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Max B. FAJARDO Jr.

SIMPLIFIED CONSTRUCTION /

C9 I I I I I U I C
Third Edition

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Ma* 3. FAJASWO Jr.

SIMPLIFIED CONSTRUCTION
•

estimate Thir i Edition

 SIMPLIFIED CONSTRUCTION

estimate
Third Edition

By

Max B. Fajardo, Jr.

Max B. Fajardo Jr. Project Manager 1 V , Regional Director, DPWH;

Author of : Simplified Methods on Building Construction; Plumbing Design
and Estimate; Electrical Layout and Estimate; Planning and Designers
Handbook and Elements of Roads and Highways .
 Philippines Copyright
1995

by

5138 Merchandising
Publisher

ISBN 971- 8589 -10 - 4

^
SELRKJ ITS RESERVED
 FOREWORD

The first edition of Simplified Construction Estimate was

unique and a significant publication. It has stimulated for the first
time the imagination and bright ideas of planners and builders as
well as laymen towards a new technique in estimating. It
established a comprehensive source of data dealing with the
functional analysis presented in an illustrative examples using a
simple arithmetical approach.

The first edition however , was written during a period that

marked the beginning of changes. A radical transformation from
English to Metrication which was considered as the beginning of a
new era of mensuration. Under such pressing condition, in order to
maintain its effectiveness, the book requires adjustment and
revisions to incorporate the new development and thinking. As a
result the second edition of Simplified Construction Estimate has
been published in 1988. Painting , wail papering and many others
were introduced as a response to the many suggestions from
those who have read the first edition.

Six years later , the author stresses the need for an urgent
transformation to the new development because of the belief that
the ways of the past are no longer adequate. Tables and formulas
were revised and improved to obtain a more intriguing results.

For this third edition, the author again reiterate that, he does
not claim that this work in itself is perfect. In fact he would welcome
suggestions from those who are better knowledgeable that may
further enrich the contents of this book.

mbf
 PREFACE TO THE FIRST EDITION

Estimate has always been regarded as a valuation based on opinion

or roughly made from imperfect or incomplete data; a calculation not
professedly exact, an appraisement; also a statement, as by a builder, in
regard to the cost of certain work.

This book is intended for the last definition. As the science of

technology advances and the demand for technologists increases
educators, architects, engineers, contractors, carpenters and other
workers in allied fields are becoming more cognizant of the importance
and value of near accuracy even in estimates.

It is at this juncture that the author, in his desire to be of some help,

prepared this edition as an introductory text to facilitate the studies of
beginning students and other interested persons who wish an instant
answerto their problems involving cost of construction materials. This book
is, by design, an outline whose purpose is to guide those persons
concerned on matters of estimation. Tables and formula have been
prepared with the hope that readers or users of the book will find it easy
to formulate solutions to their cost problems. Illustrative examples were
included to serve as visual aids for more perplexing problems.

Since this is an initial attempt on the part of the author along this line
of endeavor, he does not claim that this work is in itself perfect. In fact he
would welcome suggestions from those who are better knowledgeable that
may further enrich the contents of this book

For the present edition, the author wishes to express grateful

acknowledgment for the valuable suggestions of Dean Francisca Guevara
of UNEP, Iriga City and Edgar Tuy who read the preliminary of the entire
manuscript The author likewise, wishes to express his indebtedness to
the countless persons here unnamed , who have contributed to the
scientific and experimental background from which this book has been
based
mbf
 TABLE OF CONTENTS

C H A P T E- 1RCONCRETE

1-1 Plain and Reinforced Concrete 1

1-2 The Principles of Concrete Mixing 3
1-3 The Unit of Measure 5
1-4 Concrete Proportions 8
1-5 Concrete Slab 11
1 -6 Estimating by the Area Method 14
1-7 Concrete Column 16
1-8 Estimating Concrete Column by vhe Linear
Meter Method 17
1-9 Post and Footing 21
1-10 Rectangular Column 27
1-11 Rectangular Beam and Girder 29
1-12 Circular Column 30
1-13 Concrete Pipes 33

CHAPTER- 2 MASONRY

2-1 Concrete Hollow Blocks 37

2-2 Special Types of Concrete Hollow Blocks 58
2-3 Decorative Concrete Blocks 65
2-4 Adobe Stone 68

CHAPTER- 3 METAL REINFORCEMENT

3- 1 Steel Bars 75
3-2 Identification of Steel Bars 77
3-3 Bar Splice , Hook and Bend 78
3- 4 Reinforcement for Concrete Hollow Blocks 85
3-5 Tie Wire for Steel Reinforcement 89
 3-6 Independent Footing Reinforcement 91
3- 7 Post and Column Reinforcement 99
3-8 Beams and Girders Reinforcement 100
3- 9 Lateral Ties 102
3- 10 Stirrups for Beam and Girder 115
3- 11 Spiral and Column Ties 118
3- 12 One Way Reinforced Concrete Slab 122
3- 13 Two Way Reinforced Concrete Slab 126
3- 14 Concrete Pipe Reinforcement 129

CHAPTER- 4 LUMBER
4- 1 Wood 132
4-2 Definitions of Terms 132
4 -3 Classification of Wood 134
4 -4 Methods of Log Sawing 136
4 -5 Defects in Wood 137
4 -6 Seasoning of Lumber 138
4-7 The Unit Measure of Lumber 139
4 - 8 Wood Post 144
4- 9 Girder 146
4- 10 Floor Joist and T&G Flooring 150
4 - 11 Siding Wood Board 155
4- 12 Girts, Rafters. Truss. Purlins and Fascia Board 158
4- 13 Studs 159
4- 14 Ceiling Joist 164
4-15 Ceiling Board 166
4- 16 Door Frame 172
4 -17 Window Frame 175

CHAPTER - 5 FORMS, SCAFFOLDING AND STAGING

5- 1 Forms 177
5-2 Greasing of Forms 179
5 - 3 Scaffolding and Staging 179
 5-4 Comparative Analysis Between the T&G and
Plywood as Forms 181
5-5 Forms Using Plywood 190
5-6 Forms of Circular Column 193
5-7 Estimating the Scaffolding and Staging 200

-
CHAPTER 6 ROOFING MATERIALS

6-1 Galvanized Iron Sheet 206

6-2 Gutter , Flashing, Ridge, Hipped and
Valley Roll 222
6-3 Asbestos Roofing 228
6-4 Colorbond Klip-Lok 235
6-5 Banawe Horizontal Metal Tile 237
6-6 Marcelo Roofing System 238
6-7 Colorbond Custom Orb 239
6-8 Milano Long span Steel Bricks 240
6-9 Colorbond Trimdex Hi-Ten 241
6-10 Brick Tiles Roofing 242

CHAPTER - 7 TILEWORK

7-1 Ceramic Tiles 243

7-2 Marble Tiles 252
7-3 Vinyl and Rubber Tiles 254
7-4 Terrazzo and Granolothic 256
7-5 Cement Tiles 260
7-6 Wood Tiles 263
7-7 Pebbles and Washout Finishes 265

CHAPTER - 8 HARDWARE

8-1 Bolts 267

 8- 2 Screw 274
8-3 Nails 279

CHAPTER - 9 STAIRCASE

9- 1 Introduction 291
9-2 Stairs Layout 295
9-3 Stringer 301

CHAPTER - 10 PAINTING

10-1 Paint 309

10- 2 Ingredients of Paint 310
10-3 Essential and Specific Properties of a
Good Quality Paint 312
10-4 The Elements of a Good Painting Job 313
10-5 Specifications for Surface Preparation 315
10-6 Kinds of Pamt. Uses and Area Coverage 318
10-7 Estimating Your Paint 327
10-8 Paint Failure and Remedy 333
10- 9 Waif Papering 336

CHAPTER - 11 AUXILIARY TOPICS

11- 1 Accordion Door Cover 339

11-2 Glass Jalousie 341
11- 3 Water Tank 342
11-4 Wood Piles 346
11 -5 Bituminous Surface Treatment 348
11 -6 Filling Materials 350
11 - 7 Nipa Shingle Roofing 352
11 -8 Anahaw Roofing 356
Construction Terminologies 359
 CHAPTER 1
CONCRETE
1- 1 PLAIN AND REINFORCED CONCRETE

Concrete is either Plain or Reinforced. By definition, Plain

Concrete is an artificial stone as a result of mixing cement , fine
aggregates, coarse aggregates and water. The conglomeration of
these materials producing a solid mass is called plain concrete.
Reinforced Concrete on the other hand, is a concrete with
reinforcement embedded in such a manner that the two materials
act together in resisting forces.
The different types of cement used in the construction are:

1. The Ordinary Portland Cement

2. The Rapid Hardening Portland Cement which is preferred
when high early strength is required.
3. The Blast Furnace or Sulphate Cement which is used on
structures to resist chemical attack.
4. The Low Heat Portland Cement for massive section to
reduce the heat of hydration.
5. The Portland Pozzolan Cement with a low hardening
characteristic concrete.
6. The High Alumina Cement.

1
 The High Alumina Cement is sometimes called Aluminous
Cement or Cement Fundu. Its chemical composition is different
from that of portland cement with predominant alumina oxide
contents of at least 32% by weight. The Alumina lime ratio is within
the limit of 0.85% to 1.3 % .

This type of cement has a very high rate of strength

development as compared with the ordinary portland cement.
Aside from its rapid hardening properties , it can resist chemical
attack by sulphate and weak acids including sea water. And it can
also withstand prolonged exposure to high temperature of more
than 1.000°C.

Alumina cement however , is not advisable for mixing with any

other types of cement.

The main composition of cement are:

60 to 65% Lime
18 - 25% Silica
3 - 8% Alumina
5 - 5% Iron Oxide
2 - 5% Magnesia
1 - 5% Sulfur Trioxide

Aggregates. The aggregates used in a concrete work is

classified into two categories .

1. Coarse Aggregate such as crushed stone, crushed gravel

or natural gravel with particles retained on a 5 mm sieve .

2 . Fine Aggregate such as crushed stone , crushed gravel,

sand or natural sand with particles passing on a 5 mm sieve .

2
 Size of Aggregates. For coarse aggregate (gravel), the
maximum nominal size are usually 40 mm, 20 mm, 14 mm or 10
mm. diameter . The choice from the above sizes depends upon the
dimensions of the concrete member more particularly the spacing
of steel bar reinforcements. However , good practice demands that
the maximum size of the coarse aggregate (gravel ) should not
exceed 25% of the minimum thickness of the member nor exceed
the clear distance between the reinforcing bars and the form.

The aggregate should be small enough for the concrete mixture

to flow around the reinforcement and is ready for compaction.

1 -2 THE PRINCIPLES OF CONCRETE MIXING

The purpose in concrete mixing is to select an optimum

proportion of cement , water and aggregates to produce a concrete
that will meet the specification requirements such as:

a. Workability c. Durability
b. Strength d. Economy

The proportions which will be finally adopted In concrete mixing

has to be established by actual trial and adjustments in order to
attain the desired strength of concrete required. The processes
would be as follows:

1 The Water Cement Ratio is first determined to meet the

requirements of strength and durability .

2 . The Aggregate Cement Ratio is then chosen to satisfy the

workability requirements .

3
 Laboratory test results showed that, the water -cement content
ratio is the most important factor to consider because it influences
not only the strength and durability of the concrete but also the
workability of the fresh concrete being poured inside the forms.

The ACI requirements for concrete are enumerated

as follows :
1. Fresh concrete shall be workable. Meaning, that the fresh
concrete can freely flow around the reinforcements and fill
all the voids inside the form.
2. That the hardened concrete shall be strong enough to carry
the design load.

3. The hardened concrete could withstand the conditions to

which it is expected to perform.

4. The concrete should be economically produced.

Concrete Mixture Maybe Classified as either :

a. Designed Mixture
b. Prescribed Mixture

Designed Mixture. Where the contractor is responsible in

selecting the mixture proportion to achieve the required strength
and workability.

Prescribed Mixture. Where the designing Engineer specify

the mixture proportion. The contractor’s responsibility is only to
provide a properly mixed concrete containing the right proportions
as prescribed.

4
1 -3 THE UNIT OF MEASURE

P rior to the adoption of the Metric measure otherwise known

as System International (SI) , solid concrete structure is estimated
in terms of cubic meters although the components thereof such as
cement , aggregates and water are measured in pounds , cubic foot
and gallons per bag respectively.

TABLE 1-1 CONVERSION FROM INCHES TO METER

Number Accurate Approximate Number Accurate Approximate
value value value value
1 . 0254 . 025 21 . 5334 . 525
2 . 0508 . 050 22 . 5588 . 550
3 . 0762 . 075 23 . 5842 . 575
4 . 1016 .100 24 . 6096 . 600
5 . 1270 . 125 25 . 6350 . 625
6 . 1524 . 150 26 . 6604 . 650
7 . 1778 . 175 27 . 6858 . 675
8 . 2032 . 200 28 . 7112 . 700
9 . 2286 . 225 29 . 7366 . 725
10 . 2540 . 250 30 . 7620 . 750
11 . 2794 . 275 31 . 7874 . 775
12 . 3048 . 300 32 . 8128 . 800
13 . 3302 . 325 33 . 8382 . 825
14 . 3556 . 350 34 . 8636 . 850
15 . 3810 . 375 35 . 8890 . 875
16 . 4064 . 400 36 . 9144 . 900
17 . 4318 .425 37 . 9398 . 925
18 . 4572 . 450 38 . 9652 . 950
19 . 4826 .475 39 . 9906 . 975
20 . 5080 . 500 40 1.0160 1.000

5
 Lately however, after the adoption of the SI unit of measures,
the 94 pounds per bag cement which is equivalent to 42.72
kilograms was changed and fixed at 40 kilos per bag. This simply
means a reduction of 2.72 kilograms of cement per bag. Such
changes therefore , requires adjustment of all measurements
relative to the proportion of concrete .

The traditional measurement of a box being used to measure

the sand and gravel is 12 inches wide by 12 inches long and 12
inches high having a net volume of one cubic foot. These
measurements will be changed to 30 x 30 x 30 centimeters box
wherein the values presented in Table 1-2 is based .

Very recently , a 50 kilogram cement was released in the market

for commercial purposes. This new development was already
incorporated in our tables and illustrations which would be noticed
in the variation of values presented in Table 1-2

The values presented in Table 1-1 could be useful in two

purposes. One for the accurate conversion of distance from English
to Metric and the other is the approximate value which will be
generally used in our simplified methods of estimating .

For Instance:

A) In solving problems, the probability of committing error is

high if more number is being used.

Example:

It is easier to use .10 meter ( the equivalent of 4“ ) than .1016

the exact equivalent value of 4“ be it by multiplication or by division
processes.

6
 8 » 80 by inspection and analysis
.10

8 = 78.7 by long process of division

.1016

B) To memorize the values given in Table 1-1 is a waste of

time and not a practical approach. A simple guide will be adopted
so that anybody could easily determine the equivalent values from
the English to Metric or vice versa .

Example:

1 . To convert Meter to Feet . . Divide the length by .30

Say 6.0 m. = 20 ft .
. 30

2. To convert Feet to Meter . . Multiply by . 30

Say , 30 ft. x . 30 * 9.0 meters

3. To convert Inches to Meter, just remember the following

values of equivalent:

1 inch = .025 m.
2 inches = .050 m.
3 inches = . 075 m.
4 inches = .10 m.

Note that ail length in inches are divisible by any one of these
four numbers and could be easily converted to meters by summing
up their divisible equivalent.

7
 Example:

a) What is the meter length equivalent of 7 Inches ?

By simple analysis 7 inches could be the sum of 4 and 3
Therefore:

4 inches = 100 m.
,

3 inches = .075 m.
Answer = ,175 m.

b) How about 21 inches to meter ?

5 x 4” = 20" = .50 m.
plus 1 " = 1” « .025 m.
Answer . . . . - .525 m.

Using the above simple guide, convert any number from inches
to meter and vice versa as an exercise problems.

1-4 CONCRETE PROPORTIONS

The most common and easy way of proportioning concrete is

the volume method using a measuring box for sand and gravel as
explained in Section 1-3, the Unit of Measure . The reasons behind
its traditional acceptance and use is the convenience in measuring
and fast handling of the aggregates from the stock pile to the mixer .

This volume method of concrete proportioning however had .

long been practiced in almost all types of concrete construction and
time have proven it to be effective and successful.

8
 38 cm

Measuring Box for 40 kg. Cement Box for 50 kg. Cement

FIGURE 1-1

TABLE 1-2 CONCRETE PROPORTION

Cement Sand Gravel
Class Mixture 40 kg. / Sag^ 50 kg. / Bag cu.m. cu. m.
AA 1 : 11/2 ; 3 12.0 9.5 .5 0 1.0
A 1: 2 : 4 9.0 7.0 .5 0 1.0
B 1 : 21/2 : 5 7.5 6.0 .5 0 1.0
C 1: 3 : 6 6.0 5.0 .5 0 1.0

It is interesting to note that the volume of sand and gravel for

all classes of mixture is constant at .50 cu.m, and 1.0 cu. m,
respectively . This is true on the assumption that the cement paste
enters the void of the sand and at the same instance the
combination of these two materials fills the void of the gravel and
thereafter , forming a solid mass called solid concrete equivalent to
one cubic meter.

9
 Based from actual concreting work, one cubic meter of gravel
plus one half cubic meter sand mixed with cement and water will
obtain a little bit more than one cubic meter solid concrete. The little
excess over one cubic meter will be considered as contingencies.

Comments

In actual concreting and masonry work , there are several

factors that might affect the accuracy of the estimate. Some of
which are enumerated as follows:

1. Inaccurate volume of delivered aggregates by the supplier

is very common. Delivery truck measurements must be
checked to assure that the volume of aggregates delivered
is exactly as ordered.

2. Dumping of aggregates (sand and gravel ) on uneven

ground surface and grassy areas reduces the net volume
of the aggregates.

3. Improper measuring of the aggregates during the mixing

operation. This is a common practice when the work is on
its momentum where laborers fails to observed the right
measuring of aggregates being delivered to the mixer.

4. The cement and fine aggregate for grouting concrete joints

is often overlooked in the estimate.

5. Cement waste due to bag breakage is usually caused by

wreckless handling and hauling.

6. Pilferages of materials. This could be avoided through a

good system of construction management.

10
 Ordering coarse aggregate must be specific as to :

a ) Kind of gravel, either crushed stone or natural gravel

from the creek .

b ) The minimum and maximum size of the stone must be

specified. It shall be free from mixed sand because sand
is cheaper than gravel. Natural gravel from the creek
requires screening to obtain a well graded aggregate but
screening involves additional cost on labor .

1 -5 CONCRETE SLAB

The discussions from cement to concrete proportions plus the

Tables presented could be more meaningful and appreciated if
accompanied by illustrations or examples of actual applications.

ILLUSTRATION 1 - 1
A proposed concrete pavement has a general dimensions of 4
inches thick, 3.00 meters wide and 5.00 meters long. Determine
the number of cement in bags, sand and gravel in cubic meters
required using class ”C" mixture.

FIGURE 1- 2

11
 SOLUTION:

.
1 Determine the volume of the proposed concrete pavement
Convert 4 inches to meter = .10 m.

V = .10 x 3.00 x 5.00

V = 1.5 cu. m.

2. Refer to Table 1-2. Using 40 kg. cement class "C ” mixture;

Multiply:

Cement : 1.5 x 6.0 = 9 bags

Sand : 1.5 x .50 = .75 cu.m.
Gravel : 1.5 x 1.0 = 1.50 cu.m.

Suppose there is no available 40 kg. cement but instead what

is available is a 50 kg. per bag. How many bags will be ordered ?

SOLUTION:

1. Knowing the volume to be 1.5 cu. m.

2. Refer to Table 1-2 under 50 kg. cement class C mixture ;

Multiply:

Cement 1.5 x 5.0 = 7 5 bags

Sand : 1.5 x .50 = .75 cu. m.
Gravel : 1.5 x 1.0 « 1 . 5 cu. m.

3. Since we cannot buy 7.5 bags, order 8 bags at 50 kg. / bag

12
 ILLUSTRATION 1- 2

A barangay road 6.00 meters wide and one kilometer long after
base preparation requires concreting. Find the number of bags
cement, sand and gravel in cubic meters required using class A
concrete if the slab is 6 inches thick.

15 CfTl

FIGURE 1-3

SOLUTION:

1. Determine the volume of the concrete pavement.

Convert 6” to meter = . 15 m; 1-km. = 1000 m.

V = t x w x I
V = .15 x 6.00 x 1 ,000 m.
V = 900 cu. m.

2. Referring to Table 1-2, using 40 kg . cement;

Multiply ;

Cement : 900 x 9.0 = 8 , 100 bags

Sand : 900 x . 50 * 450 cu . m.
Gravel : 900 x 1.0 = 900 cu. m.

13
 If there is no 40 kg. cement available , a 50 kg. cement requires:

Cement : 900 x 7 = 6 ,300 bags

Sand and gravel = the same as computed above .

The solution of the preceding illustration is the volume method

which could be simplified further by the Area Method.

1-6 ESTIMATING BY THE AREA METHOD

Estimating by the Area Method is much easier than the

Volume Method, but this could be done easily with the aid of Table
1-3 which readily gives the quantity of cement, sand and gravel
per square meter depending upon the required thickness of the
slab .

-
TABLE 1 3 QUANTITY OF CEMENT, SAND AND GRAVEL
ON SLABS AND WALLS PER SQUARE METER
Thick Mixture Class Sand Gravel
cm. 40 kg. Cement 50 kg. Cement cu. in. cu. m.
A B C A B C
5.0 .450 .375 . 30 . 350 .30 .250 .0250 .050
7.5 .675 . 563 . 45 . 525 . 450 .375 .0375 .075
10.0 .900 .750 . 60 .700 .600 .500 .0500 .100
125 1.125 .938 .75 .875 .750 .625 .0630 .125
15.0 1.350 1.125 .90 1.050 .900 750 .0750 .150
17.5 1.575 1.313 1.05 1.225 j 1.050 .875 .0880 .175
20.0
22.5
1.800
2 030
1.500
1.688
1.20
1.35
1.400
1.575
11.200 1.000
' 1.350 j 1.125
.1000 .200
1125 .225
25.0 2.250 1.875 1.50 1750 1.500 1.250 1250 .250
27.5 2475 2063 1.65 1925 1650 11375 . 1380 .275
30.0 2.700 2.250 1.80 2.100 1800 11500
.
.1500 .300

14
 ILLUSTRATION 1- 3

Adopting the problem of Illustration 1-1 and 1-2 using the Area
Method with the aid of Table 1-3, the solution will be:

Solution for Illustration 1- 1

1. Solve for the pavement area.

A = width x length
A = 3.00 x 5.00 m.
A = 15 sq. m.

2. Referring to Table 1- 3 , for 10 cm. slab , class "C" mixture and

using 40 kg. cement;
Multiply:

Cement : 15 x .60 = 9 bags

Sand : 15 x .05 = .75 cu. m.
Gravel : 15 x .100 = 1.5 cu. m.

Solution for Illustration 1 -2

1. Find the area of the concrete barangay road.

A = width x length
A = 6.00 m. x 1,000 meters
A = 6 , 000 sq. m.

2. Refer to Table 1-3. Using class A mixture for a 15 cm. thick

concrete slab. Multiply:

15
 Cement : 6 ,000 x 1.350 * 8.100 bags
Sand : 6 ,000 x .075 -
450 cu. m .
Grave! : 6 , 000 x .150 =
900 cu . m.

1-7 CONCRETE COLUMN

Estimating the quantity of materials for concrete column is

done in two different ways:
1 . By the Volume Method
2. By the Linear or Meter Length Method

ILLUSTRATION 1 -4
A concrete column is 5.00 meters high with cross sectional
dimensions of 25 cm x 30 cm If there are 8 columns of the same
size in the row , find the quantity of cement , sand and gravel content
of the columns if it is poured with class TV' concrete.

r
X X
T
5 00 m

25 cm

30 cm

Cross Section x x *

FIGURE 1*4

16
 SOLUTION:

1. Find the volume of one column.

V * .25 x .30 x 5.00 m.

V = . 375 cu. m.

2. Find the total volume of the eight columns

Vt = . 375 x 8
Vt = 3.0 cu . m.

3. Referring to Table 1-2, using class A concrete;

Multiply :

Cement 3.0 x 9.0 = 27 bags at 40 kg.

Sand 3.0 x .50 = 1.5 cu . m.
Gravel 3.0 x 1 0 = 3.0 cu. m.
,

1 - 8 ESTIMATING CONCRETE COLUMN BY

TH E LINEAR METER METHOD

Another way of estimating the quantity of materials for

concrete column is by the Linear Meter Method. Under this
method, the length of the column is first determined, then with the
aid of Table 1-4, the quantity required is found by single step of
multiplication.

17
 TABLE 1-4 QUANTITY OF CEMENT, SAND AND GRAVEL
FOR POST , BEAM AND GIRDER PER METER LENGTH
Size Mixture Class Sand Gravel
cm. 40 kg . Cement 50 kg. Cement cu. m. cu. m.
A B A B
15 x 15 .203 .169 .158 .135 .011 .023
15 x 20 .270 .225 . 210 .180 .015 .030
15 x 25 .338 .281 . 263 .225 .019 .038
15 x 30 .405 .338 .315 .270 . 023 .045
15 x 35 .473 .394 .369 .315 .026 .053
15 x 40 .540 .450 .420 .360 .030 . 060

20 x 20 .360 .300 .280 .240 . 020 .040

20 x 25 .450 .375 . 350 .300 .025 .050
20 x 30 .540 .450 .420 .360 .030 .060
20 x 35 .630 .525 .490 .420 .035 . 070
20 x 40 .720 .600 .560 .480 .040 .080

25 x 25 .563 .469 .438 .375 .031 .063

25 x 30 .675 .563 .525 .450 .038 .075
25 x 35 . 788 .656 .613 .525 .044 .088
25 x 40 .900 .750 . 700 .600 .050 .100
25 x 45 1.013 .844 .788 .675 .056 .113
25 X 50 1.125 .938 .875 . 750 .063 .125
30 x 30 .810 .675 .630 .540 .045 .090
30 X 35 . 945 .788 .735 .630 .053 .105
30 x 40 1.080 .900 . 840 .720 .060 .120
30 x 45 1.215 1.013 .945 .810 .068 .135
30 x 50 1.350 1.125 1.050 .900 .075 .150

35 x 35 1.103 . 919 .858 .735 . 061 .123

35 x 40 1.260 1.050 . 980 .840 . 070 .140

18
35 x 45 1.418 1.181 1.103 . 945 .079 .158
35 x 50 1.575 1.313 1.225 1.050 .088 .175
35 x 55 1.733 1.444 1.348 1.155 . 096 .193
35 x 60 1.890 1.575 1.470 1.260 .105 .210

40 x 40 1.440 1.200 1.120 .960 .080 .160

40 x 45 1.620 1.350 1.260 1.080 .090 .180
40 x 50 1.800 1.500 1.400 1.200 . 100 .200
40 x 55 1.980 1.650 1.540 1.320 .110 .220
40 x 60 2.160 1.800 1.680 1.440 . 120 .240

45 x 45 1823 1.519 1.418 1.215 . 101 .203

45 x 50 2.025 1.688 1.575 1350 .113 .225
45 x 55 2.228 1856 1733 1.485 .124 .248
45 x 60 2.430 2.025 1.890 1.620 .135 .270

50 x 50 2.250 1.875 1750 1.500 .125 .250

50 x 55 2.475 2.063 1.925 1, 650 . 138 . 275
50 x 60 2.700 2.250 2.100 1800 . 150 .300

55 x 60 2.970 2.475 2.310 1980 . 165 .330

55 x 70 3.465 2.888 2.695 2.310 .193 .385
55 x 80 3.960 3.300 3.080 2.640 . 220 .440
55 x 90 4.455 3.713 3.465 2.970 . 248 .495
55 x 100 4.950 4.125 3.850 3.300 . 275 .550

60 x 60 3.240 2.700 2.520 2.160 . 180 .360

60 x 70 3.780 3.150 2.940 2.520 . 210 420
.

60 x 80 4.320 3.600 3.360 2.880 . 240 .480

60 x 90 4.860 4.050 3.780 3.240 . 270 .540
60 x 100 5.400 4.500 4.200 3.600 . 300 .600

19
65 x 60 3.510 2.925 2.730 2.340 .195 .390
65 x 70 4.095 3.413 3.185 2.730 .228 .455
65 x 80 4.680 3.900 3.640 3.120 . 260 .520
65 X 90 5.265 4.388 4.095 3.510 .293 .585
65 x 100 5.850 4.875 4.550 3.900 . 325 .650

70 x 70 4.410 3.675 3.430 2.940 . 245 .490

70 x 80 5.040 4.200 3.920 3.360 .280 .560
70 x 90 5.670 4.725 4.410 3.780 . 315 .630
70 x 100 6.300 5.250 4.900 4.200 . 350 .700

75 x 70 4.725 3.938 3.675 3.150 . 263 .525

75 X 80 5.400 4.500 4.200 3.600 . 300 .600
75 X 90 6.075 5.063 4.725 4.050 . 338 .675
75 X 100 6.750 5.625 5.250 4.500 . 375 .750

80 x 80 5.760 4.800 4.480 3.840 . 320 .640

80 x 90 6.480 5.400 5.040 4.320 . 360 .720
80 x 100 7.200 6.000 5.600 4.800 . 400 .800

85 x 80 6.120 5.100 4.760 4.080 . 340 .680

85 x 90 6.885 5.738 5.355 4.590 . 383 .765
85 x 100 7.650 6.375 5.950 5.100 .425 .850

90 x 90 7.290 6.075 5.670 4.860 .405 .810

90 X 100 8.100 6.750 6.300 5.400 .450 . 900

95 X 90 7.695 6.413 5.985 5.130 .428 .855

95 x 100 8.550 7.125 6.650 5.700 .475 .950
100 x 100 9.000 7.500 7.000 6.000 . 500 1.000

20
 ILLUSTRATION 1-5

Adopting the problem of illustration 1-4 where there are 8

columns at 5.00 meters high each , we have:

SOLUTION:

1. Find the total length of the 8 columns.

8 x 5.00 m. = 40 meters

2. Referring to Table 1-4 , using class A concrete for the

25 x 30 cm. column size:
Multiply:

Cement : 40 x . 675 - 27 bags

Sand : 40 x .038 = 1.5 cu. m.
Gravel : 40 x .075 = 3.0 cu. m.

1-9 POST AND FOOTING

Structurally , post is always supported by a slab called footing.

Estimating the quantity of materials could be done in two ways:

1. By the Volume Method

2. By the Linear Meter and Area Methods combined. ( Linear

for the post and area for the slab. )

21
 ILLUSTRATION 1 6 -
A concrete post 4.00 meters high with cross sectional
dimensions of 20 cm . x 25 cm . is supported by a footing slab 20
cm. thick by 80 cm. square. Using class MA" concrete, find the
quantity of concrete materials if there are 12 posts of the same
size.

4.00 m.

Li 25 cm. 80 cm .
20 cm.
i
80 cm. 20 cm.
80 cm.
FIGURE 1-5

SOLUTION:

A , By the Volume Method

1. Find the volume of the 12 posts

V * 12 x ( .20 x .25 ) x 4.00 m.

V = 2.4 cu . m.

2. Solve for the volume of the slab.

22
 V * 12 ( .20 x . 80 x .80)
V * 1.54 cu. m.

3. Total volume of post and slab

V = 2.4 + 1.54
= 3.94 cu. m.

4. Refer to Table 1-2. Using class A concrete and 40 kg. cement

Multiply :

Cement : 3.94 x 9.0 = 35.46 say 36 bags

Sand : 3.94 x .50 - 1.97 say 2 cu. m.
Gravel : 3.94 x 1.00 * 3.94 say 4 cu. m.

B. Solution by the Linear and Area Method

1 . Solve for the total length of the 12 posts .

L = 12 x 4.00
= 48 meters

2. Refer to Table 1-4 along the 20 x 25 cm. column size

class A mixture ;
Multiply:

Cement : 48 x . 450 * 21.6 bags

Sand : 48 x .025 = 1.2 cu . m.
Gravel : 48 x . 050 = 2.4 cu. m.

3 . Find the total area of the footing slab

23
 A = 12 x (. 80 x . 80 )
» 7.68 sq. m.

4 . Referring to Table 1- 3 , class A mixture , 20 cm. thick slab:

Multiply:

Cement : 7.68 x 1.800 = 13.824

Sand : 7.68 x . 100 = .768
Gravel : 7.68 x .200 = 1.540

5. Add the results of 2 and 4

Cement : 21.60 + 13.824 * 35.424 say 36 bags

Sand 1.20 + .768 = 1 97 say 2 cu. m.
Gravel : 2.40 + 1.54 - 3.94 say 4 cu. m.

30 cm.

30 cm.
r —!
! Ft t
L- j-l
3.00 m.
P
90 cm.
l
• W \\
i ii
I
i « i F 15 cm
4.00 m.
I *» 80 cm —

FIGURE 1-6

24
 ILLUSTRATION 1-7

From Figure 1-6, determine the number of 40 kg. cement, sand

and gravel required using class "A" concrete for the footing and
class "C" concrete for the flooring.

SOLUTION ( By the Volume Method )

A ) Footing Slab

1. Solve for the volume of F

V » .15 x . 80 x .80
V » .096 cu. m.

2. Total volume of 4 footing slab

V< « .096 x4 a 0.384

B) Pedestal

1. Solve for volume P

V a .30 x .30 x .90

V a 0.81 cu. m.

2. Multiply oy 4 pcs = .081 x 4 a 0.324

Total volume F & P a 0.708 cu. m.

3. Referring to Table 1-2, using class "A” concrete:

Multiply:

25
 Cement .708 x 9.0 - 6.37 bags
Sand .708 x .50 * .354 cu. m.
Gravel .708 x 1.0 « .708 cu. m.

C ) Concrete Floor Slab

1. Determine the volume of the concrete floor slab.

V = . 10 x 3.00 x 4.00
V = 1.2 cu. m.

2. From Table 1-2, using class "C" mixture;

Multiply:

Cement : 1.2 x 6.0 * 7.2 bags

Sand : 1.2 x .50 - 0.6 cu. m.
Gravel : 1.2 x 1.0 = 1.2 cu. m.

Summary of the Materials

Cement : 6.37 + 7.2 * 13.57 say 14 bags

Sand : .354 + 0.6 = .95 cu. m.
Gravel : .708 + 1.2 * 1.9 cu. m.

Problem Exercise

Using the same problem of illustration 1-7 , solve for the quantity
of cement at 50 kg . per bag , including the sand and gravel using
the linear meter and the area method of estimating.

Note: The quantity of cement that will be found in this problem

exercise will not be equal to 14 bags as computed because the

26
problem calls for 50 kg. cement. However, the result could be
checked and compared with by converting the number of 50 kg.
cement found into kilograms divided by 40. The result must be
equal to 13.7 or 14 bags cement

MO RECTANGULAR COLUMN

The estimating procedure for the square or rectangular

column is practically the same . It could be either by the volume or
linear meter methods depending upon the choice and convenience
of the estimator .

ILLUSTRATION 1 -8

A series of eight rectangular concrete column with cross

sectional dimensions of 40 x 60 centimeters is supporting a beam.
The column has a clear height of 5.00 meters from the floor line to
the bottom line of the beam. Find the quantity of cement at 50 kg.
per bag, including the sand and gravel using class "A" concrete .

SOLUTION - I ( By the Linear Meter Method )

1. Determine the total length of the eight columns.

8.00 m. x 5 m. ht. = 40 meters

2. Referring to Table 1-4 along the 40 x 60 cm. column size

and using a 50 kg . [Link] class "A ” mixture:
Multiply:

27
 Cement : 40 x 1.680 = 67.2 say 68 bags
Sand : 40 x .120 - 4.8 say 5 cu. m.
Gravel : 40 x . 240 = 9.6 say 10 cu. m.

r 1
I
5.00 m.
A

40 cm
A
I
60 cm

Cross Section A- A
FIGURE 1-7

SOLUTION -2 (By Volume Method )

1. Find the volume of the eight columns:

V = 8 ( .40 x .60 ) x 5 m. ht .
V = 9.60 cu . m.

2. Refer to Table 1-2. Using class A mixture ; 50 kg. cement

Multiply :

Cement : 9.60 x 7.0 = 67.2 say 68 bags

Sand : 9.60 x .50 = 4.8 say 5 cu. m.
Gravel : 9.60 x 1 . 0 « 9.6 say 10 cu. m.

28
1 - 11 RECTANGULAR BEAM AND GIRDER

A Beam is defined as a strong horizontal piece of reinforced

concrete for spanning and supporting weights. On the otherhand a .
beam that is carrying or supporting another beam is called Girder.
Similarly,the simplest way of estimating the materials for these type
of structures is by the volume or the linear meter method.

ILLUSTRATION 1 -9

From Figure 1-8, list down the materials required using class
“A” concrete mixture.

Girder

£ £
8.00 m. s
OQ
itu
OQ

Girder

12.00 m..
erA
A

50 cm.

75 cm.
40 cm.

Girder
Section A A
FIGURE 1- 8

29
 SOLUTION (By the Volume Method )

1. Solve for the volume of the beam.

V = 4 pcs . x .25 x .40 x 8.00 m. long

V - 3.2 cu . m.

2. Find the volume of the girder .

V = 2 pcs. x .50 x .75 x 12.00 m.

V * 9 cu. m.

3. Total volume: add 1 & 2

3.2 + 9 » 12.2 cu. m.

4 . Refer to Table 1-2. Using 40 kg. cement class "A" concrete

Multiply :

Cement : 12.2 x 9 * 109.8 say 110 bags

Sand : 12.2 x .50 - 6.1 cu. m.
Gravel : 12.2 x [Link] 12.2 cu. m.

1-12 CIRCULAR COLUMN

Estimating the quantity of materials for circular column is

typically the same as the volume method for the square and
rectangular column using Table 1 -2. However , Table 1- 5 was also
prepared for the circular column to avail of the Linear Meter Method
of estimating.

30
 TABLE 1-5 QUANTITY OF CEMENT, SAND AND GRAVEL
PER METER LENGTH OF CIRCULAR COLUMN
Size Mixture Class Sand Gravel
cm. 40 kg. Cement 50 kg. Cement cu. m. cu. m.
A B A B
25 . 442 . 368 . 344 .295 .025 .059
30 . 636 .530 . 495 .424 .035 .071
35 . 866 .722 .673 .577 . 048 .096
40 1.131 .942 .880 .754 . 063 .126
45 1.431 1.193 1.113 .954 . 080 .159
50 1.767 1.473 1.374 1.178 . 098 .196
55 2.138 1.782 1.663 1.425 .119 .238
60 2.545 2.121 1.979 1.696 .141 .283
65 2.986 2.488 2.323 1.991 .166 .332
70 3.464 2.866 2.694 2.309 .192 .385
75 3.976 3.313 3.093 2.651 . 221 . 442
80 4.524 3.770 3.519 3.016 .251 .503
85 5.107 4.256 3.972 3.405 . 284 . 567
90 5.726 4.771 4.453 3.817 . 318 .636
100 7.069 5.890 5.498 4.712 . 393 .785

6.00 m . r
A

I Section A-A
—I * 4.50 m. -}— Circular Column

FIGURE 1-9

31
 ILLUSTRATION 1 -10

A circular column with cross sectional diameter of 60 cm. has

a clear height of 6.00 meters. Find the quantity of cement , sand
and gravel required using class "A" concrete if there are 5 columns
of the same size in a row.

SOLUTION- 1 (By Volume Method )

1. Solve for the cross sectional area of the circular column.

A - TTr2
A = 3.1416 x 302 ,

A = .283 sq. m.

