When designing any facility, it is necessary to carry out work to calculate the loads that the electrical networks will carry in the future. This will help you choose the right power switching devices and select the cross-section of cable lines. The calculation of electrical loads is aimed primarily at protecting against overloads when power consumption exceeds permissible values.

The obtained calculation data allows you to select an individual wiring cross-section for each room. In order to correctly calculate the electrical load, there are several basic methods.

Calculation based on specific load

This calculation method is based on the value of the specific load, depending on the area of ​​each room. It is quite simple and does not require special knowledge. For example, the number of lamps and their power directly depend on the size of the premises. A significant disadvantage of this method is the not entirely accurate determination of loads in each individual case.

Calculation by specific power

Despite the complexity of this method, it allows you to calculate the load with high accuracy, depending on the power of all available consumers.

In order to make a correct calculation of electrical loads in the entire house or apartment, it is necessary, first of all, to establish the exact power of each electricity consumer. The resulting power indicator is multiplied by a coefficient showing the extent to which a particular device is used within an hour. In addition, the use of another correction factor allows us to take into account the uneven operation of each device.

Thus, the calculated value for each device will consist of the product of the installed power of the consumer, the utilization factor of the consumer and the coefficient allowing additional use of the equipment.

Calculation using tables

This method is most often used at the stage when constant and variable loads are taken into account in advance. Using tables, the necessary models are compiled that reflect the electrical loads of all branches.

In addition, when carrying out calculations, it is necessary to take into account changes in loads in different situations. That is, all consumers can work continuously or periodically.

Calculations are made for each category of consumers, the data obtained is summarized in a final table, where the total power consumed by all available devices is displayed. This method is the most accurate, subject to predetermined electrical equipment.

1. Load collection

Before starting the calculation of a steel beam, it is necessary to collect the load acting on the metal beam. Depending on the duration of action, loads are divided into permanent and temporary.

• own weight of the metal beam;
• own weight of the floor, etc.;

• long-term load (payload, taken depending on the purpose of the building);
• short-term load (snow load, taken depending on the geographical location of the building);
• special load (seismic, explosive, etc. Not taken into account within this calculator);

Loads on a beam are divided into two types: design and standard. Design loads are used to calculate the beam for strength and stability (1st limit state). Standard loads are established by standards and are used to calculate beams for deflection (2nd limit state). Design loads are determined by multiplying the standard load by the reliability load factor. Within the framework of this calculator, the design load is used to determine the deflection of the beam to reserve.

After you have collected the surface load on the floor, measured in kg/m2, you need to calculate how much of this surface load the beam takes on. To do this, you need to multiply the surface load by the pitch of the beams (the so-called load strip).

For example: We calculated that the total load was Qsurface = 500 kg/m2, and the beam spacing was 2.5 m. Then the distributed load on the metal beam will be: Qdistributed = 500 kg/m2 * 2.5 m = 1250 kg/m. This load is entered into the calculator

2. Constructing diagrams

Next, a diagram of moments and transverse forces is constructed. The diagram depends on the loading pattern of the beam and the type of beam support. The diagram is constructed according to the rules of structural mechanics. For the most frequently used loading and support schemes, there are ready-made tables with derived formulas for diagrams and deflections.

3. Calculation of strength and deflection

After constructing the diagrams, a calculation is made for strength (1st limit state) and deflection (2nd limit state). In order to select a beam based on strength, it is necessary to find the required moment of inertia Wtr and select a suitable metal profile from the assortment table. The vertical maximum deflection fult is taken according to table 19 from SNiP 2.01.07-85* (Loads and impacts). Point 2.a depending on the span. For example, the maximum deflection is fult=L/200 with a span of L=6m. means that the calculator will select a section of a rolled profile (I-beam, channel or two channels in a box), the maximum deflection of which will not exceed fult=6m/200=0.03m=30mm. To select a metal profile based on deflection, find the required moment of inertia Itr, which is obtained from the formula for finding the maximum deflection. And also a suitable metal profile is selected from the assortment table.

4. Selection of a metal beam from the assortment table

From two selection results (limit state 1 and 2), a metal profile with a large section number is selected.

The issue of choosing a cable cross-section for installing electrical wiring in a house or apartment is very serious. If this indicator does not correspond to the load in the circuit, then the wire insulation will simply begin to overheat, then melt and burn. The end result is a short circuit. The thing is that the load creates a certain current density. And if the cable cross-section is small, then the current density in it will be high. Therefore, before purchasing, it is necessary to calculate the cable cross-section according to the load.

Of course, you shouldn’t just randomly choose a wire with a larger cross-section. This will primarily hit your budget. With a smaller cross-section, the cable may not withstand the load and will quickly fail. Therefore, it is best to start with the question, how to calculate the cable load? And only then, based on this indicator, select the electrical wire itself.

Power calculation

The easiest way is to calculate the total power that the house or apartment will consume. This calculation will be used to select the cross-section of the wire from the power line pole to the input circuit breaker in the cottage or from the entrance switchboard to the apartment to the first distribution box. Wires in loops or rooms are calculated in the same way. It is clear that the input cable will have the largest cross-section. And the farther you are from the first distribution box, the less this indicator will decrease.

But let's get back to the calculations. So, first of all, it is necessary to determine the total power of consumers. Each of them (household appliances and lighting lamps) has this indicator marked on the body. If you can’t find it, look in your passport or instructions.

After which all powers must be added up. This is the total power of the house or apartment. Exactly the same calculation must be made for the contours. But there is one controversial point. Some experts recommend multiplying the total indicator by a reduction factor of 0.8, adhering to the rule that not all devices will be connected to the circuit at the same time. Others, on the contrary, suggest multiplying by an increasing factor of 1.2, thereby creating a certain reserve for the future, due to the fact that there is a high probability of additional household appliances appearing in the house or apartment. In our opinion, the second option is the optimal one.

Cable selection

Now, knowing the total power indicator, you can select the wiring cross-section. The PUE contains tables that make it easy to make this choice. Let's give some examples for an electric line running at 220 volts.

• If the total power is 4 kW, then the wire cross-section will be 1.5 mm².
• Power 6 kW, cross-section 2.5 mm².
• Power 10 kW – cross-section 6 mm².

There is exactly the same table for an electrical network with a voltage of 380 volts.

Current load calculation

This is the most accurate value of the calculation carried out on the current load. The formula used for this is:

I=P/U cos φ, where

• I is the current strength;
• P – total power;
• U – network voltage (in this case 220 V);
• cos φ – power factor.

There is a formula for a three-phase electrical network:

I=P/(U cos φ)*√3.

It is by the current indicator that the cable cross-section is determined according to the same tables in the PUE. Again, let's give a few examples.

• Current 19 A – cable cross-section 1.5 mm².
• 27 A – 2.5 mm².
• 46 A – 6 mm².

As in the case of determining the power cross-section, here it is also best to multiply the current indicator by a multiplying factor of 1.5.

Odds

There are certain conditions under which the current inside the wiring can increase or decrease. For example, in open electrical wiring, when the wires are laid along walls or ceilings, the current strength will be higher than in a closed circuit. This is directly related to the ambient temperature. The larger it is, the more current this cable can carry.

Attention! All of the above listed PUE tables are calculated under the condition that the wires are operated at a temperature of +25C with the temperature of the cables themselves not exceeding +65C.

That is, it turns out that if several wires are laid at once in one tray, corrugation or pipe, then the temperature inside the wiring will be increased due to the heating of the cables themselves. This leads to the fact that the permissible current load is reduced by 10-30 percent. The same applies to open wiring inside heated rooms. Therefore, we can conclude: when calculating the cable cross-section depending on the current load at elevated operating temperatures, you can choose wires of a smaller area. This is, of course, a good saving. By the way, there are also tables of reducing coefficients in the PUE.

There is one more point that concerns the length of the electrical cable used. The longer the wiring, the greater the voltage loss in the sections. Any calculations use a loss of 5%. That is, this is the maximum. If the losses are greater than this value, then the cross-section of the cable will have to be increased. By the way, it’s not difficult to independently calculate current losses if you know the wiring resistance and current load. Although the best option is to use the PUE table, which establishes the relationship between load torque and losses. In this case, the load torque is the product of the power consumption in kilowatts and the length of the cable itself in meters.

Let's look at an example in which an installed cable 30 mm long in an alternating current network with a voltage of 220 volts can withstand a load of 3 kW. In this case, the load moment will be equal to 3*30=90. We look at the PUE table, which shows that losses of 3% correspond to this moment. That is, it is less than the nominal value of 5%. What is acceptable. As mentioned above, if the calculated losses exceeded the five percent barrier, then it would be necessary to purchase and install a cable of a larger cross-section.

Attention! These losses greatly affect lighting with low-voltage lamps. Because at 220 volts 1-2 V is not reflected much, but at 12 V it is immediately visible.

Currently, aluminum wires are rarely used in wiring. But you need to know that their resistance is 1.7 times greater than that of copper ones. And that means their losses are just as many times greater.

As for three-phase networks, the load torque here is six times greater. This depends on the fact that the load itself is distributed over three phases, and this is a corresponding exponential increase in torque. Plus a double increase due to the symmetrical distribution of power consumption across phases. In this case, the current in the zero circuit must be zero. If the phase distribution is asymmetrical, and this leads to an increase in losses, then you will have to calculate the cable cross-section for the loads in each wire separately and select it according to the maximum calculated size.

