Staircases Version 2 CE IIT, Kharagpur Lesson Types and D esi Staircases Version 2 CE IIT, Kharagpur InstructionalObjectives: At the end of this lesson, the student should be able to: o classify the different types configurations, - of staircases based on geometrical name and identifythe differentelementsof a typicalflight, o state the general guidelineswhile planninga staircase, o determinethe dimensionsof trade, riser,depth of slab etc. of a staircase, o classifythe differentstaircasesbased on structuralsystems, 0 explainthe distributionof loadingsand determinationof effectivespans of stairs, o analyse differenttypes of staircasesincludingthe free-standingstaircases in a simplifiedmanner, o designthe differenttypes of staircasesas per the stipulationsof IS 456. 9.20.1 Introduction Staircase is an importantcomponent of a building providingaccess to differentfloorsand roof of the building.It consistsof a flightof steps (stairs)and one or more intermediatelandingslabs between the floor levels. Differenttypes of staircasescan be made by arrangingstairs and landingslabs. Staircase,thus, is a structure enclosing a stair. The design of the main components of a staircase-stair,landingslabs and supportingbeams or wall are already covered in earlier lessons. The design of staircase, therefore, is the applicationof the designsof the differentelementsof the staircase. Version 2 CE IIT, Kharagpur Fig. 9.2.,1£r:): £2Ipe.n=-weal staircase Fig. 9.2i}.1: Types of steimeses Fig. !i.20~,.1£-e): Hetticnisdalstaircase Central mat Fig. B&I.2l¥.1l{d}: Spéral staircase Fig. 9.2.*I: Types of steiarcases Figures 9.20.1a to e present some of the common types of staircases based on geometrical configurations: (a) Single flight staircase (Fig. 9.20.1a) Version 2 CE IIT, Kharagpur (b) Two flightstaircase(Fig. 9.20.1b) (c) Open-wellstaircase(Fig. 9.20.1c) (d) Spiral staircase(Fig. 9.20.1d) (e) Helicoidalstaircase(Fig. 9.20.1e) Architecturalconsiderationsinvolving aesthetics,structuralfeasibilityand functionalrequirementsare the major aspects to select a particulartype of the staircase. Other influencingparametersof the selectionare lighting,ventilation, comfort,accessibility,space etc. 9.203 A Typical Flight Fig. §.;.§;Yé.£i:f::ji: f!s§ef§£-slab- Fig. §.§%.2:x.';=;% frearzl-risertype §.E§.,.2{«ej;::: £s»2x%:et.ed treazzi eaten Fig. &.§.2:t: A typical Figures 9.20.2a to d present plans and sections of a typical flight of differentpossibilities.The differentterminologiesused in the staircase are given below: Version 2 CE IIT, Kharagpur (a) Tread: The horizontal top portion of a step where foot rests (Fig.9.20.2b) is known as tread. The dimension ranges from 270 mm for residential buildings and factories to 300 mm for public buildings where large number of persons use the staircase. (b) Nosing: In some cases the tread is projected outward to increase the space. This projection is designated as nosing (Fig.9.20.2b). (c) Riser: The vertical distance between two successive steps is termed as riser (Fig.9.20.2b). The dimension of the riser ranges from 150 mm for public buildings to 190 mm for residential buildings and factories. (d) Waist: The thickness of the waist-slab on which steps are made is known as waist (Fig.9.20.2b). The depth (thickness) of the waist is the minimum thickness perpendicular to the soffit of the staircase (cl. 33.3 of IS 456). The steps of the staircase resting on waist-slab can be made of bricks or concrete. (e) Going: Going is the horizontal projection between the first and the last riser of an inclined flight (Fig.9.20.2a). The flight shown in Fig.9.20.2a has two landings and one going. Figures 9.2b to d present the three ways of arranging the flight as mentioned below: (i) (ii) (iii) 9.20.4 waist-slab type (Fig.9.20.2b), tread-riser type (Fig.9.20.2c), or free-standing staircase, and isolated tread type (Fig.9.20.2d). General Guidelines The following are some of the general guidelines to be considered while planning a staircase: o The respective dimensions