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Crack Width Calculation Euro Code 2 Worked Examples

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7.3.2 Minimum reinforcenlent areas

  1. Euro Code Rocklin
  2. Euro Code 3
  3. Concrete Design To Euro Code 2 Worked Examples

(1)P If crack control is required, a minimum amount of bonded reinforcement is required to control cracking in areas where tension is expected. The amount may be estimated from equilibrium between the tensile force in concrete just before cracking and the tensile force in reinforcement at yielding or at a lower stress if necessary to limit the crack width.

(2) Unless a more rigorous calculation shows lesser areas to be adequate, the required minimum areas of reinforcement may be calculated as follows. In profiled cross sections like Tbeams and box girders, minimum reinforcement should be determined for the individual parts of the section (webs, flanges).

Calculation of crack width and crack spacing. The residual tensile strength which is used for crack width calculations should be obtained from tests. The book contains numerous examples. European Commission Joint Research Centre Institute for the Protection and Security of the Citizen Contact information Address: Joint Research Centre, Via.

(7.1)

where:

  • As,min is the minimum area of reinforcing steel within the tensile zone
  • Act is the area of concrete within tensile zone. The tensile zone is that part of the section which is calculated to be in tension just before formation of the first crack
  • σs is the absolute value of the maximum stress permitted in the reinforcement immediately after formation of the crack. This may be taken as the yield strength of the reinforcement, fyk. A lower value may, however, be needed to satisfy the crack width limits according to the maximum bar size or spacing (see 7.3.3 (2))
  • fct,eff is the mean value of the tensile strength of the concrete effective at the time when the cracks may first be expected to occur:
    fct,eff = fctm or lower, (fctm(t)), if cracking is expected earlier than 28 days
  • k is the coefficient which allows for the effect of non-uniform self-equilibrating stresses, which lead to a reduction of restraint forces
    = 1,0 for webs with h ≤ 300 mm or flanges with widths less than 300 mm
    = 0,65 for webs with h ≥ 800 mm or flanges with widths greater than 800 mm
    intermediate values may be interpolated
  • kc is a coefficient which takes account of the stress distribution within the section immediately prior to cracking and of the change of the lever arm:
    For pure tension kc = 1,0
    For bending or bending combined with axial forces:
    - For rectangular sections and webs of box sections and T -sections:
    (7.2)

    - For flanges of box sections and T -sections:
    (7.3)

    where
    • σc is the mean stress of the concrete acting on the part of the section under consideration:
      (7.4)
    • NEd is the axial force at the serviceability limit state acting on the part of the cross-section under consideration (compressive force positive). NEd should be determined considering the characteristic values of prestress and axial forces under the relevant combination of actions
    • h* = h for h < 1,0 m
      h* = 1,0 m for h ≥ 1,0 m
    • k1 is a coefficient considering the effects of axial forces on the stress distribution:
      k1 = 1,5 if NEd is a compressive force
      k1= 2h*/(3h) if NEd is a tensile force
    • Fcr is the absolute value of the tensile force within the flange immediately prior to cracking due to the cracking moment calculated with fct,eff

9.2 Beams

9.2.1 Longitudinal reinforcement

9.2.1.1 Minimum and maximum reinforcement areas

(1) The area of longitudinal tension reinforcement should not be taken as less than As,min.

Note 1: See also 7.3 for area of longitudinal tension reinforcement to control cracking.

Note 2: The value of As,min for beams for use in a Country may be found in its National Annex. The recommended value is given in the following:

As,min = 0.26·fctm/fyk·bt·d, but not less than 0.0013·bt·d

where:

  • bt denotes the mean width of the tension zone; for a T-beam with the flange in compression, only the width of the web is taken into account in calculating the value of bt
  • fctm should be determined with respect to the relevant strength class according to Table 3.1:
    fctm = 0,30 × fck(2/3), fck ≤ 50
    fctm = 2.12·Ln(1+(fcm/10)), fck > 50/60
    with fcm = fck+8 (MPa)

(2) Sections containing less reinforcement than As,min should be considered as unreinforced.

