Category: Computations of Quantities of Materials Required for Various Items of Works

Mangalore Tiled Roof

To calculate the number of Mangalore tiles and battens required to cover an area of 10 sq.m

Size of Mangalore tiles = 41 cm x 24 cm

and it can cover an area of 32 cm x 21 cm

Therefore, Area covered by one tile with overlap = 32 cm x 21 cm = 0.32 sq.m x 0.21 sq.m

= 0.672 sq.m

Therefore, Total number of tiles required = (Area to be covered / Area of one tile with overlap)

= 10 / 0.0672 = 148.81 = 149 Nos.

Length of the ridge tile = 40 cm

The size of teak wood battens is equal to 5 cm x 2.5 cm and are usually fixed at 30 cm c/c

Therefore, the length of the battens required = 36 m

Floor Finishes

Item: Plain cement tiles 20 mm thick over cement mortar (1:6) screeding, including cement float, etc.

The quantities of plain cement tiles, cement and sand required for flooring 10 sq.m area will be determined as follows:

Plain cement tiles including 5% wastage = 10.5 sq.m
Quantity of mortar for screeding, assuming 20 mm thickness

= 10 (sq.m) x 2 / 1000 (m) = 0.2 cum.

Quantity of Cement required for (1:6) proportion = 0.26 / (1+6) = 0.037 cum.

= 0.037 x 30 = 1.11 bags of cement

and Quantity of cement required for float, assuming 1.5 mm thickness = 10 x 0.0015

=0.015 cum.

=0.015 x 30 = 0.45 bags of cement

Therefore, Total quantity of cement required = 1.11 + 0.45 = 1.56 bags

and, Quantity of sand required = [0.26/(1+6)] x 6 = 0.222 cum. = 0.25 cum.

Pointing

The volume of dry mortar required for pointing depends upon the type of surface i.e. either brick work or masonry work.

For pointing ( which may be flush, struck or keyed) to the brick work, the dry volume of mortar , including wastage, required for 10 sq.m area is about 0.036 cum. and for random rubble masonry it is 0.076 cum.

Thus knowing the proportion of the mix of the mortar as 1:3, the quantity of cement and sand can be found out as follows:

For pointing to brick work

Volume of cement required = 0.036 / (1+4) = 0.009 cum.

= 0.009 x 30 = 0.270 bags

and, Volume of sand required = [0.036 / (1+3)] x 3 = 0.027 cum.

For pointing to random or coursed rubble masonry, the volume of dry mortar required = 0.076 cum.

Therefore, For pointing in cement mortar (1:3) proportion,

Quantity of cement required = 0.076 / (1+3) = 0.019 cum.

= 0.019 x 30 = 0.57 bags = 0.60 bags of cement

and, Quantity of mortar required = [0.076 / (1+3)] x 3 = 0.057 cum.

Plastering

To determine the quantity of cement (or lime) and sand, required for plastering unit square metre of the area of various thickness is to work out the volume of mortar required per sq.m of plaster by multiplying the area to be plastered by its thickness.

In order to allow extra mortar for raked out joints, cavities, uneven surfaces, etc. the above worked out quantity is to be increased approximately by 30%.

Further to convert the wet volume of mortar into its corresponding dry volume, it should be increased by about 30%.

Then the quantities of cement and sand required can be determined by dividing the total dry volume of mortar by the sum of the numerical figures of proportion or mix of the mortar and multiplying it by the individual numerical figures.

e.g. To determine the quantities of cement and sand for 12 mm thick plaster in cement mortar (1:4), the procedure would be as follows:

Considering the area to be plastered as 10 sq.m, with a thickness of 12 mm, the quantity of wet mortar required = 10 (sq.m) x 1.2 / 100 (m) = 0.12 cum.

Therefore, Adding 30% extra for filling joints, and uneven surface, etc., quantity of wet mortar required = 0.12 + (0.3 x 0.12) = 0.156 cum.

Therefore, Quantity of dry mortar required = 0.156 + (0.3 x 0.156) = 0.156 + 0.0468

= 0.2028 cum.

Further, allowing for wastage, etc.

The total quantity of dry mortar required = 0.203 cum.

Therefore, Quantity of cement required = 0.203 / (1+4) = 0.041 cum.

