Maths A - Chapter 7

7

syllabussyllabusrrefefererenceenceStrand:Applied geometry

Core topic:Linking two and three dimensions

In thisIn this chachapterpter7A Scale drawings7B Building plans7C Floor plans and elevations7D Pegging out the perimeter7E Footings and slabs7F Bracing7G The roof7H Cladding the roof7I Brickwork

Basics of construction

MQ Maths A Yr 11 - 07 Page 231 Wednesday, July 4, 2001 5:27 PM

232

M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d

Introduction

Jaclyn and Kim are newly married. Like many young couples, their aspiration is toachieve that great Australian dream of ‘owning your own home’. They realise that thisis probably the largest single financial investment they are likely to make, so it requiresvery careful planning.

Having looked at houses for sale on the market, and not finding what they reallywanted, they decided to investigate building a home. A friend introduced them to a CDproduced by G. J. Gardner Homes specially for new home buyers.

This software package is included on the CD accompanying your text book and isable to be viewed by clicking on the icon seen on the left. You should look brieflythrough the package at this stage to become familiar with its features.

Jaclyn and Kim carefully examined the range of homes on the CD and settled on theSturt design (see image below), from the Colonial series.

This chapter traces the construction of their home, and looks at the mathematicsapplied throughout the process.

© G. J. Gardner

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C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n

233

1

Change the following ratios to the form 1 :____

a

4 : 3

b

1.5 : 6

c

0.2 : 5

d

0.01 : 0.4

2

Which of the ratios in question

1

would represent an enlargement and which wouldrepresent a reduction in size, if used as a scale for a plan?

3

Complete the following:

a

1 : 100

≡

1 cm : ____ m

b

2 : 50

≡

1 mm : ____ cm

4

Determine the perimeter and area of the following shapes.

a b

5

Find the volume of the following prisms.

a b

6

Use Pythagoras’ theorem to determine the lengths of the marked sides.

a b

7

Determine the size of the marked angles (to 1 decimal place) in each of the following.

a b

8

Calculate the length of the longest stick that would fit inside the following prisms.

a b

12 m

8 m

4 m12 m

4 m

2 m

3 m

10 cm 100 mm

500 mm20 m

2400 mm

2400 mm

x 2400 mm3 m

y

2400 mm

3.5 mx

2400 mm3.5 m

y

2 m

10 m

5 m

1 m


234


9

Find the height of the square-based pyramids below.

a b

10

Determine the surface area (excluding the base) of the pyramids in question

9

.

Scale drawings

The first stage in selecting a design involves choosing a floor plan that caters for theneeds of the family. In order to be able to read plans correctly, you must first under-stand scale drawings.Plans can be represented in different ways. The following statements1 : 1001 cm

↔

1 m

all represent the same scale — that is, 1 cm on the plan represents a distance of 1 metrein the field. The scale factor is expressed as a ratio.

Scale factor

=

=

plan length : field length

In the above examples, the scale factor is , meaning that the plan measurementsare one hundredth of the field measurements, or the field measurements are 100 timesthe size of the plan measurements. Notice that in the first representation, no units arementioned. This means that the user can supply the appropriate units for the particularsituation (depending on whether the measurements are small or large). That is,

1 mm on the plan or drawing represents 100 mm in the field or original

or

1 cm on the plan or drawing represents 100 cm (1 m) in the field or original

Where the scale factor is a number greater than 1, the plan or drawing represents an

enlargement

of the original; a scale factor smaller than 1 represents a

reduction

of theoriginal.

5 m

8 m

h 3000 mm

3500 mm

h

0 1 2 3 4 5 m

plan lengthfield length----------------------------

1100---------



235

Scale drawings

1

Convert to metres.

a

8215 mm

b

350 cm

c

89 km

d

26 mm

e

4 cm

f

6.4 km

2

Measure these lengths to the nearest millimetre.

abc

A house plan is drawn to a scale of 1 : 100.a What would be the size of a bedroom measuring 3 cm square on the plan?b The patio is 3500 mm wide. What would be its measurement on the plan?

THINK WRITE

a Use cm as unit in scale because plan unit is cm.

a Scale is1 cm : 100 cm

Actual measurements are 100 times plan measurements, so multiply by 100.

3 cm : 3 × 100 cm

Calculate the actual length and convert to appropriate unit.

3 cm : 300 cmSo 3 cm on plan represents 3 m in house∴ Bedroom is 3 m square

b The plan is size of house, so divide by 100, remembering answer is in same units.

b Patio width = 3500 ÷ 100 mm

Calculate plan length, converting to appropriate unit if necessary.

Width of patio on plan = 35 mmor = 3.5 cm

1

2

3

11

100---------

2

1WORKEDExample

remember1. Scales can be represented in different ways.2. If no units are indicated on a scale representation, any unit can be inserted, so

long as the same unit is used for both the plan length and the field length.3. The scale factor compares the plan length with the field length; that is

scale factor = plan length : field length4. A scale factor greater than 1 represents an enlargement and a scale factor

smaller than 1 represents a reduction in size.5. In converting from a plan length to a field length, multiply by the scale factor.

Divide by the scale factor when progressing from the field to the plan.

remember

7A

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236 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d

3 Classify the following as enlargements or reductions.a 1 : 200b 1 mm ↔ 1 mc

d

e 10 :1f

4 Express the following in the form of a simplified ratio.a 2 cm ↔ 100 mb

c

5 A plan of a building site uses a scale of 1 : 150. What field lengths would be repre-sented by the following plan lengths? (Answer in metres.)a 12 mm b 4.5 cm c 150 mm d 2.25 cm

6 A plan of a house is drawn using a scale of 1 : 125. What would be the measurementsof the following on the plan?a Patio 3 m long and 1.75 m wideb Bedroom 3.75 m × 2.25 mc Kitchen 5.5 m × 2.5 m

7 A building block measuring 48.4 m by 41.25 m is drawn on a plan with measurementsof 8.8 cm by 7.5 cm. What scale was used to draw the plan?

8 The figure below shows a house on a block, drawn using a 1 : 250 scale.

a How far is the house from the front boundary?b The owner intends to fence this property at a cost of $32.00 per metre, plus $250 for

gates. What would it cost to fence and install gates on this property?c What is the length of the sewer line (dotted)?

0 1 2 3 4 5 mm

0 5 10 km

110 000----------------

Mathca

d

Map scales

EXCEL

Spreadsheet

Map scales 1

0 12 km6

0 1 2 3 4 mm

WORKEDExample

1a

WORKEDExample

1b

Mathca

d

Scale factors

EXCEL

Spreadsheet

Map scales 2

Fron

t bou

ndar

y

Scale 1 : 250

House


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 2379 The map below shows Mooloolaba Boat Harbour.

a How wide is the entrance to the harbour?b How far is it, as the seagull flies, from the yacht club to the water tower at Point

Cartwright?c What area (in km2) does this map cover?

Building plansDetailed building plans are necessary so that builders and other tradespersons know exactly what is required to complete a project. By the time it is completed, any building will have incorporated information drawn from many facets of construction. Drawings for a domestic structure will include a survey plan, a site plan, floor plans and elevations.

Survey planA survey plan shows all boundaries of the block of land and includes the position of roadways and nearby lots (see figure at top of next page).

Mooloolaba Boat HarbourLine of Leods 131°24

MOOLOOLABA

PARKYN PDE.AVCG

Yacht Club

Boat HarbourSupervisor

PilotStation

PublicJetty

Pilot Stn.Jetty

Dredged Mooring Basin

Dredged to 3m

Fish Board

7048000mN

RIV

ER

MOOLOOLAH

HARBOUR PDE.

BUDDINA

PAC

IFIC

B

LVD

.SOUTH

PACIFIC

OCEAN

513000mE

153°08'E

Beaconsmark channel

F Bu

Q Bu & FI R 4s

FI WR (3) 15s53m 23 / 11m

Water TowerConspic.

Point CartwrightVQR

VQR

REDWHITE

7049000mN

26° 41'S

500 metres0

Source: Reproduced with permission of Qld. Dept. of Transport from Boating safety chart "Moreton Bay – Southport to Caloundra"



Site planA site plan shows the boundaries of the lot that is to be built on, and where the struc-ture is to be situated on this lot. It may also show contour lines (see figure below).

