Materials Tribotechnolgy and Surface Engineering - Assignment Example

 Materials, Tribotechnolgy and Surface Engineering Question 1 K1c of ceramic zirconia, ZrO2= 7.5Mpa Polyethylene terephthalate K1c = 5Mpa Polyamides value (Nylons, PA) is K1c = ( GcE)0.5= 2.05Mpa Ceramic zirconia, ZrO2 K1c is larger than Polyethylene terephthalate (PET), Polyamides (Nylons, PA), or Polyethylene (PE). G1c= 6.04kJ/m2 E= 700Mpa K1c = ( GcE)0.5= 2.05Mpa The toughness is smaller. Polyethylene (PE), K1c = 4.5Mpa Toughness G1c = (K1c)2 / E G1c for Polyethylene = 50 J/m2, E= 0.8Gpa G1c for Polyethylene terephthalate= 55J/m2, E=2.27Gpa G1c for Polyamides (Nylons, PA) = 4.5J/m2, E= 2.5Gpa G1c is larger which is contrary to the expectation. There is apparent contradictions in the results as depicted in the results below. The variation is due to the structural and corresponding morphological characteristics namely quantity, orientation, size and corresponding shape of the crystalline. Question 2 Given: Carbon fibres Young’s modulus= 400GPa Epoxy resin with a Young’s modulus = 3GPa Taking Vf= 35% volume carbon fires Direction parallel to the fibers Ec= Vf Ef + (1-Vf)Em Therefore, Ec= Vf Ef + (1-Vf)Em = 0.35*400Gpa +(1-0.35)3Gpa = 140Gpa + 1.95Gpa = 141.95Gpa Direction perpendicular to the direction of the fibres Ec = 1/{Vf/Ef + (1-Vf)/Em} = 1/{0.35/400Gpa +(1-0.35)/3Gpa} = 1/{0.000875+ 0.21667)} = 1/0.217545 = 4.597Gpa Question 3 Given: Stress range Probe fail after 107 cycles= 90Mpa Stress range probe fail after 110Mpa= 110Mpa a) Stress range Δσ= σr= σmax- σmin =110Mpa- 90Mpa = 20Mpa/2 =10Mpa Thus, stress range for the probe fail at 104cycles= 10Mpa b) Δσ σm= Δσ0[1- σm/ σTS] Where Δσ σm= stress range when σm is not equal to 0 Δσ0= stress range when σm=0 σTS= tensile strength Therefore, Δσ σm= Δσ0[1- σm/ σTS] = 90[1-70/250] = 90[1-0.28] = 90* 0.72 = 64.8Mpa Hence, the stress range for 107 cycles= 64.8Mpa Question 4 a. Single shear system mainly introduces a small moment at the joint because of the axial forces in the top and the bottom plates are not axial. The axial loads will tend to align possibly resulting to the connected moments to deform. The whole load P is transferred in shear over one cross-section of the bolt. b. Double shear is prominent within pin-and-clevis connections possess merit of being balanced because of the symmetry. Load P is normally transferred over two cross-sections. Therefore, whilst the bolt transfers the same force P, the corresponding maximum shear stress is half that of the bolt of the single shear. Question 5 Given: Shear Force= 12Kn during static test Dimension of the pad: a= 170mm and b= 250mm Artificial rubber thickness = t= 60mm Force = 12Kn the top plate moves laterally by 8mm w.r.t bottom plate Required: Shear modulus of elasticity, G, of the artificial rubber Solution Shear train: ɣ= d/t =8/60 = 0.13 Ge = V/aby = 12000/[170*250*0.13*10-6] =12000/0.005525 =2.172*106Nm-2 Question 6 a) Cu2Sb b) Eutectic at 6500C : L(31% Sb) = a(12% Sb) + ᵦ(32% Sb) Eutectic at 5200C : L (77% Sb) = Cu2Sb+ δa(98% Sb) Eutectic at 4200C : ᵦ (42% Sb) = ƹ(38% Sb) + Cu2Sb c) The two-phase field at steady temperature consists of mixture of copper and Sb, with the composition of the phases at the saturation limits the values on the prevailing boundaries at the end of the tie line. Different compositions at this temperature contains diverse proportions of every phase that the entire fractions of the two elements. The underlying proportions of every phase by weight within the two-phase region such that the weight fractions are mainly fixed by the demand that the matter is conserved. d) The alloy contain 95 wt% copper and the first is austenitised, and the alloy cools slowly at room temperature at 650 degrees. At 650 degrees the alloy is liquid and it melts to 500 degrees. At the temperature of 450 degrees the alloy is pure Sb. Question 7 a) A detailed view of the arrangement of the atoms around a left-hand screw dislocation has two planes of atoms. The lattice is perfect and the filled circles are directly below the open circles. Within the lower portion of the figure, atom 1 is within the lower level, and atom 2 is above. A b) Ceramics materials are normally extremely hard and do not deform plastically under tensile stress. Nevertheless, they break suddenly after elastic deformation. Moreover, ceramics materials possess small cracks due to their prevailing processing. Compression stress normally exerts a lot of pressure on the ceramics making it to be compact and stronger to forces. Nevertheless, tensile stress leads to development of cracks on the ceramics due to dislocations and slip of the molecules making it to be weak. Since ceramics are brittle in nature, tension load leads to small cracks on them making them to fail easily. c) Fatigue limit for carbon steel is the threshold stress for non-propagation of cracks. It is the point of the load where cracks are unable to grow. Fatigue damage is normally caused by cyclic slip, which is driven by dislocation glide force. Moreover, it commonly occurs within persistent slip bands. Thus, fatigue limit is inversely