1.aci 318 code comparison with is456-2000(1)

DESCRIPTION IS 456 – 2000 ACI 318 - 2002

CLAUSE Contents 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) +1.6(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

6.2.3.1 Ec=5000 (fck) N/mm2

Es=200xE3 N/mm2

8.5.1 Ec =57000f’c psi

Es =29x10 6 psi

Mu limit Mu limit = 0.138 Fck b d2 Mulimit=

General basic comparison

'2 59.01

c

yy f

fbdf



Span to depth ratio in slab

23.2.1 Cantilever - 7

Simply supported - 20

Continuous - 26

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”

Slabs

No. 11bars – 0.75”

No. 14 – 18 bars – 1.5”



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

- ½ clear dist. Between the web

Minimum area of tension reinforcement

26.5.1.1.a

Ast(min) = 0.85 b d / Fy 3.12.5.3 For flexural members

Ast(min) = 3fc’ /fy * bw*d

And not less than

200*bw*d / fy

For slabs and foundation

Astmin = 0.14%

Grade 40-50 steel 0.2%

Grade 60 steel 0.18%

fwo

f Dbl

b 66

fwo

f Dbl

b 312



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

3.12.11.1

Flexural member

And not greater than

Side face reinforcement

26.5.1.3 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)

cs

Cf

s 5.2540

sf

3612


CLAUSE Contents 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 – 0.446 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

23.2.1 (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 / (0.225+0.00322 * fs –

0.625) 2

MF (compression) = 1.6 * Pc / Pc + 0.275 1.5

DESCRIPTION IS 456 – 2000 ACI 318 - 2002 CLAUSE Contents CLAUSE Contents

Shear reinforcement

Asv = 0.46 Sv

0.87 fy

IfV > 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’ + 2500w * 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)

DESCRIPTION IS 456 – 2000 ACI 318 - 2002 CLAUSE Contents 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)



Spacing of shear reinforcement

11.5.4.1

11.5.4.3

Not greater than d/2

When V s exceeds 4*fc’* bw*d

the above spacing is reduced to 1/2

Minimum shear reinforcement

11.5.5.1 If Vu > ½ * Vc

Av = 0.75*fc’* bw*s / fy

Not less than 500* bw*s / fy

Spacing of the shear reinforcement

11.5.6.2 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



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 11.6.1

Calculate the factored torsional moment

For solid sections

For hollow sections

'41

A'

2

cG

u

cp

cp

c fAN

Pf

'8

7.1

2

2

2

cw

c

oh

hu

w

u fdb

V

A

PT

db

V

c

db

V

A

PT

db

V

w

c

oh

hu

w

u 87.1

2

2

2



Transverse reinforcement

41.4.3 Total transverse reinforcement not less than

11.6.3.6

Longitudinal reinforcement

11.6.3.7 Additional longitudinal reinforcement shall be not less than

Minimum torsion reinforcement

11.6.5 When Tu > values of 11.6.1

Not less than

Minimum total area of longitudinal torsional reinforcement

26.5.1.7 At least one longitudinal bar in each tie. When c/s exceeds 450mm additional reinforcement to be provided

11.6.5.2

At/ s shall not be less than

y

vu

y

vusv fd

SV

fdb

STA

87.05.287.0 111

s

fAAT yvton

cot2

y

vcve

f

bS

87.0

yl

yvhtl sf

fPAA

2cot

yv

wctv f

sbfAA '75.02

yv

w

f

sb50

yl

yvht

yl

cpcl sf

fPA

f

AfA

'5min,

yv

w

f

b25



Spacing of the torsion reinforcement

26.5.1.3 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

26.5.3.1

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

4 – rectangular bars

3 – triangular bars

6- enclosed spirals

SPIRAL REINFORCEMENT

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

y

ck

k

g

f

f

A

A

136.0

y

c

c

g

f

f

A

A '145.0




26.5.3.2 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

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

- 0.5 times least lateral dimension


10.3.6.1

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”

scyccku AfAfP 67.04.0 stystgcn AfAAfP '85.085.0* max,


CLAUSE Contents CLAUSE

Contents

Footing - Critical section for shear

34.2.4.2 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 = 0.0018bh

Grade 40-50 steel 0.2%

Grade 60 steel 0.18%

Development length of bars

is taken from the table 26.2.1.1

12.2.1 For deformed bars

Not greater than 2.5

Ld not less than 12”

bd

sdL

4

bd

btr

c

yd d

bdkc

f

fL *

'*40

**3

bd

kc tr

Slender Columns with biaxial bending

Isolated footing with Moment Click

1.aci 318 code comparison with is456-2000(1)

Documents

asv sv

mu

longitudinal reinforcement

fy

clear dist

shear reinforcement 11

longitudinal bar

48 times dia