CIV 105.pdf

TECHNICAL HIGHER INSTITUTE FOR ENGINEERING AND PETROLEUM

CIV 105 SPECIFICATIONS AND BILL OF QUANTITIES

COURSE MANUAL

ISO 9001: 2008

ACKNOWLEDGMENTS

The completion of this endeavor could not have been in reality

without the undying support and immeasurable advices from the following

persons:

Engr. Mustafa Al Ghazal, General Manager of the Institute for his

advises and encouragements paving the way to make the institute the best

institute in the Kingdom.

Dr. Ali Al Gadhib, Academic Consultant of the institute for his guidance,

encouragement and unending advices for the completion of this workbook.

Engr. Ronaldo Hertez, the Department Head of the Civil Engineering

Technology valuable support and assistance.

Mr. Naguish, Institute keeper for his assistance in the reproduction of

this workbook.

His loving and beautiful wife Carmina and his sons Zadkiel Dhen and

Xian Gabriel for their love as well as understanding. This could not be possible

without them.

To all who had contributed in the development of this workbook.

Introduction of the course:

Estimating is one of the most useful math’s skills we can acquire. It

relates to all sorts of different areas of math’s – measure, handling data,

handling money, time, space, shape, number operations. Estimating and

preparing the itemized list of bill of materials required for a project to be built,

added, repaired, remodeled or renovated is an integral and no less important

part of the professional services rendered by an architect or an engineer to his

clients. An estimate is necessary to give the owner a reasonably accurate idea

of the cost to help him decide whether the work can be undertaken as proposed

or needs to be curtailed or abandoned, depending upon the availability of funds

and prospective direct and indirect benefits. Estimating is the most important

of the practical aspects of construction management, and the subject deserves

the closest attention of one aspiring to a career in the profession. It is

comparatively simple subject to understand; however, as it brings one up

against practical work, methods and procedure, knowledge of it cannot be

acquired without close application.

Course Title: Specifications and Bill of Quantities

Code: CIV 105

Course Description: This course covers quantity computations for

plain and reinforced concrete, reinforcing steel bars calculations as well as earthwork (cut and fill). This course also covers types of

project contracts, contact documentations, meaning of valued tenders, and quantities list.

Table of contents

Concrete works

Estimating concrete footings 1

Illustrative example 1 3


Student exercise 1 7






Estimating concrete beams 21





Estimating concrete columns 28





Estimating concrete slab 37





Estimating concrete wall footing 46





Masonry works

Estimating Concrete Hollow Blocks 55





CHB plastering 60





Metal reinforcement 66

Independent footing reinforcement 70





Post / Column reinforcement 75





Girder or beam reinforcement 83





Tile works

Wall and floor tiles 90

Illustrative example1 91




Painting works 99





Technical Specification sample 105

APPENDICES 110

1

SPECIFICATIONS & BILL OF QUANTITES Prepared By:

ENGR. FREDEL J. DE VERA

General objectives of the course:

This course aims at giving trainee essential skills to perform, count, and

reckon different ordinary work items from drawings and plans, or to measure

the executed work in the site.

Specific Objectives: Skills based objectives

At the end of the chapter, the students should be able to:

1. Match architectural drawing with other engineering drawing.

2. Review conditions and specifications.

3. Make sure of existing execution materials.

4. Reckon executed quantities in the field.

CONCRETE WORKS

A. Footings:

It is the bottom part of the foundation and is usually made of concrete

and reinforced with steel (rebar).

Types of Footings

The most common types of footings are:

Spot or Isolated Footings

A spot or pad footing is used to support a single point of contact, such as under a pier or post.

2



Continuous Spread Footing

A continuous spread footing is mainly used to provide a stable base around the perimeter of a building. Spread footings are often augmented

with interior spot footings. The spread footing supports the weight of the exterior or foundation walls.

Grade Beam Footing

A grade beam footing is a continuous reinforced-concrete member used to support loads with minimal bending. Grade beams are capable of spanning across non-load bearing areas, and are commonly supported

by soil or pilings.

Table 1: Concrete Proportions

CLASS MIXTURE

PROPORTION CEMENT IN BAG SAND

cu. m. GRAVEL cu. m. 40 kg. 50 kg.

AA 1 : 1 ½ : 3 12 9.5 0.5 1.0

A 1 : 2 : 4 9 7 0.5 1.0

B 1 : 2 ½ : 5 7.5 6 0.5 1.0

C 1 : 3 : 6 6 5 0.5 1.0

Procedure of estimating concrete footings:

1. Identify the number of typical footings such as column footing 1 (C1F1),

column footing 2 (C1F2), etc.

2. Compute for the corresponding volume of the column footings.

3. Compute for the number of bags, cubic meter of sand and cubic meter of

gravel by using the values in table1.

3



Illustrative Example1:

From the given isolated footing as shown, compute for the following using class

A mixture.

a. No of bags of cement (using 40 Kg. bag)

b. Cubic meter of sand

c. Cubic meter of gravel

Solution

a. Number of bags of cement

Compute for the volume of concrete:

V = 1.5m x 2.5m x 0.30m

V = 1.125 cu.m.

B = 1.125 x 9

B = 10.125 bags ≈ 11 bags


4



V = 1.125 x 0.50

V = 0.5625 cu.m.


V = 1.125 x 1.0

V = 1.125 cu.m.

5



Illustrative Example 2


A mixture.




Solution



V = 1.8m x 2.0m x 0.25m

V = 0.90 cu.m.

B = 0.90 x 9

B = 8.1 bags ≈ 9 bags

6




V = 0.90 x 0.50

V = 0.45 cu.m.