of tread and riser for all the parallel steps should be the same in consecutive floor of a building. o The minimum vertical headroom above any step should be 2 m. o Generally, the number of risers in a flight should be restricted to twelve. - The minimum width of stair (Fig.9.20.2a) should be 850 mm, though it is desirable to have the width between 1.1 to 1.6 m. In public building, cinema halls etc., large widths of the stair should be provided. Version 2 CE IIT, Kharagpur 9.20.5 Structural Systems iFiigs.. Lg§fLE§E1»B§lf§,f s;ps.a:tn.ir=;g staircases Different structural systems are possible for the staircase, shown in Fig. 9.20.3a, depending on the spanning direction. The slab component of the stair spans either in the direction of going i.e., longitudinally or in the direction of the steps, i.e., transversely. The systems are discussed below: (A) Stair slab spanning longitudinally Here, one or more supports are provided parallel to the riser for the slab bending longitudinally. Figures 9.20.3b to f show different support arrangements of a two flight stair of Fig.9.20.3a: Version 2 CE IIT, Kharagpur (i) (ii) (iii) (iv) (v) Supported on edges AE and DH (Fig.9.20.3b) Clamped along edges AE and DH (Fig.9.20.3c) Supported on edges BF and CG (Fig.9.20.3d) Supported on edges AE, CG (or BF) and DH (Fig.9.20.3e) Supported on edges AE, BF, CG and DH (Fig.9.20.3f) Cantilevered landing and intermediate supports (Figs.9.20.3d, e and f) are helpful to induce negative moments near the supports which reduce the positive moment and thereby the depth of slab becomes economic. ¬1aE:il»:§Fgt:f3=} _?_ g. ~§.2:..4{af): Beams at two ends §|&5'1:El§tt%i:_§]E :3: ii 7 1V fr: Fig.§.,2;.a¢{h): Beams atthree ends lzamteia-age Fig. 9.2.4: fSta.im:asras :[span«n§nsg imngttedinaity} and imiclings.{spanneingstransaserselyij In the case of two flight stair, sometimes the flight is supported between the landings which span transversely (Figs.9.20.4a and b). It is worth mentioning that some of the above mentioned structural systems are statically determinate while others are statically indeterminate where deformation conditions have to taken into account for the analysis. Version 2 CE IIT, Kharagpur Longitudinal spanning of stair slab is also possible with other configurations including single flight, open-well helicoidal and freestanding staircases. (B) Stair slab spanning transversely tieugmtionbarre Mainsum Fig. 9.2I.).Sfa.ft: Slabs suppormd bemeen mm strigtgiarfhaeams or walls v i HSmndmal&--saatm or R13 waii Hg. &.2.«5l;b_§: aniélever slab fmm:El5;land:r4t3a|beam orwall Fig. §i..2.5{i3}:: Dcieubsliy cziantiiesver islet}Efrem 3 central beam Fig. 9.2..5V:Trarasverselyspannirtg staircases; Here, either the waist slabs or the slab components of isolated tread-slab and trade-riser units are supported on their sides or are cantilevers along the width direction from a central beam. The slabs thus bend in a transverse vertical plane. The following are the different arrangements: (i) (ii) (iii) Slab supported between two stringer beams or walls (Fig.9.20.5a) Cantilever slabs from a spandreal beam or wall (Fig.9.20.5b) Doubly cantilever slabs from a central beam (Fig.9.20.5c) Version 2 CE IIT, Kharagpur 9.20.6 Effective Span of Stairs The stipulations of clause 33 of IS 456 are given below as a ready reference regarding the determination of effective span of stair. Three different cases are given to determine the effective span of stairs without stringer beams. (i) The horizontal centre-tocentre distance of beams should be considered as the effective span when the slab is supported at top and bottom risers by beams spanning parallel with the risers. (ii) The horizontal distance equal to the going of the stairs plus at each end either half the width of the landing or one meter, whichever is smaller when the stair slab is spanning on to the edge of a landing slab which spans parallel with the risers. See Table 9.1 for the effective span for this type of staircases shown in Fig.9.20.3a. Table 9.1 Effective span of stairs shown in Fig.9.20.3a Note: G = Going, as shown in Fig. 9.20.3a 9.20.7 Distribution of Loadings on Stairs if eerri V,, .,, . ,,«-.