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Code Comparison between

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Presentation on theme: "Code Comparison between"— Presentation transcript:

1 Code Comparison between
IS and ACI 318 Presented by K.Gokulprabhu & N.Kumaresan

2 General basic comparison
DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Loading combination 36.4.1 a) 1.5 DL +1.5 LL b) 1.5 DL l+1.5 WL c) 1.2 DL+1.2 LL+1.2wl 9.2.1 U= 1.4(D+F) U= 1.2(D+F+T) (L+H+0.5(Lr or S or R) U= 1.2D+1.6(Lr or S or R)+(L or 0.8W) U=1.2D+1.6W+L+0.5(Lr or S or R) U= 1.2D+E+L+0.2S U= 0.9D+1.6W+1.6H U= 0.9D+E+1.6H Elastic modulus for concrete and steel Ec=5000 (fck) N/mm2 Es=200xE3 N/mm2 8.5.1 Ec =57000f’c psi Es =29x10­6 psi Mu limit Mu limit = Fck b d2 Mulimit=

3 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Span to depth ratio in slab 23.2.1 Cantilever Simply supported - 20 Continuous One-way slab- 20 Continuous slab-24 Min reinforcement in slab Plain bar-0.15% High strength bar-0.12% 200/fy Nominal cover Columns Not less than 40mm When dimension is 200mm or size of bar is 12mm then cover is 25mm Footings Not less than 50mm Slabs Not less than 15mm Beams not less than 20mm Refer P-47 Table –16A of IS code to get the cover details for fire resistance   Footing Resting on mat – 2” Resting on earth – 3” Top of piles – 2” Beams and columns For dry conditions Stirrups, spirals, ties – 1.5” Main reinforcement – 2” Exposed to earth Stirrups, spirals, ties – 2” Main reinforcement – 2.5” No. 11bars – 0.75” No. 14 – 18 bars – 1.5”

4 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents T beams 23.1.1 For T beams For L beams 8.10.3 Width of slab effective as T beam flange shall not exceed one quarter of the span length of the beam Effective overhang of flange on either of web shall not exceed - 8 * slab thickness - ½ clear dist. Between the web For slab on one side - 1/12 of span of beam - 6 * slab thickness Minimum area of tension reinforcement a Ast(min) = 0.85 b d / Fy For flexural members Ast(min) = 3fc’ /fy * bw*d And not less than 200*bw*d / fy For slabs and foundation Astmin = 0.14% Grade steel 0.2% Grade 60 steel 0.18%

5 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Minimum horizontal distance between the bars 26.3.2 Greater of this following 1)larger diameter of the bars 2)Size of the aggregate + 5mm Slabs Main bars – 3d or 300mm Distribution bars – 5d or 450mm not greater than the above Beams Refer to the IS code P-46 Table-15 it depends on the percentage redistribution Flexural member And not greater than Side face reinforcement When depth >750mm Total area of reinforcement = 0.1% area of the web and spacing not exceeding 300mm 10.6.7 When d >36” Total longitudinal reinforcement not greater than half of the tensile reinforcement Spacing Ssk not greater than d/6, 12” na and 1000Ab / (d-30)

6 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Maximum spacing of the tension bars 26.3.3 1) 0% re distribution of the moment 180mm 2) For other percentage of redistribution refer table 15 Compression steel / Tension steel 1) M > Mu li mit compression steel is required Asc = M - Mu li mit / (352 – fck) * (d - d’) Ast = (Pt(limit) + Pt2) * b * d / 100 2) if M < Mu li mit compression steel is not required Ast = Pt * b* d / 100 Span / Effective depth check (a) Basic span / Effective depth = 26 Effective span / MF * d  26 MF taken from fig 4&5 using Pt and fs Fs = 0.58 *fy * (Astreq / Astpro) Or MF (tension) = 1 / ( * fs – 0.625)  2 MF (compression) = 1.6 * Pc / Pc  1.5

7 c = taken from table 19 using Pt and fck values
DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Shear reinforcement Asv = 0.46 Sv 0.87 fy IfV > c Asv = (v - Cv) b d 0.87fy d c-refer table 19  Cmax 2 V = vU / b*d c = taken from table 19 using Pt and fck values V < c shear reinforcement is not required (Asv / Sv)min = (0.4 * b / 0.87 * fy) c< v < c max / 2 (Asv / Sv) = Vu -c*b*d / 0.87*fy*d c can be derived from this formula c = 0.76 (fck*((1+5)^0.5 – 1))^0.5 / 6  = 0.8*fck / 6.89*Pt  Vn Vu where Vu – factored shear Vn – nominal shear Vn = Vc + VS Vc = 2*fc’ bw*d For members shear and flexure and for detailed analysis use Vc = (1.9*fc’ w * Vu *d / Mu ) * bw*d And not greater than 3.5*fc’ bw*d Quantity Vu *d / Mu not greater than 1 For members subj. to axial compression Vc = 2* (1+Nu / 2000*Ag) * fc’* bw*d For detailed analysis Mm = Mu – Nu (4*h – d) / 8 Vc shall not be greater than Vc = 3.5*fc’* bw*d * (1 + Nu / 500*Ag)