V = 0.90 x 1.0

V = 0.90 cu.m.

7



Student Exercise1


A mixture.






V = __________________________________ cu.m.

V = __________________________________ cu.m.

B = _________________________

B = ________________________ ≈ ________ bags

8




V = _______________

V = _______________ cu.m.


V = __________________

V = _________________ cu.m.

9



Student Exercise 2


A mixture.






V = __________________________________ cu.m.

V = __________________________________ cu.m.

B = _________________________

10



B = ________________________ ≈ ________ bags


V = _______________

V = _______________ cu.m.


V = __________________

V = _________________ cu.m.

11



Illustrative Example 3:

From the given foundation plan as shown, compute for the following using

class A mixture.




C1F1: 1.5m x 1.6m x 0.30m C1F2: 1.4m x 1.5m x 0.25m

Solution


Number of pieces of C1F1 = 9 pcs.


12



Compute for the total volume of concrete:

V1 = 1.5m x 1.6m x 0.3m x 9 pcs.

V1 = 6.48 cu.m.

V2 = 1.4m x 1.5m x 0.25m x 7 pcs.

V2 = 3.675 cu.m.

Total Volume, Vt = V1 + V2

Vt = 6.48 + 3.675

Vt = 10.155 cu.m.

B = 10.155 x 9

B = 91.935 bags ≈ 92 bags


V = 10.155 x 0.50

V = 5.0775 cu.m.


V = 10.155 x 1.0

V = 10.155 cu.m.

13





class A mixture.




C1F1: 1.2m x 1.4m x 0.20m C1F2: 1.4m x 1.6m x 0.30m

Solution




14




V1 = 1.2m x 1.4m x 0.2m x 5 pcs.

V1 = 1.68 cu.m.

V2 = 1.4m x 1.6m x 0.30m x 11 pcs.

V2 = 7.392 cu.m.


Vt = 1.68 + 7.392

Vt = 9.072 cu.m.

B = 9.072 x 9

B = 81.648 bags ≈ 82 bags


V = 9.072 x 0.50

V = 4.536 cu.m.


V = 9.072 x 1.0

V = 9.072 cu.m.

15



Student Exercise 3:


class A mixture.




C1F1: 1.2m x 1.6m x 0.30m C1F2: 1.4m x 1.6m x 0.25m

Solution


Number of pieces of C1F1 = _________________________ pcs.

16



Number of pieces of C1F2 = _________________________ pcs.


V1 = _________________________

V1 = ________________________ cu.m.

V2 = _________________________

V2 = ________________________ cu.m.


Vt = _________________________

Vt = _________________ cu.m.

B = _____________________

B = ______________ bags ≈ ________ bags


V = ___________________

V = __________________ cu.m.

17




V = __________________

V = _________________ cu.m.

18



Student Exercise 4:


class A mixture.




C1F1: 1.3m x 1.3m x 0.30m C1F2: 1.5m x 1.5m x 0.25m

Solution


Number of pieces of C1F1 = __________ pcs.

19



Number of pieces of C1F2 = __________ pcs.


V1 = _________________________

V1 = ________________________ cu.m.

V2 = _________________________

V2 = ________________________ cu.m.


Vt = _________________________

Vt = _________________ cu.m.

B = _____________________

B = ______________ bags ≈ ________ bags


V = ___________________

20



V = __________________ cu.m.


V = __________________

V = _________________ cu.m.

21



B. Beams

It is a horizontal or inclined structural member spanning a distance between

one or more supports, and carrying vertical loads across (transverse to) its longitudinal axis, as a girder, joist, purlin, or rafter.

Types of beams: (1) Simple span, supported at both ends,

(2) Continuous, supported at more than two points, and (3) Cantilever, supported at one end with the other end overhanging and free.

Procedure of estimating concrete beams:

1. Compute for the corresponding volume of the concrete beam.




From the given beam as shown, compute for the following using class A

mixture.




http://www.businessdictionary.com/definition/load.html

http://www.businessdictionary.com/definition/point.html

http://www.businessdictionary.com/definition/free.html

22



Solution



V = 5.0m x 0.25m x 0.15m

V = 0.1875 cu.m.

B = 0.1875 x 9

B = 1.6875 bags ≈ 2 bags


V = 0.1875 x 0.50

V = 0.09375 cu.m.


V = 0.1875 x 1.0

V = 0.1875 cu.m.

23





mixture.




Solution



V = 4.0m x 0.30m x 0.20m

V = 0.24 cu.m.

B = 0.24 x 9



V = 0.24 x 0.50

V = 0.12 cu.m.


V = 0.24 x 1.0

V = 0.24 cu.m.

24



Sample Exercise1:


mixture.




Solution



V = ________________________

V = _______________________ cu.m.

B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

25



V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

26



Student Exercise2:


mixture.




Solution



V = ________________________

V = _______________________ cu.m.

B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.

27




V = _______________________

V = ______________________ cu.m.

28



C. Columns

A structural member subjected principally to compressive stresses.

Concrete columns may be unreinforced, or they may be reinforced with longitudinal bars and ties (tied columns) or with longitudinal bars and spiral steel (spiral-reinforced columns).

Types of column:

1. Tied 2. Spiral 3. Composite

4. Combination 5. Steel pipe

29



Procedure of estimating concrete column:

1. Compute for the corresponding volume of the concrete column.




From the given column as shown, compute for the following using class A

mixture.




30



Solution



V = 5.0m x 0.30m x 0.30m

V = 0.45 cu.m.