Conclusion on the topic

As you can see, to calculate the cable cross-section for loads, you have to take into account various coefficients (reducing and increasing). It’s not easy to do this on your own, if you understand electrical engineering at the level of an amateur or a novice master. Therefore, my advice is to invite a highly qualified specialist, let him do all the calculations himself and draw up a competent wiring diagram. But you can do the installation yourself.

When selecting electrical equipment, it is important to know how to calculate the load on the cable. An electrical network is used to transmit electrical energy in the form of alternating current from a power source to certain electrical equipment. Electrical energy is characterized by the following parameters: current, voltage, power of the electrical network, frequency of alternating current, power of the consumer, that is, the electrical appliance.

When installing the electrical network, all existing equipment is connected only in parallel. Any electrical appliance has a certain power consumption, so the total power of the consumer is determined by the sum of the powers of all connected devices. Thus, by calculating the power of the electrical wiring of a certain room, you can select a suitable cable.

How to calculate cable load

To ensure uninterrupted operation of all electrical wiring in the room, before purchasing a cable, it is necessary to correctly calculate the loads on the cable. Since there are different types of loads, calculations are made based on all indicators. The cable cross-section is calculated based on current, power, load and voltage.

How to calculate load current

When choosing a cable, it is necessary to take into account the rated load current, which depends on the size of the cable cross-section. It is selected based on the rated current in the network, at which the cable does not heat up and the voltage in the network does not drop. Calculation of the electrical loads of the premises network is carried out according to the formula: I = P / U, where I is the current strength, P is the power of the electrical network, U is the voltage, which is 220 V.

For electrical wiring installation, various types of cables are used, which are several insulated conductors enclosed in a hermetically sealed sheath.

The maximum permissible current load on the cable directly depends on the temperature of the cable during its operation, at which its mechanical strength and elasticity are not reduced.

It is also important to determine the material of the cable core, since different metals have different conductive properties. In residential premises, copper cable is usually used, since its conductive properties are much higher than those of cable made of aluminum.

Indoor power supply systems require grounding, so the cable must be three-core. It is also worth considering the type of installation: hidden or open, as this also affects the choice of cable cross-section.

Once you have decided on the type of cable, you should take care of the safety of the electrical network. To protect against overloads in the network, electrical circuit breakers are used.

Calculation of the load on the foundation is necessary for the correct selection of its geometric dimensions and the area of ​​the foundation base. Ultimately, the strength and durability of the entire building depends on the correct calculation of the foundation. The calculation comes down to determining the load per square meter of soil and comparing it with the permissible values.

To calculate you need to know:

• The region in which the building is being built;
• Soil type and groundwater depth;
• The material from which the building’s structural elements will be made;
• Building layout, number of floors, roof type.

Based on the required data, the calculation of the foundation or its final verification is carried out after the design of the structure.

Let's try to calculate the load on the foundation for a one-story house made of solid brick of solid masonry, with a wall thickness of 40 cm. The dimensions of the house are 10x8 meters. The ceiling of the basement is reinforced concrete slabs, the ceiling of the 1st floor is wooden on steel beams. The roof is gable, covered with metal tiles, with a slope of 25 degrees. Region - Moscow region, soil type - wet loams with a porosity coefficient of 0.5. The foundation is made of fine-grained concrete, the thickness of the foundation wall for calculation is equal to the thickness of the wall.

Determining the foundation depth

The depth of installation depends on the depth of freezing and the type of soil. The table shows reference values ​​for the depth of soil freezing in various regions.

Table 1 – Reference data on the depth of soil freezing

Reference table for determining foundation depth by region

In general, the foundation depth should be greater than the freezing depth, but there are exceptions due to the type of soil; they are listed in Table 2.

Table 2 - Dependence of foundation depth on soil type

The depth of the foundation is necessary for the subsequent calculation of the load on the soil and determination of its size.

We determine the depth of soil freezing using Table 1. For Moscow it is 140 cm. Using Table 2 we find the type of soil - loam. The laying depth must be no less than the calculated freezing depth. Based on this, the foundation depth for the house is chosen to be 1.4 meters.

Roof load calculation

The roof load is distributed between those sides of the foundation on which the rafter system rests through the walls. For a regular gable roof, these are usually two opposite sides of the foundation; for a hip roof, all four sides. The distributed load of the roof is determined by the projected area of ​​the roof divided by the area of ​​the loaded sides of the foundation and multiplied by the specific gravity of the material.