(3) The cross-sectional area of tension or compression reinforcement should not exceed As,max outside lap locations.

Note: The value of As,max for beams for use in a Country may be found in its National Annex. The recommended value is 0.04·Ac.

9.3 Solid slabs

(1) This section applies to one-way and two-way solid slabs for which b and leff are not less than 5h (member for which the minimum panel dimension is not less than 5 times the overall slab thickness).

9.3.1 Flexural reinforcement

9.3.1.1 General

(1) For the minimum and the maximum steel percentages in the main direction 9.2,1,1 (1) and (3) apply.

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(2) Secondary transverse reinforcement of not less than 20% of the principal reinforcement should be provided in one way slabs. In areas near supports transverse reinforcement to principal top bars is not necessary where there is no transverse bending moment.

(3) The spacing of bars should not exceed smax,slabs.

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Note; The value of smax,slabs for use in a Country may be found in its National Annex. The recommended value is:

- for the principal reinforcement, 3·h ≤ 400 mm, where h is the total depth of the slab;
- for the secondary reinforcement, 3.5·h ≤ 450 mm

In areas with concentrated loads or areas of maximum moment those provisions become respectively:
- for the principal reinforcement, 2·h ≤ 250 mm
- for the secondary reinforcement, 3·h ≤ 400 mm.

9.5 Columns

(1) This clause deals with columns for which the larger dimension h is not greater than 4 times the smaller dimension b.

http://lepetkowsnajs1972.mystrikingly.com/blog/add-a-blog-post-title. 9.5.1 General

9.5.2 Longitudinal reinforcement

(1) Longitudinal bars should have a diameter of not less than Φmin.

Note: The value of ¢min for use in a Country may be found in its National Annex. The recommended value is 8 mm.

(2) The total amount of longitudinal reinforcement should not be less than As,min

Note: The value of As,min for use in a Country may be found in its National Annex. The recommended value is given by Expression (9.12N)

As,min = max (0.1·NEd/fyd ; 0.002·Ac)

where:

  • fyd is the design yield strength of the reinforcement
  • NEd is the design axial compression force

(3) The area of longitudinal reinforcement should not exceed As,max

Note: The value of As,max for use in a Country may be found in its National Annex. The recommended value is 0.04·Ac outside lap locations unless it can be shown that the integrity of concrete is not affected, and that the full strength is achieved at ULS. This limit should be increased to 0.08·Ac at laps.

(4) For columns having a polygonal cross-section, at least one bar should be placed at each corner. The number of longitudinal bars in a circular column should not be less than four.

9.6 Walls

9.6.1 General

(1) This clause refers to reinforced concrete walls with a length to thickness ratio of 4 or more and in which the reinforcement is taken into account in the strength analysis

9.6.2 Vertical reinforcement

(1) The area of the vertical reinforcement should lie between As,vmin and As,vmax.

Note 1: The value of As,vmin for use in a Country may be found in its National Annex. The recommended value is 0.002·Ac.

Note 2: The value of As,vmax for use in a Country may be found in its National Annex. The recommended value is 0.04·Ac outside lap locations unless it can be shown that the concrete integrity is not affected and that the full strength is achieved at ULS. This limit may be doubled at laps.

(2) Where the minimum area of reinforcement, As,vmin, controls in design, half of this area should be located at each face.

(3) The distance between two adjacent vertical bars shall not exceed 3 times the wall thickness or 400 mm whichever is the lesser.

9.6.3 Horizontal reinforcement

(1) Horizontal reinforcement running parallel to the faces of the wall (and to the free edges) should be provided at each surface. It should not be less than As,hmin.

Note: The value of As,hmin for use in a Country may be found in its National Annex. The recommended value is either 25% of the vertical reinforcement or 0.001·Ac, whichever is greater.

(2) The spacing between two adjacent horizontal bars should not be greater than 400 mm.

9.8 Foundations

9.8.1 Column and wall footings

(1) A minimum bar diameter Φmin should be provided

Note: The value of Φmin for use in a Country may be found in its National Annex. The recommended value is 8 mm.

