Estimation of service life-span of concrete structures

Exercise 11

Strength classes in by50

Workability classes

Concrete family -consept

Requires the same characteristics:• Same cement • Same aggregate• Same additional binders• Can have admixtures, as long as they don’t

substantially affect the strength• K15 – K60 (C12 – C50)• Same age at testing

In applying the concept of concrete families, a reference concrete is chosen. The reference concrete is usually the most commonly produced, or one from the mid-range of the concrete family.

Combining data into families can reduce the time taken to detect any significant changes in production quality.

All members’ compressive strength results need to be converted to that of the reference concrete. Two methods of transposing can be used:• Strength method based on a straight line relationship between

strength and water/cement ratio• Strength method based on a proportional effect

Example:The strength of the studied concrete is converted to correspond to the strength of the reference concrete.

Reference concrete K30, target strength of 36 MPaStudied concrete K45, target strength of 55 MPaConvert the strength to correspond to the strength of the reference concrete when the compressive strength results for the studied concrete was at 53 MPa.53-55 MPa = -2 MPa→36 – 2 = 34 MPa

Conformity Control (kelpoisuuden valvonta)

In the assessment of conformity control, three criterion need to be satisfied. When a family member is tested, the original compressive strength result has to conform to criterion 2 in Table 14 of SS EN 206-1. The member’s result will be converted to equivalent values of the reference concrete and assessed for conformity (Criterion 1, Table 14 of SS EN 206-1).

Criterion 1 and 2 for initial and continuous production according to SS EN 206-1: 2009

Initial Continuous

Conformity Control (kelpoisuuden valvonta)

In addition, it has also to be assessed that each individual member belongs to the family (Criterion 3, Table 15 of SS EN 206-1). In the case where a member fails to meet criterion 3, it is removed from the family and assessed individually for conformity.

Curing

- X0 and XC1 60 % of the nominal strength

- others 70 % of the nominal strength except

• XF2 and XF4 80 % of the nominal strength

- Estimation of the strength development for example with the Sadgroven maturity function

Sadgrove :

t20 = ((T + 16 oC)/36 oC) * t

Durability - against what?

• Physical erosion• Carbonation• Chlorides• Freeze/thaw resistance• Chemical durability

• Ca(OH)2 + CO2 → CaCO3

• Concrete pH 12,5 - 14,0• Carbonation pH < 9,0

- Is not dangerous in ”easy” conditions (for example, indoors)

- protective passivity layer on steel surface is broken- iron + water + oxygen → rust

- A: Fe → Fe2+ + 2e-

- C: 4e- + 2H2O + O2 → 4OH-

Carbonation

2 Fe2+ + 4 (OH)- → 2 Fe(OH)2 3 Fe(OH)2 → Fe3O4 + 2 H2O + H2

Rust demands more space than initial products → concrete breaks

Chlorides

• Are usually not harmful to concrete• Free chlorides in the pore water are

effective in initiating chloride-induced corrosion of the reinforcement

Steel

Thickness of the concrete cover• Concerns all reinforcement• Corrosion susceptible reinforcement *

presents a 10 mm additional coverage requirement

* Thickness of the reinforcement 4 mm or more* long-term stress state (in service state) over 400 MN/m2 (cold-worked steel)

Freeze-thaw resistance

• Rate of freezing and thawing• Temperature• Degree of moisture saturation • Number of cycles• Chlorides

Freeze-thaw resistance

• Air-entrained concrete• 2 % → 6 % (8 %)

• Spacing factor (huokosjako)• Specific surface area

Chemical attack• Has to be foreseen in the design process– Environment of the concrete

• Reactions with external substances • Humidity• Migration with water / drying• Acids• Sulphates SO4

2-

• SO42- + Ca(OH)2 = gypsum

• SO42- + gypsum + calsiumaluminatehydrate =

ettringite

Exposure classes

The designer must choose an appropriate exposure class for the structure in terms of the following stress or load factors:1. Corrosion caused by carbonation2. Corrosion caused by chlorides3. Corrosion caused by chlorides in sea water4. Freezing-and-thawing stress5. Chemical load

