> 2 CHAPTER 3 SLAB 3.1 INTRODUCTION 3.2 TYPES OF SLAB

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1 CHAPTER 3 SLAB 3. INTRODUCTION Reinforced concrete sabs are one of the most widey used structura eements. In many structures, in addition to providing a versatie and economica method of supporting gravity oads, the sab aso forms an integra portion of the structura frame to resist atera forces. Usuay a sab is a broad, fat pate, with top and bottom surfaces parae or neary so. It may be supported by reinforced concrete beams, by masonry or reinforced concrete was, by structura stee members, directy by coumns, or continuousy by the ground. 3. TYPES OF SLAB One-way sab : Independent of support condition. (Figure 3.a 3.b) > ; Two-way sab : Depends on support condition. (Figure 3.c) > (a) One- way sab > Figure 3.: Types of sab

2 Two-way sabs are cassified as: Two-way edge supported sab or sab with beams. ( Figure 3.d ) Two-way coumn supported sab or sab without beams. ( Figure 3.e,3.f, 3.g ) (b) One- way sab (c) Two- way sab (d) Sab with beams (Edge supported sab) (e) Fat pate (Coumn supported sab) (f) Fat sab (Coumn supported sab) (g) Grid sab (Coumn supported sab) Figure 3.: Types of sab (continued) 5

3 3.3 DESIGN OF ONE-WAY SLAB Step : Estimation of Sab Thickness (h) Sab thickness is determined according to ACI Code 9.5. as given in Tabe 3. Tabe 3.: Minimum thickness of non-prestressed one- way sab Members w c = 45 pcf f y = psi w c =90 0 pcf f y < psi or,f y > psi Rounding the thickness up Simpy supported / 0 One end continuous / 4 Both end continuous / 8 Mutipy by ( w c ) but >.09 Mutipy by f y 00,000 () h 6 in next higher ¼ in () h > 6 in next higher ½ in Cantiever / 0 # Span ength is in inches, as defined by ACI Code 8.7 given in Fig. 3.(a), (b), & (c) Step : Cacuation of Factored Load (w u ) w u =.4 D+.7 L psf Dead oad, D = w c x h psf w c = Unit weight of concrete (45 ~ 50 pcf for norma weight concrete ) 6

4 Step 3: Determination of Design Moment Design moment is determined by using ACI Moment Coefficient (ACI Code 8.3.3) as given in Tabe 3.4. Step 4 : Checking the Design Thickness d = M u f y φρf yb( 0.59ρ ) f c putting ρ = ρ max = 0.75ρ b. Where, ρ b = 0.85*β f / c * f y f y Vaues of β is given in Tabe 3. Tabe3. : Vaues of β (ACI Code ) f c / 4000 psi β = 0.85 f c / > 4000 psi β sha be reduced at a rate of 0.05 for each 000 psi of strength in excess of 4000 psi β 0.85 Tabe 3.3: Cear cover for sab ( ACI Code 7.7.) No 4 & No 8 bars / in No bar & smaer... ¾ in 7

5 If (d + cear cover) < h; Design is ok. Otherwise redesign the thickness. h t a = a + h a + t (a) Sabs not buit integray with the support (ACI Code 8.7.) h a t = a + t (b) Sabs are continuous (ACI Code 8.7.) t a (c) Sabs buit integray with support Figure 3. : Span ength 8

6 Tabe 3.4 : ACI moment coefficient Support Condition Moment Coefficent For two span:. Discontinuous ends are unrestrained Discontinuous ends are buit integray with support (spandre beam or girder) Discontinuous ends are buit integray with support (when support is a coumn ony) For continuous Span:. Discontinuous ends are unrestrained Discontinuous ends are buit integray with support (spandre beam or girder) Discontinuous ends are buit integray with support (when support is a coumn ony) Shear in end members at first interior support. Shear at a other supports.5 w w w = Tota factored oad per unit ength of beam or per unit area of sab = Cear span for positive moment and the average of two adjacent cear spans for negative moment. 9

7 Step 5 : Determination of Stea Area (A s ) Reinforcement for ft. strip towards shorter distance is cacuated by Iteration. Detais shown in Figure 3.3 (here b = in) Assume a (hints: a = 0.3d) (A s ) tria = φf y M u ( d a ) (a) corrected = A s f y 0.85 f b c (A s ) corrected Cacuate a for next tria with (A s ) corrected A s, corr No A s, tria Yes OK Figure 3.3: Iteration process to determine the stee area 0

8 Generay # 3 or # 4 bars are used for sab main reinforcement. Spacing: ACI Code specifies that Spacing 3h or 8 in, whichever is smaer but >.5 h Finding out bar spacing: Let us chose # 3 bar (0.in ) Spacing = 0.* A s in c/c Step 6 : Temperature and Shrinkage Reinforcement Reinforcement is provided norma to main reinforcements. ACI Code 7... provides required area of temperature and shrinkage reinforcement as given in Tabe 3.5. Tabe 3.5 : Minimum ratio of temperature and shrinkage reinforcement in sabs. Sabs where grade 40 or 50 deformed bars are used Sabs where grade 60 deformed bars or weded wire fabric are used Sabs where reinforcement with yied strength exceeding 60,000 psi measured at a yied strain of 0.35 percent is used ,000 f y But shoud be: ρ > Required stee area, A s = ρbh in per ft. strip

9 Spacing: ACI Code 7... specifies that Spacing 5h or 8 in, whichever is smaer Using # 3 or # 4 bar required spacing can be obtained. Step 7 : Shear Check According to ACI shear coefficient given in Tabe 3. Shear at end members at first interior support is w.5 u n Critica shear at a distance d from support, V u = (.5 W u n Wud ) Design strength for shear, φ V c = φ f c bd ; φ = 0.85 If φv c > V u, sab design for shear is OK Otherwise sab thickness shoud be revised. Step 8 : Reinforcement Detaiing Shown in Figure 3.4