B = 0.45 x 9



V = 0.45 x 0.50

V = 0.225 cu.m.


V = 0.45 x 1.0

V = 0.45 cu.m.

31





mixture.




Solution



V = 6.0m x 0.20m x 0.25m

V = 0.30 cu.m.

B = 0.30 x 9

32





V = 0.30 x 0.50

V = 0.15 cu.m.


V = 0.30 x 1.0

V = 0.30 cu.m.

33



Student Exercise1:


mixture.




Solution



V = ________________________

V = _______________________ cu.m.

34



B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

35



Student Exercise 2:


mixture.




Solution



V = ________________________

V = _______________________ cu.m.

36



B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

37



D. SLABS

A concrete slab is the foundation of a house or building in

construction, made using concrete. The characteristic of a concrete slab is a

foundation that is flat, uniform, at ground level, and is not segmented.

Types of slabs.

1. One way slab

2. Two way slab

3. Flat slab

4. Ribbed slab

5. Waffle slab

Procedure of estimating concrete slabs:

1. Compute for the corresponding volume of the concrete slab.



38




From the given slab as shown, compute for the following using class A

mixture.




Solution



V = 4.5m x 6.0m x 0.20m

V = 5.40 cu.m.

B = 5.40 x 9

B = 48.6 bags ≈ 49 bags

39




V = 5.40 x 0.50

V = 2.70 cu.m.


V = 5.40 x 1.0

V = 5.40 cu.m.

40





mixture.




Solution



V = 5.50m x 6.50m x 0.30m

V = 10.725 cu.m.

B = 10.725 x 9

B = 96.525 bags ≈ 97 bags

41




V = 10.725 x 0.50

V = 5.3625 cu.m.


V = 10.725 x 1.0

V = 10.725 cu.m.

42



Student Exercise1:

From the given slab as shown, compute for the following using class A

mixture.




Solution



V = ________________________

43



V = _______________________ cu.m.

B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

44



Student Exercise 2:


mixture.




Solution



V = ________________________

V = _______________________ cu.m.

B = _______________________

45



B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

46



E. Wall Footing

It is a portion of a foundation/substructure that transmits load unto the soil/ground.

Procedure of estimating wall footing:

1. Compute for the total length of wall footing.

2. Compute for the corresponding volume of the concrete wall footing by

multiplying the total length, the width and the thickness.



47




From the given wall footing as shown, compute for the following using class A

mixture.




Solution



Total Length = 4.20 meters

V = 4.20m x 1.20m x 0.20m

48



V = 1.008 cu.m.

B = 1.008 x 9

B = 9.072 bags ≈ 10 bags


V = 1.008 x 0.50

V = 0.504 cu.m.


V = 1.008 x 1.0

V = 1.008 cu.m.

49





mixture.




Solution



Total Length = 4.00 meters

V = 4.00m x 1.00m x 0.25m

50



V = 1.00 cu.m.

B = 1.00 x 9

B = 9 bags ≈ 9 bags


V = 1.00 x 0.50

V = 0.5 cu.m.


V = 1.00 x 1.0

V = 1.00 cu.m.

51



Student Exercise1:


mixture.




Solution



Total Length = _____________ meters

52



V = ________________________

V = _______________________ cu.m.

B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

53



Student Exercise 2:


mixture.




Solution



Total length = ______________ meters

V = ________________________

V = _______________________ cu.m.

54



B = _______________________

B = ____________ bags ≈ __________ bags


V = _______________________

V = ______________________ cu.m.


V = _______________________

V = ______________________ cu.m.

55



MASONRY WORKS

A. Concrete Hollow Blocks

Concrete hollow blocks is popularly known as CHB. A hollow or solid concrete

masonry unit consisting of Portland cement and suitable aggregates combined

with water. Lime, fly ash, air-entraining agents, or other admixtures may be

included.

Classifications of CHB

1. Load bearing blocks.

- These are used to carry loads aside from its own weight.

2. Non-load bearing blocks

- These blocks are intended for walls, partitions, fences, dividers

carrying its own weight only.

Methods of computations

1. By fundamental method

2. By area method

Formula : Area x 12.5 pcs/sq.m.

56



Steps in computing by area method:

1. Determine the total length of the wall.

2. Determine the total height of the wall

3. Compute for the area of the wall

4. Subtract wall openings if there are wall openings to get the net area.

5. Compute the net area by subtracting the wall opening.

6. Multiply the net area by 12.5 pcs/sq.m.


From the given wall as shown, compute for the number of pcs. Of CHB using

area method

Solution:

57



a. Total length of the wall = 5.79 meters

b. Total height of the wall = 3.01 meters

c. Area of the wall = 5.79m x 3.01 meters

= 17.4279 sq. meter.

d. Wall opening = NO OPENING

e. Net area = 17.4279 sq. meter

f. Total number of pcs of CHB = 17.4279 x 12.5pcs/sq.m.

= 217.84875 ≈ 218 pcs



area method

Solution:

a. Total length of the wall = 6.18 meters

b. Total height of the wall = 3.20 meters

c. Area of the wall = 6.18m x 3.20 meters

58



= 17.4279 sq. meter.

d. Wall opening = 1.66m x 1.66m = 2.7556

e. Net area = 17.4279 – 2.7556 = 14.6723 sq. meters

f. Total number of pcs of CHB = 14.6723 x 12.5pcs/sq.m.

= 183.40375 ≈ 184 pcs

Student Exercise 1


area method

Solution:

a. Total length of the wall = __________________ meters

b. Total height of the wall = __________________ meters

c. Area of the wall = __________________________

= _______________________________________ sq. meter.

