How to calculate a triangular ray

How to calculate the proper farms for canopies: drawing and assembly rules. Example of calculating a triangular farm calculation of a metal farm from an online corner calculator

Definition of Domestic Agricultural Efforts

We often do not have an opportunity to apply the usual beam to one structure or another, and we have to apply a more complex design called a farm.
However, although it is different from calculating the bar, we are not difficult to calculate. All you need is attention, an initial knowledge of algebra and geometry, and two hours of free time.
So let's get started. Before calculating the farm, let's ask a real situation that you might face. For example, you need to overlap the garage of 6 meters wide and 9 meters long, but no plate overlaid, no beams you have. Only metal corners of different profiles. Here we collect our farm!
After the farm is based on runs and professional grounds. Removing the farm on the walls of the garage - hinged.

To begin with, you need to know all the geometric dimensions and corners of your farm. This is where we need our mathematics, namely geometry. Corners are found with the cosine theorem.

Then you need to collect all the loads on your farm (you can see in the article). You can have the following download option:

Next we need to number all the elements, number the farm nodes, and set the support responses (the elements are signed in green, and the nodes are blue).

To find our reactions, write the equilibrium equation with the equilibrium equation of moments with respect to node 2.

RA + RB-100-200-200-200-100 \ u003d 0;
200 * 1.5 + 200 * 3 + 200 * 4.5 + 100 * 6-rb * 6 \ u003d 0;

From the second equation we find the RB reference reaction:

RB \ u003d (200 * 1.5 + 200 * 3 + 200 * 4.5 + 100 * 6) / 6;
Rb \ u003d 400 kg

Knowing that RB \ u003d 400 kg, from the 1st equation, we find RA:

Ra \ u003d 100 + 200 + 200 + 200 + 100-RB;
Ra \ u003d 800-400 \ u003d 400 kg;

Now that the supporting reactions are known, we need to find a node in which only a few of the unknown values ​​(each numbered element is an unknown value). From this point on we begin to divide the yard for individual nodes and find the internal efforts of agricultural smoke in each of these nodes. On these inner efforts we will pick up the cross sections of our rods.

When it happens that the efforts in the tail are directed from the center, our tail tries to straighten (return to the original position), which means that it is compressed. And when the trunk efforts are directed towards the center, the rod tries to grab, that is, it is stretched.

So let's turn to the calculation. In node 1 there are only 2 unknown values, so we consider this node (the instructions of effort S1 and S2 that we give from your deliberations, in any case we will turn out to be duly).

Consider the equations of equilibrium on the X and Y axes.

S2 * sin82.41 \ u003d 0; - on the x-axis
-100 + S1 \ u003d 0; - on the y-axis

From the first equation it can be seen that S2 \ u003d 0, that is, the second rod is not loaded with us!
From the 2nd equation it can be seen that S1 \ u003d 100 kg.

Since the value of S1 we turned out to be positive, then decided the direction of the effort, which we decided correctly! If it was negative, the direction should be changed and the sign will be changed to "+".

We can know the direction of the effort S1, we can imagine that it is the first rod.

Since one force has been sent to the node (node ​​1), the second force is directed to the node (node ​​2). So our bar is trying to stretch, which means it is compressed.
Next we consider node 2. It had 3 unknown values, but since we have already found the value and direction S1, only 2 unknown values ​​remain.

Yet again

100 + 400 - sin33.69 * s3 \ u003d 0 - on the axis
- S3 * cos33.69 + s4 \ u003d 0 - on the x-axis

The 1st equation S3 \ u003d 540.83 kg (squeeze handle 3).
From the 2nd equation S4 \ u003d 450 kg (rod No. 4 stretched).
Consider the 8th node:

We create equations on the X and Y axes:

100 + S13 \ u003d 0 - on the axis
-S11 * cos7.59 \ u003d 0 - on the x-axis

From here:

S13 \ u003d 100 kg (STEM №13 compressed)
S11 \ u003d 0 (zero bar, no effort in it)

Consider the 7th node:

We create equations on the X and Y axes:

100 + 400 - S12 * sin21.8 \ u003d 0 - on the axis
S12 * cos21.8 - S10 \ u003d 0 - on the x-axis

From the 1st equations we find S12:

S12 \ u003d 807.82 kg (stem No. 12 compressed)

From the 2nd equations we find S10:

S10 \ u003d 750.05 kg (rod No. 10 stretched)

Next, look at node number 3. As far as we can remember the 2nd rod, we have zero, so it won't draw.

