SOLVED PROBLEMS IN 45 CONSTRUCTION COST ESTIMATING Question 4.1 By referring to the drawing above, calculate the taking-off quantity for pad Footing SOLUTION 4.1 Quantity Take-Off for Footing / Pile cap/pad footing Taking off List Element: Work Below Lowest Floor Finish (WBLFF) Item: Footing / Pile cap No. Item Unit a. Excavation pit m3 b. Lean Concrete m3 c. Concrete Work m3 d. Sawn Formwork m e. Reinforcement kg a) Excavation pit Drawing No: Bill No.: Element : Pad Footing Page no: 1 Heading: Unit: Excavation of Pad Footing m 3 Description: Excavation of pad footings not exceeding 1.50m depth. Remaining excavated material is dumped and levelled. Quantity: 6.22 Excavate 4 nos of pit for pad footings 4/ 1.20 1.20 1.08 with depth not exceeding 1.50 depth Depth Lean conc 0.025 Footing 0.300 Stump 0.750 1.075 6.22
SOLVED PROBLEMS IN 46 CONSTRUCTION COST ESTIMATING b) Lean concrete Drawing No: Bill No.: Element : Pad Footing Page no: 3 Heading: Unit: Lean concrete for Pad Footing m 3 Description: Concrete grade 7 blinding for pad footing with thickness 25mm lain on earth. Quantity: 0.17 25mm thick lean concrete beneath pad footing 4/ 1.20 1.20 0.03 0.17 Lean concrete
SOLVED PROBLEMS IN 47 CONSTRUCTION COST ESTIMATING c) Concrete work Drawing No: Bill No.: Element : Pad Footing Page no: 4 Heading: Unit: Reinforced concrete for Pad Footing m 3 Description: Reinforced insitu concrete grade 20 in foundation bases. Quantity: 1.73 4/ 1.20 Width and length = 1200mm 1.20 height of foundation base = 300mm 0.30 1.73 0.30
SOLVED PROBLEMS IN 48 CONSTRUCTION COST ESTIMATING d) Sawn Formwork Drawing No: Bill No.: Element : Pad Footing Page no: 2 Heading: Unit: Formwork for Pad Footing m 2 Description: Formwork to sides of pad footing Quantity: 5.76 Pad footing girth: 1.20 + 1.20 + 1.20 + 1.20 =4.8 4/ 4.80 Fabricate formwork to sides of 4 nos of foundation with 300mm high 0.30 5.76 0.30
SOLVED PROBLEMS IN 49 CONSTRUCTION COST ESTIMATING e) Reinforcement Drawing No: Bill No.: Element : Pad Footing Page no: 5 Heading: Unit: Reinforcement bars in Pad Footing m Description: 12mm diameter steel reinforcement in straight and bent bar in pad footings. Quantity: 62.4 Concrete cover = 40mm 5Y12 12mm diameter of high yield steel as main reinforcement bar in foundation. Length 1200 Less concrete cover 2/40 1120 add 2 bends 2/300 1720 4/10/ 1.56 Less conc. Cover 2/40 62.4 Less conc. Cover 2/40 1560 62.4 m 300 1200 1200 – (2x40) 300 – 40 -40 300 – 40 -40 5 pcs 5 pcs
SOLVED PROBLEMS IN 50 CONSTRUCTION COST ESTIMATING Info: Question 4.2 By referring to the drawing above, calculate the taking-off quantity for Column Stump. SOLUTION 4.2 Taking off List Element: Work Below Lowest Floor Finish (WBLFF) Item: Column Stump No Item Unit a. Concrete m3 b. Sawn formwork m2 c. Reinforcement m Table : Example of weight measurement of reinforcement
SOLVED PROBLEMS IN 51 CONSTRUCTION COST ESTIMATING a) Concrete work Drawing No: Bill No.: Element : Column Stump Page no: 2 Heading: Unit: Reinforced concrete m3 Description: Reinforced in situ concrete grade 20 in isolated column stump Quantity: 0.27 Column stump size: 300 x 300 4/0.30 Column stump height : 750mm 0.30 Reinforced concrete grade 20 in column stump. 0.75 0.27 0.27 m3 Column Stump (image from quora.com, 2023)
