EXAMPLE 11.6 The floor of a kitchen, measuring 3.20m X 3.001, is to be finished with marble effect floor tiles each measuring 330 mm x 330mm. If there are nine tiles ina pack, calculate the number of packs required. Allow 10% for wastage. Solution: 3.20 x 3.00 = 9.60 m? 330mm X 330mm 0.33 m X 0.33 m = 0.1089 m? Area of the floor Area of one tile As there are nine tiles in a pack, each pack will cover 0.1089 x 9 or 0.98 m? of the floor. Area of the floor Number of p —— Area of tiles in one pack cks required = Oo 0.98 Wastage = 10% of 9.8 = 9.8 x 1 = 0.98 Therefore, number of packs required (including 10% for wastage) = 9.8 + 0.98 = 10.78, say 11EXAMPLE 11.7 The floor of a room, measuring 6.0m X 4.2 m, is to be covered with lamin- ate flooring. If one pack covers an area of 2.106 m2, calculate the number of packs required, Consider wastage © 10%. Solution: Om x 4.2m 5.20 m? Area of the floor Floor area Number of packs required = —_—fleor area_ f 4 Area covered by one pack 25.2 Wastage ® 10% = 5/100 x 11.97 = 0.6 Number of packs required = 11.97 + 0.6 = 12.57, say 13EXAMPLE 11.8 The 3D view of a room is shown in Figure 11.7. The internal wall and ceil- ing surfaces require a coat of emulsion paint. Calculate their areas in m?. Given: Size of the door = 2m x 1m wide Height of the skirting board = 100mmSolution: The area of each wall will be calculated separately and added later to find the total wall area. The areas of the window, door and skirting board have to be excluded. Area of wall 1 (see Figure 11.8a) = Length x Height = 4.0 * (2.9 - 0.1) = 4.0% 2.8=11.2m? 4.0m sl 4.0m le Wall 1 Wall 3 2.9m Door 2.0m pot Skirting board roam | ¢ t 7 Figure 11.8Areas (1) 109 Area of wall 2 = Length x Height 5.0 X (2.9 — 0.1) = 5.0 X 2.8 = 14.0m? = Gross area of the wall — Area of the door ~ Area of the skirting board =4.0 X 2.9 — 2.0 x 1.0— (1.5 X 0.1 + 1.5 X 0.1) (see Figure 1 1.8b) = 11.6 — 2.0 - 0.3 = 9.3m? The details of wall 4 are shown in Figure 11.9. Area of wall 3 50m | ware — | waaow | f Aly 29m am 2 ace 7 Figure 11.9 Area of wall 4 = Gross area of the wall — Area of the window — Area of the skirting board 5.0 X (2.9 — 0.1) — 1.2 * 1.2 5.0 x (2.8) — 1.44 = 12.560 Total wall area = 11.2 + 14.0 + 9.3 + 12.56 47.06m* 1g = 5.0 X 4.0 = 20m!11.5.1 Cavity walls The cavity wall of a domestic building consists of brickwork as outer leaf and blockwork as iner leaf and the cavity between the Owo filled with Rockwool. The outer leaf of brickwork is 103mm thick (also called half brick thick wall). The inner leaf of concrete blocks is 100mm thick. The measurements of bricks and concrete blocks used in area calculations are shown in Figure 11.10. The number of bricks and concrete blocks per m? are 60 and 10 respectively. EXAMPLE 11.9 A cavity wall 5.4 m long and 2.7 m high (Figure 11.10) is to be constructed using facing bricks and aerated concrete blocks. Find the number of bricks and blocks required. Allow 5% extra for wastage. Briek work cavity Block work 27m x @ Bricks Asrated concrete blocks ‘215 mm long x 65mm high /440 mmm long © 215 mm high 40mm thick mortar ©) Joints © ure 11.10 Cavity wall construction,Solution: Size of a brick without mortar joints = 215 x 102.5 x 65mm Size of a brick with mortar joints = 225 X 102.5 x 75mm Size of a block without mortar joints = 440 215 x 100mm Size of a block with mortar joints = 450 x 225 x 100mm Area of a brick = 225 x 75 = 16875mm? Area of a block = 450 x 225 = 101 250mm? We must have compatible units of all components. Therefore, change the length and the height of the wall into millimetres. 