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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
Plain and Reinforced Concrete
The Principles of Concrete Mixing
The Unit of Measure
Concrete Proportions
Concrete Slab
Estimating by the Area Method
Concrete Column
Estimating Concrete Column by vhe Linear
Meter Method
1-9 Post and Footing
1-10 Rectangular Column
1-11 Rectangular Beam and Girder
1-12 Circular Column
1-13 Concrete Pipes
1-1
1-2
1-3
1-4
1-5
1 -6
1-7
1-8
1
3
5
8
11
14
16
17
21
27
29
30
33
CHAPTER- 2 MASONRY
2-1 Concrete Hollow Blocks
2-2 Special Types of Concrete Hollow Blocks
2-3 Decorative Concrete Blocks
2-4 Adobe Stone
37
58
65
68
CHAPTER- 3 METAL REINFORCEMENT
3- 1
3-2
3-3
3- 4
3-5
Steel Bars
Identification of Steel Bars
Bar Splice , Hook and Bend
Reinforcement for Concrete Hollow Blocks
Tie Wire for Steel Reinforcement
75
77
78
85
89
Independent Footing Reinforcement
Post and Column Reinforcement
Beams and Girders Reinforcement
Lateral Ties
3- 10 Stirrups for Beam and Girder
3- 11 Spiral and Column Ties
3- 12 One Way Reinforced Concrete Slab
3- 13 Two Way Reinforced Concrete Slab
3- 14 Concrete Pipe Reinforcement
3-6
3- 7
3-8
3- 9
91
99
100
102
115
118
122
126
129
CHAPTER- 4 LUMBER
4- 1 Wood
4-2 Definitions of Terms
4 -3 Classification of Wood
4 -4 Methods of Log Sawing
4 -5 Defects in Wood
4 -6 Seasoning of Lumber
4-7 The Unit Measure of Lumber
4 - 8 Wood Post
4- 9 Girder
4- 10 Floor Joist and T&G Flooring
4 - 11 Siding Wood Board
4- 12 Girts, Rafters. Truss. Purlins and Fascia Board
4- 13 Studs
4- 14 Ceiling Joist
4-15 Ceiling Board
4- 16 Door Frame
4 -17 Window Frame
132
132
134
136
137
138
139
144
146
150
155
158
159
164
166
172
175
CHAPTER - 5 FORMS, SCAFFOLDING AND STAGING
5- 1 Forms
5-2 Greasing of Forms
5 - 3 Scaffolding and Staging
177
179
179
5-4 Comparative Analysis Between the T&G and
Plywood as Forms
5-5 Forms Using Plywood
5-6 Forms of Circular Column
5-7 Estimating the Scaffolding and Staging
181
190
193
200
-
CHAPTER 6 ROOFING MATERIALS
6-1 Galvanized Iron Sheet
6-2 Gutter , Flashing, Ridge, Hipped and
Valley Roll
6-3 Asbestos Roofing
6-4 Colorbond Klip-Lok
6-5 Banawe Horizontal Metal Tile
6-6 Marcelo Roofing System
6-7 Colorbond Custom Orb
6-8 Milano Long span Steel Bricks
6-9 Colorbond Trimdex Hi-Ten
6-10 Brick Tiles Roofing
206
222
228
235
237
238
239
240
241
242
CHAPTER - 7 TILEWORK
7-1 Ceramic Tiles
7-2 Marble Tiles
7-3 Vinyl and Rubber Tiles
7-4 Terrazzo and Granolothic
7-5 Cement Tiles
7-6 Wood Tiles
7-7 Pebbles and Washout Finishes
243
252
254
256
260
263
265
CHAPTER - 8 HARDWARE
8-1 Bolts
267
8- 2 Screw
8-3 Nails
274
279
CHAPTER - 9 STAIRCASE
9- 1 Introduction
9-2 Stairs Layout
9-3 Stringer
291
295
301
CHAPTER - 10 PAINTING
10-1 Paint
10- 2 Ingredients of Paint
10-3 Essential and Specific Properties of a
Good Quality Paint
10-4 The Elements of a Good Painting Job
10-5 Specifications for Surface Preparation
10-6 Kinds of Pamt. Uses and Area Coverage
10-7 Estimating Your Paint
10-8 Paint Failure and Remedy
10- 9 Waif Papering
309
310
312
313
315
318
327
333
336
CHAPTER - 11 AUXILIARY TOPICS
11- 1 Accordion Door Cover
11-2 Glass Jalousie
11- 3 Water Tank
11-4 Wood Piles
11 -5 Bituminous Surface Treatment
11 -6 Filling Materials
11 - 7 Nipa Shingle Roofing
11 -8 Anahaw Roofing
Construction Terminologies
339
341
342
346
348
350
352
356
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
b. Strength
c. Durability
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
. 5334
. 525
.
025
21
22
. 5588
. 0762
050
. 075
23
. 5842
. 1016
.100
. 6096
. 1270
. 125
. 1524
. 150
. 1778
. 175
. 2032
. 200
. 2286
. 225
. 2540
.
. 2794
. 275
12
. 3048
. 8128
. 800
13
14
15
16
17
. 3302
300
. 325
. 350
. 375
24
25
26
27
28
29
30
31
32
33
34
. 8382
. 825
8636
. 850
35
36
. 8890
875
. 900
. 925
. 950
. 975
1.000
1
. 0254
.
2
3
4
5
6
7
8
9
10
11
. 0508
.
18
19
20
.
3556
. 3810
. 4064
. 4318
. 4572
. 4826
5080
.
.
.
250
.
400
. 6350
. 6604
. 6858
. 7112
. 7366
7620
. 7874
.
.
. 9144
. 9398
. 450
37
38
.475
39
. 9906
500
40
1.0160
.425
5
. 9652
550
. 575
. 600
. 625
. 650
. 675
. 700
. 725
. 750
. 775
.
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.
. 30
= 20 ft .
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
=
2 inches =
3 inches =
4 inches =
.025 m.
.050 m.
. 075 m.
.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.
1” « .025 m.
plus 1 " =
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
Class
Mixture
AA
A
1 : 11/2 ; 3
1: 2 : 4
1 : 21/2 : 5
1: 3 : 6
B
C
Cement
40 kg. / Sag^ 50 kg. / Bag
12.0
9.5
7.0
6.0
5.0
9.0
7.5
6.0
Sand
cu.m.
Gravel
cu. m.
.5 0
1.0
.5 0
.5 0
1.0
1.0
.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
450 cu . m.
Sand : 900 x . 50 *
900 cu. m.
Gravel : 900 x 1.0 =
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
cm.
5.0
7.5
10.0
125
15.0
17.5
20.0
22.5
25.0
27.5
30.0
Mixture Class
50 kg. Cement
40 kg. Cement
C
C
A
A
B
B
.30
.450 .375 . 30 . 350
.250
.675
.900
1.125
1.350
1.575
1.800
2 030
2.250
2475
2.700
. 563
.750
.938
1.125
1.313
1.500
1.688
1.875
2063
2.250
. 45
. 60
.75
.90
1.05
1.20
1.35
1.50
1.65
1.80
. 525 . 450 .375
.700 .600 .500
.875 .750 .625
750
1.050 .900
1.225 j 1.050 .875
1.000
1.400
'
1.575 1.350 j 1.125
1750 1.500 1.250
1925 1650
2.100 1800 11500
11.200
11375
.
14
Sand Gravel
cu. in. cu. m.
.0250 .050
.0375 .075
.0500 .100
.0630 .125
.0750 .150
.0880 .175
.1000 .200
1125 .225
1250 .250
. 1380 .275
.1500 .300
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
: 6 ,000 x .075
450 cu. m .
Sand
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
T
5 00 m
X
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
Sand
Gravel
3.0 x 9.0 = 27 bags at 40 kg.
3.0 x .50 = 1.5 cu . m.
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
Sand Gravel
Mixture Class
cm.
40 kg . Cement
50 kg. Cement cu. m. cu. m.
A
B
A
B
.169
.158
.203
.135
15 x 15
.011 .023
15 x 20
.270
.225
.180
.015
.030
. 210
.338
.281
15 x 25
.038
. 263
.225
.019
.405
.338
.315
.270
. 023
.045
15 x 30
.473
.053
.315
.394
.369
.026
15 x 35
.540
15 x 40
.450
.420 .360
.030
. 060
20 x 20
20 x 25
20 x 30
20 x 35
20 x 40
25 x 25
25 x 30
25 x 35
25 x 40
25 x 45
25 X 50
30 x 30
30 X 35
30 x 40
30 x 45
30 x 50
35 x 35
35 x 40
.360
.450
.540
.630
.720
.300
.375
.450
.563
.469
.563
.656
.750
.844
.938
.438
.675
. 788
.900
1.013
1.125
.280
. 350
.420
.490
.560
.525
.600
.525
.613
. 700
.788
.875
.240
. 020
.040
.300
.360
.420
.480
.025
.030
.035
.040
.050
.060
. 070
.080
.375
.450
.525
.600
.675
. 750
.031
.063
.038
.044
.050
.056
.063
.075
.045
.090
.105
061
.123
.140
.810
. 945
1.080
1.215
.675
.630
.788
.900
1.013
.735
.540
.630
. 840
.720
.945
.810
1.350
1.125
1.050
.900
.053
.060
.068
.075
1.103
1.260
. 919
.858
.735
.
1.050
. 980
.840
. 070
18
.088
.100
.113
.125
.120
.135
.150
35 x 45
35 x 50
35 x 55
35 x 60
1.418
1.575
1.733
1.890
1.181
1.313
1.444
1.575
1.103
1.225
40 x 40
40 x 45
40 x 50
40 x 55
40 x 60
1.440
1.620
1.800
1.200
1.120
1.260
45 x 45
45 x 50
45 x 55
1823
2.025
2.228
2.430
45 x 60
1.980
2.160
1.350
1.500
1.650
1.800
1.348
1.470
1.400
. 945
1.050
1.155
1.260
.960
1.080
1.680
1.200
1.320
1.440
1.418
1.215
1350
1856
1.575
1733
2.025
1.519
1.688
1.540
.079
.088
.158
.
096
.193
.105
.210
.080
.090
.160
.180
100
.200
.110
.220
.240
.
.
120
.175
.203
1.485
. 101
.113
.124
1.890
1.620
.135
.270
1.500
.125
.250
1, 650
1800
. 138
. 150
. 275
.300
. 165
.330
.193
.385
. 220
.440
. 248
.495
.225
.248
50 x 50
50 x 55
50 x 60
2.250
2.475
1.875
2.063
2.700
2.250
1750
1.925
2.100
55 x 60
55 x 70
55 x 80
55 x 90
55 x 100
2.970
2.475
2.888
3.300
3.713
2.310
2.695
3.080
3.465
1980
2.310
2.640
2.970
4.125
3.850
3.300
. 275
.550
60 x 60
3.240
. 180
.360
3.780
4.320
4.860
5.400
2.520
2.940
2.160
60 x 70
60 x 80
60 x 90
2.700
3.150
2.520
. 210
.
3.600
3.360
. 240
.480
4.050
4.500
3.780
4.200
2.880
3.240
. 270
.540
3.600
. 300
.600
60 x 100
3.465
3.960
4.455
4.950
19
420
2.925
3.413
3.900
4.388
4.875
2.730
3.430
5.040
5.670
6.300
3.675
4.200
4.725
5.250
4.725
3.938
5.400
65 x 60
65 x 70
65 x 80
3.510
4.095
65 X 90
65 x 100
5.265
5.850
70 x 70
4.410
70 x 80
70 x 90
70 x 100
75 x 70
75 X 80
75 X 90
75 X 100
4.680
6.075
6.750
3.185
3.640
4.095
2.340
2.730
3.120
3.510
3.900
195
.228
. 260
.293
.