2. Find the volume of the 5 columns

V = 5 pcs. x .283 x 6 m.
V = 8.49 cu . m.

3. Refer to Table 1- 2. Using class “A” concrete .

Multiply .

Cement : 8.49 x 9.0 = 76.4 say 77 bags

Sand : 8.49 x .50 = 4.25 say 5 cu. m.
Gravel : 8.49 x 1.0 = 8.49 say 9 cu. m.

SOLUTION -2 ( By the Linear Meter Method )

1 Solve for the total length of the 5 circular columns.

32
 L - pcs . x height
L = 5 pcs. x 6.00 m
= 30 meters

2. Refer to Table 1-6 along the 60 cm diameter column;

Multiply :

Cement : 30 x 2.545 = 76.3 say 77 bags

Sand . 30 x .141 = 4.2 say 5 cu. m.
Gravel : 30 x .283 » 8.49 say 9 cu. m.

-
1 13 CONCRETE PIPES

The quantity of materials for concrete pipes is determined

through the following processes:

1. Find the net volume of the concrete. That is, by subtracting

the volume occupied by the hole from the gross volume of the pipe .
2. Knowing the volume , refer to Table 1-2 to get the quantity of
cement, sand and gravel or:
3. Use Table 1-6.

Concrete Pipe

1.00 m.

33
 ILLUSTRATION Ml

A road construction requires 12 pcs. 90 cm diameter concrete

pipe for drainage purposes. Determine the quantity of cement , sand
and gravel required for the manufacture of said pipes using class
"A" concrete, (excluding reinforcement which will be discussed
later in chapter 3)

1.10 m. d * 90 cm . »

Concrete Pipe

SOLUTION -1 (By the Volume Method )

1. Solve for the gross volume of the concrete pipe .

V = . 7854 D2 h
V = .7854 x 1.102 x 1.00 m. ht.
V ” . 95 cu. m.

2. Solve for the volume of the hole .

V = . 7854 x d2h

34
 V = . 7854 x 902 x 100 m. ht.

V = .636 cu . m.

3. Subtract the result of step 2 from step 1 to get the net

volume of the concrete pipe.

Vn - .95 - .636 = .314 cu. m.

4 . Total volume of the 12 pipes:

Vt = 12 x .314 * 3.77 cu. m.

5 . Refer volume to Table 1-2. Using class "A" concrete;

Multiply:

Cement : 3.77 x 9.0 - 33.93 say 34 bags

Sand : 3.77 x .50 = 1 88 say 2 cu. m.
Gravel : 3.77 x 1.0 * 3.77 say 4 cu. m.

SOLUTION - 2 ( By Linear Meter Method )

1 . Referring to Table 1-6 , for a 90 cm. diameter pipe;

Multiply given data by the number of pipes.

Cement : 113.10 x 12 pipes = 1, 357.2 kg.

Convert to bags of cement. Divide by 40 kg.

1,357.20 „
33.93 say 34 bags
40 kg.

2. Referring back to Table 1-6 , multiply

35
 Sand : 12 pcs. x .157 = 1.88 say 2 cu. m.
Gravel : 12 pcs. x .314 - 3.77 say 4 cu. m.

TABLE 1-6 QUANTITY OF CEMENT, SAND AND GRAVEL PER

PIPE IN KILOGRAMS
Diameter Cement in kilograms Sand Gravel
cm Class of Mixture cu. m. cu. m.
D di A B

25 15 11.31 942 .016 . 0314

30 20 14.14 11.78 .020 .0400
40 25 27.57 22.97 .038 .0770
45 30 31.81 26.51 .044 .0884

55 40 40.29 33.58 . 056 .1120

60 45 44.53 37.11 .062 .1240
65 50 48.77 40,64 .068 . 1360
80 60 79.17 65.97 .110 .2200
85 65 84 82 70.69 . 118 .2360
90 70 90.48 75.40 .126 .2510
100 80 101.79 84.82 .141 .2830
110 90 113.10 94.25 .157 .3140
120 100 124.41 103.67 .173 . 3460

145 120 187 32 156.10 . 260 .5200

175 150 229.73 191,44 .319 . 6380

The values given in Table 1-6 for cement is in kilograms not in

bags. The purpose of having it in kilogram is to make it easier to
understand and comprehend say 7 kilos than 0.175 bags. Likewise
for a problem involving only one or two small pipes, cement in
kilogram is easier to calculate for mixing.

36
 CHAPTER 2
MASONRY
2- 1 CONCRETE HOLLOW BLOCKS

Concrete Hollow Blocks are classified as bearing and

non-bearing blocks. Load bearing blocks are those whose
thickness ranges from 15 cm. to 20 cm. and are used to carry toad
aside from its own weight. Non-bearing blocks on the other hand,
are blocks which are intended for walls, partitions , fences or
dividers carrying its own weight whose thickness ranges from 7.5
cm. to 10 cm.

-
FIGURE 2 1

Concrete Hollow Blocks has three whole cells and two one half
cells at both ends having a total of four. These cells vary in sizes
as there are different manufacturers using different molds The

37
varying sizes of the cells will affect the estimated quantity of
materials. For this reason, it is recommended that the bigger cell
be adopted in the computations

Estimating the materials required for concrete hollow block

works, comprises of the following major items:

1. Quantity of the blocks.

2. Quantity of cement and sand mortar for block laying.
3. Cement , sand and gravel filler of the hollow core or cell.
4. Cement and fine sand for plastering.
5. Cement, sand and gravel for CHB footing and Pests.
6. Reinforcing steel bars.

ILLUSTRATION 2- 1

From Figure 2-2, determine the number of 10 x 20 x 40 cm.

concrete hollow blocks and the materials required for:

a) Mortar for block laying

b) Mortar filler for the hollow core or cells .
c) Plastering
d) Concrete for CHB and Post footings
e) Reinforcement

SOLUTION - 1 (By Fundamental Unit Method )

A. Concrete Hollow Blocks

1 . Divide the height of the fence by the height of one block

38
 3.00 m. * 15 pcs.
.20 -
= 3S<> /»l 2

2. Divide the total length of the fence by the length of one block

20.00 m = 50 pcs.
.40

3. Multiply the result of 1 and 2

15 x 50 * 750 pcs

3.00 m.

3.00 m.

1
Mortar ^

15 cm.
t
40 cm.

FIGURE 2- 2

39
TABLE 2-1 QUANTITY OF CEMENT AND SAND FOR MORTAR
AND PLASTER MIXTURE PER CUBIC METER
Cement in Baps Sand
Cta»» Mixture 40 kg. SO kg. cu. m.
A 1:2 18.0 14.5 1.0
B 1: 3 12.0 9.5 1.0
C 1:4 9.0 7.0 1.0
D 1:5 7.5 6.0 1.0

B. Mortar for Block Laying ( .0125 average thickness)

1. Find the volume of the mortar (one layer)

V M x w x I
V => .0125 x .10 x 20 m
V = .025 cu. m.
2. Multiply the number of layers to get the total volume of the
mortar:

Total V = .025 x 15 layers

V = .375 cu. m.

3. Refer to Table 2-1 using class B mixture 40 kg. cement

Multiply:

Cement : .375 x 12 = 4.50 bags

Sand .375 x 1.0 = 0.375 cu. m.

C. Mortar Filler for Hollow Cell

1. Find the volume of one cell

40
 V * .05 x .075 x .20
V = .00075 cu.m.
5 cm.

CHB Cell
20 cm.
-4
7.5 cm.
FIGURE 2 3 -
2. Volume of 4 cells per block

V * .00075 x 4
V » .003 cu. m.

3. Total volume of cells for 750 CHB

V =* .003 x 750
V = 2.25 cu. m.

4 . Refer to Table 2-1 using class B mortar - 40 kg. cement

Multiply:

Cement : 2.25 x 12 27 bags

Sand 2.25 x 1.0 - 2.25 cu. m.

D. Plastering (at average of 16 mm. thick )

1. Find the area of the fence (one side)

41
 3.00 x 20 — 60 sq. m.

2. Find the area of the two sides

60 x 2 = 120 sq. m.

3. Solve for the volume

V - 120 x .016
V - 1.92 cu. m.

4. Refer to Table 2-1 using class B mixture-40 kg. cement

Multiply:

Cement : 1.92 x 12 * 23.04 bags

Sand 1.92 x 1.0 = 1.92 cu. m.

E. Footing
1. Find the volume of the footing

V• t x w x L
V * .15 x .40 x 20.00
V « 1.20 cu. m.

2. Refer to Table 1-2 using class "B" concrete 40 kg . cement

Multiply:

Cement : 1.20 x 7.50 « 9.00 bags

Sand : 1.20 x .50 = .60 cu. m.
Gravel : 1.20 x 1.00 * 1.20 cu. m.

42
 Summary of the Materials

1. CHB 750 pcs.

2. 40 kg. Cement 63.54 say 64 bags
3. Sand 5.14 cu.m.
4 . Gravel 1.20 cu.m.

Note: The steel reinforcement will be discussed separately in

Sec. 3-4, Chapter 3.

TABLE 2-2 QUANTITY OF CEMENT AND SAND FOR CHB

MORTAR PER SQUARE METER WALL
Bags Cement
40 Kg. 50 Kg.
Siz ofCHB Number
; Mixture Mixture
Sand
cm. Per . m .
sq B C D B C D cu.m.
10 x 20 x 40 12.5 0.525 0.394 0.328 0.416 0.306 0.263 . 0438
15 x 20 x 40 12.5 1.013 0.759 0.633 0.802 0.591 0.506 .0844
20 x 20 x 40 12.5 1.500 1.125 0.938 1.138 0.875 0.750 .1250

The problem of illustration 2-1 can be be solved by the Area

Methods with the aid of Table 2-2 and 2-3 , thus, consider ;

SOLUTION - 2

A. Concrete Hollow Blocks

1 . Find the area of the fence

3.00 x 20.00 m. = 60 sq. m.

43
 2. Refer to Table 2-2 , multiply:

60. x 12.5 = 750 pcs. CHB

B. Mortar For Block Laying and filler of the cell

1. Referring to Table 2-2 using class "B" mixture 40 kg. cement

Multiply:

Cement : 60 x .525 * 31.5 bags

Sand : 60 x .0438 * 2.63 cu. m.

C. Plaster Mortar

1. Find the area to be plastered

60 x 2 = 120 sq. m. two faces

2. Referring to Table 2-4 using class "B" mixture 40 kg. cement

Multiply:

Cement : 120 x .192 s 23.04 bags

Sand : 120 x .016 * 1.92 cu. m.

D. Footing

1 . Determine the total length of the footing: = 20 m.

2. Referring to Table 2-3 using class "B" concrete:

For a 15 x 40 cm. Footing,
Multiply:

44
 Cement: 20 m. x .450 « 9.0 bags
Sand : 20 m. x .030 * .60 cu. m.
Gravel : 20 m. x .060 » 1.20 cu. m.

Summary of the Materials

1 . Concrete Hollow Blocks . . . 750 pcs.
2. 40 kg. Cement 63.5 say 64 bags
3. Sand 5.15 cu. m.
4. Gravel 1.20 cu. m.
-
TABLE 2 3 QUANTITY OF CEMENT, SAND AND GRAVEL FOR
CHB FOOTING PER LINEAR METER
Dimension Cement in Bags
cm. Class Mixture Aggegate
40 kg. Cement 50 . Cement
kg Sand Gravel
t w A B A B cu. m. cu. m.
10 30 .270 .225 .210 . 180 . 015 . 030
10 35 .315 .263 .245 . 210 . 018 . 035
10 40 . 360 .300 .280 . 240 . 020 . 040
10 50 .450 .380 . 350 . 300 . 025 . 050
15 40 . 540 .450 .420 . 360 . 030 .060
15 45 .608 .506 .473 .405 .034 . 068
15 50 .675 .563 .525 .450 .038 .075
15 60 . 810 .675 .630 .540 .045 .090
20 40 .720 .600 .560 . 480 . 040 .080
20 50 . 900 . 750 . 700 .600 . 050 . 100
20 60 1.080 .900 . 840 .720 . 060 . 120
 -
TABLE 2 4 QUANTITY OF CEMENT AND SAND FOR PLASTERING
PER SQUARE METER
Class Cement in Bags Sand
Mixture 40 kg. Cement 50 kg . Cement cu. m
A .288 .232 . 016
B .192 . 152 016
C .144 .112 .016
D . 120 . 096 .016
* Thickness computed at an average of 16 mm.

ILLUSTRATION 2- 2

From Figure 2-5 find the quantity of 15 x 20 x 40 cm concrete

hollow blocks , cement (40 kg. per bag ), sand and gravel required
using class B mixture by the area method .

FIGURE 2-5

SOLUTION:

A.) Concrete Hollow Blocks

46
 Solve for the net wall area :
^ .

A = Perimeter x Height
A * (20 + 20 + 15 + 10 ) x 2.00 m. ht.
A = 65.00 x 2.00 m.
A = 130 sq. m.

2. Determine the number of hollow blocks. Refer to Table 2-2

using class B mixture:
Multiply:

130 sq. m. x 12.5 pcs.

- 1 ,625 pcs.

B. Mortar for Block Laying

1. Using class B mortar , solve for cement (at 40 kg.) and sand.
R efer to Table 2-2 under size 15 x 20 x 40 CHB;
Multiply :

Cement : 130 x 1.013 = 131.7 say 132 bags

Sand : 130 x .0844 = 10.9 say 11 cu. m.

C. Plastering

1. Area of one face wall = 130 sq. m.

Area of Two faces x 2 - 260 sq. m.

2. Referring to Table 2-4 , using class B mortar;

Multiply:

Cement : 260 x .192 * 49.9 say 50 bags

Sand : 260 x .016 = 4.16 cu. m.

47
 D. CHB Footing

1. The wall perimeter is 65.00 m. Referring »•/ Table 2-3, using

15 x 40 cm. footing class B mixture;
Multiply:
Cement : 65 x .450 = 29.25 say >0 bags
Sand : 65 x .030 = 1.95 say 2 cu. m
Gravel : 65 x .050 = 3.90 say 4 cu. m.
SUMMARY
-
15 x 20 x 40 cm. Concret Hollow Block*
Cement at 40 kg.
—- 1,625 pcs.
212 begs
Sand 18 on. m.
Grave! 4 cu. m.

ILLUSTRATION 2-3

-
From Figure 2 6 prepare the bill of materials using class B
mixture.

20

s
2500
= ==- CHB

200
2500
Ground Line
500
40
1«
papi
"

* 10.00 10.00

FIGURE 2-6

48
SOLUTION (By the Area Method)

A. Concrete Hollow Blocks

1. Find the total length or perimeter of the fence

p ~ 95 m.

2. Subtract the space occupied by the posts

95 - (20 posts x .20)

= 95 - 4
- 91 m. net length

3. Solve for the net area of the fence

A. = 2.40 x 91 m.
A * 218.4 sq. m.

4 . Referring to Table 2-2, determine the number of 10 x 20 x

40 cm. CHB.
Multiply:

CHB : 218.4 x 12.5 - 2,730 pcs.

B. Cement mortar for Block Laying and Cell Filler

1. Referring to Table 2-2 using class "B" mixture 40 kg. cement

Multiply:

Cement : 218.4 x .525 = 114.66 bags

Sand : 218.4 x .0438 = 9.57 cu. m.

49
 .
C Plastering of the fence to ground line only (2.00 m.)

1. Solve for the area to be plastered (two sides)

A. = Ht. x Perimeter
A. = 2.00 x 91 m.
A. -182 sq. m. one face
Two faces: 182 x 2 = 364 sq. m.

2. Referring to Table 2-4, using class "B" mixture, soVe for

the quantity of cement and sand.
Multiply:

.
Cement : 364 x 192 = 69.88 say 70 bags
Sand : 364 x .016 = 5.82 say 6 cu. m.

D. Footing of Posts -( .60 x .60 square )

1. Solve for the volume of the footing

V= t x w L
*
V = .15 x .60 x .60 x 20 posts
V = 1.08 cu. m

2. Referring to Table 1-2 class " B" mixture

Multiply:

Cement : 1.08 x 7.50 = 8.1 bags

Sand : 1.08 x 50 = 0.54 cu. m.
Gravel : 1.08 x 1.00 = 1.08 cu. m.

50
 E. CHB Footing

1. Total length of the fence less the space occupied by the

post footings { .60 x .60 m.}

95 m. - ( .60 x 19 posts )
= 95 - 11.40
= 83 . 60 m.

Note: The number of post is only 19 instead of 20 pieces

because the two posts at the gate entrance occupies one half of
the hollow block footing, ( see figure).

2. Referring to Table 2-3 using 40 kg. cement class "8" mixture

10 x 40 cm. Footing.
Multiply:

Cement 83.6 x .300 = 25 bags

Sand 83.6 x .020 * 1.67 cu. m.
Gravel 83.6 x . 040 = 3.34 cu. m.

F. Concrete Post

1. Solve for the volume

.20 x .20 x 2.40 x 20 posts
* 1.92 cu. m.

2. From Table 1-2 using class "BM mixture 40 kg. cement.

Multiply :

Cement : 1.92 x 7.5 = 14.4 bags

Sand : 1.92 x 50 * .96 cu. m.
Gravel : 1.92 x 1.0 = 1.92 cu. m.

51
 .
G Plastering of the Post (if necessary )

1. Solve for the surface area of the post less the area occupied
by the CHB {.20 x 2.00 } see Figure 2 6 -
.60 x 2.00 m. ht. x 20 posts
= 24 sq. m.

2. Referring to Table 2-4 using class "B" mixture 40 kg. cement

Multiply:
Cement : 24 x .192 * 4.60 bags
Sand : 24 x .016 = 0.384 cu. m.
Summary

Concrete Hollow Blocks 2,730 pieces

Cement . . . . 237 bags
Sand . . . . 19.10 cu. m.
Gravel 6.34 cu. m.

i
35 70
100

1
^
4
300
120
v.
I I
1 120
270
70 B A 210 I
Natural Ground
60 60
Footing line
35 Q f 600

FIGURE 2 7 -
52
 ILLUSTRATION 2-4

From Figure 2-7 prepare the bill of materials using class B

concrete and class C mortar .

SOLUTION

A. Concrete Hollow Blocks

1. Find the area of wall "A"

A - 6.00 x ( 2.70 + . 50 + .60 ) * 22.80 sq. m.

2. Find the area of wall "B"

B * 3.50 x (3.00 + .35 + .60 ) = 13,825 sq. m.

Total area of A & B 36.625 sq. m.

3. Less Area of the windows

W -1 * 2.10 x 1.20 * 2.52 sq . m.

W -2 * .70 x 1.20 * .84 sq. m.

Total Area of W-1 and W-2 * 3.360 sq. m.

Net wall area 33.265 sq. m.

4 . Referring to Table 2-2, multiply:

CHB: 33.265 x 12.5 * 415.8 say 416 pcs.

B. Cement Mortar

53
 -
Referring to Table 2 2 using class "C” mixture 40 kg. cement
Multiply:

Cement : 33.265 x .759 = 25.24 bags

Sand : 33.265 x .084 - 2.8 cu. m.

C. Cement Plaster

1. Referring to Table 2-4 using class C Mixture:

Multiply:

Cement : 33.265 x .144 - 4.79 bags

Sand : 33.265 x .016 = 0.53 cu. m.
D. Footings

1. Total length of the wall = 9.50 m.

2. Referring to Table 2-3 for a {15 x 40} footing using class
"B" concrete
Multiply:

Cement 9.50 x .450 * 4.28 bags

Sand 9.50 x .030 = 0.28 cu. m.
Gravel 9.50 x .060 = 0.57 cu. m.

Other factors that might affect the estimated quantity

of materials

1. Improper measure of aggregates during the block laying

work. The most common attitude of the worker is to mix
sand and gravel with cement disregarding the measuring
box .

54
 2. Sometimes the mason prepares a box for measuring sand
or gravel not in accordance with the specified
measurement.

3. Addition of cement to over exposed mixed mortar not used

or applied on time.

4. The excess mortar for installation of hollow blocks are

usually dumped in a certain corner of the construction site.
This is a common practice especially after working hours
where no overtime pay is authorized.

These are considered as minor things in the construction work

which are simply overlooked, but summing them up for a months
work will surprisingly result to a figure beyond expectation affecting
the estimate.

Table 2-5 and Table 2-6 are presented to simplify further the
estimating methods in determining the quantity of materials from
block laying to the plastering work. The special feature of these
tables is that the materials like cement and sand are given per
hundred pieces not by the area method as previously presented.

-
TABLE 2 5 QUANTITY OF CEMENT AND SAND PER 100 CHB
MORTAR
Cement in Bags
Mixture
CHB Size 40 kg. Cement 50 kg. Cement Sand
cm. 8 C 0 B C D cu. m.
10 x 20 x 40 4.200 3.152 2.624 3.328 2.448 2.104 0.350
15 x 20 x 40 8.104 6.072 5.064 6.416 4.728 4.048 0.676
20 x 20 x 40 12.000 9.000 7.504 9.504 7.000 6.000 1.000

55
 TABLE 2-6 QUANTITY OF CEMENT AND SAND PER 100 CHB
PLASTER *
Cement in Bags
No. of Face Mixture
plastered 40 kg. Cement 50 kg. Cement Sand
A B C A B C cu. m.

One face 2.304 1.536 1.152 1.856 1.216 0.896 0.128

Two face 4.608 3.072 2.304 3.712 2.432 1.792 0.256

* Plaster thickness at an average of 16 mm.

ILLUSTRATION 2-5

From figure 2-8, determine the quantity of 15 cm. (6")

concrete hollow blocks, cement and sand required using class C
mixture .

60.00 m.

,
I
l 1 \
zrzi\
i CHB Wall Cement Plaster E
8
I i / l
T
K Ground Line
J»
V Footing line

FIGURE 2-8

56
SOLUTION

A. Concrete Hollow Blocks

1. Determine the area of the fence

2.00 x 60 m. - 120 sq . m.
2. Referring to Table 2-2.
Multiply:

CHB: 120 x 12.5 * 1 ,500 pcs.

.
B Cement Mortar

1. Divide:
1 ,500
"
ioo ’15
"

2. From Table 2-5, using a 40 kg. cement:

Multiply :

Cement. 15 x 6.072 = 91 bags

Sand : 15 x 0.675 = 10.125 cu . m.

.
C Cement Plaster (One Face)

1. From Table 2-6 , using a 40 kg . cement classs "C" mortar;

Multiply :

Cement : 15 x 1.152 = 17.30 say 18 bags

Sand : 15 x 0.128 = 1.92 say 2 cu . m .

57
2-2 SPECIAL TYPES OF CONCRETE HOLLOW
BLOCKS
The common and ordinary type of concrete hollow blocks are
those with three hollow cells as explained in Section 2-1 . However ,
There are concrete hollow blocks which are especially designed for
architectural and structural purposes , one of which is the concrete
blocks with two ceils.
The purpose of manufacturing the concrete hollow blocks with
two celts is to fill the hollow core with concrete not just by a mortar
but concrete which is a mixture of mortar and gravel. This is how
the special type of CHB differs from that of the ordinary CHB.

Estimating procedure for these types of hollow blocks is similar

with that of the ordinary blocks using the constant value of 12 5 pcs
per square meter. For the half block, the constant value is 25 pieces
per square meter although it is determined through direct counting.
The thickness of the blocks has no participation whatsoever in
estimating the number of pieces required because what is being
considered is the exposed side of the blocks. The thickness
however, has a direct effect on the quantity of mortar and concrete
filler for the hollow core or cells. Table 2-7 was prepared for a more
simplified methods of estimating.

Example of special types of concrete hollow blocks are:

1. 2- core Stretcher Blocks with thickness that varies from 15

cm. to 20 cm. thickness.
2. 2-core L-Corner Block
3. 2-core Single End Block
4 . Half Block
5. Beam Block

58
 X 2 Core s XI v
14 ‘
Stretcher Block
' 19
^ 2 Core
L-Comer Block

Single End 8lock

14 ,
* Half Block
39
1 19

J>
Beam Block

Half Block 1
'9

39
19

>
Beam Block Single End Block

Special Types of Hollow Blocks

FUGURE 2-9

59
 TABLE 2-7 QUANTITY OF CEMENT, SAND AND GRAVEL PER
HUNDRED SPECIAL TYPES OF CONCRETE HOLLOW BLOCKS
T
CHB Size 40 kg. Cement Mortar
in Cm. Class of Mixture Sand Gravel
B C D

Stretcher Block
2-core - 20 cm. 9.19 7.20 4.32 .67 .83
2-core - 15 cm. 6.23 4.88 4.06 . 30 25

L - Corner Block
2-core - 20 cm. 8.67 6.87 5.78 .60 . 93
2- core - 15 cm. 5.90 4.67 3.89 .41 .65

Single End Block

2-core - 20 cm. 8.85 6.98 5.82 .63 .90
2-core - 15 cm. 6.12 4.82 4.02 .44 .60
2-core - 10 cm. 4.2 3.15 2.63 .35

Half Block
20 x 20 cm , 3.98 3.15 2.63 .28 .45
15 x 15 cm. 2.70 2.14 1.78 .16 . 30

Beam Block A B C Sand Gravel

2-core - 20 cm. 8.78 7.32 5.85 . 49 . 98

2-core - 15 cm. 5.85 4.88 3.90 . 33 .65

Mortar: - A mixture of cement and sand used for block laying.

Concrete: - A mixture of cement , sand and gravel use to fill
hollow core or cells.
Values given - combination of mortar for block laying and
concrete for hollow core filler.

60
 ILLUSTRATION 2-6

From Figure 2-10 determine the number of 40 kg . cement, sand

and gravel required including the following type of blocks:
a) 2-core 15 cm. Stretcher blocks
b) 2-core 15 cm. Single end block
c) 15 cm. Half block
d) 15 cm. Beam block

* 0.00 m
4.80 m.
4.80 m.

FIGURE 2-10

SOLUTION:

A ) Find the Number of Blocks

1. Determine the total wall area:

A » 4.80 m. ht. x wall perimeter

61
 A * 4.80 x 32 m.
A * 153.6 sq. m.

2. Solve for the total total number of blocks. Refer to Table 2-2
Multiply:

153.6 x 12.5 pcs. per sq. m.

- 1 ,920 pcs.

This 1, 920 pcs. comprises at! the types of blocks from stretcher
to half blocks except the beam blocks. The next step is to find the
number of single end block and half block to be subtracted from
1,920 pcs .

3. Solve for the number of single end blocks. Add the total
height of the 4 comers.
Total length 4.80 x 4 = 19.2 meters

4. Divide by the height of one block which is 20 cm.

19.20 * 96 pcs. single end block.

.20

5. Solve for the number of half block:

4 corners at 4.80 m. = 19.20 m. divide by the height
of one block (.20m.) .20
* 96 pcs.

2 end Wall at entrance - $.60 divide by the height

of one block ( .20 ) .20
= 48 pcs. = 24 pcs.
Divide by 2 2

62
 48 pcs was divided by 2 because the half block on this end wail
is alternate with the whole block , ( see figure)

6. Get the total number of half block.

96 pcs + 24 pcs. = 120 pcs. half blocks. Convert this

into whole block size - 60 pcs.

7. Add the Single End Blocks and the converted half blocks.

96 + 60 = 156 this value now will be subtracted to

the 1920 pcs.

1,920 - 156 = 1764 pcs. is now the total number of

2- core stretcher blocks.

Note: In step no. 2 we solve for the entire quantity of hollow

blocks in the wall and we get 1,920 pcs. If we will subtract the
number of half block , it should be converted to whole size block to
have a common unit of measure. Thus, we divide 120 pcs. by 2
and we get 60 pcs.

8. Solve for the number of Beam Blocks:

Perimeter = Number of blocks

.40 m. length of one block

36.0 m - 90 pcs. beam block

.40

9. The final list of blocks will be:

1745 pcs. 15 cm. 2- core stretcher blocks

63
 96 pcs. 15 cm. Single end blocks
120 pcs 15 cm. Half blocks
90 pcs. 15 cm. x 20 x 40 cm. beam blocks

B) Solve for Cement, Sand and Gravel

Referring to Table 2-7, convert the quantity to 100 pcs. and

using class B mixture:
Multiply:

1745 * 17.5 x 6.23 = 108.7 bags for 2-core stretcher blocks

100

96 * .96 x 6.12 = 5.9 bags- Single end block

100

120 = 1.20 x 2.70 = 3.3 bags - for Half blocks

100

90 = .90 x 5.85 = 5.3 bags- for Beam blocks

100
Total . . . 123.2 say 124 bags cement

Sand Gravel

17.50 x .30 * 5.13 cu m. 17.30 x .25 = 4.30 cu. m.

.96 x .44 = .43 " M
.96 x .60 = .58 " "
1.20 x .16 = .19 " "
.90 x .33 = .30 " "
1.20
.90
x
x
.30 = .36 "
.65 * .59 " ’•
-
Total . . . . 6.05 cu m. 5.83 cu. m.

64
2-3 DECORATIVE CONCRETE BLOCKS

Decorative blocks are made out from either cement mortar or

clay . These type of construction materials had been widely used
for ventilation and decorative purposes .

TABLE 2-8 QUANTITY OF DECORATIVE BLOCKS, CEMENT AND

SAND FOR BLOCK LAYING MORTAR
Size Number 40 kg. Cement Mortar Sand
cm. per sq. m. Class Mixture per
A B 100 Blk.
5 x 10 200 . 180 . 120 .010
5 x 15 133 . 270 . 180 . 015
5 x 20 100 . 360 . 240 . 020
5 x 25 80 .450 . 300 . 025
10 x 20 50 . 720 . 480 . 040
10 x 25 40 . 900 .600 . 050
10 x 30 33 1.080 .720 . 060

TA BLE 2-9 QUANTITY OF CEMENT AND SAND FOR VARIOUS

TYPES OF BRICKS PER HUNDRED BLOCKS
Size Number 40 kg. Cement per 100 Blocks Sand
cm. per sq. m. Mixture Class per
t h xI A B 100 blks.
*
6 x 12 x 19 38.5 . 346 . 230 . 0192
10 x 14 x 19 33.3 . 612 . 408 .0340
10 x 14 x 23 27.8 684 . 456 . 0380
10 x 24 x 24 160 . 882 . 588 . 0490
10 x 14 x 39 16.7 . 972 . 648 . 0540
10 x 19 x 39 12.5 1.062 . 708 .0590

65
ITALIAN

55 x 215 x 125 mm

BOLIVIAN
100 x 160 x 180 mm

100 x 140 X 240 mm

MSpf
5or
dbti
CJ££J [ cy?& ii#?43
DO o OOOPO^
Cfetll
100 x 250 x 250 mm
CORINTHIAN

ROMAN 100 . x 230 x 250 mm

FIGURE 2- 11

66
 m mm
JOSEPHINE
9 @s100 x 250 x 250 mm

SKMS8
ASG
.
100 x 250 x 250 mm

057ZI] ayjD a a n?z3

^ .
cr /ra cr^'a I croT3
cw^a I OkVXJ
CI/MZJ ama cra -
o| POO
OOTZ3
CJ&YZ1
^- .
aw a vjsIn o,^
.
ci o' jj 0.9 13 IQ9£)|
1176 a3
.
crmra |CTAO jc7A o -
AUM 100 x 250 x 250 mm

EGYPTIAN 100 x 250 x 250 mm

S
a
3 83 S3
(Jo oLJo QUO

°O“l“0S
“?oQ2 EkoliFfcfi
PERSIAN 100 x 250 x 250 mm
FIGURE 2-12

67
 -
2 4 ADOBE STONE

Adobe Stone is commonly used for fencing materials as

substitute to concrete hollow blocks for economic reasons. Lately
however, the used of adobe stone is no longer limited to the
ordinary zocalo and fencing work but also extensively used as
finishing and decorative materials for the exterior and interior of
buildings.

Adobe Stone

T
T IS
J.. :; - re.*.- H

FIGURE 2 13 -
The use of adobe stone for buttresses, cross footings , fences
and stairs minimizes the use of mortar filler . Plastering is
sometimes disregarded specially when the design calls for
exposure of the natural texture of the stones.

68
 TABLE 2-10 QUANTITY OF CEMENT AND SAND FOR ADOBE
MORTAR PER SQUARE METER *
No. Size 40 kg. Cement 50 kg. Cement
/ sq. cm. Class Mixture Class Mixture Sand
m. B C D B C D cu. m

12 1 5 x 1 5 x 4 5 . 2808 . 2106 . 1756 2220 . 1638 . 1404

. .0234
10 1 5 x 2 0 x 4 5 .2520 .1890 . 1575 . 1995 . 1470 . 1260 . 0210
11 1 5 x 3 0 x 3 0 .2280 . 1708 . 1423 . 1803 .1328 .1139 .0190
8 1 5 x 3 0 x 4 0 .2079 . 1559 . 1299T . 1646 . 1213 . 1040 . 0173

6.5 1 5 x 3 0 x 4 5 . 1901 . 1426 . 1188 . 1505 . 1109 . 0951 . 0158

* Average thickness = 20 mm

TABLE Ml QUANTITY OF ADOBE STONE, CEMENT AND SAND

FOR BUTTRESSES AND FOOTINGS

Buttress and Footing Cement Mortar per Stone

Buttress No . Number Mixture Class
cross of of stone 40 kg. Cement SO kg. Cement Sand
Section Course per m, ht. B C D B C D cu. m .

30 x 45 2 12 .027 .0203 .017 .0214 .016 .014 .0023

45 x 45 3 18 .029 .0220 .018 .0230 .017 .015 .0025
45 x 60 4 24 027 .0203 .017
. .0214 .016 .014 .0023
45 x 75 5 30 .032 .0240 .020 .0250 .018 .016 .0026
45 x 95 6 36 .034 .0253 . 021 .0270 .020 .017 .0028

ILLUSTRATION 2- 7

From Figure 2-14 compute for the quantity of adobe stone ,

cement and sand using class C mixture.

69
 500 .. 500 500

_
15
JTF 15

30
15 15
30
15

Buttress
rf
~
>l
r
15

150
Buttress
.. — Ground Line
50
Footing

45 4”
FIGURE 2- 14

SOLUTION

A. Adobe Stone Fence

1. Determine the net length of the fence minus the space

occupied by the buttresses

15.00 m. - { . 30 x 3 post } - 14.1 m.

2. Solve for the net area

14.10 m. x 2.00 m. ht. = 28.20 sq. m.

70
3. Referring to Table 2-10, using a 1 5 x 1 5 x 4 5 cm. stone;
Multiply:

28.20 x 12 - 338.4 ssay 339 pcs.

4. Total height of the four posts

2.00 x 4 post « 8.00 m.

5. From Table 2-11 using 30 x 45 cm. buttresses;

Multiply:

8.00 m. x 12 pcs. per meter ht. ~ 96 pcs.

plus the adobe stone footing per buttress or post.

96 + ( 4 pcs. x 4 posts ) “ 112 pcs.

6. Determine the length of the fence less the area occupied

by the buttress footing.

15.00 m. - (.45 x 3 ) = 13.65 m.

7. Multiply by 6 stone per meter length

13.65 X 6 = 82 pcs.

B. Cement Mortar

1. Wall Fence Area = 28.20 sq, m.

Referring to Table 2- 10, using class C mixture 40 kg.
cement; Multiply :

Cement : 28.20 x .2106 = 5.9 say 6 bags.

Sand : 28.20 x .0234 « .66 say . 7 cu. m.

71
 2. Buttress (post) and Footing - 112 pcs.
Referring to Table 2- 11 using class ”CM mixture 40 kg.
cement; Multiply:

Cement : 112 x . 0203 * 2.3 bags

Sand : 112 x .0023 = .26 cu. m.

3. Fence footing = 82 pcs.

Referring to Table 2- 11; Multiply :

Cement : 82 x .0203 * 1.7 bags

Sand : 82 x .0023 = . 18 cu m.

.
C Cement Plaster: ( One Face )

1 . Determine the total surface area of the wall plus the surface
area of the buttress to be plastered .

(15.00 m. + .45) + {.15 x 8}

= 16.65 m.

2. Solve for the area

Length x height

* 16.65 m. x 1.50 m.
= 25 sq. m.

From Table 2-12 using class "C” mixture; Multiply :

Cement : 25 x .225 = 5.6 bags

Sand : 25 x .025 * .63 cu. m.

72
 3. Multiply the result by two if two sides are to be plastered.

Cement : 5.60 x 2 = 11.2 bags

Sand : 0.63 x 2 = 1.3 cu. m.

Summary

Adobe stone. 629 pcs.

Cement 21.2 say 22 bags
Sand 2.4 cu. m.

TABLE 2-12 QUANTITY OF CEMENT AND SAND FOR

PLASTERING PER SQUARE METER

Bag Cement and Mixture C!ass

Side 40 kg. Cement 50 kg. Cement Sand
B C D B C D cu. m.

One face . 300 . 225 . 188 .238 . 175 . 150 . 025

Two Faces .600 . 450 . 375 . 476 . 350 . 300 . 025

Problem Exercise

From Figure 2-15, Determine the quantity of the following

materials:
1.15 x 20 x 45 adobe stone
2 . Buttresses and Footings

73
3. Cement for:
a. Mortar and
b. Plastering
4 . Sand

FIGURE 2-15

74
 CHAPTER 3
METAL REINFORCEMENT
-
3 1 STEEL BARS
Steel is the most widely used reinforcing material for almost
all types of concrete construction . It is an excellent partner of
concrete in resisting both tension and compression stresses.
Comparatively , steel is ten times stronger than concrete in resisting
compression load and hundred times stronger in tensile stresses.
The design of concrete assumes that concrete and steel
reinforcement acts together in resisting load and likewise to be in
the state of simultaneous deformation , otherwise, the steel bars
might slip from the concrete in the absence of sufficient bond due
to excessive load .
In order to provide a higher degree of sufficient bond between
the two materials, steel reinforcing bars with a surface deformation
in various designs are introduced.

'
:Vl
wl ‘

i i-

Types of Deformed Bars

FIGURE 3-1

75
 TABLE 3-1 STANDARD WEIGHT OF PLAIN OR DEFORMED
ROUND STEEL BARS IN KILOGRAMS

Dtam. 5.0 m 6.0 m 7.5 m 9.0 m 10.5 m 12.0 m 13.5 m

8 mm 1.96 2.37 2.96 3.56 4.15 4.74 5.33

10 mm 3.08 3.70 4.62 5.54 6.47 7.39 8.32
12 mm 4.44 5.33 6.66 7.99 9.32 10.66 11.99
13 mm 5.21 6.25 7.83 9.38 10.94 12.50 14.07
16 mm 7.90 9.47 11.84 14.21 16.58 18.95 21.32
20 mm 12.33 14.80 18.50 22.19 25.89 29.59 33.29
25 mm 19.27 23.12 28.90 34.68 40.46 46.24 52.02
28 mm 24.17 29.00 36.25 43.50 50.75 58.00 65.25
30 mm 27.75 33.29 41.62 49.94 58.26 66.59 74.91
32 mm 31.57 37.88 47.35 56.62 66.29 75.76 8523
36 mm 39.96 47.95 59.93 71.92 83.91 95.89 107.88

-
TABLE 3 2 DEFORMATION REQUIREMENTS

T
Nominal Max. Average Height Tolerance Max Value
Diameter Spacing of luge Minimum Maximum Summation
of lug& gap

8 7.0 0.3 0.6 5.5

10 7.0 0.4 0.8 7.8
12 8.4 05 1.0 9.4
13 9.1 0.6 1.2 10.2
16 11.2 0.7 1.4 12.6
20 14 0 1.0 2.0 15.7
25 17 5 1.2 2.4 19.6
28 196 14 28 22.0
30 21 0 15 30 236
32 22.4 ! 1.6 32 25.1
i
36 24.5 18 36 27.5

76
 -
TABLE 3 2A MECHANICAL PROPERTIES

Yield Tensile Elongation Bending Diameter

Strength Strength Specimen in 200 mm Angle of Pin
Class Grade MPa MPa mm. percent Degree d^nominal
mm mm min. dia. at
specimen

Hot Rolled 230 230 390 D < 25 18 180 3d

Non Wel- D > 25 16 4d
dable def- 275 275 480 D < 25 10 180 4d
Steel Bar D > 25 8 5d
415 415 020 D < 25 8 180 5d
D > 25 7 6d
Hot Rolled 230 230 390 D < 25 20 180 3d
Weldable D > 25 18 4d
Def. or 275 275 480 D < 25 16 180 4d
Plain bar D < 25 14 5d
415 415 550 D < 25 14 180 5d
D > 25 12 6d

3-2 IDENTIFICATION OF STEEL BARS

Steel reinforcing bars are provided with distinctive markings

which identify the name of the manufacturer with its initial and the
bar size number including the type of steel bars such as:

N « For Billet
A - For Axle
Rail Sign = For Rail Steel

77
 — Main Ribs

tnitial of
Manufacturer
- -M
— Bar Size -6
N Steel Type A
Grade Mark

Grade 40
Grade 50 0 One Line - Grade SO
Two Line - Grade 75

Steel Bars Marking System

FIGURE 3-2

3-3 BAR SPLICE, HOOK ANTI BFNH

In estimating the quantity of the steel reinforcing bars, one has

to consider the additional length for the hook , the bend and the
splice whose length vanes depending upon the limitation as
provided for by the National Building Code .

Types of Reinforcement Minimum Splice length

1. Tension Bars * 25 x Bar Size + 150 mm.
2. Compression Bars * 20 x Bar Size + 150 mm.

ILLUSTRATION 3- 1
Determine the length ot the splice joint for a 16 mm. steel bars
under the following conditions:

78
 a) Tensile reinforcement of a beam
b) Compressive reinforcement of a column

SOLUTION 3- 1
a) Classification of the reinforcement is under tension. Thus,
Multiply:
25 x 16 mm + 150 mm. » 550 mm.

b) For reinforcement under compression

Multiply:
20 x 16 mm + 150 mm * 470 mm

I
4d
h * 9d
cir 2d| l

P
n = 5d

0f /

k-t Hook Bend

Hook Length - L+ h for hook
Mild Steel Minimum Hook and Bend Allowance^

g j I

d
=
- 1 4d
ri
h = 11d 6d
i I
n = 5.5d
*A
9
3d
7
Hook Bend
Hook Length » L + n for Bend
High Yield Bars Minimum Kook and Bend Allowance

FIGURE 3 3 -
79
 A

L = 2A 3B +22d

B
A

IU
L * 2 ( A B ) 20d Total Length = A B C

=
A

1 A~1
Total Length * A + B + C D
Id
B
-
Total Length * A + B 1/2 r d -
B

Length * A + 2B + C -
D 2r 4d -
FIGURE 3 4 -
80
 A —
A
Total Length * A
Total Length = A h

1
'
71 i
B 1
l
h

I
r
A
Total Length = A + B - 1/2 r - d
Total Length = A 2h

h r
A
,a
i
V
r
A
Total Length = A + h Total Length = A + 2h

%
-A

B
dlc I C
B \

Total Length = A + B + C - r - 2d Total Length = A + B C

C —^
FIGURE 3-5

81
Comments:

1. To those who have not been exposed to detail drafting work

nor actual field construction of reinforced concrete will find
it difficult to make a detailed estimate of the various
reinforcement required.

The difficulties lies not on the ability of the estimator to

compute but his inability to familiarize with the different
types and parts of the reinforcement that comprises the
footings, columns, beams, slab etc.

2. The various parts of reinforcement that an estimator should

be familiar with are

Concrete Hollow Block Reinforcement is the simplest

type of vertical and horizontal reinforcement in between
cells and layers of the blocks respectively .

Footing Reinforcements comprises the following:

1. Footing slab reinforcement for small and medium

size.
2. Beam reinforcement for large foundations
3. Dowels

Post and Column Reinforcement:

1. Main vertical reinforcement

2. Lateral ties
a. Outer ties
b. Inner ties

82
 c. Straight ties
3. Spiral ties

Beam and Girder Reinforcement

1. Main reinforcement
a. Straight bars
b. Bend bars

2. Stirrups
a . Open stirrups
b. Closed stirrups

3. Cut Bars
a. Over and across the support
b. Between supports
c. Dowels

Floor Slab Reinforcement

1. Main Reinforcement
a . Straight main reinforcing bars
b. Main alternate reinforcing bend bars

2. Temperature Bars
3. Cut additional alternate bars over support (beam)
4. Dowels.

Not until after familiarizing with these different types of

reinforcement could one make a sound and reliable estimate.

Estimating steel bar reinforcements could be done with the

following procedures:

83
 1. The main reinforcement for post , columns, beams, girders
and the like , is determined by the " Direct Counting
Method” .Thai is, by counting the number of main
reinforcement in one post , column or beam as the case
maybe then multiplied by the total number of the same
category in the plan . The additional length for hook , bend
and splices for lapping should not be overlooked as it is
always the case in ordering length.

2. For other structural member such as lateral ties, stirrups ,

spirals , dowels, cut bars and the like should be treated
separately , one at a time. The length of their cut must
include the mandatory additional length for hook and bend.

TABLE 3-3 AREAS OF GROUPS OF REINFORCING BARS

Bar
Dia. Number of Bars ( mm)2
1 2 3 4 5 6 7 6 9 10
6 28 57 85 113 141 170 198 226 254 283
8 50 101 151 201 251 302 352 402 452 503
10 79 157 236 314 393 471 550 628 707 785
12 113 226 339 452 565 679 792 905 1017 1131
16 201 402 603 804 1005 1206 1407 1608 1809 2011
20 314 628 942 1257 1571 1885 2199 2513 2827 3142
25 491 982 1473 1963 2454 2945 3436 3927 4418 4909
32 804 1608 2412 3216 4021 4825 5629 6433 7237 8042
40 1256 2513 3769 5026 6283 7539 8796 1005 1131 1257

3. After knowing the length of the lateral ties, stirrups, and

othersmilar parts, select the steel bars to be ordered from
the various commercial length of from 5.00 meters to 13.50

84
 meters long avoiding extra cuts which might be classified
as junk unless they could be of service or used on other
structural members.

4. Tie wire for reinforcement joints and intersections is a big

item of large construction work considering its cost. Cutting
of each tie wire should be done to the minimum required
length based from the diameter of the bars to be tied with.

-
3 4 REINFORCEMENT FOR CONCRETE
HOLLOW BLOCKS

Steel bars as reinforcement is a requirement in all types of

concrete and masonry structures of which concrete hollow blocks
is one. The National Building Code has promulgated guidelines on
how and what kind of reinforcement is appropriate for this type of
work depending upon the purpose it is to serve.

The size and spacing requirements for concrete hollow block

reinforcement must be indicated on the plan or specifications. The
number of steel bars for concrete hollow block work coukt be
determined in three ways:

1. By Direct Counting Method

2. By the Unit Block Method
3. By the Area Method

Under the Direct Counting Method the vertical and

horizontal reinforcements are directly counted in the plan. The
length is also determined from the the plan or elevation although,
the hook, bend and lapping splices are imaginably added to its

85
length because it is very very rare to see a plan with a large scale
detailed drawing showing this particular requirements of reinforcing
steel bars. Thus, estimators must be familiar with the hook, bend
and splicing requirements to be able to work on effectively even if
the plan is not accompanied with such details .

The Area and Unit Block Method Is the simplest method of

computing the steel bar reinforcement for CHB with the aid of Table
3-4. The values presented in this table includes the allowances
required for standard bend, hook and splices.

ILLUSTRATION 3-2

From Figure 3-6, determine the number of 10 cm.(4") concrete

hollow blocks including the 10 mm vertical and horizontal
reinforcing bars required if it is spaced at 80 cm. on center and one
horizontal bars at every after three layers respectively .

CHB CHB Vertical Relnf. © 80 cm . o. c.

i -- fteverv
rizontef Relnf
3 layers
.

2 60 m
T\

\ Natural Ground
IZE
£ -
40 cm
Footino
4 00 m

FIGURE 3-6

86
1st SOLUTION (By the Square Meter or Area Method)

1. Solve for the area of the fence

A = length x height
A = 4.00 x 3.00
A = 12 sq. m.

2. For vertical reinforcement spaced at 80 cm. on center;

Refer to Table 3-4. By the square meter or area method,
Multiply:

12 x 1.60 = 19. 2 meters

3. Convert this value to the commercial length of steel bars

ranging from 5.00 to 13.50 meters long. Select the most
economical length avoiding extra cut.

Select: 4 pcs. 10 mm. x 5.00 m. long

= 20 meters

4. Horizontal bars at every after 3 layers. From Table 3-4 ,

Multiply:

12 x 2.15 * 25.80 meters

Select: 4 pcs. at 5.00 m. and

1 pc. at 6.00 m. long

5. Order:
8 pcs. 10 mm x 5.00 m. Steel bars
1 pcs. 10 mm. x 6.00 m. Steel bars

87
 TABLE 3-4 LENGTH OF STEEL BAR REINFORCEMENT FOR
CONCRETE HOLLOW BLOCK WORK

Vertical Reinforcement Horizontal Reinforcement

Spacing Length of Bars in Meter Spacing Length of Bars in Meter
cm. Per Block Per Sq. M. Layers Per Block Per Sq. M.

40 0.235 2.93 2 0.264 3.30

60 0.171 2.13 3 0.172 2.15
80 0.128 1.60 4 0.138 1 72

2nd SOLUTION (By the Unit Block Method)

1. Solve for the area of the fence

A = 4 x 3.00
A = 12 sq. m.

2. Determine the number of CHB

12 x 12.5 = 150 pcs.

3. Referring to Table 3-4

a) Vertical Reinforcement per block spaced at .80 m. o.c.

Multiply:

150 x 0.128 = 19.2 meters

Select: 4 pcs. 10 mm. x 5.00 m. long

88
 b} Horizontal bars at every after 3 layers
Referring to Table 3-4 ,
Multiply:

150 x 0.172 = 25.8 m.

c} Convert to commercial length

Select:
4 pcs. 10 mm x 5.00 m. long and
1 pc. 10 mm x 6.00 m. long

4. Order:
8 pcs. 10 mm x 5.00 m. and
1 pcs. 10 mm x 6.00 m. Steel bars

3-5 TIE WIRE FOR STEEL REINFORCEMENT

Tie wire refers to gauge No. 16 galvanized iron wire popularly

called G.l. tie wire. Tie wire is used to secure the steel bars in its
designed position before accepting fresh concrete .

Ordering tie wire is not by feet or meter length but in kilograms

or roil. One roll is equivalent to 40 to 45 kilograms or approximately
2,285 meters or 53 meters per kilogram.

The length of each tie wire depends upon the size of the bars
to be tied on. However, tie wire is cut into length ranging from 20
to 40 centimeters for small and medium size steel bars.

This is one item of construction materials which is always

included in the bill of materials but never computed. The quantity

89
is determined through a more or less calculation. In short, it is a
quantity with uncertainty of its accuracy . The only thing that is
certain is either it is over estimated or under estimated which is as
bad as the other

The problem is how to determine the number of kilograms

required which when cut into pieces as ties will be sufficient enough
to provide adequate ties to all joints as required by the Code.

Tie wire for CHB Reinforcement.

The common size of steel reinforcement specified for concrete

hollow block work is either 10 mm, 12 mm , 13 mm or 16 mm
depending upon the plan and specifications. For these particular
size of reinforcement, a 25 cm. or 30 cm. long tie wire folded at the
center will be satisfactory.

TABLE 3-5 KILOGRAMS OF NO. 16 TIE WIRE FOR CHB

REINFORCEMENT

Vertical Horizontal Kilogram per 100 CHB

Spacing Layer Spacing 25 cm. ties 30 cm. ties

40 2 .0042 .0051
40 3 .0031 .0038
40 4 .0028 .0033
60 2 .0028 .0034
60 3 .0021 .0025
60 4 .0018 .0022
80 2 .0021 . 0025
80 3 .0016 .0019
80 4 .0014 . 0017

90
 ILLUSTRATION 3-3

Continuing the solution of illustration 3- 2 and from the

following data obtained , determine the quantity of tie wire required
in kilograms.

Vertical Reinforcement spacing = 80 cm.

Horizontal Reinforcement spacing at every 3 layers
Area of the wall “ 12 sq. m.

SOLUTION 3-3
1. Determine the number of CHB

12 x 12.5 - 150 pcs.

2. Referring to Table 3-5 , using a 25 cm. long tie wire,

Multiply :
150 x . 0016 = .24 kg . No. 16 G.l. tie wire

More of this tie wire example will be presented in the

succeeding examples of steel reinforcement.

3-6 INDEPENDENT FOOTING REINFORCEMENT

Independent footing is an Individual or Isolated footing. The

steel bar reinforcement for this type of structure is determined
through the following methods:

1. Know the actual dimensions of the footing as to its length

and width.

91
 2. Remember that the minimum underground protective
covering of concrete to the steel reinforcement is 7.5 cm.

3. If the plan does not call for a hook or bend of the footing
reinforcement , the length of the bar is equal to the length or
width of the footing minus the protective covering at both
ends.
-
s Footing Slab

hr ./ :- |
\ • • Steel Bars ReinfC~ steel Bars Reinf""*)- .:
-
v* . £§> ^ *3%
1
5
•
<

^.5 cm 7.5 cm
I
7.5 cm
-L- 1.5 cm

Length of Bar * L Length of Bar = l 2 (7.5 cm)

FIGURE 3- 7 FIGURE 3*8

4 If the plan calls for a hook or bend of the reinforcement, the

bar-cut should include the allowances for hook and bend as
presented in Figure 3-4 and 3-5 .

S. Know the spacing distance of the steel bars both ways to

determine the exact number required . As much as
possible, select the appropriate steel bar length which is
divisible by the cut length to avoid unwanted extra cuts.

Various problems in computing for the steel bar reinforcement

will be encountered because of varied measurements and
designs. Problems however , usually arise on reinforcing members
which requires cutting and bending . Others could be determined
by the Direct Counting Method specially those which does not
require hook , and bend .

92
 ILLUSTRATION 3-4

From Figure 3-9 , determine the number of 12 mm. steel bars

required if there are 6 footings with a general dimensions of 1.50
x 1.50 meters.

i . 4

1.50 m 1.35 m
7.5 cm 7.5 cm

12 mm steel bars

1.50 m.

FIGURE 3 9 -
SOLUTION 3 4 -
1 . The net length of one reinforcing cut bar is,

1.50 - (.075 + .075 ) = 1.35

2. Find the total number of cut bars in one footing .
By direct counting

13 x 2 = 26 pcs.

93
 Get the total number of bars for the 6 footings.

26 x 6 = 156 pcs. at 1.35 m. long

4 . Select the steel bars whose length is economically cut into

1.35 m. long. Say 6.00 meters.

6.00 m, = 4.44 pcs.

1.35 m.

The fractional value of .44 is inevitable , but should not be

included in the computation because it is less than one cut bar
length. Use the whole value of 4.0 thus,

5. Divide the result of step 3 by 4.0

156 = 39 pcs. of 12 mm. x 6.00 m.

4.0

The common error committed in estimating the number of steel

bars is presented below .

Using the same data we have :

1. The net length of one cut reinforcing bar is 1.35 m. long.

2. Total number of bars in one footing is

13 x 2 - 26 pcs.

3 . Total number of cut bars for 6 footings

26 x 6 - 156 pcs.

94
 4 . The total length of these cut bars In meter Is:

156 x 1.35 * 210.6 m.

5 . Converting this length to commercial steel bar length

say 6.00 m.

210.6 m. = 35 pcs. steel bars

6.00 m.

Analysis:

Comparing the results of the two estimating procedures, the

answers has a difference of 4 pieces steel bars at 6.00 meters
long. This is the result by including the fractional amount of . 44 as
divisor of the reinforcing bar length.
The second procedure could be correct if the quotient in
dividing the length of one commercial steel bar by the length of
one cut bar yields a whole value, on the contrary, if the result has
a fractional amount, the second procedure will not give a correct
answer.

The following illustration is an example where the

second procedure applies.

ILLUSTRATION 3-5

From Figure 3-10, determine the number of 12 mm steel bars

including the tie wire In kilograms if there are 20 pcs. independent
square footing with a general dimensions of 1.15 x 1.15 meters.

95
 r 12 mm Steel Bars

P~DT TT |

y 1.15 m.

L = 1.20 m.
u-U i t >

1.15 m.

FIGURE 3-10

SOLUTION 3-5

1. Determine the net length of one reinforcing cut-bar

1.00 m. + .20 m. - 1.20 m.

2. Total cut-bars in one footing

6 x 2 - 12 pcs.

3. Total cut-bars for 20 footings

12 x 20 = 240 pcs.

4. Total length of all the bars

240 x 1.20 = 288 meters

96
 5. Divide the above result by the length of one steel bar say
6.00 meters.

288 * 48 pcs.
6.00

Order: 48 pcs. 12 mm x 6.00 m. long steel bars

Solving the same problem by the 1st procedure, we

have:

1. Net length of one reinforcing cut bar = 1.20 m.

2. Total bars in one footing

6 x 2 = 12 pcs .

3. Total for 20 footings

12 x 20 = 240 pcs.

4. Divide the commercial length of one steel bar by

the length of one cut bar;

6.00 * 5 pcs .
1.20

This simply means that 5 pcs. at 1.20 meter long reinforcing

bar could be taken from a 6.00 meters long steel bar, thus,

5. Divide the total cut bars for 20 footings by 5

97
 240 = 48 pieces
5

The question now is when to use the first procedure and when
to adopt the second procedure. In determining alone what
procedure to adopt is an additional burden, to avoid such confusion,
the following rules will help in making the right choice.

.
1 Determine the net length of one reinforcing cut bar

2. Divide 6.00 m. or any chosen commercial length of steel

bar by the result of step 1.

3. If the result is a whole number (exact value) use the

second procedure.

4. If the result has a fractional value, adopt the first procedure.

Considering illustration 3-5, Tie Wire could be

determined through the following steps:

1. Find the number of steel bar intersections in one footing

6 x 6 - 36 ties

2. Total Ties for 20 footings

20 x 36 = 720 ties

3. Using 25 cm. length per tie wire

Multiply:

98
 720 x . 25 m. = 180 meters

4 . One kilo of No . 16 G.l . wire is approximately 53 meters

long . Divide

180 = 3 4 kilos
53

3- 7 POST AND COLUMN REINFORCEMENT

The reinforcement of posts and columns to be considered in

the estimates are:

1. The Main or Vertical Reinforcement

2. The Lateral Ties or
3. The Spiral Ties for Circular Column

The quantity aha length of the main reinforcement is

determined by the ” Direct Counting Method giving special
attention to the additional length for:

a . Lap joints of end splices.

b. Allowance for bending and or hook.
c. Additional length for the beam depth and floor thickness if
the height indicated in the building plan is from floor to
ceiling.
d. Distance from floor to footing.
e. Provisions for splices of i succeeding floors.

99
 Floor Slab

Add Length
r Beam
1' -•
< Column
Lap Joint

FIGURE 3-11

-
3 8 BEAMS AND GIRDERS REINFORCEMENT

The "Direct Counting Method" is the best method in

determining the main reinforcement for beams and girders.
Provided that, in determining the length of steel bars, the following
physical conditions of the beam in relation with their support must
be considered.

1. Verify from the plan if the span of the column where the
beam is resting indicates the following conditions:

a) Center to Center of the column

b) Outer to center of the column
c) Outer to Outer of the column

100
 Each physical condition of the beam must be given special
attention in determining the length of steel bars

2. Verify the splicing position of the reinforcement if it is

adjusted to the commercial length of steel bars. Remember
that 'The Lesser the splice the lesser is the cost"

3. Identify the bars with bend and hook, for adjustment of their
order length.

Beam Beam

- - Column — *

Center to Center Center to Outer Side

Beam Beam

* — Column

Inside to Inside Outer to Outer side

Span of Beam

FIGURE 3-12

101
3-9 LATERAL TIES

Tied column has reinforcement consisting of vertical bars

held in a position by lateral reinforcement called Lateral Ties.

The ACI Code so provides that: " AU non-prestressed bars

for tied column shall be enclosed by lateral ties of at least No. 3.
in size for longitudinal bars No. 10 or smaller and at least No. 4
in size for No. 11 to 18 and bundled longitudinal bars."

The Code provisions simply mean that;

a.) If the main longitudinal reinforcement of a tied column is

No. 10 bar or smaller in size, the Lateral Ties should not be
smaller than No . 3 steel bar in size.

b) If the main reinforcement of tied column is No. 11 to No.

18 and bundled bars, Lateral Ties should not be less than
No. 4 steel bar .

Steel Bars Diameter

Number Designation Inches mm. (approx . )
2 1/4 6
3 3/8 10
4 1/2 13
5 5/8 16
6 3/4 19
7 7/8 22
8 1.0 25
9 1 1/8 28
10 1 1/4 31

102
 11 1 3/8 35
12 1 1/2 38
13 1 5/8 41
14 1 3/4 44
15 1 7/ 8 47
16 2.0 50
17 2 1 /8 53
18 2 1/4 56

The ACI Code further provides that, spacing of the lateral

ties shall not exceed the following:

1.16 x the longitudinal bar diameter

2. 48 x lateral tie bar diameter or
3. The least dimension of the column

ILLUSTRATION 3-6

Determine the spacing of the lateral ties for a tied column as

shown in Figure 3-13.

SOLUTION 3-6

1. Diameter of the. main longitudinal bars is 20 mm.

103
 2. Diameter of the lateral ties is 10 mm.

3. Multiply :
16 x 20 = 320 mm.
48 x 10 = 480 mm.
Shortest side of the column = 300 mm.

4 . Adopt the least value of 300 mm or 30 cm. spacing

20 mm

10 mm Lateral Ties

30 cm
- »- Column Reinforcement

T cm
X X cm

FIGURE 3-13

ILLUSTRATION 3-7

A building has a series of 26 square columns having a cross

sectional dimensions of 30 x 30 cm. 7.00 m. long with 8 pieces
20 mm vertical reinforcing bars for each column . Make an order
of 10 mm steel bars required for making the lateral ties.

104
 30 ciTi . - ».[
Column
Reinforcement
O O
30 cm 0- Col. Reinf.
10 mm
Lateral Ties O

T
30 cm. 10 mm Lateral Ties

FIGURE 3-14

SOLUTION 3-7

1. Determine the spacing of the lateral ties.

16 x 20 mm. = 320 mm. or 32 cm.

48 x 10 mm. = 480 mm. or 48 cm.
The shortest side of the column is = 30 cm.

2. Adopt the 30 cm. spacing

3. Determine the number of lateral ties in one column

7.00 m. ht. « 23.3 pcs.

.30

105
 4. The 23 pcs. is the distance between the lateral ties, what
we need is the number of ties in one column, so, we add
one to be exact.

23 + 1 - 24 pcs.

5. Solve for the total ties of the 26 columns.

26 x 24 * 624 pcs.

6. Find the length of one lateral tie.

By inspection the tie is 120 cm. long or 1.20 m.

7. Determine the number of 1.20 m. cut from a

6.00 m. commercial length steel bar

6.00 = 5 pcs.
1.20

8. Divide the result of 5 by result of 7

624 = 124.8 say 125 pcs.

5

9. Order: 125 pcs. of 10 mm x 6.00 m. steel bars.

ILLUSTRATION 3- 8

From Figure 3- 15 prepare an order of 10 mm. steel bars for

making the lateral ties.

106
 30 cm -
Col . Reinf . I
30 cm
'

Lateral Tie*
"
T
30 cm
... 1_

FIGURE 3 15 -

SOLUTION 3 8 -
I . This problem is an improvement of illustration 3-7 Where
the outer ties have been found to be 125 pcs. 10 mm. x 6.00 m.
What is to be determined here is the inner ties.

2. By inspection the length of the inner tie is 83 cm.

3. Determine how many 83 cm. could be cut in a 6.00 m. steel

bar .
600 = 7.23 pcs .
83

107
 4. Disregard the fractional value of .23 accept 7 pcs. and
Divide:

624 pcs. by 7 pcs. (See 5 of illustration 3-7)

624 - 89 pcs. of 10mm. x 6.00 m.

7

5. Total: 125 + 89 = 214 pcs. of 10 mm x 6.00 steel bars

ILLUSTRATION 3-9

From Figure 3-16 prepare an order of 10 mm steel bars for

making the lateral ties including the tie wire required.

i r JLL -TJ.

?HF - — Outer Ties

40 cm p-5 Straight Ties

i h r
Inner Ties
au c
25 cm 25 cm
1
n V,

FIGURE 3- 16

108
Given Data:

Number of Columns = 16 pcs .

Size = 25 x 40 cm.
Clear Height = 4.60 m.
Ties Spacing = 25 cm.

SOLUTION 3-9

1. By inspection , there are 3 types of ties:

a) Outer ties = 1 2 0 cm. long

b) Inner ties = 85 cm. long
c) Straight Ties = 50 cm. long

2. Determine how many 120 cm., 85 cm. and 50 cm. long

could be made out from one 6.00 m. steel bar .

Outer ties: 600 ~ 5 pcs.

120
Inner ties: 600 = 7 pcs.
85
Straight ties: 600 = 12 pcs.
50

3. Determine the number of ties in one column

4.60 m: ht. * 18.4 say 19 pcs.

.25 m. spacing

109
 4 . Total ties for 16 columns

19 x 16 = 304 pcs.

5. Divide 304 by each type of tie (in step 2)

304 = 60.8 say 61 pcs.

5
304 = 43.4 say 44 pcs.
7
304 = 25.3 say 26 pcs
12

6. Order: 131 pcs. of 10 mm x 6.00 m. steel bars.

Note:
The values found in step 5 was rounded to the next whole
number having the sum of 131 pcs . steel bars.

7. Solving for the tie wire:

a) Number of Joints per lateral tie by direct counting = 12.

b) Number of ties in a column

12 x 19 = 228 pcs.

c) Total tie wire for the 16 columns

228 x 16 = 3 ,648 pcs.

.
d) Multiply by the length of each tie wire say 30 cm.

110
 -
TABLE 3-6 NUMBER OF LATERAL TIES IN ONE STEEL BAR AND
QUANTITY PER METER LENGTH OF COLUMN

Spacing of Number of Length of NUMBER OF CUT IN ONE

Lateral Lateral Ties Ties with STEEL BAR LENGTH
Ties cm. PerM. Ht. Hook & Bend 5.0 m. 6.0 m. 7.5 m . 9.0 m. 12.0 m.

15 6.70 60 x 10 X 15 20
70 7 x x x 17
20 5.15 80 6 x x 11 15
85 x 7 x X 14
25 4.13 90 x X 8 10 13
95 5 x X X X
30 3.43 100 5 6 x 9 12
105 x x 7 x X
35 3.00 110 x x X 8 X
115 x 5 X X X
40 2.64 120 4 5 x X 10
125 4 x 6 x x
45 2.36 130 x x X X 9
135 X x x X X
50 2.14 140 X 4 x X X
145 X 4 5 6 x
55 1.96 150 x 4 5 6 8
160 3 x X X X
60 1.81 170 x x X 5 7
180 x X 4 5 X
190 x 3 X X 6
200 x 3 x X 6

-
X Not advisable length for economical reasons.

111
 3,648 x ,30 m. = 1,094.4 m.

e}. Divide by 53 m. length of tie wire per kilogram

1 , 094.4
53
- 20.6
say 21 kilograms of No . 16 G.l wire

Table 3-6 was prepared to simplify further the estimate for

column lateral ties and stirrups for beams and girders . It will be
noted that there are x- entry in the Table which simply mean that
such length of steel bars is not recommended for economical
reasons. The main objective of this Table is to guide the estimator
in selecting reinforcing bar whose commercial length when divided
by the length of each lateral tie will avoid extra cut of unwanted
size. To use the table , consider the following example .

ILLUSTRATION 3- 10

A building has 12 columns with a cross sectional dimensions

of 30 cm, x 40 cm. each with a clear height of 7.00 meters . Prepare
an order of 10 mm steel bars for the lateral ties spaced at 20 cm.

SOLUTION 3-10

1. By inspection there are two types of lateral ties

a) outer ties * 125 cm. long with hook

b) inside ties = 80 cm. long with hook

112
 30 cm

Col. Reinf .
40 cm

10 mm
Lateral Ties 20 cm
20 mm col. Reinf.
i

FIGURE 3-17

2. Find the total length of the 12 columns

12 x 7.00 m. = 84.00 meters

3. Referring to Table 3-6 under spacing at 20 cm.

Multiply:

84 x 5.15 = 432 pcs.

4 . Referring to Table 3-6, fcr a 125 cm. outer ties

Divide:

113
 432 = 72 pcs. 10 mm x 7.50 m. bars
6
or
432 = 108 pcs. 10 mm x 5.00 m. bars
4

5 . From Table 3-6 for the 80 cm. Inside ties

Divide:

432 = 72 pcs. 10 mm. x 5.00 m. bars

6
6. Order:

72 pcs. 10 mm. x 7.50 m. steel bars

72 pcs. 10 mm x 5.00 m. steel bars

Solving for the Tie Wire, we have;

1. Total number of lateral ties = 432

2. Total number of vertical reinforcement = 8

3. Multiply:

432 x 8 = 3 ,456

4 . If the length of each tie is 40 cm.

Multiply:

3,456 x 40 m. = 1,382.4 meters

,

114
 5 . Divide by 53 meters ( the length of tie wire in one kilo) .

1,382.4 = 26 kilograms
53

3- 10 STIRRUPS FOR BEAM AND GIRDER

Stirrup is the structural reinforcing member that holds or binds

together the main reinforcement of a beam or girder to a designed
position.

The two types of stirrups commonly used are:

1 . Open Stirrups
2. Closed stirrups

ffi)
^
Open Stirrups Closed Stirrups

FIGURE 3-18

115
 The methods in estimating the number of stirrups required is
the same as that of the lateral ties as explained in Article 3-9 with
the aid of Table 3-6. However , the spacing of the stirrups could
not be determined by the linear meter method because the spacing
of stirrups become closer as it approaches the beam support. The
number of stirrups is best determined by direct counting per span
by categories according to the design as indicated in the detailed
drawings.

ILLUSTRATION 3-11

A concrete beam with a cross-sectional dimension of 25 x 40

cm. requires 10 mm. open stirrups spaced as shown in Figure
3- 19 Prepare an order of 10 mm. steel bars for stirrups of 8 beams
with the same category .

Beam ~
irum v I
- - 4 l/l i i ITT i
E3 M
¥
Bend Bars
j I
Main Reinforcement

i! Column —•“
>

.
M
I
!

FIGURE 3-19

116
 SOLUTION 3-11

1. By direct counting, there are 24 stirrups at 98 cm. say

1.00 m. long

2. Total number of stirrups

24 x 8 beams = 1 9 2 pcs.

3. Referring to Table 3-6 for a 1.00 m. long stirrups using a

5.00 m. long steel bars.

Divide:
192 = 38.4 pcs. 10 mm. x 5.00 m.
5
or using a 6.00 m. long steel bars.
Divide:
192 = 32.0 pcs. 10 mm x 6.00 m.
6

4 . Order : 32 pcs. of 10 mm x 6.00 m. steel bars

Comment;

If 5.00 m. steel bar is chosen, the order will be 39 pcs. not 38.4
because we cannot order . 4 steel bar. After cutting the stirrups
there will be an excess of .6 of 5.00 meter steel bar which is
equivalent to 3.00 meters long. Thus, in order to be exact, a 6.00
meter steel bar should be chosen.

117
3- 11 SPIRAL AND COLUMN TIES

The spiral reinforcement consist of evenly spaced continuous

spirals held firmly in place by at least three vertical bar spacers
under the following considerations:

1. That, the center to center spacing of this spiral should not

exceed 6th part of the diameter core.
2 . That, the dear spacing between the spirals should not
exceed 7.5 cm. nor less than 5.0 cm. or,
3. The clear spacing between the spirals be less than one and
one half ( 1 1/2) times the biggest size of the coarse
aggregate (gravel).

ILLUSTRATION 3- 12

A spiral column with a cross sectional diameter of 50 cm.

requires 10 mm spiral reinforcement as shown in Figure 3-20. If
there are 14 columns at 7.00 meters high find the number of 10
,

mm steel bars needed for a 5.00 cm. pitch spirals.

SOLUTION 3-12

A. Spiral Reinforcement

1 . Find the total length of the 14 columns

7 x 14 = 98 meters

118
 2. From Table 3-7 for a 50 cm. column diameter 5.00 cm.
pitch
Multiply :

98 x 3.223 = 315.8 say 316 pcs .

3. Order: 316 pcs . 10 mm x 9. CO m. steel bars

. Column Reinf.

50 cm —
— Spiral Ties
»

t 5 cm Pitch h
i

Spiral Column

FIGURE 3 * 20

B. Tic Wire

1. Find the number of vertical bars per column - 12

2 Refer to Table 3-7 . Along 50 cm column diameter ,

5 cm. pitch, the number of turn per meter height is 21.
Multiply :
12 x 21 = 252 ties per meter height.

119
 3. Total tie wire for 14 column bar intersections at 7.00 m. ht.

252 x 7.00 m. x 14 = 24 ,696 pcs

4. Total length of the wire at .30 m. long per tie wire

24,696 x .30 * 7,409 m.

5. Convert to kilogram at 53 m. long per kg.

Divide:

7409 = 139.8 say 140 kgs.

53

6 . Order: 140 kilograms of No. 16 G.l. wire

TABLE 3-7 NUMBER OF SPIRAL REINFORCING BARS PER METER

HEIGHT *

Col. Dia. Pitch No. or Turn Number of Steel Bars from a

cm. cm. per Meter Ht. 6.00 m. 9.00 m. 12.00 m.

5.00 21.0 2.604 1.706 1.269

30.0 6.25 17.0 2.108 1.381 1.027
7.50 14.3 1.778 1.165 0.866

5.00 21.0 2.894 1.896 1.410

32.5 6.25 17.0 2.342 1 535 1.141
7.50 14.3 1.975 1.294 0.962

5.00 21.0 3.183 2.085 1.550

35.0 625 17.0 2.577 1.688 1.255
7.50 14.3 2.172 1 423 1.058

120
 5.00 21.0 3.472 2.275 1.692
37.5 6.25 17.0 2.811 1.842 1.393
7.50 14.3 2.370 1.524 1.154

5.00 21.0 3.762 2.465 1 833

40.0 6.25 17.0 3.045 1.995 1.484
7.50 14.3 2.567 1 682 1.251

5.00 21.0 4.051 2.654 1.974

42.5 6.25 17.0 3.281 2.149 1.598
7.50 14.3 2:765 1.812 1.347

5.00 21.0 4.340 2.844 2.115

45.0 6.25 17.0 3.513 2.302 1.712
7.50 14.3 2.962 1.940 1.443

500 21.0 4.630 3.033 2.256

47.5 6.25 17.0 3.748 2.455 1.826
7.50 14.3 3.159 2.070 1.539

5.00 21.0 4919 3.223 2.397

50.0 6.25 17.0 3982 2.609 1.940
7.50 14.3 3.357 2.199 1.635

5.00 21.0 5.498 3.602 2.678

55.0 6.25 17.0 4.451 2.916 2.168
7.50 14.3 3.752 2.458 1.828

5.00 21.0 6.077 3.981 2.960

60.0 6.25 17.0 4.919 3.223 2.396
7.50 14.3 4.146 2.717 2.020

5.00 21.G 7.234 4.740 3.524

70.0 6.25 17.0 5.856 3.837 2.853
7.50 14.3 4.936 3.234 2.405

121
 5.00 21.0 8.391 5.498 4.088
80.0 6.25 17.0 6.793 4.451 3.310
7.50 14.3 5.726 3.752 2.790

5.00 21.0 9.549 6.256 4.652

90.0 6.25 17.0 7.730 5.064 3.766
7.50 14.3 6.366 4.171 3.101

5.00 21.0 10.706 7.014 5.216

100 6.25 17.0 8.667 5.678 4.222
7.50 14.3 7.137 4.676 3.477

* Values given includes the end-l ap or splice allowance.

3- 12 ONE WAY REINFORCED CONCRETE

SLAB

The one way reinforcement of concrete slab is adopted

when the concrete beams or girders that supports the floor slab
is almost or rectangular in shape. Solution can either be by direct
counting or by the area method.

ILLUSTRATION 3- 13

From Figure 3-21, determine the number of steel bars for a

one way reinforcement slab including the tie wires required.

SOLUTION 3- 13

1 . Given Data:
Spacing of Main Reinforcement = 150 mm. (15 . cm.)

122
 Temperature Bars Spacing = 250 mm. (25 cm.)
Size of the Reinforcement = 12 mm. diameter
Type of reinforcement = Oneway

”
Cut Bars 1 075 m .
I Bars

Wl . I l
Straight Kars
4 30 m
'
2.15 m .
Bend Bars
t
1
rfl

^ Cut Bars I ‘
1.075 m

4.70 m .

One Way Slab Reinforcement

FIGURE 3- 21

2. Determine the number of Main Reinforcements.

4.70 + 1 = 32.33 pcs . at 5.00 m. long steel bars

.15

Note: In dividing the span by the bar spacing, what is found

is the number of spacing. Add one ( 1) to get the number of steel
bars .

3. Determine the length of the alternate cut bars to be installed

in between the main reinforcement.

123
 1.075 + .175 (hook) * 1.25 meters

4. Using a 5.00 m. steel bars,divide:

5.00 = 4 pcs. (number of cut bars taken from one

1.25 steel bar)

5. Divide: Step 2 by step 4

32.33 = 8 pcs. at 5.00 m. steel bar

4

6. Total Main Reinforcing Bars

32.33 + 8 * 40.33 pcs.

7. Solve for the Temperature bars at 2.15 m. span

2.15 + 1 = 9.6 say 10 pcs.

.25
8. Temperature bars at 1.075 span

1.075 + 1 = 5.3 pcs.

. 25

9. Multiply by 2 layers at 2 sides

5.3 x 4 - 21.2 pcs.

10. Add result of 6 and 9

9.6 + 21.2 = 30.8 pcs.

124
 11. Summary total of 6 and 10

40.33 + 30.8
= 71.1 pcs. 12 mm. x 5.00 m. steel bars.

Order: 72 pcs. 12 mm x 5.00 m. steel bars.

TA BLE 3-8 QUANTITY OF STEEL BARS AND TIE WIRES IN A

-
ONE WAY REINFORCED CONCRETE SLAB

Bar Number of Steel Bars per Square Meter Length of Tie

Spacing Wire / sq. m.
cm. 5.0 m. 6.0 m. 7.50 m. 9.0 m. 12.0 m. 25 cm. 30 cm.

10.0 4.493 3.667 2.856 2.320 1.834 . 242 .291

12.5 3.911 3.186 2.483 2.015 1.593 . 197 .236
15.0 3.524 2.866 2.234 1.812 1.433 . 163 . 195
17.5 3.247 2.637 2.056 1.667 1.319 . 141 . 169
20.0 3.039 2.465 1.192 1.558 1.233 . 126 . 152
22.5 2.878 2.332 1.819 1.473 1.166 .111 .133
25.0 2.749 2.225 1.737 1.405 1.113 .101 .121
27.5 2.643 2.138 1.669 1.350 1.069 .091 . 109
30.0 2.554 2.065 1.612 1.304 1.033 .086 . 103

2nd SOLUTION ( By the Area Method )

1. Solve for the area of the floor

4.70 x 4.30 * 20.21 sq. m.

