Maths A - Chapter 7

7syllabussyllabusrrefefererenceence
Strand:
Applied geometry
Core topic:
Linking two and three
dimensions
In thisIn this chachapterpter
7A Scale drawings
7B Building plans
7C Floor plans and elevations
7D Pegging out the perimeter
7E Footings and slabs
7F Bracing
7G The roof
7H Cladding the roof
7I 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 to
achieve that great Australian dream of ‘owning your own home’. They realise that this
is probably the largest single financial investment they are likely to make, so it requires
very careful planning.
Having looked at houses for sale on the market, and not finding what they really
wanted, they decided to investigate building a home. A friend introduced them to a CD
produced by G. J. Gardner Homes specially for new home buyers.
This software package is included on the CD accompanying your text book and is
able to be viewed by clicking on the icon seen on the left. You should look briefly
through 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 the
Sturt design (see image below), from the Colonial series.
This chapter traces the construction of their home, and looks at the mathematics
applied throughout the process.
© G. J. Gardner
GJG
ardner
MQ Maths A Yr 11 - 07 Page 232 Friday, July 6, 2001 8:58 AM

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 would
represent 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 m
12 m
4 m
2 m
3 m
10 cm 100 mm
500 mm
20 m
2400 mm
2400 mm
x 2400 mm3 m
y
2400 mm
3.5 m
x
2400 mm
3.5 m
y
2 m
10 m
5 m
1 m

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 the
needs 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 statements
1 : 100
1 cm ↔ 1 m
all represent the same scale — that is, 1 cm on the plan represents a distance of 1 metre
in 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 measurements
are one hundredth of the field measurements, or the field measurements are 100 times
the size of the plan measurements. Notice that in the first representation, no units are
mentioned. This means that the user can supply the appropriate units for the particular
situation (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 the
original.
5 m
8 m
h 3000 mm
3500 mm
h
0 1 2 3 4 5 m
plan length
field length
----------------------------
1
100
---------

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.
a
b
c
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 is
1 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 cm
So 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 mm
or = 3.5 cm
1
2
3
1
1
100
---------
2
1WORKEDExample
remember
1. 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 length
4. 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
SkillS
HEET 7.1
SkillS
HEET 7.2

3 Classify the following as enlargements or reductions.
a 1 : 200
b 1 mm ↔ 1 m
c
d
e 10 :1
f
4 Express the following in the form of a simplified ratio.
a 2 cm ↔ 100 m
b
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 measurements
of the following on the plan?
a Patio 3 m long and 1.75 m wide
b Bedroom 3.75 m × 2.25 m
c 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 measurements
of 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
1
10 000
----------------
Mat
hcad
Map
scales
EXCE
L Spreadshe
et
Map
scales 1
0 12 km6
0 1 2 3 4 mm
WORKED
Example
1a
WORKED
Example
1b
Mat
hcad
Scale
factors
EXCE
L Spreadshe
et
Map
scales 2
Frontboundary
Scale 1 : 250
House

9 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 plans
Detailed 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 plan
A 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 Harbour
LineofLeods131°24
MOOLOOLABA
PARKYN PDE.AVCG
Yacht Club
Boat Harbour
Supervisor
Pilot
Station
Public
Jetty
Pilot Stn.
Jetty
Dredged Mooring Basin
Dredged to 3m
Fish Board
7048000mN
RIVER
MOOLOOLAH
HARBO
UR
PDE.
BUDDINA
PACIFICBLVD.
SOUTH
PACIFIC
OCEAN
513000mE
153°
08'E
Beacons
mark channel
F Bu
Q Bu & FI R 4s
FI WR (3) 15s
53m 23 / 11m
Water Tower
Conspic.
Point CartwrightVQR
VQR
RED
W
H
ITE
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 plan
A 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 plan
A floor plan shows the exact dimensions of the building, including dimensions and
names of all rooms, the size and position of doors and windows, the direction in which
doors open, the thickness of walls and the location of stairs.
In simple constructions, the roof plan and the locations of all electrical and plumbing
fittings are superimposed on the floor plan. For more complex constructions, plans for
these services would be made separately. A floor plan without electrical or plumbing
fittings is shown in the following figure. If units are not indicated on a plan, it is
assumed that dimensions are in millimetres.
110
630 m2
30.0
21.0
21.0
187
23.0
27.499
23.0
189
15.0
38.914
38.607
7.219
13.219 5.0
796 m2
Lot X
11 500 7 200
100014400
R.L.
31.400
LOT 8
GURNERSTREET
75 000
75 000
28000
28000
SET
OUT
POINT
30300
30600
30900
31200
31500
31800
32100
N

