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Midterm2F10Solutions (1) - UNIVERSITY OF MICHIGAN Official...

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Unformatted text preview: - _ - _ UNIVERSITY OF MICHIGAN Official Examination Booklet _ — — — - _ _ — _— Name of Student: Date of Examination: November 17, 2010 Course: CEES 11 —— Midterm #2 The University of Michigan Honor Code The Honor Code outlines standards for ethical conduct for graduate and undergraduate students, faculty members, and administrators of the College of Engineering at the University of Michigan. Policies When Taking Exam The instructor need not monitor the exam The instructor will announce the time of the exam and the instructor's whereabouts will be communicated to the class Students will allow at least one empty seat between themselves and their neighbor The instructor will inform the class prior to the exam if aids are allowed during the exam Students must write and sign the Honor Piedge on their exams The Honor Pledge is: I acknowledge that I have neither given nor received aid on this examination nor have I concealed any vioiatr‘on of the Honor Code. (Signed) QUESTION #1 (50 POINTS) —- NEAR—FIELD EARTHQUAKES (50 PTS) You are hired by a client to design a simple structure (single bay frame) for placement near a major fault in the Los Angeles metropolitan area. As the engineer of record, you are obligated by code to retain a seismologist to aid you in assessing the seismic risk posed to the structure. After hiring the seismologist, she delivers some bad news — your site is very close (within 2 km) to a particularly active fault line. As a result, you must design the structure for the near field earthquake affect. In general, near fault earthquake records are often characterized by a dominate “pulse” in the measured acceleration. The seismologist describes the characteristic pulse behavior of the fault based on its past seismic record. She provides the following information about a likely fault pulse scenario for which you should design the structure: ~mi3gI .L Assuming that you are designing the single bay frame as a simplified, Eumped mass single degree—ofufreedom (SDOF) system, determine the time history response (relative to the base) of the structure, u(t). To keep your analysis simple, please assume no damping is present in the structure. Represent the mass of the structure as m and the lateral stiffness of the frame ask. mg? 669.59 fig”? *9“??? Egg} i —-. ,n. 4' ‘ a“ E m “E” ‘3' CE 8mm”? mg m. i, “ “E mswfiz; 3mm ifie/rc‘ifll: U H" I! (33} $2 22*": 2,2522% €55); 122232? 5A3? 95213 j: ’E‘ éeawfifioét 2. 22222922 §: #22222: 22222 “E Vi; SW“: WW) («322 £224., ~—._ 2 ,mwflmmm#3%wmflfivzaw 2 . u Jam gem 5’54 Wm 323 K? a 5 522222222222 E 7; 22222222“: 22?“: 5:3? 5255.3; Wfi‘ag (WWW 2 22322 €72? E25522?" fig, :: $lhfiwefi “522222222"; ._-2_ 22:22.. Mnfre-a’wuf'nw 3:, 2” 2 22 $5??? 52,2522 5%? {$2 555% WP? ‘ 3 220 ~ '2 _ a [05 2.222%: S n WE; 2* @222 ”Z 2:222: 22.22222? 