Floor planA floor plan shows the exact dimensions of the building, including dimensions andnames of all rooms, the size and position of doors and windows, the direction in whichdoors open, the thickness of walls and the location of stairs.

In simple constructions, the roof plan and the locations of all electrical and plumbingfittings are superimposed on the floor plan. For more complex constructions, plans forthese services would be made separately. A floor plan without electrical or plumbingfittings is shown in the following figure. If units are not indicated on a plan, it isassumed that dimensions are in millimetres.

110630 m2

30.0

21.0

21.0

187

23.0

27.4

99

23.0

189

15.0

38.9

1438.6

07

7.21913.219 5.0

796 m2

Lot X

11 500 7 200

1 00

014

400R.L.

31.400

LOT 8

GU

RN

ER

ST

RE

ET

75 000

75 000

28 0

00

28 0

00

SETOUTPOINT

30 3

00

30 6

00

30 9

00

31 2

00

31 5

00

31 8

00

32 1

00

N


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 239

This questions refers to Lot X on the previous survey plan.a What shape is Lot X?b The two parallel sides measure 30 m and 33 m, while the front and back boundaries

measure 21 m and 23 m respectively. Calculate the area of Lot X.c Lot 110 is for sale for $59 850. Suggest a reasonable sale price for Lot X by comparing

its area with the area of Lot 110.

THINK WRITE

a Lot X is 4-sided with one pair of parallel sides.

a Lot X has the shape of a trapezium.

b Write the formula. b A = (a + b)h

Substitute for the variables. = (30 + 33) × 21 m2

Calculate the area. = 661.5 m2

c Calculate the price per m2 based on Lot 110 information. c Price per m2 =

= $95 per m2

Multiply by the number of square metres in Lot X.

Price = $95 × 661.5= $62 842.50

A reasonable price for Lot X would be $62 850.

112---

212---

3

1 $59 850630 m2-------------------

2

2WORKEDExample



Building plans

1 In the survey plan on page 238a What is the area in square metres of Lot 110?b What is the length and breadth of Lot 110?c What scale has been used to draw the survey plan?d Redraw Lot 110 using a 1 : 500 scale.e The area of Lot 187 is not shown. Find this area.f Before the metric system was introduced, the area of house blocks was measured in

perches (1 perch = 25.3 m2).i A block of 42 perches is advertised for sale at $61 500. Convert the area to

square metres and find the price per square metre.ii One lot is 850 m2 and another is 28 perches. Which is the larger?

g Lot 110 is for sale at $59 850, and Lot 189 is for sale at $82 000.i Which represents the best value per square metre?ii What features of a block of land might attract a purchaser even though its dollar

value per square metre may be higher than surrounding blocks? (Comparing thepositions of Lots 110 and 189 can assist in your answer, but include as manyother features as possible.)

2 The site plan on page 238 shows Lot 8 on Gurner St. All dimensions given are in milli-metres.a Find the area of Lot 8 in square metres and perches.b The shaded sketch shows the area of the dwelling proposed to be erected on this lot.

What is the area of the proposed dwelling?c What scale has been used to produce this diagram? (Note: this scale may not be a

simple ratio.)d The dotted lines are contour lines (lines of height). All points along the 31 800 line

are 31 800 mm above sea level.i Is the block rising or falling as I walk from the Gurner St entrance to the rear of

the block?ii Calculate the angle of rise or fall from the front to the rear.

3 The floor plan on page 239 shows a plan of a one-bedroom dwelling. All dimensionshave been given in mm. (Note: this scale may not be a simple ratio.)a Calculate the area of this dwelling in square metres. (Include the patio.)b The patio is to be tiled using tiles costing $22.50 per square metre. Find the cost.

Include an extra 5% for cutting.c Find the area of the bedroom, kitchen, laundry and bathroom.

remember1. Look carefully at features and measurements on plans.2. If no units are mentioned on a plan, the figures are assumed to be in

millimetres.3. Remember to give your answers in appropriate units.

remember

7BWORKEDExample

2



241

Floor plan

Jaclyn and Kim have chosen to build the Sturt home. The floor plan is shown below. You can also locate it on the CD and print a copy for yourself.

© G. J. Gardner

The measurements of the rooms are indicated in metres. The dimensions of the external length and breadth are given in millimetres.

1

Using the measurements shown, make a 1 : 100 scale drawing of the floor plan of the Sturt.

2

Choose another house style from the G.J. Gardner CD collection and make a 1 : 100 scale drawing. (Home building plans generally use a 1 : 100 scale representation.)

Changing a floor plan

Although you may at first be happy with a floor plan, when you examine it in detail you may discover features you would like to change.

Access the G.J. Gardner collection from the CD and select the floor plan of the Sturt. Note that:

1. One feature enables you to zoom in and out for greater definition.2. Another feature enables you to modify the plan using the drawing tools.

1

Experiment with erasing walls, drawing walls in different places and locating furniture at various places in rooms. The writing tool also lets you relabel rooms (bedroom 3 may become a study).

2

Assuming that you had chosen this floor plan yourself, undertake some changes to make it more suitable to your lifestyle. Explain these changes then print out a copy.

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3 Experiment with other plans that may be more to your liking. You will generally find, on closer inspection, that there are features you would like to move or change.

4 Note that, although on paper you can virtually place things wherever you want, in practice you have to remember that the roof requires support from the walls. Large expanses in a room require special beams as support for the roof. Take note of this in any modifications you make.

5 In two-storey houses, you will generally find the walls in the top storey are supported by walls in the lower storey. Print out the top and lower levels of a two-storey design and position them on top of each other.



ElevationsThe scale drawings we have considered so far are representations of what we wouldview from above, looking directly towards the ground. A three-dimensional structuremost frequently has four vertical faces and we can view these faces from four differentdirections. We can observe a front and back view, as well as two side views. Theseviews are called elevations. The side facing mainly towards the east is known as theeastern elevation (and similarly for the other sides). Earlier we saw the front elevationof the Sturt home. The elevations for a house provide an impression of how the com-pleted structure would appear.

Consider the floor plan below and its three accompanying elevations.



Study the elevations carefully, trying to match the features marked on the boundary ofthe floor plan with those depicted on the elevations. Note that the windows and doorshave been labelled for ease of identification.

A view from the southgives the South elevation

South elevation

N

A view from this direction gives West elevation

A view from this side gives East elevation



Floor plans and elevations

1 Use the floor plan and accompanying elevations on pages 243 and 244 to answer thefollowing.a How many windows and doors are on the southern exterior of the house?b How many internal doors are there in the dwelling?c There is one internal sliding door. Which two rooms does it separate?d What is the height of the ceiling above the floor?e There are four points on the floor plan indicating DP. Where are they located? What

are they? f What is the height of the bathroom window from the floor?g How wide are the external walls of the house? What is the width of the internal

walls?

Use the floor plan on page 243 and its accompanying elevations (page 244) to answer the following.a Give the window numbers on the eastern side of the house.b How many external doors are on the western side?c Where is the linen cupboard situated?

THINK WRITE

a Identify the windows on the eastern side of the floor plan and check them on the eastern elevation.

a On the eastern side are windows W6, W7,W8 and W9.

b Cross-check the western side of the floor plan with the western elevation.

b On the western side there are doors D1, D2and D3. So there are 3 external doors on thewest.

c The linen cupboard is identified by L on the floor plan.

c The linen cupboard is situated in the hallwayat the entrance to the bathroom.

3WORKEDExample

remember1. Floor plans represent a view from above.2. Elevations represent a view from the side.3. The north elevation is that side of the building which faces north. Similarly,

there are south, east and west elevations.4. A floor plan will give indications of the width of features in the elevations, but

no indication of the height of features (windows, doors).

remember

7CWORKEDExample

3



h Name all the items drawn in the bathroom.i What items are drawn in the kitchen?j Consider the northern elevation of the house.

i How many windows/doors are there? Determine their width.ii What is the total width of the slab? Determine the floor width.

k Assume that the shape of the roof is symmetrical, and that the highest point of theroof is 4 metres above the level of the floor. Also assume that any windows on thenorthern side are of the same style as those on the western side. Sketch the northernelevation of the house. Mark all features and measurements you have identified onyour sketch.

