SOLVED PROBLEMS IN First Edition Learn strategies for solving problems in step-by-step detail Prepare effectively for students’ exams and assessment problems. Basic reference in project cost estimating SEZEE GOROTOP | KRISTY WONG
SOLVED PROBLEMS IN CONSTRUCTION COST ESTIMATING FOR BUILDING First Edition by Sezee binti Gorotop Kristy Wong Politeknik Kota Kinabalu
First Edition 2023 Published in Malaysia by POLITEKNIK KOTA KINABALU No. 4, Jalan Politeknik KKIP Barat Kota Kinabalu Industrial Park 88460 Kota Kinabalu Sabah Phone No. : (088) 401800 Fax No. : (088) 499960 Website : https://polikk.mypolycc.edu.my/ e ISBN 978-967-2301-94-3 ______________________________________________________________________________ Copyright © 2023 Politeknik Kota Kinabalu All rights reserved. No parts of this publication may be copied, stored in form or by any means, electronic, mechanical, photocopying and recording or otherwise or by any means for reproduced without the prior permission of Politeknik Kota Kinabalu.
PREFACE This First edition of Solved Problems in Construction Cost Estimation for Building was produced based on the Contract Estimating course subject offered by Malaysian Polytechnic. This book is a compilation of questions and answers related to topics covered in the polytechnic syllabus and contains a straightforward, practical approach that has made it easy to understand and solve simple cost estimation problems. It was designed to provide readers with an understanding of the concepts of solving problems and this edition includes topics such as The Preliminary Estimating Method, Build-up Rate Method, and The Quantity Measurement for sub-structure and superstructure. The production of this book aimed to help readers understand how concepts apply in the real world and to make it easier for lecturers or students to understand related calculations in related topics. The calculations shown are according to the correct and structured workflow. Finally, it is hoped that this book can help students in cost estimating problems for specifically for buildings. The highest appreciation goes to all parties involved in the preparation of this book, either directly or indirectly. May all the efforts made be beneficial in educating students in related fields. Sezee Binti Gorotop Kristy Wong
TABLE OF CONTENT CHAPTER 1 THE PRELIMINARY ESTIMATING METHOD 1 - 16 CHAPTER 2 THE BUILD-UP RATE METHOD 17 - 31 CHAPTER 3 THE QUANTITY MEASUREMENT (EARTHWORKS & PILING) 32 - 40 CHAPTER 4 THE QUANTITY MEASUREMENT (SUBSTRUCTURE) 41 - 63 CHAPTER 5 THE QUANTITY MEASUREMENT (SUPERSTRUCTURE) 64 - 73
ABSTRACT SOLVED PROBLEMS IN CONSTRUCTION COST ESTIMATION FOR BUILDING Difficulty in understanding and following learning sessions in the classroom is the biggest problem as students are unable to answer test and examination questions for Contract and Estimation. Therefore, this book is written so that students are more familiar with and understand the various styles of questions with solutions in this course. The contents of this book can be used in standard lecture or self-paced classes. Keywords: cost estimation, preliminary estimates, quantity measurement, construction cost
SOLVED PROBLEMS IN 1 CONSTRUCTION COST ESTIMATING THE PRELIMINARY ESTIMATING METHOD