.390
.455
.520
.585
. 325
.650
2.940
3.360
3.780
4.200
. 245
.490
.280
.560
. 315
.630
. 350
.700
3.150
3.600
4.050
. 263
4.500
5.063
5.625
3.675
4.200
4.725
5.250
338
. 375
.525
.600
.675
.750
4.480
5.040
5.600
3.840
4.320
4.800
. 320
.640
. 360
.720
. 400
.800
4.760
4.080
4.590
5.100
. 340
.680
. 383
.765
.425
.850
80 x 80
80 x 90
80 x 100
5.760
6.480
7.200
4.800
5.400
6.000
85 x 80
85 x 90
85 x 100
6.120
6.885
7.650
5.100
5.738
6.375
90 x 90
90 X 100
7.290
8.100
6.075
95 X 90
95 x 100
100 x 100
7.695
8.550
9.000
6.750
6.413
7.125
7.500
4.550
3.920
4.410
4.900
5.355
5.950
4.500
. 300
.
5.670
6.300
4.860
.405
.810
5.400
.450
. 900
5.985
6.650
7.000
5.130
5.700
6.000
.428
.475
.855
.950
. 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
: 40 x .038 = 1.5 cu. m.
Sand
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
20 cm.
80 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
: 48 x .025 = 1.2 cu . m.
Sand
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
: 7.68 x . 100 = .768
Sand
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
1.20 + .768 = 1 97 say 2 cu. m.
Sand
Gravel : 2.40 + 1.54
- 3.94 say 4 cu. m.
30 cm.
30 cm.
r —!
! Ft
t
L- j-l
3.00 m.
l
• W \\
i
i
ii
«
P
90 cm.
I
i
F
4.00 m.
I
FIGURE 1-6
24
*»
80 cm —
15 cm
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 x 4 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
a 0.708 cu. m.
Total volume F & P
3. Referring to Table 1-2, using class "A” concrete:
Multiply:
25
Cement
Sand
Gravel
.708 x 9.0 - 6.37 bags
.708 x .50 * .354 cu. m.
.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
: 1.2 x .50 - 0.6 cu. m.
Sand
Gravel : 1.2 x 1.0 = 1.2 cu. m.
Summary of the Materials
Cement : 6.37 + 7.2 * 13.57 say 14 bags
: .354 + 0.6 = .95 cu. m.
Sand
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 . c.ement class "A ” mixture:
Multiply:
27
Cement : 40 x 1.680 = 67.2 say 68 bags
: 40 x .120 - 4.8 say 5 cu. m.
Sand
Gravel : 40 x . 240 = 9.6 say 10 cu. m.
r
A
1
A
I
5.00 m.
I
40 cm
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
£
£
s
8.00 m.
OQ
i
tu
OQ
erA
Girder
A
12.00 m..
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 I.OO
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
Mixture Class
Size
cm.
40 kg. Cement
25
A
. 442
. 636
30
35
40
45
. 866
1.131
1.431
1.767
50
55
2.138
2.545
2.986
3.464
3.976
4.524
5.107
5.726
7.069
60
65
70
75
80
85
90
100
B
. 368
.530
.722
.942
1.193
1.473
1.782
2.121
2.488
2.866
3.313
3.770
4.256
4.771
5.890
50 kg. Cement
A
B
.295
. 344
. 495
.424
.673
.577
.880
.754
.954
1.113
1.178
1.374
1.663
1.979
1.425
1.696
2.323
2.694
1.991
2.309
2.651
3.016
3.093
3.519
3.972
4.453
5.498
3.405
3.817
4.712
Sand
Gravel
cu. m.
cu. m.
.025
.059
.071
.096
.126
.159
.196
.238
.283
.332
.385
. 442
.503
.035
. 048
. 063
. 080
. 098
.119
.141
.166
.192
. 221
.251
. 284
. 318
. 393
. 567
.636
.785
r
6.00 m .
A
—I
*
4.50 m.
-}—
I
Circular Column
FIGURE 1-9
31
Section A-A
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
. 30 x .141 =
Sand
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)
d * 90 cm .
1.10 m.
»
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
40 kg.
„
33.93 say 34 bags
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
Gravel
Sand
Cement in kilograms
Diameter
cu. m.
cu. m.
Class of Mixture
cm
B
A
D
di
25
30
40
45
55
60
65
80
85
90
100
110
15
20
120
100
120
145
175
25
30
40
45
50
60
65
70
80
90
150
11.31
14.14
27.57
31.81
40.29
44.53
48.77
79.17
84 82
90.48
101.79
113.10
124.41
187 32
229.73
942
11.78
22.97
26.51
33.58
37.11
40,64
65.97
70.69
75.40
84.82
94.25
103.67
156.10
191,44
.038
. 0314
.0400
.0770
.044
.0884
. 056
. 118
.1120
.1240
. 1360
.2200
.2360
.126
.2510
.141
.157
.2830
.3140
.173
. 3460
. 260
.5200
.319
. 6380
.016
.020
.062
.068
.110
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.
Mortar
^
1
15 cm.
40 cm.
FIGURE 2- 2
39
t
TABLE 2-1 QUANTITY OF CEMENT AND SAND FOR MORTAR
AND PLASTER MIXTURE PER CUBIC METER
Cement in Baps
Cta»»
Mixture
40 kg.
SO kg.
A
B
C
D
1:2
1: 3
1:4
1:5
18.0
12.0
9.0
14.5
9.5
7.5
7.0
6.0
Sand
cu. m.
1.0
1.0
1.0
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 :
Sand
.375 x 12 = 4.50 bags
.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.
20 cm.
CHB Cell
-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 :
Sand
2.25 x 12 27 bags
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 :
Sand
1.92 x 12 * 23.04 bags
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
2. 40 kg. Cement
3. Sand
4 . Gravel
750 pcs.
63.54 say 64 bags
5.14 cu.m.
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.
Mixture
Siz ofCHB Number
Mixture
Sand
sq
Per . m .
cm.
B
C
C
cu.m.
D
B
D
10 x 20 x 40
0.525 0.394 0.328 0.416 0.306 0.263 . 0438
12.5
1.013 0.759 0.633 0.802 0.591 0.506 .0844
15 x 20 x 40
12.5
12.5
20 x 20 x 40
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
: 60 x .0438
Sand
* 31.5 bags
* 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.
63.5 say 64 bags
2. 40 kg. Cement
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
kg
40 kg. Cement
Sand Gravel
50 . Cement
t
w
A
B
cu. m. cu. m.
A
B
10
10
10
10
15
15
15
15
20
20
20
30
.270
.225
.210
. 180
. 015
. 030
35
.315
.263
.245
. 018
. 035
40
. 360
.300
.280
. 210
. 240
. 040
50
.450
40
45
. 540
.608
.675
. 810
.720
. 900
1.080
.380
.450
. 350
.420
.473
.525
. 020
. 025
. 030
.034
.038
.045
. 040
50
60
40
50
60
.506
.563
. 300
. 360
.405
.450
.630
.540
. 480
. 750
.560
. 700
.900
. 840
.675
.600
.600
.720
. 050
. 060
. 050
.060
. 068
.075
.090
.080
. 100
. 120
-
TABLE 2 4 QUANTITY OF CEMENT AND SAND FOR PLASTERING
PER SQUARE METER
Sand
Cement in Bags
Class
40 kg. Cement
50 kg . Cement
cu. m
Mixture
.232
.288
A
. 152
.192
B
C
.144
.112
. 096
D
. 120
* Thickness computed at an average of 16 mm.
. 016
016
.016
.016
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 :
Sand :
130 x 1.013 = 131.7 say 132 bags
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.
Sand
Grave!
-
pcs.
—- 1,625
212 begs
18 on. m.
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
* 10.00
40
"
10.00
FIGURE 2-6
48
1«
papi
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
Sand : 218.4 x .0438
49
= 114.66 bags
= 9.57 cu. m.
.
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
: 364 x .016 = 5.82 say 6 cu. m.
Sand
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
Sand
Gravel
83.6 x .300 = 25 bags
83.6 x .020 * 1.67 cu. m.
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.
35
i
300
60
70
1
100
^
4
v.
120
70
35 Q
B
1
120
I I
210
A
270
I
Natural Ground
60
f
600
-
FIGURE 2 7
52
Footing line
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.
36.625 sq. m.
Total area of A & B
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
Net wall area
*
3.360 sq. m.
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:
= 25.24 bags
Cement : 33.265 x .759
: 33.265 x .084
Sand
- 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
Sand
Gravel
9.50 x .450 * 4.28 bags
9.50 x .030 = 0.28 cu. m.
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
C
cm.
8
0
C
B
cu. m.
D
10 x 20 x 40
4.200 3.152 2.624 3.328 2.448 2.104 0.350
8.104 6.072 5.064 6.416 4.728 4.048 0.676
15 x 20 x 40
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
Mixture
40 kg. Cement
50 kg. Cement
C
A
A
B
C
B
No. of Face
plastered
2.304
4.608
One face
Two face
1.536
3.072
1.152
1.856
1.216
2.304
3.712
2.432
0.896
1.792
Sand
cu. m.
0.128
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.
l
, 1\
I
zrzi\
i
I
i /
CHB Wall
T
Cement Plaster
E
8
l
K
Ground Line
J»
Footing line
V
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
Stretcher Block
' 19
XI v
s
^
14
‘
2 Core
L-Comer Block
Single End 8lock
14
,
Half Block
*
39
1
19
J>
Beam Block
1
Half Block
'9
39
19
>
Beam Block
Special Types of Hollow Blocks
FUGURE 2-9
59
Single End Block
TABLE 2-7 QUANTITY OF CEMENT, SAND AND GRAVEL PER
HUNDRED SPECIAL TYPES OF CONCRETE HOLLOW BLOCKS
T
CHB Size
in Cm.
Stretcher Block
2-core - 20 cm.
2-core - 15 cm.
L - Corner Block
2-core - 20 cm.
2- core - 15 cm.
40 kg. Cement Mortar
Class of Mixture
C
D
B
Sand
Gravel
4.32
4.06
.67
.83
6.23
7.20
4.88
. 30
25
8.67
5.90
6.87
4.67
5.78
3.89
.60
. 93
.41
.65
8.85
6.12
6.98
.63
.90
.44
.60
4.2
4.82
3.15
5.82
4.02
2.63
Half Block
20 x 20 cm ,
15 x 15 cm.
3.98
2.70
3.15
2.14
2.63
1.78
.28
.16
.45
Beam Block
2-core - 20 cm.
2-core - 15 cm.
A
8.78
5.85
B
7.32
4.88
C
5.85
3.90
Sand
. 49
Gravel
. 98
.65
Single End Block
2-core - 20 cm.
2-core - 15 cm.
2-core - 10 cm.
9.19
.35
. 33
.
30
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
.20
*
96 pcs. single end block.
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
.40
- 90 pcs. beam block
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
17.30 x .25 = 4.30 cu. m.
.96 x .60 = .58 " "
1.20 x .30 = .36 "
.90 x .65 * .59 " ’•
5.83 cu. m.
17.50 x .30 * 5.13 cu m.
.96 x .44 = .43 "
1.20 x .16 = .19 " "
.90 x .33 = .30 " "
Total . . . . 6.05 cu m.
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
5 x 10
5 x 15
5 x 20
5 x 25
10 x 20
10 x 25
10 x 30
200
133
100
A
B
100 Blk.
180
. 270
. 120
.010
. 180
. 015
240
. 300
. 480
.600
.720
. 020
.
. 360
80
.450
50
40
33
. 720
. 900
1.080
.
.
025
. 040
. 050
. 060
TA BLE 2-9 QUANTITY OF CEMENT AND SAND FOR VARIOUS
TYPES OF BRICKS PER HUNDRED BLOCKS
Size
cm.
t h xI
*
6 x 12 x 19
10 x 14 x 19
10 x 14 x 23
10 x 24 x 24
10 x 14 x 39
10 x 19 x 39
Number
40 kg. Cement per 100 Blocks
per sq. m.
Mixture Class
A
B
38.5
. 346
. 230
33.3
. 612
. 408
27.8
684
. 456
160
. 882
. 588
16.7
. 972
. 648
1.062
12.5
. 708
65
Sand
per
100 blks.
