Measurements and calculations are an important part of almost every task on a construction site. In this chapter we will cover:
Have another look at the chapter titled Planning Your Work. In summary, to plan and prepare for a measurement or calculation task you need to:
The basic equipment used for measuring and calculating on a construction site includes tapes, rules, trundle wheels and calculators.
Tapes & Rules
Tapes and rules are discussed on page 53 in the chapter on Hand Tools. All the measurements you make with tapes and rules should be accurate to one millimetre.
Trundle Wheel
A trundle wheel is often used for taking long site measurements that don’t have to be perfectly accurate. It consists of a wheel and axle, handle and a counter which records the measurement. Wheels generally have a circumference of one metre. As the trundle wheel is pushed over the ground, each revolution causes the counter to add one metre to the measured distance.
Calculator
Calculations can be performed with a calculator or they can be done manually (using your mathematical skills). However, it is much more efficient to do large or complex calculations with a calculator.
Always remember to double check all measurements and calculations for accuracy.
[[ sh /n ][ Linear Measurement ]]The term linear means in a line or in one dimension only, for example a length measurement.
In the General Construction industry linear measurements are in millimetres (mm) and metres (m). There are 1000 millimetres in 1 metre.
Construction drawings are mostly dimensioned in millimetres. Have a look at the drawings in the chapter on Plans & Specifications.
You will notice that the unit symbols mm and m are not shown on the dimensioned construction drawings. Very long measurements can be shown in metres. A decimal point is used to indicate a measurement in metres, for example:
Converting millimetres to metres, or metres to millimetres simply requires a decimal point to be removed or inserted, for example:
Timber is sold in metres with standard lengths starting at 1·8m and generally increasing in increments of 0·3m, for example 1·8m, 2·1m, 2·4m, 2·7m and 3·0m.
The photograph on the right shows how standard lengths of timber are often stored in racks in a timber merchant’s yard.
Timber orders are written in a fairly standard format, however there may be some slight differences between states or geographical areas. Millimetre (mm) and metre (m) symbols are generally left out.
Timber orders should include:
Sample Order
A quantity of 75x50 rough sawn hardwood is required for a job. You have worked out that you need 8 pieces 3·6 metres long, 5 pieces 2·7 metres long, 24 pieces 2·4 metres long and 4 pieces 1·8 metres long.
This is how you would write the order:
75x50 RS Hardwood 8/3·6 5/2·7 24/2·4 4/1·8
Note that the lengths are listed in order from the longest to the shortest. If you were placing the order over the phone you would say:
75 by 50 rough sawn hardwood, 8 at 3·6, 5 at 2·7, 24 at 2·4 and 4 at 1·8.
Example 1
Calculate the total linear metres in the following timber order: 100x25 Rough Sawn Hardwood 5/3·6 4/2·7 12/2·4 10/1·8.
Linear Metres = (5 × 3·6) + (4 × 2·7) + (12 × 2·4) + (10 × 1.8)
= 18·0 + 10·8 + 28·8 + 18·0
= 75·6 m
Example 2
The photograph on the right shows a typical temporary fence which is used to provide a safety barricade around a construction site.
Find the length of temporary fencing required to erect a safety barricade on a site that is rectangular in shape and measures 34·500 × 28·750.
Perimeter of Rectangle = (Length + Width) × 2
= (34·500 + 28·750) × 2
= 63·250 × 2
Fencing req’d = 126·500 m
The photograph above shows a large circular hopper that is being installed on a construction site.
Example 3
Calculate the length of barricade mesh needed to go around a circular work area 3·5 metres in diameter. Allow an extra metre for overlap.
Before leaving this page, think of what you've just been reading, and test yourself with these questions.
[[ mc /r /f ][ Converting 2.3m to mm should result in: ][ 2300mm ][ * 230mm ][ * 23mm ][ * 0.23mm ][ The unit gets smaller by a factor of 1000, so the quantity gets bigger by the same factor. ]]
[[ mc /r /f ][ Standard lengths of timber get bigger or smaller in increments of: ][ 0.3m ][ * 0.5m ][ * 0.1m ][ * 0.6m ][ Timber is sold in 1.8m, 2.1m, 2.4m , 2.7m, 3.0m lengths. ]]
[[ mr /f ][ What info should be included in a timber order: ][ The section size and shape. ][ The type of timber. ][ The type of finish. ][ * The straightness of the timber. ][ The number of pieces. ][ Hopefully, the timber supplied will be straight enough to do the job. ]]
[[ mc /r /f ][ If you have to enclose an area 12m x 15m with temporary fencing, how much fencing do you need to order to the jobsite? ][ * 180m ][ * 90m ][ * 60m ][ 54m ][ * 36m ][ (12 x 2) + (15 x 2) = 24 + 30 = 54m. ]]
The accurate measurement and calculation of surface area is very important in the General Construction industry.