SOLVED PROBLEMS IN 2 CONSTRUCTION COST ESTIMATING UNIT METHOD EXAMPLE 1.1 Company A constructed 380 units of houses in the year 2018 which cost RM5.7 million. In 2023, this company wants to construct another housing project which will involve 500 units of houses where the design is based on the previous project. The construction cost is expected to be increased by 2.5% per annum. Estimate the total cost of the new project in 2023. SOLUTION 1.1 House A (Year 2018) = RM5,700,000.00 380 = 15,000.00 5 × 2.5% × 15,000.00 = 1,875.00 = 15,000.00 + 1,875.00 = 16,875.00 = , . × = , , . (. ) EXAMPLE 1.2 The following schools were constructed in the year 2019. A new school will be constructed in the year 2023 which can accommodate 1000 chairs. Calculate the school construction cost by taking the increasing cost of construction at 6.5% per annum. School Construction cost (RM) Nos of chair A 6.06 million 700 B 5.90 million 600 SOLUTION 1.2 ℎ ( 2019) = 6,060,000.00 700 = 8,657.14 ℎ ( 2019) = 5,900,000.00 600 = 9,833.33 =8,657.14 + 9,833.33 2 = 9,245.24 4 × 6.5% × 9,245.24 = 2,403.76 = 9,245.24 + 2,403.76 = 11,649.00 ( ) = , . × = , , . (. )
SOLVED PROBLEMS IN 3 CONSTRUCTION COST ESTIMATING EXAMPLE 1.3 Based on the table below, the school were constructed in year 2020. The proposed school will be constructed in the year 2024. The new school will be equipped with 600 chairs. Calculate the school construction cost by taking the increasing rate at 6% per annum. School Construction cost (RM) Nos of chair A 800,000 400 B 850,000 450 C 970,000 500 SOLUTION 1.3 ℎ ( 2020) =800,000.00 400 = 2,000.00 ℎ ( 2020) =850,000.00 450 = 1,888.89 ℎ ( 2020 =970,000.00 500 = 1,940.00 =2,000.00 + 1,888.89 + 1,940.00 3 = 1,942.96 4 ; 4 × 6.0% × 1,942.96 = 466.31 = 1,942.96 + 466.31 = 2,409.27 ( ) = , . × = , , . (. ) EXAMPLE 1.4 Based on table below, calculate the cost of construction of Hospital C which can accommodate 6000 patients, taking into consideration that there will be an additional cost of 25% for materials. Types Construction cost (RM) Nos of patients Hospital A 2,800,500.00 3,500 Hospital B 3,500,000.00 4,200 Hospital C ? 6000
SOLVED PROBLEMS IN 4 CONSTRUCTION COST ESTIMATING SOLUTION 1.4 = 2,800,500.00 3,500 = 800.14 =3,500,000.00 4,200 = 833.33 =800.14 + 833.33 2 = 816.74 25% × 816.74 = 204.19 = 816.74 + 204.19 = 1,020.93 = , . × = , , . (. ) EXAMPLE 1.5 Based on the information below, calculate the cost of construction for a new hostel building to accommodate about 3,200 students in 2024. Consider the cost of materials will increase 5% per annum. Types of hostel Construction cost in year 2019 (RM) Number of students Increase in labour salary (2019 – 2014) Hostel A 3,000,000.00 1650 Hostel B 2,500,000.00 1500 2% Hostel C 2,450,000.00 1300 SOLUTION 1.5 = 3,000,000.00 1650 = 1,818.18 =2,500,000.00 1500 = 1,666.67 =2,450,000.00 1300 = 1,884.62 =1,818.18 + 1,666.67 + 1,884.62 3 = 1,789.82 ( 2019 − 2024) 6 × (5% + 2%) × 1,789.82 = 757.72 = 1,789.82 + 757.72 = 2,541.54 ( ) = , . × , = , , . (. )
SOLVED PROBLEMS IN 5 CONSTRUCTION COST ESTIMATING EXAMPLE 1.6 Based on table below, calculate the estimated construction cost for Training center, which can accommodate 3500 students. Type Construction Cost (RM) No. of students Training Center A 6,500,000.00 2000 Training Center B 3,200,000.00 1500 SOLUTION 1.6 =6,500,000.00 2000 = 3,250.00 =3,200,000.00 1500 = 2,133.33 =3,250.00 + 2,133.33 2 = 2,691.67 = , . × = , , . ( . ) FLOOR AREA METHOD EXAMPLE 1.7 Based on the figure below, estimate the cost of the building using the Floor area Method. Given the price rate is RM450/m2. 6800 5000 1500 3000 Porch
SOLVED PROBLEMS IN 6 CONSTRUCTION COST ESTIMATING SOLUTION 1.7 = 5.0 × 6.8 = 34.02 ℎ = 1 2 × 3.0 × 1.5 = 2.252 = (. + . ) × . = , . (. ) EXAMPLE 1.8 Based on figure below, calculate the building area by using Floor Area Method. Given, the wall thickness is 225 mm. SOLUTION 1.8 Actual Area = Area building 1 + Area building 2 = (10.0m × 15.0m) + (7.0m × 3.5m) = 1502 + 24.52 = . (. ) 15000 10000 3500 7000