=,.~.t4.~u--.~ war T W? l "fhe teaman mgsmmmn areas to be t2its.eri as uniamalt each dire:-met Fig. E!..2=.:.Lciadir%~gee on epen»-tweet; staigreasee Version 2 CE IIT, Kharagpur Eil?lf.=3»$ti*»."&* §.:lre.at%t l Fig. 9..2¬l.?: Loading an staimases btrilatairtrtn walls Figure 9.20.6 shows one open-well stair where spans partly cross at right angle. The load in such stairs on areas common to any two such spans should be taken as fifty per cent in each direction as shown in Fig.9.20.7. Moreover, one 150 mm strip may be deducted from the loaded area and the effective breadth of the section is increased by 75 mm for the design where flights or landings are embedded into walls for a length of at least 110 mm and are designed to span in the direction of the flight (Fig.9.20.7). 9.20.8 Structural Analysis Most of the structural systems of stair spanning longitudinally or transversely are standard problems of structural analysis, either statically determinate or indeterminate. Accordingly, they can be analysed by methods of analysis suitable for a particular system. However, the rigorous analysis is difficult and involved for a trade-riser type or free standing staircase where the slab is repeatedly folded. This type of staircase has drawn special attraction due to its aesthetic appeal and, therefore, simplified analysis for this type of staircase spanning longitudinally is explained below. It is worth mentioning that certain idealizations are made in the actual structures for the applicability of the simplified analysis. The designs based on the simplified analysis have been found to satisfy the practical needs. Version 2 CE IIT, Kharagpur A \_ U T; ; 5 '1g 5!'%".7#"«,u at f._,g ' .-n'rgw:P'F X -ml..,,__,;t.t. gm skis? apt spa" rag 9.2s.acc};EM déagramst alongtreads 5FE} AlaPT E 3% : tr iii Fi§§.5§.EU.B{hf') FED attradla slab EF am3» 4 i 5 PT A 3FT 2? g it? 2:3 P W- Fig. i§}*.3i3t;8§_;g}: FEE} ofriser sigatr BE Fig. 5":l.;?§'§.=Bél:,T':li FED t3$*.f"Il£i{§ slab ED Fig. 9.33.3: Strtmtural analysis citsiampilysuppartaedtradenarisser staircase Figure 9.20.8a shows the simply supported trade-riser staircase. The uniformly distributed loads are assumed to act at the riser levels (Fig.9.20.8b). The bending moment and shear force diagrams along the treads and the bending moment diagram along the risers are shown in Figs.9.20.8c, d and e, respectively. The free body diagrams of CD, DE and EF are shown in (either compressive or tensile). The assumption is that the riser and trade slabs are rigidly connected. It has been observed that both trade and riser slabs may be designed for bending moment alone as the shear stresses in trade slabs and axial forces in riser slabs are comparatively low. The slab thickness of the trade and risers should be kept the same and equal to span/25 for simply supported and span/30 for continuous stairs. l Fig. Q.2{i.§;fb}:Support mraemen.t ielarmzr Fig. ~:i.2a-s{cf::: Fab of Ei's.»L.,§? Flg..e.2mr.fEI{djt: Fan at @E Fig. §.2&.§: S«ftructurai a:naiyysfiis at an indeterrtti:r1..at.e trade-raiserstailrizase Figure 9.20.9a shows an indeterminate trade-riser staircase. Here, the analysis can be done by adding the effect of the support moment MA (Fig.9.20.9b) with the results of earlier simply supported case. However, the value of MA can be determined using the momentarea method. The free body diagrams of two vertical risers BC and DE are show in Figs.9.20.9c and d, respectively. 9.20.9 illustrative Examples Two typical examples of waist-slab and trade-riser types spanning longitudinally are taken up here to illustrate the design. Version 2 CE IIT, Kharagpur Eiilfl H Going E7r'?EI} Vg ; _.