Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 12, Problem 12.12P

(a)

To determine

The strain where the first change in elastic modulus occurs in composite material.

(a)

Expert Solution
Check Mark

Answer to Problem 12.12P

Matrix will rupture before the fiber at the strain of 0.003 .

Explanation of Solution

Given:

Elastic modulus of Boronis 400GPa .

Tensile strength of Boronis 2.8GPa .

Elastic modulus of Aluminumis 70 GPa .

Yield strength of Aluminumis 200 MPa .

Ultimate tensile strength of aluminum is 200 MPa .

Concept used:

Write the expression for failure strain in Fiber.

  εfu=σfuEf …… (1)

Here, εfu is the failure strain in Fiber, σfu is the tensile strength of Fiber and Ef is the modulus of elasticity of Fiber.

Write the expression for failure strain in Matrix.

  εmu=σmuEm …… (2)

Here, εmu is the failure strain in Matrix, σmu is the tensile strength of Matrix and Em is the modulus of elasticity of Matrix.

Calculation:

Substitute 2.8 GPa for σfu and 400GPa for Ef in equation (1).

  εfu=2.8GPa400GPa=0.007

Substitute 200MPa for σmu and 70GPa for Em in equation (2).

  εmu=200MPa70 GPa( 10 3  MPa 1 GPa )=0.003

Conclusion:

Thus, matrix will rupture before the fiber at the strain of 0.003 .

(b)

To determine

The composite stress where the change in elastic modulus occurs.

(b)

Expert Solution
Check Mark

Answer to Problem 12.12P

The composite stress where the change in elastic modulus occurs is 800MPa .

Explanation of Solution

Given:

Volume percent of Boron fiber is 60% .

Volume percent of Aluminumis 40% .

Strain at fracture of Aluminumis 15% .

Concept used:

Write the expression for stress carried by composite.

  σc=Vfσf+Vmσm …… (3)

Here, σc is the stress in of composite, σf is the stress in fiber, Vf is the volume fraction of fiber, σm is the stress in matrix and Vm is the volume fraction of matrix.

Write the expression for stress in fiber by considering the iso-strain model for axial strain.

  σf=εEf …… (4)

Here, σf is the stress in fiber.

Calculation:

Substitute 400GPa for Ef and 0.003 for ε in equation (4).

  σf=(0.003)(400GPa)=1.2GPa

Substitute 0.6 for Vf , 0.4 for Vm , 1.2GPa for σf and 0.20 GPa for σm in equation (3).

  σc=0.6(1.2)+0.4(0.20)=0.8 GPa( 1000 MPa 1 GPa)=800 MPa

Conclusion:

Thus, the composite stress where the change in elastic modulus occurs is 800MPa .

(c)

To determine

The elastic modulus at low strain and the elastic modulus after Aluminum yields, if Aluminum yields before fiber fails.

(c)

Expert Solution
Check Mark

Answer to Problem 12.12P

The elastic modulus at low strain is 268 GPa and the elastic modulus after Aluminum yields, if Aluminum yields before fiber fail is 240 GPa .

Explanation of Solution

Concept used:

Write the expression for modulus of elasticity of composite.

  Ec=VfEf+VmEm …… (5)

Here, Ec is the modulus of elasticity in of composite, Ef is the modulus of elasticity in fiber and Em is the modulus of elasticity of matrix.

Calculation:

Substitute 0.6 for Vf , 0.4 for Vm , 400GPa for Ef and 70 GPa for Em in equation (5).

  Ec=(0.6)(400)+(0.4)(170)=268GPa

After yielding of matrix the elastic modulus of composite is the elastic modulus of fiber in it. Therefore, the elastic modulus will be equal to 60% of 400GPa which will be equal to 240GPa .

Conclusion:

Thus, the elastic modulus at low strain is 268 GPa and the elastic modulus after Aluminum yields, if Aluminum yields before fiber fail is 240 GPa .

(d)

To determine

Fracture strength of composite.

(d)

Expert Solution
Check Mark

Answer to Problem 12.12P

Fracture strength of composite is 1760 MPa .

Explanation of Solution

Concept used:

Write the expression for fracture strength.

  σts=Vfσf+Vmσm …… (6)

Here, σts is the fracture strength, σf is the maximum strength of fiber and σm is the maximum strength of matrix.

Calculation:

Substitute 0.6 for Vf , 0.4 for Vm , 2.8 GPa for σf and 0.20 GPa for σm in equation (6).

  σts=(0.6)(2.8)+(0.4)(0.20)=1.760 GPa( 1000 MPa 1 GPa)=1760 MPa

Conclusion:

Thus, the Fracture strength of composite is 1760 MPa .

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