1 m = 1000mm Length of the wall = 5.4m = 5400mm Height of the wall = 2.7m = 2700mm Area of the wall = 5400 x 2700 = 14580000 mm? Area of the wall Area of one brick 14 580 000 = eee = 864 16 875 Number of bricks required = Areas (1) Wastage @ 5% = 864 X 5/100 = 43.2, say 44 Total number of bricks = 864 + 44 = 908 Number of blocks required = eee 144 101250 Wastage @ 5% = 144 X 5/100 = 7.2, say 8 Total number of bricks = 144 + 8 = 152 12.3 Mass, volume The density of building materials is an important property that a struc and density tural engineer has to consider for calculating the forces acting on a build- The self-weight of building components is permanent and known as the dead load. The live load on the other hand keeps on changing, andtypical examples are: weight of people using a building. wind force act- ing on a building etc.: Mass of a material/component = Density x Volume EXAMPLE 12.6 A water storage cistern measuring 2.5m x 2.0m x 1.2m high is to be provided in a building: (a) Calculate the volume of water when the cistern is filled to 15. cm from the top rim (b) Use the volume of water calculated in part (a) to determine the mass of water, The density of water is 1000 kg/m? (Q Find the load |force) on the joists supporting the cistern. 1 kg = 9.8 newton (N) Solution: (a) Convert 15 cm into metres. To do this divide 15 by 100 as tem = im 100 5 15 15cm = —2 m=0.15m 100 Depth of water in the cistem = 1.20 — 0.15 = 1.05m Volume of water = wh = 2.5 X 2.0 X 1.05 = 5.25m* (b) Density of water = 1000 kg/m? This means that 1 m? has a mass of 1000 kg Mass of 5.25 m? = 5.25 x 1000 = 5250kg (©) Load (or force) on the joists = 5250 x 9.0 = 51450N 12.4 Concrete mix Concrete is one of the important materials used in the construction of and its constituents — buildings and civil engineering structures, The constituents of a con- crete mix are cement, fine aggregates (sand), coarse aggregates (gravel, blast-furnace slag ete.) and water. Water is added for the chemical reac~ tion between cement and water to take place and produce a solid con- crete from a semi-fluid state. The amounts of these constituents may be determined by considering either their volume or mass. If the propor- tions of cement, sand and gravel are measured by volume, the concrete mix is known as a nominal mix. Mixing by volume does not take into account the moisture content of the aggregates, therefore nominal mixes are used only for minor work. Typical examples of nominal mixes are1:2:4 concrete and 1:3:6 concrete. The 1:2:4 concrete means that it is prepared by mixing I part cement, 2 parts fine aggregates and 4 parts coarse aggregates. The volume of water is about 50-60% of the volume of cement. In this section only nominal mixes are considered. If we want to prepare 1m? of 1:2:4 concrete mix, the quantities of dry materials may be calculated as shown below. The volume of cement in a 25 kg bag is 0.0166 m’. Assuming the volume of water to be 55% of the volume of cement, the proportions of the concrete mix may be written as 1:2:4:0.55. The total of these proportions is 7.55: Volume of cement = — <1 = 0.132 mi 155 Volume of fine aggregates = oa X 2 = 0.265 m? Volume of coarse aggregates = i: x 4 = 0.53 m Volume of water = a X 0.55 = 0.073 mé or 73 litres ‘The coarse aggregates contain a lot of air voids which are filled by the other ingredients when they are mixed. The final volume of the concrete mix will be less than I m°, The volume of the dry materials must be increased by 50% (approximately) to get the required quantity of con- crete, The detailed calculations are given in Appendix 1.ALT 1:2:4 concrete Appendix 1 Concrete mix A concrete mix can be prepared by two methods: (a) mixing by volume and (b) mixing by mass. In the latter method the mass of cement, fine aggregates (sand) and coarse aggregates (gravel, crushed