Reinforced Concrete Structures

Transcription

1 School of Engineering Department of Civil and Bioytem Engineering Reinforced Concrete Structure Formula and Tale for SABS 0100:199 John. Roert Septemer 004

2 Although care ha een taken to enure that all data and information contained herein i accurate to the extent that it relate to either matter of fact or accepted practice or matter of opinion at the time of pulication, the Univerity of Pretoria and the author aume no reponiility for any error in or miinterpretation of uch data and/or information, or any lo or damage ariing from or related to it ue. Thi formula ook i intended a an aid to tudent during tet and exam. It therefore contain a ummary of only the mot important deign equation and doe not replace the deign code of practice SABS 0100 to which deigner hould refer.

3 Tale of Content 1. aterial Propertie 1.1 Concrete Reinforcement..... Limit State Deign 3 3. Analyi and Deign for Flexure Relationhip etween train and neutral axi depth Singly reinforced rectangular ection Douly reinforced rectangular ection Flanged eam Elatic analyi and deign Deign of Beam for Shear 9 5. Deign for Torion Bond and Anchorage 1 7. Deign for Serviceaility Cover to concrete aximum clear pacing of reinforcement inimum pacing of reinforcement inimum area of reinforcement aximum area of reinforcement Reinforcement at ide of eam exceeding 750 mm in depth Span-effective depth ratio Deign of Beam Effective pan length Analyi of continuou eam Flanged eam Beam with compreion reinforcement Curtailment of reinforcement Deign of Short Column 9.1 oment-axial force interaction diagram Axially loaded hort column Deign of Supended Floor One-way panning la Two-way panning edge upported la Flat la Punching hear in la Deign of Slender Column Braced and unraced column Effective length Slenderne oment and force in column Bi-axial ending Staircae Deflection and Crack Width Pretreed Concrete Sign convention aterial propertie Elatic tree Ultimate limit tate...46

4 Univerity of Pretoria 1 aterial Propertie 1.1 Concrete odulu of elaticity E E ct, c, 8 f cu, t f cu, 8 (1-1) Poion ratio = 0. (1-) Coefficient of thermal expanion th = ºC 1 (1-3) Unit weight c = 4 kn/m 3 (1-4) Shear modulu Ec G 04. Ec 1 (1-5) Total long-term concrete train c, tot tel () t cr() t h() t th() t (1-6) t Shrinkage train t e h() h, 1 where h, range from 0.10 to (1-7) f c Creep train cr() t fc C() t () t (1-8) E t Creep coefficient () t e c Tale 1: Characteritic cue trength f cu,8 (Pa) Secant modulu of elaticity at 8 day E c,8 (GPa) Average Typical range where range from 1.5 to 3.5. (1-9) Elatic and creep train el cr f c () t E () t eff where E eff Ec () t (1-10) 1 () t 1. Reinforcement Tale : Reinforcement type. Reinforcement type Symol inimum characteritic yield trength f y (Pa) Hot rolled mild teel R 50 Hot rolled high-yield teel Y 450 Cold worked high-yield teel Y 450 Welded wire faric FS or FD 485 odulu of elaticity: E = 00 GPa For reinforcement area ee Tale 31 and 30.

5 Reinforced Concrete Structure - SABS Limit State Deign Tale 3: Partial factor of afety for material m. Limit tate Concrete Steel Ultimate Flexure, axial load Shear Bond 1.4 Serviceaility: Deign load effect Deign load effect Q f / (-1) Characteritic trength f f Tale 4: Typical * partial factor of afety for load f. Load comination Self-weight load Self-weight load 1.5 (0.9) Self-weight load + live load Self-weight load + live load + wind load 1. (0.9) 1. (0.9) Ultimate limit tate Impoed load k Wind load f n k m m 164. (-) Serviceaility limit tate Self-weight load 1.1 (1.0) 1.6 (0) 0.5 (0) 1.1 (1.0) 1.3 (1.3) 1.1 (1.0) Impoed load * See SABS 0160 (1989) for a complete dicuion on load and their comination. 1.0 (0) 0.3 (0) Wind load 0.6 (0.6) 3 Analyi and Deign for Flexure 067. f cu m Stre E ci Paraolic curve E ci c0 55. f cu m GPa f cu m f cu in Pa c0 Strain cu = Figure 1: Paraolic-rectangular tre-train relationhip for concrete in flexure.

6 4 Univerity of Pretoria Stre f f y Tenion m yc f E yc E = 00 GPa 1 y f y E m Strain Compreion f yc f y f y f / 000 m y = Characteritic yield trength (in Pa) Figure : Stre-train relationhip for reinforcement. Tale 5: Yield tre and train for reinforcement. Reinforcement type Symol Tenion Compreion Yield trength f y Yield train y (10 3 ) 3.1 Relationhip etween train and neutral axi depth Yield trength f yc Yield train yc (10 3 ) ild teel R High yield Y Welded wire faric FS or FD The value in thi tale include m = Strain in tenion reinforcement t dx c (3-1) x Strain in compreion reinforcement xd c c (3-) x d A A d c c N.A. x Neutral axi/reinforcement depth x c d t c (3-3) Cro-ection t Strain Figure 3: Relationhip etween train and neutral axi depth.

