Structures 2 Final Exam

Structures 2 Final Exam - 1. Return to p.A:—l3 of the...

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Unformatted text preview: 1. Return to p.A:—l3 of the concrete sectio FINAL EXAMINATION May 2, 2005 CE 482 .Spluhon. ...... Name: . n of your handout. You are to substitute a wood floor for the concrete, with #2 Southern Pine joists at 16"_c.c. spanning north— sOuth in wide Southern Pine gluelam girders will replace the concrete place of the beams B4} 8%" elam girders. Assume a dead load of 10 girders. The joist ends will sit on top of the glu pounds/square foot for the sub-flooring and joists. Let CD = CM = Ct = CH = 1. 1. What live load per square foot can non-—composite 2X 12 joists_carry in addition to the dead load? You must consider bending and shear strength as well as live load deflection. 5:1 I ,F'xll #1:: L '3 I, A '5 [2A 3 6L.” M: 31.6)}.IZO: fichflin‘zhs = Ski-(£5 Fla: QWFJJJS’: HELOFSL . 2a§o= V8 xwwml a'w: gm :ts/fie saw—{Lg}: #90st —> LL = 3%.0 :5 Chafik SheuréT}en§+k3 ’25— V= GEHxQLE— H...” 1:. :53? “=5 *v 3 90rd 4» 2/3 a so. {.5‘ 11.15’2 mots: >> 537 IL: 63. Kt I z ham“. “.253 a main” E”: Lax/own. Wu: *+%’ 5 “'33 H’s/E" A 3* £4.33: Hahn :-—"v->« M -:. _ ___J__ r A 354% {.gxroéx 173 ‘00233 ‘ Has < 360 ‘9' K- 2. What STURD— I-FLOOR should you use for the suh— flooring? f5 o,<:. (1v. Aw") 3. Assume the gluelam.girders will he“8%"x:l9%" in cross section. What will be the actual self weight of the floor system.in pounds per square foot? Giulch weigh 8'iL:£‘ZI \ .E’fxcz;¥fr : 3lec/H‘ 5' i: E Zgorftsfic Eolfr we} k l-S’xfkt ' _' ' ‘ '3 “H. “5536M — moans/£1— : Lifoaulrzrsslozrsi IQ‘Q-C- Sub £409? -: alqz’lJ-Z‘“XISI$N G21“ 2 Fm 4. Select the lightest A992 steel W section for a column that is 5.5m.high, with no lateral ' support except at the ends, which are "pinned"; The column has to carry a factored axial load Pu = BOOOkN, a factored moment of SOOkNm about the strong axis, and a factored moment of BOOkNm about the weak axis; Use the simple expression and table.The frame is . "braced"; W _ k . I a D r 5 K2 =——£‘§55 = iao’ EL = 67” MM = 368§+-k.T: Hay» 23'“ 1“? Try on; RE ':. 6714+— 3g2,1.3 +-22;.zu.3:173o‘< ' CPI“) Use WJ‘h m; TM. on IS gems,“ : CONTINUED ON" OTHER 5. SIDE and bending_moments of question 4 are to be used to design a I. The length, axial load, circular cross section reinforced concrete column. Use fé = 4000 psi and fy = 60,000 psi. ' lumn. There will be zero moment at the bottom. I avoid slenderness calculations? The frame is braced. k 3 __léLLI£.. = 251g" : C(45'h4u« .25‘aA3H Use-‘26 titan co'wmn ’______,__.__.——r——-'—'_‘\_/ 6. Assume the answer to question 5 is 24". You plan to use #11 rebars as reinforcement,.and will have "interior" exposure. How many #11 rebars will you need? Ye ZLWté‘nLc—Wo 3 ‘7 ‘ ,- "’3jfif33fq"‘“ 0 7 Use 7; “atom = kmo’k = 5520"“ “(El—Er r— Sfa'o‘ ("a "7- mé—Og I‘cSi } 1T). 