The calculation of steel structures has become a stumbling block for many builders. Using the example of the simplest farms for an outdoor shed, we will tell you how to correctly calculate the loads, and also share simple methods of self-assembly without using expensive equipment.
General calculation methodology
Trusses are used where it is impractical to use a solid bearing beam. These structures are characterized by lower spatial density, while maintaining stability to perceive impacts without deformation due to the correct arrangement of parts.
Structurally, the truss consists of an outer chord and filling elements. The essence of the operation of such a lattice is quite simple: since each horizontal (conditionally) element cannot withstand the full load due to an insufficiently large section, two elements are located on the axis of the main action (gravity) in such a way that the distance between them provides a sufficiently large cross section of the entire structure ... It can be explained even more simply as follows: from the point of view of the perception of loads, the truss is considered as if it is made of solid material, while the filling provides sufficient strength, based only on the calculated applied weight.
The structure of the truss from a shaped pipe: 1 - lower belt; 2 - braces; 3 - racks; 4 - side belt; 5 - upper belt
This approach is extremely simple and is often more than enough for the construction of simple metal structures, however, the material consumption with a rough calculation turns out to be extremely high. A more detailed consideration of the existing impacts helps to reduce the consumption of metal by 2 or more times, this approach will be most useful for our task - to design a light and fairly rigid truss, and then assemble it.
The main profiles of trusses for the canopy: 1 - trapezoidal; 2 - with parallel belts; 3 - triangular; 4 - arched
You start by defining the overall configuration for your farm. It usually has a triangular or trapezoidal profile. The lower element of the belt is placed mainly horizontally, the upper one - at an angle, ensuring the correct slope of the roofing system. In this case, the cross-section and strength of the chord elements should be chosen close to such that the structure can support its own weight with the existing support system. Next, you add vertical bridges and oblique ties in an arbitrary number. The structure must be displayed on a sketch to visualize the mechanics of interaction, indicating the real dimensions of all elements. Then Her Majesty Physicist comes into play.
Determination of combined actions and support reactions
From the statics section of the school mechanics course, we will take two key equations: the balance of forces and moments. We will use them to calculate the response of the supports on which the beam is placed. For simplicity of calculations, we will consider the supports to be hinged, that is, they do not have rigid connections (terminations) at the point of contact with the beam.
An example of a metal farm: 1 - farm; 2 - lathing beams; 3 - roof covering
On the sketch, you must first mark the pitch of the roofing system, because it is in these places that the points of concentration of the applied load should be located. Usually, it is at the points of application of the load that the nodes of the convergence of the braces are located, so it is easier to calculate the load. Knowing the total weight of the roof and the number of trusses in the shed, it is easy to calculate the load on one truss, and the factor of uniformity of the coverage will determine whether the applied forces at the points of concentration will be equal, or they will differ. The latter, by the way, is possible if in a certain part of the canopy one coating material is replaced by another, there is a gangway or, for example, an area with an unevenly distributed snow load. Also, the effect on different points of the truss will be uneven if its upper beam has a rounding, in this case the points of application of the force must be connected by segments and the arc should be considered as a broken line.
When all the acting forces are marked on the truss sketch, we proceed to calculating the support reaction. With respect to each of them, the farm can be represented as nothing more than a lever with a corresponding amount of influences on it. To calculate the moment of force at the fulcrum, you need to multiply the load at each point in kilograms by the length of the arm of the application of this load in meters. The first equation says that the sum of the actions at each point is equal to the reaction of the support:
- 200 1.5 + 200 3 + 200 4.5 + 100 6 \u003d R 2 6 - the equation of equilibrium of moments relative to the node and, where 6 m is the shoulder length)
- R 2 \u003d (200 1.5 + 200 3 + 200 4.5 + 100 6) / 6 \u003d 400 kg
The second equation determines equilibrium: the sum of the reactions of the two supports will be exactly equal to the applied weight, that is, knowing the reaction of one support, you can easily find the value for the other:
- R 1 + R 2 \u003d 100 + 200 + 200 + 200 + 100
- R1 \u003d 800 - 400 \u003d 400 kg
But make no mistake: the leverage rule also applies here, so if the truss has a significant extension beyond one of the supports, then the load in this place will be higher in proportion to the difference in the distances from the center of mass to the supports.
