Reinforcement weight: table of specific gravity of 1 linear meter depending on diameter
20.05.2017
Reinforced concrete today is the most common material used in the construction of multi-story buildings, roads, tunnels, bridges and any other objects. Reinforcement is an important component of such structures - non-reinforced concrete, although it can withstand significant compressive loads, practically does not work in bending and tension, collapsing under relatively small loads.
But the use of metal rods - regular or prestressed - eliminates this drawback. Often builders find themselves in situations where they need to know the weight of reinforcement in order to calculate the required amount of material for construction. A table of reinforcement weights will help them with this.
You will find it below in the article, in the reinforcement table, the value of the mass of metal rods of all diameters is presented.
What does the mass of rods depend on?
Of course, first of all, the mass of the rod depends on the thickness. The larger the diameter, the greater the weight. Today, in construction, metal rods with a diameter of 6 to 80 millimeters are most often used.
The weight of 1 m of reinforcement, the thinnest, weighs only 222 grams, while for the thickest this figure is 39.46 kilograms. As you can see, the difference is huge.
Therefore, knowing the weight of the reinforcement will also not be superfluous when calculating the pressure of the structure on the foundation - several unaccounted tons of load can have a detrimental effect on the reliability and durability of any building.
How much does the fittings weigh?
In order to find out the reinforcement weight, the easiest and most convenient way is to use the special table presented below.
Rebar weight table
Diameter, mmWeight of 1 meter of reinforcement, kgLinear meters in ton| 6 | 0,222 | 4504,5 |
| 8 | 0,395 | 2531,65 |
| 10 | 0,617 | 1620,75 |
| 12 | 0,888 | 1126,13 |
| 14 | 1,21 | 826,45 |
| 16 | 1,58 | 632,91 |
| 18 | 2 | 500 |
| 20 | 2,47 | 404,86 |
| 22 | 2,98 | 335,57 |
| 25 | 3,85 | 259,74 |
| 28 | 4,83 | 207,04 |
| 32 | 6,31 | 158,48 |
| 36 | 7,99 | 125,16 |
| 40 | 9,87 | 101,32 |
| 45 | 12,48 | 80,13 |
| 50 | 15,41 | 64,89 |
| 55 | 18,65 | 53,62 |
| 60 | 22,19 | 45,07 |
| 70 | 30,21 | 33,1 |
| 80 | 39,46 | 25,34 |
All data indicated in this table fully comply with the current GOST. The error can be a maximum of a few percent - such errors will not cause significant trouble and will certainly not cause damage to the structure.
Having a table at hand, you can quickly calculate the weight of reinforcement, for example, with a diameter of 32 mm. Find the corresponding diameter in the first column and immediately find out that its mass is 6.32 kg per 1 m, and a ton includes 158.48 meters.
Why do you need to know the weight?
Professional builders often have a question: what is the weight of a linear meter of reinforcement. Why do they need this? The fact is that when purchasing rods for the construction of large structures, it is not purchased individually, as in individual construction, but in tons.
But it’s difficult to calculate how long a certain mass of material will last if you don’t know how much a meter of reinforcement weighs. Knowing the total mass and specific gravity of the reinforcement, 1 meter, you can make simple calculations in a matter of seconds, obtaining the total length of the metal rods.
To do this, we take the entire mass of the necessary rods and divide them by the weight of 1 linear meter.
Calculation example
To reinforce the beams, 2.5 tons of 25-diameter rods are required. We take the value of the mass of 1 meter from the table, equal to 3.85 kg. Next, we convert tons to kilograms, multiply by 1000, it will be 2500 kg, and divide by 3.85, we get 649 meters of material.
The standard length of a metal rod is 11.7 m, to find out the required number of rods, divide 649 by 11.7, we get 55.5 pieces. In this way, you can count the number of rods with any cross-section.
This will help, especially in private construction, to check that the correct amount of material has been delivered to you.
The opposite situation may also occur. The specialist knows how much material he needs, and also knows the optimal diameter. Having found out the theoretical weight of a meter of reinforcement, he just needs to multiply this number by the total length of the required metal rods to determine how much material is needed for construction.
Source: https://VseoArmature.ru/raschet/ves-armatury
Strip foundation reinforcement
Calculator Reinforcement-Tapes-Online v.1.0
Calculation of longitudinal working, structural and transverse reinforcement for strip foundations. The calculator is based on SP 52-101-2003 (SNiP 52-01-2003, SNiP 2.03.01-84), Manual for SP 52-101-2003, Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete (without prestressing) .
If this menu item is selected, the calculator will calculate the minimum content of working longitudinal reinforcement for the foundation structure in accordance with SP 52-101-2003. The minimum percentage of reinforcement for reinforced concrete products lies in the range of 0.1-0.25% of the cross-sectional area of the concrete, equal to the product of the width of the tape and the working height of the tape.
SP 52-101-2003 Clause 8.3.4 (analogous to the Manual for SP 52-101-2003 Clause 5.11, Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, clause 3.8)
Manual to SP 52-101-2003 Clause 5.11
In our case, the minimum percentage of reinforcement will be 0.1% for the tension zone. Due to the fact that in a strip foundation the stretched zone can be both the top and bottom of the strip, the percentage of reinforcement will be 0.1% for the top chord and 0.1% for the bottom chord of the strip.
For longitudinal working reinforcement, rods with a diameter of 10-40 mm are used. For the foundation, it is recommended to use rods with a diameter of 12 mm.
Manual to SP 52-101-2003 Clause 5.17
Guidelines for the design of concrete and reinforced concrete products made of heavy concrete, paragraph 3.11
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, paragraph 3.27
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, paragraph 3.94
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, paragraph 3.94
Distance between rods of longitudinal working reinforcement
Manual to SP 52-101-2003 Clause 5.13 (SP 52-101-2003 Clause 8.3.6)
Manual to SP 52-101-2003 Clause 5.14 (SP 52-101-2003 Clause 8.3.7)
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, paragraph 3.95
Structural reinforcement (anti-shrinkage)
According to the manual for the design of concrete and reinforced concrete structures made of heavy concrete, paragraph 3.104 (analogous to the Manual for SP 52-101-2003, paragraph 5.
16) for beams with a height of more than 700 mm, structural reinforcement is provided on the side surfaces (2 reinforcement bars in one horizontal row). The distance between the structural reinforcement bars in height should be no more than 400 mm.
The cross-sectional area of one reinforcement must be at least 0.1% of the cross-sectional area, equal in height to the distance between these rods, half the width of the tape in width, but not more than 200 mm.
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, paragraph 3.104 (Manual to SP 52-101-2003, paragraph 5.16)
According to calculations, it turns out that the maximum diameter of the structural reinforcement will be 12 mm. According to the calculator, it may turn out to be less (8-10mm), but still, in order to have a margin of safety, it is better to use reinforcement with a diameter of 12mm.
Example
Initial data:
- Foundation dimensions in plan: 10x10m (+one load-bearing internal wall)
- Tape width: 0.4m (400mm)
- Tape height: 1m (1000mm)
- Concrete cover: 50mm (selected by default)
- Reinforcement diameter: 12mm
Calculation:
Working height of the tape section [ho] = Tape height – (Protective layer of concrete + 0.5 * Diameter of working reinforcement) = 1000 – (50 + 0.5 * 12) = 944 mm
Sectional area of the working reinforcement for the lower (upper) chord = (Tape width * Working height of the tape section) * 0.001 = (400 * 944) * 0.001 = 378 mm2
We select the number of rods according to SP 52-101-2003 Appendix 1.