59



d. Wall opening = ___________________________

e. Net area = ______________________________ sq. meters

f. Total number of pcs of CHB = ________________________

= _____________________ ≈ _________ pcs

Sample Exercise 2


area method

Solution:

a. Total length of the wall = __________________ meters

b. Total height of the wall = __________________ meters

c. Area of the wall = __________________________

60



= _______________________________________ sq. meter.

d. Wall opening = ___________________________

e. Net area = ______________________________ sq. meters

f. Total number of pcs of CHB = ________________________

= _____________________ ≈ _________ pcs

B. CHB plastering

After knowing the number of blocks needed for a particular

masonry work, the next step is to find its work partner called plastering which

is another item to consider. Most estimators however, make their estimate of

mortar for block laying and plastering through simple guessing and

calculation, assuming the quantity of cement and sand without the pain of

computation. The reason is simple; they are just in a hurry and have no time to

do it.

Methods of plastering computation.

a. The volume method

61



b. The area method

Steps in solving plastering works by area method;

1. Compute the total surface area of the wall.

2. Solve for the cement and sand by referring data in table 2

Table 2: Quantity of Cement and Sand for Plaster per square meter

Mixture

Class

Cement in Bags Thickness of Plaster

8 mm 12 mm 16 mm 20 mm 25 mm

A 0.144 0.216 0.288 0.36 0.45

B 0.096 0.144 0.192 0.24 0.3

C 0.072 0.108 0.144 0.18 0.225

D 0.06 0.09 0.12 0.15 0.188

Sand 0.008 0.012 0.016 0.02 0.025


From the given figure below, list down the cement and sand necessary to plaster the

two faces of the fence at an average thickness of 0.20 mm. class “C” mixture.

62



Solution (By the area method)

a. Total surface area of the wall

S.A = (5 x 8) + (5 x 6.5) + (5 x 8)

S.A = 112.5 sq.meters

b. Solve for the cement and sand referring from table 2 using class “C” 20 mm

thickness of plaster.

Cement = 112.5 x 0.18 = 20.25 bags ≈ 21 bags

Sand = 112.5 x 0.02 = 2.25 ≈ 3 cu.m.






S.A = (4 x 7) + (5 x 4)

S.A = 48 sq.meters

63





Cement = 48 x 0.18 = 8.64 bags ≈ 9 bags

Sand = 48 x 0.02 = 0.96 ≈ 1 cu.m.

64



Student Exercise 1


two faces of the fence at an average thickness of 0.16 mm. class “D” mixture.



S.A = ____________________________

S.A = ____________________________ sq.meters



Cement = ____________________ = ___________ bags ≈ ________ bags

Sand = _______________________= ____________ ≈ ____________ cu.m.

65



Student exercise 2





S.A = ____________________________

S.A = ____________________________ sq.meters



Cement = ____________________ = ___________ bags ≈ ________ bags

Sand = _______________________= ____________ ≈ ____________ cu.m.

66



METAL REINFORCEMENT

Steel is the most widely used reinforcing materials 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.

Types of deformed steel bars

67



IDENTIFICATION OF STEEL BARS

Steel reinforcing bars provided with distinctive markings identifying the

name of the manufacturer with its initial brand and the bar size number

including the type of steel bars presented as follows:

N = for billet

A = for axle

Rail Sign = for rail steel

Steel bars marking system

A. Independent Footing reinforcement

Independent column footing is also referred to as individual or isolated

footing. The ACI Code provides that the minimum underground protective

covering of concrete to steel reinforcement shall not be less than 7.5

centimeters. The reinforcement for this type of structure is determined by

direct counting from the detailed plan under the following procedures.

68



Individual footing reinforcement

Procedure in computing the reinforcement:

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

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

3. Find the total cut bars for the total number of footings.

4. Find the total length by multiplying it by the net length.

5. Divide the total length by the length of one commercial steel bar which is

6 meters.

6. Compute for the total kilograms of tie wire by multiplying the number of

bars along the length and width by the length of tie wire to be used.

69



70




From the given figure, determine the number of 12mm diameter steel bars

including the tie wire in kilograms if there are 30 pieces 1.15 m x 1.15 m

independent square footing. Length of hook is 0.10m and 0.30m length of tie

wire.

Solution

1. Net length of one reinforcing cut-bar

Ln = 1 + 0.10 + 0.10

Ln = 1.20 meters

2. Total numbers of cut burs in one footing

N = 6 + 6 = 12 pcs. Per footing

3. Total cut bars

Nt = 12 x 30 = 360 pcs.

4. Total length of bars

Lt = 360 x 1.20 = 432 meters

5. Number of pcs. Of bars

Nb = 432/6 = 72 pcs 12mm x 6.0 meters RSB

6. Kilograms of tie wire

K = (6 x 6 x 0.30 x 30)/53 = 6.113 ≈ 7 kgs.

71




From the given figure, determine the number of 16mm diameter steel

bars including the tie wire in kilograms if there are 20 pieces 1.00 m x 1.60 m

independent square footing. Length of hook is 0.10m and 0.40m length of tie

wire.

Solution


Ln1 = 1 + 0.10 + 0.10 = 1.20 meters

Ln2 = 1.6 + 0.10 + 0.10 = 1.80 meters


N1 = 9 pcs.

N2 = 6 pcs.

3. Total cut bars

Nt = 9 + 6 = 15 pcs.