2 The following diagrams are representations of houses with a variety of roof types.Draw a plan of the south and east elevations of these houses. The direction of north isgiven.

a b

c d

WorkS

HEET 7.1

Gable roof Boxed gable roof

N N

Hip roof

N N

Gambrel(a roof combining the additionof small gables to a hip roof)



Below is the site plan for a block of land.

1 What is the scale of the plan?

2 In which direction does the garage face?

3 What are the dimensions of the block of land?

4 What are the dimensions of the house?

5 How wide is the driveway?

1

Scale 1:250

N

Shed

Garage

House

Driveway

Garden bed

Gardenbed


248


The house plan is shown below.

6

What is the scale of this plan?

7

What are the dimensions of the lounge room?

8

Which bedroom is the largest?

9

What are its dimensions?

10

What is the width of the hallway?

Elevations

You have previously been shown the floor plan and front elevation of the Sturt home (see pages 241 and 232)

1

Considering both diagrams, sketch the front elevation of Sturt. Note that the front view shows windows all around the verandah except for bedroom 1, which exits onto the verandah via a sliding door. You can also view both plans from the CD.

2

Note that the floor plan indicates the back of the house as having four sets of sliding windows and two sliding doors (from the laundry and the dining room). The windows in the bathroom, toilet and kitchen are likely to be higher from the floor than those in the bedroom. With this knowledge, sketch the back elevation of Sturt.

3

Choose a double-storey home from the CD and draw its front and back elevations.

Lounge Bed 3

Bed 1 Bed 2Toilet

Bathroom

Pantry

Laundry

Kitchen

Bed 4 Family

Scale 1:125

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MQ Maths A Yr 11 - 07 Page 248 Friday, July 6, 2001 2:01 PM


Preparing to buildWith an understanding of plans, we are now in a position to start looking at the processof building.

Establishing levelsOnce a suitable building block has been selected, the land must be cleared and levelledin preparation for erecting the structure. There are sophisticated electronic devicesavailable today, but the older methods of using a spirit level with a straight edge, orusing a water level are still widely used.

A spirit level is a device, usually made of aluminium, containing a vial of liquidwith an air bubble. The bubble indicates whether the surface on which the device restsis level.

Since water always finds its own level, a length of clear hose filled with water can beused to establish levels between two points. Although this method seems to be a rela-tively crude one, it is actually very accurate. This device is called a water level.

Straight edgeSpirit level

Level datum

Using a spirit level and straight edge

Water level in tubegives cut off line for stump

Keep water in line withthe top of the stump

Using the water level



Pegging out the perimeter of the house

If you contract a builder to construct a dwelling for you, it is a wise idea to visit regu-larly to keep an eye on the progress of the work. Instances have been cited when dwell-ings have even been erected on the wrong block of land.

Once the land is levelled, the shape of the house can be pegged out. It is obviouslymost important that the corners of the house, which are meant to be right angles, areexactly that. Even one or two degrees error may have dire consequences (rememberthat the error becomes magnified, the longer the wall).

Builders make use of a building square, which can be easily constructed using threepieces of timber.

LevelsIt is relatively easy to construct simple devices to determine whether surfaces or points are level.

Spirit level1 Obtain a length of clear plastic tubing about 10 cm long.

2 Block one end with a cork, then almost fill the tube with water.

3 Block the second end with a cork, leaving a small air bubble in the tube.

4 Securely tape both ends, then tie or glue the straight tubing to a piece of wood or metal (a ruler would do).

5 Place your spirit level on a surface which you know to be level and mark the plastic showing the outline of the bubble.

6 You can now use your device to determine whether other surfaces are level. Try the surfaces around you.

Water levelAs this method is generally used to determine a level between points which are some distance apart, a considerable length of clear plastic tubing is required.

1 Obtain a length of clear tubing, as long as possible.

2 Almost fill the tubing with water.

3 On one end, mark a level about 5 cm from the top. Your device is now ready for use.

4 The water level requires two people to operate. One person can stand with the level mark against an object, while a second person can determine the position on another object some distance away at the same level. This method is actually used quite often when raising houses, to ensure that the final position of the house is level.

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The lengths of these three pieces of timber are such that they form a right angle(check, using Pythagoras’ theorem). Commonly used dimensions of a building squareare 1500 mm : 2000 mm : 2500 mm.

From the site plan, the position of one corner and one side of the building is deter-mined. A corner peg and a line representing one side of the building are put in place, asshown in the following diagram. The building square is then placed under this firstbuilding line, with its apex on the corner peg. The second building line can then be pos-itioned along the other arm of the building square. In this way, builders can be assuredof constructing right-angled corners where necessary.

Using suitable ratio of sides to form a right angle

90°

2500

2000

1500

Hold armsin positionwith temporaryfixings

Cut and fix brace

Constructing a builder’s square

First building line

Corner of building

Building corner

Second building lineparallel to arm of square

Establish first building line andbuilding corner

Establish second building line at 90°

Using the builder’s square



Pegging out the perimeter

1 A builder chooses to make a builder’s square using two pieces of timber 1200 mm longto contain the right angle.a How long should the third piece be?b What would be the size of the other two angles in the device?

2 Two pieces of timber 600 mm and 700 mm long are to be used to form the right anglein a builder’s square. The builder finds another piece of timber 800 mm long. If thispiece is used, what type of angle will result between the two shorter pieces?

3 A square shed is pegged out on the ground. The builder has checked that all the cornersare right angles. The length of the diagonal is 17 m. What is the size of the shed?

4 A builder has pegged out a rectangular shed. After measuring the lengths of all sides,two opposite sides are 10 m long and the other two opposite sides are 8 m long. Can thebuilder be sure that the shed is rectangular? Explain.

A builder’s square is to be made using two pieces of timber 500 mm and 600 mm long. If these two pieces are to contain the right angle, how long should the third piece be cut to complete the device?

THINK WRITE

Draw a diagram

Use Pythagoras’ theorem to determine the length of the hypotenuse.

hyp2 = base2 + height2

l2 = 6002 + 5002

= 360 000 + 250 000= 610 000

l = = 781 mm

So the third piece should be cut 781 mm long.

1

500 mml

600 mm

2

610 000

4WORKEDExample

remember1. Horizontal levels can be obtained using a spirit level with a straight edge (for

points relatively close together) or using clear plastic tubing filled with water (for points further apart).

2. It is important to ensure that corners of buildings are right angles. A builder’s square is used to check this in the pegging-out stage.

3. Pythagoras’ theorem is used to check for right angles.

remember

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4


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 2535 After pegging out a rectangular shed, a builder measures the diagonals to be the same

length. Is it possible that the shed is in fact not rectangular? Explain.

6 A square shed has a floor area of 156.25 m2. What is the length of its diagonal?

7 A builder uses a water level to check the horizontal level of the top of the walls of arectangular shed before he starts constructing the frame for the roof. He attaches thewater level to the top of the wall at diagonally opposite corners (positions X and Y). Heobserves the level of the water to be 2 cm above point X when the water level is atpoint Y. Draw a diagram to show this situation and explain what it means.

Footings and slabOnce the perimeter of thehouse has been marked out,the positions of drainage pipes(for sinks, showers and toilets)and perhaps cabling (electricaland telephone) need to bedetermined. Trenches are dugto house these pipes andcables. Footings (in the shapeof rectangular prisms) are alsodug around the perimeter ofthe structure. They are thenfilled with concrete and rein-forcing to support the weightof the walls. Preparations arethen made for pouring theconcrete slab to form the floorof the house.

Pegging out the perimeterJaclyn and Kim are keeping a close eye on the progress of their home. When they visit the site, they find that the perimeter of their house has been pegged out. The house is to be built on a slab on the ground, so the outline of the structure is rectangular.

1 Draw a 1 : 100 scale plan of the perimeter of their house (Sturt). Take care that the corners are all right angles.

2 Calculate the length of the diagonal of the slab. Compare your answer with the measurements of the two diagonals on your plan.

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© G. J. Gardner


254


Before the slab is poured, preparations include:1. marking the position of all underground cables and pipes, so that these points are not

covered when the slab is poured2. laying a bed of sand, which has been poisoned, to prevent white-ant infestations3. covering the sand with a layer of plastic4. laying reinforcing mesh over the plastic.The concrete slab can then be poured.

Determining the quantity of concrete needed for the footings and slab simplyinvolves calculating the volume of a prism. For surfaces that are rectangular in shape,the process is simpler than for those that are composite shapes. Two methods can beused.

Early stages of construction

1

Run the G.J. Gardner CD.

2

Enter the Construction section (through the right-hand window).

3

View the first three stages of construction:(a) Site preparation(b) Footings(c) Laying the slab.

4

Note the reinforcing mesh in the footings and on the slab.

5

Note the underground cables protruding from the slab.

6

After viewing the video clip, write a short paragraph describing these first three stages of building a house.

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A house with a rectangular floor plan measuring 18 000 by 12 000 has footings 350 mm wide and 450 mm deep dug around its perimeter. What volume of concrete would the footings require?

THINK WRITE

Draw a diagram.Add measurements. (Because of large values obtained when working in mm, convert all measurements to metres.)

Calculate the outer area of the rectangular surface.

Outer area = L × W= 18 m × 12 m= 216 m2

1

12 000 mm = 12 m

18 000 mm = 18 m

350 mm= 0.35 m

450 mm = 0.45 m2

3

5WORKEDExample



For composite-shaped surfaces, it is sometimes simpler to find the total length of thefootings, then multiply by the width and depth of the footings. In this case, it is neces-sary to take care not to count corner sections twice when finding the total length of thefootings.

THINK WRITE

Calculate the inner area of the rectangular surface (remember to subtract two footing widths from each measurement).Note: Do not round any values yet

Inner length = 18 − 2 × 0.35= 17.3 m

Inner width = 12 − 2 × 0.35= 11.3 m

Inner area = 17.3 m × 11.3 m= 195.49 m2

Calculate the surface area. Surface area = outer area − inner area= 216 m2 − 195.49 m2

= 20.51 m2

Calculate the volume (round if necessary). Volume = area of surface × depth= 20.51 m2 × 0.45 m= 9.2295 m3

Volume of concrete required for footings ≈ 9.3 m3

4

5

6

The footings to be dug within the perimeter of the following floor plan are 400 mm wide and 600 mm deep. What volume of concrete would be required to fill them?

THINK WRITE

Draw a diagram showing footings.Add measurements (all in metres).Break footings into lengths (either horizontally or vertically).

Find the total length of footings, taking into account their width (start at one corner and work around the figure; for example from the top left and work clockwise).

Total length of footings (m)= 15 + (3 − 2 × 0.4) + (5 + 0.4)

+ (2 − 0.4) + 10 + (5 − 2 × 0.4) m= 38.4 m

Calculate the volume, using the length of footings, width of footings and depth of footings (round if necessary).

Volume = L × W × D= 38.4 m × 0.4 m × 0.6 m= 9.216 m3

Volume of concrete required to fill footings≈ 9.3 m3

5 m

15 m

2 m10 m

1

5 m5 m

15 m

0.4 m0.4 m

2 m

3 m

10 m

23

4

5

6WORKEDExample



The concrete in the footings and the slab is reinforced with steel mesh to increase its strength and to prevent cracking (note its presence in the video in the construction section of the CD). Reinforcing mesh is often also provided beneath internal walls as a structural support. The diagram below illustrates how the mesh is laid. Note the overlap at the corners and the 500-mm minimum overlap between the sheets.

Plan of strip footing at corner

Trench mesh to be lappedfull width at corners

500min.lap

Trench mesh is to be laid on the perimeter of a slab measuring 7.5 m by 10 m. Two layers of mesh are required and the mesh is available in 6-metre lengths. The mesh must overlap at the corners, and when two sheets meet there must be 500 mm of overlap. The 6-metre lengths may be cut if necessary. Find the number of lengths required.

THINK WRITE

Draw a diagram, marking in the measurements.

Work around the trench, laying and cutting mesh. Remember the overlap at the corners and on joins.

Each 10-m side requires6 m + 4.5 m (including overlap).

Each 7.5-m side requires6 m + 2 m (including overlap).

Find the total number of 6-metre lengths required, remembering to provide for 2 layers of mesh.

Each 10-m side requires2 lengths of mesh; that is, 4 lengths for both sides.

Each 7.5-m side requires1 length + 2 metres; that is, 3 lengths for both sides.

∴ Total number of 6-metre lengths required for 2 layers of mesh = 2(4 + 3)

= 14 lengths

1

2 m

6 m 7.5 m

6 m 4.5 m10 m

2

3

7WORKEDExample



257

Footings and slab for Sturt

Refer to the floor plan of the Sturt home which Jaclyn and Kim are building (see page 241, or the CD). Since the verandah is included as part of the house, the slab is basically rectangular in shape. Note the external measurements of the house.

1

If the footings are dug 400 mm wide and 500 mm deep within the perimeter, what volume of concrete would need to be ordered?

2

Compare this with the quantity of concrete required if the footings were dug 500 mm wide and 400 mm deep.

3

How much more concrete would be required if the footings were dug an extra 50 mm deep?

4

Applying the conditions outlined previously for the laying of trench mesh, how many 6-metre lengths of mesh would be required for two layers?

5

What volume of concrete would be required to lay a 100-mm slab?

The answers to these questions provide realistic figures, which contribute to the overall cost of house-building.

Using the floor plan in worked example 6, calculate the volume of concrete required for the slab, which is to be laid 100 mm thick.

THINK WRITE

Break the composite shape into simpler shapes.

Define shapes and measurements.

Calculate the area as the total of individual shapes.

Total area = area of 2 rectangles= (10 m × 5 m) + (5 m × 3 m)= 50 m2 + 15 m2

= 65 m2

Calculate the volume (take care with units).

Volume = area of base × depth= 65 m2 × 0.1 m= 6.5 m3

So the volume of concrete required for the slab= 6.5 m3.

1

5 m 10 m × 5 m5 m × 3 m

15 m

2 m 5 m

3 m

10 m

2

3

4

8WORKEDExample

G JGar

dner

inve


n



Footings and slab

1 Greg needs to order concrete for the footings of the following rectangular floor plans.Calculate the volume of concrete required for each. (Remember that, if no units aregiven, the measurements are assumed to be in millimetres.)a Floor: 16m by 10m width: 350 depth: 500b Floor: 1750 by 9500 width: 400 depth: 600c Floor: 19 000 by 19 000 width: 400 depth: 700

2 Calculate the quantity of concrete required for footings for the following floor plan ifthey are 350 mm wide and 550 mm deep.

3 A rectangular structure has dimensions 8.5 m by 5.5 m. Footings 400 mm wide and500 mm deep have been dug within its perimeter. Two layers of trench mesh arerequired in the footings. The trench mesh is sold in 6-metre lengths. Industry standardsdictate that the trench mesh must overlap in the corners, and, if sheets are cut, a500-mm overlap is required.a How many lengths of trench mesh would need to be ordered?b What volume of concrete would the footings require?

4 How much concrete would be required for a 100-mm slab floor for the plan in question 2?

5 Use the two floor plans a and b on the opposite page for the following questions.Notice that footings have been dug inside the perimeter (as well as around theboundary) as a load-bearing support for internal walls.

remember1. Footings are trenches dug within the perimeter of a structure. They are

sometimes also dug beneath internal walls to support their weight.2. Calculating the volume of concrete required for footings involves determining

the volume of a rectangular prism.3. Take care to convert all measurements to the same unit (preferably metres).4. Trench mesh is purchased in 6-metre lengths (which may be cut if necessary).

It must overlap at the corners and have a minimum overlap of 500 mm between sheets.

5. The process of calculating the volume of concrete required for a slab involves calculating the area of the floor plan, then multiplying by the depth of the slab (remember to keep common units).

remember

7EWORKEDExample

5

WORKEDExample

6

10 m

8 m

12 m

5 m

WORKEDExample

7

WORKEDExample

8


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 259i Determine the quantity of concrete required for the footings of each dwelling if

they are 350 mm wide and 600 mm deep.ii Estimate the number of lengths of reinforcing mesh required in each case, using

two layers, and applying the same placement conditions as in question 3.iii How much concrete would be necessary for the slab of each floor plan, if the slab

thickness was 100 mm?

