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PE Review Course Construction Engineering. Pramen P. Shrestha, Ph.D., P.E. October 5, 2009. Topics to be Covered. Construction Estimating Construction Scheduling Project Controls Engineering Economics. Construction Estimating. Construction Cost consists of Direct Cost

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## PE Review Course Construction Engineering

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**PE Review CourseConstruction Engineering**Pramen P. Shrestha, Ph.D., P.E. October 5, 2009**Topics to be Covered**• Construction Estimating • Construction Scheduling • Project Controls • Engineering Economics**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**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**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]**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**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**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**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**Adjustments**• Time • Location • Size • Complexity • Need appropriate contingency**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**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**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**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**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**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)**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**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**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**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**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**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**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**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**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**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)**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)**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**Equipment Annual Ownership Cost**• Net equipment value = $90,000 • Expected salvage value = $20,000 • Useful life = 5 years • Interest Rate = 17%**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**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**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**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**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**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**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)**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**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**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**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.**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**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**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**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**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**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**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**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**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

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