses required for the two slip systems, which are expected slip system in aluminum and the slip system (110)/[ 11¯ 1 ] in aluminum is 0.442

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Quang Nguyen

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