Exam II Solutions - Introduction to Science of Engineering Materials | MATS 2001, Exams of Materials science

Download Exam II Solutions - Introduction to Science of Engineering Materials | MATS 2001 and more Exams Materials science in PDF only on Docsity! MatS 2001 - Exam #2 Solution 1. Lines, Planes, and Crystals (20 points) (a) On the cells below, label the x, y, and z axes using the right hand rule convention. Inside the left cell draw 111   . Inside the right cell draw [121]. (2 pts) Comment: labeling x, y and z is necessary (b) On the cells below, label the x, y and z directions using the right hand rule convention. Inside the left cell draw (021). Inside the right cell draw  211 . (4 pts) comment: some students forget how to differentiate directions and planes. How to determine the intercepts for planes is very important. Page 1 of 9 FL12 Z X X Y Y Z z z y y xx (c) Your colleague claims to have invented a new super alloy of copper and aluminum. He further claims that the new structure consists of a FCC lattice of Cu atoms with Al atoms in positions (1/2, 0, 0) relative to each Cu atom. Draw the new structure below. Be sure to distinguish between Cu and Al atoms. (6 pts) comment: most got it right. Cu Al (d) Calculate the packing fraction for the new alloy. The radius of Al is 0.143 nm and the radius of Cu is 0.128 nm. (8 pts) Comment: how to determine a is very important, the majority of students did it very well.   3 3 3 2 2 0.542 nm 4 4 4 (0.143) (0.128) 3 3 P.F. = 0.528 or 52.8% 0.542 Cu Ala r r Vol atoms Vol cell              Page 2 of 9 FL12 X X X X X X X X X X X X X X 3. C Cid. working 40. points) Trang ems (TAI Gi = Ab mm l inwvial, dls Yad = 21 ram deemed yanus Vo Yd = Ql woking AnnAd. ; | esl Ae—P 8» ponstenk wren — number —[aes At Cailictor = cmtteek —wlam number | ” na wy" ea? 2 th = Gl ah PG = 835 me or ; Howeuin, Page 5 of 9 FL12 Key 4) Strengthening and Slip (21 pts) (a) List 3 of the Hume-Rothery rules for a substitutional solid solution. (6 points) 1. Similac atomic radii (aré 157) Q. Makeing cvystal Stcture 3, Similar” Clee-trontyativity 4 Similor valevec (b) Cirele True or False (Do not guess: +2 points for correct answer, 0 points for no answer, -1 point for incorrect answer). (6 points possible) a) T @ A preferred slip system for dislocation motion in BCC crystals is {111} <110 >. b) T @) Ima FCC crystal the planar atom density (atoms/em’) is lower on a (111) than ona (222). c) (£) F The critical resolved shear stress of a metal is a “material property”, and therefore not a function of the orientation of the crystal relative to the applied stress. (c) Name three methods for strengthening metals and explain them briefly. (9 points) 1 Grin Sige Reductioa — B e : 3 pts By Creating sewn lle 5 (bes dhe deusify ak \ nwae rin boudaris is merase. Didviatin , aa A. desciption y Gan badaries oul Ths The Walon, / 1g hag nai 95? Allyn | Solid Solotin - Besides tha welton of es | wane 4 Ae 2 desiite ce ee eal Sess cvealel arr ga inbestiPi! oy vestitrtin will beet disheatens aud. taht their Mobilis, ; Co i aps 8 Wore Worleimy a wmaterial vil imercare olilecatiie clewsity incase THIS chereases the alist, betuer, clulcebins Deshocahins vepel eet Other as Tuts diceatin wection 1s Iyudned, Tais leads Yo @ showper Page 6 of 10 Page 6 of 9 FL12 5. Strain Energy (10 points) If all the dislocation strain energy is converted to heat in a strain hardened piece of copper we estimate that the temperature rise of the copper will be 1C. Calculate the dislocation density C. Calculate the dislocation density that would generate this temperature rise. The heat capacity of copper is 390 J/kg-K, density = 8950 kg/m3, E = 130 GPa, Poissons ratio = ν = 0.36, RCu = 0.128 nm. Energy density for temperature rise: Eheat=ρCp∆T Dislocation energy density: Edis= LGb2 V For copper, FCC structure, burgers’ vector: (Some students could not get the magnitude of burgers vector. 2 points if student can get the correct b, because it is very important in material science.) b= a 2 (11 0 ) b=2 r Dislocation energy is turned into (internal) energy for temperature rise: Eheat=Edis (Energy conservation, 1-2 points if student can show this concept in any readable form) ρCp∆T= LGb2 V Shear modulus: G= E 2(1+ν) Dislocation density: L V = ρC p∆T Gb2 = ρCp∆T 4Gr 2 = ρC p∆T (1+ν ) 2 Er2 = 8950 kg m3 ∙ 390 J kgK ∙1K ∙ (1+0.36 ) 2∙130 ∙109Pa ∙ (0.128 ∙10−9m ) 2 ¿1.1×1015m/m3 Many people did not write down anything for this problem. If time is not enough for calculation, write down the major equations that will be used and the right equations will get you most of the points. Average score is about 5. Page 7 of 9 FL12