Composition of cement paste, concrete admixtures and mix design

of superplasticized concreteExercise 4

1

A concrete sample was extracted from a structureand dried at 105 °C in which case 80 kg/m3 of water evaporated. The degree of hydration ( ) wasdetermined at 0,5. The mix design of the concretewas 1 : 6,0 : 0,6 and air content was measured at 3 %. How much of the water had been evaporatedprior to drying? What were the amounts (in volume) of unhydrated cement, solid part of cement gel, gel water, capillary water, contractionpores and capillary pores at the time of sampling?

Mix design1 : 6,0 : 0,6 3 % air

Basic equation of concrete:

+ + + = 1000

, + , 0,6C + 30 = 1000

C , , , = 970 C = 306 kg/m3

Amount of added water:Wo = 0,6 * 306 kg/m3 = 184 kg/m3

Concrete density = (1+6,0+0,6)C = 7,6*C = 2326 kg/m3

When the sample was dried, 80 kg/m3 waterevaporated

Non evaporable water consists of chemicallycombined water !

The amount of evaporable water should have beenWo – WN

= 184 – 0,25*0,5*306= 184 – 38= 146 kg/m3

Prior to drying, water had evaporated:146 – 80 = 66 kg/m3

WN = Chemically bound water = 0,25* * C

Unhydrated cement VC.UNHYD

WC.UNHYD = 306 – 0,5*306 = 153 kg/m3

VC.UNHYD = 153/3,1 = 49,4 l/m3

Solid products of hydration Vgs= hydrated cement Vch + chemically bound water VN –contraction pores (supistumishuokoset) Vcon

Vgs = Vch + VN – Vcon Vcon = 0,25VN

= Vch + VN – 0,25VN = Vch + 0,75VN

= + 0,75 ,

= ,, + , , ,

= 78,0 l/m3

The volume of gel pores Vgh are 28 % of the total volume of the sement gel

Vgh / (Vgh + Vgs) = 0,28 (Vgs is the solid part of the cement gel)

Vgh = 0,28/0,72 * Vgs

= 0,28/0,72 * 78,0 l/m3 = 30,3 l/m3

Contraction pores Vcon= 0,25 * VN

= 0,25*0,25* *C= 9,6 l/m3

The amount of evoporable water consists of capillarywater and gel water.

The amount of evaporated water was 80 kg/m3

Wcap + Wgh = 80 kg/m3 Wgh= gel water

Wcap = 80 kg/m3 – 30 kg/m3 = 50 kg/m3

Vcap = Wcap V = 50 l/m3

The total volume of the capillary pores Vcap= Vo – VN – Vgh VN = chemically bound water

VN = 0,25* * C = 0,25 * 0,5 * 306 = 38,3 l/m3

The total volume of the capillary pores Vcap= Vo – VN – Vgh= 184 – 38 – 30 = 116 l/m3

Concrete´s cement and water amounts were 350 kg/m3

and 135 kg/m3 respectively. Calculate the degree of hydration and amount of gel pores a) without wet curing b) when wet cured.

1)1(4,1max

Maximum degree of hydration

1. Without wet curing (no outside water):

2. When wet cured

1)1(2,1max

c

wcw

cw

= = 0,386

=1

3,1= 0,323

=0,386

0,386 + 0,323= 0,544

= ,, ( , )

= 0,852 not wet cured

= ,, ( , )

= 0,994 wet cured

Total amount of gel pores ?2 ways of calculating:

1)Vgh = 0,2 * * C (see exercise 3 for details)

2)

)1(6,0gwv

Formula 1:not wet cured: Vgh = 0,2 * 0,852 * 350 = 59,6 dm3

wet cured: Vgh = 0,2 * 0,994 * 350 = 69,6 dm3

OR

Formula 2:vgh = 0,6 x (1 - 0,544) x 0,852 = 0,233vgh = 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 + 135/1) = 57,79 dm3

0,272 x (350/3,1 + 135/1) = 67,4 dm3

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?

Maximum degree of hydration

1. Without wet curing (no outside water):

2. When wet cured

1)1()6,14,1(

max

csk

1)1()9,02,1(

max

csk

cs

cw

cw

s

w

c

w

csk

4,11

1

Amount of silica: s =0,07 * 350 kg/m3 = 24,50 kg/m3

Amount of cement: c = 350 kg/m3 – 24,5 kg/m3 = 325,5 kg/m3

wc

=135

325,5= 0,4147

=10003100

= 0,323

=10002200

= 0,455

sc

=24,5

325,5= 0, 07527

k =1

1 + 1,4 0,07527= 0,9047

=wc

wc + + s

c

=0,4147

0,4147 + 0,323 + 0,455 0,07527= 0,5372

Now we can calculate:

1)1()6,14,1(

max

csk

1)1()9,02,1(

max

csk

=0,5372

0,9047 (1,4 + 1,6 0,07527) (1 0,5372) = 0,844

= ,, ( , , , ( , )

= 1,012 = 1 because max 1

OR

And how much changes the volume of unhydrated cement?

)1()1(cvVolume fraction of unhydrated cement in problem 2:

Volume fraction of unhydrated cement with silica:)1()1(kvc

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 + 135/1) = 16,6 dm3 or0,003 * (350/3,1 + 135/1) = 0,7 dm3

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 dm3

Concrete admixturesWater-reducing admixtures

• water-reducing /plasticising admixtures• superplasticizers

Air-entraining agentsAccelerating admixturesRetarding admixturesWater-proofing admixturesOther admixtures

e.g. grouting admixtures (injektointiaine) , antibacterialadmixtures …

Accelerating and retardingadmixtures

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!

Accelerating admixtures aka accelerators areused to speed up concrete setting or the earlystrength development (hardening) whenconcrete is to placed at lower temperatures, in the manufacture of precast concrete or othersituations where a rapid removal of formworkis desired.

Air entraining agents

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

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.

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)?

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

The granulometric value of H (rakeisuusluku H) of the combinedaggregate 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

- slump 120 mm- 28 compressive strength 40 MPa- cement strength 48 MPa

From the mix design form:- Cement 355 kg/m3

- Aggregate 1840 kg/m3

- Water 178 kg/m3

- air 20 l/m3

a) The amount of cement stays the same, wateramount is 10 % smaller

Composition:

cement 355 kg/m3

water 178 - 0,1*178 160,2 kg/m3

air 20 l/m3

New water-air/cement -ratio:

, 0,51

From the mix design form we can read:Ks = 45 MPaThe original design strength was 42,5 MPa, thusTHE STRENGTH WOULD INCREASE BY 2,5 MPa

New amount of aggregate can be calculated byusing the basic equation of concrete:

+ + + = 1000

1000 – 355/3,1 – 160,2/1,0 – 20 = 705,3 dm3

705,3 *2,68 kg/m3 = 1890,2 kg/m3

b) Decrease the cement amount

cement ?? kg/m3

water 160,2 kg/m3

air 20 l/m3

We wan to keep the original strength so the water-air/cement –ratio stays the same

Original ,

New , ,

cement = 323 kg/m3

Cement is saved 355 – 323 = 32 kg/m3

The new amount of aggregate can be calculatedby using the basic equation of concrete:

1000 – 323/3,1 – 160,2/1,0 – 20 = 715,6 dm3

715,6 * 2,68 = 1918 kg/m3