L1 = 1.20 x 9 = 10.80 meters

L2 = 1.80 x 6 = 10.80 meters

Lt = (10.80 meters + 10.80 meters) x 20

Lt = 432 meters


Nb = 432/6 = 72 pcs 16mm x 6.0 meters RSB


K = (9 x 6 x 0.40 x 20)/53 = 8.15 ≈ 9 kgs.

72



Student Exercise 1


bars including the number of tie wire in kilograms if there are 30 pieces 2.65 m

x 2.65 m independent square footing. Length of hook is 0.10m and 0.30m

length of tie wire.

Solution


Ln = _________________________________________

Ln = ________________________________________ meters


N = _________________________________________ pcs. Per footing

3. Total cut bars

Nt = _________________________________________ pcs.


Lt = _________________________________________ meters


Nb = ____________ = ____________________________

2.50 m

2.65 m

2.65 m

16mm Steel bars

2.50 m

73




K = ___________________________ kgs.

Student Exercise 2


bars including the tie wire in kilograms if there are 25 pieces in an independent

square footing. Length of hook is 0.10m and 0.40m length of tie wire.

Solution


Ln1 = ______________________________________________ meters

Ln2 = ______________________________________________ meters


N1 = ____________________________ pcs.

N2 = ____________________________ pcs.

3. Total cut bars

Nt = ______________ pcs.


L1 = _________________ meters

12mm Steel bars

1.20 m

0.8

0 m

74



L2 = __________________meters

Lt = __________________ meters

Lt = __________________ meters


Nb = _______________ pcs


K = _________________________________________ kgs.

75



B. Post or Column Reinforcement

A reinforced concrete column is a structural members designed to carry

compressive loads, composed of concrete with an embedded steel frame to provide reinforcement. For design purposes, the columns are separated into

two categories: short columns and slender columns.

Procedure of computation:

1. Compute for the number of pieces of vertical main bars by multiplying

the total length by the number of pieces of bars per column/post and the

number of pieces of columns.

2. Compute for the number of pieces of lateral ties by dividing the length of

the column by the spacing of the ties and multiplying it by the number of

pieces of column.

3. Compute for the kilograms of tie wires by multiplying the number of

pieces of main bars and the total number of pieces of tie wires multiplied

by the length of tie wire to be used divided by 53.

http://en.wikipedia.org/wiki/List_of_structural_elements

http://en.wikipedia.org/wiki/Compressive_stress

http://en.wikipedia.org/wiki/Steel

76




From the given column details as shown, compute for the following if there are

15 pieces of identical columns.

a. Number of pieces of 16mm diameter vertical main bars.

b. Number of pieces of 10mm diameter lateral ties.

c. Kilograms of tie wire.

77



Solution:


Lt = 6 x 2.48 x 15

Lt = 223.20 meters

Nb = 223.20/6 = 37.2 pcs ≈ 38 pcs.


Nt = (2.48 / 0.20) + 1 = 13.4 pcs. Per column

Ntotal = 13.4 x 15 = 201 pcs

Length per tie = 0.50 + 0.50 + 0.50 + 0.50

Length per tie = 2 meters

Total length = 2 x 201 = 402 meters

Nb = 402/6 = 67 pcs

c. Kilograms of tie wire

K = (201 x 6 x 0.30) / 53 = 6.83 kgs. ≈ 7 kgs.

78





20 pieces of identical columns. Length of tie wire is 0.30m.




Solution:


Lt = 6 x 4.0 x 20

Lt = 480 meters

Nb = 480/6 = 80 pcs.


Nt = (4.00 / 0.20) + 1 = 21 pcs. Per column

Ntotal = 21 x 20 = 420 pcs

Length per tie = 0.30 + 0.30 + 0.30 + 0.30

Length per tie = 1.2 meters

Total length = 1.2 x 420 = 504 meters

Nb = 504/6 = 84 pcs


4.0

0 M

0.30 0.3

0

79



K = (420 x 6 x 0.30) / 53 = 14.26 kgs. ≈ 15 kgs.

Student Exercise 1






Solution:


Lt = ____________________________________________________

Lt = ____________________________________________________ meters

Nb = ___________________________________________________ pcs.


Nt = ______________________________________ pcs. Per column

Ntotal = ___________________________________ pcs

5.0

0 M

0.25

0.2

5

80



Length per tie = ___________________________

Length per tie = ____________________ meters

Total length = _______________________________ meters

Nb = _________________________________ pcs


K = ________________________________ kgs.

81



Student Exercise 2






Solution:


Lt = _____________________________________________

Lt = _____________________________________________ meters

Nb = ____________________________________________ pcs.


Nt = ______________________________________ pcs. Per column

Ntotal = ___________________________________ pcs

7.0

0 M

0.20

0.2

0

82



Length per tie = ___________________________

Length per tie = ____________________ meters

Total length = ____________________________________________ meters

Nb = ______________________________________________________ pcs


K = _________________________________________________________ kgs.

83



C. Girder and Beam Reinforcement

The direct counting is so far the best method in determining the number

of main reinforcement of beam and girder. The length however, is determined

by the physical condition of the structures in relation with their support.

84




From the given beam details as shown, compute for the following if there are 8

pieces of identical beams. Length of tie wire is 0.30m.




Solution

a. Number of pieces of 12mm Ø main bars

Lt = 6 x 8 x 8 = 384 meters

Nb = 384/6 = 64 pcs 12mm Ø RSB

b. Number of pieces of 10mm diameter stirrups.

Ns = (8/0.20) + 1 = 41 pcs. Per beam

Ntotal = 41 x 8 = 328 pcs.