We select a section greater than or equal to the section found above.
The result was 4 reinforcement bars with a diameter of 12 mm (4F12 A III) with a cross-sectional area of 452 mm.
So, we have found rods for one belt of our tape (let's say the bottom one). For the top one it will be the same. Eventually:
Number of rods on the bottom belt of the tape: 4
Number of rods on the upper belt belt: 4
Total number of longitudinal working rods: 8
Total cross-section of longitudinal working reinforcement per strip = Cross-section of one rod * Total number of longitudinal rods = 113.1 * 8 = 905mm2
Total length of the tape = Foundation length * 3 + Foundation width * 2 = 10 * 3 + 10 * 2 = 50m (47.6m in the calculator taking into account the width of the tape)
Total length of rods = Total length of tape * Total number of longitudinal rods = 47.6 * 8 = 400m = 381m
Total weight of reinforcement = Weight of one meter of reinforcement (we find it from the table above) * Total length of rods = 0.888 * 381 = 339 kg
Volume of reinforcement per strip = Section of one longitudinal reinforcement * Total length of rods / 1000000 = 113.1 * 381 / 1000000 = 0.04m3
Design reinforcement
If this type of menu is selected, then the calculation of the longitudinal working reinforcement for the tension zone will be performed according to the formulas of the manual to SP 52-101-2003.
In our case, tensile reinforcement is installed at the top and bottom of the tape, so we will have working reinforcement in both the compressed and tension zones.
Transverse reinforcement (clamps)
Transverse reinforcement is calculated based on user data.
Transverse reinforcement standards
Manual to SP 52-101-2003 Clause 5.18
Manual to SP 52-101-2003 Clause 5.21
Manual to SP 52-101-2003 Clause 5.21
Manual to SP 52-101-2003 Clause 5.23
Manual to SP 52-101-2003 Clause 5.20
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete. Paragraph 3.105
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete. Paragraph 3.106
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete. Paragraph 3.107
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete. Paragraph 3.109
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete. Paragraph 3.111
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete. Clause 2.14
Manual to SP 52-101-2003 Clause 5.24
Manual to SP 52-101-2003 Clause 5.22
Protective layer of concrete
Manual to SP 52-101-2003 Clause 5.6
Manual for SP 52-101-2003 Clause 5.8 (Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete, clause 3.4)
Useful
Regulatory documentation
SP 52-101-2003 Concrete and reinforced concrete structures without preliminary
reinforcement stresses Manual for SP 52-101-2003 on the design of concrete and reinforced concrete structures without preliminary
reinforcement stress SNiP 2.03.01-84 Concrete and reinforced concrete structures
Guidelines for the design of concrete and reinforced concrete structures made of heavy concrete (without prestressing)
Books
Reinforcement of elements of monolithic reinforced concrete buildings I.N. Tikhonov 2007
Construction calculators
Source: https://www.gvozdem.ru/stroim-dom/kalkulyatory/armirovanie-lentochnogo-fundamenta.php
Reinforcement weight table
Reinforcement is an integral part of any construction process when it comes to high-strength concrete structures. Reinforcement is used both in the construction of foundations and in the construction of heavy loaded floors.
You can also often find less massive structures in partitions and in certain types of finishing work.
The reinforcement, or more precisely the steel from which it is made, is subject to special requirements, which are standardized by GOST, and each reinforcing steel is used strictly depending on the load prescribed by the standards and external conditions.
Reinforcement parameters and weight
Each enterprise that deals with steel rolling complies with certain technical conditions that the fittings must comply with. First of all, this concerns weight. The reinforcement weight table can be used for various purposes - both to calculate transport costs and to calculate the load that the reinforced structure will exert. The table shows the main parameters for the most common types of reinforcing bars.
The weight of reinforcing steel is calculated by multiplying the total length of all rods by the weight of one linear meter; this is why such tables exist.
How many meters are there in a kilogram and how many kilograms are in a meter?
Naturally, whenever this table is needed, it will not be at hand. Therefore, in order to accurately calculate how many kilograms are in a meter of reinforcement, you need to proceed from the volume of the material. Calculating the weight of reinforcement along its length can only be done when we know its diameter exactly.
For an accurate calculation, we use a simple formula - 1m*(3.14*D*D/4). Thus, by calculating the volume of the reinforcement, you can easily find out its weight. To do this, you need to know the specific gravity of steel, which, according to reference books, is 7,850 kg/m³. Length by weight is calculated in the same way. Knowing the specific gravity of steel and the diameter of the reinforcement, you can easily operate with both the length and weight of the rod.
Application of reinforcing steel
Reinforcement, in principle, is a system of interconnected rods, which, when interacting with the concrete mixture, give the final product sufficient strength and rigidity. Reinforcement systems take on all the tensile, bending and compressive loads that a reinforced concrete structure must withstand.
Basically, periodic reinforcing steel is used in the construction of foundations, and is also often used in the construction of buildings by constructing walls by pouring permanent wall formwork with concrete mixture. For such construction, special types of concrete are used, but the requirements for reinforcing steel are the same. They are approved according to GOST 5781-82, and are observed by all metallurgical enterprises.
Dependence of the weight of reinforcement on various factors
As we have already found out, it is necessary to know the weight of the reinforcement during calculations that are carried out to determine the load that the reinforced structure will exert on other elements, or on the soil, if the calculations concern the construction of foundations. Also, construction engineers calculate the reinforcement coefficient of a structure based on the weight of the reinforcement and its volume, so parameters such as:
- Diameter of fittings.
- Weight per linear meter of reinforcement.
- Volume of fittings.
- The ratio of the volume of reinforcement to the volume of concrete.
- Reinforcing steel grade.
Vital to know in engineering calculations.
Reinforcing steel grade
The weight of any reinforcement is presented in the table that we have posted on the page, and the weight of a linear meter of reinforcement depends on its diameter. GOST clearly defines the standard size of diameters, which we presented in the table. This table also shows the class of steel from which the reinforcement itself is made.
The weight of a linear meter of reinforcement depends not only on the diameter, but also on the shape of the surface of the reinforcing rod or reinforcing wire. The profile can be either smooth or corrugated. The corrugated profile of the reinforcement is necessary in order to improve the adhesion of steel to the concrete mixture, as well as to give additional strength to the reinforcing rod or wire. Depending on the production technology of reinforcing steel, it is divided into three types:
- hot rolled;
- cold rolled;
- cold drawn wire.
It is noteworthy that the mass of a linear meter of hot-rolled steel does not depend on the basic mechanical characteristics it possesses. These characteristics are called steel classes and are designated by the letter A and a digital index.
Imported steel brands are similar in performance, but have slightly different markings and a higher price, so they are used extremely rarely.
Source: https://stroydetali.com/tablica-vesov-armatury_/
Weight of 1 meter of reinforcement - formula and table
During the construction process, an important parameter is the weight of the steel reinforcement, which is taken into account when designing the building. To calculate the mass of a batch of reinforcing bars, you need to know their diameter, mass per linear meter and total length.
Calculation of the weight of smooth and corrugated reinforcement: options
The mass of a linear meter of reinforcing bar can be determined in three ways: using the GOST 5781-82 , which determines the range of this metal product, using a formula and using an online calculator.
