Concrete Mix Design as per IS 10262:1982.

 Concrete mix design deals with following steps.

Step 1 -

The target mean strength is calculated.

        ft = fck + K x S

where, ft = Target mean strength

           fck = The strength of concrete mix which we want to design

          K = 1.65 

          S = standard deviation

Step 2 -

The value of w/c ratio corresponding to target mean strength is taken from the graph.

Step 3 -

Maximum water content is taken from IS 10262:1982 corresponding to nominal size of coarse aggregate and slump 25 to 50 mm.

• Value of water content is increased by 3% for each additional 25 mm slump and reduced by 5-10% for plasticizer and 20% for superplasticizer.
• Requirement of water reduces with increase in normal size of aggregate.

Step 4 -

The quantity of cement is calculated.

    wt. of cement = (wt. of water)/(w/c ratio)

This value should not be less than minimum cement required corresponding to exposure condition.

Step 5 -

Proper proportion of coarse aggregate in total aggregate is taken from IS 10262 for w/c ratio 0.5.

For w/c ratio other than 0.5 above value are modified as follows -

• +0.01 in proportion for each -0.05 w/c ratio.
• -0.01 in proportion for each +0.05 w/c ratio.

Step 6 -

Weight of fine aggregate and coarse aggregate is calculated as follows :-

(a) total volume of concrete = 1 m^3

(b) vol. of concrete mass = 1 - air content

(c) vol. of cement = (wt. of cement)/(unit weight of cement)

(d) vol. of water = (wt. of water)/(unit weight of water)

(e) vol. of admixture = (wt. of admixture)/(unit wt. of admixture)

(f) vol. of aggregate = (b) - (c+d+e)

• vol. of coarse aggregate = (proportion) x ( f )
• wt. of coarse aggregate = (proportion) x (f) x (unit weight)
• vol. of fine aggregate = (1- proportion) x (f) 
• wt. of fine aggregate = (1- proportion) x (f) x (unit weight)

Thank you.