7 Reinforced Concrete Structure - SABS Singly reinforced rectangular ection cu = f / cu 045. f cu m 0.45 f cu x F cc = 0.9 x F cc / d N.A. zd A F t F t t (a) Cro-ection () Strain (c) Paraolic-rectangular concrete tre lock (d) Equivalent rectangular concrete tre lock Figure 4: Equivalent rectangular tre lock for ingly reinforced rectangular ection. Neutral axi depth x( 04. ) d05. (3-4) where oment at the ection following reditriution = (3-5) oment at the ection efore reditriution with 1 (3-6) > 0.75 under normal condition > 0.8 if the cro-ection varie along the memer > 0.9 for tructure exceeding 4 torey For tenion reinforcement only K K f d (3-7) where cu for 0. 9 K ( 0. 4) 0. 18( 0. 4) for 09. Tale 6: Redi tri. (%) oment reditriution and limit on neutral axi depth. x/d z/d K (3-8) K Internal lever-arm zd d (3-9) Required area of reinforcement A (3-10) 087. f z y

8 6 Univerity of Pretoria 3.3 Douly reinforced rectangular ection d cu = f cu 045. f cu d A F c / x = c = 0.9 x d F cc ( 0.4) N.A. z A F t (a) Cro-ection t () Strain (c) Concrete tree and reultant force Figure 5: Douly reinforced rectangular concrete ection. Compreion reinforcement mut e provided if K K Required area of reinforcement A K K d f f ( dd) A yc cu K f d f cu yc A 087. f z 087. f y y (3-11) (3-1) K where zd and x dz 09. (3-13, 14) Tale 7: Condition wherey reinforcement yield. Yield trength f y (Pa) Tenion reinforcement yield when x / d Compreion reinforcement yield when d / d d / x = 0.90 = 0.85 = 0.80 = Compreion reinforcement yield when d f yc 1 ( 0. 4) d E cu

9 Reinforced Concrete Structure - SABS A 15 d 14 SABS 0100: d f cu 30 Pa 1 f y 450 Pa A 11 d d d 10 A xd / = d (Pa) xd / = Suitale for deign A d Figure 6: Deign chart for flexure. 3.4 Flanged eam Neutral axi within the we (x > h f ): Solve w from h f w 045. fcu f h fd 045. fcu w w dhf (3-15) Simplified deign for x > h f with x = d/: A 045. f h 045. f 087. f cu f f cu w w A y 01. f d 045. dh 087. f d05. h cu w f y f (3-16) (3-17) f cu 0.45 f cu d h f x = 0.9 x N.A. F cf = 0.45 f h Fcw = 0.45 fcu ww w z 1 z cu f f A w t F =f A t Cro-ection Strain Stree and force Figure 7: T-ection with neutral axi within the we.

10 8 Univerity of Pretoria 3.5 Elatic analyi and deign Cracked rectangular ection Find x from x x E n1a ( xd) nadx where n (3-18) E Cracked tranformed econd moment of area 1 3 Icr x n1a xd na ( dx) (3-19) 3 Stree d A f cc x I, f n f n ( d x ) t ec, f n x d c (3-0) cr I cr I cr ( n 1) A d A na x + = N.A. A na na c (a) Cro-ection () Tranformed ection Figure 9: Cracked tranformed rectangular ection with compreion reinforcement. Uncracked ection Find x from x n A x d h 1 x ( ) n1adx (3-1) Uncracked tranformed econd moment of area Ico x hx n1a xd n1a ( dx) (3-) 3 3 d ( n 1) A A na d A x + = N.A. A na (a) Cro-ection A ( n 1) A () Tranformed ection Figure 8: Uncracked tranformed rectangular ection with compreion reinforcement.

11 Reinforced Concrete Structure - SABS Deign of Beam for Shear V Step 1: Ultimate deign hear tre v (4-1) v d where v = width of the eam (average width of the we elow the flange for a T-ection) Step : Check that v < v u were v u 075. f cu leer of 475. Pa Step 3: Shear capacity of the eam without hear reinforcement (4-) where m = 1.4 f cu 40 Pa v c fcu A 14 / d d m 13 / 13 / v 100 A d v 3 (4-3) A = area of properly anchored tenion reinforcement (ditance d eyond the ection under conideration) Tale 8: Shear capacity v c (in Pa) for eam without hear reinforcement for f cu = 30 Pa. 100A d v Effective depth d (mm) Step 4: Nominal hear reinforcement: A v v for f yv 50 Pa for f yv 450 Pa (4-4) Step 5: If v > v c, hear reinforcement mut e provided For ent-up ar: V A 087. f co in cot or yv A dd (4-5) vvcv 087. f coin cot yv (4-6) where A = area of a ent-up ar, = pacing of ent-up ar, = angle etween the horizontal

12 10 Univerity of Pretoria and the ar and = angle etween compreive trut in the concrete and the horizontal. For vertical link: A v v v vc with f yv 450 Pa. (4-7) 087. f yv Tale 9: A v / v (mm /mm) for link with two leg Diameter Link pacing (mm) (mm) Step 6: aximum pacing = 075. d 15. d for link for ent - up ar (4-8) Further comment: Within a ditance d from a upport, or concentrated load, the hear reitance v c may e increaed a follow d 075 f cu v c. leer of av 475. Pa (4-9) For eam carrying mainly uniformly ditriuted load, or where the principal load i applied further than d from the face of the upport, a critical ection at a ditance d, from the face of the upport i conidered. For la thinner than 00 mm the reitance of the hear reinforcement hould e reduced y 10 % for every 10 mm reduction of la thickne elow 00 mm. 5 Deign for Torion Step 1: Find A and A v to reit ending and hear. Step : Find the torional hear tre from v t h min T h h max 3 min (5-1) where h min i the maller and h max i the larger ection dimenion. 3 Step 3: Divide T- L- and I-ection into component that maximie h h each component individually, ujected to a moment min max. Conider

13 Reinforced Concrete Structure - SABS T i h 3 min hmax T 3 h h min i max i (5-) Step 4: Check that v v v (5-3) where v i the hear tre equal to V/(d). Alo check for mall ection that v v Sep 5: Provide torional reinforcement if v hear and torion refer to Tale 11. t t t tu tu y 1 550mm Tale 10: inimum and maximum tree for torion (Pa). (5-4) vt,min with v t,min from Tale 10. For comined where v f cu (Pa) v t,min v tu < v tu < 4.75 tu 071. f 4.75 Pa and v 006 f 0.36 Pa cu t, min. Thee value include m = 1.4 Tale 11: Reinforcement for comined hear and torion. cu v( v c 0. 4) inimum hear reinforcement; no torion reinforcement v( v c 0. 4) Deigned hear reinforcement; no torion reinforcement v t v,min v v t t t,min Deigned torion reinforcement ut not le than minimum hear reinforcement Deigned hear and torion reinforcement Step 6: Deigned torion reinforcement (additional to that required for hear in tep 1) A A v v T 08. x y ( 087. f ) A v v f f 1 1 Step 7: Detailing requirement: Only ue cloed link for torion. aximum pacing for link i the leer of: x 1, y 1 / or 00 mm. yv y yv (5-5) x 1 y 1 (5-6)

