GROUP 3 METAL REINFORCEMENT

[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!.

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

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..

[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..

[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..

[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..

[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..

[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..

[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..

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.

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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”..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..

[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..