[Virtual Presenter] Good evening everyone. Today we are here to discuss and learn about steel reinforcement and its different weight and design variations.Let's get started!. 

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JADE ERWIN BASUL HUDSON CEASER GEALOGO CEDRIC JOHN SARANZA

[Audio] Steel bars are a necessary material for any kind of concrete construction. Steel is much stronger than concrete, being ten times stronger in compression and a hundred times stronger in tension. To make sure the steel bars and concrete function effectively together in carrying external loads, it is imperative that they are deformed together. If not, the steel bars may be forced out of the concrete due to too much weight and inadequate grip.. 

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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

[Audio] Steel bars are commonly utilized to reinforce concrete and increase the stability of constructions. They have a surface deformation that further improves the adherence to the concrete, producing incredibly sturdy structures. The deformation can come in various shapes and designs, making it applicable for a broad range of applications.. 

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In order to provide a higher degree of 	sufficient bond between the concrete and the 	reinforcement, steel bars were provided with 	a surface deformation in various forms and 	designs.

[Audio] The table below shows the length of steel bars in meters according to the different diameters. The length of the bars increases with the increase of the diameter. For instance, the length of a 5-millimeter-diameter steel bar is 1.98 meters, that of a 7.5-millimeter-diameter is 2.96 meters and that of a 36-millimeter-diameter is 107.88 meters. This table can be helpful when reinforcing steel bars.. 

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Diam. 5 6 7.5 9 10.5 12 13.5







 8mm 1.98 2.37 2.96 3.56 4.15 4.74 5.33 10mm 3.08 3.7 4.62 5.54 6.47 7.39 8.32 12mm 4.44 5.33 6.66 7.99 9.32 10.66 11.99 13mm 5.21 6.25 7.83 9.38 10.94 12.5 14.07 16mm 7.9 9.47 11.84 14.21 16.58 18.95 21.32 20mm 12.33 14.8 18.5 22.19 25.69 29.59 33.29 25mm 19.27 23.12 28.9 34.68 40.46 46.24 52.02 28mm 24.17 29 36.25 43.5 50.75 58 65.25 30mm 27.75 33.29 41.62 49.94 58.26 66.59 74.91 32mm 31.57 37.88 47.35 56.82 66.29 75.76 85.23 36mm 39.96 47.95 59.93 71.92 83.91 95.89 107.88

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

[Audio] The table provides the areas, in square millimetres, for the most common groups of reinforcing steel bars. The bar diameter is found along the left-hand side and the area for each size can be read off.. 

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Bar Dia. mm.











 1 2 3 4 5 6 7 8 9 10










 6 28 57 85 113 141 170 198 226 254 283 8 50 101 151 201 251 302 352 402 452 503 10 79 157 236 314 393 471 550 628 707 785 12 113 226 339 452 565 679 792 905 1017 1131 16 201 402 603 804 1005 1206 1407 1608 1809 2011 20 113 226 339 452 565 679 792 905 1017 1131 24 201 402 603 804 1005 1206 1407 1608 1809 2011 32 314 628 942 1257 1571 1885 2199 2513 2827 3142

TABLE 3-4 AREAS OF GROUPS OF REINFORCING STEEL BARS

[Audio] When considering the length of steel reinforcement bars for splicing, keep in mind the following guidelines. For tension bars, the minimum splicing length should be 25 times the bar size plus 150 millimeters. For compression bars, the minimum splicing length should be 20 times the bar size plus 160 millimeters. This will ensure the proper integrity of the structure.. 

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TYPES OF REINFORCEMENT MINIMUM SPLICE LENGTH 1. Tension Bars 25 x BAR SIZE + 150mm 2. Compression Bars 20 x BAR SIZE + 160mm

HOW TO DETERMINE SPLICING LENGTH OF STEEL BARS

[Audio] The length of the splice joint for a 16mm steel bar is calculated by multiplying 25 times the bar diameter plus 150mm. This would result in 25 times 16mm plus 150mm, equal to 550mm or 55cm. Likewise, the same calculations would apply to a 20mm steel bar.. 

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Determine the length of the splice joint for a 16mm and 20mm steel bars under the following conditions:
 a.)Tensile reinforcement of a beam. b) Compressive reinforcement of a column.
 SOLUTION (For 16mm steel bars) a.) Classification of the reinforcement is under tension. Multiply: 	25 x 16mm + 150mm = 550mm or 55cm

[Audio] Multiplying 20 x 16mm plus 160mm gives a total of 480mm or 48 centimeters, so the splice joint for a 16mm tension bar should be 55 centimeters long and 48 centimeters for compression bars. For a 20mm steel bar, the length of splice for a tension bar is 25 x 20mm plus 150mm, resulting in a total of 650mm or 65 centimeters. As for compression bars, the splice length is 20 x 20mm plus 160mm, totaling 560mm or 56 centimeters.. 

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b.) For the reinforcement under compression. Multiply:  	20 x 16mm + 160mm = 480mm or 48cm
 Therefore, the splice joint for a 16mm tension bar is 55 centimeters long and 48 centimeters for a compression bars.
 SOLUTION (For 20mm steel bar)
 a.) Length of splice for tension bar. 	25 x 20mm + 150mm = 650mm or 65cm b.) length of splice for compression bars. 	20 x 20mm +  160mm =  560mm or 56cm

[Audio] Steel bars are marked with distinct markings to identify the manufacturer and their brand, the bar size number, and the type of steel bar. These markings include N for Billet, A for Axle, and Rail Sign for rail steel.. 

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Steel reinforcing bars are 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

6 N Main Ribo Initial ot Menu"cturer Bar Size Steel Type Grade 40 Gr•de 50 6 Main Ribt Bar Size Grade Mark One Line • Grade 60 Two One • Grade 75 STEEL BARS MARKING SYSTEM FIGURE 3-2. 