2‘43 A F (P' “ fl=.034’ lb : M, \75— s ero kid Aé/ 'I'T‘x ZLF- . 7 5‘}; ZLIL».03I=15.21.1 .1 1515’ | I N _ 1.5-6 z io’l raga” a U“? H #51 (They not“ fl?) s 13 #11 rebars. The footing for the column is to be lumn can resist in'the‘absence of the bending re. You are to make the footing 7. Assume the answer to question 6 i I designed'for the maximum axial load the co moments. The soil is a gravel and sand— gravel mixtu square and 1100mm.thiCk. What side length, in mm, will the footing have to have? VP = : .OLILIS —-> K» e 3.3 —-s» F;L : 3,2H—rqa‘nfi'lf: moo 14?: = nulo'kN ‘ A in: :2.u7;q=,r7.s'kPe 3’ A b; jlqo ‘ : ML _ x . A-E 03.51.66] “3 9-53 > 3W ( Assume the axial load,-Pu = 7000kN. 8. You will make the footing 3200x 3200mm in plan. llOOumn will the footing'be” Using.fér= 2819a and recalling.that h of the footing = strong enough in Eunching shear? ' ciafr iaofin _ = Hoo- ma: :ooonm ‘ 2. Q6 : —jl:’l—:a(2,t,30e5‘+ I) '= ZKOLi ML l1 Tr. (ax.305‘+:) = 5.06m = some.“ 1:. A5+5 = I§,z.ML Aa- 7000v('— x103: 837 525’» 3060 5‘23? WK“ ' ‘ CE 482 FINAL EXAMINATION December 13, 2004 Nme:... §?l“p?fi. - - a o n a o - Consider the example starting on p.7 of the timber part of your handout. You decide to use a similar floor system for a home you want to build, but will modify it‘a bit to save money. The loads per unit area and the overall dimensions will remain as in the example. The following questions pertain to your redesign. ' 1. First yon decide to place the joists a nominal 20 inches center to center. What Sturd—I—j ' Floor panels would you spéfidfy for your plywood subfloors? , s- r F ‘" ' I D Tabla? A: 33: §{:::fic.”a¥"5>‘_ 1U, NQEC) (Si/11g 40 39875 ‘5 roml‘aéle: 7or {3 “Z 0: Uggogf -rfiwo'c. I 2. To avoid some of the problems encountered in the example, you take advantage of the fact ' that the maximum loads will be primarily short term, and:design for composite designgin flexurelas well as for deflectiofi, What maximum bending stressiwould you expect in the‘ an i ‘ 11.2511? '4‘ l;5*II.ZFI(7.Zd-— )1 : 3L{L{ = ifliflcj : r 3 T7 5 mg v.3”, _, b .,*. Mtyaké’ngOZMa: rzeoomnt, M; Uni [\‘l Cf“ ‘k " — 5—“?- 00 l WEI: W ‘42,} : ‘IZOFS‘L J .‘ as. l— __ I ‘ FE - q7é—txlllli; : HZOFM, £243.?» W“ 3. Assume lhe moment of idertia of the composite section is 350inh Find the_short team "'live load deflection at mid— span of the joists. Is that acceptable according to Code? 71m :12 ,(go,,2 3 . 5 L3 - _'S. I [Z a __w__, : .14, :z _ my, '0 m 4 A“ 38915]: 3811 Lme‘ . 33—0 1 \ “filly - ._ '~ 3 r TR L fl".- \:' A A : hilkjo x zone: .Lm.mr'“1':__ 2.78: r0"3 o.\<l~ ll I ; g 4. The loads on the spine beam won't change, but it will be less expensive to use steel "beams in place of gluelam. To maintian good appearance, you decide to use structural tubes. Assuming continuous lateral support, what is the lightest structural tube7that can resist the factOred bending moment?7Recall fy of tubes = ééfisi. From 70? Furl MLi = V2 * (£ng .