. 0192
.0340
. 0380
. 0490
. 0540
.0590
ITALIAN
55 x 215 x 125 mm
BOLIVIAN
100 x 160 x 180 mm
100 x 140 X 240 mm
MSpf
5
or
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
9 @s
JOSEPHINE
100 x 250 x 250 mm
A
SKMS8
.
ASG
100 x 250 x 250 mm
057ZI]
CI/MZJ
OOTZ3
CJ&YZ1
a a n?z3
cr /ra cr^'a I croT3
cw^a I OkVXJ
ama cra
o| POO
ci o' jj 0.9 13 IQ9£)|
1176 a3
aw a vjsIn o,^
crmra |CTAO jc7A o
AUM
100 x 250 x 250 mm
EGYPTIAN
100 x 250 x 250 mm
ayjD
^
^-
.
.
-
.
.-
3 83 S3
S
a
(Jo oLJo QUO
°O“l“0S
“?oQ2 EkoliFfcfi
100 x 250 x 250 mm
PERSIAN
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
J.. :; -
T IS
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 *
Size
No.
40 kg. Cement
50 kg. Cement
cm.
/ sq.
Class Mixture
Class Mixture
Sand
B
C
D
cu. m
m.
B
C
D
1 5 x 1 5 x 4 5 . 2808 . 2106 . 1756 . 2220 . 1638 . 1404
1 5 x 2 0 x 4 5 .2520 .1890 . 1575 . 1995 . 1470 .1260
1 5 x 3 0 x 3 0 .2280 . 1708 . 1423 . 1803 .1328 .1139
8 1 5 x 3 0 x 4 0 .2079 . 1559 . 1299T . 1646 . 1213 . 1040
6.5 1 5 x 3 0 x 4 5 . 1901 . 1426 . 1188 . 1505 .1109 . 0951
12
10
11
.0234
. 0210
.0190
. 0173
. 0158
* Average thickness = 20 mm
TABLE Ml QUANTITY OF ADOBE STONE, CEMENT AND SAND
FOR BUTTRESSES AND FOOTINGS
Buttress and Footing
Buttress
cross
Section
30 x 45
45 x 45
45 x 60
45 x 75
45 x 95
No .
Cement Mortar per Stone
Mixture Class
Number
40 kg. Cement
of stone
Course per m, ht.
of
2
3
4
5
6
12
18
24
30
36
B
C
D
.027 .0203 .017
.029 .0220 .018
027 .0203 .017
.
.032
.0240 .020
.034 .0253 . 021
SO kg. Cement
B
C
D
Sand
cu. m .
.0214 .016 .014
.0230 .017 .015
.0214 .016 .014
.0250 .018 .016
.0270 .020 .017
.0023
.0025
.0023
.0026
.0028
ILLUSTRATION 2- 7
From Figure 2-14 compute for the quantity of adobe stone ,
cement and sand using class C mixture.
69
_
JTF
500
. . 500
500
15
15
15
30
Buttress
30
15
15
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
: 82 x .0023 = . 18 cu m.
Sand
.
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.
Cement
Sand
629 pcs.
21.2 say 22 bags
2.4 cu. m.
TABLE 2-12 QUANTITY OF CEMENT AND SAND FOR
PLASTERING PER SQUARE METER
Bag Cement and Mixture C!ass
40 kg. Cement
50 kg. Cement
B
C
D
B
C
D
Side
One face
Two Faces
. 300
600
.
225
. 450
.
. 188
. 375
238
. 476
.
Sand
cu. m.
.
175
. 150
. 025
.
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
i
Types of Deformed Bars
FIGURE 3-1
75
wl
‘
i-
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
8 mm
10 mm
12 mm
13 mm
16 mm
1.96
3.08
2.37
2.96
3.70
5.33
6.25
9.47
14.80
23.12
4.62
6.66
3.56
5.54
4.15
6.47
4.74
7.39
5.33
8.32
9.32
10.94
16.58
10.66
11.84
18.50
28.90
11.99
14.07
21.32
33.29
52.02
29.00
36.25
33.29
41.62
7.99
9.38
14.21
22.19
34.68
43.50
49.94
37.88
47.35
56.62
58.26
66.29
47.95
59.93
71.92
83.91
20 mm
25 mm
28 mm
30 mm
32 mm
36 mm
4.44
5.21
7.90
12.33
19.27
24.17
27.75
31.57
39.96
7.83
10.5 m 12.0 m 13.5 m
12.50
25.89
18.95
29.59
40.46
50.75
46.24
58.00
65.25
74.91
66.59
75.76 8523
95.89 107.88
-
TABLE 3 2 DEFORMATION REQUIREMENTS
T
Max. Average
Height
Tolerance
Diameter
Spacing of luge
Minimum
Maximum
8
7.0
10
7.0
8.4
0.3
0.4
05
0.6
0.7
1.0
1.2
14
0.6
0.8
1.0
1.2
1.4
15
30
1.6
18
32
Nominal
12
13
16
20
25
28
30
32
36
9.1
11.2
14 0
17 5
196
21 0
22.4
24.5
!
i
76
2.0
2.4
28
36
Max Value
Summation
of lug& gap
5.5
7.8
9.4
10.2
12.6
15.7
19.6
22.0
236
25.1
27.5
-
TABLE 3 2A MECHANICAL PROPERTIES
Yield
Strength
Class
Grade
Tensile
Elongation
Bending
Diameter
Strength Specimen
in 200 mm
Angle
of Pin
percent
Degree
d^nominal
MPa
MPa
mm
mm
mm.
min.
dia. at
specimen
Hot Rolled 230
Non Weldable def- 275
Steel Bar
415
230
390
275
480
415
020
D < 25
D > 25
D < 25
D > 25
D < 25
18
16
10
8
180
D > 25
7
20
180
390
D < 25
275
480
D > 25
D < 25
D < 25
415
550
Hot Rolled 230
Weldable
275
Def. or
Plain bar
230
415
D < 25
D > 25
180
180
8
18
16
14
14
12
180
180
3d
4d
4d
5d
5d
6d
3d
4d
4d
5d
5d
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
—
N
-
-M
Bar Size
-6
Steel Type
A
Grade Mark
0
Grade 40
Grade 50
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
1. Tension Bars
2. Compression Bars
Minimum Splice length
* 25 x Bar Size + 150 mm.
* 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
h * 9d
cir
n = 5d
0f
2d|
4d
l
P
/
k-t
Hook
Hook Length
- L+
Bend
h for hook
Mild Steel Minimum Hook and Bend Allowance^
j
g
=
ri
d
h = 11d
I
4d
1
n = 5.5d
6d
i
I
*A
3d
9
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
=
1
Total Length
=A B C
A
Id
~1
A
Total Length * A + B + C
D
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
'
1
71
i
Total Length = A h
h
B
l
Total Length = A
,a
V
r
A
2h
r
A
i
Total Length = A + h
%
A
I
Total Length = A + B - 1/2 r - d
h
r
1
Total Length = A + 2h
-A
B
dlc
I
B
\
C
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
Number of Bars ( mm)2
Dia.
6
28
2
57
8
50
101
151
201
251
302
10
79
157
236
314
393
12
113
226
339
452
565
16
402
603
804
1005
20
201
314
471
679
1206
628
942
1257
1571
1885
25
491
1963
2454
32
982 1473
804 1608 2412
3216
40
1256 2513 3769
5026
1
3
4
5
6
7
6
9
10
85
113
141
170
198
254
452
283
352
550
226
402
628
707
905
503
792
1407
2199
1017
785
1131
1608 1809
2011
2513 2827
3142
3436
3927 4418
4909
4021
2945
4825
5629
6433 7237
8042
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 Vertical Relnf. © 80 cm . o. c.
CHB
rizontef Relnf
-- fteverv
3 layers
.
i
2 60 m
T\
\
IZE
£
4 00 m
FIGURE 3-6
86
Natural Ground
40 cm
- Footino
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
Length of Bars in Meter
Spacing
Length of Bars in Meter
cm.
Per Block
Layers
Per Block
Per Sq. M.
40
60
80
0.235
2
3
4
0.264
0.172
0.138
3.30
2.15
1 72
Spacing
0.171
0.128
Per Sq. M.
2.93
2.13
1.60
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
Spacing
Horizontal
Layer Spacing
40
40
40
2
3
4
2
3
4
2
3
4
60
60
60
80
80
80
Kilogram per 100 CHB
30 cm. ties
25 cm. ties
.0042
.0051
.0031
.0038
.0028
.0033
.0028
.0021
.0034
.0018
.0022
.0021
. 0025
.0019
. 0017
0016
.0014
.
90
.0025
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.
- Footing Slab
s
hr ./ :- |
\
• •
v* .
^
Steel Bars ReinfC~ steel Bars Reinf""*)- .:
1
5
•
<
£§>
.5 cm
I
7.5 cm
7.5 cm
-L-
^ *3%
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.
.
4
i
1.50 m
1.35 m
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
7.5 cm
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
P~DT
12 mm Steel Bars
TT |
y
u-U
1.15 m.
L = 1.20 m.
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
6.00
* 48 pcs.
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
r Beam
Add Length
1'
Lap Joint
-•
<
Column
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
Outer to Outer side
Inside to Inside
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
Inches
mm. (approx . )
1/4
6
3/8
10
1/2
13
5/8
16
Number Designation
2
3
4
5
6
7
3/4
7/8
1.0
1 1/8
1 1/4
8
9
10
102
19
22
25
28
31
11
12
13
14
15
16
1 3/8
1 1/2
1 5/8
1 3/4
1 7/ 8
2.0
17
2 1 /8
2 1/4
18
35
38
41
44
47
50
53
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 cm
X
1
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
30 cm
10 mm
Lateral Ties
O
0 - Col. Reinf.
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
'
T
"
Lateral Tie*
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.
ir
40 cm
-TJ.
JLL
?HF - — Outer Ties
Straight Ties
-5
p
i h r
au c
Inner Ties
25 cm
25 cm
1
n
V,
FIGURE 3- 16
108
Given Data:
Number of Columns
Size
Clear Height
Ties Spacing
= 16 pcs .
= 25 x 40 cm.
4.60 m.
= 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:
Inner ties:
Straight ties:
600 ~
120
600 =
5 pcs.
7 pcs.
85
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
NUMBER OF CUT IN ONE
STEEL BAR LENGTH
5.0 m. 6.0 m. 7.5 m . 9.0 m. 12.0 m.
Spacing of Number of
Length of
Lateral
Lateral Ties
Ties with
Ties cm.
PerM. Ht.
Hook & Bend
15
6.70
60
70
80
x
7
85
90
95
100
105
20
25
30
35
40
45
50
55
60
5.15
4.13
3.43
3.00
2.64
2.36
2.14
1.96
1.81
10
X
15
x
x
x
x
x
11
x
x
7
x
X
X
8
10
20
17
15
14
13
5
x
X
X
X
5
6
x
9
12
x
x
7
x
X
x
x
X
8
X
x
5
5
X
X
X
x
X
10
6
x
x
x
x
x
X
X
9
X
x
X
X
X
4
4
4
x
x
X
X
5
5
6
6
x
8
x
x
X
X
X
X
5
7
6
110
115
120
125
130
135
140
145
150
160
170
180
190
x
x
x
X
4
5
X
3
X
X
200
x
3
x
X
6
6
4
4
X
x
3
-
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
i
20 mm col. Reinf.
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
2. Total number of vertical reinforcement
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
= 432
8
=
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)
Closed Stirrups
Open 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 .
irum
¥
Bend Bars
Beam ~
v
I
- - 4 l/l i i ITT
E3 M
j
i
I
Main Reinforcement
i!
Column
—•“
I
>
!
.
M
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.
— Spiral Ties
50 cm —
»
t
i
5 cm Pitch h
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.
cm.
Pitch
cm.
No. or Turn
per Meter Ht.
21.0
2.604
1.706
30.0
5.00
6.25
7.50
17.0
14.3
2.108
1.381
1.778
1.165
1.269
1.027
0.866
5.00
21.0
2.894
6.25
17.0
2.342
1.896
1 535
1.410
1.141
7.50
14.3
1.975
1.294
0.962
5.00
625
7.50
21.0
3.183
2.577
2.172
2.085
1.688
1 423
1.550
1.255
1.058
32.5
35.0
17.0
14.3
120
Number of Steel Bars from a
6.00 m.
9.00 m.
12.00 m.
37.5
5.00
6.25
7.50
21.0
17.0
14.3
3.472
2.811
2.370
2.275
1.842
1.524
1.692
1.393
1.154
40.0
5.00
6.25
7.50
21.0
17.0
3.762
3.045
1 833
1.484
14.3
2.567
2.465
1.995
1 682
5.00
6.25
7.50
21.0
4.051
3.281
2:765
2.654
2.149
1.812
1.974
1.598
1.347
5.00
6.25
7.50
21.0
4.340
3.513
2.962
2.844
2.302
1.940
2.115
1.712
500
6.25
7.50
21.0
17.0
14.3
4.630
3.033
2.256
3.748
3.159
2.455
2.070
1.826
1.539
5.00
6.25
21.0
17.0
7.50
14.3
4919
3982
3.357
3.223
2.609
2.199
2.397
1.940
1.635
5.00
6.25
21.0
17.0
7.50
14.3
5.498
4.451
3.752
3.602
2.916
2.458
2.678
2.168
1.828
5.00
21.0
6.25
7.50
17.0
14.3
6.077
4.919
4.146
3.981
3.223
2.717
2.960
2.396
2.020
5.00
6.25
7.50
21.G
17.0
14.3
7.234
5.856
4.936
4.740
3.837
3.234
3.524
42.5
45.0
47.5
50.0
55.0
60.0
70.0
17.0
14.3
17.0
14.3
121
1.251
1.443
2.853
2.405
80.0
90.0
100
5.00
6.25
7.50
21.0
5.00
6.25
21.0
17.0
7.50
14.3
5.00
21.0
17.0
14.3
6.25
7.50
17.0
14.3
8.391
6.793
5.726
5.498
9.549
7.730
6.366
6.256
10.706
7.014
8.667
7.137
5.678
4.451
3.752
5.064
4.171
4.676
4.088
3.310
2.790
4.652
3.766
3.101
5.216
4.222
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
Size of the Reinforcement
Type of reinforcement
I
= 250 mm. (25 cm.)
= 12 mm. diameter
= Oneway
”
1 075 m .
Cut Bars
Bars
Wl . I
l
4 30 m
Straight Kars
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
Spacing
Number of Steel Bars per Square Meter
Length of Tie
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
3.911
3.524
3.247
3.039
3.667
3.186
2.866
2.637
2.465
2.332
2.225
2.138
2.065
2.856
2.483
1.669
2.320
2.015
1.812
1.667
1.558
1.473
1.405
1.350
1.612
1.304
12.5
15.0
17.5
20.0
22.5
25.0
27.5
2.878
2.749
30.0
2.554
2.643
2.234
2.056
1.192
1.819
1.737
1.834
1.593
1.433
1.319
1.233
. 141
.291
.236
. 195
. 169
. 126
. 152
1.166
1.113
.111
.101
.133
.121
1.069
1.033
.091
. 109
.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
. 242
. 197
. 163
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
^
1.80 m
7.20 m.
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 are.two way - run of reinforcement
37 x 2 = 74 pcs. steel bars
127
I
Straight Bar#
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
'
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
Spacing
cm.
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
Number of Steel Bars per Square Meter
5.0 m.
6.0 m.
7.5 m.
9.0 m.
12.0 m.
Length of Tie
Wire / sq. m.
25 cm- 30 cm.
4.953
3.995
3.549
3.252
3.039
2.880
2.756
2.656
2.575
3.050
2.703
2.471
2.306
2.182
2.085
3.047
2.734
2.000
.364
.437
1.775
.279
.335
. 286
.208
. 250
. 185
. 222
.168
. 202
. 156
. 187
.146
. 175
1.892
1.626
1.520
1.440
1.378
1.328
1.288
1.254
.238
2.507
2.524
2.377
2.266
2.179
2.109
2.053
2.005
.138
. 165
4.409
4.047
3.788
3.594
3.443
3.322
3.223
3.141
2.008
1.945
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.
53.00 m.
= .44 say .50 kg. No . 16 Tie wire
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 Heartwood
Outer Bark
-
Ca mbium
Inner Bark
Cross Section of a Tree
Straight Grain
Crooked Gram
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
®
Quarter Tangential
1
o
*— Tangential
P|ain or
Bastarrd Sawing
Star Shake
Wind or Cupr
Shake
Ii
Broken
Branch
r j .v: ^—
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
r
-
3,00 m
FIGURE 4-4
SOLUTION
1. Multiply the width by the length
6 x 3.00 = 18 pulgadas (inches)
142
i
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:
14x
20 ft.
600m.
= 6.00 meters
= 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.
Girts
2.70 m .
3/4“ Wood Flooring
^
2" x 6" Joists
.20 cm
*
Girder
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.
.20
Girder
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
1 II
1
IT If
\i
FT f
11
Girder
:0
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
Girder
4F
~
ll
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 -«
H
o
o
Girder
p
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
n
i
o
Girder
«
«
o
Post
o
Post -
II
i
Girder
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
n
Floor Joist
T & G Flooring —,
©
o
Girder
o
n n r
Post
3.00 m
.60 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.
.30
This is an odd number, adjust to
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
%
Tongue
T & G Flooring
-
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
2" x 6" Floor Joists
ksL
£
8
l
.
Girder
Solid Bridging
Jim,
5.00 m.
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
12
t
= 516 bd. ft. TAG.
TABLE 4-1 QUANTITY OF T & C WOOD BOARD PER METER RUN
Size of Board
Cm
In.
3/4 x 3"
3/4" x 4"
3/4" x S'
H
3/4" X 6"
Add % forAllowance
Direct Count Method
Number of Board
Per Meter Run
.0125
16.66
.0123
12.20
.0117
9.34
.0115
7 70
2.0 x 7.5
2.0 x 10.0
2.0 x 12.5
2.0 x 15 0
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
2. Double Stone cut
3. V- cut
4. BCB cut
5 . Weather cut
Double Cut
BCB
Single Cut
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
Per Meter Run
Approximate Board Ft.
Per Sq. M.
15
7.41
13.65
20
5.55
13.55
Commercial Size
Width of Board
Cm.
Inches
6"
8"
Number of Board
5.00 m
2.10
20 cm
J
3.10 m
U- iii I
t
50 cm .
I
J
i
FFHff
iii
: ;
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
3.10 m.
Floor to ceiling
Depth of Girts
. 20 m.
Flooring & Joist
.17 m.
.25 m.
Depth of Girder
Studs
.075 m.
3.795 m. * 12.65 ft.
14 ft.
Order
Length of the wall
Less the opening
Net Wall Length
5.00 m.
2.10 m.
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
Vertreal Strut
2
|
1
Girts
Bottom Chord
King Post
FIGURE 4-17
158
Coll«r Plate
Top Chord
Diagonal Strut
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
Spacing in Centimeters (Center to Center)
Lumber
Size In.
1x 2
2x2
2x 3
2x4
2x5
2x6
30 x 30
30 x 60
40 x 40
4.230
8.460
3.208
12.688
3.256
6.513
9.769
16.920
21.146
25.375
13.026
16.282
19.539
6.417
9.625
12.833
16.042
19.250
-
40 x 60 ! 60 x60
2.771
5.445
|
i
8.312 ;
11.083
13.854
16.625
|
2.333
4.667
7.000
9.333
| 11.667
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 bd.ft. 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 .
Plywood Ceiling
—
80 cm
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
Centimeters
30 x 30
40 X 40
40 x 60
60 x 60
60 x 120
90 X 180
120 x 240
Effective Covering
per board in sq , m.
0.90
0.16
0.24
0.36
0.72
1.62
2.88
167
Number of Pieces
per square meter
11.111
6.250
4.167
2.778
1.389
0.617
0.347
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
*
40’
+1.80+1.80+1.80+
1.80
“
FIGURE 422
170
25 cm
**
-
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.
or 14.5 ft.
= 14 ' - 6".
2. Length of Header
. . . . . 3.5 ft .
( 3 ft . + 6 In.)
18.0 ft.
Total length of 1 & 2
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"
or 14.5 ft.
M
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.
1,170 bd. ft.
M
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
2. Header
3. Transom
3. Transom
4 . Window sill
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
3" x 4"
- Jamb
5 ft
— Mullion
- 3 x 6"
H
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:
3 Mullions:
Order
4
Transom:
Order
2 pcs. 3" x 6" x 8 ft. or
1 pc. 3" x 6” x 16 ft.
< 5’ - 0" ) x 2 = 10
’
: 1 pc. 3“ x r x 10 ft
( 7’ - 0 " )
: 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
2. Metal
3. Plastic
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 is a temporary structure of wooden poles and
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~
K£
71
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 x 2 )
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" x 8'
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
1.20 m .
30 cm.
30 cm
FIGURE 5-3
184
40 cm
Summary of T & G for Five footings
28 pcs. 1" x 6" x 8’
9 pcs. 2” x 2" x 8’
10 pcs. 2* x 2" x 16’
= 112 bd. ft.
= 72 In. ft.
= 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.
20 pcs. 2" x Y x ,40 m.
B
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
= 240 In. ft.
30 pcs. 2” x 2" x 8 ’
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
T & G
Bd. Ft. per Sq. M.
2 x 2 Frame
Bd . Ft . per Sq. M.
Ln. Ft.
per Sq. M.
1" x 4"
r x 6”
18.33
17.50
7.32
7.32
21.96
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
40 cm
E
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
Height
Cross Sectional Dimension
Frame
= 6
= 4 meters
* .30 x 30 cm.
* 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
Size
in M .
Number
of Pieces
x 1.80
1.20 x 2.40
. 842
. 90
*
.
Size of Frame or Ribs
Bd Ft. per Sq . M
2x2
2x 3
12.71
12.71
488
18.96
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
FIGURE 5-5
192
30 cm
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
r
/i
2" x 2" Vertical m
Frame
r
i
i
•
\
Line of Cutting
/
/
\
FIGURE 5-4
SOLUTION
1. Solve for the circumference of one column
- 3.1416 x .60 = 1.88 m.
C
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.
x 14" X 10’ = 350 bd. ft.
r
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’
1" x 14" Arc
Iill
Plain G.l . Sheet
rr
90 cm
2" x 2" Supporters
d ~ 60 cm
Circular Column Form
FIGURE 5-7
196
u
TABLE S-3 QUANTITY OF LUMBER FORM FOR CIRCULAR
COLUMN
Lumber
Size
10 Cm.
1” x T
2" x 2"
2” x 3"
1" x 8"
8.5
17.0
25.5
7.7
6.5
13.0
35.1
7.7
10.0
20.0
30.0
7.7
9.6
11.6
13.5
9.6
11.6
13.5
9.6
11.6
13.5
9.6
11.6
11.7
23.4
1" x 10"
1" x 12*
1" x 14"
Board Foor Per Plain G. i Sheet
Spacing of Frame
15 Cm.
2.5 Cm.
20 Cm.
19.5
7.7
13.5
ILLUSTRATION 5-5
Solving the problem of illustration 5-4 using Table 5-3, we have
the following data:
Number of columns
Height
Diameter
Lumber
Circumference
- 6 pcs.
- 4.50 m.
= .60 m.
- 1 x 14 and 2 x 2
- 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
Size (m.)
Number
per sq . m.
.90 x 1.80
. 81
1.20 x 2.40
.42
Lumber Frame Bd. Ft. per Plywood
2x2
2x 3
1x 2
8.0
16.0
16.0
24.0
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
Size
T x 2"
2' x 3”
2' x 4"
Column
Bd. Ft. per Meter Ht.
Vert.
Hor.
4.67
21.00 11.67
7.00
9.33
31.67 17.50
42.22 23.33
Brace
Beam
Bd Ft. / M. Ht.
Vert
Hor.
4.00
6.00
8.00
4.67
7.00
9.33
Flooring
Bd. Ft / M. Ht.
6.10
910
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 .
-ijj
Column — %
v.
