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PE Review Course Construction Engineering

PE Review Course Construction Engineering

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PE Review Course Construction Engineering

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  1. PE Review CourseConstruction Engineering Pramen P. Shrestha, Ph.D., P.E. October 5, 2009

  2. Topics to be Covered • Construction Estimating • Construction Scheduling • Project Controls • Engineering Economics

  3. Construction Estimating • Construction Cost consists of • Direct Cost • Labor, material, equipment, and sub-contractor cost • Indirect Cost • Overhead, taxes, bonds, insurance cost • Contingency Cost • Potential unforeseen work based on the amount of risk • Profit • Compensation costs for performing the work

  4. Steps for Preparing an Estimate • Review the scope of the project • Consider effect of location, security, available storage, traffic on costs • Determine quantities • Material quantity takeoff • Price material • Material cost = Quantity x Unit price of material

  5. Steps for Preparing an Estimate • Price labor • Based on labor production rates and crew sizes • Labor cost = [ (quantity)/(labor production rates)] x [labor rate] • Price equipment • Based on equipment production rates and equipment spreads • Equipment cost = [ (quantity)/(equip. production rates)] x [equip. rate]

  6. Steps for Preparing an Estimate • Obtain specialty sub-contractors’ bid • Obtain suppliers’ bid • Calculate taxes, bonds, insurance, and overhead • Contingency • Potential unforeseen work based on the amount of risk • Profit • Compensation costs for performing the work

  7. Types of Estimate • Conceptual Cost Estimate • Preliminary, feasibility, budget estimate etc. • Conducted before detail design • Conducted in planning or feasibility stage • Detailed Cost Estimate • Conducted after the detail design is complete • Basis for bid

  8. Conceptual Estimates • Prepared from completed similar projects • Size of project • No. of unit • No. of SF • No. of cars in a parking garage • Developed from unit cost • Weighting of average, maximum and minimum value

  9. Estimating Equation • Weighted Unit Cost Estimating • Equation to forecast unit cost • UC = (A + 4B + C) / 6 • Where • UC = Unit Cost • A = Minimum unit cost of previous projects • B = Average unit cost of previous projects • C = Maximum unit cost of previous projects

  10. Adjustments • Time • Location • Size • Complexity • Need appropriate contingency

  11. Weighted Unit Cost Estimate • 1. Weighted Unit Cost Estimating • Problem: Cost information from 6 previously completed housing projects are shown in the following table. These projects were completed in Las Vegas in 2004. Now a contractor has to build a house (2000 SF) in New Orleans, in 2009. Estimate the cost of that house using conceptual estimating method. • Projects Cost Square Foot Cost/ SF • 1 $500,000 2,000 $250 • 2 $351,000 1,300 $270 • 3 $371,000 1,400 $265 • 4 $550,000 2,500 $220 • 5 $600,000 3,000 $200 • 6 $200,000 1,100 $182 • Cost Indices • Years Indices • 2004 3980 • 2005 4339 • 2006 4614 • 2007 4877 • Location Indices • Location Index • Las Vegas 1205 • Austin 1000 • Los Angeles 1665 • New Orleans 1050

  12. Weighted Unit Cost Estimate • Solution: • From historical data: • Average cost of building per SF = ($250 + $270 + $265 + $220 + $200 + $182)/6 = $231.17 • Minimum SF Rate = $182 • Maximum SF Rate = $270 • Weighted Unit Cost • = ($182 + 4 x $231.17 + $270) / 6 • = $229.45 / SF • Conceptual cost estimate for 2,000 SF of building in Las Vegas, in 2004 • = 2,000 SF x $229.45/ SF • = $458,900 • Adjustment for time • Find out the average yearly interest rate • {4877 / 3980} = (1+i)n • Where is i = average yearly interest rate • n = number of years = 3 • Substituting the n value • 1.225 = (1+i)3 • i = 7% • Time Adjustment factor building in the year 2009 (n = 5 years) for Las Vegas • = (1+.07)5 • = 1.402 • Adjustment Factor for Location • = (1050 / 1205) = 0.871 • Adjusted Cost for the building • = 1.402 x 0.871 x $458,900 • = $560,382

  13. Detailed Estimate – Labor Cost • Straight time wage rate • Overtime wage rate • Workdays – more than 8 hour per day should be paid 1.5 times straight time wage rate • Weekends – all hours should be paid 2 times straight time wage rate

  14. Cost of Labor • Straight time or overtime wage • Social security tax (FICA) • Unemployment compensation tax • Worker’s compensation insurance • Public liability and property damage insurance • Fringe Benefits

  15. Labor Production Rates • Number of units of work produced by a person in a specified time (hour or day) • For example 1,000 bricks laid in 12 hours • Production rate should be calculated considering person will not work 60 min/hr • Production rates depend upon • Climatic condition • Job supervision • Complexity of the job

