HW%2012%20Solution - CE4401 Fall 2009 Homework #12 Due Mon....

Info iconThis preview shows pages 1–12. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CE4401 Fall 2009 Homework #12 Due Mon. Dec. 14 For all problems (for this homework and future homeworks) assume a #3 tie and max agg size of 3/ ” unless otherwise stated. You may make any other assumptions necessary, but be sure to state them in your solutions 1. Make a list of all the references you have seen in class or doing homework to the AC1 code this list should include the classes through Wed Dec 9 and this homework assignment. Include the code section, code page number, the topic, and a brief summary of what the code says in that section if possible ( i.e. if you can summarize it in a few words otherwise leave this column blank). Do not copy word for word out of the code. Please hand in your ENTIRE list (i.e back from HW 8) and indicate what is new on the list. 2. Determine the development length required in the column for the bars shown using both the “short method” and the “long method.” If the available development length is not sufficient to develop the tensile strength of the steel, design an anchorage using a 1800 hook and check its adequacy. fc’ = 4,000 psi. fy = 60,000 psi. The clear space between the No 10 bars is 3 in. with side cover of 2.5 in. The top cover is 2 in. (as shown) and the cover from the end of the bar is 1.5” (as shown). No. 3 stirrups spaced at 7 inches are present throughout the development length. I 2" clear __ —______ l 1.5"C1631' lén 1" 3 3 l 1 4| 3" l 3. For the simply supported beam shown, draw the moment capacity (oMn) diagram. The beam has a total length of 16’ from center—to—center of supports and is symmetric about the midspan. The effective depth, d=21.5”. Assume that the cutoff location 3 ft from the support satisfies the requirements of 12.105, and the 1211.3 is satisfied. What is the maximum uniformly distributed factored load that the beam may carry. Use f,” = 3500 psi. fy = 40,000 psi. 1’14"“ 4. An 18 ft normal—weight concrete cantilever beam is subjected to uniform factored load which produces a factored Mu = 3500 “k and a factored shear V“ = 32.4 k at the face of the support. Design the top reinforcement and the appropriate embedment of 90° hook into the concrete wall to sustain the external shear and moment. fc’ = 4,500 psi. fy = 60,000. You do not need to design the shear reinforcement. Include cut-off of reinforcement in your design. Do not worry about satisfying 12.105. CE 4401 Problem 2. 14 Dec. 2009 HW #12 KEY 2.1 Determine the development length required in the column for the bars shown using the short method and the long method. if id is not sufficient to develop the tensile strength of the steel, design an anchorage using a 180° hook and check its adequacy. fc = 4 ksi and fy = 60 ksi. Clear space between number 10 bars is 3 inches, side cover is 2.5 inches, top cover is 2 inches and end cover is 1.5 inches. No. 3 stirrups spaced at 7 inches are present throughout ld. Development length using short method 0 No. 10 bars No. 10 bar diameter db := 1.270-in Cover values amp 2: 2-in aside := 2.5-in Clear spacing ._ 3 , between No. 10 bars 561" 'm Clear cover Ccl : min(Ctop ’ Cside) Ccl : Z'in Determine short method equation to Equation 2: "Use 1/20 Equation” if SCI 2 2-db /\ Cd 2 db use for NO_ 7 bars "To Be Determined" otherwise and lar er 9 Equation = "Use 1/20 Equation" Matieral strengths fy 2: 60‘ksi fc := 4-ksi > 12" concrete below \Iit :.