Cost Estimating in Construction Course
December 11, 2016 | Author: Nong Noc | Category: N/A
Short Description
Cost Estimating in Construction Course...
Description
Cost Estimating in Construction – Part 1, Course #406
Presented by:
PDH Enterprises, LLC PO Box 942 Morrisville, NC 27560 www.pdhsite.com
This course “Cost Estimation in Construction – Part 1” provides a study of the methods and procedures needed for estimating the cost of construction projects. Students are given exercises to take out quantities from building drawings and to select the appropriate type of tools and equipment needed for estimating the project construction costs. Knowledge about the labor and equipment required and their costs are also imparted. Bond, tax, and insurance requirements are also dealt with. Part 1 of the course offers hands- on experience for estimating project costs in the areas of transport of materials, earthwork excavation, and foundations. To receive credit for this course, each student must pass an online quiz consisting of twenty-five (25) questions. A passing score is 70% or better. Completion of this course and successfully passing the quiz will qualify the student for four (4) hours of continuing education credit. Course Author: JN Ramaswamy, PhD, PE
Copyright © J.N. Ramaswamy, Ph D, PE
www.PDHSite.com
TABLE OF CONTENTS Page No I. II. III. IV. V. VI. VII.
Introduction………………………………………………………………………………………………………………………….2 Estimating Process…………………………………………………………………………………………………………………4 Conceptual Cost Estimating……………………………………………………………………………………………………6 Cost of Labor and Equipment……………………………………………………………………………………………….14 Handling and Transporting Material……………………………………………………………………………………..20 Earthwork and Excavation…………………………………………………………………………………………………….25 Foundations………………………………………………………………………………………………………………………….37 List of Figures
II.1. Estimating work progress………………………………………………………………………………………………………4 III.1. Cost capacity curve ………………………………………………………………………………………………………………10 VII.1. Cross section of a trench with shoring, wales and braces…………………………………….……………….37 List of Tables III.1. III.2. III.3. III.4. III.5. III.6. VI.1. VI.2. VI.3. VI.4. VI.5. VI.6. VI.7. VI.8. VI.9. VI.10. VI.11. VI.12. VII.1. VII.2. VII.3. VII.4. VII.5.
Cost data of previously constructed projects……………………………………………………………………………7 Unit cost per car space……………………………………………………………………………………………..……………..7 Construction cost indices for Time ………………………………………………………………………………………….8 Construction cost indices for Location ……………………………………………………………………………………..9 Equipment with factors……………………………………………………………………………………….....................12 Equipment with purchase price……………………………………………………………………………………………….13 Range of swell factors………………………………………………………………………………………………………………26 Approximate unit weight of materials………………………………………………………………………………………26 Rates of handling earth by hand………………………………………………………………………………………………28 Data on wheel‐type trenching machines……………………………………………………………… ………………..29 Data on ladder‐type trenching machines…………………………………………………………………………………29 Ideal output of draglines, cy/hr, bank measure……………………………………………………………………….30 Adjustment factors for angle of swing and other than ideal depth of cut for drag lines…………..31 Fill factors for front shovel buckets………………………………………………………………………………………….32 Time for elements in a shovel production cycle……………………………………………………………………….33 Adjustment factors for depth of cut and angle of swing for shovel………………………………………….33 Representative cycle time for backhoes……………………………………………………………………………..……35 Representative values of bucket fill factors………………………………………………………………………………36 Representative number of sheet piles driven/hour …………………………………………………………………39 Representative number of wood piles driven/hour, full penetration……….………………………………4o Approximate number of steel piles driven to full penetration/hour………………………………………...42 Representative rates of drilling shafts in soils……………………………………………………………………………44 Representative rates of drilling shafts in soft rock material………………………………………………………45
I.
INTRODUCTION
The purpose of this course is to enable the student to gain fundamental knowledge of estimating the cost of construction projects. The information contained in this course presents the basic concepts to assist the student in understanding the estimating process and procedures. Estimation is not an exact science. Knowledge of construction, common sense, and judgment are required. The estimator and his team play a vital role in preparing estimates. The estimator must review and check all parts of an estimate to ensure realistic costs. He must also document the estimate so that it can be used for cost control during the construction process. The aim of estimation is to determine the costs required to complete a project in accordance with the contract plans and specifications. For a given project, the estimator can determine the direct costs for labor, materials, and equipment. The bid price can then be determined by adding the costs for overhead (indirect costs), contingencies (costs for any potential unforeseen work), and profit to the direct cost. Because the estimate is prepared before the project is constructed, the estimate is, at best, a close approximation of the actual cost. Estimates are performed throughout the life of the project. Estimates are classified as approximate estimates and detailed estimates. Approximate estimates are also called feasibility, screening, authorization, preliminary, conceptual, order‐of‐magnitude, equipment‐factored, or budget estimates. Detailed estimates are also called final, bid/tender, or definitive estimates. Approximate estimates are sufficiently accurate for the evaluation of design alternatives or the presentation of preliminary construction estimates to the owner. Detailed estimates are prepared by determining the costs of materials, labor, equipment, subcontract work, overhead and profit. The detailed estimate is important to both the owner and the contractor. A significant amount of work is performed by subcontractors who specialize in a particular type of work. Examples include clearing, drywall, painting, roofing, guard rails, striping, signs, and fences, etc. Estimate includes both direct and indirect costs. Direct cost comprises of material, labor, equipment, and subcontractor costs. The quantity of material in a project can be accurately determined from the drawings and the different cost of material can be obtained from material suppliers. Multiplying the total quantity of work with the corresponding unit cost of the material gives the total cost of the material. If equipment is used, a similar procedure is adopted to get the equipment cost. Adding labor, material and equipment costs gives the direct cost total. Indirect cost comprises of mobilization, field office expenses, taxes, bonds, insurance, final cleanup, and overhead expenses required to complete the project. Taxes on labor and material vary depending on the location. Besides, social security tax and unemployment tax must be added. These taxes vary from year to year and they must be ascertained to include in the estimate. Bonds include bid bond, material and labor payment bond, and performance bond. The bid bond ensures the owner that the contractor will sign a contract for the bid amount. Usually it is about 5 percent of the construction cost of the project. The performance bond ensures the owner that the contractor will perform all work in accordance with the contract documents. The payment bond ensures the owner that all material labor and material will be paid. The above two bonds will be usually 2
about 100 to 200 percent. Insurance requirement includes workmen’s compensation, contractor’s public liability & property damage, and builder’s risk insurance. The base estimate is the total of direct and indirect costs. An appropriate contingency is added to the base estimate to account for risk and uncertainties such as pricing, escalation, schedule, omissions, and errors. Contingency is a real and necessary component of an estimate. Finally, a profit is added and it depends upon numerous factors such as size and complexity of project, accuracy and completeness of design documents, competition for work, availability of money, and volume of construction activity in the project area. It ranges from 5 percent for large projects to 30 percent for small high risk projects.
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II. ESTIMATING PROCESS Estimating is a process. Information must be assembled, evaluated, documented, and managed in an organized manner. People involved in preparing the estimate must know their role, responsibility, and authority. Effective communications among members of the estimating team are essential to selecting the estimate methodology, collecting project data, confirming historical cost information, organizing the estimate to the desired format, reviewing and checking the estimate, and documenting the estimate after it is complete. Figure II.1 shows Estimating work progress.
Figure II.1 Estimating work progress The first step is the kickoff meeting which provides background information about the project to the estimating team, expectations of the team, and any pertinent information that may be needed to prepare the estimate. The next step is to establish a work plan prior to starting the estimating process. The work plan identifies the work that is needed to prepare the estimate, including who is going to do it, when it is to be done, and the budget for preparing the estimate. The work plan should contain sufficient detail to enable all members of the estimating team to understand what is expected of them. Selection of the method of preparing an estimate depends upon the level of scope definition, time allotted to prepare, desired level of accuracy, and intended use of the estimate. As the estimate is being prepared, it is important to perform “reality checks” to make sure the costs developed are within reason. The quality of an estimate is governed by the following considerations: 1. Quality and amount of information available for preparing the estimate. 2. Time allocated to prepare the estimate. 3. Proficiency of the estimator and the estimating team. 4. Tools and techniques used in preparing the estimate. 4
The preparation of an estimate involves the following: 1. Development of construction methods. 2. Preparation of the construction schedule. 3. Material quantity takeoffs. 4. Assessment of risks for contingency. For accuracy, an investigation of the physical characteristics of the project site must be conducted. The schedule is an integral part of the estimate. Estimate documentation is essential as it improves communication between the estimating team and management, establishes a mechanism for estimate reviews, and forms a basis for early project cost control. Once the estimate is complete, a detailed review should be made. The number of reviews will vary depending on the size of the project, type of estimate and, the length of time allowed for preparing the estimate. For any estimate there should be at least two reviews, one during the development of the estimate and the other at or near the completion of the estimate. The final estimate review is more a structured process and the review meeting may be lengthy. A final project cost report must be prepared and it is an extremely valuable document to capture lessons learned for improving future estimates. It provides a real feed back to compare with the original cost estimate. Estimate feed back is an integral part of the estimating process.