For example, quantities of materials that cover a surface, such as plasterboard, sheet flooring, paint, floor and wall tiles, bricks, paving and roofing materials, all need to be calculated during the planning and construction of a project.
You will find many different surface shapes in the design of buildings and their surrounds.
Some of these surface shapes are simple plane figures such as squares, rectangles, triangles, quadrilaterals and circles.
More complex shapes can generally be made up from a combination of simple shapes.
The unit of area measurement used in the General Construction industry is the square metre.
If you imagine a piece of MDF or plywood that has been cut perfectly square and measures 1·000 × 1·000, its surface area is one square metre.
The symbol used to denote square metres is (m2).
Example 1
Find the area of a rectangular shaped suburban building block which measures 36·500 × 22·000.
Area of Rectangle = Length × Width
= 36·500 × 22·000
Area of the block = 803 m2
Example 2
The floor area of a building measures 15·600 × 10·000. How many square metres of sheet flooring will be required for the job, allowing 10% waste for cutting?
Area of Rectangle = Length × Width
= 15·600 × 10·000
= 156 m2
10% of 156 = 15·6 m2
Flooring required = 156 m2 + 15·6 m2
= 171·6 m2
Example 3
How many sheets of flooring measuring 3.600 × 1·200 will be required for the building in Example 2 above?
Example 4
The walls of a room which measures 7200 × 3600 are to be lined with plasterboard. The room has one door 2100 × 900 and one window 1800 × 1200. The ceiling height of the room is 2400. Calculate the net wall area (total wall area less openings). Note: When calculating square metres measurements must be converted to metres.
Total wall area = Perimeter of the room × Ceiling height
= (7·200 + 3·600 + 7·200 + 3·600) × 2·400
= 21·600 × 2·400
= 51·84 m2
Area of openings = Area of window + Area of door
= (1·800 × 1·200) + (2·100 × 0·900)
= 2·16 m2 + 1·89 m2
= 4·05 m2
Net wall area = Total wall area – Area of openings
= 51·84 m2 – 4·05 m2
= 47·79 m2
= 48 m2 (whole square metres)
Example 5
How many litres of paint would be required for one coat on the walls and ceiling of the room in Exercise 4 above? One litre of the paint to be used covers approximately 14 m2. Allow 10% for wastage and discrepancies.
Example 6
Find the area of a circular courtyard 5·600 in diameter.
Example 7
The photograph below shows a walkway that is to be paved. The feature edging is finished and the pavers for the body of the walkway are yet to be laid.
The shape of the remainder to be covered is made up of a semicircle 6·800 in diameter and a rectangle 12·000 × 2·200.
Calculate the number of square metres of pavers that will be required to complete the job. The size of the pavers and the pattern to be laid suggest that 15% should be allowed for cutting and wastage.
Area of semicircle = πR^2 ÷ 2
= 3·14 × 3·400 × 3·400 ÷ 2
= 18·15 m^2
Area of rectangle = Length × Width
= 12·000 × 2·200
= 26·4 m^2
Total area of walkway = Area of semicircle + Area of rectangle
= 18·15 m^2 + 26·4 m^2
= 44·55 m^2
Add 15% = 44·55 m^2 + 6·68 m^2
= 51·23 m^2
Pavers required = 52 m^2
Example 8
Triangles often form part of complex shapes. Think of a triangle as half a rectangle. Find the area of a triangle with a base of 9·750 and perpendicular width 6·250.
Example 9
The illustration on the right represents the shape of a block of land which is the combination of a square and triangle shown by the shaded areas. Calculate the area of the block of land.
Area of block = Area of square + Area of triangle
= Side^2 + (Base × Perp Width ÷ 2)
= (50 × 50) + (50 × 20 ÷ 2)
= 2500 + 500
= 3000 m^2
Example 10
Quadrilateral shapes (any plane figure with four sides) are made up of two triangles as shown in the illustration on the right. The area of any quadrilateral can be found by adding the areas of the triangles. The area of parallelograms can be found by multiplying the base by perpendicular width. The area of quadrilaterals with two parallel sides can be found by multiplying half the sum of the two parallel sides by the perpendicular distance between them.
For example, to find the area of the quadrilateral with two parallel sides in the adjacent illustration:
Irregular Shapes
Irregular shapes and other regular shapes such as polygons can be divided into squares, rectangles, triangles and quadrilaterals. The area of the overall shape is found by calculating the area of the constituent shapes and adding them together.
Before leaving this page, think of what you've just been reading, and test yourself with these questions.
[[ mc /r /f ][ Which of these is not a plane shape? ][ * Square. ][ * Rectangle. ][ * Triangle. ][ Cone. ][ * Circle. ][ A cone is a solid or 3-dimensional shape. ]]
[[ mc /r /f ][ What is the standard unit of area in the general construction industry? ][ Square metres. ][ * Square millimetres. ][ * Metres. ][ * Millimetres. ][ Square metres (m2) are the industry standard unit for area. ]]
[[ mc /r /f ][ If you have to clear an area 12m x 15m of topsoil during site preparation, what area are you actually clearing? ][ 180m2 ][ * 90m2 ][ * 60m2 ][ * 54m2 ][ * 36m2 ][ 12m x 15m = 180m2. ]]
[[ mc /r /f ][ A wall measuring 8.0m x 3.0m has a doorway 2.0m x 0.8m and a window 1.0m x 3.0m. What is the paintable area of the wall? ][ * 24m2 ][ 19.4m2 ][ * 28.6m2 ][ * 17.2m2 ][ (8.0 x 3.0) - (2.0 x 0.8 + 1.0 x 3.0) = 24 - (1.6 + 3) = 24 - 4.6 = 19.4m2 ]]
[[ mc /r /f ][ The end-wall of a brick building has a width of 9.0m, with a centre height of 4.0m and side heights of 3.0m. What is the area of this triangle-on-rectangle wall? ][ * 36.0m2 ][ * 27.0m2 ][ * 63.0m2 ][ 31.5m2 ][ ☐ area + △ area = (w x h) + (w x h / 2) = 9.0 x 3.0 + 9.0 x (4.0 - 3.0) / 2 = 27 + 4.5 = 31.5m2 ]]
Volume measurements in the General Construction industry are used to work out quantities of materials such as concrete, sand, fill and soil removed from excavations.
The unit of volume measurement used in the General Construction industry is the cubic metre.
Imagine a square box (like the cube shown below) which measures 1·000 × 1·000 × 1·000. The amount of material the box would hold is described as one cubic metre. The symbol used to denote cubic metres is (m^3).
In General Construction you will work out volumes mainly for box shapes and cylinders.
The basic formula for finding their volume is area of base multiplied by the perpendicular height.
Sometimes you may need to work out volumes of cone and pyramid shapes.
The formula for calculating these volumes is area of base multiplied by half the perpendicular height.
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Waste Allowance Factor
Working out quantities of concrete would probably be the most common use for volume calculations in the General Construction industry. Concrete is either poured onto the ground or into prepared forms.
Concrete poured onto the ground requires a larger allowance for discrepancies because the surface will usually be uneven. An allowance of 10% is generally applied when concrete is poured onto the ground and 5% when poured into forms.
In some of the area calculations we worked out the amount of the waste allowance, then added the waste allowance to the net amount to get the total material required. Have another look at Example 2 on page 134.
Another way to calculate the total is to multiply the net amount by a waste allowance factor. For example, the factor for a waste allowance of 15% is 1·15, 10% is 1·10 and 5% is 1·05. The numeral 1 represents the net amount and the waste allowance is included after the decimal point.
You will find multiplying the net amount by the waste allowance factor is a very convenient method, particularly when you are using a calculator.
Example 1
In example 6 on page 135 the area of a circular courtyard was found to be 25·0 m2. The courtyard is to be concreted to a depth of 100. How much concrete would be required for the job? Allow 10% for discrepancies and wastage.