Factors influencing the service life of concrete

- Strength class- Amount of cement andcement/additional binders- water/cement -ratio- Air content- Curing - Age - Service operations- Environment

Service life of concrete

• Service life requirement can be designed with either tabular data (taulukkomitoitus) or calculations

Use of tabular data50 or 100 years• Carbon dioxide• Chlorides• Sea water and thawing agents (salts)• Freeze-thaw stress• Chemical load

Tabular data

• Simple, fast• Does not enable optimization• Useful with strenght classes ≤ K40, in other

cases may lead to too thick concrete covers• Only for service lifes of 50 or 100 years

DESIGNING WITH TABULAR DATA 50 YEARSFeasible when the strength grade is close to the minimumThe requirement of minimum cement content must be fulfilledOnly option in classes XA

Minimum cover of concrete using tabular data

Permitted negative deviation generally 10 mm

Exposure classes XS and XD

Calculating the service life

• 50…200 years• Uses reference service life-span of 50 years• For all exposure classes

– Estimated seperately for each class and the shortest of these will be the determining one

A. Materials, porosityB. Design structural detailsC. Performance of workD. Interior climateE. Exterior exposure to weatherF. Working loadG. Maintenance measures

1.

Calculate the service life for a K30 foundation with regard to carbonation for which a CEM I A cement was used and the air content of the concrete was measured at 2,0 %.

Working life with regard to carbonation

Exposure classes X for foundations

XO no risk of corrosion or chemical attack -

XC carbonation +XS chlorides, sea water -XD chlorides, from other sources -XF freezing and thawing +XA chemical loads +

Exposure class for carbonation XC???

XC2 Minimum cover of concrete 20 mm(+tolerance)

The working life is calculated using the equation:

tL = tLr x A x B x C x D x E x F x G

tLr is the reference service life-span of 50 years

tL is the service life

A. Materials, porosityB. Design structural detailsC. Performance of workD. Interior climateE. Exterior exposure to

weatherF. Working loadG. Maintenance measures

A materials, porosityA1 K30 0,95

A2 CEM I A 1,00

A3 2 % 1,08

B Design, structural details

B1 30 mm concrete 1,44

B2 Coating no coating 1,0

C Curing

E Exterior exposure to weather

E1 XC2 foundations and 1,4

other undergroundstructures

E2…E4: if the structure is protected from rain,

coefficients E2, E3 and E4 shall be given the value 1 1,0

D Interior climate -1,0

F Working load -

G1 None 0,85

G Inspection and maintenance frequency

What did we get?

tL = tLr x A x B x C x D x E x F x G

= 50 x (0,95*1,00*1,08) x (1,44*1,0)x (1,0) x (1,4) x (0,85)

= 50 x 1,76 = 88 years

2

A: 80,5 years

3.

Design a foundation for a service life of 100 years using tabular data.

Foundation for a service life of 100 years using tabular dataTable 4.2 (from by50 Concrete code 2004)

Exposure classes?

Corrosion caused by carbonation, XC

Minimum amount of cement 230 kg/m3

Strength grade K35

Chemical load, XA

Minimum amount of cement 320 kg/m3

Strength grade K45Max w/c ratio 0,45 -> w = 144 kg/m3

F- and P-factors

The F-factor describes the freeze-thaw resistance in a non-saline environment:

In which w/c is the effective water/cement ratio

a is the measured air content

0,4)1(

)/(2,7;25,0max

1

1 4,0

4 5,0

a

cwF

F-factor

1,7;4min140,0;25,0max

1

0,4)10,5()320/144(

2,7;25,0max

1

14,0

45,0

F

F-factor

Calculated life span is the product of F-factor and 50 years (k x t50 years)

Using tabular data 50 years XF1 1,0 and XF3 1,5100 years XF1 2,0 and XF3 3,0

P-factor

The P-factor describes the freeze-thaw resistance in a saline environment: