GROUP 3 METAL REINFORCEMENT

GROUP 3 METAL REINFORCEMENT. JADE ERWIN BASUL HUDSON CEASER GEALOGO CEDRIC JOHN SARANZA.

[Audio] Steel is a highly valuable material in concrete construction, providing impressive strength and flexibility. It is characterized by its unbeatable tensile and compression strength, making it capable of withstanding a range of forces. Its strength and flexibility ensure that it is well-suited to be bonded together with concrete. Steel reinforcement of concrete structures ensures a strong and secure bond between the two materials, able to expand and contract in harmony, thus ensuring a longer lifespan for the structure..

[Audio] Steel bars are a critical element in strengthening concrete constructions. A range of surface deformations have been created to guarantee a solid connection between the concrete and the steel bars. This technique is necessary to strengthen the concrete and guarantee that the structure will endure..

[Audio] The table shows that the length of steel bars range from 1.98m to 107.88m, depending on their diameter. This should be considered when purchasing and using steel bars for any type of construction..

[Audio] Strength and durability are the most important attributes of metal reinforcement. Today, I'd like to present a table that demonstrates the areas of groups of reinforcing steel bars with different diameters. As you can see, the table includes the bar diameters from 6 millimeters to 32 millimeters, and the respective areas of the groups of reinforcing steel bars. This can be very helpful for those who want to ensure the strength and durability of their builds. Let's take a look at the table in detail..

[Audio] Splicing steel bars is a crucial element in construction projects. To ensure the integrity of the steel bar, it is vital to ascertain the appropriate splice length. According to the table, for tension bars the splice length should amount to 25 times the bar size plus 150 millimeters, whereas for compression bars the splice length should be 20 times the bar size plus 160 millimeters. Acknowledging the right splice length is a fundamental step to guarantee safety and quality in construction projects..

[Audio] Reinforcing steel bars are employed to enhance the resilience and capacity to hold a load of a structure. This Slide takes a glimpse at the example of a splice joint for 16mm and 20mm steel bars with a specific set of requirements. Regarding the 16mm steel bars, we can work out the length of the splice joint for tensile reinforcement in a beam by multiplying the diameter of the bar by 25, plus 150mm, resulting in a length of 550mm or 55 cm..

[Audio] Reinforcement under compression necessitates that 20 times 16mm plus 160mm equals 480mm or 48 centimeters. As a result, the splice joint for a 16mm tension bar would be 55 centimeters long and 48 centimeters for a compression bars. For a 20mm steel bar, the length of splice for tension bar would be 25 times 20mm plus 150mm which amounts to 650mm or 65 centimeters while the length of splice for compression bars would be 20 times 20mm plus 160mm, totaling 560mm or 56 centimeters..

[Audio] Steel reinforcing bars are an important structural element in reinforced concrete structures. Each bar is provided with distinctive markings that identify the manufacturer, brand, size number, and type of steel bar. These markings include 'N' for billet, 'A' for axle, and 'Rail Sign' for rail steel. Together, these markings provide a reliable way to identify the correct kind of steel bar for your construction projects..

[Audio] The steel bar marking system, as illustrated in Figure 3-2, is used to mark the steel type, grade, and bar size on the bar. This aids manufacturers in easily determining the various specifications of the bar, and helps to ensure that the requirements are met when using the steel during construction..

[Audio] In accounting for the quantity of steel reinforcing bars needed for a construction project, one must add the length associated with the hook, bend, and splice. This allowance, for mild steel, is based on the requirements stipulated in the National Building Code..

[Audio] Reinforced concrete is a significant and habitual building material. It requires scrupulous accounting and scheming to make sure that the correct types of metal reinforcements are employed. This slide has gone over the three primary categories of metal reinforcement: Concrete Hollow Block Reinforcement, Footing Reinforcement, and Post and Column Reinforcement. Every one of these types has to have exact estimations and situating of metal reinforcing bars to promise the strength of the concrete. These estimations and placements must be considered when doing construction with reinforced concrete..

[Audio] Steel reinforcement is a key element in any building structure, including columns, beams, partitions and ceilings. This slide will focus on group 3 metal reinforcement. Lateral ties can be split into three categories: outer ties, inner ties and straight ties. Spiral ties are used for reinforcing circular columns, while dowels are used for partitions and other future attachments. Beam and girder reinforcements consist of main reinforcement, straight bars, bent bars, extra cut bars for tension and compression, and dowel bars for future attachments. Stirrups are divided into open stirrups, closed stirrups and straight stirrups or ties, while cut bars are used over and across the support, between supports and for attaching dowels and hangers for ceilings and partitions..

