Costs associated with ground work, brick purchasing and roofing and cladding of the walls - Essay Example

? Assignment Production economics 2B Location Table of Contents Table of Contents 2 Introduction 3 Part 1: Groundwork 3 Part 2: Brick Work 5 Part 3: Roofing and Wall Cladding 8 Introduction This essay will be divided into three phases; the groundwork, brickwork, roofing and wall cladding. The first part deals with costs associated with ground work. This includes excavation cost and the factors to affect its success or failure to meet the set deadline. The second phase deals with cost to be associated with brick purchasing and installing of the same at the site. The last phase deals with roofing and cladding of the walls. This will be considering the cost of the sheets, the screws and the cost of installing the same at the site. This essay is about construction economics. The essay involves working out by applying reasonable assumption where applicable to assess and analyze construction activities of a six workshop unit. The plans and elevations of the plan as show in 648298_2.pdf and 648298_1.pdf are to be used in both assumption and equations. Part 1: Groundwork The ground work of a construction includes the laying down of a building’s foundation. The foundation is that part of walls, piers and a column which is in direct contact with, and transmits loads to the ground (Emmitt and Gorse, 2010). The most basic and fundamental requirement of a foundation is strength and stability; therefore the foundation should be deep enough in order to foster stability of the building. The excavated soil should be transported to another site away from the construction site in order to create space (Rasmussen, 2011). Various costs are to be incurred in the process of excavating soil for the construction of the six workshop units. The various costs to be incurred in the excavation process include; payment of the wheeled excavator, cost of hiring a lorry for transportation of unfit soil, payment for employing the loader backhoe, other general operative expenses and tipping fees (Fuss and Mcfadden,1978). Other factors such as perimeter of a building and depth to be excavated may also affect the total cost of ground work. This is because more time will be spent digging deeper, bearing in mind that all the expenses except tipping fees are paid according to time spent (Ducker, 2007). Assuming that workers are expected to work for eight hours a day, construction work is done six days in a week; the exaction work is expected to take a total of 96 hours since the contractor’s program allows maximum of two weeks of excavation. Assuming also that, on the first day of the first week, it is only the excavator and the driver were present on the site since there would have been no material for the loader and the lorry to work on the first day. That means as the excavator and general operations covered 12 days; the lorry and the loader were present in the site for 11 days in the two excavation weeks time. Assuming also, that two Lorries were used to ferry the waste material to the tipping site and as the first lorry got loaded, it was taking the same time to and fro the site and the tipping site as loading the next lorry. Since the distance from the excavation site to the tipping site is 5miles, by converting it to kilometers , its’ equal to 8 kilometers as 1 mile is equivalent to 1.6 km. Assuming the time to load the waste to be 30 minutes (0.5 hours) . Hence the lorry will travel at a speed of 32 kilometers per hour. That means in a day the Lorries will cover 16 trips, subtracting the 1 trip lost at the start of every day while loading the first lorry. The total cost to be incurred can be established as below; General operation costs (12*8*12) =1152 Euros. Lorry and Driver (11*8*38) =3344 Euros. Wheel excavator and driver (12*8*35) =3072 Euros. Loader backhoe (11*8*32) =2816 Euros. Tipping fees (15*20*11) =3300 Euros. Therefore, the total excavation cost for the two weeks will be; (1152 Euros + 3344 Euros + 3072 Euros +2816 Euros + 3300 Euros= 13684 Euros). (b) In the two weeks allocated for excavation and removal of the extracted materials, time may fail to be adequate to finish the whole work. This can be proved by working on the assumption that, the first day of excavation process will require the excavator. The excavator will spend the first day excavating the ground. The lorry that has been used can only be loaded to a maximum capacity of 20 tonnes and loading is taking half an hour. This time is same as ferrying the material to fro. The lorry will ferry (20*16) =320 tonnes in one day. This is same as what the loader will load in a day. That means the excavator will also excavates 320 tonnes in a day, as that is what it had to prepare the first day for the lorry and the loader to start work the following day. In two weeks there will be (300*12) = 3600 tonnes of waste material. In two weeks the amount ferried will be (15*20*11) =3300 tonnes. Therefore, (3600-3300) = 300 tonnes will remain unloaded and not disposed to the tipping site. It takes half an hour to load 20 tonnes as well an additional half an hour to travel the material to the tipping site. So, (300tonnes/20 tonnes per lorry) =15 trips. (15 trips*0.5 hrs)=7.5 hours+0.5 hrs to load in the first) =8 hours. That means the time deadline set out will have to be extended by another day as there will be waste left at the end of the two weeks. Part 2: Brick Work Brick work involves making of the bricks at Edenhall facing bricks and from Strawbone facing bricks. The perimeter, height and partition of the six unit workshop will determine the number of bricks to be used (Gruneberg, 1997). The overall cost of the exercise will include the cost of material, 8% on overhead cost and profit. The perimeter of the wall is (85364* 2 +25464 *2) =221656mm 60 bricks will cover an area of 1m2, so, is equivalent to 60 bricks covering 1000mm to the length of the wall and another 1000mm up the height of the wall. So, in 1000000mm2 area of the wall, there are 60 bricks. Assume that, the height covered by the bricks on the wall is 1m. Further assume that the bricks are square shaped. The area of wall will be given by perimeter*height. The height of the walls will be 1m or 1000mm. hence the areas will be (1000*221656) = 221656000mm2. If 1000mm2 has a coverage of 60 bricks, then a coverage of 221656000mm2 will be covered by; 221656000*60/1000= 13299360 bricks. For the partition, the number of bricks will be as follows; (25464*1000) = 25464000mm2 If 1000mm2 has 60 bricks, then in 25464000mm2 the number of bricks will be; (25464000*60/1000) = 1527840 bricks. So, the total number of bricks will be; (13299360+1527840) = 14,827,200 bricks. The cost of one brick from Edenhall facing bricks is; (340 pounds/ 1000 bricks) = 0.34 pounds. That means, 14,827,200 bricks will cost; (14,827,200*0.34 pounds) = 5,041,248 pounds. The number of Strawbone facing bricks will be; 1 m= 1000mm, therefore, the number of bricks around the wall will be; (221656/1000) =222 bricks since the bricks will cover an area of 3 m, it means that, there will be two sets of bricks around the wall. Hence, the number of bricks from Strawbone facing bricks will be; (222 bricks*2) = 444 bricks. The cost of strawbone facing bricks will amount to; (450/1000*444) = 199.80 pounds Assume that some of the bricks brought from Edenhall facing bricks get damaged (1 in every 20 bricks). Also, 1 brick in every 10 bricks from Strawbone will spoil. Therefore, out of 14827200 bricks Edenhall bricks used represented 95% of the bricks purchased. Hence, the actual bricks purchased from Edenhall facing bricks are; (100/95*14827200) =15607579 bricks. This increases the cost up to; (15607579*0.34) =5306576.80 pounds. In the case of Strawbone, the total bricks used represent 90% of the bricks bought. Therefore, the actual number of bricks brought in was; (100/90*444) = 493 bricks, which means the actual cost will be; (493*450/1000) =222 pounds. Assume that, the other material required in the site, that is cement, hydrated lime, building sand and colored mortar will be 2 tonnes, 4tonnes, 5 tonnes and 2 tonnes respectively. Therefore, the cost will amount to; (2*90+4*95+5*25+2*240) =1165 pounds. Assume that, it would take the contractor one year (365 days) to complete the work of installing the Edenhall bricks. Also, per day the work takes 8 hours. The labor cost will be quantified as following. General operative cost; (12*8*365) = 35040 pounds. Craft operative cost; (16*8*365) =46720 pounds. Hire of mixer; (50*12) = 600 pounds. So the total cost amounts to, 82360 pounds for installing Edenhall bricks. If it takes 50% longer to complete installing Strawbone work compared to Edenhall, then by calculation, installing Strawbone will take; It takes 50% more time to install Strawbone bricks.. The time it will take to install them will be; (365*1.5) = 548 days. The cost of labour to install Edenhall bricks excluding hire for mixer is 20160 pounds in 90 days. So, in a day it costs, (20160/90) = 224 pounds. Since the cost of labour has gone up by 6%, the cost per day then will be (224*116/100) = 260 pounds per day. In 548 days it will amount to (260*548) =142480pounds. Also, in addition will be cost of hiring the mixer in those 548 days that amounts to; (50*548) = 27400 pounds. The total cost will hence be (142480+27400) = 169880 pounds So, the total cost in the brick work phase will be; Cost of Edenhall facing bricks in total 5306576.80 pounds. Cost of Strawbone facing bricks 222 pounds. Cost of material used to be used at site 1165 pounds. Cost of walling Edenhall facing bricks 82360 pounds. Cost of installing Strawbone bricks 169880 pounds. Total cost 5560203.80 pounds. If the overhead cost is 8 %, then it amounts to (8/100*5560203.8) = 444816.30 pounds. That puts the total project cost to be (5560203+444816.30) =6005019.30pounds. By charging this part of project 95000 pounds, the contractor have made a loss of, (6005019.3-95000) = 5910019.30 pounds. The project will not be realistic to undertake since it can only result to loss making. Part 3: Roofing and Wall Cladding The lump sum price amounts to 121000 pounds while scaffolding costs 10,000. So the net amount for the rest part of work is (121000-10000) = 111000 pounds. This is the amount that will be spent on buying the materials and the cost of labor. Cost of materials delivered to the site in roofing and wall cladding phase can be found by working on this equation; (number of sheets*price per sheet) + (number of Plastic headed stainless steel screws* price per unit) i.e. (n.s*p.s) + (n.ps*p.u) Cost of labor will be equated as