10 '_zoom '_zoom SLAB A Temperature & shrinkage reinforcement Main positive reinforcement / 4 / 4 A (a) Pan of bottom reinforcement / 4 negative reinforcement at discontinuous edge negative reinforcement at continuous edge / 4 (b) Pan of Top reinforcement Temperature & shrinkage reinforcement / 4 / 3 / 3 / 3 6" 6" (c) Cross section (A- A) 3

11 Main positive reinforcement Temperature & shrinkage reinforcement Negative reinforcement at discontinuous edge Negative reinforcement at continuous edge Figure 3.4: Reinforcement detaiing in one-way sab (continued) 4

12 3.4 TWO WAY SLAB ACI Code 3.5. states that a sab system sha be designed by any procedure satisfying conditions of equiibrium and geometric compatibiity, if it is shown that the design strength at every section is at east equa to required strength, and that a serviceabiity conditions, incuding imits on defections, are met. According to ACI Code , a Two-way sab system are to be anayzed and designed either by the Direct Design Method or the Equivaent Frame Method for gravity oads ony. For atera oads, separate eastic anaysis shoud be worked out. ACI Code permits the combining to the gravity oad anaysis with the resut of atera oad anaysis. Adaptation of any one of the two methods demands fufiment of certain requirements. However, when the requirements are not met, an od procedure is sti foowed by the Engineers as specified in 963 ACI Code, named as Coefficient Method DIRECT DESIGN METHOD (DDM) Genera The design is based on equivaent rigid frame system as shown in Figure 3.5. ACI Code3.. specifies: Width of equivaent frame = Width of coumn strip = or whichever is ess The equivaent frames are considered in both ongitudina and transverse directions Limitations For DDM ACI Code 3.6. specifies that the design of Two-way sab system by DDM sha be permitted within the foowing imitations: There sha be a minimum of three continuous spans in each direction. The panes sha be rectanguar, with the ratio of the onger to the shorter spans within a pane not greater than. 5

13 The successive span engths in each direction sha not differ by more than one third of the onger span. Coumn may be offset a maximum of 0% of the span in the direction of successive coumn. Loads sha be due to gravity ony and ive oad sha not exceed times the dead oad. H.M.S Interior Equivaent Frame C.S H.M.S Exterior Equivaent Frame H.M.S C.S H.M.S = Haf Midde Strip; C.S= Coumn Strip Figure 3.5 : Equivaent frame system for DDM 6

14 If beams are used on the coumn ines, the reative stiffness of the beams in the two perpendicuar direction sha be: α 0.< α < 5.0 α = α in direction of α = α in direction of α = E E cb cs I I b s ACI Code aows the deviation from above imitations, if it can be shown that the requirements of ACI Code 3.5. (as stated in section 3.4) are satisfied DESIGN METHOD BY DDM Step : Determination of Sab Thickness ACI Code specifies the minimum thickness for two- way sab system Tabe 3.6: Minimum thickness for sab without beams (ACI Tabe 9.5.c) Yied strength f y psi 40,000 60,000 75,000 Without drops panes Exterior panes Without edge beams n 33 n 30 n 8 Without edge beams * n 36 n 33 n 3 Interior panes n 36 n 33 n 3 h > 5 in ( ACI Code a) With drop panes Exterior panes Without edge beams n 36 n 33 n 3 Without edge beams a n 40 n 36 n 34 Interior panes n 40 n 36 n 34 h > 4 in (ACI code b) 7

15 For sabs with beams aong exterior edges, the vaue of α for edge beam sha not be ess than For f y between given vaues the minimum thickness shoud be obtained by inear interpoation. Tabe 3.7 : Minimum thickness for sab with beams (ACI Code ) () α m < 0. () 0. < α m < 0.* (3) α m > 0.* The provisions of Tabe 3.4 sha appy h = n(0.8 + ) 00, β ( α 0.) h > 5 in m f y fy n(0.8 + ) 00,000 h = β h > 3.5 in For edge beam α > 0.80, otherwise h min as provided by coumn (), (3) must be increased by 0% in the pane with edge beam (3) (ACI Code d) Step : Determination of Tota Factored Static Moment (M o ) According to ACI Code M o = u n W 8 M o, of an exterior equivaent frame = Mo, of an interior equivaent frame. n = face to face of coumns or capitas or was. n 0.65 and in determining n, circuar and poygon shaped supports sha be treated as square supports with the same area, shown in Fig 3.6. (ACI Code ) 8

16 Figure 3. 6 : Exampe of equivaent square section for supporting members Step 3 : Longitudina Distribution of Moments. The tota factored static moment, M o is distributed to the negative and positive zone of a equivaent frame according to ACI Code and as given in Tabe

17 Tabe 3.8 : Distribution of tota factored static moment. Interior Span (ACI Code ) 0.35 Mo Mo 0.65 Mo 0.65 Mo Competey fixed at both ends Case -: 0.63Mo Mo Exterior edge unrestrained Case 0.57Mo 0.75Mo Mo 0.6 Mo Beams on a Coumn ines Case 3 0.5Mo 0.70Mo 0.6 Mo Mo No beams ( Fat sab, Fat pate) Case Mo Mo 0.30 Mo Edge beam ony Case - 5: 0.35 Mo 0.70Mo Exterior edge fuy restrained by monoithic concrete wa 0.65Mo 0.65Mo 30

18 Step 4 : Transverse Distribution of Longitudina Moment The ongitudina negative and positive moments are for the entire width of equivaent frame. Each of these moments is to be distributed proportionatey among coumn strip and two haf midde strips foowing ACI Code Before distribution of moment the foowing 3 parameters are to be obtained: Aspect ratio = EcbI b Stiffness ratio, α = Ecs I s EcbC Ratio β t = E I E cb = moduus of easticity of beam concrete, psi E cs = moduus of easticity of sab concrete, psi I b = moment of inertia of beam I s = moment of inertia of sab C = torsion constant cs s Evauating the three parameters, distribute the percentage of ongitudina moment into coumn strip and the remainder into two haf midde strips according to Tabe 3.9. Tabe 3.9 : Percentage of ongitudina moment in coumn strip (ACI code , , ) Aspect ratio = / / β t = o α 3