Determining the mass of reinforcing steel from the table
The table shows values calculated using nominal diameters and an average steel density of 7850 kg/m3. The nominal diameter of a grooved rod is equal to the nominal diameter of a rod with a smooth surface of the same profile number. This means that the table shows the masses per linear meter of a certain profile number, suitable for both smooth and grooved reinforcement products.
Weight table, kg, 1 linear meter of reinforcement with smooth and corrugated surface of classes A1 and A3
| Profile number (nominal rod diameter) | Theoretical mass 1 m, kg | Profile number (nominal rod diameter) | Theoretical mass 1 m, kg | Profile number (nominal rod diameter) | Theoretical mass 1 m, kg |
| 6 | 0,222 | 20 | 2,47 | 45 | 12,48 |
| 8 | 0,395 | 22 | 2,98 | 50 | 15,41 |
| 10 | 0,617 | 25 | 3,85 | 55 | 18,65 |
| 12 | 0,888 | 28 | 4,83 | 60 | 22,19 |
| 14 | 1,21 | 32 | 6,31 | 70 | 30,21 |
| 16 | 1,58 | 36 | 7,99 | 80 | 39,46 |
| 18 | 2,00 | 40 | 9,87 |
To determine the mass of a batch of metal products, the mass of 1 linear meter is multiplied by the total footage.
Calculation by formula
M = π *( D 2 /4)* ρ , in which:
M – mass of 1 m, kg,
D – nominal diameter, m,
ρ – density equal to 7850 kg/m3.
Calculation using an online calculator
Determining the weight of reinforcement (kg) by diameter and total length using an online calculator is the simplest and most convenient option.
Source: https://TreydMetall.ru/info/ves-1-metra-armatury-formula-i-tablica
Reinforcement: specific gravity of reinforcement and weight of a meter of reinforcement 12 mm, what is a linear meter, formulas
Reinforced concrete, thanks to its exceptional qualities, is successfully used in construction. The material is concrete with reinforcement in it. From this article you will learn why it is used in concrete, as well as how to calculate the specific gravity of 12 mm reinforcement in structures. The information will be useful to everyone who is engaged in construction and wants to obtain high-quality and durable building material.
It is used to strengthen concrete, giving it even greater strength. It is thanks to the strengthening of the foundation with the help of reinforced concrete blocks that many buildings, various structures and bridges have been built. There are several options that will help increase the strength of concrete using construction reinforcement. Both reinforcement without prestressing and already prestressed can be placed in the concrete body.
Reinforced concrete is a symbiosis of two different materials , which are represented: concrete of different grades, characterized by special strength in compression, but with low tensile strength, and steel reinforcement.
Reinforcement is embedded in the concrete body to provide the manufactured product with the required tensile strength. To understand what all this is for, you need to understand how the design works.
Before the actual process of filling the manufactured product with concrete, the metal structure is tensioned using a mechanical method, thereby increasing the tension of the reinforced structure. When the material has completely hardened, the force of the metal structure will be transferred to the concrete, after which it will be compressed by tension. The existing compression in the concrete itself during subsequent operation will eliminate the stress that stretches the reinforced concrete structure.
The strength of each structure made of reinforced concrete mainly depends on the correct choice of reinforcement, as well as its characteristics. That is why you should pay special attention to its quality.
Why is it necessary to accurately calculate the mass of reinforcement?
Reinforcement, as well as pouring slabs, is often used to cover private houses. The material is an innovation, but nevertheless is actively used in individual construction.
Floor slab is a common reinforced concrete product used in construction.
Reinforcement of monolithic slabs is carried out strictly adhering to technology, because the lower part of the reinforcement is the carrier of the main load . The slabs will not be able to withstand it if the reinforcement is carried out incorrectly.
The working load that acts on the floor slab itself will be directed downward. From the very point of application of the load, it will be distributed evenly throughout the monolithic slab. If the reinforcement is done incorrectly, the slab will not withstand the load.
The main load will fall on the lower part of the reinforcement, which contributes to the tensile process, which is why special strength is required. The top layer of the slab will experience compression .
Concrete will tolerate this process quite well even without additional reinforcement.
Reinforcing monolithic concrete floors can be done with your own hands, but this will require a lot of time and effort. Before starting work, you should make a calculation of the production of such an overlap. This will require special software, and possibly the help of specialists.
Correct calculation of the reinforcement of a monolithic slab has several advantages:
- monolithic flooring will have a high load-bearing capacity ;
- Accurate calculations will help you find the best option when choosing reinforcement, as well as the width of the slab and the required quantity and grade of concrete. This calculation will save money and even time;
- the use of professional calculations will allow you to use walls and columns located in the middle of the room as support for a monolithic covering;
- using calculations, you can determine not only the scope of work, but also the cost;
- it is possible to calculate a floor slab that has a non-standard geometric shape;
- The service life of the floor, which was constructed according to reinforcement calculations, can be considered practically unlimited.
What determines the weight of a linear meter of reinforcement?
The required amount of reinforcement for reinforcing the foundation is calculated in linear meters. As a result of the calculations, you can find out how many linear meters of reinforcement are required to purchase in order to build a dacha or house. In the store, the price is indicated for 1 ton, therefore, you need to convert linear meters to tons.
The number of linear meters in 1 ton will depend on the diameter of the reinforcement bars . The larger their diameter, the less meters there will be in 1 ton.
In order to convert meters into tons, as well as determine the number of kilograms in 1 m, you need to use a special table that indicates the correspondence between the diameter of the reinforcement and the weight.
Using it is quite simple: you need to find a line with the required diameter, and then see how many kilograms are in 1 m and the number of meters in 1 ton.
Method for calculating the weight of reinforcement
What methods exist for calculating the weight of reinforcement? You can calculate the weight in several ways, but for this you need, first of all, to know the linear footage, as well as the specific gravity of the 12 mm reinforcement, in order to roughly indicate the cost of building a house or other structure. Using the methods below, you can easily calculate the weight of reinforcement with a diameter of 16 mm.
There are three ways to determine the weight of a rod with a diameter of 16 mm :
Method No. 1
This method is quite effective and also simple. To do this, you need to use an online metal calculator. To carry out the necessary calculations, you only need to know the diameter of the metal rod (according to our data, this is 16 mm), as well as the length of the rod in linear meters. You need to enter the data in the appropriate columns in a special program, and you can immediately find out the result.
Method number 2
This calculation option using a table . This method of calculation can be used if it is not possible to carry out calculations on an online calculator, but the markings of the reinforcement are known.
To find out its mass, you need to use columns 1 and 2 of the table for its calculation. This table can be found in the reference literature or on this page - just below.
In the first column you need to find the diameter of the reinforcement rod you are interested in, and in the same line in the third column its mass is indicated in linear meters.
Method number 3
This calculation method is the most labor-intensive , and should only be used if you cannot use the previous two methods. First, you need to write down the volumetric mass of the material - 7856. Next, calculations should be made according to the given scheme:
- you need to look at the cross section and calculate the area (in our case, you need to find the area of a circle). The area of the circle is determined by the following formula - P*R*R, in which R is the radius of the circle of the rolled metal itself; P value = 3.14. For a rod with a diameter of 16 mm, the calculations will be as follows: R=16/2/1000=0.008 m. Area of the circle - 3.14*0.008*0.008=0.0002;
- then you should determine the volume of reinforcement , for this you need to know the length of the rod in meters (for example, 16 m). Volume - 16*0.0002=0.0032;
- Next, you need to use the previously obtained data and multiply them by the volumetric mass of the material: 0.0032 * 7856 = 25.3 kg.