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6
N
Main Ribo
Initial ot
Menu"cturer
Bar Size
Steel Type
Grade 40
Gr•de 50
6
Main Ribt
Bar Size
Grade Mark
One Line • Grade 60
Two One • Grade 75
STEEL BARS MARKING SYSTEM
FIGURE 3-2

[Audio] When estimating the quantity of steel reinforcing bars, one must keep in mind the additional length for the hook, bend, and splice. This length can differ based on the limitations specified in the National Building Code. In the case of mild steel, the minimum hook and bend allowances must always be taken into account.. 

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In estimating the quantity of steel reinforcing bars, one has to consider the additional length for the hook, the bend and the splice whose length varies depending upon the limitations as prescribed by the National Building Code

MILD STEEL MINIMUM HOOK AND BEND ALLOWANCE

h Ild
HIGH YIELD BARS MINIMUM HOOK AND
BEND ALLOWANCE
Hook Length L + n
Hook Length n for Bend
HOOK
BEND
FIGURE 3-3 HOOK AND ALLOWNCE

[Audio] Reinforcement is a vital part of any concrete construction project. Understanding the various type of metal reinforcement and how they should be applied can significantly improve the success of the project. With concrete hollow block reinforcement, such as in walls, metal is used to strengthen the individual blocks, and to connect them together in a cohesive manner that holds the structure together. When it comes to footing and post and column reinforcement, a number of elements are necessary. This could include footing slabs, dowels, cut bars, anchors, and bend bars. Beam and slab reinforcements may be needed for larger construction projects, and stirrups play an important role in providing support to the structure. Ultimately, with a mastery of the various type of metal reinforcement and its applications, it is possible to create sturdy reinforced concrete constructions that will guarantee the safety and long-term success of the project.. 

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To those who have not yet been exposed to detailed drafting work or actual field construction of reinforced concrete. Will find it difficult to make a detailed estimate of the various types of reinforcement required. The various type of metal reinforcement that an estimator should be familiarized with are: A.) Concrete Hollow Block Reinforcement. 	This is the simplest type of vertical and horizontal reinforcement placed in between the layers and hollow core of the blocks. This type of reinforcement is installed and spliced progressively with the rise of the concrete blocks.
 B. Footing Reinforcement  1. Footing slab reinforcement for small and medium size footings 2. Dowels, cut bars, anchor, and bend bars 3. Beam reinforcement for medium construction  4. Beam and slab for large construction 5. Stirrups
 C. Post and Column Reinforcement  1. Main vertical reinforcement  a. single or b. bundled bar

[Audio] It is essential to consider a range of factors when reinforcing a structure with steel. Lateral ties are used to tie the layers of the concrete together to ensure its strength, with outer, inner, and straight ties, and in circular columns, spiral ties are used in addition. Dowels are used for partitions and other future attachments. For beams and girders, main reinforcements include straight bars, bent bars, cut bars for tension and compression, and dowel bars for future attachment, and stirrups (open, closed, and straight) provide additional support. Cut bars should be included over and across the support, between supports, and for dowels and hangers for ceiling and partitions, as all of these reinforcements are necessary for the proper amount of strength and durability of the structure.. 

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2. Lateral Ties a. outer ties b. inner ties c. straight ties 3. Spiral Ties for Circular column 4. Dowels for partitions and other future attachments.

D. Beam and Girder Reinforcements 
 1. Main Reinforcement  a. Straight bars  b. Bend bars  c. Additional cut bars for tension and compression  d. Dowel bars for future attachment  2. Stirrups  a. Open stirrups  b. Closed stirrups  c. Straight stirrups or ties  3. Cut Bars  a. Over and across the support  b. Between supports  c. Dowels and hangers for ceiling and partition

[Audio] Straight main reinforcing bars are employed to span from one beam to the other in the floor slab reinforcements, while alternate main reinforcing bars with bends are used to connect across the beam support. Temperature bars are tied transverse to the main reinforcement, and extra alternate cut bars are located over the support. Furthermore, dowels and hangers are utilized for ceiling and other necessary attachments.. 

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E. Floor Slab Reinforcement 
 1. Main reinforcements 
 a. Straight main reinforcing bars extending from one beam to the other.  b. Alternate main reinforcing bars with bend between and over the beam support.  c. Main alternate bars over support(beam or girder) 
 2. Temperature bars tied perpendicular to the main reinforcement. 
 3. Additional alternate cut bars over the support(beam). 
 4. Dowels and hangers for ceiling and other attachment.

[Audio] Reinforcement of concrete hollow blocks is essential in concrete and masonry works, with its size and spacing required to be specified according to the plan or specifications, in line with the national building code. Three standard methods are available to determine the reinforcement: the Direct Counting Method, the Area Method, and the Unit Block Method.. 

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Reinforcement of Concrete Hollow Blocks
 Steel bar as reinforcement is a requirement for all types of concrete and masonry works. The national building code has promulgated guidelines on how and what kind of reinforcement is appropriate for a certain type of work depending upon the purpose for which it is to serve. The size and the spacing requirements for concrete hollow block reinforcement must be indicated on the 	plan or specifications.concrete hollow block reinforcement could be determined in three different ways; 1. By the Direct Counting Method 2. By the Area Method 3.By the Unit Block Method

[Audio] TheDirect Counting Method is employed to gauge the length of steel bars in reinforced concrete structures. This method entails counting bars individually on the plan and computing length based on the elevation. To guarantee precision, estimators must be familiar with hook, bend and splice prerequisites if they are not specified in the plan. In case the Direct CountingMethod is hard to implement, the Area Method can be used along with Table 3-5 to accurately determine the amount of steel bars needed for reinforced concrete structures.. 

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under the counting method. The vertical and horizontal reinforcements are counted individually in the plan. The length is also determined from the elevation. The hook, bend and lapping splices are imaginably calculated and added to its length because it is very rare to see a plan in a large scale 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 effectively even if the plan is not accompanied with such details.
  In the event that estimating by Direct Counting Method is somewhat difficult, one can use the Area Method with the aid of Table 3-5 prepared for this purpose.

[Audio] The area method is the simplest approach to computing the reinforcement for reinforced concrete hollow block (CHB). Table 3-5 provides allowances for standard bending, hooking, and lapping splices. Illustration 3-2 from Figure 3-6 can be used to determine the number of CHB with 10x20x40cm measurements, as well as the vertical reinforcement spacing at 80 cm, and the horizontal reinforcement required after every 3 layers.. 

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The area method is the simplest approach in computing the steel bar reinforcement for CHB with the aid of table 3-5. the values presented in the table include the allowances required for standard bend, hook and lapping splices.
 ILLUSTRATION 3-2 From Figure 3-6, determine the number of: a.) 10 x 20 x 40 cm CHB b.) Vertical reinforcement spaced at 80 cm. c.) Horizontal reinforcement at every 3 layers.

Figure 3-6. CHB METAL REINFORCEMENT CHB VerOc•l Rein' SO em. o.e. Horizont•l Reht. ot every 3 layers 2.60 m Natural Ground 40 em Footing 00 m FIGURE. 

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CHB
METAL REINFORCEMENT
CHB VerOc•l Rein' SO em. o.e.
Horizont•l Reht.
ot every 3 layers
2.60 m
Natural Ground
40 em
Footing
00 m
FIGURE

[Audio] Provided is a table displaying the lengths of steel bar reinforcement for Concrete Hollow Blocks, in meters, for different spacing layers. Vertical and horizontal reinforcement lengths are presented in meters per block and per square meter. An increase in the number of blocks used will correspond to a rise in the needed reinforcement. Therefore, it is essential to make sure the correct amount of steel bar reinforcement is used to guarantee strength and durability of the project.. 

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Spacing Vertical reinforcement lengths of bar in meter
 Spacing layer  Horizontal reinforcement lengths of bar in meter
 cm. Per Block  Per Sq. M per Per Block Per sq. m. 40 0.235 2.93  2 0.264 3.30 60 0.171 2.13 3 0.172 2.15 8-  0.128 1.60 4 0.138 1.72

TABLE 3-5 LENGTH OF REINFORCING BARS FOR CHB IN METERS

[Audio] In order to form a fence using steel reinforcing bars, the area of the fence must first be calculated. In this case, the fence has a length of 4 meters and a height of 3 meters, for a total area of 12 square meters. Using Table 3-5, which gives the lengths for reinforcement spaced at 80 centimeters, we can determine the required length of the vertical bars as 19.20 meters and the horizontal bars as 25.80 meters.. 

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1.) Solve for the area of the fence 		Area= Length x Height 		A= 4.00 X 3.00 m. 		A= 12 sq. m.
 2.) For the reinforcement spaced at 80 centimeters, refer to table 3-5. under      column length per sq. m. multiply: 		12 sq. m. x 1.60 = 19.20 m. long
 3.) Solve for the horizontal bars at every 3 layers, from table 3-5 under horizontal reinforcement per square meter, multiply: 		12 sq. m. x 2.15 = 25.80 meters long

[Audio] Calculate the number of hollow blocks by multiplying the surface area by 12.5 pieces per square meter. Refer to Table 3-5 for the vertical reinforcement, and calculate the length of steel bar per block by multiplying the number of pieces by 0.128 meters. Add the results from both calculation and convert the length to the commercial size of steel bars. In this case, 7.5 orders of 8 pieces 10mm x 6.00m long are needed.. 

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2. Determine the number of hollow blocks. Refer to table 2-2. multiply: 		CHB; = 12 sq. m. x 12.5 = 150 pieces
 3. Refer to table 3-5 a.) vertical reinforcement at 80 cm spacing, length of steel bar per block; multiply: 	150 x 0.128 = 19.2 meters long
 4. Add the result of (a) and (b) 	19.2 + 25.80 = 45.00 meters long
 5. Convert this length to the commercial size of steel bars, say 6.00 meters long. Divide: 		45.00 m ÷ 6.00 = 7.5 order 8 pieces 10 mm x 6.00 m.

[Audio] Humans have been taking advantage of the strength and durability of steel for centuries. Tie wire is one of its most popular forms and is used to secure steel bars in the right position. Usually made from galvanized iron wire, these tie wires come in a gauge number 16. Their length varies depending on the size of the bar being secured, but usually ranges from 20 – 40 centimeters for smaller and medium size steel bars. In concrete hollow block work, the most popular reinforcement sizes for steel bars are 10mm, 12mm, and 16mm in diameter. Tie wires are a key component in the reinforcement process, providing maximum strength for stability and security of structures.. 

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3-5-TIE WIRE FOR CHB  REINFORCEMENT  Tie wire refers to  gauge no.16 galvanized iron wire popularly known as G.I tie wire. Tie wire is used to secure the steel bars into its designed position. The length of each tie wire depends upon the size  of the bar to be tied on ranging from 20 to 40 centimeters from small and medium size steel bars.
 LENGTH OF TIE WIRE REINFORCEMENT The common size of steel bar reinforcement specified for concrete hollow block work is either; 10 mm, 12 mm, or 16 mm diameter depending upon the plan and specification.

