DESIGN AIDS FOR REINFORCED CONCRETE TO IS : 456-l 978 As in the Original Standard, this Page is Intentionally Left Blank DesignAids to IS : 4564978 For Reinforced Concrete BUREAU OF INDIAN STANDARDS BAHADUR SHAH ZAFAR MARC, NEW DLEHI 110 002 SP16:1980 FIRST PUBLISHED ELEVENTH (Incorporatinp SEPTEMBER MARCH No. 1980 1999 REPRINT Amendment I) 0 BUREAU OF INDIAN STANDARDS UDC 624.0 12.45.04 (026) PRICE Rs. 500.00 I'KiNTED 1N INDIA AT VlB,\ PRESS PVT. LTD., 122 DSIDC SHEDS. OKHLA INDL!STRIAL ARtA. AND PI II3LISHED BY I IS : 1139-1966t I 1 bars up to Medium tensile steel (hotrolled deformed bars) 34.5 kgf/mm* for bars over 20 mm'dia up to 40 mm iiia 33 kgf/mm" for bars over 40 mm dia 42.5 kgf/mm2 for all sizes High yield strength steel (hotrolled deformed bars) High yield strength steel deformed (cold-twisted bars) Hard-drawn steel wire fabric IS : 1139-1966t IS : 1786-1979$ 7 4 15 N/mm2 for all bar sizes 500 N/mm* for all bar sizes 49 kgf/mm* IS : 1566-19674 and IS : 432 (Part II)-19661 Nom-S1 units have been used in IS: 1786-19793; in other Indian Standards. SI units will be adopted in their next revisions. *Specification for mild steel and medium tensile steel bars and hard-drawn steel wire for concrete reinforcement: Part I Mild steel and medium tensile steel bars (second revision). tSpecification for hot rolled mild steel, medium tensile steel and high yield strength steel deformed bars for concrete reinforcement (revised). $Specificatiod for cold-worked steel high strength &formed bars for concrete reinforcement (second WlSiO#). &+eciiication for hard-drawn steel wire fabric for concrete reinforcement ijSpecification for mild steel and medium tensile steel bars concrete reinforcement: Part II Hard drawn steel wire (second revision). (#rsr revisfon ). hard-drawn steel and wire for MATERIAL STRENGTHS AND STRESS-STRAIN RELATIONSHIPS 3 Taking the above values into consideration, most of the charts and tables have been prepared for three grades of steel having characteristic strength& equal to 250 N/mm*, 415 N/mm2 and 500 N/mm2. 1.2.1 If the steel being used in a design has a strength which is slightly diflerent from the above values, the chart or table for the nearest value may be used and the area ofreinforcement thus obtained be modi$ed in proportion to the ratio of the strengths. 1.2.2 Five values of fY (includinglthe value for hard-drawn steel wire fabric) have been included in the tables for singly reinforced sections. I/ I / .I a.002 STRAIN 0'001 FIG. 1 DESIGN STRKSS-STRAINCURVE FOR CONCRETE 1.3 STRESS-STRAIN FOR CONCRETE RELATIONSHIP The Code permits the use of any appropriate curve for the relationship between the compressive stress and strain distribution in concrete, subject to the condition that it results in the prediction of strength in substantial agreement with test results [37.2(c) of the Code]. An acceptable stress-strain curve (see Fig. 1) given in Fig. 20 of the Code will form the basis for the design aids in this publication. The compressive strength of concrete in the structure is assumed to be O-67fd. With a value of l-5 for the partial safety factor ym for material strength (35.4.2.1 of the Code), the maximum compressive stress in concrete for design purpose is 0.446 fck (see Fig. I). . 200000 N/mm' ? STRAIN -- FIG. 2 STRESS-STRAIN CURVE FOR MILD STEEL 1.4 STRESS-STRAIN FOR STEEL RELATIONSHIP be adopted. According to this, the stress is proportional to strain up to a stress of 0.8 fY. Thereafter, the stress-strain curve is defined as given below: hu#aslic~srrain Stress Nil O*SOfy OQOOl 0.85 fr 0.0% 3 0*9ofy o*ooo 7 0*9sf, 0~0010 0.975 fy 0.002 0 l-O& The stress-strain curve for design purposes is obtained by substituting fYe for fY in the above. For two grades of cold-worked bars with 0.2 percent proof stress values of 415 N/mms and 500 N/mm2 respectively, the values of total strains and design stresses corresponding to the points defined above are given in Table A (see page 6). The stressstrain curves for these two grades of coldworked bars have been plotted in Fig. 3. DESIGN AIDS FOR REINFORCED CONCRETE The modulus of elasticity of steel, E,, is taken as 200 000 N/mm2 (4.6.2 of the Code). This value is applicable to all types of reinforcing steels. The design yield stress (or 0.2 percent proof stress) of steel is equal to fr/ym. With a value of l-15 for ym (3.5.4.2.2 of the Code), the design yield stress fv becomes 0#87 f,. The stress-strain relations1.tp for steel in tension and compression is assumed to be the same. For mild steel, the stress is proportional to strain up to yield point and thereafter the strain increases at constant stress (see Fig. 2). For cold-worked bars, the stress-strain relationship given in Fig. 22 of the Code will 4 1 500 so0 450 m2 `1.0 soo/ `iv' `mnl 400 UC/l1-n 350 T < 2. 300 250 200 150 100 50 0 0 0.001 o-002 STRAIN FIG. 3 0.003 o-004 0*005 STRESS-STRAIN CURVES FOR COLD-WORKED STEELE MATERIAL SrRENGTHS AND STRESS-STRAIN RELATIONSHIPS 5 TABLE A SALIENT POINTS ON THE DESlGN STRESS-STRAIN COLD-WORKED BARS ( Chse 1.4 ) CURVE GOR STRESS LEVEL f, 0 415 N/mm' f-* Strain (`1 Stress (3) N/mm* 288.1, 306.7 324.8 342.8 351.8 360.9 > ,_-.-k Strain (4) fy= 500 N/mm8 b Stress (5) N/mm* 347.8 369.6 391.3 413.0 423.9 434.8 (1) 0.80 fyd 0.85 fyd 0.90&l 0'95 fyd 0.975 fyd l'ofyd 090144 0031 63 0~00192 0032 4 I 0.002 76 0.003 80 woo174 0.001 95 0.002 26 0.002 77 0.003 12 MO4 17 NOTE -- Linear interpolation may be done for intermediate values. 6 As in the Original Standard, this Page is Intentionally Left Blank 2. FLEXURAL MEMBERS 2.2 ASSUMPTJONS . The basic assumptions in the design of flexur&lmembers for the limit state of collapse are fcivenbelow (see 37.2 of the Code): 2.2 MAXIMUM DEPTH OF NEUTRAL AXIS Assumptions (b) and (f')govern the maximum depth of neutral axis in flexural members. T& strain distribution across a member corresponding to those limiting conditions is shown in Fig, 4. The maximum depth of neutral axis x,,, - is obtained directlyfrom the strain diagram by considering similar triangles. 0.003 5 x0,,_ d (0.005 5 f 0.87 f,/&) The values of * 4 Plane sections normal to the axis of the member remain plane after bending. This means that the strain at any point on the cross section is directly proportional to the distance from the neutral RXiS. W Ihe maximum strain in concrete at the outermost compression fibre is 0903 5. for concrete is taken as indicated in Fig. 1. for three grades of d The design stress-strain relationship reinforcing steel are given in Table B. TABLE B VALUES OF F FOR d) The tensile strength of concrete is ignored. e), Tbt in reinforcement are derived from the strains using given -in f,, design stresses DIFFERENT GRADES OF STEEL (Cfuu.re 2.2) the stress-strain relationship N/mms 250 0531 415 0.479 500 O-456 Fig. 2 and 3. f) The strain in the tension reinforcement is to be not less than 7 2.3 RECTANGULAR SECTIONS This assumption is intended to ensure ductile fail&e, that is, the tensile reinforcementhas to undergo a certain degree of inelastic deformation before the concrete fails in compression. The compressive stress block for concrete is represented by the design stress-strain curve as in Fig. 1. It is. seen from this stress block (see Fig. 4) that the centroid of compressive force in a rectangular section lies 0*0035 t X u,m*a f !zzx + 0*002 E* STRAIN OIAGRAM FIQ. 4 SINOLY REINFQRCSDSECTION FLEXURAL MRMM3R.S O= 87 f-, STRESS DIAGRAM at a distance or U-416 xu (wnlcn nas oecn rounded off to 0.42 xu in the code) from the extreme compression fibre; and the total force of compression is 0.36 fck bxu. The lever arm, that is, the distance between the centroid of compressive force and centroid of tensile force is equal to (d - 0.416 x,). Hence the upper limit for the moment of resistance of a singly reinforced rectangular section is given by the following equation: M u,lim = O-36& bxu,,, x(d - 0.416 ~u,mu) Substituting for xu,from Table B and transposing fdr bd2, we get the values of tie limiting moment of resistance factors for singly reinforced rectangular beams and slabs. These values are given in Table C. The tensile reinforcement percentage, pt,lim corresponding to the limiting moment of resistance is obtained by equating the forces of tension and compression. TABLE E MAXIMUM PERCENTAGE OF TENSILE REINFORCEMENT pt,lim FOR SINGLY REINFStRmTNSRE!aANGW (c%u.w 2.3) fdr, N/mm* 15 4 /y, Nhm' r 250 415 ;g l.43 u)o "0% b 1.32 1.76 220 2% . YE 2.3.1 Under-Reinforced Section Substituting for xu,mPx from Table B, we get the values of Pt,lim fYj& as given in Table C. TABLE C LIMITING MOMENT OF RESISTANCE AND REINFORCEMENT INDEX FOR SINGLY REl;~&FOR~N~ RECTANGULAR (Clause 2.3) j& N/mm* -250 0.149 415 W138 500 0.133 Under-reinforced section means a singly reitiorced section with reinforcement percentage not exceeding the appropriate value given in Table E. For such sections, the depth of neutral axis xu will be smaller than x",,,,~. The strain in steel at the limit state of collapse will, therefore, be more than 0.87 fy + 0902 and, the design stress in E. steel will be 0937fy. The depth of neutral axis is obtained by equating the forces of tension and compression. `G (0.87 fr) - 0.36 fdrb xu M*,lhl - Lk bd' Plrllrn fy / ck 21.97 19.82 18.87 The moment of resistance of the section is equal to the prdduct of the tensile force and the lever arm. Mu = pG (@87f,) (d - 0,416 xu) The values of the limiting moment of resistance factor Mu/bd2 for different grades of concrete and steel are given in Table D. The corresponding percentages of reinforcements are given in Table E. These are the maximum permissible percentages for singly reinforced sections. TABLE D LIMITING MOMENT OF RESISTANCE FAVOR Mu,,im/bd', N/mm' FOR SINGLY REINFC);&yE$sECTANGULAR (Clause 2.3) /CL, fy, & l- 0.4165 bd2 ) ( )( Substituting foi $ we get =O*87fy _ _ x C 1- 1.005 &$]bda N/-Y 500 2.00 Is: 3.45 414 2.66 3.33 3.99 N/mm' 15 rK-----2.24 2.98 3.73 4.47 3: 30 2.3.Z.Z Charts 1 to 28 have been prepared by assigning different values to Mu/b and plotting d versus pt. The moment values in the charts are in units of kN.m per metr$ width. Charts are given for three grades of steel and, two grades of concrete, namely M 15 and M 20, which are most commonly used for flexural members. Tables 1 to 4 cover a wider range, that `is, five values of fy and four grades of concrete up to M 30. In these tables, the values of percentage of reinforcement pt have been tabulated against Mu/bd2. DESIGN AIDS FOR WNFORCED CONCRETE 10 2.3.2.2 The moment of resistance of slabs, Retbrring to C/r& 6, corresponding to with bars of different diameters and spacings h&,/b - 567 kN.m and d = 5625 cm, are given in Tables 5 to 44. Tables are given for concrete grades M 15 and M 20, with Percentage of steel pt - lOOAs M = 0.6 two grades of steel. Ten different thicknesses ranging from 10 cm to 25 cm, are included. 0.6 bd 0.6~30~5625 __O1 ,,* These tables take into account 25.5.2.2 .*. A,= -jijiy 100 of the Code, that is, the maximum bar diameter does not exceed one-eighth the thickness of the slab. Clear cover for reinforcement has been taken as 15 mm or the bar For referring to Tables, we need the value diameter, whichever is greater [see 25.4.1(d) Mu of the Code]. Jn these tables, the zeros at w the top right hand comer indicate the region where the reinforcement percentage would 170 x IO' M" exceed pt,lim; and the zeros at the lower bd' - -3m6.25 x 56.25 x IO' left hand comer indicate the region where I 1.79 N/mm' the reinforcement is less than the minimum according to 25.5.2.1 of the Code. From Table 1, Percentage of reinforcement, pt = 0.594 of Example 1 Singly Reinforced Beam .