Curing concreteConcrete must be kept damp for at least 7 days after it is poured. This process is known as curing. If it dries too quickly, the concrete is weakened. A longer period of drying produces concrete with a greater strength. This, in turn, produces greater structural support in terms of load bearing. The strength of the concrete results from its slow chemical reaction with water. If the concrete dries too quickly, this chemical reaction does not go to completion, and weaker concrete results.

The diagram below compares the strength of the resulting concrete with no curing, a 3-day curing period and continuous curing.

1 Read values for the strength of the concrete over 7-day intervals for each of the three curing times.

2 Set up a spreadsheet, and enter these values.

3 Compare the graphs obtained from your spreadsheet with those above.

4 Compare the percentage increase in strength after 28 days, of the continuously cured concrete with that of the concrete with no curing and 3-day curing.

Footings aroundperimeter andinternally

8.5 m

15 m

perimeter andinternally

10 m

20 m

6 m11 m

a b

inve


n

Age (days)Concrete strength gain with time

Nocuring

3 dayscured

Continuouslycured

Com

pres

sive

str

engt

h (%

of

28 d

ay m

oist

cur

ed s

tren

gth)

07 14 21 28

20

40

60

80

100

120

140



The frameAfter the footings and slab have been completed, the frame can be erected. Before com-mencing this section, review the Construction section on the CD, looking particularly atthe two parts, ‘Erecting the frame’ and ‘Roofing’. You will notice that these days thewooden frames are generally prefabricated in a factory, then transported to the sitewhere they are assembled.

Testing that a wall is verticalIt is obviously very important that all walls be vertical (the exception, perhaps, beingthe Leaning Tower of Pisa). Two simple devices commonly used today are:1. the spirit level and straight edge2. the plumb bob.

The spirit level works on the same principle as described earlier when the device wasused to test horizontal levels. If the surface is vertical, it is said to be ‘plumb’.

A plumb bob consists of a metal object with a piece of string attached. When theend of the string is held, the heavy metal causes the string line to be perfectly vertical(plumb). Builders use this device as shown in the diagrams on the next page. The stringis firmly attached at the top, a short horizontal distance from the surface to be tested.The distance between the surface and the line is measured at the top of the work, and iscompared with the distance measured at the base of the work. If these two measure-ments are the same, the surface is plumb (vertical).

Straight edge



261

Plumb bob

To make a plumb bob, all that is required is a line (string, fishing line etc.) and a heavy object (sinker, bolt etc.).

1

Construct a plumb bob by attaching a heavy object to a line.

2

Use your plumb bob as described above to test that doors, walls etc. are vertical.

Plumb line

Edge of gauge block to beparallel to the wall with thegauge nail hard against endof the plate

Read

Gauge nail

Gauge block

Compare top and bottom measurements

inve


n


262


Bracing

Most walls are rectangular. Because a rectangular frame does not retain its shape whenplaced under stress (it can be pushed into the shape of a parallelogram), it requires

bracing

. This procedure involves attaching sheets of plywood, strips of timber or metalstraps. (Note the metal strap on the frame on the CD.) The straps are placed on an angleto the vertical, and attached to the frame. House frames are commonly constructed of

studs

(which run vertically) and

noggings

(which are the horizontal separators betweenthe studs).

Lintels

and

sills

are usually placed above and below windows, while

topplates

and

bottom plates

support the top and bottom of the frame. The diagrams belowillustrate the use of bracing in a variety of situations.

Bottom plate

SillLintel

Angle brace

Noggings

Studs

Top plate

Diagonalbracing



263


264


Building standards prescribe that the angle which the bracing makes with the bottomof the frame must lie within the range 37

°

to 53

°

. Houses are commonly constructedthese days with a ceiling height of 2400 mm, so most of our calculations will useframes of this height.

Bracing

1

Obtain 4 paddle pop sticks and form a rectangle by joining them a short distance from their ends with thumb tacks. Notice how the frame can very easily be pushed out of shape to form a parallelogram.

2

Obtain another paddle pop stick and brace your frame by attaching this stick at an angle to the top and bottom of the frame. Notice how this brace produces stability in the frame.

3

Placing the brace vertically does not prevent movement of the frame. Show that this is so.

4

Observe what happens when the angle which the brace makes with the bottom of the frame is moved through 37

°

to 53

°

. Which angle requires the longest brace?

inve


n

Find the length of the longest brace which could be used to provide stability in the frame below.

THINK WRITE

The longest brace would be the diagonal, but this might be outside the angle range of 37°–53° with the base of the frame.

Test this first using trigonometry. Test to see whether the diagonal is within the angle range (37° – 53°)

tan θ =

=

θ = tan−1

= 33.7°This is outside the range.

3600

2400

1

3600X

Y

θ

2400

2

oppadj---------

24003600------------

24003600------------

9WORKEDExample



265

THINK WRITE

The longest brace will then be at an angle of 37° to the horizontal.

So the longest brace is 37° to the base

Use trigonometry to find the length of the brace.

sin 37° =

sin 37° =

b sin 37° = 2400

b =

= 3988So the longest brace is 3988 mm.

3

37°

b2400

4opphyp---------

2400b

------------

2400sin 37°-----------------

Will a metal strap 3 m long provide the longest brace for the frame below?

Continued over page

THINK WRITE

The metal strap obviously cannot pass through the window. Find the angle of the brace from the top corner to the bottom of the window, using trigonometry.

tan θ =

=

θ = tan−1

= 39.8°Check to see whether this angle is within the limits of 37° to 53°.

This angle is within the limits for a brace(37° to 53°)

1200 1200 1200

1000

2400Window

11000

1200θ

oppadj---------

10001200------------

10001200------------

2

10WORKEDExample


266


Bracing

1

The frame at right has the dimensions shown. It is proposed to put a brace across the diagonal of the frame.

a Calculate the length of this brace.b Show that the angle which this brace makes with

the base of the frame is not within industry standards.c Find the length of the longest brace that does

comply with industry standards.

2 The frame at right has a full-length glass panel to be inserted on the right (that is, a brace cannot pass through it). Will the brace indicated on the frame comply with industry standards? Explain.

THINK WRITE

Extend the brace to the base. Calculate the length of this brace using the base angle of 39.8° and the trigonometric ratio sin.

sin 39.8° =

sin 39.8° =

b sin 39.8° = 2400

b =

= 3749So the longest brace is 3749 mm

Compare the answer with 3 m. A metal strap of 3 m would not provide the longest brace.

3

2400

b

39.8°

opphyp---------

2400b

------------

2400sin 39.8°---------------------

4

remember1. A spirit level and a plumb bob are two devices which can be used to test the

verticality of a wall.2. Bracing is necessary in frames to provide strength and rigidity. In wall frames

the angle which the brace makes with the horizontal must lie within the range 37° to 53° (remember to check for this). Take care that a brace does not pass through a window or door in the frame.

remember

7FWORKEDExample

9 2.4 m

6 m

4750 mm

5810 mm

1060 mm

GLASS


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 2673 The simple bookcase at right has been constructed using

five pieces of 200-mm × 1200-mm timber. To strengthen the structure, a brace is to be attached along the diagonal at the back.a What would be the length of this brace?b Find the angle it would make with the floor.c Is this within the required limits?

4 Calculate the length of the longest brace to provide structural strength for the frame at right. Remember that the brace cannot pass through the window. What angle does the brace make with the base?

5 The frame at right requires two braces as shown.a Find the base angles of each.b What total length of bracing would be required

for the job?

6 The metal frames below have bracing as indicated. Calculate the total length of bracingrequired for each.a

b

The roofBefore commencing this section, view the ‘roofing’ unit of the Construction section ofthe CD. Note the bracing which provides strength for the structure.

Roof trusses and pitchRoof trusses are often used in structures where large, clear spaces are necessary,without obstruction from the posts that are usually needed to support the roof. Theytransfer the load of the roof from the internal walls to the external walls.

These trusses are assembled in factories, then delivered to the site. They are oftenused in domestic constructions (rumpus rooms, entertainment areas) as well as in largecommercial projects (factories, halls). Some examples of truss designs are shown in thefollowing diagrams.

WORKEDExample

10

2400

1200Window

4000

2500

2400

25005300

4 m

12 m

Bracing

8 m

30°

Bracing

30°



Consider a section of a truss roof as drawn below. Note the terms rafter, king postand tie beam.

