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: > Ultimate Strength Design (USD) meétnod Ss strain io 9-003 concrete pe iy 0-865 / / Pe TL. «=| 7%. A LEeeee aw d | Na. by rare ania ti od igi L e —_ T TASTY OF ASEL SS a Rd roaheaiait ‘ -—— b——_} : \. | paleo] = STRAIN STRESS | EQUIVALENT STRESS DIAGRAM) | pDiseram BLOCK DARA) . mse ; | palanced = under reinforced condition Zr over -rcinpocrd = > i 7y | | 1 4 9 o e|. tf x steel yieldsx [| | wan wisttel oes >| —»>) - under cessed = sudden failure i not yield - excessive deflection | = most efFiciem, ~ corereté will fail (no =») - ctacks Cs iiizer puke moment resistance ) = pas warning Signs | capacity - oe steel (under stressed) rd - yield point istee) - concrete is under : ase ” wtressed | E % py 5 whitey’ ‘conctaant Ue strain ee qe. mPa pa : Net Tenaite cai" g wo4 fe 228 0.85, Lob Lea = ety, 0-05 np < ge < 5 0.5 2O2(fe5262-| ery < e4xo-e0s | o-e5 +025 an fic 2 5S. 0-05 Et Z 0-008 0-90 x Stept for eegular vections (Under /over reinporces ) ~ aa assume cee! Yields (C=T), 3. muse Gmai - po-B5fcab (a-4) | Adee Pd | | 2 9% Jestcb “ids oP TA B. Accume cbtel doer not yield a | 0.8550 (cB)b> pc [eetae) a 07 Br | im | op . © : | 9. a oP! ~)) Goo(d-c) Pi ete Ha t 7 = | f 0(d-c + §6 C to. fo = CUSED 5 Gr ay F sip FS 7 FY > correct assumption (#) 1 Repeat @-3 fe < $Y, wrong ascumption (6) ep es 2 HH : 7 oO. &t (<9 1 compare yor p t= €¢ = goe(a-e) ene * @, reclucton factor (phi) Role oF reduction Factor - imperfection in material ~ stress canied = assumptions iA design ‘calculations - design loadings # actual loads - workmanship ( Fron plans to actual! conctuction) pit PROBLEM 1 Determine the ultimate motnent capacity of the wean section chown pelow: Use Fic = 2IMPA H FY = ais MPa- : 2 if i | acs FED (3) 420] L NA.) fe} | As 7 SOB = 1844-25048 mn? i | 3-28eng in ‘g = 21mPO : fy > 415 nea aan | Bi > 0:85 ALSume SLeel Wielac ( Fé > FY) [ 3 Cnt L : | | I | [ as ASY seb (415), oh hag A “| 9.85¢¢b 0°85(21/) (320) | fl 038s a {34-21 mm JA) 2 157. 89 9m | pee GaLHO) , goo(see | c tT BB | fo = 995/228 mPa > assmpa /correct, ascumption | . i if oo f— gnome BL | ! (0054 oes OP ='0. 898 [FY p nus om b (0-865 av (4--*)) | :B 0-05 (21) (320) ( 4¢0--B) | x10°° [mus BBioa wT | 4 PRET T TRANS rV we a (i | PROBLEM 3. Determine tne Ultimate Moment eapacity @ balanced condition se* Z21MPA, fy¥7 415 MPA, 3-ZEmmgy dz 420mm , oF te beam in Provlem 1. Uce AS = 1587 @ Balanced condition | ( pe FY) fo = fy Wa c) ays c Ge0(420- 2) gig ec c= 248 2%emm |A\ b + Baomm , 6, = 0-85 Ey or esl # ety -. B= 0-65 qe ef > 0-85[4) = 2) 024mm Mu= 0-65 | p-85/al) [BI(220) (420 -B) Mu = 4. FO IeN= ! I : i 7] ¥ Stepy for Balanced - Condition les 1 fr Fy iTS woo( d-€). Fy 2. a= epi 3 Eg = € = ety B41 Gor 4. Mu = > aon mu = 2 (0 5 scab (4-%)) 4 Ht x Stepr re Irregular. Cectionc For T~ ‘beam CT (Assume Steel ‘Yielde) 0-BSfe Aconc > ACFY — Aeone = DESpe Ap phe mate (@ plange) “AR SAC f. a 74s (@ wev) — be alo.