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 12.8P

(a)

To determine

The elastic modulus of composite in axial direction.

(a)

Expert Solution
Check Mark

Answer to Problem 12.8P

The elastic modulus of composite in axial direction is 99.5GPa .

Explanation of Solution

Given:

Volume percent of Kevlar fiber is 75% .

Volume percent of epoxy is 25% .

Elastic modulus of Kevlar is 131×109 Pa .

Elastic modulus of epoxy is 5×109Pa .

Concept used:

Write the expression for elastic modulus of composite.

  EC=EmVm+EfVf …… (1)

Here, EC is the modulus of elasticity of composite, Em is the modulus of elasticity of matrix, Vm is the volume fraction of matrix, Ef is the modulus of elasticity fiber and Vf is the volume fraction of fiber.

Calculation:

Substitute 131GPa for Ef , 0.75 for Vf , 5GPa for Em and 0.25 for Vm in equation (1).

  EC=(0.25)(5 GPa)+(0.70)5(131GPa)=99.5 GPa

Conclusion:

Thus, the elastic modulus of composite in axial direction is 99.5GPa .

(b)

To determine

The elastic modulus of composite in transverse direction.

(b)

Expert Solution
Check Mark

Answer to Problem 12.8P

The elastic modulus of composite in transverse direction is 18 GPa .

Explanation of Solution

Concept used:

Write the expression for elastic modulus of E-glass composite in the direction perpendicular to fiber axis.

  EC=EfEmVfEm+VmEf …… (2)

Here, EC is the transverse elastic modulus of material.

Calculation:

Substitute 131 GPa for Ef , 0.75 for Vf , 5 GPa for Em and 0.25 for Vm in equation (2).

  EC=( 131)(5)( 0.75)(5)+( 0.25)( 131)=18 GPa

Conclusion:

Thus, the elastic modulus of composite in transverse direction is 18 GPa .

(c)

To determine

The axial strainin composite due to axial stress.

(c)

Expert Solution
Check Mark

Answer to Problem 12.8P

The axial strain in composite due to axial stress is 0.005 .

Explanation of Solution

Given:

Axial stress is 0.5×109Pa .

Concept used:

Write the expression for axial strain in composite.

  ε=σEC …… (3)

Here, ε is the axial strain in composite and σ is the average composite stress.

Calculation:

Substitute 0.5×109N/m2 for σ and 99.5 GPa for EC in equation (3).

  ε=0.5× 109 N/m299.5 GPa( 10 9 N/m 2 1GPa )=0.005

Conclusion:

Thus, the axial strain in composite due to axial stress is 0.005 .

(d)

To determine

The stresses in the fiber and in matrix of composite due to axial stress.

(d)

Expert Solution
Check Mark

Answer to Problem 12.8P

The stresses in the fiber and in matrix of composite due to axial stress is 0.665GPa and 0.025 GPa respectively.

Explanation of Solution

Given:

Transverse stress is 4×106 Pa .

Tensile strength of Kevlar is 4.0×109Pa .

Concept used:

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.

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

  σm=εEm …… (5)

Here, σm is the stress in fiber.

Calculation:

Substitute 0.005 for ε and 131 GPa for Ef in equation (4).

  σf=(0.005)(131)=0.655 GPa

Substitute 0.005 for ε and 5 GPa for Em in equation (5).

  σm=(0.005)(5)=0.025 GPa

Conclusion:

Thus, the stresses in the fiber and in matrix of composite due to axial stress is 0.665GPa and 0.025 GPa respectively.

(e)

To determine

Whether each of the component can withstand the axial stress or not, if not what is the fracture stress.

(e)

Expert Solution
Check Mark

Explanation of Solution

The calculated stress for fiber is very less than the minimum tensile strength of Kevlar. This indicates that the fiber is sufficiently string to withstand the stress.

Matrix can also hold the tensile stress as the minimum tensile strength of epoxy resin is larger than the calculate stress in it.

(f)

To determine

The transverse fracture stress.

(f)

Expert Solution
Check Mark

Answer to Problem 12.8P

The transverse fracture stress is 1.1×106 Pa .

Explanation of Solution

Given:

Tensile strength of epoxy is 50×106Pa .

Concept used:

Write the expression for transverse tensile strength of composite.

  σ2u=σmu(12( v f π )0.5) …… (1)

Here, σ2u is the transverse tensile strength of composite, σmu is the tensile strength of composite and vf is the volume fraction of fiber.

Calculation:

Substitute 50×106 Pa for σmu and 0.75 for vf in equation (1).

  σ2u=(50× 106 Pa)(12 ( 0.75 π ) 0.5)=(50× 106 Pa)(10.98)=1.1×106 Pa

Conclusion:

Thus, the transverse fracture stress is 1.1×106 Pa .

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