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

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

2 1 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 () was determined at 0,5. The mix design of the concrete was 1 : 6,0 : 0,6 and air content was measured at 3 %. 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?

3 Mix design 1 : 6,0 : 0,6 3 % air Basic equation of concrete: = 1000, + 0,6C + 30 = 1000, C, = 970 C = 306 kg/m3,,

4 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

5 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 N = Chemically bound water = 0,25* * C Wo WN = 184 0,25*0,5*306 = = 146 kg/m 3 Prior to drying, water had evaporated: = 66 kg/m 3

6 Unhydrated cement V C.UNHYD W C.UNHYD = 306 0,5*306 = 153 kg/m 3 V C.UNHYD = 153/3,1 = 49,4 l/m 3 Solid products of hydration V gs = hydrated cement V ch + chemically bound water V N contraction pores (supistumishuokoset) V con V gs = V ch + V N V con V con = 0,25V N = V ch + V N 0,25V N = V ch + 0,75V N = + 0,75, =,, = 78,0 l/m 3 +,,,

7 The volume of gel pores V gh are 28 % of the total volume of the sement gel Vgh / (Vgh + Vgs) = 0,28 (V gs is the solid part of the cement gel) Vgh = 0,28/0,72 * Vgs = 0,28/0,72 * 78,0 l/m 3 = 30,3 l/m 3 Contraction pores V con = 0,25 * VN = 0,25*0,25**C = 9,6 l/m 3

8 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 W cap = 80 kg/m 3 30 kg/m 3 = 50 kg/m 3 V cap = W cap V = 50 l/m 3 W gh = gel water The total volume of the capillary pores V cap = V o V N V gh V N = chemically bound water V N = 0,25* * C = 0,25 * 0,5 * 306 = 38,3 l/m 3

9 The total volume of the capillary pores V cap = V o V N V gh = = 116 l/m 3

10 Concrete s cement and water amounts were 350 kg/m 3 and 135 kg/m 3 respectively. Calculate the degree of hydration and amount of gel pores a) without wet curing b) when wet cured.

11 Maximum degree of hydration 1. Without wet curing (no outside water): max 1,4 (1 ) 2. When wet cured 1 w c w c w c max 1,2 (1 ) 1

12 = = 0,386 = 1 3,1 = 0,323 = 0,386 0, ,323 = 0,544 = =,,(,),,(,) = 0,852 not wet cured = 0,994 wet cured

13 Total amount of gel pores? 2 ways of calculating: 1) 2) V gh = 0,2 * * C (see exercise 3 for details) v gw 0,6 (1 )

14 Formula 1: not wet cured: Vgh = 0,2 * 0,852 * 350 = 59,6 dm 3 wet cured: Vgh = 0,2 * 0,994 * 350 = 69,6 dm 3 OR Formula 2: vgh = 0,6 x (1-0,544) x 0,852 = 0,233 vgh = 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

15 Exercise 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?

16 Maximum degree of hydration 1. Without wet curing (no outside water): 2. When wet cured 1 ) (1 ) 1,6 (1,4 max c s k 1 ) (1 ) 0,9 (1,2 max c s k c s c w c w s w c w c s k 1,4 1 1

17 Amount of silica: s =0,07 * 350 kg/m 3 = 24,50 kg/m 3 Amount of cement: c = 350 kg/m 3 24,5 kg/m 3 = 325,5 kg/m 3 w c = ,5 = 0,4147 = = 0,323 = = 0,455 s c = 24,5 325,5 = 0, k = ,4 0,07527 = 0,9047

18 = = w c w c + + s c 0,4147 0, , ,455 0,07527 = 0,5372 Now we can calculate: max k (1,4 1,6 s c ) (1 ) 1 max k (1,2 s 0,9 ) (1 ) c 1 = 0,5372 0,9047 (1,4 + 1,6 0,07527) (1 0,5372) = 0,844 OR =,,(,,,(,) = 1,012 = 1because max 1

19 And how much changes the volume of unhydrated cement? Volume fraction of unhydrated cement in problem 2: v c ( 1 ) (1 ) Thus the volume fraction of unhydrated cement in problem 2 is: c = (1-0,544)*(1-0,852) or c = (1-0,544)*(1-0,994) 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 Volume fraction of unhydrated cement with silica: k ( 1 ) (1 ) v c Thus the volume fraction of unhydrated cement with silica is: 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

20 Concrete admixtures Water-reducing admixtures water-reducing /plasticising admixtures superplasticizers Air-entraining agents Accelerating admixtures Retarding admixtures Water-proofing admixtures Other admixtures e.g. grouting admixtures (injektointiaine), antibacterial admixtures

21 Accelerating and retarding admixtures Accelerating admixtures aka accelerators are used to speed up concrete setting or the early strength development (hardening) when concrete is to placed at lower temperatures, in the manufacture of precast concrete or other situations where a rapid removal of formwork is desired. Retarding admixtures aka retarders generally slow down also the hardening of the concrete paste. Retarders are useful in concreting in hot weather when the normal setting time is shortened by the higher temperature and in preventing the formation of cold joints. In general they prolong the time during which concrete can be transported, placed and compacted. It is important to notice that retarders cannot prevent the loss of slump and it does not decrease the maximum hydration temperature in a structure it can only postpone it!

22 Air entraining agents Air entraining admixtures comprise a group of surfactants (pinta-aktiiviset aineet) which act at the water air interface in cement paste, thereby stabilising air entrapped during the mixing process in the form of tiny discrete bubbles

23 Water-reducing admixtures water-reducing /plasticising admixtures superplasticizers As their name implies, the function of water reducing admixtures is to reduce the water content of the mix, usually by 5 or 10 per cent. These admixtures also comprise of a group of surfactants (pinta-aktiiviset aineet) which act at the cement water interface.

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27 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 attain the same strength (40 MPa)?

28 Calculate the proportioning strength (suhteituslujuus) K s K s = 1,2*K*42,5/N N is the test strength of the cement 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

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30 - slump 120 mm - 28 compressive strength 40 MPa - cement strength 48 MPa From the mix design form: - Cement 355 kg/m 3 - Aggregate 1840 kg/m 3 - Water 178 kg/m 3 - air 20 l/m 3

31 a) The amount of cement stays the same, water amount is 10 % smaller Composition: cement 355 kg/m 3 water 178-0,1* ,2 kg/m 3 air 20 l/m 3 New water-air/cement -ratio:, 0,51

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33 From the mix design form we can read: K s = 45 MPa The original design strength was 42,5 MPa, thus THE STRENGTH WOULD INCREASE BY 2,5 MPa New amount of aggregate can be calculated by using the basic equation of concrete: = /3,1 160,2/1,0 20 = 705,3 dm 3 705,3 *2,68 kg/m 3 = 1890,2 kg/m 3

34 b) Decrease the cement amount cement?? kg/m 3 water 160,2 kg/m 3 air 20 l/m 3 We wan to keep the original strength so the waterair/cement ratio stays the same Original New,,, cement = 323 kg/m 3

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