Case Studies Chapter 1

Design using Carbon Nanotubes

Given• We will calculate the deflection of a simply supported

beam of PLA with dimensions of 6 inch width, 1/8 inch thick, 4 feet long. The distributed load of 0.25 pounds will be applied to the beam. We will do the calculation with multiple reinforcing fibers at 3 different fiber loadings.

Reinforcing materials– Carbon Nanotubes (CNT)– Carbon Fiber– Glass Fiber– No Reinforcement

Loadings– .5%– 5%– 20%

Moduli

• Pure PLA: 3.3 Gpa• Glass Fiber: 70 Gpa• Carbon Fiber: 400 Gpa• CNT: 1000 Gpa

Note: Gpa= Gigapascal= 1billion Pascals

Proportion

v

v ff

Fiber Fraction Calculation

Calculation for .5% fibers– νf= .005– ν= 1

• So the proportion at:– .5% is .005– 5% is .05– 20% is .2

v

v ff

005.1

005.f

Composite Modulus

• E1= Composite Modulus

• ϕf = Proportion of fiber volume to overall volume• Ef = Modulus of fibers• Em= Modulus of matrix (polymer)

mfff EEE *)1(*1

Formula to calculate the modulus with perfect adhesion at the interface

Composite Modulus Calculation

Glass Fiber at .5% loading• ϕf = .005• Ef = 70 GPa

• Em= 3.3 GPa

mfff EEE *)1(*1

GPaE

GPaGPaE

6335.3

3.3*)005.1(70*005.

1

1

Beam Calculations

• Now that we have all the moduli for the different reinforcements and loadings we must calculate the moment of inertia of the beam in order to make the deflection calculations

Moment of Inertia

• I= moment of inertia

• b= base

• h= height

6in

.125in

Moment of Inertia Calculation

6in

.125in

43

3

0001.012

125.*6

"125.0

"0.612

*

inI

h

b

hbI

Deflection

• ymax= max deflection• w = load• L = unsupported length• E = modulus• I = moment of inertia

4ft

.125in

Force: 0.25lbs distributed

IE

Lwy

**384

**5 4

max

iny

ininlb

ininlb

y

IE

Lwy

inLin

lbinlbsw

inI

psiE

833.6

0001.0*996,526*384

)48(*00521.0*5

**384

**5

48

00521.048/25.0

0001.0

996,526

max

42

4

max

4

max

4

1

Glass Fiber at .5%