[Audio] Metal reinforcement is essential for ensuring the stability and safety of a structure. This slide presents a set of metal reinforcements for a floor slab. Straight main reinforcing bars span between two beams, and alternate reinforcing bars with bends are placed both in between and above the beams. Temperature bars are then tied perpendicularly to the main reinforcement, followed by additional alternate cut bars over the support. Lastly, dowels and hangers are utilized to firmly secure the ceiling and other attachments. All these components combine to create a secure and dependable structure..

[Audio] Concrete hollow blocks are an important element in any construction project, as the use of reinforcement improves the strength and reliability of the structure. The national building code provides guidance for determining the appropriate type and size of reinforcement for a particular construction, depending on its purpose. For this type of block, there are three common methods to determine the size and spacing of reinforcement: the Direct Counting Method, the Area Method, and the Unit Block Method. Following these guidelines will help to ensure the quality and safety of the construction..

[Audio] For this slide, let's discuss estimating the amount of steel reinforcement needed for a certain project. The direct counting method is one way to do this, as it necessitates carefully examining a plan and calculating the vertical and horizontal reinforcements independently. It is also necessary to take into consideration hooks, bends, and splices, and to calculate and add their length to the total. To use this method successfully, an estimator should possess knowledge of the necessary requirements for hooks, bends, and splices. If this method is not feasible, an alternative could be the area method, which is assisted by Table 3-5..

[Audio] We discuss the Area Method for estimating steel bar reinforcements for CHB in this slide. As depicted in Figure 3-6, we need to calculate the number of 10 x 20 x 40 cm CHB, vertical reinforcement spaced at 80 cm, and horizontal reinforcement at every 3 layers. To do so, we can look up Table 3-5 for the required allowances for standard bends, hooks, and lapping splices. This computation lets us determine the steel bar reinforcements for CHB precisely and expeditiously..

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] Table 3-5 shows that the vertical reinforcement length of bar in meter increases as the spacing layer increases. For example, at 40 cm. per Block, the vertical reinforcement lengths of bar is 0.235 m. per Sq. M compared to 0.171 m. per Sq. M at 60 cm. per Block. Conversely, the horizontal reinforcement lengths of bar in meter decreases with increasing spacing layer. For instance, at 2 spacing layer, the horizontal reinforcement lengths of bar is 0.264 m. per Sq. M while at 4 spacing layer, it is 0.138 m. per Sq. M..

[Audio] We calculated the area of the fence to be 12 square meters and then consulted Table 3-5 to look up the required amount of longitudal and horizontal bars. The total length of the longitudal bars came out to 19.2 meters and the total length of the horizontal bars was 25.8 meters..

[Audio] For this slide, let’s take a look at the calculations needed to determine the number of reinforcement bars required. Starting with 12 square meters of hollow blocks, multiplying that by 12.5 gives us a total of 150 pieces. Then there’s the vertical reinforcement at 80 cm spacing, which comes out to 19.2 meters. That combined with the additional 25.8 gives us a total of 45 meters. Since that amount of steel has to be purchased in 6-meter lengths, we get a total of 7.5 orders of 8 pieces of 10mm x 6m..

[Audio] Gauge no. 16 galvanized iron wire, or G I tie wire, is the most common type of tie wire used to reinforce metal in concrete hollow block works. The length of the tie wire varies depending on the size of the steel bar, ranging from 20 to 40 centimeters for small and medium-sized bars. When working with 10 mm, 12 mm or 16 mm diameter steel bars, it is important to use the appropriate length of tie wire to ensure the highest safety standards. Knowing the right amount of reinforcement is pivotal for any job..

[Audio] The table presents data for the vertical and horizontal spacing between steel reinforcement bars, as well as the weight of the steel required per square meter. It can be observed that the weight of the steel for the reinforcement decreases with the increase of both vertical and horizontal spacing. This can be a useful reference for designers when specifying the steel reinforcement for their projects..

[Audio] The wall area is 4 meters by 3 meters, or 12 square meters. Referring to Table 3-6, 0.29 kilograms of 16-G I wire is needed for tie wires, with 80 centimeter vertical spacing and 3 layers of 30 centimeter horizontal spacing of reinforcement..

