Prepared by: Engr. Ranz Gabriel D.
Turingan
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 concrete with reinforcement
embedded in such a manner that the two materials act together in resisting
forces
▪ The Ordinary Portland Cement
▪ The Rapid Hardening Portland Cement which is preferred when high early
strength is required
▪ The Blast Furnace of Sulphate Cement which is used on structures to resist
chemical attack
▪ The Low Heat Portland Cement for massive section to reduce the heat of hydration
▪ The Portland Pozzolan Cement with a low hardening characteristic concrete
▪ The high Alumina Cement
▪ 60 – 65% Lime
▪ 18 – 25% Silica
▪ 3 – 8% Alumina
▪ 5% Iron Oxide
▪ 2 – 5% Magnesia
▪ 1 – 5% Sulfur Trioxide
▪ Coarse Aggregates
- crushed stone, crushed gravel or natural gravel with particles retained on a
5mm sieve
▪ Fine Aggregates
- Crushed stone, crushed gravel, sand or natural sand with particles passing on a
5mm sieve
The most common and easy way of proportioning in concrete is the volume
method using a measuring box for sand and gravel. The reason 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
A proposed concrete pavement has a general 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
1. Determine the volume of the proposed concrete pavement.
Convert 4 inches to meter = .10m
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 0.50 = .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?
Refer to Table 1-2. Using 50 kg. cement class “C” mixture;
Multiply:
Cement : 1.5 x 5.0 = 7.5 bags
Sand : 1.5 x 0.50 = .75 cu.m.
Gravel : 1.5 x 1.0 = 1.50 cu.m
Since we cannot buy 7.5 bags, order 8 bags at 50 kg./ bag
Concrete hollow blocks are classified as bearing and non- bearing
blocks. Load bearing blocks are those whose thickness ranges from 15cm to
20cm. And are used to carry load aside from its own weight. Non- bearing
blocks on the other hand, are blocks which are intended for walls, partitions,
fences or dividers carrying it own weight whose thickness ranges from 7.5cm
to 10 cm.
From the figure, determine the number of 10 x 20 x 40cm. Concrete Hollow Blocks and
the materials required for:
a. Mortar for block laying
b. Mortar filler for the hollow core cells
c. Plastering
d. Concrete for CHB and post footings
A. Concrete Hollow Blocks
1. Find the area of the fence
A = 3.00 x 20.00m = 60 sq.m
2. Refer to Table 2-2. Multiply:
60.00 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 0.525 = 31.5 bags
Sand = 60 x 0.0438 = 2.63 cu.m
C. Plaster Mortar
1. Finde 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
Cement = 120 x 0.192 = 23.04 bags
Sand = 120x 0.016 = 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.
Cement = 20m x 0.450 = 9.0 bags
Sand = 20mx 0.030 = 0.60 cu.m
Gravel = 20mx 0.060 = 1.20 cu.m
Summary of Materials
1. Concrete Hollow Blocks– 750 pcs.
2. 40 kg. Cement – 63.5 say 64 bags
3. Sand – 5.15 cu.m
4. Gravel – 1.2 cu.m
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 time stronger in
tensile stresses
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 varies
depending upon the limitation as provided for the National Building Code.
Determine the length of the splice joint for a 16mm steel bars under the following
conditions:
a. Tensile reinforcement of a beam
Multiply:
25 x 16mm + 150mm = 550mm
b. Compressive reinforcement of a column
Multiply:
20 x 16mm + 150mm = 470mm
Tie wire refers to gauge NO. 16 galvanized iron wire popularly called
G.I. tie wire. Tie wire is used to secure the steel bars in its designed position
before accepting fresh concrete.
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 a
small and medium size steel bars.
From the Figure 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
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.I. tie wire
▪ Determine the number of 12 mm steel bars required 6 meters long if there are 6
footings with a general dimensions of 1.5 x 1.5 meters and tie wires using 25 cm
length per tie wire.
▪ Solve for the net length
1.5 -2(0.075)=1.35 m
▪ Find the total number of cut bars in one footing by direct counting
13 x 2 = 26 pcs.
▪ Get the total number of bars for the 6 footings.
26 x 6 = 156 pcs. @ 1.35 m. long
▪ Select the steel bars whose length is economically cut into 1.35 m long.
6 m/1.35 m= 4.44 pcs.
▪ The fractional value of 0.44 is inevitable, but should not be included in the
computation because it is less than one cut bar length. Use the whole value 4.0
thus,
▪ Divide the total number of bars to be used by 4.0.
156/4 = 39 pcs. Of 12 mm x 6 m
▪ Find the number of intersections of steel bars in one footing.
▪ 13 x 13 =169 ties
▪ Total ties for 6 footings
169 x 6 = 1014 ties
▪ Using 25 cm length per tie wire
1014 x 0.25 =253.5 meters
▪ One kilo of No.16 G.I wire is approximately 53 meters long.
253.5/53 = 4.783 kilos