SOLVED PROBLEMS IN 7 CONSTRUCTION COST ESTIMATING EXAMPLE 1.9 Based on Figure below, calculate the cost of the building using the Floor Area Method. Given the rate per unit is RM480.00 / m2. SOLUTION 1.9 Floor Plan Porch A B
SOLVED PROBLEMS IN 8 CONSTRUCTION COST ESTIMATING ℎ = [ + ] + ℎ = [10.5 × 6.0]2 + [9.4 × 3.5]2 + [ 1 2 × 6 × 3]2 = 104.92 = . × . = , (. ) EXAMPLE 1.10 Based on the figure below, calculate the cost of the building using the Floor Area Method. Given the price rate is RM525.00/m2 and the wall thickness is 115mm. All the dimensions are measured from the center line to the center line. SOLUTION 1.10 = [ + + 2 + ℎ] + ℎ = [(10.5 + 0.12) × (6.25 + 0.12)] + [ 1 2 × (3.0 + 0.12)] = [10.62 × 6.37]2 + 1.562 = 204.482 = . × . = , . (. ) Floor Plan
SOLVED PROBLEMS IN 9 CONSTRUCTION COST ESTIMATING EXAMPLE 1.11 Based on the Figure below, estimate the building cost by using the Floor Area Method. Assume, the rate per m2 is RM 550.00. SOLUTION 1.11 = [1 + 2 + ] − [ℎ] = [9.5 × 7.0] − [6.1 × 3.5] = 66.52 − 21.352 = 45.152 = [45.152 × 550.00] + ��1 2 × 21.352 × 550.00�� = 24,832.50 + 5,871.25 = , . (. ) ∗ 1 2 ℎ
SOLVED PROBLEMS IN 10 CONSTRUCTION COST ESTIMATING CUBIC CONTENT METHOD EXAMPLE 1.12 Based on figure below, calculate the cost of the building using the Cubic Content Method, assuming that the price rate is RM 550.00 / m3. 6800 1500 800 7500 4100 2000 8000 3200 Floor Plan Side View
SOLVED PROBLEMS IN 11 CONSTRUCTION COST ESTIMATING SOLUTION 1.12 1 = 4.1 × 2.0 = 8.22 : ℎ > 600 ℎℎ: + ( ℎℎ ) ℎℎ: 7.5 + 0.8 = 8.3 2 = 3.2m × 8.0m = 25.62 : ℎ ℎℎ: + ℎ 2 ℎℎ: 6.8 + 1.5 2 = 7.6 = 1 + 2 = [(8.22 × 8.3) + (25.62 × 7.6)] × 550.00 = , . (. ) EXAMPLE 1.13 According to figure below, calculate the cost of the building using the Cubic Content Method. The price rate given is RM 525.00/m3. Area A Area B 230mm thick brickwall 11,200 mm 3800 mm 2500 mm 7500 mm Floor Plan
SOLVED PROBLEMS IN 12 CONSTRUCTION COST ESTIMATING SOLUTION 1.13 = + = (11.2 × 10.0)2 + (3.8 × 7.5)2 = 140.52 : ℎ ℎℎ: + ℎ 2 ℎℎ: (1.8 + 5.5) + ( 2.1 2 ) = 8.35 = = (140.5 × 8.35)3 × 525.00 = , . ( . ) 5500 2100 Ground Level 1800 Full pitch roof Cross Section
SOLVED PROBLEMS IN 13 CONSTRUCTION COST ESTIMATING EXAMPLE 1.14 According to Figure below, calculate the cost of the building using the Cubic Content Method assuming the price rate is RM450.00/m3 and there should be an additional cost about 30%. 6800 8000 4000 5000 3500 2000 Floor Plan Side View
SOLVED PROBLEMS IN 14 CONSTRUCTION COST ESTIMATING SOLUTION 1.14 ( ) 1 = 5.0 × 4.0 = 20.02 ℎ ℎ: + 0.6 ℎ ℎ: 8.0 + 0.6 = 8.6 ( ℎ ) 2 = 3.5m × 2.0m = 7.02 ℎ ℎ: + ℎ 2 ℎ ℎ: 6.8 + 1.5 2 = 7.6 = 1 + 2 = [(20.02 × 8.6) + (7.02 × 7.6)] × 450.00 = RM101,340.00 (Ans. ) = 30% × RM101,340.00 = RM30,402.00 = , . + , . = , . (. )
SOLVED PROBLEMS IN 15 CONSTRUCTION COST ESTIMATING EXAMPLE 1.15 Mr Eric proposed to build a food stall in Tuaran, Sabah. The building should be completed with mechanical ventilation. Based on the figure below, estimate the cost of the proposed building by using Cubic Content Method. Given the price rate/m3 is RM400.00 and wall thickness is 115mm. All the dimensions are measured from center to center of the wall. Floor Plan Side View
SOLVED PROBLEMS IN 16 CONSTRUCTION COST ESTIMATING SOLUTION 1.15 : ℎ = [ × × ]3 × 400.00 = [(10.0 + 0.12) × (8.0 + 0.12) × �(0.8 + 5.0) + 1.75 2 �] × 400.00 = , , . (. ) EXAMPLE 1.17 Based on the table below, the given price for a 1-meter cube building is RM620. Estimate the price of the school building if the construction cost increases at the rate of 15% using cubic method. Building Length mm Width mm Height mm School 6200 3500 4500 Assume all roofs are flat roof. SOLUTION 1.17 600mm parapet wall for flat roof. Thus, Building height = 4.5 + 0.6 = 5.1 = [ × × ]3 × 620.00 = [6.2 × 3.5 × 5.1}3 × 620.00 = RM68,615.40(Ans. ) s 15% × 68,615.40 = 10,292.31 = 68,615.40 + 9,081.45 = , . (. )
