Bored excavation for 300mm diameter piers
Total volume is calculated by considering bulking factor of soil (Peterson, 2014).
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
1 |
Bored excavation volume for 53 Ø300mm and 2.5m deep piers Cross sectional area of pier = πr2 = π x 0.152 = 0.07m2 Volume of pier = cross sectional area x depth = 0.07 x 2.5 = 0.18m3 Volume of 53 piers = 0.18m3 x 53 = 9.54m3 Assume bulking factor of 1.4 (clay soil), total volume of soil material to be excavated for boring of piers = 1.4 x 9.54 = 13.6m3 |
m3 |
14 |
Summary:
A total of 53 53 Ø300mm and 2.5m deep pier bores shall be excavated. For all the 53 piers, a total volume of about 14m3 of soil material will be excavated for the piers to be bored.
Question 2: Bulk excavation for general site preparation (DelPico, 2012)
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
2 |
Bulk excavation |
m3 |
150 |
||
|
Overall bulk area of excavation (plus working space) = (12.83 + 4.08 + 2.00) x (11.87 + 2.00) = 18.91 x 13.87 = 262.28 Estimated depths at corners: NE – 0.69m, NW – 0.55m, SE – 0.50 and SW – 0.45) Average depth = (0.690 + 0.54 + 0.45 + 0.40)/4 = 0.55m Volume of bulk excavation = area x depth = 262.28m2 x 0.55m = 144.25 Assume that only one side is battered = 18.91m x 1.0m wide x 0.55m = 10.40m3 Since battering is done at 45 degrees, battered volume is = 0.5 x 10.40 = 5.20 Total volume of bulk excavation = 144.25 + 5.20 = 149.45 |
A total of 150m3 of shall be excavated from the site during general preparation of the site. This bulk volume includes working space, which extends outside the building towards the boundary by 1m. The longest side of the building shall also be battered at an angle of 45 degrees and a width of 1m from the building. The total bulk volume of 150m3 to be excavated includes the actual site of the house, working space and battered volume. The subcontractor has to use this value to determine the most suitable equipment to use for bulk excavation and if the excavated material will have to be cart away.
Question 3: N25 concrete for the ground slab (beams, patio slab and pad footings)
Thickness of slab = 150mm = 0.15m
Thickness of beam = 450mm = 0.35m
Width of beam = 100mm = 0.1m
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
3 |
N25 Concrete for ground slab |
m3 |
|||
|
Slab on ground = length x width x slab thickness = 12.83m x 11.87m x 0.15m = 22.84 |
m3 |
23 |
|||
|
Patio slab = length x width x thickness = 6.35m x 1.67m x 0.15m = 1.59 |
m3 |
2 |
|||
|
Family room = length x width x thickness = 5.51m x 4.08m x 0.15m = 3.37 |
m3 |
4 |
|||
|
Beams on slab = total length of beams x width of beam x depth of beam = 82.82m x 0.1m x 0.45m = 3.73 |
m3 |
4 |
|||
|
Pad Footings Area = 10(1.2 x 0.9) + (0.35 x 2.4) = 10.80 + 0.84 = 11.64m2 Volume = area x depth = 11.64 x 0.35 = 4.074 |
m3 |
5 |
|||
|
Total |
m3 |
38 |
Summary:
A total volume of 38m3 of N25 concrete shall be required for the whole ground slab, including beams, patio slab and pad footings but excluding piers. Depending on the mix design recommended by the engineer, it noes becomes easier to determine the volume of concrete ingredients (concrete, coarse and fine aggregates and sand) needed for the ground slab.
Question 4: SL72 reinforcing mesh for the ground slabs
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
4 |
Number of SL72 Reinforcing Mesh Sheets |
||||
|
4 (a) |
Area of ground slab to be reinforced with mesh (including 50mm cover) = length x width = 11.98m x 11.02m = 132.02m2 Area of family room to be reinforced with mesh = length x width = 4.66m x 3.68m = 17.15m2 Total area of slab to be covered = 132.02 + 17.15 = 149.17m2 |
m2 |
150 |
||
|
4 (b) |
Number of sheets = total area of slab/ effective area of SL72 Total area of slab to be reinforced with SL72 mesh = 150m2 Effective area of SL72 sheet = 5.8m x 2.2m = 12.76m2 Number of SL reinforcing mesh sheets = 150/12.76 = 11.76 |
No |
12 |
Summary:
The total area of slab that requires to be covered by the SL72 reinforcing mesh is 150m2. Based on this area (and putting into consideration the required overlap length and cover at the edges), the number of 6.0m x 2.4m with 400mm lap SL72 reinforcing sheets required to complete the slab is 12.