19 Cacuation of Three Parameters Cacuation of Aspect Ratio = Cacuation of Stiffness Ratio (α) Ratio of fexura stiffness is reated to sab with beams either on a sides or on edge ony. For sab without beams e.g. fat pate or fat sab, α = 0. α = E E cb cs I I b s Determination of Moment of Inertia of the Beam 3 b I b = k w h 3

20 Definition of b e is shown in Figure 3.7 ( ACI code 3..4) h w 4h f b w +h w b w + 8h f Figure 3.7 : Exampe of portion of sab to be incuded with beam or definition of b e The Moment of Inertia of Sab Section 3 bh I s = h = sab thickness b = for interior equivaent frame, or = / for exterior equivaent frame It shoud be determined in both directions. Cacuation of Ratio β t β t = E cb E cs C I s 33

21 Torsion Constant ACI Code specifies that torsiona members sha be assumed to have a constant cross section throughout their ength consisting of the argest of (a), (b) and (c) as shown in Figure 3.8. ACI Code 3.0 defines torsion constant as:- x x C = 3 y 0.63 y 3 x = shorter dimension of a component rectange y = onger dimension of a component rectange The component rectange shoud be taken in such a way that the argest vaue of C is obtained. Exampe for sab with beam & fat pate shown in Figure 3.9. Step 5 : Determination of Effective Depth (d) Where two stee ayers (aong two directions, perpendicuar with each other) are in contact, the arger d is assigned to the stee of greater moment ( i.e. stee for greater moment sha be paced near to either top or bottom face ). Large d = h cear cover (min ¾ in) - d b + ½ d b in Short d = h cear cover - ½ d b in 34

22 t Fat pate direction of moment t h f Torsiona member t (a) Torsiona member (ACI Code a) beam Sab Sab with beam Coumn t t (b) Torsiona member (ACI Code b) be=(bw+hw) (bw+4hf) be=(bw+hw) (bw+8hf) hf h h w 4h f h w 4h f h w 4h f hw bw Edge beam bw Interior beam (c) Torsiona member ( ACI Code c) Figure 3.8 : Torsiona member 35

23 y y X X y y C C X C=Larger of C and C X (a) Sab with beam (Edge section) y y X x Y C C y X C=Larger of C and C X (b) Sab with beam (Interior section) t t Imaginary beam t t x h f x Y Short direction Y Long direction (c) Fat pate Figure 3.9 : Determination of C 36

24 Step 6 : Determination of Stee Area (A s ) Tota stee area for the strip is obtained by iteration process as shown in Figure 3.3. b = width of strip or, b= width of drop pane in the direction of moment (For sab with drop pane in negative moment zone) Step 7 : Check for Minimum Stee A s, min = ρbh ρ = minimum stee ratio for temperature and shrinkage as shown in Tabe 3.5 A provided s, > A s, min (OK) Otherwise provide A s, min Step 8 : Tota Number of Bar A A s N =, A b = cross section area of bar used b Spacing = N b h; b = width of strip Step 9 : Check for Defection Contro According to ACI Code contro of defection is achieved by providing the sab thickness in accordance to Tabe 3.6 and Tabe 3.7. For detais see section

25 Step 0 : Reinforcement Detaiing Pacing: Fexura reinforcement in two- way sab system is paced in an orthogona grid, with bars parae to the sides of the panes. Straight Bars: Straight bars are generay used throughout, athough in some cases positive moment stee is bent up where no onger needed, in order to provide for part or the entire negative moment requirements. Spacing: Maximum spacing h. Figure 3.0.a (ACI Code 3.3.). Concrete Cover: Minimum concrete cover = ¾ in. Figure.0.a. (ACI Code 7.7.). Effective Depth: When bars are paced in perpendicuar ayers either on top or bottom together, stacking probem arises. The inner stee wi have an effective depth - bar diameter ess than the outer stee. For reativey arger moment bars in one direction are provided with greater d. Detais in Figure 3.0.a & Figure 3.0.b. Embedment for Positive Moment: Positive moment reinforcement perpendicuar to a discontinuous edge sha extend to the edge of sab and have embedment, straight or hooked, at east 6 in. in spandre beams, coumns or was. Detais in Figure 3.0.c. (ACI Code 3.3.3). Embedment for Negative Moment: Negative moment reinforcement perpendicuar to a discontinuous edge sha be bent, hooked or anchored in spandre beams, coumns or was. Detais in Figure 3.0.c. (ACI Code 3.3.4). Cantiever Sab: Where sab is not supported by a spandre beam or wa at a discontinuous edge or where a sab cantievers beyond the support, anchorage of reinforcement sha be permitted within the sab. Detais in Figure 3.0.d. (ACI Code 3.3.5). Corner Reinforcement: In sabs with beams between supports, with vaue α >.0, specia top and bottom reinforcement sha be provided at exterior corners. Detais in Figure 3.. (ACI Code 3.3.6). Sab with Drop Pane: Detai dimensions are shown in Figure 3.. (ACI Code 3.3.7). Detais of Reinforcement in Sabs without Beams: In addition to other requirements as mentioned through paragraph - 0, detaiing shown in Figure 3.3 shoud be observed. (ACI Code 3.3.8) 38

26 Maximum spacing < h ( ACI Code 3.3.) h 3 4 " min (a) Section aong short span ( ) 3 4 " min (b) Section aong ong span ( ) Figure 3.0 : Detais of reinforcement in two-way sab 39

27 6" (c) Embedment of positive moment reinforcement (ACI Code 3.3.3) Anchorage (d) Anchorage of reinforcement in cantiever sab (ACI Code 3.3.5) Figure 3.0 : Detais of reinforcement in two-way sab (continued) 40

28 L/5 L/5 L/5 L/5 5 Diagona type (Providedin band) 5 Top bar Bottom bar Gride type (Provided in two ayers) 5 5 Figure 3. : Spira reinforcement at exterior corner 4