* . As- 0.594 x 30 x 56.25 _ ,omo2,,* Determine the main tension reinforcement required for a rectangular beam section with the following data: 3ox6Ocm Sixe of beam M 15 Concrete mix 415 N/mm' Characteristic strength of reinforcement 170kN.m *Factored moment *Assuming 25 mm dia bars with 25 mm clear cover, Effectivedepth I 60 - 2.5 -2;5625 cm From Table D, for fr P 415 N/mm' and fcrc - 15 N/mm* MWliUJM' p 2.07 N/mm: 100 Example 2 Slab Determine the main reinforcement required for a slab with the following data: Factored moment 9.60 kN.m E%etre 10 cm M 15 a) 4 15 N/mm2 b) 250 N/mm* Depth of slab Concrete mix Characteristic strength of reinforcement h&l-HODOF REPERRING TO TABLES FOR SLABS v$g$ :. &am x (1000)' e; 2.07 x 101kN/m* - 2.07 x 1O'W I 2-07 x 10' x fa x 30 Referring to Table 15 (for fy - 415 N/mmz), directly we get the following reinforcement for a moment of resistance of 9.6 kN.m per metre width: 8 mm dia at 13 cm spacing or 10 mm dia at 20 cm spacing Reinforcement given in the tables is based on a cover of 15 mm or bar diameter whichever is greater. MFXHOD OF RFNRRJNG TO FLBXURB CHART I 1965 kN.m $%ua] moment. of. 170 kN.m is less *than The sectton 1stherefore to be destgned as u'~mm'singly reinforced (unde&einforced) rectangular section. GTOFU3XURECHART fVfM'HOD OF RBFQIRIN Assume 10 mm dia bars with 15 mm cover, d - 10 - 1.5 - 9 =8cm = 2.07 N/mm* For referring to Chart, we need the value of moment per metre width. Mu/b-g = 567kN.m per metre width. a) For fy= 415 N/mm' From Table D, Mu,tidb# :. J%lirn - 2.07 x lOa x z x (A)' ' _' = 13.25kN.m *The term `factoredmoment'means the moment Actual bending moment of 960 kN.m is less due to characteristic loads multiplied by the appropriate value of p&rtialsafety factor yf. than the limiting bending moment. FLExuRALmMBERs 11 Referring to Chart 4, reinforcement centage, pt 6 0.475 Referring to Chart 90, provide 8 mm dia at 13 cm spacing or 10 mm dia at 20 cm spacing. Alternately, per- A, = O-475 x 100 x &J = 3.8 cm* per metre width. From Table %, we get the same reinforcement as before. b) Forf, = 250 N/mm* From Table D, Mu&bd" M u&m = The lever arm for the additional moment of resistance is equal to the distance between centroids of tension reinforcement and compression reinforcement, that is (d-d') where d' is the distance from the extreme compression fibre to the centroid of compression reinforcement. Therefore, considering the moment of resistance due to the additional tensile reinforcement and the compression reinforcement we get the following: Mu, - Asts (0*87f,) (d - a,) also, Mu, =&Us-fQC)(d-J') where A1t2 is the area of additional tensile reinforcement, AK is the area of compression reinforcement, I= is the stress in compression reinforcement, and fee is the compressive stress in concrete at the level of the centroid of compression reinforcement. Since the additional tensile force is balanced by the additional compressive force, A, (l;c - fee)= At, (0*87&j Any two of the above three equations may be used for finding Alt, and A,. The total tensile reinforcement Ast is given by, Ast = Pblim = 2.24 N/mm2 2.24 x 10' x 1 x(h) `---' = 14.336 kN.m Actual bending moment of 9.6 kN.m is less than the limiting bending moment. Referring to Chart 2, reinforcement percentage, pt = 0.78 Referring to Churf PO, provide 10 mm dia at 13 cm spacing. 2.3.2 Doubly Reinforced Sections - Doubly reinforced sections are generally adopted when the dimensions of the beam have been predetermined from other considerations and the design moment exceeds the moment of resistance of a singly reinforced section. The additional moment of resistance needed is obtained by providing compression reinforcement and additional tensile reinforcement. The moment of resistance of a doubly reinforced section is thus the sum of the limiting moment of resistance Mu,lim of a singly reinforced .section and the additional moment of resistance Mu,. Given the values of Mu which is greater than M",lim, the value of Mu, can be calculated. Mu, = Mu - Muslim m bd$ Asc, It will be noticed that we need the values of frc and J& before we can calculate Al. The approach, given here is meant for design of sections and not for analysing a given section. The depth of neutral axis is, therefore, taken as equal to x,,,-. As shown in Fig. 5, strain at the level of the compression reinforcement will be equal to O-003 5 1- XU,UWZ > ( d' STRAIN FIG. 5 OlAGRkM DOUBLY REINKIRCED SECI-ION DESIGN AIDS FOR REINFORCED CONCRIXE 12 For values of d'/d up to 0.2, fee is equal to 0446 fck; and for mild steel reinforcement fz would be equal to the design yield stress of 0.87 fY. When the reinforcement is coldworked bars, the design stress in compression reinforcement fw for different values of d'/d up to 0.2 will be as given in Table F. TABLE F STRESS IN COMPRESSION REINFORCEMENT ftc, N/mm* IN DOUBLY REINFORCED BEAMS WITH COLDWORKED BARS (Clause 2.3 2) The values of pt and d'jd up to 0.2 have MU/bd2 in Tables 45 for three grades of of concrete. Example 3 pc for four values of been tabulated against to 56. Tables are given steel and four grades Doubly Reinforced Beam Determine the main reinforcements required for a rectangular beam section with the following data: Size of beam Concrete mix Characteristic strength of reinforcement Factored moment 30 x 6Ocm M 15 415 N/mm2 320 kN.m f Y9 N/mm' 415 500 -A 0.0s 355 424 0.10 353 412 d'ld O-15 342 395 , 0.20 329 370 2.3.2.2 Astz has been plotted against (d -d') for different values of MU, in Charts 19 and 20. These charts have been prepared for fs = 217.5 N/mm2 and it is directly applicable. for mild steel reinforcement with yield stress of 250 N/mm*. Values of Aat? for other grades of steel and also the values of A, can be obtained by multiplying the value read from the chart by the factors given in Table G. The multiplying factors for A=, given in this Table, are based on a value of fee corresponding to concrete grade M20, but it can be used for all grades of concrete with little error. TABLE G MULTIPLYING USE WITH CHARTS FACTORS FOR 19 AND 20 Assuming 25 mm dia bars with 25 mm clear cover, d = fj0 - 2.5 - 225 = 56*25 cm From Table D, for fy = 415 N/mm2 and fck = 15 N/mm2 Mu,linJbd"=2.07 N/mm2 = 2.07 x IO2kN/m" -2.07 x 103bd2 .*. Mu,lim -2.07 56.25 56.25 30 x 10"x loo x Tsx -100- = 196.5 kN.m Actual moment of 320 kN.m is greater than Mu,lim .*. The section is to be designed as a doubly reinforced section. Reinforcement from Tables Mu $$ = O-562 320 5)2 x 103~~~~~ N/mm2 `Clause 2.3.2.1) N&P 250 415 500 f FACTOR FOR FACTOR FOR A, FOR d'jd c-- A at* 1.00 0.60 0.50 0.05 1.04 0.10 1.04 0.63 0.54 0.15 1.04 0.65 0.56 0.2 1.04 0.68 0.60 d'/d c 2.5 + 1.25 i o,07 5625 > 0.63 0.52 Next higher value of d'/d = 0.1 will be used for referring to Tables. Referring to Table 49 corresponding = 0.1, to 2.3.2.2 The expression for the moment of resistance of a doubly reinforced section may also be written in the following manner: Mu = Mu bj2 = Mu,lim Mu,lim ___bd" + MU/bd2 = 3.37 and $ . .. %(0*87fy) + (d-d') I;> Pt = 1.117, pc = 0.418 At - 18.85 cm2, A, = 7.05 cm2 REINFORCEMENTFROM CHARTS -&(0*87f,)( where ptz is the additional reinforcement. Pt = phlim + (d-d') = (56.25 - 3.75) - 52.5 cm Mu2 - (320 - 196.5) = 123.5 kN.m percentage of tensile pt2 Chart is given only for fy = 250 N/mm2; therefore use Chart 20 and modification factors according to Table G. Referring to Chart 20, Art2 (for fY = 250 N/mm2) = 10.7 cm2 13 PC =P"[-L-] FLEXURAL MEMBERS usia Jl¶odibrion factors given in Table G Y= 415 N/nuns, for B I& - 10.7 x 0.60 r! 6-42 cm* ,& I 10.7 x 0.63 = 674cm' Referring to ruble E, pt,nm - 072 5625 x 30 - 1215cm' ,oo .*. Ast,u,n -0.72 .x E 12.15+ 642 = 18.57cm' These values of At and AE are comparable to the values obtained from the table. A*: distribution in the flanp would not be uniform. The expression Bven in E-22.1 of the Code is an approximation which makes allowance for the variation of stress in the flange. This expression is obtained by substitutin# JYfor &in the equation of E-2.2 of the CO& yf being equal to (0.15 X,,m&+ 065 or) but not greater than Dr. With this m&&ation, Mudin~~T9 Mu,lirn,web f 0446 f& Mr-WY+ f ) Dividing both sides by&k bw P, 2.4 T-SECTIONS The moment of resistance of a T-beam can be considered as the sum of the moment of resistance of the concrete in the web of width b, and the contribution due to ,flanges of width br. The maximum moment of resistance is obtained when the depth of neutral axis is x,,,~. When the thickness of flange is small, that is, less than about 0.2 d, the stress in the flange will be uniform or nearly uniform (see Fig. 6) and the centroid of the compressive force in the flange can be taken at Df/2 from the extreme compression fibre. Therefore, the following expression is obtained for the limiting moment of resistance of T-beams with small values of Dfjd. x(br-bw)h( d-$) x(& l)$(l+$) xu;= + 0.65 !$ where but .f < $ Using the above expression, the ~2: of the moment of resistance Mu,lim,T~ck b,# for different values of h/b* and &/d have been worked out and given in Tables 57 to 59 for three grades of steel. 2.5 CONTROL OF DEFLECTION whereMll,llltniiveb 30.36 fd bwxu,,,,.x (d-O.416 x0,,,,&. The equation given in E-2.2 of the code is the same as above, with the numericals rounded off to two decimals. When the flange thick-, ness is greater than about 0.2 d, the above expression is not corre4ztbecause the stress 2.5.2 The deffection of beams and slabs would generally be, within permissible limits if the ratio of span to effective depth of the member does not exceed the values obtained in accordance with 22.2.1 of the Code. The following basic values of span to effective depth are given: En\!;;;Eorted Cantilever 20 "4 0.0. -_b -* 047 f" 0.002 0.87 f, E, STRAIN DIAGRAM ho. 6 T-SECTION DBSIGN AIDS FOR REINPORCED CDNCRETE STRESS DIAORAM 14 Further modifying factors are given in order to account for the effects of grade and percentage of tension reinforcement and percentage of compression reinforcement. 2.5.2 In normal designs where the reinforcement provided is equal to that required from strength considerations, the basic values of span to effective depth can be multiplied by the appropriate values of the modifying factors and given in a form suitable for direct reference. Such charts have been prepared as explained below : 4 The basic span to effective depth ratio for simply supported members is multiplied by the modifying factor for tension reinforcement (Fig. 3 of the Code) and plotted as the base curve in the chart. A separate chart is drawn for each grade of steel. In the chart, span to effective depth ratio is plotted on the vertical axis and the tensile reinforcement percentage is dotted on the horizontal axis. b) When the tensile reinforcement ex.ceeds ~I,II,,, the section will be doubly reinforced. The percentage of compression reinforcement is proportional to the additional tensile reinforcement @t - PM,,) as explained in 2.3.2. However, the value of Pt,lim and pc will depend on the grade of concrete also. Therefore, the values of span to effective depth ratio according to base curve is modified as follows for each grade of concrete: 1) For values of pt greater than the appropriate value of pt,lim, the value of (pt - pt,lim) is calculated and then the percentage of compression reinforcement p= required is calculated. Thus, the value of pc corresponding to a value of pt is obtained. (For this purpose d'/d has been assumed as 0.10 but the chart, thus obtained can generally be used for all values of d'/d in the normal range, without significant error in the value of maximum span to effective depth ratio.) 2) The value of span to effective depth ratio of the base curve is multiplied by the modifying factor for compression reinforcement from Fig. 4 of the Code. 3) The value obtained above is plotted on the same Chart in which the base curve was drawn earlier. Hence the span to effective depth ratio for doubly reinforced section is plotted against the tensile reinforcement percentage pt without specifically indicating the value of pc on the Chart. FLBXURAL MEMBERS 25.3 The values read from these Charts are directly applicable for simply supported members of rectangular cross section for spans up to 10 m. For simply supported or continuous spans larger than 10 m, the values should be further multiplied by the factor (lo/span in me&es). For continuous spans or cantilevers, the values read from the charts are to be modified in proportion to the basic values of span to effective depth ratio. The tn.l~G$ing factors for this purpose are as . . conned; spans & In the case of cantilevers which are longer than 10 m the Code recommends that the deflections should be calculated in order to ensure that they do. not exceed permissible limits. 2.54 For flanged beams, the Code recommends that the values of span to effective depth ratios may be determined as for rectanEoeons, subject to the followmg modi. . 4 The reinforcement be,bcz&zm percentage should the area brd while referrmg b) The value of span to effective depth ratio obtained as explained earlier should be reduced by multiplying by the following factors: Factor b&v >:::3 For intermediate values, linear interpolation may be done. Nom --The above method for flanged beams alay sometimes give anomalous mwlts. If the fhges arc ignored and the beam is considered as a rectangular section, the value of span to effective depth ratio thus obtained (percen of rciaforcemcnt being based on the area l&) Y s ould always be oa the safe side. 2.5.5 In the case of tw way slabs supported on all four sides, the sPorter span should be considered for the purpose of calculating the span to effective depth ratio (see Note 1 below 23.2 of the Code). 2.5.6 In the case of flat slabs the longer span should be considered (30.2.2 of the Code). When drop panels conforming to 30.2.2 of the Code are not provided, the values of span to effective depth ratio obtained from the Charts should be multiplied by 0.9. Example 4 Control of Deflection Check whether the depth of the member in the following cases is adequate for controlling deflection : a) Beam of Example 1, as a simply supported beam over a span of 7.5 m 15 b) Beam of Example 3, as a cantilever beam Cl .