July 19, 2016

HomeHOW TOHow To Calculate the Quantity of Materials (Sand and Cement) Required for 1 Cubic Meter of Cement Mortar

# How To Calculate the Quantity of Materials (Sand and Cement) Required for 1 Cubic Meter of Cement Mortar

Sometimes on site, we are faced with the problem of calculating for the amount of materials (sand and cement) required in plastering or rendering! This Post is made to tackle this issue!

Today, i want to talk about how you can calculate the quantity of sand and cement required in Mortar

As we all know, Mortar used in Plastering is mixed or produced in different proportions which could include: 1:1, 1:2, 1:3, etc. But here i would like to adopt the 1:6 proportion and do analysis based on this proportion.

**QUANTITY OF CEMENT AND SAND NEEDED IN BRICKLAYING OR RENDERING (Plastering)**

For the

**calculation of cement mortar**, lets Assume we are to mix for a cubic meter (1m^{3}): I would then provide the steps on calculating these below:**STEPS TO CALCULATE THIS IS FOUND BELOW**

1. Calculate the dry volume of materials required for 1m

^{3}cement mortar. Considering voids in sands, we assume that materials consists of 60% voids. That is, for 1m^{3}of wet cement mortar, 1.6m3 of materials are required.2. Now we calculate the volume of materials used in cement mortar based on its proportions.

Letâ€™s say, the proportion of cement and sand in mortar is 1:X, where X is the volume of sand required.

Then, the volume of sand required for 1:X for 1m

^{3}Cement Mortar will be3. The Volume of Cement required will be calculated as:

Since the volume of 1 bag of cement is 0.0347 m

^{3}, so the number of bag of cement will be calculated as:Therefore, if we were to render a cubic meter of Mortar (1m3) with a mix ratio 1:6 (Meaning One Bag Of Cement for 6 Parts Of sand),

The quantity of sand needed here would be

= 1.371m

= 1.371m

^{3}(Volume of sand needed)Lets convert this in

**Headpans and Wheelbarrow**(For Small projects)Volume of 1 headpan =0.0175m

^{3}Therefore there will be 1.371/0.0175 = 78.5 Headpans

But 4 Headpans = 1 wheelbarrow

Therefore 78.5/4 =

**19.6 or 20 wheelbarrows****Quantity of cement**= = 0.22857m

^{3}

Now lets find the number of

**Bags of Cement**since we have known the total Volume required for 1m^{3}(NOTE:

**Volume of 1 bag of cement = 0.0347m**)^{3}But total volume for 1m

^{3}is 0.22857m^{3}Therefore, the number of bags would be

BAGS = Calculated volume/fixed Volume Per Bag Of Cement = (0.22857/0.0347)

Bags = 6.58

So about 6.6 bags would be required for a 1m

Please feel free to Share with Friends and criticize this work as it will help me make it better for other readers

^{3}of mortar along with 20 wheelbarrow of sharp sand or 78.5 Head pans of sharp sand!Please feel free to Share with Friends and criticize this work as it will help me make it better for other readers

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#### About Author

##### MrPinner

Pro Nigerian Blogger, freelancer with Higher National Diploma in Civil Engineering Technology. I help people build their homes without wasting money

One Comment

Good job. Educating. More grace.