Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
Question
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Chapter 12, Problem 12.14P

(a)

To determine

The strain in composite when stress in matrix is 60 MPa .

(a)

Expert Solution
Check Mark

Answer to Problem 12.14P

The strain in composite when stress in matrix reaches 60 MPa is 0.025.

Explanation of Solution

Given:

Elastic modulus of polycarbonate matrix is 2.4×109 Pa .

Maximum allowable stress in matrix is 60 MPa .

Concept used:

Write the expression for strain in composite.

  εC=σmEm …… (1)

Here, εC is the strain in composite, σm is the stress in matrix and Em is the elastic modulus of matrix.

Calculation:

Substitute 60 MPa for σm and 2.4×109 Pa for Em in equation (1).

  εC=60 MPa( 10 6 Pa 1MPa )2.4× 109 Pa=0.025

Conclusion:

Thus, the strain in composite when stress in matrix reaches 60 MPa is 0.025.

(b)

To determine

The fiber stress for continuous fiber composite and identify if this is less than tensile strength of fiber.

(b)

Expert Solution
Check Mark

Answer to Problem 12.14P

The fiber stress for continuous fiber composite is 3.25 GPa this is close to tensile strength of fiber.

Explanation of Solution

Given:

Elastic modulus of Fiber is 130×109 Pa .

Concept used:

Write the expression for fiber stress.

  σf=εCEf …… (2)

Here, σf is the fiber stress and Ef is the modulus of elasticity of fiber.

Calculation:

Substitute 0.025 for εC and 130×109 Pa for Ef in equation (2).

  σf=(0.025)(130× 109 Pa)=3.25×109 Pa( 1 GPa 10 9  Pa)=3.25 GPa

Conclusion:

Thus, the fiber stress for continuous fiber composite is 3.25 GPa this is close to tensile strength of fiber.

(c)

To determine

Critical fiber length

(c)

Expert Solution
Check Mark

Answer to Problem 12.14P

Critical fiber length is 6.1×104m .

Explanation of Solution

Given:

Diameter of aramid fiber is 12×106 m .

Shear strength of Polycarbonate-aramid interface is 32×106 Pa .

Concept used:

Write the expression for critical aspect ratio.

  lCdf=σf2τmf* …… (3)

Here, lC is the critical fiber length, df is the diameter of fiber and τmf* is the maximum shear stress in interface.

Calculation:

Substitute 12×106 m for df , 32×106 Pa for τfm* and 3.25 GPa for σf in equation (3).

  lC12× 10 6 m=3.25 GPa( 10 9  Pa 1 GPa )2( 32× 10 6  Pa)lC=6.1×104m

Conclusion:

Thus, the Critical fiber length is 6.1×104m .

(d)

To determine

Average stress in chopped aramid fiber.

(d)

Expert Solution
Check Mark

Answer to Problem 12.14P

Average stress in chopped aramid fiber is 1.6 GPa .

Explanation of Solution

Given:

Length of aramid fiber is 6×104 m .

Concept used:

Write the expression for average fiber stress.

  σ¯f=σf(1lC2l) …… (4)

Here, σf¯ is the average fiber stress and l is the length of fiber.

Calculation:

Substitute 3.25 GPa for σf , 6.1×104m for lC and 6×104 m for l in equation (4).

  σ¯f=3.25 GPa(1 6.1× 10 4  m 2( 6× 10 4  m ))=1.6 GPa

Conclusion:

Thus, the average stress in chopped aramid fiber is 1.6 GPa .

(e)

To determine

Composite material stress.

(e)

Expert Solution
Check Mark

Answer to Problem 12.14P

Composite material stress is 753 MPa .

Explanation of Solution

Given:

Volume percentage of aramid chopped fiber is 45% .

Volume percent of polycarbonate matrix is 55% .

Tensile strength of aramid fiber is 3.6×109 Pa .

Tensile strength of Polycarbonate is 80 MPa .

Concept used:

Write the expression for composite material stress.

  σC=Vfσ¯f+Vmσm …… (5)

Here, σC is the composite material stress, Vf is the volume fraction of fiber and Vm is the volume fraction of matrix.

Calculation:

Substitute 0.45 for Vf , 0.55 for Vm , 1.6 GPa for σ¯f and 60 MPa for σm in equation (5).

  σC=(0.45)(1.6 GPa)( 10 9  Pa 1 GPa)+(0.55)(60 MPa)( 1 0 6  Pa 1 MPa)=720×106 Pa+33×106 Pa=753×106 Pa( 1 MPa 10 6  Pa)=753 MPa

Conclusion:

Thus, the composite material stress is 753 MPa .

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