Calculating the parameters of the fence and the material for its manufacture is of great importance - if construction standards are violated, the owner of the site, due to complaints from neighbors, may be fined. In addition, if gross violations were committed during the design and installation of the fence, then such a fence must be demolished. Calculating the parameters is not as difficult as it seems at first glance.
General criteria and rules
There is no separate regulatory act regulating the construction and parameters of fencing between sites in the Russian Federation. When designing a fence, you should adhere to SNiP, which was adopted by local authorities.
According to SNiP 30–20–97, each dacha or garden plot may have a fence, fence or fence. The maximum height of the fence should not exceed 1.5 m, and only mesh or lattice materials should be used for its construction.
By mutual decision of the members of the dacha association, it is allowed to install blind fences on the side of the street and roadway with a height of up to 2.2 m. These standards were established in order to maintain normal illumination of neighboring areas and the roadway.
In practice, the height and material from which the fence is made are not subject to state standards. This is due to the following provisions:
This dacha association has adopted other rules regulating the height and parameters of fences;
The owners of neighboring plots agreed on mutually beneficial terms to construct a fence of a different type and height;
When increasing the height of the fence, an indentation was made from the border of the land plots in order to prevent shading;
The high fence was installed without the consent of the members of the partnership.
The above requirements apply only to fences between suburban areas. For individual housing construction and subsidiary farming, these standards apply only if they are described in territorial rules and agreed with the regional administration. In other cases, it is allowed to install fences of any type and height, if they do not violate the legal rights of the owner of the neighboring plot.
Linear meter
The named quantity is a universal unit of measurement used when calculating a material or product of long length. It is equal to one meter of material, regardless of its width and height .
Most often, the value is used when calculating the material for continuous fencing made of corrugated board, wood or polycarbonate. It is worth noting that the height of the structure will not always be equal to the size of the material used. Therefore, when purchasing, you should coordinate the parameters of the material and the enclosing structure.
Dependence of fence length on plot area
The length of the fence is equal to the perimeter of the site or the sum of its sides. The perimeter of the plot directly depends on the shape - for two plots of the same area, but different shapes, the perimeter will be different.
As an example, let's calculate the length of the fence for a summer cottage with an area of 6 acres (600 m2), which can have different shapes:
A square is a plot with sides of 24.5 m. The perimeter of the plot is: 24.5 * 4 = 98 m.
A rectangle is a plot with sides of 30×20 or 15×40 m. The perimeter of the plot will be equal to: (30+20) *2 = 100 m, (15+40) * 2 = 110 m.
Trapezium - plot 23.5x30x20x24.5 m. The perimeter of the plot is: 23.5 + 30 + 20 + 24.5 = 98 m.
Triangle - plot 36x40x36 m. The perimeter of the plot is: 36 + 40 + 36 = 112 m.
From the calculations it can be seen that the length of the structure for a square-shaped section will be the smallest, and for a triangular-shaped section it will be the largest. A similar trend continues for plots of 9, 11, 20 or more acres.
Before calculating building materials for the fence, it is recommended to find a cadastral plan of the site, in which the shape of the site is outlined in detail, indicating the length of each side. This will allow you to more accurately calculate the estimated length of the structure. Typical parameters and perimeter of plots of various sizes are shown in the table above.
How to calculate the material for a fence
Various materials are used to make fencing and fencing. The most popular include brick, wooden boards, imitation metal boards, corrugated sheets, polycarbonate and chain-link mesh.
The calculation is carried out immediately before its purchase and delivery to the work site. Calculation operations can be performed in two ways: an online calculator and arithmetic formulas. We recommend using both methods. Manual calculation gives a more accurate and guaranteed result, and online programs allow you to verify data and find errors, if any.
As an example, we will calculate the material for a rectangular plot of 10 acres (1000 m2).
Brick
A brick fence is a monolithic structure consisting of support pillars and spans. It is erected on a pre-equipped concrete foundation.
Red, facing or frost-resistant bricks are used for masonry. The amount of brick required depends on the design of the fence and the thickness of the masonry. On average, 1 m2 of single masonry will require 100 bricks, and for the construction of one row of a support column of 1.5 bricks (380 × 380 mm.) - 4 pieces.
The sequence of calculations is as follows:
Fence area - rectangular plot 25x40 m. Perimeter of the plot: (25+40) * 2 = 130 m. Maximum height area: 130 * 2.2 = 286 m2.