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.
1
1
2
---
2
1
2
---
3
1 $59 850
630 m2
-------------------
2
2WORKEDExample

Building plans
1 In the survey plan on page 238
a 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 the
positions of Lots 110 and 189 can assist in your answer, but include as many
other 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 dimensions
have 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.
remember
1. 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
7B
WORKED
Example
2

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.
GJG
ardner
inv
estigat
ioninv
estigat
ion

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.

Elevations
The scale drawings we have considered so far are representations of what we would
view from above, looking directly towards the ground. A three-dimensional structure
most frequently has four vertical faces and we can view these faces from four different
directions. We can observe a front and back view, as well as two side views. These
views are called elevations. The side facing mainly towards the east is known as the
eastern elevation (and similarly for the other sides). Earlier we saw the front elevation
of 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 of
the ﬂoor plan with those depicted on the elevations. Note that the windows and doors
have been labelled for ease of identiﬁcation.
A view from the south
gives 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 the
following.
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, D2
and D3. So there are 3 external doors on the
west.
c The linen cupboard is identified by L on
the floor plan.
c The linen cupboard is situated in the hallway
at the entrance to the bathroom.
3WORKEDExample
remember
1. 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
7C
WORKED
Example
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 the
roof is 4 metres above the level of the floor. Also assume that any windows on the
northern side are of the same style as those on the western side. Sketch the northern
elevation of the house. Mark all features and measurements you have identified on
your 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 is
given.
a b
c d
Work
SHEET 7.1
Gable roof Boxed gable roof
N N
Hip roof
N N
Gambrel
(a roof combining the addition
of 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
Garden
bed

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 2
Toilet
Bathroom
Pantry
Laundry
Kitchen
Bed 4 Family
Scale 1:125
GJG
ardner
inv
estigat
ioninv
estigat
ion
MQ Maths A Yr 11 - 07 Page 248 Friday, July 6, 2001 2:01 PM

Preparing to build
With an understanding of plans, we are now in a position to start looking at the process
of building.
Establishing levels
Once a suitable building block has been selected, the land must be cleared and levelled
in preparation for erecting the structure. There are sophisticated electronic devices
available today, but the older methods of using a spirit level with a straight edge, or
using a water level are still widely used.
A spirit level is a device, usually made of aluminium, containing a vial of liquid
with an air bubble. The bubble indicates whether the surface on which the device rests
is level.
Since water always ﬁnds its own level, a length of clear hose ﬁlled with water can be
used 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 edge
Spirit level
Level datum
Using a spirit level and straight edge
Water level in tube
gives cut off line for stump
Keep water in line with
the 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 obviously
most important that the corners of the house, which are meant to be right angles, are
exactly that. Even one or two degrees error may have dire consequences (remember
that the error becomes magnified, the longer the wall).
Builders make use of a building square, which can be easily constructed using three
pieces of timber.
Levels
It is relatively easy to construct simple devices to determine whether surfaces or
points are level.
Spirit level
1 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 level
As 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.
investigat
ioninvestigat
ion

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 square
are 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, as
shown in the following diagram. The building square is then placed under this ﬁrst
building 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 assured
of constructing right-angled corners where necessary.
Using suitable ratio of sides to form a right angle
90°
2500
2000
1500
Hold arms
in position
with temporary
fixings
Cut and fix brace
Constructing a builder’s square
First building line
Corner of building
Building corner
Second building line
parallel to arm of square
Establish first building line and
building 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 long
to 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 angle
in a builder’s square. The builder ﬁnds another piece of timber 800 mm long. If this
piece 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 corners
are 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 the
builder 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 mm
l
600 mm
2
610 000
4WORKEDExample
remember
1. 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
7D
SkillS
HEET 7.3 WORKED
Example
4