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III. CONCEPTUAL COST ESTIMATING Always owners want to know the approximate cost of a project in order to evaluate the economic feasibility for proceeding with the project. A conceptual cost estimate provides such approximate cost of the project. A conceptual cost estimate is identified by the information from which the estimate is compiled and is classified into three levels as shown below: 1. Level 1 –It is an estimate prepared from the description of the project scope where there is little or no design. The accuracy of this estimate falls between +40 and ‐10 percent. 2. Level 2 – It is an estimate prepared upon completion of preliminary design and the accuracy of which falls between + 25 and ‐5 percent. 3. Level 3 – It is an estimate prepared upon completion of the final design and the accuracy of which falls between +10 and ‐3 percent. The conceptual estimate is generally prepared by the owner as part of economic feasibility analysis or by the designer during the design phase for selecting the design alternatives or by the contractor for negotiated work between owner and a contractor. The preparation of conceptual cost estimates requires knowledge and experience with the work required to complete the project. Cost information from previous projects of similar type and size is essential. From the cost records of previous projects, the estimator can develop unit costs to forecast the cost of future projects. The unit cost is developed from weighting of the data that emphasizes the average value, yet it accounts for the extreme maximum and minimum values. The following equation is used for weighting cost data from previous projects: UC = (A + 4B + C)/6 ………………………………………Eq.(III.1) where UC = forecast unit cost A = minimum unit cost of previous projects B = average unit cost of previous projects C = maximum unit cost of previous projects Example III.1 illustrates the weighting of the cost data from previous projects to forecast the unit cost of a proposed project. 6
Example III.1: Cost data from eight previously constructed parking garage projects is shown in Table III.1. Table III.1 Cost data of previously constructed projects Project Total Cost No. of car spaces 1 $466,560 150 2 290,304 80 3 525,096 120 4 349,920 90 5 259,290 60 6 657,206 220 7 291,718 70 8 711,414 180 Use the weighted unit cost to determine the conceptual cost estimate for a proposed parking garage that is to contain 135 parking spaces. Solution: From the data provided in the Table III.1, the unit cost per parking space is calculated for each project by dividing the total cost by the respective number of spaces and the result is shown in Table III.2 below: Table III.2 Unit cost of previously constructed projects Project Unit cost per car space 1 $3,110.40 2 3,628.80 3 4,375.80 4 3,888.00 5 4,321.50 6 2,987.30 7 4,167.40 8 3,952.30 Total $30,431.50 The average cost per car space = $30,431.50 ÷ 8 = $3,803.94 7
From Eq. (III.1), the forecast unit cost (UC) will be = 2,987.30 + 4(3,803.94) + 4,375.80 6 = $3,763.14 The conceptual cost estimate for 135 car parking spaces will be = 135 x 3,763.14 = $508,023.00 The cost information of previously completed projects must be adjusted for time, location, and size before applying the cost for the proposed project because of the difference in cost due to time, location and size between the proposed project and the previous projects. III.1 Time adjustment – Since projects are constructed at different times, an adjustment representing relative inflation or deflation of costs due to factors such as labor rates, material costs, interest rates etc. must be made. Every quarter, the Engineering News Record (ENR) publishes indices of construction industry representing the economic trend of the industry with respect to time. The index can be used to adjust previous cost information for use in preparing a conceptual cost estimate. The change in value of an index between any two years can be used to calculate an equivalent compound interest rate. This equivalent interest can be used to adjust the past records to forecast future project costs. Example III.2 illustrates the time adjustment by using the indices. Example III.2: A project was constructed at a cost of $843,500 last year. A new project is proposed for construction 3 years from now. The construction cost indices for Time, as obtained from ENR, are as shown in Table III.3: Table III.3 Construction cost indices for Time Year Index 3 yr ago 358 2 yr ago 359 1 yr ago 357 Current yr 378 Determine the conceptual cost estimate of the proposed project. Solution: The cost index has increased from 358 to 378 in 3 years. An equivalent compound interest can be calculated from this change, that is, 378 = 358(1 +i)3 or (378/358) = (1 + i)3 or 1.05586 = (1 + i)3 or 1.01828 = 1 + i or i = 0.01828 or 1.828 % Using this i value and the 4 year difference in construction 8
of the two projects, the cost of the proposed project is adjusted to be = $843,500 x (1 + 0.01828)4 = $906,960. III.2 Location adjustment – The cost of construction varies from one location to the other due to the difference in costs of materials, equipment, and labor. Hence an adjustment has to be made that represents the cost difference between the locations. The ENR publishes indices of construction costs with respect to geographic locations. Using these indices, an adjustment can be made to the previous cost information to determine the conceptual cost of the proposed construction. Example III.3 illustrates the location adjustment by using construction cost indices. Example III.3: A project was constructed in City A at a cost of $387,200 and a new project is proposed for construction in City D. The location indices for construction costs of some cities are as shown in Table III.4: Table III.4 Construction cost indices for Location Location Index City A 1,025 City B 1,170 City C 1,260 City D 1,105 City E 1,240 Using the information from the above Table, determine the cost of the proposed project. Solution: The cost of a project is proportional to the cost index of the city where the project is constructed. In this example, (cost of project in City A/ city A index) = (cost of project in City D/city D index) Assume the cost of the proposed project in City D as X and rewriting the above equation with numbers, ($387,200/1,025) = (X/1,105) or X = ($387,200 x 1,105)/1,025 or X = $417, 200 The cost of the proposed project is $417,200 III.3 Size adjustment – In general, the cost of a project is directly proportional to its size. So an adjustment has to be made to represent the size of the proposed project with respect to the previously constructed project whose construction cost is available. The adjustment is a simple ratio of the sizes. Example III.4 illustrates the size adjustment. Example III.4: A project containing 62,700sf of floor area is proposed for construction. A project constructed earlier containing a floor area of 38,500sf at a cost of $2,197, 540 is to be used for determining the conceptual cost of the proposed project. What is the cost of the proposed project? 9
Solution: Assume the cost of the proposed project as X Since the cost is proportional to the size, ($2,197,540/38,500) = ($X/62,700) or X = ($2,197,540 x 62,700)/38,500 = $3,578,850 For a given project, all the three adjustments have to be made. III.4 Conceptual Cost Estimating for Industrial projects – For industrial projects, the commonly used methods include: 1. Cost capacity curve. 2. Capacity ratios, raised to an exponent. 3. Plant cost per unit of production. 4. Equipment‐factored estimates. III.4.1. Cost capacity curve – A cost capacity curve is simply a graph that plots cost on the vertical axis and capacity on the horizontal axis. These curves are developed for a variety of individual process units, systems and services and updated utilizing return cost data from completed jobs. The estimated cost is determined by locating the capacity on the horizontal axis, then following a straight line up to the point of intersection with the curve. The estimated cost is then read from the vertical axis by a straight line from the horizontal axis point of intersection with the curve to the vertical axis. The total installed cost derived from the curve may be adjusted for time and location. Example III.5 illustrates the procedure for preparing a cost capacity curve estimate. Example III.5: Cost capacity curves for different process units in a chemical plant are shown in Figure III.1 Estimate the cost for a project that has a process unit C with a capacity of 6,000 barrels per day?
