Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
Question
Book Icon
Chapter 12, Problem 12.16P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 12.16P

The magnitude and direction of thermal stresses in matrix and fibers are 26.1×106Pa and 26.1×106Pa respectively.

Explanation of Solution

Given:

Volume fraction of uniaxial glass fiber is 50% .

Volume fraction of epoxy resin is 50% .

Initial temperature is 150°C .

Final temperature is 25°C .

Elastic modulus of glass fiber is 76GPa .

Thermal expansion coefficientof glass fiberis 5×106°C1 .

Elastic modulus of Epoxyis 4GPa .

Thermal expansion coefficientof 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 coefficientof fiber, αm is the thermal expansion coefficientof 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 , 0.5 for Vf , 0.5 for Vm in equation (1).

  0=(0.5)σf+(0.5)σmσm=σf

Substitute σf for σm in equation (2).

  εm=σfEm

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

  (αfαm)(TfTi)+(σfEf+σfEm)=0

Substitute 5×106°C1 for αf , 60×106°C1 for αm , 25°C for Tf , 150°C for Ti , 76GPa for Ef and 4 GPa for Em in above expression.

  [( 5× 10 6 ° C 1 60× 10 6 ° C 1 )( 25°C150°C)+ σ f( 1 76GPa + 1 4 GPa )]=06.875×103+σf(0.263  GPa 1( 1 GPa 10 9  Pa ))=0

Simplify above expression for σf .

  σf=6.875× 10 30.263× 10 9  Pa 1σf=26.14×106 Pa

Calculate the value of stress in matrix.

  σm=σfσm=26.14×106 Pa

Conclusion:

Thus, the magnitude and direction of thermal stresses in matrix and fibers are 26.1×106Pa and 26.1×106Pa respectively.

(b)

To determine

Composite strain in cooling the composite.

(b)

Expert Solution
Check Mark

Answer to Problem 12.16P

Composite strain in cooling the composite is 0.00097 .

Explanation of Solution

Given:

Initial temperature is 150°C .

Final temperature is 25°C .

Concept used:

Write the expression for stress in composite.

  εC=αm(TfTi)+εm …… (4)

Here, εC is the stress in composite

Calculation:

Substitute 26.1×106Pa for σm and 4 GPa for Em in equation (2).

  εm=26.1× 106 Pa4 GPa( 10 9  Pa 1 GPa )=0.00653

Substitute 0.00653 for εm , 60×106°C1 for αm , 25°C for Tf and 150°C for Ti in equation (4).

  εC=(60× 10 6°C 1)(25°C150°C)+0.00653=0.00097

Conclusion:

Thus, Composite strain in cooling the composite is 0.00097 .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
7 Has the same composition at any point? a. Isotropic b. Composite c. Prismatic d. Homogeneous 8. The composite material exhibits elastic properties in one direction different from that in the perpendicular direction. Orthotropic b. Anisotropic Isotropic d. Aeolotropic а. с. Materials having the same properties in all directions. a. Isotropic b. Anisotropic c. Aeolotropic d. Orthotropic 10 The term for the value above which the stress is no longer proportional to the strain. a. Proportional limit b. Plastic Range Rupture Stress d. Elastic Range с.
Determine the product of inertia (lxy) for the composite 100 mm 50 mm 60 mm
A discontinuous fibre reinforced composite consists of 20mm long carbon fibres in a polymer matrix. The volume fraction of the composite is 0.5, interfacial shear strength 5MPa and the critical fibre length 2.5mm. The fibre modulus is 250GPA and strain to failure 1.5%. The matrix modulus is 3.5GPa and failure strain 10%. If the ultimate fibre strength = 3500MPa, calculate the average failure stress in the composite.
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Materials Science And Engineering Properties
Civil Engineering
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning