ENGR
BCEE_345_First_Midterm_exam_solution

BCEE_345_First_Midterm_exam_solution - O FBCEE 545"...

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Unformatted text preview: # O) FBCEE 545 " STRUCTURAL 335““ TE— CPEmFb/acao chaeve'DfésKaN) 4,5251» MIDT’EIZM EXAM SOLLlTnON‘ WmTee Zoo? (Dr. K. GALAL [Ombbm/ No (1):. ’ F: 2-29 KN a} [I] gel? we\3\n‘\’(w) = 0.4x0515x2q __._ 7'1 WNW 2. _ wLZ 1 12 12) .,. 2. mm WWW, MD “ "a" ("8" ”’6‘ L=Izm. 4/ Lf 2 Mg _._. (.25 MD + (—5 ML 3 [.26 * 1211.4 +1.5 was =II'52.‘<~.M [2] MAXKMMM amwm‘k‘j +w515n 915‘ “Hot/W311 V3 Aquiwccd M +01, Carmemdifl mmean Wacib ‘5 Hrbalmd M. = 0.313 , ,3 = «>908 (£0, #9: 25 M95) Cb == :::+$j *A : 7:0:Hoo 7k 660 = [1120 mm. germ "Ht (flu‘thbnmvw Q 2: ”I" 7% NI 186 fi‘Cb b = 4% Asbgj M 5 Co 'r , c 0‘ -o‘wmzo Vb r (cl (6%.) g: $5'k5‘12‘7'fi-H00 (560 2 ) H qJ—US’! : X M;3 Nam" = ‘?L\5{7’ KNMx [3] Mg > H”: :> WWW @513»er WIN be U434. (2) M4 = W2 — Hg: 1152-0 — emera— = 2063}; KNM WW we (Wsj 30M w J—mS’am/x 554C _3‘, 30+H51— ¥=Eézm 6 NAT: AM\ “w, sngézw CJ—c‘ £577,564, A} = AT = W! = Mm”; {£le (2:) -‘ 4%. D“ pc‘ 0‘3544400 — 065 Z Fo(C€S =' 0 :1) CC. ‘f C5 '-' ’T- A; = quQCLMM 1; momma £3 9 £3 >% 3 4:5 =Q§ =33) 0° \ :V \ o 91% oénacéejc 13%,}ch As {j : gig/$515 A; :- Asb + As‘ ., = 50127 + [01443-2 a . l ‘ ° 0 g5*‘433*25%[3.9p8* CHM 12°] ; 66175.2, W3- r c 046% ”loco 4!: I400 -—- 0.85% [200 kqoo , gctcuCcA 10—30 M :7ooomm 0,0 C = LIN-{‘23 WM‘ ; 925.1% _,_ 2q5.5 .: H71“? 3 0.45 * 053:3 ->& 25 [0.903 + a~85 * [200-31- HM (6573- 564-) rKN.m > M$ = H52 (5) Kw — Mr 7 \2 How 3 9,. I , 0 6am (Ha/l1) : ammr (Homage/’2) MW '5 T W1 ' 760/2 -: H2 5 KN M Mr : Wm WM >I‘1Mcr= ”“25 ”35 WM —~—-—@ ’ ASW‘IW : O‘Zjaqua Lt I" : 0"Z—-——"’¢—Z—5_ it 400% 750 =:: 750 mm1 #00 A5 flan/{Jeni > AZWB‘ _ ”m @ 0 OMUK mM'uvnum Loafi— Sch'mj ._- Sud- : LL100~1*30—1*H.3-—67(20l-cl:Hl‘qjsmm lO—BOM Sm >/Iu al.9— IH¥297=HI~3( mm M. 7/ ‘HGKM = [q-k25z5’5 MW‘"OK W 7/ 3° W -m, M A=/~' Io _. lo dc30+|13+r¥z 5625’mm Hes=6§1ej V A A =( Hum) Q5415 x7367: '22 = 178% 2 N/W < 30 N/w Umtww) ~-—~., a 1’1, :— 750 mm © 96:4 rtlrézw’ce/mod’ is H0+ 7165.454- f fczzr Mflo‘ ? 3;: 4’00 // w? .. ‘F-‘fi—“fi WV“ K ‘ 2w +(fn)1 ~fa ! j o o o o 4-7/5‘144 .71.. BGUMM h” 5; 2001000 ’ Z? = 2 7”! CC 4300/? s 1.32 2, 42$“? 50‘ X: 00/21 is 600.30. Hrsv ZS‘ 300 $54{,'L - 7- 54-{1 Mh—s Z, “M + “A; (oh-(Rod) l H Rd : 0.3??x 54.5.2 :: (95.0 Mk— 500 (1015)? z + m (zaoo) ($46.2 ~ IQS‘) 223,07.— MPA ”£2 gnaw? 7436.0! MP“ <£9 Problem 3: Answer briefly the following questions: [15 points] A. What are the five main groups of limit states? Acccwdhé? t}; A (23.3 -011 CL 8 J #99. L‘mi’r s’i‘al‘es are 1.— o Durabi Liz] . $7M: {6,535me . Ll Lit-mate, UMH’ stares . Sax/rectum?) \{mn-Smx-g . '3?chme iniaari’lj aY‘ 3910‘“ limi‘i' Starfi- B. What are the three types of RC sections (with respect to the amount of reinforcement) and their corresponding failure modes? What is the favorable type of failure? How to ensure that a well- designed beam will experience this type of failure? The ‘WWEL 1:39;; (”i QC ggchms am 1, . Unclearctnjewcgd “chm .4 4706M: (n Steal _-‘ - - BO‘QMCJLA sat/hm .4, failwre m Canc.6bv\el st—eel szr. reimfwfltd SLC‘TWL __._.:,. @filwz in (“($11 MW rei‘nsporcgcl , use Can, \ &ZO.OD35 _ H‘fié’ifi' fan/o racble. In WW 1% emsm Mi- 0‘ SLCJTM ‘3 CMC/K ei-reuw 2—“ C < C5 0;]: C GP A5 < ASMAX i W 25 > 2—3 T: 43 Ape; C. For the case of RC section that has a specific amount of As, draw on the following diagram the expected change in the strain distribution when having As'. What would be the expected change (approximately) in the moment capacity of the beam when having As'? 8CU K A . S wx—rg‘ A; \ watemud' A; MA;‘,A5‘ ~ 1_00..7!.|5 MAS ° ‘ A5 A5 ' , 85 The End Best wishes ...
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