[Audio] In slide 25, three-six independent footing reinforcement is a popular way to reinforce concrete structures, as per the American Concrete Institution, a bare minimum underground covering of 7.5cm is essential for steel bar reinforcement. It is essential to calculate the necessary quantity and length of bar reinforcement and weigh out the best business lengths to limit cuts. Doing this will ensure that your structure has the best reinforcement to offer support..

[Audio] For reinforcing 24 footings, the amount of 12mm steel bars required can be calculated by taking the net length of one cut bar - 1.35m - and multiplying it by 10 pieces or 2 pieces per footing. This amounts to 480 pieces. Since a 6.00-meter long steel bar is used, divide 480 by 4.0 to get 120 pieces of 12mm steel bars..

[Audio] We need to reinforce the footings with metal. According to the figure, each footing requires 10 intersection of steel bars to be tied, so for 24 footings, there would be a total of 2400 ties. Each tie requires 30 cm of tie wire, making the total amount of tie wire 720 meters. One kilogram of no.16 tie wire is equivalent to 53 meters, meaning we need 14 kg #16 kg G I wire in total..

[Audio] When it comes to concrete reinforcement, it is important to consider the length and quantity of main reinforcement. The Direct Counting Method can be used to do this, taking into account additional length requirements for lap joints of end splices, allowance for bending and or hook, additional length for beam depth and floor thickness, distance from floor to footing slab, and provision for splices and succeeding floors. This method guarantees precision in the reinforcement and ensures that the structure has sufficient strength..

[Audio] To begin, we need to calculate the total length of the main steel bar reinforcement for our metal reinforcement of 10 columns from the footing to the second floor. This includes a bend at the base footing, a length from the footing to ground floor, a height from the ground line to beam, the depth of the beam, the thickness of the floor slab, and a dowel for the second floor of 20 by 20 millimeters, totaling 5.95 meters. Therefore, we select a 6.00-meter long steel bar and multiply it by ten to get the number of bars needed, 80 pieces. To best suit our needs, we order 80 pieces of 20-millimeter-long steel bars, 8-millimeter dowels, .40-meter-squared masonry, and .15-meter concrete slabs..

[Audio] Determining the right number of main reinforcement for reinforcing a beam or girder is best done by direct counting. The length of reinforcement should be based on the physical structure and support. Ensure the span or distance between columns is suitable for the specific situation. Additionally, bear in mind that the number of splices needed can have a significant impact on the overall cost, so minimizing the number of splices can help reduce this cost..

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

[Audio] 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 order to hold the reinforcement of vertical bars in position. Technology has changed the way we stay connected in the past few years, so let's take a look at how this works..

[Audio] According to code provisions, the size of the lateral ties should be no smaller than No. 3 bar or 10 mm diameter, depending on the size of the main longitudinal reinforcement bar. When reinforcing a tied column, if the main reinforcement is No. 11 to No. 18 (35 mm to 66 mm) and bundled bars, the lateral ties should be No. 4 or 12 mm diameter. Lateral ties spacing should not exceed 16 times the diameter of the longitudinal bar, 48 times the lateral tie bar diameter, or the least dimension of the column, which should be determined accordingly..

[Audio] Stirrups are an important part of any beam or girder construction. They help to hold the main reinforcement in a fixed and designed position. Open and closed stirrups are the two most commonly used variations. The type of stirrup used depends on the size and load of the beam or Girder..

[Audio] For this reinforced concrete beam, our team has made a calculation to determine which materials are required for the stirrups. Using Table 3-8, we found that the stirrup length is 6 meters and there are 272 stirrups total. Using this, we determined that the material needed was 46 pieces of 10mm x 6m steel bar..

[Audio] In order to reinforce the 12 girder design, 180 12mm stirrups are needed. This can be determined by calculation; 15 stirrups per span multiplied by 12 girders equals 180 stirrups. Furthermore, inspection can be used to measure the length of one stirrup at 150 cm. Table 3-8 shows the availability of 5, 6, 8 and 12 meter steel bars, from which four cuts can be derived. In order to use a 6 meter steel bar, 45 12mmx 6m bars and 36 12mmx 7.5m bars are needed if a 7.5 meter steel bar is chosen..