associated with grain size. Tenacious cyclic slip results to intrusions and extrusions on the metal surface. Surface intrusion offers the original notch for the crack nucleation and corresponding growth. d) Aircraft grade titanium alloy (Ti-6/4) has greater fracture toughness due to the beta-processed type microstructure intrinsic in castings after the post HIP heat treatment. It also composes of aluminum, vanadium and iron that make it to be considerably stronger with stiffness and corresponding thermal properties exclusive of thermal conductivity. Moreover, it is capable of operating at high temperature. Conversely, armour grade silicon carbide have fracture cracks making it relatively weaker. e) BCC metal possesses numerous slip systems making the movement of dislocation to happen solely in the line of atom thus enhancing application of stress and thermal activation. Moreover, it has planes that are not close-packed and any underlying slip imply that the corner atom moves to the centre of the cube thus resulting to the kink nucleation and propagation within the BCC dislocation movement. BCC metals are shortages dislocation at high temperature making it ductile. Therefore, BCC dislocations association is thermally activated whilst the FCC dislocations movements requires expressively smaller activation hence making BCC materials to be brittle at low temperatures whilst FCC remains ductile regardless of the temperature. Moreover, FCC activation energy slips at lower kbT. f) FCC metals are stronger as it stems from the underlying grain flow due to forging. As the underlying metals are normally pounded on the grain deform to follow the prevailing shape of the section thus, the grains are unbroken throughout the metal. Cold rolling and forging makes that metal to have high strength to weight ratio. Moreover, when metal are cold worked via the forging and rolling, its underlying shape is permanently deformed due to the dislocations within the grain structure, which moves via the entire crystal structure. These prevailing dislocations within the grains structure permit the entire change within the shape of the metal, which increases its strength. g) At room temperature, prevailing copper atoms normally separate through diffusion to form copper rich zone commonly known as Guinier Preston zones, which are in two dimensions and logical with the underlying lattice. This results to the existence of variation within the atomic size of copper and aluminium strains. Thus, hardening occur due to the escalated work demanded to move the dislocations via the underlying strains lattice and corresponding work demanded for the dislocations to pass via cutting stress. This is counteracts the decrease within the solid solution strengthening as the prevailing copper concentration within the aluminium decreases. Over-aging creates the equilibrium tetragonal phase (Al2Cu) is disjointed and lacks strong strengthening impact. This microstructure is conspicuous and the underlying crystallographic associations amidst the precipitous and the corresponding aluminium lattice are obvious from the corresponding arrangement of the precipitates. h) The room temperature atomic size of pure metal varies due to strains. Strains result to hardening that emanates from dislocations of the strain lattice and cutting stress. Hardening and increase in yield strength of the increase in grain size of the material from 10microns to corresponding 40 microns decreases by two due to the shrinkage of the lattice structure. References Raghavan, V. (2004). Materials science and engineering: a first course. New Delhi, Prentice- Hall of India. Callister, W. D., & Rethwisch, D. G. (2012). Fundamentals of materials science and engineering: an integrated approach. Hoboken, N.J., Wiley. Bergman, T. L., & Incropera, F. P. (2011). Fundamentals of heat and mass transfer. Hoboken, NJ, Wiley. Ageorges, C., & Ye, L. (2002). Fusion bonding of polymer composites: [from basic mechanisms to process optimisation]. London [u.a.], Springer. Shackelford, J. F. (2008). Introduction to materials science for engineers. Upper Saddle River, N.J., Prentice Hall. Swab, J. J. (2011). Advances in Ceramic Armor VII Ceramic Engineering and Science Proceedings. Hoboken, John Wiley & Sons. http://www.123library.org/book_details/?id=31718. Medvedovsk, E. (2012). Ceramic Armor And Armor Systems Proceedings Of The Symposium Held At The 105th Annual Meeting of The American Ceramic Society, April 27-30, 2003, in Nashville, Tennessee, Ceramic Transactions. Hoboken, John Wiley & Sons. http://public.eblib.com/choice/publicfullrecord.aspx?p=892308. Swab, J., Zhu, D., & Kriven, W. M. (2009). Advances in Ceramic Armor a Collection of Papers Presented at the 29th International Conference on Advanced Ceramics and Composites, January 23-28, 2005, Cocoa Beach, Florida, Ceramic Engineering and Science Proceedings. Hoboken, John Wiley & Sons. http://www.123library.org/book_details/?id=21522. Donachie, M. J. (2000). Titanium a technical guide. Materials Park, OH, ASM International. http://app.knovel.com/web/toc.v/cid:kpTATGE013. Read More