= Acone + (a7 tf) tw = Ae - AF ig ie ak ay Tm es ler 2 { || F tafost itey o 7} d ety ==> rr msg [age ae yal-3)] For Trang uler Aco = 7 OA) 3 ** ‘my = aC ‘BSF Acong,| ( (a- #2) oe. PPTTETC Nga at rrr 4 i? nd Bo@ | act AB lad ina le Mel et alla tastes aie 8 # IMregular. sections: : \ | ' T-beam (beam, then sla) in consiuction True~T-beam (lab & beam , sabay) - Rectangular to Tocam Rectangular beam (From beam design) | - T-veam to rectangular in| ot ye not okauy | PROBLEM 1 Determine tne ultimate moment, capacity ie T-beam chown below. Use s'o 3Z24mPa, fy= 415 MPS, fr 0-85 yf we eeuuuuuE i Goo" Zin Peonoreaey f'c 24 mea - 7 ahh CIEL TLL | po | |A%E AIS, mPa | br 70° 85 Dee dF 825 mm z Noa q JLejyy ft \\ As = -Zopomm 2 & 2s" 3 “ $= 200mm? 7 ; ce ae < 3 e +5 me yy goo"™ > ea assume steel vies eee rail | 5 | { 0-B55'¢ Acone = ASFY ebay A _ 2000415). | I conc 9-85 (24) | maa Acone * 40,080. 27451 mn* Bl Ag = G@00(100) = G0, COD mm * At > Aconc -- aQ<|0o (@ Flange) 3 p= @oo™ iT vevevevuvur Acont = ao gy Fe in , _ Aask Fe ce e - a= Goo =o , as @F- 810mm Bi | C299-F74nm 6) | | | . | oe sank) 7 (nee EE Bas 7 0:009205 = ra 1} pr 4 fate, fer 1844 . 219 MPS 7 HIsimea oe ey it EE = 0S, 0.002075 B | 4 mu 9 | oesge heoce ( 4-4) k - 0.90 | 089(24) B Mee Mu = 217-448 kN-m " €4 2 0:005 -- g 20.90 aS sf] % SS Pe ere eee @ Stee) [tension AYY = Aytnaya Se @0,000(50) + 200 63) (‘100 4 0-5fE3) | (0000 + 300[E] Ni ¥ = 70.193 mm {x} | = Basss(a-y) Mu nee | son (22102) \es-w a Mu = 2UB- 489 KN-m M if Any oer PROBLEM 3 Determine te required arto oF conccte BNA AC in valence condition of the veam choo in Probiem 1% 2. {i | @ balanced \ i pond Acones Ar + tw(¥) | " on G00 (d-c) _| = 1@@, 0m +, 3° gee tare? sy | [Rewes_ 79,990 (48 me), [B) G00( 325-C) ns 0-85,5'c Acone > Atval FY erst a Acbal C= 192-118 By a uu 4 Ta* GB Ae pai > 3862-869 am™ | = 0-85 /5) A = 103 300 mm y= 3 - 300mm a ; N ‘ i | { yh A a Tachi s a er ae a a DESIGN OF RECTANGULAR BEAM Rus coerricient of redictance - MONA ve 1 toG@ Mudes = designing factored moment = - 2-8 (best since = 0.00 (under tension controlled) ductile eaiture) [ 2 AR ae = pod] P+ steel ratio ba * Ductility requirement pyri frin = Prega = Pmax — usepPreadl Preq'd < Prin whe p> pmin jncréawe the «ection or ch Preqia = pmax design ar doubly reinporced ( DRB) Prega > 0.06 42 [ to fy Bem Teg) iz MG fy. 1 | Brie Prin = ae out not ea thar a ad inas madalec mag_govern hes het : I pmax - | 00st 5 Es 0.004 5 cr pa Derivation For (max = =| : 6.003 7 To Co dee | eet. IS e bos 0004 | 9.854 (.