125
 2. Referring to Table 3-8 using a 5.00 m. bars at 18 cm.
spacing .
Multiply :

20.21 x 3.524 * 71.2 pcs

3 . Order: 72 pcs. 12 mm deformed steel bars.

Solving for the tie wires at 25 cm. o.c.

1. Referring to Table 3- 8
Multiply :
20.21 x .163 = 3.29 kgs.

2. Order: 4 kgs. No . 16 G .I . wire

3- 13 TWO WAY REINFORCED CONCRETE

SLAB

A two way reinforced concrete slab is adopted when the beam

or girder that supports the concrete floor slab is almost or square
in shape or position.

ILLUSTRATION 3- 14

From Figure 3-22, determine the number of 13 mm steel bars

and tie wire required .

126
1.80 m

7.20 m

^ Straight Bar#

1.80 m

7.20 m.
I
I
Two Way Slab Reinforcement

FIGURE 3-22

SOLUTION 3 14 -
1. Solve for the main reinforcement

3.60 + 1 = 37 pcs.
.10

2. There [Link] way - run of reinforcement

37 x 2 = 74 pcs. steel bars

127
3. For cut bars at 1.87 m. long, 4 pcs . can be obtained in a
7.50 m. commercial steel bar.
Divide :

74 = 18.5 pcs. steel bars

'

4. Temperature bars at 1.80 m. span

1.80 + 1 = 8.2 pcs.

. 25

5 . Multiply by 4 sides at 2 layers

8.2 x 8 - 65.6 pcs.

6 . Summary total of step 2.3 and 5

74 + 18.5 + 65.6 = 158 pcs.

2nd SOLUTION ( By the area Method )

1. Area of the floor slab:

7.20 x 7.20 = 51.84 sq . m.

2. Referring to Table 3- 9 , for a 7.50 m. steel bars at t o cm.

spacing ; Multiply :

51.84 x 3.050 = 158 pcs.

126
 3. Order:
158 pcs, 13 mm x 7.50 m. steel bars

Solve for No. 16 G.l. tie wire cut at 30 cm.

1. Referring to Table 3-9;

Multiply:

51.84 x . 437 = 22.65 say 23 kilos .

TABLE 3-9 QUANTITY OF STEEL BARS AND TIE WIRES

ON A TWO WAY REINFORCED CONCRETE SLAB

Bar Number of Steel Bars per Square Meter Length of Tie

Spacing Wire / sq. m.
cm. 5.0 m. 6.0 m. 7.5 m. 9.0 m. 12.0 m. 25 cm- 30 cm.

10.0 4.953 3.995 3.050 3.047 2.000 .364 .437

12.5 4.409 3.549 2.703 2.734 1.775 .279 .335
15.0 4.047 3.252 2.471 2.524 1.626 .238 . 286
17.5 3.788 3.039 2.306 2.377 1.520 .208 . 250
20.0 3.594 2.880 2.182 2.266 1.440 . 185 . 222
22.5 3.443 2.756 2.085 2.179 1.378 .168 . 202
25.0 3.322 2.656 2.008 2.109 1.328 . 156 . 187
27.5 3.223 2.575 1.945 2.053 1.288 .146 . 175
30.0 3.141 2.507 1.892 2.005 1.254 .138 . 165

3-14 CONCRETE PIPE REINFORCEMENT

From Figure 3-21, determine the quantity of steel bar

reinforcement using 10 mm diameter and number 16 Tie Wire .

129
 Concrete Pipe
ReJnf. Ring

1.00 m. Temp . Bars

FIGURE 3-23

SOLUTION

1. Solve for the circumference of the circle at midpoint of the

concrete thickness T.

C = ¥d
C = 3.1416 x (.90 + .10 )
= 3.1416 x 1.00 m.
= 3.1416 m.

2. Total length of one ring plus 15 cm. splice

3.1416 + .15 = 3.29 m.

3. Find the total number of ring at .20 cm. o.c. spacing.

1.00 m. ht.
.20 cm.

130
 = 5 + 1 to get the total number of ring.
= 6 pcs.

4 . Temp, bars at .25 m. o.c

3.1416 = 12.56 say 13 pcs. at 100 long

. 25

Tie Wire

1. The total number of ring multiplied by the number of

temperature bars.

6 x 13 = 78 pcs. at . 30 m. long per tie wire

2. Total length: 78 x .30 = 23.40 m.

3. Divide by 53.00 meters to find the weight in kg.

23.40 m. = .44 say .50 kg. No . 16 Tie wire

53.00 m.

131
 CHAPTER 4
LUMBER
4- 1 WOOD

Wood is that fibrous substance which composes the trunk

and branches of the tree that lies between the pith and the bark.
The versatility of using wood in the construction has lifted it to its
present importance and high demand in almost all types of
construction.

Even with the introduction and acceptance of new materials

and methods of construction , wood is evidently much in use. Wood
because of its strength , light in weight , durability and ease of
fastening has become one of the most important building material.

4-2 DEFINITION OF TERMS

Lumber is the term applied to wood after it is sawed or sliced

into boards, planks, timber , etc.

Rough Lumber is the term applied to unplaned or undressed

lumber.

132
 Surface or Dressed lumber is a planed lumber having at least
one smooth side.

S 2S and S 4S are dressed lumber wherein the number

connotes the number of smooth sides such as s2s means smooth
on two sides and s4s on four sides.

Slab is a kind of rough lumber whichis cuttangent to the annual

rings running the full length of the log and containing at least one
flat surface.

Timber is a piece of lumber five inches or 13 cm. or larger

in its smallest dimension.

Plank is a wide piece of lumber from 4 to 13 cm. thick.

Board is a piece of lumber less than 4 cm. thick with at least

10 cm. wide.

Flitch is a thick piece of lumber.

Fine Grained , when the annua) rings are small, the grain or
marking which separates the adjacent rings is said to be fine
grained. When large, it is called Coarse Grained.

Straight Grained when the direction of the fibers are nearly

parallel with the side and edges of the board, it is said to be straight
grained.

Crooked or Cross - Grained is a lumbertaken from a crooked

tree .

133
4-3 CLASSIFICATION OF WOOD

Wood that are used in building construction are those which

grows larger by addition of a layer on the outer surface each year
known to botanists as Exogens.

Wood are classified according to:

.
1 Mode of growth

a) Exogenous are those outward growing trees which are most

preferred for lumbering .

b) indigenous are those inside growing trees which are not

preferred for lumbering because they produce a soft center
core

2. Density
a) Soft
- Density is either:
b) Hard

-
3. Leaves The leaves of the tree is either:
a) Needle shape
b) Broad shape

4« Shade or color
a) White
b) Yellow
c) Red
d) Brown
e) Black , etc.

134
5. Grain
a) Straight
b) Cross
c) Fine
d) Coarse

6. Nature of the surface when sawed

a) Plain
b) Grained
c) Figured or marked

Sapwood Modular Rays

Pitch - Outer Bark

Heartwood
-
Ca mbium
Inner Bark

Cross Section of a Tree

Crooked Gram Straight Grain

FIGURE 4-1

135
4-4 METHODS OF LOG SAWING

Lumbering is the term applied to the operations performed

in preparing the wood for commercial purposes. Logging is the
process or operation from cutting of trees, hauling and delivering
of wood to the sawmill for sawing. Sawing on the other hand , is
the operation of cutting the logs into commercial sizes of lumber.

Combined Radial and

Tangential - — Radial

bj
® 1

o
Quarter Tangential *— Tangential P|ain or
Bastarrd Sawing

Star Shake
Wind or Cupr
Shake

Ii
Broken

r j .v: ^—
Branch
r
r
- Heajt Shake '
(

FIGURE 4 2 -
136
 The methods and manner of log sawing are:

1. Plain or Bastard Sawing

2. Quarter or Rift Sawing

a) Radial
b) Tangential
c) Quarter Tangential
d) Combined Radial and Tangential

4-5 DEFECTS IN WOOD

Defects are irregularities found in wood. The most common

defects in wood are:

1. Caused by Abnormal Growth such as:

a) Heart Shakes are radical cracks originating at the heart of

the logs commonly found in old trees.

b) Wind Shakes or Cup Shakes are cracks or breaks across

the annual rings of the wood during its growth caused by
excessive bending of the tree due to strong wind.

c) Star Shakes are composed of several heart shakes which

radiate from the center of the log in star like manner.

d) Knots occurs at the starting point af a limb or branch of the

wood.

137
 2. Due to Deterioration

a) Dry Rot is caused by fungi in a seasoned lumber due to the

presence of moisture.

b) Wet rot takes place sometime in the growth of the tree

caused by water saturation.

4-6 SEASONING OF LUMBER

By nature, trees contain moisture in their cell layers. These

moisture has to be expelled thoroughly in order to preserve the
wood from shrinkage or decay. Experiments have proven that
wood which are immersed in water immediately after cutting into
flitches is less subject to splitting and decay . It reduces warping
but becomes brittle and less elastic. Soaking of wood in liquid is
the oldest method of seasoning lumber introduced and practiced
by the ancient Roman builders.

The various methods of seasoning lumber are:

1. Natural or Air Seasoning is considered as one of the best

method of seasoning lumber although the period involved is
relatively longer.

2. Artificial Seasoning is a process wherein the lumber is

stacked in a drying kiln and then exposed to steam and hot air.
Wood from this process undergoes quick drying and is classified
as quite inferior in quality as compared to those lumber seasoned
by the natural or air seasoning method.

138
 The artificial methods of seasoning wood are:

a) Forced Air Drying

b) Kiln Drying
c) Radio Frequency Dialetric Drying

Good seasoning is the primary consideration for a successful

preservation of wood. Wood does not decay naturally through age,
nor will it decay if \i is kept constantly dry or continuously
submerged in water.

The common causes of decay in wood are:

1.) Alternate moisture and dryness

2.) Fungi and Molds
3.) Insects and Worms
4.) Heat and Confined Air

The process of preserving wood are:

1. External - The wood is coated with preservative ( as paint)

which penetrates the fiber.

2. Internal -A chemical compound is impregnated at a

prescribed pressure to permeate th ^ wood thoroughly.

4-7 THE UNIT MEASURE OF LUMBER

Board foot is the unit of measure used in computing volume

of lumber. Despite the adoption of the Metric System (SI ) measure

139
board foot for lumber is still in use for convenience and practical
use of lumber sizes .

One board foot simply mean, one square foot by one inch thick
lumber or an equivalent of 144 cu. inches. The width and thickness
of commercial lumber are expressed in inches except the length
which are in feet of even numbers.
Board foot is found by dividing the product of the thickness, the
width and the length by 12.

ILLUSTRATION 4- 1

Find the total board foot of :

5 pieces 2" x 6" x 14 ft . lumber .

SOLUTION

5 x 2 x 6 x 14 = 70 bd. ft.
12

The number of board foot in a log could be determined using

the following formula:

Bd. Ft. Volume = ( D - 4 )2 L

16
Where:
D * is the smaller diameter of log
L = the length of the log
4 = slab reduction allowance

140
 ILLUSTRATION 4-2

Determine the total board foot of lumber which could be derived

from a log 28 inches diameter by 20 feet long.

6.00 m ( 20' )

FIGURE 4-3

SOLUTION

( 28 - 4 )2 x 20 = 720 bd. ft.

16

Prior to the introduction of sawmills , the manner of sawing

lumber is by manual hand sawing. Two persons are involved . One
at each end of the hand saw pulling alternately at their own direction
until the log or lumber is sliced into the desired sizes.

Labor cost is computed under the following manner:

1. By the board foot or

2. By the puigada system

141
 The Board Foot method is simpfy finding the total board feet
of sawed lumber multiplied by the agreed unit price.

The Pufgada System is computed by multiplying the width

in inches by the length of lumber in meter, the result is then
multiplied by the unit price.

ILLUSTRATION 4 3 -
How much will it cost to slice a 6" x 6" x 3.00 m. lumber into a
2" x 6" x 3.00 m. lumber if the unit price is 50 centavos per pulgada?

Lin© of Cutting ig

3,00 m - i

FIGURE 4-4

SOLUTION

1. Multiply the width by the length

6 x 3.00 = 18 pulgadas (inches)

142
 2. Multiply by the number of run or slice

18 x 2 = 36 pulgadas

3. Finding the cost, multiply:

36 x P 0.50 = P 18.00

ILLUSTRATION 4-4

How much will it cost to convert a 42" x 14" x 20 ft. fumber

toasize of 2" x 6" x 20 ft. at a price of P 0.50 per pulgada?

r— Line of Cutting

FIGURE 4-5

SOLUTION

1. The first run of the saw along 14 inches

Convert: 20 ft. = 6.00 meters

14x 600m. = 84 pulgada

143
 2. The succeeding run along the 12” is

12 x 6.00 m. = 72 pulgada per run

3. Total run : 72 x 6 * 432 pulgadas

4. Total cost : 432 + 84 = 516 x P 0.50 = P 258.D0

This pulgada system of slicing log or flitches into small pieces

of lumber is already obsolete with the introduction of sawmills,
circular saw and the modem chain saw. The introduction of this
modern chain saw have tremendously destroyed the forest
rapidly. However, with the enforcement of the log ban law, this
modern chain saw is now on its final momentum of wiping
out the coconut tree as substitute to wood lumber.

4-8 WOOD POST

In estimating wooden post for building structure, there are only

three things to consider:

a) The size of the post

b) The quantity or number of posts
c) The length or height

The size of the post is already indicated in the plan. The

quantity or number of the post is determined through direct
counting based from the detailed plan of the building. The length
is determined under the following considerations:

a) For a one storey building verify if the elevation height

144
indicates from floor to ceiling, if the ceiling is below the girts, then,
add the depth of the girts, and the bottom chord or the rafters to the
height of the post.

b) For a two storey building verify if the height indicates from

floorto floor, if so, then consider the additional length for the girders ,
floor joist and the flooring. For the second floor, add the depth of
the girts, bottom chord or rafters to the height of the post.

c) Remember that the commercial length of lumber is always

of even number , if the computed length is odd number adjust the
order to the next even number or length.

ILLUSTRATION 4- 5

From Figure 4-6, determine the length of the wooden post

required.

.20 cm
Girts *
3/4“ Wood Flooring 2.70 m .
^

Girder
2" x 6" Joists
2.70 m .
i

FIGURE 4 6 -
145
 SOLUTION

1. Find the total height from floor to ceiling

2.70 + 2.70 = 5.40 meters

2. Determine the depth of girder, floor joist, flooring and girts.

Girder .20
Floor Joist * .15
Flooring = . 025
Girts = .20
.575 m.

3. Add the result of 1 and 2

5.40 + .575 = 5.975 m.

4. Convert to feet

5.975 m. = 19.9 ft.

.30

5. Order length = 20 feet

-
4 9 GIRDER

The Girder is the structural member of a building that carries

the floor joist and the flooring . It is determined by Direct Counting
Method from the floor framing of the building plan . The length
however , is determined under the following considerations.

146
1. If the span or distance of the post is indicated from center
to center , the length of the girder is equal to the span plus
the width or one side of the post.

Floor Joist
T & G Flooring

±:0
1 II \i
1
IT If
Girder
11 FT f

i Post
L - Span + 2 ( 1/2 Side of Post )
ffl
FIGURE 4-7

2. If the span of the post indicates from outer to outer side of

the post, the girder length is equal to the span of the post.

a
Floor Joist

T & G. Flooring —*
~

II Hi 4F ll
\ Girder o

-— Post
L = Span of Post

FIGURE 4-8

3. If the span or distance indicates from center to outer side

of the posts, the length of the girder is equal to the span
plus one half the width of one post.

147
 Floor Joist —
T & G Flooring -«

o
H o
p
Girder o

Post

T L = Span + 1/2 Side of Post

FIGURE 4-9

4 . If the span or distance of the post indicates inside

measurement , the length of the girder is equal to the span
plus two width or sides of the post.

Floor Joist
T & G. Flooring

Girder
n o «
i II
Girder
i
« o o

Post
Post - Overhang
I
L * Span + 2 Sides of Post

FIGURE 4-10

5. If the second storey has a floor overhang wherein the

girder has to carry the floor joist, the length of the girder is
equal to the span as stated above plus the overhang length .

148
ILLUSTRATION 4-6

From Figure 4-11 what girder length shall be ordered?

a ©
o
Floor Joist
T & G Flooring —,

Girder
Post
n o
n n r

.60 m.
3.00 m ^
Overhang

L = Span + Overhang + 2 ( 1 /2 Side of Poet )

FIGURE 4 11-

SOLUTION

1. Determine the span and the overhang.

3.00 + .60 *= 3.60 m.

2. Add two widths of the post.

3.60 + 2 (. 15) = 3.90 m.

3. Convert to feet

3.90 = 13 ft. This is an odd number, adjust to

.30 the next even number.

4. Order : 14 ft. long

149
4-10 FLOOR JOIST AND T & G FLOORING

Floor Joist is the structural member of a building that carries

the wood flooring. The best way of estimating floor joist is by direct
counting from the detailed floor framing plan ofthe building wherein
the actual number and length ofthe joist is indicated.

Girder

Ir

I — Solid Bridging

Floor Joist

Girder
:

FIGURE 4-12

Groove
T & G Flooring
%
Tongue

FIGURE 4 13 -
150
 TheT & G is the popular name for Tongue and Groove wooden
board used for flooring, ceiling, forms etc. The thickness of the
board varies from 18 mm to 25 mm., while its width also varies from
5 to 15 centimeters. Other sizes for Architectural purposes are
obtained through special order.

There are two methods presented on how to determine the

required number of pieces and board feet of T & G for a given
area.
1. Direct counting method
2. The number of pieces per meter run

ILLUSTRATION 4-7

From the floor framing plan as shown in Figure 4-14, determine

the number and board foot of floor joist and the T& G flooring.

SOLUTION

A. Floor Joist

1. By Direct Counting

5.00 = 16.66
.30

2. Add one to get the actual number of joist

16.66 + 1 - 17.66 say 18 pcs.

151
 3. For two spans

18 x 2 = 36 pcs.

4. Determine the span of the joist ( in feet )

3.50 m. = 11.66 say 12 ft.

Order:

36 - 2" x 6” x 12' = 432 bd. ft

12

T
£ ksL
2" x 6" Floor Joists

8 Solid Bridging

l Girder

5.00 m.
. Jim, 5.00 m.

FIGURE 4-14

B. Solid Bridging

1. Span of girder

5.00 * 16.66 ft.

.30

152
 2. Total number of joist

18 x 2 inches thickness
* 36 inches or 3 feet

3. Subtract from step 1

16.66 - 3 ft. « 13.66 or 14 ft.

4. For solid bridging, order:

2 pcs. 2" x 6" x 14 ft. = 28 bd. ft.

C. T & G Flooring

1st Solution: ( By Direct Counting )

1. The length of the girder is the length of the T & G flooring

5 , 00 * 16.66 say 18 ft.

.30

2. Determine the number of 4" T & G board

3.50 x 2 span = 70 pcs

. 10

3. Referring to Table 4-1

Multiply:
70 x 1.222 = 85.54 say 86 pcs.

153
 4. Order:
,,
86 - 1" x 4 x 18 t
= 516 bd. ft. TAG.
12

TABLE 4-1 QUANTITY OF T & C WOOD BOARD PER METER RUN

Size of Board Add % forAllowance Number of Board

In. Cm Direct Count Method Per Meter Run

3/4 x 3"
H
2.0 x 7.5 .0125 16.66
3/4" x 4" 2.0 x 10.0 .0123 12.20
3/4" x S' 2.0 x 12.5 .0117 9.34
3/4" X 6" 2.0 x 15 0 .0115 7 70

2nd Solution: (By the number of board per meter run)

1. Determine the length of T & G flooring

5.00 - 16.66 say 18 ft.

”
30

2. Referring to Table 4-1 using 4” ( 10 cm.) TAG

Multiply:

7.00 x 12.20 * 85.4 say 86 pcs.

86 - 1" x 4” x 18* « 516 bd. ft.

12

154
4- 11 SIDING WOOD BOARD

The common type of commercial siding wood boards are:

1. Stone cut 4. BCB cut
2. Double Stone cut 5 . Weather cut
3. V- cut

Single Cut Double Cut BCB

V-Cut Weather Cut

FIGURE 4-15

The thickness of these boards varies from 13 , 16 or 20 mm.

The width is from 15 to 20 cm. of even length from 6 to 20 feet.
The procedure in estimating the quantity of siding wood board is
the same as that of T & G wood flooring with the aid of Table
4-2. However, unlike the T & G wood flooring , estimating the
siding board has to consider the following :

1 . The area of the opening such as windows, doors and the

like should be deducted from the area of the wall to be
covered by the siding boards.

2. The length of the siding wood boa u should be specified in

(

155
 the order to prevent joints of the board in between the
height.

3. The common error in determining the length or height of

the board is the omission of the additional length for the
girts, flooring .floor joist and girder depth.

TABLE 4-2 QUANTITY OF SIDING WOOD BOARD

Commercial Size Number of Board Approximate Board Ft.

Width of Board Per Meter Run Per Sq. M.
Inches Cm.

6" 15 7.41 13.65

8" 20 5.55 13.55

5.00 m
2.10
I
20 cm J i
J
3.10 m
FFHff
: ;
50 cm .
U- iii I iii
t Double Cut
1

FIGURE 4-16

156
 ILLUSTRATION 4-8

From Figure 4-16 solve for the required board foot and number
of 200 mm. ( 8" ) double cui siding board.

SOLUTION

1 . Determine the total length of the board

Floor to ceiling 3.10 m.
Depth of Girts . 20 m.
Flooring & Joist .17 m.
Depth of Girder .25 m.
Studs .075 m.
3.795 m. * 12.65 ft.
Order 14 ft.

Length of the wall 5.00 m.

Less the opening 2.10 m.
Net Wall Length 2.90 m.

3. Referring to Table 4-2 for a 200 mm . board

Multiply :

2.90 x 5.55 = 16 pcs .

4 . Order: 16 pcs. 3/4" x 8" x 14" double cut

* 149 bd. ft.

2nd Solution: ( By the Board Feet per Square Meter)

1 . Solve for the wall area

157
 A * 3.795 x 2.90
A = 11 sq. m.

2. Referring to Table 4-2

Multiply:

11 x 13.55 = 149 bd. ft.

4-12 GIRTS, RAFTERS, TRUSS, PURLINS, AND

FASCIA BOARD

These items in building construction are determined by direct

counting and measuring method . Members with shorter length like
collar plate , struts , blocks for splice of joints, sway bridging etc. are
computed according to their sizes combined together and adjusted
to the commercial length of lumber. For accuracy of estimating
these items, a detailed drawing indicating their sizes and length
shall be made as basis in finding the unit length of every parts

Coll«r Plate
Top Chord
Vertreal Strut | Diagonal Strut

2 Girts 1 Bottom Chord

King Post

FIGURE 4-17

158
4-13 STUDS

Stud is the structural member in building construction where

the siding or partition boards are fastened. It is sometimes referred
to as ribs of wooden walls or partitions. Lumber intended for studs
shall be straight and uniform in width of either S2S or S4S for
uniformity of wall thickness.

The advantages of using S2S or S4S lumber are:

1. Lumber are straight, uniform in thickness and of good

quality.
2. It is economical in terms of labor cost.
3. The work progress is not affected or delayed.

There are two methods presented in this Chapter on how to

find the quantity of studs at a given vertical and horizontal spacing.

1. The Direct Counting Method

2. The Square Meter Method

The Direct Counting Method is done by counting the

number of vertical and horizontal member from a detailed plan
including its length. In the absence of a detailed drawing plan, an
imaginary counting through mathematical computation will do as
an alternative.

The Square Meter or Area Method is by simply finding the

area of the wail multiplied by the values given in Table 4-3
corresponding to the size and spacing of the studs.

159
 ILLUSTRATION 4-9
A wall partition 6.00 meters long by 2.60 meters high specify
a 2" x 4" studs spaced at .60 m. o.c. both ways. Find the total
board feet required.

SOLUTION- 1 ( By Direct Counting )

1. Find the number of vertical studs

6.00 « 10 + 1 « 11 pcs. at 10 ft.

.60
2. Horizontal Studs

2.50 * 4.2 say 5 pcs. at 20 ft.

.60
3. Order:
Vertical Studs 11 pcs. 2" x 4" x 10’ * 73.3 bd. ft.
Horizontal Studs 5 pcs. T x 4" x 20’ « 66.6 bd. ft.
Total 139.9 bd. ft.

Solution-2 (By the area Method)

1. Solve for the area of the wall partitions

6.00 x 2.50 = 15 sq. m.

2. Referring to Table 4-3, using 2 x 4 at .60 m. spacing

Multiply:
15 x 9.333 * 140 bd ft.

160
 TABLE 4-3 NUMBER OF BOARD FOOT OF STUDS AND NAILING
JOIST PER SQUARE METER

Lumber Spacing in Centimeters (Center to Center)

Size In.
30 x 30 30 x 60 40 x 40 -
40 x 60 ! 60 x60

1x 2 4.230 3.256 3.208 2.771 | 2.333

2x2 8.460 6.513 6.417 5.445 i 4.667
2x 3 12.688 9.769 9.625 8.312 ; 7.000
2x4 16.920 13.026 12.833 11.083 | 9.333
2x5 21.146 16.282 16.042 13.854 | 11.667
2x6 25.375 19.539 19.250 16.625 I 14.000
!
Comment:

Comparatively , computation by the Area Method with the aid

of Table 4-3 instantly gives a result in board foot. Unlike the first
solution wherein the number of pieces and length were known
outright ahead of the board foot. However, as to which method will
be adopted, depends upon the choice and purpose of the estimator

There are instances where small discrepancies arises between

the results of the two methods. This is due to the adjustment of
lumber from odd to even length. Naturally, if length is adjusted
to the next even length the number of board foot will also increase
while the area covered remains the same. Under this circumstances
small discrepancies between the result of the direct counting
method and the area method cannot be avoided but to a very
.eligible amount .

161
 ILLUSTRATION 4- 10

A partition wall measures 8.00 m. long by 2.70 meters high

specify the use of 2" x 3" studs with a general spacing at 40 cm.
for vertical and 60 cm. for horizontal center to center distance.
Prepare the order list of 2" x 3" lumber for wall studs.

Vertical Strip Q .40 m. o.c.

2.70 m,

8.00 m.
Hor. Strip Q .60 m. o.c.
FIGURE 4-18

1st SOLUTION (By Direct Counting)

1. Find the number of vertical studs

8.00 c 20 spacing
.40
Add one spacing to get the actual number of studs ( see Fig 4-18)

20 + 1 = 21 studs

Order : 21 pcs. 2" x 3" x 10‘ = 105 bd. ft.

2. Find the number of horizontal studs

162
 2.70 = 4.5 Add 1 to get actual number of studs
.60

4.5 + 1 ~ 5.5 pcs. at 4.00 m. long

Since the wall is 8.00 m. long (see Figure 4-18)

multiply:

5.5 x 2 = 11 pcs.

Order:
11 pcs. 2" x 3” x 14’ = 77 bd. ft.

Summary
21 pcs. 2" x 3" x 10’ * 105 bd. ft.
11 pcs. T x 3" x 14’ = 77 bd. ft.
Total « •
182 bd. ft.

2nd SOLUTION ( By the Area Method)

1. Solve for the area of the wail partition

8 x 2.70 = 21.6 sq m . .
2. Referring to Table 4-3, using 2 x 3 studs at .40 x .60
m. on center spacing
Multiply:

21.6 x 8.312 = 179.54 say 180 bd. ft.

3. Note the difference of 2 bd. ft. between the two solutions

which is negligible.

163
4- 14 CEILING JOIST

Ceiling Joist is the structural member in building construction

that holds the ceiling board. It is otherwise known as the nailing
strip. The common size used for ceiling joist are V x 2"; 2" x 2"
and 2” x 3" lumber spaced to suit the size of the ceiling board. In
short, the ceiling board dimension governs the spacing of the
ceiling joist for economy. The methods adopted in computing the
quantity of ceiling joist is the same as that of the studs with the aid
of Table 4-3.

2" x 2 @ ,40 m oc.

N

Bothways

4.00 m.

7.00 m.

FIGURE 4-19
*
ILLUSTRATION 4- 11

Find the total board foot required for a 7.00 by 4.00 meters
bedroom using 2” x 2" ceiling joist spaced at .40 x .40 m. on
center.

164
SOLUTION ( By Direct Counting )

1 . Find the number of joist perpendicular to 7.00 meters.

7.00 = 17.5 + 1 = 18.5

.40
= say 19 pcs. at 4.00 m. (14‘)

2. Find the number of joist perpendicular to 4.00 m . span

4.00 = 10 + 1 * 11 pcs. at 7.00 m. or 23 feet

. 40

A combination of:

11 pcs. 2" x T x 14’ and

11 pcs. 2" x 2" x 10 ’ or
22 pcs. 2” x T x 12'

3. Order:
19 pcs. 2" x T x 14’ = 88.66 bd. ft.
22 pcs. 2" x T x 12’ = 88.00 bd. ft.
* 176.66 bd . ft.
say 177.0 bd ft

SOLUTION ( by the Area Method)

1. Area of the ceiling

7.00 x 4.00 = 28 sq . m.

2. Referring to Table 4-3, multiply:

165
 28 x 6.417 = 179.7 say 180 bd. ft.

3. Note that the 3 [Link]. difference is negligible.

4-15 CEILING BOARD

There are numerous kinds of ceiling board of different brand ,

quality and dimensions available for building construction .
However, the simplest way in finding the number of boards required
is to divide the total ceiling area by the effective covering of one
ceiling board or by the square meter area method with the aid of
-
Table 4 4.

ILLUSTRATION 4 12 -
A bedroom with a general dimensions of 4.00 m. x 5.00 meters
specifies the use of 1/4 x 4’ x 8’ plywood for ceiling on 2" x 2" ceiling
joist spaced at 40 x 60 centimeters o.c. Determine the number of
plywood and ceiling joists required .

—
80 cm
Plywood Ceiling
2.40 m.

80 cm.

1.20 1.20 1 1.20 1.20 ’

10 10

FIGURE 4*20

166
 SOLUTION ( By the effective covering method )

A. Ceiling Joist

1. Find the area of the ceiling

4.00 m. x 5.00 m. = 20 $q. m.

2. Refer to Table 4-3. Along 2" x 2" at 40 x 60 spacing

Multiply:
20 x 5.445 = 106.90 say 109 bd. ft.

B. Ceiling Board

1. Find the area of the ceiling.

4.00 m. x 5.00 m. - 20 sq, m.

2. Refer to Table A~A . Along 120 x 240 ( 1/4 x 4’ x 8’ )

Plywood board, effective covering is 2.88. Divide:

20
2 8g = 6.9 say 7 pcs. plywood board

TABLE 4-4 QUANTITY OF CEILING BOARD PER SQUARE METER

Size Effective Covering Number of Pieces
Centimeters per board in sq , m. per square meter

30 x 30 0.90 11.111
40 X 40 0.16 6.250
40 x 60 0.24 4.167
60 x 60 0.36 2.778
60 x 120 0.72 1.389
90 X 180 1.62 0.617
120 x 240 2.88 0.347

167
 ILLUSTRATION 4- 13

An office room with a general dimensions of 6.00 m. x 9.60

meters specify the use of a .60 x 1.20 m. ceiling board. Find the
number of pieces required.

9.60 m.

E 2' x 4 Plywood Ceiling

8
6
(

FIGURE 4-21

1 st SOLUTION (By the Effective Covering Area Method )

a . Find the area of the ceiling

6.00 x 9.60 m. = 57.6 sq. m.

b. Referring to Table 4-4 using a .60 x 1.20 board

57.6 = 80 pcs.
. 72

2nd SOLUTION ( By the Number of Pieces per Square

Meter )

168
 1. Find the area of the celling

6.00 x 9.60 * 57.6 sq. m.

2. Referring to Table 4-4 using a .60 x 1.20 board

Multiply :

57.6 x 1.389 = 80 pcs.

3rd SOLUTION ( By the Direct Counting Method )

1. Find the number of boards along the 6.00 m.

6.00 = 10 pcs.
.60

2. Find the number of board along the 9.60 m.

9.60 = 8 pcs.
1.20

3 . Multiply results of 1 and 2

10 x 8 * 80 pcs.

Comment:

The results of the three methods as presented are correct and

satisfactory if the ceiling area falls under the following conditions:

1. That the quotient in dividing the area of the ceiling by the

169
 effective area covering of one board yields an exact number
or value ( no fraction ).

2. Thai the ceiling design is plain and not interrupted by beams,

girders , rafters , partitions , openings etc.

3. That the ceiling has no intricate design or decorations that

requires more cutting of the ceiling board.

4. When cutting of the ceiling boards could not be avoided,

wastage is also inevitable but could be replenished by an
allowance factor of about 2 to 5 % depending upon the
design.

ILLUSTRATION 4- 14

A living room measures 6.80 m. x 8.00 m. specify the use of

a .90 x 1.80 m. ceiling board. Find the number of pieces required.

25 cm

6.30 m.
90 1.80 petting Board
*

25 cm

40’
+1.80+1.80+1.80+
1.80 “ **

FIGURE 422

170
 -
SOLUTION 1: ( By the Effective Covering Area Method )

1. Find the area of the ceiling

6.80 x 8.00 = 54.40 sq. m.

2. Referring to Table 4-4 using a . 90 x 1.80 ceiling board

54.40 = 33.58 say 34 pcs.

1.62

-
SOLUTION 2: (By The Number of Pieces per sq. m .)

1. Find the area of the ceiling

6.80 x 8.00 = 54.40 sq. m

2. Referring to Table 4-4

Multiply :

54.40 x .617 = 33 .56 say 34 pcs.

SOLUTION 3: ( By the Direct Counting Method . )

*

1. Find the number of boards along the 6.80 m. side

6.80 = 3.78
1.80
2. Find the number of boards along the 8.00 m. side

8.00 = 8.89
.90

171
 3. Multiply 1 and 2

3.78 x 8.89 = 33.60 say 34 pcs.

4-16 DOOR FRAME

Estimating the materials for fabrication of door frame is simply

determining the size and length of the lumber to suit the size of door
panel whose width varies from .60 m. to 1.00 meter wide Door
frame bigger than one meter opening is considered as special
design and order
In ordering lumber for door frame, the estimator has two
options:

1. Ordering one length of 18 ft for each door jamb or a

2. Combination of headers or jamb and header to suit the
commercial length of lumber for economic reasons

3 ft. Header

7 ft .

3" x 6"

FIGURE 4 23-
172
 ILLUSTRATION 4 15 -
A 20 classroom school building with 2 doors per room specify
the use of 3" x 6" door jamb. Prepare an order list of lumber for
fabrication of the door jamb.

SOLUTION

A. Ordering one length for each jamb

f . Determine the total length of the jamb.

Jamb = ( 7’ + 3" ) x 2 pcs.

= 14 ' - 6". or 14.5 ft.

2. Length of Header

( 3 ft . + 6 In.) . . . . . 3.5 ft .
Total length of 1 & 2 18.0 ft.

3. Order: 40 - 3" X 6HX 18’ = 1 ,080 bd. ft.

B . Combination of Headers and Jambs

1 . One header is 3’ - 6” or 3.5 ft.

2 Four Headers * 3.5 ft. x 4 = 14 ft.

3 For 40 Headers
Order: 10 pcs. 3" x 6" x 14 '

173
 4. Jambs = ( 7" + 3 ) x 2 sides = 14’ - 6"
M

or 14.5 ft.

5. Lumber length is even number, for 40 jambs

Order : 40 pcs. 3" x 6” x 16’

6. Summary:
Header: 10 pcs. 3‘* x 6” x 14’ =* 210 bd. ft.
Jamb : 40 pcs. 3 x 6" x 16’ = 960 bd. ft.
M

1,170 bd. ft.

Comment:

Comparing the result of the two procedures, it will he noted

that the second procedure is 90 board feet more than the result of
the first Considering the present cost of lumber , one has to
choose the order of 40 pcs. at 18 ft long. However, in ordering
lumber length from 18 feet and above the following disadvantages
might be encountered:

1. Due to the scarcity of lumber, length from 18 feet and above

might not be available in the market and this might cause delay of
the construction work.

2 The price of lumber varies and is considerably higher as

length goes longer. Thus, the cost of the 90 board feet difference
between the two procedures might be more than the difference in
cost if longer length is ordered.

In adjusting the ordered length from 14' • 6" to 16 feet ( step

4 ) there is an excess length of 40 pcs. one and one half feet or

174
45 cm. which could be used on other parts of the construction or
they could be sliced for studs or ceiling joist which then could not
be totally considered as waste.

4-17 WINDOW FRAME

The different parts of a window frame to be considered in

estimating are:
1 . Jamb 3. Transom
2. Header 4 . Window sill
3. Transom 5 . Mullion

Estimating Procedure :

1. To find the length of the jamb, add the thickness of the sill,
the mullion and the header.

2. The length of the sill and header shall include the thickness
of the two jambs and the mullion.

3. The transom is equal to the length of the sill or header less

the thickness of the two jambs.

4. The length of the -million is equal to the length of the jambs

less the thickness of the head and sill.

ILLUSTRATION 4- 16

From Figure 4-24 prepare the list of the lumber materials to be

ordered.

175
 Header
r
Tranec .n

- Jamb 3" x 4"

5 ft
— Mullion

-3 H
x 6"
s-izr
Sill
7 ft .
3"

FIGURE 4-24

SOLUTION

1 . Jamb: ( 5 ' + 6" ) x 2 - 1 1 feet

Order: 1 pc. 3" x 6" x 12 ft = 18 bd ft ..

2. Header & Sill: (7' + 6" ) x 2 = 15 ft.

Order: 2 pcs. 3" x 6" x 8 ft. or

1 pc. 3" x 6” x 16 ft.

3 Mullions: < 5’ - 0" ) x 2 = 10’

Order : 1 pc. 3“ x r x 10 ft

4 Transom: ( 7’ - 0 " )

Order : 1 pc. 3’* x 6" x 8 ft.

176
 CHAPTER 5
FORMS,SCAFFOLDING
AND STAGING
5- 1 FORMS

Form is a temporary boarding, sheathing or pan used to

produce the desired shape and size of concrete. The structural
members of a building are built-up into its desired shape and
dimension through the use of forms which serve as mold for the
mixed concrete.

Concrete mixture is generally semi-fluid that reproduces the

shape of anything into which it is poured. Concrete forms should
be watertight, strong enough and rigid to sustain the weight of the
concrete. It should be simple and economically designed in such
a manner that they are easily removed and re- assembled without
damage to themselves or to the concrete.

Selection of forms are based on:

1. Cost of the materials.

2. The construction and assembling cost.

177
 3. The number of times it could be used
4 . Strength and resistance to pressure and tear and wear.

Classification of Forms:

A . Materials
1 . Wood 3. Plastic
2. Metal 4 . Composite

B. Shape
1. Straight
2 . Circular , etc .

C. Solid or Hollow Cast

1 . Single
2. Double

D. Methods of Construction
1 . Ordinary
2. Unit

E. Uses
1. Foundation and column
2. Wall
3. Steps
4 . Beams and girders
5 . Slabs
6 Sidewalks, etc.

F. Construction of Forms consist of

1 . Retaining board
2. Supporters or studs

178
 3. Braces
4 Spacer
5 Tie Wire
6 Bolts and nails

G. Tjpcvof WJIII Fnrm

1 Continuous
2. Full unit
3. Layer unit
a Continuous
b Sectional

5- 2 GREASING OF FORMS

The purpose of greasing the form is to make the wood water

proof, thus, preventing the absorption of water in the concrete which
causes swelling and warping Greasing of forms also prevent
adherence of concrete to the pores of the wood.

