# PARAM VISIONS

All about civil construction knowledge- PARAM VISIONS

### Calculating the Plastering & flooring for a shed / Estimation & costing of a shed ( part 4 ).

1.  Let us plaster the shed internally, with a single coat of 12 mm. thickness in a 1: 4 ratio.

From the above drawings, the area of the wall that should be plastered

= front wall + back wall + (2 nos.× sidewall ) - ( deduction of a door + window.)

= (9 ft.× 8 ft.) +( 9 ft. × 9 ft.) + (2 nos. × 7 ft.× 8.5 ft.) - (( 6.5 ft.× 2.5 ft.) +( 3 ft. × 3.5 ft.))

(Height of side wall = (9ft + 8 ft) ÷ 2 = 8.5 ft.)ðŸ‘†

= 72 + 81 + 119 - (16.25 + 10.5 )

= 272 - 26.75

=245.25 sq.ft.

The wet volume of the plastering mortar

= plastering area × plaster thickness

= 245.25 sq.ft. × (12 × 0.00328084) ft.

(1mm = 0.00328084 ft.)

= 9.6555 cu ft.

The number of cement bags required for plastering

= 0.217 bags × ( 9.6555 cu ft. ÷ 1 cu ft.)

= 2.0952 bags.

The volume of sand required for plastering

= 1.064 cu ft. × ( 9.6555 cu ft. ÷ 1 cu ft.)

= 10.2734cu ft.

Note: The ratio taken above for cement & sand is from the article "Calculating the quantity of materials in 1 cum. & 1cu ft. of plastering mortar in different mix ratios".

2. Let us fix vitrified tiles of 2ft × 2 ft. size for the flooring.

Area of the flooring

= 9ft × 7 ft.

= 63 sq.ft.

Area of a tile

= 2ft × 2ft

=  4 sq.ft.

Number of tiles required

= area of flooring ÷ area of 1 tile

= 63sq.ft.÷ 4sq.ft

= 15.75 nos.

By rounding off, the number of tiles required = 16 nos.

Let us fix these tiles in 1: 4 mortar having 40 mm. thickness.

The number of cement bags required

= 2.84 bags ×( shed area in sq ft. ÷ 100 sq ft. area)

= 2.84 bags × ( 63 sq.ft ÷ 100 sq ft.)

= 1.789 bags

The volume of sand required for flooring.

= 13.9624 cu ft. × ( 63 sq.ft ÷ 100 sq ft.)

= 8.7963 cu ft.

Note: The ratio of cement & sand in-floor mortar is taken from "Calculating the quantity of materials for the 100 sq. ft. of tile flooring".

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