Differential force calculation
We pass from the general to the particular: now it is necessary to establish the quantitative value of the efforts acting on each element of the farm. To do this, we list each belt segment and filling inserts with a list, then we consider each of them as a balanced flat system.
For the convenience of calculations, each connecting node of the truss can be represented as a vector diagram, where the action vectors run along the longitudinal axes of the elements. All that is needed for calculations is to know the length of the segments converging at the node and the angles between them.
You need to start from the node for which the maximum possible number of known quantities was established during the calculation of the support reaction. Let's start with the extreme vertical element: the equilibrium equation for it says that the sum of the vectors of converging loads is equal to zero, respectively, the counteraction to the force of gravity acting along the vertical axis is equivalent to the reaction of the support, equal in magnitude, but opposite in sign. Note that the obtained value is only a part of the overall reaction of the support acting for a given node, the rest of the load will fall on the horizontal parts of the belt.
Knot b
- -100 + S 1 \u003d 0
- S 1 \u003d 100 kg
Next, we move on to the extreme lower corner node, in which the vertical and horizontal segments of the chord converge, as well as the inclined brace. The force acting on the vertical segment, calculated in the previous paragraph, is the pressing weight and the reaction of the support. The force acting on the inclined element is calculated from the projection of the axis of this element onto the vertical axis: subtract the action of gravity from the support reaction, then divide the "pure" result by the sin of the angle at which the brace is inclined to the horizontal. The load on a horizontal element is also found by projection, but already on the horizontal axis. We multiply the just obtained load on the inclined element by the cos of the angle of inclination of the brace and obtain the value of the impact on the extreme horizontal segment of the chord.
Knot a
- -100 + 400 - sin (33.69) S 3 \u003d 0 - equilibrium equation per axis at
- S 3 \u003d 300 / sin (33.69) \u003d 540.83 kg - rod 3 compressed
- -S 3 cos (33.69) + S 4 \u003d 0 - equilibrium equation per axis x
- S 4 \u003d 540.83 cos (33.69) \u003d 450 kg - rod 4 stretched
Thus, successively passing from node to node, it is necessary to calculate the forces acting in each of them. Note that the oppositely directed action vectors compress the bar and, vice versa, stretch it if directed oppositely from each other.
Determination of the section of elements
When all the acting loads are known for the truss, it is time to determine the section of the elements. It does not have to be equal for all parts: the belt is traditionally made of rolled products with a larger section than the filling parts. This ensures a safety margin of the design.
where: F tr - cross-sectional area of \u200b\u200bthe stretched part; N - effort from the design loads; R y γ with
If everything is relatively simple with breaking loads for steel parts, then the calculation of compressed bars is not made for strength, but for stability, since the final result is quantitatively less and, accordingly, is considered a critical value. It can be calculated using an online calculator, or it can be done manually, having previously determined the length reduction factor, which determines at what part of the total length the rod is capable of bending. This coefficient depends on the method of fastening the edges of the bar: for butt welding it is a unit, and in the presence of “ideally” rigid gussets it can approach 0.5.
where: F tr - cross-sectional area of \u200b\u200bthe compressed part; N - effort from the design loads; φ - coefficient of longitudinal bending of compressed elements (determined from the table); R y - design resistance of the material; γ with - coefficient of working conditions.
You also need to know the minimum radius of gyration, defined as the square root of the quotient of dividing the axial moment of inertia by the cross-sectional area. The axial moment is determined by the shape and symmetry of the section; it is better to take this value from the table.
where: i x - radius of inertia of the section; J x - axial moment of inertia; F tr - cross-sectional area.
Thus, if you divide the length (taking into account the coefficient of reduction) by the minimum radius of gyration, you can get a quantitative value of the flexibility. For a stable bar, the condition is met that the quotient from dividing the load by the cross-sectional area should not be less than the product of the permissible compressive load and the buckling coefficient, which is determined by the value of the flexibility of a particular bar and the material of its manufacture.
where: l x - estimated length in the plane of the truss; i x - the minimum radius of inertia of the section along the x axis; l y - estimated length from the plane of the truss; i y - the minimum radius of gyration of the section along the y-axis.
Please note that it is in the compressed bar stability analysis that the whole essence of the truss operation is displayed. If the cross-section of the element is insufficient, which does not allow ensuring its stability, we have the right to add thinner connections by changing the fastening system. This complicates the configuration of the truss, but allows for greater stability with less weight.
Manufacturing of parts for the farm
The accuracy of the assembly of the truss is extremely important, because we carried out all the calculations by the method of vector diagrams, and the vector, as you know, can only be absolutely straight. Therefore, the slightest stresses arising from curvatures due to improper fit of the elements will make the truss extremely unstable.
First you need to decide on the dimensions of the parts of the outer belt. If everything is quite simple with the lower beam, then to find the length of the upper one, you can use either the Pythagorean theorem, or the trigonometric ratio of sides and angles. The latter is preferable when working with materials such as angle steel and shaped tube. If the angle of the truss slope is known, it can be made as a correction when trimming the edges of parts. The right angles of the belt are connected by trimming at 45 °, inclined - by adding to the 45 ° the angle of inclination on one side of the joint and subtracting it from the other.
The filling details are cut out by analogy with the belt elements. The main catch is that a farm is a strictly unified product, and therefore precise detailing is required for its manufacture. As in the calculation of impacts, each element must be considered individually, determining the angles of convergence and, accordingly, the angles of undercut edges.
Farms are often made with radius ones. Such structures have a more complex calculation method, but greater structural strength, due to a more uniform perception of loads. It makes no sense to make filling elements with rounded elements, but this is quite applicable for belt parts. Typically, arched trusses consist of several segments that are connected at the convergence of the filling braces, which must be taken into account when designing.
Assembly with hardware or welding?
In conclusion, it would be nice to outline the practical difference between the methods of assembling a truss by welding and using detachable joints. To begin with, drilling holes for bolts or rivets in the body of an element practically does not affect its flexibility, and therefore is not taken into account in practice.
When it came to the method of fastening the elements of the truss, we found that in the presence of gussets, the length of the section of the rod that can bend is significantly reduced, due to which its cross-section can be reduced. This is the advantage of assembling the truss on gussets, which are attached to the side of the truss elements. In this case, there is no particular difference in the assembly method: the length of the welds will be guaranteed to be sufficient to withstand the concentrated stresses in the nodes.
If the truss is assembled by joining elements without kerchiefs, special skills are needed here. The strength of the entire truss is determined by its least strong knot, and therefore a defect in welding at least one of the elements can lead to the destruction of the entire structure. If you do not have enough welding skills, it is recommended to assemble with bolts or rivets using clamps, angle brackets or cover plates. In this case, the fastening of each element to the node must be carried out at least at two points.
The forerunner for the construction of a stationary shed are calculations. The calculation of the canopy is necessary for the structure to be reliable, to withstand its own weight, as well as the loads created by wind and snow. Within the framework of this publication, we will only talk about the drawing and calculations of various parts of the structure using the example of a polycarbonate car canopy. The entire package of project documentation is much larger and a separate article will be devoted to it.
What do you need to keep in mind when preparing a project?
Before making a drawing of a polycarbonate canopy, you need to decide on the general design and design concept, namely, how the structure will look, what shape it will have, and what it will be intended for. Next, you need to draw a sketch of the structure, where you indicate the overall dimensions of the polycarbonate canopy (length, width and other parameters) and its main elements. At the next stage, you can prepare a drawing of a carport made of polycarbonate, while you need to remember.
For your information! When preparing a drawing of a structure, it is necessary to find and attach to it technical data on the materials used.
We calculate an arch-type farm
We have a sketch of a large carport made of metal, designed for 2 cars with an arched roof (arc) covered with sheets of cellular polycarbonate. The width of the canopy from support to support is 5.8 meters, the width of the arched truss (arc) should be 6 m. Let's calculate the cross-section of the profile that will be used in the manufacture of the arched floor.
ɒ pr \u003d (ɒ 2 + 4t 2) 0.5 ≥R / 2, we decipher this formula:
- ɒ - standard voltage;
- R - iron strength C235, about 2440 kgf / cm 2;
- t - tangential stress.
Now, consistently selecting the indicators, we can calculate the profile of a suitable section so that it can withstand the required loads. We take a square shaped pipe 30x30x3.5 mm with a cross section of 35 mm 2 with a moment of inertia of 3.98 cm 4, a load coupling factor of 0.5, the estimated load on the locking part of the arch is 914.82 kgf.
All the necessary data for the calculation has been collected, the formula is there, now it remains to substitute the data into the formula and obtain the calculation of the load on the arched truss (arc) of a polycarbonate car canopy.
ɒ pr \u003d ((914.82 / 3.5) 2 +4 (919.1 * 1.854 / ((0.35 + 0.35) 3.98) 2) 0.5 \u003d 1250.96 kg / cm 2 ...
What does it mean? And this means that if we weld or twist a six-meter arch from a 30x30x3.5 mm profile, it will fully withstand its own weight and the weight of the roofing material, that is, cellular polycarbonate. There is even a decent stock.
We calculate the supporting part of the structure
Next, you need to calculate what will be the supports for the carport made of polycarbonate. There is a special technique by which it is customary to calculate steel columns; without it, an adequate calculation of the canopy is impossible. Let's apply the formula:
F \u003d N / ϕR y. Let's decipher the formula:
- F is a cross-section of a square pipe that can be used as a support;
- ϕ - coefficient determining buckling;
- R y - the value of the resistance of the material.
In order to make calculations, you will have to find data on the strength of materials. In our case, the resistance of steel square pipes 70x70, 80x80, 100x100 mm, the values \u200b\u200bfound will need to be compared with the results of the calculations and draw conclusions. We make calculations:
F \u003d 3000 / (0.599 * 2050)
As a result, we get a value of 2.44 cm 2, which must be rounded up. As a result, the value on which we should rely when searching for a suitable profile of 2.5 cm 2. These indicators correspond to a square steel pipe 70x70x2 mm, there is even a small margin.
Roof loads from snow and wind
It is possible to answer the question of how to calculate a carport for a car only if you calculate the load-bearing structures of the structure and the load on the roof from snow and wind. With the calculation of supporting structures, we figured out in general terms. Now we need to solve the problem with the loads from wind and snow.
To obtain the data needed for the calculation, you need to refer to the average wind and snow load in your area. You can find such information in the corresponding SNiP.
For example, let's take a wind load value of 23kg / m2. But in our case, this value will not work because 23kg / m 2 is determined for buildings and structures that have walls. A carport has supports, arcs, lintels, a purlin and a roof, so pressure will only be exerted on these. We determine the average wind effect on the canopy, we get 0.34 with a support height of more than three meters, a value from 0.34 to 0.75 kg / m 2. We calculate the maximum wind load on the entire structure: arches, supports, girder, roof.
W m \u003d 23 * 0.75 * 0.34. The result is a value of 5.9. Now let's calculate the load created by the snow cover. These loads differ in different regions of the country, and they differ significantly. In mountainous areas, such a load can be more than 600 kg / m2, but we will take as an example a more modest figure of 180 kg / m2 (Moscow region).
To calculate the maximum load on the canopy, you need to multiply 180 by the value of the transition coefficient, which has yet to be obtained. The figure below shows the calculation of the snow load on the canopy.
The maximum snow load on the canopy was calculated. Now it remains for us to find out the inertia index for the roofing material we have chosen. Such data cannot be found in the usual commercial description of the material, but in the technical description it is. For example, cellular polycarbonate with a thickness of 12 mm has an inertia of 3.41 cm 4. Find a material with a calculated value or more and you can safely let it on the roof of a car shed. You can read more about what you can make a roof for a canopy in the article.
In conclusion, we note that the structures of carports are not so complicated, nevertheless, one cannot relate freely to the construction of such structures. First, the general structure of the canopy must be drawn on a sketch, indicating the length of the structural elements, their diameter and other simple parameters. After that, you can start calculating and making a drawing. In the process, you will have to calculate the parameters of the arched truss (arc) and much more. If you feel that this work is beyond your strength, contact a specialist. Good luck!
To figure out how to calculate a polycarbonate canopy, you need to clearly imagine the structure and draw up a plan or drawing of the building. By and large, polycarbonate panels are just a covering that determines the total area, but besides this, there are also racks and a rafter system. In addition, the required materials will include joining, corner and end profiles, fastening material and (possibly) lighting. It is important to calculate every detail in order to obtain a strong and durable structure.
What parameters to take into account when calculating polycarbonate for a canopy
Folded roof in the garden
Please note that polycarbonate is much more durable than glass (200 times), plastic and polyvinyl chloride. But not all panels can be bent, so their structure should be taken into account (sheets with triangular cells are not bent).