* . The section is satisfactory for control over a span of 4.0 m Slab .of Example 2, as a continuous slab spanning in two directions the shorter and longer spans being, 2.5 m and 3.5 m respectively. The moment given in Example 2 corresponds to shorter spa'n. ratio of Span Eflective depth = 13.33 tension reinforcement of deflection. c) Actual ratio of Span Effective depth Actual a> = (56.;5;,oo) Percentage required, pt = 0.6 of =-=2.5 31.25 0.08 (for slabs spanning in two directions, the shorter of the two is to be considered) (i) Forfv = 415 N/mm2 pt = 0,475 Referring to Chart 22, Max Span = 23.6 (-> d For continuous slabs the factor obtained from the Chart should be multiplied by 1.3. :. Max "7 for continuous slab Referring to Char1 22, value of Max corresponding to Pt = 0.6, is 22.2. ( Span T > Actual ratio of span to effective depth is less than the allowable value. Hence the depth provided is adequate for controlling deflection. Span b) Actual ratio of Etfective depth `(d&J = 7.11 reinforcement, = 23.6 x 1.3 F 30.68 Actual ratio of span to effective depth is slightly greater than the allowable. Therefore the section may be slightly modified or actual deflection calculations may be made to ascertain whether it is within permissible limits. (ii) F0r.j; = 250 N/mm2 pt = 0.78 Referring to Chart 21, Span = 31.3 (-1 d :. For continuous slab, Max %% d Percentage of tensile pr = 1.117 Referring to Churl 22, Max Max value of %!a? = 21.0 ( Cl 1 For cantilevers, values read from the Chart are to be multiplied by 0.35. :. Max value of 1 I/d for ) =21.0x0*35=7.35 cantilever J = 31.3 x 1.3 = 40.69 Actual ratio of span to effective depth is less than the allowable value. Hence the section provided is adequate for controlling deflection. 16 DESIGN AIDS FOR REINFORCED CONCRETE As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank TABLE 1 FLEXURE - REINFORCEMENT PERCENTAGE, REINFORCED SECTIONS pc FOR SINGLY 240 250 415 480 500 : 7ck 25 O-398 i;: 0.423 fY N/mm= b 250 0.141 O-166 . 8'E Q240 415 Oa5 @lOO . 8'::; O-144 0.159 0.175 i-E Oh O-237 0.244 O-250 O-257 0'263 O-270 0.276 0283 O-289 0.2% 0.303 EJ O-323 O-329 Ei O-350 O-357 O-364 0.371 0.378 0.385 0.392 O-399 480 0.074 OQ86 o&E O-125 0.138 0.151 0.164 O-178 0'191 Q-205 0.211 O-216 o-222 O-227 0993 l-007 1*021 1935 l-049 0.262 o-267 O-273 O-279 O-285 O-291 O-297 O-303 O-309 O-315 0.321 0.327 0.333 O-339 O-345 l-136 l-151 l-166 1.181 l-197 l-212 1228 1,243 1.259 1.275 0.685 O-693 O-703 0.712 i' . O-503 O-510 O-518 O-526 8:Z 0'550 0.558 O-566 0'574 O-582 O-590 O-448 0.455 O-461 0468 O-475 0482 0489 0.496 0'503 o-510 O-517 0.525 O-532 0.539 0.546 0554 0.561 0.569 O-576 O-584 O-592 E 040 O-45 O-50 O-276 O-302 O-329 O-356 O-383 0.410 O-421 0.433 IZ 8::;; 0489 O-500 O-512 0.523 O-535 0.546 0.558 O-570 1.10 l-12 l-14 l-16 l-18 1.20 1.22 ::z l-28 1.30 l-32 1.34 I:: O-581 0393 @605 . 8':;; 0.641 0.653 O-665 O-678 O-690 8E O-727 O-740 0.752 0.765 O-778 x:z 0.816 NOTE-Blanks O-265 0.290 E O-368 O-394 O-405 O-415 O-426 O-437 0448 O-458 0469 0.480 0.491 O-502 0.513 0.524 8:::; 0.558 0370 0.581 0.592 O-604 O-615 0.627 O-639 O-650 O-662 I:G O-698 O-710 O-722 O-734 o-747 0.759 0.771 O-784 O-430 O-436 O-443 8::: o-463 O-469 O-476 0483 0.490 O-497 o":?i 0.518 0.525 O-532 0.539 0442 X:t:Y O-465 0.472 O-382 O-389 0.395 x:z indicate inadmissible reinforcement percentage (see Table E). FLeXURAL MEMBERS 240 250 415 480 f -- TABLE 2 FLEXURE -REINFORCEMENT PERCENTAGE, REINFORCED `SECTIONS pt FOR SINGLY ck f& - zO.N/m' 20 "0% O-188 O-213 O-237 O-073 Efl 0.111 0'123 izi! FEJ . O-131 O-143 XE O-181 1.253 :`26: 1% LOZ l-323 E . :`:z . :`z l-423 l-438 l-452 1467 1~482 l-497 :::;; 1.542 l-558 ::z l-604 l-203 l-216 :z 1.256 O-627 O-633 I:: 0.661 pj . O-690 . x'% O-615 0.621 0.628 O-635 xzz 0.655 0.662 O-201 ::E 0.242 0.255 l-05 1.10 1'15 . :`z E E l-50 1.55 ;:g l-70 l-75 l-80 I-85 E 200 3g E 210 . f':: . f':; 220 0.538 O-566 OO'E * O-650 O-678 O-707 0.736 O-765 0.795 O-825 8$;: . O-916 0947 * %! 1'041 1.073 lm6 l-119 :::z l-159 l-172 l-185 1*199 . :`E l-239 X:E O-521 o-537 0.553 0.517 O-543 0.570 0.597 @624 O-651 O-679 0.707 0.735 0'763 ::a:; O-297 O-311 0'325 0.339 O-354 . EL! 0397 O-193 Ffg O-245 0.258 8:;:; 0.298 0312 l-338 l-352 1966 1380 1'394 o-697 0.704 o-711 o-719 O-726 O-734 O-741 O-748 0.756 0.764 1481 1'495 :::1': l-540 ;:gj 0.500 O-515 o-531 0.537 0'543 O-550 O-556 O-562 1.782 . :`E l-833 1.632 l-647 l-663 _g;; . l-711 . :`% l-760 Num -Blanks lndk$teinadmissible reinforcement pcmntagc (see Table J% 48 DESIGNAIDS FOR REINFORCED CGNCRRI'E TABLE 3 FLEXURE - REINFORCEMENT PERCENTAGE, REINFORCED SECTIONs pt FOR SINGLY /ck = 25 N/mms 240 250 4'1 5 480 500 7ck 25 fY Mlw2, 250 0.30 0.35 g 0.146 O-171 O-195 :z O-271 O-296 0?321 O-347 O-373 0.80 O-85 8C lfl0 l-05 1.10 1.15 ::z l-30 l-35 :z 1.50 0.399 t-I.425 0.451 O-477 O-504 O-530 8% O-611 O-638 0.666 0.693 0.721 0.749 O-777 0.509 6535 O-561 0.140 0.164 0188 O-211 0.236 O-260 !% 0.333 O-358 415 0.084 O-099 0.113 Of27 O-142 O-156 0.171 0.186 O-201 O-216 0231 X% 0.276 O-291 0.307 O-322 0.338 0353 0.369 O-385 O-401 w417 0.433 O-449 O-466 O-482 0.499 0.515 O-532 O-549 O-566 O-583 O-601 O-618 O-635 O-653 0.671 0.689 O-707 0.725 O-743 0.762 0781 o-799 0.333 0.347 8% O-388 0403 x:::: 8:% O-415 O-489 fj:$ 480 500. O-070 :%z O-106 O-118 0.130 0.142 O-154 0167 0.179 0191 O-204 0.216 0229 O-242 0.255 0.267 0.280 0.293 0.306 0.320 O-333 0.346 O-359 0373 O-387 x%Y 0.428 O-442 0.456 0.470 0.484 O-498 o-513 O-527 0542 O-557 O-572 0.587 O-602 0.617 O-632 0.648 0.663 3.60 3.62 ::z 3.68 3.70 3-72 3.74 % 290 2-95 3.00 3.05 3.10 3.15 :::: 33% :::t 3-38 N/mms fu,N/mm2 7' 1.415 1448 1482 1.515 l-549 l-584 1.618 1.653 l-689 l-724 l-760 1.797 1.834 l-871 1909 1.947 l-962 l-978 1993 2QO9 2.025 Z:% 2.072 2-088 2.104 2.120 z::: 2.170 2.186 2.203 2.219 ;:22:; 2.270 2.287 2.304 2.099 2.115 2.131 2.147 2.163 2.179 2196 L --Y 250 1.358 l-390 l-422 l-455 1487 l-520 :::z l-621 l-655 ::% l-760 l-796 1.832 l-869 l-884 l-899 l-914 l-929 l-944 415 O-818 0.837 ::::z O-896 ::;:z O-956 o-977 0.997 1.018 1.039 l-061 l-082 1.104 ::::; :::z 1.162 500 O-679 :z: O-727 O-744 8% O-794 O-811 0.828 0.845 O-863 0.880 0.898 O-916 0.935 0942 . z-;;: 0.639 X:% @719 0.746 o-773 8::: 0.856 O-883 . :`zc l-989 2.005 1.80 l-83 1.90 1.95 2-00 2-05 210 2.15 $12": 0.949 0979 l-009 l-038 1,068 :z l-160 l-191 1.222 l-254 13283 l-317 l-350 l-382 O-911 O-940 0968 . YZ 1.055 : :% l-143 1.173 l-204 l-234 l-265 : :% 8% O-580 0.596 0.611 O-627 O-643 ZR O-691 NOTE - Blanks indicate inadmissible reinforcement percentage (see Table E). m3xuRAL MEMBERS 49 240 250 415 480 500 fck 30 'Y TABLE 4 FLEXURE - REINFORCEMENT PERCENTAGE, pi FOR SINGLY REINFORCED SECTIONS fd = 30 N/mm2 250 0.140 500 0.070 0.082 0.093 0.105 0.117 MUW, N/mm2 r240 255 fZ 270 275 c8": tz 3.00 1.374 ::zi 1467 1.498 1.530 1.562 1.594 1.626 1.659 1.691 1.725 1.758 1.791 1.825 1.859 1.893 zi 1998 2z 2105 2.142 2.178 2215 2253 2291 2329 2367 f:iz 2485 2.525 2566 A. N/mm2 * 250 1.319 ::zi 1.408 1.438 ~~~ 1.530 1.561 1.592 1.624 1.656 1.687 1.720 1.752 1.785 1.818 1.851 I.884 1.918 1.952 1.986 2021 2056 2091 0.978 zz3 1.036 1.055 1.075 1.095 1.115 1.135 1.156 l-176 1.197 1.218 1.239 1.260 1.281 1.303 1.325 1.347 1.369 i::z 2386 2.424 2463 1.391 1.414 415 0.794 0.812 0.830 0.848 0.866 480 @687 0.702 0.718 0.733 0.749 is; 0.797 0813 0.829 0.846 0.862 0.879 0.896 0.913 0.930 0.947 0.964 0.981 0.999 KG; 1.053 1.071 1.089 1.108 -7 500 . 8'::; @211 0235 0.734 0.750 0.765 0.781 0.796 0.812 0.828 0.844 0.860 0.876 0380 0405 0.429 0.454 0.479 0.525 0552 0.578 . x'z ~~i?i 0.712 0739 0.766 8';1;: oi49 x:zi 0.932 0961 0.989 1.018 1946 1.075 . :x 1.163 1.192 1'173 ::i!f 1.260 1289 @631 x::: 0.709 0'735 0.762 0.788 xz:: 0.868 0.895 0.922 0.950 0977 1.005 0.252 0.265 Ct.277 0.290 0.303 0316 xz 0.355 0.368 @381 0.394 x:z 0.434 oo:z 0.475 0488 0502 3.05 3.10 3.15 33:g 3.30 3.35 :z 3.50 3.55 3.60 3.65 3.70 3.75 3.80 3.85 it:z 4.00 0976 zf: 1.028 1.046 KS 1.099 1.118 . :`:z 1.164 1.184 NOTE - Blanksindicate inadmissible reinforcement pyceniage(see TableE). so DESIGN AIDS FOR REmFORCJiD CONCREIB As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank 3. COMPRESSION MEMBERS 3.1 A$MU\;,OADED COMPRESSION lower sectioqs would eliminate tho need for any calculation. This is particularly useful as an aid for deciding the sizes of columns at the preliminary design stage of multisforeyed buildings. Example 5 Axially Loaded Column All compression members arc to be designed for a minimum eccentricity of load in two oribcioal directions. Clause 24.4 of the Code specifiks the following minimum eccentricity, eminfor the design of columns: 1 i-D 3o subject to a minimum of emin=ggg I 2 cm. where 1 is the unsupported length of the column (see 24.1.3 of the Code for definition of unsupported length), and D is the lateral dimension of the column in the direction under consideration. After determining the eccentricity, the section should be designed for combined axial load and bending (see 3.2). However, as a simplification, when the value of the mininium eccentricity calculated as above is less than or equal to 0*05D, 38.3 of the Code permits the design of short axially loaded compression members by the following equation: P,=@4f,k where PU is the axial load (ultimate), A, is the area of concrete, and Asc is the afea of reinforcement. The above equation can be written as P" = 0.4 f& PA, Determine the cross section and the reinforcement required for an axially loaded columc with the following data: Factored load Concrete grade Characteristic strength of reinforcement Unyoyior;d length of 3000kN M20 415 N/mm' 3.0 m The cross-sectional dimensions required will depend on the percentage of reinforcement. Assuming 1.0 percent reinforcement and referring to Chart 25, Required cross-sectional area of column, A, - 2 700 cm* Provide a section of 60 x 45 cm. Area of reinforcement, A, - 1.0 x m~$,j2 1: 27 cm8 We have to check whether the minimum eccentricity to be considered is within 0.05 times the lateral dimensions of the column. In the direction of longer dimension, D --I &in 500 +30 AC-i-0.67fY Ar A, - +$) t- 0.67fy loo where As is the gross area of cross section, and p is the percentage of reinforcement. Dividing both sides by A,, 3*0x 102 60 + jg P 0.6 j-2.0 - 2.6 cm 500 or, e&D = 26160 = O-043 = In the direction of the shorter dimension, 3.0 x 102 45 emrn = + = 0.6 + 1.5 500 30 = 2.1 cm or, e,i,/b = 2*1/45 = @047 The minimum eccentricity ratio is less than @05 in both directions. Hence the design of the section by the simplified method of 38.3 of the Code is valid. 3.2 COMBINED AXIAL LOAD UNIAXIAL BENDING AND PU = @4&( 1 - j&) +"`67fy $j Charts 24 to 26 can be used for designing short columns in accordance with the above equations. In the lower section of these charts, P./A, has been plotted against reinforcement percentage p for different grades of concrete. If the cross section of the column is known, PU/Al can be calculated and the reinforcement percentage read from the chart. In the upper section of the charts, PU/As is plotted against PUfor various values of AS. The combined use of the upper and COMPRESSION MEMBERS As already mentioned in 3.1, all compression members should be designed for 99 minimum eccentricity of load. It should always be ensuredthat the scotionis designed for a moment which is not lessthan that due to the prescribedtinimum eccentricity. 3.2.1 Amanptio&Assumptiom (a), (c), (d) and (e) for flexural members (see 2.1) are also applicable to members subjcoted to combined axial load and bending. `The assumption (b) that the maximum strain in concrete at the outermost eom ression fibre is 04N35 is also applicable wi en the neutralaxis k within the seotionand in the limitingcase when the neutralaxis lies along one edge of the section; in the latter oasc the strain varies from 0@035 at the highly compressed edge to zero at the opposik ed~. For purely axial compression, the strain is assumed to be uniformly equal 00002 acxossthe seotion[see 38.l(a) of the Code]. The strairidistributionlines for these ~' oases intersecteaeh other at a depth of ~ffom the highly compressed edge. This point is assumedto act as a fulcrum for the strain distribution line when the neutral axis lies outsidethe motion (see Fig. 7). This leads to the assumption that the strain at the highly compresseded~ is 00035 minus 0?5 times the strain at the least compressed edge [see 38.Z(b) of the Cole], "-i t- `* I b I q q q 0 q : q ---1-" 1 * i q q q q c ! * t I 0 : q q 00 I J: I HIWilmY C6MPRE S SE II EOOE CENTRO13AL AXIS +'+ ikh ROW OF REINFORCEMENT STRAIN 0035 DIAGRAMS Neutral axis wlthln the scctlon -30/7 -1 --- ----- Neutral axis outside the sect ion FIG. 7 Cmramm Am- LOAD AND UNIAXIAL BENDING No DESIGNAIDSIK)R REINFORCED CONCR81E 3.2.2 Stress Block Parameters Wh&n the Neutral Ax& Lies O&side the Section - When the Area of stress block - 0446f,D-5 g cated in Fig. 8. The stress is uniformly compressed edge because the strain is more than 0402 and thereafter the stress diagram is parabolic. the shape of the stress block will be as. indi- neutral axis lies outside the section, ( 4 ,-D > 0446fd for a distance of Ly from the highly = 04461&D +gD - 0446fdr D [l-&&J] The centroid of the stress block will be found by taking moments about the highly compressed edge. Moment about the highly compressed edge pO1446fckD D i (1 -$ gD STRAIN DIAORAM i t The position of the centroid is obtained by dividing the moment by the area. For diierent values of k, the area of stress block and the position of its centroid are given in Table H. TABLE H STRESS BLOCKPARAhUTTERS WHENTHE NETmmtA&mA?N LIES OUTSIDE (Clause 3.2.2) O-446 1, BTRESS OIAORAW FIG. 8 STRBSS BLOCK Am h¶ WHEN THE NEUTRAL oUT?3IDE THE SECTION kD and let g be the ditference Let x0between the strxs at the highly compressed edgo and the stress at the least compressed edge. Considering the geometric properties of a parabola, Nom-Values of stress block parameters have been tabulated for valuesof k upto4'00 for infomtion only. For constructionof interactiond@cams it b merally adaquata to consider values of k up to about 1.2. 