14 1 Univerity of Pretoria Longitudinal torion reinforcement mut e ditriuted evenly around the inide perimeter of the link o that the maximum clear ditance etween ar i le than 300 mm. Each corner of a link hould contain at leat one longitudinal ar. Torion reinforcement may e included at level of exiting flexural reinforcement y increaing the diameter of the flexural reinforcement appropriately. Torion reinforcement hould extend for at leat a ditance equal to the larget ection dimenion eyond the point where it i theoretically required. For T-, L- and I-ection the reinforcement hould e detailed o that they interlock and tie the component rectangle together. If vt vt,min for a maller component rectangle, torion reinforcement may e omitted for that component. 6 Bond and Anchorage Anchorage ond length Bearing tre inide a end Lapping of ar: f L (6-1) 4 f u Ft f cu r 1 / a (6-) inimum lap length mut e the greater of 15 or 300 mm for ar and 50 mm for faric. Lap length for ar of different diameter can e aed on the maller diameter. Compreion lap mut e 5% greater than deign anchorage length in compreion. When oth ar in a lap are 5 mm or greater in ize, and the cover i le than 1.5 time the maller ar ize, then tranvere link of at leat 1/4 of the maller ar ize hould e provided at a maximum pacing of 00 mm. If ar are placed in a undle, only one ar at a time may e lapped. The maximum numer of ar in a undle, including lap, hould not e more than 4. Tale 1: Deign ultimate ond tre f u in Pa (SABS 0100). Bar type Concrete grade Plain ar in tenion Plain ar in compreion Deformed ar in tenion Deformed ar in compreion Reduce thee value y 30% for deformed top ar and y 50% for plain top ar in element where the depth exceed 300 mm.

15 Reinforced Concrete Structure - SABS Tale 13: Ultimate anchorage ond length a multiple of ar ize. Bar type r in Concrete grade Plain ar 1 in tenion Plain ar 1 in compreion Deformed ar in tenion Deformed ar in compreion ild teel f y = 50 Pa High yield teel f y = 450 Pa Reduce thee value y 30% for deformed top ar and y 50% for plain top ar in element where the depth exceed 300 mm. 1.4 Cover f F t r / / r Bar diameter = F t / = Diameter of main reinforcement a a (a) Equilirium of force () Definition of a Figure 10: Bearing tre inide a end. Figure 11: Increaing lap length Straight anchorage length L (a) Straight ar B r 4 L B () Bend Effective anchorage B = leer of 4 r and 1 B 4 r inimum radiu: ild teel ( f y = 50 Pa) High yield trength ( = 450 Pa) f y r min = r min =3 L H (c) Hook Effective anchorage H = leer of 8 r and 4 Figure 1: Equivalent anchorage of a hook and a end.

16 14 Univerity of Pretoria 7 Deign for Serviceaility 7.1 Cover to concrete Tale 14: inimum cover (in mm) for variou expoure condition. Concrete Condition of expoure ild oderate Severe Very evere Extreme Normal denity concrete Low-denity concrete Concrete with a denity in the range 00 to 500 kg/m 3. Concrete with a denity < 000 kg/m 3 made with low denity aggregate. 7. aximum clear pacing of reinforcement clear pacing (in mm) f ut 300 mm (7-1) where f i the tre in the reinforcement (in Pa) under ervice load and i given y f Areq f y 1, A 3 4 prov, (7-) with 1 = partial afety factor for elf-weight load at SLS (typically = 1.1) = partial afety factor for impoed load at SLS (typically = 1.0 or 0) 3 = partial afety factor for elf-weight load at ULS (typically = 1. or 1.5) 4 = partial afety factor for impoed load at ULS (typically = 1.6 or 0) Sla: If any one of the following three condition apply, the maximum clear pacing i the leer of 3 d and 750 mm: (a) For high yield trength teel (f y = 450 Pa) the la depth h 00 mm. () For mild teel (f y = 50 Pa) the la depth h 50 mm. (c) 03. % where 100 A d If neither of the aove condition apply, the maximum clear pacing given in Tale 15 are adjuted a follow: If 1 %, maximum clear pacing i taken from Tale 15 If < 1 %, maximum clear pacing i the value from Tale 15 divided y.

17 Reinforced Concrete Structure - SABS Tale 15: aximum clear pacing (in mm) etween ar (SABS 0100). Characteritic trength of reinforcement f y (Pa) 7.3 inimum pacing of reinforcement Percentage reditriution to or from ection conidered * * If the percentage reditriution i unknown, a value of 15 hould e aumed for moment at upport and zero for pan moment. Tale 16: inimum clear pacing etween ar. Configuration Orientation inimum clear pacing of ar Single ar Horizontal (h agg + 5 mm) Vertical 3 h agg Bar in pair Horizontal (h agg + 5 mm) Vertical 3 h agg for ar in the pair on top of each other (h agg + 5 mm) for ar in the pair ide y ide Bundled ar Horizontal and vertical (h agg +15mm) 7.4 inimum area of reinforcement (See Tale 17.) 7.5 aximum area of reinforcement 100 A h or 100 A c 4% (7-3) h Where ar are eing lapped the um of the reinforcement diameter in a particular layer hould not e greater than 40 % of the ection width at that level. For column the following limit apply: 100 A h c 6 % for column cat vertically 8 % for column cat horizontally 10% at lap for oth of the cae aove (7-4)

