# Crack Width Calculation As Per Aci 318 14

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

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The crack width, `w _{k}`, may be calculated as follows:

Calculate the required reinforcement to resist the moment at the stem base: Mu t Use #8 bars with 2.0 in. Concrete cover per ACI 318-14 (Table 20.6.1.3.1). The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below: d. Jiang Y S, Liang S S, Chen D W. Research on anti-crack and crack width calculation method of high strength concrete beams with high strength reinforcement. In: Proceedings of the 2nd Seminar of High Strength Concrete and Application. Nanjing, Jiangsu, China, 1995, 147–154.

w = _{k}s⋅(_{r,max}ε - _{sm}ε) _{cm} | (7.8) |

where:

`s`_{r,max}- is the maximum crack spacing
`ε`_{sm}- 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
`ε`_{cm}- is the mean strain in the concrete between cracks.

(7.9) |

where:

`σ`

_{s}see application for a rectangular section or application for a T-section

`E`

_{s}`α`

_{e}`E`/

_{s}`E`

_{cm}with

`E`_{cm}- the secant modulus of elasticity of concrete

`f`

_{ct,eff}`f`=

_{ct,eff}`f`or lower, (

_{ctm}`f`(

_{ctm}`t`)), if cracking is expected earlier than 28 days

`ρ`

_{p,eff}= (A + _{s}ξ⋅_{1}A')/_{p}A_{c,eff} | (7.10) |

with

`A`

_{s}`A'`

_{p}`A`

_{c,eff}`A`

_{c,eff}`h`, where

_{c,ef}`h`is the lesser of 2,5(

_{c,ef}`h`-

`d`), (

`h`-

`x`)/3 or

`h`/2 (see Figure 7.1)

`ξ`

_{1}ξ = _{1} | (7.5) |

with

`ξ`- the ratio of bond strength of prestressing and reinforcing steel, according to Table 6.2
`Φ`_{S}- the largest bar diameter of the reinforcing steel
`Φ`_{P}- the diameter or equivalent diameter of prestressing steel:
`Φ`= 1,6⋅√_{p}`A`_{P}for bundles, where`A`_{P}is the area of a prestressing steel,`Φ`= 1,75⋅_{p}`Φ`for single 7 wire strands,_{wire}`Φ`= 1,20⋅_{p}`Φ`for single 3 wire strands, where_{wire}`Φ`is the wire diameter._{wire}

`k`

_{t}`k`= 0,6 for short term loading,

_{t}`k`= 0,4 for long term loading.

_{t} • 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 `s _{r,max}` may be calculated as follows:

s = _{r,max}k_{3}c + k_{1}k_{2}k_{4}Φ / ρ_{p,eff} | (7.11) |

where:

`Φ`

`Φ`, should be used.

_{eq}`c`

`ρ`

_{p,eff}*the difference of the mean strains*above

`k`

_{1}`k`= 0,8 for high bond bars,

_{1}`k`= 1,6 for bars with an effectively plain surface (e.g. prestressing tendons).

_{1}`k`

_{2}`k`= 0,5 for bending,

_{2}`k`= 1,0 for pure tension.

_{2}Intermediate values of

`k`should be used for cases of eccentric tension or for local areas:

_{2}## Crack Width Calculation As Per Aci 318 14 Online

k = (_{2}ε + _{1}ε)/(2_{2}ε) _{1} | (7.13) |

where `ε _{1}` is the greater and

`ε`is the lesser tensile strain at the boundaries of the section considered, assessed on the basis of a cracked section.

_{2}`k`

_{3}`k`

_{4} • 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 `s _{r,max}` may be calculated as follows:

s = 1,3(_{r,max}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 `w _{k}` from your inputs. Intermediate results will also be given.

First, change the following option if necessary: