Module Reinforced Concrete Slabs Lesson Oneway Version 2 CE IIT, Kharagpur instructional Objectives: At the end of this lesson, the student should be able to: state the names of different types of slabs used in construction, identify oneway and twoway slabs stating the limits of I, /l,, ratios for one and twoway slabs, explain the share of loads by the supporting beams of one- and twoway slabs when subjected to uniformly distributed vertical loads, explain the roles of the total depth in resisting the bending moments, shear force and in controlling the deflection, state the variation of design shear strength of concrete in slabs of different depths with identical percentage of steel reinforcement, assume the depth of slab required for the control of deflection for different support conditions, determine the positive and negative bending moments and shear force, determine the amount of reinforcing bars along the longer span, state the maximum diameter of a bar that can be used in a particular slab of given depth, decide the maximum spacing of reinforcing bars along two directions of oneway slab, design oneway slab applying the design principles and following the stipulated guidelines of IS 456, draw the detailing of reinforcing bars of oneway slabs after the design. Version 2 CE IIT, Kharagpur 8.18.1 Introduction Fag a_L13_«§{a}; mg 393;; .. . .;.; Fig. 3.1331:2]:nnnums in .......,...,..,.... ...».~,...~....... .......,. ....; diramimns . Fiégi, a.13.1(b]:Cannunus in ans:direrzm Mates: 1 Bnamay silt ii L,:~« EL; 2. asnd'sc=a&«as thai mt:suppsnriis needed E1I3,:- EL;and is needed 3. End Eigl may be simpty cardamped I, *4;EL 8.13.1: Hari2:.mtal slabs Hmmnm gilab LV_ :9}/M hv ga r2LtViEvasm§} Em:._l_ig*1s5d smh Fig. 8.18.2: Stair Ease Version 2 CE IIT, Kharagpur _;; ;..:@' 4 Fig. lttrrsiimd radii Slabs, used in floors and roofs of buildings mostly integrated with the supporting beams, carry the distributed loads primarily by bending. It has been mentioned in sec. 5.10.1 of Lesson 10 that a part of the integrated slab is considered as flange of T or L-beams because of monolithic construction. However, the remaining part of the slab needs design considerations. These slabs are either single span or continuous having different support conditions like fixed, hinged or free along the edges (Figs.8.18.1a,b and c). Though normally these slabs are horizontal, inclined slabs are also used in ramps, stair cases and inclined roofs (Figs.8.18.2 and 3). While square or rectangular plan forms are normally used, triangular, circular and other plan forms are also needed for different functional requirements. This lesson takes up horizontal and rectangular /square slabs of buildings supported by beams in one or both directions and subjected to uniformly distributed vertical loadings. The other types of slabs, not taken up in this module, are given below. All these slabs have additional requirements depending on the nature and magnitude of loadings in respective cases. (a) horizontal or inclined bridge and fly over deck slabs carrying heavy concentrated loads, (b) horizontal slabs of different plan forms like triangular, polygonal or circular, (c) flat slabs having no beams and supported by columns only, (d) inverted slabs in footings with or without beams, (e) slabs with large voids or openings, (f) grid floor and ribbed slabs. Version 2 CE IIT, Kharagpur 8.18.2 Oneway and Two-way Slabs if ;§§'f3{i%§mEv§i;§F§¬rf§ $:& §Ol34m»92«¢§v2z.¢lol§i%:».¢2c+-Fmk siuads wrsie-*3by Ftszumrrf L2 """-~«.... leresbsostiic $&«tI3§.lJ|¬Zt~l!§l almrtgzj slrrgmti sgzsarit Fig.ia.*ta.a.:a;}: "l"§%'»W§'f' stab. 3e'3 Eln -ifs ' Fig. 3.3% mt: Fig. T*wfo»w:ar§« wars igit, :2 3;: Sfharingof loads Figures 8.18.4a and b explain the share of loads on beams supporting solid slabs along four edges when vertical loads are uniformly distributed. It is evident from the figures that the share of loads on beams in two perpendicular directions depends upon the aspect ratio ly /IX of the slab, IX being the shorter span. For large values of ly, the triangular area is much less than the trapezoidal area (Fig.8.18.4a). Hence, the share of loads on beams along shorter span will gradually reduce with increasing ratio of I, /IX. In such cases, it may be said that the loads are primarily taken by beams along longer span. The deflection profiles of the slab along both directions are also shown in the figure. The deflection profile is found to be constant along the longer span except near the edges for the slab panel of Fig.8.18.4a. These slabs are designated as one-way slabs as Version 2 CE IIT, Kharagpur they span in one direction (shorter one) only for a large part of the slab when I, /IX > 2. On the other hand, for square slabs of I, /IX= 1 and rectangular slabs of I, /IX up to 2, the deflection profiles in the two directions are parabolic (Fig.8.18.4b). Thus, they are spanning in two directions and these slabs with I, /I,, up to 2 are designated as twoway slabs, when supported on all edges. It would be noted that an entirely oneway slab would need lack of support on short edges. Also, even for I, /I,, < 2, absence of supports in two parallel edges will render the slab oneway. In Fig. 8.18.4b, the separating line at 45 degree is tentative serving purpose of design. Actually, this angle is a function of I, /IX. This lesson discusses the analysis and design aspects of one-way slabs. The twoway slabs are taken up in the next lesson. 8.18.3 Design Shear Strength of Concrete in Slabs Experimental tests confirmed that the shear strength of solid slabs up to a depth of 300 mm is comparatively more than those of depth greater than 300 mm. Accordingly, cl.40.2.1.1 of IS 456 stipulates the values of a factor k to be multiplied with r, given in Table 19 of IS 456 for different overall depths of slab. Table 8.1 presents the values of kas a ready reference below: Table 8.1 Values of the multiplying factor k Overall 300 or depth of slab more 275 250 225 200 175 150 or less mm WT Thin slabs, therefore, have more shear strength than that of thicker slabs. It is the normal practice to choose the depth of the slabs so that the concrete can resist the shear without any stirrups for slab subjected to uniformly distributed loads. However, for deck slabs, culverts, bridges and fly over, shear reinforcement should be provided as the loads are heavily concentrated in those slabs. Though, the selection of depth should be made for normal floor and roof slabs to avoid stirrups, it is essential that the depth is checked for the shear for these slabs taking due consideration of enhanced shear strength as discussed above depending on the overall depth of the slabs. Version 2 CE IIT, Kharagpur 8.18.4 Structural Analysis As explained in sec. 8.18.2, oneway slabs subjected to mostly uniformly distributed vertical loads carry them primarily by bending in the shorter direction. Therefore, for the design, it is important to analyse the slab to find out the bending moment (both positive and negative) depending upon the supports. Moreover, the shear forces are also to be computed for such slabs. These internal bending moments and shear forces can be determined using elastic method of analysis considering the slab as beam of unit width i.e. one metre (Fig.8.18.1a). However, these values may also be determined with the help of the coefficients given in Tables 12 and 13 of IS 456 in c|.22.5.1. It is worth mentioning that these coefficients are applicable if the slab is of uniform crosssection and subjected to substantially uniformly distributed loads over three or more spans and the spans do not differ by more than fifteen