Table 3 - Specific gravity of different types of roofing

Reference table - Specific gravity of different types of roofing

1. Determine the projection area of ​​the roof. The dimensions of the house are 10x8 meters, the projection area of ​​the gable roof is equal to the area of ​​the house: 10·8=80 m2.
2. The length of the foundation is equal to the sum of its two long sides, since the gable roof rests on two long opposite sides. Therefore, we define the length of the loaded foundation as 10 2 = 20 m.
3. Area of ​​a 0.4 m thick foundation loaded with a roof: 20·0.4=8 m2.
4. The type of coating is metal tile, the slope angle is 25 degrees, which means that the calculated load according to Table 3 is 30 kg/m2.
5. The load of the roof on the foundation is 80/8·30 = 300 kg/m2.

Snow load calculation

The snow load is transferred to the foundation through the roof and walls, so the same sides of the foundation are loaded as when calculating the roof. The area of ​​snow cover equal to the area of ​​the roof is calculated. The resulting value is divided by the area of ​​the loaded sides of the foundation and multiplied by the specific snow load determined from the map.

Table - calculation of snow load on the foundation

1. The length of the slope for a roof with a slope of 25 degrees is (8/2)/cos25° = 4.4 m.
2. The roof area is equal to the length of the ridge multiplied by the length of the slope (4.4·10)·2=88 m2.
3. The snow load for the Moscow region according to the map is 126 kg/m2. We multiply it by the roof area and divide by the area of ​​the loaded part of the foundation 88·126/8=1386 kg/m2.

Calculation of floor loads

The floors, like the roof, usually rest on two opposite sides of the foundation, so the calculation is carried out taking into account the area of ​​these sides. The floor area is equal to the area of ​​the building. To calculate the floor load, you need to take into account the number of floors and the basement floor, that is, the floor of the first floor.

The area of ​​each floor is multiplied by the specific gravity of the material from Table 4 and divided by the area of ​​the loaded part of the foundation.

Table 4 - Specific gravity of floors

1. The floor area is equal to the area of ​​the house - 80 m2. The house has two floors: one made of reinforced concrete and one made of wood on steel beams.
2. We multiply the area of ​​the reinforced concrete floor by the specific gravity from Table 4: 80·500=40000 kg.
3. We multiply the area of ​​the wooden floor by the specific gravity from Table 4: 80·200=16000 kg.
4. We sum them up and find the load per 1 m2 of the loaded part of the foundation: (40000+16000)/8=7000 kg/m2.

Wall load calculation

The load of the walls is determined as the volume of the walls multiplied by the specific gravity from Table 5, the result obtained is divided by the length of all sides of the foundation multiplied by its thickness.

Table 5 - Specific gravity of wall materials

Table - Specific gravity of walls

1. The area of ​​the walls is equal to the height of the building multiplied by the perimeter of the house: 3·(10·2+8·2)=108 m2.
2. The volume of the walls is the area multiplied by the thickness, it is equal to 108·0.4=43.2 m3.
3. We find the weight of the walls by multiplying the volume by the specific gravity of the material from Table 5: 43.2·1800=77760 kg.
4. The area of ​​all sides of the foundation is equal to the perimeter multiplied by the thickness: (10 2 + 8 2) 0.4 = 14.4 m 2.
5. The specific load of the walls on the foundation is 77760/14.4=5400 kg.

Preliminary calculation of the foundation load on the ground

The load of the foundation on the ground is calculated as the product of the volume of the foundation and the specific density of the material from which it is made, divided by 1 m 2 of the area of ​​its base. The volume can be found as the product of the depth of the foundation and the thickness of the foundation. The thickness of the foundation is taken during preliminary calculation to be equal to the thickness of the walls.

Table 6 - Specific density of foundation materials

Table - specific density of materials for soil

1. The area of ​​the foundation is 14.4 m2, the depth is 1.4 m. The volume of the foundation is 14.4·1.4=20.2 m3.
2. The mass of the foundation made of fine-grained concrete is: 20.2·1800=36360 kg.
3. Ground load: 36360/14.4=2525 kg/m2.

Calculation of the total load per 1 m 2 of soil

The results of previous calculations are summarized, and the maximum load on the foundation is calculated, which will be greater for those sides on which the roof rests.

The conditional design soil resistance R 0 is determined according to the tables of SNiP 2.02.01-83 “Foundations of buildings and structures”.

1. We sum up the weight of the roof, the snow load, the weight of the floors and walls, as well as the foundation on the ground: 300+1386+7000+5400+2525=16,611 kg/m 2 =17 t/m 2 .
2. We determine the conditional design resistance of the soil according to the tables of SNiP 2.02.01-83. For wet loams with a porosity coefficient of 0.5, R0 is 2.5 kg/cm2, or 25 t/m2.

From the calculation it is clear that the load on the soil is within the permissible limits.