= 0.042 x 30 = 1.23 bags

and Quantity of sand required = [0.203 / (1+4)] x 4 = 0.162 cum.

Approximate method

The approximate method of determining the volume of dry mortar required is to multiply the quantity of wet mortar required by a factor 1.8

i.e. Volume of wet mortar required = 0.12 cum.

Therefore, Volume of dry mortar required = 0.12 x 1.8 = 0.216 cum.

Therefore, Quantity of cement required = 0.216 / (1+4) = 0.043 cum.

= 0.043 x 30 = 1.29 = 1.30 bags

and, Quantity of sand required = [0.216 / (1+4)] x 4 = 0.0173 cum. which is same as determined above.

Neeru finish coat of 1.5 mm thickness.

If the inside face is to be plastered further with a Neeru finish of 1.5 mm thickness then the quantity of Neeru required for 10 sq.m surface

= 10 (sq.m) x 1.5 / 1000 (m)

= 0.015 cum.

i.e. 0.015 x 30 = 0.45 bags = 0.5 bags of Neeru which is available as ‘Sagol’ or ‘Sunala’ as brand name in the market.

10 cm Thick Brick Work in partition Walls in Cement Mortar (1:4)

In order to calculate the quantities of cement and sand required for 10 cm thick brick partition wall, with I.S. size (i.e. modular) bricks, the procedure would be as follows.

Considering 10 sq.m area of the brick work to be constructed with 10 cm thick wall.

The quantity of brick work in partition = 10 x (10/100) = 1 cum.

Therefore, Number of bricks required (considering the thickness of joints on 10 cm)

= 100/(0.20 x 0.1 x 0.1)

=500

Adding 5% extra for wastage = 25

Therefore, Total number of bricks = 525

Therefore, Quantity of mortar required = (10 x 0.1) – 500(0.19 x 0.9 x 0.9) = 1 – 0.77 = 0.23

Adding 10% extra for frog, bonding, wastage, etc. = 0.023

Therefore, Wet volume of mortar required = 0.253 cum.

Therefore, Dry volume required = 1.25 x 0.253 = 0.316 cum. = 0.30 cum.

i.e. for 10 sq.m area of brick work thickness 10 cm, the quantity of dry mortar required = 0.3 cum.

Further, knowing the proportion of mix, the quantities of cement and sand can be found out by dividing the total quantity of dry volume of mortar by the sum of the numerical figures of the proportion or mix of the mortar and then multiplying it by the individual numerals.

Reinforcement for Reinforced Cement Concrete

It is usual practice to express the steel required for reinforced cement concrete as percentage of volume of concrete e.g. 1% steel in R.C.C. slab indicates that the quantity of steel required will be equal to 1% of the volume of concrete i.e. for every one square metre sectional area of the slab cut, there will be 0.01 sq.m of steel bar utilized.
As the steel weighs 7850 kg per cubic metre, the 1% steel reinforcement means there will be 78 kg of steel per cubic metre of volume of concrete.
The values of usual percentage of steel assumed for various items will be as shown below:
The quintal of binding wire for reinforced steel is usually assumed as 1 to 1.3 kg per quintal of steel reinforcement.
Due allowance is to be made for wastage of steel which is 5 to 10%

Main Brick Work in Superstructure

In order to calculate the quantities of materials required for brick work in cement mortar, it is necessary to decide the size of the bricks to be used in the masonry.
The size of the conventional or traditional bricks vary from 8 3/4″ x 3 3/16″ x 2 5/8″ (i.e. 22.23 cm x 10.64 cm x 6.67 cm) to 9″ x 4 1/2″ x 3″ (i.e. 22.86 cm x 11.43 cm x 7.62 cm)
The new I.S. size brick i.e. modular brick is actual 19 cm x 9 cm x 9 cm with a frog of 10 cm x 4 cm x 1 cm size.
Normally, the mortar joint is taken as 1 cm throughout, therefore, the normal size of brick will be 20 cm x 10 cm x 10 cm.

Therefore, Volume of one I.S. size brick, with thickness of joint as 1 cm = 0.20 x 0.1 x 0.1 = 0.002 cum.

Therefore, For 1 cum. of brick work, the total number of I.S. size (i.e. modular) bricks required = 1 cum. /0.002 = 500 Nos.

