Composition of cement paste, concrete admixtures and mix design

of superplasticized concreteExercise 4

INPUT:• A concrete sample was extracted from a structure and dried at 105 °C in

which case 80 kg/m3 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 %.

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

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

Basic equation of concrete:

+ + + = 1000

, + , 0,6C + 30 = 1000

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

Cem : Agg : Water

Q = amount= density

QL = Air content (ilma)

W/C = Water Cement Ratio

How much of the water had been evaporated prior to drying?

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

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/m3 water evaporatedNon evaporable water consists of chemically combinedwater !

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= Hydr. degree

How much of the water had been evaporated prior to drying?

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

= Hydr. degree

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

Vgh = 0.28*Vgh + 0.28*Vgs0.72*Vgh = 0.28*Vgs

chemically bound water VN

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

v = density of water 1 kg/dm³VN = chemically bound water

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

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

and 135 kg/m3 respectively.

Q?Calculate the degree of hydration and amount of gel pores

a) without wet curing b) when wet cured.

Exercise 4/02

1)1(4,1max

Maximum degree of hydration

1. Without wet curing (no outside water):

2. When wet cured

1)1(2,1max

c

w

cw

cw

Initial porosity

= = 0,386

=1

3,1= 0,323

=0,386

0,386 + 0,323= 0,544

= ,, ( , )

= 0,852 not wet cured

= ,, ( , )

= 0,994 wet cured

c

w

cw

cw

Initial porosity

Total amount of gel pores ?

2 ways of calculating:

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

2)*)1(*6,0gwv

= Initial porosity(1 - ) = solid part of the pasteC = Cement

= Hydr. degree

Formula 1: Vgh = 0,2 * * C

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

*)1(*6,0gwv

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

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

Initial porosity (cem. + silica)

Coefficient based on the effect of (cem + silica)

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 c

sk4,11

1

Initial porosity (cem. + silica)

Coefficient based on the effect of (cem + silica)

cs

cw

cw

s

w

c

w

=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 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 admixturesiv. Miscellaneous admixtures include:

» Latexes» Corrosion inhibitors» Expansive admixtures

Concrete admixturesBeneficial effects of admixtures on concrete properties

Concrete admixtures

SP = Super plastisizerDCI = 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):

SuperplasticizerReduce 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 formsi. Solid powerii. Liquid --- 40% - 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 bleedingiii. entraining too much air.

Air-entraining admixtures• Entrained air:

– On purpose– Size: 50 to 200 m

• 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) KsKs = 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 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

INPUT- H (aggregates) 418- slump 120 mm- 28 compressive strength 40 MPa- Ks (Design strength) 42.5 MPa- cement strength 48 MPa

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

- Aggregate 1840 kg/m3

- Water + air 198 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

W/C–ratio 0.51

From the mix design form we can read:Ks = 45 MPa (from W/C 0.51) The 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

water amount is 10 % smaller

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