Calculating the mass of reinforcement using these methods is not difficult at all. All calculation methods will show the same result.
The weight of the reinforcement plays a big role in construction, so correct calculation will help you build a strong foundation for a cottage or house that will serve you for a long time and withstand heavy loads. When choosing a material for reinforcement, you need to pay attention to its quality, since the strength of the poured foundation depends on this, and cheap rolled metal can negatively affect the entire foundation of your building.
Source: https://plita.guru/raboty/armirovanie/ispolzovanie-armatury-i-raschet-ee-vesa-na-pogonnyy-metr.html
Reinforcement: weight per linear meter, formulas and calculation examples
In the total cost of construction, a significant portion falls on the reinforcement of reinforced concrete structures. In retail sales, the price is per linear meter. However, when constructing a foundation, a large amount of reinforcement is required, so it is cheaper to purchase it at wholesale stores. And the wholesale price is indicated in rubles/ton of products. This means that linear meters must somehow be converted into tons.
State standards indicate the weight of one linear meter of reinforcement of a specific diameter. To calculate the required kilograms or tons, you need to multiply the weight of one meter by the total length of all rods of the same diameter. By weight, by the way, the percentage of reinforcement of a reinforced concrete structure is also determined (the ratio of the mass of metal and the volume of concrete).
How to use the table
The table shows:
- Reinforcement diameter
- Sectional area of the rod
- Weight of one linear meter
- Steel grade
First, in the “diameter of reinforcement” column, you find the product with which you are going to reinforce the structure, then, moving along the horizontal line of the table, you look for the weight of a linear meter.
Table - weight per linear meter of reinforcement
If you don't have a table at hand
Not everyone has the necessary GOST, but each of us studied at school. To independently calculate the weight of a linear meter, basic knowledge of mathematics and physics is sufficient. Everyone knows that mass is equal to the volume of a body multiplied by the specific gravity of the material. The volume is calculated using the formula:
V = F x L, where
- V – body volume, m3
- F – its cross-sectional area, m2
- L – body length, m
The cross section of the reinforcement is a circle. Its area is easy to calculate, knowing the diameter of the rod:
F = 3.14 x D2/ 4 = 0.785 x D2, where
- D – reinforcement diameter (in meters)
- 3.14 – the well-known constant value π (it is dimensionless)
As you can see, calculating the cross-sectional area and volume of the rod is not difficult. Now you can calculate the weight of a linear meter. This is also done simply, using the formula:
M = V xp, where
- p – specific gravity of steel. It is equal to 7850 kg/m3.
There is some inaccuracy in these calculations: the reinforcement is not a smooth rod, and we do not take into account the dimensions of the corrugations. But if you calculate the weight of a meter rod in this way and compare it with the tabular data, you will see that your result is not much different from them.
Calculation examples
As examples, let us consider calculating the weight of a linear meter of reinforcement with the most common diameters of 6 mm and 12 mm of class A III. This material has a periodic profile (longitudinal ribs and transverse protrusions are applied to its forming surface). Special steel is used to manufacture A3 reinforcement. Thanks to the complex surface, metal and concrete form a single monolith.
Reinforcement with a diameter of 6 and 12 mm is used in the construction of dachas, private houses - for reinforcing strip foundations.
- Let's calculate the weight of a linear meter of A3 reinforcement with a diameter of 6 mm:
- Sectional area F = 3.14 x 0.006 x 0.006/ 4 = 0.000028 m2
- Volume per linear meter V = 0.000028 m2 x 1m = 0.000028 m3
- Weight M = 0.000028 m3 x 7850 kg/m3 = 0.221 kg
- Let's calculate the weight of a linear meter of reinforcement with a diameter of 12 mm:
- Sectional area F = 3.14 x 0.012 x 0.012/ 4 = 0.000113 m2
- Volume per linear meter V = 0.000113 m2 x 1 m = 0.000113 m3
- Weight M = 0.000113 m3 x 7850 kg/m3 = 0.887 kg
According to the GOST table, weight is 1 linear. m of reinforcement 6 is 0.222 kg, reinforcement 12 is 0.888 kg. As you can see, the discrepancy in numbers is small. It must be admitted, however, that the data presented in the table also does not claim to be highly accurate. They were also calculated theoretically.
In fact, the actual weight of a linear meter may differ from the table by 0.2 - 3%, both plus and minus.
on how to calculate the weight of building reinforcement
The video shows examples of calculating the weight of reinforcement, and also shows the calculation process in an online calculator.
How to calculate the weight of reinforcement in the Masmat program. You can calculate the weight of a linear meter of reinforcement of 16 mm, 32 mm and any other diameter sizes.
Source: http://MegaBeaver.ru/materials/armatura/armatura-ves-pogonnogo-metra-formuly-i-primery-rascheta.html
Reinforcement weight, weight table (weight) per 1 meter of reinforcement
When carrying out construction and installation work, calculating the mass of metal products is extremely important, since it allows you to estimate the final parameters of the structures being built and determine the cost of the material (for this, the weight of the reinforcement is taken to be 10 mm per meter).
To carry out calculations, you can use special tables that indicate the parameters of the rods and their estimated weight, as well as popular online calculators, for the use of which you need to know the exact data on the technical characteristics of rolled metal.
Knowing the exact weight of rolled materials, you can save a lot by choosing the right vehicle to transport them. If you are not sure that you can make the calculations correctly, the Regional House of Metal company will help you find out the weight of 12 mm reinforcement per meter with extreme accuracy, since they will calculate it using a special formula. View available types of foundation reinforcement.
Rebar weight table
You can find out what mass the product has - reinforcement 12 weight 1 meter - from the tables that indicate:
- weight of one linear m of product;
- number of rolled meters in one ton;
- rolled diameter in millimeters;
- cross-sectional area of the rods in square centimeters;
- class of steel used in production.
| Assortment | Weight 1 meter | |
| Weight (theoretical), kg. | Limit deviations, % | |
| 6 | 0,222 | +9 / -7 |
| 8 | 0,395 | |
| 10 | 0,617 | +5 / -6 |
| 12 | 0,888 | |
| 14 | 1,21 | |
| 16 | 1,58 | +3 / -5 |
| 18 | 2,0 | |
| 20 | 2,470 | |
| 22 | 2,980 | |
| 25 | 3,850 | |
| 28 | 4,830 | |
| 32 | 6,310 | +3 / -4 |
| 36 | 7,990 | |
| 40 | 9,870 | |
| 45 | 12,480 | |
| 50 | 15,410 | +2 / -4 |
| 55 | 18,650 | |
| 60 | 22,190 | |
| 70 | 30,210 | |
| 80 | 39,460 |
In most cases, using a table, you will be able to find the desired value. If the table did not help determine the weight of 16 mm reinforcement per meter, you can resort to using an online size calculator to make calculations. To use it, you need to know the following parameters: rolled diameter, rod length and number.
The calculator will calculate the total mass for one rod, the total length of the rods, and the volume in cubic meters. There are also calculators that rely on reference data for calculations. To use them, you need to know the GOST according to which the rolled products are manufactured, the material of manufacture and the assortment (name of the rolled products).
There are also products for which this tool is not suitable, one of such products is masonry mesh, the page of which can be found here.
Weight of reinforcement
What to do if you don’t have an online calculator at hand, and you don’t really trust the data from tables on the Internet? It’s simple - you can determine the weight of 8 mm reinforcement per meter yourself using the most common calculator.
