CHECK FOR CRACK WIDTH AS PER IS 456-2000

BEAM SECTION :-

WIDTH (b) = 250 mmDEPTH (D)= 350 mm

d-xEFFECTIVE DEPTH (d) = 312 mm d DCLEAR COVER = 30 mm

716 mm²

TOTAL COVER = 38 mm

M 25 Strain Diagram

25000

200000MODULAR RATIO (m) = 8

4.33 T/m

FROM DIAGRAM, ……………………….( 1 )

Now, FROM THE METHOD OF THE TRANSFORMED SECTION, So,

x = 98.83 mm

###

FROM THE EQUATION (1) , 0.0013 mm

TAKING, D = a = 350 mm e2 = 0.000001 mmSO THERE WILL BE CRACK. (REF. APPENDIX F OF IS 456)

(REF. APPENDIX F OF IS 456)

0.001276 mm (REF. APPENDIX F OF IS 456)

(REF. APPENDIX F OF IS 456)###

FROM THE EQUATION (2) , BAR DIA = 16 mm

CRACK WIDTH = 0.18 mm < 0.2, Hence ok

Ast =

fc/Ec

GRADE OF CONCRETE (fck ) =

ELASTIC MODULUS OF CONCRETE (Ec) = 5000X SQRT(fck) =

ELASTIC MODULUS OF CONCRETE (Es) =

MAXm MOMENT (Ms) =

e1 =(fs/Es)x(D-x)/(d-x)

b*x*x/2 = m*A

(PUTTING THE VALUE OF b, m , As ,& d )

fs = Ms/(As*(d-x/3)) =

e1 =

e2 (DUE TO STRAIN HARDENING) = 1.5*b*(D-x)*(a-x)/{3*Es*As*(d-x)}

CRACK WIDTH = 3*acr*ξm/{1+2*(acr-cmin)/(D-x)}

HERE, ξm = e1-e2 =

acr = Sqrt[(R/4)²+ec²]-dia/2

R

D-x

x

Strain Diagram

N/mm²

N/mm²

……………………….( 1 )

45.824 ###57188.352

###

< e1

(REF. APPENDIX F OF IS 456) ……. ( 2 )

(REF. APPENDIX F OF IS 456)

(REF. APPENDIX F OF IS 456)

46 mm

e1

fs/Es

b*x*x/2 = m*As*(d-x)

(PUTTING THE VALUE OF b, m , As ,& d )

ec =

cmin+dia/2

R =b-2*cmin-dia

158 mm