1. Cement chemistry; reactions and adsorption of water, volume of hydration products and porosity of cement paste Rak-82.3131 Concrete technology 2 Exercise 3

2. Recap from last week:

4. Reactions and adsorption of water

8. Characteristics of hydration of the cement compounds Amount of Contribution to cement CompoundsReaction rateheat liberated StrengthHeat liberation C3SC3S Moderate High C2SC2S Slow Low Low initially,Low high later Fast Very high Low Very high Moderate Low Moderate

9. Hydration Tricalcium silicate (alite): 2C 3 S + 6H → C 3 S 2 H 3 + 3CH 1g + 0,24g → 0,75g + 0,49g Dicalcium silicate (belite): 2C 2 S + 4H → C 3 S 2 H 3 + CH 1g + 0,21g → 0,99g + 0,22g

11. Tetracalcium aluminoferrite: C 4 AF + 2CH + 10H → C 3 AH 6 + C 3 FH 6 1g + 0,30g + 0,37g → 0,78g + 0,90g

12. Water requirement of the hydration reactions The total water requirement of the hydration reactions is the sum of the water requirements of the cement minerals.

13. w = w (SP) + w (AP) + w (FP) + w (OC) in which, SP = silicate phase AP = aluminate phase FP = ferrite phase OC = other components Water requirement w:

15. In Exercise 2 concrete´s compound composition was calculated as: {C 3 S, C 2 S, C 3 A, C 4 AF, S ̅} = {52.3, 21.0, 9.3, 8.2, 4.9} %. Calculate the water requirement of the cement (for complete hydration). Take into account the amounts of CaO free (0.96 %) and MgO (1.08 %). 1

16. Problem 1

17. Water requirement of the hydration reactions

18. MONOSULPHATE From the reactions it can be seen that 0,63g gypsum binds 1g C 3 A and 0,67g water.

21. C 4 AF (tetracalcium aluminoferrite) w(C 4 AF) = 0,37 * 0,082 = 0,03034 g/g In addition we need to take into account the chemically bound water by MgO and CaO free and thus we get: w(CaO free ) = 0,32 * 0,0096 = 0,00307 g/g w(MgO) = 0,45 * 0,0108 = 0,00486 g/g in which case w tot = 0,12552 + 0,0441 + 0,06231 + 0,01162+ 0,03034 + 0,00307 + 0,00486 = 0,25146 g/g CALCIUMCILICATES w(C 3 S) = 0,24 * 0,523 = 0,12552 g/g w(C 2 S) = 0,21 * 0,21 = 0,0441 g/g

22. How does the water requirement of the cement in exercise 1 change when {S ̅} = {1,9} % ?

23. Exercise 2

25. 1 st, the amount of gypsum added to cement clinker is expressed as the mass of SO 3 present; this is limited by European Standards. 2 nd, an excess of gypsum leads to an expansion and consequent disruption of the set of cement paste. Optimum gypsum content leads to a desirable rate of early reaction and prevents local high concentration of products of hydration. In consequence the size of pores in hydrated cement paste is reduced and strength is increased.

26. Volume of hydration products and porosity of cement paste

27. Properties of concrete, p.31

28. Calculating the volumes of hydration products INITIAL DATA & ABBREVIATIONS (=lyhenteet) The quantity and volume of cement C and V c The quantity and volume of water W o and V o Water/cement ratio w/c = W o /C Degree of hydration α Air L

33. The cement content of a concrete is 375 kg/m 3 and the w/c ratio is 0,46. At a specific moment the degree of hydration α = 0,80. Calculate how much (l/m 3 ) there is a) chemically bound water, b) gel water and d) capillary water. Problem 3

35. 3b V gh = 0,28*(V gs +V gh ) = 0,28/0,72 V gs V gs = V ch + V N – V con = α*C/ρ C + 0,25*α*C/ρ V – 0,25*0,25*α*C/ρ V = αC(1/ρ C + 0,1875/ρ V ) → V gh = 0,28/0,72* α C(1/ρ C + 0,1875/ρ V ) = 0,2αC = 0,2*0,8*375 = 60 l/m³ A MORE SPECIFIC WAY : If the clinker mineral composition is known the formula W g.h = 3*k*W N can be used In which k = 0,23*0,523+0,32*0,210+0,317*0,093+0,368*0,082 = 0,247 Because W N was calculated at 75,44 kg/m 3 in part a, we get: W N = W gh = 3* 0,247*75,44 = 55,93 kg/m³

36. 3c Excess water from the initial water amount (W o ), forms the capillary water Original water amount W o = c*w/c = 375*0,46 = 172,5 kg/m³ V cap = V o – V N – V g.h = 172,5 – 75 – 60 = 37,5 l/m³ A MORE PRECISE WAY V cap = 172,5 - 75,44 - 55,93 = 41,13 l/m³

37. Unreinforced concrete ducts were manufactured at a factory with a recipe: Aggregate 835 kg/m³ Cement 150 kg/m³ Water 57 kg/m³ A series of ducts were weighed right after manufacture and again 24 hours after. An average of 2,8 % weight loss caused by concrete drying was measured. What meaning does this water loss have? Problem 4

38. Weight of the batch = 835 + 150 + 57 = 1042 kg Loss of water = 0,028*1042 = 29,176 kg Thus, from the added water over 50 % has evaporated. In order to demonstrate the effect of this water loss we can calculate the maximum hydration degree (α MAX ) which could be achieved with this remaining amount of water.

39. Total volume of the cement paste V paste (In this type of examination, the amount of air can be excluded)

40. The volume of the gel pores (gel water) are 28 % of the total volume of the sement gel V gh / (V gh + V gs ) = 0,28 V gh = 0,28/0,72 * V gs V gh = 0,2*α*C[dm³] V N = 0,25*α*C[dm³]

41. The maximum degree of hydration of the original mix design (without evaporation) can be calculated as: V capv = 0 = V o – V N –V gh = 57 – 0,25*α*C - 0,2*α*C = 57 – 0,45*α*CC=150 = 57 – 67,5* α → α MAX = 0,84 And the maximum degree of hydration for the evaporated batch: ∆V = 29,2 V kapv = 0 = V o – ∆V – V N – V gh = 57 – 29,2 – 0,45*α*150 = 27,8 – 67,5α → α MAX = 0,41

42. 5 The composition of a concrete is such that it can theoretically achieve full hydration. Calculate the w/c ratio of such concrete.

44. V gh = 0,28/0,72*0,51 l = 0,198 l TOTAL WATER AMOUNT IS THUS: 250g + 198 g = 448g AND THE WATER CEMENT RATIO: w/c = 0,448