Figure III.1 cost capacity curve 10
Solution: Adopt the following steps: 1. Locate the 6,000 barrels per day capacity along the horizontal axis. 2. Draw a vertical line upward until it intersects the Process Unit C cost curve. 3. Draw a horizontal line from the above intersection until it cuts the vertical axis. 4. Read the number where the horizontal line cuts the vertical axis. By following the above steps, the estimated cost is found to be $151,000. III.4.2. Capacity Ratios Raised to an Exponent – This method takes into account the economy of scale on the total installed cost. The cost of a process unit is determined by means of the following equation: Cost of Process Unit A = {Capacity of process unit A}X Cost of Process Unit B {Capacity of Process Unit B}x Eq. (III.2) where Cost of Process Unit A is to be determined Cost of Process Unit B is known Capacity of the two process units are known x is an exponent derived mathematically from historical records from completed projects Typically, the value of exponent x ranges from 0.55 to 0.88 depending upon the type of process unit. Generally a value of 0.6 is used. The use of the equation is illustrated in Example III.6 Example III.6: The cost of a 320 ft3/hr process unit is $675,000. From historical cost records, the capacity ratio exponent of the process unit is 0.72. Estimate the cost of a similar process unit with a capacity of 450 ft3/hr. Solution: Cost of Process Unit to be determined = {Capacity of Process Unit whose cost is to be determined }0.72 Cost of Process Unit already determined {Capacity of Process Unit whose cost is already determined}0.72 Substituting numbers in the above equation, (Cost of process Unit to be determined) = {450}0.72 (675,000) {320}0.72 11
Cost of Process Unit is determined to be = $675,000 x {450}0.72 = $862,797. {320}0.72 III.4.3. Plant cost per Unit of production – This method is used to estimate the total plant cost based on the average plant costs per unit of production obtained from previously completed projects. This technique assumes that the relationship between plant cost and production capacity is linear. This method is similar to the square foot estimating method used for building projects. III.4.4. Equipment‐factored Estimates – This technique involves developing a factor for each equipment. This factor is the ratio between the total installed cost to the base equipment cost. Total installed cost includes the base cost of the equipment plus costs associated with transportation, labor, insurance etc. which are necessary to install and operate the equipment. This method of estimation is illustrated in Example III.7 Example III.7: Table III.5 shows different types of equipment and their equipment factors: Table III.5 Equipment with factors Equipment Factor 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
Condensers 2.4 Control instruments 4.1 Compressors 2.5 Fans 2.7 Furnaces 2.0 Generators 1.7 Heat exchangers 4.5 Motors 1.8 Pumps 5.3 Reactors 4.0 Power vessels 3.5 Tanks 2.5
Using the information from Table III.5, estimate the cost of a plant where the following types of equipment are to be installed. The name of the equipment and its purchase price is provided in Table III.6 12
Table III.6 Equipment with purchase p[rice Equipment Price Factor Plant cost Condenser $15,000 2.4 $36,000 Control instruments 22,000 4.1 90,200 Compressors 85,000 2.5 212,500 Fans 15,000 2.7 40,500 Furnaces 140,000 2.0 280,000 Generators 25,000 1.7 42,500 Heat exchangers 95,000 4.5 427,500 Motors 55,000 1.8 99,000 Pumps 18,000 5.3 95,400 Reactors 120,000 4.0 480,000 Tower vessels 325,000 3.5 1,137,500 Tanks 140,000 2.7 378,000 Total 1,055,000 3,318,900 Solution: The equipment factor from Table III.5 for all items is posted against the respective item of equipment in Table III.6 (Equipment with purchase price) for all items. The equipment base price is multiplied by the corresponding equipment factor and the amount is shown in the last column of Table III.6. All such costs are added to give the total plant cost of $3,318,000
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IV. COST OF LABOR AND EQUIPMENT People are the most important resource on a project. The cost to hire a laborer includes the straight‐time wage plus any overtime pay, worker’s compensation insurance, social security tax, unemployment compensation tax, liability and property damage insurance, and fringe benefits. Fringe benefits include apprentice plans, pension plans, and health and welfare insurance. Straight‐ time wage normally applies to work done during the 40‐hr work week, 8 hrs/day and 5 days/week. For work in excess of 8 hr/day or 40 hr/week, the straight‐time wage rate is generally increased to 1.5 to 2 times the straight‐time rate. IV.1. Cost of labor ‐ The hourly rate of construction laborers is determined by one of the following three means: 1. Union wage ‐ construction workers who are members of a labor union are paid a wage rate established by a labor contract between their union and the construction contractor’s management. Union wage rates usually include fringe benefits, which are paid directly to the union. 2. Open‐shop wage ‐ construction workers who are not members of a union are paid a open‐shop wage agreed to by each individual employee and the employer. 3. Prevailing wage ‐ for construction workers who are employed on projects funded with state or federal money, their wage rate is established by the prevailing wage at the project location. The federal government and many states have government‐established wage for each construction craft for each geographic location. Example IV.1 illustrates the calculation of the hourly cost when a worker works overtime. Example IV.1: An iron worker works 10 hr/day, 6 days a week. A base wage of $20.97/hr is paid for all straight‐time work, 8 hr/day, 5 day/week. An overtime rate of time and one half is paid for all hours over 8 hr/day, Monday through Friday and double time for Saturday work. The social security tax is 7.65 percent and the unemployment tax is 3 percent of actual wages. The rate for workmen’s compensation insurance is $12.50/$100.00 of base wage, and public liability and property damage insurance rate is $3.25/#100.00 base wage. Fringe benefits are $3.15/hr. Calculate the average hourly cost of the iron worker.
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Solution: Actual work hours = 10 x 6 = 60 hr Pay hours = {(5 x 8)@ 1.0} + {(5 x 2)@1.5} + {(10 x 1)@2.0} = 40 + 15 + 20 = 75 hr Base wage = $20.97 Average hourly pay = (75/60) x $20.97 = $26.2125 Taxes are paid on actual wage while insurance is paid on base wage. Average hourly pay = $26.2125 Social security tax: 7.65% x $26.2125 = 2.0053 Unemployment tax: 3% x $26.2125 = 0.7864 Workers’ compensation: ($12.50/$100) x $20.97 = 2.6213 Public liability and property damage insurance: ($3.25/$100) x $20.97 = 0.6815 Fringe benefits: 3.1500 Average hourly cost = sum of all the above = $35.4570/hr From the hourly rate, the daily rate is calculated by multiplying $35.4570 by 10 = $354.57 the weekly cost can be calculated by multiplying $35.4570 by 60 = $2,127.42 the yearly cost is calculated by multiplying the weekly cost by 52 = $2,127 x 52 = $110,625.84 and the monthly cost is calculated by dividing the yearly cost by 12 = $110,625.84/12 =$9,218.82 IV.2 Cost of Equipment – All projects involve use of construction equipment to some extent. Construction equipment can be purchased or rented. The equipment is purchased when extensive use is required and it is rented if it is used sparingly. When equipment is purchased, it is necessary to determine (a) the cost of owning (ownership cost) and (b) the cost of operating (operating cost) the unit. (a) The ownership cost (cost of owning) includes the following: 1. Investment on money required to purchase the equipment. The money to purchase the equipment will be borrowed from a lender, or it will be taken from the reserve funds of the purchaser. In either case, there will be interest cost to the buyer. That is, either the lender will charge an interest for the borrowed money, or the buyer will lose any interest income on the money taken out of his reserve funds. Therefore, the interest expense or the loss of the interest income, is part of the ownership cost. Interest rate varies depending upon the economic 15
situation. An additional appropriate amount must be added to account for the risk involved in purchasing the equipment. The method used to estimate the probable ownership cost uses the capital recovery and sinking fund equations. Capital Recovery Equation is as follows: A = P{ i(1 + i)n } Eq. (IV.1) {(1 + i)n‐1} where P = purchase price A = equivalent annual value I = annual interest rate n = useful life, yrs The above equation gives the equivalent annual value (A) of the purchase price (P) assuming an annual interest rate of (i) during the useful life of (n) years. Sinking Fund equation is as follows: A = F { __i___} Eq. (IV.2) {(1 + i)n‐1} where A = equivalent annual value F = future salvage value i = annual interest rate n = useful life, yrs The above equation gives the equivalent annual value (A) of the future salvage value (F) assuming an annual interest rate (i) during the useful life of (n) years. In addition to the interest rate for the borrowed money, additional rate of interest must be added to cover the risk and an equivalent interest rate for taxes, insurance and storage must be added. The sum of all the three interest rates is called the minimum attractive rate of return (MARR). Example IV.2 illustrates the determination of annual ownership cost. Example IV.2: The purchase price of equipment is $145,000. The estimated salvage value is $25,000 after the end of its expected useful life of 6 years. Assume interest for borrowing money as 9 percent, for risk as 5 percent, and the equivalent interest rate for taxes, insurance and storage as 3 percent. Determine the annual ownership cost. Solution: 16
