How to Calculate and Solve for Randomly Oriented Composite Modulus of Elasticity | Composites

The randomly oriented composite modulus of elasticity is illustrated by the image below.

To compute for randomly oriented composite modulus of elasticity, five essential parameters are needed and these parameters are Fibre Efficiency Parameter (K), Elastic Modulus of the Fibre (Ef), Elastic Modulus of the Matrix (Em), Volume Fraction of the Fibre (Vf) and Volume Fraction of the Matrix (Vm).

The formula for calculating randomly oriented composite modulus of elasticity:

E = KEfVf + EmVm

Where:

E = Randomly Oriented Composite Modulus of Elasticity
K = Fibre Efficiency Parameter
Ef = Elastic Modulus of the Fibre
Em = Elastic Modulus of the Matrix
Vf = Volume Fraction of the Fibre
Vm = Volume Fraction of the Matrix

Let’s solve an example;
Find the randomly oriented composite modulus of elasticity when the fibre efficiency parameter is 14, the elastic modulus of the fibre is 10, the elastic modulus of the matrix is 16, the volume fraction of the fibre is 12 and the volume fraction of the matrix is 8.

This implies that;

K = Fibre Efficiency Parameter = 14
Ef = Elastic Modulus of the Fibre = 10
Em = Elastic Modulus of the Matrix = 16
Vf = Volume Fraction of the Fibre = 12
Vm = Volume Fraction of the Matrix = 8

E = KEfVf + EmVm
E = (14)(10)(12) + (16)(8)
E = (1680) + (128)
E = 1808

Therefore, the randomly oriented composite modulus of elasticity is 1808 Pa.

Continue reading How to Calculate and Solve for Randomly Oriented Composite Modulus of Elasticity | Composites

How to Calculate and Solve for Elastic Modulus on Transverse Direction | Composites

The elastic modulus on transverse direction is illustrated by the image below.

To compute for elastic modulus on transverse direction, three essential parameters are needed and these parameters are Elastic Modulus of the Fibre (Ef), Elastic Modulus of the Matrix (Em) and Volume Fraction of the Fibre (Vf).

The formula for calculating elastic modulus on transverse direction:

Ec = EmEf/(1 – Vf)Ef + VfEm

Where:

Ec = Elastic Modulus on Transverse Direction
Em = Elastic Modulus of the Fibre
Ef = Elastic Modulus of the Matrix
Vf = Volume Fraction of the Fibre

Given an example;
Find the elastic modulus on transverse direction when the elastic modulus of the fibre is 4, the elastic modulus of the matrix is 8 and the volume fraction of the fibre is 6.

This implies that;

Em = Elastic Modulus of the Fibre = 4
Ef = Elastic Modulus of the Matrix = 8
Vf = Volume Fraction of the Fibre = 6

Ec = EmEf/(1 – Vf)Ef + VfEm
Ec = (4)(8)/(1 – 6)8 + (6)(4)
Ec = (32)/(-5)8 + (24)
Ec = (32)/(-40) + (24)
Ec = (32)/(-16)
Ec = -2

Therefore, the elastic modulus on transverse direction is -2 Pa.

Continue reading How to Calculate and Solve for Elastic Modulus on Transverse Direction | Composites

How to Calculate and Solve for Ratio of Load of Fibre to Matrix | Composites

The ratio of load of fibre to matrix is illustrated by the image below.

To compute for ratio of load of fibre to matrix, four essential parameters are needed and these parameters are Elastic Modulus of the Fibre (Ef), Elastic Modulus of the Matrix (Em), Volume Fraction of the Fibre (Vf) and Volume Fraction of the Matrix (Vm).

The formula for calculating ratio of load of fibre to matrix:

F(f/m) = EfVf/EmVm

Where:

F(f/m) = Ratio of Load of Fibre to Matrix
Ef = Elastic Modulus of the Fibre
Em = Elastic Modulus of the Matrix
Vf = Volume Fraction of the Fibre
Vm = Volume Fraction of the Matrix

Let’s solve an example;
Find the ratio of load of fibre to matix when the elastic modulus of the fibre is 14, the elastic modulus of the matrix is 10, the volume fraction of the fibre is 6 and the volume fraction of the matrix is 2.

This implies that;

Ef = Elastic Modulus of the Fibre = 14
Em = Elastic Modulus of the Matrix = 10
Vf = Volume Fraction of the Fibre = 6
Vm = Volume Fraction of the Matrix = 2

F(f/m) = EfVf/EmVm
F(f/m) = (14)(6)/(10)(2)
F(f/m) = (84)/(20)
F(f/m) = 4.2

Therefore, the ratio of load of fibre to matrix is 4.2.

Calculating the Elastic Modulus of the Fibre when the Ratio of Load of Fibre to Matrix, the Elastic Modulus of the Matrix, the Volume Fraction of the Fibre and the Volume Fraction of the Matrix is Given.

Ef = F(f/m) x EmVm / Vf

Where:

Ef = Elastic Modulus of the Fibre
F(f/m) = Ratio of Load of Fibre to Matrix
Em = Elastic Modulus of the Matrix
Vf = Volume Fraction of the Fibre
Vm = Volume Fraction of the Matrix

Let’s solve an example;
Find the elastic modulus of the fibre when the ratio of load of fibre to matrix is 20, the elastic modulus of the matrix is 4, the volume fraction of the matrix is 2 and the volume fraction of the fibre is 12.

This implies that;

F(f/m) = Ratio of Load of Fibre to Matrix = 20
Em = Elastic Modulus of the Matrix = 4
Vf = Volume Fraction of the Fibre = 12
Vm = Volume Fraction of the Matrix = 2

Ef = F(f/m) x EmVm / Vf
Ef = 20 x (4)(2) / 12
Ef = 20 x 8 / 12
Ef = 160 / 12
Ef = 13.33

Therefore, the elastic modulus of the fibre is 13.33

Continue reading How to Calculate and Solve for Ratio of Load of Fibre to Matrix | Composites

How to Calculate and Solve for Elastic Modulus on Longitudinal Direction | Composites

The elastic modulus on longitudinal direction is illustrated by the image below.

To compute for elastic modulus on longitudinal direction, three essential parameters are needed and these parameters are Elastic Modulus of the Matrix (Em), Elastic Modulus of the Fibre (Ef) and Volume Fraction of the Fibre (Vf).

The formula for calculating elastic modulus on longitudinal direction:

Eu = Em(1 – Vf) + EfVf

Where:

Eu = Elastic Modulus on Longitudinal Direction
Em = Elastic Modulus of the Matrix
Ef = Elastic Modulus of the Fibre
Vf = Volume Fraction of the Fibre

Let’s solve an example;
Find the elastic modulus on longitudinal direction when the elastic modulus of the matrix is 12, the elastic modulus of the fibre is 10 and volume fraction of the fibre is 4.

This implies that;

Em = Elastic Modulus of the Matrix = 12
Ef = Elastic Modulus of the Fibre = 10
Vf = Volume Fraction of the Fibre = 4

Eu = Em(1 – Vf) + EfVf
Eu = (12)(1 – 4) + (10)(4)
Eu = (12)(-3) + (40)
Eu = (-36) + (40)
Eu = 4

Therefore, the elastic modulus on longitudinal direction is 4 Pa.

Continue reading How to Calculate and Solve for Elastic Modulus on Longitudinal Direction | Composites