Length per stirrup = 0.20 + 0.30 + 0.20 + 0.30

Length per stirrup = 1.00 meter


Nb = 328/6 = 54.67 ≈ 55 pcs 10mm Ø RSB

85




K = (6 x 328 x 0.30)/53 = 11.14 ≈ 12 Kgs







Solution


Lt = 6 x 7.5 x 10 = 450 meters

Nb = 450/6 = 75 pcs 12mm Ø RSB


Ns = (7.50/0.20) + 1 = 38.5 pcs. Per beam

Ntotal = 38.5 x 10 = 385 pcs.

Length per stirrup = 0.40 + 0.30 + 0.40 + 0.30

Length per stirrup = 1.40 meter


86



Nb = 539/6 = 89.83 ≈ 90 pcs 10mm Ø RSB


K = (6 x 385 x 0.30)/53 = 13.075 ≈ 14 Kgs.

Student Exercise 1






Solution


Lt = ____________________________________

Nb = ___________________________________ 12mm Ø RSB


Ns = ____________________________________ pcs. Per beam

Ntotal = _________________________________ pcs.

87



Length per stirrup = ________________________________

Length per stirrup = ________________________________ meter

Total length = _________________________________ meters

Nb = __________________________________________10mm Ø RSB


K = ___________________________________________ Kgs.

88



Student Exercise 2






Solution


Lt = ____________________________________

Nb = ___________________________________ 12mm Ø RSB


Ns = ____________________________________ pcs. Per beam

Ntotal = _________________________________ pcs.

Length per stirrup = ___________________________

89



Length per stirrup = ___________________________ meter

Total length = _________________________________ meters

Nb = __________________________________________10mm Ø RSB


K = ___________________________________________ Kgs.

90



TILE WORKS

Ceramic tile is one of man’s oldest unique building materials

continuously in use because of its durability, functional, aesthetic

properties. It is practically indestructible and offers unlimited choices not

only in design pattern but also in color that does not fade.

Decorative ceramic tile was extensively used as early as

period of medieval Islamic Architecture from Persia to Spain. Its

popularity and used was extended up to the period of contemporary

architecture.

Quantity of tiles per square foot and meter

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From the given figure, determine the quantity of the following materials:

a. 10 cm. x 20 cm. glazed tiles

b. 20 cm. x 20 cm. unglazed floor tiles.

c. Cement for mortar.

d. White cement for joint filler.

e. Tile adhesive

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Solution

a. Solving for 10cm. x 20 cm. glazed tiles

Wall area = 1.50 x (5.00 + 3.00) = 12 sq.m.

Area of one tile = 0.10 x 0.20 = 0.02 sq.m.

Number of tiles = 12/0.02 = 600 pcs.

b. Solving for 20cm. x 20 cm. unglazed floor tile

Solving for the floor area = 5.00 x 3.00 = 15 sq.m.

Area of one tile = 0.20m. x 0.20m. = 0.04 sq.m.


c. Solving for cement mortar

Total area of wall and floor = 12 + 15 = 27 sq.m.

Referring from table 4:

Cement mortar = 27 x 0.086 = 2.3 ≈ 3 bags

d. White cement filler = 27 x 0.50 = 13.5 ≈ 14 bags.

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e. Tile adhesive = 27 x 0.11 = 2.79 ≈ 3 bags







e. Tile adhesive

94



Solution


Wall area = 5 x (6.00 + 6.00) = 60 sq.m.

Area of one tile = 0.10 x 0.10 = 0.01 sq.m.



Solving for the floor area = 6.00 x 6.00 = 36 sq.m.

Area of one tile = 0.20m. x 0.20m. = 0.04 sq.m.



Total area of wall and floor = 60 + 36 = 27 sq.m.


Cement mortar = 96 x 0.086 = 8.256 ≈ 9 bags

d. White cement filler = 96 x 0.50 = 48 bags.

e. Tile adhesive = 96 x 0.11 = 10.56 ≈ 11 bags

95



Student Exercise 1:






e. Tile adhesive

Solution


Wall area = ___________________________________________ sq.m.

Area of one tile = _____________________________________ sq.m.

96



Number of tiles = _____________________________________ pcs.


Solving for the floor area = _________________________________ sq.m.

Area of one tile = __________________________________________ sq.m.

Number of tiles = ___________________________________________ pcs.


Total area of wall and floor = _______________________ sq.m.


Cement mortar = ________________________________ bags

d. White cement filler = ____________________________ bags.

e. Tile adhesive = __________________________________ bags

97



Student Exercise 2:






e. Tile adhesive

Solution

a. Solving for 10cm. x 10cm. glazed tiles

Wall area = ____________________________________________ sq.m.

Area of one tile = _______________________________________ sq.m.

98



Number of tiles = ______________________________________ pcs.


Solving for the floor area = __________________________ sq.m.

Area of one tile = ____________________________________ sq.m.

Number of tiles = ____________________________________ pcs.


Total area of wall and floor = _______________________ sq.m.


Cement mortar = ________________________________ bags

d. White cement filler = ____________________________ bags.

e. Tile adhesive = __________________________________ bags

99



PAINTING WORKS

Paint is commonly referred to as a “Surface Coating”. It is defined

as “a coating applied to a surface or substrate to decorate, to protect, or

to perform some other specialized functions.

Almost everybody knows the word paint, its uses colors, including the

brand name rated as poor, good, and durable. There are those who have

little knowledge of paint but rated a brand based on how it is advertised

while others on cost.

Estimating your paint

Paint manufacturer specifications include the estimated area

coverage per gallon having 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 30 to 40

square 40 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 as presented:

Table 3: Paint coverage surface area

Type of surface Coverage area,

sq. m.