Equations on the X and Y axes:

200 + 540.83 * sin33.69 - S5 * cos56.31 + S6 * sin7.59 \ u003d 0 - on the y-axis
540.83 * cos33.69 - S6 * cos7.59 + s5 * sin56.31 \ u003d 0 - on the x-axis

And here we need an algebra. I will not describe in detail the methodology to find unknown values, but the essence is such - from the 1st equation Express S5 and replace it in the 2nd equation.
After the result we get:

S5 \ u003d 360.56 kg (rod No. 5 stretched)
S6 \ u003d 756.64 kg (rod No. 6 compressed)

Consider node number 6:

We create equations on the X and Y axes:

200 - S8 * sin7.59 + s9 * sin21.8 + 807.82 * sin21.8 \ u003d 0 - on the axis
S8 * cos7.59 + s9 * cos21.8 - 807.82 * cos21.8 \ u003d 0 - on the x-axis

As in the 3rd node, we find our unknown.

S8 \ u003d 756.64 kg (shaft No. 8 compressed)
S9 \ u003d 0 kg (rod No. 9 zero)

Consider node number 5:

Make an equation:

200 + S7 - 756.64 * sin7.59 + 756.64 * sin7.59 \ u003d 0 - on the axis
756.64 * cos7.59 - 756.64 * cos7.59 \ u003d 0 - on the x-axis

From the 1st equations we find S7:

S7 \ u003d 200 kg (stem No. 7 compressed)

As an inspection of our calculations, let's consider the 4th knot (efforts in the bar # 9 no):

We create equations on the X and Y axes:

200 + 360.56 * sin33.69 \ u003d 0 - on the axis
-360.56 * cos33.69 - 450 + 750.05 \ u003d 0 - on the x-axis

In the 1st equation it appears:

In the 2nd equation:

This error is permissible and is most likely with the angles (2 characters after the comma instead of 3-ex).
After the result we get the following values:

I decided to check all of our calculations in the program and got exactly the same values:

Selection of the cross-section of the agricultural elements

To the farm bill made of metal After all the internal efforts are found in the rods, we can proceed to the selection of the cross-section of our rods.
For convenience, all values ​​are in the table.

2.6.1. General concepts.

A flat rod system that after the hinges are turned on, all the nodes that turn on all the nodes remain geometrically immutable, called a farm.

Examples of farms are shown in Fig. 2.37 ..

In real bar structures, which are suitable for defining the "farm", the bars in the nodes are not connected by hinges, but bars, rivets, welding or evil (in reinforced concrete structures). In the calculated schemes, however, such structures can be entered in hinge nodes, but this is provided

· Tails are perfectly straight;

· The bar axis crosses in the middle of the node;

· Focused forces are only applied to nodes;

· The dimensions of the cross-sections of the bars are significantly smaller than their length.

Fig.2.37 .. Static, definable flat farms.

Under these conditions the agricultural rods only work when they are stretched or compressed, only longitudinal forces arise in them..

This fact greatly simplifies the calculation of the rod system and enables you to obtain results with a sufficient degree of accuracy.

To determine the efforts in the farm staffs by the departmental method, it is necessary:

1) Cross section to do that

· Crossed the axis of the rod, which determines the effort;

· Crossed as no more than three poles;

· Common farm in two parts.

2) Longitudinal efforts in the bars directly in the positive direction, i.e. H. from the knot.