Support column - height 2.2 m, thickness 380 mm. Laying one row requires 4 bricks. Total number of rows: 220 / 8.8 = 25. Total number of bricks per pillar: 25 * 4 = 100 pcs. Number of pillars: 130 / 3 = 43 pcs.
Span - height 2 m, length 3 m, thickness 166 mm. To lay one row you will need: 300 / 25 = 12 pcs. Total number of rows: 200 / 12 = 22.7. Total number of bricks per span: 22.7 * 12 * 2 = 545 pcs. Number of spans: 130 / 3 = 43 pcs.
The total number of bricks is 43 * 100 + 43 * 545 = 27735 pcs. To build according to the parameters described above, without taking into account the thickness of the mortar joints and the minimum height, you will need 27,735 bricks.
The final result is an approximate value, but since the calculations did not take into account the height of the span, the size of the wicket and the gate, then when purchasing, you don’t have to make an allowance for overruns.
To check the calculation logic, you should use an online program. As a result, it turned out that, according to more accurate data, 26,400 bricks would be required, which corresponds to reality and confirms the logic of our calculations.
Fence
A picket fence is a type of enclosing structure with a vertical wooden board that is attached to horizontal slats. There is a small gap between the board.
To make such a fence, an ordinary planed or unplaned wooden board made of pine or spruce is used. For a picket fence, a board 20x100x3000 or 20x110x3000 mm is better suited. The thickness of the board varies from 20 to 25 mm.
To calculate the material for the fence, the standard formula is used: N = L / (S + d), where L is the length of the fence, S is the width of the board used, and d is the width of the gap between the vertical slats.
As an example, let's calculate the number of boards (20x100x3000 mm) for a fence 1.5 m high with 40 mm gaps:
The length of the fence is a rectangular plot of 25×40 m. The perimeter of the plot: (25+40) * 2 = 130 m.
Number of strips - 130 / (0.10 + 0.04) = 928.5 pcs. Number of boards in cubic meters: 928.5 * 1.5 = 1392.7 cubic meters Number of boards: 1392.7 / 3 = 464 pcs.
Number of board cubes - there are 168 boards in 1 m3 cube, taking into account our parameters. The result is: 464 /168 = 2.8 m3.
To build a 1.5 m high picket fence, you will need 2.8 m3 of planed boards 20x100x3000, as well as 260 m of boards 45x140 mm.
To calculate a fence made from a European picket fence, a similar formula and calculation logic are used. First, the perimeter of the fence is calculated. Next, the parameters of the European picket fence and the distance between the slats are determined. The data is then plugged into the formula.
Corrugated sheet
Profiled sheet is a wall cladding material used for the construction of external fences. It is made of galvanized steel, onto which paint and polymer coating are applied.
For the construction of fences, corrugated sheets marked “C” or “HC” with a thickness of 0.7 mm or more are used. The sheet parameters depend on the profile type. The most popular profile types: C8, HC 13, HC 20.
As an example, let’s calculate the amount of corrugated sheet with parameters 1200×2000 mm:
The length of the fence is a rectangular plot of land 25×40 m. Perimeter of the site: (25+40) * 2 = 130 m.
Number of sheets - for a sheet with a width of 1200 mm, the useful width is 1150 mm. The number of sheets is: 130 / 1.15 = 113 pcs.
Very often, profiled sheets are sold in linear or square meters. Therefore, calculating the number of sheets is not always relevant. When choosing corrugated sheeting, check its cost taking into account the units of measurement and only then carry out the calculation.
Polycarbonate
Polycarbonate is a transparent material characterized by high strength and reliability. Previously it was used only for the construction of floors and greenhouses. Nowadays many parts are used as fencing material. Available in sheets up to six meters long. The thickness of polycarbonate varies from 6 to 12 mm.
Manufacturers mainly sell polycarbonate for fencing at a price per 1 m.p. When making calculations, it is enough to calculate the perimeter of the site - this will be the amount required for the construction of a polycarbonate structure.
For example, taking into account the examples described above, it turns out:
The perimeter of the plot is an allotment of 25×40 m. The perimeter of the plot is: (25+40) * 2 = 130 m.
Polycarbonate for a solid fence - to build a fence 2 meters high, you will need 130 m of cellular polycarbonate, 260 m of steel angle, 10x250 mm pipes in the amount of 52 pieces.
Chainlink mesh
Chain-link mesh is the most popular material for creating fencing, temporary fences, enclosures, cages, etc. It is made from low-carbon steel wire. For the construction of fences for private housing, galvanized chain-link is most often used.
