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ss_quiz_2 - E-344 Fall 2005 Name: Quiz 2 7 10 November,...

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Unformatted text preview: E-344 Fall 2005 Name: Quiz 2 7 10 November, 2005 ID#: Honor Pledge: Use the backs of the pages if you need more space. Clearly label work there with an appropriate question number. 1. (A) (12 points) (S 1.1) Using the stress-strain data presented below for the SAE 1340 steel water quenched and tempered, determine the elastic modulus, the yield stress, the ultimate tensile stress, and the percent elongation to failure. . . z. so; mew . v was: M379 6;; ZLKS 15». 0—; z b3 I5 E . /\‘7 MI {0.2% WJ\ (B) (3 points) ($1.4) During a tensile test of 1340 steel, at What stress level does dislocation motion begin? Briefly justify your answer. vb\b\vwmom Mvfim [p mmukfig mam-mtg exams» M"- W wads sweats. \l Warm-3‘6 RWV’J'LBW ems»; nmtmmfima‘ we we LDCszWm-Js News“: WWW haeowwrsm w Gar/“($Wkkiujz Mme-tats. (C) (5 points) (81.6) Suppose Rockwell indentation tests are used to measure the hardness of the 18—8 stainless steel and 17-7PH stainless steel sheet. Which of these will have the larger indent? Why? W ($3,, 5N\m%&% um“ M TM when?» Yet-ems)sz \T' mass m. LC)le mars. grammar/sue SAE 1340 steel water-quenched and. —-"” tempered at 700° F Iii‘x,‘ Stainless steel sheet 17—7PH / Z29 ‘ It, \ Stainless stee1(18 '- / - fit O‘ooe— ,1" \ Annealed titanium alloy sheet (6A1-4V) ‘ I fit \ — .— éé) 0'03 _’-’ \ Nickel alloy 'teel ' l .i'" \ - flwwww‘wmm...-..,m Annealed N—155 alloy sheet Stress a, .1000 psi Structural steel (mild steel} w“. . an «n and.” Magnes in In ' ‘ ,__l_____l_._._..L___1____ 0 1‘; 0.04 0.08 0.12 0.16 0.20 , e being considered for possible use as a turbine blade in a jet engine with an anticipated operating temperature of 1250 0C. Their fracture toughness, yield , and melting temperature are given in the table below. C. (4 pts; 81.13) Which ofthe ca 11d1date materials would have the greatest creep resistance? Briefly justify. ' ._ SW35 T o \ v 4— Material KIC (MPa—mm) 0,, (MPa) tune) PWPW'M Z»:- A 76 1310 1450 Vsbvhth' Vmo Mk B 12 1500 2650 2\L\ My?“ 44.. \goO “‘5”. LC: 0' M 632,. = ' ‘C Q T I N'Ld.mm may. Y2 \HWEW ® Memtm k \s We Eem LEMmC‘L V2.90“ W vieseacnvfz 09 Sammie. WWW; Waxes-kw . \T cw yam To “(JAN k CW3; vex—mus \k UQzC’e Esoqu w P55: ERA tier! one km Wovas $529M 5:92ch we newscast-v. Mm‘i‘i’chm E 09 M omi CAN suF‘ffiJL. C-kmsflr—neml Harm: Marika, Hwh cmwg m was we {seem mesa: “as Wu; \5 Wfifccfléfit$eyakjl $052 “5301-0: k R¥%L\C¥hm. j WW2. @aemusrz T: 9990C \‘= um» Ewe-fit “T 0 " t #9? ‘7’0‘ ‘ k- mxk WW1»: “(K MW Eh HEW \M Vim Puts—Wake 5 B h mi) ‘5‘ $USCXPT\%KE ‘T'D dam. TM >>Th F» a; x ‘80 0J1. woods ‘EMXLCX' mwme\ 3?: *0 “Mg, N OVmede C P—w M\<5Nw.&. sum/FM 9 \UWZNAK CWULX cm kt}? PS. 3E3 com. , _ W 2 W———'————“———1 3. The monomer unit for poly(acrylic acid) [PAA] is given in the adjacent diagram. ‘ CHE—pH . ([32:31 _ OI—I A. (5.8; 5 pts) Suppose a sample of FAA has an average molecular weight of 6000 g/mole. Estimate the average degree of polymerization. Mum: Eula?) \[fi on“ = shag-7'.“ e = m Wm n: 6000 :4 “47,2, --*__....._ _ polyacrjrlil: acid B. (5.1; 5 pts) Estimate the total length of the average PAA molecule in this sample. 5% 1,1: Emma‘s ‘6: "473%. TOM 0:955“ o€~ M swamps MR NDLUZZVLFL \0 “‘3 WEN-z. \3 “NUS ~ @s\i..osm§-— 23% C. (5.1; 5 pts) Sketch a possible conformation of an average PAA molecule in this sample. D. (5.5; 5 pts) Assuming one had a fully dense solid specimen of PAA with a volume of 1 01113, approximately how much would this sample weigh? W awn“! a? veg“ how‘s \s Meow r—_—_——_—__ '1 4. Consider the diffusion of aluminum into a wafer of silicon 1 mm thick that initially contains 1018 atoms/cm3. An Al—containing gaseous atmosphere in a furnace fixes the surface concentration of Al on the wafer at 10is atoms/cm? The furnace is held at 1100°C where the A1 diffusivity in Si is 4x10'13 cmz/sec. ‘ (A) (32.2; 8 pts) Sketch a series of composition profiles that describe the Al concentration as a function of depth into the silicon for t=0 and for two times thereafter. Baa“ Woe-M: bases par M \Q \‘Tfi‘k TNME , (B) (32.3; 12 pts) The furnace temperature is lowered to 1000 °c where the diffusivity is 2 x1015 ch/sec and then further lowered to 900 °C. What is the diffusivity of Al in Silicon at 900 °C? . -S' " Ti: We“. \> = mxe-‘é' (WW Ts mu bf 2*‘0‘ “M 9 . re. Negaw‘ we he; m; A“ =‘“’©~h*"’"¥ 2%: 3+1 \ . - 3- 45.. \ ...