33.3 -o+Mf& ( & 1 1 and bending are given in the form of interaction diagmms in which curyes for PJbDfd versus MdbD* fb are plotted for different values of p/f&, where p is the reinforcement percentage. 101 Construction of InteractionDiagram Design charts for combined axial compression COMPRESSlONMEMBERS 3.2.3.2 For the case of purely axial compression, the points plotted on the y-axis of the charts are obtained as follows: P,= 0446f,rbd + `g (A 0.446 fek) The above expression can be written as n Taking moment of the forces about centroid of the section, where the fr is the compressive stress in steel corresponding to a strain of 0.002. The second term within parenthesis represents the deduction for the concrete replaced by the reinforcement bars. This term is usually neglected for convenience. However, a9 a better approximation, a constant value corresponding to concrete grade M20 has been used in the present work so that the error is negligibly small over ;he range of concrete mixes normally used. An accurate consideration of this term will necessitate the preparation of separate Charts for each grade of concrete, which is not considered worthwhile. 3.2.3.2 When bending moments are also acting in addition to axial load, the points for plotting the Charts are obtained by assuming different positions of neutral axis. For each position of neutral axis, the strain distribution across the section and the stress block parameters are determined as explained earlier. The stresses in the reinforcement are also calculated from the known strains. Thereafter the resultant axial force and the moment about the centroid of the section are calculated as follows: a) When the neutral axis lies outside the section li i-1 where + xg i- 1 (.Ai - fci).Yi where C,D is the distance of the centroid of the concrete stress block, measured from the highly compressed edge; and Yi is the distance from the centroid of the section to the ith row of reinforcement; positive towards the highly compressed edge and negative towards the least compressed edge. sides of the (O-5-Cd Dividing both fck bD", c, equation by n +X*(&i -&i)(s) i- 1 b) When the neutral axis lies within the section In this case, the stress block parameters are simpler and they can be directly incorporated into the expressions which are otherwise same as for the earlier case. Thus we get the following r;xpressions: =@36k+ c i-1 &j&&d Cl - coefficient for the area of stress Pi fii - block to be taken from Table H (see 3.2.2); Ad where A,i is the area of reinbx forcement in the ith row; n fci n stress in the ith row of reinform ment, compression being positive and tension being negative; - stress in concrete at the level of the ith row of reinforcement; and - number of rows of reinforcement. where kDepth of neutral axis D DESIGN AIDS FOR REINFORCED CONCRETE 102 An approximation is made for the value Offci for M20, as in the case of 3.2.3.1. For circular sections the procedure is same as above, except that the stress block parameters given earlier are not applicable; hence the section is divided into strips and summation is done for determining the forces and moments due to the stresses in concrete. 3.2.3.3 Chartsfor compression with bending Charts for rectangular sections have been given for reinforcement on two sides (Charts 27 to 38) and for reinforcement on four sides (Charts 39 to 50). The Charts for the latter case have been prepared for a section with 20 bars equally distributed on all sides, but they can be used without significant. error for any other number of bars (greater than 8) provided the bars are distributed equally on the four sides. The Charts for circular section (Charts 51 to 62) have been prepared for a section with 8 bars, but they can generally be used for sections with any number of bars but not less than 6. Charts have been given for three grades of steel and four values of d'/D for each case mentioned above. The dotted lines in these charts indicate the stress in the bars nearest to the tension face of the member. The line for fs, I; 0 indicates that the neutral axis lies along the outermost row of reinforcement. For points lying above this line on the Chart, all the bars in the section will be in compression. The line for fSt = fYd indicates that the outermost tension reinforcement reaches the design yield strength. For points below this line, the outermost tension reinforcement undergoes inelastic deformation while successive inner rows may reach a stress of fyd. It should be noted that all these stress values are at the failure condition corresponding to the limit state of collapse and not at working Ioads. 3.2.3.4 These Charts for tension with bending - only; they do not take into account crack control which may be important for tension members. Example Bending 6 Square Column with Uniaxial Determine the reinforcement to be provided in a. square column subjected to uniaxial bending, with the following data: Size of column 45 x 45cm Concrete mix M 25 Characteristic strength of 415 N/mm% reinforcement Factored load 2500kN (characteristic load multiplied by yr) Factored moment 200 kN.m Arraugement of reinforcement: (a) On two sides (b) On four sides (Assume moment due to minimum eccentricity to be less than the actual moment). Assuming 25 mm bars with 40 mm cover, d = 40 + 12.5 OP52.5 mm z 5.25 cm d'/D = 5.25145 - 0.12 Charts for d'/D = 0.15 will be used f& = 25 x 2 45 500 xx 45103 x lo2 _ = 0.494 200 x 106 25x45~45~45~103 _ a) Reinforcement on two sides, Referring to Chart 33, p/fck = 0.09 - = 0.088 b) Percentage of reinforcement, p = 0.09 x 25 - 2.25 As = p bD/lOO = 2.25 x 45 x 45/100 = 45.56 cm2 Reinforcement on four sides from Chart 45, p&k = 0.10 p p. 0.10 x 25 = 2.5 As = 2.5 x 45 x 45/100 = 50.63 cm" Charts are extensions of the Charts for compression with bending. Points for plotting these Charts are obtained by assuming low values of k in the expressions given earlier. For the case of purely axial tension, Pu g (O-87fy) (@87fy) Example Bending 7 Circular Column with Uniaxial hk Charts 66 to 75 are given for rectangular sections with reinforcement on two sides and Charts 76 to 85 are for reinforcement on four sides. It should be noted that these charts are meant for strength calculations COMPRESSION MEMBERS Determine the reinforcement to be provided in a circular column with the following data: Diameter of column 50 cm Grade of concrete M20 Characteristic strength 250 N/mm2 for of reinforcement bars up to 20 mm+ 240 N/mm2 for bars over 20mm# 103 Factored load 16OOkN Factored moment 125 kN.m Lateral reinforcement : (a) Hoop reinforcement (b) Helical reinforcement (Assume moment due to minimum eccentricity to be less than the actual moment). Assuming 25 mm bars with 40 mm cover, d' = 40 x 12.5 = 52.5 mm m 5.25 cm d'/D - 5.25150 = 0.105 Charts for d'/D = 0.10 will be used. (a) Column with hoop reinforcement 1600 x 103 50 x 50 x ioa - o'32 125 x 10 20 x 50 x 50 x 50 x 103 - 0.05 20 Core_diie;; = 50-2 (4-O - 0.8) AI/AC IP 5O'/43*6' = 1.315 0.36 (A,,& - 1)falJlr I egg; 0.315 x 201250 Volume of helical reinforcement Volume of core Aarc .(42*8) --_------= ;(43*6%) Q, 0.09 A,J, a, x where, Ash is the area of the bar forming the helix and sh is the pitch of the helix. In order to satisfy the coda1 requirement, 0.09 Art&k > O*OO91 For 8 mm dia bar, Ati = O-503 cm2 sh ( 0.09 x 0.503 ' 0.0091 `__ < 4.97 cm 3.3 COMPRESSION MEMBERS JECT TO BIAXIAL BENDING SUB- Referring to Chart 52, for fy I 250 N/mm1 p/fck = = A: = = o-87 0.87 x 20 = 1.74 pnD2/400 1.74 x nx50x50/400=34*16cm2 Forf, I 240 N/mm2, AS = 34.16 x 2501240 = 35.58 cm2 (b) Column with Helical Reinforcement Exact design of members subject to axial load and biaxial bending is extremely laborious. Therefore, the Code permits the design of such members by the following equation : According to 38.4 of the Code, the $rength of a compression member with hehcal re-inforcement is 1.05 times the strength of a similar member with lateral ties. Therefore, the, given load and moment should be divided by 1.05 before referring to the chart. Hence, lhere M,,, M,, are the moments about x and y axes respectively due to design loads, M "Xl, MUYl are the maximum uniaxial moment capacities with an axial load P,, bending about x and y axes respectively, and ozn is an exponent whose value depends on Pu/Puz (see table below) where P uz = 0.45 fck A, + 0*75fy As: PUIPUZ `an From Chart 52, for fy = 250 N/mm2, p,$_k = 0.078 p = 0.078 x 20 = 1.56 As = 1.56 x x x 50 x 50/44X = 30.63 cm2 For fy = 240 N/mm%, A, = 30.63 x 2501240 = 31.91 cm2 go.2 )0*8 1.0 2.0 According to 38.4.1 of the Code the ratio of the volume of helical reinforcement to the volume of the core shall not be less than 0.36 (A,/Ac - 1) fck Ify where A, is the gross area of the section and Ac is the area of the core measured to the outside diameter of the helix. Assuming 8 mm dia bars for the helix, 104 For intermediate values, linear interpolation may be done. Chart 63 can be used for evaluating Puz. For different values of Pu/Puz, the appropriate value of azn has been taken and curves for the. equation (!$)"' + (z)=" = 1.0 have been plotted in Chart 64. DESIGN AIDS FOR REINFORCED CONCRETE . ExampIe Be?tdi?lg 8 Rectangular colrmu, with Biaxial DeWmine the reinforcement to be provided in a short column subjected to biaxial bending, with the following data: size of column Concrete mix Characteristic strength of reinforcement Factored load, P,, Factored moment acting parallel to the larger dimension, M,w Factored moment acting parallel to the shorter dimension, Mu, EPcm 415 N/mm' 1600kN 120 kN 90 kN of ns,, Referring to Churn 64, the permissible v&a MBa qrmsponding to the above v&see qual to 0.58. The actual value of 0.617 is only sli&ly higher than the value read from the Chart. This can be made up by slight increase in reinforcement. 100 12 bars of 18 mm will give A.130.53 c& Reinforcement percentage provided, p _ 30.53 x 100 6o x40 - 1.27 With this percentage, the section may be rechecked as follows: p/f& - l-27115= 0.084 7 Referring to Chart 44, A6 - Moments due to minimum eccentricity are less than the values given above. Reinforcement is distributed equally on four sides. As a iirst trial assume the reinforcement percentage, p= 1.2 P&k - 1*2/l 5 - 0.08 Uniaxial moment capacity of the section about xx-axis: d'/D 5.25 6. - 0.087 5 1-2x40x60 _2&8,.& Chart for d'/D = 0.1 will be used. 1600 x 10s p&k bD = 15 x 40 x 60 x 10"- 0444 Referring to Chart 44, - 0,095 f$ &li Mw z 0.;9; . - 0.085 x 40 x 60' x 10*/10* . Referring to Chart 45 f+ M WI z W&354 x&52 60 x 40' x M&k bD= = 0.09 l :. MUX, - 0.09 x 15 x 40 x 60' x loylo~ 10a/lO' - 194.4 kN.m Uniaxial moment capacity of the section about yv_axis: 5.25 d'JD = 400.131 Referring to Char; 63, PUZ= 10.4 N/mm2 Al Puz - 10.4 x 60 x 40 x lO'/lO~ Chart for d'/D - 0.15 will be used. Referring to Chart 45, M&k bD' - 0.083 :. MuYl - 0.083 x 15 x 60 x40*x 10a/lO' I 119.52kN.m Calculation of P,,: Referring to Chart 63 corresponding to p = 1.2,fu= 415 and fck= 15, - 2 496 kN Referring to Chart 64, Corresponding to the above values of PU Muy the permissible value of MuY, and z' - 10.3 x 40 x 108/10S kN 2 472 kN COMPRESSION, MEMBERS 60X MUX, MUX is 0.6. Hence the section is O.K. 105 3.4 SLENDER COMPRESSION MEMBERS of In accordance with 38.7.1.1 of the Code, the additional moments may be reduced by the multiplying factor k given below: &?. or # When the slenderness ratio D a compression member exceeds 12, it is considered to be a slender compression member (see 24.2.2 of the Code); In and i, being the effective lengths with respect to the major axis and minor axis respectively. When a compression member is slender with respect to the major axis, an additional moment Mu given by the following equation (modified as indicated later) should be taken into account in the design (see 38.7.1 of the Code) : where P,, = 0.45 &k Ac + 0.75 fy A, which may be obtained from Chart 63, and Pb is the axial load corresponding to the condition of maximum compressive strain of 0.003 5 in concrete and tensile strain of O%Ml2in outermost layer of tension steel. Though this modification is optional according to the Code, it should always be taken advantage of, since the value of k could be substantially less than unity. The value of Pb will depend on arrangement of reinforcement and the cover ratio d'/D, in addition to the grades of concrete and steel. The values of the coefficients required for evaluating Pb for various cases are given in Table 60. The values given in Table 60 are based on the same assumptions as for members with axial load and uniaxial bending. The expression for k can be written as follows : Similarly, if the column is slender about the minor axis an additional moment M.,, should be considered. M ay 2000 = Pub &" b (1 The expressions for the additional moments can be written in the form of eccentricities of load, as follows: Mu where P, eu Chart 65 can be used for finding the ratio of k after calculating the ratios P,/Pu, and pb/&z. Table 1 gives the values b or 3 for Example 9 Slender Column (with biaxial bending) Determine the reinforcement a column which is restrained with the following data: Size of column Concrete grade Characteristic strength of reinforcement Effective length for bending parallel to larger dimension, Z, Effective length for bending parallel to shorter dimension, ly Unsupported length Factored load Factored.moment in the d!"";f larger required for against sway, different values of slenderness ratio. TABLE I ADDITIONAL ECCENTRICITY FOR SLENDER COMPRESSION MEMBERS (Chuxe 3.4) 40 x 30 cm M 30 415 N/mm1 6-O m 5.0 m 70m 1500kN 40 kN.m at top and 22.5 kN.m at bottom DESIGN AIDS FOR RRINFORCED CONCRKI-E Factored moment in the direction of shorter dimension 30 kN.m at top ~tdJOcN.xn Pb (about yy-axis) i 0.184 -+ O-028x 3 30 The column is bent in double curvature. Reinforcement will be distributed equally on four sides. 