29 t t min 4 h b b 6 t h Figure 3. : Detais of drop panes 4

30 Figure 3.3 : Minimum extension for reinforcement in sabs without beams (ACI Figure 3.3.8) 43

31 3.4. EQUIVALENT FRAME METHOD (EFM) Genera The EFM is an aternate method to the DDM for computing ongitudina moments and shear for gravity oads in sabs, supported on coumn or was. ACI Code Commentary R3.7 states that EFM invoves the representation of the three dimensiona sab systems by a series of twodimensiona frames that are then anayzed for oads acting in the pane of the frames. ACI code 3.7. defines the equivaent frame as in Figure 3.4 Figure 3.4 : Eements of equivaent frame system 44

32 3.4.. Moment Of Inertia of Sab Beam (I s ) Considering gross area of concrete (ACI Code ) 3 I s = h Variation in I s aong axis of sab-beam sha be taken into account. The first change from midspan I s occurs at the edge of drop panes, the next occurs at the edge of the coumn or capita. (ACI code ) I I s from center of coumn to face coumn = s ; I s at face of coumn C (ACI Code ) The Equivaent Coumn ACI Code Commentary R3.7.4 estabishes a concept of an equivaent coumn that combines the stiffness of the sab- beam and torsiona member into a composite eement. The coumn fexibiity is modified to account for the torsiona fexibiity of the sab- to- coumn connection that reduces its efficiency for transmission of moment. The equivaent coumn is shown in Figure 3.7. Figure3.5 : Equivaent coumn 45

33 Moment of Inertia of Coumn (I c ) ACI code defines the moment of inertia of coumn as shown in Figure 3.6. I = 3 c c I c = Variab e c I = Figure 3.6 : Coumn area for moment of inertia Design Method by EFM Step : Determination of Factored Load Step : Determination of Sab Thickness Minimum required sab thickness s obtained from Tabe 3.6 & Tabe 3.7. Step 3 : Fexura Stiffness of Actua Coumn (K c ) K c = κ ce cc c I cc κ c = Coumn stiffness coefficient ( to be obtained from Appendix C-) E cc = moduus of easticity of coumn concrete 46

34 I c = moment of inertia of coumn = 3 C C c = ength of coumn (c/c) Step 3 : Torsiona Stiffness of Transverse Torsiona Member (K t ) 9E cs C K t = 3 C C can be determined as mentioned in step 4 (c) of section ` Sign impies that K t of the transverse member in each side of interior coumn is computed separatey and added. For exterior coumns, there is ony one transverse member. For beam aong center ine of coumn K t shoud be corrected.(aci Code ) I sb K t, corrected = K t * I s I sb = Moment of inertia of sab with a beam I s = Moment of inertia of sab without such beam Step 5 : Fexura Stiffness of Equivaent Coumn (K ec ) K ec = K c + K t K c = K c + K c Step 6 : Fexura Stiffness of Sab (K s ) K s κ E = s cs I s κ s = coefficient of sab stiffness ( to be obtained from Appendix C- and C-3) 3 I s = h 47

35 Step 7 : Distribution Factor (D.F) k s k K s S s3 + k ec K S + K S 3 + Kec K S + K S 3 + Kec K DF DF DF DF DF k s k s k s 3 k ec K ec k k ec ec s + kec ks + ks3 + kec k Figure 3.7 : Distribution factors for sab-coumn joints Step 8 : Carry Over Factors and Moment Coefficient (M) Carry over factors (C.O.F) and moment coefficient (M) for sab beam are obtained from Appendix C-,, 3) Step 9 : Moment Anaysis The ongitudina moments of equivaent frames are obtained by Moment Distribution Method. For different oading conditions distributed negative and positive moments are computed. Maximum moments are taken as design moment. Live oading pattern is known, frame sha be anayzed for that oad. ( ACI Code ) 3 Variabe LL, but LL < DL, then maximum factored moment occur at a sections 4 with fu LL on entire sab system. ( ACI Code ) 48

36 Variabe LL, but LL > 4 3 DL, three oading case to be considered : (ACI Code ) Tota oad (w u ) on a panes DL on a panes and 4 3 LL on midspan of a pane. DL on a panes and 4 3 LL on adjacent panes. Tota pane moment (Mp) is computed using equation: wun M p = 8 F.E.M are computed using equation: F.E.M = M w u Anayzing by Moment Distribution Method fina negative moments at the supports are computed. Positive moments at midspan is obtained by M(+) = M p - [sum of M(-) in a pane after distribution] Reduction in Negative Moments: The negative moment as obtained is appicabe for centerines of support. Since the support is not a knife edge but rather a broad band, ACI Code specifies a reduction in negative moment at critica section. When a sab system satisfy the six imitations of DDM, but are anayzed by EFM, Mo further reduction in computed moments are permitted to the proportions of as such, Design moments < Mo (ACI code ) M T = tota pane moment Wun Mo = 8 M T 49

37 Step 0 : Transverse Distribution of Longitudina Moment According to ACI Code the distribution of ongitudina moments to coumn strip and haf midde strips to be done as mentioned in step 4 of section For Step to Step 6 foow Step5 to Step0 of section Step : Determination of Effective Depth (d) Step : Determination of Stee Area (A s ) Step 3: Check for Minimum Stee Step 4: Tota Number of Bar Step 5: Check for Defection Contro Step 6: Reinforcement Detaiing 50

38 3.4.3 COEFFICIENT METHOD Genera The method makes use of tabes of moment coefficient for a variety of conditions. These coefficients are based on eastic anaysis but aso account for ineastic redistribution. This method was recommended in 963 ACI Code for the specia case of two-way sabs supported on four sides by reativey deep, stiff, edge beams. b a 4 ; C. a a ;M a 4 ;CS b 4 b ;C ; 4 b ;C S C.S = coumn strip; M.S = midde strip Figure 3.8 : Eements of two- way sab with beam by coefficient method a = ength of cear span (face of support to support) in short direction b = ength of cear span (face of support to support) in ong direction 5

39 The moments in the midde strips in two directions are: M a = C a w u a M b = C b w u b C a, C b = tabuated moment coefficients DESIGN BY COEFFICIENT METHOD Step : Seection of Stab Thickness P h = in, P = pane perimeter Step : Cacuation of Factored Load w u =.4 D +.7L D = dead oad = w c * h psf ; w c = 50 b Step 3 : Determination of Moment Coefficient a m = b Case type is identified from end conditions. Using the vaue of m corresponding moment coefficients are obtained for respective case type : C a, neg and C b, neg are obtained from Appendix D-. C a, d, pos and C b, d, pos are obtained from Appendix D-. C a,, pos and C b,, pos are obtained from Appendix D-3. 5

40 Step 4 : Cacuation of Moment Midde strip moment Positive moment M a, pos = C a, d w u a + C a, w u a M b, pos = C b,d w u b + C b, w u b Negative Moments for continuous Edge M a, neg,cont = C a, neg w u a M b, neg,cont = C b, neg w u b Negative Moments for Discontinuous Edge M a, neg, discont = 3 Ma, pos M b, neg, discont = 3 Mb, pos Coumn strip moment The moments in coumn strips shoud be taken as /3 rd of midde strip s moment in respective directions. Step 5 : Check the Design Thickness d = M u f y φρf yb( 0.59ρ ) f c If (d + cear cover) < h ; design is ok. Otherwise redesign the thickness. (For detais see step 4 of section 3.3) 53

41 Step 6 : Reinforcement for Midde Strip Required reinforcement can be determined by Iteration process as given in Figure 3.3. Reinforcement sha be determined for short direction and ong direction separatey as foows: Short Direction Midspan Continuous Edge Discontinuous Edge Long Direction Midspan Continuous Edge Discontinuous Edge Check for Minimum Reinforcement: According to ACI Code 3.3. the minimum reinforcement in each direction sha be as mentioned in Tabe 3.5. Spacing: Using # 3 or # 4 bar required spacing is determined. Maximum spacing < h (ACI Code 3.3.) Step 7 : Reinforcement for Coumn Strip Bars seected for midde strip are used in coumn strips, with the spacing 3/ times that in the midde strip, but spacing < h. Step 8 : Check for Shear Percent of tota oad as transmitted in each direction is obtained from Appendix D-4 Load per foot on the beams are determined. The shear to be transmitted by the sab to these beams is = beam oads Shear at critica section at a distance d from beam face = V u 54

42 Shear strength of the sab, φ V c = f c bd φv c > Vu design is ok, otherwise thickness shoud be redesigned. 3.5 CONTROL OF DEFLECTION 3.5. GENERAL ACI Code Commentary R 9.5. estabishes two methods for controing defections: For non-prestressed two-way construction, minimum thickness as required by Tabe 3.4 & Tabe 3.5 wi satisfy the requirements of the code. When there is need to use member depths shaower than are permitted by Tabe 3.4 & Tabe 3.5 or when members support construction is ikey to be damaged by arge defections, defections shoud be cacuated and compared with ACI Code imiting vaues as given in Tabe IMMEDIATE DEFLECTION Immediate defection is aso termed as Short-Term defection and cacuated using the formua given in Tabe

43 Tabe 3. : Cacuation of immediate defection. Live oad defection 3M bb Δ = 3E I c e # Both ends continuous or equay restrained or, One or both ends discontinuous, but monoithic with beam. # M b = ive oad +ve moment # Sab supported by masonry was # M b = ive oad +ve moment. Live oad defection 5M bb Δ = 48EcI e 3. Dead oad defection M bb Δ d = 6EcI e 4. Dead oad defection 5M bb Δ d = 48EcI e b = Cear span in ong direction M b, pos, d M b, pos, or M b = Unfoctored moment =.4. 7 in ong direction Defection can be cacuated in short direction aso in the same way. I e = Effective moment of inertia for computation of defection # Both ends continuous and fuy fixed # M b = maximum dead oad +ve moment # Both ends free of restraint (Supported on masonry wa) # M b = maximum dead oad +ve moment b = Cear span in ong direction; I e = effective moment of inertia M b pos, d M b, or M b = Unfoctored moment =.4. 7, pos, in ong direction Determination of I e Where, M cr = f r y I t g (ACI Code ) 56

44 y t = distance from centroida axis of gross section, negecting reinforcement, to extreme fibre in tension, in. f r = moduus of rupture of concrete, psi. For norma weight concrete: f r = 7.5 / f c For ight weight concrete one of the foowing modifications sha appy: When average tensie strength, f ct is specified f ct f f r = 7.5 =. f ct, ct / f c When f ct is not specified f ct f r = 0.75 * 7.5 ; for a ightweight concrete 6. 7 f ct f r = 0.85 * 7.5 ; for sand- ightweight concrete 6. 7 I e for Continuous Spans ( ACI Code ) I e = 0.50 I em ( I e + I e ) I em = effective moment of inertia for the midspan section I e, I e = negative moment sections at the respective bean ends LONG TERM DEFLECTION Initia defections increase significanty if dead oads sustain over a ong period of time, due to the effects of shrinkage and creep According to ACI Code

45 Δ ong = Δ d,short * λ ξ Where, λ= / + 50ρ / ρ = vaue at midspan for simpe and continuous span = at support for cantiever ξ = time- dependent factor (Tabe 3. or Figure 3.9) Tabe 3. : Vaues of ξ ( ACI Code ) 5 years or more.0 months.4 6 months. 3 months.0 Figure 3.9 : Vaues of ξ TOTAL DEFLECTION Δ tota = Δ ong + Δ, short Defection shoud be cacuated aong both direction and maximum vaues wi be considered. (ACI Code ) 58

46 Desired vaue : Δ tota < imiting vaue given in Tabe 3.3 Tabe 3.3 : Maximum permissibe computed defection (ACI Tabe 9.5.b) Type of member Defection to be considered Defection imitation Fat roofs not supporting or attached to nonstructura eements ikey to be damaged by arge defection Foors not supporting or attached to nonstructura eements ikey to be damaged by arge defection Immediate defection due to ive oad L 80 Immediate defection due to ive oad L 360 Roof or foor construction supporting or attached to nonstructura eements ikey to be damaged by arge defection Roof or foor construction supporting or attached to nonstructura eements not ikey to be damaged by arge defection That part of the tota defection which occurs after attachment of the nonstructura eements, the sum of the ong- time defection due to a sustained oads, and the immediate defection due to ive oad L STRIP METHOD FOR SLABS Introduction: The strip method is a ower bound approach, based on satisfaction of equiibrium requirements everywhere in the sab. By the strip method a moment fied is first determined that fufis equiibrium requirements, after which the reinforcement of the sab at each point is designed for this moment fied. The strip method gives resuts on the safe side, which is certainy preferabe in practice, and differences from the true carrying capacity wi never impair safety. The strip method is a design method, by which the needed reinforcement can be cacuated. It encourages the designer to vary the reinforcement in a ogica way, eading to an economica arrangement of stee as we as a safe design. 59

47 Choice of Load Distribution Condition-: The simpest oad distribution is obtained by setting k = 0.5 over the entire sab, as shown in figure beow. The oad on a strips in each direction is then w/, as iustrated by oad dispersion arrows in figure. This gives maximum design moments m x = m y = wa 6 Y Simpe supports 4 sides a wa 6 A A a (a) Pan view X (d) m x across X=a/ w/ (b) w x aong A-A (c) m x aong A-A Figure 3.0: Square sab with oad shared equay in two directions 60

48 Condition-: An aternative, more reasonabe distribution is shown in figure beow. Here the regions of different oad dispersion, separated by dash-dotted discontinuity ines, foow the diagonas, and a of the oad on any region is carried in the direction giving the shortest distance to the nearest support. The soution proceeds, giving k vaues of either 0 or, depending on the region, with oad transmitted in the direction indicated by the arrows in figure. For a strip A-A at a distance y a/ from the X-axis, the design moment is m x = wy Y Simpe supports 4 sides a A A y a (a) Pan view X Wa / (d) m x across X=a/ w w y (b) w x aong A-A Wy / (c) m x aong Figure 3.: Square sab with oad dispersion ines foowing diagonas 6

49 Condition-3: A third aternative distribution is shown in figure beow. Here the division is made so that the oad is carried to the nearest support, as before, but oad near the diagonas has been divided, with one-haf taken in each direction. Thus k is given vaues of 0 or aong the midde edges and vaue of 0,5 in the corner and center of the sab. For an X direction strip aong section A-A, the maximum moment is wa w a a m x = x x = And for a strip aong section B-B, the maximum moment is a a w a m x = w x x + x x a 8 = 5wa 64 6

50 Y a/4 a/ a/4 Wa /64 a/4 w/ w/ a/ B w w/ w B a a/4 A w/ w/ A X a (a) Pan view 5Wa /6 (d) m x across x=a/ w/ w/ Wa /64 (b) w x and m x aong A-A w w/ w (c) w x and m x aong B-B 5Wa /6 Figure.: Square sab with oad near diagonas shared equay in two directions 63

51 Condition-4: The preferred arrangement, shown in figure beow, gives design moment as foows: In the X direction: Side strips: wb w b b m x = x x = Midde strips: wb b b m x = w x x = In the Y direction: Side strips: wb w b b m x = x x = Midde strips: wb b m x = w x b x =

52 b b a b a a b b a b/ b b/ a Figure.3: Rectanguar sab with discontinuity ines originating at the corners b/4 A-b/ b/4 Wa /64 w/ w/ b/4 b b/ w/ w/ b/4 a Figure.4: Discontinuity ines parae to the sides for a rectanguar sab 65

53 Condition-5: For sab strips with one end fixed and one end simpy supported, the due goas of constant moment in the unoaded centra region and a suitabe ratio of negative to positive moments govern the ocation to be chosen for the discontinuity ines. Figure a shows a uniformy oaded rectanguar sab having two adjacent edges fixed and the other two edges simpy supported. The moment curve of figure b is chosen so that moment is constant over the unoaded part, i.e., shearing force is zero. The maximum positive moment in the X direction midde strip is then m xf = αwb x αb 4 α wb = 8 Accordingy, the distance from the right support, figure c, to the maximum positive moment section is chosen as α b. It foows that the maximum positive moment is m yf = αwb x αb α wb = With the above expressions, a the design moments for the sab can be found once a suitabe vaue for α is chosen. The vaues of α from 0.35 to 0.39 give corresponding ratios of negative to positive moments from.45 to

54 b ( α ) b a b α B w/ w/ b α b A b/ A w/ w/ b ( α ) w B a (a) Pan view w ( α ) wb 8 wb α 8 Figure.5: w x and m x aong A-A 67

55 wb α ( α ) wb Figure.6:-w y and m y aong B-B 3.6. DESIGN BY STRIP METHOD: Step-: Seection of Sab Thickness From tabe-3.7 (ACI Code ) p

56 Step-: Cacuation of Factored Load W u W u =.4 D+.7 L (ACI Code-00) =. D+.6 L (ACI Code-0) Where, D =50 x h (psf) Step-3: Seection of Load Distribution From choice of oad distribution. Step-4: Cacuation of Moment From the equations of oading condition moments are cacuated. Step-5: Check For Design Thickness d = M u f φρf y ( 0.59ρ f y c' ) If (d + cear cover) h; design is OK. Step-6: Reinforcement Cacuation Reinforcement cacuation is done by iteration method from figure-3.3 but compare of moment shoud be done with minimum stee requirement. φm n =φρf y bd f y ( ( 0.59ρ ) f If φm n <M; then ony minimum reinforcement. If φm n >M; then iteration from figure-3.3. c' 69

57 ρ min from Tabe-3.3 (ACI Code-7.) Spacing: Using #3 and #4 bar. Maximum spacing h (ACI Code-3.3.) Cut-off points can be cacuated from moment diagrams and deveopment ength shoud be provided. 70

58 Tabe 3.4: Choice of Load Distribution Case- Case- m x = m x = m y = wy wa 6 Case-3 For an X direction strip aong section A-A, the maximum moment is w a a wa m x = x x = And for a strip aong section B-B, the maximum moment is a a w a 3a 5wa m x = w x x + x x = In the X direction: Side strips: wb w b b m x = x x = Case-4 Midde strips: In the Y direction: Side strips: wb b b m x = w x x = w b b m x = x x = wb Midde strips: wb b m x = w x b x = 8 8 Case-5 The maximum positive moment in the X direction midde strip is then αwb αb α wb m xf = x = 4 8 The maximum positive moment is αb α wb m yf = αwb x = 7

59 3.7 EXAMPLE FOR DESIGN OF SLAB 3.7. EXAMPLE: DESIGN OF SLAB BY DDM Probem: A pan of a market buiding is given in Figure 3.0. Necessary data are furnished beow: Live oad = 60 psf Story height = 9 ft Sab thickness = 5 in. f c = psi f y = 50,000 psi No edge beam. Design the sab as Fat pate by DDM. A Coumn = 60 = in.x0 in. 5 = 75 Figure 3.0 : Foor pan of the buiding of Exampe

60 Soution: The probem is soved with reference to section Step : Cacuation of Factored Load Thickness of the sab = 5½ 5. 5 DL = 50 x = 69 psf and LL = 60 psf W=.4D +.7L = (.4 * 69) + (.7 * 60) = 98 psf = 0.98 ksf Step : Check for Sab Thickness Cear span n.ong = (5-) = 4 and n,short = (-0/) =.7 For α = o; n = 5- = 4 Using Tabe 3.6 by interpoation for f y = 50 ksi For Exterior Pane t min = n + n = *68 + = in For Interior Pane t min = n + n = *68 + = in According to ACI Code (a) the min m thickness for fat pate is 5 So, given thickness of sab = 5½ (Ok) 73

61 Step 3 : Determination of Tota Factored Static Moment M o = u n W 8 M o,ong = 8 (0.98)* *4 = 58. ft kips M o,short = 8 (0.98) *5 *.8 = 46.3 ft kips Step 4 : Longitudina Distribution of Moment From Tabe 3.8 for Fat Pate (Case 3 and Case 5) M o for A = 58. ft kips M o for B = (58.) = 9. ft - kips M o for C = 43.3 ft kips M o for D = 3. ft kips 74

62 Step 5 : Transverse Distribution of Longitudina Moment Cacuation of Aspect Ratio For A & B : = = : 5 For C & D : = =. 5 Cacuation of α Since no edge beam α = o for a Cacuation of β t I s in β t I s = 3 bh For A & B : I s = (x)(5.5) 3 = 000in 4 For C & D : (5x)(5.5) Torsiona Constant, C 3 = 500in 4 Since no actua edge beam, use Figure.8 (b) for cacuation of the torsiona member x x C = 3 y 0.63 y 3 For ong direction C = x0 0.63* = 363in

63 For short direction: C = β t = x 0.63* = 474in 3 C I s For A & B : β t = = 0. 8 x For C & D : β t = = x 500 Finding out α A B C D β t α α Percentage of Longitudina Moment in Coumn Strip For Exterior Negative Moment A B C D 98.8% 98.8% 99.3% 99.3% Expanation for A and B = 0.80: α = 0; β t =

64 From Tabe 3.9: α = β t = β t = β t > Interpoation in both directions For β t = 0.8 % of moment = x =.% decrease % of moment in coumn strip = (00.) = 98.8% Expanation for C and D =.5: α = 0 ; β t = From above Tabe, by interpoation in both direction For, βt = 0.073, % of moment decrease = ( ) * = 0.70%.50 0 % of moment in coumn strip = ( ) = 99.3% For Positive Moment For α = 0 A B C D 60% 60% 60% 60% 77

65 For Interior Negative Moment For α = 0 and = 0.80 and =.5 A B C D 75% 75% 75% 75% Tabe 3.5 : Summary of Cacuation Transverse Distribution of Longitudina Moment Ser. Equivaent Rigid Frame A B C D. Transverse width (in) Coumn strip width (in) Haf midde strip width C (in 4 ) I s (in 4 ) in β t β t α α Exterior ve moment, percent to coumn strip. 98.8% 98.8% 99.3% 99.3%. Positive moment percent to coumn strip 60% 60% 60% 60%. Interior negative moment, percent to coumn strip. 75% 75% 75% 75% 78

66 Tabe 3.6 : Distribution of factored moment in Coumn Strip & Midde Strip For equivaent Rigid Frame A Ser. Moments at Vritica section (ft.- kips) -ve moment Exterior span +ve moment -ve moment -ve moment Interior span +ve moment -ve moment. Tota moment in equivaent rigid frame A. Percentage to coumn strip 3. Moment in coumn strip 4. Moment in midde strip % 60% 75% 75% 60% 75%

67 3.7. EXAMPLE: DESIGN OF SLAB BY EFM Probem: A muti-story market buiding is panned using a fat pate foor system as shown in Figure 3.. Necessary data are given beow: Live Load = 00 psf Foor finish = 0 psf Foor to foor height = ft. fc = 4000 psi f y = 60,000 psi Coumn size = 8 in. x 8 in. Design the Interior Pane C by EFM. A B ft B C ft Figure 3. : Foor pan for Exampe

68 Soution: The probem is soved with reference to section The EFM is used to determine the ongitudina moments ony. As mentioned earier the transverse distribution process of ongitudina moments and reinforcement cacuation are simiar to DDM (Section , step 4). Hence, this probem is soved upto determination of ongitudina moments. The structure is identica in each direction, permitting the design for one direction to be used for both. Step : Determination of Sab Thickness Minimum thickness h for a fat pate is obtained from Tabe 3.6.For an exterior pane: n h= 30 = 0.5* 30 = 8.0 in in. Step : Determination of Factored Load Sab DL = w c * h = (50*8.50)/ = 06 psf ; w c = weight of concrete= 50 pcf Super imposed DL = 0 psf Tota DL = (06=0) 6 psf Factored Load: DL=.4D =.4* 6= 76psf LL=.7L =.7*00= 70 psf Step 3 : Determination Fexura Stiffness of Actua Coumn K c = k c E cc c I cc Obtain k c from Appendix C- 8