—~ 5.11mi. - . e...— .. ~— "5 ‘a r x 0"" ' {$5.59) 65%] \‘S’r's \hs =5 €.%§= (E\00.66‘%q\ =3 3: sin 29 i (:3ng Wicket: t: SWQMOABB J» «353A '2. “1%vg 3" 642‘}? 6‘) Q3333 - so 0 Lea-mt . 4351\— =_) __ ‘ EMS; Ewe? El-09*\5°\W lg.eme“)§m§\ \s, a oer M :-. \.Zi~ \0‘\€C+~FZM F, Yuais’ ,_ “£554 '\ _________..—- bkcl00°c:\\WSV-\= OLDEMOWIB stsfieflo-s)h\}-g\ -= tenor-em 4 shes «— were 4 r———‘“———“————1 In the following multiple-choice problems, circle the one answer which BEST completes the sentence. 3 points per problem. No partial credit. S. (82.4) A vacancy corresponds to a missing: , 5 I. ‘ fl “iii a. electron from the valance band; i g fif'w’g’ E l w m WM b. electron from the conduction band; {m £3 to @ atom from a substitutional lattice position; . atom from an interstitial lattice position; a? k as a W . 2157 e. none of the above (a? g f 6. (81.5) Two samples of stainless steel are identical except that sample 1 has a finer grain size than sample 2. @Sample 1 has a higher yield stress than sample 2. b. Sample 2 has a higher yield stress than sample 1. c. Thetwo samples have identical yield stresses. (1. The yield stresses may be different depending on the size of each sample. 6. none of the above «creek? 7. (81.9) Suppose a ceramic rod 2 inches in diameter with a circumferential surface scratch 0.1 inches deep is alternately loaded in tension and then in compression to a stress level 0.2 of 0),. After one million loading cycles the rod breaks. Failure most likely occurred: a. by creep; b. by brittle fracture; c. during the compression part of the loading cycle; C9 during the tensile part of the loading cycle; e. none of the above. 8. (SL4) Dislocations play an important role in: a elastic deformation of crystalline solids; lastic deformation of crystalline solids; 0) primary bonding; d) viscosity; e) none of the above. 9. (82.3) The rate at which a typical solute atom moves (diffuses) in a crystalline solvent: increases with increasing temperature; . decreases with increasing temperature; c. does not depend on temperature; d. depends on the Fick’s Second Law; e. none of the above. r*—‘“‘*——“—“n 10. (5.6) A key difference between a thermoplastic polymer and an e'lastomer like rubber is: rubber has a small number of covalent crosslinks and thermoplastic polymers have none; . therrnoplastics don’t melt when heated but rubbers do; 0. rubber can be subjected to extensive plastic deformation whereas typical thermoplastics have little plasticity; d. rubber is soluble in water but thermoplastics are only soluble in organic solvents; e. all of the above. 11. (5.4) Recycling a thermoplastic polymer is possible, because: different polymer chains in a thermoplastic are held together by relatively weak Van der Waals bonds which can be broken at relatively low temperatures; b. thermoplastics are only lightly crosslinked in comparison to thermosets; c. thermOplastics dissolve in water; d. a and b; e. none of the above. 12. (81.2) Suppose the stress-strain data ' described in the adjacent diagram were collected from a specimen of poly(ethylene) (PE), a specimen of 6063 Aluminum alloy; and a silicon nitride (Si3N4) ceramic. Which curve corresponds to which material? _ a. 1= PE; 2 = 6063 Al; 3 = (Sl3N4) @1 = (Si3N4), 2 = 6063 A1; 3 = PB; 0. 1 = 6063 Al; 2 = (Si3N4), 3 = PE; (1. 1 = 6063 Al; 2 = PE; 3 = (Si3N4), e. none of the above. 13. (S 1 . 12) In the context of materials engineering, fatigue refers to failure due to: a. corrosion—assisted deformation; b. high—energy impact; yclic loading; d. a constant load at elevated temperature; e. none of the above. ...
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ss_quiz_2 - E-344 Fall 2005 Name: Quiz 2 7 10 November,...

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