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Crack Width Calculation

Indian or British Codes

IS 456 - 2000 Annex F and BS 8110 Part II, Clause 3.8 include the same method for determining the nominal crack width for design. This procedure is based on work by Andrew Beeby. The crack width determined by this method represents a best-fit for experimental data where the stress in the tension steel is limited to 0.8x Fy .

The following assumptions are made for this method:

  • Sm = 1.5 acr
  • Wmean is a funciton of Sm and εm
  • Wmax = 2x Wmean

The following procedure is then used by the program to evaluate the crack width for design:

  1. Calculate the modular ration, m as:

    m = Ec/Es

    Where:

    Es = 20,000 N/mm2 (IS 456 - 2000, Annex F)

    Ec = 5,000·√(fc) (Cl. 6.2.3.1 of IS 456 - 2000)

  2. Calculate the depth of the neutral axis, x, per the limit state method (IS 456 - 2000 Annex G and BS 8110 Part I Cl. 3.4.4.4)

  3. Calculate the moment of inertia of the equivalent section, Ieff.
  4. Calculate the effective modulus of elasticity of concrete, Eeff.

    E e f f = E c 1 + c

    Where:

    c = safety factor taken as 1.0.

  5. Calculate the average steel strain at the depth considered (e.g., the centerline of rebar), ε1, based on the absolute value of the critical bending moment.
  6. Calculate the average strain, εm, at the point of interest as:

    ϵ m = ϵ 1 b ( h x ) ( a 1 x ) 3 E s A s t ( d x )
  7. Calculate Crack Width W_cr from IS 456 -20000 BS 8110 Part II

    W c r = 3 a c r ϵ m 1 + 2 ( a c r C min ) h x

    Where:

    acr = distance from the point of interest to the surface of the nearest longitudinal bar

    h = total depth of the section

    Cmin = clear cover to longitudinal steel

  8. Check whether the crack width value is safe or not depending upon the maximum allowed range defined by codes (IS 456 2000 & BS 8110 Part II)