The chain-link mesh is produced in the form of rolls 10 m long. To calculate the required amount of mesh for the fence, you will need to calculate the perimeter of the site. Next, the perimeter must be divided by the length of one grid. For example, taking into account our parameters it turns out:
The perimeter of the plot is a plot of 25×40 m. The perimeter of the plot is: (25+40) * 2 = 130 m.
The number of rolls of mesh netting is 130 / 10 = 13 pcs.
In addition to the mesh itself, installation will require metal poles 10x2.5 meters, steel hooks in the amount of 5 pcs. for 1 column. The pillars are buried in the ground every 2–2.5 m. For greater strength, the pillars are concreted with a cement-sand mixture before tensioning the mesh.
| Part 1 | Part 2 >>
So, the fence is subject to forces of its own weight and forces caused by external factors, the main of which is, of course, the wind. If the weight force acts (almost) along the axis of the post and (almost) does not cause a bending moment, then the wind tends to blow across the fence, and sometimes causes a very significant load, many times greater than the load from its own weight.
As a result, a bending moment acts on the fence post, reaching a maximum at the point where the fence post emerges from the ground. It is this that is the main reason for the destruction (fall) of the fence. Let us immediately note that all further calculations are based on determining the maximum permissible bending moment acting on the pole at the place of its rigid embedding (near the ground).
The moment from its own weight (the one that is “almost”...) is neglected as negligible. Calculating the column for compression (under the structure’s own weight) does not make sense, because even for a 60x2 mm square pipe, the maximum load is more than 7.5 tons, which is practically unattainable with any fence design and any reasonable materials. For the same reason, we do not consider the calculation of a fence post for shear (or pure shear).
In addition, we ignored the calculation of the pillar for stability, i.e. violation of the geometry of the walls and curvature of the column as a whole due to loss of stability. We neglected the calculation due to its excessive complexity and inexpediency within the framework of this article. Frankly speaking, assessing the stability of structures is the most difficult part of the design engineering course... In practice, there is no reason to worry, because... Science knows no cases of fence destruction due to pillars losing stability. The fundamental relationship here is simple: the larger the standard size of the pillar and the thickness of the wall, the more stable the pillar!
In this article, we hope not just to present you with bare numbers demonstrating the advantages of our poles, but to provide a tool for performing such calculations yourself.
"Wind, wind, you are powerful..."
First, let's determine what force the wind exerts on the fence. This force depends, firstly, on the wind speed, as well as on temperature and atmospheric pressure. The last two factors determine the air density, i.e. its mass in one cubic meter. For normal conditions it is 1.22 kg/m3.
From the school physics course we remember that with increasing pressure and decreasing temperature, the density of a substance increases and vice versa. So in winter and at high pressure the wind load is a little more, and in summer, at low pressure, a little less.
All of the above applies to dry, clean air. If there is a snow or sand storm or heavy rain with gusts of wind, then the density of the environment increases significantly by an undetermined amount. In our calculations, we will provide for such an increase in load by introducing a coefficient, which in the discipline “Strength of Materials” is called a safety factor.
The situation with wind speed is somewhat more significant. If the dependence of wind force on air density is linear, then on speed it is quadratic, that is, it changes proportionally to the square of the speed. Thus, wind speed is a determining factor in the load on the fence.
A simplified formula for calculating the force acting on an area of 1 sq.m. looks like this:
F=0.61V 2 /9.8
Where F is force in kgf; 0.61 is 1/2 the density of air (under normal conditions), V is the wind speed in m/s, and 9.8 is the “g” or acceleration of gravity to convert obscure Newtons (N) into understandable kilogram-forces ( kgf).
For different wind speeds, its force F acting on 1 sq.m of fence is shown in the table and graph:
V, m/s | |||||
5 | 10 | 15 | 20 | 25 | |
F, kgf | 1,56 | 6,22 | 14,01 | 24,90 | 38,90 |
To be fair, it should be noted that wind speeds of 25 m/s are observed extremely rarely in the Moscow region, but this happens once every 5-10 years. Therefore, we will still take it as a criterion for calculating the strength of a fence post. After all, we want to build a fence not for 10 years, but for 50-100, right? Wind gusts of 20 m/s are quite common and occur almost every year. Moreover, they can happen relatively suddenly, without weather services announcing a storm warning.
Forecasts of wind speeds of more than 20 m/s are necessarily accompanied by the announcement of a storm warning in all media because There is a real danger of human casualties from flying objects (twigs and branches, parts of building roofs...) as well as from falling trees and light building structures.