5 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 a
rectangular shed before he starts constructing the frame for the roof. He attaches the
water level to the top of the wall at diagonally opposite corners (positions X and Y). He
observes the level of the water to be 2 cm above point X when the water level is at
point Y. Draw a diagram to show this situation and explain what it means.
Footings and slab
Once the perimeter of the
house has been marked out,
the positions of drainage pipes
(for sinks, showers and toilets)
and perhaps cabling (electrical
and telephone) need to be
determined. Trenches are dug
to house these pipes and
cables. Footings (in the shape
of rectangular prisms) are also
dug around the perimeter of
the structure. They are then
filled with concrete and rein-
forcing to support the weight
of the walls. Preparations are
then made for pouring the
concrete slab to form the floor
of the house.
Pegging out the perimeter
Jaclyn 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.
inv
estigat
ioninv
estigat
ion
© G. J. Gardner

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 poured
2. laying a bed of sand, which has been poisoned, to prevent white-ant infestations
3. covering the sand with a layer of plastic
4. laying reinforcing mesh over the plastic.
The concrete slab can then be poured.
Determining the quantity of concrete needed for the footings and slab simply
involves 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 be
used.
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.
GJG
ardner
inv
estigat
ioninv
estigat
ion
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 m
2
3
5WORKEDExample

For composite-shaped surfaces, it is sometimes simpler to find the total length of the
footings, 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 the
footings.
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 m
10 m
1
5 m
5 m
15 m
0.4 m
0.4 m
2 m
3 m
10 m
2
3
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 lapped
full width at corners
500
min.
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 requires
6 m + 4.5 m (including overlap).
Each 7.5-m side requires
6 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 requires
2 lengths of mesh; that is, 4 lengths for both sides.
Each 7.5-m side requires
1 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 m
10 m
2
3
7WORKEDExample

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 m
5 m × 3 m
15 m
2 m 5 m
3 m
10 m
2
3
4
8WORKEDExample
GJGardner
inv
estigat
ioninv
estigat
ion

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 are
given, the measurements are assumed to be in millimetres.)
a Floor: 16m by 10m width: 350 depth: 500
b Floor: 1750 by 9500 width: 400 depth: 600
c Floor: 19 000 by 19 000 width: 400 depth: 700
2 Calculate the quantity of concrete required for footings for the following floor plan if
they 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 and
500 mm deep have been dug within its perimeter. Two layers of trench mesh are
required in the footings. The trench mesh is sold in 6-metre lengths. Industry standards
dictate that the trench mesh must overlap in the corners, and, if sheets are cut, a
500-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 the
boundary) as a load-bearing support for internal walls.
remember
1. 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
7E
WORKED
Example
5
WORKED
Example
6
10 m
8 m
12 m
5 m
WORKED
Example
7
WORKED
Example
8

i 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 concrete
Concrete 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 around
perimeter and
internally
8.5 m
15 m
perimeter and
internally
10 m
20 m
6 m
11 m
a b
inv
estigat
ioninv
estigat
ion
Age (days)
Concrete strength gain with time
No
curing
3 days
cured
Continuously
cured
Compressivestrength(%of
28daymoistcuredstrength)
0
7 14 21 28
20
40
60
80
100
120
140

The frame
After 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 at
the two parts, ‘Erecting the frame’ and ‘Rooﬁng’. You will notice that these days the
wooden frames are generally prefabricated in a factory, then transported to the site
where they are assembled.
Testing that a wall is vertical
It is obviously very important that all walls be vertical (the exception, perhaps, being
the Leaning Tower of Pisa). Two simple devices commonly used today are:
1. the spirit level and straight edge
2. the plumb bob.
The spirit level works on the same principle as described earlier when the device was
used 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 the
end 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 string
is 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 is
compared 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