  16. Labor Cost Calculation • Brick layers Cost • Base wage = $22.50/hr • FICA Tax = 7.65% • Unemployment Tax = 3.00% • Worker’s Compensation Insur. = $15 per $100 • PL & PD Insurance = $2.50 per $100 • Fringe Benefits = $3.50 per hour Work 10 hours and 6 days per week (Mon – Saturday)

  17. Labor Cost Calculation • Actual Hours and Pay Hours Calculation: • Actual hours = 10 hours/day X 6 days • = 60 hours • Pay hours: • Normal hours = 8 hours/day X 5 days • = 40 hours • Overtime hours in weekdays • = 1.5 X 2 hours/day X 5 days • = 15 hours • Overtime hours in weekend • = 2 X 10 hours/day x 1 day • = 20 hours • Total Pay hours = 40 hours + 15 hours + 20 hours • = 75 hours

  18. Labor Cost Calculation • Average Hourly Pay Rate Calculation: • Average actual hourly pay = $22.50/ hour X (75hr / 60 hr) • = $28.12/ hour • 3. FICA Tax Calculation: • FICA tax per hour = 7.65% of actual hourly pay • = 0.0765 X $28.12/hour • = $2.15 /hour • 4. Unemployment Tax Calculation: • Unempl. tax per hour = 3.00% of actual hourly pay • = 0.03 X $28.12/hour • = $0.84 /hour • 5. Worker’s Compensation Insurance Calculation: • Work Comp. Ins. per hour = $15 per $100 of base hourly pay • = ($15/$100) X $22.50/hour • = $3.37 /hour • 6. PL & PD Insurance Calculation: • PL & PD Ins. per hour = $2.50 per $100 of base hourly pay • = ($2.50/$100) X $22.50/hour • = $0.56 /hour • 7. Fringe Benefits: = $3.50 /hour

  19. Labor Cost Calculation • Total cost per hour = $28.12 + $2.15 + $0.84 + $3.37 + $0.563 + $3.50 = $38.54 • Total cost per day = 10 hours x $38.54 /hours = $385.40 • Total cost per week = 60 hours x $38.54 /hours = $2,312.40 • Total cost per month = $2,312.40 /week x (52 weeks / 12 months) = $10,020.40 • Total cost per year = $10,020.40 /month x 12 months = $120,244.80

  20. Equipment Costs • If Equipment is rented: • Rental Cost • Monthly rent fixed by the rental company • Rate obtained from Rental Rate Blue Book by Dataquest Inc., DOT etc. • Does not include operating cost • Include fuel, oil and lubricants • If Equipment is purchased • Ownership Costs • Operating Costs

  21. Ownership Costs • Cost associated with equipment whether it is used or not • Useful life – expected duration the equipment can be used • It includes • Investment cost • Depreciation • Taxes and insurance

  22. Investment Costs • Cost of purchasing equipment • Money borrowed from bank or lender • Interest cost associated with borrowed money • Interest cost will depend upon economic condition • Most contractor will use current interest rate and add interest for risk in buying equipment

  23. Depreciation Costs • Loss of value due to use and age • Salvage value – equipment value at the end of useful life • If P is purchase amount and F is salvage value Depreciation = P - F

  24. Depreciation Methods • Straight line depreciation method • Yearly depreciation = (P – F) / No. of years • Depreciation cost is determine for • Estimate equipment cost for the project • Depreciate cost for tax purpose • Straight line depreciation is used for estimate • Double-declining-balance and sum of years digit method are used for tax purpose

  25. Hourly Ownership Cost • Ownership cost is calculated on hourly basis method • This method used time-value-of-money • Generally provided in Engineering Economic analysis book • This method used two equations • Capital Recovery Equation • Sinking Fund Equation

  26. Capital Recovery Equation A = Equivalent Annual Value P = Purchase Price i = Annual Interest Rate n = Useful Life in Years Economic Analysis Table A = P x (A/P, i%, n)

  27. Sinking Fund Equation A = Equivalent Annual Value F = Future Salvage Value i = Annual Interest Rate n = Useful Life in Years Economic Analysis Table A = F x (A/F, i%, n)

  28. Engineering Economic Table

  29. Steps for Determining Annual Ownership Cost • Obtain Purchase Price (P) • Estimate Salvage Value (F) • Estimate Useful Life (n) • Estimate Interest Rate • First estimate interest rate for borrowing money • Estimate taxes, insurance, and storage and convert these in equivalent interest rate • Add these two rates to get Minimum Attractive Rate of Return (MARR) • Use Capital Recovery and Sinking Fund equations to find out annual ownership cost

  30. Equipment Annual Ownership Cost • Net equipment value = $90,000 • Expected salvage value = $20,000 • Useful life = 5 years • Interest Rate = 17%