Net volume of concrete = Area of courtyard × Depth of concrete
= 25·0 m2 × 0·100
= 2·5 m3
Apply waste allowance = 2·5 m3 × 1·10 (waste allowance factor)
= 2·75 m3
Example 2
The photograph on the right shows formwork for a rectangular column being constructed with reinforced concrete. In total there are 12 of these columns in the building. The finished size of each column is 450 × 300 and the length of each column is 6·800. Calculate the volume of concrete required for the job allowing 5% for waste and discrepancies.
Volume of one column = 0·450 × 0·300 × 6·800
= 0·918 m3
Volume of 12 col’s = 0·918 m3 × 12
= 11·016 m3
Apply waste allowance = 11·016 m3 × 1·05
Concrete required = 11·6 m3 (rounded to one decimal place)
Example 3
The photograph below shows a backhoe excavating a trench on a construction site. The trench is to be 600 deep and 450 wide. The total length of the trench will be 42·500.
Calculate the number of cubic metres of soil that will be excavated. Remember that all measurements must be in metres.
Applying the basic formula (Volume = Area of Base × Perpendicular Height) to the excavation:
Volume of soil to be excavated = Length of the trench × Width × Depth
= 42·500 × 0·450 × 0·600
= 11·5 m3 (rounded to one decimal place)
Example 4
The drawing on the right shows two views of a concrete base for a large modern sculpture that is to be erected in a park.
Specifications for the project say that concrete is to be poured to a depth of 300 in the ground. The excavation has been completed and formwork constructed above ground. Your task is to calculate the quantity of concrete to order.
Note 1: The volume of concrete below ground should be calculated with 10% waste allowance while the volume of concrete in the form can be calculated with 5% waste allowance.
Note 2: The shape of the concrete contained in the form is like the lower part of a rectangular based pyramid that has had a smaller pyramid removed from the top. The volume of concrete above ground will be the difference between the volume of the large pyramid and the volume of the small pyramid.
Total volume of concrete = Concrete above ground + Concrete below ground
Volume above ground = Volume large pyramid – Volume small pyramid
= (3·150 × 2·400 × 3·000 ÷ 2) – (2·100 × 1·600 × 2·000 ÷ 2)
= 11·34 m3 – 3·36 m3
= 7·98 m3
Apply waste allowance = 7·98 m3 × 1·05
Concrete above ground = 8·38 m3 (rounded to two decimal places)
Volume below ground = 3·150 × 2·400 × 0·300
= 2·268 m3
Apply waste allowance = 2·268 m3 × 1·10
Concrete below ground = 2·50 m3 (rounded to two decimal places)
Vol. concrete required = Vol. conc. above ground + Vol. conc. below ground
= 8·38 m3 + 2·50 m3
= 10·9 m3 (rounded to one decimal place)
Before leaving this page, think of what you've just been reading, and test yourself with these questions.
[[ mm /f ][ Match the solid shape with the way to calculate its volume: ][ Cube ~ Square base area x height. ][ Cylinder ~ Circular base area x height. ][ Cone ~ Circular base area x height / 3. ][ Pyramid ~ Square base area x height / 3. ][ Cube => Square x height; Cylinder => Circle x height; Cone => Circle x height / 3; Pyramid => Square x height / 3. ]]
[[ mc /r /f ][ A trench 12.0m long, 300mm wide and 600mm deep needs to be dug. How much soil will be removed in this dig? ][ 2.16m3 ][ * 2160m3 ][ * 21600m3 ][ * 21.6m3 ][ * 216m3 ][ 12.0 x 0.3 x 0.6 = 2.16m3 ]]
[[ mc /r /f ][ How much concrete should you order to construct a small pyramid with a base 1.2m square and a height of 600mm? Allow an extra 10% for wastage ][ * 0.29m3 ][ 0.32m3 ][ * 1.8m3 ][ * 7.2m3 ][ base x height / 3 = (1.2 x 1.2) x 0.6 / 3 = 0.288m3 => 0.288 x 1.10 = 0.3168m3 ]]
Without being too technical, the terms weight and mass mean much the same thing. When you calculate quantities of reinforcing steel, mass is used for the ‘weight’ per linear metre and the term weight is used for the total tonnage required.
Units of measurement are the kilogram (kg) and the tonne (t). There are 1000 kilograms in 1 tonne.
The basic formula used for calculating the amount of reinforcing steel required is:
The table on the right shows the mass per linear metre for three common sizes of reinforcing steel.
