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Case Studies Chapter 1 Design using Carbon Nanotubes.

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Presentation on theme: "Case Studies Chapter 1 Design using Carbon Nanotubes."— Presentation transcript:

1 Case Studies Chapter 1 Design using Carbon Nanotubes

2 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%

3 Moduli Pure PLA: 3.3 Gpa Glass Fiber: 70 Gpa Carbon Fiber: 400 Gpa CNT: 1000 Gpa Note: Gpa= Gigapascal= 1billion Pascals

4 Proportion The proportion of fibers and matrix in the composite are expressed as fractions of the total volume they occupy. ϕ f = Proportion of fibers ν f = Volume of fibers ν = Total volume

5 Fiber Fraction Calculation Calculation for.5% fibers – ν f =.005 – ν= 1 So the proportion at: –.5% is.005 – 5% is.05 – 20% is.2

6 Composite Modulus E 1 = Composite Modulus ϕ f = Proportion of fiber volume to overall volume E f = Modulus of fibers E m = Modulus of matrix (polymer) Formula to calculate the modulus with perfect adhesion at the interface

7 Composite Modulus Calculation Glass Fiber at.5% loading ϕ f =.005 E f = 70 GPa E m = 3.3 GPa

8 Composite Modulus Calculation Modulus (GPa)PLA Modulus (GPa)ProportionComposite Modulus (GPa)Composite Modulus (psi)% Change Over PLA (%) PLA3.3 1 4786250.0 Glass Fiber 703.30.0053.652699610.1 703.30.056.6962327101.1 703.30.216.62413432404.2 Carbon Fiber 4003.30.0055.376630860.1 4003.30.0523.13355454601.1 4003.30.282.6119859402404.2 CNT 10003.30.0058.31201422151.0 10003.30.0553.177065941510.2 10003.30.2202.6293905006040.6

9 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

10 Moment of Inertia I= moment of inertia b= base h= height 6in.125in

11 Moment of Inertia Calculation 6in.125in

12 Deflection y max = max deflection w = load L = unsupported length E = modulus I = moment of inertia 4ft.125in Force: 0.25lbs distributed Glass Fiber at.5%

13 Deflection Calculation Proportion Composite Modulus (psi) Load (lb/in) Unsupported Length (in) Moment of Inertia (in 4 )Deflection (in) % Change Over PLA (%) PLA 14786250.00521480.0001-7.5240.0 Glass Fiber 0.0055269960.00521480.0001-6.833-9.2 0.059623270.00521480.0001-3.742-50.3 0.224134320.00521480.0001-1.492-80.2 Carbon Fiber 0.0057663080.00521480.0001-4.699-37.5 0.0533554540.00521480.0001-1.073-85.7 0.2119859400.00521480.0001-0.300-96.0 CNT 0.00512014220.00521480.0001-2.997-60.2 0.0577065940.00521480.0001-0.467-93.8 0.2293905000.00521480.0001-0.123-98.4


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