[Audio] The table shows that the kilograms per square meter are impacted by the vertical spacing. If the vertical spacing is 25 centimeters, the kilograms per square meter for a horizontal layer spacing of 2, 3 and 4 are .054, .039 and .024 respectively. As the horizontal layer spacing increases, the kilograms per square meter increases too. For a horizontal layer spacing of 4, the kilograms per square meter are .066, .063 and .039 for vertical spacings of 25, 30 and 40 centimeters respectively, demonstrating the relation between the layers' spacing and the vertical spacing on the kilograms per square meter.. 

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Vertical Spacing

 Horizontal Layer  Spacing 
 Kilograms per Square Meter



 25 cm 30 cm  40 cm 40 40 40 2 3 4 .054 .039 .024 .065 .047 .029 .066 .063 .039 60 60 60 2 3 4 .036 .026 .020 .044 .032 .024 .057 .042 .032 80 80 80 2 3 4 .027 .020 .015 .033 .024 .018 .044 .032 .024

TABLE 3-6 NO. 16 G.I TIE WIRE FOR CHB REINFORMENT  PER SQUARE METER

[Audio] To solve Illustration 3-2, the wall area must be determined to be 12 square meters. By referring to Table 3-6, when reinforcement is spaced 80 centimeters vertically and 30 centimeters horizontally between each 3 layers, the necessary tie wire in kilograms is 0.29 kilograms for 16 gauge galvanized iron wire.. 

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Illustration 3-3 Continuing  the solution of illustration 3-2, find the required tie wire in kilograms if the reinforcement are spaced at: Vertical bars spaced at 80 centimeters and one horizontal bars at every after 3 layers of the block. Tie wire in kilograms

SOLUTION:
 1. Solve for the wall area.         Area = 4.00 × 3.00              A = 12 sq.m
 2. Refer to Table 3-6. Along 80 cm. vertical spacing and 3 layers horizontal spacing of reinforcement  at 30 cm. long tie wire  12sq.m × 0.024 = 0.29 kgs                    # 16- G.I wire

CHB
METAL REINFORCEMENT
CHB Veroc•l Reint SO em. o.e.
Horizon
•t ever€
2.60m
Reht
•yen
Ground
40 cm
— Foot"tg
00 m
FIGURE 34

[Audio] When designing a structure with individual footings, it is essential to bear in mind the properties of the steel reinforcement for protection. According to the American Concrete Institution (ACI) code, the minimum underground protective covering for concrete to steel reinforcement must not be below 7.5 centimeters. Additionally, it is necessary to take into account the length and quantity of the needed reinforcement. Longer commercial length bars should be chosen in order to minimize the number of unwanted cuts required to get the required size. For instance, if five 1.2 meter-long pieces are needed, then the most suitable order would be 6.00 meter-long steel bars, as this would reduce the amount of cuts needed to achieve the required length.. 

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3-6 INDEPENDENT FOOTING REINFORCEMENT
 Also referred to as individual or isolated footing. The ACI( American Concrete Institution)  code provides that the minimum underground protective covering of concrete to steel reinforcement shall not be less done 7.5 centimeters. 






 One important consideration in estimating steel bar reinforcement is to find the required length and quantity of a  particular reinforcement then choose a commercial length bars which when cut into the required size will minimize unwanted cuts. For instance. If 5 pieces at 1.20 meters long is needed then the most ideal order is  6.00 meters long steel bars.

[Audio] For this example, the calculation for the number of steel bars required is fairly straightforward. To start, the net length of one cut reinforcing bar is 1.5 meters minus two 0.075cm, resulting in a length of 1.35 meters. For each footing, two pieces of 12mm steel bars at a 1.35 meter length is required, giving a total of 20 pieces. Multiplying this number by the 24 footing for the project gives us 480 pieces. To determine the number of 6.00 meter steel bars we will need, we divide 6.00 meters by 1.35 meters, resulting in 4.44 or 4 pieces. Dividing the total cut bars from step three by 4 gives us the total number of steel bars required, which is 120 pieces.. 

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Example-1  From the figure, find the number of 12mm steel bars required if there are 24 footings having a general dimensions of 1.5 x 1.5 meters.

nunnunu
13m
P4N
12 m
7.5
L: 13m.
CROSS SECTION

Solution:  1. Find the net length of one cut reinforcing  bar.                    Net length = 1.5 m -2(0.075cm)                                   L  = 1.35m  2. Find the number of cut bars per footing.            10 × 2 = 20 pieces of 12 mm at 1.35 m. long 3. Total number of cut bars in 24 footings.                  Total Cut bars = 24(20) = 480 pieces 4. If 6.00 meters long steel bar will be used, we obtain the following cuts.                   6.00/1.35  =  4.44 or 4.0 pieces 5. Divide the result of step 3 by 4.0 to get  the number of steel bars required.                   480/4.0  = 120 pcs of 12mm steel bars

[Audio] Estimating the right amount of tie wire for a footing requires some calculation steps. First, you need to calculate the number of ties for one footing. This can be done by looking at the cross-section of the steel bars that need to be tied together. For example, if the cross-section shows 10 (10) steel bars to be tied together, then the number of ties for one footing will be 100. Then you need to calculate the total number of ties for all the footings, which in this case, with 24 footings, will be 2400 ties. Next, calculate the total length of the tie wire needed by multiplying the number of ties by the length of each tie. In this case, that would be 2400 (30 cm length per ties) = 72000 cm, or 720 meters. Finally, convert the length to kilogram. To do this, divide the length by the number of meters in a kilogram of #16 tie wire, which is 53 meters. In this case it would be 720m/53m = 14 kg #16 kg G I wire.. 

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Estimating the footing tie wire 1. Find the number of ties for one footing.  2. Find the total number of ties for all the footings.  3. Find the total length of the tie wire needed.  4. Convert the length to kilogram.

