Composition of cement paste, concrete admixtures and mix design of superplasticized concrete Exercise 4

INPUT: A concrete sample was extracted from a structure and dried at 105 °C in which case 80 kg/m 3 of water evaporated. The degree of hydration (α) was determined at 0,5. The mix design of the concrete was 1 : 6,0 : 0,6 and air content was measured at 3 %. Q? How much of the water had been evaporated prior to drying? What were the amounts (in volume) of: – unhydrated cement, – solid part of cement gel, – gel water, – capillary water, – contraction pores and – capillary pores at the time of sampling? Exercise 4/01

Cem : Agg : Water Q = amount  = density Q L = Air content (ilma) W/C = Water Cement Ratio How much of the water had been evaporated prior to drying?

Amount of added water: W o = 0,6 * 306 kg/m 3 = 184 kg/m 3 Concrete density = (1+6,0+0,6)C = 7,6*C = 2326 kg/m 3 W/C = Water Cement Ratio Cem : Agg : Water How much of the water had been evaporated prior to drying?

When the sample was dried, 80 kg/m 3 water evaporated Non evaporable water consists of chemically combined water ! The amount of evaporable water should have been W o – W N = 184 – 0,25*0,5*306 = 184 – 38 = 146 kg/m 3 Prior to drying, water had evaporated: 146 – 80 = 66 kg/m 3 W N = Chemically bound water = 0,25* α * C α = Hydr. degree How much of the water had been evaporated prior to drying?

α = Hydr. degree

The volume of gel pores V gh are 28 % of the total volume of the Cement gel → V gh / (V gh + V gs ) = 0,28 (V gs is the solid part of the cement gel) → V gh = 0,28/0,72 * V gs = 0,28/0,72 * 78,0 l/m 3 = 30,3 l/m 3 Contraction pores V con = 0,25 * V N = 0,25*0,25*α*C = 9,6 l/m 3 V gh = 0.28*V gh *V gs 0.72*V gh = 0.28*V gs chemically bound water V N

The amount of evoporable water consists of capillary water and gel water. The amount of evaporated water was 80 kg/m 3 W cap + W gh = 80 kg/m 3 Wgh= gel water W cap = 80 kg/m 3 – 30 kg/m 3 = 50 kg/m 3 V cap = W cap /ρ V = 50 l/m 3 The total volume of the capillary pores V cap = V o – V N – V gh V N = chemically bound water V N = 0,25* α * C = 0,25 * 0,5 * 306 = 38,3 l/m 3  v = density of water 1 kg/dm³ V N = chemically bound water

The total volume of the capillary pores V cap = V o – V N – V gh = 184 – 38 – 30 = 116 l/m 3

INPUT: Concrete´s cement and water amounts were 350 kg/m 3 and 135 kg/m 3 respectively. Q? Calculate the degree of hydration and amount of gel pores a)without wet curing b)when wet cured. Exercise 4/02

Maximum degree of hydration 1. Without wet curing (no outside water): 2. When wet cured Initial porosity

Total amount of gel pores ? 2 ways of calculating: 1) V gh = 0,2 * α * C (see exercise 3 for details) 2)  = Initial porosity (1 -  ) = solid part of the paste C = Cement α = Hydr. degree

Formula 1: V gh = 0,2 * α * C not wet cured: V gh = 0,2 * 0,852 * 350 = 59,6 dm 3 wet cured: V gh = 0,2 * 0,994 * 350 = 69,6 dm 3 OR Formula 2: v gh = 0,6 x (1 - 0,544) x 0,852 = 0,233 v gh = 0,6 x (1 - 0,544) x 0,994 = 0,272 !!! Formula 2 gives the volume fraction of pores in cement gel. Thus, this is only just the proportional share (suhteellinen osuus) of the whole volume!!! Therefore, 0,233 x (350/3, /1) = 57,79 dm 3 0,272 x (350/3, /1) = 67,4 dm 3

How does the degree of hydration change, when 7 % of cement is replaced with silica powder? And how much changes the volume of unhydrated cement? Exercise 4/03

Maximum degree of hydration 1. Without wet curing (no outside water): 2. When wet cured Initial porosity (cem. + silica) Coefficient based on the effect of (cem + silica)

Initial porosity (cem. + silica) Coefficient based on the effect of (cem + silica)