7-y
Q
2 x 2 Horizontal and
Diagonal Support
S
2 x 3 Vertical Support
Perspective
rffft
' 4 oo * 7nr
\
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
1 , 176 bd. ft. of
2" x 3" x 14 ft.
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‘
14
22 36
20.25
35.78
40.25
44.72 49.19 53.66
28.35
32.40
36.45
40.50 44.55 48.60
16
.163
18.14
25.39
29.02
32.64
17
18
.147
16.43
26.83
24.30
27.76
19.72
31.30
15
.203
.180
14.73
17.67
19
.117
13.03
15.63
20
. 102
11.32
13.58
15 85
26.29
23.56
20.84
18.11
29.58
.132
23.00
20 62
18.24
36.27 39.90 43.52
32.86 36.15 39,43
21
.094
14.60
16.69
22
.086
.079
.071
13.46
12.22
15.38
23
10.43
9.62
8.73
23.45
20.38
18.78
17.31
13.96
15.71
11.07
12.66
14.24
9.84
11.24
8.66
9.90
8 06
9.22
12.65
11.14
10.37
9.59
8.82
8.06
24
25
26
27
28
29
30
.064
7.03
.056
6.19
.051
.048
.043
5.76
12.52
11.54
10.47
9.49
8.43
7.43
6.91
5.33
6.39
7.46
490
5.88
6.86
8.52
7.84
.041
4.48
5.37
6.27
7.16
7.91
Iff
26.51
1V
12'
29.45 32.40 35.34
26.05 28.66 31.27
22.64 24.90 27.17
20.86 22.95 25.03
19.23 21.15 23.08
1745 19.20 20.94
15.82 17.40 18.29
14.05 15.45 16.86
12.38 13.62 14.86
11.52 12.67 13.82
10.65 11.72 12.78
9.80 10.78 11.76
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.
Length
Feet - Meter
1.80
2.10
2.40
6
7
8
9
10
12
2.70
3.00
3.60
Effective Width Covering
Side Lapping
2 1/2
1 1/2
Number of
Purlins
Distance Nails or Rivets
(cm. )
per Sheet
. 70
.60
. 75
.70
.70
. 70
. 70
.70
.60
. 60
. 70
. 60
. 67
. 66
.60
60
.
.60
.60
14
18
18
22
22
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
Number of Pieces per Kilogram
Materials
G.l. Rooring nails
G. I. Rivets
G.l. Washers
Lead Washers
Umbrella Nails
120
180
126
75
120
TABLE 6-4 SIZE AND QUANTITY OF STRAPS IN
ONE PLAIN G. I. SHEET
Size of Purlins
In
2" x 3" 50 x 75
2" x 4" 50 x 100
T x 5" 50 x 125
rx 6" 50 x 150
Size of G.I. Strap
in Cm.
Number of Strap
in One Plain Sheet
2.5 x 22.5
2.5 x 25 0
2.5 x 27.5
2.5 x 30.0
384
210
342
312
288
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
180
= 5.33 say 5.5 kg.
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
126
= 15.24 say 15.5 kg.
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 sheets.Referring 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
For 5.40 m. rafter combine
3.60 and 2.70 m. sheets
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 =
For the 3 .00 m. sheet: 60 x 22 =
For the 2.70 m. sheet : 175 x 22 =
Total =
2 ,990 pcs.
1,320 pcs.
3,850 pcs.
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
No. of
Rafter
Length Sheets
3.00
3.30
3.60
3.90
4.20
4.50
4.80
5.10
5.40
5.70
6.00
6.30
1
1
1
2
2
2
2
2
2
2
2
2
6.60
2
6.90
7.20
7.50
7.80
8.10
2
3
3
3
3
8.40
3
8.70
9.00
3
3
Combination of Roof Sheet Length
Meters (Feet)
3.00 (10‘)
3.30 (11‘)
3.60 (12* )
2.10 (7‘)
2.40 (8‘)
2.40 (8' )
3.00 (10' )
3.00 (10')
3.00 (10* )
3.00 (10' )
3.60 (12* )
3.60 (12‘)
3.60 (12' )
3.60 (12' )
3.00 (10' )
3.00 (10‘)
3.00 (10*)
3.60 (12' )
3.60 (12' )
3.60 (12' )
3.60 (12' )
and
and
and
and
and
and
and
and
and
and
and
and
and
and
and
and
and
and
221
2.10 (7' )
2.10 ( 7' )
2.40 (8‘)
2.10 (7' )
2.40 (S' )
2.70 (9‘)
3.00 (10' )
2.70 9‘)
3.00 (10‘ )
3.30 (11* )
3.60 (12* )
2.40 ( 8' )
3.00 (10’ )
3.00 (10' )
3.00 (10' )
3.00 (10' )
3.60 (12‘ )
3.60 (12' )
<
and
and
and
and
and
and
and
2.40 (8' )
2.10 (7' )
2.40 (8' )
2.10 ( 7‘ )
2.40 ( 8‘)
2.10 ( 7* )
2.40 ( 8' )
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 .l.strap 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
00
1
2.5 cm
as
7.50 m .
Cross Section of Gutter
FIGURE 6-7
223
(N
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 .
Ijr
22 5 cm. „
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
9 pcs. 90 cm. x 2.40 m. plain G .l. sheets
Total
TABLE 6-6 ROOF ACCESSORIES
Effective Length in Meter
Item
2.35
2.30
Gutter
Flashing
Ridge Roll
Valley Roll
Hipped Roll
Soldering Lead
Muriatic Acid
2.20
2.35
2.20
1/4 bar ( .25 ) per Solder Joint
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
Effective Width Covering
Corrugation
1 1/2
2 1/2
Commercial Size
In . / ft
Meter
26" x 8 '
29" x 8 '
650 X 2.40
. 725 x 2.40
46 m
. 53 m.
.
.
.31 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
= 22 26 pcs
52.00
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
End lapping: Below 20 degrees
Above 20 degrees
Effective width
.30 m.
.15 m.
.838 m.
.838 m.
Ridge Roll effective length
Gutter effective length
Outside Flashing
Hip Roll
.838
1.20 x 3.00 m..
2.336 m.
2.286 m.
1.676 m.
m.
Lap
.98 m.
STANDARD CORRUGATED SHEET
FIGURE 6-10
231
TABLE 6-9 CORRUGATED SHEETS TECHNICAL DATA
2.438 m.
.965 m.
. 965 m.
Standard Length
Effective width
Ridge Roll effective length
Outside Flashing effective length
2.286 m.
965 m
Lap
v/WVM
1.CX) m.
CORRUGATED SHEET
FIGURE 6-11
.177 m
.177 m.
.244 m
Lap
.058 m
KANALETAS
FIGURE 6- 12
232
TABLE 6*10 KANALETAS
Length in Meter
Items
7.315
Length
. 885
Effective Width
. 885
Eaves Flashing
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
812
Standard Length
. 600
Effective Length
1.180
Standard Width
1 100
Effective Wdth
200
End Lap
080
Side Lap
1 100
Ridge Roll Effective Length
Outside Flashing Effective Length
2 286
1 100
Ridge Flashing Effective length
1 100
Eaves Flashing Effective length
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
Lapping
Effective Length
Standard Width
Effective Width
Outside Flashing
Minus Lapping
Ridge Roll
2.44
.15 or .30
2.29 or 2.14
.748
.675
1.50 to 3.00
. 15
.953
Other accessories , estimate by direct count .
.07 m.
.45 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
m
Female Rib
Male Rib
203 mm
1b
203 mm
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:
= 0.60 mm.
Total Coated thickness
* 0.63 mm.
Weight per meter length of panel = 2.66 kg.
= 6.55 kg/m2
Weight per covered area
- 15 m.
Length available up to
Longer length through special
Steel base thickness
order up to
Overall width
Effective Width Coverage
- 35 m.
= .427 m.
= .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)
(2.5 mm) thick
wafer head self drilling
wafer head self drilling
and tapping screw
and tapping screw
No . 10-24 x 5/8" ( 16 mm )
No . 10- 24 x 7/8" (22 mm)
wafer head self drilling
wafer head self drilling
and tapping screw
and tapping screw
Steel 3/32" to 3/ 16"
-
(2.5 5 mm) thick
236
M
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
Hardwood
Softwood
2“ x 9G(50 x 3.75 mm)
screw
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
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
Effective Width Coverage
Length
Longer Length
Minimum Roof Slope
228 m.
. 204 m
12.19 m.
Special Order
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
Average No. of fasteners
10 pcs. per sheet
12 DCS. per sheet
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.
1.35 m.
Special Order
Length:
Longer Length
Maximum recommended
length for continuous sheet
without expansion joints
24 m.
239
6-7 MILANO LONGSPAN STEEL BRICKS
38 cm
i3 cfC
rT tiffin
30 cm
-
—
.
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 ridge and gutter line
* Nailing position for intermediate
tile presses.
tile presses
FIGURE 6 *20
Technical Data
0.40 mm (No. 26)
0.46 mm
4.53 kg.
3.44 kg.
.67 m.
6.00 m.
Special Order
10° min.
Steei base thickness
Total coated thickness
Weight per sq m.
Weight per length
Effective coverage
Length up to
Longer length
Recommended roof slope
-
6 8 COLORBOND TRIMDEX HI -TEN
240
Notch and turn aown toga or
RIDGE CAPPING
copping bat ween ribs.
Fatten at alternate rib .
Fatten at alternate rib .
*
Turn up tray between rib .
*
*
Turn up tray between rib .
TRANSVERSE FASCIA CAPPING
*
Notch and turn down edga of
capping between rib
*.
1 1/8 in
( 29 mm )
I
. +
2 3/ 8 in
S 1/8 in
( 130 m ti)
(60 mm)
30 in (760 mm) COVER
FIGURE 6-21
Technical Data
Steel base thickness
Total coated thickness
Weight:
Per unit area
Per unit length
Normal width
Effective width coverage
Available length up to
Lenger length
Minimum slope:
Single sheet
Roof with end lap
0.40 m.
0.46 m.
4.28 kg./ sq. m.
3.26 kg / sq. m.
0.83 m.
0.76 m.
15 m.
Special Order
3°
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:
Description
Marceille Type(flat)
Ondula Type (wavy )
Sr. (Standard ridge )
Half Marceille
JRT- 1
JRT- 2
JRT- 3
JRT -4
JRT- 5
SRT- 1
SRT- 2
SRT- L
SRT- R
JSR- 1
JSR -2
SRT- U
SRT- T
Weight per
Piece in kg
3.60
400
280
200
3.00
3.75
3.20
3 20
220
400
2.20
2.25
2.25
2.00
3.00
1 75
1.50
242
Total Number Required
14.0 per sq. m.
15.0 per sq. m.
2.5 per In. m.
1.0 per In. m.
13.0 per sq. m.
4.0 persq. m.
1.0 per In m
1.0 per In. m.
4.0 per In m.
13.0 persq . m.
1.0 per In m.
1.0 per In. m
10 per In. m
10 per In. m.
4.0 per In. m.
25.0 per sq. m.
25.0 per sq. m.
7
CHAPTER
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.
.10 x .20
=
12
.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
Size
Classifications
in.
cm.
12 x 12
30 x 30
1.0
1076
7.5 x 7.5
3x3
4x4
10 x 10
4 1/4 x 4 1 /4 10.62 x 10.62
10 x 20
4x8
15 x 15
6x6
20 x 20
8x8
30 x 30
12 x 12
16
9
7.97
177.8
Mosaic Tiles
Glazed and
Unglazed
Tiles
Number of Pieces Per
Square Ft Square M.
5 Point Hex. Tiles
4 Point Hex Tiles
5
2
2
5
85.7
50.0
4.5
4
2.25
1.0
25.0
10.76
61 / ft
4.9 / ft
20 / m.
16 / m.
External Comer Bead - Direct count
Internal Comer Bead - Direct count
Portland cement paste mortar - . 076 bags per sq. m
- . 5 klg per sq. m.
White Cement Filler paste
246
100.0
44.5
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
White Cement
27 sq. m. x .076 = 2.0 bags
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 *
Number
per sq. m.
Size
cm.