< :.c. + §OOxLL)x 201,1; : Swim—MP mp Ejf‘ . s , l“ ‘- J f, 2 aqutqg = 19:7.”4 USE tIZ'r LI-LE/H'. (5:19»G1n3) LUTI: [9.4 ‘5“ W” L!“ 7m (2: 25.3.:1413) JV: 21:2 Hag/ft. - ,m C: 4/ fl: , .l. :. éiiflif I _ m ' pONTINUED ON OTHER SIDE Continuing with changes, you decide toruse pipe columns? in place of the gluelam posts and to increase column height to 1.2 feet to permit duct work above the ceiling. For safety you also decide to forgo the live load reduction for purposes of the pipe Column design, using the actual AOpsf live load. What is the lightest pipe column you can use? Assumétkss 1 ‘1' ’1 I {H‘LIOOf/(LIUI r. e + 15“ 1- a] ’ 19 = 3‘23 Rip: Fm r~ W7 4-64 - b": I'rgL-Il'nl L: z; " / 6. You decide to use a 9 inch diameter circu'lar'h'aSe plate uner“ your pipe column and to set it directly on the slab. You note that the service load on the pipe column is 22 hips. How' thick muSt you make the slab to safely carry this load in the middle of the slab? hath-1‘ The subgrade is uniformly graded sand, and you should be conservative in “choosing the subgrade modulus. Assume fé = 4,000 psi. Anumek=200 Psifn ;)Pfic5€'3 E:- 303L011; » {53”}: — Tr)! 11: a” g: 2.11“ 2,30 — 1—! 11>ng A (—Z.i7+g,fl[8)l,glg : #68135: no 500d 5-1.: 2.2000 6‘! i2” it = US$12." Halo ML Jill-i -29~ar30* 4153“, (#3231 + mu (- I.q1+ egg) a em = 220?»; Tlxole‘: WL7 columns Shmulri newer he flamed on ‘Hu; sl'oxln / L; r95“, (Jig r1 + h?- — .-5 75 k) (J31.L|+r:LI—5.Ll>= *Z‘J’] r: LII-F" _ = " - P . assume Allowuiale = Z» l 755%} Ltosokj // : 237?:1’. \l 7. Assume the answer to question 6 is 10 inches. If you omit all control joints in the 7 slab to save money, ivhat spacing of'i'M rebars'would you have to use in each direction to control shrinkage cracking?- In 60 51'. other-non A Heed *2 LN? 12'? 901 x [(900 :: IFSMM : (,23” (In. > 110 xZSTLixJOZEF-KZJS'» Q0: {2x253 21170 : SOBMML/vq .331mz/51‘ In V0 (Ursa—Ion head 2/3 as math -> Use 5H1 @Cl'rcc or ZZE‘MM on.» M Lg" 25L: : [+57 “MHz/M < EQAE’OS‘ Doe: maul—coma! QJfL‘DTFdI’r'Q(TT°y‘ (hack A: > goc|glnjn = 8. As an engineer you prefer the appearance of a W section column.) Since small W section columns are not included in the axial load tables, you decide to use the lightest A36 steel W section which has aweakaxis radius'of gyration ‘4 that of the pipe column. Which W section will you have to use? See question 5. r of? Flirt col. hm", uni WG w or fl}: "(Z/ll { .kh .9 4! 1‘ Have a great vacation, cbulue cmefiwflflg and. )LQ/tuftll 66162591. Happy New Yea/t! CE #82 FINAL EXAMINATION December 15, 2006 Name=. . .3elytlieo. 1. You are to design a column 40ft tall, with pinned" end conditions. The ends are braced in all directions, and there is lateral bracing at mid—height in the east-west direction, but no bracing in the north- south direction. The architect can‘t tell you what the vertical load on the column will be, but he does want a 10% inch wide by 16 lame deep Southern Pine glulam column that is to be oriented in the most advantageous direction. Load duration is normal and service condition is dry. How should the column be oriented, and what axial service load, in kips, can it carry? F); 3 F3”..- E I: ’23:: {cyan/£5 _— :53“ i :I f \ 'c. ‘ ,J' «2» ‘. z \‘2, Elo‘Il-li :: I 3_LiQ"‘na\" _. 201in "\ fl iflxll' __ k m J "c3 ‘\ arr} “ L474 {Mums-J "- 52; ( Wins) ' 2090 . J, N ~' hr“ 4 w J :, ( rtdgéggged (or-5) E? F —_ .Lugxlsao A ., fl FF/ % : MW _ “5 SL7. “ LL“? > a; 4 1,7 ““ «3147 (PA) C .-, L847 ‘ _ —‘ ‘ ‘EQZ " 1.2 P = ,eq: x Huey. 22 272‘C 2. Architects are notorious for changing their minds about materials. Yours now wants to use a steel W section column. Assume the answer to question 1 was P — 255kips service load. Half the load is dead load, and half is live load. Select the lightest A992 steel W section in the braCed frame. r E; = 1,q x 255': 35"?Trq orich wifb vdeok axé: Lracsd gbwud-kfifikT“ KLy : 20’ W onsa' : 375* Q4,=13| #E-= £3.U'v»30%k mo 3095L (fi.u—a0j 7 W12:58:3Q_h n/vzzlio Kthjq)<20 O’H' (Pfisngg) U” WWW” (Maxi J: its? 3: was-Am) 3. You decide to use a W12X 65 which can carry Pu * 470 kips. The footing concrete will have fé = QOOOpsi, and you will use ah" diameter anchor bolts. Find the horizontal dimensions and necessary thickness of the A36 steel base plate needed for the 470k factored load. :24 u 105' = l2” From-{$.70 af- :tTcEI' seamen Cl +3»: H Minimum Slie.‘ NS mfg" =i'?”-> ‘: 2%an C <- A P r ’1‘“; ' on? T: ; L35) 2 new <<1¢H a: ya"('qy*52rl—.€'<f2)= .Wa e N: we Tuna = $1.3" < m" use Nr-HI/g” E: :7“ (ram) X = [ wixzn.;~ } x “To _ _‘““" {11.1 +12)?‘ macs} "7910” “1:; “gins-71‘": “Jr —l _ u M tum-=5“ “v /:.(w-.s_u>~3.7 w: y. 2 3.01" 3‘: WZF—HW‘W’ = "N7 An) m '4' IE—ah’ULi - . Ji= 3.10" £> 3'7 /_2.:u7_o____‘ 3 a ’ £13611?an "'7 '3 y. I Use F: l 714‘ >‘17"! I’-5“/g" = FE Bo a L432 « trsH'mm CONTINUED ON OTEER SIDE 4. The architect has again changed his mind about the material f or the column. He now wants to use reinforced concrete. The factored load remains as given in question 3, and the length and bracing conditions remain as in problems 1 and 2. Note the end moments — 0. cross section and its reinforcement using Design and sketch a minimum size. rectangular, You are to size the cross section so that fé = 4000psi concrete and grade 60 #11 rebars. slenderness will not have to be considered. Show_the bar layout on your sketch, as well as the ties needed. P .