Now we can calculate : OR

And how much changes the volume of unhydrated cement? Volume fraction of unhydrated cement in problem 2: Volume fraction of unhydrated cement with silica: Thus the volume fraction of unhydrated cement in problem 2 is: ν c = (1- 0,544)*(1-0,852) or ν c = (1-0,544)*(1-0,994) ν c = 0,067 or 0,003 So the volume is: 0,067 * (350/3, /1) = 16,6 dm 3 or 0,003 * (350/3, /1) = 0,7 dm 3 Thus the volume fraction of unhydrated cement with silica is: ν c = 0,9047* (1- 0,537)*(1-0,844) = 0,065 So the volume is: 0,065 * (325,5/3,1 + 24,5/2,2+ 135/1) = 16,3 dm 3

Concrete admixtures Material other than water, aggregates, cement and reinforcing fibers that is used in concrete as an ingredient and added to the batch immediately before or during mixing. i.Air-entraining agents (ASTM C260) ii.Chemical admixtures (ASTM C494 and BS5075) iii.Mineral admixtures iv.Miscellaneous admixtures include: » Latexes » Corrosion inhibitors » Expansive admixtures

Concrete admixtures Beneficial effects of admixtures on concrete properties

Concrete admixtures SP = Super plastisizer DCI = Darex Corrosion Inhibitor

Water reducing admixtures Water-reducing admixture lowers the water required to attain a given workability. Mechanism: – separate the cement particles – Release the entrapped water

Water reducing admixtures Two kinds of water-reducing admixture: i.The normal range (WR):  Reduce 5 – 10% of water ii.The high range water reducing admixture (HRWR):  Superplasticizer  Reduce water in a range of 15-30%

Water reducing admixtures Superplasticizer – Superplasticizers are used for two main purposes: i.To produce high strength concrete at w/c ratio in a range of 0.23 – 0.3 (60 – 150MPa) ii.To create “flowing” concrete with high slumps in the range of 175 to 225mm. Self compacting concrete: for beam-column joint and footing (heavy reinforced) – Two forms i.Solid power ii.Liquid % - 60% of water – Normal dosage of superplasticizer for concrete is 1%-2% by weight of cement. – drawbacks of superplasticizer are: i.retarding of setting (especially at large amount addition) ii.causing more bleeding iii.entraining too much air.

Air-entraining admixtures Entrained air: – On purpose – Size: 50 to 200 μm Entrapped air: – By chance – As large as 3mm

Air-entraining admixtures

Advantages of adding air entraining admixtures Improved workability --- air bubble as lubricant Improved ductility --- more deformation from small hole Reduced permeability --- isolated air bubble Improved impact resistance --- more deformation Improved durability --- freezing and thawing(release ice forming pressure)

Disadvantages of adding air entraining admixtures Strength loss of 10-20%

We require a concrete mix with a 28 day compressive strength of 40 MPa and a slump of 120 mm, ordinary Portland cement being used with cement strength of 48 MPa. Grading of the aggregate is presented in the forms. Proportioning is to be done by using a superplasticizer in which case the required water amount can be reduced by 10 %. How much does the strength of the concrete increase when water is decreased (assuming that the cement content stays the same)? By how much could the cement content be decreased in order to abtain the same strength (40 MPa)? Exercise 4/04

Calculate the proportioning strength (suhteituslujuus) K s K s = 1,2*K*42,5/N N is the test strength of the cement Ks = 1,2*40*(42,5/48) = 42,5 The granulometric value of H (rakeisuusluku H) of the combined aggregate has already been calculated Use the mix design form to specify the amounts of water, cement and aggregate Export the material data to the “Concrete composition” form, i.e. BETONIN KOOSTUMUS

INPUT - H (aggregates)418 - slump120 mm -28 compressive strength 40 MPa -Ks(Design strength)42.5 MPa - cement strength 48 MPa

From the mix design form: - Cement 355 kg/m 3 - Aggregate1840 kg/m 3 - Water + air198 kg/m 3 - Water 178 kg/m 3 - air20 l/m 3

W/C–ratio 0.51

water amount is 10 % smaller

Cement is saved 355 – 323 = 32 kg/m 3 The new amount of aggregate can be calculated by using the basic equation of concrete: 1000 – 323/3,1 – 160,2/1,0 – 20 = 715,6 dm 3 → 715,6 * 2,68 = 1918 kg/m 3