69 For fat pate structure it is assumed a members are prismatic, negecting the increase in stiffness within the joint region. Take k c = 4 Consider E c = constant Step 4 : Determination Torsiona Stiffness of Transverse Torsiona Member C = x x 3 y 0.63 y 3 h f = 8.5 in. C = 8 in C = * = 590 in 4 9E K cs C t = 3 C 9E * 590 c K t = 3 8 / 64 = 09 E c Step 5 : Determination of Fexura Stiffness of Equivaent Coumn K ec = K c + K t 8

70 K c = * 43 E c K t = * 09 E c = + K ec = 5 E c K 486E 8E ec c c Step 6 : Determination of Fexura Stiffness of Sab κ E K = s s cs I s Obtain vaue of κ s from Appendix C-. For fat pate assume κ s = 4 3 h I s = = 3 64 *8.5 =350 K = s E 4 c * = 05 E c Step 7 : Cacuation of Distribution Factor Distribution factors at each joint are cacuated according to step 7 of section Step 8 : Determination of Carry Over Factors and Moment Coefficient C.O.F and Moment coefficient for sab- beam are obtained from Appendix C-.Moment coefficient COF=.503(for both cases). 83

71 Step 9 : Moment Anaysis Since LL = 70 psf > 4 3 DL=3 psf three oading cases shoud be considered: Tota oad on a pane DL oad on a panes and ¾ LL on midspan of a pane DL on a panes and ¾ LL on adjacent pane Tabe 3.7 : Longitudina moment in fat pate foor Pane B C B Joint DF Fixed end moments Fina moments Span moment in C 3 (b) 76 psf panes B 304 psf pane C Fixed end moments Fina moments Span moment in C 5 (c) 304 psf panes B(eft) & C &76 psf pane B (right) Fixed end moments Fixed end moments Span moment in C 34 84

72 3.7.3 EXAMPLE: DESIGN BY COEFFICIENT METHOD Probem: A pan of a residentia buiding is given in Figure 3.. Necessary data are given beow: Live Load = 40 psf fc = 3000 psi fy = 60,000 psi Coumn size = x Design the corner pane A as two-way sab with beam by coefficient Method. 6 A Coumn size: Figure 3. : Foor pan for exampe Soution: The probem is soved with reference to section and Appendix D-,, 3, 4. 85

73 Step: Determination of Minimum Thickness h = p 80 = = 0.36 ft. = 4.7 in. whwre, P = (6+6) = 6 Seect h = 5 in. as tria depth Step : Cacuation of Factored Load DL = w c * h psf DL = 50 x LL = 40 psf 5 = psf ; where w c = 50 pcf W =.4D +.7 L = (.4 * * 40) = 36 psf Step 3 : Determination of Moment Coefficient a 6 Length ratio, m = = = 6 b From the end condition case type is Case 4 From Appendix D- C a, neg = 0.05;C b, neg = 0.05 From Appendix D- C a,d,pos = 0.07; C b,d,pos = 0.07 From Appendix D-3 C a,,pos = 0.03; C b,,pos =

74 Step 4 : Cacuation of Moment Midde Strip Moment: Positive Moments at Midspan M a, pos = C a, d W a + C a, W a M a, pos = 0.07 * 36 * * 36 * 6 = 494 ft-b M b, pos = C b, d W b + C a, W b M b, pos = 0.07 * 36 * * 36 * 6 = 494 ft-b Negative Moments at Continuous Edge M a, neg = C a, neg W a = 0.05 * 36 * 6 = 473 ft b M b, neg = C b, neg W b = 0.05 * 36 * 6 = 473 ft b Negative Moment at Discontinuous Edge M a, neg, discontinuous = 3 * Ma, pos = 3 * 494 = 64 ft b M b, neg, discontinuous = 3 * Mb, pos = 3 * 494 = 64 ft b Coumn Strip Moment: Coumn strip moments are /3 of corresponding midde strip s moments in respective direction. 87

75 Step 5 : Check the Design Thickness d = φρf y M u f ( 0.59ρ f y c' ) Here ρ = ρ max = 0.75ρ b = 0.75 * 0.85*β f / c * f y f y = 0.06 d = 494* *0.06*60,000*( 0.059*0.06* ) 3 =.4 in. h required = (d + cear cover= in ) = 3.4 in. h required < h design, design is OK Step 6 : Cacuation for Reinforcement for Midde Strip In Short Direction: Midspan M u = 494 * b- in. By Iteration process as given in Step 5 of section 3.3 find: A s = 0.30 in. /ft Using # 3 bar required spacing: Spacing = x0. = c / c

76 Continuous Edge M u = 473 * b- in. By Iteration process as given in Step 5 of section 3.3 find: As = 0.5 in /ft. Using # 3 bar required spacing: Spacing = x0. = c / c 0.5 Discontinuous Edge The negative moment at discontinuous edge is one third of positive moment in the span. It woud be adequate to bend up every third bar from the bottom to provide negative moment stee at discontinuous edge. However the spacing woud be = But maximum aowabe spacing = h = 0 in. So, using # 0 in c/c. In Long Direction: Being equa moments, the reinforcement in ong direction wi be equa to short direction in this case. Step 7 : Cacuation for Reinforcement for Coumn Strip The average moments in coumns being two-third of the corresponding moments in the midde 3 strips, adequate stee wi be furnished if the spacing of this stee is times that in the midde strip. Using # 3 bar spacing for coumn strip 89

77 Midspan = 4 x 3 = 6 c/c Continuous edge = 5x 3 = 7.5 c/c Discontinuous edge = 8x 3 = c/c But maximum aowabe spacing = h = 0 in. Use 0 c/c. Step 8: Detaiing B A B A Figure 3.3 : Detaiing for exampe

78 # 0 in c/c # 4 in c/c # 5 in c/c Section A-A Section B-B Figure 3.3 : Detaiing for exampe (continued) 9