3) Select such equilibrium equations for a part of the farm that would only include a desired effort. Such equations are for example

· The sum of the moments relative to the point at which the efforts in the rods of the fumes cut into the cross-section cuts into the cross-section; such points are invoked;

· The amount of protrusions for the vertical axis for dividing farms with parallel belts.

4) To determine the effort in the racks, cut knots if there are no more than three bars.

5) To simplify the definition of the shoulders of the internal force relative to the moment, the point in time in making the equations of the moments is, if necessary, to replace the desired efforts with their protrusions on the single vertical axis.

2.6.2. Determination of the efforts in the rods of the farm.

To determine efforts in the agricultural rods it is necessary:

· Determine the reactions of the carriers;

· Determine the effort required;

· Review the results obtained.

The responses of carriers in simple jet farms shown in Fig. 27 are also determined as in frames with equations of species

To check the reactions of the carriers we use the equation

Consider the calculation algorithm on a specific example.

There is a calculated farm scheme (Fig.2.38).

It is necessary to determine the efforts in bars 4-6, 3-6, 3-5, 3-4, 7-8.

The solution of the problem.

1) Determine the reactions of the support.

To do this, use the equilibrium equation:

Record equations with the adopted control rule:

Solve equations

Check the reactions of the carriers by equation.

2) Determine the efforts in the agricultural rods.

a) Effort in rods 4-6, 3-6, 3-5.

To determine the effort in the specified bars, cut the farm with a cross section a-A There are two parts and take into account the balance of the left part of the farm (Fig 2.39.

On the left side of the farm we apply the wearer's reaction, the force acting in node 4 and the desired efforts in the agricultural rods. These efforts are guided by the knot along the appropriate rods, that is, in a positive direction.

To find the effort, you can use the following system of equations:

In this case, however, we get a common system of equations in which all the desired efforts are included.

To simplify the solution to the problem, it is necessary to use the equilibrium equations in which only one would be unknown.

To find the effort is this equation

d. H. The sum of the moments in relation to node 3, in which the lines of action of the effort and, since the moments of these forces relative to node 3 are zero. For effort is this equation

d. H. The sum of the moments relative to node 6 crossing the lines of effort and.

To determine the effort, the equation of the sum of the moments against the point O, where the lines of effort and its, i.e. H.

In recording these equations, mathematical difficulties arise in determining the shoulders of the forces relative to the corresponding points. To simplify the solution to this problem, it is recommended to decompose the desired force along the X, Y axes and use the projection of the effort in plotting the equilibrium equation.

Let's show this for an example of an effort (Fig. 2.40).

We write the equation:

Solving the equation, we get:

In this example, the projection of the force on the X-axis has a moment around an equal zero point as its line passes through point O.

3) We determine the exertion in rod 3-4.

To determine the exertion, cut into node 4 of the cross-section of the farm b-b. (Fig. 2.41.A).

4) We determine the effort in the bar 7-8.

Cut into section 8 c-s. (Fig.2.41.b). We make two equations of equilibrium

To determine the effort, we have two equations with three unknowns. As a result, one of these unknowns (or) has to be determined beforehand.

Once the effort is known, the equation can be used to find the effort:

the amount of protrusions on the line of force perpendicular to the node at the node on the X-axis.

It should be noted that the efforts in the farm bars can be determined by alternately considering balance of its nodes and forming two equations for each node

It is necessary to start with a knot in which only two rods converge and then consistently consider the knots in which only two unknown efforts are inside. See example (Fig.2.42).

1) We consider node 1, in which only two bars converge. We make and solve the equation

2) We consider node 2, in which 3 bars converge, but the effort is known:

Solving the system of equations, we find:

Then node 4 and so on are considered.

This method of determining the efforts in the agricultural rods has the following disadvantages:

· The error made in the calculation process is applied to subsequent calculations.

· It is not rational to determine the efforts only in individual agricultural rods.

