Foundation reinforcement: calculation of reinforcement, laying and knitting

For the correct reinforcement of the foundation of a private house, it is necessary to calculate the reinforcement, its competent laying and knitting. Incorrect calculation will lead to damage to the foundation or to unnecessary costs. We will discuss the reinforcement of foundations of various structures and the principle of calculating steel reinforcement, accompanied by diagrams and summary tables.

Reinforcement of the foundation requires the study of the structure of the frame from the reinforcement, the selection and calculation of the section, length and mass of profiled rolled products. Insufficiency of reinforcement leads to a decrease in strength and a possible violation of the integrity of the building, and its excess leads to unreasonably high costs at this stage.

What you need to know about fittings

When strengthening a concrete base, two types of building reinforcement are used:

• class A-I - smooth;
• class A-III - ribbed.

Smooth reinforcement is used in unloaded areas. It only forms the framework. Ribbed reinforcement, due to the developed surface, provides better adhesion to concrete. Such rods are used to compensate for the load. Therefore, the diameter of such reinforcement, as a rule, is larger than that of smooth reinforcement within the same foundation.

The diameter of the rod depends on the type of soil and the mass of the structure.

Table No. 1. Minimum standard diameters of reinforcement

If it is planned to build a wooden one-story building on dense soil, tabular values ​​\u200b\u200bof the diameters of the reinforcement can be taken. If the house is massive, and the soil is heaving, the diameters of the longitudinal reinforcement are taken in the range of 12-16 mm, in exceptional cases - up to 20 mm.

In the calculations, you will need information about the reinforcement from GOST-2590-2006.

Table number 2

Consumption of reinforcement for various types of foundation

Foundations of various designs differ in the area over which the load from the structure is distributed. For each type, the calculation of the amount of reinforcement is carried out according to its requirements. For a correct comparison, we will calculate all the foundations for the following house sizes:

• width - 6 m;
• length - 8 m;
• the length of the bearing walls is 14 m.

Calculation of reinforcement for slab foundation

This is the most material-intensive type of foundations. Two levels of reinforcing bars are placed in concrete, located below the upper and above the lower border of the slab by 50 mm. The laying step depends on the perceived loads. For houses made of stone / brick, the frame cell is usually 200x200 mm. At the intersection points of the reinforcement, the upper and lower levels of the frame are connected by vertically arranged bars.

Reinforcing frame of slab foundation

Let's calculate the reinforcement for our reference house (see above).

1. Horizontal reinforcement, Ø 14 mm, corrugated.

• 8000 mm / 200 mm + 1 = 41 pcs. 6 m long.
• 6000 mm / 200 mm + 1 = 31 pcs. 8 m long.
• Total: (41 pieces x 6 m + 31 pieces x 8 m) x 2 = 988 m - for both levels.
• Weight 1 linear m rod Ø 14 mm - 1.21 kg.
• The total weight is 1195.5 kg.

2. Vertical reinforcement, Ø 8 mm, smooth. For a plate thickness of 200 mm, the length of the bar will be 100 mm.

• Number of crossings of horizontal reinforcement: 31 x 41 = 1271 pcs.
• Total length: 0.1m x 1271pcs = 127.1 m.
• Weight: 127.1 m x 0.395 kg/m = 50.2 kg.

3. Heat-treated wire Ø 1.2-1.4 mm is usually used as a knitting wire. Since the place of one connection, as a rule, is tied up twice - first when laying horizontal bars, then vertical ones, the total amount of wire doubles. For one connection, approximately 0.3 m of thin wire is needed.

• 1271 pcs. x 2 x 0.3 m = 762.6 m.
• The specific weight of wire Ø 1.4 mm is 12.078 g/m.
• Wire weight: (762.6 m x 12.078 g/m) / 1000 = 9.21 kg.

Since thin wire can break / get lost, you need to purchase it with a margin.

The total amount of materials for reinforcing the slab frame is shown in Table No. 3.

Table No. 3

Calculation of strip foundation reinforcement

The strip foundation is reinforced concrete beams located under all load-bearing walls. It contains straight sections, corners and "tees". The calculation is performed for straight sections with a small margin for corner reinforcement. We accept the width of the tape - 400 mm, depth - 700 mm.

Schematic representation of a straight section of a strip foundation

The junction of load-bearing internal and external walls

External or internal corner of external walls

Reinforcement of strip foundations is also two-level. For longitudinal sections, a class A-III bar is used, and for vertical and transverse (clamps) - a bar of class A-I. The reinforcement cross section is taken for strip foundations somewhat lower than for slab foundations, under the same construction conditions.

Let's calculate the reinforcement for the reference building chosen as an example (see above).

1. Horizontal longitudinal reinforcement, Ø 12 mm, corrugated. For a tape width of 400 mm, it is enough to lay two rods in each of the two levels. For a wider tape, 3 rods should be laid.