Question 5: Lengths of SL12TM for ground slab construction
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
5 |
Length and number of SL12TM Trench Mesh |
||||
|
5 (a) |
Length of trench mesh (including 200mm lap) = length of one trench x number of trenches = 13.23m x 3 = 39.69m = length of one trench x number of trenches = 12.27m x 3 = 36.81m = length of one trench x number of trenches = 4.48m x 2 = 8.96m = length of one trench x number of trenches = 5.91m x 1 = 5.91m Total length of trench mesh = 39.69 + 36.81 + 8.96 + 5.91 = 91.37 |
m |
92 |
||
|
5 (b) |
Number of SL12TM trench mesh 16 sheets = Length of trenches/length of trench sheets = 92/6 = 15.33 |
No |
16 |
Summary:
The total length of SL12TM trench mesh needed for ground lab (including overlap of 200mm at each end) is 92m. Since each trench measure comes in 6m lengths, the number of lengths of SL12TM trench mesh needed is 16. Therefore the contractor will have to supply 16 sheets of 6m long SL12TM trench mesh.
Question 6: Structural steel
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
6 |
Structural steel |
tonnes |
1.78 |
||
|
B1 Weight = 25.7kg/m x 12.83m = 329.73kg |
|||||
|
B2 Weight = 32kg/m x 12.83m = 410.56kg |
|||||
|
B3 Weight = 14kg/m x 11.87m = 166.18kg |
|||||
|
B4 Weight = 14kg/m x 11.87m = 166.18kg |
|||||
|
B6 Weight = 14kg/m x 11.87m = 166.18kg |
|||||
|
B5 Weight = 31.6kg/m x 12.83m = 405.43kg |
|||||
|
L1 Weight = 14.6kg/m x 4.08m = 59.57kg |
|||||
|
L2 Weight = 14.6kg/m x 5.51m = 80.45kg |
|||||
|
Total weight = 329.73 + 410.56 + 166.18 + 166.18 + 166.18 + 405.43 + 59.57 + 80.45 = 1784.28kg |
Summary:
The total weight of structural steel needed for the project is 1.78 tonnes.
Question 7: Plasterboard
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
7 |
Plasterboard |
||||
|
7 (a) |
Area of plasterboard Assume that the height of plasterboard wall is 2.4m Area of wall = length of wall x height of wall A1 = 11.93m x 2.4m = 28.63m2 A2 = 11.93m x 2.4m = 28.63m2 A3 = 11.93m x 2.4m = 28.63m2 A4 = 11.93m x 2.4m = 28.63m2 A5 = 10.97m x 2.4m = 26.33m2 A6 = 10.97m x 2.4m = 26.33m2 A7 = 10.97m x 2.4m = 26.33m2 A8 = 10.97m x 2.4m = 26.33m2 A9 = 3.63m x 2.4m = 8.71m2 A10 = 3.63m x 2.4m = 8.71m2 A11 = 4.61m x 2.4m = 11.06m2 A12 = 4.61m x 2.4m = 11.06m2 Total area of walls = 4(28.63) + 4(26.33) + 2(8.71) + 2(11.06) = 259.38m2 Deduct area of doors and windows Total area of doors = 5(2m x 0.9m) = 9.00m2 Area of windows = 6(1.2m x 0.8m) = 5.76m2 Total area of doors and windows = 9.00 + 5.76 = 14.76m2 Net measure = 259.38 – 14.76 = 244.62m2 |
m2 |
245 |
||
|
7 (b) |
Number of plasterboard sheets The selected size of plasterboard sheets is 1200mm wide and 2400mm long Area of one plasterboard sheet = 1.2m x 2.4m = 2.88m2 Number of sheets = net measure of wall/area of one sheet = 245m2/2.88m2 =85.07 |
No |
86 |
Summary:
The total area of wall that will be covered with plasterboard is 245m2 and a total of 86 plasterboard sheets measuring 1200mm wide and 2400mm long shall be required for the project.