15 x 30
20 X 20
20 x 40
30 x 30
30 x 60
40 x 40
60 x 60
;
22.3
25.0
12.5
11.1
5.6
6.3
2.8
Sand
Cement Bags
Mixture
I
|
A
B
C
cu . m.
.45
. 30
.225
.025
.45
. 30
. 30
.225
.225
.225
.225
. 225
, 225
.025
.025
.45
.45
. 30
.45
. 30
.
45
. 30
.
45
.30
.025
.025
. 025
.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.
I
- 12 00 m
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 selfdealing
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
3 mm thick
. 20 x
20
. 225 x .225
.25 x .25
. 30 x . 30
.40 x 40
.60 x .60
Number Per
Sq. M.
Gallons of Adhesive
per Sq. M.
25.00
19.75
16.00
.042
11.11
6.25
2.78
. 042
. 042
042
. 042
. 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
8 00 m
Granolithic Floor
•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
stock
Pieces
Size
m.
per
sq. m.
.20 x .20
.225 x .225
.25 X .25
.30 x .30
.35 X .35
.40 x .40
25.00
1&.75
16.00
11.11
8.16
6.25
Cement bags
per sq. m.
Mixture Class
A
B
Sand
cu. m.
.338
.225
.0188
.338
.225
.225
.225
.0188
.0188
. 0188
.225
.225
. 0188
.0188
.338
.338
.338
.338
sq. m.
Brass Divider
meter / sq. in-
10.8
10.0
8.9
8.0
6.0
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
Brass Divider
Floor Plan
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” )
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' )
H
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
FIGURE 7-6
260
Cement Tiles
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
— - 8.00 m
B
18.00 m
FIGURE 7*7
262
8 00 m
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
For Class B mixture multiply by
For Sand or Pebbles "
.216
.144
.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
countersunk hole.
-
Has a beveled head fitting into a
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
In.
1/2
12
5/8
16
10.0
1.44
1 69
2.45
2.85
3.24
364
4.03
4.43
12.5
15.0
17.5
20.0
225
25.0
27 5
30.0
32.5
35.0
1.94
2.19
2.45
270
2.95
3.20
3.46
3.71
3.96
482
522
5.61
6.01
640
D i a m e t e r of B o l t s
3/4
7/8
1
20
25
22
1 1/8
28
1 1/4
31
3.64
4.21
9.76
11.06
12.60
14 18
15 76
17.35
4.78
5.35
5.92
6.49
7 06
7.63
8.20
8.77
9.34
524
6.01
678
7.55
832
9.09
9.86
10,63
1140
12.17
1294
268
7.23
824
9.26
10.27
11.29
12.30
1331
14.33
15.34
1636
17.37
12.33
13.61
14.89
16.17
17.44
18.72
20.00
21 27
22.55
18.93
20.51
22.09
23.67
25.26
26.84
28 42
Machine Bolts
FIGURE 8- J
Bevel Head
Countersunk Head
Turned Oval Head
Bastard Head
Carriage Bolts
FIGURE 8-2
Nut End j
Body
Stud Bolt
FIGURE 8-3
263
Attachment Head
ffillHmPjJ
Railroad Track Boft
^
safi
Welded Eye Bolt
Step Bolt
Plain and Shouldered
Forged Eye Bolt
Expansion Bolt
Various Types of Bolt
FIGURE 8- 4
270
Estimating Procedure in Determining Length of the
Bolt
Post
I"
E*
W.l. Post Strap
Bolts
E
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
1/4 5 / 16 3/8
9
6
of Bolts
7
No of Thread 20 18
16
per Inch
5
8
Diameter of
6
13
5
1
Top Drill
4
64
16
Diameter
7/16 1/2
10
12
13
14
9/16
5/8
14
16
11
9
11
23
27
64
64
12
15
64
12
13
17
32
3/4 7/8 1
19 22 25
9
9
10
18
32
64
19
3
4
22
55
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
JSr- 4r
~
TT
5 cm.
5 cm.
S cm.
[I
liTSf
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.
Size of Host
cm
10 X 10
13 x 13
15 x 15
18 x 18
20 X 20
23 x 23
25 x 25
28 x 28
TABLE 8-3 POST ROUGH HARDWARE
No. Bolt
W.!. Post
W. l. Splice
Strap (mm)
Strap (mm)
No.
Bolt
6 x 45 x 600
2
12
6 x 37 x 600
4
12
6 x 50 x 600
2
16
6 x 50 x 600
4
16
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 length.Each
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
2. Steel
5 . Bronze
6 . Aluminum
3. Brass
4. Copper
Shape of the Head
1. Flat
2. Round
3. Fillister
4. Oval
5. Winged
6 . Bung
7. Punched
8. Headless
9. Slotted (wood screw)
10. Square (lag screw)
11 . Hexagonal
12. Clove
13. Grooved
Shape of the Point
1.Standard
2. Full length
3. Coarse
Duty
1. Wood (light duty)
2. Lag (heavy duty)
Finish
1. Bright
2. Blued
3. Nickel Plated
4 . Silver Plated
5. Brass
6 Bronzed
7. Coppered
8. Japanned
9. Lacquered
10 . Galvanized
276
TABLE 8- 4 STANDARD WOOD SCREW AND NUMBER PER
KILOGRAMS
5
5
Inches
1/2
1
1 1/2
2
21/2
3
31/2
4
mm
12
25
37
50
62
75
87
100 112 125 150
4 1/2
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
Round Head
Flat Head
Counter Sunk Fillister Head
Oval Head
Felloe
Closed Head
Hexagonal
Round Bung Head
CO
Headless
Grooved
Oowel
Pinched Head
Winged
Square Bung Head
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.
Cone
Gimlet
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
b) Sprigs
c) Brads
d) Nails
3. Materials
a) Steel
b) Brass
4. Finish
a) Plain
b) Coated
e) Spike
c) Copper
c) Galvanized
d) Blued
5. Service
c) Finishing
d) Roofing
a) Common
b) Flooring
e) Boat , etc.
Types of Nail Points
UF
¥ FT
Blunt
Chisel Point
Diamond
Front
Side
Sheared Bevel
FIGURE 8-12
280
Long Diamond
Needle
Cut Nail
I
-c
i
I
1
1
J
-
i
i
Ci
==9
=4
2d
3d
4d
=4 5d
-“4 7
isfl
31
6d
I
d
t
i
\
.iL
=3
*
j
l
I
8d
j
9d
10d
12d
I
16d
*
>
31/2"
i
i
<
i
1 1/2"
2 1/2"
3"
*
2"
M
1
Length of Nails { Actual Size )
FIGURE 8- 13
2d to 20d For Cabinetwork Furnitures
Common Nail
Box
Casing
Finishing Nail
3/ 16" to 3" For Light Work Mouldings .
Wire Brad
2d to 20d Flooring Construction .
6" to 12" For Heavy Construction . .
Cut
Spike
2d to 60d For General Construction
2d to 40d For Light Construction Household Use.
2d to 40d For Interior Trim
281
8
2d to 60d . . . .Common Nail
'
luiii
8
6
*
2d to 40d. . . Box
2d to 40d. . .. Casing
2d to 20d. . . Finishing Nail
&
c*
3/16" to 3" . Wire Brad*
>
8^1
2d to 20d
. . Cut
itfrr
6" to 12“ . .
Spike
Types of Nails and Their Uses
-
FIGURE 8 14
Kil
Flat
Common
Large Flat
I
Large Flat
Reinforced
Oval
Wire Spike
3
Round
Countersunk
Round
Countersunk
Offset
W
Hook
£1
Ud
Non Leak
Cone
Curved
Cut Nail
Types of Nail Heads
-
FIGURE 8 15
282
Brad
Diamond
Barge Spike
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
3/16
1/4
5/16
3/8
3
35,200
23, 465
17,600
14,080
11, 732
8,800
5,865
4,400
10, 120
7/16
9/16
5/8
11/16
3/4
13/16
5
6
8
10
12
14
7/8
15/16
16
17
19
20.5
22
24
1
1 1/16
1 1/8
25
27
28
2.930
2, 514
1, 200
1, 953
1, 760
1,599
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
No.
6d
50
11
7d
32
37
11
10
44
10
75
81
87
9
6
7
100
6
8d
9d
10d
12d
16d
20d
Gauge
No .
No. of Nails
per kg.
12
12
11
14.500
12.500
173
131
95
11
10
76
57
8
7,800
5, 900
4 , 300
3,450
2,600
Dia. Head Approx.
Gauge
No. / kg.
6
6
5
5
4
3
2
1
322
277
200
9
7
9.000
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
2d
3d
4d
5d
6d
7d
8d
9d
10d
12d
16d
20d
40d
50d
60d
Approximate Number
Length
Gauge
No.
Inches
mm
15
1 1/4
25
14
12.5
12.5
11.5
11.5
10.25
1 1/ 2
31
1 1/2
37
44
50
56
63
10.25
9
2 3 /4
3
3 1/4
3 1 /2
4
1 3/4
2
21/4
2 1/2
9
8
6
4
3
5
5 1/ 2
6
2
69
75
Per Kilogram Per Keg
1 , 831
1 , 177
666
580
382
344
208
188
138
124
81
100
112
125
93
58
45
34
150
20
88
82,400
53, 000
30 , 000
26 , 100
17,200
15, 500
9,400
8, 500
6 , 250
5 ,600
4 , 200
2, 625
2,040
1,540
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
Inches
2d
3d
1
1 1/4
4d
1 1/2
5(t
1 3/4
6d
2
21/4
2 1/2
2 3/4
7d
8d
9d
10d
12d
16d
20d
30d
40d
50d
3
33/4
3 1/2
4
4 1 /2
5
5 1 /2
Approximate Number
Per Kilogram Per Keg
25
31
37
44
50
56
63
69
75
81
88
100
112
125
137
1, 904
1,206
662
566
397
29,600
25,500
17,900
340
15.300
224
197
146
137
108
68
53
10,100
40
28
286
85,700
54.300
8.900
6,600
6,200
4.900
3,100
2,400
1,800
1 , 300
TABLE H-') ORDINARY SPIKES TECHNICAL DATA
Size
Length
Inches
10d
12d
16d
20d
30d
40d
50d
60d
175 mm
3
3 1/4
3 1/2
4
4 1 /2
5
5 1/2
6
200 mm
225 mm
250 mm
7
8
9
10
300 mm
12
Approx . No.
per Kilogram
75
90
81
88
83
100
112
50
37
125
137
150
175
28
200
225
9
8
250
300
7
66
22
19
15
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
Size
Kind of
Nails
.17
20d
CWN
15
. 09
6d
6d
Flooring brad
Flooring brad
sq m
sq. m.
. 08
6d
06
6d
Casing brads
Casing brads
sq. m.
sq. m
. 08
8d
8d
CWN
Unit Per
Approx kg.
Required
Floor joist and
bridging at 30 o.c
sq. m
T & G Flooring
1x 4
1x6
sq. m
sq. m.
Siding Wood Board
on studs at 60 o. c.
1x6
1x 8
Studs:
at .40 o. c
at .60 o. c.
.
. 05
CWN
20d
CWN
2d
Finishing nails
20d
CWN
and Cleats
Meter ht.
. 73
of post
Per sheet
at . 15 o.c on
. 055
40 x 60 joist
Per sq. m.
.20
purlins at 70 o.c.
Base Board
Meter length
03
6d
CWN
Fascia Board
Ceiling Joist
at 40 x 60 o c.
Meter length
.
048
8d
CWN
Sq. m
.05
8d
CWN
Scaffolding
Plywobd Wall
and Ceiling
Rafters. Purlins
,
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
Multiply:
60 x .06 = 3.6 say 4 kilos 6dCWNail
M
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
Gauge
No. of Nails/ Kg.
Penny
Length
1"
1 1/4
1 1/2"
1 3/4"
2"
2 1/4"
2 1/2"
2 3/4"
3"
3 1/4"
31/2"
4"
4 1/2"
5"
5 1/2"
6"
M
2d
3d
4d
5d
6d
7d
8d
9d
10d
12d
16d
20d
30d
40d
50d
60d
15
14
12 1/2
121/2
11 1/2
11 1/2
10 1/4
101/4
9
1863
1195
9
149
103
64
48
37
31
24
8
6
5
4
3
2
290
651
559
367
330
222
202
145
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
riser.