following; (labour cost of cladding the wall +labour cost of roofing, i.e. ( c.cw+c.r). Given that, the labour cost of cladding the wall is more expensive than labour cost for roofing, then, the equation can be put as follows; (c.r*1.15+c.r). Where c.r represents the labour cost of roofing Total cost will be; 10,000 pounds + (n.s*ps) + (n.ps*p.u) + (c.r1.15+c.r) =121,000 pounds. Length of the roof is the length of the wall which is 85364 mm. The spreading of the sheets along the length of the roof will be its corresponding width (Ducker, 2007) The width of one sheet is 900 mm. So, the number of sheets required to cover a length of 85364 mm will be; (85364/900) = 94.85, that will be 95 sheets along the length of the roof on one side. Assuming that the segmentation shown in 64828_1.pdf, on the part of roofing represents the number of sheets running up the roof, then, the number of sheets used up the roof will be 11 pieces of sheets. So, the number of sheets to be required on each side of the roof will be; (95*11) = 1045 sheets. If the cost of one coated steel sheet, of the size calculated above costs 35 pounds. Then, the total amount the sheets used for roofing will sum up to; (1045*35) = 36575 pounds. Since the roofing will be carried out on both two sides, the total cost of the roofing sheets will be; (36575*2) = 73150 pounds. The number of sheets required for cladding will depend on the height, length and width which is the perimeter of the wall to be cladded (Yanagida and Ching, 1985). The length of the wall and its width amounts to the perimeter of the wall. The perimeter of the roof as calculated above is 221656 mm. So, the number of sheets required to cover this perimeter will be; (221656/900) = 246 sheets. Whereby, 900mm is the width of one sheet. Assuming that, the larger segmentation shown in 648298_1pdf, represents the number of sheets up the wall, then the number of sheets will be 4 up the wall. So, the total number of sheets used in the cladding will be; (246*4) = 985 sheets. Given that the cost of one sheet is 35 pounds, then the total cost of cladding sheets will be; (985*35) = 34480 pounds. Assume that, in spreading the roof sheets, the steel screws were used in a ratio of 1:1000, that is, in every 1000 mm, a screw is used to attach the sheet (Gruneberg, 1997). Since the number of sheets up the roof as assumed in the assessment are 11 sheets and, one sheet has a length of 3000 mm, then the total width will be; (3000*11) = 33000 mm. So, the number of screws to be applied will be; (33000/1000) = 33 screws. Using the above ratio, the screws applied along the length of the roof will be; (85364/1000) =85.364 screws, that is approximately 86 screws. So, the total number of screws to be applied in the roofing part will be; (33*86)2 = 5676 screws. Since the cost of screws is 12 pounds per 100 screws, then the cost of the screws will be; (5676*12/100) = 681.12 pounds in the whole roofing work. The same ratio applies in attaching the screws in the cladding process (Fuss and Mcfadden,1978). Since the number of sheets is four, up the wall and the length of one sheet is 3000mm. Then the number of screws used in the wall will be; (3000*4/1000) = 12 screws. Also, to find the number of screws used along the perimeter of the wall, the same formula is used; (221656/1000) = 221.7screws, which is approximately to 222 screws. So, the total number of screws used in cladding will be; (12*222) = 2664 screws. Given that, the cost of screws is 12 pounds per 100 screws, then the cost of the cladding screws will be; (2664*12/100)= 319.68 pounds. The cost of roofing material will amount to; (73150+681.12) = 73831 pounds. The cost of cladding material will amount to; (34480+319.68) = 34799.68 pounds. The total cost of materials used will amount to; (73150+34480+681.12+319.68) = 108630.80 pounds The summation of lump sum price amount will now be; 108630.80+ (c.r1.15+c.r) =121,000 pounds. The total labour cost will be; (c.r1.15+c.r) = 12369.20 pounds To amount of each labour cost will be; c.r (1.15+1) = 12369.20 c.r=12369/2.15. c.r= 5753.11 pounds. c.r is the labour cost of roofing. So, the labour cost of roofing is 5753.11 pounds. Since the labour cost of cladding the wall is 15% more expensive than that of roofing, the labour cost of cladding will be; (1.15*5753.11) = 6616 pounds Key: n.s represents the number of insulated plastic coated steel sheet. p.s represents the price of each insulated plastic coated steel sheet. n.ps represents the number of plastic headed stainless steel screws. p.u represents the price of plastic headed stainless steel screws. c.cw represents the cost of cladding the wall c.r1.15 represents the cost of cladding the wall. c.r represents the cost of roofing. Reference Ducker, 2007. The structure of Production. New York University press: New York. Emit, Stephen and Christopher, 2010. Construction of materials,2nd ed. West Sussex: Wiley-Blackwell. Fuss and Mcfadden,1978. A dual approach to theory and application. North Holland. Gruneberg, 1997. Construction economics, Macmillan building and surveying series, London. Rasmussen, 2011. The basic theory of production optimization, Institute of food and Resource production. Copenhagen: University of Copenhagen. Yanagida and Ching 1985. Mathematical Development and Application. New Jersey: New Brunswick. Read More