Question 8: 66x18mm finished DAR primed pine skirting
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
8 |
66x18mm finished DAR primed pine skirting |
||||
|
8 (a) |
Length of skirting L1 = 11.93m L2 = 11.93m L3 = 11.93m L4 = 11.93m L5 = 10.97m L6 = 10.97m L7 = 10.97m L8 = 10.97m L9 = 3.63m L10 = 3.63m L11 = 4.61m L12 = 4.61m Total length of skirting = 4(11.93) + 4(10.97) + 2(3.63) + 2(4.61) = 108.08m Plus 5% wastage Total length of skirting (including wastage) = 1.05 x 108.08 = 113.48 |
m |
114 |
||
|
8 (b) |
Number of skirting lengths The selected length of 66x18mm finished DAR primed pine skirting is 2.7m Number of skirting lengths = length of skirting/ selected length of skirting L1: 11.93m/2.7m = 4.42 = 5 sheets L2: 11.93m/2.7m = 4.42 = 5 sheets L3: 11.93m/2.7m = 4.42 = 5 sheets L4: 11.93m/2.7m = 4.42 = 5 sheets L5: 10.97m/2.7m = 4.06 = 5 sheets L6: 10.97m/2.7m = 4.06 = 5 sheets L7: 10.97m/2.7m = 4.06 = 5 sheets L8: 10.97m/2.7m = 4.06 = 5 sheets L9: 3.63m/2.7m = 1.34 = 2 sheets L10: 3.63m/2.7m = 1.34 = 2 sheets L11: 4.61m/2.7m = 1.71 = 2 sheets L12: 4.61m/2.7m = 1.71 = 2 sheets Total sheets = 8(5) + 4(2) = 48 |
No |
48 |
Summary:
The total length of skirting is 114m and a total of 48 lengths of 66x18mm finished DAR primed pine skirting measuring 2.7m long shall be required for the project.
Question 9: Ceramic floor tiles
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
9 |
Ceramic floor tiles |
||||
|
9 (a) |
Amount of ceramic floor tiles Family room = 4.08m x 5.51m = 22.48m2 Plus 10% wastage = 1.10 x 22.48m2 = 24.73m2 ≈ 25m2 Meals room = 2.43m x 4.68m = 11.37m2 Plus 10% wastage = 1.10 x 11.37m2 = 12.51m2 ≈ 13m2 Kitchen Floor tiles: 4.59m x 6.11m = 28.04m2 Plus 10% wastage = 1.10 x 28.04m2 = 30.85m2 Wall tiles (under cupboards) = 2(4.59m x 1.8m) + 2(6.11m x 1.8m) = 38.52m2 Plus 10% wastage = 1.10 x 38.52m2 = 42.37m2 Total kitchen = 30.85 + 42.37 = 73.22m2 ≈ 74m2 Foyer/stair area = 1.71m x 1.09m = 1.86m2 Plus 10% wastage = 1.10 x 1.86m2 = 2.05m2 ≈ 3m2 Total = 25 + 13 + 74 + 3 = 115 |
m2 |
115 |
||
|
9 (b) (i) |
Number of boxes of floor tiles Number of boxes of floor tiles = area of floor tiles/1.2m2 = (25 + 13 + 31 + 3)/1.2 = 72m2/1.2m2 = 60 |
No |
60 |
||
|
9 (b) (ii) |
Number of boxes of wall tiles Number of wall tiles = area of wall tiles/1.4m2 = 43m2/1.4m2 = 30.71 |
No |
31 |
A total area of 115m2 (including family room, meals room, kitchen and foyer/stair area) shall be tiled. 60 boxes of floor tiles and 31 boxes of wall tiles shall be required for the project.
References
DelPico, W., 2012. Estimating Building Costs for the Residential and Light Commercial Construction Professional. New York: John Wiley & Sons.
Peterson, S., 2014. Estimating in Building Construction. Chicago: Pearson Education.
Bored excavation for 300mm diameter piers
Total volume is calculated by considering bulking factor of soil (Peterson, 2014).