7.5
Isom.
1.35m
75 m.
12 mm stel bn crcmwise
7.5
L = 1.sn
CROSS SECTION

Example: 1. Looking at the figure the number of Ties per footing               10(10) = 100 intersection of steel bars to                                be tied per footing 2. Solve for total ties of the 24 footings.              100(24) = 2400 ties  3. Solve for total length of tie wire needed.               2400(30 cm length per ties) = 72000 cm                                                                  = 720 meters 4. Convert length to kilogram (1 kilogram of no.16 tie wire is 53 meters long)                 720m/53m = 14 kg #16 kg G.I wire.

[Audio] The quantity and length of main reinforcement can be determined through the “Direct Counting Method”. This method requires special attention to the additional length needed for lap joints of the end splices, any allowance for bending or hooking, additional length for beam depth and floor thickness, distance from the floor to the footing slab, and provisions for splices and succeeding floors.. 

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3-7 POST AND COLUMN REINFORCEMENT  The reinforcement of post and column to be considered in this study are: (a) The main vertical reinforcement, (b) The lateral ties or (c) Spiral ties for circular column.

Floor Slab
Beam
CS Scanne± Carn.Scanner
Add Length
Add Depth of Floor and Beam
Re"rce
Fbor line

The quantity and length of main reinforcement is determined by “Direct Counting Method” giving special attention to the additional length for: 
 Lap joints of end splices  Allowance for bending and or hook  Additional length for beam depth and floor thickness if the height indicated in the plan is from floor to ceiling Distance from floor to footing slab  Provisions for splices and succeeding floors

[Audio] This slide provides an example of how to calculate the amount of steel reinforcement required for a two-story building with 10 columns. To start, you need to determine the total length of the main steel bar reinforcements. This includes a bend at the base footing, the length from the footing to the ground floor, the height from the ground line to the beam, the depth of the beam, and the thickness of the floor slab. The dowel for the second floor requires a 20mm by 20mm steel bar, which adds 0.4 meters to the total length. According to the figure, the total length of the reinforcement is 5.95 meters. Once you have the total length, you then need to select a 6.00 meter long steel bar and multiply it by the number of bars in one post, which is 8 bars multiplied by 10 posts. The result is 80 pieces of 20mm by 6.00 meters long steel bars. Put together this solution summarises as 80 pieces of 20mm by 6.00 meters long steel bars.. 

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Example: From the following figure, list down the main reinforcement from the footing to the second floor using 20mm if there are 10 columns in the plan. Solution: 1. Determine the total length of the main steel bar reinforcements.  Bend at the base footing……………………20  Length from footing to ground  floor…1.20 Height from ground  line to beam…….3.50 Depth of beam………………………………….0.50 Thickness of the floor slab…………………0.15 Dowel for second floor(20×20 mm)…_0.40 _                 Total length of the Reinforcement……5.95 M
 2. Select a 6.00 meters long steel bar 
 3. Multiply by number of bars in one past × 10 post.                 8 BARS × 10 = 80 pieces 4. Order: 80 pcs.20 mm × 6.00 meters long steel bars

[Audio] Maintaining the integrity of a structure such as a beam or girder is an essential part of building construction. To ensure the strength and security of the structure, more than just steel is needed. The direct counting method is usually the go-to process when calculating the number of main reinforcements for beams and girders. The length of the reinforcement, however, is determined by the physical condition of the structure and its relation to the support. It is important to pay particular attention to the plans in order to verify the span of the columns where the beam is resting, such as center to center of the column, outer to center of the column, outer to outer side of the column, and inside to inside span. When looking at the splicing position of reinforcement, it must be adjusted to the commercial length of steel bars. It is also worth remembering that “the lesser the splice, the lesser is the cost”.. 

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3-8 BEAM AND GIRDER REINFORCEMENT The direct counting so far is 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 structure in relation of the support.
 Verify the plan if the span or distance of the column where the beam is resting indicates the following conditions: Center to center of the column  Outer to center of the column Outer to outer side of the column Inside to inside span
 Verify the splicing position of the reinforcement if it is adjusted to the commercial length of steel bars. Take note that “the lesser the splice the lesser is the cost”.

[Audio] This slide displays various steel bars which can be utilized as a reinforcing material in construction. The steel bars featured from left to right are 'Colum', 'Cow', 'Beam' and 'Centert00frndt'. All steel bars possess a distinctive length and area which can be tailored to any project.. 

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

bnm
Colum
to Center
Slab
bam
Cow
Irde to
Beam
Centert00frndt
Slab
Bum

[Audio] Lateral ties are employed in reinforced tied columns to hold the vertical reinforcement in position. Generally, they are constructed of a steel of size No. 3 or 4 for longitudinal bars No. 10 or smaller, and No. 11 to 18 for bundled bars. According to ACI Code, all non-prestressed longitudinal bars must be encased by lateral ties to ensure the reinforcement stays stable and that the desired strength is attained.. 

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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 provides that “ All non pre- stressed bars for tied column shall be enclosed by lateral ties of at least No.3 in a seize for longitudinal bars No. 10 or smaller and at least No4 in size for No.11 to 18 and bundled longitudinal bars”

[Audio] For a tied column, the lateral ties should not be less than 10 millimeters in diameter if the main reinforcement consists of No. 11 to No. 18 bars or bundled bars with a diameter of 35 to 66 millimeters. The lateral ties should not be less than 12 millimeters in diameter. The spacing of the lateral ties should not exceed 16 times the longitudinal bar diameter, 48 times the lateral tie bar diameter, or the least dimension of the column. For example, if the longitudinal bar diameter is 20 millimeters, the lateral ties should be spaced no further than 320 millimeters, or 32 centimeters apart. If the lateral ties are 10 millimeters in diameter, they should be no further than 480 millimeters, or 48 centimeters apart. The lowest value however, is the least dimension of the column, in this case 30 centimeters. Therefore, the lateral ties should be spaced no further than 30 centimeters.. 

