Quantities of Cement Sand and Aggregate in Cubic Meter, Kg, Cubic Feet for M25, M20, M15, M10, M7.5, M5

Hello, friends in this article you will learn to calculate the Quantity of Cement, Sand, and Aggregate In Kilograms, Cubic Meter and Cubic Feet. Also, you find the Table where you will get the Quantity of Cement, sand, and aggregate of different Grade of Concrete Such as M 5, M 10, M 15, M 20, and M 25. As we know that the Quantity of Cement is measured in bags so in this post you will also learn to calculate the Quantity of cement in bags and also get the table where you find the Number of cement bags in different Grade of Concrete. For easy understanding, I have calculated the Quantity of Material for the M 15 Grade of Concrete.

Let’s Find

Volume of Concrete Work =1 cum .

Note:  for Concrete work: Dry volume =wet volume x 1.54

Consider,

Grade of concrete (M 15) =1:2:4

Where 1= Part of cement, 2 = Part of sand, 4 = Parts of aggregates

Sum of Ratio = 1+2+4 =7

Dry volume = 1 x 1.54 m3 = 1.54 m3

Quantity of Cement, Sand & Aggregate in Cubic Meter

Cement Quantity in Cum

Quantity of cement = Dry Volume x (Ratio of cement / Sum of Total ratio)

                                              = 1.54 x (1/7) =0. 22 m3

Quantity of Sand in Cum

Quantity of sand in cum = Dry Volume x (Ratio of sand/ Sum of Total ratio)

                                             =1.54 x (2/7) =0.44 m3.

Quantity of Aggregate in Cum

Quantity of aggregate in cum = Dry Volume x (Ratio of aggregate / Sum of Total ratio)

=1.54 x (4/7) =0.88 m3

Quantity of Cement Sand and Aggregate in cum for 1 cum Concrete Work

S. No.Grade of ConcreteCement (Cum)Sand (Cum)Aggregate (Cum)
1M 25(1:1:2)0.390.39 Cum0.77 Cum
2.M 20 (1:1.5:3)0.280.42 Cum0.84 Cum
3.M 15 (1:2:4)0.220.44 Cum0.88 Cum
4.M 10 (1:3:6)0.150.46 Cum0.92 Cum
5.M 7.5 (1:4:8)0.120.47 Cum0.95 Cum
6.M 5 (1: 5:10)0.100.48 Cum0.96 Cum

Quantities of Cement Sand and Aggregate in Kilograms

Cement Quantity in kg

Quantity of cement in cum = Dry Volume x (Ratio of cement / Sum of Total ratio) x Density of Cement

We Know the Density of Cement = 1440 Kg/ m3.

= 1.54 x (1/7) x 1440 = 316.8 kg

Quantity of Cement in Bags

Weight of 1 bags of Cement = 50 kg

Quantities of cement in bags = (316.8/ 50) =6.336 bags (1 bags = 50 kg).

 Quantity of Sand in Kg

Quantity of sand in kg = Dry Volume x (Ratio of sand/ Sum of Total ratio) x Density of Sand

The Density of Dry Clean Sand as per IS 875 Part-1 =1540 -1600 Kg/m3

Quantity of sand in kg = 1.54 (2/7) x 1600 =704 kg

Aggregate Quantity in Kg

Quantity of aggregate in kg = Dry Volume x (Ratio of aggregate/ Sum of Total ratio) x Density of aggregate

The Density of Dry Aggregate as per IS 875 Part-1 =1600-1870 Kg/m3

Quantity of aggregate in kg = 1.54 (4/7) x 1870 =1645.6 kg

Quantities of Cement Sand and Aggregate in Kg for 1 cum Concrete Work

Grade of ConcreteCement (Kg)Cement (Bags)Sand (Kg)Aggregate (Kg)
M 25(1:1:2)554.4 Kg11.09 Bags6161439.9
M 20 (1:1.5:3)403.2 Kg8.06 Bags6721570.8
M 15 (1:2:4)316.8 Kg6.34 Bags7041645.6
M 10 (1:3:6)221.76 Kg4.44 Bags739.21727.88
M 7.5 (1:4:8)170.58 Kg3.41 Bags758.151772.18
M 5 (1: 5:10)138.6 Kg2.77 Bags7701799.88

Quantities of Cement Sand and Aggregate in Cubic Feet

Quantity of Cement in Cubic Feet

If we multiply the Number of cement Bags by 1.226, then we get the value in Cubic Feet.

Number of Cement Bags = 6.336 Bags

                                              = 6.336 x 1.226 = 7.77 ft3

Quantity of Sand in Cubic Feet

Quantity of sand in cft = Dry Volume x (Ratio of sand/ Sum of Total ratio) x 35.3147

                                             =1.54 x (2/7) x 35.3147 = 15.54 ft3.

Quantity of Aggregate in Cubic Feet

Quantity of aggregate in cft = Dry Volume x (Ratio of aggregate / Sum of Total ratio) x 35.3147

=1.54 x (4/7) x 35.3147 =31.08 ft3

S. No.Grade of ConcreteCement  (CFT)Sand (CFT)Aggregate(CFT)
1M 25(1:1:2)13.5913.627.19
2.M 20 (1:1.5:3)9.8914.8329.66
3.M 15 (1:2:4)7.7715.5431.08
4.M 10 (1:3:6)5.4416.3232.63
5.M 7.5 (1:4:8)4.1816.7333.47
6.M 5 (1: 5:10)3.41733.99

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