Crude oil is the most economical and satis factoiy material for
this purpose. The etude oil »s mixed with No. 40 motor oil to a
proportion of 1:3 mixture with varying viscosity according to the
temperature. Thicker mixture :s necessary on warm weather.
_
However. greasing of forms should not be ajtcwed after the steel
,

bars have been, set torts position

,

5-3 SCAFFOLDING AND STAGING

Scaffoiding a temporary structure of wooden poles and

is
planks providing platform fc. working men to stend on while
'
erecting or repairing a building. It is further defined as a temporary
framework for other purposes.

Staging on the otherhand , is a more substantial framework

progressively built up as tail building rises up. The term staging is
applied because it is built up in stages one storey at a time.

Numerous accidents in building construction usually

happened because of faulty construction method and insufficient
supports and braces. One tragic incident happened in the
construction of the Film Palace in Metro Manila where several lives
including the Supervising Engineer were buried alive in cement and
rubbles when the forms and staging swayed and rammed down in
total coHapse.
Staging is not as simple as others think of it . It requires special
skill, and experienced men to do the work. Incidentally , the primary
cause of accidents and failure of the framework is the use of inferior
quality lumber, inadequate supports and braces , nails and others
for economy sake or negligence . Definitely , out lumber has no
place in horizontal scaffolding or staging work if the builder is
aware of the value of life and property involved in building
construction. Lumber intended for temporary structure to support
heavy load such as concrete should be selected from straight grain
of wood free from shakes or knots and other defects.

The use of coconut lumber is gaining wide acceptance due to

the scarcity and the prohibitive price of wood lumber. However ,
extra care should be exercised in selecting coco-lumber. Those
with or near the bark is considered harder and better quality. Closer
spacing of supports and braces should always be in mind when
using coco-lumber as scaffolding or staging. Do not rely too much
on coco lumber fastened with nails , remember the principles of post
and lintel structures .

180
 The different parts of staging to be considered are:

1 . Vertical support
2 Footing base { as need arises )
3 . Horizontal braces
4 . Blocks and wedges support
5 . Nails

- jr
^ ^f=
.
T
r~

71K£

5- 4 COMPARATIVE ANALYSIS BETWEEN THE

T & G AND PLYWOOD AS FORMS

Cost is the primary consideration in selecting the kind of

materials to be used as forms. Cost is a broad term in construction
which under this particular item refers to:

1 . Initial investment on materials.

181
 2. Assembling cost .
3. The number of times it could be used.
4. Durability of the materials to resist pressure and tear and
wear.

ILLUSTRATION 5- 1

A residential house has 10 wooden posts resting on a

concrete footing as shown in Figure 5-2. Prepare a comparative
bill of materials using T & G and plywood forms.

2" x 2" Frame

i T& G

60 cm
30 cm.

30 cm

30 cm .

40 cm
1.20 m.

T
1.20 m.
i
Footing

FIGURE 5- 2

182
SOLUTION

a ) Using a l” x 6" T & G Board

1 . Solve for the lateral width of one footing.

( . 30 x 2 ) + ( . 40 x2)
.6 + .8 = 1.40 m.

2. Referring to Table 4-1, multiply:

1.40 x 7.70 = 10.78 say 11 pcs.

3. For 10 posts prepare 5 forms only , thus:

11 x 5 = 55 pcs.

4 . Determine the height of the form

1.20 m. " 4 feet

5 . Order:
55 pcs. 1" x 6" x 4’ or
8 pcs. 1" x 6" x 8’ = 112 bd. ft.

b ) Form Ribs and Frame @ .40 m. distance

Form A
.30 x 8 pcs. - 2.40 or 8 ft .
1.20 x 4 pcs. = 4.80 m. or 16 ft .

For 5 forms , Order:

183
 5 pcs. 2" x 2" 8'x
5 pcs. T x 2" x 16’

Form B
.60 x 8 pcs. = 4.80 or 16 ft.

For 5 forms Order:

5 pcs . 2" x 2" x 16’

c) Brace Holder and Stake

1. By direct counting from the figure we have:

4 pcs. 2” x T x 8’

2" x 2" Frame

Plywood -

T 30 cm.
1.20 m .

30 cm 40 cm

FIGURE 5-3

184
 Summary of T & G for Five footings

28 pcs. 1" x 6" x 8’ = 112 bd. ft.

9 pcs. 2” x 2" x 8’ = 72 In. ft.
10 pcs. 2* x 2" x 16’ = 160 In. ft .

2nd SOLUTION

a) Using plywood forms

.
1. The total lateral width of one forrr = 1.40 m.

2. Length or Height of the form = 1.20 m.

3. Area of one form

1.40 x 1.20 - 1.68 sq. m.

4. Total area of 5 forms

1.68 x 5 = 8.40 sq. m.

5. Referring to Table 4 -4, using 1.20 x 2.40 plywood by

effective covering method;
Divide ;

8.40 = 2.92 say 3 pcs. plywood

2.88

b) Forms or Ribs

Form A
1. By direct counting

185
 6 pcs. 2" x 2" x 1.20 m. ( 4 ft.)
4 pcs. 2" x 2" x .30 m. ( 11t.)

2. For 5 forms, we have:

30 pcs. 2" x 2” x 4 ft.
20 pcs. 2" x 2" x 1 ft.

3. Order:
15 pcs. 2" x 2" x 8 ft.
2 pcs. T x 2” x 10 ft.

Form B

1. By direct counting

6 pcs. Y x 2* x 1.20 m. ( 4 ft.)

.
4 pcs 2" x 2“ x .40 m.

2. For 5 forms , we get:

30 pcs. 2” x 2 x 4 ft.
B

20 pcs. 2" x Y x ,40 m.

3. Order:
15 pcs. 2" x 2“ x 8 ft.
5 pcs. Y x 2“ x 1.60 or 6 ft.

c) Brace Holder and Stake

1. By direct counting
4 pcs. 2* x 2" x 8 ft.

186
 Summary for 5 Footings

3 pcs. 6 mm. x 1.20 x 2.40 plywood

30 pcs. 2” x 2" x 8 ’ = 240 In. ft.
2 pcs. T x 2” x 10’ = 20 In. ft.
5 pcs. 2" x 2" x 6' = 30 In . ft.
290 In. ft .

From the above results, knowing the materials required for T

& G and plywood form, canvass the prices and make your choice
as to what materials will be used.

Comment:

The common material used for forms in all types of construction

during the time of lumber abundancy is the T & G . Unfortunately,
with the present condition of our forest where the price of wood is
highly prohivitive , using a T & G . lumber as form is very costly
unless extremely necessary. Presently, the materials being used
as form is either plywood or metal sheet The use of plastic form is
the next alternative after the wood considering its weight , durability
and recycling properties

TABLE 5.1 QUANTITY OF TAG WOOD FORM FOR PIERS AND

COLUMNS

Materials Bd. Ft. per Sq. M. 2 x 2 Frame Ln. Ft.

T & G Bd . Ft . per Sq. M. per Sq. M.

1" x 4" 18.33 7.32 21.96

r x 6” 17.50 7.32 21.96

187
 Note:
To convert board ft. to Linear Ft. - Multiply by 3
To convert Linear ft. to Board Ft. - Divide by 3

ILLUSTRATION 5-2

Prepare Bill of Materials for 8 columns at 4 meters high with

uniform cross-sectional dimensions of 30 cm. x 40 cm. using 1"
x 6" T & G form with 2" x 2" frame .

—- — ^ - 2" x 2" Frame —

30 cm I
E 40 cm
T&G 8
T & G. Form
I
r
40 cm

FIGURE 5-4

SOLUTION

1 . Find the lateral surface area of one column

188
 ( .30 x 2 ) + (. 40 x 2 ) x 4.00 m. ht.
(.60 + . 80 ) x 4.00 m. = 5.6 sq. m.

2. Referring toTable 5-1 using 1" x 6" T & G Board

Multiply :

5.6 x 17.50 = 98 bd . ft .

3. For 8 columns
Multiply:

98 x 8 = 784 bd. ft. 1 ” x 6" x 14’

4 . Find the 2" x 2" frame for one form

Referring to Table 5- 1
Multiply:

5.6 x 7.32 = 41 bd. ft.

5. For 8 columns
Multiply:
41 x 8 = 328 bd. ft.

Comment

1. Illustration 5- 1 is the direct counting meihod. One will notice

the intricacies involved in itemizing the materials for the
different parts of the form including the adjustment of the
lumber to the commercial length for that simple wooden
footings . What if the estimate calls for the entire forms
necessary to construct a multi- storey reinforced concrete
building? Table 5- 1 is prepared for this purpose and
presented in illustration 5-2.

189
 2. T h e 2 x 2 f r a m e a s found in illustration 5-2 does not
include yet the vertical and horizontal support and the
diagonal braces which will be discussed later in Section 5-8
under the scaffolding and staging.

5-5 FORMS USING PLYWOOD

Plywood is a versatile construction material not only used for

walls, partitions and cabinets but also for furnitures as well as forms
for reinforced concrete constructions. The plywood thickness
varies from 4 mm. 6mm. 12 mm. and 20 mm. with a commercial
size of .90 x 1.80 m. and 1.20 m. x 2.40 meters.

Plywood as form has the following advantages:

1. It is economical in terms of labor cost.

2 . Light weight and handy .
3 . Smooth surface which requires less plaster or no plastering
at all.
4. Less consumption of nails.

ILLUSTRATION 5- 3
Six concrete posts 4.00 m. high with a uniform cross sectional
dimensions of 30 x 30 cm. specify the use of 12 mm plywood on
a 2" x 2" frame . Prepare the bill of materials .

190
SOLUTION

Data:
Number of post = 6
Height = 4 meters
Cross Sectional Dimension * .30 x 30 cm.
Frame * 2" x 2"

1. Solve for the lateral surface area of one column.

.30 x 4 sides = 1.20 m.

2. Multiply by the height.

1.20 x 4.00 m. = 4.8 sq. m.

3. Total area of 6 columns

4.8 X 6 * 28.8 sq. m.

4. Referring to Table 5-2 using 1.20 x 2.40 plywood

Multiply:

28.8 x .488 * 14 pcs.

5. Solve for the frame, from Table 5-2 using 2 x 2 lumber:

Multiply:

28.8 x 12.71 = 366 bd. ft.

6. The height of the post is the length of the frame

Order:
366 bd. ft . T x 2" x 14 ft.

191
 -
TABLE 5 2 QUANTITY OF PLYWOOD FORM AND ITS FRAME FOR
COLUMNS PER SQUARE METER LATERAL AREA

Plywood Number Size of Frame or Ribs

Size of Pieces Bd Ft. per Sq . M
in M . 2x2 2x 3

. 90x 1.80 . 842 12.71 18.96

1.20 x 2.40 . 488 12.71 18.96

* The values given under the frame or rihs column are computed
from a longitudinal rib type considering its economical
advantages.

Plywood Form

30 cm

2” x T Frame

C? -
'••••
30 cm.

30 cm.
40 cm 30 cm

FIGURE 5-5

192
5-6 FORMS OF CIRCULAR COLUMN

The galvanized iron sheet otherwise known as G.l. sheet is

the most common material used as form for circular , oval or
elliptical structure considering its versatility in forming any shape
of various forms and design. Where G.l. sheet is specified as form
for a cylindrical column, wood board and built up supporters are
necessary to form the circumferential arc frame . The standard
dimension of the plain G .l sheet is 90 cm. wide by 2.40 meters
long.
Estimating procedure:

t . Find the circumference of the circle

where C = 3.1416 x diameter .
2. Multiply the circumference by the column height.
3. Divide the result found in step 2 by 2.16 , the effective
covering area of one G.l. sheet.
4 . Where extra cuts could not be avoided, an allowance of 5
to 10 % for waste is satisfactory .
5 . Solve for the number of supporter or ribs. Divide the
circumference by . 10 if the spacing is 4" or by .15 if the
spacing is 6 inches.
6 . Provide 2 pieces circumferential supporter for every joint of
the unit form which is equivalent to 90 cm. die width of one
G.l . sheet.

ILLUSTRATION 5-4

Determine the required plain G .l .sheet form for 6 circular

columns 4.50 m. high with a uniform cross-sectional diameter of

193
.60 m. using 1" x 14" and T x 2" supporters.

I" Wood Board

Circumferential
Supporter

/i r r 2" x 2" Vertical m

i Frame
i • Line of Cutting
\ /
\
/

FIGURE 5-4

SOLUTION

1. Solve for the circumference of one column

C - 3.1416 x .60 = 1.88 m.

2. Multiply by the column height

Area = 1.88 x 4.50 = 8.46 sq . m.

3. Area of the 6 columns

8.46 x 6 * 5076 sq. m.

4. Find the number of sheets required . Divide by the effective

194
 covering of one G.I sheet 2.16

50.76 = 23.5 pcs.

T 16
5 . Consider 10 % allowable factor

23.5 x 1.10 = 25.85

6 . Total G. l . sheets

* 25.85 say 26 pcs.

7. Solve for the 2" x 2" frame of one column @ 10 m. o.c.

1.88 = 18.8 say pcs.

.10

8. For 6 columns

19 x 6 = 114 pcs.

9 . Height of the column is 4.50 m. or 16 ft. Thus, the order will

be:
114 pcs. 2" x T x 16 ' = 608 bd . ft.

10. Circumferential arc supporter

4.50 m. ht. = 5 pcs.

. 90

11. Multiply by 2; 5 x 2 = 10 pcs.

195
 The value of 2 is the circumferential supporter frame per G.l.
sheet form at 90 cm. high having 2 frames per joint (see figure 5-7).

12. Total for 6 columns

1 0 x 6 = 60 pcs. 1" x 14" x 5" or

30 pcs. r
x 14" X 10’ = 350 bd. ft.

Note
The circular column diameter is .60 m. or 2. ft. thus, a 5 ft.
board length divided by two will be satisfactory for the
circumferential arc. ( See Figure 5-7 ).

Summary

26 pcs. .90 x 2.40 m. (36" x 8‘) plain G.l. sheet

114 pcs. 2” x 2" x 16’
30 pcs. 1" x 14” x 10’

Iill
1" x 14" Arc

Plain G.l . Sheet rr90 cm

2" x 2" Supporters u

d ~ 60 cm

Circular Column Form

FIGURE 5-7

196
 TABLE S-3 QUANTITY OF LUMBER FORM FOR CIRCULAR
COLUMN

Lumber Board Foor Per Plain G. i Sheet

Size Spacing of Frame
10 Cm. 2.5 Cm. 15 Cm. 20 Cm.

1” x T 11.7 10.0 8.5 6.5

2" x 2" 23.4 20.0 17.0 13.0
2” x 3" 35.1 30.0 25.5 19.5
1" x 8" 7.7 7.7 7.7 7.7
1" x 10" 9.6 9.6 9.6 9.6
1" x 12* 11.6 11.6 11.6 11.6
1" x 14" 13.5 13.5 13.5 13.5

ILLUSTRATION 5-5

Solving the problem of illustration 5-4 using Table 5-3, we have

the following data:

Number of columns - 6 pcs.

Height - 4.50 m.
Diameter = .60 m.
Lumber - 1 x 14 and 2 x 2
Circumference - 1.88 m.

SOLUTION

1 . Find the lateral surface area of the 6 circular columns

1.88 x 4.50 m. x 6 columns.

= 50.76 sq. m

197
 2. Divide by the effective covering area of one G.l Sheet

50.76 = 23.50 pcs.

2.16

3. Consider 10% allowance

23.22 x 1.10 — 25.54 say 26 pcs.

4. Solving for the supporter frame or ribs @ .10 m. ox.
From Table 5-3;
Multiply:

26 x 23.39 = 608 bd. ft.

.
5. Circumferential supporter, using T' x 14” board Table 5-3
Multiply:

26 x 13.47 = 350 bd. ft.

TABLE 5-4 FORM FOR BEAMS AND GIRDERS

Plywood Number Lumber Frame Bd. Ft. per Plywood

Size (m.) per sq . m. 1x 2 2x2 2x 3

.90 x 1.80 . 81 8.0 16.0 24.0

1.20 x 2.40 .42 16.0 32.0 48.0

ILLUSTRATION 5-6

Six concrete beams with cross-sectional dimensions of .30 x

.40 m. has a uniform clear span of 4.50 meters. If 6 mm x 1.20 x

198
2.40 plywood form will be used on 2 x 2 wood frame , prepare the
bill of matedats.

30 cm
Plywood Form
40 cm
4.50 m.

2“ X 2” Frame
40 cm —

FIGURE 5-8

SOLUTION

1. Determine the total length of the two sides (depth) and the
bottom width of the beam.

( .40 x 2) + . 30 - 1.10 m.

2. Multiply by the clear span

1.10 x 4.50 - 4.95 sq. m .

3. Referring to Table 5-4 , solve for the plywood form.

Multiply:

199
 4.95 x .42 = 2.1 pcs.

4 . Solve for the frame or ribs. From Table 5-4,

Multiply :

2.1 pcs. x 32.0 * 64 67.2 bd. ft

Summary for 6 Beams

12 pcs. 6 mm. x 1.20 x 2.40 m. plywood

384 bd . ft. T x 2“ x 16’ lumber .

TABLE 5-5 QUANTITY OF LUMBER FOR SCAFFOLDING AND

STAGING

Lumber Column Beam Flooring

Size Bd. Ft. per Meter Ht. Bd Ft. / M. Ht. Bd. Ft / M. Ht.
Vert. Hor. Brace Vert Hor.

T x 2" 4.67 21.00 11.67 4.00 4.67 6.10

2' x 3” 7.00 31.67 17.50 6.00 7.00 910
2' x 4" 9.33 42.22 23.33 8.00 9.33 12.10

5-7 ESTIMATING THE SCAFFOLDING AND

STAGING

Estimating the quantity of the materials required for

scaffolding or staging is somewhat difficult considering the volume
of the materials involved. The computation requires time and

200
imagination in counting the various parts of the structure such as
the vertical and horizontal support, the diagonal braces plus the
blocks and wedges which are not shown even on a detailed plan
of the building .

The usual practice of most estimator is to to make an estimate

by either the quantity of the materials or by a lump sum amount for
forms and staging item in the bill of materials. Table 5-5 is specially
prepared for for this purpose .

Column —% -ijj
v.

7-y

Q 2 x 2 Horizontal and
S Diagonal Support
2 x 3 Vertical Support

Perspective

rffft
' 4 oo *7 nr\

Plan
FIGURE 5-9

201
 ILLUSTRATION 5- 7

A reinforced concrete building has 9 columns with a clear

height of 4.00 meters as shown on Figure 5-9. Determine the
required staging under the following specifications:

Vertical Support use 2" x 3” lumber

Horizontal Support use 2" x 2" lumber
Diagonal Braces use 2” x 2" lumber

SOLUTION

A . ) Staging for Columns

1. Find the total length of the 9 columns

4.00 m. x 9 = 36 m.

2. Referring to Table 5-5 and using 2** x 3" vertical support;

Multiply:

36 x 7.00 = 252 bd. ft.

If the height is 4.00 , order:

252 bd. ft. 2" x 3" x 14 ft.

3. For the horizontal support , refer to Table 5-5; using 2 x 2

Multiply :

36 x 21.00 = 756 bd . ft .

202
 4 . Diagonal Braces . From Table 5-5
Multiply :

£6 x 11.67 = 420 bd . ft.

Summary:

252 bd. ft. of 2" x 3" x 14 ft.

1 , 176 bd. ft. of 2” x 2" x 16 ft.

Note:
If 2” x 2" is to be computed in linear ft .
Multiply : by 3

1 , 176 x 3 = 3,528 In. ft.

B.) Staging for Beams

1. Find the total length of the beams

<4.50 x 6) + (4.00 x 6)
27 + 24 = 51 meters

2. Referring to Table 5-5 , we have;

a) For vertical support using 2” x 3";

Multiply:

51 x 6.00 = 306 bd. ft.

b) For horizontal support using 2" x 2";

Multiply :

203
 51 x 4,67 = 238 bd. ft.

c) Total:

306 bd. ft . 2" x 3” x 14 ft .

238 bd. ft. T x 3" x 14 ft .

C.) Staging for Concrete Floor Slab

1. Find the area of the floor

4.50 x 4.00 x 4 units = 72 sq. m.

2. Referring to Table 5-5 using 2" x 3”;

Multiply :

72 x 9.10 = 655 bd. ft .

3. Order:
655 bd . ft. 2" x 3” x 14’

Comments:

In the construction of multistorey building, the Transfer of the

forms, scaffolding or staging from one floor to the next floor is an
inevitable normal operation wherein waste of materials could not
be avoided due to tear and wear . The percentage of waste varies
depending upon the following factors.

1. The difference in height between the 1st and the 2nd floor
naturally requires adjustment of the vertical support.

204
 2. The difference in sizes of beams and girders also requires
form adjustments .

3. The tear and wear of forms and scaffolding are caused by

the dismantling, transferring and re -assembling.

4. Reckless use and handling of the materials. This includes

making firewood and pilferages during the period of
construction.

The use of inferior quality lumber will only result to high

percentage waste and risk. To use poor quality lumber for
economic reasons should be carefully evaluated because the final
accounting result might be to the contrary as expected.

As previously mentionedjhe percentage of wasfe in

transfening the staging from one floor to the next varies from 10
to 20 % per floor depending upon the physical condition of the
structure, the quality of materials and the manner of how they are
handled.

205
 CHAPTER 6
ROOFING MATERIALS
6- 1 GALVANIZED IRON SHEET

Galvanized iron sheet is either Plain or Corrugated. Plain

G .l. Sheet is widely used for roofing, gutter , flashing , downspout,
ridge , valley and hip roll etc. Plain G.l. sheet standard commercial
size is .90 m. wide by 2.40 meters long (3‘ x 8‘). Corrugated G.l .
Sheet on the otherhand, is widely used for roofing and siding
material having standard width of 80 centimeters with varying
length from 1.50 meters to 3.60 meters at an interval length of 30
centimeters.

Corrugated G. I. Roof Sheet

-
FIGURE 6 1

206
 -
TABLE 6 1 STANDARD WEIGHT OF GALVANIZED IRON SHEET
IN KILOGRAM

Gauge Thick 1.5 m. 1.8 m. 2.1 m. 2.4 m. 2.7 m. 3.0 m. 3.3 m . 3.6 m.
No. cm. 5' 6' T 8‘ Iff 1V 12'

14 .203 22 36 26.83 31.30 35.78 40.25 44.72 49.19 53.66

15 .180 20.25 24.30 28.35 32.40 36.45 40.50 44.55 48.60
16 .163 18.14 27.76 25.39 29.02 32.64 36.27 39.90 43.52
17 .147 16.43 19.72 23.00 26.29 29.58 32.86 36.15 39,43
18 .132 14.73 17.67 20 62 23.56 26.51 29.45 32.40 35.34
19 .117 13.03 15.63 18.24 20.84 23.45 26.05 28.66 31.27
20 . 102 11.32 13.58 15 85 18.11 20.38 22.64 24.90 27.17
21 .094 10.43 12.52 14.60 16.69 18.78 20.86 22.95 25.03
22 .086 9.62 11.54 13.46 15.38 17.31 19.23 21.15 23.08
23 .079 8.73 10.47 12.22 13.96 15.71 1745 19.20 20.94
24 .071 7.91 9.49 11.07 12.66 14.24 15.82 17.40 18.29
25 .064 7.03 8.43 9.84 11.24 12.65 14.05 15.45 16.86
26 .056 6.19 7.43 8.66 9.90 11.14 12.38 13.62 14.86
27 .051 5.76 6.91 8 06 9.22 10.37 11.52 12.67 13.82
28 .048 5.33 6.39 7.46 8.52 9.59 10.65 11.72 12.78
29 .043 490 5.88 6.86 7.84 8.82 9.80 10.78 11.76
30 .041 4.48 5.37 6.27 7.16 8.06 8 95 9.85 10.74

The thickness of galvanized iron sheets is measured in terms

of Gauge number from 14 to 30. The sheet becomes thinner as the
gauge number becomes larger. Gauge number 26 is the most
extensively used in various tinsmithing work although gauge 24 is
specified for gutters and flashing .

How to distinguish the difference in thickness of G.l. sheets

between the consecutive gauges from 14 to 30 is difficult even with

207
the aid of a caliper. The gauge is expressed in terms of hundredth
of an inch or centimeter. The only way by which one could be sure
that he is buying the right thickness of the sheet is by weight
measure . Table 6- 1 is presented for this purpose;

Procedures in estimating corrugated G .I . sheets

1 . Verify the specifications for side lapping if it is 1 1/2 or 2 1/2

corrugations.

2. If it is 1 1 /2 corrugations the effective width covering per

sheet is 70 cm. , or 60 cm. for 2 1/ 2 corrugations .

3. The standard end lapping joint is from 25 cm. to 30 cm.

4 . The distances or spacing between purlins should be

proportionally spaced and adjusted to the length of the G.I .
sheets to avoid unnecessary cutting of the sheets. In short,
the length of the roofing sheet shall govern the distances or
spacing of the purlins. (Referto Table 6-2.)
5. As much as possible, minimize the end-lapping joint of the
roof sheets. Always specify longer length for economical
reasons .

& i - ,
_ 2 1 /2 Corr .
_
- G I. Sheets

Puffins

1 1/2 Corr .
End Lapping Side Lapping

FIGURE 6-2

208
 TABLE 6-2 EFFECTIVE COVERAGE OF CORRUGATED G.L
SHEETS, ROOF ACCESSORIES AND PURLINS DISTANCES.

Effective Width Covering Purlins Number of

Length Side Lapping Distance Nails or Rivets
Feet - Meter 1 1/2 2 1/2 (cm. ) per Sheet

6 1.80 . 70 .60 . 75 14
7 2.10 .70 .60 . 60 18
8 2.40 .70 .60 . 70 18
9 2.70 . 70 60
. . 60 22
10 3.00 . 70 .60 . 67 22
12 3.60 .70 .60 . 66 26

Procedures in estimating the quantity of corrugated

G.I. roofing and its accessories.
1. Determine the length of the purlins along the gutter line. This
distance is perpendicular with the roof direction.

2. Divide this length with the effective width covering of one

sheet which is .70 m. for 1 1/2 corrugations or .60 m. for 2
1/2 corrugations (See Table 6-2 ) the result is the number
of G.I . sheets in one row.

3. Determine the length of the rafter or the top chord. Choose

the right combination of G.I . roofing sheets that will satisfy
this length considering the 30 cm. end lapping joint.

4. Multiply the result found in step 2 by each length of G.I .

sheet combination found in step 3.

5. Determine the number of G.I. nails or rivets and washers in

209
 kilogram using Table 6-2 and Table 6-3.

6. Take note that the number of anchor G.l. strap and lead
washer is the same as the quantity of the rivets. The G.l.
washer is double the quantity of the rivets (see Table 6-3).

7 . Solve for the number of plain G.l. sheets required for anchor
strap with the aid of Table 6-4 .

-
TABLE 6 3 QUANTITY OF ROOF ACCESSORIES IN KILOGRAMS

Materials Number of Pieces per Kilogram

G.l. Rooring nails 120

G. I. Rivets 180
G.l. Washers 126
Lead Washers 75
Umbrella Nails 120

TABLE 6-4 SIZE AND QUANTITY OF STRAPS IN

ONE PLAIN G. I. SHEET

Size of Purlins Size of G.I. Strap Number of Strap

In in Cm. in One Plain Sheet

2" x 3" 50 x 75 2.5 x 22.5 384

2" x 4" 50 x 100 2.5 x 25 0 342
T x 5" 50 x 125 2.5 x 27.5 312
rx 6" 50 x 150 2.5 x 30.0 288

210
 ILLUSTRATION 6- 1
From Figure 6-3, find the number of corrugated G.l. sheets
and its accessories required if the side lapping specify 1 1/2
corrugations with 30 cm . end lapping on a 50 mm. x 75 mm.
purlins.

Corr . G. I . Roofing
5.00

3 .o
GQ
% ^°

FIGURE 6 3 *

SOLUTION

A,) For Corrugated G.L Sheets

1. Divide the length of the gutter by the effective covering of

one sheet . Referring to Table 6- 2
Divide:

14.00 m. = 20 pcs.
.70 m.

211
 2. The length of the rafter is 6.00 m., thus, a combination of
3.60 m. and 2.70 m. long G.I. sheets.

3. Order:

20 pcs. .80 x 3.60 m. G.l. Sheets

20 pcs. .80 x 2.70 m. G . l . Sheets

B.) Rivets

1. Referring to Table 6-2 for a 3.60 m. and 2.70 m . sheet .

Multiply :

For 3.60 m. long: 20 pcs. x 26 = 520 pcs.

For 2.70 m. long: 20 pcs. x 22 « 440 pcs.
Total number rivets = 960 pcs.

-
2. Convert to kilograms. Referring to Table 6 3.
Divide:

960 = 5.33 say 5.5 kg.

180

C.) G.L Washers

1. Double the number of rivets

960 x 2 * 1,920 pcs .

2. Convert to kilograms. Referring to Table 6-3

1,920 = 15.24 say 15.5 kg.

126

212
D.) Plain G. I . Strap on 50 % 75 mm . purlins

1. Total number of rivets = 960 pcs.

2. Referring to Table 6-4 using 50 x 75 mm. purlins

Divide:

960 - 2.5 pcs. Plain G.I. Sheets

384

£.) Lead Washers

1. The number of rivets is equal to the number of lead washers

= 960 pcs.

2. Referring to Table 6-3

Divide:

960 = 12.8 say 13 kgs.

75
Summary
20 pcs. 80 cm. x 360 cm. Corr. G.I. sheets
20 pcs. 80 cm. x 270 cm. Corr. G.I sheets
5.5 kg. G.I. Rivets
15.5 Kg. G.I. Washers
13 kg. Lead Washers
2.5 pcs. Plain G.I sheets

ILLUSTRATION 6-2

From Figure 6-4 find the number of corrugated G.I . sheets

213
including the umbrella nails required if the G .I . sheets are laid at 2
1/2 side corrugations and .30 m. end lapping joint.

Gu%r

FIGURE 6 4-
SOLUTION

A.) Corrugated G. l . Sheets

1. Determine the number of roof [Link] to Table 6^2

Divide:

18.00 m. = 30 pcs.
.60 m.

2. The length of the rafter is 6.00 m. or a combination of 3.60

and 2.70 m. sheets with 30 meter end lapping ( See Table
6-5 ) .
Multiply each sheet length by the result of step 1, thus;

214
 30 pcs. - .80 m. x 3.60 m. Corr. G.l . Sheets
30 pcs. - .80 m. x 2.70 m. Corr . G .l . Sheets

B.) Umbrella Nails

1. Determine the number of umbrella nails for the 3.60 m.

and 2.70 m. roof sheets. Referring to Table 6- 2
Multiply:

30 sheets x 26 = 780 pcs.

30 sheets x 22 = 660 pcs.
Total = 1,440 pcs.
2. Convert to kilograms. Referring to Table 6-3;
Multiply:

1 ,440 = 12 kg.
120

.
Corr. G l. Roofing

300
*».
.

FIGURE 6-5

215
 ILLUSTRATION 6- 3

From Figure 6-5 , find the number of corrugated G .l. sheets,

roof nails, washers and lead washers required if the side lapping
specify 11/2 corrugations .

SOLUTION

A .) Corrugated G . l . Roofing

1. Determine the number of corrugated roof sheets . Referring

to Table 6-2;
Divide:

12.90 m. = 18.43 pcs .

.70 m.

2. Determine the length of the rafters ( see figure 6-5 ).

4.80 m. or 16 ft .

3. Referring to Table 6- 5 , 4 80 m. rafter requires a combination

of 3 00 m. and 2.10 m. G.l. sheets.
Therefore:

18.43 x 2 = 36 86 say 37 pcs. @ 3.00 m long .

18 43 x 2 5 36.86 say 37 pcs. @ 2.10 m. long

B.) Roof Nails

1. Solve for the quantity of roof nails . Referring to Table 6- 2

216
 Multiply:

Fora 3.00 sheet 37 x 22 = 814 pcs.

Fora 2.10 sheet 37 x 18 = 666 pcs.
Total = 1 ,480 pcs

2. Convert 1 ,480 pcs . to kilograms Referring to Table 6-3,

Divide:

1 ,480 = 12.33 say 13 kilograms roof nails.

120 .

1 ,480 = 11.75 say 12 kg. G.l. washers

126

1,480 = 19.73 say 20 kg . Lead washers

-
FIGURE 6 6

217
 ILLUSTRATION 6-4

From Figure 6-6, determine the number of corrugated G.I.

sheets , rivets, washers, lead washers including the plain G.l. straps
required if the roof sheet is laid at 2 1/2 corrugations side lapping
and 30 cm. end lapping on a 50 x 100 mm ( 2” x 4”) purlins.

SOLUTION

A .) Corrugated G . l. Sheets

1. Determine the number of corrugated sheets. Referring to

Table 6-2 for 2 1/2 side lapping;
Divide:
34.50 - 57.50 pcs.
.60

18.00 = 30 pcs.
.60

2. For a 6.00 m. and 5.40 m. length of rafter, refer to Table

6-5; Combine length:

For 6.00 m. rafter combine 3.60 and 2.70 m. sheets

For 5.40 m. rafter combine 3.00 and 2.70 m. sheets

3. Determine the number of G .l. sheets

For 3.60 m. sheet: 57.50 x 2 run * 115 pcs. ( 12’)

For 2.70 m. sheet: 57.50 x 2 run = 1.15 pcs . ( 9’ )
For 3.00 m. sheet: 30 x 2 run - 60 pcs. (10')
For 2.70 m. sheet: 30 x 2 run = 60 pcs . ( 9 ’ )

218
B.) G.I. Rivets:

1. Determine the number of rivets. Referring to Table 6-2;

Multiply:

For the 3.60 m. sheet: 115 x 26 = 2 ,990 pcs.

For the 3 .00 m. sheet: 60 x 22 = 1,320 pcs.
For the 2.70 m. sheet : 175 x 22 = 3,850 pcs.
Total = 8,160 pcs .

2. Convert to kilograms. Referring to Table 6-3;

Divide:

a) Rivets: 8 , 160 = 45.3 say 46 kg.

180

b) G .I . Washers ( Double the Rivets ), from Table 6-3;

Divide:

8,160 x 2 = 129.52 say 130 kg.

126

c) Lead Washer ( same number as the rivets),

from Table 6-3;
Divide:

8, 160 « 108.8 say 109 kg.

75

C.) Plain G. I. Strap

1 . Size of G.I . straps on a 50 x 100 mm. purlins

- 2.5 by 25 cm.

219
 2. Number of strap is equal to number of rivets
= 8,160 pcs.

3. Solve for the number of plain G.l. sheets.

Referring to Table 6-4

8,160 *= 23.86 say 24 pcs.

342

4. Common wire nails for the anchor straps

8,160 x 3 nails per strap

= 24,480 pcs. 32 mm. CW Nail
Note;
1 kilo of 4a, 32 mm. CW Nail is Approximately 695 pcs.
Convert to kilograms.
Divide:

24,480 * 35.2 say 36 kg.

695

D.) Ridge Roll , Fascia and Gutter( to be discussed

later)

Summary

115 pcs. .80 m. x 3.60 m. .

Corr. G I. Sheets
175 pcs. ,80 m . x .
2.70 m. Corr. G l. Sheets
60 pcs. .80 m. x 3.00 m. Corr. G.l. Sheets
46 kgs. ..
G l rivets
130 kgs. .
G l. washers

220
 109 kgs. Lead washers
24 pcs. 90 m. x 2.40 m. Plain G .l. sheets
36 kgs . 4 d, 38 mm. 32 mm. CW Nail

-
TABLE 6 5 COMBINATION OF CORRUGATED G.L ROOF SHEETS
ON A GIVEN RAFTER LENGTH

Rafter No. of Combination of Roof Sheet Length

Length Sheets Meters (Feet)

3.00 1 3.00 (10‘)

3.30 1 3.30 (11‘)
3.60 1 3.60 (12* )
3.90 2 2.10 (7‘) and 2.10 (7' )
4.20 2 2.40 (8‘) and 2.10 ( 7' )
4.50 2 2.40 (8' ) and 2.40 (8‘)
4.80 2 3.00 (10' ) and 2.10 (7' )
5.10 2 3.00 (10') and 2.40 (S' )
5.40 2 3.00 (10* ) and 2.70 (9‘)
5.70 2 3.00 (10' ) and 3.00 (10' )
6.00 2 3.60 (12* ) and <
2.70 9‘)
6.30 2 3.60 (12‘) and 3.00 (10‘ )
6.60 2 3.60 (12' ) and 3.30 (11* )
6.90 2 3.60 (12' ) and 3.60 (12* )
7.20 3 3.00 (10' ) and 2.40 ( 8' ) and 2.40 (8' )
7.50 3 3.00 (10‘) and 3.00 (10’ ) and 2.10 (7' )
7.80 3 3.00 (10*) and 3.00 (10' ) and 2.40 (8' )
8.10 3 3.60 (12' ) and 3.00 (10' ) and 2.10 ( 7‘ )
8.40 3 3.60 (12' ) and 3.00 (10' ) and 2.40 ( 8‘)
8.70 3 3.60 (12' ) and 3.60 (12‘ ) and 2.10 ( 7* )
9.00 3 3.60 (12' ) and 3.60 (12' ) and 2.40 ( 8' )

221
 Comments:

The estimating procedure for a hipped-roof is the same as that

of the lean-to or gable type roofing, considering the effective
covering of one sheet to be constant . However, a little variation
might occur in actual construction under the following conditions:

°
1. If the hipped roof is not patterned at 45 extra cutting of the
GJ . roofing is inevitable .
2 . Errors might be committed in cutting and /or lapping of the
corrugated G .l . sheets .

Under these circumtances, an allowance of 5% to 10% will be

satisfactory.

6-2 GUTTER , FLASHING , RIDGE, HIPPED AND

VALLEY ROLL

Estimating these kind of roof- accessories is simply

determining the number of plain G. l . sheets needed to fabricate the
gutter , flashing and the different rolls according to the shape as
shown in the plan. In building construction , this is categorized
under the tinsmithing work.

The primary consideration in tinsmithing job is economy which

simply mean "To utilize every inch of the tin sheet ’'. As much as
possible , any unnecessary or unwise cutting of the tin sheet
should be avoided . Cutting shall start from the widest to the
narrowest part of roof accessories. Experienced tinsmith start
cutting from the gutter , then the flashing down to the smallest plain
G .[Link] thereby avoiding waste .

222
Estimating procedure:

1. Determine the total length of the gutter.

2. Divide this length by 2.35 m. to find the number of gutter

required. ( 2.35 is the effective length of one gutter )

3. Find the total width of one gutter (see the detailed plan)
,

4. Divide .90 m. (36 inches) width of one plain G .i. sheets by

the result of step 3 to find how many gutter could be made
out from one plain G .I. sheet. The fractional value or extra
cut shall be reserved for other smaller parts.

5. Divide the result of step 2 by the result of step 4. The result

is the required number of plain G.I sheets.

34.50 m

m
E
S
27 00 m

as
(N
00
2.5 cm
1
7.50 m .

Cross Section of Gutter

FIGURE 6-7

223
 ILLUSTRATION 6- 5

Reproducing Figure 6-6 of illustration 6-4 , find the number of

plain G .l . sheets required to fabricate the gutter as illustrated.

SOLUTION

1 . Find the total length of the roof gutter.

34.50 + 27.00 + 2(7.50) + 18 + 25.50 = 120 meters.

2. Divide the length by 2.35 (effective length of one gutter «

8‘)
120 = 51 pcs .
2.35

3. The total width of one gutter is 55 cm. (see figure);

Subtract :

90 cm. - 55 cm. = 35 cm. extra cut

This extra cut of 35 cm. from one plain G.l . sheet could be
set aside momentarily to be considered in making other roof
accessories. Thus, only one gutter at 2.40 m. ( 8' ) long could
be made out from one sheet

4 . Therefore , order 51 pcs. plain G .l . sheet for fabrication of

gutters.