22.22222 222225222 95$ _ :‘Z‘vi‘ «Whefivrww-mrv »: 2:25.. mummwm $22 $2 22222222? Ma 22223222 2 E2 ’2 E 2 f?" S§V§§E§Wa§ “5 ”é". :2 h QUESTION #2 (50 POINTS) -- EARTHQUAKE DESIGN You are provided the blueprints of a concrete frame structure designed for a state agency in California. The structure is located in a seismic zone (Sacramento, California) where it is highly probable (e.g., 80% probability in the next 20 years) that the next earthquake will have a 0.2 g peak ground acceleration. WM WOW?“ Spandrel Beam: 300 x SUD mm Columns: A )fl 300 x 300 mm The frame is built with reinforced concrete with a 30,000 MPA elastic modulus. The column and beam have a 300 x 300 mm square cross-section. Furthermore, the slab supported on the top of the frame is 150 mm thick. The total mass at the roof level is estimated to be 18 x 103 kg. Based on the International Building Code (IBC), you can safely assume that reinforced frame structures of this type are 2% of critical damping. Also provided by the design code is an appropriate design spectrum (see attached spectrum) for the Sacramento area. a. Determine the anticipate peak displacement, velocity and acceleration of the frame. b. Determine the maximum base shear of the structure. 0. Determine the maximum overturning moment of the structure. A. é‘ééb’fi‘f Ofiééfiaz ggaweafi {gag w“) w {mamw} EZézai [<5 3 “7““ ‘ @w’oé ”/w E : 3 mam: wag; iiigwg I ,5. g «56* £353 a“? 22 % “E" ”5:2 a J“ 53:29 wa§ {fisggwg'fi} Mm Pseudo—velocity, 81, (cm/s) 2 egg, v 1’4» 0 1 0 0 g: _? (m ' \W/ s, at ‘h‘r .v ‘1 V V r "7" rr"" ! mmmm '7'} fifigyyfir‘ , 3y 1“. v «9'- . MAfiEM'J ’ v V» a, my" . 5 9.. _ . r—x‘mmv-v h, _ . 4 ,A" L W? '1' A, mfifi ,0 him? ’43; ”Image?” - gunman?!“ w ‘0} 523”“ L Mk3? ‘ whqum- Y‘! ( L 7 .4“. I}... 5‘ ”AIyA‘KMIdWfiggmyzqggngafi'. ‘ zgzumgwg/Angmz ‘ , ‘ ‘W'Idaklfifi'wNV/J'IA'M . 3wzzzy <wvaa<§ mm. 1'1 Wigs , I o W __, 'qgammmmflm}; AflifikQ-fl . ‘ V 7 ‘ V K 7 V17 “3254‘ 4‘ ,A! J, ’4!” 7' E as r‘ [A E 3“? ii )4 IL '7 a! m ‘1'!!! 3:59} «as. ;. 'i u )- 9. I A‘R \ w v“ a H.k . ~ A,” ‘ ’ m' 1' . m:w.fi‘%m§% / 21 0 . -A mam ‘fi la n m9 Aw‘ NR‘Y L. q ‘1 ”lg sum. ,1 r A». L In 12' a . *1 ._. . _.. I n! ' ’4 I ‘hfidfix'lk K4, fflflfifiixm U; Jugaggmgwmgmggmwmznmg w ‘aa‘wwvw wwwv’w Vi AL. AL). $1092“. A ALA; L .93.:10241 ALA; ' VW‘OVVV 7"! ‘V'V'vvvv " ‘ 0.04 0.00 0.1 0 2 0.4 0.0 0.01 2 4 0 3 10 \ *1. “5%”: M. 5 «I 5:3”... .5me9 :35595; .3 o .omcucfimhui I In?“ "35. "4mm“ wk ,3: p.35; I gamma”. “5 I . _. Qumomoocmr VN my. 37 “amen. 3+ flcEv€_I a mimfim cw H miomouo msmwmmifi INN m\m$:_m :wn “w Q? Em am mil N... away a + cw u I mgoImSIfi New" Mr. mgflmmnoomt+ never? a awn «Wang £8 I. m 2: SE“? .5 w... L cmfirfic,n a. # I m m a Lila“. 3 ave , IIEImMII! II a m . 3!... I: x a... IE I w a... I: I 4 . I . I . , I 3+ 33:1 x |E + $5 I It. 3 In 33 x. {E I fix. I «I TAMI 9&1. ”QUE i m TAMI Qtw+ “3. «I: N c _ .I qIY A a ff LI a T? 3% I. QIWI Qmmh imiIifi? .3 SET. on J m \Nmn maismriu £73. a: I? é. If If -EI c _ am I. III N_\ATN+UN.IH$H_3 3 .6\I&._H I»... . Ii NEH «N _ I 5;“ H . H Ila. . . I I I 3. In I I. méaLa ET 3%.... I 4?. m IVA I I: M mi I1 m 9%: LINE. I 3%? a a Ei .3. I I? aqfiw II EIQIIcQETI "$54 I 1%.. 