6-o x 100 Lx `40 = 15.0 > 12 D P cy= I b 5.0 x 100 = 16-7 > 12 30 is slender about x40x30x30 x 10*/1os .pby - 672 kN p -;b; I `;ym17r .*. k, I = k .y = oG5 Puz - Pu p "7. - pby - 2700-1500 2 700 - 672 = o-592 Therefore the column both the axes. From Table I, For For Lz The additional moments calculated earlier, will now be multiplied by the above values of k. :: = 67.8 x O-625 = 42.4 kN.m = 63.0 x 0.592 - 37.3 kN.m P 15, eJD = 0.113 lay = 167, e,/b = O-140 b Additional moments: M 1x = Puex = 1 500 x0-1 13 x & -67.8kN.m May = Pue, = 1 500 x0*14x &=63*0 kN.m The additional moments due to slenderness effects should be added to the initial moments after modifying the initial moments as follows (see Note 1 under 38.7.2 of the Code) : M,,=(O*6 x 40 - 0.4 x 22.5) = 15-OkN.m KY= (0.6 x 30 - 0.4 x 20) = 10.0 kN.m The above actual moments should be compared with those calculated from minimum eccentricity consideration (see 24.4 of the Code) and greater value is to be taken as the initial moment for adding the additional moments. ex = - The above moments will have to be reduced in accordance with 38.7.1.1 of the Code; but multiplication factors can be evaluated only if the reinforcement is known. For first trial, assumep p: 3.0 (with reinforcement equally on all the four sides). &-=40x 30= 1200cm2 &+$-7g + 40 3o= 2*73cm = 2.4 cm From Chart 63, Puz/As = 22.5 N/mm2 .-. Pu = 22.5 x 1200 x 102/10s =2 700 kN Calculation of Pb : Assuming 25 mm dia bars with 40 mm cover d'/D (about xx-axis) cs g = 0.13 Muy - Both e, and e, are greater than 20 cm. Moments due to minimum eccentricity: Mux = 1 500 x `g = 41.0 kN.m > 15.0 kN.m 1500x2'4 100 = 36.0 kN.m > 10.0 kN.m :. Total moments for which the column is to be designed are: - 41.0 + 42.4 = 83.4 kN.m MUX iu Uy= 36.0 + 37.3 = 73.3 kN.m checked for biaxial Chart or Table for d'/d P O-15 will be used. 5.25 d'/D (about yy-axis) = 3. = 0.17 Chart or Table for d'/d = 0.20 will be used. From Table 60, Pb (about xx-axis) = (k, + k2 &rbD 0.196 + 0.203 X o 3 The section is to be betiding. Pul& bD = x 30 x 30 x 40 x 102/103 = -779 kN COMPRESSION MEMBERS 1500 x 10s 30x 30 x40 x 102 = 0.417 107 i = 0.10 Plfck - .g M - MUY UYl 73.3 Ic -103.7 = 0.71 = 0.104 x 30 x 30 x 40 x 40 x MUX1 103/10' = 149.8 kN.m Refep; to ihart 46 (d'/D P O-20), v cka 2 = 0,096 :. M"Yl =0*096 x 30 x 40 x 30'~ 30 x :. 103/106 = 103.7 kN.m 83.4 ic e 0.56 PulPu, = - 1500 = 0.56 2700 Referring to Chciit 64, the maximum allowable value of M,,/M,,, corresponding to the above values of M,,/M,,, and PuIPuz is 0.58 which is slightly higher than' the actual value of 0.56. The assumed reinforcement of 3.0 percent is therefore satisfactory. A s = pbD/lOO - 3.0 x 30 x 40/100 M",, M", - 149.8 L= 36.0 cm2 108 250 415 - 500 f ck 'Y Chart 63 VALUES OF Puz for COMPRESSION MEMBERS 15 20 25 30 ti i i i i i i i i i i i i i i i i i i IWi 148 ._ DESIGN AIDS FOR REINFORCED CONCRETE Chart64 BIAXIALBENDINGIN COMPRESSION MEMBERS 0 0.1 O-2 O-3 0.4 0.5 %/L O-6 O-7 0.8 0.9 1.0 COMPRFSSIONMEMBERS Chart 65 SLENDER COMPRESSION MEMBERS Multiplying Factor k for Additional Moments P k+ ur-pu PUZ -Pe a 04 04 O-3 0*2 o-1 0 150 DESIGN AIDS FOR REINFORCED CONCRl3-b TABLE 60 lktMgdW_: SLENDER COMPRESSION MEMBERS-VALUES OF A Valora of k, &ID r 005 0219 0172 WlO om7 @MO 015 01% 0149 OQQ 0184 0.138 9 V@mof4 fr N,`mrn~ #ID r O-05 0.10 &lS 020 \ COMPRESSION MEMBERS 171 As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank 4. SHEAR AND TORSION 4.f DESIGN SHEAR CONCRETE STRENGTH OF For a series of inclined stirrups, the value of Vup/d for vertical stirrups should be multiplied by (since i- coscc) where cc is the angle between the inclined stirrups and the axis of the member. The multiplying factor works out to 1.41 and 1.37 for 45" and 60" angles respectively. For a bent up bar, VuI= 0*87fY ASvsince where g =0.8 fck/6*89pl, but not less than 1.0, and Pt = 100 A&&. The value of `F~corresponding to pl varying from 0.20 to 3.00 at intervals of 0.10 are given in Table 61 for different grades of concrete. 4.2 NOMINAL SHEAR STRESS Values of V,,, for different sizes of bars, bent up at 45" and 60" to the axis of the member are given in Table 63 for two grades of steel. 4.4 TORSION Separate Charts or Tables are not given for torsion. The method of design for torsion is based on the calculation of an equivalent shear force and an equivalent bending moment. After determining these, some of the Charts and Tables for shear and flexure can be used. The method of design for torsion is illustrated in Example 11. *The design shear strength of concrete is given in Table 13 of the Code. The values given in the Code are based on the following equation: The nominal shear stress 7" is calculated by the following equation: 7" = - VU bd where V,, is the shear force. When rv exceeds 7c, shear reinforcement should be provided for carrying a shear equal to Vu- Q bd. The shear stress rv should not in any case exceed the values of T~,~, given in Table J. (If T"> T~,~~, the section is to be redesigned.) TABLE J CONMAXIMUM Ml5 25 SHEAR STRESS w,mu M20 M25 M30 M35 M40 2% 3-l 3.5 3-l 40 Example 10 Shear Determine the shear reinforcement (vertical stirrups) required for a beam section with the following data: `30 x 60 cm Ream size acrn Depth of beam M 15 Concrete grade 250 N/mma Characteristic strength of stirrup reinforcement Tensile reinforcement 0.8 percentage Factored shear force, Vu 180 kN Assuming 25 mm dia bars with 25 mm cover, d = 60 -T - 2.5 = 56.25 cm -30 ~8p,$o,",, = l-07 N/mm* From Table J for M15, 7c,max= 2.5 N/mm2 T" is less than 7c,mu From Table 61, for P1=0.8, ~~20.55 N/mm* Shear capacity of concrete section = Q bd = 0*55 x 30 x 56.25 x 102/103=92*8 kN 175 GRADE Q., mu, N/mm' 4.3 SHEAR REINFORCEMENT The design shear strength of vertical stirrups is given by the following equation: v "I _ - @87f,A,vd Shear stress, 7" =i g sv where A," is the total cross sectional area of the vertical legs of the stirrups, and sv is the spacing (pitch) of the stirrups. The shear strength expressed as Vu/d are given in Table 62 for different diameters and spacings of stirrups, for two grades of steel. SHEAR AND TORSION Shear to be carried by stirrups, VU,== V, -~bd = 180 - 92.8 = 87.2 kN 87.2 = V", -EL-1.55 kN/cm d 5625 Referring to Table 62, for steelf, -250 N!mme. Provide 8 mm diameter two legged vertical stirrups at 14 cm spacing. Referring to Table I, corresponding to Mujbdz .= 2.05 PI = 0.708 A,, = O-708 x 30 x 5625/100 = 1 l-95 cm* Example I I Torsion Provide 4 bars of 20 mm dia (A*= 12.56 cm*) on the flexural tensile face. As Mt is less than MU,we need not consider Me, according to 40.4.2.1 of the Code. Therefore, provide only two bars of 12 mm dia on the compression face, one bar being at each corner. As the depth of the beam is more than 45 cm, side face reinforcement of 0.05 percent on each side is to be provided (see 25.5.1.7 and 25.5.1.3 of the Code). Providing one bar at the middle of each side, Spacing of bar = 53.412 = 267 cm 0.05 x 30 x 26.7 Area required for each bar= .,oo = 040 Determine the reinforcements required for a rectangular beam section with the following data : Size of the beam Concrete grade Characteristic strength of steel Factored shear force Factored torsional moment Factored bending moment 30 Y 6Ocm M 15 415 N/mm2 95 kN 45 kN.m 11S kN.m cm* Assuming 25 mm dia bars with 25 mm cover, d = 60 - 2.5 - `G Equivalent shear, Vc = V-I- 1*6(f , 56.25 cm Provide one bar of 12 mm dia on each side. Transverse reinforcement (see 40.4.3 of the Code) : Area of two legs of the stirrup should satisfy the following: 595-t 45 1*6x m = 95-l-240 = 335 kN W---30 cm- Equivalent shear stress. 335 x 101 V, %e = Fd = 3. x 56.25 >rlo2 = 1.99 N/mm* From Table J, for M 15, `F~,,,,.~ = 2.5 N/mm" ~~~is less than sc.,,,-; hence the section does not require revision. From Table 61, for an assumed value of pt = 0.5, T. = 0.46 N/mm* c T"=. Hence longitudinal ments are to be reinforcement (see Equivalent bending Me,.= M,C Mt and transverse reinforccdesigned Longitudinal 40.4.2 of the Code): moment, * b, = 23 cm Y&6cm 1 I 7 k ,-FLEXURAL TENSION FACE 6C1 cm d, 953-4 cm -m = 115-J-79.4 = 194.4 kN.m M,Jbd2 194.4x 10" = 30 x (56.292 x 103 = 2.05 N/mma - 1 * 176 DESIGN AIDS FOR REINFORCED CONCREFE Assuming diameter of stirrups as 10 mm da = 60 - (2.5 + l-O)-(2*5+0%)-53.4 cm b1=30-2(25+1.0)=23cm 45 x 10' Aav(0*87&J S" -23 x 53.4 x lOa +25 x 9553.4 x IO? x 10-366.4-l-71.2 = 437.6 N/mm P 438 kN/cm Area of all the legs of the stirrup should satisfy the condition that A& should not From Table 61, for tensile reinforcement percentage of @71, the value of o is O-53 N/mm' - (I.99 - 0.53) 30 x 10 II 438 N/mm -438 kN/cm Nom-It is only a coincidence that the values of Aav (@87/rllSv cdcuhted by the hvo cqlmtions 8rc the srmc. Referring T&e 62 (forfr - 415 N/mm'). Provide 10 mm + two legged stirrups at 12.5 cm spacing. According to 25.5.2.7(u) of the Code, the spacing of stirrupa shall not exceed xl, (x, C yJ4 and 300 mm, where x1 and 1 arc the short and long dimensions of tL stirrup. xl - 30 - 2(2.5 - O-5)= 26 cm y,r60-2(25-05)=56cm (xl f y&/4 - (26 i- 56)/4 - 20.5 cm 10 mm + two legged stirru at 12.5 cm spacing will satisfy all the cod$ requirements. SHEAR AND TORSION 177 f ck 15 20 25 30 35 40 TABLE 61 SHEAR I IS DESIGN SHEAR STRENGTH fck, OF CONCRETE, TC)N/mm2 , pt 20 0.33 0.39 8:Z 0.51 0.55 0.57 8:Z ~~~ 0.68 0.70 072 073 0.75 076 077 0.79 0.80 0.81 0.82 082 0.82 0.82 0.82 @82 0.82 0.82 - N/mm* --. 25 0.33 0.39 0.45 0.49 0.53 056 0.59 0.62 0.64 0.66 0.69 0.71 072 0.74 076 0.77 0.79 0.80 0.82 0.83 0.84 0.86 0.87 0.88 X:E 091 0.92 0.92 30 0.33 0.40 045 0.50 0.54 0.57 iE 066 0.68 0.70 0.72 0.74 076 0.78 0.80 0.81 0.83 084 086 087 088 ;I: 0.92 0.93 0.94 0.95 0.96 35 0.34 0.40 046 0.50 0.54 0.58 0.61 8:; 069 0.72 0.74 076 0.78 0.80 0.81 0.83 085 0.86 0.88 0.89 0.91 092 0.93 094 0.96 097 0.98 0.99 40 034 0.41 046 0.51 0.55 059 062 0.65 0.68 0.70 0.73 0.75 0.77 0.79 0.81 0.83 0.85 086 088 090 0.91 0.93 0.94 0.95 097 0.98 0.99 l*oO 1.01 032 0.38 0.43 0.46 0.50 0.53 0.55 0.57 O%O 0.62 0.63 065 0.67 068 0.69 0.71 0.71 0.71 0.71 0.71 071 0.71 0.71 0.71 0.71 0.71 x::: 0.71 0.80 % 1.10 1.20 1.30 1'40 1.50 1*60 1.70 :z 2.00 210 220 2.30 f% E E 3.00 178 DESIGN AIDS FGR RHNFGRCBD CGNCmB 'v 250 415 TABLE 62 SHEAR -VERTICAL STIRRUPS Values of VW/d for two legged stirrups, kN/cm. STIRRUP J, = 250 N/mm2 DIAMEIZR, mm /x = 415 N/mm* DIAMETER, mm -7 SPmNO, em 5 , 6 8 4373 3644 3.124 2733 2429 2186 1.988 1.822 1682 1.562 1.458 ::E 1.215 1.151 1.093 0.875 0.729 0625 0.547 0.486 10 6833 6E 4271 3.796 3,416 3.106 2847 2628 ::z 2135 2010 1.898 1.798 1.708 1.367 1.139 0.976 0.854 0.759 12 ' 6 4083 3403 t:;:: 2269 2042 8 7.259 f :% 4537 4.033 3.630 3.299 3.025 2792 2593 2420 2269 2135 10 11.342 9452 8.102 7089 6302 5.671 5.156 4726 4363 4051 3.781 3.545 3.336 3.151 2985 2.836 :z 1.620 1.418 1.260 7424 6806 L% 5445 5.104 4804 t:;:;: 4083 1537 l-367 -1.230 4472 4100 3.784 Et 3.075 2894 2733 2589 2460 I.856 1.701 1.571 1.458 1.361 1.276 1.201 1.134 1.075 1.020 0.817 0681 0'583 0.510 0.454 . :8:8 1.815 l-452 1.210 1.037 0907 0.807 TABLE 63 SHEAR- BENT-UP BARS Values of Vu, for singal bar, kN BAR Dm, mm :; :B" 20 /i - 250 N/mm* I , a = 45' 1739 1208 30.92 3914 48.32 5846 75.49 94.70 123.69 15654 a=60° 21.30 1479 47.93 37.87 5918 7160 9246 115.98 151.49 191.73 fy = 415 N/mm2 I Q i= 45" 28.87 20.05 51.33 64.97 SO.21 97.05 _. __ 125.32 15720 205.32 25986 a=60° > 2456 35.36 %E 98.23 118.86 . :;;*z 251.47 318-27 NOTE- a is the angle between the bent-up bar and the axis of the member. 8HEAR AND TOIlsION 179 As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank 5. DEVELOPMENT LENGTH AND ANCHORAGE 5.2 DEVELOPMENT LENGTH OF BARS The development length Ld, is given by Ld = ~+ es 4 Tbd where 4 is the diameter of the bar, a, is the stress in the bar, and 7bd is the design bond stress given in 25.2.1.1 of the Code. The value of the development kngth corresponding to a stress of 0937 fY in the reinforcement, is required for determining the maximum permissible bar diameter for positive moment reinforcement [see 25.2.3.3(c) of the Code] and for determining the length of lap splices (see 25.2.5.2 of the Code). Values of this development length for different grades of steel and concrete are given in Tables 64 to 66. The tables contain the development length values for bars in tension as well as compression. 