18 16 Univerity of Pretoria Tale 17: inimum percentage of reinforcement (SABS 0100). Tenion reinforcement Situation Definition f y = 50 Pa f y = 450 Pa Section ujected mainly to pure tenion 100A /A c Section ujected to flexure (a) () (c) Flanged eam, we in tenion (1) w / < A / w h () w / A / w h Flanged eam, flange in tenion over a continuou upport (1) T-eam 100A / w h () L-eam 100A / w h Rectangular ection (in olid la thi reinforcement hould e provided in oth direction) Compreion reinforcement (where uch reinforcement i required for the ultimate limit tate) 100A /A c General rule 100A c /A cc Simplified rule for particular cae: (a) Rectangular column or wall 100A c /A c () Flanged eam: (1) Flange in compreion 100A c /h f () We in compreion 100A c / w h (c) Rectangular eam 100A c /A c Tranvere reinforcement in flange of flanged eam (provided over the full effective flange width near top urface to reit horizontal hear) 100A t /h f A c = total area of concrete = width of ection A cc = total area of concrete in compreion w = width, or effective width of the ri * A c = minimum area of compreion reinf. h = total depth of ection A = minimum area of tenion reinf. h f = depth of flange A t = minimum area of tranvere reinf. in the flange = pan of eam * For a ox, T-, or I-ection, w i taken a the average width elow the flange.

19 Reinforced Concrete Structure - SABS Reinforcement at ide of eam exceeding 750 mm in depth f y w (7-5) 3 h 1 h > 750 mm where 50 mm. Ue w 500 mm in aove equation. 7.7 Span-effective depth ratio Tale 19: Baic pan-effective depth (L/d) ratio for eam. odification factor for tenion reinforcement: F A 477 f d 0. (7-6) odification factor for compreion reinforcement: where Support condition F A (7-7) 100 A, d prov Tale 18: L/d modification factor for compreion reinforcement. Rectangular ection Flanged ection w 0.3 Truly imply upported Simply upported with nominally retrained end One end continuou Both end continuou 8.4 Cantilever (7-8) w Figure 13: Reinforcement at ide of eam exceeding 750 mm in depth. 100 Aprov, odification d factor Intermediate value may e determined y interpolation If the aic L/d ratio i applied to a eam with pan longer than 10 m, the aic L/d ratio hould e multiplied y 10/L

20 18 Univerity of Pretoria Tale 0: L/d modification factor for tenion reinforcement. Steel ervice tre f (Pa) / d

21 Reinforced Concrete Structure - SABS Deign of Beam 8.1 Effective pan length Tale 1: Effective length in eam. Beam Simply upported Continuou eam Cantilever Effective pan length The leer of: ditance etween centre of earing, and clear ditance etween upport plu an effective depth. Ditance etween centre of upport. For an emedded end the centre of upport hould e taken a half an effective depth from the face of the upport. The length to the face of the upport plu half an effective depth. If the cantilever form part of a continuou eam the effective length hould e taken a the clear length plu the ditance to the centre of the upport. 8. Analyi of continuou eam Condition for the ue of the implified method: (a) Q n 15. G n () The load on the eam mut e utantially uniformly ditriuted load. (c) There mut e 3 or more pan. (d) The pan may not vary y more than 15% in length with regard to the longet pan. Tale : Ultimate ending moment and hear force for continuou eam (implified method). Poition oment Shear Outer upport F Near centre of end pan FL eff /11 - Firt internal upport FL eff /9 0.6F Centre of interior pan FL eff /14 - Interior upport FL eff /1 0.55F Thee moment may not e reditriuted. Aume = 0.9. F = Total load on pan (in kn) = 1. G n Q n L eff = Effective pan

22 0 Univerity of Pretoria 8.3 Flanged eam Effective flange width: T-ection L-ection eff eff Lz w leer of 5 actual flange width (8-1) L z w leer of 10 actual flange width (8-) where L z i the ditance etween zero moment. A a implified approach for continuou eam, L z can e aumed to e 0.7 of the effective pan. 8.4 Beam with compreion reinforcement Containment of compreion reinforcement: A A A A prov req prov req,,,, (8-3) Link hould pa around outer ar and each alternate ar The link hould e at leat 1/4 the ize of the larget compreion ar The maximum longitudinal pacing of link i 1 time the diameter of the mallet compreion ar For the containment to e effective, the link hould pa around the ar with an inide angle not more than 135º No compreion ar hould not e more than 150 mm from a contained ar 8.5 Curtailment of reinforcement The curtailment anchorage length mut e the greater of: 1. the effective depth d of a memer, or. twelve time the ar ize (1 ). For ar in tenion, the mallet ditance from one of the following condition mut alo apply: 3. The ar mut extend the ultimate anchorage ond length L ua eyond the theoretical cut-off point (TCP). For the ultimate anchorage ond the tre in the ar i taken a 0.87 f y. 4. At the phyical cut-off point (PCP) the hear capacity i twice the actual hear force. 5. At the PCP the flexural capacity of remaining ar i twice the actual ending moment. If the condition to the ue of Tale applie, the implified rule for curtailment may e ued.

23 Reinforced Concrete Structure - SABS C L Support C L Support 1 (a) Equivalent anchorage 1 () Greatet of up /3 and 30 mm C L Support Solid la v < v c 1 Equivalent d/ anchorage d d/ 1 d up (e) (c) (d) Figure 14: Curtailment of reinforcement imply upported end. C L Support 1 d/+1 Condition 1: A hook or end may not tart efore the centre line of a upport (a) Simple upport Condition : A hook or end may not tart cloer than d/ from the face of the upport d 0.5 A1 A 1 d d/ 0.08 L L = Effective pan () Simply upported eam Greater of L/ en 45 A A 1 d d/ L = Effective pan (c) Cantilever 0. A 0.5 L ( 45 ) 0.15 L ( 45 ) 0.6 A A A A A 1 0.1L 0.15 L d/ L = Effective pan (d) Continuou eam Figure 15: Simplified curtailment rule for eam.

