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midterm1_solutions

midterm1_solutions - Middle East Technical University...

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Unformatted text preview: Middle East Technical University Department of Civil Engineering CE486 Structural Design : Concrete Structures 2003—2004 Spring Term Midterm Exam I April 26, 2004 Duration : 120 min Dr. Ergin Atimtay Name and Surname: S O L. O 0 Indicate your answer clearly. 0 Circle the closest answer to what you have found. 0 Show 21le calculations clearly. 0 You can—use your text books and speciﬁcations. (TS — 500 and EQ Code) 0 You can not use your lecture notes. 0 No credit will be given if there is no solution. 0 All questions have equal marks. 1. The beam is subjected to a distributed load of 3 t/m from A to C. Estimate the position of the inﬂection point at 022(1) from support C. Calculate the moment at support C. A) 16.02 tm B) 7.81 tm C) 9.26 tm @ 10.19 tm E)12.31tm K: J: (3)0...68) =-. \$01 if”, f -1 . x K = 344, lm l g T 1}th Jim“) .4112, ’rm 4.4.3 M: \$390.31) + J7: (3)0511) :@ 2. The beam is subjected to Pd= 4 t/m ﬁ'om A to C and P= 2 tons at C. Calculate the shear force at “(1” distance away from face of support B. Take d = 40 cm. ® 8.4t B) 11.44: C) 6.53 t D) 7.02 t. E)13.40 t1 P= 2 tons 3. For the under-reinforced concrete beam cross section shown, calculate the ratio of ultimate steel strain to yield strain. Use design strengths. Use TS 500, As: 12 cm2. C16 / S220 (according to TS 500) A) 18.05 B) 9.13 C) 12.77 D) 8.13 (E) 10.47 a: AS z Mm~m :qigCM 0.25 CA 10“ 0.85 (0.141(25) cl 8 d—SOcm 0‘85 AS 0.003 . . . “.541 14—4 “ 1" «.3 ‘ bW=25 cm iJEO—Jkskw givém \‘E : O‘OOZ" lgieéw -; O‘O‘l 0 Cl 27’: (ﬁlo- : 0.000155 O‘OOO‘ISS / 2110‘ 4. Calculate the design moment capacity of the beam cross section shown. C16 / 5220. Use TS 500 As=6 cmz. A) 414.11 t.cm B) 506.9 t.cm C) 286.7 t.cm D) 395.3 t.cm ® 338.8 t.cm Egg/£74 —: oM/lﬁt (Hm?) Fr, ASQYJ:LJ(1.64):M.L,HM For W 200m 1 1: —. 0.<3s(o,14)(Jaxlano)=\%.ogt>ttl16£ 1 c . d=40cm SO, d<20 cm FC 1 F)E 03510.10141‘q1a): Hlaé 20cm a :15.éécM <20 (m 014/ M( = l\.L,(, (’10—- 1:.115114) ""/ 5. The continuous beam shown is subjected to a design moment of M = 1300 t.cm at support B. The cross section is shown. Calculate the total tension steel area such that no deﬂection problems will occur. Concrete cover = 5 cm at top and bottom. C16 / 8220 d = 45 cm, 5') = 0.86, p]: 0.0135, K1: 44.9 cmzlt. @ 17.51 cm2 B) 18.93 cm2 C) 14.09 cm2 D) 20.83 cm2 E) 19.33 cm2 (4: 35%}.— 218.41 < KL: 993"“75 00 b=80cm l“‘——" A A A A B C 1<———>1<—————>1 h=50 cm 4 m 6 m d=45 cm " DQQ\QLL)0(‘ Fro\2\1m "-9 Deodo‘e KainQochA 68.9"? |4___.| Asp. y((t2w1):o.ons(15,195):15.H<m" 25 cm M) a 15.1%,: 1311(01341115‘) = 1413 10”“ M1; 1300,1113 : \ﬁ—Lcm 1?? m 7- A a; _ ﬂ... _ 1.11m Lal(%s_s) A32. A5‘1- A51: lSJGl—k- 1‘37. =/ 6. Calculate the moment at suppon line AB of the slab shown. Support width is 25cm. Pd: 1 t/m2 ® 31.0 t.cm B) 49.0 t.cm C) 53.0 t.cm D) 60.2 t.cm E) 67.8 t.cm Momtf\\’ £00013 ‘0“: ATreLx'Ton is independent B C , 0Q, vhan Atmenhoﬁk 7.21.»