M30 Material calculations; bricks 1m3, Plastering 1m3 material calculations

Bricks calculations :-

Brick size= 190mmX90 mmX90mm

Volume of 1 no. brick=0.001539

No. of bricks required 1 m3= 650 bricks

Plastering required:-

If we have to carry out plastering of 12 mm then volume of plastering for 1m2 area plastering is 1X1X0.012= 0.012 m3 of mortar is required.

Density of mortar is 2200 kg/m3

Ratio of mortar=1:6

Sum of proportion=7

Therefore cement required=1/7*2200=315 Kg

Sand=1885 Kg

This is for 1 m3 of mortar

Cement required= 0.012*315=3.78 Kg

Sand required= 0.012*1885=22.62 Kg

Weight calculations of Steel bars:-

If there is requirement of calculation of steel bars weight which has been installed during RCC works. Then below are the calculations:-

Formula for weight calculations= D^2/162.5

If there are 12 mm bars installed during RCC work having length of 3 meter then weight will be= ((12)^2/162.5)*3=2.65 Kg

Fe415 is uncommon name mostly we hear about Fe500 and Fe500D. Fe415 D and Fe415 S are also available where Fe415 means it has carbon content of 0.30% and Fe415D and Fe415S has maximum carbon of 0.25%.

Fe415 TMT has maximum 0.060% Sulphur content and Fe415D and Fe415S has maximum 0.045% content. Similarly for Phosphorous content. Max. Combined phosphorous and Sulphur content should be 0.110% while in Fe415S and Fe415D have Sulphur and Phosphorous content of 0.085%.

RCC Mixture ratios:-

M15 grade- Nominal Proportion will be 1:3:6 cement(bag)/cum=4.5 bags, sand(in cft)=16.04, Aggregate(in cft)= 32.07

M20 Grade- Nominal Proportion will be 1:2:4, cement(bag)/cum=6 bags, sand(in cft)= 14.98, aggregate(cft)=29.96

M25 Grade- Nominal Proportion will be 1:1.75:3.5, Cement(bag)/cum= 6.75, sand (in cft)= 14.75, Aggregate=29.14

M30 Grade- Nominal Proportion will be  1:1.5:3, Cement (bag)/cu,= 7.5-8 , sand (in  cft)=18.06, Aggregate(in Cft)=32.11

Density of cement= 1440 Kg/ Cum (Approx)

Dry loose bulk density of Sand= 1600 Kg/cum

Dry loose bulk density of aggregate= 1450 Kg/cum

Volume of 1 Kg of cement= 1/1440= 0.00069 cum

Volume of 50 kg bag= 50*0.00069=0.0347 cum

Now in M30 grade ration of sand with respect of cement 1:1.5 and for Aggregate is 1:3

So volume of sand required= 0.0347*1.5= 0.0521 cum

Volume of Aggregate required= 0.0347*3= 0.104 cum

Now Sand weight required= 0.0521*1600= 83.33 Kg

Weight of Aggregate required= 0.104*1450=151.04 Kg

So one bag of cement has to be mixed with 83.33 kg of sand and 151.04 kg of aggregate to produce M30 grade.

For producing 1 cum of M30 concrete quantity of cement, sand and aggregate required are as below:-

Density of concrete= 2400 Kg/cum

Weight of concrete produced with 1 bag of cement= 50+83.33+151.04+27.5= 311.87 Kg

So 1 bag of cement and other ingredients will produce =312/2400=0.13 Cum of concrete

Cement required for 1 cum= 1/0.13= 7.7 bags

Sand required= 83.33/0.13=641 Kg

Aggregate required= 151.04/0.13=1161.84 Kg

There is an alternative method of arriving the Kg required for Cement, Sand and aggregates:-

Dry weight to Wet weight of cement= 1.54

Now total aggregate of ratios= 1+1.5+3=5.5

So cement required =1.54*(1/5.5)*1440= 403.2 Kg

Each Bag weight= 50 Kg so no. of bags required= 8 bags

Similarly for Sand , sand weight required= 1.54*(1.5/5.5)*1600=672 Kg

Aggregate required= 1.54*(3/5.5)*1450=1218 kg