Minimum attractive rate of return = 9+5+3 = 17 percent or i = 0.17 Useful life n = 6 yrs Initial investment (P) = $145,000 Salvage value at the end of useful life = $25,000 Substituting the numbers in Eq. (IV.1) A = $145,000 x {0.17(1+ 0.17)6} { (1 + 0.17)6 ‐1} = $145,000 x 0.2786 = $40,397. This is the equivalent annual cost. Since there is a salvage value, its equivalent annual value must be determined and this value must be subtracted from $40,397 to get the net equivalent annual cost. Using Eq. (IV.2) and substituting numbers, A = $25,000 x { 0.17___} {(1+ 0.17)6‐1 } = $25,000 x 0.1086 = $2715 and net annual ownership cost = $40,397 ‐ $2,715 = $37,684. 2. Depreciation of the equipment over its useful life The depreciation of equipment is normally assumed to be linear over its useful life. In other words, annual depreciation is calculated by dividing the purchase price by the useful life of the equipment. 3. Taxes 4. Insurance, and 5. Storage when not in use. For items 3 to 5 above, an equivalent interest rate is added to the interest rate on the borrowed money. (b) The cost of operating the unit (operating cost) includes the following: 1. Cost of fuel – construction equipment is driven by internal combustion engines which require fuel. The fuel may be gasoline or diesel. The equipment is seldom used for 60 minutes in an hour. Most machines normally operate for 45 minutes in an hour. Moreover, the machine is not operated at its full capacity all the time. It may work at its full power only during heavy load conditions. At normal temperature and pressure, a gasoline engine will consume approximately 0.06 gallons of fuel for each actual horse‐ power developed and a diesel engine will consume approximately 0.04 gallons of fuel for each horse‐power developed. 17
For an engine that works at its full power only for a fraction of its cycle of operation and with partial hour operation, an operating factor must be established first. The operating factor consists of two parts: engine factor and time factor. The operating factor is a product of these factors. To calculate the fuel needed, the rated horse‐power must be multiplied by the operating factor and the required fuel/hp. Example IV.3 illustrates the calculation for fuel requirement: Example IV.3: A shovel is used in a digging operation. Its rated horse‐power is 160. During an operating cycle of 20 seconds, the engine is operated at full power while filling the bucket in tough ground, requiring 5 seconds. During the balance of the cycle, the engine is operated at 50 percent of its rated power and the shovel is operated only for 45 minutes in an hour. Calculate the diesel required per hour. Solution: Engine factor: Filling the bucket: (5/20) x 100% = 0.250 Rest of cycle: (15/20) x 50% = 0.375 Engine factor = 0.625 Time factor: 45/60 = 0.750 Operating factor = 0.625 x 0.750 = 0.470 Fuel required = 0.47 x 160 x 0.04 = 3.0 gallons/hour 2. Cost of lubricating oil‐ the quantity of lubricating oil required by an engine will vary with the size of the engine, the capacity of the crankcase, the conditions of the pistons, and the number of hours between oil changes. The quantity of oil consumed can be estimated by means of the following equation: Q = {(hp x 0.6 x 0.006)/7.4} + c/t Eq. (IV.3) where Q = quantity of oil required, gal/hr hp = rated horsepower of the engine c = capacity of crankcase, gal t = time between oil changes, hrs and 0.6 = operating factor (generally used) Example IV.4 illustrates the use of Eq. (IV.3). 18
Example IV.4: A shovel fitted with a 100‐hp motor is used in a digging operation. The capacity of crankcase is 4 gal and oil has to be changed every 100 hours of operation. Assuming an operating factor of 0.6, calculate the lubricating oil required for the engine. Solution: Substituting numbers in Eq. (IV.3), Q = {(100 x 0.6 x 0.006)/7.4} + 4/100 = 0.526 gal/hr In addition, engines require grease which is normally 0.5 lb/hr 3. Cost of maintenance, and repairs. The annual cost of maintenance and repairs is often expressed as a percentage of the purchase price or as a percentage of the straight‐line depreciation costs (P – F)/n
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V. HANDLING & TRANSPORTING MATERIAL Construction materials have to be transported from the storage yard of the material supplier to the job site and also from the stockpiles on the jobsite to the location where the material will be permanently installed. This involves a cost that must be included in the estimate for the project. The time required by a truck for transport of materials is divided into the following four elements: 1. Load. 2. Haul, loaded. 3. Unload. 4. Return, empty. These four elements define the cycle time for transporting material. The time required for each element should be estimated. Time for elements 2 and 4 can be determined from the speed of the vehicle and the distance traveled. Speed depends upon the vehicle, traffic congestion, condition of the road, and other factors. For determining time for elements 1 and 3, production rate must be known. Production rate is defined as the number of units of work produced by a unit of equipment or a person in a specified unit of time. The unit of time is usually I hr. The production rate will be a maximum if the work is performed at the same rate during the 60 min of the hour. But this is seldom possible due to interruptions and delays. A machine or a worker may work only for 45 min in an hr, in which case, the actual production rate is 45/60 or 0.75 of the maximum production rate. The ratio 45/60 or 0.75 is defined as the efficiency factor. Production rates are crucial to estimate the time and cost of projects. The time that labor and equipment will be on the job can be calculated by dividing the total quantity of work by the production rate. After the time to perform the work is calculated, the cost of labor and equipment can be determined by multiplying the total time by the hourly rate of labor and equipment. Example V.1 illustrates the above process: Example V.1: 175 tons of sand with a density of 100 lb/cf must be transported 7 mi using a 12‐cy dump truck. Two laborers and a driver each will load the truck at a rate of 1.5cy/hr. The haul speed is 30 mph and return speed is 40 mph. It takes 3 min to unload the truck. The cost of the truck is $25/hr, the driver is $18/hr, and the laborer is $15/hr. Determine the total time, total cost, and the cost/unit of transporting the sand if the actual working time is 45 min in one hr. 20
Solution: 1. Quantity of work: Volume of sand (175 x 2,000 x 100)/27 = 130cy 2. Cycle time: Load = 12/(3 x 1.5) = 2.667 hr Haul = 7/30 = 0.233 “ Dump = 3/60 = 0.050 “ Return = 7/40 = 0.175 “ Total cycle time = 3.125 hr/trip 3. Production rate: Number of trips/hr = 1/3.125 = 0.32 Quantity hauled/trip = 12cy x 0.32 = 3.84 cy/hr Production rate = 3.84cy x 45/60 = 2.88 cy/hr 4. Time: Using I truck and 2 laborers, time taken to transport 130cy of sand = 130/2.88 = 45.1 hrs. 5. Cost: Truck =45.1hrs x 1 x $25 = $1,127.50 Driver = 45.1hrs x 1 $18 = 811.80 Laborers = 45.1hrsx 2 x $15 = 1,53.00 Total cost $3,292.30 6. Unit cost: Cost/cy = $3,292.30/130 = $25.33 Cost/ton = $3,292.30/175=$18.81 In Example V.1, the load time (2.667 hr) is significantly greater than the travel and dump time (0.458 hr) which indicates an imbalance between loading and hauling. The load time can be reduced by using a tractor loader instead of labor. Example V.2 illustrates the use of a loader to calculate the cost of transportation of the sand mentioned in Example V.1 21
Example V.2: To load the truck with sand, a small tractor loader is rented at a cost of $50/hr. The production rate of the loader is 95cy/hr. The cost of transporting the loader to the site is $400. The loader operator cost is $21/hr. Determine the total time, cost, and cost/cy for transporting the 175 tons of sand. Solution: 1. Quantity of work: Volume of sand (175 x 2,000 x 100)/27 = 130cy 2. Cycle time: Load = 12/95 = 0.126 hr Haul = 7/30 = 0.233 “ Dump = 3/60 = 0.050 “ Return = 7/40 = 0.175 “ Total cycle time = 0.584hr/trip 3. Production rate: Number of trips/hr = 1/0.584 = 1.71 Quantity hauled/trip = 12 x 1.71 = 20.5 cy/hr Production rate = 20.85 x 45/60 = 15.4 cy/hr 4. Time: Using I truck and I loader, time taken to transport 130cy of sand = 130/15.4 = 8.4 hrs 5. Cost: Truck =8.4 hrs x 1 x $25 = $210.00 Driver = 8.4 hrs x 1 $18 = 151.20 Loader = 8.4 hrs x 1 x $50 = 420.00 Operator 8.4 hrs x 1 x $21 = 176.40 Total = $957.60 Transporting loader = 400.00 Total cost = $1,357.60