The pitch of the roof is the angle the roof makes with the horizontal. The roof abovehas a pitch of 12.5°. Often the pitch is expressed as a pitch ratio, written in the form1 : x, where 1 : x represents the tangent ratio of the pitch angle.

King post

Standard type A

Standard type B

Scissor

Fink

Scissor parallel

Half

Parallel

Pitched warren

Cantilever

Bow string

King post

Tie beam12.5° 12.5°

RafterRafter



Express a pitch angle of 12.5° in the form of a pitch ratio.

THINK WRITE

Draw a diagram, labelling the features.

The pitch ratio represents the tangent ratio.

tan 12.5° =

x tan 12.5° = 1

x =

x = 4.5

Write the pitch ratio in the form 1 : x. So the pitch ratio for a pitch angle of 12.5° is 1 : 4.5

1

12.5°1

x

21x---

1tan 12.5°----------------------

3

11WORKEDExample

Express a pitch ratio of 1 : 6.3 in the form of a pitch angle.

THINK WRITE

Draw a labelled diagram. The pitch ratio represents the tangent of the pitch angle.

Use the tangent ratio to determine the pitch angle.

tan θ =

tan θ =

∴ θ = tan−1

= 9°∴ Pitch angle = 9°

1

6.3θ

1

2oppadj---------

16.3-------

16.3-------

12WORKEDExample



The roof

1 Complete the following table, converting to pitch angle or pitch ratio as required.

Pitch ratio Pitch angle

1 : 22.9

1°

1 : 7.6

10°

1 : 1

20°

35°

1 : 1.08

A symmetrical roof has a pitch of 12.5°. If the building is 14 metres wide, find the height of the king post in the roof truss.

THINK WRITE

Draw a labelled diagram. The roof is symmetrical so the king post is directly above the middle of the base.Use the tangent ratio to determine the height of the king post. tan 12.5° =

tan 12.5° =

7 × tan 12.5° = h1.552 m = h

Give the answer to the nearest millimetre.

So the king post is 1552 mm high.

1

12.5°14 m7 m

h2

3 oppadj---------

h7---

4

13WORKEDExample

remember1. The pitch of a roof is the angle that the roof makes with the horizontal. It can

also be expressed as a pitch ratio in the form 1 : x.2. The pitch ratio represents the tangent of the pitch angle in the form 1 : x.3. Trigonometry is used to calculate truss heights and building widths.

remember

7G

SkillSH

EET 7.4

WORKEDExample

11

WORKEDExample

12


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 2712 A simple symmetrical roof truss is drawn below.

Complete the following table.

3 Find the angle of pitch in each of the following roof structures.

4 Calculate the pitch ratio and the angle of pitch for the following roof trusses.

5 The structures below are in the shape of a gable roof. Using the measurements indi-cated, calculate the angle of pitch for each.

Roof pitch Building width King post height Rafter length

10° 8 m

12° 10 m

12 m 2 m

25° 15 m

3 m 16 m

14 m 8.5 m

WORKEDExample

13

Building width

Roofpitch

Rafter Rafter

King post

θ

12 000 mm

2850 mm

23 000 mm

3500 mm

4000 mm

5000 mm

12 000 mm

7500 mm

3 m

1 m

10 m

3 m

PR = 9000 mmPQ = 9000 mmQR = 13 000 mm

Q

P

R

ST = 6000 mmSV = 6000 mmVT = 10 000 mm

S

V

T



The gable and hip roofThe difference in shape between a gable roof and a hip roof is shown in the sketches below.

To illustrate the differences more clearly, we can construct a paper model of each.

1 Draw a scale enlargement of the models, or use a photocopier to enlarge them to a more manageable size.

2 Cut along the unbroken lines, and fold along the dotted lines. Fold each roof into shape.

3 Notice the difference in shape of each roof. How many flat surfaces does each have? What are their shapes?

4 Observe the effect of changing the angle of pitch on the height of the roof and the width of the structure.

5 Draw a 3-D sketch of the hip roof shown below.

6 If we were to draw a plan of the roof (looking down from above), it would appear as below. Note the position of the apex of the hip with respect to the centre of the roof and its distance from the edge of the roof.

inve


n

Gable roof Boxed gable roofHip roof

Boxed gableroof

Hiproof

Hip Hip

Ridge

Pitch angle

Y X

Ridge

Half widthof roof

Half widthof roof

Y X



Cladding the roofThe most common cladding for a roof these days is concrete tiles or some type oftreated metal sheeting (colorbond). To compare the costs of a concrete tiled roof and ametal roof is not simply a matter of considering the cost of the materials. Tiles requirea stronger roof frame with more structural supports to which they are tied. Metalsheeting, on the other hand, does not require as many supports. Laying tiles is morelabour intensive than attaching long metal sheets, which sometimes wrap from one sideof the roof to the other. These and other factors influence the cost.

However, house builders will advise that there is little difference in cost between pro-viding a metal roof and providing a concrete tile roof when all factors are considered.If we were to purchase the materials from a supplier and construct our own roof, wewould find a concrete tile roof dearer than a colorbond roof. This anomaly existsbecause building contractors are able to negotiate special deals with suppliers whoobviously wish their product to be used rather than a competitor’s product.

a Identify the types of roof structure on the house pictured above.b The top storey roof is 6 m wide and 12.5 m deep and has a pitch of 25°. Determine the

cost to supply concrete tiles at $35/m2.

Continued over page

THINK WRITE

a The one on the right front has a flat vertical front face, so it is a gable roof. The others have a front face which slopes backwards, so they are hip roofs.

a There is one gable roof on the front right.The front left roof, the upper storey roof andthe roof on the lower storey are each of hiproof style.

b Draw a diagram of the roof, marking the measurements. Remember that the apex of a hip roof is roof width in from the edges.

b1

12---

25°

3 m

3 m6 m

12.5 m

h

p

14WORKEDExample



When a roof is to be given metalcladding, the material is purchased insheets which may be cut to anyrequired length. The corrugations ofadjacent sheets overlap to seal the rooffrom the weather. We talk about theeffective width of a sheet taking intoaccount the overlap. The sheets are alllaid from the apex of the roof towardsthe gutter.

THINK WRITE

Draw the shape of each face of the roof. The roof consists of 4 faces.

Determine the area of the two trapeziums. First calculate the perpendicular height using the pitch angle and the base length of the triangle.

For trapeziums,

cos 25° =

cos 25° =

h cos 25° = 3

h =

h = 3.3 mArea of 2 trapeziums

= (12.5 + 6.5) × 3.3 × 2 m2

= 62.7 m2

Determine the area of the triangles. Note: the perpendicular height of the triangles (p) is the same as the perpendicular height of the trapeziums (h).

Area of 2 triangles = × 2 m2

= 19.8 m2

Find the total area of the roof. Total area of roof = 62.7 m2 + 19.8 m2

= 82.5 m2

Calculate the cost of tiles from the roof area.

∴ Cost to supply concrete tiles at $35/m2

= $35 × 82.5= $2887.50

2

2 trapeziums 2 triangles

3

12.5 m 3 m

hh25°

adjhyp---------

3h---

3cos 25°------------------

12---

4

6 m

p = 3.3 m

6 3.3×2

----------------

5

6



The carport above has a storeroom at the rear. The gable roof is 6 m wide and 9 m deep and has a pitch of 25°. The cladding for the roof is to be colorbond. The metal roofing is supplied in sheets which may be cut to any length. The effective width of each sheet is 762 mm, which allows for a minimum overlap of one and one-half corrugations (about 80 mm). Calculate the cost of supplying the colorbond if the manufacturer quotes $8.50 per linear metre.

Continued over page

THINK WRITE

Draw a diagram of the shape of the roof. A gable roof has two rectangular areas. Mark in the known measurements.

Find the width of the rectangular section of the roof (w) using the pitch angle and the base of the triangle. This will be the length of the colorbond sheets.

cos 25° =

cos 25° =

w cos 25° = 3

w =

w = 3.3 m∴ The colorbond sheets are each 3.3 m long.

Determine the number of colorbond sheets down the depth of the roof.

Effective width of colorbond sheet = 762 mm∴ No. of sheets required for 9 m

= 9 m ÷ 762 mm= 9 m ÷ 0.762 m= 11.8 sheets

This means that 12 sheets are required for each side of the roof.

