Every time we calculate the concrete or cement mortar calculation, We use a 1.54 constant to get a dry volume of ingredients.

But why exactly 1.54 why not 1.75? What is the reason behind that?

Its simple.

**Adding Bulking of Sand & Void Ratio Compressibility**

**1. Bulking of Sand â€“** Up to the M25 **grade of concrete**, we make nominal mixes by volumetric measurement. In the design mix, we weigh the materials.

If the moisture is present in the sand, then it makes the sand look bulkier, which could result in inadequate sand proportion in the concrete ratio.

**For Example,**Â If we need to add 1m^{3}Â sand in the concrete mix ratio, we take 1.3 m^{3 }(**30% more**). The reason for that is the moisture content present in the sand makes it a little bulkier. The 5% to 8% of surface moisture will increase the bulking of sand up to 20% to 30%. When we add more water (more than 8%) to the sand, the thin film will disappear, and volume decreases.

**2. Void Ratio â€“** If you fill the shuttering area with blue metal alone it has lots of voids in it which also need to be filled by the other materials such as sand & cement. Such void gap is known as **void ratio compressibility** which is 20% for coarse aggregate.

So if we need 1 m3 of concrete in the wet condition we have to take consideration of the bulking of sand & void ratio compressibility for both fine & coarse aggregate (30%+20%+4%) = 54%

Adding 4% as wastage it comes as 54%.

Thus every time you need to calculate the volume of concrete you need to add the above percentage with dry volume

Wet Volume of concrete = Dry Volume of concrete + 54% of the dry volume of concrete or

Wet Volume of concrete = Dry Volume of concrete x 1.54

So whatever volume we get from the above formula, we need to multiply the value with 1.54%

So to get 1 m3 of wet concrete, we have to multiply the dry volume by 1.54.

Consider M20 Mix Ratio (**1 : 1.5 : 3)**

So total part of mortar = 1+1.5+3 = 5.5

Therefore,

- Required Volume of Cement = 1/5.5 x 1.54 (Bulkage & Wastage) x Mortar Volume

Â = 1/5.5 x 1.54= 0.28 m^{3}

To convert into cement bagsÂ â€“ (0.28 x 1440)/50 = 8.06 Bags

Unit weight of Cement â€“ 1440 Kg/m^{3}Â Â 1 bag cement â€“ 50 Kg

- Required Sand Volume = 1.5/5.5 x 1.54 (Bulkage & Wastage) x Mortar Volume

= 1.5/5.5 x 1.54Â = 0.42 m^{3}

- Required Volume of Coarse Aggregate = 3/5.5 x 1.54 (Bulkage & Wastage) x Mortar Volume

= 3/5.5 x 1.54Â = 0.84 m^{3}

Happy Learning ðŸ™‚

## 1 Comment

Thank You for Guidance on civil engineering basics knowledge