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6. Unit cost: Cost/cy = $1,357.60/130 = $10.44 Cost/ton = $1,357.60/175= $7.67 The time, total cost, and cost/unit using the loader are significantly lower than those using the laborers as in Example V.1. It is therefore justified to use the loader even though there is a charge for transporting it to the site. Use of the loader becomes economical if large quantity of work is involved. For smaller quantities, it may prove to be more expensive. Therefore it becomes necessary to calculate the quantity of material over which it is economical to use the loader. The determination of this quantity is made as shown below using the data from Examples V.1 and V.2. Hourly cost using laborers to load the truck = $25 + $18 + 2 x $15 = $73/hr Hourly cost using loader to load the truck = $50 + $21 + $25 +$18 = $114/hr Let ‘X’ cy be the quantity over which loader is economical It costs $73/hr using laborers and produces 2.88 cy/hr Therefore cost/cy = $73/2.88 and for ‘X’ cy the cost = (73/2.88) x ‘X’ ………………….. (1) Likewise it costs $114/hr using the loader and produces 15.4 cy/hr + $ 400 additionally for transportation of the loader to the job site. Therefore cost/cy = $114/15.4 and for X cy the cost = ($114/15.4)x ‘X’ + $400…….. (2) Expressions (1) and (2) must be equal {73/2.88}’X’ = {114/15.4}’X’ + 400 23.53’X’ = 7.4’X’ + 400 and ‘X’ = 22.3cy. Thus, if it is less than 22.3 cy of sand, it is more economical to use the two laborers and one truck to transport the sand. If it is more than 22.3cy, it is more economical to use the loader and one truck. It is seen in Example V.2, the load time is much less than the transport time by 1 truck and thus the loader is kept idle while the truck is transporting the material. Additional trucks must be added such that the load time and transport time balance. Example V.3 illustrates the use of multiple trucks with the loader and calculates the cost of transportation. Example V.3: Using the data in Example V.2, determine the economical number of trucks such that the load time and transport time balance. Also determine the cost/unit for transporting the material. Solution: The number of trucks required to balance the loader can be calculated by dividing the total cycle time by the load time. No. of trucks required = total cycle time/load time = 0.584/0.126 =4.63 23
Alternative 1 – consider using 4 trucks. With 4 trucks, there are fewer trucks than needed and hence the production rate is governed by the truck production rate. Quantity hauled by one truck = 20.5 cy/hr Quantity hauled by 4 trucks = 4 x 20.5 = 82.0 cy/hr Time required for transporting 130 cy of sand = 130/82 = 1.6 hrs Cost: Loader = 1.6 hrs x$ 50 = $80.00 Loader operator = 1.6 hrs x $21 = 33.60 Trucks = 1.6 hrs x 4 x $25 = 160.00 Truck drivers = 1.6 hrs x 4 x $18 = 115.20 Total labor and equipment = 388.80 Transporting loader = 400.00 Total cost = $788.20 Cost/cy = $788.20/130 = $6.07 Alternative 2 – consider using 5 trucks. With 5 trucks, there are more trucks than needed and hence the production rate is governed by that of the loader. Production rate of loader = 95 cy/hr Time required to transport 130cy of sand = 130/95 = 1.4 hrs Cost: Loader = 1.4 hrs x $50 = $470.00 Loader operator = 1.4 hrs x$ 21 = 29.40 Trucks = 1.4 hrs x 5 x $25 = 175.00 Truck drivers = 1.4 hrs x 5 x $18 = 126.00 Total labor and equipment= 400.40 Transporting loader = 400.00 Total cost = $800.40 Cost/cy = 800.40/130 = $6.16 Comparing the two alternatives, alternative 1, which is using 4 trucks, is better because it cost less for transport by $0.09/cy.
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VI. EARTHWORK AND EXCAVATION Most projects involve excavation to some extent. Usually machinery (excavating equipment) is employed in excavation. To calculate the cost of excavation, the time taken for excavation must be ascertained. The time taken by a machine depends upon its production rate which depends upon job factors such as type of soil, extent of water present, weather conditions, freedom of workers and equipment to operate on the job, size of job, and length of haul for disposal. Management factors such as organizing the job, maintaining good morale among workers, selecting and using suitable equipment and construction methods, exercising care in servicing equipment, establishing good field supervisory personnel etc. also affect the production rate. Methods of excavating vary from hand digging and shoveling for small jobs to that done by machines such as backhoe, front shovel, dragline, clamshell, scraper, bulldozer, trenching machine, and dredge. The size of the hauling unit must be balanced with the excavating equipment as described in Section V. The earth material can be measured as excavated, or hauled, or compacted. Earth that is to be excavated is called bank measure. Earth that is placed in a hauling unit for transportation is called loose measure. Earth that is to be compacted is called compact measure. Earth after being loosened during excavation and placed in the hauling unit will occupy a larger volume, with a corresponding reduction in weight/unit volume. This increase in volume is called swell. The amount of swell depends upon the type of soil and the amount of loosening during excavation. When the earth is placed in a fill area and compacted with compaction equipment, it occupies a smaller volume than in its natural state in the bank measure. This decrease in volume is called shrinkage. The amount of shrinkage depends upon the type of soil, moisture content of soil, type of the compaction equipment, and the number of passes of the compaction equipment. The correlation between unit weights, volume, swell, and shrinkage are shown in the following equations: L = {1 + Sw//100}B……………..Eq.(VI.1) and C = {1 – Sh/100}B……………………..Eq.(VI.2) where L = volume of loose soil B = volume of undisturbed soil C = volume of compacted soil Sw = percentage of swell Sh = percentage of shrinkage Equations VI.1 and VI.2 give the relationship between the volumes of different measures. Equations VI.3 and VI.4 give the relationship between the unit weight of the different measures. 25
L1 = B1/(1 + Sw/100)………..Eq.(VI.3) and C1 = B1/(1 – Sh/100)……………………..Eq.(VI.4) where L1 = unit weight of loose soil B1 = unit weight of undisturbed soil C1 = unit weight of compacted soil Sw = percentage of swell Sh = percentage of shrinkage In estimation, it is important to convert one measure into another and the above equations are important for this purpose. Table VI.1 gives the range of swell factors for some materials: Table VI.1 Range of swell factors Material Swell, % Sand or gravel 10 ‐15 Loam 15 ‐20 Common earth 20‐ 30 Hard clay 25 ‐40 Blasted rock 50 ‐60 Table VI.2 below gives the approximate unit weight of some materials: Table VI.2 Approximate unit weight of materials Material Bank weight, lb/cf Loam 80 Sand, dry 95 Sand, wet 100 Clay, dry 100 Clay, wet 110 Earth, dry 105 Earth, wet 115 Earth and gravel 120 Gravel, dry 105 Gravel, wet 125 Limestone 160 Rock, well blasted 155 Shale 130
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The relationship between the different volume and unit weight using Equations VI.1 to VI.4 is illustrated in Example VI. Example VI.1: Clay soil with a bank unit weight of 105 lb/cf is excavated, hauled by 22cy (loose measure) trucks, and placed in a fill and compacted. The empty weight of the truck is 67,500 lb. The swell factor and shrinkage factor respectively are 30 and 15 percent. Calculate the following: 1. Equivalent bank measure volume and compacted measure volume for the 22‐cy truck load of soil, and 2. The total vertical weight of the hauling unit with the pay load of the soil. Solution: Equivalent loose measure volume: By Eq. (VI.1), L = (1 + Sw/100)B 22 cy = (1 + 30/100)B B = 22/(1 + 0.30) = 22/1.3 = 16.9 cy bank volume. BY Eq. (VI.2), C = (1‐Sh/100)B = {1‐(15/100)} 16.9 = 0.85 x16.9 = 14.4 cy compacted volume Weight of soil pay load, by Eq.(VI.3), L1 = {B1/1 + (Sw/100)} = 105/{1 + (30/100)} = 105/1.3 = 80.8 lb/cy loose weight of soil Soil pay load = 80.8 x 27 x 22 = 47, 995 lb Empty weight of truck = 67,500 lb Total weight with pay load = 67,500 + 47, 995 = 115,495 lbs VI.1. Excavating by hand – Numerous types and sizes of excavating equipment is generally used in construction except in very small projects where using equipment will be more expensive as shown in Section V . Even if the job is big, sometimes there may not be sufficient room to operate the equipment such as for a motor‐pump foundation for a unit located in a confined space in a refinery. Also, if numerous underground utilities exist, it may be difficult to use equipment and so, under these circumstances, manual excavation has to be resorted to. The rate of excavation by a laborer varies with the type of material, extent of digging required, height to which the material must be lifted, and climatic conditions. Table VI.3 gives the representative rates of earth excavation by hand. 27