24 Univerity of Pretoria 9 Deign of Short Column 9.1 oment-axial force interaction diagram Stre and train in reinforcement f 087. f y for y ( ) E for y yc f yc for yc (9-1) dx x (9-) Tenion Stre f y f yc Compreion E yc 0.87 f y Figure 16: Sign convention Strain From equilirium: N 045. fcu fc A f A (9-3) 045. f cu xp fc A xp d fadxp (9-4) Fcc ( h/ ) Fc df d oment aout top of ection yield platic centroid: x p (9-5) F F F h N x p A d d x cu = c N.A. cc c (< h) 0.45 f cu F c F cc A Cro-ection (a) Small axial force F HG t Strain h x 09. I K J F t Stree and force d cu = f cu A d x c (= h) F c F cc A t F t Cro-ection () Large axial force F HG Strain h x 09. I K J Stree and force Figure 17: oment and axial force acting on a ection.

25 Reinforced Concrete Structure - SABS Defining point: (a) Pure flexure, N =0 () Balance point = y x al d 1 y (9-6) (c) Pure compreion =0 (d) F t =0, =0,x = d (e) Compreion reinforcement yield c = yc. Point (e) can e aove or elow (). N (c) N al 0.00 > yc c > yc =0 (d) Compreion failure () Tenion failure (e) < y (a) al x al = y < Figure 18: Point defining the N-interaction diagram. y c = yc 9. Axially loaded hort column inimum eccentricity e min = 0.05 h 0 mm (h i meaured perpendicular to axi of ending). Nominal eccentricity moment min = Ne min (9-7) For moment le than min the axial capacity i N u 040. fcuac067. f yac (9-8) 50 SABS 0100: 199 N h (Pa) A c h 8 xh / =0.9 xh / =0.8 xh / =0.7 h d xh / =0.6 A c A c f y = 450 Pa f cu =30Pa d =0.h d =0.8h d xh / =0.5 xh / = h (Pa) Figure 19: Typical N-interaction diagram.

26 4 Univerity of Pretoria 10 Deign of Supended Floor 10.1 One-way panning la The la hould e deigned to pan in one direction if the long pan exceed 3 time the hort pan. The ingle load cae of maximum deign load on all pan may e ued when: (a) area of a ay 30 m () Q 1. 5G (10-1) n n (c) Q n 5kPa (10-) (d) Reinforcement mut e curtailed according to the implified rule (Fig. 0). When uing thi ingle load cae in the analyi of a continuou la, a reditriution of moment hould e applied y reducing the upport moment y 0% and increaing the pan moment accordingly. Nowhere hould the reditriuted moment e le than 70% of the elatic moment. 1 or equivalent anchorage For v < v c / C L Greatet of /3 and 30 mm 0.5 A1 A 1 d d/ 0.08 L L = Effective pan (a) Simply upported la 0.3 L 0.5 A Greatet of 0.15 L and 45 A A 1 Greater of L/ en A 1 A A 1 0. L L = Effective pan () Continuou la d d/ L = Effective pan (c) Cantilever la Tenile anchorage length Greatet of 0.15 L and 45 *0.5 A 1 * But more than minimum reinforcement *0.5 A A A 1 0.1L A 1 0.1L L = Effective pan (d) Retrained end where zero moment were aumed in the analyi Figure 0: Simplified detailing rule for one-way panning la.

27 Reinforced Concrete Structure - SABS If the pan adjacent to a cantilever i le than 3 time the length of the cantilever, the load cae where the cantilever carrie the maximum load and the adjacent pan carrying a minimum load hould alo e conidered. The implified method given in Tale 3 may ued if: (a) The aove condition for the implified load arrangement apply () The pan are approximately equal (c) There are 3 or more pan. Tale 3: Ultimate ending moment and hear force in one-way panning la. Poition oment Shear Outer upport: 0 0.4F Near centre of end pan FL eff - Firt interior upport FL eff 0.6F Centre of interior pan FL eff - Interior upport FL eff 0.5F Thee value may not e reditriuted. F = 1. G n Q n, = Effective pan L eff Concentrated load The width of eam upporting the load i the um of the load width and the following width on each ide of the load x 1. x1 03. L (10-3) L Load where x i meaured from the nearet upport to the load. x L Unupported edge Load width F HG x 1. x 1 L I K J 0.3L Effective la width upporting the load Figure 1: Effective width of la upporting a concentrated load.

28 6 Univerity of Pretoria 10. Two-way panning edge upported la Simply upported la For a rectangular la, imply upported along all four edge o that lifting of the corner are not prevented, the maximum moment per unit width in the centre of the la i given y m m x x n x y y n x (10-4a, ) where x = hort pan y = long pan n = total deign load (kn/m ) Tale 4: Bending moment coefficient for imply upported two-way panning la. = 1. g n q n Detailing rule are given elow. 1 or equivalent anchorage 0.5 A1 A 1 d d/ 0.1 L L = Effective pan = or Figure : Simplified detailing rule for two-way panning imply upported la Sla with retrained edge x y y / x x y The implified method elow may e ued if: (a) The nominal dead and impoed load on adjacent panel hould e approximately the ame a on the panel under conideration. () In the direction of the pan eing conidered, the adjacent pan length mut e approximately equal to that of the pan under conideration. m m x x n x y y n x where x and y are ending moment coefficient from Tale 5. (10-5a, )

29 Reinforced Concrete Structure - SABS Tale 5: Bending moment coefficient for rectangular panel upported on four ide with proviion for torional reinforcement in corner. Type of panel and moment conidered Short pan coefficient x for y / x Long pan coefficient y for all y / x 1. Interior panel Negative moment at continuou edge Poitive moment at midpan. One hort edge dicontinuou Negative moment at continuou edge Poitive moment at midpan 3. One long edge dicontinuou Negative moment at continuou edge Poitive moment at midpan 4. Two adjacent edge dicontinuou Negative moment at continuou edge Poitive moment at midpan 5. Two hort edge dicontinuou Negative moment at continuou edge Poitive moment at midpan Two long edge dicontinuou Negative moment at continuou edge Poitive moment at midpan Three edge dicontinuou (One long edge continuou) Negative moment at continuou edge Poitive moment at midpan Three edge dicontinuou (One long edge continuou) Negative moment at continuou edge Poitive moment at midpan Four edge dicontinuou Poitive moment at midpan