stone, crushed bricks, blast-furnace slag ete.) are calculated for preparing the concrete mix. In this section, only the former method is described, If'we want to prepare Im’ of 1:2:4 concrete mix, the quantities of dry materials may be calculated as shown below: The volume of cement in a 25kg bag is 0.0166m', Assuming the volume of water to be 55% of the volume of cement, the proportions of the concrete mix may be written as 1:2:4:0.55. The total of these proportions is 7.55: Comentifine aggregatesicoarse aggregatesiwater IL 2 4 0.55 Volume of cemer Volume of fine aggregates = od 2 = 0.265 m? Volume of coarse aggregates = —!_ x 4 = 053m) 735 Volume of water = —1 x 0,55 = 0,073 m* or 73 litres 755 The coarse aggregates contain a lot of air voids which are filled by cement, water and fine aggregates when they are mixed. The mass of each ingredient can be calculated by multiplying its volume by density: Mass of 0.132 m* of cement = 0.132 1506 = 198.8kg Mass of 0.265 m" of sand = 0.265 X 1650 = 437.25kg Mass of 0.53:m? of gravel = 0.53 1600 = $48 kg Mass of 0.073 m’ of water = 0.073 1000 = 73 ke Total = 1557 kgim*The density of fresh concrete is 2350kg/m' The ratio in which the volumes of the dry materials should be inereased is 23501557 or 1.5 The volumes of dry materials that will give 1m’ of concrete after mixing are: Volume of cement = 0.132 x 1.5 = 0.198m* Volume of sand = 0.265 * 1.5 = 0.397m* Volume of gravel = 0.53 x 1.5 = 0.8m" Volume of water = 0,073 X 1.5 = 0,109m' or 109 litres Cement:fine aggregates:coarse aggregates: water 13 6 0.55 Volume of cement = * 1 = 0.095 m* 10.55 1 in 5 = 3 = 0.284 m? Volume of fine aggregates = 7-55 1 Volume of coarse aggregates = —— x 6 = 0.569m? olume of coarse azerezates = T= m Volume of water = * 0.55 = 0.052 m} or 52 litres 1 10.55 ‘The mass of each ingre volume by density nt can be calculated by multiplying its Mass of 0.095 m* of cement = 0.095 * 1506 = 143.1 kg. Mass of 0,284 m? of sand = 0.284 X 1650 = 468,6kg ‘Mass of 0.569:m" of gravel = 0.569 X 1600 = 910.4kg, Mass of 0.052 m' of water = 0.052 x 1000 = 52kg Total = 1574 kgfm* The density af fresh concrete is 2350kg/m’ (approx.) The ratio in which the volumes of the dry materials should is 2350/1574 or 1.49 (say 1.5) ‘The volumes of dry materials that will give Im! of concrete after mixing are: increased, Volume of cement = 0.095 * 1,5 = 0.143 m* Volume of sand = 0.284 1.5 = 0.426m* Volume of gravel = 0.569 x 1.5 = 0.854m* Volume of water = 0,052 x 1.5 = 0.078m’ or 78 litresEXAMPLE 12.7 A concrete drive to a garage is to be 10 m long, 3 m wide and 150 mm thick. Calculate: (a) The volume of concrete required to construct the drive (b) The quantities of cement, sand and gravel required, if 1:2:4 concrete is to be used Solution: (a) Convert 150mm into metres, as the units of all measurements should be the same: 150 mm = 12° = 9.15 m 1000 The shape of the drive is a rectangular prism. Volume of concrete required = lwh 10 X 3X 0.15 = 4.5m? (b) The quantities of dry materials to make 1 m? are (from Appendix 1) Cement = 0.198m’; sand = 0.397m’; gravel = 0.8m’; water = 109 litres The quantities of the above materials to make 4.5 m? are: Cement = 4.5 X 0.198 = 0.891 m? Sand = 4.5 X 0.397 = 1.787m* Gravel 5X08 = 3.60m' Water = 4.5 X 109 = 491 litres EXAMPLE 12.8 Figure 12.9 shows the plan of the proposed extension to a building. Find: (a) The volume of the earth to be excavated (b) The volume of concrete required to construct the deep strip foundationPlen m 8] Extension Take down existing wall 0.260 m+} } 39m _,) |__2.sm_) Existing building (a) 260 mm 260 mm 7 J, oattywal SLY a | DPC r 50mm GL. TTR SAY be 0.120.m Concrete 750 mm 0.5m Section aa Section b-b Oo © Figure 12.9 Solution: Figure 12.9 shows the sectional details of the deep strip foundation. The foundation may be considered to be a rectangular prism; hence its volume will be the product of the cross-sectional area and the depth. The width of the trench will be equal to the width of the foundation, i.e. 500 mm (or 0.5 m). The foundation plan, shown in Figure 12.10, has been divided into parts A, B and C to simplify the calculations.0.260 + 3.74 + 0.260 + 0.120 = 4.380 m 3 260 mm § 500 mm wide foundation trench 260 mm. ‘sl | -260 mm thick 500 mm wide cavity wall foundation’ Extension trench 500 mm 3.740 m «120 mm Existing building Figure 12.10 The total cross-sectional area will be the sum of the areas of parts A, B and C. Area of part A = 2.88 X 0.5 = 1.44m? Area of part B = 4.38 x 0.5 = 2.19m? Area of part C = Area of part B = 1.44m? Total area = 1.44 + 2.19 + 1.44 = 5.07 m? Depth of the trench = 900 mm or 0.9m Volume of earth = 5.07 X 0.9 = 4.563 m* Depth of the foundation = 750mm = 0.75m Volume of concrete = 5.07 X 0.75 = 3.803 m*15.1 Introduction 15.2 Foundations 15 Costing — materials and labour LEARNING OUTCOMES: (2) Calculate the quantities af materials jar a range of construction activities tb) Work out the cast of matertals {e) Calculate the cos! of labour for 3 range of construction activities ‘One of the important branches of building technology is to prepare the estimates of the cost involved in constructing a building. The overall cost of a project determines its viability and enables the client to arrange capital for financing the construction. The main components of the cost ‘of constructing a building are the cost of materials, labour, plant and con- actor's profit This chapter gives some examples on how to prepare approximate estimates. The costs of materials and labour are never constant and they also differ from one region to another. For latest information on the cost ‘of materials, labour and plant reference should be made to publications which update this information at regular intervals. ‘Conerete is by far the most commenly used material in the construction ‘of foundations of buildings. For plain concrete 1:3:6 concrete mix is usually used, but for reinforced concrete 1:2:4 of a stronger mix is used. Asdescribed in Chapter 12, the quantities of cement and aggregates for preparing a concrete mix may be determined by considering either their volume or mass. In this section we will consider how to work out the mass of the cement and the aggregates. This gives a better quality of concrete as we can take into account the moisture content of the-aggre- gates if they are not dry.172 Construction Mathematics Consider a 1:3:6 concrete mix. The density of concrete is approxi- mately 2400 kg/m’ or in other words, | m* of concrete has a mass of 2400kg. To prepare | m* of concrete the quantities of cement and the aggregates are: a % 2400 = 240kg (the quantity of cement is | part out of 10) = 720 kg 2 x 2400 10 Coarse aggregates 1440 kg ‘The amount of water depends on where the conerete is 10 be used. For strip foundations the amount of water could be about 50% of the amount of cement, giving a water cement ratio of 0.5. EXAMPLE 15.1 Calculate the cost of materials and labour required to construct the 1:3:6 conerete strip foundation shown in Figure 15.1, given that: Cost of materials Cement = £3.60 per 25 kg bag Fine aggregates = £32.00 per 850 kg jumbo bog, Coarse aggregates = £32.00 per 850 kg jumbo bag Labour Approximate labour hours = 2 per m? for mixing and placing concrete Hourly rate = £12.00 Solution: Area of foundation = 2[17.3 776m 7.76 X 0.2 0.6 + 0.6) x 0.6 + (6.3 x 0.6i] Volume of concrete 3.552 m3 The quantities of materials to prepare 1 m* of