The advantages of the method relate to the possibility of using it when compiling programs for calculating computers.

2.6.3. Check the calculation results.

To check the results of the calculation, you need to use equilibrium equations that involve the greatest number of efforts. For example, for testing, (Fig.3.3) with such equations are

Metal farms from the profile pipe - metal structures, the arrangement of which is made by means of lattice metal bars. They are quite a complicated and arduous process to make, but the result usually lives up to expectations. An important dignity can also be called the economy of the resulting design. In the production process, steam metal and jams are often used as connecting metal parts. The further assembly process is based on riveting or welding.

Advantages of metal structures

The metal farm has many advantages. With their help, you can slightly overlap the span of the length. However, it is understood that the correct installation involves the primary competent calculation of the farm from the profile pipes. In this case, it is possible to be confident as the created metal structure. It is also worth sticking to the planned plans, drawing, and markup so that the product meets the requirements.

This benefit of the product doesn't end there. The following benefits can be assigned:

1. Shelf life of the metal product.
2. Light weight compared to other similar structures.
3. Endurance.
4. Resistance to damage and negative environmental factors.
5. Strong knots that help withstand all types of loads.
6. The ability to save finances by self-assembly as the finished metal product is expensive.
7. Farm structural features

The profile tube has distinctive features that should be remembered in advance. Based on the division, you can select certain parameters. The main value is the number of straps. The following types can be distinguished:

The second important parameter, without which the drawing of the farm will not work, is contours and shape. Depending on the latter, you can assign direct, double or one-sided, arched farms. The contour can also be used to divide metal structures into several options. The first is the designs with a parallel belt. They are considered the optimal solution for creating a soft roof. Metal support is extremely simple, and its components are identical, in size the grill coincides with the bars, making installation a simple operation.

The second option is individual metal structures. Rigid nodes in their base that provide resistance to external loads. The creation of such a design is characterized by the engineering of the material and, accordingly, in small editions. Third view - polygonal farms.They stand out for a long time and quite complicated installation, and the advantage becomes the ability to withstand a lot of weight. The fourth variant is the triangular farms of the profile pipe. They are used when it is planned to create a metal farm with a large angle of inclination, but a minus is the presence of waste after construction.

The next important parameter is the angle of inclination. Depending on the situation, metal farms made of profile tubes are divided into three main groups. Metal structures with an angle of inclination at 22-30 degrees fall into the first group. In this case, the length and height of the product are represented by a 1: 5 ratio. Among the advantages of such a metal structure can be distinguished by a small weight. Metal triangular farms are most commonly created.

At the same time, it may be necessary to use the dividers mounted from top to bottom if the height of the spans exceeds 14 meters. The upper belt is located 150-250 cm panel. As a result, a design with two straps and one even panel is obtained. Assuming that the span is more than 20 meters, it is necessary to assemble the subcentating metal structures, tying it to the supporting columns.

The second group includes farms of square pipes or of profdres and other species, if the angle of inclination is 15-22 degrees. The ratio of height and length to each other reach 1: 7. The maximum frame length should not exceed 20 meters. If there is a need to increase the height, then additional procedures will be required, for example, a broken belt is created.

The third group includes metal structures with an angle of inclination less than 15 degrees. A trapezoidal rafter system is used in these projects. They also have a short rack. This allows you to increase the opposition to the longitudinal deflection. When a single table roof is mounted, the angle of inclination is 6-10 degrees, it is necessary to think about an asymmetrical shape. The separation of the span can vary depending on the design features and can reach seven, eight or nine parts.

Separately assign the farm Polonso, assembled by hands. It is represented by two triangular farms connected by a tightening. This makes it possible to eliminate the need to install long divosers that would have to be in the middle panels. As a result, the weight of the design is optimal.

