Question

In: Civil Engineering

Example of Mix design proportion calculation • You need to design the mix proportion of a...

Example of Mix design proportion calculation
• You need to design the mix proportion of a grade 40 for a column with workability of 50
mm slump and w/c not more than 0.55. The maximum size of gravel aggregate is 10 mm
and the fine aggregate is quarry blast sand. The moisture content of the aggregate is as
follow:
• The gravel aggregate is in moist condition with moisture content of =3.5%, dry density
= 1700 kg/m3
• The sand is in moist condition with saturated surface dry density = 2600 kg/m3
. Percent
passing at 600m =35%,water absorption = 0.5%,
• Aggregate relative combined density (saturated surface dry) = 2.60
• Cement used is Sulfate Resistance because part of column is in soil identified to contain
sulfate

Solutions

Expert Solution

Ans) Let the total volume of trial mix be 1 m3 then according to ACI 211.1-91 table 6.3.3 for 50 mm slump and nominal aggregate size of 10 mm , amount of water required per cubic meter of concrete is 207 kg

=> Amount of water = 207 kg per m3 concrete

Since ,air content for 10 mm aggregate size is 3%

Given water cement ratio = 0.55

Hence, amount of cement = 207 / 0.55 = 376.36 kg

Now, according to table 6.3.6 for nominal aggregate size of 10 mm and 35% passing 600 micron , volume of coarse aggregate is 0.48 m3

=> Amount of coarse aggregate = dry density x volume = 1700 x 0.48 = 816 kg

Volume of fine aggregate = Total volume of concrete - Volume of water, cement, coarse aggregate and air

=> Fine aggregate volume = 1 - [(207/1000) + (376.36 / 3.15 x 1000) + (816/ 2.6x 1000) + 0.03]

=> Fine aggregate volume = 1 - 0.67 = 0.33 m3

=> Amount of fine aggregate = volume x density = 0.33 x 2600 = 858 kg

Now, since aggregates has moisture and has absorption capacity , amount of mixing water needs to be corrected

Water provided by coarse aggregate = 0.035 x 816 = 28.56 kg

Water absorbed by fine aggregate = - 0.005 x 858 = - 4.29 kg

=> Actual amount of water to be added = 207 - 28.56 + 4.29 = 182.73 kg

Hence, composition for trial mix for 1 m3 concrete is as follows :

Component Amount (kg)
Water 182.73
Cement 376.36
Coarse aggregate 816
Fine aggregate 858

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