30 8 Univerity of Pretoria Reinforcement in x -direction Reinforcement in -direction y y y Edge trip x /8 x Edge trip iddle trip Edge trip y /8 3/4 y y /8 x iddle trip Edge trip 3/4 x x /8 Figure 3: Ditriution of la into middle and edge trip x y 0. x Edge eam x 3/4 A x 3/8 A x 7 x A x x 3/8 Ax None Figure 4: Support condition for la panel in Tale 5. Edge eam Interior eam Figure 5: Placing and quantitie of torional reinforcement. 0.1L 0.5 A 0.3L 0.15L A 0.5 A A A A L 0.15L Dicontinuou edge L = Effective pan = or x y 0.5L Continuou edge Figure 6: Simplified detailing rule for two-way panning la with retrained edge.

31 Reinforced Concrete Structure - SABS The method illutrated in Fig. 7 i propoed when moment at the mutual upport to two adjacent la differ. A B B B3 = O B3 B A (a) Bending moment from Eq. 9-5 B = + D or = D B B BA O B B3 BC O A C () Adjuted ending moment 1 A B A 0.5 A 0.5 x x (c) Anchorage of reinforcement Figure 7: Adjutment to moment in edge upported la. A y Edge eam B C 45º 60º x D 60º 45º E Edge eam Figure 8: Load on upporting eam.

32 30 Univerity of Pretoria 10.3 Flat la Notation 1 = length of the panel, meaured etween column centre line in the direction under conideration = width of the panel, meaured etween column centre line perpendicular to the direction under conideration x = horter pan y = longer pan = effective pan = 1 h 3 c (10-1) h c = effective diameter of the column head h 4 h for round column head 1 4 for quare column head h = effective dimenion of a column head ho = leer of h,max cdh40 mm 1 (10-) (10-3) Analyi ho = actual dimenion of column head d h = depth of column head elow offit (or elow drop, if preent) c = dimenion of the column (meaured in the ame direction) Following the analyi of an equivalent frame, the ending moment hould e divided in column and middle trip (ee Fig. 31) a hown in Tale 6. The maximum deign moment at a upport can e taken a the moment a ditance h c / from the column centre line, provided the um of the poitive pan moment and the average of the upport moment i greater than the following n h c (10-4) The maximum moment that can e tranferred to a column i given y t e cu, max d f (10-5) where e = width of a trip depending on the ditance etween the column and the free

33 Reinforced Concrete Structure - SABS Tale 6: Diviion of moment in trip (SABS 0100). Column trip iddle trip Negative moment 75% 5% Poitive moment 55% 45% For the cae where the width of the column trip i taken a equal to that of the drop and the width of the middle trip therefore increae, the deign moment reited y the middle trip hould e increaed in proportion to it increae in width. The deign moment to e reited y the column trip may then e decreaed y an amount uch that the total negative and total poitive moment reited y the column and middle trip together are unchanged. h,max h,max d h 45º 40 mm d h 45º 40 mm h c ho = h,max h,max (a) Flared column head h c ho = h,max ho d h 45º 40 mm d h 45º 40 mm h c ho = h,max () Contant column head h c ho = ho Figure 9: Effective dimenion of a column head. C x C y y e = C x = C + y e x e = Cx+ Cy y y x y e = Cx+ y column trip e = Cx+ y/ e = x+ y/ Figure 30: Width of trip e to tranfer moment at la-column connection.

34 3 Univerity of Pretoria = Column trip hort pan = x x /4 x / x /4 hort pan = x x /4 y x/ long pan = y x /4 (a) Sla without drop Note: Ignore drop if < x /3 drop drop drop / drop x drop / drop / y drop long pan = y drop / drop () Sla with drop drop Figure 31: Diviion of panel into column and middle trip.

35 Reinforced Concrete Structure - SABS Tale 7: Ultimate ending moment and hear flat la. Outer upport: Poition oment Shear Total column moment Column Wall 0.04 F 0.0 F 0.45 F 0.4 F 0.04 F Near centre of end pan F Firt internal upport F 0.6 F 0.0 F Centre of interior pan F Interior upport F 0.5 F 0.0 F Thee moment may not e reditriuted. Aume = 0.8. F = Total load on pan (in kn) = 1. G n Q n = Effective pan in the direction under conideration 1 3 h c edge of the la (ee Fig. 30). For an internal column e hould not e greater than the width of the column trip. d = effective depth of top reinforcement in the column trip f cu = characteritic concrete trength If the condition for the implified load arrangement (ee ection 10.1) apply, the implified method given in Tale 7 may e ued to otain the ending moment and hear force in the la. The following additional requirement mut alo e met efore the implified method may e ued: (d) The tructure i raced o that ideway taility of the frame doe not depend on the la-column connection. (The tructure i raced). (e) (f) There hould e at leat three pan in the direction under conideration. The pan length hould e approximately equal. (It i aumed here that the pan will e approximately equal if they do not differ y more than 15% from the longet pan). (g) The curtailment rule for olid one-way paning la (Fig. 0, ec. 10.1) hould e ued Deflection For flat la with drop the aic allowale /d-ratio for eam i ued ut multiplied y a factor of 0.9. If the plan dimenion of the drop are at leat a one-third of the repective pan in each direction, the 0.9 factor can e omitted. The /d-ratio hould alway e conidered in the critical direction, which i uually the long-pan direction in flat la.