According to the Beaufort scale, adopted by the World Meteorological Organization for estimating wind speed, a storm is considered a wind speed in the range from 20.8 to 24.4 m/s. For reference, the wind speed of a hurricane exceeds 32.4 m/s. It causes a pressure of more than 64 kgf per m 2. It is almost impossible for a person to move and even stand still.
Fence posts must withstand loads acting on the entire fence, so there are quite stringent requirements for their strength.
Choosing materials for poles “by eye” can lead to destruction and significant financial losses.
Materials for pillars can be brick, wood, reinforced concrete, asbestos-cement and wooden pipes.
- Brick supports have sufficient strength, durability, attractive appearance, but require the construction of foundations of significant size, the use of high-quality bricks, and their laying takes a lot of time. All this increases the cost of construction and is not always justified.
- pillars They are installed quickly, are not subject to corrosion, but have significant weight and their installation often requires lifting and transport equipment.
- Wood in some areas it is the most affordable material, is easy to process, looks great in combination with the fence and the surrounding nature, but is susceptible to rotting. To increase their service life, wooden poles must be impregnated with special compounds.
- Asbestos cement pipes They are good for everyone, but they do not tolerate mechanical shocks well. It is difficult to attach the veins to the posts. Punching holes in the supports is not recommended due to a decrease in load-bearing capacity, and clamps can ruin the appearance.
- Metal pipes have relatively light weight and sufficient strength. Installation of metal posts and veins by welding is quite simple and does not take much time. A large selection of rolled metal allows you to select materials for any conditions, and ease of processing allows you to produce structures of the required dimensions at the installation site. Metal is susceptible to corrosion and requires protective coatings.
The pillars are subject to loads from their own weight, the weight of the fence and pressure. The supports easily withstand axial loads directed from top to bottom, but poorly - lateral (bending). Since the forces from the weight of the fence are axial, they can be neglected. Please note that this statement does not apply to poles on which a gate or wicket is cantilevered.
Wind loads are lateral, they try to bend the supports and turn them out of the ground, and the amount of effort can be quite significant. It depends on wind speed and air mass. One cubic meter contains 1.22 kg of air. This value depends on pressure, humidity and increases greatly during showers and snow storms. Changes in mass are taken into account in the formulas by introducing a safety factor.
The force acting on one square meter of fence surface is calculated by the formula:
F=0.61V 2 /9.8,
where F is the wind force in kgf, V is the air speed in m/sec, 0.61 is half the specific gravity of the air, 9.8 is the coefficient for converting newtons to kilograms of force. Substituting the speed values into the formula, we can determine that at a wind speed of 5 m/sec a force of 1.56 kgf is exerted on a square meter of fence, at 15 m/sec - 14.01 kgf, at 25 m/sec - 38.9 kgf.
You should know that when wind speeds exceed 25 m/sec, a storm warning is issued, and wind speeds of 30 m/sec are considered hurricane-force.
Recommendation: For reliable operation of the poles, not only their strength is important, but also the reliability of their attachment to the soil. When the foundations are shallow, hurricane winds can tear out the strongest supports. Therefore, experts recommend burying the supports into the soil to a depth greater than half the span height.
Before calculating the pillars, it is necessary to determine the height and span length. If the dimensions change, the calculation will have to be repeated. For example, let’s calculate a pillar for a span of 2.5 meters long and 2 meters high. We take the estimated wind speed to be 25 m/sec. Then the force will act on the pillar
F=2x2.5x38.9=194.5 kgf.
The point of application of force is 1.25 meters above ground level(half the span height plus the distance from the soil to the bottom edge of the canvas). Then at the zero mark the pole will be affected by a torque
M = FLk = 194.5 x 1.25 x 1.5 = 364.7 kgf m,
where k=1.5 is the safety factor, L =1.25 m is the length of the force application arm.
Maximum permissible pipe bending moment must be greater than the torque acting on the column. It is calculated by the formula:
M = σW/1000,
G where σ is the yield strength of the metal (for steel 10 it is equal to 20 kgf/sq. mm), W is the moment of resistance of the section (cubic mm), 1000 is the coefficient for converting millimeters into meters.
The moment of resistance for a round pipe is determined by the formula:
W=π(D 4 -d 4)/32D,
where n is pi equal to 3.14, D and d are the outer and inner diameters of the pipe in millimeters.
For a square pipe the formula is:
W=(H 4 -h 4)/6H,
where H and h are the external and internal width of the section face.