SOLVED PROBLEMS IN 17 CONSTRUCTION COST ESTIMATING THE BUILD-UP RATE METHOD
SOLVED PROBLEMS IN 18 CONSTRUCTION COST ESTIMATING EXAMPLE 2.1 A concrete mixture of (1:3:6 - 20mm aggregates) will be mixed manually for the construction of a slab for a guard house. Calculate the material cost for 1 cubic meter of concrete by referring to the following particular. Portland cement (1m3 = 28.7bags) = RM22.00/bag Sand = RM15.00 Aggregates = RM35.00 Wastage = 50% SOLUTION 2.1 (: : ) . : 28.7 × 22.00 = 631.40 . : 33 × 15.00 = 45.00 . : 63 × 35.00 = 210.00 Total cost of materials = RM886.40 (%) 50% × RM886.40 = RM443.20 Cost for 103 = 1,329.60 . = . (. ) EXAMPLE 2.2 Four (4) numbers of columns will be built in the construction of a garage. The size of the column is 400mm x 400mm and 3m in height. The concrete mixture of (1:3:6 -20mm aggregates) will be mixed manually. Calculate the total price of all columns needed in the construction. Particular; 1m3 of concrete = RM135.60 Labor charge = RM60.00/day Total hours to mix 1m3 = 2 hours
SOLVED PROBLEMS IN 19 CONSTRUCTION COST ESTIMATING SOLUTION 2.2 (: : ) 1.03 = 135.60 2 8 × 60.00 = 15.00 1.03 = + = 150.60 * = % 15% × 150.60 = 22.59 Total cost rate for 1.03 = 150.60 + 22.59 = 173.19 ∶ − × (. × . × ) × . = . (. ) EXAMPLE 2.3 Based on the information given in Table below, calculate the construction cost for lm3 concrete work (1 :3: 6-20mm agg.) mixed manually. Material Cost (RM) Cement RM30/bag Sand RM35/m3 Aggregate RM 45/m3 Labour Output Mixing concrete 2.5 hour / m3 Placing concrete 1.5 hour / m3 Compacting concrete 1.0 hour / m3 Others Labour wages (General labour) RM 120/day Profit & Overhead cost 10% 1m3 cement 28 bags
SOLVED PROBLEMS IN 20 CONSTRUCTION COST ESTIMATING SOLUTION 2.3 (: : ) Take 1.03 = 28 Cement = 28 bags × RM30.00 = RM840.00 Sand = 33 × 35.00 = 105.00 Aggregate = 63 × 45.00 = 270.00 Total cost for materials = RM1,215.00 % 50% × RM1,215.00 = RM607.50 Cost for 103 = 1,822.50 Cost for 1.03 = 182.25 . =2.5 8 × 120.00 = 37.50 . =1.5 8 × 120.00 = 22.50 . =1.0 8 × 120.00 = 15.00 Total cost of labour = RM75.00 Cost for 1.03 = + = 185.25 + 75.00 = 260.25 % 10% × RM260.25 = RM26.03 . = . (. )
SOLVED PROBLEMS IN 21 CONSTRUCTION COST ESTIMATING EXAMPLE 2.4 Calculate the construction cost for excavation work for pad foundation done manually based on the information given in Table below. Details of pad foundation Type size Depth Nos F1 1.30m x 1.30m x 0.50m 1.50m 5 F2 1.20m x 1.20m x 0.50m 2.0m 5 Labour Constant Excavation of soil 3.0 hour/m3 Dispose soil 1.5 hour/m3 Backfill soil 1.2 hour/m3 Others Labour wages (General labour) RM80/day Profit & Overhead cost 15% SOLUTION 2.4 i. Excavation of soil = 3.0hrs ii. Dispose of soil = 1.5 hrs iii. Backfill soil = 1.2 hrs 0.30m x 0.30m column stump D T Original ground level
SOLVED PROBLEMS IN 22 CONSTRUCTION COST ESTIMATING S ( ∗ %) 25% × 1.5hrs = 0.375hrs = 6.075 ℎ 8 × 80.00 = 60.75 % 15% × RM60.75 = RM9.11 . = . 1 = 5 × (1.30 × 1.30 × 2.00) × 69.86 = 1,180.63 2 = 5 × (1.20 × 1.20 × 2.50) × 69.86 = 1,257.48 = + = , . (. ) EXAMPLE 2.5 Based on the following information, calculate the cost of concrete work (1 : 2 : 4 – 19 mm aggregates) where concrete is mixed using machine. Material Cement : RM 9.80/bag Sand : RM 21.00/m3 Aggregates : RM 17.50/m3 Labour Mixing concrete : 2 hours/m3 Transport and pouring concrete : 2.25 hours/m3 Compacting concrete : 0.75 hours/m3 Levelling concrete : 0.5 hours/m3 Labourer : 3 person Labour wages : RM32/day Concrete mixer (Machine) Machine productivity = 5m3/hr Machine rental rates = RM 45/hr Others: Profit : 10% Wastage : 50%