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
1 |
Bored excavation volume for 53 Ø300mm and 2.5m deep piers Cross sectional area of pier = πr2 = π x 0.152 = 0.07m2 Volume of pier = cross sectional area x depth = 0.07 x 2.5 = 0.18m3 Volume of 53 piers = 0.18m3 x 53 = 9.54m3 Assume bulking factor of 1.4 (clay soil), total volume of soil material to be excavated for boring of piers = 1.4 x 9.54 = 13.6m3 |
m3 |
14 |
Summary:
A total of 53 53 Ø300mm and 2.5m deep pier bores shall be excavated. For all the 53 piers, a total volume of about 14m3 of soil material will be excavated for the piers to be bored.
Question 2: Bulk excavation for general site preparation (DelPico, 2012)
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
2 |
Bulk excavation |
m3 |
150 |
||
|
Overall bulk area of excavation (plus working space) = (12.83 + 4.08 + 2.00) x (11.87 + 2.00) = 18.91 x 13.87 = 262.28 Estimated depths at corners: NE – 0.69m, NW – 0.55m, SE – 0.50 and SW – 0.45) Average depth = (0.690 + 0.54 + 0.45 + 0.40)/4 = 0.55m Volume of bulk excavation = area x depth = 262.28m2 x 0.55m = 144.25 Assume that only one side is battered = 18.91m x 1.0m wide x 0.55m = 10.40m3 Since battering is done at 45 degrees, battered volume is = 0.5 x 10.40 = 5.20 Total volume of bulk excavation = 144.25 + 5.20 = 149.45 |
A total of 150m3 of shall be excavated from the site during general preparation of the site. This bulk volume includes working space, which extends outside the building towards the boundary by 1m. The longest side of the building shall also be battered at an angle of 45 degrees and a width of 1m from the building. The total bulk volume of 150m3 to be excavated includes the actual site of the house, working space and battered volume. The subcontractor has to use this value to determine the most suitable equipment to use for bulk excavation and if the excavated material will have to be cart away.
Question 3: N25 concrete for the ground slab (beams, patio slab and pad footings)
Thickness of slab = 150mm = 0.15m
Thickness of beam = 450mm = 0.35m
Width of beam = 100mm = 0.1m
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
3 |
N25 Concrete for ground slab |
m3 |
|||
|
Slab on ground = length x width x slab thickness = 12.83m x 11.87m x 0.15m = 22.84 |
m3 |
23 |
|||
|
Patio slab = length x width x thickness = 6.35m x 1.67m x 0.15m = 1.59 |
m3 |
2 |
|||
|
Family room = length x width x thickness = 5.51m x 4.08m x 0.15m = 3.37 |
m3 |
4 |
|||
|
Beams on slab = total length of beams x width of beam x depth of beam = 82.82m x 0.1m x 0.45m = 3.73 |
m3 |
4 |
|||
|
Pad Footings Area = 10(1.2 x 0.9) + (0.35 x 2.4) = 10.80 + 0.84 = 11.64m2 Volume = area x depth = 11.64 x 0.35 = 4.074 |
m3 |
5 |
|||
|
Total |
m3 |
38 |
Summary:
A total volume of 38m3 of N25 concrete shall be required for the whole ground slab, including beams, patio slab and pad footings but excluding piers. Depending on the mix design recommended by the engineer, it noes becomes easier to determine the volume of concrete ingredients (concrete, coarse and fine aggregates and sand) needed for the ground slab.
Question 4: SL72 reinforcing mesh for the ground slabs
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
4 |
Number of SL72 Reinforcing Mesh Sheets |
||||
|
4 (a) |
Area of ground slab to be reinforced with mesh (including 50mm cover) = length x width = 11.98m x 11.02m = 132.02m2 Area of family room to be reinforced with mesh = length x width = 4.66m x 3.68m = 17.15m2 Total area of slab to be covered = 132.02 + 17.15 = 149.17m2 |
m2 |
150 |
||
|
4 (b) |
Number of sheets = total area of slab/ effective area of SL72 Total area of slab to be reinforced with SL72 mesh = 150m2 Effective area of SL72 sheet = 5.8m x 2.2m = 12.76m2 Number of SL reinforcing mesh sheets = 150/12.76 = 11.76 |
No |
12 |
Summary:
The total area of slab that requires to be covered by the SL72 reinforcing mesh is 150m2. Based on this area (and putting into consideration the required overlap length and cover at the edges), the number of 6.0m x 2.4m with 400mm lap SL72 reinforcing sheets required to complete the slab is 12.
