Composition of cement paste,
concrete admixtures and mix design
of superplasticized concrete
Exercise 4
1
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 (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
%. 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?
Mix design
1 : 6,0 : 0,6
3 % air
Basic equation of concrete:
𝑄𝑐
𝜌𝑐
+
𝑄𝑎
𝜌𝑎
+
𝑄𝑤
𝜌𝑤
+ 𝑄𝐿 = 1000
C
6∙C
+
+ 0,6C
3,1
2,67
C
1
6
+
+0,6
3,1 2,67
+ 30 = 1000
= 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 water
evaporated
Non evaporable water consists of chemically
combined water !
The amount of evaporable water should have been
WN = Chemically bound water = 0,25* α * C
Wo – WN
= 184 – 0,25*0,5*306
= 184 – 38
= 146 kg/m3
Prior to drying, water had evaporated:
146 – 80 = 66 kg/m3
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
0,25∗α∗C
= α∗C
+
0,75
∗
ρ
ρ
C
V
0,25∗0,5∗306
= 0,5∗306
+
0,75∗
3,1
1
= 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 (V 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
gs
Contraction pores Vcon= 0,25 * VN
= 0,25*0,25*α*C
= 9,6 l/m3
The amount of evoporable water consists of capillary
water 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/ρV = 50 l/m3
The total volume of the capillary pores Vcap
= Vo – VN – Vgh
VN = chemically bound water
VN = 0,25* α * 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.
Maximum degree of
hydration
1. Without wet curing (no
outside water):
 max 

1,4  (1   )
1
2. When wet cured
 max 

1,2  (1   )
1

w
c
w w

c c
w
c
=
135
350
= 0,386
ρv
1
=
= 0,323
ρc 3,1
0,386
𝜃=
= 0,544
0,386 + 0,323
αmax =
0,544
1,4∙(1−0,544)
= 0,852 not wet cured
αmax =
0,544
1,2∙(1−0,544)
= 0,994 wet cured
Total amount of gel pores ?
2 ways of calculating:
1)
Vgh = 0,2 * α * C
2)
v
gw
 0,6 (1   )  
(see exercise 3 for details)
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 = 0,233
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 + 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):
 max 

s
k  (1,4  1,6 )  (1   )
c

1
2. When wet cured
 max 

s
k  (1,2  0,9 )  (1   )
c
1
w
c
w w w s



c c  s c
k
1
s
1  1,4 
c
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
w
135
=
= 0,4147
c
325,5
ρv
1000
=
= 0,323
ρc
3100
ρv
1000
=
= 0,455
ρs
2200
s
24,5
=
= 0, 07527
c
325,5
k sil
1
=
= 0,9047
1 + 1,4 ∙ 0,07527
w
θ= w ρ c ρ s
+ w+ w∙
c
ρc
ρs c
0,4147
=
= 0,5372
0,4147 + 0,323 + 0,455 ∙ 0,07527
Now we can calculate:
 max 

s
k  (1,4  1,6 )  (1   )
c
𝛼𝑚𝑎𝑥
1
 max 

s
k  (1,2  0,9 )  (1   )
c
1
0,5372
=
= 0,844
0,9047 ∙ (1,4 + 1,6 ∙ 0,07527) ∙ (1 − 0,5372)
OR
αmax =
0,5372
0,9047∙(1,2+0,9∙0,07527)∙(1−0,5372)
= 1,012 = 1 because αmax ≤ 1
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:
νc = (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 + 135/1) = 16,6 dm3 or
0,003 * (350/3,1 + 135/1) = 0,7 dm3
Volume fraction of unhydrated cement with silica:
v  k  (1   )  (1   )
c
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 dm3
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 …
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!
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
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) Ks
Ks = 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
- slump
- 28 compressive strength
- cement strength
From the mix design form:
- Cement
- Aggregate
- Water
- air
120 mm
40 MPa
48 MPa
355 kg/m3
1840 kg/m3
178 kg/m3
20 l/m3
a) The amount of cement stays the same, water
amount is 10 % smaller
Composition:
cement
water 178 - 0,1*178
air
355 kg/m3
160,2 kg/m3
20 l/m3
New water-air/cement -ratio:
160,2+20
355
≈ 0,51
From the mix design form we can read:
Ks = 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:
𝑄𝑐
𝜌𝑐
𝑄𝑎
+
𝜌𝑎
+
𝑄𝑤
𝜌𝑤
+ 𝑄𝐿 = 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
?? kg/m3
160,2 kg/m3
20 l/m3
cement
water
air
We wan to keep the original strength so the waterair/cement –ratio stays the same
Original
178+20
=0,558
355
New
160,2+20
=0,558
cement
→ cement = 323 kg/m3
Cement is saved 355 – 323 = 32 kg/m3
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 dm3
→ 715,6 * 2,68 = 1918 kg/m3