Cement volume is 0.25m3, Sand volume is 0.25m3, Coarse aggregate volume is 0.5m3. Concrete grade is M25.(mix proportion is 1:1:2). What will be the concrete volume in m3?

Dry shrinkage or drying shrinkage occurs due to loss of moisture in the concrete. There are different types of water in a hydrated cement paste. Dry shrinkage occurs due to the loss of capillary water. This water can be found in voids of sizes ranging from 5 nm to 50 nm in the hydrated cement paste.Read more

Dry shrinkage or drying shrinkage occurs due to loss of moisture in the concrete. There are different types of water in a hydrated cement paste. Dry shrinkage occurs due to the loss of capillary water.

This water can be found in voids of sizes ranging from **5 nm to 50 nm **in the hydrated cement paste. Due to the small size of the voids, the water is held inside by **surface tension or capillary tension**. Due to the bond created by the surface tension, evaporation, or any other form of removal of this water causes significant volume changes to the structure. This volume changes causes dry shrinkage to the structure.

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## Gopal Mishra

Volume of fully compacted concrete is given by following formula: where Vc is the volume of concrete. W, C, Fa and Ca are the masses of water, cement, fine aggregates and coarse aggregates respectively. Density of cement = 1500 kg/m3, fine aggregates = 1700 kg/m3 and coarse aggregates = 1650 kg/m3.Read more

Volume of fully compacted concrete is given by following formula:

where V

_{c}is the volume of concrete. W, C, F_{a}and C_{a}are the masses of water, cement, fine aggregates and coarse aggregates respectively.Density of cement = 1500 kg/m

^{3}, fine aggregates = 1700 kg/m^{3}and coarse aggregates = 1650 kg/m^{3}.So, Weight of cement C = 0.25 x 1500 = 375kg

Weight of Fine aggregates F

_{a}= 0.25 x 1700 = 425 kgWeight of Coarse aggregates C

_{a}= 0.5 x 1650 = 825 kgConsider w/c ratio as 0.45, so weight of water = 0.45×0.25×1000 = 112.5 L or kg

S

_{c}, S_{fa}and S_{ca}are the specific weight of cement, fine aggregates and coarse aggregates which are 3.15 for cement and 2.6 for fine and coarse aggregates.Therefore, putting these values in the formula above, we get

V

_{c}= (112.5/1000)+(375/(1000×3.15))+(425/(1000×2.6))+(825/(1000×2.6)) = 0.712m^{3}So, Volume of concrete = 0.712m

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