per cent of the longer span. It is also important to note that the average of the two values of the negative moment at the support should be considered for unequal spans or if the spans are not equally loaded. Further, the redistribution of moments shall not be permitted to the values of moments obtained by employing the coefficients of bending moments as given in IS 456. For slabs built into a masonry wall developing only partial restraint, the negative moment at the face of the support should be taken as WI/24, where W is the total design loads on unit width and I is the effective span. The shear coefficients, given in Table 13 of IS 456, in such a situation, may be increased by 0.05 at the end support as per cI.22.5.2 of IS 456. 8.18.5 Design Considerations The primary design considerations of both one and two-way slabs are strength and deflection. The depth of the slab and areas of steel reinforcement are to be determined from these two aspects. The detailed procedure of design of oneway slab is taken up in the next section. However, the following aspects are to be decided first. (:3) Effective span (cI.22.2 of IS 456) The effective span of a slab depends on the boundary condition. Table 8.2 gives the guidelines stipulated in cI.22.2 of IS 456 to determine the effective span of a slab. Table 8.2 Effective span of slab (cI.22.2 of IS 456) Su o ort condition Simply supported not built integrally with its supports Effective san Lesser of (i) clear span + effective depth of slab, and (ii) centre to centre of su o orts Version 2 CE IIT, Kharagpur 2 suContinuous when the width ofthe ort is <1/12" of clear san 3 Continuous when the widthof the supportis > lesserof 1/12"of clear (i) Clear span betweenthe supports span or 600 mm (i) for end span with one end fixed and the other end continuous or for intermediate spans, (ii) Lesser of (a) clear span + half the effective depth of slab, and (b) clear span + half the width (ii) for end span with one end free and of the discontinuous support the other end continuous, (iii) spans with roller or rocker (iii) The distance between the centres of bearings beannos. 4 5 Cantilever slab at the end of a Length up to the centre of continuous su slab Cantilever span ort Length up to the face of the support + half the effective deth HE Centre to centre distance (b) Effective span to effective depth ratio (cls.23.2.1a-e of IS 456) The deflection of the slab can be kept under control if the ratios of effective span to effective depth of oneway slabs are taken up from the provisions in c|.23.2.1ae of IS 456. These stipulations are for the beams and are also applicable for oneway slabs as they are designed considering them as beam of unit width. These provisions are explained in sec.3.6.2.2 of Lesson 6. (c) Nominal cover (cl.26.4 of IS 456) The nominal cover to be provided depends upon durability and fire resistance requirements. Table 16 and 16A of IS 456 provide the respective values. Appropriate value of the nominal cover is to be provided from these tables for the particular requirement of the structure. (d) Minimum reinforcement (cl.26.5.2.1 of IS 456) Both for one and twoway slabs, the amount of minimum reinforcement in either direction shall not be less than 0.15 and 0.12 per cents of the total crosssectional area for mild steel (Fe 250) and high strength deformed bars (Fe 415 and Fe 500)/welded wire fabric, respectively. Version 2 CE IIT, Kharagpur (e) Maximum diameter of reinforcing bars (cl.26.5.2.2) The maximumdiameterof reinforcingbars of one and two-way slabs shall not exceed one-eighthof the total depth of the slab. (f) Maximum distance between bars (cI.26.3.3 of is 456) The maximum horizontaldistance between parallel main reinforcingbars shall be the lesserof (i) three times the effectivedepth, or (ii) 300 mm. However, the