How to calculate the canopy

The calculation and manufacture of farms from the profile line should be based on the basic requirements that are written in Snip. When calculating, it is important to build up the drawing of the product, without the impossibility of subsequent installation. First of all, it should be ready to prepare a scheme that will indicate the basic dependencies between the slope of the roof and the construction length in general. In particular, the following should be taken into account:

1. Support contour bands. They will help determine the appointment of metal structures, the angle of inclination and the type of roof.
2. When choosing, it is necessary to follow the principle of savings, if the requirements do not assume the opposite.
3. The size calculation is done with the burdens of the design in mind. It is important to remember that the corners of the rafter may differ, but the panel must match them.
4. The last calculation concerns the gap between the nodes. Most often it is chosen to match the width of the panel.

It should be remembered that the height of the height leads to an increase in storage performance. In this case, the snow cover is not kept on the roof. In order to further improve the metal structures, it is necessary to install the rigid ribs. To find out the gabarits of the farm, it is necessary to be guided through such data:

• constructions up to 4.5 meters wide are assembled from parts of parts of 40x20x2 mm dimensions.
• 5.5 meter wide products are made from the components of 40x40x2 mm.
• if the width of the design exceeds 5.5 meters, optimally choose the parts 40x40x3 mm or 60x30x2 mm.

Next it is necessary to calculate the step, because this takes into account the distance from one to the next skeletal support. Often it is standard and reaches 1.7 meters. If it is broken, the structural strength may be a little broken. After all the necessary parameters are calculated, it is necessary to obtain a design scheme. To do this, use the program to achieve the required strength. Most programs have a name similar to the process they are running. You can choose the farm calculation program, "Farm Calculation 1.0" and similar.

When calculating the cost of a ton of metal in procurement, be sure to consider, as well as the cost of making most metal structures, that is, the cost of welding, anti-corrosion protection and installation. Now it remains to cook with the cooking of a farm from the profile pipes.

In order for the welding of farms to be of high quality, it is necessary to follow a number of recommendations. Among them are the following:

In order for the design to meet the requirements, it is important to adhere to a certain working algorithm. Originally do the marking of the site. For this, vertical supports and mortgage parts are installed. If necessary, metal profile pipes can be immediately placed in the pits and concrete. The installation of vertical supports is selected with a deletion to check the parallelism, stretch the cable.

There is an outdoor area with dimensions of 10x5 m near the house and I want to make this platform closed so that the summer could drink tea on the street, not on the weather conditions or rather loosely, but from a reliable canopy and so that you The car can be put under the carport to save the garage, and in fact was protection against solar heat on a summer day. That's only 10 meters - the span is a big one and a beam for such a span, it's heavy, and too massive this very beam is boring and generally reminiscent of the factory workshop. In such cases, the optimal option is to create beams instead of the farm, and then toss the crate on the farms and make the roof. Of course, the shape of a farm can be any, but then the calculation of the triangular farm will be considered as the simplest option. The problems of calculating the columns for a similar canopy are considered separately, the calculation of two or bolts, with which farms are described here, is also not given.

It is assumed that the farms are in 1 meter increments, and the load on the farm from the box is transferred only to the farm node. Roofing material serves professional flooring. The height of the farm can theoretically be any, that is only if it is a canopy, adjacent to the main building, the main delimiter is a roof shape if the building is one-story, or the second floor if the floors are more, but in in any case, the pawn height more than 1 m is hardly emphasized, but taking into account the fact that there is also a riglel between columns, then 0.8 m will not always come out (anyway we will use this figure for calculations). Based on these assumptions, a farm can be designed:

Figure 272.1. Overall preliminary farm cavity scheme.