SOLVED PROBLEMS IN 23 CONSTRUCTION COST ESTIMATING SOLUTION 2.5 Material Cost Concrete ratio (1 : 2 : 4) Let 1m3 cement = 28.7 bags of cement So, 1m3 cement = 28.7 bags x RM9.80 = RM281.26 2m3 Sand = 2 x RM21.00 = RM42.00 4m3 Aggregates = 4 x RM17.50 = RM70.00 Total cost of 7m3 concrete = RM281.26 + RM42.00 + RM70.00 = RM393.26 Rate for 1m3 concrete = RM56.18/m3 Add wastage (50%) = RM56.18 x 1.5 = RM84.27/m3 Labour Cost Total hours of working = (2+2.25+0.75+0.5)hrs/m3 = 5.50 hrs /m3 Labour fees/hr Labour fee for 5.5hrs = RM4.00/hr x 5.50hrs/m3 = RM22.00/m3 For 3 person = RM22.00 x 3 = RM66.00/m3 Machinery Cost In 1 hour, machine can mix 5m3 with rent RM45/hr, so the machine rate is = = RM9.00/m3 Total cost of concrete work = Material Cost + Labour Cost + Machinery cost = (RM84.27 + RM66.00 +RM9.00)/m3 = RM159.27/m3 Add 10% profit = (10% x RM159.27/m3) + RM159.27/m3 = RM175.20/m3
SOLVED PROBLEMS IN 24 CONSTRUCTION COST ESTIMATING EXAMPLE 2.6 Calculate the build-up rate for 1 cubic meter of concrete ( 1: 3 : 6 – 20mm aggregates) to construct the beam and column, which are mixed using a machine. The calculation is based on the following information. Construction Material Price Portland Cement) 1m3 = 28.4bag) = RM18.00/bag Sand = RM 21.00 Aggregates (20mm) = RM 30.00 Diesel = RM 1.70/L Lubricant = RM 4.50/L Labour constant Transporting concrete = 1.80 hrs/m3 Compacting concrete = 1.00hr/m3 Levelling concrete = 0.35 hrs/m3 Concrete Mixer Machine Price = RM 10,000.00 Productivity = 4m3/hr Bank Interest = 6% per year Machine Service charge for 5 years = 10% of machine price Cost of transporting machine for 5 years = 5% of machine price Life expectancy of machine = 5 years Machine working period = 280 days/year Diesel = 15 liters/day Lubricant = 2 liters/day Driver = 1 person Labour = 1 person Others: Driver wages = RM70/day Labour wages = RM40/day Total time of working (Driver) = 9 hrs/day Total time of working (Labourer) = 8 hrs/day Company profits = 10%
SOLVED PROBLEMS IN 25 CONSTRUCTION COST ESTIMATING SOLUTION 2.6 Material Cost Concrete ratio (1 : 3 : 6) Given 1m3 cement = 28.4 bag So, 1m3 Portland cement = 28.4 bags x RM18.00 = RM511.20 3m3 Sand = 3 x RM21.00 = RM63.00 6m3 Aggregates = 6 x RM30.00 = RM180.00 Total cost of 10m3 concrete = RM511.20 + RM63.00 + RM180.00 = RM754.20 Rate for 1m3 concrete = RM754.20 / 10m3 = RM75.42/m3 Add wastage 50% = 1.5 x RM75.42 = RM113.13/m3 Labour cost Total working hours = (1.80+1.00+0.35)hrs/m3 = 3.15 hrs/m3 Driver wages/hr = RM70/9hrs = RM7.78/hr Labour wages/hr = RM40/8hrs = RM5.00/hr Rate of labour cost = (RM7.78 + RM5.00)/hr x 3.15hrs/m3 = RM 40.26/m3 Machinery cost Diesel = RM1.70 x 15L/day = RM25.50/day Lubricant = RM4.50 x 2L/day = RM9.00/day Total fuel cost/hour = (RM25.50 +RM9.00)/8hrs = RM4.31/hr Machine cost = RM10000 Bank Interest for 5 years = RM10000 x 5 x 6% = RM3000 Machinery sercice for 5 years = 10% x RM10000 = RM1000 Transporting cost for 5 years = 5% x RM10000 = RM500