same for secondary/distribution bars for temperature,shrinkageetc. shall be the lesserof (i) five times the effectivedepth, or (ii) 450 mm. 8.18.6 Designof One-way Slabs The procedure of the design of oneway slab is the same as that of beams. However, the amountsof reinforcingbars are for one metre width of the slab as to be determinedfrom either the governingdesign moments (positiveor negative)or from the requirementof minimumreinforcement.The differentsteps of the designare explainedbelow. Step 1: Selection of preliminary depth of slab The depth of the slab shall be assumed from the span to effectivedepth ratiosas given in section3.6.2.2 of Lesson6 and mentionedhere in sec.8.18.5b. Step 2: Design loads, bending moments and shear forces The total factored (design) loads are to be determined adding the estimated dead load of the slab, load of the floor finish, given or assumed live loads etc. after multiplyingeach of them with the respectivepartialsafety factors. Thereafter, the design positiveand negative bending momentsand shear forces are to be determinedusingthe respectivecoefficientsgiven in Tables 12 and 13 of IS 456 and explainedin sec.8.18.4 earlier. Step 3: Determination/checking of the effective and total depths of slabs The effectivedepth of the slab shall be determinedemployingEq.3.25 of sec.3.5.6 of Lesson5 and is given below as a ready referencehere, M,,,,-,,, = R,,,-,,, bdz (3.25) where the values of R),-m for three different grades of concrete and three differentgrades of steel are given in Table 3.3 of Lesson5 (sec.3.5.6). The value of b shall be taken as one metre. Version 2 CE IIT, Kharagpur The total depth of the slab shall then be determined adding appropriate nominal cover (Table 16 and 16A of c|.26.4 of IS 456) and half of the diameter of the larger bar if the bars are of different sizes. Normally, the computed depth of the slab comes out to be much less than the assumed depth in Step 1. However, final selection of the depth shall be done after checking the depth for shear force. Step 4: epth of the slab for shear force Theoretically, the depth of the slab can be checked for shear force if the design shear strength of concrete is known. Since this depends upon the percentage of tensile reinforcement, the design shear strength shall be assumed considering the lowest percentage of steel. The value of r, shall be modified after knowing the multiplying factor k from the depth tentatively selected for the slab in Step 3. If necessary, the depth of the slab shall be modified. Step 5: Determination of areas of steel Area of steel reinforcement along the direction of oneway slab should be determined employing Eq.3.23 of sec.3.5.5 of Lesson 5 and given below as a ready reference. Mu = 0.87 fy As;d{1 (As,)(fy)/(fck)(bd)} (3.23) The above equation is applicable as the slab in most of the cases is underreinforced due to the selection of depth larger than the computed value in Step 3. The area of steel so determined should be checked whether it is at least the minimum area of steel as mentioned in c|.26.5.2.1 of IS 456 and explained in sec.8.18.5d. Alternatively, tables and charts of SP-16 may be used to determine the depth of the slab and the corresponding area of steel. Tables 5 to 44 of SP-16 covering a wide range of grades of concrete and Chart 90 shall be used for determining the depth and reinforcement of slabs. Tables of SP-16 take into consideration of maximum diameter of bars not exceeding one-eighth the depth of the slab. Zeros at the top right hand corner of these tables indicate the region where the percentage of reinforcement would exceed p,,,,-m. Similarly, zeros at the lower left and corner indicate the region where the reinforcement is less than the minimum stipulated in the code. Therefore, no separate checking is needed for the