To find out the mass of a linear meter of rolled metal, you need to determine the total length of the rods, and then multiply the specific gravity of a linear meter of the product by the number of meters. The formula used for calculation is: 1 m x (3.14 x D x D/4). By performing the actions in brackets, we obtain the geometric area of a circle with a given diameter.
Did not you find what you were looking for? You may be interested in the page with pipeline valves, which can be found here: http://rdmetall.ru/truboprovodnaya-armatura/zatvory/.
Thus, we obtain the weight of a linear meter of reinforcement by multiplying the volume by the specific gravity of the product, equal to 7850 kilograms per cubic meter. An example of calculations for one m of a rod with a diameter of 8 millimeters.
Metal volume: 1 m x (3.14 x 0.008 m x 0.008 m/4) = 0.00005024. Specific gravity: 0.00005024 cubic meter x 7850 kilograms per cubic meter = 0.394384 kilograms.
You can substitute any value of D into the formula and obtain accurate data for any rolled metal, which will allow you to determine the cost of structures for construction.
Source: http://RDMetall.ru/armatura-dlya-fundamenta/ves-armatury/
ANSWER: 0 meters. (0 rods per millimeter)
The reinforcement is manufactured in accordance with GOST 5781-82 “Hot-rolled steel for the reinforcement of reinforced concrete structures. Technical conditions" and GOST R 52544-2006 "Rolled welded reinforcement of periodic profile of classes A500C and B500C for reinforcement of reinforced concrete structures. Technical specifications"
Depending on the mechanical properties, reinforcing steel is divided into classes A-I (A240), A-II (A300), A-III (A400), A-IV (A600), A-V (A800), A-VI (A1000 ). In the designation A500C and B500C, the letter A means hot-rolled or thermomechanically strengthened reinforcing bars, the letter B means cold-deformed reinforcing bars, and the letter C means weldable.
The number in the designation shows the rounded value of the yield strength in N/mm 2. The yield strength is a mechanical characteristic of a material that characterizes the stress at which strain continues to increase without increasing the load.
What else to read on the site:
Equal-flange steel angle weight calculator. Angle meter weight table. Number of meters of angle in a ton. Dimensions of the metal corner.
Weight calculator for steel unequal angle. Table weight per linear meter of corner. Number of meters of angle in a ton. Metal corner, unequal dimensions.
domatut.rf is a site for professional builders and those who build with their own hands. Articles about the procedure for registration, composition and maintenance of as-built documentation in construction. Examples of correct completion of work logs and acts. Articles about the design and technology of construction and installation works. Calculators for calculating the amount of materials and other useful information for professional builders and those who build with their own hands.
Reinforcement weight calculator
Reinforcement weight calculator. Weight of a meter of reinforcement. Number of meters of reinforcement per ton. Reinforcement diameters table.
Source: https://postroifundament.ru/massa-pogonnogo-metra-armatury.html
Calculation of the quantity and weight of reinforcement for a strip foundation
To make a foundation that will serve you for a long time and without problems: cracks, breaks and other troubles, you need to calculate and lay a sufficient amount of reinforcement in advance. It's better to report than not to report.
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Many people often wonder how to calculate reinforcement for a strip foundation of a house or bathhouse. It’s good if you are doing the design of your house with some professional who knows how to count and has practice in this type of work. But many people lay the foundation for a bathhouse without a design and even themselves do not even approximate the minimum maintenance of the fittings. Not to mention the fact that they do not take into account the properties of soils. The result: a rupture of the foundation after the first winter.
There is another option when designers make mistakes because they try to take the minimum, so to speak, to reduce the cost of the foundation. From my own experience, I can say that if your project includes 12ASh reinforcement, feel free to increase it to 14ASh.
The other day, an acquaintance approached me, asking me to look at the foundation drawings for a house.
This is the summary I gave him: increase the cross-section of the reinforcement.
He tried to convince me that the designer swore to him that the calculations were correct.
Therefore, let's look in more detail using this example to test the calculation of the reinforcement section for the foundation.
In our example there were several more dubious points, but in this article we will consider only the longitudinal (working) reinforcement.
The project contained the following data:
According to the project, the cross-section of the foundation is 400x700mm
- According to the project, the cross-section of the monolithic strip reinforced concrete foundation is assumed to have the following dimensions: width 400 mm, height 700 mm.
- The dimensions of the house foundation along the width of the house are 10.8 m and along the length of the house - 13.8 m. Although the configuration of the foundation in plan is complex, in this example we will consider it simpler.
- The longitudinal reinforcement according to the project is adopted with a section of 12 mm periodic profile
- The pitch of the longitudinal reinforcement along a height of 700 mm was taken by the designer to be equal to 600 mm. That is, the longitudinal reinforcement is located in 2 levels: in the lower and upper zones of the concrete foundation mass. So to speak, the “extra” 100mm of section height is distributed over 50mm into the protective layers of concrete for the reinforcement.
Calculation of the amount of reinforcement for the foundation
In order to check or do your own calculation of the amount of reinforcement for the foundation, you must first decide on the number of longitudinal rods for the foundation section you have adopted.
If you adhere to building codes, you can independently determine the quantity and diameter of the required reinforcement.
First, in accordance with clause 8.3.6 SP52-101-2003, for a foundation width of up to 400 mm there must be two reinforcement bars.
The distance between the axes of the longitudinal reinforcement bars of no more than 400 mm ensures:
- effective joint work of concrete and reinforcement
- uniform distribution of stress and strain
- limiting the width of cracks between reinforcement bars
If the foundation width is more than 400 mm, the distance between the axes of the longitudinal reinforcement must still comply with the standard, which means that with a foundation width of 600 mm, the longitudinal rods should no longer be 4 pieces, but 6 pieces.
We check the accepted reinforcement cross-section:
- In accordance with this point, it is necessary to first determine the cross-sectional area of our foundation: 700 x 400 = 280000 mm2 = 2800 cm2 Therefore, the minimum cross-section of the reinforcement will be required equal to 2800 cm2 x 0.001 = 2.8 cm2, where 0.001 is 0.1%.
- Using the table you can select the reinforcement cross-section
Now you need to select the reinforcement area not lower than this calculated value. There is a table of reinforcement areas, but in it, in accordance with the diameter of the reinforcement, the area of one rod is indicated. Therefore, first we will discuss that there must be at least 4 rods in the foundation: two in the lower zone and two in the upper zone of the foundation section. Let's go to the table and see that the calculated area of 2.8cm2 corresponds to column 4 rods and a row of rod diameter 10mm-3.14cm2
- but Therefore, we take not 2.8 cm2, but 5.6 cm2 and select for this value the number of rods and their diameter in accordance with their cross-sectional area. It turns out that if the number of rods is 4, you need to take reinforcement with a diameter of 14 mm (6.16 cm2).
- but again! In the project, the foundation section is 700 mm high and there is reinforcement in the lower and upper zones, that is, the distance between the longitudinal reinforcement is 600 mm. I have already mentioned this in the initial data for the project. But, this means that a foundation with a height of 700 mm according to the project cannot be reinforced only in the lower and upper zones. This means that it must also have reinforcement in the middle. In our case, when reinforced not with 4, but with 6 rods, the distance between them will be 300 mm, which corresponds to the standard. Therefore, we select 6 rods in the table and get a diameter of 12 mm with a cross section of 6.79 cm2.