?SM 3m Fof ‘4 XLO'OL" A "L F——’| D -ng): o.oc,\ (1.0)(135) a 0.34 +.m 5‘“ — / 7. Calculate the design moment at support line C — D of the slab system shown. Take support width as 25 cm. Pd = 1 t/mz, Use TS 500. s A) 90.4 t.cm B) 73.3 t.cm © 59.0 t.cm D) 68.2 t.cm E) 46.0 tom {30: 9.0.25; 335M 3, 4m M‘: 0.0 ¢,1(Lo)(3}g) : 0‘53 L”. MIL: 0‘0“ (Lo)(3.?5) ; 0.35 1mm M\ M1. >02 No .5u€?,pr'rv1c~monl" Corre (Al-Ton Ms ‘ 0‘5 5‘ “mgitmjl/ 8. Calculate the equivalent distributed beam load on beam C — D of the slab system shown. Pd = 1 t/m2, Use TS 500. A) 3.83 t/m B) 5.12 t/m C) 3.26 t/rn D) 2.13 t/m @ 2.95 t/m mwcbkcl): )7n0 WAVWW‘7 2 .5 M i;- mm "' = ms + no :(Las HM’K/ 9. Calculate the additional top steel area at support C — D of the slab system shown. Pd = 10ml, Md(CD) = 68 t.cm, Slab: h=100m, d=8.50m, C16/ 3220 Span reinforcement: D101: ¢8/28 (str) + ¢8l28 (bent—up) D102: ¢8/30 (str) + 418/30 (bent-up) @11961112 B) 0.86 cm2 C) 1.80 cm2 D) 1.43 cm2 E) 2.21 cm2 ale/5220 «W's 3:1 £94,; 1.». owl/4.44 (Jr/m?) M‘L (CD) .. 68 1.8m B [AS(,CD)=..-€£——Mwa 4.65882)", (HMO-3821.55) 6m 1808:1801 wrntpmmmx dag/281101004: 42/1001”) A ‘00 4 0.5,; l” —. 34,; m7,“ AS: 015* 2 8 3.0 AAS: 0.65—3.94 =WQCN3/M/l/ 2.5m 4m 10. Calculate the cross section area of column (C) as required by the Turkish Earthquake Code (AY — 97). Slab design load including the weight of the beams is Pd = 1 t/mz. Assume a column cross section of 40x40 cm and consider weight of columns. Use TS 500 (1.4g + 1.6q). Number of ﬂoors: 8, Floor height: 3m C20 / 8220. A) 2002.5 cm2 B) 1305.8 cm2 C) 1703.0 cm2 D) 1646.7 cm2 63) 1754.4 cm2 TH‘UULO'Y Arcaa 1015 Mt N; CC): “944th 10.13 mxs 8 s—lortes +1.L1(0.L,,<0.4-,*’L.5+{N2)x 8,113,“ NACC)‘=162+U.% c 185.6919,” 1 _ 6.5144 0.8092) / l l 11. Calculate the maximum moment at centerline of span AB using PCA Method. Distribution factors: DF (AB) = 0.4, DF (BA) = O. gd=2t/m, qd= lt/m, Pd: 3t/m. A)3.l3t.m @ 4.6t.m C) 4.89 t.m D)3.40t.m E) 3.77 m ,Mﬁziﬁmxs): e15 4m - A M 1M?-1,7.54. TM(,/>xcs)-.2.vzs + ((+o_¢,)+ «.1; (H03) +M(Ag) s Z‘l’LS + 01258 + am: +M(AB)= than“ 12. Calculate the maximum beam moment at support B using the PCA Methodl Distribution factors; DF(AB)=0.4, DF(BA)=0.3, DF(BC)=O.35, DF(CB)=1.0 gd=2t/ m, qd= lt/m. @705tm B) 6.63 t.m C) 8.33 t.m D) 9.09m E) 8‘71t.m EPWM MSCAE):—\i{ (’l)(5)::‘r.\7'-l.m MS(@C)=T\{(1)(H} LLH‘M W(A%)=—3;<2)(s) = 415% MAE“): ﬁCng) = Holman AM 775 AM—.\.s Ealam +096 0.325 “Total -105 6.53 / 13. Calculate the maximum moment on column BD by considering the proper live load arrangement Use PCA method. Distribution factors; DF(AB)=O.4, DF(BA)=O.3, DF(BC)=0.45, DF(CB)=0.4 Column heights; l(BD)=3m, l(BE)=4m gd=2t/m, qd= ltlm. A) 1.36 t.m B) 0.58 t.m C) 0.75 t.m ® 0.61 t.m E) 0.88 t.m MP: 11; (m5) = 615 4..“ D .1. x u 1 M3: nuns): \ 1‘ \m A C OF N M 0134. (/3 8 2‘11}; W Ale \ ‘ ‘ “’ N“ D) l/3+l/LI galonce. To‘l'al AM:l.O:I>.LM 14. Calculate the minimum slab thickness h as required by TS 500. Beam width is 25cm. A) 13.03 cm 14.37 cm C) 11.89 cm D) 12.26 cm E) 10.43 cm 15¢Z£L " M 4m D‘O‘: M: i»:\.0 6 6m 15“. 400-25: 545m _ (Xt-OJS k) 5‘”; 1-9.2) 5m 6m / IS+£C ‘I (.o 15. Calculate the maximum span moment +M(CD) on span (CD), g¢=2t/m, qd= 1t/m. A)5.09t.m B)3.862t.m C) 2.08 t.m D) 4.09m @3375’Lm A B C ng'Sl-Im D l E 5m 1.5m 3m 2m _ F L: ~+ M v 4 8 16. Calculate the spacing of bars at midspan in the short direction AB. Use (1)8 bars. Pd=lt/m2, d=10.5cm, C16 / 8220, Use TS 500. (19 10.0 cm B) 9.03 cm C) 11.26 cm D) 13.03 cm E) 14.38 cm B 'CCA/QYJ ’—‘ OM flail 045—. 0.ol—.o 13“; 5.0.7.5: 9.15m +M : o.oeoCLo)(L..°rs) c 0.90%“ A z: ‘30 LCM + AS = go -— 9.6m uX/m ...
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