Coarse to rough surface 30

Fine to coarse surface 35

Smooth surface 40

100



Paints and other coatings are estimated using the formula:

(surface area) x (No.of coats)

coverage of paint = Quantity of paint required

101




The computer area and the number of coats and coverage of paint to be

used for a painting job are as follows:

Area of masonry surface = 750 square meters Primer coat to be used = 1 coat with coverage of 30 sq.m. per 4 liters Topcoat to be used = 1 coat with coverage of 40 sq.m. per 4 liters

Area of wood surfaces = 900 square meters

Primer coat to be used = 1 coat with coverage of 25 sq.m. per 4 liters

Topcoat to be used = 1 coat with coverage of 35 sq.m. per 4 liters

Solution:

A. Paints required for masonry surfaces: 1. Converting the coverage primer coat to its equivalent rate per liter:

Coverage = 30 sq.m. / 4 liters = 7.5 square meters per liter

Primer required = 750/7.5 = 100 liters Number of cans @ 4Liters = 100/4 = 25 cans

2. Converting the coverage of Topcoat paint to its equivalent rate per liter:

Coverage = 40 sq.m. / 4 liters = 10 sq. m. per liter Topcoat paint required = 750/10 = 75 liters Number of cans @ 4Liters = 75/4 = 18 cans

B. Paints required for wood surfaces:

1. Converting the coverage of primer coat to its equivalent rate per liter: Coverage = 25 sq.m. / 4 liters = 6.25 square meters per liter Primer required = 900/6.25 = 144 liters

Number of cans @ 4Liters = 144/4 = 36 cans

2. Converting the coverage of Topcoat paint to its equivalent rate per

liter: Coverage = 35 sq.m. / 4 liters = 8.75 sq. m. per liter

Topcoat paint required = 900/8.75 = 102.8 liters Number of cans @ 4Liters = 102.8/4 = 25.7 ≈ 26 cans

102






Area of masonry surface = 800 square meters Primer coat to be used = 1 coat with coverage of 40 sq.m. per 4 liters Topcoat to be used = 1 coat with coverage of 50 sq.m. per 4 liters

Area of wood surfaces = 1000 square meters

Primer coat to be used = 1 coat with coverage of 30 sq.m. per 4 liters


Solution:

A. Paints required for masonry surfaces: 1. Converting the coverage primer coat to its equivalent rate per liter:

Coverage = 40 sq.m. / 4 liters = 10 square meters per liter

Primer required = 800/10 = 80 liters Number of cans @ 4Liters = 80/4 = 20 cans

2. Converting the coverage of Topcoat paint to its equivalent rate per liter:

Coverage = 50 sq.m. / 4 liters = 12.5 sq. m. per liter Topcoat paint required = 800/12.5 = 64 liters Number of cans @ 4Liters = 64/4 = 16 cans

B. Paints required for wood surfaces:

1. Converting the coverage of primer coat to its equivalent rate per liter: Coverage = 30 sq.m. / 4 liters = 7.5 square meters per liter Primer required = 1000/7.5 = 133.33 liters

Number of cans @ 4Liters = 133.3/4 = 33.33 ≈ 34 cans


liter: Coverage = 40 sq.m. / 4 liters = 10 sq. m. per liter

Topcoat paint required = 1000/10 = 100 liters Number of cans @ 4Liters = 100/4 = 25 cans

103



Student Exercise 1:



Area of masonry surface = 600 square meters

Primer coat to be used = 1 coat with coverage of 20 sq.m. per 4 liters Topcoat to be used = 1 coat with coverage of 30 sq.m. per 4 liters

Area of wood surfaces = 550 square meters Primer coat to be used = 1 coat with coverage of 20 sq.m. per 4 liters


Solution:

A. Paints required for masonry surfaces:

1. Converting the coverage primer coat to its equivalent rate per liter: Coverage = _______________________________ square meters per liter

Primer required = ________________________________ liters

Number of cans @ 4Liters = ______________________ cans


liter: Coverage = ______________________________________ sq. m. per liter

Topcoat paint required = __________________________ liters

Number of cans @ 4Liters = _______________________ cans

104



Student Exercise 2:



Area of masonry surface = 950 square meters Primer coat to be used = 1 coat with coverage of 35 sq.m. per 4 liters


Area of wood surfaces = 1500 square meters Primer coat to be used = 1 coat with coverage of 25 sq.m. per 4 liters


Solution:

A. Paints required for masonry surfaces:

1. Converting the coverage primer coat to its equivalent rate per liter:

Coverage = _______________________________ square meters per liter Primer required = ________________________________ liters

Number of cans @ 4Liters = ______________________ cans


liter:

Coverage = _______________________________________ sq. m. per liter

Topcoat paint required = ___________________________________ liters Number of cans @ 4Liters = _______________________ cans

105



TECHNICAL SPECIFICATIONS

For the proposed residence of:

OWNER: __________________________________

PROJECT TITLE: __________________________________

LOCATION: __________________________________

1.1. Nature and Scope of Construction Works The work to be done shall generally consist of but not necessarily

limited to provisions of all materials, labor, plant, equipment,

supervisions required for the completion of the

___________________________________________ in strict accordance

with this technical specifications and other related contract

documents unless stated otherwise in the contract.

1.2. National Laws, Local Ordinances, and Building Rules and Regulations

Construction of the proposed residential building shall be in conformity with national laws, local ordinances, and building rules

and regulations.