36 34 Univerity of Pretoria = Column trip 0.4 x x B col B col 0.1 B col 0.4 y y (a) Opening common to two interecting middle trip 0.1 B col () Opening common to two column trip B col 0.5 B col (c) Opening common to a column trip and a middle trip Figure 3: Opening in panel Detailing of reinforcement If the implified method have een ued to otain the ending moment (ection 10.1), the curtailment rule for olid two-way paning la (ee ection 10.1.) hould e ued. The column trip reinforcement that pae over the column mut e placed o that two-third of thi reinforcement i placed within half the width of the column trip, centrally over the column Punching hear in la The maximum hear tre at the edge of the loaded area hould not exceed v u 075. f cu leer of 475. Pa (10-6)

37 Reinforced Concrete Structure - SABS = Failure zone 0.75d 1.5d 1.5d 1.5d 1.5d 1.5d 1.5d 1.5d * * * *At leat et of reinforcement within a failure zone (a) Firt failure zone () Second failure zone (c) Third failure zone Figure 33: Conidering ucceive failure zone for punching hear. The deign proce i ummarized a follow: (a) The firt perimeter i conidered at a ditance 1.5 d from the loaded area. If v v c no hear reinforcement i required and no further check are neceary. Reinforcement ued for v c mut extend at leat an effective depth d or 1 diameter eyond the zone on either ide. () If v > v c the following hear reinforcement i required: v vcud For v v v A c 16. c : v 087. f yv with vv 04. Pa where: vvc ud For 1. 6 vc v vc: Av 087. f u = outide perimeter of the failure zone A v = area of hear reinforcement c yv (10-4) f yv = characteritic trength of the hear reinforcement (not exceeding 450 Pa) Thee equation apply only when: v <v c Link are ued a hear reinforcement. The la i at leat 00 mm thick. For every 10 mm le than 00 mm, a 10 % lo of efficiency hould e aumed for the hear reinforcement.

38 e e 36 Univerity of Pretoria l p l p l p l p l p lp lp l p l p l p i a multiple of 0.75 d l p lp l p (a) Perimeter for a general hape of loaded area Sla edge Sla edge <6d Perimeter a Perimeter a Perimeter Perimeter () Effective perimeter allowing for opening (c) Effective perimeter cloe to la edge (d) Effective perimeter cloe to la corner Figure 34: Perimeter for punching hear. (c) Shear reinforcement mut e ditriuted into at leat two perimeter within the failure zone under conideration (ee Fig. 3-33). (d) Shear reinforcement placed for a previou failure zone may e included in the failure zone under conideration where uch zone overlap. (e) (f) The firt perimeter of hear reinforcement hould e approximately 0.5 d from the face of the loaded area and hould contain at leat 40 % of the required area of hear reinforcement. Shear reinforcement mut e anchored around at leat one layer of tenion reinforcement. (g) Shear tree are checked on the next perimeter a ditance 0.75 d from the current perimeter. If v v c no hear reinforcement i required and no further check are neceary, otherwie repeat the proce from tep () aove. The effective hear force in flat la can e calculated from Fig. 36 and 35. The implified equation (hown with *) may e ued when the tructure i raced, the ratio of pan doe not exceed 1.5 and the maximum deign load i applied on all pan adjacent to the column under conideration. l p Corner column V eff = 1.5 V t C y C x l p x Edge column V V eff eff V eff = 1.5 V e = Vt t Vx t = 1.4 V * t t or Internal column V V eff eff e = Vt t Vx t = 1.15 V * t or t V t Perimeter eing conidered Bending moment Shear force Figure 35: Effective hear force. Figure 36: Definition of V t and t.

39 Reinforced Concrete Structure - SABS Deign of Slender Column A vertical load-earing memer i defined a a column when h 4 and a wall when h 4 where h i the larger and the maller cro-ection dimenion, repectively Braced and unraced column A tructure can e conider raced if the ratio S /S u i greater than 5, where S i the lateral tiffne of the raced tructure and S u i the way limne of the unraced tructure. 11. Effective length The effective length i determined from: e (11-1) where o = clear height etween end retraint and = factor otained from Tale 8. Tale 8: -Value for effective length. o End condition top Braced column -Value for effective length Unraced column Definition of end condition: 1 The end of the column i connected monolithically to eam on either ide which are at leat a deep a the overall dimenion of the column in the plane eing conidered. The end of the column i connected monolithically to eam or la on either ide which are hallower than the overall dimenion of the column in the plane eing conidered. 3 The end of the column i connected to memer which, while not deigned to provide retraint to rotation, will neverthele, provide ome retraint. 4 The end of the column i unretrained againt oth lateral movement and rotation.

40 38 Univerity of Pretoria Alternatively, the effective length can e determined from Braced column: Unraced column: e o c, 1c, leer of o c,min e leer of o c,min o c, 1 c, o (11-) (11-3) where c,1 = ratio of the um of the column tiffne to the um of the eam tiffnee at the lower end of the column c, = ratio of the um of the column tiffne to the um of the eam tiffnee at the upper end of the column c,min = leer of c,1 and c, If a ae have een deigned to reit the moment, c can taken a 1, otherwie c hould e taken a 10. For a very large tiff ae, c can taken a 0. For imply upported eam framing into a column c hould e taken a Slenderne A column hould e conidered lender when l h e / for raced column 1 for unraced column (11-4) where l e = effective height = l o ( from Tale 8) l o = clear height etween end retraint 1 = maller initial end moment due to ultimate deign moment (negative for ending in doule curvature) = larger initial end moment due to ultimate deign moment Limit: Braced column: l o <60 and 0.5h (11-5) Unraced column: l o <5 and 0.5h (11-6) 11.4 oment and force in column inimum eccentricity All column hould therefore e deigned for a minimum moment reulting from eccentric loading where e min = 005. h 005. min N e (11-7) min for ending aout the x - axi 0 mm (11-8) for ending aout the y - axi

41 Reinforced Concrete Structure - SABS Additional moment in lender column add = Na u (11-9) a u = a Kh (11-10) a = K = 1 l e 000 h (11-11) N uz N N uz N i al al vir K 1 (11-1) vir K 1 N uz = 0.45 f cu A c f y A c ( m included) (11-13) For ymmetrically reinforced rectangular ection al = f cu d f yc A c (d d) (11-14) N al = 0.5 f cu d (11-15) SABS 0100: N 0.5 h f cu hn min = Ac 1 ( Pa) h fcu K = 1.0 h d A c A c f y = 450 Pa d = 0. h d = 0.8 h d h f cu Figure 37: Interaction diagram with K-value.