Table VI.3 Rates of handling earth by hand
Example VI.2 illustrates estimating the cost of earth excavation by hand. Example VI.2: A trench in a confined area in a refinery has to be excavated by hand. The trench is 15 ft long, 3 ft wide and 4 ft deep and the soil is sandy loam. Calculate the cost of excavation. Cost of labor is $15.56/hr. Solution: Quantity of work: Volume of earth = (3 x 4 x 15)/27 = 6.7cy Production rate: Using data from Table VI.1, Loosening earth = 6.7 x 2 hrs = 13.4 hrs Shoveling loose earth from trench = 6.7 x 1hr = 6.7 hrs Shoveling back to the trench = 6.7 x 0.5 hr =3.4 hrs Backfilling trench = 6.7 x 0.5 hr = 3.4 hrs Total labor hours = 26.9 hrs Cost: Total cost = 26.9 hrs x $15.56 = $418.56 Cost/cy = $418.56/6.7 = $62.47 Cost/ft = $418.56/15 = $27.90 VI.2. Excavating by machine – For larger jobs, the cost of excavation by machine is considerably less than the cost by hand once the machine is transported to the job site. The savings in excavating cost must be sufficient to offset the cost of transporting the machine to the job and back to storage after the job is completed. Otherwise, hand labor is more economical. VI.2.1. Trenching machines‐ There are two types of trenching machines: (1) Wheel‐type and (2) Ladder‐ type. Wheel‐type machine is frequently used for water mains, gas lines, and oil pipe lines. The wheel rotates at the rear of the machine, which is mounted on crawler tracks. A combination of teeth and buckets attached to the wheel loosens and removes the earth from the trench as the machine advances. The 28
earth is cast into a windrow along the trench. The machine can be used to excavate trenches 16 to 20 in. wide and depths up to 6 ft. Table VI.4 provides data on this type of machines. Table VI.4 Data on wheel‐type trenching machines
Ladder‐type machine is used for deeper trenches such as those required for sewer pipes and other utilities. Inclined or vertical booms are mounted at the rear of the machine. Cutter teeth and buckets are attached to endless chains that travel along the boom. The depth of cut is adjusted by raising or lowering the boom. By adding side cutters, the width of the trench can be increased. This machine is used to excavate trenches from 16 to 36 in. wide and depths up to 12 ft. Table VI.5 provides data on this type of machines. Table VI.5 Data on ladder‐type trenching machines
Example VI.3 illustrates earth excavation using a wheel‐type trenching machine. Example VI.3: Estimate the total cost and cost/linear foot for excavating 2,940 ft of trench in common earth using a ladder‐type trenching machine. The trench will be 30 in. wide and the average depth will be 7 ft. The machine will work for 45 min. in an hour. A machine operator, a laborer and a foreman will be employed on the job with a pickup truck. The rate for trenching machine is $87.50/hr. The machine operator is paid at $21.67/hr, the laborer is paid at $25.56/hr and the foreman is paid at $25.00/hr. The rate for pickup truck is $12.00/hr. In addition, transporting the machine to and from the job cost a lump sum of $1,500.00 29
Solution: From Table VI.5, the digging speed is found to vary between 25 and 50 ft/hr for a width of 30 in and depth of 7 ft. Use the average digging speed of 37.5 ft/hr. Time for digging trench = 2,940/37.5 = 78.4 hrs Adjusting for 45 min/hr, time required = 78.4 x (60/45) = 104.5 hrs Cost: Trenching machine : 104.5 hrs @ $87.50/hr = $9,143.75 Machine operator 104.5 hrs @ $21.67 = 2,264.50 Laborer 104.5 hrs @ $15.56 = 1,626.02 Foreman 104.5 hrs @ $25.00 = 2,612.50 Pickup truck 104.5 hrs @ $12.00 = 1,254.00 Transporting machine to and from job = 1,500.00 Total cost = $18,400.77 Cost/linear foot = $18,400.77/2,940 = $6.26 VI.2.2 Drag line – Drag lines are used for excavating earth for drainage channels and building levees where water is present. It can operate on wet ground and can dig earth out of water‐logged pits because they do not have go into pits or hole to excavate. They cannot excavate rock. Frequently they are used with a long boom to dispose of the material along the canal or near the pit. This eliminates the need for hauling units. The size of the drag line is indicated by the size of the bucket expressed in cubic yards. Most drag lines can handle more than one size bucket depending on the length of the boom and class and weight of the material excavated. Greatest output can be achieved if the job is planned to permit excavation at the optimum depth of cut. Table VI.6 gives ideal output, in bank measure, Table VI.6 Ideal output of draglines, bank measure, cy/hr
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for short‐boom drag lines when excavating at optimum depth of cut with an angle of swing of 900, based on a 60‐min hour. The upper figure is the optimum depth in feet and the lower number is the ideal output in cubic yards. Adjustments have to be made for outputs at other depths and angle of swing. Table VI.7 provides the adjustment factors. Table VI.7 Adjustment factors for angle of swing and other than ideal depth of cut for Drag lines
Example VI.4 illustrates calculation of excavation cost using a dragline. Example VI.4: A drainage ditch 10 ft wide at the bottom, 36 ft wide at the top, 12 ft deep, and 15,100 ft long is to be excavated using a dragline. The soil is good common earth and the excavated earth can be deposited on one or both sides of the ditch. A 1.5cy dragline with an average angle of swing of 1200 will be used. It will require 2 days to set up and remove the dragline. Two laborers will assist the operation which will be supervised by a foreman. The dragline will operate an average of 45 min/hr. The following cost data will apply: dragline $87.69/hr., machine operator $21.67/hr., laborer $15.56/hr. In addition, it will cost $2,500 to move the dragline to and from the job. Estimate the total cost and cost/cy for excavating the ditch. Solution: Quantity of material: Volume of earth = (10 + 36)/2 x 12 x 15,110 = 4,170,360cf = 4,170,360/27 = 154,458cy Production rate: From Table VI.6, the ideal output = 190 cy/hr, bank measure, and the optimum depth of cut = 9.0 ft Percent of optimum depth= 12/9 = 1.33 or 133% From Table VI.7, for 133% optimum depth and 1200 angle of swing, the depth‐swing adjustment factor = 0.89 For a 60‐min/hr operation, production rate = 0.89 x 190 = 169.1 cy/hr 31
For a 45‐min/hr operation, production rate = 169.1 x (45/60) = 126.8 cy/hr bank measure. Time required for full excavation = 154,458/126.8 = 1,218 hrs Cost: Dragline 1,218 hrs@ $87.69/hr = $106,806.42 Operator 1,218 hrs @ $21.67/hr = 26,394.06 Laborers 1,218 hrs x 2 @ $15.56/hr = 37,904.16 Foreman 1,218 hrs @ $25.00/hr = 30,450.00 Cost to set up and remove the dragline: Operator 16 hrs @ $21.67/hr = 346.72 Laborers 16 hrs x 2 @ $15.56/hr = 497.92 Cost to move the dragline to and from the job = 2,500.00 Total cost =$204,999.28 Cost/cy = $204,999.28/154,458 = $1.33 VI.2.3. Shovel – A shovel is a hydraulic excavator. There are two kinds of shovels: one having the digging action in an upward direction and the other having the digging action in a downward direction. The former is called front shovel or simply shovel and the latter is called hoe or backhoe or trackhoe. Shovels are mostly used for pit excavation where the bucket load is obtained from the vertical face of the excavation pit above and in front of the excavator. These machines can handle all classes of earth without loosening. The output of a shovel depends upon the class of earth to be excavated, the height of the cut, the ease with which hauling equipment can approach the shovel, the angle of swing from digging to emptying the bucket, and the size of the bucket. The size of the shovel is designated by the size of the bucket, expressed in cubic yards, loose measure. A bucket can be rated as struck capacity or heaped capacity. The struck capacity is defined as the volume in the bucket when it is filled even with, but not above, the sides. The heaped capacity is defined as the volume that a bucket will hold when the earth is piled above the sides. The heaped capacity will depend upon the depth of the earth above the sides and the base area of the bucket. The heaped slope usually is 1:1 or 2:1 above the sides of the bucket. The equipment has a fill factor which is a percentage to be used to multiply the heaped capacity to obtain the average payload of the bucket. Table VI.8 provides fill factors for shovels. Table VI.8 Fill factors for front shovel buckets
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There are four elements in the production cycle of a shovel: load bucket, swing with load, dump load, and return swing empty. Adding times for these elements provides the cycle time for the shovel. Table VI.9 provides the range of element times for shovels with bucket sizes ranging from 3 to 5cy. Table VI.9 Time for elements in a shovel production cycle
The optimum height of cut for a shovel is that depth at which the bucket comes to the surface of the ground with a full load without overcrowding or under crowding the bucket. The optimum depth varies with the class of soil and the size of the bucket and varies from 30 to 50 percent of the maximum digging height. The output has to be adjusted for the percentage of optimum heights of cut and angles of swing. Table VI.10 provides the adjustment factors. Table VI.10 Adjustment factors for depth of cut and angle of swing for a shovel