Question 5: Lengths of SL12TM for ground slab construction
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
5 |
Length and number of SL12TM Trench Mesh |
||||
|
5 (a) |
Length of trench mesh (including 200mm lap) = length of one trench x number of trenches = 13.23m x 3 = 39.69m = length of one trench x number of trenches = 12.27m x 3 = 36.81m = length of one trench x number of trenches = 4.48m x 2 = 8.96m = length of one trench x number of trenches = 5.91m x 1 = 5.91m Total length of trench mesh = 39.69 + 36.81 + 8.96 + 5.91 = 91.37 |
m |
92 |
||
|
5 (b) |
Number of SL12TM trench mesh 16 sheets = Length of trenches/length of trench sheets = 92/6 = 15.33 |
No |
16 |
Summary:
The total length of SL12TM trench mesh needed for ground lab (including overlap of 200mm at each end) is 92m. Since each trench measure comes in 6m lengths, the number of lengths of SL12TM trench mesh needed is 16. Therefore the contractor will have to supply 16 sheets of 6m long SL12TM trench mesh.
Question 6: Structural steel
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
6 |
Structural steel |
tonnes |
1.78 |
||
|
B1 Weight = 25.7kg/m x 12.83m = 329.73kg |
|||||
|
B2 Weight = 32kg/m x 12.83m = 410.56kg |
|||||
|
B3 Weight = 14kg/m x 11.87m = 166.18kg |
|||||
|
B4 Weight = 14kg/m x 11.87m = 166.18kg |
|||||
|
B6 Weight = 14kg/m x 11.87m = 166.18kg |
|||||
|
B5 Weight = 31.6kg/m x 12.83m = 405.43kg |
|||||
|
L1 Weight = 14.6kg/m x 4.08m = 59.57kg |
|||||
|
L2 Weight = 14.6kg/m x 5.51m = 80.45kg |
|||||
|
Total weight = 329.73 + 410.56 + 166.18 + 166.18 + 166.18 + 405.43 + 59.57 + 80.45 = 1784.28kg |
Summary:
The total weight of structural steel needed for the project is 1.78 tonnes.
Question 7: Plasterboard
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
7 |
Plasterboard |
||||
|
7 (a) |
Area of plasterboard Assume that the height of plasterboard wall is 2.4m Area of wall = length of wall x height of wall A1 = 11.93m x 2.4m = 28.63m2 A2 = 11.93m x 2.4m = 28.63m2 A3 = 11.93m x 2.4m = 28.63m2 A4 = 11.93m x 2.4m = 28.63m2 A5 = 10.97m x 2.4m = 26.33m2 A6 = 10.97m x 2.4m = 26.33m2 A7 = 10.97m x 2.4m = 26.33m2 A8 = 10.97m x 2.4m = 26.33m2 A9 = 3.63m x 2.4m = 8.71m2 A10 = 3.63m x 2.4m = 8.71m2 A11 = 4.61m x 2.4m = 11.06m2 A12 = 4.61m x 2.4m = 11.06m2 Total area of walls = 4(28.63) + 4(26.33) + 2(8.71) + 2(11.06) = 259.38m2 Deduct area of doors and windows Total area of doors = 5(2m x 0.9m) = 9.00m2 Area of windows = 6(1.2m x 0.8m) = 5.76m2 Total area of doors and windows = 9.00 + 5.76 = 14.76m2 Net measure = 259.38 – 14.76 = 244.62m2 |
m2 |
245 |
||
|
7 (b) |
Number of plasterboard sheets The selected size of plasterboard sheets is 1200mm wide and 2400mm long Area of one plasterboard sheet = 1.2m x 2.4m = 2.88m2 Number of sheets = net measure of wall/area of one sheet = 245m2/2.88m2 =85.07 |
No |
86 |
Summary:
The total area of wall that will be covered with plasterboard is 245m2 and a total of 86 plasterboard sheets measuring 1200mm wide and 2400mm long shall be required for the project.