5.2 ANCHORAGE AND BENDS VALUE OF HOOKS In the case of bars in tension, a standard hook has an anchorage value equivalent to a straight length of 16# and a 90" bend has an anchorage value of 84. The anchorage values of standard hooks and bends for different bar diameters are given in Table 67. DEVIXWMENT LRNGTH AND ANCHORAGE 183 'v 250 415 f ck TABLE 64 DEVELOPMENT LENGTH FOR FULLY STRESSED PLAIN BARS 15 20 25 30 DIAMETER, f-A----, Jj = 250 N/mm* for bars up to 20 mm diameter. = 240 K/mm' for bars over 20 mm diameter. Tabulated values are in ccatimatns. TENSIONBARS COMPRESSION BARS GRADE OF CONCRETE BAR mm x 10 12 :: z 25 ;; 36 GRADE OF CONCRETE Ml5 43.5 326 544 65.3 97.9 87.0 108.8 114.8 1305 146.2 :z:x NOTE -The M20 % 453 544 725 81.6 z.76 108.8 1392 121.8 1566 M25 23.3 31.1 38.8 46.6 621 699 820 777 93.2 1193 lW4 1342 M30 21.8 290 363 435 653 58.0 725 766 870 111.4 974 125.3 , Ml5 261 348 43.5 522 696 78.3 91.9 870 104.4 133.6 1169 150.3 M20 21.8 z:: 435 65.3 5&O 766 725 87.0 974 f:::: M25 18.6 -M301 174 23.2 290 348 zt 58.0 61.2 696 % 10&2 497 559 65.6 621 746 95.5 835 107.4 development lengths given above ara for a stress of @87/y in the bar. _ TABLE 65 DEVELOPMENT LENGTH FOR FULLY STRESSED DEFORMED BARS fy PI 415 N/mm* Tabulated values are in cantimetrcs. BAR DIAMETER, C- TENSIONBARS GRADE OF CONCRETE A- COMPRESSION BARS GRADE OF CONCRETE -- Ml5 33'8 45.1 564 677 1% 112.8 124-l 141.0 1580 180.5 203.1 Nora-The M20 28.2 37.6 47-o 564 75.2 846 940 103.4 117.5 M25 242 322 M30 22.6 391 37.6 45.1 z:: I Ml5 271 361 45.1 542 72.2 81.2 z*: 112.8 126.4 144.4 162.5 M20 M25 193 25.8 322 38.7 51.6 58.0 %Z - M30 18.1 241 30.1 361 48.1 542 %"z 75.2 2:: 108.3 SC 827 1:A:: 169.3 development 100.7 1128 128.9 145.0 940 105.3 120.3 135.4 94.0 105.3 120.3 135.4 in the bars. 806 1E 1161 lengths given above are for a stress of 087fy 184 DESIGN AIDS FOR REINFORCED CONCRETE TABLE 67 ANCHORAGE VALUE OF HOOKS AND BENDS Tabulated values are in centimetres. BAR DIAMETER, mm Anchorage Value of hook Anchorage Value of 90" bend 6 9.6 4.8 8 128 6.4 10 160 8-O 12 192 9.6 16 25.6 12.8 18 28.8 14.4 20 320 16.0 22 35.2 17.6 25 40.0 20-O 28 448 22.4 32 51.2 25.6 36 57-6 28.8 4 cb mir _L_- -I--- STANDARD HOOK STANDARD 90' BEND STANDARD HOOK AND BEND Type of Steel Mild steel Cold worked steel NOTE 1 -Table No,rli 2 Hooks Min Valye of k 2 4 bars. given above. is applicable to all grades of reinforcqment and binds shall conform to the details 186 DESIGN AIDS FOR REINFORCED CONCRETE As in the Original Standard, this Page is Intentionally Left Blank 6. WORKING STRESS DESIGN 6.1 FLEXURAL MEMBERS Reinforcement percentage Pt,bal for balanced section is determined by equating the compressive force and tensile force. a,h _ kdb _ PI,~I bd as1 2 100 hbal = Design of flexural members by working stress method is based on the well known assumptions given in 43.3 of the Code. The value of the modular ratio, m is given by 93.33 280 m E-E3 acbc acbc Therefore, for all values of acb we have m acbc = 93.33 b;bc 50 k.a,a a t . The value of pt,w for different values of a,bc and a,t are given in Table L. TABLE L PERCENTAGEOF TENSILE REINFORCEMENTP..,,., FOR SINGLY REINFORCEDBALANCEDSECTION .,"_. 7 -. l-k T-1 7.0 8.5 10.0 (Clause 6.1.1) $&ma Gil 5.0 0.71 l-00 1.21 1.43 esl N/mm* L 230 0.31 0.44 053 0.63 , 275 O-23 0.32 0.39 O-46 FIG. 9 BALANCED SECTION (WORKING STRESS DESIGN) 6.1.1 Balanced Section (see Fig. 9) 6.1.2 Under Reinforced Section Stress in steel = ast =maLbc(+-1) The position of the neutral axis is found by equating the moments of the equivalent areas. bkdz 2 pt bd = loo m (d - kd) = bd2 `$ 1 -= k 93.33 as( f- 93.33 The value of k for balanced section depends only on qt. It is independent of a,bc. Moment of resistance of a balanced section is given k= bd2 7 (1 - k) - k) k2 = k2 + p$(l by hfbal = yach k( 1 - f );The values k = !$- !!!=o. of Mbal/bd2 for different values of U&c and asI are given in Table K. The positive root of this equation is given by p2,m2 + ptm -- Ptm + - 50 100 J F (100)" FACTOR TABLEK MOMENTOF RESISTANCE This is the general expression for the depth M/b@,N/mm*FOR BALANCED RECTANGULARSECTION of neutral axis of a singly reinforced section. Moment of resistance of an under-reinforced `%I, N/mm* =cbc section is given by N/mm' c +`140 230 275 5.0 0.87 1.21 147 1.73 0.65 0.91 1.11 1.30 0.58 O-81 0.99 l-16 7-o 8.5 10-o M/bd2 Values of the moment of resistance factor have been tabulated against pt in 189 WORKING STRm DESIGN x (1 -;)a& To& tensile reinforcement A,l is given by Ast = AM, -f- Astt FIG. 10 DOUBLY R~r~~onci?n SECTION (WORKING Smm.s D~SGN) Tables 68 to 71. The Tables cover four grades of concrete and five values of uu. 6.1.3 Doubly Reinforced Section - Doubly reinforced sections are adopted when the bending moment exceeds the moment of resistance of a balanced section. M=b&i-M' The additional moment M' is resisted by providing compression reinforcement and 0 additional tensile reinforcement. The stress in the compression reinforcement is taken as I.5 m times the stress in the surrounding concrete. Taking moment about the centroid of tensile reinforcement, bd(1.5 m - 1) ucbc M' = PC -ldc where and Atu = pl,bPI &f 100 A,Q = The compression reinforcement can be expressed as a ratio of the additional tensile reinforcement area Altp. = Qcbc (1.5 m - 1) (l-d'/kd) Values of this ratio have been tabulated for different values of d'/d and ucbcin Table M. The table includes two values of ust. The values of pt and pc for fear values of d'/d have been tabulated against' M/bd' in Tables 72 to 79. Tables are given for four grades of concrete and two grades of steel. USI 1 TABLE M VALUESOF THE RATIO A&,,, (Clause6.1.3) I +& (1.5m - 1) uck d'ld "cbc %t N/mm' N/mm' m-- 0.15 1.38 % l-44 l-66 I.68 1.70 1.72 0.20 2.07 2.11 213 2.15 5.0 x(1 -$)(I-;)bd' 140 I ;:y i 10.0 1.19 1.20 l-22 1.23 Equating the additional tensile force and dditional compressive force, ;:g 214 2.12 2.65 2.61 2.71 2.68 3.60 3.55 ;I$ 5.63 5.54 . :*:I: Xi k$! (1.5 m - l)U&( olt l-2) 6.2 COMPRESSION MEMBERS or (pt - pt,bd =pc (1*5m-l)ucbc 190 (l-ii) Charts 86 and 87 are given for determining the permissible axial load on a pedestal or short column reinforced with longitudinal bars and lateral ties. Charts are given for two vrdues of 0%. These charts have been made in accordance with 45.1 of the Code. DESIGN AIDS FOR REINFORCED CONCRETE According to 46.3 of the Code, members subject to combined axial load and bending designed by methods based bn elastic theory should be further checked for their strength under ultimate load conditions. Therefore it would be advisable to design such members directly by the limit state method. Hence, no design aids are given for designing such members-- by elastic theory. 6.3 SHEAR AND TORSION The method of design for shear and torsion by working stress method are similar to the limit state method. The values of Permissible shear stress in concrete are given in Table 80. Tables 81 and 82 are given for design of shear reinforcement. 6.4 DEVELOPMENT LENGTH AND ANCHORAGE The method of calculating development length is the same as given under limit state design. The difference is only in the values of bond stresses. Development lenaths for plain bars and two grade; of deformed bars are given in Tables 83 to 85. Anchorage value of standard hoolcs and bends as given in Table 67 are applicable to working stress method also. WORKlNC3 STRESS DESlCiN 191 As in the Original Standard, this Page is Intentionally Left Blank %t WORKING I STRESS METHOD TABLE 68 FLEXURE - MOMENT OF RESISTANCE FACTOR, SINGLY REINFORCED SECTIONS iU/bda, N/mm* FOR 130 140 190 230 275 est. N/mma 130 0,146 O-158 O-170 O-181 O-193 0.205 0.216 O-228 . 8'::: O-262 ::z o-297 O-308 O-319 O-331 O-342 O-353 0'364 O-376 O-387 O-398 :z 0431 o-443 ez O-476 O-487 O-498 0.631 0.647 O-663 O-679 0.695 O-711 0.728 O-77 z; O+O 190 230 O-258 O-321 O-341 275 pt 0.47 0.48 O-49 0.50 O-51 @52 O-53 O-54 0.55 0.56 f 130 140 0.583 0.595 O-607 O-619 0.630 O-642 O-654 O-665 0x77 O-689 O-700 O-712 0.724 0.735 O-747 O-758 O-770 O-781 O-793 O-804 X:iE 0'839 O-850 O-862 190 230 275 O-265 O-282 O-299 O-316 O-333 . x'::; O-383 0400 0.417 0.433 O-450 0.467 O-483 Oao @516 @533 x:2: . O-248 Ei 0.542 O-553 O-564 O-574 0.585 0.5% O-607 0,618 0.629 OTi40 O-650 O-661 0:672 O-683 O-693 L- . 50 O-362 O-383 . Es 0444 0.62 0.63 x:s O-66 0.704 O-715 0.726 O-736 0.747 O-758 0.768 O-779 X:E O-811 O-821 . :ii; O-853 O-864 . 8'0;: 0.895 %t 130 140 190 230 275 %bc 1 WORKING STRESS METHOD 1 TABLE 69 FLEXURE - MOMENT OF RESISTANCE FACTOR, M/hP, SINGLY REINFORCED SECTIONS N/mm* FOR . 70 us:, N/mm* 130 O-242 O-266 O-289 . :::: O-358 O-381 :zi O-449 . 8Z iFg . O-539 O-551 O-562 O-573 O-584 O-581 O-593 @605 O-617 0.629 O-641 O-653 :zE OS89 O-651 . :z :z 0706 O-717 O-728 0.739 O-750 O-761 K& . O-804 0.815 . X'E Om8 O-859 O-878 . t'E O-913 O-925 @701 O-713 . x! O-748 O-760 :z O-795 0807 O-951 O-967 O-983 SE O-523 O-557 . x'z!i O-657 O-690 O-723 0.739 . X'Ei :z O-821 0837 O-854 O-96 O-97 O-98 099 1.00 140 190 230 0.428 0.470 0.511 @552 O-593 O-633 O-674 z:: O-795 O-835 x::;: O-757 O-806 275- pt O-76 O-77 0.78 O-79 0.80 O-81 O-82 0.83 8:: O-86 O-87 O-88 O-89 O-90 130 O-869 @880 0.891 8ZZ 140 O-936 0948 O-960 0.971 O-983 o-994 190 230 27i F% 1.029 l-041 l-064 __. l-052 l-075 l-087 l-099 l-031 l*!Ml l-052 :iE l-084 ::z: 1.116 l-127 ::::I 1.158 l-169 l-180 l-110 l-122 l-133 l-145 l-156 1.16 1.179 l*l!bl 1.202 1% DElION AIDS FOR RRINKRkCRD CDNCRR'I'B 1 WORKING STRESS METHOD ( TABLE 70 FLEXURE-MOMENT OF RESISTANCE FACTOR, hf/bd¶, N/mm* FOR SINGLY REINFORCED SECTIONS 130 140 190 230 275 %bc ' a, N/mm' pt ' 130 O-244 8::;: O-314 O-337 O-30 O-32 0.34 O-36 O-38 0.361 O-394 O-407 O-430 O-453 O-476 O-498 O-521 O-544 O-567 O-50 0.52 O-54 056 O-58 O-589 O-612 O-634 x:::3 0.701 0.723 0.746 O-768 O+Ml 0.70 O-72 O-74 O-76 0.78 O-80 O-82 O-83 O-84 0.85 8:: O-88 O-89 090 0.812 O-834 0.856 0.878 0900 XE O-955 O-966 0.977 0.987 0998 ltM9 l-020 l-031 l-042 E: . l-074 l-085 1.122 l-134 1.145 l-157 l-169 O-875 O-898 0.922 0946 0.969 0993 . :% :zE 140 0.262 0.288 O-313 Ei O-388 O-413 tee @488 O-512 x:::: O-586 0.610 O-634 O-659 0%83 0.707 0.731 l-025 l-057 l-090 ::i:: l-187 l-219 190 x::;: O-425 O-459 O-493 O-527 O-561 O-595 O-628 O-662 230 O-431 O-473 @514 O-556 0.597 O-638 0.679 O-720 0.761 O-801 Ow2 ::tiz O-962 l-002 1.042 l-082 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 E 1.26 ::f; i-29 1.30 1.31 :::z . :*:: 1.36 1.37 1.310 l-321 1.332 ::E: 1.364 1.374 :::;: 1406 1.417 1.427 1.438 1448 1.459 1469 1,480 1.491 1a501 I.512 1.522 1.533 l-411 l-423 1.434 . :`% 1468 275O-515 O-565 0.615 O-664 O-714 O-763 O-812 O-861 E pt 0.96 0.97 0.98 O-99 1.00 ' 130 1.096 x: 1.128 1.139 ::E 1.171 1.182 1.193 1.203 ::z 1.236 1.246 140 1.180 1.192 1.203 1.215 1.227 1.238 1.250 190 230 275 . 85 WORKING STRESS DESIGN 197 %t 130 140 190 230 275 %bc I TABLE 71 FLEXURE - MOMENT OF RESISTANCE FACTOR, SINGLY REINFORCED SECTIONS M/bd2, N/mm2 FOR WC = 10.0N/mm' USI,N/mm* 230 O-433 0.475 O-517 X:E OTi42 O-683 O-724 ALE 0.847 . x'% O-969 1*009 1.049 ;:g l-210 1.250 l-289 1'30 1'31 l-32 l-33 1'34 l-35 :::4 1.38 l-39 ;:g . 1.43 l-44 1.45 1.46 1.47 l-48 1.49 1.50 1.51 1.52 1.53 1.54 135 1.56 . :z 1.59 l-60 l-632 1.643 1.653 l-664 1.675 1.685 EE . l-717 l-727 1.738 l-749 1.759 1.770 l-780 l-791 . 100 Pl I.10 1.12 1.14 I.16 1.18 ' 130 1.257 l-279 1.301 1.322 l-344 1.365 1.387 1.408 l-430 l-451 l-473 l-483 l-494 l-505 l-515 1.526 l-537 1.547 l-558 1.569 140 %i 1.401 1.424 1'447 1.470 1.494 E . l-563 1.586 l-597 :z 1,632 1.643 l-655 t%i . l-689 1'701 t ::i: 190 230 27; DESIGNAIDS FOR RBINFORCED CON- 1 WORKING STRESS METHOD 1 TABLE 73 FLEXURE - REINFORCEMENT PERCENTAGES REINFORCED SECTIONS FOR DOUBLY 7.0 N/mm2 6dc = qt = 140 N/mm2 MJbd a N/mm2 1.22 1.25 :::i 140 1.45 150 1.55 l-60 l-65 1.70 ::ii l-85 1.90 1.95 2.00 2.05 2.10 2.15 z 2.30 2.35 240 245 2.50 255 E 2-70 2'75 280 285 290 295 ::z 3.10 3.15 ::z :::: 340 d'Jd is 0.05 I d'/d pi O-10 d'ld = 0.15 r- pt 1a05 l-028 KG l-140 1.178 1.216 1.253 1.291 l-328 ::z 1441 1.479 l-516 3 PO . oaK O-033 O-078 O-124 O-169 pt PO- d'/d = 0.20 -A-------\ pt lTM6 l-031 l-073 1*:15 l-157 PC p, 1.006 1.033 l-077 l-122 1.167 1.211 1.255 l-301 1.345 1.390 1.435 1.479 1.524 l-568 l-613 1.658 1.702 1.747 1.792 1.836 1.881 1.926 1.970 2.015 2.060 2.104 2.149 2.193 2.238 2283 `2.327 2.372 ::t :: 2506 2.551 2-595 ;zg 2.774 2818 2.863 2908 2-952 PC0.013 oa69 O-163 O-257 0.351 0.445 0539 0.633 0.727 0.821 0.915 1009 l-103 1.197 l-291 1.385 1.479 1.573 1.667 1.761 1.855 1.949 2.043 2.137 2.231 2.325 2.419 2.513 2.607 2701 2.795 2.888 2.982 3.076 3.170 3.264 3.358 3.452 3.546 3-540 3.734 3.828 3.922 4.016 4.110 X:E O-123 O-193 0.264 0.335 :% 0.547 O-618 0.689 O-760 0.830 0901 0.972 1a43 1.113 1.184 1.255 I.326 I.396 1.467 l-538 l-609 1.680 1.750 1.821 l-892 l-963 2.033 2.104 2.175 2.246 2.316 2387 2.458 2-529 2.599 2.670 2741 2.812 2883 2.953 3.024 3.095 1.188 1228 l-267 ::z:: 0.264 O-319 O-375 0.431 O-486 :::ii 0,653 0.709 0.765 O-821 0.876 O-932 O-988 l-043 1.099 1.155 ::ti: 1.322 1.378 1.433 1.489 1.545 l%OO 1.656 1.712 :34: 1.879 1.934 ::Ez 2.102 2.157 2.213 2.269 2324 2.380 2.436 I.199 1.241 1.283 l-325 1.367 1409 1.451 1.493 1.535 l-577 1.619 1.661 l-703 l-745 1.787 1.829 1.871 1.913 l-955 l-997 2.039 2.081 2123 2165 2-207 2.249 2.291 2.333 2375 2-417 2.459 2501 2.543 2585 2.627 2.669 2.711 2.754 2796 2.838 ::z O-531 0.576 0.621 O-666 0.712 