SOLVED PROBLEMS IN 26 CONSTRUCTION COST ESTIMATING Total cost for machine in 5 years = RM(10000+3000+1000+500) = RM14500 For I year = RM14500/5yr = RM2900/yr For 1 day = RM2900/280days = RM10.36/day For 1 hour = RM10.36/8hours = RM1.30/hr Fuel cost + Machine cost = RM4.31/hr + RM1.30/hr = RM5.61/hr Productivity of machine = 4m3/hr which means within 1 hour, machine can produce 4m3 concrete with capital cost of RM5.61 So, rate of machinery cost = RM5.61/4m3 = RM1.40/m3 Total cost/m3 = Material cost + Labour cost + Machinery cost = RM113.13 +RM40.26 +RM1.40 = RM154.79/m3 Add 10% profit = (10% x RM154.79/m3) + RM154.79/m3 Build-up rate = RM170.27/m3 EXAMPLE 2.7 Calculate the total cost of reinforced concrete (1:11/2 :3 – 19mm aggregates) for five (5) pad footings with size 1350 x 1350 x 225 (mm) and mixed using hands. Given, Labour constant Mixing concrete = 4 hrs/m3 Transport and pouring concrete = 21/2 hrs/m3 Compacting concrete = 3/4 hrs/m3 Levelling surface = 1/4 hrs/m3 Rate of semi-skilled worker = RM22.00/day Material constant Portland Cement = RM18.00/bag Sand = RM22.50/m3 Aggregates = RM31.00/m3
SOLVED PROBLEMS IN 27 CONSTRUCTION COST ESTIMATING Other Profits = 15% 1m3 concrete = 28.40 bags SOLUTION 2.7 Material cost Concrete ratio ( 1: 11/2 : 3) Given 1m3 cement = 28.4bags 1m3 portland cement = RM18.00 x 28.4 = RM511.20 11/2 m3 Sand = RM15.00 x 3/2 = RM33.75 3m3 Aggregates = RM31.00 x 3 = RM93.00 Total cost for 5.5 m3 concrete = RM511.20 + RM33.75 + RM93 = RM637.95 Rate for 1m3 concrete = RM637.95/5.5m3 = RM116.00/m3 Add wastage (50%) = 1.5 x RM116.00/m3 = RM174.00/m3 Labour cost Total hours to mix 1m3 concrete = (4+5/2+3/4+1/4)hrs/m3 = 7.5hrs/m3 Labour fee per hour = RM22/8hrs = RM2.75/hr Build up rate for 1m3 = RM2.75/hr x 7.5hrs/m3 = RM20.63/m3 Total cost for 1m3 = RM174 + RM20.63 = RM194.63/m3 Add 15% profit = (15% x RM194.63) + RM194.63 = RM223.82/m3 Given, Volume of 1 pad footing= 1.35 x 1.35 x 0.225 x 5 = 2.05m3
SOLVED PROBLEMS IN 28 CONSTRUCTION COST ESTIMATING For 1m3 = RM223.82 For 2.05 m3 concrete (5 footings) = RM223.82 x 2.05 = RM458.84 EXAMPLE 2.8 As a Supervisor in a construction company, you are assigned by your manager to calculate the cost of excavation work for the foundation by using machine. Size of footing : 1.20m x 1.20 m x 600mm Excavation depth : 1.60m Number of footings : 8 Type of soil : ordinary soil Machinery Rent fees : RM250.00/hr Productivity : 3.00m3/hr Operator : 1 person Assistant : 2 person Working period : 8hrs/day Others Driver wages : RM80.00/day Assistant wages : RM40.00/day Diesel (RM1.70/L) :4.55L/hr Lubricant (RM5.50/L) : 0.14 L/hr Profits and Overhead : 20% SOLUTION 2.8 Labour cost Rate of wages = (RM80 + (RM40 x 2) )/8hrs = RM20/hr Machinery Cost Rent fees = RM250.00/hr Fuel cost Diesel = RM1.70/L x 4.55L/hr = RM7.74/hr
SOLVED PROBLEMS IN 29 CONSTRUCTION COST ESTIMATING Lubricant = RM5.50L x 0.14 L/hr = RM0.77/hr Total = RM7.74/hr + RM0.77/hr = RM8.51/hr Total cost per hour = RM20/hr + RM250.00/hr + RM8.51/hr = RM278.51/hr Given, productivity = 3.00m3/hr For 1 hour, volume of soil excavated = 3m3 and cost = RM278.51 So, cost of 1 m3 = RM278.51/3 = RM92.84/m3 Add 20% profit and overhead = (20% x RM92.84) + RM92.84 = RM111.41/m3 Volume of soil excavated = 1.2 x 1.2 x 1.6 = 2.30 m3 Volume of soil excavated (8 footings) = 2.30 x 8 = 18.40 m3 Cost of excavation work = RM111.41/m3 x 18.40m3 = RM2019.84 EXAMPLE 2.9 Based on Table B2, estimate the price rate/m3 for excavation work manually done by using hand tools. The type of soil is ordinary soil. Labour Output Excavation soil 3.0 hrs/m3 Dispose soil 1.5 hrs/m3 Backfill soil 1.2 hrs/m3 Others Labour wages RM70/day Profit & Overhead 15%