- as you can see in the drawing there are 5 rods, one of which is in the middle zone. Firstly, for some reason the project uses 4 rods with a diameter of 12 mm and this one, in the middle zone of the foundation mass, has a diameter of 10 mm. The total is 4.52+0.785=5.3 cm2. There's a little shortage, but it's better to always have a spare! And I don’t like one rod, there should be a number of them in pairs, because who knows how the foundation will behave, especially since in this particular case there are a lot of unaccounted facts and complicating elements, such as a very large difference in ground level marks, the presence the basement is only on one side of the building, hence the uneven settlement, which needs to be calculated in advance, etc.
So, conclusions from all of the above:
- calculation of the minimum reinforcement of the foundation showed that it is necessary to lay 6 rods with a diameter of 12 mm
- in order to provide some reserve in the reinforcement, it is enough to increase the diameter by one amount, that is, take not the 12th diameter, but 14 mm
- if you want to decide on reinforcement for the foundation of a bathhouse, for which you do not need to make a high foundation, then 4 rods will be enough, 2 each in the lower and upper zones. And the calculation of the diameter of the reinforcement can be done using this method, depending on the adopted cross-section foundation. For example, your cross-section of the foundation for a bathhouse is 500 mm in height and 300 mm in width, then the minimum reinforcement area will be 3 cm2, which corresponds to 4 rods with a diameter of 10 AS (3.14 cm2). You also add a reserve and get 4 rods 12Ash (4.52 cm2).
How much reinforcement is needed for a foundation?
It is advisable to know how much reinforcement is needed for the foundation, at least in order to purchase the required amount for your construction project or so that you can control the consumption of reinforcement by a hired team of builders.
Above, you determined for yourself how many longitudinal reinforcement bars there should be across the width and height of the foundation. Now all that remains is to calculate the total footage by multiplying the quantity by linear meters. That is, if according to calculation you got 4 reinforcement bars with a diameter of 12 mm, 2 each in the lower and upper zones, then the total footage will be 196.8 linear meters, so the perimeter of the building in our example: 10.8m*2+13.8m*2= 49.2m.
If you consider that 12mm reinforcement is sold in rods 11.90 m long, then you need to buy 17 pieces (17 * 11.90 m = 202.30 m).
Based on rolled metal, the reinforcement can be immediately cut to the length that you need for installation or for transportation to the construction site. Additional cutting costs little money, but it is quick and convenient for loading onto the car.
Calculation of reinforcement weight
weight of fittings depending on diameter
Often it is necessary to calculate not only the total length of the longitudinal reinforcement required for the foundation, but also to convert the length into tonnage, that is, to calculate the weight of the reinforcement:
- in order to determine the cost in aggregate
- in order to figure out which car to hire for transportation
For example, you decide to transport in your passenger car, but you can put no more than 80 kg on the trunk. So we determine that only 90 m of reinforcement with a diameter of 12 mm can be loaded onto the trunk. The weight calculation is as follows: 90m*0.888kg/m=79.92kg
And the cost of 202.3 meters of reinforcement with a diameter of 12 mm for the project considered in this article is calculated by multiplying the tonnage of the reinforcement, determined from the weights table, by the cost of a ton:
- weight of 1 linear meter of diameter 12mm = 0.888 kg. This means the weight of 202.30m will be 179.64 kg.
- The cost of 1 ton of 12mm reinforcement in our region is 26,000 rubles, which means for the purchase you will need to take 4,671 rubles with you.
Assortment weight program
built-in calculator allows you to quickly calculate the weight of the assortment
In order not to endlessly search on the Internet for the weight of rolled metal, not only fittings, but also pipes, channels, etc., I use a program that has the weight of any assortment. You can use the built-in calculator. I recommend downloading.
I hope that the method for calculating reinforcement for a strip foundation will be useful to you both for laying the foundation for a bathhouse yourself and for checking the reliability of the future foundation of the house.
Source: http://banjstroi.ru/fundament/raschet-armaturyi-dlya-lentochnogo-fundamenta.html
Reinforcement weight
» Directory » Reference information » Reinforcement » Reinforcement weight
Reinforcement is the most important part of the construction of multi-story buildings, cottages or industrial premises. It is important to correctly calculate the total weight of the reinforcement for building a foundation or reinforcing structures.
In addition, accurate knowledge of the weight of the reinforcement will allow you to calculate the exact cost of the material, not to buy too much and not to buy more later because there was not enough.
The weight of the reinforcement is calculated separately for each structure; as a rule, it is indicated by specialists in the project, but if it is necessary to replace one diameter of the reinforcement with another, then recalculations cannot be avoided.
There are smooth , round and periodic profile reinforcement. In other words, there are rods with a smooth surface, longitudinal ribs or with a transverse pattern on the rod.
Managers of the company Metal Service Center YugMetCenter will advise you, calculate the weight of the reinforcement, as well as the required quantity for your construction. You can calculate the theoretical weight of reinforcement to understand how much reinforcement is needed for each specific object. The weight of a meter of reinforcement is multiplied by the sum of the lengths of all rods.
Let's look at an example of calculating the weight of reinforcement:
You need reinforcement with a diameter of 12 mm, 25 meters. The table says that 1 meter of reinforcement with a diameter of 12 mm weighs 0.888 kilograms, multiply it by the number of meters: 25 x 0.888 = 22.2 kg. In other words, 25 meters of reinforcement weigh just over 22 kilograms.
Reinforcement weight table
| Diameter (mm) | Weight kg/meter |
| 5.5 mm | 0,187 |
| 6 mm | 0,222 |
| 8 mm | 0,395 |
| 10 mm | 0,617 |
| 12 mm | 0,888 |
| 14 mm | 1,210 |
| 16 mm | 1,580 |
| 18 mm | 2,000 |
| 20 mm | 2,470 |
| 22 mm | 2,980 |
| 25 mm | 3,850 |
| 28 mm | 4,830 |
| 32 mm | 6,310 |
| 36 mm | 7,990 |
| 40 mm | 9,870 |
| 45 mm | 12,480 |
| 50 mm | 15,410 |
Reinforcement calculation
If for some reason you cannot use reference information, then the weight of a meter of reinforcement can be calculated using the formula:
V (m3) = 1m x (0.785 x D x D),
Weight = K x 7850 (kg/m3),
where D is the diameter of the reinforcement, 7850 is the specific gravity of the metal (in kg/m3), 0.785 is the coefficient.
For example, let’s determine the weight of a meter of reinforcement with a diameter of 16 mm.
V = 1m x (0.785 x 0.016 m x 0.016 m) = 0.00020096 m3,
Weight = 0.00020096 m3 x 7850 kg/m3 = 1.577536 kg (compare with the value in the table).
It is important to remember: the thinner the reinforcement, the weight of which you need to know, the more it is needed to reinforce the structure.
To find out the actual weight of the reinforcement, it is weighed on special scales, which also have their own error. Experts advise checking the weight of the reinforcement both theoretically and actually. The results will be slightly different, but that's okay. There should be no significant differences. Southern Metal Center specialists will help you make the correct calculations.
Source: http://www.ugmc.ru/ves-armaturyi
Reinforcement calculation
get a discount
Nowadays, reinforcement is used on all construction sites, be it low-rise buildings or high-rise buildings. To prepare the foundations of one- or two-story private cottages, it is necessary to calculate the number and type of reinforcing products.
The foundation of any house must be durable and strong - the service life of the entire facility will depend on its proper construction. Correct calculation of reinforcement plays a huge role in increasing the service life of a structure. To do this, it is necessary to correctly determine the type and volume of material.