concrete are given in see- tion 15.2, For preparing 3.552 m’ of concrete the quantities af material and their costs are: Material Mass Cement 35% 25kg bags Fine aggregates 720 ¥ 3.552 3 ¥ 850kg jumbo bags Coarse aggregates 1440 * 3,552 6 x B50kg jumbo bagsCosting ~ materials and labour 173 ees 7 | i cutee @: ooo : screens wo soni Figure 15.1 15.3 Cavity walls Cost of cement = £3.60 x 35 = £126.00 Cost of fine aggregates = 3 = £96.00 Cost of coarse aggregates = £32 x 6 = £192.00 Total cost of materials = £414.00 Labour hours = 2 x 3.552 = 7.104 Labour cost = 7.104 X £12.00 = £85.25 Total cost (materials and labour) = £414.00 + £85.25 = £499.25 ‘The cavity walls in dwelling houses consist of a 102.5mm, outer leaf of bricks and a 100mm thick inner leaf of lightweight concrete174 Constuiction Mathematics blocks, The space between the (wo leaves, called the cavity, is 50 10 75mm wide and provided with 50mm thick Rockwool or other insula- tion material suitable for the purpose. Figure 15.2 shows a brick and a block as they are laid in the con- struction of walls. The dimensions shown are without the mortar joints, With 10mm thick mortar joints their dimensions are: Bricks: 225 > 102.5 x 75mm Concrete blocks: 450. 100 2 225 mm i (@) Brick 215 mm (8) Concrete block Figure 15.2 Surface area of | brick = 0.225 x 0.075m = 0.0169 m? I Number of bricks per m: = 59.2 say 60) 0.0169 Surface area of | concrete block = 0.450 * 0.225 m = 0,10125m™ L Number of concrete blocks per m® = = 9.88 say 10 0.10125Costing ~ materials and labour 175 ‘The quantity of mortar required for Im? of brickwork is 0.026 m' and for Lm’ of blockwork is 0.012m', EXAMPLE 15.2 Calculate the number of bricks and 100 mm thick aerated concrete blocks, and the quantities of cement and sand, tw construct a 4,0 m % 2.8m high cavity wall with a 2.0 m * 1.2 m high window. Allow for 5% extra bricks and blocks and 10% extra mortar Labourrates are: £15,00/m? for brickwork and £1 1.30/m? for blockwork Solution: Area of the wall = 4.0 X 2.8 — (2X 1.3) 8.8 x 60 = 528 Number of bricks with 5% extra allowance Number of blocks = 88 X 10 = 88 Number of blocks with 5% extra allowance = 88 % 1.05 Mortar required for brickwork = 0.026 % 8.8 = 0.2288m? Quantity of mortar with 10% extra allowance = 1.1 X 0.2288 = 0.252 m? Mortar requited for blockwork = 0.012 x 88 = 0.1056m' Quantity of mortarwith 10% extra allowance = 1.1% 0.1056 = 0.116 m? BX 105 = 555 3 Brickwork Assume the density of 1:3 cement/sand mortar to be 2300 kgim3 Mass of 0.252 m! of mortar = 0.252 x 2300 = 579.6kg ‘Mass of cement = tx 579.6 = 144.9 kg Mass of sand = + 579.6 = 434.7 kg Blockwork: 300 kgimn? 116 % 2300 = 266.8kg Assume the density of 1:6 cement/sand mortar to be ‘Mass of 0.116 m? of mortar 1 Mass of cement = 2X 266.8 = 38,1Tkg, Mass uf sand = X 266.8 = 228.69 kg Total quantity of coment = 144.9 + 38.11 = 183.01kg Total quantity of sand = 434.7 + 228.69 = 663.39kg, Cost of labour Brickwork 15,00 X 8.8 = £132.00 £11.50 X 8.8 = £101.20 Blockworke176 Consuuction Mathematics 15.4 Flooring Softwood boards and chipboard sheets are used as floor covering mater- ials in dwelling houses. The thickness of these materials depends on the spacing between the floor joists. For a spacing of 400mm between the joists, 18mm thick floorboards and chipboard sheets are satisfactory. ‘The coverage provided by tongued and geooved floor boards/chipboard is slightly Tess than their actual surface area as the tongue of one board fits into the groove of the next one. A floorboard measuring 2400 x 121 x 18mm provides coverage of 0.272 n°, Figure 15.3 shows the joists and floorboards of an upper floor. Tongued and groowed Figure 15.3. Tongued ane grocved softwood flooring.