-:0;I "' a 53 iii? 5 . t, g 3- ' E Q; 65 §; l § ;: E; i The effectivespan (cls. 33.1b and c) = 750 + 2700 + 1500 + 150 = 5100 mm. The depth of waist slab = 5100/20 = 255 mm. Let us assume total depth of 250 mm and effectivedepth = 250 20 6 = 224 mm (assumingcover = 20 mm and diameter of main reinforcingbar = 12 mm). The depth of landing slab is assumedas 200 mm and effectivedepth = 200 20 6 = 174 mm. Step 2: Calculation of loads (Fig.9.20.11, sec. 1-1) (i) Loadson going (on projectedplan area) (a) Se|fweight ofwaists|ab= 25(0.25)(313.85)/270 = 7.265 kN/m2 (b) Se|fweight ofsteps= 25(0.5)(0.16)= 2.0 kN/m2 (c) Finishes (given)= 1.0 kN/m2 (d) Liveloads(given)= 5.0 kN/m2 Total = 15.265 kN/m2 Totalfactored loads= 1.5(15.265)= 22.9 kN/m2 (ii) Loadson landingslab A (50% of estimatedloads) (a) Se|fweight of landing slab = 25(0.2) = 5 kN/m2 (b) Finishes (given)= 1 kN/m2 (c) Liveloads(given)= 5 kN/m2 Total = 11 kN/m2 Factored loadsonlanding s|abA (iii) 0.5(1.5)(11)= 8.25 kN/m2 Factored loadsonlanding slabB = (1.5)(11) = 16.5 kN/m2 The loadsare drawn in Fig.9.20.11. Step 3: Bending moment and shear force (Fig. 9.20.11) Total loads for 1.5 m width of flight 16.5(1.65)} = 142.86 1.5{8.25(0.75) + 22.9(2.7) + kN Version 2 CE IIT, Kharagpur v0 = 1.5{8.25(0.75)(5.1 0.375) + 22.9(2.7)(5.1 0.75 1.35) +16.5(1.65)(1.65)(0.5)}/5.1 = 69.76 kN VD = 142.8669.76 The distance xfrom = 73.1 kN the left where shear force is zero is obtained x = {69.761.5(8.25)(0.75)+1.5(22.9)(0.75)}/(1.5)(22.9) The maximum bending moment at x: = 69.76(2.51) (1.5)(8.25)(0.75)(2.51 (1.5)(22.9)(2.51 0.75)(2.51 from: = 2.51 m 2.51 m is 0.375) 0.75)(0.5) = 102.08 kNm. For the landing slab B, the bending moment at a distance of 1.65 m from = 73.1(1.65)1.5(16.5)(1.65)(1.65)(0.5) = 86.92 kNm Step 4: Checking of depth of slab Fromthe maximummoment,we get d = {102080/2(2.76)}/2 = 135.98 mm < 224 mm for waist-slab and < 174 mm for landing slabs. Hence, both the depths of 250 mm and 200 mm for waist-slab and landing slab are more than adequate for bending. Forthe waist-slab,rv = 73100/1500(224) = 0.217 N/mm2.Forthe waistslab of depth 250 mm, k = 1.1 (cl. 40.2.1.1 of IS 456) and from Table 19 of IS 456, rc = 1.1(0.28)= 0.308N/mm2. Table20 of IS 456, rcm = 2.8 N/mm2. Since rv < rt < ram , the depth of waist-slab as 250 mm is safe for shear. For the landingslab, rv = 73100/1500(174) = 0.28 N/mm2.For the landing slab of depth 200 mm, k: 1.2 (cl. 40.2.1.1 of IS 456) and from Table 19 of IS456,rc = 1.2(0.28) = 0.336N/mm2 andfromTable20of IS456, rcm = 2.8 N/mm2. Herealso rv < rt < rum, so thedepthof landingslabas 200mmis safe for shear. Step 5: Determination of areas of steel reinforcement Version 2 CE IIT, Kharagpur 1.1:-.a A .. . .&,...M,_, A5 .5at , _s -1aft .2 gr. _ . .: 1, .kg! 9 itare:1 iii W>tmx->v.1r.»..v. "wsm 3.20.12: Retit"ttntm=lr'tktg bars atExairttpule Ext,eats; twtnet 9.29550 (i) Waist-slab:M./bdz= 102080/(1.5)224(224) = 1.356N/mm2.Table2 of SP16 gives p = 0.411. The areaof steel = 0.411(1000)(224)/(100) = 920.64 mm2.Provide12 mmdiameter@ 120mmc/c (= 942mm2/m). (ii) Landingslab B: Mu/bdzat a distanceof 1.65 m from VD(Fig. 9.20.11)= 86920/(1.5)(174)(174) = 1.91 N/mm2.Table2 of SP-16gives: p = 0.606.The areaof steel = 0.606(1000)(174)/100= 1054mm2/m.Provide16 mm diameter @ 240 mm c/c and 12 mm dia. @ 240 mm c/c (1309 mm2)at the bottomof landing slab B of which 16 mm bars will be terminated at a distance of 500 mm from the end and will continue up to a distance of 1000 mm at the bottom of waist slab (Fig. 9.20.12). Distribution steel: The same distribution steel is provided for both the slabs as calculated for the waist-slab. The amount is = 0.12(250) (1000)/100 = 300 mm2/m.Provide8 mmdiameter@ 160mmc/c (= 314 mm2/m). Step 6: Checking of development length and diameter of main bars Development length of 12 mm diameter bars = 47(12) = 564 mm, say 600 mm and the same of 16 mm dia. Bars = 47(16) = 752 mm, say 800 mm. (i) For waist-slab Version 2 CE IIT, Kharagpur M, for 12 mmdiameter@ 120mmc/c (= 942mm?)