The Lennard-Jones Potential Curve for Carbon fiber and Copper Fiber

The Lennard-Jones Potential Curve for Carbon fiber and Copper Fiber

Task 1

1. Carbon fibres are considered as a material for composite part in a wind turbine structure.
   1. Calculate the theoretical strength of the carbon fibres by assuming a carbon-carbon bond length of 0.15 nm, a material surface free energy of 50 mJ/m2 and an elastic modulus of 1 TPa.                                                                                             [5 Marks]

 

Carbon – Carbon bond distance:-0.15nm

Material Surface Energy:- U = 50Mj/m^2

Elastic Modulus: - E = 1 Tpa

b) Sketch a Lennard-Jones curve for the carbon-carbon bonds in the carbon fiber and show a comparable Lennard-Jones curve for copper fibers where the strength and elastic

Modulus of the copper is lower than for carbon. Clearly indicate the differences between

the two materials. [9 Marks]

 

 

The Lennard-Jones Potential Curve for Carbon fiber and Copper Fiber  ;

As we know more energy is require to increase the distance between two molecule of the fiber. And as the curve is become more sharper the potential energy is increases.

If we compare the curve of carbon fiber and copper fiber, it could be conclude that the amount of energy required to separate the material is higher in copper fiber then in the carbon fiber.

C) Sketch a diagram to show carbon fiber production. Your diagram should clearly indicate the three temperature treatments required in the production and name each of these temperature treatments.

 

Diagram of Carbon fiber Production

 

·Spinning

Wet spinning is used to produce the PAN fibers .The Powder of plastic is added into the coagulation bath and mix well. Then after it is extruded through the spinneret. Spinneret is made of metal and work as a die. The spinneret consisting of no of hole, which is determined by the count of the Fiber. This process is very fragile so the rollers draw it and deep into the wash to remove extra coagulant .the after it is dried for some time. Afterwards the fishing oil is apply on the fiber to prevent tangle. The color of this PAN fiber is white .In the last it is again dried and wound on to bobbins and stored for some time.

·Oxidation

In this process the PAN fibers is passing through no of a series of oven. This process is very time consuming process of carbon fiber production process. Now the Bobbins are loaded on the mechanism so called creel, which is used for the feed the wire in the production. Now the fibers are spread flat with use of the tow band. The working temperature of oxidation generally lies between 200°C to 300°C. In this Process, the reaction is take place between the Pan Fibers and oxygen from the air. Because of this reaction the cross-linking is start in the polymer. This process increase the density from 1.18 g/cc to 1.38 g/cc. There are usually the 5 to 6 ovens are place in the series and the time takes per oven is 70 to 130 mins. In this process the mainly 10 to 12 pass of the fiber in the oven .At the end process the oxygen is oxidized into fiber and stabilize. At the end of the process Pan Fiber consisting of 55 to 65 % carbon and other element in the balance form.

Carbonization

Carbonization process is normally operate in the oxygen free atmosphere or in the inert gas condition. In this process, normally the temperature is increasing progressively. In this process at the starting and exit of the chamber there is purge chamber to prevent the oxygen to get inside the oven. This process prevent the loss of carbon fiber at such high temperatures. In this process other element like VOCs or cyanide which can be removed .This process start in low temperature furnace with 700-800 and completed at the 1200-1500 .In the whole process the fiber must be in the tension condition. At the end of the process the fiber consisting of 90 % carbon. In subsequent process of graphitization at temperature 1315°C, this fiber contain 94 to 95 %. At last the fiber contain 99 % carbon at 1950 to 2475 °C.As the process completion of carbonization process weight is reduced .the length and diameter is contracts up 10 %.   

Surface treatment and sizing

In this process the reinforcing of the carbon fibers can be enhance. This process is start with pulling the fibers in to the electrolyte bath. In this process the chemical is react with surface of fiber and rough surface. This can increase the surface area of filament for interfacial fiber. In the next stage, the coating is applying on the whole surface of the fiber. Which is 0.5 to 5 % weight of fiber. When the sizing dries, the long process is complete.

(b) The composite uses the carbon fibres with 50% in a 45° and 50% in a -45° orientation to improve

fatigue resistance. An epoxy with an elastic modulus of 1 GPa and a low-grade carbon fibre with an

elastic modulus of 200 GPa are used in the composite.

 

  1 .State the krenchel equation that can be used to to predict the elastic modulus of the composite.