O-757 @802 O-847 1.386 1.426 1.466 1.505 1.545 1.585 1.624 Kz I.743 l-783 1.823 l-862 1902 l-942 1.981 Et 2101 2140 2180 3% 2.299 2339 2378 3% 2497 2-537 2.577 2.616 2.656 2696 2-735 : ::;3 ::i:: 1.892 t:;zt 2005 2.043 2080 2-118 2.155 2193 2231 2268 2306 2343 2381 2.419 2.456 2.494 2531 2569 f:fg Ef 0.983 l-028 l-073 1.119 1.164 l-209 1::;; l-345 1.390 l-435 l-480 1.526 l-571 l-616 1.661 1a707 1.752 1.797 I.842 f:E 1.978 ' 200 DESIGN AIDS FOR REINFORCED CONCRETE 1 WORKING !3T@ESS METHOD 1 TABLE 80 SHEAR - PERMISSIBLE SHEAR STRESS IN CONCRETE, sc, N/mm* M30 o-21 O-25 E 033 O-36 O-38 0.39 o-41 0.43 E gz 0.49 8:: O-52 O-53 O-54 0.54 0.55 @56 o-57 x::: O-59 O-59 M35 O-21 O-25 LE O-34 XG 040 O-42 O-43 O-45 E O-49 o-50 O-51 0.52 0.53 O-54 O-55 O-56 O-57 iE: O-59 la40 O-21 0.25 iz :3$ O-39 O-41 O-42 O-44 O-45 0.47 O-48 O-49 851 O-52 0.53 O-54 0.55 O-56 @57 O-58 0.59 O-60 8:: :::i TABLE 81 SHEAR - VERTICAL. STIRRUPS Valuesof 9 for hvo leggedstirrups, kN/cm Gv 3: MN/mm* Dmmnm, mm UN - 230N/mma I)l-. mm STIRRUP sPAuNo, col : 7 x 16 13 l-979 1.863 I -759 1.667 l-583 1.267 l-056 0905 :z :: 2081 1a734 f'ii! 1:156 WOltKlNO STRBSS DSlCiN 207 1 WORKING SfRESS METHOD 1 TABLE 82 SHEAR Valuewf BENT UP BARS V, for single bar, kN %V = 140 N/mm* up to 20 mm diameter = 130 N/mm' over 20 mm diameter , 0=4S0 a=60° 9.52 b r- %v=230 a=450 1277 N/mm' 778 1120 lP90 25-19 31.10 13.71 2438 3086 38.09 iii*%3 3270 41.39 51.09 61.82 7Pa3 100-14 130-80 16554 . E 93.57 NOTE- a is the an6 6P32 PO-54 11460 between the bent up bar and the axis of the member. TABLE 83 DEVELOPMENT LENGTH FOR PLAIN BARS us,= 140 N/mm* for bars up to 20 mm diameter = 130 N/mm* for bars over 20 mm diameter 0IE = 130 N/mm' for all diameter Tabulated values are in centimetres. TENSION BARS Gnrse OF CONCRETE D&R. mm 6 8 f8 :; z 25 28 :: , Ml5 35-o z: 70-O 105.0 93-3 1167 llP2 135-4 151.7 195.0 173.3 M20 263 35.0 :s:: 78-a 70.0 875 w-4 101.6 113-a 146.3 130-O T 23.3 31.1 46.7 38-P 622 70.0 77.8 7P4 1Z 1300 115.6 21.0 28-O 42.0 35.0 63.0 56-O 71.5 70-O 91-o 81.3 117-o 104-o `M1S z:(: 43.3 520 78-o 6P3, 993 867 108.3 121.3 1560 138.7 MU) lP-5 E3 3PO ;:3 E M2S 17.3 23-l z:; 46.2 520 63.6 57-a za 1E tikXlPRE%SlON BARS GRAIX OF CONCRETEi 3 M30 15.6 208 26-o 31.2 41.6 46.8 :27p . z:: 93.6 83.2 - 81.3 91-o 1170 104-o 208 DESIGN AIDS FOR REINFORCED CONCRBIg WORKING STRESS METHOD I TABLE 84 DEVELOPMENT LENGTH FOR DEFORMED BARS Tabulated values are in centimetres. %t = 230 N/mm* SC = 190 N/mm' BAR DIAMETER, mm : 10 12 16 18 TENSIONBARS GRADE OF CONCRETE PA_ M25 M20 30.8 41.1 51.3 61.6 82-l 92.4 1027 1129 128.3 143.8 164.3 184.8 27-4 36.5 45.6 548 73.0 82.1 91.3 100.4 114.1 127.8 146.0 164.3 COMPREWON BARS GRADE OF CONCRETE M30 24.6 329 41.1 49.3 65.7 73.9 82.1 90.4 1027 115*0 131.4 147.9 xi----41.1 54.8 68-5 82-l 109.5 123.2 136-9 150.6 171'1 191.7 219.0 2464 ' TiiY 27.1 36.2 45.2 M20 20.4 27.1 33.9 40.7 54.3 61.1 67-9 74.6 84.8 95.0 108.6 122.1 M25 18.1 24-l 302 36.2 48.3 54-3 EC: 75.4 84.4 96.5 108.6 M3d 16.3 21-7 27.1 32-6 43.4 48-9 54-3 597 67-9 L!&! 97.7 ,54*3 724 81.4 z*: 113.1 126.7 144.8 1629 TABLE 85 DEVELOPMENT LENGTH FOR DEFORMED BARS Tabulated values are in centimetres. Qll = 275 N/mm* 0,=19ON/mm BAR DLAMEI-ER, ,_-__h_ Ml5 mm 49-l :::; 98.2 16 18 ;: 131-o 147.3 163.7 180.1 % f TENSIONBARS _.___.___ GRADE OF CONCRETE M20 36.8 491 61.4 73.7 98.2 ::I; 135.0 171.9 153.5 196.4 221.0 M25 327 43,7 54.6 65.5 87.3 98.2 109*1 120.6 152.8 136.4 174.6 1964 M30 29.5 49.1 39.3 58.9 78,6 88.4 98.2 108.0 1228 z: 176.8 > 27.1 36.2 45.2 54-3 72.4 81.4 E: 113.1 126.7 144.8 162.9 COMPRESSION BARS GRADE OF CONCRETE M20 20.4 27.1 33.9 40.7 z: 67.9 74-6 848 95.0 108.6 1221 M25 18.1 241 30.2 36.2 48-3 z:: 66.3 75.4 84.4 96.5 108.6 \ M30 16.3 21.7 27-l 326 43.4 E 597 67.9 76.0 86.9 97.7 261.9 2946 WORKXNQ STRBSS DESIGN 209 As in the Original Standard, this Page is Intentionally Left Blank As in the Original Standard, this Page is Intentionally Left Blank 7. DEFLECTION CALCULATION 7.1 EFFECTIVE INERTIA MOMENT OF The. chart takes into account the condition 4 I > 1. After finding the value of Zd it has A method of calculating the deflections is given in Appendix E of the Code. This method requires the use of an effective moment of inertia I& given by the following equation Z* I&r 1.2-s; 1-2 +_ ( ) but, Whrn Ir to be compared with Z* and the lower of the two values should be used for calculating the deflection. For continuous beams, a weighted average value of Z~lr should be used, as given in B-2.1 of the Code. 7.2 SHRINKAGE AND CREEP DEFLECTIONS Ir < Za < b is the moment of inertia of the cracked section ; Mr M z d x b, b where fa is the modulus of rupture of concrete, Zmis the moment of inertia of the gross section neglecting the reinforcement and yt is the distance from the centroidal axis of the gross section to the extreme fibre in tension ; is the maximum moment under service loads; is the lever arm; is the effective depth; is the depth of neutral axis; is the breadth of the web; and is the breadth of the compression face. fcllll be calculated in accordance with clauses B-3 is the cracking moment, equal to and B-4 of the Code. This is illustrated in Yt Example 12. Example 12 Check for deflection Calculate the deflection of a cantilever beam of the section designed in Example 3, with further data as given below: Span of cantilever 4.0 m Re\didimoment at service 210 kN.m Sixty percent of the above moment is due to rmanent loads, the loading being distri% uted uniformly on the span. BP ZE =-i-T= 300 x @O)* _ 5.4 x 10' mm' 12 Deflections due to shrinkage and creep can The values of x and z are those obtained by elastic theory. Hence z = d - x/3 for rectangular sections; also b = b, for rectangular sections. For flanged sections where the flange is in compression, b will be equal to the flange width br. The value of z for flanged beams will depend on the tlange dimensions, but in order to simplify the calculations it is conservatively assumed the value of z for ganged beam is also d - x/3. With this assumption, the expression e&ctive moment of inertia may be written as follows : From clause 5.2.2 of the Code, Flexural tensile strength, fcr = 0.74 z N/mm9 fcr P O-7 t/E = 2.71 N/mm' Yt -D/2=~=3OOmm 2.71 x 5.4 x 10' - 488 x 10' N.mm - O-067 a'/d II 005 will be used in referring to Tables. From 5.2.3.Z of the Code, EC = 5700 q/fck N/mm* I 5 700 d/13= 22-l x 10' N/mm* A?& P 200 kN/mm* = 2 x 10s N/mm% > 1 r and Zen< Zm Chorr 89 can be used for finding the value of F I in accordance with the above equation. but, F DEFLECTION CALCULATION 213 From Example 3, p, = 1.117.p, =0.418 p,(m - I)/@, m) = (0.418 X 8.05)/ (1.117 X 9.05) = 0.333 PJ?? = 1.117 x 9.05 = 10.11 Referring to Table 87, I,/(bd'/ 12) = 0.720 .. I, = 0.720 X 300 X (562.5)`/ 12 = 3.204 X IO9 mm4 Referring to Table 91, Deflection due to creep, a,, (pcrm) = a,,, (p,r,nj- a, ,,,cmr, In the absence of data, the age at loading is assumed to be 28 days and the value of creep coefficient, 8 is taken as 1.6 from 5.2.5.1 of the Code. EC, = E, 1 +e 22.1 x IO3 1 + 1.6 = 8.5 X 10' N/mm2 E, 2x IO5 = 23.53 = J = 0.338 Moment at service load, M = 210 = 21.0 X 10' N.mm 4.88 X 10' Mr/ M = 21.0 x lo'= o.232 Referring to Chart 89. I,,,/ I, = 1.0 . Ierr = I, = 3.204 X IO9 mm' For a cantilever with uniformly load, 2 kN.m m = z = 8.5 X lo3 p, = 1.117, pc (m - l)/(p,m) pL = 0.418 = 0.418(23.53 - I)/ (1.117 X 23.53) = 0.358 Referring distributed to Table 87, Elastic deflection cll 2 1 .o X 10' x (4000)? z -__--4 x 22.1 X lo3 x 3.204 X 10" = I I.86 mm = f .g t,/(bd'/ 12) = I,.497 I, = 1.497 X 300 (562.5)3/ 12 = 6.66 X lo9 mm" I, < Lrr d I$q 6.66)X 10" d I,,, < 5.4 x IO9 .* . Ierr = 5.4 X 10' mm4 alcc (,,rr,,r,= Initial plus creep deflection due to permanent loads obtained using the Above modulus of elasticity 1 Ml2 --= 4 E&r . ..( 1) Deflection due to shrinkage of the Code): IILo= k+ Vv, I` ki = 0.5 for cantilevers (see clause B-3 p, = l.l17,p, = 0.418 pi--p<= 1.117-0.418=0.699< ... ~4=0.72Xy& Pt = 0,72 x (1.1.17 - 0.418) fii-iT ==0.476 ' = $X 1.0 (0.6 X 21 X 10') (4 000)2 8.5 X IO3 X 5.4 X IO9 = 10.98 mm In the absence of data, the value of the ultimate shrinkage strain &, is taken as 0.000 3 as given in 5.2.4.1 of the Code. L)=6OOmm .'. Shrinkagecurvature q\Ir,,= k4 g aI (pwn\ = Short term deflection due to permanent load obtained using EC 1 (0.6 X 21 X 10') (4 000)' =i-x- 22.1 X 10' X 3.204 X IO9 = 7.12 mm = 10.98 - 7.12 = 3.86 . ..(3) ... a`r(pc.,m) . . Total deflection (long term) due to initial load, shrinkage and creep = 1 1.86 + 1.90 + 3.86 = 17.62 mm. According to 22.2(a) of the Code the final deflection should not exceed span/2SO. . Permlsslble deflection = = 0.476 X 0.000 3 = 2 38 x 1o-7 600 a,, = 0.5 X 2.38 X 10e7 X (4 000)2 `&!I?!!? 250 = 16 mm. = 1.90 mm . ..(2) The calculated deflection is only slightly greater than the permissible value and hence the section may not be revised. 214 DESIGN AIDS FOR REINFORCED CONCRETE i w t ii L 1.0 10 RATIO bf/bw DEFLECTION CALCULATION 215 Chart 89 EFFECTIVE MOMENT OF INERTIA FOR CALCULATING DEFLECTION f i- b-0 3i 1.0 1.1 L_;______;-__ _;-c_$ I I I IY I I I I I I I I I hi-I i I I I I Al I I I I - 1 %++-II"8 I l-4 l-5 1.6 1.7 1.8 l-9 2~0"""""""'~""""""~ 216 DESIGN AIDS FOR REINFORCED CONCRETE Chart 90 PERCENTAGE, AREA AND SPACING `OF BARS IN SLABS 29 24 23 22 21 20 19 10 9 0 1 2 J b s 9 1 D 9 10 11 12 13 14 19 AREA OF REINFORCEMENT cm' PER METER WIOTH +USf ECfCCtlVL OCfTM OR OVERALL WHICHEVER IS USLO POR CALCUlATING p DBFLECTION CALCULATION 217 Chart 91 EFFECTIVE LEhlGlH OF COLUMNSFrame Restrained Against Sway FUED 0 0.6 P 0*7 0.8 0.9 l*O ii E Pa Kb z r 8 BXand Paare the valuesof 19 at the top and bottom of the column when, pdone for the members framin8 into r joint; KC and arc the fkxural sKc the summation being r;K, + tKb ' stiffacoses of column and &m mpiwfy. 218 DESk3N AIDS FOR REINFORCED CONCRHI chrrt 92 EFFECTIVE LENGTH OF COLUMNS Frame Wiiut Restraint to Sway 0*9 04 0.6 P1 FIXED 0 o\ 0.1 0.2 0.3 04 03 0*6 0.7 0.8 0.9 1-O x ii wp P2 b'~Kc~sKb, tbe of &Umn hand @,are the values of b at the top and bottom of the column. where done for tbo mombem framing into a joint; KC and swnmation being Kb are the llex~rd dfhessu and beamrespeCtivclY. DFMACMON CALCULATION 219 TABLE 86 MOMENT OF INERTIAd. cm 10 VALUES OF M/12 000 b, cm 15 1.2 l-7 $; 20 25 30 35 40 45 50 4-2 5.5 ;`2 11.4 14-l 17-l z 28.6 33.3 38-6 iii:: 57-6 ;; ;:; 46 ::: 5-8 ;:; ::g 6-5 1:: :::: 18.4 21.9 25.7 g:`: 10-2 12-l 143 10-O :::4 15.2 17.3 19.5 220 24-6 27-4 30-5 33.8 41.0 49.1 58.3 68.6 53-3 61.7 E 922 lW2 117-2 131.2 146-3 1626 180-O %:X 351.6 4267 511.8 607-s 7145 833.3 it: :z3 138-2 :f:; 196.8 219-5 243-9 270-O 343.3 428.7 527-3 640.0 767.7 911'3 1071.7 1250.0 45.0 546 65.5 77'8 91-5 106-7 123.5 1420 1622 :843 208-3 234.3 i;::`: 325.2 360.0 ::::7' 703.1 853.3 1023.5 %X 1666-7 326 36-6 41-o 45-7 50-8 z:: 81.9 97.2 1143 ::::: 177-5 202-8 23@4 260-4 2929 ~%I: 406.5 450.0 5721 7146 878.9 1066-7 1279-4 1518.8 1786-2 2083.3 g 146 17-l 1rt 14-3 17-O 20-O 23.3 27-o 31.1 35.5 40.3 45.6 51.3 z:t 71-l 11.3 z 19-4 m9 52-l :!:f 73-2 81.3 90.0 109.2 131.0 :z . 213.3 E8 324-5 368.6 416-7 iii:; 585-4 650-4 ES . tz:: 1706-7 2047-l ii% . 3333'3 58.6 65.9 i::!! 91.5 101.3 122.9 147-4 175.0 205.8 2400 277.8 3194 365.0 4147 E: 590-5 658.6 731.7 8AO.O 10298 1286-2 15820 192@0 65-l 73-2 82-O 91.5 101.6 1125 %5 . 194-4 228.6 266.7 :::I: 405-6 460-8 5208 i%; 731.7 813-O 1E :%-ii 2133:3 %T . 3572'4 4166-7 67-5 81.9 98.3 fK4 78.8 95.6 114.6 136-l 160.0 186.7 216-l 248.5 283.9 322.6 3125 351.5 393.7 439-o 487.8 5400 686.6 857.5 10547 1280-O z:: 2143.4 2500.0 364.6 410-l 459-3 5122 569-l 630.0 801.0 fzz 1493'3 1791.2 2126-3 K:: 220 DEMON AIDS FOR REINFORCED CONCRETE TABLE 87 MOMENT OF INERTIA OF CRACKED SECTION VALUES OF &/(R") . 0-O O-3 O-100 11::;1;: O-226 O-264 0.8 O-100 Ki 0229 0269 1.6 O-298 :::z 0398 O-430 iz . x':;; O-601 O-628 @653 O-678 O-703 O-472 FiTi O-596 O-625 O-654 O-682 O-710 0,738 0765 O-792 O-818 O-844 O-870 0.839 O-860 O-881 :zi 1.121 l-179 . :-zz l-351 l-123 l-156 l-188 1.220 l-250 ;:g l-575 l-630 l-685 l-739 ::z 1902 O-490 O-525 0.559 zi . 