The choice of polycarbonate by thickness
First of all, in order to calculate the polycarbonate canopy, you need to take into account the possible mechanical load (snow, wind), on which the thickness of the panels depends. For monolithic panels, the thickness is 2, 3, 4, 5, 6. 8, 10 and 12 mm, they are called "vandal-proof", since the sheets are difficult to break mechanically.
The difference in the structure of cellular polycarbonate
The honeycomb structure implies not only the thickness, but also the configuration of the cell:
- SX is a five-layer 25mm sheet with angled stiffeners. The thickness can also be 32 mm. Triangular mesh panels are not suitable for curved roofs;
- SW - the sheet also consists of five layers, only the honeycomb looks like a rectangle (the edges are arranged vertically). The thickness is 16 to 20 mm;
- 3X - the sheet has 3 layers, the thickness is 16 mm, and the stiffeners are adjustable in density:
- 3H - made of 3 layers with a rectangular structure. The panel is produced in 6, 8 and 10 mm;
- 2H is the simplest sheet with square cells. Sheets are made in 4, 6, 8 and 10 mm.
Solid standard polycarbonate sheet
The thickness of the honeycomb polycarbonate changes only by 2 mm. That is, if the thinnest cellular sheet is 4 mm, and the thickest is 32 mm, then all intermediate sizes will be multiples of two.
Dimensions of polycarbonate sheet around the perimeter
The standard calculation of a monolithic polycarbonate canopy is made according to the dimensions of 3050 × 2050 mm. If you wish, you can agree with the manufacturer to change the perimeter of the panel, but a special order is usually more expensive.
Standard size cellular polycarbonate
Standards for cellular polycarbonate vary in two parameters, they are 210 × 600 cm and 210 × 1200 cm. Long sheets are useful for wide awnings, for example, in collective car parks with curved roofs, where joints are made only along the side edges. Also, by order at the factory, they cut from 1 m to 9 m, but this is only for colored panels.
There is also a profiled sheet, where the thickness does not exceed 1.2 mm, but, thanks to the wave, the height of which reaches 5 cm, the strength increases and precipitation is easily carried out. The standard width is 126 cm and the length is 224 cm.
Profiled (corrugated) polycarbonate sheets
Calculation of materials by types of canopies and types of roofs
To calculate a canopy made of corrugated board, polycarbonate or any other material, you need to take into account the configuration of the roof and the type of supporting frame. Such awnings are made of three types - single-slope, gable and bent (oval). The most difficult is the bent type, but the whole problem lies only in manufacturing, not in operation.
Shed sheds with an adjoining house
In cases where one side of the frame is held on the wall of the house, the calculation of the canopy from a rectangular pipe will be minus half of the vertical supports. That is, one side of the batten is held against the wall of the building. In any case, there must be a profile at the joints of the sheets, therefore, the distance between them is kept 126 cm, 210 cm or 205 cm, but this does not mean that the entire crate consists only of these profiles.
One side is attached to the wall of the house
In any case, the width of the roof must correspond to the parameters of the car and it is not less than 3 m in order to have a free passage. But such a length of the profile will cause its deformation (deflection), and this should be avoided, therefore, a rafter system will have to be made for the canopy.
When calculating the canopy to the house, 6 vertical supports will be needed - only on one side, but if the structure is autonomous, then twice as many risers will be required - 12 pieces. The principle here is as follows - for each rafter leg, supports should be installed on both sides, but if one side is attached to the building, then risers are not needed there.
In addition, beams are installed along the length, and for a 6-meter width they will need 6 pieces - 2 at the edges of the overhangs, 2 along the pillars and 2 in the middle of the roof. If the canopy length is 10.5 m, then 10.5 * 6 \u003d 63 m or 63/6 \u003d 11 pieces of profiles. The ends of the cellular polycarbonate are muffled with an end profile.
Dimensional drawing for a lean-to building
Free-standing canopy calculations
To calculate a canopy in the yard, one should take into account not only its width and length, but also the amount of precipitation that falls in winter. The fact is that the snow exerts a strong mechanical load and it will have to be contained in some way. The most optimal option for stiffening the frame is a triangle - this is the only geometric figure that does not provide for backlash.