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The Code Provisions Simply Mean that: a.) If the main longitudinal rNo. 10 bars or smaller in size (lower than 30 mm), the Lateral Ties should not be smaller than No, 3 bar or 10 mm diameter. Reinforcement of a tied column is  b.) If the main reinforcement of a tied COIumn is No. 11 to No. 18 (35 mm to 66 mm) and bundled bars, the Lateral Ties should not be less than No. 4 or 12 mm diameter,          The Code further provides that Lateral Ties Spacing shall not exceed the following: 	1. 16 x the longitudinal bar diameter 	2, 48 x the lateral tie bar diameter or 	3. The least dimension of the column

iumn Rein%rcement 20 mm.
em
10 mm Lateral
30 em.
ELEVATION SECTION
30
CROSS SECTION
3-13 SPACING DISTANCE OF LATERAL TIES

SOLUTION 1. Determine the spacing distance of the lateral ties. a. 16 x 20 mm = 320 mm or 32 cm. b. 48 x 10 mm = 480 mm or 48 cm. c. The least Side of the column is = 30 cm. 2. The lowest value is 30. Therefore, adopt 30 centimeters spacing of the lateral ties

[Audio] Stirrups are essential for reinforcing beams and girders as they enable the main reinforcements to be held in position. There are two types of stirrups used for this purpose - open stirrups and closed stirrups. Open stirrups have one vertical leg and one horizontal leg, while closed stirrups have two vertical legs connected by a cross-bar. The type of stirrup used depends on the structure’s requirements such as weight, length and size. Selecting the right stirrup is essential for effective beam and girder design.. 

Scene 34

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 stirrup commonly used are the open stirrups and the closed stirrups

[Audio] The materials required for the stirrups for this reinforced concrete beam design include 46 pieces of 10 mm steel bars that are 6.0 meters in length. This amount was calculated as 272 stirrups were needed for the 16 beams, each with a length of 1.00 meter, and Table 3-8 was used to calculate the length of steel bar to be used.. 

Scene 35

ILLUSTRATION 3-12      A reinforced concrete beam with a cross-sectional dimensions as shown above specify the use of 10mm open stirrups spaced as shown in Figure 3-19 if the 16 beams of the same design, find the materials required for the stirrups.

SOLUTION:  1. By direct counting there are 17 stirrups at 99cm. Long say 1.00 meter 2. Find the total number of stirrups         17x16 beams = 272 3. Refer to Table 3-8. For a 1.00m long stirrups the choice is either 6.0m, 9.0m or 12m long steel bar. For easy handling use 6.0 meters long. Divide:; 272/6 = 45.3 say 46 pcs – 10mm x 6.0 m

TABLE NUMBER OF t.AtERAL nes ONE STEEL BAR
Weal
Ties
Sp•ang
15
20
45
X
ANO QUANTITY PtR MCTER LENGIH COLUMN
Nurnber ot ot
ties Ties wth
M. ht. Hook & Bend
5M • M Its M 9M
670
S. 15
40
343
300
264
196
10
12M
20
17
14
13
12
10
100
IOS
110
Its
120
125
130
135
140
145
150
160
170
180
190
— Not advisable length for economical reasons.

[Audio] For this slide, the voiceover script would be: Figure 3-20 is used to calculate the number of 12mm stirrups needed for a series of 12 girders. By direct counting, it is determined that each span requires 15 stirrups, for a total of 180 stirrups. Table 3-8 provides information on the length of steel bars required to generate the necessary 4, 5, 6, and 8 cuts of stirrups. If 6.00m long steel bars are chosen, 45 bars are needed. If 7.50m long steel bars are chosen, 36 bars are needed.. 

Scene 36

ILLUSTRATION 3-13      From Figure 3-20, compute the number of 12 mm stirrups required if there are 12 girder of the same design.

SOLUTION 1. By direct counting there are 15 stirrups per span. If there are 12 girders; multiply: 	15x12= 180 2. By inspection the length of one stirrups is 150 cm . Refer to table 3-8 along 150 cm length of ties, 4,5,6,and 8 cuts could be derived from 6.00, 7:50; 9.00 and 12 meters steel bars respectively. If we chose 6.00m long 	180/4 = 45 12mmx6.00 steel bars 3. If we chose 7.50 meters long steel bars we get; 	180/5 = 36 and 12 mm x 7.50 m

"UMBER OF LATERAL ONE STEEL EAR
ANO QUANTITY PER H COLUMN
Number
6, 70
4.13
343
3.00
2.64
o' Che
10
45
100
1 os
120
125
130
13S
140
145
150
170
190
17
14
13
12
10
Not advisable length for economical reasons

[Audio] Spiral ties consist of continuous spirals of reinforcement held firmly in place with at least three vertical bar spacers. The center-to-center spacing of the spiral must not exceed one-sixth of the diameter of the core and the clear spacing between the spirals should not exceed 7.5 centimeters, or be less than 5.0 centimeters. Furthermore, the clear spacing between the spirals should be less than 1-1/2 times the size of the largest coarse aggregate used.. 

Scene 37

3-11 SPIRAL AND COLUMN TIES The spiral reinforcement consist of evenly spaced continuous spirals held firmly place by at least three vertical bar spacers under the following consideration.

That the center to center spacing of the spiral should not exceed 6th part  of the diameter core. 2.That, the clear spacing between the spirals should not exceed 7.5 centimeters nor  less than 5.0 centimeters 3.That the clear spacing between the spirals be less than 1-1/2 times the biggest  size of the coarse aggregate.

[Audio] The illustration depicts a spiral column with a cross sectional diameter of 10 mm and figure 3-1 with spiral reinforcement. There would be a total length of 98 meters if there were fourteen columns, each 7.00 meters high. Thus, the spiral reinforcement can be calculated.. 