42 40 Univerity of Pretoria Braced lender column The deign moment i the greater of (a) () i + add (11-16a c) (c) Ne min where i = (11-17) add / add / + add = Larger moment add / add / i + add = max 1 Smaller moment add / 1 + add / (a) Braced frame () End condition for column (c) Initial moment from analyi (d) Additional moment caued y lenderne (e) oment envelope Figure 38: oment in raced lender column Unraced lender column The deign moment i the greater of (a) V H1 add, unr V H () add, raced (11-18a c) where (c) Ne min

43 Reinforced Concrete Structure - SABS add,raced = additional moment from Eq. (11-3), ut uing the raced effective length in Eq. (11-5) 1, = maller and larger end moment including the effect of way V + H add, unr F HG H H V I KJ V H F HG 1 add, unr H V I KJ + = Le tiff end joint Smaller moment add may e reduced in proportion to the tiffne ratio of the le tiff to the tiffer joint 1 + = (a) Unraced frame Stiffer end joint () End condition for column V + H Larger moment (c) Initial moment from analyi add, unr F HG H H (d) Additional moment caued y lenderne V I KJ V F HG, H 1 H (e) oment envelope add unr V I KJ Figure 39: oment in unraced lender column (SABS 0100) Slender column ent uniaxially A column which i lender aout oth axe ut i ent uniaxially, mut e deigned to reit the additional moment aout oth axe eparately. If the column i lender aout one axi only, the additional moment only have to e conidered in one plane. Enure that i min Bi-axial ending For For h h x x y then y then where i given in Tale 9. h h x x x y y y (11-19a ) Tale 9: -Value for iaxial ending. N hf cu

44 4 Univerity of Pretoria 1 Staircae The unit weight of the wait, meaured horizontally, i determined y multiplying the unit weight meaured along the lope of the tair with R G (1-1) G The unit weight of the tair (without the wait), meaured horizontally, i determined y approximating it a a la with thickne R/. If a tair panning in the direction of the flight i uilt in at leat 110 mm into a wall along the length, a width of 150 mm adjacent to the wall may e deducted from the loaded area. The effective width of the tair may then include /3 of the emedded width, up to a maximum of 80 mm. If a tair i upported y element panning at right angel to the pan of the tair, the effective pan of the tair may e taken a the clear ditance etween upport plu half the width of the upporting element, up to a maximum ditance of 900 mm at oth end. The allowale /d may e increaed y 15% if the tair make out 60% or more of the pan. 13 Deflection and Crack Width Deflection are calculated from i K (13-1) EI c e R N T G W G = Going R = Rie T = tread N = Noing W = Wait Figure 40: Definition of term where I e cr a 3 cr I g 3 1 Icr Ig (13-) a f r f f cu cu cr fi r g (13-3) y t for unretrained ection (13-4) for retrained ection where cracking can occur efore loading Value for K are given in Figure 41. The total deflection (elatic and creep) i given y i (13-5) where 1x i and xi x/ d (13-6) If compreion reinforcement i preent, replace with ( 1/ ) (13-7) where A / A

45 Reinforced Concrete Structure - SABS Loading Bending moment diagram K 0.15 a W Wa( 1a) 34a 48( 1a) 1 If a, 1 K W/ W/ a a Wa 015. a 6 q q q q a a q W q A Wa qa C B K F I HG K J and A End deflection K a ( 3 a ) 6 If a1, K End deflection K a ( 4 a ) 1 If a1, K05. C B a a A B C W a 4 ( 34 ) K F I HG K J 1 80 ( 54a ) 34a A C B Figure 41: Value for K for different ending moment diagram. Deflection caued y hrinkage are determined from where KK c 1 for cantilever 1 K 8 for imply upported memer for one end continuou 1 16 for oth end continuou (13-8) h (13-9)

46 44 Univerity of Pretoria en 0 for uncracked memer K c ( ) 1 en 03. for cracked memer (13-10) 100A d 100A d 1 3 The maximum crack width i determined from w max where m 3acrm acr c 1 hx 1 min t ( hx)( a x) 3EA( dx) (13-11) (13-1) 1 = concrete train at the level under conideration a = ditance from the compreion edge to the level under conideration a cr = ditance from the point under conideration to the nearet longitudinal ar t = width of the ection at the level of the tenion reinforcement c min = minimum cover to tenion reinforcement To calculate x and 1 the cracked tranformed ection hould e aed on E c / to account for the effect of creep in the concrete.

47 Reinforced Concrete Structure - SABS Pretreed Concrete 14.1 Sign convention Tenile tree and force are poitive. Hogging ending (concave curvature) i poitive. The eccentricity e of the pretreing force i meaured from the centroid of the ection and i taken poitive elow the centroidal axi. The ign of the ection modulu Z=I/y with repect to a particular fire i determined y the ditance y of the fire meaured from the centroidal axi. Thi ditance i taken poitive for fire located elow the centroidal axi. 14. aterial propertie odulu of elaticity: E p = 05 GPa for high tenile teel wire (14-1) = 195 GPa for 7-wire trand = 165 GPa for high tenile alloy ar. Stre f p f py = / f pu m f p1 = 0.8 f py E p f py py f p1 p1 f f E p p1 p p p1 d i E p p py Strain p Figure 4: Stre-train relationhip for pretreing teel Elatic tree Concrete tree f P Pey y A P Pe I I A Z Z (14-)