Eurocode 2 part 1-1: Design of concrete structures 7.3 Crack control

Euro code 3

The crack width, wk, may be calculated as follows:

wk = sr,max⋅(εsm - εcm) (7.8)

where:

sr,max
is the maximum crack spacing
εsm
is the mean strain in the reinforcement under the relevant combination of loads, including the effect of imposed deformations and taking into account the effects of tension stiffening
εcm
is the mean strain in the concrete between cracks.
(7.9)

where:

σs
is the stress in the tension reinforcement assuming a cracked section,
see application for a rectangular section or application for a T-section
Es
is the design value of the modulus of elasticity of the reinforcing steel, see § 3.2.7 (4)
αe
is the ratio Es/Ecm

with

Ecm
the secant modulus of elasticity of concrete
fct,eff
is the mean value of the tensile strength of the concrete effective at the time when the cracks may first be expected to occur:
fct,eff = fctm or lower, (fctm(t)), if cracking is expected earlier than 28 days
ρp,eff
= (As + ξ1A'p)/Ac,eff(7.10)

with

As
the cross sectional area of reinforcement
A'p
the area of pre or post-tensioned tendons within Ac,eff
Ac,eff
the effective area of concrete in tension surrounding the reinforcement or prestressing tendons of depth, hc,ef, where hc,ef is the lesser of 2,5(h - d), (h - x)/3 or h/2 (see Figure 7.1)
ξ1
the adjusted ratio of bond strength taking into account the different diameters of prestressing and reinforcing steel:
ξ1 = (7.5)

with

ξ
the ratio of bond strength of prestressing and reinforcing steel, according to Table 6.2
ΦS
the largest bar diameter of the reinforcing steel
ΦP
the diameter or equivalent diameter of prestressing steel:
Φp = 1,6⋅√AP for bundles, where AP is the area of a prestressing steel,
Φp = 1,75⋅Φwire for single 7 wire strands,
Φp = 1,20⋅Φwire for single 3 wire strands, where Φwire is the wire diameter.
kt
is a factor dependent on the duration of the load:
kt = 0,6 for short term loading,
kt = 0,4 for long term loading.

• Where the bonded reinforcenlent is fixed at reasonably close centres within the tension zone (spacing ≤ 5(c + Φ/2), cf. Figure 7.2), the maximum crack spacing sr,max may be calculated as follows:

sr,max = k3c + k1k2k4Φ / ρp,eff(7.11)

where:

Φ
is the bar diameter. Where a mixture of bar diameters is used in a section, an equivalent diameter, Φeq, should be used.
c
is the cover to the longitudinal reinforcement
ρp,eff
see the difference of the mean strains above
k1
is a coefficient which takes account of the bond properties of the bonded reinforcement:
k1 = 0,8 for high bond bars,
k1 = 1,6 for bars with an effectively plain surface (e.g. prestressing tendons).
k2
is a coefficient which takes account of the distribution of strain:
k2 = 0,5 for bending,
k2

Euro Code Rocklin

= 1,0 for pure tension.
Intermediate values of k2 should be used for cases of eccentric tension or for local areas:
k2 = (ε1 + ε2)/(2ε1) (7.13)

where ε1 is the greater and ε2 is the lesser tensile strain at the boundaries of the section considered, assessed on the basis of a cracked section.

k3
is a Nationally Determined Parameter, see § 7.3.4 (3)
k4
is a Nationally Determined Parameter, see § 7.3.4 (3).

Euro Code 3

• Where the spacing of the bonded reinforcement exceeds 5(c + Φ/2) (cf. Figure 7.2), or where there is no bonded reinforcement within the tension zone, the maximum crack spacing sr,max may be calculated as follows:

sr,max = 1,3(h - x) (7.14)

where:

h
is the overall depth of the section (see Figure 7.1)
x
is the neutral axis depth of the section (see Figure 7.1).

This application calculates the crack width wk from your inputs. Intermediate results will also be given.

First, change the following option if necessary:

Concrete Design To Euro Code 2 Worked Examples


Output
(7.5)
(7.10)
(7.9)
mm(7.11)
mm(7.8)
Comments
Recipients




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