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 be
parallel to the wall with the
gauge nail hard against end
of the plate
Read
Gauge nail
Gauge block
Compare top and bottom measurements
inv
estigat
ioninv
estigat
ion

Bracing
Most walls are rectangular. Because a rectangular frame does not retain its shape when
placed 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 metal
straps. (Note the metal strap on the frame on the CD.) The straps are placed on an angle
to the vertical, and attached to the frame. House frames are commonly constructed of
studs (which run vertically) and noggings (which are the horizontal separators between
the studs). Lintels and sills are usually placed above and below windows, while top
plates and bottom plates support the top and bottom of the frame. The diagrams below
illustrate the use of bracing in a variety of situations.
Bottom plate
Sill
Lintel
Angle brace
Noggings
Studs
Top plate
Diagonal
bracing

Building standards prescribe that the angle which the bracing makes with the bottom
of the frame must lie within the range 37° to 53°. Houses are commonly constructed
these days with a ceiling height of 2400 mm, so most of our calculations will use
frames 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?
inv
estigat
ioninv
estigat
ion
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
opp
adj
---------
2400
3600
------------
2400
3600
------------
9WORKEDExample

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 =
= 3988
So the longest brace is 3988 mm.
3
37°
b
2400
4
opp
hyp
---------
2400
b
------------
2400
sin 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
2400
Window
1
1000
1200
θ
opp
adj
---------
1000
1200
------------
1000
1200
------------
2
10WORKEDExample

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 =
= 3749
So 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°
opp
hyp
---------
2400
b
------------
2400
sin 39.8°
---------------------
4
remember
1. 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
7F
WORKED
Example
9 2.4 m
6 m
4750 mm
5810 mm
1060 mm
G
L
A
S
S

3 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 bracing
required for each.
a
b
The roof
Before commencing this section, view the ‘roofing’ unit of the Construction section of
the CD. Note the bracing which provides strength for the structure.
Roof trusses and pitch
Roof 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. They
transfer 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 often
used in domestic constructions (rumpus rooms, entertainment areas) as well as in large
commercial projects (factories, halls). Some examples of truss designs are shown in the
following diagrams.
WORKED
Example
10
2400
1200
Window
4000
2500
2400
2500
5300
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 post
and tie beam.
The pitch of the roof is the angle the roof makes with the horizontal. The roof above
has a pitch of 12.5°. Often the pitch is expressed as a pitch ratio, written in the form
1 : 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 beam
12.5° 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
2
1
x
---
1
tan 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
6.3
θ
1
2
opp
adj
---------
1
6.3
-------
1
6.3
-------
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° = h
1.552 m = h
Give the answer to the nearest
millimetre.
So the king post is 1552 mm high.
1
12.5°
14 m
7 m
h
2
3 opp
adj
---------
h
7
---
4
13WORKEDExample
remember
1. 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
SkillS
HEET 7.4
WORKED
Example
11
WORKED
Example
12

2 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
WORKED
Example
13
Building width
Roof
pitch
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 mm
PQ = 9000 mm
QR = 13 000 mm
Q
P
R
ST = 6000 mm
SV = 6000 mm
VT = 10 000 mm
S
V
T

The gable and hip roof
The 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.
investigat
ioninvestigat
ion
Gable roof Boxed gable roof
Hip roof
Boxed gable
roof
Hip
roof
Hip Hip
Ridge
Pitch angle
Y X
Ridge
Half width
of roof
Half width
of roof
Y X

Cladding the roof
The most common cladding for a roof these days is concrete tiles or some type of
treated metal sheeting (colorbond). To compare the costs of a concrete tiled roof and a
metal roof is not simply a matter of considering the cost of the materials. Tiles require
a stronger roof frame with more structural supports to which they are tied. Metal
sheeting, on the other hand, does not require as many supports. Laying tiles is more
labour intensive than attaching long metal sheets, which sometimes wrap from one side
of 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, we
would find a concrete tile roof dearer than a colorbond roof. This anomaly exists
because building contractors are able to negotiate special deals with suppliers who
obviously 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 and
the roof on the lower storey are each of hip
roof 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
1
2
---
25°
3 m
3 m6 m
12.5 m
h
p
14WORKEDExample

