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.)👆
= 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".
PART 3 👈 BACK CONTINUED 👉 PART 5
No comments:
Post a Comment