¢ Br) B= ASmax " te +o7d- 0-85 f¢(SaP) Br * Prax ¥A fy | | = at | ‘ d= er Pmax = [os 6] czy (transition) 0-004 0- 004 ( transition) 0-005 (under jencin controlted) 3} - C = 3A (tension controited ) minimum pacing Requirement 50m" $ 2 ade 4 dogg. PROBLEM: 2- Determine the Fackored derigning moments op the \restrainad wean snown. The beam rection i¢ 4oomm x OO mm. \ D= lorN/m (excl. beam wt) p = Wokn/n L = [4 kN/m (uniform) ee Ee ae oe ee ee m | an @ lert support, 5 4 j Wor = lot 24(0-4)(0-6) + 15. FORN/M SX Py RN] | 3 i yw fee ECOG xr FS | L? e2 | 2 7 af pe ee =-13099125 FF sm ge XS 14 x 3x i be aext - | 15.96(8)? , yacey? | Mp 7 - 85- 30114583 kN [Bl muive =~ 74.007 kN-m [El mu = 44D > 4461 = = pa.spsuodz KN-m Mu = 420 + 4GL = 4-288) + UIE). =.-221-9000 417 RNTM v Mui = 7221 90 kN-M @ righ pport | Shai jeg e* * ay (xt 5)?(8-x) | age, (RD, mp = 90-0B02083 kN-m [FE] M+ F4COT eNom El | mu = -4D os aE = 120-0812292 kN-m i Mu = -2D & Pol = /1-2{E} +) OG) + 227-S302917 pn-m - ~- [eo Migs) = 227-5362017 Nm | TET — 4 | ” @ mid Support re t= 48)?! 24 24 ML = 97-393 KN-M PUTT RTPA usirg Tree- Moment Equation P okN/n Poy m wy |, el a7 5t* 5” x 3 bt 37x - 5 ~ GAG | ( 2X59) [ 92 _ (osx)? | dx = 100.9375 L lb 8 } | — 3 t Gab of x(a) 2_ sx)" = 6B.9625 fee [s coma" | 68.96 15.F0(8)? $ iuMa 4 8mMB + + 08-9025 7O lwMa + mg = - 2080.2425 — eg#T | 15. (8)? BMA 4 (OMG + —a + (06.5375 70, BMA + IOMB = - 2123-8195 —w eg#7Z MA = - BS- 20N45OS KN-m FAL y postive Mpei= -90-03802083 kN-™ |B) @ whole section / beam Me -O | Ra (B) + MO-ma - 15-FH(BX4) - T(4)(F7( 13 x9) =O i - Bl + Bl + 5-70 (04) + Y2(@)(3)(1) A > g Ra = 03-577 kN [Cl 2Pv 70 Ra t Re ~ 15-46 ( B)- Z(G)(3) =O pe > #1502 IN Bl , DESIGN OF ONE-WAY SLAB ) Minimum thickness oF Slab bs - Simply supported 7. “Ye 22 oe way ) - one-erd continuous Lay bye 52 4wo wary - Botn-end continuous = Y/2g f } - cantiever “ho w) cone. cover = 20mm ) “ y Wzownee , moc ip x For sy otner than 42omPpa, modipy by multiplying w/ (0-40 =) ¥, For temperature bare | fy < 420 mPa Asmin 7 0-002 0% 3 DteAg fi 2 420 mPa Asin 20-0018 ( “7°/4y.) 24 PROBLEM 1: Design a cinpty supporied 2.@0mM one-way slab carrying super - imposed loads . 