Essentials of Materials Science and Engineering, SI Edition
Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
Question
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Chapter 17, Problem 17.29P
Interpretation Introduction

(a)

Interpretation:

Density of composite is to be calculated.

Concept introduction:

Rules of mixtures is:

  ρc=( f i ρ i )for{i=1..........n}ρc=f1ρ1+f2ρ2...................fnρn

Where

  ρc= density of composite

  fi= volume fraction of i component

  ρi= density of i component

Volume fraction is defined as:

  f=V1V1+V2f=volumefractionV1=volumeoffirstcomponentV2=volumeofsecondcomponentWhereV1andV2V1=weightoffirstcomponentdensityoffirstcomponentV2=weightofsecondcomponentdensityofsecondcomponentintermsofweightanddensity

  f=V1=weightoffirstcomponentdensityoffirstcomponentV1=weightoffirstcomponentdensityoffirstcomponent+V2=weightofsecondcomponentdensityofsecondcomponent

Expert Solution
Check Mark

Answer to Problem 17.29P

The requiredvalue of volume fraction of composite = 2.558g/cm3

Explanation of Solution

Given information:

Weight of boron fiber in unidirectional orientation = 5 kg

Weight of aluminum fiber in unidirectional orientation = 8 kg

Based on given information:

Applying rule of mixing,

  ρc=f1ρ1+f2ρ2

  ρc = density of composite

  f1 = Volume fraction of boron

  f2 = Volume fraction of aluminum

  ρ1 = Density of boronis 2.36g/cm3

  ρ2 = Density of aluminum is 2.7g/cm3

  ρc=f1ρ1+f2ρ2

Calculation of volume fraction of boron is defined as the ratio of volume of boron to total volume:

  fboron=VboronVboron+VAl

Calculation of volume of boron, expressed as the ratio of mass of boron to density of boron:

  Vboron=mBρB

Conversion of mass in gram, therefore multiplying the units by 1000 grams.

  Vboron=5×10002.36Vboron=2118.644g/cm3

Calculation of volume of aluminum:

  VAl=m Alρ AlVAl=8×10002.7VAl=2962.96g/cm3

Calculation of volume fraction of boron and aluminum on substituting the value of volume for boron and aluminum.

  fB=V boronV boron+V AlfB=2118.6442118.644+2962.96fB=0.41692fAl=V AlV boron+V AlfAl=2962.962118.644+2962.96fAl=0.5830

Volume fraction of boron = 0.41692

Volume fraction of aluminum = 0.5830

Applying rule of mixing

  ρc=f1ρ1+f2ρ2

  ρc=(2.36×0.41692)+(2.7×0.5830)ρc=0.9839+1.5741ρc=2.558g/cm3

Interpretation Introduction

(b)

Interpretation:

Modulus of elasticity parallel to fiber is to be calculated.

Concept introduction:

Modulus of elasticity is defined as the ratio of shear stress to shear strain.

Relation for modulus of elasticity is given as:

  EC=fBEB+fAlEAl

  fBEB are the volume fraction and modulus of elasticity for boron.

  fAlEAl are the volume fraction and modulus of elasticity for aluminum.

  f=V1V1+V2f=volumefractionV1=volumeoffirstcomponentV2=volumeofsecondcomponent

Expert Solution
Check Mark

Answer to Problem 17.29P

The required value of modulus of elasticity parallel to fibers is EC=198239.68MPa

Explanation of Solution

Calculation of volume fraction of boron is defined as the ratio of volume of boron to total volume:

  fboron=VboronVboron+VAl

Calculation of volume of boron, expressed as the ratio of mass of boron to density of boron:

  Vboron=mBρB

Conversion of mass in gram, therefore multiplying the units by 1000 grams:

  Vboron=5×10002.36Vboron=2118.644g/cm3

Calculation of volume of aluminum:

  VAl=m Alρ AlVAl=8×10002.7VAl=2962.96g/cm3

Calculation of volume fraction of boron and aluminum on substituting the value of volume for boron and aluminum:

  fB=V boronV boron+V AlfB=2118.6442118.644+2962.96fB=0.41692fAl=V AlV boron+V AlfAl=2962.962118.644+2962.96fAl=0.5830

Volume fraction of boron = 0.41692

Volume fraction of aluminum = 0.5830

Substituting the following values in the formula of modulus of elasticity:

  EC=fBEB+fAlEAlEB=379×103MPaEAl=69×103MPafB=0.41692

  fAl=0.5830

  EC=0.41692×379×103+0.5830×69×103EC=198239.68MPa

The required value of modulus of elasticity is 198239.68 MPa.

Interpretation Introduction

(c)

Interpretation:

Modulus of elasticity perpendicular to fiber is to be calculated.

Concept introduction:

Modulus of elasticity is defined as the ratio of shear stress to shear strain

Relation for modulus of elasticity perpendicular to fiber is given as:

  1EC=fBEB+fAlEAl

  fBEB are the volume fraction and modulus of elasticity for boron

  fAlEAl are the volume fraction and modulus of elasticity for aluminum

  f=V1V1+V2f=volumefractionV1=volumeoffirstcomponentV2=volumeofsecondcomponent

Expert Solution
Check Mark

Answer to Problem 17.29P

The required value of modulus of elasticity perpendicular to fibers is EC=104719.40MPa

Explanation of Solution

Calculation of volume fraction of boron is defined as the ratio of volume of boron to total volume:

  fboron=VboronVboron+VAl

Calculation of volume of boron, expressed as the ratio of mass of boron to density of boron:

  Vboron=mBρB

Conversion of mass in gram, therefore multiplying the units by 1000 grams:

  Vboron=5×10002.36Vboron=2118.644g/cm3

Calculation of volume of aluminum:

  VAl=m Alρ AlVAl=8×10002.7VAl=2962.96g/cm3

Calculation of volume fraction of boron and aluminum on substituting the value of volume for boron and aluminum:

  fB=V boronV boron+V AlfB=2118.6442118.644+2962.96fB=0.41692fAl=V AlV boron+V AlfAl=2962.962118.644+2962.96fAl=0.5830

Volume fraction of boron = 0.41692

Volume fraction of aluminum = 0.5830

Substituting the following values in the formula of modulus of elasticity

  EB=379×103MPaEAl=69×103MPafB=0.41692

  fAl=0.5830

  (1 E C )=( 0.41692 379× 10 3 )+( 0.5830 69× 10 3 )EC=104719.40MPa

The required value of modulus of elasticity is 104719.40 MPa.

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