SOLVED PROBLEMS IN 52 CONSTRUCTION COST ESTIMATING b) Sawn Formwork Drawing No: Bill No.: Element : Column Stump Page no: 1 Heading: Unit: Formwork m 2 Description: Formwork to sides of column stump Quantity: 3.60 Column stump girth: 4/0.300 = 1.20m 4/ 1.20 formwork to sides of 4 nos of 750 mm high column stump ( 4 sides). 0.75 3.60 3.60 m2 Column Stump Formwork
SOLVED PROBLEMS IN 53 CONSTRUCTION COST ESTIMATING C) Reinforcement work i. Main bar Drawing No: Bill No.: Element : Column Stump Page no: 2 Heading: Unit: Reinforcement bars m Description: Reinforcement bars Y20 in column stump Quantity: 27.04 4Y20 stump length Foundation = 0.300 Column stump height = 0.750 Laps in column = 0.500 1.550 4/4/ Less 1.69 concrete cover = 0.040 foundation bars = 2/0.012 = 0.024 27.04 1.486 Add bend = 0.200 1.686 27.04 m * Reinforced concrete with 40mm concrete cover
SOLVED PROBLEMS IN 54 CONSTRUCTION COST ESTIMATING ii. Links/Stirrups Drawing No: Bill No.: Element : Column Stump Page no: 2 Heading: Unit: Reinforcement Link m Description: 8mm diameter reinforcement in link in isolated column stump Quantity: 25.68 6R8-225 Link girth Width : 0.300 4/6/ ddt conc. Cover 2/0.040 0.080 1.07 = 220 Multiply by 4 sides : 4/220 25.68 = 880 Add hooks: 24d = 24/0.008 = 0.192 1.072 8mm reinforcement bar (mild steel) in column stump 29.75m
SOLVED PROBLEMS IN 55 CONSTRUCTION COST ESTIMATING Question 4.3 By referring to the drawing above, prepare taking off quantity for the slab with the thickness of 300mm. The dimension sheet shall include concrete work and reinforcement (bottom only). The nominal concrete cover = 25mm SOLUTION 4.3 Taking off List Element: Work Below Lowest Floor Finish (WBLFF) Item: Ground Slab No Item Unit a. Concrete work m3 b. Reinforcement work m c. Formwork m2 6700mm 7000mm T10-100 T12-200 Nominal cover 25mm
SOLVED PROBLEMS IN 56 CONSTRUCTION COST ESTIMATING a. Concrete Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Concrete work m3 Description: Reinforced insitu concrete grade 20 in ground slab not exceeding 300mm thickness Quantity: 14.07 Concrete work 7 Length: 7 m 6.7 Width: 6.7 m 0.3 Thickness: 0.3 m 14.07 b. Reinforcement i. 12 mm reinforcement bar Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 12mm reinforcement in ground floor slab Quantity: 240.72 Reinforcement (Bottom -Y direction) Length: 7 less(-) cover 2 x 0.025 = 0.050 6.95 T12-200
SOLVED PROBLEMS IN 57 CONSTRUCTION COST ESTIMATING Add (+) bend 2 x 5.5d = 2 x 5.5 x 0.012 = 0.132 7.08 Total length required steel (6700 – 25 – 25) = 6650 No. of rebar: (6650/200) + 1 = 34 34 7.08 240.72 240.72m, T12 ii. 10mm reinforcement bar Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 10mm reinforcement in ground floor slab Quantity: 473.20 Reinforcement (Bottom -Y direction) Length: 6.7 less(-) cover 2 x 0.025 = 0.050 6.65 Add (+) bend 2 x 5.5d = 2 x 5.5 x 0.010 = 0.11 6.76 No of rebar = (Length / C-C) + 1 T10-100
SOLVED PROBLEMS IN 58 CONSTRUCTION COST ESTIMATING Total length required steel (7000 – 25 – 25) = 6950 No. of steel (6950/100) + 1 = 70 70 6.76 473.20 473.20m, T10 c. Formwork Drawing No: Bill No.: Element : Ground Slab Page no: Heading: Unit: Formwork m2 Description: Formwork to sides of ground slab Quantity: 8.22 Formwork Length: 6.7 m 27.40 Width: 7.0m 0.30 Perimeter of slab: 6.7+7.0+6.7+7.0 = 27.40 slab thickness: 0.30m 8.22 No of rebar = (Length / C-C) + 1