‘Costing ~ materials and labour 17 EXAMPLE 15,3 Find the cost of providing and fixing: {a) sofnwoud floorboards in a room measuring 4.5m x 3.9m {b) chipboard flooring in a raom measuring 5.0m % 4.2m Floorboards measuring 3000 ¥ 121 X 18mm thick cust £16.99 for @ pack of five, One pack covers an area of 1,71, Cost of labour is £9.00hm", Qne sheet of chipboard mezsuring 2400 * 600 x 18mm thick costs £5.30, One sheet covers an area of 1.44 m?, Labour cost is £4.50 fm! Allow wastage @ 10%. Solution: fal Area of the floor = Number of packs of floorboards = Wastage © 10% = 10.26 x 22 = 1.026 100, Tolal number of packs required = 10.26 + 1.026 = 11.286 or 12 Cost of material = £16.99 % 12 = £203.88 Labour cost = £9.00 X 17.55 = £157.95 Total cost = £203.88 + £157.95 = £361.83 tb) Area of the room = 3.0% 4.2 = 210m? ‘Area of 1 sheet of chipboard = 2.4 x 0.6 = 1.4m? Number of sheets required a 14.58 Wastage @ 10% = 14.58 x 14 = 1.458 : Too Total number of chipboard sheets = 1 Cost of material = £5.30 x 16 = £84.80 Labour cost = 21% £4.50 = £94.50 Total cust = £84.80 + £94.50 = £179.30 + 1.458 = 16.038 or 16 15.5 Painting, Painting is necessary to give doors, windows, walls ete. a finish that enhances their appearance and provides protection from dust, dirt and ‘other harmful substances. In this section only the external surfaces of178 Constuetion Mathematics Walls are considered, Some of the main factors that affect the amount of paint required are the type of surface, the number of caats and the type ‘of paint. EXAMPLE 15.4 The front wall of a house, shown in Figure 15.4, s tinished with roughcast rendering and needs (wo coats of masunry paint, The spreading rale of paint is 3m? per litre, Find the cost of paint and labour if one 3 litre can ‘of masonry pain{ costs £15.89, and the labour cost is £4.30 per m? ,- 8, zo, | g at BI | 8 g} of “ dpe 3 | 7 000 (All imensions in mm} Figure 15.4 Front elevation, Solution: The net area of the wall is calculated by subtracting the atea of the deor and window from the total wal! area; Area of roughest finish on the wall 9.0 x (3.0 = 0.15) = 20% 1.2 = 21% 10 = 25.65 — 2.42.1 =21.15m? Area of wall Volume of paint required = — ————__ Paint req Spreading rate of paint = 2115 — 7.05 bites 3 Two cuats of paint require 2% 7.05 of 14,1 litres of paint 14a Number of 5 litre cans 82 say 3Costing materials and labour 179 Cost of paint = £15.89 x 3 = £47.67 Labour cost = 2 (ovals) X 21.15 X £4.30. Total cost = £47.67 + £181.89 = £229.56 181.89 EXERCISE 15.1 ‘The answers to Exercise 15.1 can be found in Appendix 2 For exercises 1 to 4 refer to the relevant example for material and labour costs 1. Caleulate the cost of materials and labour required to construct the 13:6 concrete strip foundation shown in Figure 15.5 50 e600 3 I (Allimensions in mma) (@) Foundation plan 750 Deep strip foundation |/- deer sta {All dimensions to mim) (0) Section, Figure 15.5180) Construction Mathomatcs Brickwork 450m wide deop stp foundation Figure 15.5. (Contimed) 2. Calculate the number of bricks and 100mm thick aerated concrete blocks, and the quantities of cement and sand, to construct @ 5.51% 2.7m high cavity wall with Ow 1.801m x 1.2m high windows, Allow for 5% extra bricks and blocks and 10% extra mortar 2. Find the cost of providing and fixing {a) Softwood flocrboards in a rourn measuring 5.0m x 42am {} Chipboard flooringin a room measuring 4.3m X 3.9m Allow wastage @ 10% 4 The walls of a house, shown in Figure 15.6, ate finished with rougheast rendering and need lwo coals of masonry aint, The spreading rate of paint is 3m? per lie. Find the ‘cox of paint and labour if ene 5 litre can of masonry paint ‘cons £15.89, and the labour cus is £4.30 per a”