33% I 23$. _ 3:33 .3 $35.5 N 323ka .3 wsmcxé $.ng N r.._§ 3....an nn...‘ “Gnu F1. £12.53 «553% 3.3. $55:me 8 3.1a u Sui 51 STATICS DERIVATIVES AND INDEFINITE INTEGRALS in these fommlas, u, v, and w represent functions (:fx. Also. a, c, and r: represent constants. A11 arguments of the trigonometric functions are. in radians. A constant ofintsgration should be added to the integrals. To avoid telminolugy difficuity, the following definitions are foliowed: arcsin u t: sin“1 1:. (sin at)"i = l/sin u. I Q’s/cit: : 0 2 (iv/(Lt = 1 3 flaw/cit 2 C du/dx 4. d(u + v m tad/(ix = du/a’x + dv/dx — dw/dx 5 d(uv)/’dx = u dv/dx + v (in/air 6 durum/d): = 11W dw/dx + mv dv/dx + vw dIr/dr 7 (Kn/v) _ v du/cir — u dv/dx dx v2 8. d(u")/dx 3 mt“ dart/dz: 9. d[f(u)]fdx = {dU(-u)]fdu} dm’dr 10. du/dx = 1/(cix/du) d(logau)_ 1 d_u 11. dx (logac)— 1; (Tc 011111 11) LE 12. 6b: — M a" q. d1a“)_ u d_u 1:. 4" -—.(En (1)51 d—x 14. d(e")/dx = e“ dIr/dt 15. d(u”)/dx =5 WM du/dx + (In a) u"dv/dx l6. d(si11 z:)/dv " cos a du/dx 17 610305 u)/dx — —sin 1: dn/dv 1.8. d(tan await ~= seczu du/dr 19. ({(cot u)/a'x = ~csc2u dzr/dt 20. d(sec LO/dx = sec 2: tan arr du/dx 21. ((051: z:)/dx = ~—csc u cot z: (in/ah; - .«1 7 (215111 111% 1 fl _ < -—| < 1 2_. dx 1 ”2 cit 1 “Jr/2 _ sm 1.: _ 1/2) 23 d1cos"‘zr) =_ 1. Egg” (0 < cos‘lu < ?r) ' dx 1 __ ”3 dx 'fl . _ d1tan‘lu) _ 1. fl _ . —1 24. ~—afic — m1 + “2 (iv 1 arr/2 < tan :1 < 7:112) 25. Mom—I”) 2.. 1 Q 10 < cot—1t: < 7:) dx 1 + u‘ six 26. d1scc' In] _ 1 dz: 27. __._........_‘ dx = (0 < cscflu E 7:12) (—7: < ass-"11.1 S — m’Z) I I Mr) =m) 2. fdxux 3. jaflx) dx = a jflx) ab: 4 f [u(x) i v(x}] dx 2 j u(x) 51x :1: ,1" v(x) dx m -=- i J}: _ x ' _ 5. f.» dr— m + 1 (m ¢ 1) 6. f ”(x) dv(x) = “(1') v(x) — f v (x) duh”) - dx “1 7. jm: + h_ Ir1111M + 171 I - dx 5. ~— fl 2 . J ‘5 «f; x __ a" 9' [a d): "111:: 10. fsinx dx = — 005x 11. fcosxdx “4 sinx - - 2 _ E _ sian 12. j sm xdx — 2 4 2x 13. jcos2 xdxw 2 -£- Sin4 l4. fx sin x air = sin x —x (:03 x 15. _[x cosxaivxcosx-E-xsinx 16. f sin x cos x dx 2 (sinzxyil 17~ fsinaxcosbxdx=—W—W1ag$bz) 18. ftanxdxmmln1cosxl=1n lsecxl 19. fentxair=~1nlcscx1wm lsinxl 20. _[ tanzx dx=tanxwx 2.1. fcntzx dxm—cotx—x 22. [emit-J (1/a) 0‘“ 23. fxe‘“ dx L" (c‘Lr/aZXch — 1) 24. [111x dx =x [111(x)—— E] (x > 0) dx _ l ~1£ 25. f a2 + x2 m a tan a (a % 0) dx 1 —1 a s 26. = ¢ . — , > 0. ‘t' 0 szrx + 6 ac (Zr/:1 (a 'C ) dx I 2cm: + b 27 1_———-= ———~mtan —— a " ax + bx + C ‘f4a02 — 15' “4116—sz (4m: — 52 > 0) 20x+bmvlb2—4a(‘ - dx 1 —,~ummm=—;_—1.n W mtg-+1.4... m 2...+b+m 1b'—4ac>0) dv 2 -2 27C. w—m— , (b —4ac:0) Jaxz +bx+c. 25‘9“"1‘17 MATHEMATICS 27 Identities csc G = l/sin 8 sec 8 = l/cos 6 tan 6 = sin films 9 cot: B = liter: 9 sinZG + c0338 = 1 {mile + 1 = seczfl c0128 + E = cscEB sin (ow B) a sin 0t cos {3 + cos ct sin [3 cos (ct + [3) = cos ct cos [3— sin 0t sin B sin 2a ~“~"~ 2 sin on cos 01 cos 2a = coszot — sinzu 2 l — 2 SinZOt. 