- The front edge of the steps that project beyond the
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 :
—— -
Width of Tread
25 cm.
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
t
-
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
7
Cut Type
Cleated
Built-Up
FIGURE 9-5
302
Rabbeted
4
-
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:
/
SL = /(3.50) 2 + (2.50) 2
Stringer length = (Run ) 2 + (Rise) 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
.18
Riser
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 )
No.of
Steps
Length of Stringer
Tread Width
30 cm.
25 cm.
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1, 05
1.23
.80 l 1,00
1.31
1.54
1.00
1.25
1.57
1.84
2.10
1.85
2.16
2.47
2.78
3.08
3.39
3.70
4.00
4.31
4.62
1.20
1.40
1.60
1.80
1.50
1.75
4.93
5.24
5.55
5.85
3.20
3.40
6.16
4.00
2.36
2.62
289
3.15
3.41
3.67
3.94
4.20
4.46
4.73
5.00
5.62
Length of Run
Tread Width
25 cm. , 30 cm.
2.00
2.20
2.40
2.60
2.80
3.00
3.60
3.80
68
.85
.72
1.02
1.08
1.19
1.36
1.26
1.44
2.25
2.50
2.75
1.53
1.70
1.62
1.87
1.80
1.98
3.00
3.25
3.50
2.04
2.21
2.16
2.34
2.38
2.52
3.75
4.00
4.25
4.50
475
5.00
2.55
2.72
2.70
2.89
3.06
3.24
2.00
Stairs Inclination Angie
Ladder
Ramp
Stairs
Ideal Stairs
- 50° to 90°
- 1° to 20°
- 20° to 50°
- 30° to 35°
305
Height of Rise
Riser Height at
18 cm.
17 cm.
.
3.06
3.23
3.40
.90
2.88
3.42
3.60
-
TABLE 9 2 SPIRAL STAIRS
Open Riser Treads
No. of T read
Tread Degrees
in Circle
22° - 30‘
28° - 0*
30° 0*
Cantilever Treads
Riser
Head Room
Meter
cm.
16
17.5
18.0
12 - 13
12 13
-
-
2.10
2.00
2.00
20.0
M
Elevation
Cantilever Treads
5
(
Step
m%
J
%
Plan
¥
EE
Elevation
Open Riser Treads
-
FIGURE 9 7
306
30°
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
Kind of Paint
Uses
Thinning
Drying Time
Coverage
per 4 Its.
PRIMERS*
1. Interior Primer
and Sealer
For interior wood
surfaces
Paint
Thinner
2 hours, allow
overnight before
25 to 30
sq. m.
recoating
2. Exterior wood
primer
For exterior wood
surfaces
Paint
Thinner
6 hours, allow
overnightbefore
30 to 40
sq. m.
recoating
3. Prepakote
Red Oxide
Primer
Primer for ferrous Paint
and non-ferrous
Thinner
materials
4. Zinc Chromate
Primer
For exterior and
interior metal
surfaces exposed
to normal indus
Paint
Thinner
-
trial environment
318
3 hours
35 to 40
sq. m.
3 hours
30 to 40
sq. m.
5. Red Lead
Primer
Rust preventive
primer for ferrous
Paint
Thinner
3 hours
30 to 40
sq. m.
6 hours, allow
overnight before
recoating
30 to 40
sq. m.
surfaces
For steel, aluminum, galvanized
iron
6. Epoxy Primer
Epoxy
Reducer
* WATER BASE MASONRY PAINT *
1. Acrylic Latex
Paint
Exterior & Interior
1/2 Its7
masonry surfaces
4 Its
.
water
30 minutes.
Allow 6 hrs.
before recoat
30 to 40
sq. m.
2. Acrylic-SemiGloss Latex
Exterior & interior
masonry surfaces
water
30 minutes.
Allow 6 hrs.
before recoat
30 to 40
sq. m.
3. Acrylic Gloss
Latex paint
Exterior & Interior
masonry surface
water
30 minutes.
Allow 6 hrs
before recoat
30 to 40
sq. m.
4. Acrylic Clear
Gloss Emulsion
paint
For chalky surface water
to improve adhesion of new coats
of latex paint
1 hour. Allow
30 to 40
sq. m.
5. Latex Hi-gloss
Enamel
For furniture, cabi- Usse as
is
nets, doors, windows, toots, toys
wrought iron, and
primed metals
1 hour. Allow
4 hours before
30 tO 40
sq. m.
Primer to old and
30 min. Allow
6. Acrylic
water
319
.
recoating
30 to 40
concrete sealer
7. Masonry
-
6 hrs. before
recoating
new concrete, ex
terior & interior
Primer for old
surface
conditions
chalky paint film
8. Tinting Colors
Acrylic colors
Use as
is
sq. m.
24 hours. Allow 30 to 40
sq. m.
overnight before
recoating
water
* ROOF PAINT *
1. All weather
acrylic root
shield
2. Portland
cement paint
3. Davies Roof
paint
For gafvanized
iron sheet, as
bestos bricks
concrete and
stucco
For G.i. sheet
CHB , stucco,
concrete, brick
and other zinc
coated metal
surface
-
G.I. roofs and
other metal
such as aluminum and steel
water
4 hours. Allow
overnight before
40 to 50
sq. m.
recoating
Paint
Thinner
6 hours. Allow
overnight before
30 to 40
sq. m.
recoating
Paint
Thinner
6 hours. Allow
overnight before
30 to 40
sq. m.
recoating
* ENAMEL AND GLOSS PAINT *
1. Quick Drying
Enamel
For Exterior &
Interior wood
and metal surf.
Paint
Thinner
5 hours. Allow
8 hrs. before
recoating
320
30 to 40
sq. m.
2. Interior semi-
Gloss Enamel
3. Flat Wall
Enamel
Interior wood
and metal
surfaces
Paint
Thinner
6 hours. Allow
overnight before
recoating
25 to 30
sq. m.
For Interior
wall & ceilings
Paint
Thinner
3 hours. Allow
overnight before
40 to 50
sq. m.
recoating
4. Exterior
Gloss paint
.
For ext wood
and properly
primed metal
surfaces.
6 hours. Allow
48 hrs. before
recoating
30 to 35
sq. m.
Use as
is
10 minutes
30 minutes
for dry hard
20 to 25
sq. m.
Paint.
Thinner
12 min. Allow
Paint
Thinner
* VARNISHING *
1. No. 48 Davies
For dark wood
wood bleach
land 2
to be changed
to light natural
2. No. 77 Dawes
Lax-Tire Plastic wood Dough
3. Wood filler
paste
4. Non-Grain
raising wood
finish & making
old wood color
uniform
For patching up
wood defects
like knots,nail
holes and cracks
Sealer for open
grain of interior
wood
For wood sur
faces
Use as
is
25 to 35
overnight before sq. m.
recoating
-
30 minutes
stain
321
30 to 40
sq. m
.
5. Oil Woodstain
For panelling,
cabinets, floors,
furnitures, door
jambs, and other
Allow 24 hours
before recoating
30 to 40
sq. m.
Overnight
35 to 40
sq. m.
24 hours
40 to 50
sq. m.
woodworks
6. Finishing Oil
To seal and finish
interior wood surf,
such as furnitures
wood & panelling
7 Valsparor
Spar Varnish
.
For floors, sidings Paint
furnitures, deck of Thinner
8. Daxpar
Varnish
For interior and
exterior wood
surfaces, nautical
& aeronautical
varnish.
Paint
Thinner
24 hours
40 to 50
sq. m.
9. Hi- solid sanding sealer
For Interior new
wood furnitures
and fixture, cabi
net, doors, etc
Lacquer
10 minutes.
Allow 30 min.
before recoat
40 to 45
sq m.
10. Hi- Solid
clear gloss
Lacquer
For furnitures,
cabinets, fixture
door panelling
and trim.
Lacquer
Allow 30 min.
before recoat
30 to 40
sq. m.
11. Hi- Solid
Semi-gloss
For Interior
wood furnitures,
cabinets, doors
shelves etc.
Lacquer
Allow 30 min.
before recoat
30 to 40
sq. m.
boats etc.
.
Lacquer
-
Thinner
Thinner
Thinner
322
.
12. Hi-Solid dead
flat lacquer
13. Water white
gloss lacquer
For Interior wood Lacquer
furniture, cabinets Thinner
door jambs, trim
panelling etc.
For furnitures,
cabinets, doors,
panel and interior walls
Lacquer
Thinner
Allow 30 min.
before recoat
30 to 40
sq. m.
Allow 30 min.
before recoat
30 to 40
sq. m.
* ENAMEL AND GLOSS PAINT *
1. Quick drying
Enamel
2 Interior Semi
gloss Enamel
For Exterior and
Interior wood &
metal surfaces
Paint
Interior wood &
metal surfaces
Paint
Thinner
Thinner
5 hours. Allow
8 hrs. before
30 to 40
sq. m
.
recoating
6 hours. Allow
overnight before
25 to 30
sq. m.
recoating
3. Flat Wall
For Interior walls
Enamel
and ceiling
4. Exterior gloss
Exterior wood
Paint
and properly
primed metal
Paint
Thinner
Paint
Thinner
surfaces
5. Tinting Colors
Oil base Tinting
color
323
3 hours. Allow
overnight before
recoating
6 hours. Allow
48 hrs. before
recoating
40 to 50
sq. m
.
30 to 35
sq. m.
AUTOMOTIVE FINISHING
1. Lacquer
Enamel
For properly
Lacquer
Allow 30 min.
primed metal
Thinner
before recoat
Lacquer
Thinner
30 minutes
40 to 50
sq. m.
40 to 50
sq. m.
and wood
2. Automotive
Lacquer
For Extenor &
Interior metal
or wood surf.
3. Lacquer
Primer
Surfacer
For metal and
wood surface
Lacquer
30 minutes
40 to 50
sq m.
4. Lacquer Putty
For Exterior and
Interior metal &
wood surfaces
Lacquer
Thinner
10 minutes
dry to sand
in 1 hour
20 to 30
sq. m.
Automotive fin.
for residential &
commercial
Paint
Thinner
2 hours. Dry
hard in 10 hrs.
30 to 40
sq. m,
For Exterior and
Interior wood &
metal surfaces
Paint
2 hours. Allow
8 hours before
30 to 40
sq. m.
_
5. PnH ux Auto
Enamel
6. Pro-Lux
Enamel Primer
-
7. Pro Lux
Glazing Putty
For Exterior and
Interior metal &
wood surfaces
Thinner
recoating
Use as
is
2 hours
25 to 35
sq. m.
* INDUSTRIAL PAINTS *
1. Silver fiwsh
Aluminum
For steel tanks,
Ext. & Int. metal
wood & masonry
Use as
is
1 hour. Allow
24 hrs. before
recoating
324
40 to 50
sq. m.
2 Heat Resisting
For Interior and
Exterior surfaces
like radiators,
boilers, pipes
& general Ind
equipment.
Use as
is
1 hour. Allow
24 hrs. before
40 to 50
sq. m.
3 Hi-Heai Re sisting paint
For superheated
steam lines,
boiler casings,
drum and rocket
launchers
Use as
is
1 hour at
450° F max.
40 to 50
sq. m
4 Asphalt Base
Aluminum
For asbestos ce ment composition & metal
Use as
1 hour. Allow
24 hrs. before
recoating
20 to 30
sq. m.
5. Traffic Paint
For Asphalt and
masonry surface
Use as
is
1 hour. Allow
24 hrs. before
recoating
20 to 30
sq. m.
6. Blackboard
For wood or
metal surface
Paint
Thinner
12 hrs. Allow
35 to 40
overnight before
sq. m.
Slating
is
recoating
7 Davies AntiCorrosive
Marine paint
8. Marine Bott
Topping Paint
For hulls and
below water
line of ships
For properly
primed surfaces
b/w the light &
deep load lines
of ships
Paint
Thinner
12 hrs. Allow
overnight before
recoating
30 to 40
sq. m.
