Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337670845
Author: ASKELAND
Publisher: Cengage
Question
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Chapter 13, Problem 13.13P
Interpretation Introduction

Interpretation:

The interference in 1080 between ferrite and cementite and percent reduction in the surface needs to be determined.

Concept Introduction:

Pearlite is composed by adding layers of ferrite containing 87.5wt%, and cementite contain 12.5wt% in contribution. Pearlite was identified by Sorby and named it sorbate, pearlite is formed by the cooling of austenite. Pearlite is one in the strongest structure.Stainless steels may be attracted into slender cables using a perlitic structure or a near-pearlitic microstructuring. Such cables, often packed into cords, have been used extensively for piano cables, suspension bridge cables, and tire reinforcement steel cable.

Expert Solution & Answer
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Answer to Problem 13.13P

Area of interference of pearlite is 11,000cm2 and percentage reduction is 98.4%.

Explanation of Solution

Given:

Thickness of cementite = 4×10-5cm

Ferrite = 14×10-5cm

Diameter of cementite sphere = 4×10-3cm

Density of ferrite = 7.87 g/cm3

Density of cementite= 7.66 g/cm3

Calculations:

The formula for the lever is given as

  Xa=c-ba-b×100

Where,

   c= 0.8= Entire alloy

  a=6.67 =

  α Phase weight percentage

  b= 0.0218 = θ Phase weight percentage

Therefore,

  %Fe3c=0.80-0.02186.67-0.0218×100=0.77826.6482×100=11.705%

In 1080 steel alloy weight percent of ferrite,

  %Fe=100-%Fe3c=100-11.705=88.295%

Now, percent of volume fraction of Fe3c is given by,

  %volume fraction Fe3c   =[weightofFe3cdensityofFe3c][weightofFe3cdensityofFe]+[weightofFe3cdensityofFe]×100 Substitute, Weight of Fe3c= 11.705Density of Fe3c= 7.66Weight % of Fe= 88.295Density of Fe = 7.78

Therefore % volume fraction becomes

  =[ 11.705 7.66[ 11.705 7.66]+[ 88.295 7.87]]×100=1.52811.5281+11.219×100=11.987

Calculate the number of the interface in the thickness of ferrite and cementite

Number of interface per cm = Number of interface[thickness of ferrite + thickness of cementite]

Where,

   Number of interface= 2 Thickness of ferrite = 4×10 -5 cm Thickness of cementite =14×10 -5 cm  

Therefore,

Number of interface per cm=

  24×10-5cm+14×10-5

  =218×10-5

  =1.1×104

  interface /cm

Now for the area of interface (A) in pearlite

A= number of interface

  ( 1 .1×10 4 interface/cm) ( 1cm 3 )        =11,000cm2 

Calculate the volume of cementite sphere with radius 2×10-3cm

  v=43πr3=43π(2×10-3cm)3=3.35×10-8cm3

Now, the number of cementite in the sphere,

  N=volume fraction of cementite Volume fraction of cementite sphere       =0 .11987cm33 .35×10 -8 cm3=3.58×106

Calculate the total surface area of a sphere,

Sphere = (4πr2)(N)

Where,

  N=3.58×106

Interface,

  %

  =180cm2

  interface/cm3

Now, the percentage reduction in the surface during anodizing pearlite steel

%reduction in the interface area =A-sphereA×100

Where,

  A=11000cm2sphere=180cm2

  % Reduction

  =11000-18011000×100=1082011000×100=98.4%

Conclusion

Area of interference of pearlite is 11,000cm2 and percentage reduction is 98.4%.

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