The calculation of reinforcement for the foundation occurs already at the design stage and is its most important component. It is produced, taking into account SNiP 52 – 01 – 2003 in the selection of the class of reinforcement, its quantity and section. Reinforcement of monolithic structures is made in order to improve the tensile strength of the concrete structure. After all, unreinforced concrete can collapse when swelling the ground.
Calculation of reinforcement for the foundation type slab
Slab foundation is used for the construction of cottages and country houses, as well as other buildings without a basement. This base is a monolithic concrete slab, which is reinforced with a rod in two perpendicular directions. The thickness of this foundation is more than 20 cm, and the mesh is knitted both from above and below.
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First determined with the type of rod bars. For slab monolithic basement, which is performed on solid, dense and non-fossil soils, which have a very low probability of horizontal shear, it is possible to allow the use of a ribbed rebar with a diameter of 10 mm and class A-I. If the ground is rather weak, heaving or the building is projected on a slope – the reinforcement must be taken at least 14 mm thick. Vertical connections between the lower and upper rows of the reinforcing mesh will be sufficient to use a smooth 6 mm bar class A-I.
The material of the future walls of the building is also very important. After all, the load on the foundation has significant differences in the frame, as well as wooden houses and buildings of brick or aerated concrete blocks. As a rule, for light buildings, it is possible to use an armature rod with a diameter of 10-12 mm, and for walls made of brick or blocks – at least 14-16 mm.
The gaps between the rods in the reinforcing mesh are usually about 20 cm in the longitudinal, as well as in the transverse direction. This circumstance assumes the presence of 5 reinforcing bars per 1 meter of the length of the basement wall. Between themselves, the intersections of the perpendicular rods are tied with soft wire using a device such as a crochet hook.
Helpful advice! If the volume of construction is very large, then a special gun can be purchased for knitting fittings. It is able to automatically interconnect rods at very high speeds.
Suppose that we need to perform the calculation of the reinforcement for the foundation of a private house of aerated concrete light blocks. Its installation on a slab foundation, which has a thickness of 40 cm, is projected. The data of geological surveys suggest that the soil under the foundation is loamy with medium heaving. House dimensions – 9×6 m:
- since we conceived a sufficiently large thickness of the foundation, we will need to fill it with two horizontal grids. The block structure on medium-to-soil soils requires horizontal bars with a diameter of 16 mm and ribbing, and vertical rods can be smooth with a thickness of 6 mm;
- to calculate the required number of longitudinal reinforcement, take the length of the largest side of the basement wall and divide it by the grating step. In our example: 9 / 0,2 = 45 thick reinforcing bars, which have a standard length of 6 meters. We calculate the total number of bars, which equals: 45×6 = 270 m;
- in the same way, we find the number of reinforcement bars for transverse ligaments: 6 / 0.2 = 30 pieces; 30×9 = 270 m;
- multiplying by 2, we obtain the required number of horizontal reinforcement in both grids: (270 + 270) x 2 = 1080 m;
- vertical ligaments have a length equal to the entire height of the foundation, that is, 40 cm. Their number is calculated from the number of perpendicular intersections of the longitudinal rods with transverse ones: 45Х30 = 1350 pcs. Multiplying 1350х0,4, we get the total length of 540 m;
- it turns out that for the construction of the required foundation you will need: 1080 m rod A-III D16; 540 m rod A-I D6.
Helpful advice! In order to calculate the mass of all valves, it is necessary to use GOST 2590. According to this document, 1 rm. rebar D16 has a weight of 1.58 kg, and D6 – 0.22 kg. Based on this, the total mass of the whole structure: 1080×1.58 = 1706.4 kg; 540×0.222 = 119.9 kg.
For the construction of reinforcement requires also knitting wire. Its number can also be calculated. If you knit with a regular crochet, then one knot will take about 40 cm. One row contains 1350 connections, and two – 2700. Therefore, the total wire consumption for knitting will be 2700х0.4 = 1080 m. At the same time, 1 m of wire with a diameter of 1 mm weighs 6.12 g. So its total weight is calculated as follows: 1080×6.12 = 6610 g = 6.6 kg.
Features of the strip foundation are such that its rupture is most likely in the longitudinal direction. Based on this, the need for reinforcement for the foundation is calculated. The calculation here is not very different from the previous one, which was made for a slab-like foundation. Therefore, the thickness of the rod can be for longitudinal mounting 12-16 mm, and for the transverse, as well as vertical 6 – 10 mm. In the case of a tape base, a pitch of no more than 10-15 cm is chosen in order to avoid a longitudinal gap, since the load in it is much greater.
For example, we calculate the foundation tape type in application to a wooden house. Suppose that its width is 40 cm and its height is 1 m. The geometrical dimensions of the structure are 6×12 m. The soil is sandy heaving:
- in the case of a strip footing, the installation of two reinforcing meshes is mandatory. The lower one prevents the physical rupture of the monolithic tape during ground subsidence, and the top one when heaving the ground;
- the grid step of 20 cm seems to be optimal. Therefore, for the correct arrangement of the tape of such a foundation, 0.4 / 0.2 = 2 bars of the longitudinal reinforcement in both layers of reinforcement are needed;
- for a wooden house, the diameter of the rebar is 12 mm. To perform a two-layer reinforcement of the longest sides of the base, you need 2x12x2x2 = 96 m of the rod. Short sides require 2x6x2x2 = 48 m;
- for transverse crossbars we take a 10-millimeter bar. The pitch is 50 cm.
The perimeter of the building: (6 + 12) x 2 = 36 m. We divide it into a step: 36 / 0.5 = 72 reinforcing cross bars. Since their length is equal to the width of the foundation, the total need is 72×0.4 = 28.2 m;
- for vertical connections, the rod D10 is also applicable. Since the height of the vertical component of the reinforcement is equal to the full height of the foundation (1 m), the required number is determined by the number of intersections. To do this, multiply the number of transverse rods by the number of longitudinal: 72×4 = 288 pcs. For a height of 1 m, the total length will be 288 m;
- that is, to complete the reinforcement of our strip foundations, you need: 144 m A-III rod D12; 316.2 m rod A-I D10.
Helpful advice! In accordance with the same GOST 2590, it is possible to determine the mass of all reinforcement on the basis of the fact that 1 rm. bar D16 has a weight of 0.888 kg; D6 – 0.617 kg. Hence the total weight: 144×0.8 = 126.7 kg; 316.2 x0.62 = 193.5 kg.
The carried out examples of the calculation of reinforcement for the foundation will help you navigate the needs of materials in any case. To do this, you only need to substitute your data in the formulas.