[Audio] Spiral and Column ties can be used to reinforce concrete columns and walls. The spiral reinforcement should have evenly spaced continuous spirals held in place by three vertical bar spacers. The center-to-center spacing of the spirals should not exceed 1/6th of the core's diameter, and the clear spacing between the spirals should not exceed 7.5 centimeters nor be less than 5.0 centimeters. Additionally, the clear spacing between the spirals should be less than 1.5 times the size of the biggest piece of the coarse aggregate..

[Audio] Metal reinforcement is an essential component of modern building designs. This slide discusses a spiral column with a cross-sectional diameter of 10 mm and 14 columns at 7.00 meters high. To address its reinforcement needs, the total length of the 14 columns must be calculated, which is 98 meters..

[Audio] This solution involves finding the number of 10mm bar spirals needed to reinforce a 50 cm column of 5 cm pitch. According to Table 3-9, 483 x 10mm bar spirals are needed for 6.00 meters of steel bars. If 9.00 meters are used, the entry is 3.223, and multiplying by 98m gives us 316 pieces of 10mm x 9.00m..

[Audio] In this case, our task is to determine the number of ties wire, needed for 14 columns that are 7m tall. To do this, we first need to find the number of vertical bars per column which is 12. We then refer to Table 3-9 under 50 cm. column diameter and S cm pitch, the number of turns per meter height is 21. We multiply 12 by 21 to get 252 ties per meter height. Next, we multiply 252 ties by 7 meters tall by 14 columns which adds up to 24,696 tiles. Following that, we find the total length of the wire at 30 m long per tie which is 740,880/100 or 7,409 meters. Finally we convert this to kilograms by dividing 7,409 by 53 which ends up being around 140 kilograms..

[Audio] A one-way reinforced concrete slab is supported by beams on two opposite sides to carry loads along one direction. This type of slab is characterised by a ratio of its longer span to shorter span being equal to or greater than 2, meaning that it will bend in the direction of the shorter span. As such, a one-way slab is a great choice for flat roofs, upper stories or any other slab that needs to bend in one direction..

[Audio] Reinforcement of metals is essential for this engineering project. To determine the number of 12 mm steel bars, including tie wire, required for a one-way reinforced concrete slab as represented in Figure 3-22, we use the given data of the main reinforcement spacing at 15 cm and the temperature bar spacing at 30 cm. Through direct counting, an exact number of 12 mm steel bars can be established for successful completion of the project..

[Audio] Using 81 pieces of 12mmx6 steel bars spaced 15 cm apart and 8kg of #16 G.1 wire is the optimal solution for reinforcing the floor area of 5.70x5.50. This ensures that the floor area has the strength to withstand the expected stress..

[Audio] Reinforced concrete slab systems provide a strong and reliable structure for modern buildings. Two-way reinforcement is a design strategy consisting of two sets of steel bars placed perpendicularly across the slab which share the forces and spread the load more evenly. This prevents potential cracking and increases the life span and durability of the slab..

[Audio] We uncovered an effective solution to our challenge for Slide 45 of our presentation. To solve this problem, we found the area of the floor slab and referred to Table 3-12. This revealed we needed 173 pieces of 12mm x 7.50 meter steel bars at 10cm spacing of main reinforcement. This plan of action should ensure we have the necessary reinforcement for the job..

[Audio] Referring to Table 3-13, for a 10 cm bar spacing and 30 cm long tie wire, the amount of material needed is calculated by multiplying the area by 51.84 and .648, resulting in a total of 33.6. Consequently, No.16 G L wire must be ordered in an amount of 37 kilograms..

[Audio] When it comes to reinforcing concrete pipes, we need to consider the circumference of the circle at the midpoint of the concrete thickness. Calculating this circumference, π multiplied by the diameter of 0.90 plus 0.10 yields a result of 3.1416. From this, we can determine the quantity of 10 mm steel bars and tie wire required for the reinforcement..

[Audio] Without greetings and without beginning with Today: In this slide, we'll look into the total number of rings and shrinkage and temperature bars that we need for our reinforcement. To find the total number of rings, we need to find out the spacing distance of 15 centimeters. After doing the calculations, we come to the conclusion that there are 8 pieces at 3.29 meters. Secondly, to find the number of shrinkage and temperature bars, we need to find out the spacing distance of 20 centimeters. After doing the calculations, we come to the conclusion that there are 17 pieces at 1.00 meter. Finally, to solve for the tie wire, we need to multiply the total number of ring by the number of shrinkage and temperature bars. That's it for this slide. Thank you for your attention..