Materials: A Stainless Steel 316L (SS-316L) - Assignment Example

MATERIALS By Materials (Couse 10th February Materials Question A Stainless Steel 316L (SS-316L) tube having an external diameter of 20 mm and an internal diameter of 10mm has a Copper-Zinc (CZ) alloy rod of 10mm diameter inside it, the two materials being joined rigidly at their ends when the ambient temperature is 18°C. Determine the stresses in the two materials: (a) When the temperature is raised to 100°C
 (b) Use SolidWorks software to validate your results obtained in case (a). Draw the bespoke part, apply the boundary conditions and calculate the error between the theoretical and simulation outcomes. For CZ: Modulus of elasticity = 80GPa; Coefficient of expansion = 17 μstrain /°C For SS-316L: Modulus of elasticity = 210GPa; Coefficient of expansion = 11 
 μstrain / °C. Solution. (a) Since the relation for stresses due to change in temperature as mentioned by Askeland & Fulais (2006) is given as ; … (1) Where; … (2) … (3) (° C) Substituting the values (2) and (3) of and in equation (1); … (2) Alternatively as for given Stainless Steel 316L (SS-316L) tube having; External diameter , Internal diameter Hence initial circumference of steel tube is given by relation; … (4) … (5) Similarly; … (6) … (7) … (8) As the strain due to change in temperature is; … (9) … (10) Substituting the values we have; … (11) Similarly for Copper-Zinc (CZ) alloy rod having 10 mm diameter as, … (12) … (13) Hence applying equation (1) we have; … (14) Question 2 A Thermal Barrier Coating (TBC) is required for a Nickel super- alloy turbine blade. Using CES level 2 to choose a material that might be suitable for the high temperature application (1200°C). Discuss your selection methodology with the current Laser Shock Peening (LSP) coating and include details of any CES work you carry out. Answer. The possible materials for Thermal Barrier Coating for Nickel Turbine blade include Alumina, Aluminum Nitride, Boron Carbide, Zirconia, Silicon Nitride and Silicon Carbide. Here, the focus is on the application temperature and the materials costs. The six materials were checked on the CES database and the results are as shown in the graph overleaf. From the CES results, the best material for the job is Zirconia. This is due to the fact that Zirconia has high service temperature. Its service temperature is far much higher than 1200°C, which is the required temperature. Also, the cost of this material is relatively low as compared to other materials. Question 3 You are required to select a light reasonable material for intermediate case in Trent- 900 Gas Turbine Engine. Using appropriate materials indices to select a sheet of material, this has: 1. The best performance in terms of Young’s modulus to density ratio 
 2. The best performance in terms of Young’s modulus to cost-density ratio 3. The best performance in terms of Stiffness to recycle ratio of maximum 100 GBP/Kg. Solution The basic criterion to a satisfactorily design is to create products that can perform their function effectively, safely and at acceptable cost. For designing an object the selection of material plays a critical role. The key factors involved while selecting the most appropriate material are properties and functions of material, manufacturability, cost, failure mode and it environmental considerations. In general all such factors are arranged in groups of material properties known as Material Indices (MIs) as given in 1below. The form of the MI depends on the functional requirements of an object and its geometry. In material selecting procedure material properties including cost are compared and material having higher MI is selected to use in manufacturing the specific object. Table1. MIs used for material selection For given case similar approach can be applied to select an appropriate light reasonable material that can be used to manufacture the intermediate case in Trent- 900 Gas Turbine Engine. Comparing the properties of various materials given in table 1 then; 1) For material having the best performance in terms of Young’s modulus to density ratio comparing the material properties given in 4 and 5 as given in table 1 then comparing the properties of various metals and Composite Fiber Reinforced Polymers (CFRP) – the best suited material for intermediate case in Trent- 900 Gas Turbine Engine, the lightest appropriate material having least value of density (ρ, Mg/m3) is CFRP Laminate (graphite) hence it is the best light material in terms of Young’s modulus to density i.e. … (1) for intermediate case in Trent- 900 Gas Turbine Engine. Again comparing the Young’s modulus to cost-density ratio the best material for for intermediate case in Trent- 900 Gas Turbine Engine is Aluminum alloy (7075- T6) as compared to composite and metal it quite cheaper (1.8 $/kg) with light weight (. For 3.the best performance in terms of Stiffness to recycle ratio of maximum 100 GBP/Kg as composite materials have not been properly recycled, due to their inherent nature of heterogeneity, in particular for the thermoset-based polymer composites while on the other light hand metals and matels alloys are like Aluminum (7075-T6) can be the material with the best performance in terms of Stiffness to recycle ratio of maximum 100 GBP/Kg that can be used to manufacture the intermediate case in Trent- 900 Gas Turbine Engine (Yang et al., 2012). Question 4 You are working as a mechanical engineer involved in the design of various structural components. Using the information in Tables 1 and 2, calculate the highest stiffness and strength metal (Steel, Aluminum, Titanium, and Magnesium) in case of a tie rod, a beam and a panel. Arrange to a new table and put your final results and discuss why the geometry plays important role in your finding. Table1. Table 2 Solution The required calculated values for stiffness (N/m) and strength (Pa) for a tie rod, a beam and a panel manufactured from Steel, Aluminum, Titanium and Magnesium are given in table 4.1 below. S. No. Object Material Youngs Modulus of elasticity (E) GPa Yield Strength (σy) MPa Density (ρ) kg/mᶾ Stiffness (N/m) Strength (Pa) formula Value formula Value 1 Steel 210 600 7850     Aluminum 70 380 2700   Titanium 115 850 4600 Magnesium 45 240 1750   2 Steel 210 600 7850  1.846   Aluminum 70 380 2700 3. 100 Titanium 115 850 4600  2.331 Magnesium 45 240 1750  3. 833 3 Steel 210 600 7850  0.757  3.120 Aluminum 70 380 2700  1.526  7.220 Titanium 115 850 4600  1.0576   6.337 Magnesium 45 240 1750  2. 033   8.852 Table 4.1 Experimental studies have proved that the variation in geometry of objects also affect material properties such as poisson ratio, compressive strength, ductility, yield strength and others (Rothenberg, 2015; Pilky & Pilkey, 2008; Vorobyev & Guo, 2015). For given case stiffness and strength of three different geometry (tie, beam and panel) objects made from Steel, Aluminum, Titanium and Magnesium are compared. Since for beam and panel geometry objects both the Youngs Modulus of elasticity E (GPa) and Yield Strength σy (MPa) vary exponentially in contrast to tie shape object with linear variation. Therefore, compared to beam and panel geometry objects, the tie shape object is more stiffer and stronger as obvious from values given in stiffness and strength columns. References ASKELAND, D. & FULU, P. (2006). The Science & Engineering of Materials. Toront Thomas Canada Ltd. VOROBYEV, A. Y. & CHUNEI, C. (2015). ‘Multifunctional surfaces produced by femtosecond laser pulses’, Journal of Physics, vol. 117, no. 3 pp. 033103; doi:10.1063/1.4905616vol. ROTHENBERF, B. (2015). Geometry & Material Performance. Available from: ˂http://www.bradleyrothenberg.com/geometry-material-performance/>. P I L K E Y, W. D. & P I L K E Y, D. (2008). Peterson’s Stress Concentration Factors. (3ed). New York: John Wiley & Son. YANG, Y. et. al. (2012). ‘Recycling of composite materials’ Chemical Engineering and Processing: Process Intensification. [Online]. 51.53– 68. Available from: ˂ http://www.sciencedirect.com/science/article/pii/S0255270111002029> Read More