Crack width calculation due to flexure.

b

x/3x

h dz

Section Strain stress

Depth of section h = 300 mmwidth of section b = 1000 mmcover = 50 mmtention bar dia Ф = 16 mm

= 242 mmAs, tention = 1005Bar spacing s = 200 mmYoung's modulus of steel = 200Young's modulus of concrete Ec = 27

Ec/2 = 13.5Fcu = 35fy = 460

= 0.004153

Modular ratio = 14.8

Depth of neutral axis, x = 71.3 mm

lever arm z=d-x/3 = 218 mmz = 218 mm

Applied service moment = 44 kNm/m

Steel tensile stress = 201Satisfy

Concrete comp stress = 5.66Satisfy

Ɛc fc

Ɛs fs/αe

deffmm²/m

Es kN/mm²kN/mm²kN/mm²N/mm²N/mm²

ρ =As/bd

αe=Es/Ec

x=d.αe.ρ.[(1+2/αe.ρ)^0.5-1]

fs=Ms/z.As < 0.8fy N/mm²

fcb=2.Ms/z.b.x N/mm²

Ɛ1=(h-x).fs/(d-x).Es = 0.001344

a' , distance from comp.face to crack point = 300 mm

limitted crack width,

for 0.2 mm for 0.1 mm

Ɛ2 =b.(h-x).(a'-x)/3.Es.As.(d-x) Ɛ2 =1.5.b.(h-x).(a'-x)/3.Es.As.(d-x)

Ɛ2 = 0.000508 Ɛ2 = 0.000762

Ɛm = Ɛ1-Ɛ2 = 0.000836 = 0.000582

s s =Ф =

=

=

Crack width,

= 0.18 mm = 0.12

cmin cmin

acracr

w = 3.acr.Ɛm/(1+2.(acr-cmin)/(h-x))

= 229.9 mm0.95deff

Ɛ2 =1.5.b.(h-x).(a'-x)/3.Es.As.(d-x)

200 mm16 mm50 mm

107.60 mm

mm