- k _ ZOxEL _ .. u 3 ‘ 2'43... U #70 ha/_ 'Bxaq ‘123539 4H hws: mar (95? P*~ r Ag: 11502.61 FrouijfO ? I [ LII—ID _ Jr, H31: f) (éonasy Ll) Jasnawnm ' Nil 5 W3“ we“; use £81m A; = LOI‘ 1:50: Mia-n: —-——> need 8 a H EFL; “ES 22”“, l -: t 1/2 CIE‘Argawr'er 5. Columns need footings. The soil is rather poor in your location, consisting of stiff clay (p.Ar-3 of the steel section). You will use a square footing. How large and how thick will it have to be in order to carry the steel column and base plate of question 3, with = 470k? (Hints: See the middle of p.65 of the concrete section, and try a footing Pu thickness = 2%: first.) Assume the answer to question 3 was that the base plate is 18x 18?. STGQ! C?t:5:r-T§a'cn : i and; : org” x .‘ifs 11.1 .2 {1.57" use 5:: L! k5; Eeo‘vnjemTcanc"* 49 : I3+9.& N r T fi‘r H Fm " I" ma = t3:8 x m‘ghfl‘y ma” (350.3175 mm} r7 5oon; 22 (a SGOMm} "fatal; F7; riggéomm hr23S0+HéFJ “e ' ‘ P f L k l_ ' J’ 375W“) 3399M G. = (.33i-LIGJXL.375+.LM) = .676 M" u. ‘ I70“ H'L") = zoqo ltN 1/3 : Llhsf :: W2 J-‘Pa. > inqo A I : 3-6‘33’”; use 3"“ “3m Fooling c. at; Zoqo{1~ ‘ azmszqowm— : WWW“ < “60mm OrK. C 3009:.WLnn or Wu :0" W» 20" or 2J~Gu " 600nmh and a 3m X313 footing. Also assume 6. Assume you will use a footing thickness that you found the equivalent concrete column of question 5 to be 366x 360nmn How many grade 60 #8 rebars will be required as flexural reinforcement in the footing? i=i§€el,3gm isfi'OOMH/l M = 30°10 MBZl a c u q x “fies. X 3 —— €07 [sally—fl 3m . a, A 2 Mi; _ L. a r. _ at} so S “W * fool-85' 3 HOW?“ P" 300035“ch ‘ ‘00 N EEGL : v! E l x 5| - .1 507 "a “F51Wflfifihlmrechy. : assognm' Howe a great hofiidag and New yea/r. Drums cmc-fiufity and Ire/twin {SELL/32x61}. FINAL EXAMINATION A May ‘5, 2004 I l I u— I - u u . Name=. 5.013139?) . . "3 "Youfiyorklas a constructioneng'ineer for a contractor. .An.8” thick reinforced concrete 7 floor slab is to‘he cast onto forms which will be supported, by vertical #2 Southern 'Pine 4:: 45. The form's'weig'h .25 pounds/ft2 service load. The vertical supports will lie 10' “tall, with no .bracing between bottom and top. They may he considered "pinned-'rgktop and bottom. In addition to the weight of the forms and the concrete, allow for 30 pounds/ft2 for the weight of workers and finishing machinery. What is the largest horiiontal spacing? equal in both directions}? at which youcan place the 4Xf_§_2__¥pu had better assume they ' “ill-‘be'we-t’ but” temp'étaiuwilh-se‘neeeleue ' a = 36 _.._. if: K” U—‘OOB ' 239‘}! ME UP W'SJ' tun c4511 51$ *" ’1 = More CD?“ Case-8 CH1,” _. * Ff = J'Hs—o.1.2:..é.r.15’-=(t7oF:L=r.51L.L-\(3.6) “ ‘ " " “ r E’ = MOO-.235. “Md L51 14:; rue moo. \‘ 2’“ m !G'\ 1:“? 3 My FCC—I. 3t? “A 3,, lg}? 39.3 Ea= "‘ L”) = ,367 kg; F? : j’g zazeo I! . _ /..: Ax :- ‘ i P M: = 63579 i z E: ‘zoq’f'c7 Cit-{i . E'2 \ ' t " 1\ Fallowqwd 0’1 on: ‘1' L ,zg kw“ OF (OLUMAIS - c . _‘ Flue/L WQGD (‘0 dr-W fl___,_,- 2. Suppose you weffifsfih‘s'm' diameter Piper standard steel pipe- for the 42: As. At what spacing‘ficould you put the pipes? «WW— “WWEB 3 L Act” "STEEL )Izg’u)4 LL) fig 7 r. ' ' J’J’ZEG“M5=C£{5A.J% Area='%,=7G/f§r"'_t SFQC‘IHE 22% 1,6,, H 3_. Acircular cross section“ reinforced concrete column is 406m in diameter and is rein- forced with 9 #11 'reher'fs and a #3 spiral. Factored applied axial load Pu = 1-670 RN. What factored moment Mu”in kNm can the column resist in addition to that axial load? Let fy = 601051 and f' = Alwi. Cover = 135”. ,. ’3 ,«e ,. P rm: I'Liaihz' fl: 2’0"! ; .07 W10 firZ-\_ A a T}. 