When a roof is to be given metal
cladding, the material is purchased in
sheets which may be cut to any
required length. The corrugations of
adjacent sheets overlap to seal the roof
from the weather. We talk about the
effective width of a sheet taking into
account the overlap. The sheets are all
laid from the apex of the roof towards
the 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 m
Area 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
hh
25°
adj
hyp
---------
3
h
---
3
cos 25°
------------------
1
2
---
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 to
front and back gables
Roller shutter door
to store RM
Carport
Rear elevation
Hardiplank cladding
Selected alum
glazing
Face
brickwork
1
9 m
6 m
25°w
25°
3 m
w
2
adj
hyp
---------
3
w
----
3
cos 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 m
Determine the cost. ∴ Cost of colorbond = $8.50 × 79.2
= $673.20
4
5
6
remember
1. 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
WORKED
Example
14
E
D C
H
G
FA
B
SkillS
HEET 7.5
SkillS
HEET 7.6

3 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 the
lengths 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-m
by 20-m shed if the pitch of the gable roof was 22° and the metal cladding was $12.50
per 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 the
Construction section. Once the roof cladding has been fixed in place, the external walls
can be bricked around the wall frame. Notice that the bricks are not laid one on top of
the other, but each brick straddles two bricks in the course below. This provides greater
structural strength. The bricks are held together with (and separated by) mortar
(a mixture of cement and sand).
P V
T
S
Q
R
WORKED
Example
15

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 are
laid 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 mm
The effective height of a brick
(including mortar) = 86 mm
∴ effective surface area of 1 brick
= 240 mm × 86 mm
Convert 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
=
= 48.4
Allow for breakages, etc. There are approx. 50 bricks required per m2
1
240 mm
86 mm
2
3
1
0.020 64
--------------------
4
16WORKEDExample

When we are constructing a brick wall, we often refer to a course of bricks. The term
course 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
3
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
remember
1. 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 household
bricks measuring 230 mm by 110 mm by 76 mm and that the mortar allowance is
10 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 of
length:
a 12 m b 600 cm c 4500 mm
4 What height would walls consisting of the following number of courses (rows) of
bricks 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 be
required for the wall? Compare your answer with that obtained in part (c). Com-
ment on any difference.
7I
WORKED
Example
16
900
2400
2000
Door
2500
2400
18000
Glass
panel
DoorWindow
500
900
910
1200
2000
WORKED
Example
17
WORKED
Example
18

7 The floor plan below depicts an open garage with a workshop at the rear. All walls are
brick, and entry to the workshop is through a 900 by 2000 sliding door. The window in
the workshop is 900 by 1200. If the walls are 3 m high, allowing for openings, estimate
the 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).
Work
SHEET 7.2
6000
20006000
900
900
Sliding
door
Window
Workshop
Garage
GJG
ardner
inv
estigat
ioninv
estigat
ion

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.
CHAPTER
review
© 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 residence
c the features of the main bedroom d the size of the garage
e the position of the smoke alarms f the floor area of the house
g 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

4 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 23
b using the estimate of 50 bricks/m2
.
7E
7E
7F
7F
600
2400
1400
1000 1000
7G
7G
7G
Footings and slab to be in accordance with
wall detail and engineer’s specifications
Engineer designed prefabricated timber roof
trusses to manufacturer’s specification @ 900 CRS
2440fromtopof
slabtou/softruss
7510
2100
600
Joinery
21° Pitch 21° Pitch
Ceiling
Floor
300
2400
7H
7H
7I
7I
7I
testtest
CHAPTER
yyourselfourself
testyyourselfourself
7

Maths A - Chapter 7

Recomendados

Recomendados

Más contenido relacionado

La actualidad más candente

La actualidad más candente (18)

Similar a Maths A - Chapter 7

Similar a Maths A - Chapter 7 (20)

Más de westy67968

Más de westy67968 (20)

Último

Último (20)

Maths A - Chapter 7