For calculations, take the conditional width of the roof 6 m, length 10.6 m and polycarbonate with a width of 2100 × 600 mm. The rafters can be made from a pipe profile of 60 × 40 mm or from a wooden board 100 × 50 mm. Of course, a metal profile is better than wood and its service life has practically no restrictions in the foreseeable future.
The principle of the rafter structure
The drawing above shows a structure where the upper part of the ramp is 240 cm, and the rafter device consists of 11 triangles - this is the best option. Given the fact that metal profiles are usually 6 m in length, the width will turn out to be slightly smaller, but for each rafter leg 6 profiles are required, taking into account the vertical and inclined lintels. In total, you will need 6 rafters and 5 sheets of polycarbonate.
Of course, you can save on metal and make only 2 triangles, as shown in the top photo. In this case, the calculation of the canopy frame will be reduced by at least 2 profiles for each rafter leg, but if there are 6 of them, then this is already 12 profiles. However, for an average amount of precipitation, this is quite enough - it is possible to calculate a lean-to canopy in a budget mode, saving on metal.
Single-slope freestanding design
Gable carports
For gable roofs, the calculation of the metal frame of the canopy is very similar to the pitched roofs, that is, the rigidity is created by the same triangles. Such awnings are usually made for large parking lots, the width of which exceeds 6 m, that is, there is an opportunity for parking several cars or buses.
The principle of installing polycarbonate does not change - at each joint there must be a profile, and in this case, these are rafter legs. The number of triangles directly affects the rigidity of the structure - the more, the better. The most optimal option is as follows - each running meter is divided by a vertical profile, and this figure is divided diagonally into two triangles.
The principle of installing a gable canopy
To calculate the metal canopy, you need to immediately determine the dimensions of the roof, and for example, you can consider the same option 10.6 × 6 m.To cover here, you will also need 5 sheets, but they will have to be cut in half, connecting them in the center with a ridge profile. The number of metal vertical supports is twice the number of rafters, if there are 6 of them, then 12 risers will be required.
More longitudinal beams are needed here - 7 pieces - a ridge beam is added. Total:
- 2 profiles at the edges of the overhangs;
- 2 along the pillars;
- 2 between the supports and the ridge;
- 1 - on the skate.
Diagram of a gable building
If we translate the longitudinal beams into pieces, then 10.5 * 7/6 \u003d 12.25 or 13 six-meter profiles. The section for such beams is the same as for rafters (usually 60 × 40 mm), but for risers, a pipe of 80-100 mm or a pipe profile of a similar section is used.
The advantage for a gable roof is that the calculation of the metal structures of the canopy will turn out to be more economical. Two rafter legs with a jumper already form a triangle, which can be divided into two parts in the middle. As a result, two figures with horizontal (lower) sides of 3 m will be learned.
Calculation of materials for a curved canopy
It is more difficult to calculate a canopy with a curved roof on your own, since here much depends on its convexity, that is, the steeper the bend, the more materials are consumed. But you can proceed from the same dimensions: 10.5 m long and 6 m wide, although the width here will be reduced due to bending.
Curved carport
A clear advantage of this design is the saving of material when assembling the rafter system. For a given size, you can do with only two or three rafter systems, along the edges and in the middle - all the other legs are simply made in the form of an arc without a lower lintel, as in the photo. A curved metal profile, fixed on two supports, in itself is a rigid figure and the only question here is a good fastening of the risers.
In this case, the calculation of the carport will consist of 6 bent six-meter profiles, two or three of which are supplied with a jumper and are divided into several triangles. Supports are also required for each arc, which means there will be 12 of them. 6 longitudinal beams are enough:
- 2 at the edges of the overhangs;
- 2 along the pillars;
- 2 along the roof.
Arched canopy
In total, you get 12 * 10.5 / 6 \u003d 21 and 4 more profiles for the jumpers.
It is quite natural that for narrower canopies less material is consumed, but here it is important to take into account the length of the polycarbonate. That is, if you work with 6-meter sheets, then they should be used either whole or cut in half so that there is no waste. In this case, the roof will turn out to be 6 m or 3 m wide, and the length is already adjusted as necessary.
As a result, we can say that the most economical calculation of the canopy will turn out with a curved type roof, although this is the most difficult option. Nevertheless, in such designs, it is possible to save on metal profiles, so the benefits are obvious here.