37-: 201;"; A :. q..u -an _, ‘22 - ' ,. REDLA' lg .67? Zr mmmey LICTkJ-egy‘ J; .5._7‘. S : r ' IDLL . ‘ Cadwffif-SugdAu-iyfé' q Hl/ 0 GO 5' 3’ Am /conver:{on foamy- ' A h = ’ ' . . I. 5" = .7 ‘ I _ x z . P~A~Ir “ta/J = 0.75} W a He 7:" "75' 3m- 15 ~“3 329 MW AL ,- - A 7" ‘ 3 . ' . j‘ 3 V L . " {fiduh‘flw ’1 . lg»? ‘/ _ M). 3Kpr .i CONTINUED ON OTHER SIDE u: 11g 3 {9°} '5 1 Assess the answer to qnestion 3 is Mu = 23lkfim. Pu remains ‘ ' ht_5 14 ft and'the ends can be assmed.pinned.5Using the e divalent axial load method 3; Select the lightest A.992 W Sectiog that can resist/that moment about its strong axis 7 When also loaded with that Pu. ;; . Q 66' . Hui: 1'70 Inf-Etc?) “PW = 375' fur:- -.~ ER]? Try WY F“; 2 '315‘ + n14: r10 =- as k '—» wwx 7m " A thriq‘y; '70 = (530“ n> Uzi ,x 65’ <—U:¢ (p. H-ZS‘) LWOHJ '11:}? 3?3'+ rifle 51.70 = GBI 12—) was, 77 5. A_painters' scaffold is made of three #2. Son-them Pine 2:: 12.p1anks' arranged as shown. One painter plus paint bucket can be assumed to 'weigh 220 pounds, and to occupy 2 feet longitudinally- and about . 3 feet transversely, occupying ,the _ ’ A full width and a part of the length ' ' - . of the scaffold. If the painters and their buckets are crowded together near midspan as cl-osely as possible, so that each occupies just the assmedZ' 2L3", 'how many painters? and their buckets can the scaffold safely carry?37'essume the planks act in unison, that bending moment is the critical forceyfignore dead load and shear strength, but do "consider CD, C511, and. Cr" [3* - 'no Shaun F6 a 97:75: CD a has" C5”: 1.2 ct: NE 7»,- F' = [63'0st 5 .—. 3a “'25” "52' s. {2, 7‘ 3/fluiwb1 A” A .- My _/ g g I m aqum: = Tax JGSO~ 12.7: 1780 ytlbs. \ t __ _ Z I ,. , , _.‘___1 72 2'0 = “O H]:/ {gm-1:}; . ‘ R=—“—”‘_ M=LR.;3_J_ X1-13 no 2'- r” x 3' z_ . w —-- — ——-‘ HOx - .___. R Z l (2.) L{ a X I R r730 = 335—35: 43le - - ’ x‘j—zaxdw'zeso . A X = '/z_ [26- flac— u-m'] = 6.7:‘1‘ = 3'EG-J'htcr: + 3 theta} I l (x 7' M“! 3:12.“— ”‘W “i '6. Consider the floor system of par—lit of .the concrete part of the handout. What positive and hega‘tive factored moments,"in ENE, can beams 1133 resist?.«?=let.,f film-{Pa and f; = 28mm. ‘3 - Y _OL=(106:-:fr—‘gg—‘k‘jvzcu = 202mm fl 1 AL”: 6-333: 2330 ms- A: :. a: 339= 3100 as? f _ meow; M_:=I8f-\Z330‘H1szog:HIE—C: M I I I Hf: his—atmoo- um uzog- Jo“ = r-zootuw RN HF 1‘- SWCCT} telliv . t ...
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Structures 2 Final Exam - 1. Return to p.A:—l3 of the...

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