  31. Operating Costs • Maintenance and Repair Costs • Replacement cost, labor, oil & lubricating services • Varies depending upon the equipment • Cost is expressed in % of purchase cost or depreciation cost • Fuel Costs • Consumption per hour calculated from the equation • Lubricating Oil Costs • Consumption per hour • Cost of Rubber Tires • % of depreciation cost

  32. Fuel Consumption Cost • Fuel Consumed per hour = OF x Engine HP x [0.04 gal/(hp-hr)] for Diesel engine = OF x Engine HP x [0.06 gal/(hp-hr)] for Gasoline engine OF = Operating Factor = Time Factor x Engine Factor Time Factor = % time the engine runs in one hour Engine Factor = % of time the engine operated at a full power

  33. Lubricating Oil Consumption Cost • Oil Consumed per hour = [HP x OF x 0.006 lb/(hp-hr) / 7.4 lb/gal] + c/t Where, OF = Operating Factor = 0.6 c = capacity of crankcase in gallons t = hours between oil change For example Engine 100 hp, crankcase capacity 4 gal and oil change every 100 hrs Oil consumed =(100 x 0.6 x 0.006/7.4) + (4/100) = 0.089 gal/hr

  34. Handling and Transporting Material • Loading • Material being loaded in truck • Haul (Loaded) • Material being transported to site • Unload • Material being dumped on site • Return (Empty) • Truck returns back to loading site

  35. Cycle Time • Total time to complete these four activities is called cycle time • Hauling and Returning time can be combined if they require same time • Distance between material site and jobsite • Effective speed of the vehicle • Vehicle • Traffic congestion • Condition of road • Other factors

  36. Production Rate • Total quantity of work done per hour • Labor and Equipment Production rate • Labor and equipment can be involved together • Only equipment can be involved • Only labor can be involved • Analysis process is very important • Estimate must be done with different alternatives

  37. Earthwork Measurement • Excavated: Cut or bank measure ( Unit weight 95 lb/cf to 105 lb/cf) • Hauled: Loose measure ( Volume increases due to swell but unit weight reduces- 80 to 95 lb/cf) • Compacted: Fill or compacted measure (Volume decreases but unit weight increases – 110 to 120 lb/cf)

  38. Soil Report • Soil classification • Unit weight • Moisture content • Swell factor ( % of swell) - % gain in volume compared to original volume • Shrinkage factor- When soil is compacted, volume decreases

  39. Correlation between Volume, Swell, and Shrinkage • L = (1 + Sw /100) B • C = (1 – Sh /100) B Where L = Volume of loose soil (Loose measure) B = Volume of undisturbed soil (Bank measure) C = Volume of compacted soil (Fill or Compacted measure) Sw = % of swell Sh = % of shrinkage

  40. Correlation between Weight, Swell, and Shrinkage • L = B / (1 + Sw /100) • C = B / (1 – Sh /100) Where L = Unit weight of loose soil (Loose measure) B = Unit weight of undisturbed soil (Bank measure) C = Unit weight of compacted soil (Compacted measure) Sw = % of swell Sh = % of shrinkage

  41. Earthwork Excavation Estimation • Example Volume of earthwork = 60,000 cy bank measure Production rate of backhoe = 300 cy / hr bank measure Swell factor = 25% Capacity of truck = 20 cy loose measure = (20/1.25) bank cy = 16 bank cy Hauling distance = 4 miles Average speed = 30 mph Dump time = 4 minutes Truck wait time = 3 minutes Efficiency 45 minute per hour • Cost data: Cost of backhoe = $75.00/ hr Cost of operator = $35.00 /hr Cost of trucks = $45.00 /hr Cost of drivers = $30.00 /hr Cost of supervisor = $45.00 /hr Find out production rate in bank cubic yard, cost per cubic yard of bank measure, and total cost.

  42. Production Rate of Hauling • Solution: • Cycle Time • Loading time: (16 cy / 300 cy) = 0.05 hrs • Hauling time and return ( 4+4) mi / 30 mph = 0.27 hrs • Dump Time = 4 min = 4 min / 60 min/hr = 0.07 hr • Waiting to load = 3 min = 3 min/ 60 min/hr = 0.05 hr • Total cycle time = 0.05 hr + 0.27 hr + 0.07 hr + 0.05 hr • = 0.44 hrs • Production Rate • Number of trips per hour = 1/ 0.44 hrs / trip • = 2.3 trips • Quantity hauled per hour = 16 cy/ trip x 2.3 trip/hr bank measure • = 36.8 cy/ hr bank measure • Assuming operating factor as 45 min/ hr • Production rate = (45 min / 60 min/hr) x 36.8 bcy/hr • = 27.6 bcy/hr