Question 8: 66x18mm finished DAR primed pine skirting
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
8 |
66x18mm finished DAR primed pine skirting |
||||
|
8 (a) |
Length of skirting L1 = 11.93m L2 = 11.93m L3 = 11.93m L4 = 11.93m L5 = 10.97m L6 = 10.97m L7 = 10.97m L8 = 10.97m L9 = 3.63m L10 = 3.63m L11 = 4.61m L12 = 4.61m Total length of skirting = 4(11.93) + 4(10.97) + 2(3.63) + 2(4.61) = 108.08m Plus 5% wastage Total length of skirting (including wastage) = 1.05 x 108.08 = 113.48 |
m |
114 |
||
|
8 (b) |
Number of skirting lengths The selected length of 66x18mm finished DAR primed pine skirting is 2.7m Number of skirting lengths = length of skirting/ selected length of skirting L1: 11.93m/2.7m = 4.42 = 5 sheets L2: 11.93m/2.7m = 4.42 = 5 sheets L3: 11.93m/2.7m = 4.42 = 5 sheets L4: 11.93m/2.7m = 4.42 = 5 sheets L5: 10.97m/2.7m = 4.06 = 5 sheets L6: 10.97m/2.7m = 4.06 = 5 sheets L7: 10.97m/2.7m = 4.06 = 5 sheets L8: 10.97m/2.7m = 4.06 = 5 sheets L9: 3.63m/2.7m = 1.34 = 2 sheets L10: 3.63m/2.7m = 1.34 = 2 sheets L11: 4.61m/2.7m = 1.71 = 2 sheets L12: 4.61m/2.7m = 1.71 = 2 sheets Total sheets = 8(5) + 4(2) = 48 |
No |
48 |
Summary:
The total length of skirting is 114m and a total of 48 lengths of 66x18mm finished DAR primed pine skirting measuring 2.7m long shall be required for the project.
Question 9: Ceramic floor tiles
|
Item |
Particulars |
Unit |
Quantity |
Rate |
$ C |
|
9 |
Ceramic floor tiles |
||||
|
9 (a) |
Amount of ceramic floor tiles Family room = 4.08m x 5.51m = 22.48m2 Plus 10% wastage = 1.10 x 22.48m2 = 24.73m2 ≈ 25m2 Meals room = 2.43m x 4.68m = 11.37m2 Plus 10% wastage = 1.10 x 11.37m2 = 12.51m2 ≈ 13m2 Kitchen Floor tiles: 4.59m x 6.11m = 28.04m2 Plus 10% wastage = 1.10 x 28.04m2 = 30.85m2 Wall tiles (under cupboards) = 2(4.59m x 1.8m) + 2(6.11m x 1.8m) = 38.52m2 Plus 10% wastage = 1.10 x 38.52m2 = 42.37m2 Total kitchen = 30.85 + 42.37 = 73.22m2 ≈ 74m2 Foyer/stair area = 1.71m x 1.09m = 1.86m2 Plus 10% wastage = 1.10 x 1.86m2 = 2.05m2 ≈ 3m2 Total = 25 + 13 + 74 + 3 = 115 |
m2 |
115 |
||
|
9 (b) (i) |
Number of boxes of floor tiles Number of boxes of floor tiles = area of floor tiles/1.2m2 = (25 + 13 + 31 + 3)/1.2 = 72m2/1.2m2 = 60 |
No |
60 |
||
|
9 (b) (ii) |
Number of boxes of wall tiles Number of wall tiles = area of wall tiles/1.4m2 = 43m2/1.4m2 = 30.71 |
No |
31 |
A total area of 115m2 (including family room, meals room, kitchen and foyer/stair area) shall be tiled. 60 boxes of floor tiles and 31 boxes of wall tiles shall be required for the project.
References
DelPico, W., 2012. Estimating Building Costs for the Residential and Light Commercial Construction Professional. New York: John Wiley & Sons.
Peterson, S., 2014. Estimating in Building Construction. Chicago: Pearson Education.
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First, you will need to complete an order form. It's not difficult but, in case there is anything you find not to be clear, you may always call us so that we can guide you through it. On the order form, you will need to include some basic information concerning your order: subject, topic, number of pages, etc. We also encourage our clients to upload any relevant information or sources that will help.
Complete the order form
Once we have all the information and instructions that we need, we select the most suitable writer for your assignment. While everything seems to be clear, the writer, who has complete knowledge of the subject, may need clarification from you. It is at that point that you would receive a call or email from us.
Writer’s assignment
As soon as the writer has finished, it will be delivered both to the website and to your email address so that you will not miss it. If your deadline is close at hand, we will place a call to you to make sure that you receive the paper on time.
Completing the order and download