Front elevation

Hardiplank cladding tofront and back gables

Roller shutter doorto store RM

Carport

Rear elevation

Hardiplank cladding

Selected alumglazing

Facebrickwork

1

9 m

6 m

25°w

25°3 m

w

2adjhyp---------

3w----

3cos 25°------------------

3

15WORKEDExample



Cladding the roof

1 Draw sketches to illustrate the difference in shape between a hip roof and a gable roof.Identify the shapes of the faces on each roof.

2 Copy this diagram of a hip roof and complete the labelling from the following information and calculations:a The roof is 18 m long and 6 m wide (label EF

and FH).b It has a pitch of 20° (label angle CAB).c Label the length of AB (half the width of the roof).d Use the right-angled triangle ABC to determine the lengths of AC and BC. Mark

their values on the diagram.e Label the length of the ridge CD.f Calculate the area of the trapezium section of the roof CDEF.g The length CG (the perpendicular height of triangle CFH) is the same as CA. Deter-

mine the area of triangle CFH.h What is the total area of the roof?i If concrete tiles cost $45/m2 to supply and fit, what would it cost to clad the roof?

THINK WRITE

Calculate the number of sheets for both sides of the roof.

∴ No of sheets required for both sides of roof= 12 × 2= 24 sheets

Determine the total length of colorbond.

Each sheet of colorbond is 3.3 m long∴ Total length of colorbond = 3.3 × 24 m

= 79.2 mDetermine the cost. ∴ Cost of colorbond = $8.50 × 79.2

= $673.20

4

5

6

remember1. Two common roof shapes are the gable roof and the hip roof.2. In the gable roof, the end triangular shapes are vertical. These two triangles are

not normally classed as part of the roof. They are usually clad in a different material.

3. A hip roof consists of two triangular shapes and two trapeziums.4. The apex of the hip section of a hip roof is half the roof width in from the

edges of the roof.5. The most common claddings for a roof are concrete tiles and metal sheeting.6. Metal sheeting is laid from the ridge of the roof towards the gutter.7. The effective width of a metal sheet is 762 mm allowing for the overlap.

remember

7H

WORKEDExample

14

E

D C

H

G

FA

B

SkillSH

EET 7.5

SkillSH

EET 7.6


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 2773 Repeat the process in question 2 for a roof the same

size in a gable style.a Label the diagram with the length (QR), width (PQ)

and pitch (angle TQV).b Label the length VQ (half the width of the roof).c Use the right-angled triangle VTQ to calculate the

lengths TV and TQ. Mark these on the diagram.d Determine the area of the rectangular section of the roof QRST and hence deter-

mine the total area of the roof.e What would it cost to cover the roof with the same concrete tile cladding at $45/m2?

4 A hip roof with a pitch of 18° covers a shed 8 m by 12 m. Draw a sketch of the roof,labelling the lengths of all sides.

5 If the roof covering the shed in question 4 was a gable style instead of hip, compare thelengths of the sides by drawing a labelled sketch.

6 Compare a hip and a gable roof for a square floor plan.a Draw a hip roof with a 15° pitch for a 12-m-square floor plan.

i What is special about the shape of the roof?ii Calculate the roof area.

b Draw a gable roof with a 15° pitch for a 12-m-square floor plan. Calculate the roof area.

7 What would be the cost to supply colorbond metal sheeting to cover the roof of a 10-mby 20-m shed if the pitch of the gable roof was 22° and the metal cladding was $12.50per linear metre. (The sheeting has an effective width of 762 mm.)

8 A gable roof with a pitch of 20° covers a shed 5 m by 10 m.a Draw a 3-D sketch of the roof.b Label the lengths of all sides.c Draw a 2-D net of the roof to scale.

Bricking the walls

The CD shows brickworking in action in the ‘brickwork and electricals’ unit of theConstruction section. Once the roof cladding has been fixed in place, the external wallscan be bricked around the wall frame. Notice that the bricks are not laid one on top ofthe other, but each brick straddles two bricks in the course below. This provides greaterstructural strength. The bricks are held together with (and separated by) mortar(a mixture of cement and sand).

P V

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WORKEDExample

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A common house brick measures 230 mm by 110 mm by 76 mm.

There is usually 10 mm of mortar between the bricks. This means 10 mm between each brick in a row, and 10 mm of mortar between each row. Thus, the effective size of a brick is 240 mm by 110 mm by 86 mm when bricks arelaid in rows as for a wall of single brick thickness.

230 mm

110 mm 76 mm

A normal house brick measures 230 mm by 110 mm by 76 mm. If 10 mm is the allowance for mortar, give an estimate of the number of bricks required per square metre when bricking a wall.

THINK WRITE

Find the effective surface area exposed by 1 brick including its surrounding mortar.

The effective length of a brick(including mortar) = 240 mmThe effective height of a brick (including mortar) = 86 mm∴ effective surface area of 1 brick

= 240 mm × 86 mmConvert the area to m2; that is divide by 1 000 000.

= 20 640 mm2

= 0.020 64 m2

Calculate the number required for 1 m2. No. of bricks/m2 =

= 48.4Allow for breakages, etc. There are approx. 50 bricks required per m2

1

240 mm

86 mm

2

31

0.020 64--------------------

4

16WORKEDExample



When we are constructing a brick wall, we often refer to a course of bricks. The termcourse is just another word for row.

How many bricks would be needed per row for a wall of common house bricks (230 mm by 110 mm by 76 mm) stretching 4 m? Allow 10 mm for mortar between the bricks.

THINK WRITE

The question involves only the length of a brick. Make allowance for mortar.

Effective length of 1 brick= 230 mm + 10 mm (mortar)= 240 mm

Find the number of bricks of this length in a total length of 4 m. Convert to the same units before dividing.

No. of bricks in a 4-m length= 4 m ÷ 240 mm= 4000 mm ÷ 240 mm= 16.7 bricks

Round up. So 17 bricks would be needed per row for a wall that is 4 m long.

1

2

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17WORKEDExample

What height would the wall in the previous example reach, if it consisted of 8 courses of bricks?

THINK WRITE

This question involves only the height of the brick. Make allowance for mortar.

Effective height of 1 brick= 76 mm + 10 mm (mortar)= 86 mm

Find the height of 8 rows of bricks. ∴ Height of 8 rows of bricks= 8 × 86 mm= 688 mm or 68.8 cm

1

2

18WORKEDExample

remember1. Common house bricks measure 230 mm by 110 mm by 76 mm.2. The allowance for mortar is generally 10 mm between bricks and between

rows.3. The effective length of a brick is 240 mm, while its effective height is 86 mm.4. As an estimate, there are 50 bricks per square metre in a wall.5. When providing estimates of numbers of bricks required for a job, remember to

allow for areas occupied by windows and doors.

remember



Brickwork

In the following questions, assume that the bricks referred to are normal householdbricks measuring 230 mm by 110 mm by 76 mm and that the mortar allowance is10 mm.

1 Using the estimate for the number of household bricks per square metre in a brick wall(50), how many bricks would be required for the following jobs:

a a wall 14 m long and 2.4 m high

b a fence 30 m long and 2 m high

c a shed 5 m square and 3 m high (ignore door and windows)

d the side of a house 16 000 by 2400.

2 Give an estimate of the number of bricks required for the following walls. Make allow-ance for any doors and windows indicated.a b

3 How many bricks would be required per row (don’t forget the mortar) for a wall oflength:

a 12 m b 600 cm c 4500 mm

4 What height would walls consisting of the following number of courses (rows) ofbricks reach?

a 5 b 22 c 55

5 Calculate the number of courses (rows) of bricks in walls of the following heights(don’t forget the allowance for mortar).

a 2.15 m b 860 mm c 1032 cm

6 A wall measures 8400 mm long and 1290 mm high.

a How many courses of bricks are in the wall?

b How many bricks are in each course?

c Calculate the total number of bricks in the wall.

d Using the estimate of the number of bricks per square metre, what number would berequired for the wall? Compare your answer with that obtained in part (c). Com-ment on any difference.

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900

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7

The floor plan below depicts an open garage with a workshop at the rear. All walls arebrick, and entry to the workshop is through a 900 by 2000 sliding door. The window inthe workshop is 900 by 1200. If the walls are 3 m high, allowing for openings, estimatethe number of bricks in the building’s construction.