= 942(102.08)/920.64 = 104.44 kNm. With V (shear force) = 73.1 kN, the diameter of main bars 3 {1.3(104440)/73.1}/47 s 39.5 mm. Hence, 12 mm diameter is o.k. (ii) For landing-slab B M1 for 16 mm diameter @ 120 mm c/c (= 1675 mm2) = 1675(102.08)/1650.88 = 103.57 kNm. With V (shear force) = 73.1 kN, the diameter of main bars 3 {1.3(103570)/73.1}/47 = 39.18 mm. Hence, 16 mm diameter is o.k. The reinforcing bars are shown in Fig.9.20.12 (sec. 1-1). (B) Design of landing slab A Step 1: Effective span and depth of slab The effective span is lesser of (i) (1500 + 1500 + 150 + 174), and (ii) (1500 + 1500 + 150 + 300) = 3324 mm. The depth of landing slab = 3324/20 = 166 mm, < 200 mm already assumed. So, the depth is 200 mm. 3:2,from one ?[figf1 0.336N/mm2 Theabovevalueof 7, = 0.336N/mm2 for landingslabof depth200mmhas been obtained in Step 4 of A. However, here rt is for the minimum tensile steel in the slab. The checking of depth for shear shall be done after determining the area of tensile steel as the value of r, is marginally higher. Step 5: Determination of areas of steel reinforcement @.2v..¬a:l: eiaiersiug barsrzti Example:ea: IiifFig. For M./bdz = 75060/(1.5)(174)(174) = 1.65N/mm2,Table2 of SP-16 gives p = 0.512. The areaof steel = (0.512)(1000)(174/100 = 890.88 mm2/m.Provide 12 mm diameter @ 120 mm c/c (= 942 mm /m). With this area of steel p = 942(1oo)/1ooo(174) = 0.541. Version 2 CE IIT, Kharagpur Distribution steel = The same as in Step 5 of A i.e., 8 mm diameter @ 160 mm c/c. Step 6: Checking of depth for shear Table19andcl. 40.2.1.1gives: r6 = (1.2)(o.493) = 0.5916N/mm2.xv= 0.347N/mm2 (seeStep3 of B) isnowlessthan rc (= 0.5916N/mm2). Since,zv < rt < rcm , the depth of 200 mm is safe for shear. The reinforcing bars are shown in Fig. 9.20.14. Example 9.2: if...7i F1ig.9.2.."l§: Eluzampiaa 9.2 Design a trade-riser staircase shown in Fig.9.20.15 spanning longitudinally. Landing slabs are supported on beams spanning transversely. The dimensions of riser and trade are 160 mm and 270 mm, respectively. The finish loadsand live loadsare 1 kN/m2and 5 kN/m2,respectively.Use M 20 and Fe 415. Solution: The distribution of loads on landings common to two spans perpendicular to each other shall be done as per cl. 33.2 of IS 456 (50% in each direction), since the going is supported on landing slabs which span transversely. The effective span in the longitudinal direction shall be taken as the distance between two centre lines of landings. (A) Design of going Version 2 CE IIT, Kharagpur Step 1: Effective span and depth of slab ETILZI H 21% aw1.43e~%1:3.5g A M35 Fig. lrfangema~nt of loadings and Gairnsgg fE.1¢.alf-]:rE& Figure 9.20.16 shows the arrangement of the landings and going. The effective span is 4200 mm. Assume the thickness of trade-riser slab = 4200/25 = 168 mm, say 200 mm. The thickness of landing slab is also assumed as 200 mm. Step 2: Calculation of loads (Fig. 9.20.17) Total lim-2;s=.». a EI;l3.?Ekw ~ I 13335Wm (.Sr3%}- 2L ~"'fgf11:\x"a *2 g 2 *§E.;3?3 kt-lin1g*Es£}% E 52.39RN al KN Fig...9.29.i?:.l%...oad3 of Seotlon"l-l. Fig. 9.2l3..i.5,.{Example 92} Version 2 CE IIT, Kharagpur The total loads including self-weight, finish and live loads on projected area of going (1500 mm x 2465 mm) is first determined to estimate the total factored loads per metre run. (i) Se|fweight of going (a) Nine units of (0.2)(0.36)(1.5) @ 25(9) = 24.3 kN (b) One unit of (0.27)(0.36)(1.5) @ 25(1) = 3.645 kN (c) Nine units of (o.o7)(o.2)(1.5) @ 25(9) = 4.725 kN (ii) Finishloads@1 kN/m2= (1.5)(2.465)(1)= 3.6975kN (iii) Liveloads@ 5 kN/m2= (1.5)(2.465)(5)= 18.4875kN Total = 54.855 kN Factored loads per metre run = 1.5(54.855)/2.465 = 33.38 kN/m (iv) Se|fweight of landing slabs per metre run = 1.5(0.2)(25) = 7.5 kN/m (v) Live loads on landings = (1.5)(5) = 7.5 kN/m (vi) Finish loads on landings = (1.5)(1) = 1.5 kN/m Total = 16.5 kN/m Factored loads =1.5(16.5) = 24.75 kN/m Due to common area of landings only 50 per cent of this load should be considered. So, the loads = 12.375 kN/m. The loads are shown in Fig.9.20.17. Step 3: Bending moment and shear force Total factored loads = 33.38(2.465) + 12.375(0.85 + 0.885) = 103.75 kN V0 = {12.375(0.85)(4.2 0.425) + 33.38(2.465)(0.885 + 