• The length of all tapes: (8 m + 6 m) x 2 + 14 m = 42 m.
• Total length of reinforcement: 42 m x 4 = 168 m.
• Reinforcement weight: 168 m x 0.888 kg = 149.2 kg.
• Taking into account the strengthening of the corners, the weight of the bars will be 160 kg.

2. Vertical reinforcement Ø 8 mm, smooth. For a tape depth of 700 mm, the length of the bar will be 600 mm. The distance between the vertical bars along the length of the tape is 500 mm.

• Total length of bars: 85 pcs. x 0.6 m = 51 m.
• Weight of bars: 51 m x 0.395 kg/m = 20.1 kg.

3. Horizontal transverse (yoke) reinforcement Ø 6 mm, smooth. For a tape width of 400 mm, the length of the bar will be 300 mm. The distance between the transverse bars along the length of the tape is taken to be 500 mm.

• Number of bars: 42 m / 0.5 + 1 = 85 pcs.
• Total length of bars: 85 pcs. x 0.3 m = 25.5 m.
• Weight of bars: 25.5 m x 0.222 kg/m = 5.7 kg.

4. Knitting wire. Calculation when tying each connection with one wire Ø 1.4 mm:

• Number of nodes: 85 x 4 = 340 pcs.
• Total length: 340 pieces x 0.3 m = 102 m.
• Total weight: (102 m x 12.078 g/m) / 1000 = 1.23 kg.
• When knitting knots in two times, the mass of the wire will be 2.5 kg.

The total amount of materials for reinforcing the tape frame is given in table No. 4.

Table No. 4

Consumption of metal elements for a columnar foundation

Such a foundation consists of supports, the lower part of which is below the freezing zone, and a strip foundation resting on them. For a freezing depth of 1.5 m, the height of the pillars is 1300 mm (see Fig.), That is, their base is 1700 mm below the soil level.

The location of the reinforcement in the columnar foundation, side view: 1 - sand cushion; 2 - fittings Ø 12 mm; 3 - pile reinforcement

Poles are installed in the corners of the building and along the tape every 2-2.5 m.

Let's calculate the number of rods for the configuration of the house, taken as an example (see above). To do this, you need to calculate the amount of reinforcement for the pillars and sum it up with the result of the calculation for the strip foundation.

In the pillars, only vertical bars are loaded, the horizontal ones serve to form the frame. A column with a diameter of 200 mm is reinforced with four vertical reinforcements. Number of poles: 42 m / 2 m = 21 pcs.

1. Vertical reinforcement Ø 12 mm, corrugated.

• Total length of fittings: 21 pcs. x 4 pcs. x 1.3 m = 109.28 m.
• Reinforcement weight: 109.29 m x 0.888 kg = 97.0 kg.

2. Horizontal reinforcement Ø 6 mm, smooth. For dressing, it is necessary to place horizontal clamps at a distance of no more than 0.5 m. For a depth of 1.3 m, three dressing levels are sufficient. Vertical sections are located from each other at a distance of 100 mm. The length of each horizontal segment is 130 mm.

• Total length of horizontal bars: 21 pcs. x 3 pcs. x 4 pcs. x 0.13 m = 32.76 m.
• Weight of bars: 32.76 m x 0.222 kg/m = 7.3 kg.

3. Knitting wire. There are three levels of horizontal bars in each column, which are tied around four vertical ones.

• Binding wire length per pole: 3 pcs. x 4 pcs. x 0.3 m = 3.6 m.
• Wire length for all poles: 3.6 m x 21 pcs. = 75.6 m.
• Total weight: (75.6 m x 12.078 g/m) / 1000 = 0.9 kg.

The total amount of materials for reinforcing the columnar foundation, taking into account the strip frame, is given in table No. 5.

Table No. 5

Methods and techniques for connecting reinforcement

To connect the crossed rods, welding and wire knitting are used. For foundations, welding is not the best method of installation, as it weakens the structure due to the violation of structural integrity and the risk of corrosion. Therefore, as a rule, the reinforced frame is “knitted”.

This can be done manually with pliers or hooks, as well as with a special gun. With the help of pliers, unannealed wire of large diameter is knitted.

Techniques for manual knitting of reinforcement using pliers: 1 - knitting with wire in bundles without pulling; 2 - knitting corner knots; 3 - two-row knot; 4 - cross knot; 5 - dead node; 6 - fastening of the rods with a connecting element; 7 - rods; 8 - connecting element; 9 - front view; 10 - rear view

For thin annealed wire, it is more convenient to use hooks: simple or screw.

Video: A visual lesson in knitting reinforcement with a homemade crochet

knitting gun

For large volumes of work, a knitting gun is used. The mating speed is much higher than traditional methods, but there is a dependence on the power source. In addition, it is for foundations that the gun can not be used everywhere - some areas are hard to reach for it.