Question

A concrete mix design is needed for a residential street paving project in Detroit, Michigan. Pavement thickness is 8”, unreinforced, subjected to freezing and thawing and deicing agents are anticipated. The specified compressive strength is 4000 psi. Assume both Fine Aggregate (FA) and Coarse Aggregate (CA) are in saturated surface dry condition.

Specific Gravity of Fine Aggregate (FA) = 2.71

Fineness Modulus (FM) of Fine Aggregate (FA) = 2.70

Specific Gravity of Coarse Aggregate (CA) = 2.65

Bulk Density of Coarse Aggregate (CA) = 105 pcf

Nominal Maximum Size of Coarse Aggregate = ¾”

Specific Gravity of Cement = 3.15

Fill in ALL boxes in the tables below:

Slump (Table 9-6)	3''
Maximum size of aggregate (Page 152, 2nd Paragraph)	3/4
Air entrainment (Yes/No)	yes
Air content (Table 9-5)	6%
Required Compressive Strength (Table 9-11)	5200
w/c ratio (Table 9-3)	0.42

Material	Weight (lbs)	Volume (ft3)
Water (Table 9-5)	295
Cement
CA (Table 9-4)	1956.15
FA
Air
Total

Answer 1

Ans) Let the total volume of trial mix be 1 cubic yard (27 )  then according to ACI 211.1-91 , Table 6.3.1 slump required for pavement and slab is 1 to 3 in

Ans) Maximum size of aggregate is equal to nominal maximum size = 3/4 in

Ans) Since, concrete is exposed to freezing and thawing and deicing agents, yes air entrainemet should present

Ans) According to ACI 211.1-91 ,table 6.3.3 for severe exposure , recommended air content is 6%

Ans) According to ACI 318, in the absence of statical data, required compressive strength , f'cr

f'cr = f'c + 1200

where, f'c is specified compressive strength = 4000 psi

=> fcr = 4000 + 1200

=> fcr = 5200 psi

Ans) According to ACI 211.1-91, table 6.3.4(a) , for required compressive strength of 5200 psi, required w/c ratio is 0.38

Ans) According to ACI 211.1-91, table 6.3.3 , for 3/4 in aggregate and 3 in slump , amount of water required per cubic yard of concrete is 305 lb/cy or 4.88 cu.ft

Ans) Amount of cement = Amount of water / w-c ratio

=> Amount of cement required = 305 lb / 0.38 = 802.6 lb

Volume of cement = Weight / (Specific gravity x 62.4) = 802.6 / (3.15 x 62.4) = 4.08 cu.ft

Ans) Now, according to ACI 211.1-91 table 6.3.6 , for nominal aggregate size of 3/4 in and fineness modulus of 2.7, volume of coarse aggregate is 0.63 cy or 17.01 ft3

=> Amount of coarse aggregate = Dry rodded density x volume

=> Amount of coarse aggregate = 105 x 17.01 = 1786 lb

Ans) Volume of fine aggregate(sand) = Total volume of concrete - Volume of water,cement,coarse aggregate and air

=> Fine aggregate volume = 27 - [4.88 + 4.08 + (1728/ 2.65 x 62.4) + 0.06(27)]

=> Fine aggregate volume = 27 - 21.03 = 5.97 ft3

=> Amount of fine aggregate = Volume x Specific gravity x Water unit weight = 5.97 x 2.71 x 62.4 = 1009.5 lb

Hence, composition for trial mix for 1 cy concrete is as follow :

Material	Weight (lb)	Volume (ft3)
Water	305	4.88
Cement	802.6	4.08
Coarse aggregate	1786	10.45
Fine aggregate	1009.5	5.97
Air	-	1.62