Scene 38

ILLUSTRATION 3-1	4      A spiral column with a cross sectional diameter of quires 10. mm spiral reinforcement as figure. If. there are 14 columns at 7.00 meters high,

SOLUTION  Solving the spiral Reinforcement  Total length of 14 columns: 7.00 m. x 14 = 98

[Audio] The number of 10mm bar spirals must be calculated for this slide. Refer to Table 3-9 which applies to a 50cm column diameter with a 5cm pitch. 482.1 pieces of 10mm bars is the result of multiplying 98m by 4.919. For a 9m spiral along the 50cm column, the entry is 3.223 and multiplying 98m by 3.223 yields 316 pieces of 10mm bars.. 

Scene 39

SOLUTION  2. Find the number of 10 mm bar spirals. Refer  {o Table 3-9. For a 50 cm. column diameter, 5.0 cm, pitch  6.00 meters steel bars, multiply 98 m. x 4.919 = 482.1 say 483 x 10 mm  3. If 9.00 meters will be used as spiral, along 50 cm column diameter, 5 cm. pitch under 9m steel bars the entry is 3.223. Multiply: 98 m. x 3.223 = 316 pieces. 10 mm x 9.00 m.

TABLE 3-9 NUMBER OF BARS
PER
47. S
ss.o
7 SO
5.00
6.25
5.00
6 as
7.50
5 00
7 so
S 00
6.25
S_OO
e_2S
500
62S
No Turn
17.0
14.3
21.0
17.0
14.3
170
14.3
21.0
17.0
17.0
14.3
21.0
170
14.3
21 .o
170
14.3
170
4.3
21.0
17.0
14.3
17.0
14.3
21 .0
170
143
N•ume• Of Bars
2604
204
Z 342
1 _97S
3.183
ZS77
2172
3472
2811
2370
3.762
3045
3281
3S13
Y 748
3.1 so
4919
4S2S
3.7S2
4.919
2 08S
1.423
.842
2_6S4
2' 49
Z 302
940
3 033
2070
2.009
2.199
3.602
2.S24
2717
.027
_410
.393
'.1S4
1.97a
1712
2097
2678
2.253
2.020

[Audio] To reinforce steel, the correct tie wire is essential. Refer to Table 3-9 to calculate the number of turns per meter height, then multiply that result by the number of columns to determine the total ties for the construction. Calculate the length of the ties by adding 30 cm per tie, then divide the overall length by 53 to get the kilograms of tie wire needed. In this example, 140 kilograms of tie wire is required.. 

Scene 40

B. Finding the Tie Wire Find the number of vertical bars per column = 12 pieces
 2. Refer to Table 3-9. under 50 cm. column diameter; S cm pitch, the number of turn per meter height is 21. Multiply: 	12 x 21 = 252 Ties per meter height
 3. Total Ties for 14 column at 7:00 m. high is: 	252 x 7.00 x 14 = 24,696 tiles
 4. Total length of the wire at 30.m long per tie 	24,696 x 30 = 740,880/100 = 7,409 meters
 5. Convert to kilograms. Divide by 53 	7,409/53 = 139.8 say 140 kilograms

[Audio] A one-way reinforced concrete slab is supported by beams on two opposite sides, with a ratio of the longer span to shorter span being two or more. This causes the slab to bend in the direction of the shorter span, enabling it to carry the load in one direction. This is a useful and economical method for structurally reinforcing a concrete slab.. 

Scene 41

3-12 ONE WAY REINFORCED CONCRETE SLAB One way slab is a slab which is supported by beams on the two opposite sides to carry the load along one direction. The ratio of longer span (l) to shorter span (b) is equal or greater than 2, considered as One way slab because this slab will bend in one direction i.e in the direction along its shorter span.

[Audio] Steel is a key factor in reinforcing concrete structures as shown in this illustration. The given data allows one to calculate the number of 12 mm diameter steel bars necessary for the construction. This exemplifies the significance of using the right steel reinforcements and considering various aspects to develop strong, lasting structures.. 

Scene 42

ILLUSTRATION 3-1	6 A spiral column with a cross sectional diameter of quires 10. mm spiral reinforcement as figure. If. there are 14 columns at 7.00 meters high,

From Figure 3-22 of a one-way reinforced concrete slab, determine the number  of 12 mm. steel bars including the tie wire required

A. SOLUTION (By Direct Counting) Given Data: Spacing of main reinforcement = 15 cm

20 mm
r—illki i
Str•Oht b•r•
u—uuUUt11Ä
11
3-22 ONE WAY SUB REDGORCEMENT

[Audio] A solution to reinforce the floors of a structure utilizing steel as a reinforcing material can be calculated by taking the floor area of 5.70 x 5.50 meters and the given spacing of 15 cm into consideration. According to Table 3-10, 81 pieces of 12 mm x 6 meter bars are required. Furthermore, Table 3-11 indicates that 8 kg of #16 G.1 wire must be utilized.. 

Scene 43

TABLE 2-11 OF TIE ON A ONE WAY
CONCRETE SLAB PER SQUARE METER
Length of Steel
Mfifiåü
2.CO
10.0
12 s
17 s
20.0
22 s
250

SOLUTION  Determine the floor area 	5.70 x 5.50 = 31.35
 2. Refer to Table 3-10. Using a 6.00 meters long steel bars at 15 cm spacing distance, multiply 31.35 x 2.584 = 81 pieces 12mm x6

Refer to Table 3-11 along bar spacing at 15cm using a 30cm tire wire multiply 31.35 x .228 = 7.15 say 8 kg #16 G.1 wire

[Audio] Reinforced concrete slabs designed for two-way stresses typically consist of two rebar layers arranged in each direction, regardless if beams are present or not. The strength of these systems is determined by factors such as the size, spacing, and grade of the rebar used. Enhancing the concrete with this two-way reinforcement system creates structures that are strong and reliable.. 