8 For members subjected to axial tension
DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Shear reinforcement For members subjected to axial tension Vc = 2*(1 + Nu / 500*Ag) *fc’* bw*d but not less than zero Where Nu is negative for tension Nominal shear strength provided by concrete when diagonal cracking results results from the combined shear and moment Shear strength Vci = 0.6*fc’* bw*d + Vd + Vi * Mcr / Mmax and not less than 1.7 *fc’* bw*d Mcr = I / t * (6*fc’ + fpe – fd)

9 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Spacing of shear reinforcement Not greater than d/2 When V­s exceeds 4*fc’* bw*d the above spacing is reduced to 1/2 Minimum shear reinforcement If Vu > ½ * Vc Av = 0.75*fc’* bw*s / fy Not less than 500* bw*s / fy Spacing of the shear reinforcement Vu >  * Vc Vs = Vu -  * Vc /  For straight bars Vs = Av*fy*d / s Inclined stirrups Vs = Av*fy (sin + cos)*d / s Group of bars all bent up Vs = Av*fy*sin not greater than 3* fc’* bw*d Spacing are calculation for straight bars Av / s = Vu - *Vc / *fy­*d

10 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Torsion 41.4.2 The longitudinal reinforcement shall be designed to resist an equivalent BM Me1 = Mu + Mt Mt = Tu (1+D/b / 1.7) If Mt > Mu Then longitudinal reinforcement shall be provided on flexural compression member to withstand the moment Me2 11.6.1   11.6.3    Torsion can be neglected when For non prestressed members Tu < f*fc’*(A2 cp / Pcp) For non prestressed members subjected to axial tensile or compressive force If it exceeds the condition as per Calculate the factored torsional moment For solid sections For hollow sections

11 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Transverse reinforcement 41.4.3 Total transverse reinforcement not less than  Longitudinal reinforcement Additional longitudinal reinforcement shall be not less than Minimum torsion reinforcement 11.6.5 When Tu > values of Not less than Minimum total area of longitudinal torsional reinforcement At least one longitudinal bar in each tie. When c/s exceeds 450mm additional reinforcement to be provided At/ s shall not be less than

12 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Spacing of the torsion reinforcement Rectangular closed stirrups. Spacing shall not exceed x1, x1+y1 / 4 , 300mm 11.6.6 Spacing shall not exceed  12” or PA/8 Column reinforcement  39.4.1 0.8 to 6% of gross c/s area 4 – rectangular bars 6- circular bars Minimum dia. 12mm Spacing of bars not greater than 300mm SPIRAL REINFORCEMENT Ratio of volume of helical reinforcement to volume of core shall not be less than 10.9 Minimum longitudinal reinforcement 0.01% to 0.08% gross cross sectional area Minimum bars 3 – triangular bars 6- enclosed spirals Minimum diameter 3/8” no.3 bar LATERAL TIES Diameter not less than 0.02 times greater dimension of the member No.3 to No. 5 bar Vertical spacing of bar Not greater than -16 times dia. Longitudinal bar - 48 times dia. Of tie bar

13 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Column reinforcement d Maximum pitch not greater than 75mm LATERAL TIES Pitch of the transverse reinforcement shall not be more than the least of the following - least lateral dimension - 16* dia. Longitudinal bar - 300mm dia. Of the lateral ties not less than ¼ of the largest longitudinal bar Minimum diameter 3/8” no.3 bar Diameter not less than 0.02 times greater dimension of the member No.3 to No. 5 bar Vertical spacing of bar Not greater than -16 times dia. Longitudinal bar - 48 times dia. Of tie bar - 0.5 times least lateral dimension Foundations Footing thickness 34.1.2 Thickness at the Edge Footing on soil not less than 150mm Piles – 300mm 15.7 Depth of the footing above the bottom reiforcement should not be less than - Footing on soil 6” - footing on piles 12” 

14 DESCRIPTION IS 456 – 2000 ACI CLAUSE Contents Footing - Critical section for shear D / 2 distance alround the face of the column 15.5.4 dp / 2, alround the face of the column Minimum reinforcement in footings 34.5.2 Nominal reinforcement for concrete thickness greater than 1m shall be 360mm2.this provision does not supercede the requirement of the min. tensile reinforcement based on depth calculation  Longitudinal reinforcement atleast 0.5% of the c/s area of the supported column Astmin = bh Grade steel 0.2% Grade 60 steel 0.18% Development length of bars is taken from the table 12.2.1 For deformed bars Not greater than 2.5 Ld not less than 12”

15 Slender Columns with biaxial bending
Isolated footing with Moment Click


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