1.2325) +12.375(0.885)(0.885)(0.5)}/4.2 = 52.09 kN VD = 103.7552.09 The distance x = 51.66 kN from the left support where shear force is zero is now determined: Version 2 CE IIT, Kharagpur 52.09 or 12.375(0.85) 32.38(x 0.85) X = {52.09 = 0 12.375(0.85) + 33.38(0.85)}/33.38 = 2.095 m Maximum factored bending moment at x = 2.095 m is 52.09(2.095) 0.85)(0.5) = 65.69 12.375(0.85)(2.095 0.425) - 33.38(2.095 0.85)(2.095 kNm Step 4: Checking of depth of slab From the maximum bending moment, we have d= {(65690)/(1.5)(2.76)}1/2 = 125.97mm < 174mm From the shear force V,, = VA,we get rv = 52090/(1500)(174) = 0.199 N/mm2. Fromcl. 40.2.1.1andTable19 of IS 456,we have rc = (1.2)(0.28) = 0.336N/mm2. Table20 of IS 456gives rum: 2.8N/mm2. Since, rvi;ze4i»:¢-¢6si:wavu4» -- T1.21:: 135155lI¥£§"~J liig 9.2.22: Ealmiiatipin:5116565 SEElent«f Fig. 9.30.2/'f, Example 9.5 Step 2: Calculation of loads (Fig.9.20.22) (i) Loadson going (on projectedplan area) (a) Selfweightof waistslab=25(0.25)(313.85/270) = 7.265 kN/m2 (b) Selfweightof steps= 25(0.5)(0.16)= 2.0 kN/m2 (c) Finishloads(given)= 1.0 kN/m2 (d) Liveloads(given)= 5.0 kN/m2 Total = 15.265 kN/m2 So,the factored loads= 1.5(15.265)= 22.9 kN/m2 (ii) Landing slab A (a) Selfweightof slab = 25(0.25)= 6.25 kN/m2 (b) Finishloads= 1.00 kN/m2 (c) Liveloads= 5.00 kN/m2 Total = 12.25 kN/m2 Factored loads 1.5(12.25)= 18.375kN/m2 (iii) LandingslabB = 50 percentof loadsof landingslabA = 9.187 kN/m2 Version 2 CE IIT, Kharagpur The total loads of (i), (ii) and (iii) are shown in Fig.9.22. Total loads (i) going = 22.9(1.96)(2) = 89.768 kN Total loads (ii) landing slab A = 18.375(2.15)(2) = 79.013 kN Total loads (iii) landing slab B = 9.187(1.0)(2) = 18.374 kN Total loads = 187.155 kN The loads are shown in Fig. 9.20.22. Step 3: Bending moment and shear force (width = 2.0 m, Fig. 9.20.22) Vp = {79.o13(5.11 = 98.97 98.97 or 3.13) + 18.374(0.5)}/5.11 kN VJ = 187.15598.97 The distance 1.075) + 89.768(5.11 xwhere 79.013 = 88.185 kN the shear force is zero is obtained from: 22.9(2)(x 2.15) = 0 x = 2.15 + (98.9779.013)/22.9(2) = 2.586 m Maximum bending moment at x: 2.586 m (width = 2 m) = 98.97(2.586) Maximum shear 79.013 force (22.9)(2)(0.436)(0.436)(0.5) = 161.013 kNm = 98.97 kN Step 4: Checking of depth Fromthe maximummoment d = {161.013(103)/2(2.76)}/2 = 170.8mm < 224 mm. Hence o.k. Fromthe maximum shearforce, av= 98970/2000(224) = 0.221N/mm2. For the depth of slab as 250 mm, k = 1.1(cl. 40.2.1.1 of IS 456) and r6 = 1.1(0.28) = 0.308N/mm2 (Table19of ls 456). ram,= 2.8N/mm2 (Table20of ls 456). Since, rv < rt < ram , the depth of slab as 250 mm is safe. Step 5: Determination of areas of steel reinforcement Version 2 CE IIT, Kharagpur M,/bdz = 161.013(103)/2(224)(224) = 1.60 N/mm2.Table2 of SP16 gives p = 0.494,to have A5,= 0.494(1000)(224)/100 = 1106.56mm2/m.Provide 12 mmdiameterbars@ 100mmc/c (= 1131mm2/m)bothfor landingsandwaist slab. Distributionreinforcement= 0.12(1000)(250)/100 = 300 mm2/m.Provide 8 mmdiameter@ 160mmc/c (= 314mm2). Step 6: Checking of development length i l l l l Leading.15. . . Fig. B.;213.2t3t: Rsei.nf0rEingtsnars,sec l»=-11% of Fig E_,:::amp|=e i;'3t;..% Development length of 12 mm diameter bars 7(12) = 564 mm. Provide L0;= 600 mm. For the slabs M1 for 12 mm diameter @ 100 mm c/c = (1131)(161.013)/1106.56 = 164.57 kNm. Shear force = 98.97 kN. Hence, 47 ¢ 3 1.3(164.57)/98.97 3 2161.67 mm or the diameter of main bar ¢ S 45.99 mm. Hence, 12 mm diameter is o.k. The reinforcing bars are shown in Fig.9.20.23. (B) Design of landing slabs B and C and going (sec. 22 of Fig.9.20.21) Step 1: Effective span and depth of slab The effective span from the centre line of landing slab B to the centre line of landing slab C = 1000 + 1960 + 1000 = 3960 mm. The depths of waist slab and landing slabs are maintained as 250 mm like those of sec. 11. Version 2 CE IIT, Kharagpur Step 2: Calculation of loads (Fig.9.20.24) *£3'@'.§:f?5$*f 75~~.-f?%§ tiff ......,................ .-3. ,m&«,,_ ~ l§}l.s..aa...m vé;.vi:yu«».\ my: "$26.24; itaimiiaaiimof icssas. 