If difficulties arise in the settlement process, you can use special programs and services of professionals.
To arrange a canopy over the porch, recreation area, playground or parking lot, structures made of shaped pipes are often used. Frameworks made of professional pipes are easy to manufacture, aesthetic and suitable for the construction of small architectural forms for various purposes and sizes. For the device of a durable and reliable canopy with your own hands, it is necessary to correctly calculate the configuration of the frame.
A canopy is a small architectural form, part of a covered area that has a functional or decorative purpose. The main function of this structure is to protect the site from precipitation and the sun.
The metal profile compares favorably with other materials in resistance to biological, chemical, mechanical stress. Square or rectangular cross-sections are easy to design and install. They are equally well suited for the installation of a small canopy for the porch and the device of a spacious covered area.
The canopy consists of:
- foundation;
- supports or suspensions - vertical and inclined load-bearing elements of the frame;
- side connectors - slopes and trusses;
- rafter system, consisting of rafter legs, trusses, lathing;
- roofs.
In the case of erecting a canopy over a small area, for example, a porch or a sandbox, you can do without farms - the structure will withstand snow and wind, since there will be no long horizontal sections in it.
If it is planned to equip a canopy over a parking place or a swimming pool, the horizontal floors and rafter legs will be of greater length. Such extended elements are vulnerable to stress. To distribute it and give the structure rigidity, in these areas, not separate pipes are used, but trusses (reinforced structural parts consisting of two pipes and vertical and inclined elements connecting them - racks and braces).
The foundation for the canopy is most often made columnar, since its equipment does not require large excavation work. The roof is made of polycarbonate, profiled sheet, ondulin and other sheet materials.
Scope of application
The ease of installation and reliability of the structure provided the canopies from the profile pipe with a wide range of applications - they are erected for shading and protection from rain and snow:
- parking lots,
- children's and sports grounds,
- individual game and sports equipment,
- summer cafes,
- recreation areas,
- entrance groups of residential buildings and non-residential buildings,
- places for storing equipment and implements in personal plots.
Types of outbuildings
Sheds from a professional pipe are classified according to the type of support and rafter systems.
The type of support system depends on the location of the canopy relative to other buildings. Separate attached, console and free-standing structures.
A canopy made of a profile pipe is a very common design that can be found in almost every yard. From shaped pipes, you can make both a small canopy over the porch and a large roof for a parking lot - and the structure in any case will be quite strong, beautiful and easy to equip. This article will consider the calculation of a canopy from a profile pipe and its installation.
Calculation and drawing of the canopy
Competent calculation and creation of a good drawing imply compliance with a number of standards and requirements for structures made of shaped pipes. However, small single-pitched sheds do not need to be calculated so precisely - a small visor from a profile pipe does not differ in large weight, so this kind of structure does not pose any danger. Large carports or swimming pools must be sized to avoid problems.
A drawing of a canopy from a professional pipe always begins with a sketch - a simple outline, which indicates the type of structure, its main features and approximate dimensions. In order to accurately determine the size of the future canopy, it is worth taking measurements on the site where the structure will be located. In the event that the canopy will be attached to the house, then it is also necessary to measure the wall in order to accurately know the dimensions of the profile pipe for the canopy.
You can consider the calculation method using the example of a structure located on a 9x7 m site located in front of a house with dimensions of 9x6 m:
- The length of the canopy may well be equal to the length of the wall (9 m), and the outreach of the structure is one meter shorter than the width of the site - i.e. 6 m;
- The lower edge may well have a height of 2.4 m, and the high one should be raised to 3.5-3.6 m;
- The angle of inclination of the slope is determined depending on the difference in the heights of the lower and upper edges (in this example, it turns out about 12-13 degrees);
- To calculate the loads on the structure, you need to find maps showing the level of atmospheric precipitation in a given region, and build on them;
- When the size of the structure and the expected loads are calculated, it remains to draw up a detailed drawing, select materials and start assembling the canopy.
Drawings of trusses from a profile pipe for a canopy should be displayed separately with all the details. It is also worth remembering that the minimum canopy slope is 6 degrees, and the optimal value is 8 degrees. Too little slope will prevent the snow from sliding on its own.
Having finished with the drawings, the appropriate material and its quantity are selected. The calculation must be carried out accurately, and before purchasing it is worth adding about 5% of the tolerance - during work very often small losses occur, and marriage is not uncommon. According to similar calculations, it is possible to make a garage frame from a profile pipe, which is quite in demand.
