Materials, Tribology and Surface Engineering - Math Problem Example

MATERIALS, TRIBOLOGY AND SURFACE ENGINEERING ASSIGNMENT Student Name: Course: Professor: Institution Name: Date: Materials, Tribology and Surface Engineering Assignment Question One (a). Brass (CW614N - CZ121) has a Young’s modulus of 97GPa and a Poisson’s ratio of 0.34.Calculate the effective modulus at the interface when two brass plates are loaded together. (b) Copper has a modulus of 140 GPa and a hardness of 150 MPa. Estimate the elastic strain in copper at the yield stress (10 marks) SOLUTION: (a) (b) Strain, ε is given by σ/ E = (150x103)/ (140x106) Strain =1.07x10-3 Question Two A steel shaft with a Ra value of 0.4μm is rotating in a brass bush with a Ra value in its inner diameter of 0.7μm. The shaft and bush are immersed in oil and, during operation; an oil film thickness of 5μm is developed. In which lubrication regime is the sliding interface operating? Your answer should be supported by suitable examples and references. (10 marks) SOLUTION: Boundary lubrication Since the value of film parameter obtained falls in the range: 5≤4.55≤10, the lubrication regime can be classified as Elastohydrodynamic lubrication[Jam13]. Question3. Using appropriate examples or references discuss four limitations of liquid lubricants. (10 marks) Discussion: Liquid lubricant lubricants are manufactured from vegetable oils, animal fats and mineral oils. Limitations: i. Vegetable oils are rapidly oxidized and less stable at high temperature environments. They also contain high levels of natural boundary lubricants. Examples of these oils include Castor oil and Rapeseed oil. ii. Compared to solid and semi-solid lubricants, liquid lubricants, especially from animal fats, are easily contaminated by presence of debris. In addition, they easily contaminate surfaces that they come into contact with. Examples include of animal fat oils include tallow oil and fish oil. iii. These lubricants usually have challenges in controlling loss of lubricant by creep, leakages, absorption by porous surfaces, evaporation and chemical reactivity of the lubricant and contact surfaces. Loss of lubricant from the required area of working makes it lose its function as a lubricant. iv. Liquid lubricants extracted from vegetable oils and animal fats generally exhibit variations in temperature in their viscosity cause variation on system efficiency or torque on the load bearing component. Question4. It is found that a polymer-based bearing supporting a rotating steel shaft has a depth wear rate of 0.25mm in 1000 hours both at a bearing pressure of 10MPa and speed of 10-1ms-1, and at a pressure of 1 MPa and a speed of 1ms-1. (a) Show that the bearing is operating in the range where the specific wear rate is constant. SOLUTION: (6 marks) Assumption: The volume of the wear material is a unit. Case I: Normal load, F= 10MPa Speed = 10-1ms-1 Wear rate = 0.25mm in 1000 hours= 2.5x10-4 mm/hr Specific wear rate, k = (volume of material wear)/ (Normal force x sliding distance) Sliding distance= (10-1ms-1x1000x3600) = 3.6x105m k1=1/ (10x3.6x105m) = 2.8 x 10-7 Case II: Normal load, F= 1MPa Speed = 1-1ms-1 Wear rate = 0.25mm in 1000 hours = 2.5x10-4 mm/hr Specific wear rate, k = (volume of material wear)/ (Normal force x sliding distance) Sliding distance= (1ms-1x1000x3600) = 3.6x106m k2=1/ (1x3.6x106m) = 2.8 x 10-7 In both case I and case II, the specific wear rate k1=k2, i.e. the specific wear rate is constant. (b) Calculate the time taken to reach a wear depth of 0.25mm if the pressure and speed were 2 MPa and 0.2 ms-1. (4 marks) Solution: Pressure =2MPa Speed =0.2 ms-1 k=2.8 x 10-7 sliding distance = volume/ (pressure x specific wear rate) =1/ (0.2 ms-1x (2.8 x 10-7) = 5.6x106 m sliding distance= speed (ms-1) x time (sec) Therefore, time= (sliding distance)/ (speed) = (5.6x106 m)/ 0.2 ms-1 = (2.8x107) sec= 7777.8 hrs. Hence, it will take 7777.8 hrs to reach a wear depth of 0.25mm if the pressure is and speed were 2 MPa and 0.2 ms-1 respectively[Jam13]. (c) Given this information, decide if you could safely calculate the wear rate at a pressure of 10MPa and a speed of 1ms-1, and give reasons for your decision. (4 marks) It is not possible to obtain the wear rate of the shaft for the reasons stated below: i. We need to measure the radius of the shaft to enable us obtain the area of the cross-section. ii. We also need to know the length of the bearing length, as this will assist in calculating the volume of the shaft. (d) The test results were obtained using a polished steel shaft in a laboratory environment and at room