Example VI.5 illustrates the calculation of cost of excavation using a shovel. Example VI.5:
A 4‐cy shovel will be used to excavate and haul 58,640cy, bank measure, of common earth with a swell factor of 0.25. The maximum digging height for the shovel is 34 ft and the average height of cut is 15 ft. The shovel swings 1200 to load the haul units. The earth will be hauled 4 mi by 20cy, loose measure trucks, at an average speed of 30 mph. The expected time at the dump is 4 min. The truck time waiting at the shovel to move into loading position will average 3 min and the trucks will work 45 min in an hr. In addition to the shovel operator and truck drivers, 2 laborers and a foreman will be employed on the job. The rate of renting a shovel will be $135.00/hr and that for the truck is $55.00/hr. The wage rates for the workers will be as follows: shovel operator @ $21.67/hr, truck driver @ $18.17/hr, laborer @ $15.56/hr, and foreman @ $25.00. Calculate the total cost and cost/cy based on using sufficient trucks to balance the production rate of the shovel. 33
Solution: Volume of earth = 58,640cy, bank measure Volume of truck = 20/(1 + 0.25) = 16cy, bank measure Cycle time of shovel: From Table VI.9 using average values, Load bucket = 8 sec Swing with load = 5 sec Dump bucket = 3 sec Return swing = 4 sec Total time = 20 sec Cycles/min = 60/20 = 3 Production rate of shovel: Bucket fill factor for common earth from Table VI.7 = 105% Average bucket payload = 4 x 105% = 4.2cy Ideal production rate = 4.2 x 3 = 12.6cy/min Optimum height of cut = 40% x 34 = 13.6 ft Average height of cut = 15 ft Percent optimum height of cut = 15/13.6 = 1.1 or 110% Adjustment factor for 110% optimum depth of cut and 1200 angle of swing from Table VI.9 = 0.87 Probable output = 12.6 x 0.87 x 45 = 493 cy/hr, loose measure Swell factor for common earth is 25% Probable production rate = 493/(1 + 0.25) = 395 cy/hr, bank measure Cycle time for trucks: Loading truck: 16/395 = 0.04 hr Traveling: 8/30 = 0.26 hr Dump time: 4/60 = 0.07 hr Waiting to load: 3/60 = 0.05 hr Total cycle time = 0.42 hr/trip Production rate of trucks: Number of trips/hr = 1/0.42 = 2.4 Number of trips in 45 min = 2.4 x 45/60 = 1.8 trips/hr Volume hauled/trip = 1.8 x 16 = 28.8 cy/hr, bank measure Time to complete the job: 34
Number of trucks required = 395/28.8 =13.7 Use 14 trucks – there will be more trucks than required and hence the shovel production rate of 395 cy/hr will govern the time required = 58,640/395 =148.5 hrs Cost: Shovel 148.5 hrs @ $135.00/hr = $20,047.50 Trucks 145.5 hrs x 14 x $55.00/hr = 112,035.00 Shovel operator 145.5 hrs @ $21.67/hr = 3,152.99 Truck drivers 145.5 hrs x 14 x $18.17/hr = 37,012.29 Laborers 145.5 hrs x 2 x $15.56/hr = 4,527.96 Foreman 145.5 hrs @ $25.00/hr = 3,637.50 Transporting the shovel to and from the job = 2,850.00 Total cost = $183,263.24 Cost/cy = $183,263.24/58,640 = $3.13/cy bank measure VI.2.4. Backhoe –Backhoes are used below natural surface such as for trenches and basements requiring precise control of depths. It may be a wheel type or track type. The latter is called trackhoe. A backhoe is superior to the machines already described due to its convenient loading into dump trucks, its greater tooth pressure and its capacity for digging utility trenches requiring no shoring. The production rate depends upon bucket payload, average cycle time, and job efficiency. The cycle time depends upon the difficulty in loosening the soil, the angle of swing, the size of the truck the backhoe must load, and the skill of the operator. Table VI.11 provides representative cycle time for backhoes under average conditions and Table VI.12 provides representative values of the bucket fill factor for backhoes. Table VI.11 Representative cycle time for backhoes
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Table VI.12 Representative values of bucket fill factors
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VII. FOUNDATIONS Foundations that support structures include footings, piles, and drilled shafts. Footings are shallow while piles and drilled shafts are deep. Footings are placed at shallow depths, usually less than 5 ft. Footings may be isolated or continuous. Isolated footings are placed at one location to support a column. Continuous footing provides support for a wall of a building. Construction of a footing includes excavating the soil to the required depth, erecting formwork, setting reinforcing steel, placing concrete, removing formwork, and backfilling soil above the footing to the surface of the ground. A trench is excavated into the soil to construct a continuous footing. If the sides of the excavated soil are stable enough to support itself without caving in, there is no need for providing formwork. If the soil is unstable, it will be necessary to install a system of shores, braces, and solid sheeting along the excavated walls to hold the earth in position. The sheeting is typically constructed with 2‐,3‐ or 4‐in thick lumber, placed side by side or overlapping along the whole length of the trench. Wales consisting of 4 x 6‐in thick lumber can be placed in a horizontal direction to provide additional support for the sheeting. Depending upon the stability of the earth, it may be necessary to install braces at suitable intervals. Figure VII.1 shows a cross section of a trench with sheeting, wales and braces.
Figure VII.1 Cross section of a trench with sheeting, wales and braces 37
Example VII.1 illustrates the cost estimation for a trench with sheeting, wales and braces. Example VII.1 Estimate the cost of installing and removing solid sheeting and bracing for a trench 100 ft long and 7 ft deep. The sheeting will be 2 x 12‐in lumber, 8 ft long. Two horizontal rows of 4 x 6‐in wales will be placed on each side of the trench for the full length of the trench. Trench braces will be placed 4 ft apart along each row of wales. The sheeting will be driven one plank at a time by using an air compressor and a pneumatic hammer. Two laborers combined will install the sheeting at 4/hr and wales and braces at 3/hr. The following material cost will apply: sheeting $0.73/bf, wales $0.92/bf and braces $3.50 each. The following equipment cost will apply: air compressor $8.75/hr and air hammer $1.75/hr. Labor costs are $21.67/hr for driving sheet piles and $18.17 for installing wales and braces. Solution: Quantity of materials: Sheeting, using actual dimension of 2 x 12 lumber as 1.5 in x 11.25 in Number of pieces for two sides= 100/(11.25/12) x 2 = 214 pieces. = Quantity of sheeting = 214 x 2 x (12/12) x 8 = 3,424 bf Wales: Number of pieces for one row = 100/8 12.5., use 13 pieces = Quantity of lumber = 13 x 2 x 2 x 4 x (6/12) x 8 = 832 bf Trench braces = (100/4) x 2= 50 Time to perform the job: 2 laborers installing sheeting = 214/4 = 53.5 hrs 2 laborers installing wales and braces = (50 + 13)/3 = 21 hrs Labor cost: Laborers driving piles 53.5 hrs x 2 x $21.67 = $2,318.69 Laborers installing wales and braces = 21hrs x 2 x $18.17 = $763.14 Total labor cost = $3,081.83 Material cost: Sheeting 3,424 bf @ $0.73 = $2,499.52 Wales 832 bf @ $0.92 = 765.44 Braces 50 @ $3.50 = 175.00 Total material cost = $3,439.96 38
Equipment cost: Air compressor 53.5 hrs x $8.75 = $468.13 Air hammer, hose etc 53.5 hrs x $1.75 = 93.63 Total equipment cost = $561.76 Summary of costs: Labor = $3,081.83 Material = 3,439.96 Equipment = 561.76 Total cost = $7,083.55 Cost/linear foot of trench = $7,083.55/100 = $70.84 Cost/sf of sheeting = $7,083.55/(100 x 7 x 2) = $5.06 VII.1 Pile foundation – The following type of piles are in use: VII.1.1 ‐ Sheet piles ‐ Sheet piles are made in Z and flat sections and they are driven two simultaneously. The flat section is designed to interlock with each other to provide strength. They are suitable for construction of cellular structures. The Z section is designed for bending and hence it is more suitable for construction of retaining walls, bulkheads, cofferdams, and for supporting excavation. Table VII.1 gives representative number of sheet piles driven per hour. Table VII.1 Representative number of sheet piles driven per hour
From Table VII.1, knowing the length of pile and the type of soil, the number of piles that can be driven in an hour can be calculated. From this information, the total time required for the job can be