Scene 44

3-13 TWO WAY REINFORCED CONCRETE SLAB The Building Structural Standard defines a two-way slab system as a concrete slab system in which two rebars are arranged in two directions regardless of the presence or absence of a beam that transmits a load to a column.

[Audio] To determine the floor slab solution, start by calculating the area of the floor slab by multiplying the length and width. In this instance, the floor area is 51.84 sq. meters. Next, referring to Table 3-12, and using a 7.50 meters steel bars at 10 cm spacing of the main reinforcement, the total quantity of pieces required is 173 pieces 12 mmx 7.50 meters.. 

Scene 45

ILLUSTRATION 3-1	7 From Figure 3-24, determine the number of 12 mm steel bars spaced at 10 cm, including the   tie wire required.

132 m•
- 720m
87
720m
PLAN
Concrete
Bars
SECTION
1.87

2, Refer to Table 3-12. Using a 7.50 m steel bars at 10 cm. spacing of main reinforcement, multiply: 	51 84 x 3.337	173 pieces 12mm x 7.50 meters

TABLE 3-12 QUANTITY OF STEEL BARS IN A TWO
REJNFORCED CONCRETE SLAB PER SQUARE METER
aar Spacåig
in
Centimeters
128
15.0
17.5
200
228
25.0
Number of Steel Bars per Square Meter
6.00M
4.359
3.603
3221
2.647
2.360
2.168
1.977
3B37
2.662
2.296
2025
1775
1.601
1 466
9.00 M
2772
2245
1 650
1 461
1 312
1 217
2022
1k35
1 382
1196
1.056
0 907
0 862
• The ACI Code pro•åes tat the center to cente spacing of temperature
must not be geater than frvetimes the slab thickness

[Audio] To determine the amount of No.16 G.l. wire needed for the project, one should refer to Table 3-13. The area should be multiplied by 51.84 times .648 when the bar spacing is 10 cm and the tie wire length is 30 cm. This yields a result of 33.6. Consequently, 37 kilograms of No.16 G.l. wire should be ordered.. 

Scene 46

1. . Refer to Table 3-13, along 10 cm bar spacing and 30 cm. long tie wire, multiply by the area:
 51.84 x .648 = 33.6
 2. Order 37 kilograms of No.16 G.l. wire.

130m'
iiiiii'
. Jar.
87
720m
PLAN
Concrete
Ban
secti0N

TABLE 3-13 TIE WIRE IN A TWO-WAY REINFORCED CONCRETE
SLAB IN KILOGRAMS PER SQUARE METER
Length of Steel Bars
Bar Spacng 600 M
750M
10.0
126
150
175
200
225
25.0
.592
.416
.350
247
.174
.168
.142
.89€
554
466
330
232
.224
.189
.648
420
307
.231
.171
.150
.134
863
560
409
308
227
200
, 178
648
431
.257
.223
.173
,148
.114
864
574
343
298
231
.198
.152
596
, 421
297
.2C6
.10
.13C
.116
794
561
396
.275
.217
-173
.155

[Audio] The figure in the slide shows how to calculate the quantity of 10mm steel bars and tie wire needed for the concrete pipe reinforcement. The circumference of the circle is found by multiplying Pi by the diameter of the circle at the middle point of the concrete thickness. In this case, the circumference of the circle is 3.1416.. 

Scene 47

3-14  CONCRETE PIPE REINFORCEMENT From the following Figure, determine the quantity of 10 mm steel bars and the tie wire required.

1. Solve for the circumference of the circle at midpoint Of the concrete thickness “t”

C = πd C = 3.1416 x (0.90 +. 10) = 3.1416

[Audio] Divide the total length of the structure by the spacing distance of the rings to find the total number of rings. In this case, the length is 1.00 meter, and the spacing distance is .15 meter which gives us 6.7 meters of rings. Add one to get the total number of rings, resulting in 7.7 or 8 pieces at 3.29 meters. Determine the number of shrinkage and temperature bars at the spacing of .20 meter, which is 16.45 or 17 pieces. Calculate the total number of rings and shrinkage and temperature bars to solve for the tie wire. This can then be used to create a steel reinforcing structure. Thank you.. 

Scene 48

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

3. Find the total number of ring at 15 cm. spacing distance.      1.00 m / .15 m = 6.7 meters

4. Add one to get the number of ring: 6.7 + 1 = 7.7  say 8 pieces at 3.29 meters.

5. Find the number of shrinkage and temperature  bars at say .20 m. on center. 3.29/.2 = 16.45 say 17 pieces

8 pcs. 10 mm. x 3,29 m, long 17 pcs. 10 mm. x 1.00 m. long

8 x 17 = 136 ties x .30 cm. long per tie 41 meters

1. Total number of ring multiplied by the number of shrinkage and temperature bars.

2. Convert this length to kilograms. Divide by 53. 41/53 = .77 say 1 kilograms #16 G.I wire

GROUP 3 METAL REINFORCEMENT


Scene 0

Steel bars are a necessary material for any kind of concrete construction

Steel is much stronger than concrete, being ten times stronger in compression and a hundred times stronger in tension

To make sure the steel bars and concrete function effectively together in carrying external loads, it is imperative that they are deformed together

If not, the steel bars may be forced out of the concrete due to too much weight and inadequate grip.

avatar

Steel bars are commonly utilized to reinforce concrete and increase the stability of constructions

They have a surface deformation that further improves the adherence to the concrete, producing incredibly sturdy structures

The deformation can come in various shapes and designs, making it applicable for a broad range of applications.

The table below shows the length of steel bars in meters according to the different diameters

The length of the bars increases with the increase of the diameter