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 12.17P

(a)

To determine

The magnitude and direction of thermal stresses in matrix and fibers.

(a)

Expert Solution
Check Mark

Answer to Problem 12.17P

The magnitude and direction of thermal stresses in matrix and fibers are 171.31×106 Pa and 57.10×106 Pa respectively.

Explanation of Solution

Given:

Volume fraction of carbon fiberis 75% .

Volume fraction of epoxy resin is 25% .

Initial temperature is 22°C .

Final temperature is 260°C .

Elastic modulus of carbon fiber is 380GPa .

Ultimate tensile strength of carbon fiber is 2.4GPa .

Thermal expansion coefficient of carbon fiber is 0.7×106K1 .

Elastic modulus of Epoxy is 4GPa .

Ultimate tensile strength of epoxy resin is 0.07 GPa

Thermal expansion coefficient of Epoxy is 60×106°C1 .

Concept used:

Write the expression for stress carried by composite.

  σc=Vfσf+Vmσm …… (1)

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 strain in Fiber.

  εf=σfEf

Here, εf is the strain in Fiber, σf is the tensile strength of Fiber and Ef is the modulus of elasticity of Fiber.

Write the expression for strain in Matrix.

  εm=σmEm …… (2)

Here, εm is the strain in Matrix, σm is the tensile strength of Matrix and Em is the modulus of elasticity of Matrix.

Write the expression for the total strain in the composite.

  (αfαm)(TfTi)+(εfεm)=0 …… (3)

Here, αf is the thermal expansion coefficient of fiber, αm is the thermal expansion coefficient of matrix, Ti is the initial in temperature of composite and , Tf is the final in temperature of composite.

Calculation:

The composite stress is zero because there is no application of stress to the composite material in the axial direction.

Substitute 0 MPa for σC , 0.75 for Vf , 0.25 for Vm in equation (1).

  0=(0.75)σf+(0.25)σmσm=3σf

Substitute 3σf for σm in equation (2).

  εm=3σfEm

Substitute 3σfEm for εm and σfEf for εf in equation (3).

  (αfαm)(TfTi)+(σfEf+3σfEm)=0

Substitute 0.7×106°C1 for αf , 60×106°C1 for αm , 260°C for Tf , 22°C for Ti , 380GPa for Ef and 4 GPa for Em in above expression.

  [( 0.7× 10 6 ° C 1 60× 10 6 ° C 1 )( 260°C22°C)+ σ f( 1 380GPa + 1 4 GPa )]=00.014447+σf(0.253  GPa 1( 1 GPa 10 9  Pa ))=0

Simplify above expression for σf .

  σf=0.0144470.253× 10 9  Pa 1σf=57.10×106 Pa

Calculate the value of stress in matrix.

  σm=3σfσm=171.31×106 Pa

Conclusion:

Thus, the magnitude and direction of thermal stresses in matrix and fibers are 171.31×106 Pa and 57.10×106 Pa respectively.

(b)

To determine

Composite strain in cooling the composite.

(b)

Expert Solution
Check Mark

Answer to Problem 12.17P

Composite strain in cooling the composite is 0.02855 .

Explanation of Solution

Given:

Initial temperature is 22°C .

Final temperature is 260°C .

Concept used:

Write the expression for stress in composite.

  εC=αm(TfTi)+εm …… (4)

Here, εC is the stress in composite

Calculation:

Substitute 171.31×106 Pa for σm and 4 GPa for Em in equation (2).

  εm=171.31× 106 Pa4 GPa( 10 9  Pa 1 GPa )=0.04283

Substitute 0.04283 for εm , 60×106°C1 for αm , 260°C for Tf and 22°C for Ti in equation (4).

  εC=(60× 10 6°C 1)(260°C22°C)0.04283=0.02855

Conclusion:

Thus, Composite strain in cooling the composite is 0.02855 .

(c)

To determine

(c)

Expert Solution
Check Mark

Answer to Problem 12.17P

The component of design which can cause problem and justify your answer.

Explanation of Solution

The tensile stress in the fiber is less than the ultimate tensile strength of fiber and the stress in matrix is compressive in nature whereas the tensile strength of epoxy matrix is 0.07 GPa . For the material to prevent failure the ultimate compressive strength of matrix must be greater than the applied compressive stress on it.

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