= 1.3 Assume uncoated bars \Ire :: 1.0 Assume normal . >\ = 1.0 weight concrete Not more than 1.7 Factorl := min(1.7,\IIt-II‘6) Factorl = 1.3 Not more than 100 FactorZ : 63.2-psi 1cc Factor2z= psi~min 100, ——. p51 fyFaCtorl Development length 1 := max ———————~d ,12~in 1 = 78.3~i dshort ZOAFactorz b dshort Development length using long method No. 7 and larger bars \IIS z: 1.0 On center spacing coc := sc1+ db coc = 4.27~in No. 10 bars Minimum cover fl cover := mjn(ct0p,cside) + cover = 2.635‘in distance to bar center 2 CE 4401 14 Dec. 2009 HW #12 KEY 2.2 Problem 2: (Continued) Factor Cb := nfin(cover,£§£) ch = 2.135-in 80 K U will be determined by a horizontal splitting plane Area of transverse _ ( . 2) . 2 reinforcement Atr '= 2' 0-11'111 Atr 2 0.22m Number of bars ._ 3 being developed n '— Spacing of transverse ._ 7 . reinforcement S " '1“ Transverse K '_ 40A“ _ 041 . reinforcement factor tr " Sin ' Ktr — . 9~1n Cb + Kn. Not more than 2.5 Factor3 z: mn[__.é_ ’25) Factofi : 201 b 3-fy-Factor1-II's Development length ldlong I: max m-db, 12-111 ldlong = 58.41 Available distance L := 4.1“: + 3~in — 1.5‘in L = 49.5-in Determine if available distance is sufficient to develop full tensile strength of bars Check 2: "Length Not Sufficient" if L < ldlomy 5 "Length Sufficient" otherwise Check = "Length Not Sufficient" We see the available distance is not sufficient to develop the full tens/e strength of the No. 10 bars so a 1800 hook will be designed Hook development length Hook development 002me'fy . length ldh:= Web 1dh=24.1-1n Cover factor Factor4 z: if(cside 2 2.5-in,0.7,1.0) Factor4 = 0.7 Spacing factor FactorS := if(s > 3-db,1.0,0.8) FactorS = 1 Minimum Fact0r6 := max(8-db,6.in) Factor6 = 10.16-in Modified hook _ . development length ldhook .= max(ldh~Factor4-Factor5 ,Fact0r6) ldhook = 16.9-1 11 8 424362 1 42389 2 fiaé‘iamflMwfi s3 [04% mm. E‘wyzéa CE 4401 V Problem 3. Development length using short method and #8 bars only which is conservative 14 Dec. 2009 HW #12 KEY 3.1 Draw the chn diagram for the simply supported beam. The beam is 16 feet from center to center of supports and is symmetric about its midspan. d = 21.5". Assume cutoff locations 3 ft from the supports satisfy 12.105 and 1211.3. What is the maximum uniformly distributed factored load that the beam may carry. f0 = 3.5 ksi fy = 40 ksi. No. 8 bar diameter db := l-in Assume conidtions are sufficient to use 1/20 equation for No. 7 bars and larger Matieral strengths < 12" concrete below q] _ and assume uncoated t' Assume normal weight concrete Not more than 1.7 Not more than 100 Development length 1d := max[ fy := 4O~k51 = 1.0 >\:= 1.0 Factorl := min(1.7,\Pt-\Ile) fc Factor2 := psirmin 100, —-7 1381 f -Factor1 Y Determine capacity of (2) #8 bars and (1) #7 bar Area of individual bars Area of all bars Beam width and effective depth Depth of compression block assuming f5 = f Concrete strength factor and ultimate strain Depth to neutral axis Steel modulus of elasticity A7 := 0.60-in2 Aloft i: 2A8 + A7 b := 14-111 a ._ Aleft'fy 15f" 0.85-fc-b pl 1: 0.85 C ._ aloft left 61 E := 2900mm -db, 12in] 20- X Factor2 fC :2 3.5-ksi ‘l'e := 1.0 Factorl = 1 FactorZ = 59.2 psi 1d = 33.811 A8 := 0.794112 . 2 Aleft = 111 d := 21.5-in alefi = in ecu :2 0.003 Cleft = in CE 4401 Problem 3: (Continued) Steel yield strain Steel strain Check assumptions of steel yielding and tension controlled section Nominal moment capacity Strength reduction factor Design moment capacity Area of all bars Depth of compression block assuming is = fy Depth to neutral axis Steel strain Check assumptions of steel yielding and tension controlled section Nominal moment capacity Strength reduction factor Design moment capacity 14 Dec. 2009 HW #12 KEY 3.2 fy 5y 2: E 8y 2 0.00138 _ ecu'(d _ Cleft) eleft .