SOLVED PROBLEMS IN 30 CONSTRUCTION COST ESTIMATING SOLUTION 2.9 Labour cost; RM70/8hrs = RM8.75/hr Excavation constant = 3hrs/m3 Excavation cost /m3 = 3hrs/m3 x RM8.75/hr = RM26.25/m3 Add profit & overhead (15%) = 1.15 x RM26.25 = RM30.19/m3 EXAMPLE 2.10 Based on the information given in Table and figure below, prepare the construction cost for concrete work (1:2:4 – 20mm agg) mixed manually for footings. Footing Type Size (L x W x T) Depth (D) Nos F1 1.5m x 1.5m x 0.30m 1.80m 10 F2 1.20m x 1.20m x 0.30m 2.10m 6 Material Cost Cement RM20 / bag Sand RM35 /m3 Aggregate RM45/m3 Labour Output Mixing concrete 2.5 hrs/m3 Placing concrete 1.5 hrs/m3 Compacting concrete 1.0 hrs/m3 Others Labour wages RM80/day Profit & Overhead cost 20% 1m3 cement 28 bags
SOLVED PROBLEMS IN 31 CONSTRUCTION COST ESTIMATING SOLUTION 2.10 Material cost Concrete ratio 1 : 2 : 4 1m3 cement = 28bags 1m3 portland cement = RM20.00 x 28 = RM560 2m3 Sand = RM35.00 x 2 = RM70.00 4m3 Aggregates = RM45.00 x 4 = RM180.00 Total cost for 7m3 concrete = RM560 + RM70.00 + RM180.00 = RM810.00 Rate for 1m3 concrete = RM810/7 = RM115.71/m3 Add wastage 50% = 1.50 x RM115.71 = RM173.57/m3 Labour cost Total hours to process 1m3 concrete = 2.5 hours + 1.5 hrs + 1 hr = 5 hrs/m3 Labour fee per hour = RM80/8hrs = RM10/hr Labour cost in 5 hours = RM10/hr x 5hrs/m3 = RM50/m3 Build-up rate for 1m3 = RM173.57 + RM50 (material cost + labour cost) = RM223.57/m3 Add profit and overhead (20%) = 1.20 x RM223.57/m3 = RM268.28/m3 Total volume of footings, Volume of F1 = 1.5m x 1.5m x 0.30m x 10 nos = 6.75m3 Volume of F2 = 1.20m x 1.20m x 0.30m x 6 nos = 2.59m3 Total volume of concrete for all footings = 9.34m3 Rate for 1m3 concrete = RM268.28 Thus, for 9.34m3 concrete = RM268.28/m3 x 9.34m3 = RM2506.27 Total cost of concrete work for footing is RM2506.27
SOLVED PROBLEMS IN 32 CONSTRUCTION COST ESTIMATING THE QUANTITY MEASUREMENT (EARTHWORKS & PILING)
SOLVED PROBLEMS IN 33 CONSTRUCTION COST ESTIMATING EXAMPLE 3.1 Based on the following figure and table, calculate the quantity of soil using the Triangle Method. Particular; Interval = 10 meter Formation Level = 85.5 m SOLUTION 3.1 Location Ground Level Formation Level Depth (A) Weighing (B) Weighing Depth Cut (AB) (+) / Fill (-) No of triangle A1 87.4 85.5 1.9 2 3.8 A2 84.3 85.5 -1.2 3 -3.6 A3 81.6 85.5 -3.9 1 -3.9 B1 86.5 85.5 1 3 3 B2 88.4 85.5 2.9 6 17.4 B3 87.4 85.5 1.9 3 5.7 C1 90.5 85.5 5 1 5 C2 86.7 85.5 1.2 3 3.6 C3 89.6 85.5 4.1 2 8.2 TOTAL: 24 39.2 POINT ORIGINAL LEVEL (m) A1 87.40 A2 84.30 A3 81.60 B1 86.50 B2 88.40 B3 87.40 C1 90.50 C2 86.70 C3 89.60 AI BI C1 A2 A3 B2 B3 C2 C3
SOLVED PROBLEMS IN 34 CONSTRUCTION COST ESTIMATING EXAMPLE 3.2 Refer to the grid layout in figure below, calculate the volume of soil using Square Method. Given: i. Interval = 10 meter ii. Formation Level = 97.5 meter Point Original Level (m) A1 99.50 A2 96.20 A3 94.30 B1 98.20 B2 101.80 B3 98.00 C1 102.20 C2 102.50 C3 100.10 A1 A2 A3 B1 B2 B3 C1 C2 C3 Average Height = Total Weighing Depth (AB) Total Weighing (B) = 39.2 24 Average Height = 1.63 (cut) Area of Site = 20m x 20m = 400m2 Volume of soil = Area x Average Height = 400m2 x 1.63m = 652m3 (Cut)
SOLVED PROBLEMS IN 35 CONSTRUCTION COST ESTIMATING SOLUTION 3.2 Location Ground Level Formation Level Depth (A) Weighing (B) Weighing Cut (+) / Fill (-) No of Depth (AB) Rectangle A1 99.5 97.5 2 1 2 A2 96.2 97.5 -1.3 1 -1.3 A3 94.3 97.5 -3.2 1 -3.2 B1 98.2 97.5 0.7 2 1.4 B2 101.8 97.5 4.3 4 17.2 B3 98 97.5 0.5 2 1 C1 102.2 97.5 4.7 1 4.7 C2 102.5 97.5 5 2 10 C3 100.1 97.5 2.6 1 2.6 TOTAL: 15 34.4 Average Height = Total Weighing Depth (AB) Total Weighing (B) = 34.4 15 Average Height = 2.29 (cut) Area of Site = 20m x 20m = 400m2 Volume of soil = Area x Average Height = 400m2 x 2.29m = 916m3 (Cut)