allowable maximum diameter of the bars or the computed area of steel exceeding the minimum area of steel while using tables and charts of SP-16. The amount of steel reinforcement along the large span shall be the minimum amount of steel as per c|.26.5.2.1 of IS 456 and mentioned in sec.8.18.5d earlier. Version 2 CE IIT, Kharagpur Step 6: Selection of diameters and spacings of reinforcing bars (cls.26.5.2.2 and 26.3.3 of IS 456) The diameterand spacingof bars are to be determinedas per c|s.26.5.2.2 and 26.3.3 of IS 456. As mentionedin Step 5, this step may be avoided when usingthe tables and chartsof SP-16. 8.18.7 Detailing of Reinforcement Fig. mwa_.5:a:;: Saar:M 3.13.5: Reinforcement of one-way sigh Version 2 CE IIT, Kharagpur Figures 8.18.5a and b present the plan and section 1-1 of oneway continuous slab showing the different reinforcing bars in the discontinuous and continuous ends (DEP and CEP, respectively) of end panel and continuous end of adjacent panel (CAP). The end panel has three bottom bars B1, B2 and B3 and four top bars T1, T2, T3 and T4. Only three bottom bars B4, B5 and B6 are shown in the adjacent panel. Table 8.3 presents these bars mentioning the respective zone of their placement (DEP/CEP/CAP), direction of the bars (along x or y) and the resisting moment for which they shall be designed or if to be provided on the basis of minimum reinforcement. These bars are explained below for the three types of ends of the two panels. Table 8.3 Steel bars of oneway slab (Figs.8.18.5a and b) Bars Panel 1 Alon o x 2 3 Resistin B1, B2 DEP 4 B B3 DEP B4, B5 CAP moment + 0.5 MXfor each, Minimum steel + 0.5 M for each, CAP 6 6 CEP T1, T2 x T3 DEP X T4 DEP y Minimum steel 0.5 M for each, + 0.5 M Minimum steel Notes: (i) DEP = Discontinuous End Panel (ii) CEP = Continuous End Panel (iii) CAP = Continuous Adjacent Panel (i) Discontinuous nd anal (DEP) Bottom steel bars B1 and B2 are alternately placed such that B1 bars are bars are continuous up to a distance of 0.1lx1 from the centre of support at the discontinuous end. o Top bars T4 are along the direction of y and provided up to a distance of 0.1lx1 from the centre of support at discontinuous end. These are to satisfy the requirement of minimum steel. (ii) Continuous - End Panel (CEP) Top bars T1 and T2 are along the direction of x and cover the entire ly. They are designed for the maximum negative moment Mx and each has a capacity of -0.5MX. Top bars T1 are continued up to a distance of 0.3Ix1,while T2 bars are only up to a distance of 0.15Ix1. o Top bars T4 are along y and provided up to a distance of 0.3lx1 from the support. They are on the basis of minimum steel requirement. (iii) Continuous Adjacent Panel (CAP) - Bottom bars B4 and B5 are similar to B1 and B2 bars of (i) above. - Bottom bars B6 are similar to B3 bars of (i) above. Detailing is an art and hence structural requirement can be satisfied by more than one mode of detailing each valid and acceptable. 8.18.8 Numerical Problems (a) Problem 8.1 Design the oneway continuous slab of Fig.8.18.6 subjected to uniformly distributedimposedloadsof 5 kN/m2usingM 20 and Fe 415. The loadof floor finishis 1 kN/m2.The spandimensionsshownin the figureare effectivespans. The width of beams at the support = 300 mm. jl~5ig.. B.