Calculation of reinforcement for foundation

Calculation of reinforcement for the foundation is an important stage of its design, therefore it must be carried out taking into account the requirements of SNiP 52-01-2003 for choosing the class of reinforcement, section and its required quantity.

First you need to understand why a monolithic concrete base needs metal fittings. Concrete after the recruitment of industrial strength is characterized by high compressive strength, and significantly lower tensile strength. A non-reinforced concrete base during expansion of the soil is prone to cracking, which can lead to deformation of the walls and even destruction of the entire building.

Calculation of reinforcement for slab foundation

A slab foundation is often used in the construction of cottages and country houses, as well as other buildings without a basement. It is a concrete slab reinforced with a rod in both perpendicular directions, with a foundation thickness of more than 20 cm, the mesh is made in the upper and lower layers.

Before calculation, it is necessary to determine the brand of reinforcing bar. For a slab foundation, performed on strong, non-heaped soils, where the probability of horizontal displacement of the building is negligible, it is allowed to use a ribbed reinforcing bar of class A-I with a diameter of 10 mm. If the ground is weak, puchinisty or the building is on a slope - the bar must be chosen not less than 14 mm in diameter. For vertical connections between the lower and upper reinforcing mesh, a sufficiently smooth rod with a diameter of 6 mm in class A-I is sufficient.

The material of the walls also has a value, since the load of the building is significantly different for frame or wooden houses and buildings made of bricks or aerated concrete blocks. In general, for light small structures, it is allowed to use a rod with a diameter of 10-12 mm, for brick or block - a reinforcement of 14-16 mm in diameter.

The distances between the bars in the grid are usually 20 cm in both the longitudinal and transverse directions. This means that for 1 meter length of the house you need to lay 5 reinforcing bars. Between each other, perpendicular intersecting bars are connected by a soft annealed wire using a hook for binding or a knitting pistol.

Example of calculation:

A house made of aerated concrete blocks, installed on a 40 cm thick foundation plate on medium-loamy loam. The overall dimensions of the house - 9x6 meters.

1. Since the thickness of the foundation is significant, you need two reinforcement grids, as well as vertical connections. Horizontal grids for block structure on medium-pitch soil are made of reinforced rod with a diameter of 16 mm, vertical grids are made of a smooth rod with a diameter of 6 mm.
2. The number of bars of the longitudinal reinforcement is calculated as follows: the length of the greater side of the foundation is divided by the grid pitch: 9 / 0,2 = 45 longitudinal reinforcing bars 6 meters in length, and the total amount of the rod is 45 · 6 = 270 m.
3. Similarly, the number of bars for transverseconnections: 6 / 0.2 = 30 rods;30 · 9 = 270 m.
4. The total number of bars per two reinforcing mesh is:( 270 + 270) · 2 = 1080 m.
5. Vertical ties have a length equal to the height of the foundation. Their number is found from the number of intersections of longitudinal and transverse reinforcing bars: 45 · 30 = 1350 pieces. Their total length is 1350 · 0.4 = 540 meters.
6. Thus, to perform the foundation you need:
7. 1080 meters of A-III D16 class bars;
8. 540 meters of bar class A-I D6.
9. According to GOST 2590 we find its mass. The running meter of rod D16 weighs 1.58 kg;the meter of rod D6 is 0.222 kg. We calculate the total mass: 1080 · 1.58 = 1706.4 kg;540 · 0.222 = 119.88 kg.
   

   The total cross-sectional area of ​​the rod armature

10. The calculation of the knitting wire depends on the tool used. When crocheting, the average wire consumption is 40 cm per connection. The number of connections in one row is 1350, in two - 2700. The consumption of wire will be 2700 · 0.4 = 1080 meters. The weight of 1 meter of wire with a diameter d = 1.0 mm is 6.12 g. For the binding of foundation reinforcement, 1080 · 6.12 = 6610 g = 6.6 kg of wire is required.

Calculation of reinforcement for strip foundation

In the ribbon foundation, the main tensile load is along the belt, that is, it is directed longitudinally. Therefore, for longitudinal reinforcement, a rod with a thickness of 12-16 mm is selected depending on the type of soil and material of the walls, and for bars and vertical joints, it is allowed to take a rod of smaller diameter - from 6 to 10 mm. In general, the calculation principle is similar to the calculation of the reinforcement of the slab foundation, but the pitch of the reinforcement grid is chosen to be 10-15 cm, since the efforts to break the basement can be much larger.

Calculation example:

Belt foundation of a wooden house, foundation width 0,4 m, height - 1 meter. The house is 6x12 meters in size. The ground is the sandy sandy loam.

1. Two reinforcement grids are required for the execution of the strip foundation. The lower reinforcing mesh prevents the rupture of the basement rupture during subsidence of the ground, the upper one - with its punching.
2. The grid spacing is selected 20 cm. For the foundation tape device, 0.4 / 0.2 = 2 longitudinal bars are required in each layer of the reinforcement.
3. The diameter of the longitudinal rod for a wooden house is 12 mm. To perform double-layer reinforcement of two long sides of the foundation, 2 · 12 · 2 · 2 = 96 meters of rod are necessary.
4. For short sides 2 · 6 · 2 · 2 = 48 meters.
5. For cross-links, choose a rod with a diameter of 10 mm. Laying step - 0,5 m.
6. Calculate the perimeter of the strip foundation:( 6 + 12) · 2 = 36 meters. The resulting perimeter is divided into a laying step: 36 / 0.5 = 72 transverse rods. Their length is equal to the width of the foundation, therefore, the total quantity is 72 · 0.4 = 28.2 m.
7. For vertical connections, we also use the D10 rod. The height of the vertical reinforcement is equal to the height of the foundation - 1 m. The number is determined by the number of intersections, multiplying the number of transverse rods by the number of longitudinal bars: 72 · 4 = 288 pieces. At a length of 1 m, the total length will be 288 m.
8. Thus, to perform the reinforcement of the strip foundation you will need:

• 144 meters of A-III D12 class bars;
• 316.2 meters of rod class A-I D10.
• According to GOST 2590 we find its mass. The running meter of rod D16 weighs 0.888 kg;meter of rod D6 - 0,617 kg. We calculate the total mass: 144 · 0.88 = 126.72 kg;316.2 · 0.617 = 193.51 kg.

Calculation of knitting wire: the number of connections can be calculated from the number of vertical reinforcement, multiplying it by 2 - 288 · 2 = 576 connections. The consumption of wire per connection is 0.4 meters. The consumption of the wire will be 576 · 0.4 = 230.4 meters. The weight of 1 meter of wire with a diameter d = 1.0 mm is 6.12 g. For the mating of the foundation foundation, 230.4 · 6.12 = 1410 g = 1.4 kg of wire is required.