2&2of Ffg.s;i.2i:ia.,2'f.mg (i) Loadson going(Step2(i) of A) = 22.9 kN/m2 (ii) Loadson landingslab B (Step2(iii)) = 9.187kN/m2 (iii) Loadson landingslabC (Step2(iii)) = 9.187kN/m2 Total factored loads are: (i) Going = 22.9(1.96)(2) = 89.768 kN (ii) Landing s|abA = 9.187(1.0)(2) = 18.374 kN (iii) Landing slab B = 9.187(1.0)(2) = 18.374 kN Total = 126.506 kN The loads are shown in Fig.9.20.24. Step 3: Bending moment and shear force (width = 2.0 m, Fig.9.20.24) The total load is 126.506 kN and symmetrically placed to give VG= VH= 63.253 kN. The maximum bending moment at x: 1.98 m (centre line of the span 3.96 m = 63.253(1.98) 18.374(1.98 0.5) 22.9(2)(0.98)(0.98)(0.5) = 76.05 kNm. Maximum shear force = 63.253 kN. Since the maximum bending moment and shear force are less than those of the other section (maximum moment = 161.013 kNm and maximum shear force = 98.97 kN), the depth of 250 mm here is o.k. Accordingly, the amount of reinforcing bars are determined. Step 4: Determination of areas of steel reinforcement Version 2 CE IIT, Kharagpur §_aI:?éd%2rI§ E mac E2 if 2. LimitStateDesignof ReinforcedConcrete,2 Edition,by P.C.Varghese, PrenticeHa|I 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 McGraw-Hill Publishing Company Limited, New Delhi, 2003. 5. Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam, Oxford & I.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004. 6. ReinforcedConcreteDesign, 13 RevisedEdition,by S.N.Sinha,Tata McGraw-Hill Publishing Company. New Delhi, 1990. 7. ReinforcedConcrete,6" Edition,by S.K.Ma|Iickand A.P.Gupta,Oxford & IBH Publishing Co. Pvt. Ltd. New Delhi, 1996. 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 9.20.12 Test 20 with Maximum Marks = 50, 1978, BIS, New Delhi. Solutions Maximum Time = 1 hour Answer all questions. TQ.1: Draw a typical flight and show: (a) trade, (b) nosing, (c) riser, (d) waist and (e) going. (5 marks) A.TQ.1: Figures 9.20.2a, b and c (sec. 9.20.3) Version 2 CE IIT, Kharagpur TQ.2: Draw schematic diagrams of different types of staircases based on different structural systems. (10 marks) A.TQ.2: Figures 9.20.3a, b, c, d, e; Figs.9.20.4a and b; Figs.9.20.5a, b and c (sec.9.20.5). TQ.3: Illustratethe simplifiedanalysis of longitudinallyspanning free-standing staircases. (10 marks) A.TQ.3: See sec. 9.20.8. TQ.4: Designthe staircaseof illustrativeexample 9.1 of Fig.9.20.10 if supported on beams along KQ and LR only making both the landingsA and B as cantilevers.Use the finishloads= 1 kN/m2,live loads= 5 kn/m2,riserF? = 160 mm, trade T: 270 mm, grade of concrete = M 20 and grade of steel = Fe 415. (25 marks) A.TQ.4: ii}:permistbletoads alt span ;i(§U. an alt spans xi .z«> ; {iii} LL canlandingslab .6!-1 {ivj Ll. an {V} LL mt larw/ding slabs A 8..3 Fig. El.2.23.: Fi«.ee ofloadings lj :1 Version 2 CE IIT, Kharagpur Solution: The general arrangement is shown in Fig.9.20.26. With F?= 160 mm and T= 270 mm,the inclinedlengthof eachstep= {(160)2+ (270)2}/2 = 313.85mm. The structural arrangement is that the going is supported from beams along KQ and LR and the landings A and B are cantilevers. Step 1: Effective span and depth of slab As per cl. 33.1a of IS 456, the effective span of going = 3000 mm and as per cl. 22.2c of IS 456, the effective length of cantilever landing slabs = 1350 mm. The depth of waist slab and landing is kept at 200 mm (greater of 3000/20 and 1350/7). The effective depth = 200 20 6 = 174 mm. Step 2: Calculation of loads (i) Loads on going (on projected plan area) (a) sell weightof waists|ab = 25(0.20)(313.85)/270 = 6.812kN/m2 (b) seli weightof steps= 25(0.5)(0.16)= 2.0 kN/m2 (c) Finishes(given)= 1.0kN/m2 (d) Liveloads(given)= 5.0 kN/m2 Total: 14.812 kN/m2 Totalfactoredloads=1.5(14.812)= 22.218kN/m2 (ii) Loads on landing slabs A and B (a) sell weightof landingslabs= 25(0.2) = 5 kN/m2 (b) Finishes(given)= 1 kN/m2 (c) Liveloads(given)= 5 kN/m2 Total: 11 kN/m2 Totalfactoredloads = 1.5(11) = 16.5kN/m2.Thetotal loadsare shown in Fig.9.20.27. Version 2 CE IIT, Kharagpur