SOLVED PROBLEMS IN 59 CONSTRUCTION COST ESTIMATING Question 4.4 By referring to the drawing above, prepare taking off quantity for the R.C Beam as shown in Figure below. SOLUTION 4.4 Taking off List Element: Work Below Lowest Floor Finish (WBLFF) Item: Ground Beam No Item Unit a. Concrete m3 b. Reinforcement m c. Formwork m2 150mm 500mm R6 – 150mm 2T18 2T12 300 x 300 column 6700 mm 2T18 2T12 CROSS-SECTION OF BEAM (Concrete cover = 25mm)
SOLVED PROBLEMS IN 60 CONSTRUCTION COST ESTIMATING a. Concrete Drawing No: Bill No.: Element : Ground Beam Page no: Heading: Unit: Concrete work m3 Description: Reinforced insitu concrete grade 20 in ground beam Quantity: 0.48 1/ Concrete work 6.4 Length: 6.7 m 0.15 less(-) column: 2x0.150 = 0.30 0.50 6.40 0.48 b. Formwork Drawing No: Bill No.: Element : Ground Beam Page no: Heading: Unit: Formwork m2 Description: Formwork to sides of ground beam Quantity: 6.40 Formwork Length: 6.7 m 1.00 less(-) column: 2x0.150 = 0.30 6.40 6.40 Girth: 2 x 0.5 = 1.0 m 6.40
SOLVED PROBLEMS IN 61 CONSTRUCTION COST ESTIMATING c. Reinforcement bar i. T12 rebar Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 12mm reinforcement in ground beam Quantity: 14.36 Reinforcement (top) Length: 6.7 + 0.3 = 7.0 less(-) cover 2 x 0.025 = 0.050 6.95 Add (+) bend 2 x 9.5d = 2 x 9.5 x 0.012 = 0.228 7.178 say 7.18 2/ 7.18 14.36 14.36m, T12 2T12
SOLVED PROBLEMS IN 62 CONSTRUCTION COST ESTIMATING ii. T18 rebar Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 18mm reinforcement in ground beam Quantity: 14.58 Reinforcement (bottom) Length: 6.7 + 0.3 = 7.0 less(-) cover 2 x 0.025 = 0.050 6.95 Add (+) bend 2 x 9.5d = 2 x 9.5 x 0.018 = 0.342 7.292 say 7.29 2/ 7.29 14.58 14.58m, T18 2T18
SOLVED PROBLEMS IN 63 CONSTRUCTION COST ESTIMATING iii. R6 stirrups Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 6mm stirrup reinforcement in ground beam Quantity: 54.56 Stirrups; Length = 0.500 Stirrups ground beam girth Width 0.150 Less Con.cover 2/0.025 0.050 0.100 Height 0.500 Less con. Cover 2/0.025 0.050 0.450 Girth = 2 times ( width + height) 2/(0.10 + 0.450) = 1.10 Add hooks 24d -> 24/0.006 = 0.144 = 1.244 no of stirrups: 44/ 1.24 Length / C-C +1 = 6.4 / 0.15 +1 = 43.67 say 44 54.56 Total length: 54.56m, R6 R6 – 150mm
SOLVED PROBLEMS IN 64 CONSTRUCTION COST ESTIMATING THE QUANTITY MEASUREMENT (SUPERSTRUCTURE)
SOLVED PROBLEMS IN 65 CONSTRUCTION COST ESTIMATING NOTES; The length of the placed bars must be extended by an additional length called ‘lap length’, which has to be equal or greater than the length required for the lapping of corresponding rebars between two successive storeys. This length is equal to the rebar’s diameter multiplied by the ‘contact coefficient’ (its value varies from 45 to 60). Eg: 45D to 60D
SOLVED PROBLEMS IN 66 CONSTRUCTION COST ESTIMATING QUESTION 5.1 By referring to the drawing given, prepare taking off quantity for the R.C Column as shown in Figure below. SOLUTION 5.1 Taking off List Element: FRAME Item: COLUMN No Item Unit a. Concrete work m3 b. Reinforcement work m c. Formwork m2 Particular; Column Height: 2700 Column Size : 300 x 300 Beam Size : 300 x 500 Concrete cover: 25 Reinforcement Bar (main): 4Y20 Stirrup : R8@100 c/c No. of column: 9
SOLVED PROBLEMS IN 67 CONSTRUCTION COST ESTIMATING a. Concrete Work Drawing No: Bill No.: Element : Frame Page no: 1 Heading: Unit: Reinforced concrete for Column m 3 Description: Reinforced concrete Grade 30 for column Quantity: 2.19 Size of column: Length = 300 9/ 0.30 Width = 300 0.30 Height = height of wall = 2700 2.70 2.19 2.19m3 b. Formwork Drawing No: Bill No.: Element : Frame Page no: 2 Heading: Unit: Column Formwork m 2 Description: Formwork fabrication for column Quantity: 29.16 Size of column: Length = 300 Width = 300 9/4/ 0.30 Height = height of wall = 2700 2.70 29.16 29.16m2