In the General Construction industry lengths of reinforcing steel are bought by the tonne. Reinforcing can be supplied cut to length or in stock lengths of 6·000 and 12·000.
Example 1
Calculate the total weight (tonnes) of the reinforcing steel order listed below.
20Ø, 60/6·000 16Ø, 25/12·000 12Ø, 20/6·000
Note: The top line of the formula gives the weight in kilograms. Divide by 1000 to give the weight in tonnes. Refer to the table on page 141 for mass per linear metre.
Example 2
The drawing below illustrates that part of a building block has an average grade or slope of 1 in 5. This means that over a horizontal distance of five metres the ground rises one metre. The level part of the block is to be increased by 17·500. Find the height of the retaining wall (x) that will need to be built after the earthworks have been completed.
The ratio between the height of the retaining wall (X) and 17·500 will be the same as the grade (1 in 5). This can be expressed as:
The height of the retaining wall (X) can be found by cross multiplying.
5X = 17·500
X = 17·500 ÷ 5
Height of retaining wall = 3·500
Alternative Method: You could have found the height of the retaining wall by making an accurate scale drawing of the grade. The illustration below shows the actual measurements used to make the drawing scaled to 1:100. The 1 in 5 grade line is set up by drawing the shaded right angled triangle with base 100 and perpendicular height 20.
The measurement of 175mm on the drawing represents 17·500 scaled 1:100. The height of the retaining wall is represented by the vertical line drawn from point P. This measurement on the drawing would be 35mm. The height of the retaining wall is found by scaling this measurement 100:1 (0·035 x100 = 3·500).
Sometimes you may need to convert material quantities to different units of measurement. For example, because bricks are purchased by the thousand it is necessary to convert the net wall area of a building (m2) to the number of bricks required.
As another example, large quantities of timber are purchased by the cubic metre (m3), so it may be necessary to convert the number of lengths required (linear metres) to cubic metres.
Example 3
The photograph below shows a typical brick veneer house. Brick veneer construction consists of an internal frame of wood or steel, usually lined with plasterboard and an external skin or veneer of brick.
Note: There are approximately 50 standard bricks in a square metre of finished brickwork. Net wall area is found by subtracting the area of openings from total wall area. Have a look at Example 4 on page 134 for a similar calculation.
Find the number of bricks required for a brick veneer house with net wall area of
128 m2. Allow 5% for cutting and wastage. Round up to the next 100 bricks.
Bricks required = Net wall area × Number of bricks per m2 × Waste factor
= 128 × 50 × 1.05
= 6800 (rounded up)
Example 4
Calculate the number of cubic metres in the following timber order:
100x50 RS Hardwood, 250/3·6
Cubic metres = Section × Length × Number
= 0·100 × 0·050 × 3·600 × 250
= 4·5 m3
Example 5
The illustration on the right shows a frame that has been constructed using 75x50 DAR pine. The timber list for the frame consists of five studs 2·400 long and three pieces 1·800 long for the plates and nogging.
Your task is to construct eight similar frames that measure 3550 × 2400. Studs should be spaced at no more than 450 centres. Calculate material required and write your timber order.
Studs in 1 frame = (Length of frame ÷ Maximum spacing) + 1
= (3550 ÷ 450) + 1
= 8 + 1 = 9
Studs in 8 frames = 9 × 8
= 72
Timber Order: 75x50 DAR Pine, 24/3·600 72/2·400
Note1: There is always one more stud than the number of spaces.
Note 2: The number of spaces is always rounded up, e.g. 7.88 is rounded up to 8 spaces.
Before leaving this page, think of what you've just been reading, and test yourself with these questions.
[[ mc /r /f ][ A block of land has a slope of 1 in 8. If a 12.0m distance has to be levelled for a building foundation, what depth of dirt has to be dug out at the back of block to achieve this? ][ * 1.0m ][ * 1.2m ][ 1.5m ][ * 2.0m ][ * 2.4m ][ 12.0 x 1 / 8 = 1.5m ]]
[[ mc /r /f ][ A small house has a net brick wall area of 80m2. If there are 50 bricks to the square metre, how many bricks will you need for this house? Allow 5% for cutting and wastage. ][ * 800 ][ * 4000 ][ 4200 ][ * 2000 ][ * 2200 ][ 80 x 50 = 4000; 4000 x 1.05 = 4200. ]]