Therefore, Adding 5% towards wastage = 25

Therefore, Total number of I.S. bricks required = 525 Nos.

Now,

Quantity or volume of wet mortar required = (Total volume of brick work) – (Volume occupied by 500 bricks of 19 cm x 9 cm x 9 cm size)

= (1 – 500 x 0.19 x 0.09 x 0.09) cum

= 1 – 0.7695

=0.2305 cum.

In order to allow for mortar for filling the frog, bonding and wastage during its use, 10% is to be added.

Therefore, Volume of wet mortar required = 0.2305 + 0.10 x 0.2305

= 0.235 cum.

Therefore, Dry volume of mortar required = 1.25 x 0.235

= 0.316 = 0.30

i.e. Approximately for 1 cum. of brick work, 30% of the dry mortar will be required.

Calculations of materials for 1 cum. of brick work in C.M. (1:6) with traditional size bricks 9″ x 4.375″ x 2.75″ (i.e. 22.86 cm x 11.11 cm x 6.985 cm)

Assuming thickness of joint as 1 cm throughout, the nominal size of traditional bricks = 23.86 cm x 12.11 cm x 7.985 cm

Therefore, Volume of one traditional size brick with 1 cm as thickness of joint = (0.2386 x 0.1211 x 0.07985) cum. = 0.002307 cum.

Number of traditional bricks required for 1 cum. of brickwork = 1/0.002307

= 433 Nos.

Add 5% towards wastage = 22

Therefore, Total number of traditional bricks required = 455 Nos.

Now, Value of wet mortar required = 1 – 433 x (0.2286 x 0.1111 x 0.06985)

= 1 – 0.77

= 0.23 cum. which is practically same as derived above

Therefore, Adding 10% extra for wastage = 0.023 cum.

Therefore, Wet volume of mortar required = 0.253 cum.

Therefore, Dry volume required = 1.25 (wet volume)

= 1.25 x 0.253

=0.316 cum. = 0.30 cum.

i.e. approximately 30% of dry volume of mortar is required for constructing 1 cum. of brick work. Further, knowing the proportion of cement mortar the quantities of cement (in bags) and sand can be worked out as usual.

e.g. knowing the proportion of the cement mortar, the quantities of cement and sand required can be determined as follows.

For cement mortar (1:6) proportion,

Quantity of cement required = (Dry volume of mortar)/(1+6) = (0.316/7)

= 0.045 cum. = 0.045 x 30 = 1.35 bags of cement

and, Quantity of sand required = (0.316/7) x 6 = 0.2708 cum. = 0.27 cum.

Approximate method:

The above quantities can be determined by an approximate method as follows:

For 1 cum. of brick work divide 0.3 by the sum of the proportion of the material to obtain the quantity of cement in cubic metre

i.e. Quantity of cement required = (0.3)/(1+6) = 0.3/7 = 0.043 cum.

But as certain amount of cement will be required to fill the voids in the sand, add 0.002 cubic metre extra.

Therefore, Quantity of cement required = 0.043 + 0.002 = 0.045 cum. which is same as above

Therefore, Number of cement bags required = 0.045 x 30 = 11.35 bag

and, Quantity of sand required = 0.045 x 6 = 0.27 cum.

Plain Cement Concrete (P.C.C)

The materials required for preparation of cement concrete are cement, sand (i.e. fine aggregates) and ballast (i.e. coarse aggregates) which are to be mixed in the predetermined proportion.
The voids in the coarse aggregates are filled by fine aggregates and that in the fine aggregates are filled with by cement paste (i.e. cement and water).
Thus the wet volume of the cement concrete (i.e. when water is added to the dry cement concrete mix) will always be less than its corresponding dry volume (i.e. sum of total volume of each ingredient added together).
It has been observed that in order to prepare 1 cum. of wet cement concrete, the corresponding dry volume required is about 1.52 cum.
Knowing the mix of the cement concrete (i.e. 1:4:8 or 1:3:6 or 1:2:4 etc.) the ingredient materials required can be determined as follows:

To determine the materials required for 1 cum. of (wet) concrete of 1:4:8 proportion, the dry volume of concrete required will be 1.52 cum. (which shrinks to 1 cum. after adding of water to it)