Paint
Thinner
12 hrs. Allow
24 hrs. before
30 to 40
sq. m.
recoating
325
9. Anti-Fouling
Paint
For properly
primed surface
below the water
line of ships.
Paint
Thinner
8 hrs. Allow
12 hrs. before
recoating
10. Hull, Deck,
Mast & Topside Paint
For use above the
water line of sea
Paint
12 hrs. Allow
24 hrs. before
Thinner
vessels, equip ment and struc ture s near sea.
30 to 40
sq m.
i
30 to 40
sq. m.
recoating
1 hour. Allow
24 hrs. before
recoating
11, Machinery
Engine
Enamel
Marine engine
Machineries &
equip’t. casing
Paint
12. Epoxy
Enamel
For Steel, alum,
galv. iron, wood
and concrete
Epoxy
reducer
30 to 40
6 hrs. Allow
overnight before
sq. m.
recoating
13. Epoxy Glazing putty
For body repair
aircraft, cars &
equipments
Epoxy
recucer
2 hours
14. Epoxy Glue
Multi-purpose
thermosetting
plastic material
for cementing or
bonding rigid
materials
Use as
8 hours full
strength in
96 hours
Use as thinner
exterior house
Use as
is
15. Pure Pale
Boiled Linseed oil
Thinner
is
30 to 40
sq. m.
30 to 40
sq. m.
t
i
f
paint
i
16. Concrete
Treatment to
.
i
Use as
326
24 hours
30 to 40
Neutralizer
neutralize masonry surface
is
17. Rust
Remover
Paint stripper
Use as
Overnight
25 to 35
sq. m.
18. Paint
Remover
Paint stripper
Use as
Overnight
25 to 35
sq. m.
19. Mildewcide
Solution
Destroy molds
mildews on new
and previously
Water
sq . m.
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.
30 sq . m.
= 12 gallons Acrylic concrete primer
4. Solve for the Acrylic Gloss Latex Paint final coating;
Divide:
360 sq. m.
30 sq. m. / gal.
= 12 gallons
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
Divide:
- 35;
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
, 38 gallons
Total
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:
6. Fading
1. Blistering or Peeling
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
Cm.
Length / Roll
Meter
Effective Covering
Per Roll ( sq m. )
Adhesive
Box per Roll
52
54
71
10.05
10.05
13.70
5.22
5.42
9.72
. 17
338
. 18
. 32
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
G.l sheet blade 15 cm. x 2.40 m.
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
6 mm x 32 mm x 2.00 m. Flat steel bar ( rail )
1 pc.
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 .
50 cm. G. l . pipe 10 mm diameter
1 pc.
12 pcs.
13 pcs.
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
Li J
Elevation
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
of Blade
Height
of Glass
Square Ft.
per Blade
Number
of Blade
Height
of Glass
4
5
14 7/8"
4 x 22 = 0.61
4 x 2 4 = 0.67
4 x 26 = 0.72
4 x 28 = 0.78
4 x 3 0 = 0.83
4 x 32 = 0.89
4 x 34 = 0.94
4 x 36 = 1.00
4 x 38 = 1.06
4 x 4 0 = 1.11
4 x 42 = 1.17
4 X 44 = 1.22
4 X 46 = 1.28
4 x 48 = 1.33
13
45 3/8"
49 7/8"
53 3/8"
56 7/8"
60 3/8"
63 7/8"
67 3/8"
70 7/8"
74 3/8"
77 7/8"
6
7
8
9
10
11
12
18 3/8
21 7/8"
25 3/8"
28 7/8"
32 3/8"
35 7/8"
39 3/8"
42 7/8"
14
15
16
17
18
19
20
21
2?
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
60 min.
a 50.7 gallons per minute.
TABLE 11-3 WATER CONSUMPTION IN OFFICE BUILDING
Consumption in Gallons
Per Hour per Person
Building Type
Commercial, no Airconditioning
Commercial, with Airconditioning
Owner occupied with kitchen and
laundry. No Airconditioning
Owner occupied with kitchen and
laundry with Airconditioning
343
3.8
7.3 - 9.2
7.3
9.0
TABLE 11-4 HOT WATER CONSUMPTION IN APARTMENT
BUILDING
Consumption
Public Housing
Gallons
Apartment Hotel
Gallons
Average Daily
Per Apartment
Per Person
Per Room
79
28
22
50
41
Maximum Daily
Per Apartment
Per Person
Per Room
92
33
26
69
8.9
7.6
3.2
2.5
5.3
4.9
Maximum Hourly
Per Apartment
Per Person
Per Room
36
48
43
TABLE 11-5 COLD WATER CONSUMPTION IN BUILDINGS
Type of Budding
Gallons per Day per Person
Residence - Average
Residence - Large
Apartment - Low Rent
Apartment - High Rent
50
100
Hotels
Office Buildings
100
75
100
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 cubic.meter.
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.
d2 x 1.568
d2 =*
d2
-
- 11.5 cu. m.
- 11.5
11.5
1.568
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
in Meter
Under 12 meters
t2.0 m. to 18.0 m.
Over 18 meters
Diameter from Butt
Min. Cm. Max. Cm.
30
32
35
45
45
50
Minimum Tip
Diameter Cm.
20
18
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
Use and Type
General Use
Marine Use
Type of Processing
Empty Cell Process
Full Cell Process
321 kg.
321 kg.
193 kg.
1Q3 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
2. Soft Clay
3. Wet Clay and Soft Sand
4. Clay and Sand in alternate layers
5. Finn and Dry Loom or Clay,
Hard dry day or Fine Sand
6. Confined Sand
7. Compact Coarse Sand or
Stiff Gravel
8. Sand and Gravel well Cemented
9. Good Hard Pan or Hard Shale
10. Rock
347
5.38
8.07
10.76
32.28
21.50
43.04
5.38
21.50
16.14
21.50
21.50
10.76
43.04
43.04
32.28
32.28
32.28
53.80
53.80
53.80
64.56
107.60
107.60
43.04
86.08
86.08
161.40
10.76
10.76
269.00
-
TABLE 11 9 RANGE OF SKIN FRICTION FOR VARIOUS SOIL
Type of
Soil
1. Silt and Soft Mud
2. Silt Compacted
3. Clay and Sand
4. Sand with some Clay
5. Sand and Gravel
Value in Kilograms per Square Meter
Minimum
Maximum
489
1,712
3,913
4,891
8,804
244
587
1,956
2,446
2,935
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
Application
First Spreading
First Application
Bituminous
Materials
Liter / Sq. M.
Aggregate
Coarse Key
Kg.
Liter / Sq. M.
90
4.0
Second Spreading
Second Application
Third Spreading
Third Application
Fourth Spreading
Total
13
1.8
11
1.4
8
7.2
348
122
TABLE 11-11 BITUMINOUS MACADAM PAVEMENT
Hot Asphalt Type Approximately 21/4 or 5.7 cm. thick
Activity
First Spreading
Second Spreading
First Application
Third Spreading
Second Application
Fourth Spreading
Third Application
Fifth Spreading
Total
Bituminous
Materials
Liter / Sq. M.
Aggregate
Coarse Choker Key
Kg.
Liter / Sq. M.
90
10
5.5
10
3.5
8
2.0
8
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 )
Operations
First Spreading
First Spreading
Second Application
Second Spreading
Third Application
Third Spreading
Bituminous
Materials
Liter / Sq. M.
Aggregate
Coarse
Choker
KgLiter / Sq. M.
1.0
22
1.3
6.5
0.7
4.5
Total
3.0
Values given with bulk specific gravity of 2.65.
33
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
m
B
15.00 m .
- 4
50 cm
1.30 m
-
"
M
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
Lap
Cm.
10 (4")
7.5 (T )
5.0 (T )
2.5 (1")
Effective Covering
in Square Meter
Class A Class B
. 1300
.0975
.0650
.0325
(
1090
. 081B
.0545
.0273
No. of Pieces
per Sq. Meter
A
B
7.70
10.27
15.38
30.77
9.18
12.23
18.31
36 62
No. of Rattan
per Sq Meter
*
1.8
2.0
2.6
40
36
4.0
56
80
* Rattan ties for every other bamboo split ribs.
** Rattan ties for every bamboo spirt ribs.
Nipa shingle 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
Bamboo
Class
Approx. Length
in Meters
Number of
Splits
9.0
B
8
6
C
4
6.0
A
CM, STICK
Coverage per
Square Meter
12
9
4
7.5
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
*
Bamboo Split
/
; 7.5 cm to 12.5 cm
It
'
AW
MW m
*
-*i
b•
7* 9* Spacing
-
FIGURE 11 4
355
11-8 ANAHAW ROOFING
-
TABLE 11 16 ANAHAW ROOF TECHNICAL DATA
End Lapping
Cm
Number of Leaves
Per Sq. Meter
Number of 3.0m. Rattan
Per Sq. Meter
7.5
10.0
84
7
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
Halige, Poste
Gililan
Post
Girder
Joist
Flooring
Soleras
Sahig, Suwelo
Sepo
Girt
Blga
Kilo
Barakilan
Tahilan
Reostra
Beam
Truss
Bottom Chord
Top Chord
Purlins
Collar Plate
Fascia Board
Esternal Siding
Vertical Stud
Horizontal Stud
Sinturon
Senepa
Tabike
Pilarete
Pabalagbag
Ceifing Joist
Window Sill
Window Head
Window or Door Jamb
Open Stringer
Kostiiyahe
Pasamano
Sombrero
Hamba
Hardinera
Closed Stringer
Madre de Escalera
Tread
Riser
Baytang
Takip Siipan
Gabay
Muldora
Handrail
Moulding
Eave
Sibe
Bolada
Projection
359
Framework
Gutter
Conductor
Baiangkas
Kanal
Aiulod
Wrought Iron Strap
Plantsuwela
Pierno
Plantsa
Estaka
Bolt
Scaffolding
Stake
Plastered Course
Kusturada
Stucco or Plaster
Scratch Coat
Picfcwork on Masonry
Varnish Finish
Spacing or Gap
Monyeka
Biyento
Concrete Stab (rough)
Alignment
Plumb Line
Cement Tiles
Cement Brick
Larga Masa
Asintada
Hulog
Baldosa
Ladrilyo
Palitada
Rebokada
Piketa
Door Fillet
Batidora
Groove
Wood Grain
Pattern or Schedule
Hinge
Kanal
Haspe
Plantilya
Bisagra
Paneled Door
De Bandeha
Eskumbro
Earthfill
Masonry Fill
Adobe Anchor
Solder
Lastilyas
Liyabe
Hinang
Soldering Lead
Temper (metal work)
Estanyo
Suban, Subuhan
360
Diagonal Brace
Nail Setter
Wiring Knob
Cabinet Hinge
Piye de Galyo
Punsol , Punson
Poleya
Espolon
MENSURATION FORMULA
Base x Altitude
2
Base x Altitude
Triangle
Area
Parallelogram
Area
Trapezoid
Area
s;
Circle
Area
S
Circumference
=
s
Ellipse
Area
Cone
Surface
(curved only)
Volume
Cylinder
0.7854 x Diameter 2
3.1416 x radius 2
3.1416 x Diameter
6.2832 x radius
0.7854 x Short Diameter x
Long Diameter
SS
Surface
Volume
Sum of Parallel Side x Altitude
2
SS
Slant height x Circumference
of Base
Area of Base x height
3
Length x Circumfernce plus
Area of Ends
0.7854 x Length x Diam. 2
361
Sphere
Surface
«
m
Volume
«
3.1416 x Diameter 2
Circumference x Diameter
0.5236 x Diameter 3
Circumference x Diameter 2
6
Vol. of Circumscribing Cylinder
3
362
TIME
-B. FajardoALL
.J
YOUR BOO:<r
By
;
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
. 6
6138 Mttk hmH ring
d- A Fisheries St.
Quwxi City, ^hiiit
Te >. 924 7101 o- 7:9 61 / /
.
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IS9N 97I85- 99!0-4