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