= -—-——— slefi = 0.023 Cleft AssumpCheck :2 if (Eleft < 8y v Eleft < 0.005, "Not OK" , "OK" ) AssumpCheck = "OK" ‘ a1ch . Mnleft ~= Atefrfy' ‘1 “ T Mnleft = 149 klp‘ft (heft := if (AssumpCheck = "OK" , 0.9, "TBD") ¢left = 0'9 ¢1eern1efi = 133-8 kiP' Determine capacity of (4) #8 bars and (1 i #7 bar . 2 := + = 3.76111 Afight' fy , F . =3.61in afight 0.85_fc.b afight aright Cri ht : g 51 Cfight = in Eright 1: —_ efight = 0.012 right < Ey V Eri AssumpCheckZ = "OK" . afight _ . Mnright .2 d — — AssumpCheck2 := if (E 0ht < 0.005, "Not OK" , "OK") D dpfioht := if (AssumpCheckZ = "OK" ,0.9, "TBD") (bright = 0.9 D D 5 SQUARES 5 SQUARES m m z < 3 a m m o 4268 WWW! Wm; W 57%; 1&9? H“; W §Z/ 34‘fo :é‘mfi 59'? XI fiexgg :4 3?; ’!;_$fé>§¥i> h: , ‘MQMJQ’E {yflggfipifi'ifix‘E’xq 40;“ £8 {a :1 Mg? mire/rom- Fkl’isb ‘g" 9‘»ch @5§’w‘£€a\q $wém§€u§5§ 43F?“ ¢ :3 . ,4: gram fiLgfi {kc La,“ ‘ { :7{f+9 +: €++é¥ Q) % MM??? {)7 iv; gaff V“; 3%, aqf§42m U y “figmfi‘yg? DAL7 3&6 Cgagawgcvémgd! 'éfgfflm l 4 ‘ i .3, a; 249 Erna; Edi-9i" [pg/474:?) “>494 ‘33 *WMWV L 2/ 2.. /\ 9 > ‘ fl :9 {a 45;;“5; W“, Max wg my!“ 36% ‘3’ ma& ; M 3" a; ‘ ” j MMQW‘; 3L0 3?) : Mamgzigwag :ziag”>px:g’+§é§; 5.70; gf. M : Mag fxw3g§( 2 : 1'29? —, in? 53619:,” f g a, . L '(X'}\ « )a6 -9 ‘~ g m g z w g 75 «3.? x 2 “2,, \V C > c?) 143% 39%,? ‘3 H”? Conserxgméflffiéy 'me é daxfiw i we walk/{(67% (jg/Lack _%M.S ifififij Wag}? 5,5/gMfigaf Qfihabgjiy ATL wzngégavx W «L, ,_ H 2“ g .' Mww; a “WM : ng~% $222 M+ re $54§g%’@n $.m EmEoE .0853? >538 EwEoE summon-0| >5. 8+ "£9.21; a: x Wm =_>_ tam :_>_9 . : ... E 5; Nat >>I om a OCH 0 m H 59T46QOJ g omN NNN u 2%: £2.26 meow .umo va a n E3: 5:29 (NH) wawow 00 |\ LD Ln <1” m N H o o o m o mm H E 3. HOS» mu 42-381 50 SHEETS EVE—EASE”) » 5 SQUARES A2482 100 SHEETS EYE-EASE“ A 5 SQUARES 42—389 200 SHEETS EYE-EASEm < 5 SQUARES Egg“ fiafiiwafi ®§mm i 1 v Do me‘} MW?" \ i ‘ A\ -r‘ (1° rifle/\dwdaw gw‘%c “9 $5 fibukTXOld> THUR $224313 3" 3; {i9 {m{@§ ) E~ A , {3%‘336\A\Y $€~Eflé~§w§ f“ ywr’gyg, 12:1)993 WWWWW mwmmwmaMimfinVflflmwWWW-mww-xmwgmm my; I u / 4) R Z 23’;§@2Q {fi\ ’ Z"!\§’§h “:2 274 m 5/ , fi’fifix 9A. 3/154“ N\ {MM : gaggfiz :D 14:27:“ ~ é : igf‘fikgflgf‘ gfiagafixuf‘em $95 a: /’ SQUARES SQUARES OUAHES j: w o c N no ‘1’ N e I I w w o o m c N _ m :3 a: m N 9 e “2% fiaifimai “fliémmfi a 952,. E) éfi-Pwrf /S":?}”?&/L mafiaCzaae! 343.523» ‘ : mmg’wfl 3%“ = Mtg ” a) Lwég—QLQ a} ggtzgm » may/,1; :: 13,24; >2> b‘é‘a “gt?” €,en%ruc~}~&i¢§ g (‘O’XMI-c‘az 51551;; sgfiiflngw? ‘ \i 10;; g; - g f, . -Lgowgom“ S 7' x 5 abybg’a 10$!) ' Inga” (Qgé’wax fl’g"? rs 541§(%)’2;55 4}? “j ‘ a g s f — 15’4. 3%? + zizg’yg :- Z‘W" "zgfiwzg'fi \) ‘0 :57"? f Q r W' 3e. v: “39:? as»?!va -~ 92» (i a? 55 ‘31? L 37, 3} 0913’ 5&6 jr’mls a"; ;,[g,ari7 3511’"$£:~fig fat/‘2 WWWmmmmWAN,mWWmW.wW.mmWW._MMhWWnWM_w.H..WMW.w_W._.mmMWW.um.m.m,..,-mwmmMWWWMMWWWW?MWW_WWWWMWW MWWWWWW 5 bOUARES — 5 SQUARES 5 SQUARES can moo -N .Am woos: WW: gum 5: <r B 5; k: y 133,. w $257 :3? E w big: a , await, ...
View Full Document

This note was uploaded on 06/09/2010 for the course CIVIL 4401 taught by Professor Shield during the Fall '09 term at University of Minnesota Crookston.

Page1 / 12

HW%2012%20Solution - CE4401 Fall 2009 Homework #12 Due Mon....

This preview shows document pages 1 - 12. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online