In this post, you will learn to calculate many bags of cement sand and gravel are in 1 cubic meter? Have you ever thought about why does the quantity of sand and aggregate in concrete add up to be more than the quantity of concrete itself?
The volume of dry material required for making concrete is always higher than the volume of finished hardened concrete. This is because the placement of wet concrete shrinks after hardening.
This means that the volume of dry concrete is lesser than the volume of wet concrete. so some shrinking factor is required to be incorporated. Also during mixing and placing wastage of material like cement sand aggregate is there.
Another factor is that placed concrete is well-compacted concrete thus the volume of air in finished concrete is less.
However, the dry material contains a higher amount of entrapped air e.g, in dry sand. After mixing and placement volume of these materials reduces. From a safety point of view factor of 1.54 to 1.57 is required.
This means that for 1 cum of hardened concrete, 1.54 to 1.57 cum of wet concrete is required.
Steps in Calculating Quantity of Cement, Sand and Coarse Aggregate in 1 Cubic Meter of any given grade of Concrete
Choose the grade of concrete.
The grade of concrete is defined by its compressive strength. M5, M7.5, M10, M15, and M20 are nominal grades of concrete. For these grades, concrete design is not required. Such grade can be obtained by choosing the required ratio of Cement:sand: aggregate.
M25, M30, M35, M40, M45 are standard grades of concrete, and M 50, M55, M60, M65, M70 are high-strength concrete grades.
Add the ratio of Cement, Sand, and Aggregate
Suppose we have chosen the M 20 Concrete of ratio 1:1.5:3 which means 1 part cement 1.5 part sand and 3 part coarse aggregate. The next step is to add these parts 1+1.5+3 = 5.5. 1 Part of cement + 1.5 Part of Sand + 3 Part of Coarse aggregate = 5.5 Parts.
Calculate dry volume of material required
For 1 Cum of wet concrete, 1.54 to 1.57 cum of dry concrete material is required. Thus shrinking factor of 1.54 to 1.57 is used.
Calculate Quantity of Cement
Total cement required = (1/5.5)x 1.57. Out of 5.5 Parts of total concrete 1Part is the Cement (thus 1/5.5). This one part has to be multiplied with the shrinking factor (1.54 to 1.57) which will give the total cement required in cum.
Calculate Quantity of Sand
Total Sand Required = 1.5/5.5 x 1.57. Out of a total of 5.5 Parts, 1.5 parts are of Sand which is multiplied by the shrinking factor to account for the shrinking of the dry volume of concrete on becoming wet or mixing with water.
Calculate Quantity of Coarse aggregate
Total Coarse aggregate required= 3/5.5 x 1.57. In ratio 1:1.5:3, 3 part is of the coarse aggregate. Thus the total quantity of coarse aggregate is obtained.
Let us understand it through example.
How Many Bags of cement sand and gravel are in 1 cubic meter for M20 Concrete (1:1.5:3)
For M 20 Concrete = (1:1.5:3) sum = 1+1.5+3 = 5.5
A shrinking factor or safety factor of 1.54 to 1.57 can be used. You can choose any value between 1.54 to 1.57.
Total cement required = 1/5.5 x 1.57 =0.285 cum =0.285/0.0348 = 8.20 bags of cement.
Quantity of sand = 1.5/5.5 x 1.57 = 0.428 cum
Quantity of coarse aggregate = 3/5.5 x 1.57= 0.856 cum
Calculations have been summarised in the table below:-
|Type of Concrete||Cement (Cum)||Cement (bags)||Sand (cum)||Aggregate (cum)|
|M 15 (1:2:4)||0.2225||6.4||0.445||0.89|
|M 10 (1:3:6)||0.1567||4.4||0.470||0.89|
|M 7.5 (1:4:8)||0.1175||3.4||0.470||0.89|
|M 5.0 (1:510)||0.094||2.6||0.470||0.89|
How Many Bags of cement sand and gravel are in 1 cubic meter for M15 Concrete (1:2:4)
Let us consider another example. This time we will calculate the material in Cubic Foot.
For M 15 Concrete sum of the ratio =1 + 2+ 4 =7
Shrinking facto = 1.54 to 1.57 is used.
Vol of Cement in cubic foot = 1/7 x (1.57×35.31) (1cum = 35.31 cubic foot)
= 7.92 cubic foot
Density of Cement = 1440 Kg/Cum = 40.78 Kg/cft
Weight of cement = 7.92 x 40.78 =322.98 Kg =6.45 Bags
Quantity of sand = (2 /7)x (1.57×35.31) = 15.83 cft
Quantity of Coarse aggregate = (4/7) x (1.57×35.31) =31.68 cft
|Type of Concrete||Cement (cft)||Cement (bags)||Sand (cubic foot)||Aggregate (cubic foot)|
|M 15 (1:2:4)||7.86||6.4||15.71||31.43|
|M 10 (1:3:6)||5.53||4.4||16.60||31.43|
|M 7.5 (1:4:8)||4.15||3.4||16.60||31.43|
|M 5.0 (1:510)||3.32||2.6||16.60||31.43|
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