Strip base reinforcement scheme
In order to correctly calculate the reinforcement in a reinforced concrete strip, we will consider typical cases of its location in such foundations.
When constructing private low-rise buildings, two main reinforcement options are used:
- six reinforcing elements;
- four products.
Which option is better?
In accordance with the requirements of SP 52-101-2003, when adjacent rods are located, the maximum distance should be no more than 40 cm (400 mm). When calculating the reinforcement, 5–7 cm (50–70 mm) are retreated between the extreme rod and the side wall of the base. If the width of the building's supporting structure is more than 50 cm, use a reinforcement scheme with six rods.
The optimal location of the rods has been selected; now it is necessary to determine their other parameters.
Calculation of reinforcement diameter
Determination of parameters of vertical and transverse reinforcing elements. To make the right choice, use the information from the table:
Conditions for using fittings Minimum diameter of fittings, mm| Vertical reinforcement with a cross-sectional height of the tape less than 80 cm | 6 mm |
| Vertical reinforcement for tape heights greater than 80 cm | 8 mm |
| Transverse reinforcement | 6 mm |
When building low-rise cottages (up to 2 floors), rods with a diameter of 8 mm are used for vertical and transverse piping. This indicator is enough to lay a strong strip foundation.
Calculation of the diameter of longitudinal reinforcement
In accordance with the requirements of SNiP 52-01-2003, the minimum cross-sectional area of reinforcing bars in a strip base must be 0.1% of the total transverse size of the reinforced concrete strip.
The cross-sectional area of a reinforced concrete structure is determined by multiplying the width by the height. For example, with tape parameters of 40 x 100 cm, the calculation results in 4000 cm². The reinforcement area is 0.1% of the foundation cross-section, so 4000 cm²/1000 = 4 cm².
To avoid calculating the indicator for each rod, use the table. There are minor inaccuracies due to rounding of numbers that do not affect the final result.
Important! If the length of the tape is less than 3 m, the minimum diameter of the reinforcement is 10 mm. When the structure size is more than 3 meters, rods with an indicator of 12 mm are chosen.
When calculating the reinforcement, we obtained the minimum cross-sectional area of the rods in the section of the strip base - it is equal to 4 cm² (taking into account the number of longitudinal elements).
If the width of the foundation is 40 cm, it is sufficient to use a reinforcement scheme with four rods. Let's return to the table to find out the value for 4 rods and select the indicator.
During the calculation, we determine that for a base 40 cm wide and 1 m high, reinforcement with a diameter of 12 mm will be most suitable, since the cross-sectional area of 4 elements is 4.52 cm².
For a structure with six rods, all actions are performed similarly. You just need to use the values from the corresponding column.
The longitudinal reinforcing elements for the strip base must have the same diameter. If for some reason the rods have different diameters, then the rods with a larger indicator are used in the bottom row.
How to calculate the amount of reinforcement for the foundation?
It often happens that reinforcing bars are delivered to the site, but when tying the frame, a lack of material is discovered. You have to purchase the required volume, pay for delivery, and incur additional expenses, which lead to an increase in the cost of building a private house.
For example, we have the following plan:
Let's try to calculate the reinforcement for a structure of this type.
Determination of the number of longitudinal bars
Let's do some rough calculations. To do this, find the length of all foundation walls:
6 x 3 + 12 x 2 = 42 m,
multiply the resulting parameter by 4:
4 x 42 = 168 m.
We have obtained the total length of the longitudinal rods. To correctly calculate the reinforcement, you need to take into account several more factors. When calculating the volume of material, take into account the launch of reinforcement products when joining, because the length of one element can be 4–6 meters, and to fill the distance of 12 m it is necessary to connect several sections. The rods are joined with an overlap with a margin of at least 30 diameters. To calculate the reinforcement (with a diameter of 12 mm), we determine the launch 12 x 30 = 360 mm (36 cm).
To take stock into account, two methods are used:
- a plan for the placement of rods is drawn up and the number of joints is calculated;
- add 10–15% to the obtained value.
Determination of the amount of vertical and transverse reinforcement
According to the plan for one “rectangle” you need:
2 x 0.35 + 2 x 0.90 = 2.5 m
When calculating the reinforcement, we take the values with a margin (and not 0.3 and 0.8) so that the harness is slightly larger than the resulting rectangle.
Important! When assembling the frame in a prepared trench, vertical reinforcing rods are installed at the bottom, sometimes they are deepened into the ground to increase the stability of the structure. Then, when calculating the reinforcement, you need to take the length not 0.9 m, but increase it by 10–20 cm.
We find such parts in the entire structure, taking into account the location of 2 “rectangles” at the junctions of the walls and corners.
To calculate the reinforcement, we draw a diagram of the foundation and determine the number of resulting fragments.
We take the long side (12 meters), on it there are 6 “rectangles” and two sections of the wall of 5.4 m each, where there are 10 lintels. The result is:
6 + 10 + 10 = 26 pcs.
You can calculate the number of jumpers on a 6-meter section in a similar way, we get 10 pieces. Multiply the value by the number of walls:
2 x 26 + 10 x 3 = 82
Previously it was calculated that each part required 2.5 meters of reinforcement, therefore:
82 x 2.5 = 205 m
Total amount of material
When calculating the reinforcement, it was determined that the longitudinal reinforcing elements have a diameter of 12 mm, and the vertical and transverse ones have a diameter of 8 mm. 184.4 m of rods of the first type are required, and 205 m of the second type.
Often, when knitting a frame, small scraps remain that cannot be used. Therefore, having calculated the reinforcement, it is necessary to purchase material with a reserve. You need to buy about 190–200 meters of 12 mm rods, as well as 210–220 m of products with a diameter of 8 mm. Thanks to such simple calculations, it is easy to determine the required volume of reinforcing bars.
Source: https://as-tim.ru/raschet-armatury/
Reinforcement 12: Calculation, weight, price
Calculation of the amount of reinforcement. Cost of one meter. Characteristics and GOSTs.
The choice of diameter of rod reinforcement for reinforced concrete structures depends on the load and operating conditions of the object. Starting from 12 millimeters and above, rods are used for load-bearing supports, walls, foundations, building frames, floors, highways, and bridges with significant loads. It is reinforcement 12 that is most in demand in construction. This is due to its high-quality physical characteristics and favorable price.
Steel grades for 12 mm GOST bars
Production is based on the hot-rolled method of producing shaped metal products. According to their technical properties, products are divided into prestressed and non-tensioned. Prestressed ones make it possible to achieve high bending strength, which is important in structures with pronounced lateral pressure. This type of rod is most often made from the following grades of carbon and low-alloy steel:
- st3, 32G2Rps, 25G2S, 35 GS.
Requirements for steel and assortment for rods d=12 are regulated by: GOST 10884-94, GOST 5781-82. In a concrete structure, reinforcement 12 enhances the strength properties of the finished building.
At the same time, the rods themselves take certain loads and protect the frozen block, for example, a foundation, from concrete cracking during its hardening and operation. This is the main purpose of using rods.
And this type is advantageous due to its high-quality physical characteristics, suitable for the loads of low-rise residential buildings, and an attractive price (compared to higher and more expensive brands).