 

= proportion of total fibre content

 ? = angle of fibers

 = composite efficiency factor (Krenchel)

2. Calculate Orientation Factor For Composite.

Efficiency factor

Layup = (+45/-45)

η = Cos4θ

                                                 = Cos4 45°

η = 0.25

Laminate in X-direction

                                                             = (1× 0.25)

                                                             = 0.25

Laminate in Y-direction

                                                             = (1× 0.25)

                                                             = 0.25

3.(ili) Calculate the resultant elastic modulus of the composite assuming the length efficiency

factor is 0.8. [4 Marks]

Ef = 200 GPa

 Em= 1 GPa

Vf = 0.8

Ex = (0.25 × 200×0.8) + (1×0.2)

Ex = 40 + 0.2 = 40.2 GPa

 Ey = (0.25 × 200×0.8) + (1×0.2)

Ey = 40 + 0.2 = 40.2 GPa

4. Suggest how the resultant composite elastic modulus would vary compared to your answer

in (iii) for fibre orientation approaching Voigt and Reuss conditions.

[6 Marks]

                                                                    

                                                                   Ec= 4.9 Gpa

Voigt analysis is used for the linear analysis. There is no orientation effect on the results. But the Krenchel allow us to do the calculation using orientation of the ply.

Task 2

The simply supported beam in Fig. T2 carries two concentrated loads.

(a) Derive the expressions for the shear force and the bending moment for each segment of the beam.

 (b) Sketch the shear force and bending moment diagrams.

Now,

S.F. Calculations –

S.F. at A ( Just LHS ) = 0.0

-- S.F. at A ( Just RHS ) = + 18 kN

-- S.F. at B ( Just LHS ) = +  18 kN

-- S.F. at B ( Just RHS ) = + 18.0 – 14  = 4 KN

-- S.F. at C ( Just LHS ) = +4 KN

-- S.F. at C ( Just RHS ) = + 4 – 28.0 = - 24 KN

-- S.F. at D ( Just LHS ) = -24 KN

-- S.F. at D ( Just RHS ) = -24 + 24 = 0.0 KN S.F.D. will be as shown in fig.

B.M. Calculation –

-- B.M. at A = 0.0 KN-m.

-- B.M. at B = + ( 18 x 2 ) = + 36 kN-m.

-- B.M. at C = + ( 18 x 5 ) – ( 14 x 3 ) = +  48 kN-m.

-- B.M. at D = 0.0 kN- m. The B.M.D. will be as shown in fig.

 

 

 

18 KN V KN

 

+

04 KN

 

A

 

_

B

 

 

 

24 KN

 

 

 

48 KN.m

 

M   KN.m

36 KN.m

 

+

 

D

A

 

C

B

 

 

 

Task 3

The main components of the artificial hip are the femoral stem that is fitted into the bone, the ball attached to the stem and the acetabular cup that is fixed into the pelvis. The property constrains on the materials to be used for these elements are very strict because of chemical and mechanical complexity of the hip joint.

(a) What are the main material requirements for hip implants? Briefly justify each point. [10 Marks]

< >High biocompatibility:-Nonmagnetic:-The material should be non-Magnetic. In the many time of the life human can come in interaction with the Magnetic Resonance Imaging (MRI). If a person done the transplant before. And this transplant is magnetic, then this situation can create the problem .Because of MRI the transplant gets heated and may change its position, so the material should be nonmagnetic.The surface femoral stem should be bioactive in terms of osteointegration :- It is a important that when transplant has done ,the living bone should accept the transplant and make a strong bond. So the material has to be osteointegration in nature Adequate mechanical :-  The material has to good mechanical properties like high young modulus. Good yield strength, The material should be optimum amount of ductile. The tensile strength should be good enough. Very low toughness value. The material has high hardness so that the wear and tear should be low. The fatigue strength is good.The surface of the layer has high modulus of elasticity. The structure  can be a porous structure.The layer of the implant or free surface does not have surface defects like scratches ,cracks, or any other defect.The surface morphology of the implant should be fine. If the surface  is expose to the load then  it is very fine .Or if it is for stem then the rough surface can be used.The implant should be chemically stable. There is not defect at the microscopic level like interstitial defect of point defect.The implant should have high quality but it should not be deal with the price. The price should be economical.Materials selection of femoral stem componentMaterials selection of femoral head and acetabular cup componentsMetal-on-metalCeramic-on-ceramic

 

(a) The effective (von Mises) stress for this implant. [11 Marks]

 

 

 

 

 

 

Mohr Circle

 

σx

 

σ2=-

σ1

 

 

 

 

τ

 

 

 

 

 

 

 

 

Material compressive yielding Strength = 30 MPa ,The Tresca stress is 20.4 Mpa so Material Safe in compressive yielding.

 

 

 

b) The factor of safety against yield. [3 Marks]

The Factor of safety can be found by yielding compressive strength of material to Tresca stress.

F.O.S =