How much coarse aggregate do I need for 1m3 concrete?
Table of Contents
How much coarse aggregate do I need for 1m3 concrete?
The quantities of materials for 1 m3 of concrete production can be calculated as follows: The weight of cement required = 7.29 x 50 = 364.5 kg. Weight of fine aggregate (sand) = 1.5 x 364.5 = 546.75 kg. Weight of coarse aggregate = 3 x 364.5 = 1093.5 kg.
How many bags of cement will cast?
Unit weight of cement is 1440 kg per cubic meter. one bag contains 50 kg. so approximately 29 bags per cubic meter (1440/50=28.8) .
How many bags of sand are in a cubic meter?
Bulk bag of sand:- a bulk bag of sand yields volume about 0.5 cubic metres (800/ 1600 = 0.5), 17.66 cubic feet (0.5 ×35.32 = 17.66) or 31.25 Liters (50÷1.6 = 31.25) and it require 32 bags of sand to make 1 cubic meter of sand.
How much cement do I need for 1 cubic meter concrete?
Example calculation Estimate the quantity of cement, sand and stone aggregate required for 1 cubic meter of 1:2:4 concrete mix. Ans. Materials required are 7 nos. of 50 kg bag of cement, 0.42 m 3 of sand and 0.83 m 3 of stone aggregate.
What is the volume of sand required for 50 kg of cement?
Volume of 01 bag (50 kg) of cement = 50 X 0.000694 = 0.035 cubic meter (cum) Since we know the ratio of cement to sand (1:2) and cement to aggregate (1:4) Volume of Sand required would be = 0.035*2 = 0.07 cubic meter (cum) Volume of Aggregate required would be = 0.035*4 = 0.14 cubic meter (cum)
How many bags of cement do I need to build a fence?
1 Cement 2 Sand 3 Aggregate (gravel) is too strong. 16bags (costly) 1 Cement 2.5 Sand 4 Aggregate is OK for fence posts. 13bags of cement per cubic metre. The more cement the more costly of course so don’t over do it. Note: These are 20kg bags of General Purpose cement.
What is the ratio of cement to aggregate in concrete mix?
Considering a concrete mix proportion (by volume) of 1:2:4 i.e., Cement: Fine aggregate (Sand) : Coarse Aggregate is in the ratio of 1:2:4 by volume. Cement required = 01 bag = 50 kg ~ 36 liters or 0.036 cum Fine Aggregate required = 2*0.036 ~ 0.07 cum = 0.072*1600 = 115 kg