SOLVED PROBLEMS IN 68 CONSTRUCTION COST ESTIMATING c. Rainforcement work i. Main bar Drawing No: Bill No.: Element : Frame Page no: 3 Heading: Unit: Column reinforcement bars m Description: Main bar Y20 in column Quantity: 154.80 Main Bar Y20 Length = height of column + beam thickness + 1 lap(55D) = 2700+ 500 + (55 x20) = 4300 9/4/ 4.30 154.80 154.80 m ii. Stirrup Drawing No: Bill No.: Element : Column Page no: Heading: Unit: Reinforcement Link m Description: 8mm diameter reinforcement in link in isolated column stump Quantity: 133.28 R8-100 Link girth Width : 0.300 ddt conc. Cover 2/0.025 0.050 = 0.250 Multiply by 4 sides : 4/0.250 = 1.000
SOLVED PROBLEMS IN 69 CONSTRUCTION COST ESTIMATING Add hooks: 24d = 24/0.008 = 0.192 4/28/ 1.192 1.19 no of stirrups: 133.28 Length / C-C +1 = 2.7 / 0.1 +1 = 28 8mm reinforcement bar (mild steel) in column, 133.28m QUESTION 5.2 By referring to the drawing above, prepare taking off quantity for the First Floor Beam as shown in figure below. SOLUTION 5.2 Taking off List Element: Frame Item: First Floor Beam No Item Unit a. Concrete m3 b. Reinforcement m c. Formwork m2
SOLVED PROBLEMS IN 70 CONSTRUCTION COST ESTIMATING a. Concrete Drawing No: Bill No.: Element : First floor Beam Page no: Heading: Unit: Concrete work m3 Description: Reinforced insitu concrete grade 20 in ground beam Quantity: 0.48 1/ Concrete work 6.4 Length: 6.7 m 0.15 less(-) column: 2x0.150 = 0.30 0.50 6.40 0.48 150mm 500mm R6 – 150mm 2T18 2T12 300 x 300 column 6700 mm 2T18 2T12 CROSS-SECTION OF BEAM (Concrete cover = 25mm)
SOLVED PROBLEMS IN 71 CONSTRUCTION COST ESTIMATING b. Formwork Drawing No: Bill No.: Element : First floor Beam Page no: Heading: Unit: Formwork m2 Description: Formwork to sides and soffit of first floor beam Quantity: 7.36 Formwork Length: 6.7 m less(-) column: 2x0.150 = 0.30 6.40 1.15 Perimeter: (2 x 0.5) + 0.15 = 1.15 m 6.40 7.36 c. Reinforcement bar i. T12 rebar Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 12mm reinforcement in ground beam Quantity: 14.36 Reinforcement (top) Length: 6.7 + 0.3 = 7.0 less(-) cover 2T12
SOLVED PROBLEMS IN 72 CONSTRUCTION COST ESTIMATING 2 x 0.025 = 0.050 6.95 Add (+) bend 2 x 9.5d = 2 x 9.5 x 0.012 = 0.228 7.178 say 7.18 2/ 7.18 14.36 14.36m, T12 ii. T18 rebar Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 18mm reinforcement in ground beam Quantity: 14.58 Reinforcement (bottom) Length: 6.7 + 0.3 = 7.0 less(-) cover 2 x 0.025 = 0.050 6.95 Add (+) bend 2 x 9.5d = 2 x 9.5 x 0.018 = 0.342 7.292 say 7.29 2/ 7.29 14.58 14.58m, T18 2T18
SOLVED PROBLEMS IN 73 CONSTRUCTION COST ESTIMATING iii. R6 stirrups Drawing No: Bill No.: Element : Slab Page no: Heading: Unit: Reinforcement work m Description: 6mm stirrup reinforcement in ground beam Quantity: 54.56 Stirrups; Length = 0.500 Stirrups ground beam girth Width 0.150 Less Con.cover 2/0.025 0.050 0.100 Height 0.500 Less con. Cover 2/0.025 0.050 0.450 Girth = 2 times ( width + height) 2/(0.10 + 0.450) = 1.10 Add hooks 24d -> 24/0.006 = 0.144 = 1.244 no of stirrups: 44/ 1.24 Length / C-C +1 = 6.4 / 0.15 +1 = 43.67 say 44 54.56 Total length: 54.56m, R6 R6 – 150mm
SOLVED PROBLEMS IN 74 CONSTRUCTION COST ESTIMATING