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calculated. Once the total time is determined, the total cost can be calculated by multiplying the respective labor, material and equipment cost with the number of hours. VII.1.2 Wood piles – Wood piles are classified as load‐bearing and friction piles. If the pile is driven such that it rests on the hard ground and supports the superimposed load, it is called load a bearing pile. If the soil resistance (skin friction – the friction between the pile and the adjoining soil) is adequate to oppose the superimposed load on the pile, the pile need not be driven to the hard stratum. It can be stopped at such depth where the soil resistance developed is more than the load on the pile. Such a pile is called a friction pile. Wood piles are tapered from an end point diameter of 6 to 8 in to a butt diameter of 12 to 14 in. Lengths vary from 30 to 50 ft. A preservative treatment of creosote or pentachlorophenol is generally applied. A steel boot is placed on the pile end point (tip) to reduce the tendency of breakage during driving of the pile. A reasonable allowance for extra piles should be included in the estimate since the piles may be broken during the driving process. While costing the equipment, use of crane, hammer, leads for hammer, and air compressor must be considered. Table VII.2 gives approximate number of wood piles that can be driven/hour to full penetration for various lengths in two different soil resistances. Table VII.2 Representative of wood piles driven per hour, full penetration
From the information from Table VII.2, for a given length of the pile and the nature of soil friction, the number of piles that can be driven per hour can be obtained. Then the total time can be calculated by dividing the total number of piles by the piles/hr obtained form Table VII.2. Once the total hours for the job is known, the total cost can be calculated as explained under Section VII.1.1. Example VII.2 illustrates the method of estimating the cost for driving wood piles. Example VII.2: Estimate the cost of furnishing and driving 160 wood piles 36 ft long into a soil having normal frictional resistance. The piles will be purchased in 40‐ft lengths and the tops will be cut to the required elevation after driving. The piles will be approximately 14 in. in diameter at the butt, and 7 in. at the tip. The piles will be treated with creosote preservative at a rate of 2.5 lb/cf, and breakage should be considered. The average weight of the pile will be 37 lb/ft. One crane operator, 3 laborers and one foreman will be on the job. The following cost information will apply: piles = $10.85/ft., pile points = $64.78.,
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crane = $82.40/hr., leads for hammer = $14.63/hr., hammer = $10.80/hr., air compressor = $14.65/hr., other equipment and supplies = $9.35/hr., foreman = $25.00/hr., crane operator = $21.67/hr., and general laborers = $15.56/hr. Solution: Quantity of materials: Number of piles to be driven = 160 Add 5% for breakage = 8 Total = 168 Number of linear feet of piles to be purchased = 168 x 40 = 6,720 Number of linear feet of piles to be driven = 160 x 36 = 5,760 Number of boots required = 168 Number of piles that must be cut at top = 160 Time to drive piles: From Table VII.2, for 36‐ft long pile to be driven in normal friction to full penetration, the rate of driving will be 3 piles/hr and time to drive 168 piles = 168/3 = 56 hrs. Cost of driving: Material cost: Treated piles = 6,720 ft x $10.85 = $72,912.00 Pile points = 168 x $64.78 = 10,883.04 Total material cost = $83,795.04 Equipment cost: Crane = 56 hrs x $82.40 = $4,614.40 Leads for hammer = 56 hrs x $14.63 = 819.28 Hammer = 56 hrs x $10.80 = 604.80 Air compressor = 56 hrs x $14.65 = 820.40 Other equipment and supplies 56 hrs x $9.35 = 523.60 Total equipment cost = $7,382.48 Labor cost: (Add 8 hrs for the crew to setup and take down equipment) Foreman ( 56 + 8)= 64 hrs x $25.00 = $1,600.00 Crane operator = 64 hrs x $21.67 = 1,386.88 General laborers = 64 hrs x 3 x $15.56 = 2,987.52 Total labor cost = $5,974.40 41
Summary of costs: Material = $83,795.04 Equipment = 7,382.48 Labor = 5,974.40 Total cost = $97,152.92 Cost/driven pile = $97,151.92/160 = $607.20 Cost/linear foot of driven pile = $97,151.92/5,760 = $16.86 VII.1.3 Cast‐in‐place concrete piles – These piles are suitable for use on projects where the soil conditions are such that the length of penetration is not known in advance and the depth varies among the piles driven. Tapered steel shells or steel pipes are driven to the required depth and later filled with concrete. The shells are left in place but the pipes are withdrawn as the concrete is deposited. Any length of pile, up to approximately 125 ft, can be obtained by welding extensions to the shell. In estimating the cost of cast‐in‐place concrete piles, it is necessary to determine the cost of the shells, the cost of equipment and the cost of labor to drive the shells, and the cost of concrete placed in the shells. The rate of driving varies with the length of the piles, topography of the site, class of soil into which they are driven, type of driving equipment used, spacing of the piles, and weather conditions. VII.1.4 Steel piles – These are used where it is necessary to drive piles through considerable depths of poor soil to reach solid rock or another formation having high load‐bearing properties. Standard rolled steel HP sections or wide flange beams are most frequently used. Driving operations are similar to those of other piles. A cap can be installed on the top of piles to reduce the potential for damage during the driving operation. Table VII.3 gives approximate number of steel piles driven to full penetration per hour. Table VII.3 Approximate number of steel piles driven to full penetration per hour
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From the information from Table VII.3, for a given length of the pile and the weight/linear foot of pile, the number of piles that can be driven per hour can be obtained. Then the total time can be calculated by dividing the total number of piles by the piles/hr obtained form Table VII.3. Once the total hours for the job is known, the total cost can be calculated as explained under Section VII.1.1. Example VII.3 illustrates the method of estimating the cost for driving steel piles. Example VII.3: Estimate the cost of furnishing and driving 180 steel piles, 40 ft long. The piles weigh 73 lbs/ft and are driven to full penetration into soil having high frictional resistance. A foreman with one crane operator and three laborers will be on the job. The following cost information will apply: piles = $0.32/lb., pile caps = $87.00., crane = $82.40/hr., leads for hammer = $14.63/hr., hammer = $10.80/hr., air compressor = $14.65/hr., other equipment and supplies = $9.35/hr., foreman = $25.00/hr., crane operator = $21.67/hr., and general laborers = $15.56/hr. Actual working time will be 45 min in one hour. Solution: Quantity of materials: Linear feet of piles = 180 x 40 = 7,200 Total weight of piles = 7,200 x 73 = 525,600 lbs Time to drive piles: From Table VII.3 for a 75 lb/ft pile and 40 ft long, the driving rate is 2.25piles/hr Total time for driving piles = 180/2.25 = 80 hrs With only 45 min working in one hour, actual time = 80 x (60/45) = 107 hrs Rounding this to full 8‐hr day, the total time = 112 hrs Cost to drive piles: Material cost: Piles = 525,600 lbs x $0.32 = $168,192.00 Pile caps = 180 x $87.00 = 15,660.00 Total material cost = $183,852.00 Equipment cost: Crane = 112 hrs x $82.40 = $9,928.80 Leads for hammer hrs = 112 hrs x $14.63 = 1,638.56 Hammer = 112 hrs x $10.80 = 1,209.60 Air compressor = 112 hrs x $14.65 = 1,640.80 Other equipment and supplies 112 hrs x $9.35 = 1,047.20 Total equipment cost = $15,464.96 43
Labor cost: (Add 8 hrs for the crew to setup and take down equipment) Foreman (112 + 8 )= 120 hrs x $25.00 = $3,000.00 Crane operator = 120 hrs x $21.67 = 2,600.40 General laborers = 120 hrs x 3 x $15.56 = 5,601.60 Total labor cost = $11,202.00 Summary of costs: Material = $183,852.00 Equipment = 15,464.96 Labor = 11,202.00 Total cost = $210,518.96 Cost/ pile = $210,518.96/180 = $607.20 Cost/linear foot of pile = $210,518.96/7,200 = $29.24 VII.2 Shaft foundation – Shaft foundations are installed by placing reinforced concrete in holes that have been drilled into the soil. Drilling rigs mounted on trucks can drill holes from 12 to 96 in diameter and up to approximately 100 ft depth. The bottom of the drilled hole can be under‐ reamed with a belling tool to provide an increased capacity of the foundation. When drilling into unstable soils or where water may be present, it is usually necessary to install cylindrical steel casing to hold the sides of the shaft until the reinforcing steel and concrete are placed. The casing is pulled from the hole during placement of the concrete. Table VII.4 shows representative rates of drilling various diameters of shafts into soils and Table VII.5 gives approximate rates for drilling into soft rock, such as shales. Table VII.4 Representative rates of drilling shafts in soils
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Table VII.5 Representative rates of drilling shafts in soft rock material
Vertical reinforcing steel is tied together by horizontal ties or spiral hoops, and lowered into position in the drilled shaft. Lugs attached to the reinforcing cage hold them into position at the center of the hole until concrete is placed. The cost of shaft foundations will include mobilization of equipment, drilling of the shaft, casing (if required), fabrication and placement of the reinforcement steel, and depositing concrete.
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