The sequence of calculations is as follows:
- calculate the torque acting on the pole;
- select a pipe and determine the moment of resistance based on its geometric dimensions, then the maximum permissible bending moment;
- if the maximum bending moment is much greater than the torque, repeat the calculation for a pipe with a smaller diameter, if, on the contrary, a larger one. Calculations are carried out until the optimal option is selected.
According to calculations, for the span chosen as an example, a round pipe with an outer diameter of 89 mm and a wall thickness of 3.5 mm, the maximum moment for which is 386 kgf m, or a square pipe 60x60 mm with a wall thickness of 3 mm and a maximum moment of 457 kgf, is suitable. m.
Conclusion
The given calculation method makes it possible to only approximately assess the suitability of materials for the manufacture of pillars, however, when using it, gross errors in the design of the fence are eliminated. More accurate and comprehensive calculations require knowledge of strength of materials in the scope of the institute course.
Independent calculation of corrugated sheets for a fence
Stage 1. Materials
Before we begin, let's look at the materials. For installation, you need supports, crossbars, logs, gates, and wickets. Typically the poles are installed 2.5 meters apart. For a two-meter fence you will need two logs, from 2.2 m - three.
![](https://i2.wp.com/prof-zabory.ru/images/myblog/raschet-profnastila.jpeg)
A little about corrugated sheeting: it comes in different types. We offer 3 options - inexpensive C8, rigid and more expensive C20 and C21. The cost also depends on whether the profiled sheet is painted, and if so, on one side or both. You can order any options from us.
Read the article: Which corrugated sheet to choose for a fence
Stage 2. Calculation of materials for a corrugated fence
1) Find out the length of the future fence. The length equals the perimeter, and the perimeter is the sum of the sides multiplied by two:
If the plot is a rectangle with sides 40 m and 30 m, then the perimeter will be equal to: (40 + 30) × 2 = 140 m.
Please note: the natural topography of the earth distorts remote calculations, and rarely does any area have a perfect rectangular shape. All the numbers are a little arbitrary, although close to reality. For accurate information, order the services of a surveyor - a specialist will come and measure the territory for free.
2) We calculate how many sheets of corrugated board are needed for the fence. The easiest way is to divide the perimeter by the length of the sheet. Let's say you chose C8 standard length 1.15 m. We calculate:
140 / 1.15 = 121.7, rounded up - 122 pieces.
It’s better to buy more than to make a mistake and not get enough.
3) We calculate the number of pillars. They are installed every 2.5 meters, so the perimeter must be divided by this interval:
140 / 2.5 = 56 pillars.
4) We calculate the number of lags. For a fence height of up to two meters, the perimeter simply needs to be multiplied by two. If its height is from 2.2 - by three.
The beams are welded to the support posts. The mechanism is as follows: pillars are dug into the ground to a depth of one and a half meters, crossbars are attached to them, everything is thoroughly primed with anti-corrosion compounds, then corrugated sheets are attached to the structure. Next - cosmetics: plastic plugs that hide the unaesthetic edges of the pillars and protruding screw heads.
140 × 2 = 280 m - the total length of the logs needed for our fence.
The standard log length is 3 m, although dimensions may vary. Then:
280 / 3 = 94 lags.
![](https://i1.wp.com/prof-zabory.ru/images/myblog/raschet-profnastila-2.jpeg)
So, how many sheets of corrugated sheets are needed to fence an area of 30 × 40 m?
- 122 sheets;
- 56 pillars;
- 94 lags.
Calculator - the easiest way to calculate!
Calculations can be done automatically using a calculator. Moreover, the system will provide an approximate budget - an ideal way to plan your expenses. The calculator has fields for gates and gates; it takes into account the cost of primer, screws, and the work of installation crews.
Try it now!
Almost every owner of a suburban area has to face the problem of installing fences around their property. There are many ways to solve this problem - for example, many wealthy owners of mansions prefer to build permanent brick or stone fences. But the average owner of a suburban area still prefers a more “budget” option - fencing made of picket fence or sheet materials (corrugated sheets, slate, polycarbonate and the like). In this case, metal pipes are most often used as support posts.
Such supports can be purchased ready-made - there is a fairly wide range of similar products in stores. It won’t be difficult for a good owner to make them himself by purchasing the required number of round or profile pipes. The process of installing and concreting is not too complicated. But since such work is planned, some planning will be required, including the consumption of concrete for the installation of each of the supports. And since this amount is small, the owner will almost certainly mix the solution himself, and he needs to know the amount of the original ingredients.