SOLVED PROBLEMS IN 36 CONSTRUCTION COST ESTIMATING EXAMPLE 3.3 Based on the figure below, calculate the earthworks volume by using Triangle Method. Given; Interval = 14.00m Formation Level = 151.00m Excavation of topsoil = 150 mm SOLUTION 3.3 Location Ground Level Top Soil Reduce Level Formation Level Depth (A) Weighing (B) Weighing Depth Cut (AB) (+) / Fill (-) No of triangle A1 152 0.15 151.9 151 0.85 2 1.7 A2 151.3 0.15 151.2 151 0.15 3 0.45 A3 151.4 0.15 151.3 151 0.25 1 0.25 B1 151.5 0.15 151.4 151 0.35 3 1.05 B2 151.2 0.15 151.1 151 0.05 6 0.3 B3 150.2 0.15 150.1 151 -0.95 3 -2.85 C1 150.4 0.15 150.2 151 -0.8 3 -2.4 C2 153.5 0.15 153.4 151 2.37 6 14.22 C3 149.6 0.15 149.5 151 -1.53 3 -4.59 D1 151.5 0.15 151.3 151 0.3 1 0.3 D2 152.5 0.15 152.4 151 1.35 3 4.05 D3 150.9 0.15 150.7 151 -0.3 2 -0.6 TOTAL 36 11.88 152.00 151.50 150.35 151.45 151.30 151.20 153.52 152.50 151.40 150.20 149.62 150.8 A B C D 1 2 3
SOLVED PROBLEMS IN 37 CONSTRUCTION COST ESTIMATING EXAMPLE 3.4 Based on the figure below, calculate the quantity of earthwork by using Square Method. Data given; Interval = 10.00m Formation level = 100.00m Excavate topsoil = 150 100.09 101.50 100.00 101.30 101.90 100.52 101.40 101.80 149.62 A B C 1 2 3 Average Height = Total Weighing Depth (AB) Total Weighing (B) = 11.88 36 Average Height = 0.33 (cut) Area of Site = 42m x 28m = 1176m2 Volume of soil = Area x Average Height = 1176m2 x 0.33m = 388.08m3 (Cut)
SOLVED PROBLEMS IN 38 CONSTRUCTION COST ESTIMATING SOLUTION 3.4 Location Ground Level Top Soil Reduce Level Formation Level Depth (A) Weighing (B) Weighing Depth Cut (+) / (AB) Fill (-) No of Square A1 100.1 0.15 99.94 100 -0.06 1 -0.06 A2 101.3 0.15 101.2 100 1.15 2 2.3 A3 101.4 0.15 101.3 100 1.25 1 1.25 B1 101.5 0.15 101.4 100 1.35 2 2.7 B2 101.9 0.15 101.8 100 1.75 4 7 B3 101.8 0.15 101.7 100 1.65 2 3.3 C1 100 0.15 99.85 100 -0.15 1 -0.15 C2 100.5 0.15 100.4 100 0.37 2 0.74 C3 149.6 0.15 149.5 100 49.47 1 49.47 TOTAL 16 66.55 Average Height = Total Weighing Depth (AB) Total Weighing (B) = 66.55 16 Average Height = 4.16 (cut) Area of Site = 20m x 20m = 400m2 Volume of soil = Area x Average Height = 400m2 x 4.16m = 1664m3 (Cut)
SOLVED PROBLEMS IN 39 CONSTRUCTION COST ESTIMATING EXAMPLE 3.5 Refer to the grid layout in figure below, calculate the quantity of soil by using Square Method. Data given; Interval = 10.00m Formation level = 100.00m SOLUTION 3.5 Location Ground Level Formation Level Depth (A) Weighing (B) Weighing Depth (AB) Cut (+) / Fill (-) No of triangle A1 100 100 0 1 0 A2 103.2 100 3.2 2 6.4 A3 99.4 100 -0.6 1 -0.6 B1 105.4 100 5.4 2 10.8 B2 108.9 100 8.9 4 35.6 B3 101.2 100 1.2 2 2.4 C1 107.8 100 7.8 2 15.6 C2 103 100 3 4 12 C3 98.5 100 -1.5 2 -3 D1 109.6 100 9.6 1 9.6 D2 102.9 100 2.9 2 5.8 D3 99.5 100 -0.5 1 -0.5 TOTAL: 24 94.1 A1 A2 A3 B1 B2 B3 C1 C2 C3 D1 D2 D3 100 103.20 99.40 105.4 108.9 101.20 107.8 103.0 98.50 109.6 102.9 99.50
SOLVED PROBLEMS IN 40 CONSTRUCTION COST ESTIMATING Average Height = Total Weighing Depth (AB) Total Weighing (B) = 94.10 24 Average Height = 3.92 (cut) Area of Site = 30m x 20m = 600m2 Volume of soil = Area x Average Height = 600m2 x 3.92m = 2352 m3 (Cut)
SOLVED PROBLEMS IN 41 CONSTRUCTION COST ESTIMATING THE QUANTITY MEASUREMENT (SUBSTRUCTURE)
SOLVED PROBLEMS IN 42 CONSTRUCTION COST ESTIMATING Quantity Take-off for Work Below Lowest Floor Finish (WBLFF) Components of WBLFF; • Footing / pilecap • Ground beam • Ground Slab • Column Stump • Lift pit • Cable Trench All questions in this section are based on the following layout drawing. Substructures
SOLVED PROBLEMS IN 43 CONSTRUCTION COST ESTIMATING 4Y20 6R8 - 225 300 300 Column Stump Details Layout Plan
SOLVED PROBLEMS IN 44 CONSTRUCTION COST ESTIMATING Section of pad footing Plan view of pad footing