Comment

It will be noted that the standard commercial width of one plain

224
G .I . sheet is 90 cm . ( 36” ). The total width of one gutter is 55 cm. ,
subtracting 55 from 90 will result to an extra cut of 35 cm . This
simply mean that only one gutter could be taken from one plain G.I.
sheet with an excess cut of 35 centimeters .

The 35 cm . excess does not necessarily mean to be

considered as waste because there are several parts in the
tinsmithing work such as flashing, downspout and straps that only
requires smaller cut or dimensions.

ILLUSTRATION 6-6

From Figure 6-8 , find the number of plain G.I. sheets required
for gutter and flashing.

4.50 m .

22 5 cm. „
Ijr
5 cm .
2.5 cm.
115 cm.
7.5 cm.

Flashing Gutter
FIGURE 6-8

225
 SOLUTION

A .) Gutter

1 . Find the total length of the roof gutter . From figure 6-8 the
total length is = 20 meters.

2. Divide this length by 2.35 m. effective length of one gutter.

20_ = 8.5 say 9 pcs. gutter at 2.40 m. long .

2.35
3. Determine the total width of one gutter . From figure 6-8
the total width is = 45 cm.

4 . Divide the width of the plain G. l . sheet by the width of one

gutter .

JK) = 2 pcs. Gutter per sheet

.45

5 . Divide the result of step 2 by the result of step 4

8.5 = 4.25 say 5 pcs. plain sheet .

2

B.) Flashing

1 . Determine the total length of the flashing. From figure 6- 8

length is « 18 meters.

2. Find the number of flashing

226
 18 m. * 7.8 say 8 pcs.
2.30 m.

3 . Width of the flashing is = 45 cm.

Divide:

.90 m. = 2 pcs. per G.l. sheet

.45

4 . Divide the result of 2 by step 3

8.0 = 4 pcs. plain G .l . sheet .

2

Summary

For Gutter: 5 pcs. 90 cm. x 2.40 m. plain G.l. sheets

For Flashing: 4 pcs. 90 cm. x 2.40 m. plain G .l . sheets
Total 9 pcs. 90 cm. x 2.40 m. plain G .l. sheets

TABLE 6-6 ROOF ACCESSORIES

Item Effective Length in Meter

Gutter 2.35
Flashing 2.30
Ridge Roll 2.20
Valley Roll 2.35
Hipped Roll 2.20
Soldering Lead 1/4 bar ( .25 ) per Solder Joint
Muriatic Acid 10 cc per Soldering Lead

227
 C.) Ridge, Valley and Hipped Rolls
The estimating procedure for this type of roof accessories is
the same as that of the gutter and the flashing .

TABLE 6-7 CORRUGATED PLASTIC ROOFING SHEET

Commercial Size Effective Width Covering

Corrugation
In . / ft Meter 1 1/2 2 1/2

26" x 8 ' . 650 X 2.40 . 46 m .31 m.

29" x 8 ' . 725 x 2.40 . 53 m. . 38 m .

6-3 ASBESTOS ROOFING

Unlike galvanized iron roof sheet , estimating the asbestos

roofing material is simpler because all the roof accessories and
parts to be used such as gutter , ridge , hip and valley rolls are all
factory made for installation . Unlike galvanized iron roofing where
all the accessories are made on the site out from the standard
dimension of plain G.l. sheets.

Estimating procedure for asbestos roofing is

enumerated as follows:

1 . The number of corrugated sheet is determined by dividing

the gutter length by the effective width covering of one
sheet .

228
 2. In finding the number of flashing , gutter , ridge , hip and valley
rolls, divide the total length by the effective length of the
accessories.

3. Other parts such as ridge, end cap, apron flashing , gutter

corner, downspout and fittings are determined by direct
counting, they are all readily made according to factory
standard sizes.

ILLUSTRATION 6-7

From Figure 6-9 , Find the standard asbestos roofing sheets

and accessories required.

FIGURE 6 9-
SOLUTION

A . ) Corrugated Sheets

1. Find the number of corrugated sheets. Refer to Table 6-8;

Divide;

229
 26.00 * 31.02
“
838

2 Length of rafter is 3.00 m x 2 sides .

Order .
62 pcs. 3.00 m standard asbestos sheets

B.) Gutter

1. Divide the total length of roof gutter by the effective length

of one gutter. Referring to Table 6-8;

26 x 2 3 52.00 meters total length

52.00 = 22 26 pcs
2 336 m.

2. Total length of ridge roll - 26.00 m Referring to Table 6-8;

Divide:
26.00 = 31.02 pcs. ridge roll
.838

C.) Flashing

1. Find the total length of the flashing (see figure)

3.00 x 4 sides = 12.00 m.

2 Refemng to Table 6-8

Divide
12.00 = 5.249 pcs.
2.286 m.

230
D,) Ridge End Cap - By actual counting

DIFFERENT KINDS OF ASBESTOS ROOFING

1. Standard Corrugated Sheet

2 . 4-V Corrugated Sheet
3. Kanaletas
4. Placa Romana
5. Tencor Corrugated Sheet
6. Ardex lightweight corrugated sheet
a) Standard Ardex
b) Super Ardex

TECHNICAL DATA FOR ESTIMATING PURPOSES

TABLE 6-8 STANDARD CORRUGATED SHEETS

Length 1.20 x 3.00 m..

End lapping: Below 20 degrees .30 m.
Above 20 degrees .15 m.
Effective width .838 m.
Ridge Roll effective length .838 m.
Gutter effective length 2.336 m.
Outside Flashing 2.286 m.
Hip Roll 1.676 m.

.838 m.
Lap

.98 m.

STANDARD CORRUGATED SHEET

FIGURE 6-10

231
 TABLE 6-9 CORRUGATED SHEETS TECHNICAL DATA

Standard Length 2.438 m.

Effective width .965 m.

Ridge Roll effective length . 965 m.

Outside Flashing effective length 2.286 m.

965 m
Lap

v/WVM
[Link]) m.

CORRUGATED SHEET
FIGURE 6-11

.177 m .177 m.

Lap .244 m
.058 m

KANALETAS
FIGURE 6- 12

232
 TABLE 6*10 KANALETAS
Items Length in Meter
Length 7.315
Effective Width . 885
Eaves Flashing . 885

Outside Flashing effective length 2.40 to 3.00

Lap

1.18 m.

PLACA ROMANA
FIGURE 6-13

-
TABLE 6 11 PLACA ROMANA

Item Length in Meter

Standard Length 812

Effective Length . 600
Standard Width 1.180
Effective Wdth 1 100
End Lap 200
Side Lap 080
Ridge Roll Effective Length 1 100
Outside Flashing Effective Length 2 286
Ridge Flashing Effective length 1 100
Eaves Flashing Effective length 1 100
Other accessories, estimate by direct counting

233
 .67 m Lap

.75 m.

TENCOR
FIGURE 6- 14

TABLE 6-12 TENCOR CORRUGATED SHEETS

Items Length in Meter
Standard Length 2.44
Lapping .15 or .30
Effective Length 2.29 or 2.14
Standard Width .748
Effective Width .675
Outside Flashing 1.50 to 3.00
Minus Lapping . 15
Ridge Roll .953
Other accessories , estimate by direct count .

.45 m. .07 m.

rVWWNA;
Lap

ARDEX
FIGURE 6 15 -
234
 TABLE 6-13 ARDEX CORRUGATED SHEET
Items Measurement
Standard Super

Standard Width 52 cm. 105 cm.

Effective Width 45 cm. 97 cm.
Nominal Length 75 to 315 cm. 240 to 360 cm.
Ridge Roll Eff. Length 95 cm. 95 cm.
Outside Flashing 1.50 - 200 cm. 150 to 300 cm.
Side Lapping 15 cm. 15 cm.

6-3 COLORBOND KLIP LOK -

Colorbond is a corrosion resistant zinc coated steel sheet
prepainted steel ribbed tray roofing and walling with the following
special features:

27 mm
Mean Rib Width
41 mm

Female Rib

203 mm 203 mm
m
Male Rib

1b
406 mm Coverage
( 427 mm| Overall Width

FIGURE 6-16

235
 1. Concealed fastening
2. Lock action rib design
3. Attractive flutted trays
4 . Near flat roof slopes
5. Less supports- wider spaced
6. Strong lightweight steel
7. Custom cut long lengths

Technical Data:

Steel base thickness = 0.60 mm.

Total Coated thickness * 0.63 mm.
Weight per meter length of panel = 2.66 kg.
Weight per covered area = 6.55 kg/m2
Length available up to - 15 m.
Longer length through special
order up to - 35 m.
Overall width = .427 m.
Effective Width Coverage = .406 m.

-
TABLE 6 14 RECOMMENDED FASTENERS - (TWO FASTENERS
REQUIRED PER CLIP)

Support Member Normal Fastening Fastening over Insulation

up to 100 mm (4")

Steel up to 3/32" No . 10-16 x 5/8" (16 mm) No . 10-16 x 7/8 (22 mm)
M

(2.5 mm) thick wafer head self drilling wafer head self drilling
and tapping screw and tapping screw

Steel 3/32" to 3/ 16" No . 10-24 x 5/8" ( 16 mm ) No . 10- 24 x 7/8" (22 mm)

-
(2.5 5 mm) thick wafer head self drilling wafer head self drilling
and tapping screw and tapping screw

236
Steel over 3/1 * 6" No 10- 24 x 5/8" {16 mm) Pre-drill 11/64" (4.5 mm)
(5mm) thick wafer head thread cutting hole for No. 10-24 x 7/8"
screw . Drill 11/64" (4.5mm) ( 22 mm) wafer head

self drilling and tapping

screw
Hardwood 2“ x 9G(50 x 3.75 mm) 2 1/2" x 9G (60 x 3.75 mm)
counter sunk head G.l. counter sunk head G.l.
spiral nail, or No. 10-12 spiral nail or No. 10-12 x
x 1" (25 mm) type 17 1 3/4"{ 45 mm) type 17
wafer head self drilling wafer head self drilling
wood screw. wood screw
Softwood No . 10-12 x 1 3/4" (45 mm) No . 10- 12 x 1 3/4" (45 mm)
type 17 wafer head self type 17 wafer head self
drilling wood screw. drilling wood screw

6-4 BANAWE HORIZONTAL METAL TILE

L*’ Mtn. Roof Inclination

Cross Section of Banawe Metal Tile

FIGURE 6-17

237
 Technical Data

Nominal Width 228 m.

Effective Width Coverage . 204 m
Length 12.19 m.
Longer Length Special Order
Minimum Roof Slope 15°

6-5 MARCELO ROOFING SYSTEM

1.35 m. /

FIGURE 6-18

Technical Data:
Width 1.14 m
Length 1.11 m.
Effective Width Coverage . 95 m
Effective Length . 96 m.
Effective Area Coverage per Sheet . 92 sq m.
Number of Fastener per Sheet
First row 15 pcs. per sheet

238
 Succeeding Rows 10 pcs. per sheet
Average No. of fasteners
per sheet 12 DCS. per sheet

6-6 COLORBOND CUSTOM ORB

Crest Fastening to Steel

I CUSTOM ORB

Crest Fastening to Wood

FIGURE 6 19 -
Technical Data

Nominal Width . 80 m.
Efective Coverage 76 m.
,

Length: 1.35 m.
Longer Length Special Order
Maximum recommended
length for continuous sheet
without expansion joints 24 m.

239
 6-7 MILANO LONGSPAN STEEL BRICKS

30 cm
38 cm -
i3 cfC
rT tiffin
—.

A = 5 cm to 43 cm dependingupon the length of the top tile

Fastened to supports through the crest of the corrugations

Direction of A
sheet laying/

4-
1 1/2 corrugation sKle lap 2 1/2 corrugation side lap
* Nailing position for intermediate * Nailing position for ridge and gutter line
tile presses tile presses.

FIGURE 6 *20

Technical Data

Steei base thickness 0.40 mm (No. 26)

Total coated thickness 0.46 mm
Weight per sq m. 4.53 kg.
Weight per length 3.44 kg.
Effective coverage .67 m.
Length up to 6.00 m.
Longer length Special Order
Recommended roof slope 10° min.

-
6 8 COLORBOND TRIMDEX HI -TEN

240
Notch and turn aown toga or
copping bat ween ribs. RIDGE CAPPING
Fatten at alternate rib . Fatten at alternate rib .
* Turn up tray between rib .
*
*

Turn up tray between rib .

* Notch and turn down edga of
TRANSVERSE FASCIA CAPPING
*.
capping between rib

1 1/8 in
( 29 mm )

I . +
S 1/8 in
( 130 m ti)

30 in (760 mm) COVER

2 3/ 8 in
(60 mm)

*
FIGURE 6-21
Technical Data
Steel base thickness 0.40 m.
Total coated thickness 0.46 m.
Weight:
Per unit area 4.28 kg./ sq. m.
Per unit length 3.26 kg / sq. m.
Normal width 0.83 m.
Effective width coverage 0.76 m.
Available length up to 15 m.
Lenger length Special Order
Minimum slope:
Single sheet 3°
Roof with end lap 5°

241
 Fasteners:
1 For Hardwood : Use Type 17 self drilling wood screw No. 12 x 50
mm hexagonal head with neoprene washer.
2 For Soft Wood Add 12 mm to length of screw.
3 There should be four fasteners per sheet at all supports.
4 For side lap fasteners, use type S point self drilling screw No.10 x
16 mm hexagonal head with neoprene washers.
5 Teks self drilling screw to steel supporters up to 4.5 mm thick use
No 12 x 45 mm hexagonal head with neoprene washer .

6- 9 BRICK TILES ROOFING

Technical Data:
Weight per
Description Piece in kg Total Number Required
Marceille Type(flat) 3.60 14.0 per sq. m.
Ondula Type (wavy ) 400 15.0 per sq. m.
Sr. (Standard ridge ) 280 2.5 per In. m.
Half Marceille 200 1.0 per In. m.
JRT- 1 3.00 13.0 per sq. m.
JRT- 2 3.75 4.0 persq. m.
JRT- 3 3.20 1.0 per In m
JRT -4 3 20 1.0 per In. m.
JRT- 5 220 4.0 per In m.
SRT- 1 400 13.0 persq . m.
SRT- 2 2.20 1.0 per In m.
SRT- L 2.25 1.0 per In. m
SRT- R 2.25 10 per In. m
JSR- 1 2.00 10 per In. m.
JSR -2 3.00 4.0 per In. m.
SRT- U 1 75 25.0 per sq. m.
SRT- T 1.50 25.0 per sq. m.

242
 CHAPTER 7
TILEWORK
7-1 CERAMIC TILES

Ceramic tiles is one of man’s oldest building material

continuously in use due to its unique , functional and decorative
properties. Ceramic tiles offer an almost unlimited choice of
patterns and colors which does not fade and is practically
indestructible.

Decorative ceramic tiles were widely used during the period of

the Medieval Islamic Architecture from Persia to Spain and
extended up to the period of the contemporary Architecture.

Ceramic tiles are classified as:

1 . Glazed Tiles are principally used for walls and light duty
floors.

2. Unglazed tiles are hard, dense and homogeneous

composition, primarily used for floors and walls.

Various Types of Tiles

1. Porcelain tiles are made from the pressed dust processed

243
 into fine grain, smooth, dense and shapely formed face,

.
2 Natural clay tiles are made from either the pressed method
or the plastic method from dust clay that produce a dense
body with distinctive slightly textured appearance.

3. Ceramic mosaic tiles are mounted on a 30 cm. x 30 cm.

paper as binder of the tiles to facilitate its laying or setting.

4. Quarry tiles are made through the plastic extraction

process from natural clay or shale.

5. Faience mosaic Tiles are tiles less than 15 square

centimeters in facial form.

.
6 Special purpose ceramic tiles :

a) Nonslip tiles
b) Ship or Galley
c) Frost proof tiles
d) Conductive tiles

10 x 20 Glazed Tiles 10 x 20 Glazed Tllea

Zoo 1.50 m.

20 x 20 Floor Tilee

FIGURE 7-1

244
 ILLUSTRATION 7-1

From Figure 7- 1 , determine the quantity of the following

materials:
a) 10 x 20 cm. Glazed wall tiles
b) 20 x 20 cm. Unglazed floor tiles
c) Cement and sand mortar
d) White cement paste filler

SOLUTION- 1 ( By Fundamental Method )

1. Solve for the wall area .

A = 1.50 x { 5.00 + 3.00 )

A = 1.50 x 8
* 12 sq. m.

2. Solve for the wall glazed tiles. Divide the wall area by the
area of one tile .

12 sq. m. = 12
.10 x .20 .02

= 600 pcs 10 x 20 cm (4" x 8") glazed tiles.

3. Solve for the floor tiles. Floor area divided by area of one
tile.

A = 5.00 x 3.00 floor area

a (.20 x .20) area of one tile

245
 = 15
.04

= 375 pcs. 20 x 20 cm. floor unglazed tiles

Total Area - 1 2 + 1 5 = 27 sq. m. x .076 = 2.0 bags

White Cement joint filler 27 sq. m. x .50 - 13.5 say 14 kgs.

5. For breakage allowance , 5 to 10% is satisfactory .

TABLE 7-1 QUANTITY OF CERAMIC TILES PER SQUARE METER

Classifications Size Number of Pieces Per

in. cm. Square Ft Square M.

Mosaic Tiles 12 x 12 30 x 30 1.0 1076

Glazed and 3x3 7.5 x 7.5 16 177.8

Unglazed 4x4 10 x 10 9 100.0
Tiles 4 1/4 x 4 1 /4 10.62 x 10.62 7.97 85.7
4x8 10 x 20 4.5 50.0
6x6 15 x 15 4 44.5
8x8 20 x 20 2.25 25.0
12 x 12 30 x 30 1.0 10.76

5 Point Hex. Tiles 2 5 61 / ft 20 / m.

4 Point Hex Tiles 2 5 4.9 / ft 16 / m.

External Comer Bead - Direct count

Internal Comer Bead - Direct count
Portland cement paste mortar - . 076 bags per sq. m
White Cement Filler paste - . 5 klg per sq. m.

246
SOLUTION -2 (By the Area Method )

1 Solve for the wall area

A = 1.50 m. x 8.00 m.
- 12 sq. m.

2. Refer to Table 7-1, ysing a 10 x 20 cm. glazed tiles;

Multiply:

12 sq. m. x 50 pcs. per sq. m.

600 pcs.

3. Solve for the 20 x 20 cm. floor tiles. Referring to Table 7-1

Multiply:

Floor area = 5.00 x 3.00 m.

- 15 sq. m.
15 sq. m. x 25 pcs. * 375 pcs.

4. Solve for cement mortar and paste filler.

Total Area :
12 + 15 = 27 sq. m.

5. Referring to Table 7-1:

Multiply:

Cement mortar 27 sq. m. x .076 = 2.0 bags

White Cement 27 sq. m. x .50 = 13.5 say 14 kgs.

5. For breakage allowance , 5 to 10% is satisfactory .

247
 ILLUSTRATION 7-2

From Figure 7-2, determine the quantity of the following

materials:

a. 12" x 12" mosaic floor tiles

b. 4" x 4" wall glazed tiles
c. internal bead and capping
d. Internal comer bead
e. External corner bead
f. Ordinary Portland cement
g. White cement

1.50 m

FIGURE 7-2

248
 SOLUTION

A.) Mosaic Tile Flooring

1. Solve for the floor area

1.50 x 2.00 * 3.0 sq. m.

2. Referring to Table 7 1 using 12" x 12” mosaic floor tiles;

*

Multiply:

3.0 sq. m x 10.76 * 32.3 say 33 pcs.

B.) Glazed Tiles for Walls

1. Solve for the lateral area of the wall.

2(2.00) + (1.50 + .75) x 1.50 m. ht

{ 4 + 2.25 ) x 1.50 m.
* 9.37 say 9.4 sq. m.

2. Referring to Table 7-1, using 4 " x 4" glazed tiles:

Multiply:

9.4 x 100 - 940 pos.

3 Add 2% to 5% allowance for breakage

C. Internal Bead

1. Solve for the length or perimeter of inside corners.

249
 2(2.00) + (1.50 + 1.50) = 7.00 m.

2. Solve for the length of 4 vertical wall corners :

4(1.50 m. ht.) + 7.00 m.

6.00 + 7.00 = 13.00 m.

3. Divide this length by the length of internal bed or the tile.

13.00 m = 130 pcs. internal bead.

.10 m.

D.) Capping

1. Solve for the perimeter of the wall tiles

2(2.00) + 150 + .75 = 6.25 m.

2. Add capping along door jamb : 1.50 x 2 * 3.00 m.

Total * 9.25 m

3. Divide by one capping or tile length.

9.25 « 92.50 say 93 pcs.

.10

4 . Add 1 to 2% allowance for breakage.

E.) Internal Corner Bead

By direct counting there are 4 comers , order 4 pcs .

250
F.) External Corner Bead

By direct counting, order 4 pcs.

G.) Ordinary and White Portland Cement for Paste

1 . Total floor and wall area * 3.0 + 9.4 » 12.4 sq. m.

2. Solve for the ordinary cement for paste.

Referring to Table 7-1, multiply:

12.4 x . 076 = .94 say 10 bag cement

3. Solve for the white cement filler.

Referring To Table 7-1, multiply:

12.4 x .5 kg. per sq. m. = 6.2 say 7 kgs. white cement.

Summary

33 pcs. 12" x 12” (30 x 30) mosaic tiles

940 pcs. 4" x 4 " (10 x 10) glazed tiles
130 pcs. 4" ( 10 cm.) Internal Bead
93 pcs. 4” ( 10 cm.) capping
4 pcs. 4" ( 10 cm.) Internal comer bead
4 pcs. 4” ( 10 cm.) External corner bead
1 bag at 40 kg. cement
7 kg. White cement

251
7-2 MARBLE TILES

Marble is a hard metamorphic limestone white or colored ana

sometimes streaked or mottled in crystalline or granular state and
capable of taking high polish. It is used in sculpture , furnitures
topping for slabs and floor etc .
Marble as construction materials have been extensively used
from the ancient time of the Geek to the Roman Empire down to
the modern and contemporary Architecture.

TABLE 7- 2 QUANTITY OF MARBLE TILES AND MORTAR PER

SQUARE METER *

Size Number Cement Bags Sand

cm. per sq. m. Mixture
A B C cu . m.

15 x 30 ; 22.3 .45 . 30 .225 .025

20 X 20 25.0 .45 . 30 .225 .025
20 x 40 12.5 .45 . 30 .225 .025
30 x 30 11.1 I .45 . 30 .225 .025
30 x 60 5.6 | .45 . 30 .225 .025
40 x 40 6.3 . 45 . 30 . 225 . 025
60 x 60 2.8 . 45 .30 , 225 .025

* Cement mortar computed at an average thickness of 2 5 centimeters.

ILLUSTRATION 7 -3

From Figure 7-3 , solve for the number of 30 x 60 marble tiles

required including the cement and sand for class B mortar .

252
 1.00 m.
6 00 m

4.00 m.

- 12 00 m
I
Floor Plan

FIGURE 7-3

SOLUTION:

1 . Solve for the floor area .

A * 12.00 x 5.00
A = 60 sq. m.

2. Referring to Table 7-2 for a 30 x 60 marble tiles class B

mortar mixture; Multiply:

60 sq. m. x 5.6 pcs. * 336 pcs.

3 . Cement at 40 kg. per bag using class B mixture , from Table

7-2; Multiply:

60 x .30 = 18 bags

4. Sand: 60 x .025 = 1.5 say 2 cu. m.

?53
 5 . Add 5% for cutting allowance and breakage .

Summary

336 + 16 (allowance) - 352 pcs . 30 x 60 marble tiles

18 bags cement at 40 kg.
2 cu. m. sand

Polymir Liquid .035 gal. per sq. m.

Hardener .030 quarts per sq. m.
Calsomine Powder 020 kg. per sq. m.

7-3 VINYL AND RUBBER TILES

The standard specifications for vinyl and rubber tiles provides

that:
"It shall be non-fading, odorless and nonslip even when wet
and shall be strong enough to withstand the ordinary tear and wear,
cleaning and moving of furnitures without damage and shall be self-
dealing

Tiles shall be laid to conform with the manufacturer’s

specifications which partly states that

a) Adhesive cement shall be applied to the floor every after the

tiles are laid on the surface.
b ) Tiles are pressed with linoleum roller to avoid blisters.
c) After completion, aH work shall be cleaned of cement , dirt
and other substances,
d) Apply two coats wax and poGsh to smooth shiny finish.

254
 TABLE 7 *3 VINYL AND RUBBER TILES

Stock Size Number Per Gallons of Adhesive

3 mm thick Sq. M. per Sq. M.

. 20 x 20 25.00 .042
. 225 x .225 19.75 . 042
.25 x .25 16.00 . 042
. 30 x . 30 11.11 . 042
.40 x 40 6.25 . 042
.60 x .60 2.78 . 042

ILLUSTRATION 7-4

An office room with a general dimensions of 7.00 m. x 9.00 m

is undergoing renovation. Determine the number of 30 cm. square
vinyl tiles required including its adhesive for installation.

SOLUTION

1. Solve for the floor area

7.00 x 9.00 * 63 sq. m.

2. Determine the number of 30 cm. vinyl tiles. Referring to

Table 7-3, multiply:

63 sq. m. x 11.11 - 700 pcs.

3. Finding the required adhesive cement, refer to Table 7-3;

Multiply :

255
 63 x .042 = 2.65 say 3 gallons .

7-4 TERRAZO AND GRANOLITHIC

Terrazo and granolithic are types of marble mosaic floor

finish that uses portland cement as base material. It has a
characteristics of durability , great beauty and variety . Installation
of this type of floor topping is done by either:

1. Monolithic or Cast-in Place

2. Pre-Cast

Monolithic or Cast-in Place - means massively , solid, single

and uniform floor finish cast in place . A mixture of cement and
marble chips to a proportion of 1: 3 is casted on top of a rough floor
slab surface to an average thickness of 1.25 cm The floor is then
grinded after it has attained sufficient hardness to withstand
abrasion and vibration caused by the grinding machine . Grinding
of the floor surface shall not be allowed earlier than 48 hours after
casting

Pre-Cast refers to granolithic tiles in various dimensions

hydraulically pressed and molded in a factory . The distinctive
difference between the cast in place and the pre cast installation
is the manner and place of casting or molding. The former being
installed on site and the latter at the factory site . Thus , pre cast is
installed in a tile form while cast in place is installed in a fresh mixed
form.
Normally , a dividing hard brass strips with alloy zinc are
installed in between tiles to control and localize any shrinkage or
flexure cracks. The dividing strip thickness ranges from 1.56 mm.
to 3.12 mm. or thicker depending upon the design .

256
 Monolithic or Cast-in Place Estimating Procedure

1 . Solve for the total floor area in square meters .

2. Multiply the area by . 172 to get the number of bags of white
or colored cement required.
3. Multiply the floor area by 12.5 to get the weight of the marble
chips in kilogram.
4. Multiply the quantity found by the unit cost.

ILLUSTRATION 7 5 -
For an 8.00 m. by 10.00 m. room that specify cast 4 n place
granolithic floor, list down the materials required .

pp
Granolithic Floor 8 00 m
•V

ii
10.00 m.

Floor Plan

FIGURE 7-4

SOLUTION

1. Find the floor area

257
 8.00 x 10.00 m. * 80 sq. m.

2. Determine the quantity of white cement required.

80 x .172 * 13.76 say 14 bags.

3. Determine the quantity of marble chips required.

80 x 12.5 * 1,000 kg.

-
TABLE 7 4 TERRAZO AND GRANOLITHIC FLOORING

Cement bags Sand

stock Pieces per sq. m. cu. m. Brass Divider
Size per Mixture Class sq. m. meter / sq. in-
m. sq. m. A B

.20 x .20 25.00 .338 .225 .0188 10.8

.225 x .225 1&.75 .338 .225 .0188 10.0
.25 X .25 16.00 .338 .225 .0188 8.9
.30 x .30 11.11 .338 .225 . 0188 8.0
.35 X .35 8.16 .338 .225 . 0188 6.0
.40 x .40 6.25 .338 .225 .0188 5.8

Pre Cast Installation

ILLUSTRATION 7-6

A room with a general floor dimensions of 10.00 m. x 20.00

meters specify a 40 x 40 cm. granolithic floor tiles. Using class
6 mortar base, determine the quantity of 40 cm. square tiles,
cement, sand, and brass divider required.

258
 40 x 40 cm Tiles

E 40 x 40 cm Tiles
i:
§ -
i
o

20.00 m

Floor Plan Brass Divider

FIGURE 7- 5

SOLUTION

1 . Find the total floor area .

10 m. x 20 m. = 200 sq. meters

2. Solve for the number of tiles required. Referring to Table 7-4

Multiply:

200 x 6.25 = 1,250 pcs. 40 x 40 cm. tiles.

3 . Using class B mortar , solve for cement. From Table 7-4

Multiply:

200 x 225 = 45 bags cement

4 Determine the quantity of sand. From Table 7-4 ;

Multiply:

200 x 0188 = 3.7 say 4 cubic meters.

259
 5. Determine the quantity of brass divider in between tiles.
Referring to Table 7-4 , multiply :

200 x 5.8 ~ 1 ,160 meters

7-5 CEMENT TILES

Cement tiles are hydraulic pressed, locally manufactured in

the following commercial sizes.
25 mm x 15 cm. x 15 cm. ( 1" x 6" x 6" )
25 mm x 20 cm. x 20 cm. ( 1 x 8" x 8” )
H

25 mm x 25 cm. x 25 cm. ( 1” x 10“ x 10" )

25 mm x 30 cm. x 30 cm. ( T' x 12” x 12' )
1

Estimating the number of cement tile required includes the

mortar assumed at an average thickness of 20 mm. {3/4”). The
methods applied in estimating cement tile work is either :
1. The unit measure method
2. Square meter area method

7.00 m.
20 x 20 cm. Cement Tiles

m
9.00 m,

Floor Plan
3 Cement Tiles

FIGURE 7-6

260
 ILLUSTRATION 7- 7

Find the number of 20 m x 20 m. cement tiles required for

a school classroom with a general dimensions of 7.00 m. x 9.00 m
using class B mortar mixture .

SOLUTION - I ( By ihc Unit Measure Method )

1 . Divide the length and width of the classroom by the width

of one tile .

7.00 = 35 pcs .
20

9.00 - 45 pcs.
20

2. Multiply :

35 pcs. x 45 pcs. * 1 ,575 pcs .

3 . Solve for Cement Mortar Referring to Table 7-4

Multiply :

7 00 x 9 00 x 0188 * 1.18 cu. m.

4 . Solve for the quantity of cement and sand required

mortar Referring to Table 2- 1 , multiply:

Cement : 1.18 x 12 = 14.16 say 15 bags.

Sand : 1.18 x 1.0 » 1.18 cu. m.

261
 SOLUTION -2 ( By the Square Meter Area Method)

1. Sotve for the floor area

7.00 x 9.00 = 63 sq . m.

2. Solve for the number of tiles .Referring to Table 7- 4

Multiply :

63 x 25 s 1 ,575 pcs.

3. Using class B mortar mixture , refer to Table 7- 4

Multiply:

Cement : 63 x .225 = 14.17 bags

Sand : 63 x .0188 * 1.18 cu. m.

ILLUSTRATION 7- 8

From Figure 7-7, find the number of 25 cm. x 25 cm. cement

tiles including the mortar using class "A" mixture.

15.00 m. A B 8 00 m

— - 8.00 m 18.00 m

FIGURE 7*7

262
 SOLUTION

1. Find the total area of A and B

A - 15.00 x 8.00 = 120 sq. m.

B = 18.00 x 8.00 = 144 sq. m.
Total Area = 264 sq. m.

2. Solve for the number of 25 cm. tiles. Referring to Table 7- 3

Multiply:

264 x 16 — 4 ,224 pcs

3. Solve for the mortar materials: From Table 7- 4;

Multiply

Cement : 264 x .338 = 89.20 say 83 bags

Sand : 264 x .0188 = 4.96 say 5 cu. m.

7-6 WOOD TILES

Wood tiles is a combination of wood pieces in various

dimensions with thickness that ranges from 6 mm to 8mm. Wood
tiles are carefully laid one at a time on smooth surface concrete
floor slab applied with good kind of white glue . The tiles is then
grinded with No.300 and 100 sand paper 24 hours after setting to
produce a fine and smooth even surface. Sandpaper dust is mixed
with wood glue and used as filler of tile joints.

Estimating Wood Tile is no more than the following simple

steps:

263
 1 . Solve for the net floor area to be covered with wood tiles in
square meter or square foot .

2. Multiply the area found in square meter by 10.76 to get the

equivalent in square foot. ( Wood tiles is sold in square foot
not In square meter.)

3. Multiply the floor area by . 165 to get the number of wood

glue in gallons per square meter .

ILLUSTRATION 7-9

An office room measuring 8.00 m. wide x 12.00 meters long

specify wood tiles flooring. List down the materials required.

SOLUTION

1. Find the area of the floor

A = 8.00 x 12.00
A = 96 sq. m.

2. Convert to feet ( wood tiles are purchased in sq. ft. )

96 x 10.76 = 1 ,033 sq. ft.

3. Add 5% allowance for cutting and edging = say 50 sq. ft.

4 . Order :
1, 033 + 50 = 1 , 083 sq . ft . wood tiles.

5 . Determine the white wood glue in gallons:

264
 ( Take note that 1- sq. m. floor area consumes . 165 gallons
of white wood glue. )

96 sq. m x . 165 * 15.84 say 16 gallons.

7-7 PEBBLES AND WASHOUT FINISHES

Pebbles are small roundish stone used for wails and floor
finishes called washout and pebbles respectively . The pebble
stone is mixed with pure cement to a proportion of either 1:2 or 1:3
mortar mixture then applied to a well prepared wall or flooring slab.
With the use of water sprinkler , the pebble mortar applied on the
wall or floor slab is then washed with water to a desired texture
before the concrete finally set . Twenty four hours later , the pebble
surface is then scrubed with steel brush and muriatic acid to obtain
the desired natural stone finishes.

ILLUSTRATION 7-10

A wall roughly plastered has a general dimensions of 2 meters

high and 10 meters wide requires stone pebble washout finish.
Prepare the bill of materials.

SOLUTION

1. Solve for the wall area

2.00 m. x 10.00 m. = 20 sq. m.

2. Determine the thickness of the stone pebbles say 12 mm or

.012 meter

265
3. Multiply by the wall area

20 x .012 = .24 cu. m.

4. Referring to Table 2-1, using Class "A" mixture:

Multiply:
.24 x 18 = 4.32 say 5 bags cement
.24 x 1.0 e .24 cu. m. Stone pebble.
SOLUTION -2

1. Solve for the wall area « 20 sq. m.

2. For Class A mixture multiply by .216

For Class B mixture multiply by .144
For Sand or Pebbles " .012
3. Using the above data, multiply:

20 sq. m. x .216 * 4.32 say 5 bags

20 sq. m. x .012 * .24 cu. m. stone pebbles.

266
 CHAPTER

HARDWARE
8- 1 BOLTS

Bolt ts a pin or rod with a head at one end threaded at the

other end to receive a nut.

The different kinds of bolts used in building construction

are:

1. Machine Bolt - Has a head at one end and a short thread

at the other end.

2. Countersunk Bolt - Has a beveled head fitting into a

countersunk hole.

3. Key Head Bolt - Has a head shaped end fitted to a groove

that will not turn when the nut is screwed into the other end.

4. Stud Bolt - A headless bolt threaded at both ends.

Size and Length of Bolts

267
 The length of bolt is the sum of the thickness of both pieces
being connected plus 12 mm.. With respect to the size or diameter
of bolt , depends upon the thickness of the object to be bolted.

a) Lumber up to 5 cm. thick , use . . 6 mm (1/4") diameter.

b) 7.5 cm. (3") thick lumber use . . 10 mm (3/8") diameter.
c) 10 cm. (4”) thick lumber use . . 12 mm (1/2") diameter.
d) Drill hole is 1.5 mm (1/16") larger than bolt diameter unless
snug fit is necessary.
e) Use washers under head and nut of machine bolt.
f) For carriage bolt use washer under nut only .
g) Use toggle bolt for attaching fixture to plaster wall.
h) Use expansion bolt for fastening to masonry.
i) For outdoor exposure, use brass or cadmium plated finish.

TABLE 8- 1 U.S. STANDARD THREAD OF BOLTS

Length D i a m e t e r of B o l t s
In. 1/2 5/8 3/4 7/8 1 1 1/8 1 1/4
12 16 20 22 25 28 31

10.0 1.44 2.45 3.64 524 7.23 9.76 12.60

12.5 1 69 2.85 4.21 6.01 824 11.06 14 18
15.0 1.94 3.24 4.78 678 9.26 12.33 15 76
17.5 2.19 364 5.35 7.55 10.27 13.61 17.35
20.0 2.45 4.03 5.92 832 11.29 14.89 18.93
225 270 4.43 6.49 9.09 12.30 16.17 20.51
25.0 2.95 482 7 06 9.86 1331 17.44 22.09
27 5 3.20 522 7.63 10,63 14.33 18.72 23.67
30.0 3.46 5.61 8.20 1140 15.34 20.00 25.26
32.5 3.71 6.01 8.77 12.17 1636 21 27 26.84
35.0 3.96 640 9.34 1294 17.37 22.55 28 42

268
 Machine Bolts

FIGURE 8- J

Bevel Head
Countersunk Head

Turned Oval Head

Bastard Head

Carriage Bolts

FIGURE 8-2

Nut End j Body Attachment Head

Stud Bolt
FIGURE 8-3

263
ffillHmPjJ
Railroad Track Boft ^ safi
Welded Eye Bolt

Plain and Shouldered Step Bolt

Forged Eye Bolt

Expansion Bolt

Various Types of Bolt

FIGURE 8- 4

270
 Estimating Procedure in Determining Length of the
Bolt
Post

E * I"
W.l. Post Strap

E Bolts

Footing
5 i
/ 1
ii
s>
FIGURE 8-5
WOODEN POST ANCHORED BY POST STRAP

The Length of the Bolt is equal to the width of the post plus
the two thickness of the post strap plus 20 mm allowance for the
thread and nut.
L = w + 13 mm + 20 mm.

Beam
Bolt
Washer
20 mm dap

FIGURE 8 6 -
POST AND SINGLE BEAM

The Length of Bolt is equal to the width of the post plus the
thickness of the beam .
L =w+t
271
 TABLE 8- 2 WEIGHT OF BOLTS WITH SQUARE HEADS AND
HEXAGONAL NUTS PER 10 BOLTS

Diameter 1/4 5 / 16 3/8 7/16 1/2 9/16 5/8 3/4 7/8 1

of Bolts 6 7 9 10 12 14 16 19 22 25
No of Thread 20 18 16 14 13 12 11 10 9 9
per Inch
Diameter of 5 6 8 9 11 12 13 18 19 22
Top Drill 13 1 5 23 27 15 17 32 3 55
64 4 16 64 64 64 32 64 4 64

Beam

20 mm dap

FIGURE 8-7

POST WITH TWO BEAMS OFTHE SAME THICKNESS

Length of the Bolt is equal to the width of the post plus 2

thickness of the beam minus 20 mm dap. ( There are two dap
opposite the column sides but only one is subtracted because the
other 20 mm dap is reserved for the thread that will receive the
nut . )

L = w + 21 - 20 mm

272
 Beam
Bolt
Bolt Washers

20 mm dap

FIGURE 8-8

POST AND TWO BEAMS OF DIFFERENT THICKNESS

Length of bolt is equal to the width of the post plus ti + t2

minus 20 mm.