1 DL = 2 kPa \ wor [2 4 24 (0-108) ltr }- 4-52 kN/m } LL = 4.80«Pa Wi > 4B0Ka (Im) 4-BOkN/m ’ fc 21 MPa 2 ot i A : Veonc 7 24 KN/m? (2 Mudes 7 PH Wy = 1-209 + We (Ho) ’ fy 7 295 mPa | | Wu > 19. j04 EN/M. (Factored, joad ) Main Par P > [2mm + Muges + (h073kN-m Al c i : ) Temp: Bar D = jomm tod c Mu des [A x10% as DP baz . 0-9(1000) (19)? r ‘ea pat temp bat hey ge 2.30RU mn fag 0 4 [+ [sR | 2) | 2 sonelae][r-fr- 28] , ; ] freq * 0:00903903@ (Cl bt 20 [0-4 + bo ee, 295 pri! 7 =. 0:004@S937 __ 21900 [240+ 300 AY EAD : =\—goT hoy alt ’ 140 |. b-00s0909_—* We t= 10 OF mm gay 1O5me fy + 4 fel | minimum design max * S| oes ; P| thickness (vys) | 8 of clab te [ oes S (09s) | d= j08 -20- _— diz yolem fax * 0°023645 ) b= 1000mm ins = fina; ) p * 0-85 erin pnt? fr " "P= Freq * 0-007 [C) Main Par: A 2 ba | a __ Ale AS (reg | + Atregd = B10) (79) be Zz ) AS = G03 53mm” 5 > ——#— (i008) fe) AL min * 0:002 bt | 6 = 1B. 3920560 mm = 0-002 (1000))(105 ) AGmin 7 210 men 7 i i vay UF (8S am i( bund dobn, by 5) code! SZ BbHo} 1385 4 Aregd 2 'ASmin| Wy iy / td Sz 450 4so ' ro mG Use jamn B main Car Spaced tesmer 0.C) i poy rey 4g [ Temperatufe: BAM fs { fY < 420MPA 5 Anin = 0-0%ot > Svemp | _Ale | Py Atemp AS temp: > o.002bh kl de | nto)? | 0-002 ((000) (105) f Stemp = — Es (1000) Aiemp = 210 mn? | 210° i I | Stemp > 374m! gay Syemp = 37tOmm | | . [kab | fal Code Sxemp 450 so sie a S vem Se es oy \w ob ‘bo bhot tate cec 1 papers =.» [FS USE [1omm B Temp- Bar] Wace GBH Design / Derails # T , 1200mM, + e ] ~ : weet | 2 3990 | be | 4 ; Ly jonim temperature TON zomm | A: bar spaced @. + fe eb 370mm |0-C. iy | ‘| \- i aL roy [2mm p je parr = i | } spaced @ lasmm) OC. vedoul ia s D D DD) rd ID SHEAR IN BEAMS = ( STIRRUPS > Vn *?Vo + VS Yp > Nominal Shear capacity Vo 2 Shear capacity oF the concrete waar Stiefups Tre peam section | ive? Pa. Ve + Shear capacity oF the ctiups | vu = Ultimate entar Copaciby | [vue 7045 _| 7 | | Suen Av fyd AV = Area jn Snear | vs (too leg! oF sctimaps) —- - default Dervatio + oe {Po 2) Orgy WUges = Vucap A | Pa ¥s) + P(ve t¥) | +) Als ET aisvegara | _ Nudes iy, vd | Ys (actual) gpg % ji Wen Vic edt . 1 5Y Aas | | Zz BYC = Stirrups are needed. { 2 Ae vi vu 4 Be (0- 5) Stiups are not | needed! | $°3.s PS Tey KI | Ls Shear Crack __ Earthquake Load i = Cole: a | LA | ai @ a distance | 2d From the T we | pace OF cuppor’ d | fafa I: | | & 896 \ i 2 | 24.Yeneeves @ vetigh, aways) 1|@ Pomm start! 