*lfE..: Prtjifem 8.1 Version 2 CE IIT, Kharagpur Solution of Problem 8.1 Step 1: Selection of preliminary depth of slab The basic value of span to effectivedepth ratio for the slab havingsimple supportat the end and continuousat the intermediateis (20+26)/2 = 23 (c|.23.2.1 of IS 456). Modification factorwith assumedp = 0.5 and f = 240 N/mm2isobtained as 1.18 from Fig.4 of IS 456. Therefore,the minimumeffectivedepth = 3000/23(1.18) = 110.54 mm. Let us take the effectivedepth d = 115 mm and with 25 mm cover, the total depth D = 140 mm. Step 2: Design loads, bending moment and shear force Dead loadsof slab of 1 m width = 0.14(25) = 3.5 kN/m Dead load of floor finish =1.0 kN/m Factoreddead load = 1.5(4.5) = 6.75 kN/m Factoredlive load = 1.5(5.0) = 7.50 kN/m Total factored load = 14.25 kN/m Maximum momentsand shear are determined from the coefficientsgiven in Tables 12 and 13 of IS 456. Maximumpositivemoment = 14.25(3)(3)/12 = 10.6875 kNm/m Maximumnegative moment = 14.25(3)(3)/10 = 12.825 kNm/m Maximumshear V,, = 14.25(3)(0.4) = 17.1 kN Step 3: Determination of effective and total depths of slab From Eq.3.25 of sec. 3.5.6 of Lesson5, we have M,,,,-,,, = R,,,-,,, bdz where H,,,-,,, is2.76N/mm2 fromTable3.3 of sec.3.5.6 of Lesson 5. So, d = {12.825(106)/(2.76)(1000)}°'5 = 68.17mm Version 2 CE IIT, Kharagpur Since, the computed depth is much less than that determined in Step 1, let us keep D: 140 mm and d: 115 mm. Step 4: Depth of slab for shear force Table19of IS 456gives 76= 0.28N/mm2 for the lowestpercentage of steel in the slab. Further for the total depth of 140 mm, let us use the coefficient k of cl. 40.2.1.1of IS 456as 1.3to get 7, =k To= 1.3(0.28)= 0.364N/mm2. Table20of IS456gives ram = 2.8N/mm2. Forthisproblem rv =Vu/bd = 17.1/115 = 0.148N/mm"'. Since,rv 30.8025 kNm/m. Fifty per cent of the bars should be curtailed at a Version 2 CE IIT, Kharagpur distance of larger of L0;or 0.5IX.Table 65 of SP-16 gives Ld of 10 mm bars = 470 mm and 05],, = 0.5(1850) = 925 mm from the face of the column. The curtailment distance from the centre line of beam = 925 + 150 = 1075, say 1100 mm. The above, however, is not admissible as the spacing of bars after the curtailment exceeds 300 mm. So, we provide 10 mm @ 300 c/c and 8 mm @ 300 c/c. The moment of resistance of this set is 34.3 kNm/m > 30.8025 kNm/m (see Table 44 of SP-16). Fig. 3."f$.§: Be-ta§lEnlgw:sfiibars1.rofProblem Qwl-t lse»t::. E~iof Fig. 8.13.3} Figure 8.18.9 presents the detailing of reinforcing bars of this problem. 8.18.10 References 1. ReinforcedConcreteLimit State Design,6" Edition,by Ashok K. Jain, Nem Chand & Bros, Roorkee, 2002. 2. LimitStateDesignof ReinforcedConcrete,2 Edition,by P.C.Varghese, PrenticeHa|l of India Pvt. Ltd., New Delhi, 2002. 3. Advanced Reinforced Concrete Design, by P.C.Varghese, Prentice-Hall of India Pvt. Ltd., New Delhi, 2001. 4. ReinforcedConcreteDesign,2 Edition,by S.UnnikrishnaPillai and Devdas Menon, Tata McGrawHill Publishing Company Limited, New Delhi, 2003. 5. Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam, Oxford & l.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004. 6. ReinforcedConcreteDesign, 13 RevisedEdition,by S.N.Sinha,Tata McGrawHill Publishing Company. New Delhi, 1990. 7. ReinforcedConcrete,6" Edition,by S.K.Ma|lickand A.P.Gupta,Oxford & IBH Publishing Co. Pvt. Ltd. New Delhi, 1996. Version 2 CE IIT, Kharagpur 8. Behaviour, Analysis & Design of Reinforced Concrete Structural Elements, by |.C.Sya| and R.K.Ummat, A.H.Whee|er & Co. Ltd., Allahabad, 1989. 9. ReinforcedConcreteStructures,3 Edition,by |.C.Sya|and A.K.Goe|, A.H.Whee|er & Co. Ltd., Allahabad, 1992. 10.Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited, New Delhi, 1993. 11.Designof ConcreteStructures,13" Edition,by Arthur H. Nilson,David Darwin and Charles W. Dolan, Tata McGraw-Hill Publishing Company Limited, New Delhi, 2004. 12.Concrete Technology, by A.M.Neville and J.J.Brooks, ELBS with Longman, 1994. 13.Propertiesof Concrete,4 Edition, 13 Indian reprint, by A.M.Neville, Longman, 2000. 14.Reinforced ConcreteDesignersHandbook,10"Edition,by C.E.Reyno|ds and J.C.Steedman, E & FN SPON, London, 1997. 15.lndianStandardPlain and ReinforcedConcrete Code of Practice(4" Revision), IS 456: 2000, BIS, New Delhi. 