£ IT T
II
TT
[I 5 cm.
5 cm.

JSr- 4r
~
liTSf S cm.
Wood Block
Machine Bolts

FIGURE 8-9
BOLTS ON TRUSSES

Length of bolt is equal to the thickness of the membei s in layer

plus 20 mm.

3 x 50 mm + 20 mm = 170 mm. or 17 cm

273
 Example: 3 x 50 mm = 150 mm + 20 mm
= 170 mm.

TABLE 8-3 POST ROUGH HARDWARE

Size of Host W.!. Post No. Bolt W. l. Splice No. Bolt
cm Strap (mm) Strap (mm)

10 X 10 6 x 45 x 600 2 12 6 x 37 x 600 4 12
13 x 13
15 x 15
18 x 18
20 X 20 6 x 50 x 600 2 16 6 x 50 x 600 4 16
23 x 23
25 x 25
28 x 28 6 X 62 x 600 2 20 6 x 62 x 900 4 20

8-2 SCREW

In carpentry work, screw is sometimes used instead of nails

due to the following advantages:

1. Greater holding power

2. Neat in appearance
3. Less chance of injuring the materials
4 . Ease of removal in case of repair

How to Choose Screw

1. Select one that is long enough wherein one half to two thirds

274
 of its length will enter the base in which threads are
embedded.

2 , The length of the screw should be 1/ 8" or 3mm less than

the combined thickness of the boards being joined.

3. Use fine thread screw for hard wood and coarse for soft
wood .

How to Use the Screw

1. Always drill lead hole for the screw .

2. Hole in top board should be slightly larger than the shank,
in second board slightly smaller than the threaded portion.
3. In soft wood , bore to depth half the length of the thread.
4 . In hard wood, bore nearly as deep as the length of the
screw .
5 . For lag screw , drill hole two-thirds its length then drive in
with hammer , and finally tighten with wrench.

How to buy screw

1 . Screws are classified by gauge (thickness) and [Link]

gauge has a variety of different lengths which maybe
obtained up to 12 cm. (5 inches)
2. When ordering screw , specify head shape (e.g., round
head) , finish (brass) , gauge (number 5) and length (2" to 5
cm.).
3. Square-headed lag screws come in diameters of 6 mm. to
25 mm (1/4 to 1inch) with its length from 4 cm. to 30 cm.
(1 1/2" to 12 inches.)

275
 Wood Screw - Is a screw nail with handed coarse thread to
give a grip.

Materials Used
1. Iron 3. Brass 5 . Bronze
2. Steel 4. Copper 6 . Aluminum

Shape of the Head

1. Flat 8. Headless
2. Round 9. Slotted (wood screw)
3. Fillister 10. Square (lag screw)
4. Oval 11 . Hexagonal
5. Winged 12. Clove
6 . Bung 13. Grooved
7. Punched

Shape of the Point

[Link]
2. Full length
3. Coarse

Duty
1. Wood (light duty)
2. Lag (heavy duty)

Finish
1. Bright 6 Bronzed
2. Blued 7. Coppered
3. Nickel Plated 8. Japanned
4 . Silver Plated 9. Lacquered
5. Brass 10 . Galvanized

276
 TABLE 8- 4 STANDARD WOOD SCREW AND NUMBER PER
KILOGRAMS

Inches 1/2 1 1 1/2 2 21/2 3 31/2 4 4 1/2 5 5

mm 12 25 37 50 62 75 87 100 112 125 150
Number 6,211 3,443 2 , 329 1, 779 1 , 414 1 , 166 1 ,126 910 739 655 515

Ordinary Lag S crew

Coach Lag Screw

Effective Length of Screw

Standard Wood Screw Point

FIGURE 8- 10

277
 Flat Head Round Head

Oval Head Counter Sunk Fillister Head

Felloe Closed Head

Hexagonal Round Bung Head

CO
Grooved Headless

Oowel

Pinched Head Square Bung Head

Winged Winged Winged Head

Various Types of Wood Screw

FIGURE 8-11

278
 The Three Shapes of Screw Point are:

1 . Gimlet point - used on wood and coach screws.

2 Diamond point - is used when more driving is done before

turning as in drive and tag screw.

3. Conical point - same as diamond point.

Gimlet Cone
Types of Wood Screw as to the Point
FIGURE 3- 11

8- 3 NAILS

The first handmade nails were used in the United States which
lasted up' to the end of the Colonial Period. In France, light nails
for carpenters were made by hand and hammer out of steel wire
as early as the days of Napoleon 1. In the United States, the wire
nail was first introduced by William Herser of New York in 1851.
Twenty five years later in 1876, Father Goebel introduced the
manufacture of wire nails and at the last part of the 18th century,
twenty three patents for nail making machine were approved in the
United States which was later introduced in England.

Kinds of Nail as to :

279
 1. Cross-Sectional Shape
a) Cut (rectangular)
b) Wire (circular)

2. Size
a) Tacks c) Brads e) Spike
b) Sprigs d) Nails

3. Materials
a) Steel b) Brass c) Copper

4. Finish
a) Plain c) Galvanized
b) Coated d) Blued

5. Service
a) Common c) Finishing
b) Flooring d) Roofing e) Boat , etc.

Types of Nail Points

UF Blunt Diamond Long Diamond Needle

¥ FT
Chisel Point Front Side
Sheared Bevel
Cut Nail

FIGURE 8-12

280
 I
2d
-c ==9
I
3d
1
=4
i 1
4d
J

=4 5d

- Ci
i

I
i

-“4 7
isfl
6d
d

31
t \
8d
i .iL
9d

*
j j =3 10d

12d
I l I

16d
i i < i
>
* *
2 1/2" 1 1/2"
31/2"
3" 2" 1 M

Length of Nails { Actual Size )

FIGURE 8- 13

2d to 60d For General Construction Common Nail

2d to 40d For Light Construction Household Use. Box
2d to 40d For Interior Trim Casing
2d to 20d For Cabinetwork Furnitures Finishing Nail
3/ 16" to 3" For Light Work Mouldings . Wire Brad
2d to 20d Flooring Construction . Cut
6" to 12" For Heavy Construction . . Spike

281
8 luiii '
2d to 60d . . . .Common Nail

8 2d to 40d. . . Box
2d to 40d. . .. Casing
6
* 2d to 20d. . . Finishing Nail
&
c* > 3/16" to 3" . Wire Brad*

. . Cut
8^1 2d to 20d

itfrr 6" to 12“ . . Spike

Types of Nails and Their Uses

FIGURE 8 14 -

Kil
Flat
Common
Large Flat
I
Large Flat
Reinforced
Wire Spike Oval

W
Round Round Offset Hook
Countersunk Countersunk

£1
Ud
Non Leak Cone Curved Cut Nail Brad Diamond
Barge Spike
Types of Nail Heads

FIGURE 8 15 -
282
 A) Tacks

Are small, sharp pointed nails with tapering sides and a thin flat
head. Tacks are nails chiefly used in fastening carpets and flashing
of any thin materials.

Tacks

FIGURE 8 16 -
-
TABLE 8 5 NUMBER OF WIRE TACKS PER KILOGRAM

Inches Length ( mm) Number per Kilogram

1 /8 3 35,200
3/16 5 23, 465
1/4 6 17,600
5/16 8 14,080
3/8 10 11, 732
7/16 12 8,800
9/16 14 5,865
5/8 16 4,400
11/16 17 10, 120
3/4 19 2.930
13/16 20.5 2, 514
7/8 22 1, 200
•

15/16 24 1, 953
1 25 1, 760
1 1/16 27 1,599
1 1/8 28 1, 465

283
 B) Sprigs

Are small headless nails sometimes called barbed dowel pins.

The regular size of sprigs ranges from 12 mm. to 5.0 cm. gauge
No. 8 wire or 4 mm. diameter.

C ) Brads

Are small slender nails with small deep heads. The common
variety is made in sizes from 2.5 cm. (2d) to 15 cm. (6d) in length
while the flooring brads- from 5 cm. to 10 cm. length.

TABLE 8-6 FLOOR BRADS TECHNICAL DATA

Size Length Gauge Dia. Head Approx. Gauge No. of Nails

No. Gauge No. / kg. No . per kg.

6d 50 11 6 322 12 14.500
7d 32 11 6 277 12 12.500
8d 37 10 5 200 11 9.000
9d 44 10 5 173 11 7,800
10d 75 9 4 131 10 5, 900
12d 81 6 3 95 9 4 , 300
16d 87 7 2 76 8 3,450
20d 100 6 1 57 7 2,600

D) Nails

Is a popular name for all kinds of nail except those extreme

sizes such as Tacks and Spikes. The most generally used are
called Common Nails of sizes from 2.5 cm. to 15 cm.

284
 TABLE 8-7 COMMON WIRE NAILS TECHNICAL DATA

Size Gauge Length Approximate Number

No. Inches mm Per Kilogram Per Keg

2d 15 1 1/4 25 1 , 831 82,400

3d 14 1 1/ 2 31 1 , 177 53, 000
4d 12.5 1 1/2 37 666 30 , 000
5d 12.5 1 3/4 44 580 26 , 100
6d 11.5 2 50 382 17,200
7d 11.5 21/4 56 344 15, 500
8d 10.25 2 1/2 63 208 9,400
9d 10.25 2 3 /4 69 188 8, 500
10d 9 3 75 138 6 , 250
12d 9 3 1/4 81 124 5 ,600
16d 8 3 1 /2 88 93 4 , 200
20d 6 4 100 58 2, 625
40d 4 5 112 45 2,040
50d 3 5 1/ 2 125 34 1,540
60d 2 6 150 20 910

Rat Head Diamone Point

4 ^ —tTritimwajftgfrCD
i

Oval Head Chisel Point

FIGURE 8-17

E) Spikes

An ordinary spike is a stout piece of metal from 7.5 cm. to 30

cm. in length, much thicker in proportion than a common nail.

285
 Spike is much used in attaching railroad rails, construction of
docks, piers and other work that uses large timber.

There are Two Kinds of Spikes, namely:

a) Flat Head, diamond point

b) Oval Head, chisel point

F) Boat Spikes

Are small kind of nail driven mostly in hard timber with a clear
cut sharp chisel point.

TABLE 8-8 COMMON BRADS TECHNICAL DATA

Size Length Approximate Number

Inches Per Kilogram Per Keg

2d 1 25 1, 904 85,700
3d 1 1/4 31 1,206 54.300
4d 1 1/2 37 662 29,600
5(t 1 3/4 44 566 25,500
6d 2 50 397 17,900
7d 21/4 56 340 15.300
8d 2 1/2 63 224 10,100
9d 2 3/4 69 197 8.900
10d 3 75 146 6,600
12d 33/4 81 137 6,200
16d 3 1/2 88 108 4.900
20d 4 100 68 3,100
30d 4 1 /2 112 53 2,400
40d 5 125 40 1,800
50d 5 1 /2 137 28 1 , 300

286
 TABLE H-') ORDINARY SPIKES TECHNICAL DATA

Size Length Approx . No.

Inches per Kilogram

10d 3 75 90
12d 3 1/4 81 83
16d 3 1/2 88 66
20d 4 100 50
30d 4 1 /2 112 37
40d 5 125 28
50d 5 1/2 137 22
60d 6 150 19
175 mm 7 175 15
200 mm 8 200 9
225 mm 9 225 8
250 mm 10 250 7
300 mm 12 300 6

Other Important Facts and Uses of Nails

1. Use a nail that is three times as long as the thickness of the board
2. Nails with sharp point holds better than blunt ones but tend to split
wood.
3. Flatten the point with hammer before driving into an easily split
wood
4. Use thinner nails for hardwood than for soft wood.
5. For weather resistance, use copper , aluminum or galvanized nails .
6. Zinc or cement coatings increase resistance to withdrawal
7. Barbed nails hold best in green wood.
8. Sizes of nails are indicated by "penny" abbreviated as " d '.
Example ; 20 penny nail is known as 20 d nail .

287
 -
TABLE 8 10 USES AND APPROXIMATE QUANTITY OF NAILS

Materials Unit Per Approx kg. Size Kind of

Required Nails

Floor joist and

bridging at 30 o.c sq. m .17 20d CWN

T & G Flooring
1x 4 sq. m . 15 6d Flooring brad
1x6 sq. m. . 09 6d Flooring brad
Siding Wood Board
on studs at 60 o. c.
1x6 sq m . 08 6d Casing brads
1x 8 sq. m. 06 6d Casing brads
Studs:
at .40 o. c sq. m. . 08 8d CWN
at .60 o. c. sq. m . 05 8d CWN

Scaffolding Meter ht. . 73 20d CWN

of post
Plywobd Wall Per sheet
and Ceiling at . 15 o.c on . 055 2d Finishing nails
40 x 60 joist
Rafters. Purlins Per sq. m. .20 20d CWN
and Cleats purlins at 70 o.c.

Base Board Meter length

, 03 6d CWN

Fascia Board Meter length 048

. 8d CWN
Ceiling Joist
at 40 x 60 o c. Sq. m .05 8d CWN

288
 ILLUSTRATION 8- 1

A wall partition 16.00 meters long by 3.00 meters high specify

the use 1" x 8” stone cut wood board on studs spaced at .60 m.
o . c. Determine the nails required for the studs and the wood board.

SOLUTION

A ) Nails for studs

1. Find the area of the wall

A = 20.00 x 3.00 m.
= 60 sq . m.

2. Referring to Table 8-11 along stud at .60 m. o.c.

Multiply:

60 x .05 = 3.0 kilos 8d CW Nail

B ) Nails for I " x 8" Wood Board

1. Area = 20 00 x 3.00 m.
A - 60 sq. m.

2. Referring to Table 8-11 along 1” x 8 board

M

Multiply:
60 x .06 = 3.6 say 4 kilos 6dCWNail

289
 Cement Coated Nails Measured here
Bright Rat Headed Nails Measured Here

Concrete Nails

FICURE 8* 18

TABLE 8-11 QUANTITY OF CEMENT COATED NAILS PER

KILOGRAM
Length Penny Gauge No. of Nails/ Kg.

1" 2d 15 1863
M
1 1/4 3d 14 1195
1 1/2" 4d 12 1/2 651
1 3/4" 5d 121/2 559
2" 6d 11 1/2 367
2 1/4" 7d 11 1/2 330
2 1/2" 8d 10 1/4 222
2 3/4" 9d 101/4 202
3" 10d 9 145
3 1/4" 12d 9 149
31/2" 16d 8 103
4" 20d 6 64
4 1/2" 30d 5 48
5" 40d 4 37
5 1/2" 50d 3 31
6" 60d 2 24

290
 CHAPTER 9
STAIRCASE
9- 1 INTRODUCTION

Carpenters who have tried to build stairs have found it to be

an art in itself. However, building staircase requires technical of
structural carpentry and craftsmanship of cabinet making. It is like
constructing an enclined bridge between two points on different
floors with uniform, well proportional treads and risers that are
safe and comfortable to climb and descend. And to start with the
construction and study of estimate, it is important to know and
familiarize with the technical terms comprising staircase.

.
Baluster A small post supporting the handrail or a coping.

Balustrade is a series or row of balusters joined by a handrail

or a coping as the parapet of a balcony.

Bearers - A support for winders wedged into the walls secured

by the stringer

-
Carriage That portion which supports the steps of a wooden
stairs.

Close String A staircase without open well as in a dog stairs.

-
291
Cocktail Stair - Is a term given to a winding staircase.

Circular Stair - A staircase with steps winding in a circle or

cylinder.

Elliptical Stairs - Those elliptical in plan where each tread

assembly is converging in an elliptical ring in a plan.

Flight of Stairs -Is the series of steps leading from one landing
to another.

Front String - The string on the side of stairs where handrail

is placed.

Flyers - Are steps in a flight that are parallel with each other.

Geometrical Stairs - Is a flight of a stair supported by the wall

at the end of the steps.

Half- Space - Is the interval between two flights of steps in

staircase.
Handrail - A rail running parallel with the inclination of the
stairs that holds the balusters.

Hollow Newel - An opening in the middle of the staircase as

distinguished from solid newel wherein the ends of steps are
attached.

Housing - The notches in the string board of a stair for the

reception of stairs.

Knee - Is the convex bend at the back of the handrail.

292
 -
Landing Is the horizontal floor as resting place in a flight.

-
Newel The central column where the steps of a circular
staircase wind.

Nosing - The front edge of the steps that project beyond the
riser.

Pitch - The angle of inclination of the horizontal of the stairs.

Ramp - A slope surface that rises and twists simultaneously

Rise -The height of a flight of stairs from landing to landing

or the height between successive treads or stairs.

-
Riser The vertical face of a stair step.

-
Run The horizontal distance from the first to the last riser of
a stair flight.

Spandril - The angle formed by a stairway.

Staircase- Is the whole set of stairs, the structure containing

a flight of a stairs.
Stair Suilders Truss - Crossed beams which support the
landing of a stair.

Stairhead - The initial stair at the top of a flight of stair or

staircase.

Stair Headroom - The clear vertical height measured from the

the nosing of a stair tread to any overhead obstruction.

293
Stairwell - The vertical shaft which contains a staircase.

Step - A stair which consists of one tread and one riser.

Steps - The assembly consisting of a tread and a riser.

String - The part of a flight of stairs which forms its ceiling or

soffit.

Soffit - The underneath of an arch or molding.

String Board - The board next to the wall hole which receives
the ends of the steps.

Tread ‘ The horizontal part of a step including the nosing.

Tread Run - The horizontal distance between two consecutive

risers or, on an open riser stair, the horizontal distance
between the nosing or the outer edges of successive treads all
measured perpendicular to the front edges of the nosing or
tread.

Tread Length -The dimension of a tread measured

perpendicular to the normal tine of travel on a stair.

Tread Width - The dimension of a tread plus the projection of

the nosing.

Wall String - The board placed against the wall to receive the
end of the step.

Well - The place occupied by the flight of stairs.

294
 Well Hole - The opening in the floor at the top of a flight of
stairs.

Well Staircase - A winding staircase enclosed by walls

resembling a well .
-
Winders Steps not parallel with each other.

Wreath - The whole of a helically curved hand rail.

Tread

Stringer
Flight
Riser
o<

rj
Right
X Nosing

*
-
FIGURE 9 1

9-2 STAIRS -LAYOUT

The fundamental procedures in laying out a staircase are

enumerated as follows:

295
1. Determine the clear height of the riser in meter. Normally ,
the standard comfortable rise per step is from 17 to 18 cm.
The maximum height of a step riser is up to 20 cm. but is
only allowed on special considerations where physical
conditions dictate. However , this height is understood to be
not comfortable for both ascending and descending

2. Determine the number of steps from the firsttothe next floor

by dividing the total height of the rise by the chosen step
riser of either 17cm. or 18 cm .

3. Divide the run distance by the effective width of the tread

which normally measured as follows :

25 cm.
——
Width of Tread

30 cm. ---
35 cm.
Effective Width
- .20 cm.
- .25 cm.
- .30 cm.
Angle Post

-
FIGURE 9 2

296
 The effective width of the tread is equals to the width minus
the nosing.

4. If the result of step 3 Is less than that of step 2 , adjust the

length of the run or the width of the tread to obtain an equal
distances and proportional steps.

5. The height of the risers should be equal and uniform from

the first to the last step of the stair, hence , there shall be
no fractional value in dividing the rise by the riser per step.
However , if fractional value could not be avoided in dividing
the rise by the riser, adjust the fractional value in equal
proportion to the number of risers but in no case shall the
riser per step be greater than 19 cm. nor less than 17 cm.
otherwise, the stairs will not be considered ideal and
comfortable one.

Floorline O

.183 m Rise
, 5 cm Nosing

Effective Width of Tread

Floorline

FIGURE 9 3 -
297
 ILLUSTRATION 9-1

From figure 9- 3 determine the number of steps and the height

of the riser if the total height of the rise is 2.20 m. using a 30 cm.
width of the tread.

SOLUTION:

1. The height of the rise is 2.20 meters . Assume a 17 cm.

riser.

2. Divide:
Rise * No. of risers
Riser

2.20 * 12.94
.17

3. The answer has a fractional value of .94. The rule says,

"there should be no fractional value in dividing the rise by
the riser. Thus, adjust to have an equal height per riser.

4. From the result of step 2, use the whole value of 12

disregarding the decimal amount of .94 .

2.20 = .183 m. or 18.3 cm.

12

5. The 18.3 cm. is now the height of the risers per step instead
of 17 cm. as assumed. This value is within the range of 17
and 19 cm. considered as ideal and comfortable stairs.

298
 6. Determine the distance of the run using the formula.

Run = No. of Steps - 1 x Effective width of the tread.

Where:
Effective Width = Tread Width - Nosing
Nosing is from 2 to 5 cm.

Run * 12 - 1 x 25 cm.
R = 2.75 meters

There are instances however, where the length of the run and
the height of the rise are known or given. The question is, how to
determine the width of the tread and the height of each risers.

ILLUSTRATION 9- 2

Determine the height of the riser and the width of the tread
when the rise is 2.65 m. and the run is 2.75 m.

SOLUTION:

1. Assume that the riser height is 18 cm.

2. Divide the rise height by the riser 18 cm.

2.65 - 14.72 say 15 steps

.18

3. Assuming that there are 15 steps instead of 14.72

determine the final height of the riser.

299
 2.65 * 17.7 cm.
15

This value is between 17 to 19 cm. which is acceptable

4. Assume that the tread width is 30 cm. the effective width

ofthestepis 30 - 05 nosing * 25 cm.

5 . If there are 15 steps, multiply by the effective width of the

tread .

Run = No . of steps - 1 x . 25 (see figure 9-4)

Run = (15 - 1) x 25 = 350 cm. or 3.50 m.

Note:
The 3.50 meters is longer than the 2.75 m . distance of the run
as specified in the problem , therefore , adjustment of the tread
width is necessary, thus:

a . From step 4 let us assume that the tread width is 25 cm.

not 30 cm.

b. The effective width is = 25 cm. minus 5 cm. nosing

= 20 cm.

c . Check distance by trial multiplication:

No. of steps x effective width = Run

14 x .20 = 2.80 m.

This value is acceptable since the existing distance of the run

is 2.75 meters with a difference of 5 centimeters which could
be adjusted in the construction.

300
 Well Hole

Floor

Ceiling

E Run of Step

a
Stringer a:

Tread
Nosing
Rise per Step
Riser

J1 Floor Line

Run

Cross Section and Different Parts of stairs

FIGURE 9-4

9-3 STRINGER

Stringer is the inclined plane that supports or holds the tread

and the riser of a stair. The length of the stringer is determined by

301
either the use of the Pythagorean Formula or by actual
measurement using a meter rule or tape.

The steel square is a useful and effective tool in staircase

framing. Know its functions and a satisfactory result will be
obtained.

There are several forms of stringer classified according to the

methods of attaching the risers and the tread, they are:

1. Cut type
2. Cleated type
3. Built-up
4 . Rabbeted ( Housed )

Cut type stringers - are popularly used in most modem and

contemporary house designs.

Cleated type stringer - is used for very rough work.

Built-up type stringer - is adopted on a wide stairs that

requires a center stringer.

L Stringer

/ Tread
4
7

Cut Type Cleated Built-Up Rabbeted

FIGURE 9-5

302
 -
Rabbeted type stringer is adopted on a fine work and
usually made at the mill. The risers and treads are held in the
rabbets by wedges that are set in with glue.

ILLUSTRATION 9 3 -
Determine the length of an open wood stringer with the
following data:

Run distance * 3.50 m .

Height of the Rise * 2.50 m.

Floor Line
1

E
8
ci

'
o X
i<r
Floor Lino

Run * 3.50 m.

-
FIGURE 9 6

303
SOLUTION:

1 . Using the Formula:

/
Stringer length = (Run ) 2 + (Rise) 2

SL = /(3.50) 2 + (2.50) 2

SL = 7 (12.25 + 6.25

/
SL * 1830

SL = 4.30

2. Convert to feet;

4.30 m. = 14.33 ft .
.30

3. Determine the number of steps:

Assuming 18 cm. riser height.
Divide:

Riser - 2.50 * 13-.88 say 14 steps

Riser .18

This problem can be solved by referring to Table 9- 1.

See the value of 4.31 m. along the 14 steps;
3.50 m . length of run below the 30 cm. tread and the 2.52 m.
rise below the 18 cm . riser column.

304
 TABLE 9-1 HEIGHT OF RISE, LENGTH OF STRINGER AND RUN
OF STAIRWAY (in . Meters )

[Link] Length of Stringer Length of Run Height of Rise

Steps Tread Width Tread Width Riser Height at
25 cm. 30 cm. 25 cm. , 30 cm. 17 cm. 18 cm.

4 1, 05 1.23 .80 l 1,00 .68 .72

5 1.31 1.54 1.00 1.25 .85 .90
6 1.57 1.85 1.20 1.50 1.02 1.08
7 1.84 2.16 1.40 1.75 1.19 1.26
8 2.10 2.47 1.60 2.00 1.36 1.44
9 2.36 2.78 1.80 2.25 1.53 1.62
10 2.62 3.08 2.00 2.50 1.70 1.80
11 289 3.39 2.20 2.75 1.87 1.98
12 3.15 3.70 2.40 3.00 2.04 2.16
13 3.41 4.00 2.60 3.25 2.21 2.34
14 3.67 4.31 2.80 3.50 2.38 2.52
15 3.94 4.62 3.00 3.75 2.55 2.70
16 4.20 4.93 3.20 4.00 2.72 2.88
17 4.46 5.24 3.40 4.25 2.89 3.06
18 4.73 5.55 3.60 4.50 3.06 3.24
19 5.00 5.85 3.80 475 3.23 3.42
20 5.62 6.16 4.00 5.00 3.40 3.60

Stairs Inclination Angie

Ladder - 50° to 90°

Ramp - 1° to 20°
Stairs - 20° to 50°
Ideal Stairs - 30° to 35°

305
 -
TABLE 9 2 SPIRAL STAIRS

Open Riser Treads Cantilever Treads

Tread Degrees No. of T read Riser Head Room
in Circle cm. Meter

22° - 30‘ 16 17.5 2.10

28° - 0* 12 - 13 18.0 2.00
-
30° 0* -
12 13 20.0 2.00

Elevation M
Cantilever Treads

m%
5
( Step
J 30°

Plan %

EE ¥
Elevation
Open Riser Treads

-
FIGURE 9 7

306
The National Building Code on Stairs provides that:

1. The minimum width of any stair slab and the minimum

diameter -dimension of any landing should be at least 110
centimeters.

2. The maximum rise of stairs step should be between 17 and

19 centimeters. A rse less than 16 cm. nor more than 19
cm. is not considered as ideal stair.

3. The minimum width of a tread exclusive of the nosing shall

be 25 centimeters.

4 The maximum height of a straight flight between landing is

generally 3.60 meters except those serving as exit from
places of assembly where a maximum height of 2.40
meters is normally specified.

5. The number of stairway in a building depends upon the

number of probable occupants per floor, the width of
stairway and the building floor area. The distance from any
point in an open floor area to the nearest stairway shall not
exceed 30 meters and that the corresponding distance
along corridors in a particular area shall not exceed 38
meters.

6. The combined width of all the stairway in any floor shall

accommodate at one time the total number of persons
occupying the largest floor area under the condition that one
person for each .33 sq. m. floor area on the landing and
halls within the stairway enclosure.

307
7. In building of more than 12 meters high and in ail mercantile
buildings regardless of height , the required stairways must
be completely enclosed by fireproof partitions and at least
one stairway shall continue to the roof.

308
 CHAPTER

PAINTING
10- 1 PAINT

Paint is commonly referred to as a Surface Coating. It is

defined as "a coating applied to a surface or subtrate to decorate,
to protect, or to perform some other specialized functions."

Almost everybody knows the word paint, its uses, color,

including the brand which are rated as good, best and durable.
There are those who have little knowledge of paint but rated a
brand based on how it is advertised. Others on the cost of the paint.

Generally, a good quality paint is costlier than that of a poor

one. However, in terms of surface area coverage, ease of work and
durability, a good quality paint is cheaper than a poor one, thus,
never have a second thought of having the best from a reputable
brand, otherwise to think of saving a few cents for your paint might
turn out later to be more expensive.

Obtaining good quality paint from a reputable brand however,

.
is not a guarantee that you have a long lasting paint There are
numerous kinds of paint as there are different kinds of surface to

309
be painted. Applying a premium quality paint to a surface not
suitable for such type of paint is considered a technical failure which
cannot be guaranteed by the cost neither the brand of the paint. It
is therefore imperative to know which kind of paint for what kind of
surface to be applied

10-2 INGREDIENTS OF PAINT

The ingredients of paint are:

1. Vehicle
2. Solvent
3. Pigments
4. Additives

Vehicle - is that substance in the paint which gives a film

continuity and provides adhesion to the surface or subtrate. It is
called vehicle because it carries the ingredients to the subtrate
which will remain on the surface after the paint dries.

The vehicle contains the film former which is the combination

of the following:

a) Resins
b) Plasticisers
c) Drying oil, etc.

The Vehicle is divided into the following:

1. Solid Thermoplastic Film Formers - The solid resin is

melted for application and then solidifies after application.

310
 -
2. Lacquer Type Film Formers The vehicle dries by solvent
evaporation.

3. Room Temperature Catalyzed Film Formers - Chemical

agents blended into the coating before application cause
cross-linking into a solid polymer at room temperature.

4. Oxidizing Film Formers - Oxygen from the air enters the

film and cross links it to form a solid gel.

5. Heat-Cured Film Formers - Heat causes cross-linking of

the film former or activates a catalyst that is not active until
heat has been applied.

-
6. Emulsion-Type Film Formers The solvent evaporation
and the droplets of plastic film former floating in it flows
together to form a film.

Solvents - are low viscosity volatile liquid used in coating to

improve application properties.

-
Pigment Paint pigments are solid grains or particles of
uniform and controlled sizes which are generally insoluble in the
vehicle of the coating.

The paint pigment contributes to the following properties

a) For tfie decoration of function - it contributes opacity,

color and gloss control.

-
b) For the protective function it contributes specific
properties such as hardness, resistance to corrosion, and
rapid weathering, abrasion, and improved adhesion.

311
 c) It makes sanding easier , retard flame and serves as
insulation against electricity.

d) Pigments serve to fill spaces in paint films.

Additives - are ingredients formulated in the paint to modify

the properties of either the vehicle or the pigmentation or both. They
give the wet paint or dried paint film properties not present in the
vehicle and pigmentation system. Additives improve a certain
properties of vehicle such as speed drying, pigment resistance to
fading or the entire paint such as the ease of application.

10-3 ESSENTIAL AND SPECIFIC PROPERTIES

OF A GOOD QUALITY PAINT
A good quality paint must have the following essential and
specific properties.

1. Adhesion - coating must stick to the surface or subtrate to

bring other properties into work.

2. Ease of Application - paint must be easy to apply as

utilized by the user or through the method prescribed by
the manufacturer. The paint must go into the subtrate in the
specified film thickness. It should be dried in the specified
time with the desired appearance and possessing the
necessary specific properties.

3. Film Integrity - the cured or dried film of paint must have

ail the film properties as claimed by the manufacturer. There
should be no weak spots in the film caused by imperfect film
drying or curing .

312
 4. Consistent quality - paints must be consistent in quality
such as color, viscosity , application properties and durability
from can to can, batch by batch, shipment to shipment.

5. Specific Properties - The paint should be considered for

the particular use, such as :

a) Kitchen Enamel - must resist grease; heat and repeated

cleaning.

b) Stucco or f- atex Paint - must resist water, alkali and

sunlight and must permit passagt' of water vapor.

c) Swimming Pool Paints - must have a specific chlorine,

water and sunlight.

d) Exterior Commercial Aircraft Finishes must resist ultra

violet degradation, erosion by air toss of adhesion at high
speed, rapid change of temperature, chemical attack by the
hydraulic fluids of the aircraft, and film rupture from the
flexing of the film by the denting of the surface.

10 4 THE ELEMENTS OF A GOOD PAINTING

JOB
Painting is the final touch in the construction work. It is where
all the construction defects, ugliness and roughness from masonry ,
carpentry, tinsmithing etc . are corrected, smoothened and
beautified.

The Elements of a Good Painting Job are :

1. Correct Surface Preparation: The primary and essential

313
 prpperty necessary for of paint is Adhesion. Good
adhesion demands proper surface preparation.

2. Choice of the Proper Paint System . Apply the right kind

of paint on the right surface. For instance, water base paint
for masonry and concrete, oil base paint for wood and
equivalent surface etc

3. Good Application with the right technique and tools.

a) Uniform wet and dry film thickness.
b) Correct number of sequence of application.
Verify manufacturers specification.
c) The right tools and their use

4. Correct Drying Cycle - The final properties of the dried

.
coating develops during the drying cycle Unless conditions
are favorable, correct film properties will never develop.

5. Protection Against Water. The primary cause of paint

failure is Moisture which is considered a menace to the
best of paint jobs. It is very frustrating to see your lovely
and newly painted house deteriorating so soon specially if
you bought good paint, used good tools and spend a lot for
its labor.

Water is the hidden enemy of paint. It is a pervasive element

of deterioration and it causes the following:

a) Rusting and other corrosion

b) Paint peeling
c) Masonry efflorescence and spalling
d) Wood rot
e) Corrosive water solution (staining sea water)

314
10-5 SPECIFICATIONS FOR SURFACE
PREPARATION

The quickest way to achieve paint failure is through improper

surface preparation. It is just as important to qualify a surface
preparation as it is to specify a painting system.

The following specifications give a general overview of surface

preparation, thus:

SPECIFICATIONS

A* GENERAL

Surface Examination - No exterior painting or interior

finishing shall be done under conditions which may jeopardize the
quality or appearance of the painting or finishing.

Preparation. Ail surfaces shall be in proper condition to

receive the finish. Woodwork shall be sandpapered and dusted
clean. All knotholes, pitch pockets, or sappy portions shall be
shellacked or sealed with knot-sealer. Nail holes, cracks, or defects
shall be carefully puttied after the first coat with putty matching the
color of the stain or paint.

Interior woodwork. Finishes shall be sandpapered between

coats. Cracks, holes, or imperfections irt plaster shall be filled with
patching plaster and smoothed off to match adjoining surfaces.

Plaster or masonry. Plaster or masonry shall be dry before

any sealer or paint is applied. After the primer-sealer coat is dry,
all visible suction spots shall be toughed up before succeeding

315
coats are applied. Work is not to proceed until all spots have been
sealed. In the presence of high alkali conditions, surfaces should
be washed to neutralize the alkali.

Metals . Metals shall be clean, dry and free from mill scale and
rust. Remove all grease and oil from the surfaces. Unprimed
galvanized metal shall be washed with metal etching solution and
allowed to dry.

Concrete and brick surfaces. These surfaces shall be

wire-brushed clean. Surfaces which are glazed or have traces of
patching compound shall be sandblasted or acid etched.

B. Cleaning Methods

Sandblasting. There are three general methods applied:

1. Conventional dry sandblasting The sand is not recycled .

However, dust respirators and other safety precautions
should be observed since environmental restrictions on dry
blasting are becoming increasingly severe.

2 . Vacuum sandblasting . This method reduces health

hazards since the sand is recovered. However, It is costlier
and less efficient than dry blasting, but its efficiency can be
increased by holding the vacuum cone at a slight distance
from the surface. The vacuum method is useful inside
shops and in areas where dust might damage machineries.

3. Wet sandblasting. This method reduces the dust hazard

and may be required by legal restrictions . The wet sand and
paint residues however, accumulate on ledges and other
flat areas, thus, rinsing operation is necessary.

316
 Wire-brushing and scraping. Power and hand wire-brushing
are used mainly on small jobs, in cleaning small areas after
sandblasting, and on surfaces for which sandblasting is not
feasible. Hand scraping is used on small areas, in places where
access is difficult, and for final clean-up after other methods have
been employed

.
Power toots Power tools such as rotary wire and disc tools,
rotary impact chippers , and needle sealers maybe used if
sandblasting is not feasible

.
Water blasting Water blasting is an effective method in
cleaning and removing old paint from large masonry surfaces. It is
generally used and acceptable for health and environmental
requirements. Water blasting method is preferred for underwater
or marine work.

.
C Chemical methods

.
1 Acid-etching. Is the use of an acid solution, with or without
a detergent , to roughen a dense glazed surfaces. Rinse
thoroughly the acid-etched surfaces to remove the residual
soluble reaction of calcium and magnesium chloride which
affect the adhesion and stability of latex paints in particular.

.
2. Paint removers Both the conventional solvent-based and
the water rinsable types of paint removers maybe used to
remove old paint. Most paint removers contain wax. This
wax must be removed completely before painting because
it destroys adhesion and inhibits the drying of paint.

3. Steam cleaning . Steam cleaning with or without

detergents is frequently used in food-packing plants. A

317
 mildewcide is usually added. Low-pressure steam cleaners
are use on homes and offices walls.

4. Alkali cleaning. Alkali cleaners should not be used on

masonry surfaces adjacent to aluminum, stainless steel or
galvanized metal. Surfaces which are cleaned with alkali
cleaners shall be thoroughly rinse and clean with water.
Residual alkali and detergents can cause greater damage
to paint if they are not removed completely .

10-6 KINDS OF PAINT, USES AND AREA

COVERAGE
Coverage
Kind of Paint Uses Thinning Drying Time per 4 Its.

PRIMERS*

1. Interior Primer For interior wood Paint 2 hours, allow 25 to 30

and Sealer surfaces Thinner overnight before sq. m.
recoating

2. Exterior wood For exterior wood Paint 6 hours, allow 30 to 40

primer surfaces Thinner overnightbefore sq. m.
recoating

3. Prepakote Primer for ferrous Paint 3 hours 35 to 40

Red Oxide and non-ferrous Thinner sq. m.
Primer materials

4. Zinc Chromate For exterior and Paint

Primer interior metal Thinner 3 hours 30 to 40
surfaces exposed sq. m.
to normal indus -
trial environment

318
5. Red Lead Rust preventive Paint
Primer primer for ferrous Thinner 3 hours 30 to 40
surfaces sq. m.

6. Epoxy Primer For steel, alumi- Epoxy 6 hours, allow 30 to 40

num, galvanized Reducer overnight before sq. m.
iron recoating

* WATER BASE MASONRY PAINT *

1. Acrylic Latex Exterior & Interior 1/2 Its7 30 minutes. 30 to 40

Paint masonry surfaces 4 Its. Allow 6 hrs. sq. m.
water before recoat

2. Acrylic-Semi- Exterior & interior water 30 minutes. 30 to 40

Gloss Latex masonry surfaces Allow 6 hrs. sq. m.
before recoat

3. Acrylic Gloss Exterior & Interior water 30 minutes. 30 to 40

Latex paint masonry surface Allow 6 hrs . sq. m.
before recoat

4. Acrylic Clear For chalky surface water 1 hour. Allow 30 to 40

Gloss Emulsion to improve adhe- sq. m.
paint sion of new coats
of latex paint

5. Latex Hi-gloss For furniture, cabi- Usse as 1 hour. Allow 30 tO 40

Enamel nets, doors, win- is 4 hours before sq. m.
dows, toots, toys recoating
wrought iron, and
primed metals

6. Acrylic Primer to old and water 30 min. Allow 30 to 40

319
 concrete sealer new concrete, ex - 6 hrs. before sq. m.
terior & interior recoating

7. Masonry Primer for old Use as 24 hours. Allow 30 to 40

surface chalky paint film is overnight before sq. m.
conditions recoating

8. Tinting Colors Acrylic colors water

* ROOF PAINT *

1. All weather For gafvanized water 4 hours. Allow 40 to 50

acrylic root iron sheet, as- overnight before sq. m.
shield bestos bricks recoating
concrete and
stucco
2. Portland For G.i. sheet Paint 6 hours. Allow 30 to 40
cement paint CHB , stucco, Thinner overnight before sq. m.
concrete, brick recoating
and other zinc
coated metal
surface

3. Davies Roof G.I. roofs and Paint 6 hours. Allow 30 to 40

paint other metal Thinner overnight before sq. m.
such as alumi- recoating
num and steel

* ENAMEL AND GLOSS PAINT *

1. Quick Drying For Exterior & Paint 5 hours. Allow 30 to 40

Enamel Interior wood Thinner 8 hrs. before sq. m.
and metal surf. recoating

320
2. Interior semi- Interior wood Paint 6 hours. Allow 25 to 30
Gloss Enamel and metal Thinner overnight before sq. m.
surfaces recoating

3. Flat Wall For Interior Paint 3 hours. Allow 40 to 50

Enamel wall & ceilings Thinner overnight before sq. m.
recoating

4. Exterior .
For ext wood Paint 6 hours. Allow 30 to 35
Gloss paint and properly Thinner 48 hrs. before sq. m.
primed metal recoating
surfaces.