  43. Cost Estimation of Hauling • Let us find out number of trucks to balance the production rate. Loading time = 0.05 hr Total cycle time = 0.44 hr Number of trucks required = 0.44 hr / 0.05 hr / truck = 8.8 trucks • Consider using 8 trucks: If 8 trucks are used, then there are less trucks used than required. Therefore, the production rate will be governed by the production rate of the trucks. Production rate of backhoe = 300 cy/hr x (45 min / 60 min/hr) = 225 cy/ hr bank measures Production rate of 1 truck = 27.6 cy/ hr bank measures Production rate of 8 trucks = 8 x 27.6 bcy / hr = 220.8 bcy/ hr Use production rate of 225 cy/ hr Cost per hour: = $75.00 + $35.00 + 8 ($45.00 + $30.00) + $45.00 = $755/ hr Cost per cubic yard bank measure = $755.00 /hr / 220.8 bcy/hr = $3.42 per cy bank measure

  44. Cost Estimation of Hauling • Consider using 9 trucks: If 9 trucks are used, then there are more trucks used than required. Therefore, the production rate will be governed by the production rate of the loader. Production rate of backhoe = 300 cy/hr x (45 min / 60 min/hr) = 225 cy/ hr bank measures Production rate of 1 truck = 27.6 cy/ hr bank measures Production rate of 9 trucks = 9 x 27.6 bcy / hr = 248.4 bcy/ hr Use production rate of 225 cy/ hr Cost per hour: = $75.00 + $35.00 + 9 ($45.00 + $30.00) + $45.00 = $830/ hr Cost per cubic yard bank measure = $830.00 /hr / 225cy/hr = $3.34 per bcy bank measure • Therefore using 9 trucks is more economical • Total Cost = 60,000 bcy x $3.34/bcy = $200,400

  45. Sample Question from NCCES • Find the productivity of truck in bank measure: • Maximum vehicle weight = 37,800 lb • Empty vehicle weight = 10,800 lb • Heap capacity = 12 yd3 • Struck capacity = 10 yd3 • Load + haul + return + dump times = 17 min • Delay time = 5 min/ hr • Bulk density = 110 pcf • Loose density = 100 pcf

  46. Solution for productivity • Find the volume of truck in bank cubic yard • Weight of earth in each truck = 37,800–10,800=27,000 lbs • Volume (bank) = Wt/ Bank Density = 27,000 / 110 x 27 = 9.09 yd3 • Find no. of trips per hour • Trips per hour (considering 55 min.) = 55/ 17 = 3.24 trips • Multiply no. of trips per hour and volume measured in bank cubic yard • Productivity = 3.24 trips/ hr x 9.09 yd3/ trip = 29.45 yd3

  47. Sample Question from NCCES • Size of concrete wall = 72 ft (L) x 12 ft (B) x 1 ft (thick) • Build in three equal pours. Include 10% waste on concrete • Labor rate: Carpenter= $32.73/hr, Laborer = $26.08/hr Supervisor = $35.37/hr • Productivity: Erect forms = 5.5 ft2/LH, Strip forms = 15 ft2/LH, Place concrete = 2.2 yd3/LH • Crews: Erect and strip : 4 Carpenters, 2 Laborers, 1 Supert. • Place concrete: 3 Laborers, 1 Carpenters, 1 Supervisor • Materials: Formwork Initial erection: $2.66/ft2, Reuse (2 times) = $0.34/ft2, Concrete = $97.20/yd3, Reinforcing sub contract = $120/ yd3

  48. Cost Estimation • Volume of concrete = 1.10(72*12*1 /27) = 35.2 cubic yard • Area of Wall = 2 sides ( 72*12) = 1728 ft2 • Material Cost • Form work (initial) = (1728*2.66)(1/3) = $1,532 • Reuse formwork = (1728*0.34)(2/3) = $392 • Concrete = 35.2* 97.20 = $3,421 • Reinforcing subcontract = (35.2/1.1) x $120 = $3,840 Total material cost = $9,185

  49. Cost Estimation (contd.) • Labor Cost: • Erect hours = 1728/5.5 = 314.2 hrs • Strip hours = 1728/15 = 115.2 hrs • Total erect and strip hours = 429.4 hrs • Concrete placing hours = 35.2/2.2 = 16 hrs • Erect and strip cost / LH = (4*32.73 + 2*26.08 + 35.37)/7 =$31.21/LH • Concrete place cost/hr = (3*26.08 + 32.73+35.37)/5 = $29.27/ LH

  50. Cost Estimation (contd.) • Labor Cost • Erect and strip cost = 429.4 LH x $31.21 = $13,402 • Concreting placing cost = 16 LH x $29.27 = $468 Total Cost = $13,870 • Total Cost Material cost = $9,185 Labor cost = $13,870 Total cost = $23,055