1.3. Quality Assurance Work cited under these technical specifications should be executed

according to generally accepted standard practices.

This technical specification shall be interpreted in conjunction with

the drawings for the satisfactory completion of the works involved.

Any discrepancies, if any, between the technical specifications and

drawings shall be verified with the Engineer and accordingly

adjusted immediately.

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2. Site Preparation

2.1. Clear and prepare the site within the lot. 2.2. Execute all excavation works to the required lines and grades as

required and specified in the contract documents (architectural.)

2.3. All fittings shall be done in layers not exceeding six (6) inches in thickness. Each layer being thoroughly completed to 95% density.

2.4. Execute all soil treatment works as early as allowable, but shall have been completed prior to the pouring of concrete.

3. Concrete Works

3.1. General Notes

All concrete works shall be done in accordance with the standard

specifications for concrete and reinforced concrete.

3.2. Proportioning of Concrete 3.2.1. Use Class “A” concrete for all footings, beams, and columns.

3.2.2. Use Class “C” concrete for slab on fill, mortar, and driveway pavement.

3.2.3. All slabs on ground shall not be less than 75.0 millimeters

(mm.)

4. Masonry Works

4.1. Use 4-inch (in.) concrete hollow blocks for interior partitions unless otherwise specified or noted in the drawings. Use 10 mm.

dia. deformed steel bars (DSB) vertical/ horizontal bars spaced @ 0.60 M. O.C.

4.2. Use 4-inch (in.) concrete hollow blocks for exterior walls unless

otherwise specified or noted in the drawings. Use 10 mm. dia. deformed steel bars (DSB) vertical/ horizontal bars spaced @ 0.60 M. O.C.

107



5. Wood and Plastics

5.1. General Requirements

5.1.1. Lumber shall be of good quality, well-seasoned, thoroughly dry, and free of knots, sap, shakes, or other imperfections, which may impair strength, durability or appearance.

5.1.2. Kinds of lumber required for the various parts of the work shall be as indicated in the drawings.

5.1.3. Any lumber equally good for the purpose intended may be

substituted to the kinds specified subject to the Engineer’s approval.

5.2. Door Framing

Doorframes shall be done with carefully fitted joints as much as

possible. Framing lumber shall be of the rough dimensions as

specified in the drawings.

6. Roofing 6.1. Roofing

Rib Roof type (color red), or approved equivalent.

6.2. Gutters/ Flashings Ga. 24 pre-formed, gutters, ridge rolls, and flashings (or approved

equivalent.)

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Installation shall be as per manufacturer’s standards.

7. Doors

7.1. Doors hollow core type as shown in the drawings, unless otherwise specified.

7.1.1. Hollow core (flush) type doors shall have a veneer of ¼” thick marine plywood veneer. The veneer shall be protected from peeling off at the edges by rabetting it around it around into

the doorframe. 7.1.2. All doors shall have keyed locksets of “Amerilock” brand, or

approve equivalent, satin finish. 7.1.3. Provide and set in place hinges. Use 3 ½” x 3 ½” loose pin,

butt hinges.

8. Finishes For extent of application and specific finishes, refer to the drawings.

8.1. Exposed CHB 8.1.1. Concrete blocks shall be laid plumb and leveled. 8.1.2. After each grade, all joints shall be leveled and grooved with

a suitable device. 8.1.3. Repair imperfections such as cracks, holes, and notches, to

match CHB surface. 8.1.4. Leveling plaster to match CHB surface as close as possible to

allow for further surface treatment in the future.

8.2. Trowelled Concrete Finish 8.2.1. Concrete topping mixtures shall be composed of Portland

cement ASTM C150 Type 1 (gray), fine and coarse aggregates, clean, hard, and free from deleterious matters.

Graded by weight to pass sieves according to accepted standards. Fine aggregates shall consist of sand and crushed screenings. Coarse aggregates shall consist of

gravel or crushed stones. 8.2.2. Remove dirt, loose materials, oil, grease, paint, or other

contaminants, leaving a clean surface. When base slabs are

109



unacceptable for good bonding, roughen surface by chipping or scarifying before cleaning.

8.2.3. Thoroughly dampen slab surface prior to placing topping mixture.

8.2.4. Provide float finish. Spread topping mixtures evenly over prepared base to the required elevation. Use highway straight edge, full float, or Darby, to level surface. After the

topping has stiffened sufficiently to permit the operation and water sheets have disappeared, float the surface at least twice to a uniform sandy texture. Re-straighten where

necessary with highway straight edge. 8.2.5. After floating, begin first trowel finish operations using

power-driven trowels. Continue trowelling until surface is ready to receive final trowelling. Begin final trowelling when a ringing sound is produced as trowel is moved over

trowelled surface.

9. Electrical Works 9.1. Installations of wiring in concrete, masonry, CHB walls, slabs, and

ceiling, shall be done in corrugated conduit pipes. 9.2. All sizes of wires shall follow the specifications of the Professional

Electrical Engineer designer (see electrical drawings.) 9.3. Wall switches and receptacles are on approved equivalent. 9.4. Lighting shall be surface-mounted omni bulb (18W) fixtures.

110



Appendices:

Problem:


class A mixture.

1. For the footing:




2. For the column:




3. For the beam:




4. For the slab:




5. For the 6” CHB wall:




d. Number of pcs. Of CHB

111



Footings:

C1F1: 1.2m x 1.6m x 0.30m C1F2: 1.4m x 1.6m x 0.25m

Beam: 400mm x 300mm

Column: 400mm x 400mm, Ht = 3.5meters

Documents

CIV 105.pdf