Features of a periodic profile
Rods of this parameter are produced with a smooth or corrugated surface. Continuous profile rods (corrugated) have undeniable advantages when protrusions are located in a spiral or along the axis of the rod. In the marking, the designation of the diameter of such rods means the average data of 2 values:
- Internal (implies the diameter of the rod body only);
- External (indicates the size of the entire diameter, with protrusions).
Parameters: diameters, area, weight, length
Rods of 12 millimeters have a rod body d (internal) of 11.3 mm; d external 13.7 mm. The cross-sectional area is 1.131 cm2. The weight of 1 meter of profile is 0.888 kg. 1 ton contains 1126.13 m of product.
Slicing and packaging
Production is possible in 2 ways: hot-rolled and cold-formed. Rods are produced from ferrous rolled products using the hot-rolled method, and from non-ferrous metals - using the cold-rolling method. The price of the rods depends on this parameter, regardless of the method of cutting rolled metal:
- Rods of measured length;
- Profile of unmeasured length in packs.
12 mm reinforcement is packaged in bundles, the range of rod lengths is 6-12 meters. However, by agreement, it is possible to cut the rods into segments of any length from 3 to 25 m. Standardly, 1 rod of reinforcement is cut to a length of 11.75 m. From here you can calculate its weight, which at this length will be 10.43 kg. At the same time, a ton of profile contains 1126 meters.
Reinforcement calculation 12
To calculate how many meters of rods are needed, the total mass is divided by the weight of one meter. It is important to know that rods have theoretical and actual weights, which differ due to the fact that the values are standardized within certain tolerance ranges. That is why the weight when releasing products may differ slightly, up or down from that stated in GOST 5781-82 for one linear meter and the entire order as a whole.
In order to find out the actual weight of the released rods, a control weighing of the total tonnage of the batch is carried out directly during loading. It should be noted that the maximum deviations that regulate weight are also indicated in the standards. If 12 mm reinforcement is weighed, these deviations can range from -6% to +5% according to GOST 5781-82.
Smooth profile
Smooth rods for construction are produced by the hot-rolled method, hardness class A1. Products are cut, as agreed with the customer, either standardly or to custom lengths. This type of profile has no protrusions or ribs. Welded steel reinforcement has the ability to be welded efficiently without loss of tensile and bending strength. This also applies to corrugated products.
Price of fittings dn=12 mm
The price of rod reinforcement is indicated per linear meter and per ton; more often, the manufacturer indicates the cost per ton based on the finished length of the segments. In this case, the customer can receive cuts of any length. Knowing how much 1 ton will cost and the number of meters in it, you can calculate what the price is for 1 meter of rods.
In the Moscow region, 1 meter of reinforcement with a nominal diameter of 12 costs from 24 rubles (periodic profile).
Smooth profile, price per 1 meter:
- From 29 rubles (retail);
- From 28 rub. small wholesale;
- From 27 rub. large wholesale
The price of reinforcement per ton is from 25,900 rubles (grade A500/35GS) with a standard cut of 11.7 m. With a rod length of 6 meters - from 26 thousand rubles. subject to the purchase of a wholesale batch of 20 tons.
The cost of one meter with a rod length of 3 m is 28 rubles. The lowest price per meter is 21.5 rubles. with standard cutting 11.7 m (A3, A500C). Reinforcement with a nominal diameter of 12 millimeters is used mainly for reinforcing high-grade concrete.
Source: https://the-master.ru/armatura/armirovannaya-stal-12-mm
Reinforcement weight 12 mm per meter table
Appearance of fittings.
When knitting frames, meshes, as well as when constructing a foundation, the main element is reinforcement. As for private construction, one of the most in demand here is rolled metal with a diameter of 12 millimeters. The favorable ratio of strength and affordable price allows you to use 12 mm reinforcement in the construction of a private house.
Why do you need to know the weight of rolled metal? This value will be needed to estimate the cost of construction work at different stages. Usually the weight is already calculated in the project for each structure where rolled metal A12, A3 or any other grade is used.
If you plan to do the construction calculations yourself or just want to understand this point in detail, then this material will answer all your questions.
After studying the article, the reader will be able to independently carry out calculations and find out the weight of reinforcement 12 mm, A3 or another brand.
Weight calculation
The calculation is performed in linear meters - special quantities usually used in construction work. The table also shows the weight of one linear meter. At the same time, reinforcement is sold by weight, and not by length. The builder's task is quite simple: find out how many meters are required for all structures, and then convert them into units of mass. Below is a detailed and simple table that will help you find out the weight of one linear meter. Weight calculation
In this table you need to find the required diameter (D), in this case it is 12 mm. The second column shows D - this data is not particularly needed, and converting 12 mm is quite simple (you need to divide 12 mm by 100, the result is 0.12 m).
The third column of the table is the most important - the mass m per kg is indicated here. A meter of rolled metal equals 12 millimeters to 0.888 kilograms. Also, as an example, you can take 10 mm rods, the weight of which is 0.617 kg.
The last column shows how many meters there are in one ton.
Calculator
Self-calculation
Now the reader knows how much one meter weighs. But in order to better understand the work, you need to understand the scheme by which the calculation is carried out. Having understood the essence, the builder will be able to calculate the weight of one linear meter of rods with a diameter of 12 or 10 mm.
To perform the calculation, you must proceed according to the following scheme:
The volume of one linear meter can be obtained using the following formula: 1m x (0.785 x D x D). Here the letter "D" represents the diameter of the circle. The total mass is multiplied with the specific gravity of the rods; in all cases it will be 7850 kg/m3.
To find out how much a meter weighs, you need to know the volume.
For example, you can independently calculate the mass of one meter of 10 mm reinforcement. The first step is to obtain the volume - 1m x (0.785 x 0.010 x 0.010) = 0.00010124 m3. The mass of rods is 10 mm – 00010124 m3 x 7850 = 0.616 kg. If you look at the table, one meter of reinforcement 10 weighs 0.617 kg. How much a meter of rods 14 or 16 weighs can be found using the same scheme.
Number of meters in one ton
A more detailed table, where the steel class is also present.
Above is the calculation for 10 mm. The number of meters per ton can also be calculated without using specialized tables. Here it is worth referring to the building regulations, which state that the strip base must contain at least 0.1% of rods in relation to the reinforced concrete structure. This formulation looks quite complicated.
To understand how this works, it’s worth looking at an example:
- A strip base is taken, the area of which is 2400 square cm.
- Next you will need a coefficient, for this formula it is 0.001.
- The resulting volume is multiplied with a coefficient - 2400 x 0.001 = 2.4 cm2.
- At the next stages, you won’t be able to do without reference information. Here you will need a manual that indicates the required number of rods.
For reinforcement with a diameter of 10 and 12 mm, two rods are sufficient.
What you need to know about A12 fittings
Rods are made from steel, the grade of which depends on the requirements for strength, wear and other parameters. Typically, builders choose rods made of low-alloy metal. It cannot be said that this is the most reliable and durable steel, but it does have an important advantage - low-alloy metal can be processed using arc welding.
Grade A12, like reinforcement with a diameter of 10 mm, is usually used to add strength to a structure made of reinforced concrete. Also, these rods are the main element in the construction of frame structures. In addition to this parameter, you also need to pay attention to rental, it differs by class:
- Periodic profile – A3. Class A3 fittings have transverse corrugation.
- Smooth profile – A1. Unlike A3, class A1 fittings come without corrugation.
You can purchase reinforcement, regardless of diameter or class A3, in coils or rods.
Source: http://jsnip.ru/stroitelnye-materialy/ves-armatury-12-mm-za-metr.html