Step 3: Bending moments and shear forces Here, there are two types of loads: (i) permanent loads consisting of selfweights of slabs and finishes for landings and se|fweights of slab, finishes and steps for going, and (ii) live loads. While the permanent loads will be acting everywhere all the time, the live loads can have several cases. Accordingly, five different cases are listed below. The design moments and shear forces will be considered taking into account of the values in each of the cases,. The different cases are (Fig.9.20.28): (i) Permanent loads on going and landing slabs (ii) Live loads on going and landing slabs (iii) Live loads on landing slab A only (iv) Live loads on going only (v) Live loads on landing slabs A and B only The results of V0, V3, negative bending moment at Q and positive bending moment at T (Fig.9.20.27) are summarized in Table 9.2. It is seen from Table 9.2 that the design moments and shear forces are as follows: (a) Positive bending moment = 25.195 kNm at T for load cases (i) and (iv). (b) Negative bending moment = -22.553 kNm at Q for load cases (i) and (ii). (c) Maximum shear force = 83.4 kN at Q and R for load cases (i) and (ii). Table 9.2 Values of reaction forces and bending moments for different cases of loadings (Example: TQ.4, Figs. 9.20.26 to 9.20.28) VQ(kN) V;q(kN) Negative moment Permanent atQ Positive moment atT kNm kNm +32.06 10.251 +2.404 -3.417 -10.251 loads landins Live loads on going and +32.06 landin 0 s Live loads landin on +18.605 -5.13 A onl +16-875 +16-875 n Version 2 CE IIT, Kharagpur Live loads on +15.1875 +15.1875 -1 0.251 -1 0.251 +68.215 +68.215 -12.302 +25.195 +83.40 +83.40 -22.553 +14.939 landings A and B onl Step 4: Checking of depth of slab Thedepthis checkedfor the positivemomentof 25.195kNmas thetwo depthsare the same.The effectivedepthof slab d = {25.195(106)/1500(2.76)} = 78 mm < 174 mm. Hence o.k. Thenominalshearstress rv = 83400/1500(174) = 0.3195N/mm2. Using the value of k = 1.2 (cl. 40.2.11 of IS 456) and from Table 19 of IS 456, we get rc = 1.2(0.28)= 0.336N/mm2and rcm = 2.8 N/mm2fromTable20 of IS 456. Since, rv < r6 < ram, the depth of 200 m is safe against shear. Step 5: Determination of areas of steel reinforcement at he text em 4» fi 313%} 5' §T@¢l§ erg: 3T't7.7EP1§UIlslLi t? é 3?s5fi}* 151:!t:t'r;: + ST .m:,;«.4mm~.~w~m 36;?f emcee wwww err @ stat: -«tease» e...z,....» esM...;,,e Fig. *9.211¥.29: Reittfercing have :31E_x.ampl.e Tt Version 2 CE IIT, Kharagpur (i) Waistslab: M,/ba = 25195/1.5(174)(174) = 0.555N/mm2.Table2 of SP16 gives: p = 0.165. Accordingly, As, = 0.165(1000)(174)/100 = 287.1 mm2/m.Provide8 mmdia.bars@ 150mmc/c (= 335mm2). (ii) Landing slab: Since the difference of positive and negative bending moments is not much, same reinforcement bars i.e., 8 mm diameter @ 150 mm c/c is used as positive and negative steel bars of waist and landing slabs. Distributionbars:0.12(200)(1000)/100 = 240 mm2/m.Thesamebar i.e.,8 mm diameter @ 150 mm c/c can be used as distribution bar. The extra amount is useful to take care of change of ending moments due to different cases of loadings. The reinforcement bars are shown in Fig.9.20.29. 9.20.13 Summary of this Lesson This lesson explains the different types of staircases based on geometrical considerations. The different terminologies commonly used in a typical flight are mentioned. Important guidelines to be considered at the planning stage of the staircase are discussed. The classification of staircases based on structural system is explained. The distribution of loadings, determination of effective spans and selection of preliminary discussions of trade, riser and depth of slabs are illustrated. Simplified analysis procedures of staircases including free-standing staircase are explained. Several numerical problems are solved in the illustrative examples, practice problem and test, which will help to understand the applications of all the guidelines and the design of different types of staircases. Version 2 CE IIT, Kharagpur