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ECE 302 Fall 2017
HOMEWORK #3
Due October 20, 2017
Portland Cement Concrete (67 points)
HW3-1. Design the concrete mix according to the following conditions. Please use ACI method
which has been taught in our class. Use our handout. The sentence that “only air entrainer is
allowed’ indicates “mild exposure condition.” (20 pts)
Answer
Solution
Sample Computation Conditions
Components S.G. F.M A.C. (%) MC (%)
Max. Size
(mm)
Dosage
(g) per kg
of cement
Type V Air-entraining cement 3.00
Fly ash 2.60
Slag 2.90
Coarse Aggregate 2.68 0.5 2.0 25
Fine Aggregate 2.64 2.8 0.7 6.0
Air entrainer 0.5
Retarding water reducer 3
Plastizer 30
Shrinkage reducer 15
28-day Compressive Strength 40.0 Mpa
Standard deviation 2.0 MPa
DRUW of Coarse Aggregate 1600
kg/m3
Environmental Condition Severe cold weather, high sulfate soil
Air content
Mix Design Proceduere
Step 1. Check condition (requirement)
From Tables 1 & 2, min. compressive stength: 35 Mpa
From Tables 1 & 2, max. w/c = 0.4
For a standard deviation of 2.0 Mpa, the f’cr must be grater of 43.5 Mpa
f’cr = f’c +1.34S = 40+1.34(2) = 42.7 Mpa or
f’cr = 0.9f’c +2.33S = 36+2.33(2) =
40.7 Mpa
Therefore, f’cr = 42.7 MPa
Step 2. Determine slump
Slump: Pavement & Slabs (Table 4) = 25 ~75 mm
Thus, min. slump value should be choosen. 75 mm
Step 3. Determine Norminal Max. Size of Aggregate
Nominal MSA = 19 mm
Step 4. Estimate mixing water and air content
From Table 5, air content: severe expsoure condition = 6% + 2% =
8 %
From Table 5, water content based on slump (75-100mm) = 184
kg/m3
Because crushed subangular rock reduces the water contents to 10
kg/m3
water content = 184-10 = 174
kg/m3
In addition, retarding water reducer plus super plasticizer will reduce water demand by 10%.
additional water reduction content = 174*0.10 = 17.4
kg/m3
Thus, the estimated water content = 174 – 17.4 = 156.6
kg/m3
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ECE 302 Fall 2017
Step 5. Select water to cement ratio
From table 6, the recommended w/c for an f’c = 43.5 Mpa is interpolated.
45 0.3
42.7 x
40 0.34
Thus, x = 0.32
Step 6.Estimate cement content
w/c = 0.3184; c = 156.6/0.3184 = 491.8
kg/m3
Fly ash: 15% of 556.1 = 73.8
kg/m3
Slag: 30% of 556.1 = 147.6
kg/m3
So, cement: 556.1 – 72.1 – 144.2 = 270.5
kg/m3
Step 7. Estimate coarse aggregate content
From Table 7, the bulk volume of coarse aggregate recommended when using sand with FM of 2.8 is
0.62
m3
DRUW = 1600
kg/m3
Thus, WtOD = 1600 x 0.62 = 992.0
kg/m3
Step 8. Calculate fine aggregate content
Vol.fine aggr. = 1 – [Vwater + Vcement + Vfly ash + Vslag + Vcoarse aggr. + Vair]
Water = 156.6/(1*1000) = 0.157
m3
Cement = 270.5/(3.0*1000) = 0.164
m3
Fly ash = 73.8/(2.60*1000) = 0.028
=
m3