Creating a canopy from a profile pipe
The design of the canopy is not particularly complicated. If the drawing of the canopy and the materials necessary for its assembly are already there, then you can proceed directly to the arrangement of the structure.
The production of a canopy from a profile pipe is carried out according to the following algorithm:
- First, a plot for a canopy is marked and prepared. You need to find a place for the foundation holes and dig them up, and then fill the bottom of all the holes with rubble. Embedded elements are installed in the pits, after which the foundation is poured with cement mortar.
- Steel parts of a square shape are welded to the lower parts of the canopy posts, the size of which coincides with the dimensions of the embedded parts, as well as the diameter of the bolt holes. When the solution hardens, the posts for the canopy from the profile pipe are screwed to the embedded parts.
- The next step is to assemble the frame. The profile pipe at this stage is marked and cut into the necessary pieces, and only after that trusses can be made from the profile pipe for the shed. First, the side trusses are fastened with bolts, then the front lintels, and the latter, if necessary, equip the diagonal grids. The assembled frame is installed on the racks and fixed in the chosen way.
Before installing the roof, the canopy must be painted or coated with an anti-corrosion compound to prevent possible destruction of the material - during assembly, the base coat is damaged and metal parts as a result lose their resistance to corrosion. In addition, you need to understand that external processing does not protect the structure from destruction from the inside, so the edges of the pipes must be closed with plugs.
Types of fastenings for canopy elements and their sizes
Various methods can be used to assemble canopy elements from a profile pipe:
- One of the most common ways of fixing canopies from professional pipes is a bolted connection. The quality of such a connection is quite high, while it does not differ in complexity. For work, you will need a drill with a drill for metal, as well as bolts or screws, the diameter of which depends on the section of the pipe.
- Another way that canopy elements are attached is a welded connection. Welding work requires certain skills, and equipment will be more expensive than for a bolted connection. However, the result is worth it - welding provides high structural strength without weakening it.
- For fixing small canopies made of pipes with a diameter of up to 25 mm, you can use the crab system, which is a special clamps of various shapes (in more detail: ""). Most often, when installing canopies, T-shaped and X-shaped clamps are used, ensuring the connection of three or four pipes, respectively. To tie the clamps, bolts with corresponding nuts are required, which often have to be purchased separately. The main disadvantage of crab systems is the ability to assemble the structure only at a 90-degree angle.
The choice of shaped pipes for the manufacture of trusses
When choosing pipes for arranging a large-sized canopy from a profile pipe, it is necessary to study the following standards:
- SNiP 01.07-85, which describes the relationship between the degree of loads and the weight of the constituent elements of the structure;
- SNiP P-23-81, describing the method of working with steel parts.
You can consider the arrangement of the structure using the example of a wall shed with dimensions of 4.7x9 m, resting on the outer posts in front, and attached to the building at the back. When choosing the angle of inclination, it is best to stop at an 8-degree indicator. By examining the standards, you can find out the level of snow load in the region. In this example, a pitched roof made of a shaped pipe will be subjected to a load of 84 kg / m2.
One 2.2-meter post made of a profile pipe has a weight of about 150 kg, and the degree of load on it is about 1.1 tons. Given the degree of load, you will have to select strong pipes - a standard round shaped pipe with 3 mm walls and a diameter of 43 mm will not work here. The minimum dimensions for a round pipe should be 50 mm (diameter) and 4 mm (wall thickness). If a pipe with a diameter of 45 mm and a wall thickness of 4 mm is used as a material. Using such material, a gate can also be made from a profile pipe with your own hands, which will be quite reliable and durable.
When choosing trusses, it is worth stopping at a structure of two parallel contours with a diagonal lattice. For a truss with a height of 40 cm, you can use a square profile pipe with a diameter of 35 mm and a wall thickness of 4 mm (read also: ""). Pipes with a diameter of 25 mm and a wall thickness of 3 mm are well suited for the manufacture of diagonal grids.
Conclusion
Assembling a canopy from a professional pipe with your own hands is not so difficult. For successful work, it is necessary to correctly design the future structure and take a responsible approach to each stage of the project implementation - and then the result will be a reliable structure that can stand for many years.