16. Design Aids for Reinforced Concrete to IS: 456 8.18.11 Test 18 with Maximum Marks = 50, 1978, BIS, New Delhi. Solutions Maximum Time = 30 minutes Answer all questions. TQ.1: (a) State the limit of the aspect ratio of Iy/IXof one- and two-way slabs. (b) Explain the share of loads by the supporting beams in one- and two- way slabs. (10 marks) A.TQ.1: (a) The aspect ratio ly/IX (IXis the shorter one) is from 1 to 2 for two-way slabs and beyond 2 for oneway slabs. (b) See sec. 8.18.2. TQ.2: How to determine the design shear strength of concrete in slabs of different depths having the same percentage of reinforcement? (10 marks) A.TQ.2: See sec. 8.18.3. TQ3: Determine the areas of steel, bar diameters and spacings in the two directions of a simply supported slab of effective spans 3.5 m x 8 m (Figs.8.18.10a and b) subjectedto live loadsof 4 kN/m2and the load of Version 2 CE IIT, Kharagpur floorfinishis 1 kN/m2.Use M 20 and Fe 415. Drawthe diagramshowing the detailing of reinforcement. (30 marks) A.TQ.3: isnun s ET@ we 4_ W V} _» rfrz ji ET@ Eaimin(32 } §::ig..5_?Ei.1l3|:_§«}I S-iaizzaicnra 1 ii riewwaayshah Fig. 3.18.1e: DetaiiinigBf bars inf Prahiem TEL 1 Version 2 CE IIT, Kharagpur This is oneway slab as I,/I = 8/3.5 = 2.285 > 2. The calculations are shown in different steps below: Step 1: Selection of preliminary depth of slab Clause 23.2.1 stipulates the basic value of span to effective depth ratio of 20. Using the modification factor of 1.18 from Fig.4 of IS 456, with p = 0.5 per centand f5= 240 N/mm2,we havethe spanto effectivedepthratio= 20(1.18)= 23.6. So, the minimum effective depth of slab = 3500/23.6 = 148.305 mm. Let us take d: 150 mm and D: 175 mm. Step 2: Design loads, bending moment and shear force Factored dead loads of slab = (1.5)(0.175)(25) = 6.5625 kN/m Factored load of floor finish = (1.5)(1) = 1.5 kN/m Factored live load = (1.5)(4) = 6.0 kN/m Total factored load = 14.0625 kN/m Maximum positive bending moment = 14.0625(3.5)(3.5)/8 = 21.533 kNm/m Maximum shear force = 14.0625(3.5)(0.5) = 24.61 kN/m Step 3: Determination/checking of the effective and total depths of slab Using Eq.3.25 as explained in Step 3 of sec. 8.18.6, we have d = {21.533(106)/(2.76)(103)}°'5 = 88.33mm < 150mm,as assumedin Step 1. So, let us keep d=150 mm and D=175 mm. Step 4: Depth of the slab for shear force With the multiplying factor k = 1.25 for the depth as 175 mm (vide Table 8.1 of this lesson)and r, = 0.28N/mm2fromTable19 of IS 456,we have rc = 1.25(0.28)= 0.35N/mm2. Table20 of IS 456gives em = 2.8 N/mm2.Forthis problem: 7, = 24.61(1000)/(1000)(150) =0.1641N/mmz. Thus, the effective depth of slab as 150 mm is safe as r, < r, < rcm. Version 2 CE IIT, Kharagpur Step 5: Determination of areas of steel Table 41 of SP-16 gives 8 mm diameter bars @ 120 mm c/c have 22.26 kNm/m > 21.533 kNm/m. Hence, provide 8 mm T @ 120 mm c/c as main positive steel bars along the short span of 3.5 mm. The minimum amount of reinforcement (cl. 26.5.2.1 of IS 456) = 0.12(175)(1000)/100 = 210 mm2.Provide6 mm diameterbars @ 120 mm c/c (236mm2)alongthe largespanof 8m. Figure 8.18.1Obshows the detailing of reinforcing bars. 8.18.12 Summary of this Lesson This lesson mentions the different types of slabs used in construction and explains the differences between one and two-way slabs. Illustrating the principles of design as strength and deflection, the methods of determining the bending moments and shear forces are explained. The stipulated guidelines of assuming the preliminary depth, maximum diameter of reinforcing bars, nominal covers, spacing of reinforcements, minimum amount of reinforcing bars etc. are illustrated. The steps of the design of oneway slabs are explained. Design problems are solved to illustrate the application of design guidelines. Further, the detailing of reinforcement bars are explained for typical one-way slab and for the numerical problems solved in this lesson. Version 2 CE IIT, Kharagpur