temperature. Suggest changes in these conditions that could cause the specific wear rate to be different. (6 marks) Changes that may cause difference in specific wear rate: The applied load will definitely vary when the material is used elsewhere in real applications. The change in the applied load will change the specific wear rate of the polished steel material. The more load applied, the lower the specific wear rate of the material. Another significant condition that is likely to vary the specific wear rate is the change in sliding speed and temperature. The wear rate significantly decreases with increase in temperature over relatively small range of temperatures[Jam13]. (Total 20 marks) Question5. (30 marks) “To reduce wear on a steel component, a hard wear resistant ceramic coating is applied” (a) Explain using suitable examples, what is meant by this statement. [Hint] why using ceramic coatings The widest application for ceramic material coatings is wear resistance. The coating helps in prevention of particle erosion, wear, and sliding wear, rubbing, galling and fretting. The three most outstanding performance features of ceramic coating materials are; improved wear and thermal resistance, and electrical resistance. Ceramic coatings can also resist corrosion from oxidizing agents, alkalis and strong acids. The coatings have wear resistance superior to hard metal coating, hardened steel or stainless steel and can transform ordinary metals into high performance materials. (b) With the aid of diagrams, illustrate three different techniques by which this could be done. Explain why a ceramic coating and discuss the advantages of the three techniques. You may use appropriate examples and references. Techniques of Metal Ceramic Coating: i. Plasma Flame spray technology: This technology uses a wide range of ceramic materials that can be provided in powder form. The powder passes through pressurized ionized spray of gas at elevated temperatures of up to 30,000 degrees (F). The gas enhances molten particles to the substrate where bonding to surface takes place. Using this technology, it is possible to virtually achieve any desired finish with a strongly adhering and relatively high-density coating. Figure 1: Plasma Flame Spray Detonation Gun: This technology is the most effective for specific ceramic materials such as tungsten carbide that are needed for highly dense coats on metal surfaces. The process creates an explosion of oxy-acetylene gas at temperatures as high as 6,000 degrees (F). The high temperatures melt the ceramic and fire it towards the substrate at a high speed. Figure 2: Detonation gun spray ii. Oxygen Acetylene Powder: This technology involves heating of the ceramic material powder in a flame at 5,000 degrees (F), compressing the gas and spraying the coat onto the target metal surface. This method creates a porous coat with relatively reduced adhesion strengths. Figure 3: Oxygen Acetylene Powder Another method related to this is the oxygen Acetylene Rod in which a fused ceramic material rod is passed in a torch burning at temperatures of 5,000 degrees (F). The pressurized gas is then applied to spray molten material onto the metal surface. The method produces a high cohesive bonding metal coat[Jam13]. Question 6 A Tungsten carbide ball 50mm in diameter is loaded under an increasing normal force against a stainless steel plate with a hardness of 200 GPa, as shown below. At what force does the material of the plate first yield; what is the corresponding contact width; what are the mean and maximum pressures in the contact zone; and what are the magnitude and the location of the maximum shear stress? The relevant Hertz equations for this contact are: Radius of the circle of contact, a = (3PR/4E*)1/3 (equation 1) Maximum pressure, p0 = 3P/2πa2 = (6PE*2/π3R2)1/3 (equation 2) Where P is the normal load, R is the radius of the ball and E* is the composite modulus at the contact. SOLUTION: (20 marks) Radius of Tungsten carbide ball=25mm Composite modulus, E*; 1/E* = {(1-v12)/E1} + {(1-v22)/E2} = {(1-0.242)/700} + {(1-0.272)/200} =0.001346+0.004636=0.005982 E*=167 GPa V=4/3πr3 =65.48cm3 15.63g/cm3X65.48 cm3=1023.4g Force exerted by the ball =10.03N Normal load, P=10N Radius of the circle of contact, a = (3PR/4E*) 1/3 = {(3x10x0.25)/ (4x167)} 1/3 =0.22mm Alternatively: Radius of the circle of contact, a = Contact depth={9x102/16x0.25x1672}1/3=0.2mm Maximum pressure at the contact zone is given by: P0 =3x10/2x22/7 x0.222 = 98.61N/ mm2 Maximum shear stress = =1/π (6X10X1672/0.252)1/3 = 95.2N/mm2 Reference Jam13: , (Takadoum, 2013), Read More