Quantity of bricks in Sturt

Now that you have some method of estimating the number of bricks required for a job, we’ll use this knowledge to estimate the quantity of bricks required for the Sturt home.

Because the floor plan does not indicate window and door widths, you’ll have to access the Sturt plan via the CD. (If you don’t have a floor plan, print one out to annotate it with measurements.)

1

Position the Sturt plan on site.

2

Note the measuring-tool option on the left of the screen. Using this, measure the widths of all windows and doors and write them on your floor plan.

3

Because elevations are not displayed, you are unable to measure the heights of the openings. For the purpose of this exercise, you can assume an average height of the windows as 1200 mm and a door as 2000 mm.

4

Calculate the total surface area of the external walls, and subtract the total area of the openings (work in metres). This gives an estimated area of brickwork in square metres.

5

Multiply this brick area by 50 (remember there are 50 bricks to each square metre of brickwork) to give an estimate of the total number of bricks required for the job.

6

Choose a two-storey design, and repeat the exercise (don’t forget the extra bricks for the top level).

WorkS

HEET 7.2

6000

2000

6000

900

900Slidingdoor

Window

Workshop

Garage

G JGar

dner

inve


n



Scale drawings and plans• Scales are represented by comparing the measurement on a plan with that in the

field. The scale factor = plan length : field length.• Survey, site and floor plans are reductions of field measurements.• The absence of units on a plan indicates that the measurements are in millimetres.• Floor plans are aerial views, whereas elevations are side views.• The floor plan, together with the elevations, gives an indication of the external

features of the completed structure.

Preparing the foundations• Horizontal levels may be checked using a spirit level or a water-filled clear plastic

tube.• Right angles are ensured using a builder’s square.• Footings are dug around the perimeter of a structure, and sometimes under load-

bearing internal support walls. Strength is provided with trench mesh.• The slab provides the basis for the frame.

Erecting the structure• A spirit level or a plumb bob may be used to check the vertical nature of a wall.• Bracing provides rigidity in frames. The brace angle must lie in the range 37° to

53°.• The pitch of a roof is expressed either in angle form or as a ratio.

Cladding the structure• The most common roof shapes are gable and hip.• Common cladding for a roof is either tiles or sheet metal.• The external walls of a structure are generally brick or timber.• There is an estimated number of 50 bricks per square metre.

summary



These problems draw together all the techniques practised in this chapter and focus on interpreting and applying mathematics to the plans and elevations of a typical residence. Consult the following plans or diagrams when directed to do so.

CHAPTERreview

© G. J. Gardner



© G. J. Gardner



© G. J. Gardner



1 Refer to the floor plan to determine:a the scale of the floor plan b the number of bedrooms in the residencec the features of the main bedroom d the size of the garagee the position of the smoke alarms f the floor area of the houseg the meanings of the abbreviations VB, BRM, WM, WC, WIR, W/O, R HOOD, HP, DP.

2 Make a front elevation scale drawing of the garage.

3 Make a scale drawing of the floor plan of the section that comprises bedrooms 2 and 3, including the wardrobes.

7A © Creedon Reid & Associates Pty Ltd


C h a p t e r 7 B a s i c s o f c o n s t r u c t i o n 2874 Answer the following with reference to the site plan.

a In which direction does the house face?b What is the shape of the land?c What are the dimensions of the building block?d Which street does the house face?e What is the closest distance the house approaches to the southern boundary line?f How close is the garage to the front boundary?g What percentage of the land does the house cover?

5 Consult the elevation drawings, floor plan and site plan to answer the following.

6 Refer to the footing plan to answer the following.a What is the thickness of the slab around the edge of the house?b How thick is the slab in the centre of the house?c What is the depth of the piers for the pergola?d What does the instruction ‘50 fall’ in the garage mean?e What are the dotted path lines throughout the floor?f Is the garage at the same level as the house?g What is the fall on the pergolas?h What symbol is used to indicate the shower?i Where is the storm water drain? How deep is it?

7 Through answering the previous questions you should be aware of the features of the house. Imagine that this residence was the ‘House of the week’ to be published in a magazine. Write about a page describing the house and promoting its features.

8 The living–dining room measures 6.4 m by 4 m.a Using a scale of 1 : 100, what would be its measurements on the plan?b If the scale was altered to 1 : 50, what would the plan measurements be?

9 In pegging out the rectangular pergola north of the kitchen, the builders measured the two adjacent sides to be 3.6 m and 2.67 m. If these two sides were at right angles to each other, what should be the length of the diagonal?

10 The footings around the edge of the pergola in the previous question are 150 mm wide and 200 mm deep. What would be the cost of concrete for the footings at $130/m3?

11 The scale of the footing plan is 1:160. Give an estimate of the volume of concrete required for the internal footings in the area comprising bedroom 1, the living room and the dining room. The footings are 300 mm wide and 400 mm deep.

7B

7Ca The two long windows on the left-hand side of the front elevation are in which room?

b In which direction does the opening of the garage face?

c In which corner of the house is the kitchen?

d How many windows are there in bedroom 1?

e What type of door is on the garage?f How many windows face the south?g What type of cladding is on the roof?h What is the pitch of the roof?i What type of roof structure is above the

garage section which faces the street?

j What type of cladding is on the external walls?

k What is the width of the hallway leading to bedrooms 2 and 3?

l How far does the roof overhang the external walls?

m How high is the ceiling?n What type of truss is pictured in the

section plan?o What size is the pergola north of the

kitchen?p What is the thickness of the internal

walls?q How thick are the external walls?

7A

7D

7E

7E



12 Find the number of 6-metre lengths of trench mesh required in the footings beneath the three exterior walls of bedroom 1 (including the ensuite area). The mesh must overlap at the corners and when two sheets meet there must be 500 mm of overlap. The 6-metre lengths can be cut if necessary. Allow for two layers of mesh.

13 Calculate the cost of the concrete ($130/m3) required for the slab for the garage and storeroom area if it is laid 100 mm thick

14 Consider the diagram showing the bracing of the frame. If the frame is 2400 mm high, determine the length of the longest brace (within the acceptable angle constraints) that could be used to provide rigidity to the frame.

15 The wall section between the windows in the front elevation of the garage is as shown. Find the length of the longest brace which could be used to strengthen this section. (It is possible for the brace to extend from the top of the frame through the section beneath the windows.)

16 The pitch angle of the roof is 21°. Express this as a pitch ratio.

17 If the pitch ratio in the previous question was twice as large:a would this double the pitch angle?b what would the pitch angle be?

18 This section represents the front elevation of the area comprising bedroom 1, the living room and the dining area. What is the height of the ridge of the roof above the ceiling?

19 A work shed in the back yard is 6 m by 10 m. It has a hip roof with a pitch angle of 21°. Calculate the cost to supply concrete tiles as a roof coverage at $45/m2.

20 If the roof of the shed in the previous question had been a gable style covered with colorbond, calculate the cost to supply the colorbond at $10.50 per linear metre. (The metal sheets have an effective width of 762 mm.)

21 Using the estimate that there are approximately 50 common bricks in each square metre, how many bricks would be required for a brick wall 10 m long and 2 m high?

22 A house brick measures 230 mm by 110 mm by 76 mm. With an allowance of 10 mm for mortar, how many bricks would be required per row to build a brick feature wall in the living–dining room backing on to the WIR in bedroom 1? The length of the wall is 2.6 m.

23 How many rows of bricks would be required for the wall in the previous question if it extended from the floor to the ceiling?

24 Compare the total number of bricks necessary for constructing the wall in question 23:a using your answers from questions 22 and 23b using the estimate of 50 bricks/m2.

7E

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7F

7F

600

2400

1400

1000 1000

7G7G

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Footings and slab to be in accordance withwall detail and engineer’s specifications

Engineer designed prefabricated timber rooftrusses to manufacturer’s specification @ 900 CRS

2440

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p of

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to u

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f tr

uss

7510

2100

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Joinery

21° Pitch 21° PitchCeiling

Floor300

2400

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CHAPTERyyourselfourself

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Maths A - Chapter 7

Education

gardner mq

scale factor

clear plastic

common roof

supply concrete

square oor

pitch ratio

metre lengths