3 Us¢0mm code * Vue > -2D4 LoL t LOE or 0.5E vs 7B Yfe A Vs 4 Hy fe ba vs > $fe- wel : 2 4 go a (a) increase Ne beam Sregd = Z Syegd <= a SECTION (b) increase concrete Sreqd & GOOmm Sreqd = sadmm sirength (c) combination oF Cal) F (ep) PROBLEM 1. Determine the theoretical | | spacivg oF (omm | straps), FY 2 27SMPa he “ge > 24mPA . For ‘a 3m can firever beam jared as spon’. ie aku ye Po » |OkN 4 ae We, = @kN/m } Pi? BEN PET ryy yy yyy 325 | Festa S|" jb tads dp a pj L | Lt 4 wuz b2(L4) 7 $O(@) = 20-4 N/m) Py= $2(10) + 1@(B) = 24.8KN H @a-q Ru = 26 4(9)+ 24.8 ary = Ru = 104 kN 1 ; Ru- 5 4(0-325) + Yue 70 1 { _| Vue > 104 - 20- 4(0-325) | | HOH) Yug, | okt “Vue = 98-42 KN | eS a 0 Piet tt || sheer Diagram | | cr P(vetvs) | | lip required cpecitg oF tiers Is headed * Wc hy _ Vee i | L Vs = + ~ Ve ++¥e—= te bd z 4, [aa (225)(250) x10 3 | skewed? | Ja (250)(325) x10"? Ch me Vs =: wp 29 Al AV {yd Stnieo = apy uw Reta (235)(328) x10” Ay, [Fines = 230-58mm | l Fife od 60-89 < 13268 j | df2 = 3285/2 7 (oz-Simm 84 = econ [ea aR vr aeign PIJIISAALISILELEE CCC PROBLEM 3- Same W/ Proplem 2 , but w/ an additional earthquake load: LLELUGGGUGGN op) kN Jn Wu = 50) ( oy een O rr e mt eS | ww 7 (Ruz 23-32 KN [A tT pow | | j ia | || 283% Wie des = Wao.» + Wee) We. | | i > 191.25 4 29-99 oO T } We des 7 154.5, KN [Bl Wied _ | | (+ governs ) _ We des TF aa Mein Ee = ee ke av syd eC) : : Sineo = —Ve - 2T (amy say xi? ante] (295)(378) "64S Give fs | en | 7 ha Wey won't govern ~ xo? | vg = 135-02 EN Wey, =15D- 50075) Stheg = V72Fomm | Vugy 7 1125 EN —y from Spear diagram oF proot2 © 2A @ code veel | @ distance 2A + 70mm! ve $$ fe ba | | Ay = 37°/4 = 99.45 —¥ USE | | tafest | | i {35 < 312.794mm section ic okay! Va < 156: 296mm | 2 7 Ve £ 5 (OF (239)( 378) x10"? | | 300mm i dj: = 295 /2 = 1873 : 450-50 ; : Whe 2 798 x 8 Sregd Zap mm | | Qo | I pes % lao (238375) x10" i | BBme | = 0 L 24D or = R412) = 288m | | ‘ | | USE (99 7Smm , ay 90mm Use | 2mm straps) 1@ somm , B@ Oomm, rest @lesnm oC. From the Face op Support to midspan beam PPPPP PPP OP eeeee NS Y—IEE'S OE DOUBLY). REINFORCED .BEAM (ORB) a PPPS oss eeeeeeseuyyeEyE > 4 >» _ a © le7, al | Cl orc, ai C's or C2 AS _ Dte-a = % . + rd 4 AS Aisa. Pee ea Bp Ary SAY Asi Fy Asa ey, a an b ; } mu = mu, % Notes : ] | f r ae = Mur + Mug ie r+ 4 Mun= B Aca sy (a-a') | - AcE (Od me MUL G0. 8S¢cab (a7) Ci = 0-895 ab | rf fis C2 = ANSE) onAsss | i Th > AS FY | Tes AG gy $b — TENSILE STEEL Aways “vieLo! bid egy deep 32 i 1 uaedbsk eT tC, +6g —» er | equation | C2 =T2 bof | | ici) > Th | + mucap > ¢ [> ete 20 (a) Rep late | comcrete + Mucap = B[ awry (d- 45) ’ ning (2a) | see Cention) Ni AG FY a +a a at So? oes OB, El pace i Lat 4, | bg. rade att + gs + elena he peer ' Hee 5.00% (ed 3 — 2x10F | 4 s Filio | 3 pl\ { i § 2 gente | d | ! | =e Ne Phat mAb Gb ig , Tr BS, Feb 7 I Acfy = p.B5fe (cpio + +s [emcee Ly Ags (cel comps does not yield’) bei fbi key roe } 