8':;; O-723 O-755 O-787 O-818 0.850 0.880 O-911 0942 0.310 O-348 O-386 O-424 0460 0.4% 0332 O-567 o":zz O-670 O-704 O-738 0.771 0.804 O-837 x:ii; 0.934 0.966 0998 l-030 l-061 1.093 l-124 1.155 l-217 I.278 l-340 1400 1461 l-521 l-581 ::z l-758 l-817 l-876 1.934 1993 DEFLECTION CALCULATION 221 TABLE 88 MOMENT OF INERTIA OF CRACKED SECTION VALUES OF h,(z) 0.0 @l E O-185 O-224 O-262 O-3 O-100 O-144 O-185 O-225 O-263 O-300 0.335 0.370 0403 O-436 0468 O-499 O-530 O-560 O-589 O-618 ::zE O-701 O-727 04 &lOO 0.144 0.185 0225 O-263 O-300 @336 O-371 0405 O-438 0.470 O-502 8:::: O-593 0.622 o-651 0680 O-708 0735 8:::; O-815 O-841 0867 O-6 0.100 O-144 O-186 O-226 0264 0.302 0338 O-373 zz O-8 O-100 O-144 Q186 ::z O-303 O-340 O-376 O-411 O-445 O-479 O-512 8:::: 0.609 1.0 E! O-366 O-398 O-430 0.298 O-333 O-367 Et! O-463 O-493 0523 O-551 O-580 O-607 O-634 ;:z o-71 1 F$$ 0.413 O-448 0.483 o-517 0.550 0.583 O-616 O-648 ::tE! 0.743 O-774 O-632 z: O-720 0.749 O-792 z:: . Ez I:E 8E l-022 :z: 0.993 l-021 lft49 l-077 l-131 l-185 l-239 1.292 0.727 X:E O-795 O-818 O-839 O-860 0881 Ez o-942 0.980 1.018 1:E l-123 l-156 I*188 f:E 12$ 1.337 :::9": O-854 O-876 0.898 :E 0962 EZ l-082 1*120 1.157 1.193 ::E! l-296 l-329 :::: 1.425 l-455 0999 l-045 :d~ l-176 1.218 I.260 1301 ::z: 0.805 O-836 O-866 f:E 0956 O-986 l-015 fZ 1.103 1.160 l-217 1.274 I.330 1.386 1441 f I.606 l-247 1.291 :::z I.419 1461 ::E l-583 l-623 l-298 1.347 ::z l-489 l-535 1582 . t-E l-718 l-344 1.3% ::g I.551 lTjO1 1.652 :E 1*801 :Zf 222 DESIGN AIDS FOR RRINFORCED CONCRETE TABLE 89 MOMENT OF INERTIA OF CRACKED SECTION VALUES OF It/ t s ) r o-0 O-100 0.143 O-185 x:z O-298 EZ O-398 0.430 x:z 0519 0547 O-575 0.601 O-628 O-653 O-678 o-703 O-727 0750 0773 0.795 0.818 0.839 O-860 O-881 0902 O-922 O-942 O-980 l-018 l-054 1 a89 1.123 l-156 I.188 1.220 l-250 1.280 1.308 I.337 l-364 1.391 0.1 0.100 O-143 0.185 O-224 0.262 O-298 O-333 0.367 O-399 0.43 1 O-462 X:iE 0.550 O-578 O-605 O-632 O-658 O-683 o-708 O-733 0.757 0.781 O-804 O-827 0.849 0871 0.893 0914 O-935 0956 0997 l-036 1.075 J-112 1.148 1.184 l-219 I *252 J-286 1.318 J-350 l-381 I-411 1441 O-2 O-100 O-143 O-185 O-224 O-262 O-298 O-333 O-367 z! 0.463 @494 O-523 O-552 O-581 O-608 O-635 0.662 O-688 0.714 O-739 O-764 O-788 O-812 @836 O-859 O-882 ;z 0.970 I-012 I .054 I.094 I.134 I.172 l-210 1.247 I.283 I.319 1.354 1.388 I.422 1455 1488 0.3 @JO0 O-143 0.185 O-224 O-262 O-298 0334 O-368 0401 O-433 O-465 O-495 O-525 O-555 O-583 0.611 0639 8:E O-719 0,745 @770 O-795 0820 O-844 O-868 0.891 O-915 O-938 O-960 0.983 1.027 l-070 1.112 J-154 J-194 1234 I.274 1.312 1.350 1.387 J-424 J-461 1.4% 1.532 O-4 o-100 0.143 0.185 O-224 0262 0.299 O-334 O-368 O-402 0,434 zz! 0.527 O-557 O-586 O-614 O-643 0.670 0.697 0724 O-750 0.776 O-802 0827 O-852 O-876 0901 0.925 0,949 0.972 E 1.086 l-130 1.173 I.216 I.257 I.299 l-339 1.379 1.419 : ::z: l-535 1.573 o-6 O-100 @I43 0.185 OQ24 0.262 x:E O-369 0403 0.436 0468 Ei O-561 O-591 0.620 O-649 0.678 O-706 O-733 0,761 O-788 0814 O-841 0.867 O-893 O-918 0943 O-969 0.993 :::t l-115 l-162 1 a208 O-8 O-100 0.143 xi!: O-263 0.300 FL::: O-405 O-438 0.471 o-503 O-534 O-565 0.596 O-626 O-655 O-685 O-713 O-742 O-770 0.798 O-826 O-853 O-880 X:K 0.960 0.987 l-013 J-039 l-090 I.141 l-191 l-240 J-0 O*lOO 0.143 . 8`::: O-263 0473 O-505 O-537 O-569 O-600 O-631 0.661 O-691 O-721 O-750 O-779 P808 O-836 ::8896: 0.92 I O-948 O-976 l-003 1 a30 l-057 1~111 1.164 1.216 l-268 I.320 1.371 I -422 I.473 I.523 I.476 zzi l-605 1 a7 I:627 l-574 1E I.712 1.573 `I:x:3 I.721 1 a770 DEFLMXION CALCULATION 223 TABLE 90 MOMENT OF INERTIA OF CRACKED SECTION VALUES OF Jr/ $ ( ) d'/d-O-20 Am--l)/(m) 0-l O-2 0.3 O-4 O-6 O-8 19' ;; O-399 O-430 O-298 O-333 !E! O-298 O-333 EZ O-431 O-462 O-492 :::: O-579 a298 O-333 @367 X% 8% 0.523 0.552 @580 0o:g O-663 O-689 O-715 O-741 0.766 O-791 O-815 O-839 0.863 O-887 0910 z:: 0.874 O-898 O-922 ii% O-993 1so39 l-085 l-129 l-173 l-217 l-259 1:E l-384 1.425 ::zz l-545 l-584 yg O-554 O-583 0.298 O-333 O-367 . %i Ei O-368 O-401 O-434 :%! 0528 O-558 @588 O-603 zz O-681 O-706 O-730 0754 Ei Of23 O-605 O-632 O-658 Ei O-734 O-758 0.782 EE O-607 0.634 iT% 0.712 @614 O-643 O-671 O-699 O-726 0.753 O-779 O-806 O-832 O-857 0.883 Ei @958 O-983 l-007 1955 1.103 :z l-241 l-287 l-331 . :*:z 1463 l-506 :::;; l-633 @617 ic O-731 . x'::: O-812 0839 O-865 Ei 1931 :%i 1141 l-176 . :-iii l-276 1.308 l-339 . :%i l-429 o-961 :izz l-082 l-120 0969 l-012 :z!z l-135 l-174 l-212 . :% l-324 O-978 l-022 la65 . xi 1.189 :zz 1.307 1.345 l-383 1,420 :1E 1.528 ST% 1.119 l-168 1:216 Et 1358 1a5 1.451 1.497 l-542 l-587 l-632 1.677 l-334 1368 ::zi 1464 224 DESIGN AIDS FOR REINFORCED CONCRETE TABLE 91 DEPTH OF NEUTRAL AXES BY ELASTIC THEORY VALUES OF x/d d'/d=@OS O-132 O-159 O-181 EY O-131 O-158 O-180 O-198 O-215 o-131 O-157 0.178 0.197 O-213 0227 E!i O-263 O-274 0225 O-238 O-249 z: EJ 0.305 O-312 @325 E O-345 @351 @358 O-363 0.369 0374 O-380 0.319 O-326 O-333 O-339 O-345 0.351 EZ O-367 O-372 0.130 O-155 O-176 O-194 O-209 O-223 0235 i!Lg 0.276 O-284 O-292 Ez O-314 0:321 O-327 O-333 O-339 O-128 O-153 O-173 @190 0205 0.218 tz! O-251 O-260 O-268 0.276 ;ij 0.304 O-310 O-316 O-321 C+326 :::t 0.341 0.345 O-350 O-214 x:z O-245 O-253 0.261 0.269 0.276 0.282 0.289 O-294 O-300 O-305 ::::: ::E 0328 0332 0.336 0.287 @297 Ei O-323 O-336 8E O-358 O-365 0.330 O-338 0345 0.352 O-358 MO2 Oar7 O-413 O-418 0,423 O-393 O-398 O-403 z! O-418 O-427 E O-451 O-385 O-390 O-394 O-399 O-404 :iti 0.425 O-432 O-439 xO-459 0465 0.471 O-399 :zi O-421 0428 EZ O-378 O-382 @386 O-383 @388 0392 O-396 0400 ::g O-411 0.414 O-417 O-507 O-513 0519 O-525 O-531 O-491 0497 0503 8E Ez 0488 O-493 0.498 DEFLECTlON CALCULAflON 225 TABLE 92 DEPTH OF NEUTRAL AXIS BY ELASTIC THEORY VALUES OF x/d 01 0.132 @158 @MO 0.199 0.215 0230 odz 0.268 0.278 0.2 0.131 0.158 0.179 0.198 0.214 0228 0242 0254 0.265 0275 0285 ii . 0.319 0.4 0.131 0.157 0.178 0.196 0211 0.225 0.238 0249 0.260 0.270 0.279 0288 812Z 0.311 0.318 0.325 0.331 0.337 0.343 0.349 0.354 0.359 0.364 0.369 0.6 0.130 0.156 ml76 0.194 0209 XE 0245 0.255 0.265 0.273 0.282 0289 0296 0303 0.310 0316 0322 "0:::: 0.338 0.343 0.348 0.352 0.357 0.361 0365 8% 0377 0.8 0.130 0.155 P175 0.192 0.206 0.219 0.231 0241 0.251 0.260 1.6 0.130 0.154 0.174 0.190 0.204 0263 0.270 x:18': 0289 X:E 0.313 @318 0.323 8% ::::: 0.345 0.349 0.353 0.357 0.360 0.363 0.367 0.373 8% 0.389 0354 0.360 O-365 0.370 0.375 0.295 0.300 0.305 0310 0.314 x:::: "0:::: 0.335 0331 0.339 %: 0.359 0.327 0.334 0.341 0347 0.354 8% 0.371 0.377 0.382 0.394 x:z 0.410 0.414 0.419 0.428 0.437 0445 0.453 fii$ 0481 0487 0.493 Ei 8% 0.387 0.392 0.397 ;:z 0.411 00::;; 0.435 0442 0449 0456 E; 0.475 0.395 0403 X% 0.423 0.380 O-387 @394 0400 0406 0.411 0.416 0.421 0.426 0.431 226 DESIGN AIDS FOR REINFORCBD CONCRBTB TABLE 93 DEPTH OF NEUTRAL AXIS BY ELASTIC THEORY VALUES OF x/d ' o-0 0.1 0.132 O-159 O-181 0.199 O-216 O-2 O-132 O-159 . 8':: O-215 0.3 0.132 O-159 O-180 O-198 O-214 0229 O-242 0.254 O-265 O-275 o-4 0.132 @159 @180 O-198 O-214 O-228 OatI 0.252 0.263 O-273 0.282 O-291 O-299 o-307 o-315 0322 O-329 O-335 O-341 o-347 0.353 O-358 0.364 O-369 0.374 0.378 0.383 O-387 0.392 O-396 OaO 8::; 0422 0.429 0.435 O-441 O-447 ::t:: O-463 0468 ::t:`7 0.481 0.6 o-133 0.158 :::;Y 0.212 0.8 0.133 0.158 O-179 O-196 O-211 0224 0.236 0.247 0.257 0266 1-d 0.133 O-lS8 0.178 O-195 O-209 O-222 0.234 0.244 O-254 0.262 0.270 O-278 O-285 8:ZE O-304 8Z 0.320 O-324 0.329 O-333 O-338 O-342 O-345 0.349 O-353 0.356 "o::zi O-366 O-372 0.377 O-382 O-387 ZE O-401 0405 O-408 0.412 O-415 0.419 O-422 0.425 z-z 0:258 O-270 O-281 O-292 0301 O-311 ii::: O-336 0.344 o-351 O-358 0.365 0372 O-378 O-384 O-390 O-396 0402 O-407 O-413 O-418 O-423 O-428 O-437 Et 0.463 O-231 0.244 0.257 O-268 0.279 O-289 O-299 @308 0.316 O-324 0.332 O-340 O-347 0.354 O-360 0.367 O-373 O-379 O-385 O-390 0.396 O-401 0406 0.411 O-416 O-420 0.429 0.438 iEi 0462 O-469 O-476 O-482 O-489 0495 O-501 O-506 0.512 O-517 O-287 O-296 0.305 O-313 O-321 O-329 O-336 0.343 O-349 O-356 0.362 O-368 O-374 0.379 0.384 0.390 O-395 :`iz O-409 0.413 O-422 O-430 0.438 O-445 O-452 0.459 O-466 8:::; @484 0.489 0.494 x:z 0.285 0294 @302 O-310 0.318 O-325 0.332 fF339 O-345 0.351 8:::: O-369 0.374 O-379 0384 0.389 O-393 0.398 0402 O-407 O-415 O-422 O-430 O-437 O-444 O-450 0.456 O-462 O-468 0.473 0.478 X:E 0493 0.278 0.287 0.294 x:z O-315 O-322 O-328 x:::: O-344 0.349 x::::: 0.364 O-368 O-372 0.376 O-380 0.384 O-388 O-395 O-401 0.408 O-414 0419 0.425 0.430 O-435 O-440 ::I$ 0.453 0.457 O-460 . O-274 O-282 O-290 @297 0.303 0309 O-315 O-321 O-326 O-332 0.336 O-341 O-346 0.350 O-354 O-358 0.362 O-366 O-369 O-373 O-376 O-383 "0::;; O-un 0405 O-410 0.415 O-419 O-423 0.427 0431 O-435 0438 O-442 DEFLECTION CALCULATION 227 TABLE 94 DEPTH OF NEUTRAL AXIS BY ELASTIC THEORY VALUES OF x/d d'jd=0.20 PC@+-lMptm) ptm 1.0 1.5 ;:; 3-o 3.5 9:; 5.0 5.5 :I': 3:: 8-O . 0.0 0.1 O-132 O-159 O-181 0.200 O-217 0.2 0.133 0.160 O-181 0.200 0.216 O-231 x:z O-268 O-279 x:4:: O-307 0.315 0.323 O-331 0.338 O-345 O-352 0.358 o-364 O-370 0.376 O-382 0387 O-392 0.397 0402 0407 0.411 0.416 O-425 0.433 O-441 O-448 O-455 0462 0.469 O-475 0.481 0.3 0.133 O-160 O-182 0.200 O-216 0.231 0.244 0.256 0.267 O-277 0.287 0*29., 0305 0.313 O-321 O-328 0.335 O-342 O-348 O-355 0.4 0134 O-160 O-182 8:% O-230 O-243 @255 0.266 0.276 0.286 0.295 ii:::: 0319 0.326 0.333 0339 0.345 0.351 0.357 0.363 O-368 :%I 0.6 O-135 0.161 0.182 8% 0230 0.242 O-254 O-264 0.274 ,0.283 0.291 0.300 0.307 0.314 0.321 O-328 O-334 0.340 0.345 0.351 0356 0361 0.366 O-370 0375 0.379 k% 0391 O-395 0402 %z O-422 O-428 a433 0.439 :zi 0.453 @458 @462 0.466 O-470 0.8 0.135 O-161 0.182 ::z 0229 O-241 0.252 0.262 O-272 0280 8:;;: 0303 0.310 0317 0.323 0329 0.334 0.340 1.6 0136 0162 O-183 0.200 0215 0228 0.240 0.251 O-261 0.270 O-278 0.286 0.293 ~:~ 0.313 0.318 0.324 0.329 O-334 0.339 O-343 0.348 0352 0.356 0.232 0246 0.258 O-270 0.281 0.292 0.301 @311 0.319 O-328 0.336 O-344 O-351 O-358 0365 O-231 8:;;; 0.269 O-280 0290 O-300 O-309 x:::: O-333 0.341 O-348 0.355 0362 O-368 O-374 0.380 0.386 0.391 11-o 11'5 :I:; 13.0 13.5 140 14-5 15.0 15-5 16-o 17.0 18-o 19.0 20.0 21-O 3:*8 24-o 25-O 8% O-413 O-418 0.423 o-428 0.437 0446 t%: 0471 0.479 O-486 @493 O-500 0.397 O-402 0407 O-412 O-417 8% 0.439 O-448 O-456 0388 0.392 0.397 O-402 O-406 O-410 O-419 0.427 O-434 0441 0448 O-454 0461 :z: @478 0.483 O-488 tz;ii O-383 8% 0397 0401 0405 0.413 f-b.421 0.428 0434 0441 0447 O-453 0.459 O-464 0469 X% O-484 0488 8% 0.375 0.379 O-382 O-386 0.393 O-399 0405 0410 O-416 0.421 O-426 x:1:`: 0.439 O-443 O-447 O-450 0.454 0.360 0364 ::z: O-374 0377 0384 0.389 0.395 0400 O-405 8:Z O-418 O-422 0426 O-429 @433 x:1:9" 228 DESlGN AIDS FOR REINFORCEDCONCRET'E TABLE 95 AREAS OF GIVEN NUMBERS OF BARS IN Cm) BAR DIANKIER. mm OPBAW NUMBER -6 10 ::i; 235 :z: 12 2.26 1.13 339 452 5.65 6.78 7.91 9.04 10.17 11.31 14 3.07 I -54 461 769 6.15 923 10.77 1231 13.85 15.39 1693 18.47 20.01 21.55 23.09 24.63 -. __ 2617 27.70 2924 30.78 16 4.02 2o1 6.03 10.05 8.04 12% 14.07 l&O8 18.09 20.10 211 24.12 26.13 28.14 30.15 3217 3i*iS 36.19 38.20 40.21 18 z 763 10.17 12.72 15.26 17.81 20.35 2290 2544 2799 30.53 33.08 35.62 38.17 40.71 43.26 45.80 48.34 50.89 20 3.14 i% 1256 15.70 l&85 2199 25.13 iK 34.55 37.69 z% 47.12 50% 53.40 56.54 5969 6283 22 3.80 . 1% 15.20 1900 22.80 zz 34.21 38.01 41.81 45.61 4941 :;g . 25 491 l:% 2454 1963 29.45 ii::; 44.17 49.08 5399 5890 63.81 g*;; 28 6.16 1231 18.47 2463 30.78 36.94 43.10 49.26 55.41 61.57 67.73 73.89 80.04 86.20 92.36 98.52 104.67 1 IO.83 116.99 123.15 32 8 01 1608 24.12 32.17 40.21 48.25 56.29 6434 7238 80.42 88.46 96.5 I 101.55 : g:;; 128.68 136.72 144.76 152.80 160.85 36 , IO.18 2035 30.53 40.71 50.89 61.07 71.25 81.43 91 a 101.78 111.96 122.14 132.32 142.50 152.68 162.86 f ;;:;; 193.39 203.57 8.63 9.42 10.21 1099 11.78 1244 13.57 1470 15.83 16.96 18.09 i9*i2 20.35 21.48 22.62 1:::: 5.65 10.05 1413 1492 15.70 z% 68.42 ;;g ix 88.35 ;;g DEFLECTION CALCULATlON 229 TABLE 96 AREAS OF BARS AT GIVEN SPACINGS Values in cm2 per Meter Width BAR DIAMEIER. mm 1 6 5.65 471 % 3.14 10 t:. :: 15 16 17 18 19 2.83 257 z7" 2.02 1.88 l-77 l-66 1.57 1.49 1.41 1.35 l-28 1.23 1.18 l-13 1.09 l-05 1 *Ol 0.97 30 32 34 36 :: O-94 0.88 0.83 0.78 0.74 071 8 10.05 8.38 7-18 6.28 5.58 5.03 4.57 4.19 ZJ 3.35 3.14 2.96 2.79 2.65 2.51 239 2-28 218 2-09 2.01 1.93 1.86 1.79 1.73 1.68 1.57 1.48 :z 1.26 10 15.71 13.09 11-22 9.82 8.73 7.85 7.14 6.54 4:: 5.24 4.91 4-62 4.36 4.13 3.93 3-74 3.57 3'41 3.27 3.14 ;:Ff 2.80 271 2.62 245 2.31 2.18 2.07 1.96 12 22.62 18.85 16.16 14-14 12-57 11.31 10'28 9.42 a.70 S-08 % f:E 5.95 5.65 5.39 5-14 4-92 4'71 4.52 4.35 419 4.04 3-90 3.77 3-53 3.33 3.14 2.98 2.83 14 30-79 25.66 2199 19-24 17.10 16 40-21 33.51 f 22.34 20-11 18.28 1675 15.47 1436 1340 1257 11-83 11.17 IO-58 10.05 9-57 9.14 8.74 8.38 8.04 7-73 7.45 7.18 6-93 6-70 6-28 5.91 5.58 5.29 5.03 8.48 7.95 7.48 7.07 6.70 6.36 18 50.89 4241 36.35 31.81 28.27 25.45 23.13 21.21 19-57 18.18 1696 15?w 14-97 1444 13.39 12.72 12.12 11.57 11.06 10-60 20 6283 52.36 44.88 39.27 3491 31.42 28.56 26.18 24.17 22.44 20.94 19.63 18.48 17.45 16.53 15.71 14.96 4.28 13'66 13.09 12.57 1208 11.64 1 l-22 10.83 10-47 9-82 5.24 8.73 8.27 7.85 22 76.03 63.36 54.30 47.52 4224 38.01 34.56 31.68 29.24 27.15 25.34 23.76 22.36 21.12 20-01 19.01 18.10 17.28 16-53 15.84 15.20 14.62 1408 13.58 13.11 12.67 11.88 11-18 10-56 `!?z . 25 98.17 81.81 70.12 61.36 54-54 49-09 44.62 40-91 37.76 35.06 :z 28.87 27.27 25.84 24.54 23.37 22.31 21.34 20.54 19.63 18.88 18.18 17.53 16-93 16.36 15.34 14.44 13.63 12.92 12.27 28 123'15 102%8 87.96 76-9 68'42 61.57 55.98 51-31 47-37 43.98 41.05 38.48 36.22 2421 32-41 30-79 29.32 2799 26.77 25.66 24.63 f 21.99 21.23 20.52 19.24 18.11 17.10 16.20 15.39 32 16085 :4qs 10053 89-36 80-42 73.11 6702 61.86 57.45 53.62 50-27 47.3 1 44.68 42.33 40.21 3830 36-56 34.97 33.51 g:;; 29.79 28.76 27.73 26.81 25.13 23.65 22.34 21.16 20.11 cm . E * ::-sz 1283 11.84 11-00 10.26 9-62 9.05 8-55 8.10 7.70 7.33 ::z! 6.41 6.16 5.92 5-70 5.50 5-31 5.13 4.81 4.53 4,28 4.05 3.85 E 230 DESICSN AIDS FOR REINFORCED CONCRETE Table 97 FIXED END MOMENTS FOR PRISMATIC BEAMS LOAD TYPE M rr Mm Pab' (' Pdb 1' I- PI % w, 12 1' [' 12 ad+ s2 (I-3b)l -7 121' 2 I (31~e4+3s') `,f+-3S) w I' +12 w I2 -12 W 5Wl' A -96 WI' +20 w I2 -30 t-----`----l IXSf'LECT'ION CALCULATION 231 Table 98 DEFLECTION FORMULAE FOR PRISMATIC BEAMS &it I+`/2 I P 48EI Pl' P 1' 192 L&! L ( 23PI' 6= 5 PI' 646 EI w 1 , , , P ,' , rrrrcrrl --I 6E1 H!t P 1' 3 L,MI 16 2 EI Note:- W is total distributed load DESIGN AIDS FOR REINEORCED CONCRETE