2 2 c0330: — 1 tan 20L 5 (2 tan 00/(1 — tanzo.) cot 20'. = (cotza — i)/(2 cot on) ion (ct -i- [3) = (tan ct + tan [3)!(1 m tan 0t tan [5) cot (0!, + [3) w- (cot oz cot [i — l)f(cot 0t + cot B) sin (o'.— [3) = sin 01 cos [3— cos 0: sin [3 cos (0:— [3) = cos 0t cos [5+ sin oc sinB tan (cc m £3) = (tan ct — tan {3)/(1 + tan Of. tan B) cot (ct — [5) I: (cot oz cot [3 + 1)/(cot [3m octet) sin (m2) = i (1 — cos ell/2 cos (oz/2) : L/(i + cos all2 tan (on?) = i (l —— cos {ti/(1 + cos a) cot (042) = ifll + cos o)/(l — cos Ci) sin 0! sin [3:(112)[cos(ocw [3)“ cos (a "2- [3)] cos (1 cos B = (1/2)[cos (ct ~ [3) + cos (ot + [3}] sin 0: cos [3: (1/2)[sin (01 + [5) + sin (0t m [3)] sin 0: + sin [3: 2 sin [(UZXOL + [3)] cos [(1 f2)(0t — [3)] sin 0: w sin [3: 2 cos [(1/2)(0t + [3)] sin [(1/2)(0t — [3)] cos 0'. + cos [5 2 2 cos [(1/2)(oa + [3)] cos [(1/2)(0t - [3)] cos 0; m cos = ~ 2 sin [(1/2)(ot + [3)] sin [(l/Z)(cc — [3)] COMPLEX NUMBERS Definition i: [:71 (a + ib) + (c + it!) = (a + c) + i (b + d) (a+ib)—(c+ ici’)=(cr-~c) +i(bwd) (a + ich + M) = (ac — bd) + 1' (ad + be) a + if; = (a + ib)(c — id) = (ac + bd) ~l~ iibc — ad) C‘i'id (c+id)(c—-id) (-24.51?2 Polar Coordinates x = 1' cos 9;): = r sin 9; 6 = at‘ctan (Wit) 1" = Ix + iyl = m x + iy = r (cos B + i sin 8) = re" [rl(cos 31+ 1' sin 81)]{r2(cos 62 + i sin 82)] = rirz[cos (GI -i- 63) + 1' sin (81+ 82)] (x + iy)“ x {1' (cos 8 -t~ i sin 9)]” : r"(cos of) + i sin r19) Moos 01 + isinfia) rl . . mm = m ~ .~ 9 — - ‘ — 15(005 32 + isin 92) r2 [Lo'ai I 82) [- Isinifll 82)] El Euler’s Identity 6‘93 cos 8+1” sin 8 9"“: cos 8 ~ 1‘ sin 8 ,rt) 40 i9 é'.._—|:2_e____’ Sin 6 g e __ e—tl} cost? = 2’. Roots if k is any positive integer, any complex number (other than zero) has 117 distinct roots. The I: roots of 1‘ (cos 9 + i sin 9) can he found by substituting successively n x 0, 1, 2, ..., (k-— l) in the formula w = k/F cos(-9- —i— mm) + isinfi + 113600)] k k_ L k Also, see Algebra of Complex Numbers in the ELECTRICAL AND COMPUTER ENGINEERING section. MATRICES A. matrix is an ordered rectangular array of numbers with in rows and :2 columns. The element “it refers to row i and co'iumnj. Multiplication lfA = (aw) is an m X :2 matrix and B = (by) is an n X 3 matrix, the matrix product AB is an m X 3 matrix C = (as) = (limbs where n is the common integer representing the number of columns oi’A and the number of rows oil? (I and km 1, 2, ..., :1). Addition If A x (cry) and B = (bfj) are two matrices of the same size n: x n, the sum A + B is them X 1: matrix C = (cg) where ca- 3 021' + by. Identity The matrix. [ w: (or?) is a square :2 x n identity matrix where oh; E for 1' == 1, 2, ..., n and cry-2 0 for r' if, ’l‘ranspose The matrix B is the transpose of the matrix A if each entry bfi in B is the same as the entry ab. in A and conversely. In equation form, the transpose is B =A7‘. MATHEMATICS 23 ...
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