* VARNISHING *

1. No. 48 Davies For dark wood Use as

wood bleach to be changed is
land 2 to light natural
finish & making
old wood color
uniform
2. No. 77 Dawes For patching up Use as 10 minutes 20 to 25
Lax-Tire Plas- wood defects is 30 minutes sq. m.
tic wood Dough like knots,nail for dry hard
holes and cracks

3. Wood filler Sealer for open Paint. 12 min. Allow 25 to 35

paste grain of interior Thinner overnight before sq. m.
wood recoating

4. Non-Grain For wood sur - 30 minutes 30 to 40

raising wood faces sq. m.
stain

321
5. Oil Woodstain For panelling, Allow 24 hours 30 to 40
cabinets, floors, before recoating sq. m.
furnitures, door
jambs, and other
woodworks

6. Finishing Oil To seal and finish Overnight 35 to 40

interior wood surf, sq. m.
such as furnitures
wood & panelling

.
7 Valsparor For floors, sidings Paint
Spar Varnish furnitures, deck of Thinner 24 hours 40 to 50
boats etc. sq. m.

8. Daxpar For interior and Paint 24 hours 40 to 50

Varnish exterior wood Thinner sq. m.
surfaces, nautical
& aeronautical
varnish.

9. Hi- solid san- For Interior new Lacquer 10 minutes. 40 to 45

ding sealer wood furnitures Thinner Allow 30 min. .
sq m.
and fixture, cabi - before recoat
net, doors, etc.
10. Hi- Solid For furnitures, Lacquer Allow 30 min. 30 to 40
clear gloss cabinets, fixture Thinner before recoat sq. m.
Lacquer door panelling
and trim.

11. Hi- Solid For Interior Lacquer Allow 30 min. 30 to 40

Semi-gloss wood furnitures, Thinner before recoat sq. m.
Lacquer cabinets, doors
shelves etc.

322
12. Hi-Solid dead For Interior wood Lacquer Allow 30 min. 30 to 40
flat lacquer furniture, cabinets Thinner before recoat sq. m.
door jambs, trim
panelling etc.

13. Water white For furnitures, Lacquer Allow 30 min. 30 to 40

gloss lacquer cabinets, doors, Thinner before recoat sq. m.
panel and inte-
rior walls

* ENAMEL AND GLOSS PAINT *

1. Quick drying For Exterior and Paint 5 hours. Allow 30 to 40

Enamel Interior wood & Thinner 8 hrs. before sq. m.
metal surfaces recoating

2 Interior Semi Interior wood & Paint 6 hours. Allow 25 to 30

gloss Enamel metal surfaces Thinner overnight before sq. m.
recoating

3. Flat Wall For Interior walls Paint 3 hours. Allow 40 to 50

Enamel and ceiling Thinner overnight before sq. m.
recoating

4. Exterior gloss Exterior wood Paint 6 hours. Allow 30 to 35

Paint and properly Thinner 48 hrs. before sq. m.
primed metal recoating
surfaces

5. Tinting Colors Oil base Tinting

color

323
 AUTOMOTIVE FINISHING

1. Lacquer For properly Lacquer Allow 30 min. 40 to 50

Enamel primed metal Thinner before recoat sq. m.
and wood

2. Automotive For Extenor & Lacquer 30 minutes 40 to 50

Lacquer Interior metal Thinner sq. m.
or wood surf.

3. Lacquer For metal and Lacquer 30 minutes 40 to 50

Primer wood surface sq m.
Surfacer

4. Lacquer Putty For Exterior and Lacquer 10 minutes 20 to 30

Interior metal & Thinner dry to sand sq. m.
wood surfaces in 1 hour

_
5. PnH ux Auto Automotive fin. Paint 2 hours. Dry 30 to 40
Enamel for residential & Thinner hard in 10 hrs. sq. m,
commercial
6. Pro-Lux For Exterior and Paint 2 hours. Allow 30 to 40
Enamel Primer Interior wood & Thinner 8 hours before sq. m.
metal surfaces recoating

-
7. Pro Lux For Exterior and Use as 2 hours 25 to 35
Glazing Putty Interior metal & is sq. m.
wood surfaces

* INDUSTRIAL PAINTS *

1. Silver fiwsh For steel tanks, Use as 1 hour. Allow 40 to 50

Aluminum Ext. & Int. metal is 24 hrs. before sq. m.
wood & masonry recoating

324
2 Heat Resis- For Interior and Use as 1 hour. Allow 40 to 50
ting Exterior surfaces is 24 hrs. before sq. m.
like radiators,
boilers, pipes
& general Ind
equipment.

3 Hi-Heai Re - For superheated Use as 1 hour at 40 to 50

sisting paint steam lines, is 450° F max. sq. m
boiler casings,
drum and rocket
launchers

4 Asphalt Base For asbestos ce - Use as 1 hour. Allow 20 to 30

Aluminum ment composi- is 24 hrs. before sq. m.
tion & metal recoating

5. Traffic Paint For Asphalt and Use as 1 hour. Allow 20 to 30

masonry surface is 24 hrs. before sq. m.
recoating

6. Blackboard For wood or Paint 12 hrs. Allow 35 to 40

Slating metal surface Thinner overnight before sq. m.
recoating

7 Davies Anti- For hulls and Paint 12 hrs. Allow 30 to 40

Corrosive below water Thinner overnight before sq. m.
Marine paint line of ships recoating

8. Marine Bott For properly Paint 12 hrs. Allow 30 to 40

Topping Paint primed surfaces Thinner 24 hrs. before sq. m.
b/w the light & recoating
deep load lines
of ships

325
9. Anti-Fouling For properly Paint 8 hrs. Allow 30 to 40
Paint primed surface Thinner 12 hrs. before sq m.
below the water recoating
line of ships.

10. Hull, Deck, For use above the Paint 12 hrs. Allow i 30 to 40
Mast & Top- water line of sea Thinner 24 hrs. before sq. m.
side Paint vessels, equip - recoating
ment and struc -
ture s near sea.

11, Machinery Marine engine Paint 1 hour. Allow 30 to 40

Engine Machineries & Thinner 24 hrs. before sq. m.
Enamel equip’t. casing recoating

12. Epoxy For Steel, alum, Epoxy 6 hrs. Allow 30 to 40

Enamel galv. iron, wood reducer overnight before sq. m.
and concrete recoating

13. Epoxy Gla- For body repair Epoxy 2 hours

zing putty aircraft, cars & recucer
equipments

14. Epoxy Glue Multi-purpose Use as 8 hours full 30 to 40

thermosetting is strength in sq. m.
plastic material 96 hours
for cementing or
bonding rigid t
i
materials
f
15. Pure Pale Use as thinner Use as
Boiled Lin- exterior house is
seed oil paint
i
.
16. Concrete Treatment to i Use as 24 hours 30 to 40

326
 Neutralizer neutralize ma- is sq . m.
sonry surface

17. Rust Paint stripper Use as Overnight 25 to 35

Remover sq. m.

18. Paint Paint stripper Use as Overnight 25 to 35

Remover sq. m.

19. Mildewcide Destroy molds Water

Solution mildews on new
and previously
painted surfaces

•Source: Davies Paint Manual of Information

10-7 ESTIMATING YOUR PAINT

Paint manufacturer specifications includes the estimated area

coverage per gallon with a net content of 4 liters. Generally, the
estimated area coverage is in the range interval of 10. For instance ,
one gallon of Quick Drying Enamel covers 40 to 50 square meters
surface area which simply means a minimum of 40 and a maximum
of 50 square meters depending upon the texture of the surface to
be painted.

The problem therefore is, what amount for which surface

texture will be used? To simplify our estimate, surface texture will
be classified into three categories such as:

a.) Coarse to Rough — 40 sq. m. per gallon

b.) Fine to Coarse 45 sq. m. per gallon
c.) Smooth to Fine 50 sq. m. per gallon

327
 ILLUSTRATION 10 - 1

A concrete fire wafl measures 30 meters long and 12 meters

high. Determine the number of gallons ( 4 liters content) required
using Acrylic Gloss Latex Paint if the wall is :
a.) Wooden trowelled finish ( coarse to rough )
b.) Paper finished ( Fine to coarse )
c.) Fine to Smooth finished

SOLUTION

A) For Coarse to Rough Surface

1. Solve for the wall area .
30 x 12 = 360 sq. m.

2. Deteimine the quantity of concrete primer.

Referring to Sec. 10-4 under Masonry Water Base Paint
using Acrylic Concrete Sealer as primer the area coverage
per gallon is 30 to 40 sq. m.

3 , For a Coarse to Rough Surface

Divide:

360 sq. m. = 12 gallons Acrylic concrete primer

30 sq . m.

4. Solve for the Acrylic Gloss Latex Paint final coating;

Divide:

360 sq. m. = 12 gallons

30 sq. m. / gal.

328
 5 . Solve for the concrete neutralizer .
If one quart of neutralizer is mixed with 21/2 gallons of water
Divide:

12 = 4.8 say 5 quarts concrete neutralizer

2.5

B) Solution for Fine to Coarse Surface

1. Solve for the net area of the wall « 360 sq. m.

2. Solve for the Concrete Primer Sealer: ( use 35 sq . m.)

Divide:

360 - 10.28 gallons concrete primer sealer

35

3. Solve for Acrylic Gloss Latex Paint.

Same as in No. 2 - 10.28 gallons.

C) Solution For Fine to Smooth Surface

1. Divide the wall area by 40 sq. m.

360 • 9 gallons as surface primer

40

2. For final coating, the same = 9 gallons.

ILLUSTRATION 10-2

A 10 class room elementary school building with a general

329
dimensions of 3.00 x 90.00 meters requires painting of its G.l.
roofing and plywood ceiling. The pian specifies two coatings of
Acrylic Roof Shield and Quick Drying Enamel paint for the roof and
ceiling respectively. Prepare an order list of the following materials

a) Roof Paint
b) Wood primer for ceiling
c) Quick Drying enamel paint
d) Paint thinner

FIGURE 10- t

SOLUTION:

A) Roof Faint

1. Find the total roof area:

330
 A = 8.0 mx 82.0 m.
= 656 sq. m.

2. Refer to Sec. 10-4 ; under roof paint, the area coverage of

Acrylic Roof Shield Paint per gallon is 40 to 50 sq. m.
Use 45 as average .

656 sq. m. 14.6 gallons

45

3. For two coatings, multiply by 2 = 29.2 say 30 gallons

4. Paint thinner is not required because the base of this paint

is water.

B ) Enamel Paint for the Ceiling

1. Solve for the total ceiling area including the eaves.

A = 8 x 82 m.
= 656

2. Solve for the primer paint. See Sec. 10-4 . The area coverage
of Exterior wood primer is 30 to 40 sq. m.

3. For a plywood ceiling use average value of 35.

Divide:

656 = 18.7 say 19 gallons

35

4 . Solve for the Quick Drying Enamel final coat . Refer to Sec

331
 10-4. the area coverage is 30 to 40. Use the average - 35;
Divide:

656 = 18.7 say 19 gallons

35

5. Solve for the paint thinner: at an average of 1/2 liter per

gallon of paint.

Primer 19 gallons
Quick Dry Enamel . . 19 gallons
Total , 38 gallons

Multiply: 38 x .50 (1/2) « 19 liters

Summary

30 gallons All weather Acrylic Roof Shield

19 gallons Wood Primer
19 gallons Quick Drying Enamel
19 liters paint thinner

Comment

1. Paint thinner of any type is considered as the most abused

materials in ail painting job. It is used for cleaning of hands, paint
brushes, tools and sometimes as fuel and torches.

2. Frequent washing of paintbrush will consume large amount

of paint thinner; thus, sufficient number of paint brush is necessary
and that no washing shall be allowed until after completion of the
work .

332
 3. After the days work, paint brushes should be wrapped with
paper then soak into a can with water to avoid cleaning and
hardening of the paint.

4. Your paint brush should be protected from damages and

shall be thoroughly cleaned immediately if any of the following
paint is used:

a. Epoxy paint mixed with Catalyst- Clean with Epoxy

thinner or Lacquer thinner.
b. Water Base Paint- Wash thoroughly with water and soap.
c. Lacquer Paint or Varnish- Clean with lacquer thinner

3. Lacquer thinner estimate for varnishing woik should be

doubled to anticipate frequent thinning and multiple rubbing
purposes including evaporation.

10-8 PAINT FAILURE AND REMEDY

The different types of paint failures identified are:

1. Blistering or Peeling 6. Fading
2. Chalking 7. Bleeding
3. Flaking 8. Mildew
. .
4 Cracking and Alligatoring 9 Staining
5. Peeling or Cracking of 10. Checking and Flaking
Paint on G.l. Sheets

Blistering or Peeling - Occurs when the moisture trapped in

the wood evaporates when exposed to sun pushes the paint out of
the surface.

333
 Remedy:
1. Locate and eliminate the sources of moisture.
2. Scrap off old paint around the blistered area. Let dry and
apply good primer then final paint of good quality paint

Chalking - Means that the paint was too thin for the required
paint film.

Remedy:
Be more generous to your paint. Spend a little for two coatings.

Flaking - The result of inadequate or poor surface preparation.

The paint flakes off in scales or powders or powders and chalk off.

Remedy:
Remove the paint on affected area by wire brush. Seal all
cracks against moisture, then apply final coat of paint.

Cracking and Alligatoring- Results when paint was applied

in several heavy coats not observing the sufficient drying time
between coats. The primer or undercoat may not be compatible
with the final coat. For instance , using a quick drying enamel as
final coat over a flat wall paint, or a lacquer paint over an ordinary
oil base paint.

Remedy:
Remove the paint . Clean the surface properly, apply good
primer then final coat.

Peeling or Cracking of Paint on G.I. Sheet.- Indicates the

use of improper metal primer or no primer applied. The paint film
has no adhesion on the surface .

334
 Remedy :

Strip off the paint Clean with solvent Dry, then apply
.
galvanized paint ( see roof paint Sec 10-4)

- .
Fading. Fading is a normal behavior of paint But, if fading is
too fast and excessive, that means you applied a poor kind of paint .
This is what usually happened when for a few cents of difference
in cost, the quality is sacrificed.

Remedy:

Repaint Next time be sure to buy the best of paint brand.

Remember that a good kind of paint contains more and
better pigment

Bleeding - is the result of inadequate sealing of the surface at

the first application of paint.

Remedy:
Scrape off the surface then,- Repaint

Mildew - thrives on high humidity and temperature. The

fungus are stimulated and grows on the paint film. If covered with
new coat of paint, just the same, it will grow through the new coat.

Remedy:
Wash the surface with mildew wash solution dilluted with
water. Scrub the surface. Rinse with clean water and dry for
48 hours then, apply final coat

-
Staining is an effect of wood preservatives or rust of nails.

335
 Remedy:
1. Remove paint on affected area.
2. Remove rust on nails then apply lead primer to metal and
wood primer.
3. Apply final coat with good quality paint

Checking and Flaking - is caused by expansion or

contraction of wood.

Remedy:
See remedy for Blistering.

10 9 WALLPAPERING

The term wall paper refers not only to paper substances

that are pasted on walls and ceiling but also includes vinyl, cork,
fabrics, grass cloth , foils and many other surface covering
materials.

Estimating your wall paper requires extra rolls in anticipation

of the following:

1. For replacement of ruined or damaged material in the

process of working and handling .

2. For additional areas which are not included in the plan or

overlooked in actual surface measuring.

3. For future repair which requires the same pattern color ,

texture and design.

336
 4. Trim can be used as a decorative boarder. It is sold by yard
or meter

Vinyl Wall Paper is classified into three kinds:

1. Vinyl laminated to paper

2. Vinyl laminated to cloth
3. Vinyl impregnated cloth on paper backing. This is extremely
durable, easy to clean and resistant to damage.

Caution in Buying Vinyl Wall Paper

1 . Examine the label if it is vinyl coated only.

This kind is not wear or grease resistant nor washable type .
2. Never confuse them with vinyl wall paper.
3. In buying your vinyl wall paper, always use and specify Vinyl
Mildew Resistance Adhesive only.
4. Vinyl wall paper stretches if pulled. Hair line cracks will
appear at seams as wall paper shrinks when it dries. Thus,
avoid stretching your vinyl wall paper.

Foils
Foil is another wallpaper simulated metallic finish or aluminum
laminated to paper. Do not fold or wrinkle the foil. There is no
remedy to creases. Smooth surface is required to avoid reflective
surfaces as foil magnify any imperfections on the surface to which
it is attached. Remember to specify mildew resistant vinyl adhesive
only.
Grass Cloth, Hemp, Burlap, Cork

Thes8 are mounted on paper or backing which could be

weakened from oversoaking with paste. Paste one strip at a time.

337
 Flocks
Flocks is made of nylon or rayon available on paper, vinyl or
foil wall papers Use paint roller or squeegee for best result .

Wall Paper Estimating Procedure

1 . Determine the surface area to receive wall paper .

2. Subtract the area opening such as doors , windows etc.

3. Divide the net wall area by the effective covering of the wall
paper size as presented on Table 10-1 to find the number of
roll.

4. Add 5 to 10% allowance depending upon the design pattern.

5. Multiply the number of rolls by the corresponding amount of

adhesive to get the number of boxes required.

TABLE 10- 1 WALL PAPER TECHNICAL DATA

Width Length / Roll Effective Covering Adhesive

Cm. Meter Per Roll ( sq m. ) Box per Roll

52 10.05 5.22 . 17
54 10.05 5.42 . 18
71 13.70 9.72 . 32

338
 CHAPTER 11
AUXILIARY TOPICS
11-1 ACCORDION DOOR COVER

The standard height of accordion door cover is 2.40 meters

or 8 feet Order could be lower than this height but computed at
2.40 m. standard height.
Estimating the quantity of materials for any given length of door
would be easy by using the following items comprising one meter
length of the accordion door.

-
TABLE 11 1 QUANTITY OF ACCORDION DOOR MATERIALS PER
METER WIDTH
Pieces Materials

12 pcs. G.l sheet blade 15 cm. x 2.40 m.

13 pcs. 6 mm x 2.40 m. Steel Pin
14 pcs. 5 mm x 25 mm x 2.40 m. Flat steel bars
24 pcs. 3 mm x 12 mm x 60 cm. Flat steel bars
1 pc. 6 mm x 32 mm x 2.00 m. Flat steel bar ( rail )
2 pcs. 38 mm dia. Roller Bearing with rivets and bushing.
35 pcs. 6 mm x 38 mm Rivets
52 pcs. 6 mm x 16 mm Rivets
112 pcs. 6 mm hole x 20 mm Washers .
1 pc. 50 cm. G. l . pipe 10 mm diameter

339
 To find the required material for a given door opening , multiply
each item by the span of the door opening in meters.

ILLUSTRATION 11-1

A 4.00 meters wide door requires an accordion cover for

security purposes. The accordion door is to open at the center
folded at both sides direction. List down the materials.

1.00 m
1

Black Sheet JLUL - .

Flat Bars - .|

\\

I
Elevation
Li J

Accordion Door

Plan

FIGURE 11 1 -
340
 SOLUTION:

1. Determine the span of door = 4 meters

2. Referring to Table 11-1

Multiply:

4 x 12 * 48 pcs. 15 cm. x 2.40 m. G .l Sheet Blade

4 x 13 = 52 pcs . 6 mm x 2.40 m. Steel Pin
4 x 14 = 56 pcs . 5 mm x 25 mm. x 2.40 m. Flat Bars
4 x 24 - 96 pcs . 3 mm x 12 mm. x .60 m. Flat bars
4 x 1 = 4 pcs. 6 mm x 12 mm. x 2.00 m. Flat Bars
4 x 2 = 8 pcs. 38 mm dia. Roller Bearing with bushings.
4 x 35 = 140 pcs. 6 mm. x 38 mm. Rivets
4 x 52 = 208 pcs. 6 mm. x 16 mm. Rivets
4 x 112 = 448 pcs. 6 mm hole x 20 mm washers
4 x 1= 4 pcs. 10 mm dia. x 50 cm. G.l. pipe (bushing)
,

11 -2 GLASS JALOUSIE

One important consideration in estimating glass jalousie is

the clear height of the window opening. If a window is to
accommodate glass jalousie, its height must be adjusted to the
number of blade intended to install. Table 11-2. provides the
standard height of jalousies corresponding to the number of blade.
With respect to the width of the window jalousie, it does not
present any problem because the glass blade can be adjusted to
the design length. The glass blade however ,should not be longer
than 90 centimeters.

341
 TABLE 11- 2 GLASS JALOUSIE STANDARD HEIGHT

Number Height Square Ft. Number Height

of Blade of Glass per Blade of Blade of Glass

4 14 7/8" 4 x 22 = 0.61 13 45 3/8"

5 18 3/8 4 x 2 4 = 0.67 14 49 7/8"
6 21 7/8" 4 x 26 = 0.72 15 53 3/8"
7 25 3/8" 4 x 28 = 0.78 16 56 7/8"
8 28 7/8" 4 x 3 0 = 0.83 17 60 3/8"
9 32 3/8" 4 x 32 = 0.89 18 63 7/8"
10 35 7/8" 4 x 34 = 0.94 19 67 3/8"
11 39 3/8" 4 x 36 = 1.00 20 70 7/8"
12 42 7/8" 4 x 38 = 1.06 21 74 3/8"
4 x 4 0 = 1.11 2? 77 7/8"
4 x 42 = 1.17
4 X 44 = 1.22
4 X 46 = 1.28
4 x 48 = 1.33

11-3 WATER TANK

The size and capacity of water tank for building structure

depends upon the volume of waterto be stored in. Volume of water
on the otherhand , depends upon the number of persons occupying
the building. Thus , our problem is how to determine the volume of
water and the size of the water tank which could serve efficiently
to an specified number of occupants.

ILLUSTRATION 11-2

342
 An airconditioned commercial office building has a total
occupancy of 800 persons. Determine the volume of water in cubic
meter to supply the building needs.

SOLUTION

1 . Determine the hourly water consumption of 800 persons.

Refer to Table 11-3. For airconditioned commercial type
building:
Multiply:

800 x 3.8 = 3,040 gallons per hour

2 . The tank must always have 1/2 hour supply thus;

3,040 gal. = 1,520 gal per 1 /2 hour supply

2
3. The pump therefore must supply 3 ,040 gallons / hour or:

3,040 a 50.7 gallons per minute.

60 min.

TABLE 11-3 WATER CONSUMPTION IN OFFICE BUILDING

Consumption in Gallons
Building Type Per Hour per Person
Commercial, no Airconditioning 3.8
Commercial, with Airconditioning 7.3 - 9.2
Owner occupied with kitchen and
laundry. No Airconditioning 7.3
Owner occupied with kitchen and
laundry with Airconditioning 9.0

343
 TABLE 11-4 HOT WATER CONSUMPTION IN APARTMENT
BUILDING

Public Housing Apartment Hotel

Consumption Gallons Gallons

Average Daily
Per Apartment 79 50
Per Person 28 41
Per Room 22 36

Maximum Daily
Per Apartment 92 69
Per Person 33 48
Per Room 26 43

Maximum Hourly
Per Apartment 8.9 7.6
Per Person 3.2 5.3
Per Room 2.5 4.9

TABLE 11-5 COLD WATER CONSUMPTION IN BUILDINGS

Type of Budding Gallons per Day per Person

Residence - Average 50
Residence - Large 100
Apartment - Low Rent 75
Apartment - High Rent 100
Hotels 100
Office Buildings 25

344
 ILLUSTRATION 11-3
Assuming that the water consumption of a commercial office
building is 3,040 gallons . Determine the size and volume of the
tank in cubic meters to contain 3,040 gallons.

SOLUTION

1. Byconversion:
1 cu. m. * 264.2 gallons

2. Convert 3,040 gallons to [Link].

Divide:

3,040 gal. = 11.5 cu. m.

264.2

Note: This volume is the inside dimension of the tank

3. The formula for finding the size of a cylindrical tank is:

d2 x 0.784 x height = volume

4 . Assuming that the height of the tank is 2.00 m.

d2 x 0.784 x 2.00 m. - 11.5 cu. m.

d2 x 1.568 - 11.5
d2 =* 11.5
1.568
d2 - 7.33

345
 d * 7 7.33
d •2.72 meters, diameter of the tank

5. The size of the water tank is

2.72 m. diameter by 2.00 meters high.

For more detail of this problem in plumbing , refer to the book

Plumbing Design and Estimate by the same author

11-4 WOOD PILES

Specifications. Piles shall be peeled removing all the rough

bark at least 80% of the inner bark. Not less than 80% of the
surface on any circumference shall be clean wood. No string of
inner bark remaining on pile shall be over 2 cm. wide and 20 cm.
long. All knots shall be trimmed closed to the body of the pile.

TABLE 11-6 WOODPILES

Length of Pile Diameter from Butt Minimum Tip

in Meter Min. Cm. Max. Cm. Diameter Cm.

Under 12 meters 30 45 20
t2.0 m. to 18.0 m. 32 45 18
Over 18 meters 35 50 15

The diameter of piles shall be measured in their peeled

condition. When the pile is not exactly round, the average of 3

346
measurements may be used. The butt diameter for the same length
of pile shall be as uniform as possible. All piles shall retain
preservative of at least the amount given in the following table:

-
TABLE 11 7 MINIMUM PRESERVATIVE PER CUBIC METER OF
WOOD

Type of Processing
Use and Type Empty Cell Process Full Cell Process

General Use 193 kg. 321 kg.

Marine Use 1Q3 kg. 321 kg.

-
TABLE 11 8 ALLOWANCE BEARING POWER OF DIFFERENT
SOIL

Kind of Soil Value in Tons per Sq. M.

Min. Max. Usual

1. Quick Sand and Alluvial Soil 5.38 10.76 5.38

2. Soft Clay 8.07 32.28 21.50
3. Wet Clay and Soft Sand 10.76 21.50 16.14
4. Clay and Sand in alternate layers 10.76 43.04 21.50
5. Finn and Dry Loom or Clay,
Hard dry day or Fine Sand 21.50 43.04 32.28
6. Confined Sand 10.76 43.04 32.28
7. Compact Coarse Sand or
Stiff Gravel 32.28 64.56 43.04
8. Sand and Gravel well Cemented 53.80 107.60 86.08
9. Good Hard Pan or Hard Shale 53.80 107.60 86.08
10. Rock 53.80 269.00 161.40

347
 -
TABLE 11 9 RANGE OF SKIN FRICTION FOR VARIOUS SOIL

Type of Value in Kilograms per Square Meter

Soil Minimum Maximum

1. Silt and Soft Mud 244 489

2. Silt Compacted 587 1,712
3. Clay and Sand 1,956 3,913
4. Sand with some Clay 2,446 4,891
5. Sand and Gravel 2,935 8,804

Lumber shall be treated by pressure method with creosoted

coal solution or creosote petroleum solution.

11-5 BITUMINOUS SURFACE TREATMENT

Hot Asphalt type - Approx . 1/2 or 1.25 cm. thick.

-
TABLE 11 10 BITUMINOUS MACADAM WEARING COURSE

Bituminous Aggregate
Application Materials Coarse Key
Liter / Sq. M. Kg. Liter / Sq. M.

First Spreading 90
First Application 4.0
Second Spreading 13
Second Application 1.8
Third Spreading 11
Third Application 1.4
Fourth Spreading 8
Total 7.2 122

348
 TABLE 11-11 BITUMINOUS MACADAM PAVEMENT

Hot Asphalt Type Approximately 21/4 or 5.7 cm. thick

Bituminous Aggregate
Activity Materials Coarse Choker Key
Liter / Sq. M. Kg. Liter / Sq. M.

First Spreading 90
Second Spreading 10
First Application 5.5
Third Spreading 10
Second Application 3.5
Fourth Spreading 8
Third Application 2.0
Fifth Spreading 8

Total 11.0 126

Open Graded Plan Mix Surface Course

The approximate amount of materials per square meter of the

open graded plant mix surfacing course and the sequence of
placing shall be as folows:

Plant - mixed aggregate - 80 kilos

Choker aggregate 3 kilos
Bituminous Materials 45 liters
Choker aggregate 3 to 5 kilos

The weight above are those of aggregate having a bulk

specified gravity of 2.65

349
 TABLE 11-12 BITUMINOUS SURFACE TREATMENT
( Hot Asphalt Type Approximately 1.6 cm . Thick )
Bituminous Aggregate
Operations Materials Coarse Choker
Liter / Sq. M. Kg- Liter / Sq. M.
First Spreading 1.0
First Spreading 22
Second Application 1.3
Second Spreading 6.5
Third Application 0.7
Third Spreading 4.5
Total 3.0 33
Values given with bulk specific gravity of 2.65.

11-6 FILLING MATERIALS

Estimating Procedure:

1. Compute for the volume to be filled up = L x W x H

2. Determine the kind of filling materials (see Table 11-13 )

3. Multiply the result found in step one by the corresponding

percentage of additional volume (Table 11-13 )

4. Add results to obtain the compact volume desired.

TABLE 11-13 FILLING MATERIALS

Materials ( Loose Volume ) % Additional to Obtain Compact Volume
Earth Fill 23%
Earth and Sand 16%
Selected Borrow 15%

350
 ILLUSTRATION 11-4

A residential lot as shown in Figure 11-2 requires filling.

Compute the materials required.

20.00 m. 20.00 m

X A 50 cm
m
15.00 m .
B 1.30 m

- -
"
M
-
4 SECTION X - X
PLAN

FIGURE 11-2

SOLUTION:

1. Find the volume of A

15.00 x 20.00 x 5.0 * 150 cu. m.

2. Find the volume of B

15.00 x 20.00 x .65 = 195 cu. m.

Total volume = 345 cu. m.

3. From Table 11-13 , consider 23% compaction allowance.

345 x 23% = 79.35 say 80

Volume required = 345 + 80
= 425 cu. m.

351
 -
11 7 NIPA SHINGLE ROOFING

Bamboo Split

y 7 -9" Spacing

Nipa Shingles

FIGURE 11-3

-
TABLE 11 14 NIPA SHINGLES TECHNICAL DATA

End Effective Covering No. of Pieces No. of Rattan

Lap in Square Meter per Sq. Meter per Sq Meter
Cm. Class A Class B A B *

10 (4") . 1300 1090 7.70 9.18 1.8 36

7.5 (T ) .0975
(
. 081B 10.27 12.23 2.0 4.0
5.0 (T ) .0650 .0545 15.38 18.31 2.6 56
2.5 (1") .0325 .0273 30.77 36 62 40 80

* Rattan ties for every other bamboo split ribs.

** Rattan ties for every bamboo spirt ribs.

Nipashingle is a native local product commonly used as

roofing materials. The nipa palm tree is the source of nipa shingles
which normally grows on swamp areas. One nipa stalk contains
approximately 174 to 180 nipa leaves.

352
 Nipa shingle is classified into two: class "A" and "B". The
nominal length is 150 cm. and 120 cm. respectively. The former
contains approximately 82 pieces nipa leaves while the latter has
approximately 68 pieces .

Roof Slope

The slope of the roof is an important consideration if nipa

shingle is to be used. The durability and life span of the nipa shingle
depends upon the slope or pitch of the rcof. The higher the slope,
the longer is the life, the lower the slope the more frequent you buy
nipa shingles to change your roof.
Slope of rafters less than 45 degrees is not advisable for the
following reasons:

a) Spacing of the nipa shingles would be relatively far from the

succeeding rows. Thus, the roof is considerably thin.
b) To install the nipa shingles to a closer spacing will only invite
rain water to flow back inside the house.
c) Rain water is hard to drain on a lower pitch roof. Moisture is
the number one enemy of nipa leaves, bamboo and even
wood.
d) Not only nipa shingles but even galvanized sheet deteriorate
faster on low pitch roof.

ILLUSTRATION 11-5

The area of the roof framing is 30 sq . m. having a general

dimension of 3.00 by 10.00 meters. Find the number of nipa
shingle required adopting 7.5 cm.(3") end lapping tied on the
bamboo split ribs at every other intervals using Class "A" nipa
shingle.

353
 SOLUTION:

1. Determine the area of the roof. Refer to Table 11- 14 under

class A * 7.5 cm. (3") lapping.
Multiply:

30 sq. m. x 10.27 * 308 pcs.

2. Nipa shingle is sold in bundles of 25 pcs.

Divide:

308 - 12.32 say 13 bundles

25

4. Solve for the required rattan splits. Referring to Table

11- 14 For rattan split.
Multiply:

30 sq. m. x 2 pcs./ sq. m. 60 pcs.

5. Determine the number of bamboo poles required.

Refer to Table 11-15 . using class B bamboo;
Divide

30 sq.m. = 3.3 say 4 pcs. or

9
if class A bamboo is used:

30 sq. m. * 2.5 say 3 pcs.

12

Bamboo poles are also classified into three. They are : Class
"A", "B" and "C", which are then divided into splits approximately

354
3.8 cm. to 5 cm. ( 1 1 /2" to 2" ) nailed on the purlins at an intervals
of 20 to 25 cm. apart.

-
TABLE 11 15 BAMBOO POLES DIVIDED INTO 4 CM, STICK

Bamboo Number of Approx. Length Coverage per

Class Splits in Meters Square Meter
A 8 9.0 12
B 6 7.5 9
C 4 6.0 4

Too fresh or too dried bamboo is brittle. Care should be

exercise to avoid cracks or separation of the bamboo grain in
fastening with nails. Cracks and splitting weakens your structure
but this could be avoided if the following procedures are observed.
a. Divide the newly cut fresh bamboo into the desired split
sizes.
b. Do not install nor fasten the bamboo splits with nails until
after sundried fora at least 8 hours.
c . Too dried bamboo is subject to cracks and splitting if nailed
on the purlins. Soak in water overnight before fastening with
nails.

Anahaw Leaves
uyiMj
m ; 7.5 cm to 12.5 cm

Bamboo Split
*
/ AW It '

MW
* -*i
m
b•
7* 9* Spacing
FIGURE 11 4-
355
11-8 ANAHAW ROOFING

-
TABLE 11 16 ANAHAW ROOF TECHNICAL DATA

End Lapping Number of Leaves Number of 3.0m. Rattan

Cm Per Sq. Meter Per Sq. Meter

7.5 84 7
10.0 60 6
15.0 45 5

Note:
1. Add 10 pcs. anahaw leaves per meter length along the gutter
tine.
2. Add 10 pcs. anahaw leaves per meter length of the ridge
and hip-line.
3. Add 5% allowance for damaged leaves. Dishonest supplier
insert damaged leaves in each bundle which could not be
detected until after it is opened for installation. However, if
your supplier is dead-honest, disregard no. 3.

ILLUSTRATION 11-6

From Figure 11-5, find the number of anahaw leaves and

rattan splits required.

SOLUTION:

1. Solve for the area of roof A and roof B.

Area A = 30 sq. m.

356
 Area B = 30 sq. m.
Total * 60 sq. m.

.
2. Refering to Table 11-16 adopting 10 cm. (4” ) end lapping:
Multiply:

60 x 84 = 5 ,040

3. Determine the length of:

a) Ridge line - 10.00 meters

b) Gutter line = 20.00 meters
Total - 30.00 meters

4. Referring to notation of Table 11-16 ;

Multiply:

10 pcs. x 30 meters - 300 leaves

Gutter Line

FIGURE 11 5 -
357
5. Add result of 2 and 4: Order:

5,040 + 300 = 5 ,340 pcs. anahaw leaves

6. Determine the number of rattan required. Refer to Table

11-16 Multiply:

60 x 6 - 360 pcs.

358
 CONSTRUCTION TERMINOLOGIES
ENGLISH PIPIPINO

Post Halige, Poste

Girder Gililan
Joist Soleras
Flooring Sahig, Suwelo
Girt Sepo

Beam Blga
Truss Kilo
Bottom Chord Barakilan
Top Chord Tahilan
Purlins Reostra

Collar Plate Sinturon

Fascia Board Senepa
Esternal Siding Tabike
Vertical Stud Pilarete
Horizontal Stud Pabalagbag

Ceifing Joist Kostiiyahe

Window Sill Pasamano
Window Head Sombrero
Window or Door Jamb Hamba
Open Stringer Hardinera

Closed Stringer Madre de Escalera

Tread Baytang
Riser Takip Siipan
Handrail Gabay
Moulding Muldora

Eave Sibe
Projection Bolada

359
 Framework Baiangkas
Gutter Kanal
Conductor Aiulod

Wrought Iron Strap Plantsuwela

Bolt Pierno
Scaffolding Plantsa
Stake Estaka
Plastered Course Kusturada

Stucco or Plaster Palitada

Scratch Coat Rebokada
Picfcwork on Masonry Piketa
Varnish Finish Monyeka
Spacing or Gap Biyento

Concrete Stab (rough) Larga Masa

Alignment Asintada
Plumb Line Hulog
Cement Tiles Baldosa
Cement Brick Ladrilyo

Door Fillet Batidora

Groove Kanal
Wood Grain Haspe
Pattern or Schedule Plantilya
Hinge Bisagra

Paneled Door De Bandeha

Earthfill Eskumbro
Masonry Fill Lastilyas
Adobe Anchor Liyabe
Solder Hinang
Soldering Lead Estanyo
Temper (metal work) Suban, Subuhan

360
 Diagonal Brace Piye de Galyo
Nail Setter Punsol , Punson
Wiring Knob Poleya
Cabinet Hinge Espolon

MENSURATION FORMULA

Triangle Area Base x Altitude

2
Parallelogram Area Base x Altitude

Trapezoid Area s; Sum of Parallel Side x Altitude

2

Circle Area S 0.7854 x Diameter 2

3.1416 x radius 2
Circumference = 3.1416 x Diameter
s 6.2832 x radius

Ellipse Area 0.7854 x Short Diameter x

Long Diameter

Cone Surface Slant height x Circumference

(curved only) of Base
Volume SS Area of Base x height
3

Cylinder Surface Length x Circumfernce plus

Area of Ends
Volume SS 0.7854 x Length x Diam. 2

361
Sphere Surface « 3.1416 x Diameter 2
m Circumference x Diameter
Volume « 0.5236 x Diameter 3
Circumference x Diameter 2
6
Vol. of Circumscribing Cylinder
3

362
 YOUR BOO:<r
By
-B. FajardoALL
.J;
TIME

Construct* '' i Estimate

* PUimbirg D shn r ,u Estimate
^
* planning .: id D Jgners Hanaiosk
^
* Elattrical Layout and Estfe
* Simplified Methods on Builc,1 ’rt ;onstruct;i i
* Eler . . nis of Roads and Highways
. -

- -
iMU 971 33* 3 10 -
Published c? :i Disnoufc - ' by
6138 Mttk hmH ring
d- A Fisheries St.
. 6
!
J
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;!

.
Quwxi City, ^hiiit
IS9N 97I85- 99!0-4
-
Te >. 924 7101 o- 7:9 61 / /