Figure 272.1 Blue color shows the rays of the boxes, Blue - a farm that should be calculated, the calculated, liliable rays or farms, for the pillars are based on the fact that the color change from light blue to dark purple in this case an increase in the calculated load, the mean for darker structures means more powerful profiles are required. Farms in Figure 272.1 are shown in dark green due to a completely different nature of the load. Thus, the calculation of all design elements separately, like this:

The beams of the box (the beams of the boxes can be considered multiplet beams if the length of the beams is about 5 m, if the beams are about 1 m long, that is, between farms, then these are ordinary burglary beams on hinge supports)

Roof colors (it is enough to determine normal stresses in cross-sections of the bars over which it is below)

Bays or farms under roof colors (calculated as single break bars or farms)

pose no particular problems. However, the purpose of this article is to show an example of triangular farm calculation, which we are going to do. Figure 272.1, You can consider 6 triangular farms while the load is twice less on the extreme (front and rear) agricultural farm than the rest of the farms. This means that if there is a continuing desire to save material, these two farms should be counted separately. However, for aesthetic and technological considerations, it is better to make all farms identical, which means that it is enough to calculate everything just one farm (shown in blue in Fig. 272.1). In this case the farm will be a console; H. Farm supports are not located at the ends of the farm, but in the nodes shown in Figure 272.2. Such a design scheme allows you to distribute the load more evenly, which means that the profiles of a smaller cross-section for the manufacture of farms. For the manufacture of farms, it is planned to use square profile pipes of the same type and choose the desired cross-section of the profile pipe to calculate further.

When the beams of the boxes are based on the farms, the load from the carport from the professional flooring and the snow lying on that professional flooring can be considered concentrated at the farm nodes. The bars of the farm are welded together, while the bars of the upper belt are most likely the continuous length of about 5 feet. However, we assume that all agricultural nodes are hinged. These clarifications seem like a minor title, but they allow you to maximize and simplify the calculation of the calculation for the reasons set out in another article. The only thing left for us to do further calculations is to determine a focused load, but it is not difficult to do so when the baselations or the beams of the boxes have already been calculated. When calculating the professional flooring, we found that the sheets of professional flooring with a length of 5.1-5.3 m are a perennial continuous beam with a console. This means that the support reactions for such a beam, and, accordingly, the load for our yard will not be the same, but the changes in support reactions for 5 transit beams will not be so significant and simplify the calculations that can be assumed The load from snow, professional flooring and boxes are transferred evenly, as with single break bars. Such an assumption only leads to a small stock through strength. As a result, we get the following calculation scheme for our farm:

Figure 272.2.. Calculation scheme for a triangular farm.

Figure 272.2 A) presents the entire design scheme of our farm, which is the calculated load Q \ u003d 190 kgWhat follows from the calculated snow load of 180 kg / m 2, the weight of the wave net and the possible weight of the bars of the box. Figure 272.2 b) shows the sections thanks to what efforts can be calculated in all agricultural rods, taking into account the fact that the farm and agricultural load are symmetrical and it means that it is not enough to count on the agricultural rods, But a little more than half. And so that they are not confused in numerous poles when marking the calculation, rods and farm knots. The marking shown in Fig. 272.2 c) means that the farm:

Lower strap bars: 1-A, 1-B, 1-D, 1-F, 1-and;

Upper strap bars: 2-A, 3-B, 4-g, 5th, 6-s;

Draps: AA-B, B-B, V-G, Mr., Dr., E, ZH, Z - and.

When calculating each loading bar, it is desirable to make a table in which all the bars should be made. Then this table is useful for establishing the resulting compressive or tensile stresses.

Well, the calculation itself is not particularly difficult if the farm is welded from 1-2 kinds of closed section profiles. For example, the total computation of the operation can be reduced to compute efforts in bars 1- and 6-S and H. To do this, it is enough to take into account the longitudinal forces that are taken into account when cutting the part of the farm along the line IX-IX (Fig. 272.2 g).

But let's get sweetened on the third, and see how it's done on simpler examples than this consideration

I-I cross-section (Fig. 272.2.1 d)

If in the indicated way cut off the additional part of the farm, then you need to determine the efforts only in two rods of the farm. This uses static equilibrium equations. Since in the hinge farm nodes the value of the bending moments in the farm nodes is zero, based on the same conditions of static equilibrium, the value is the sum of all forces relative to the axis H. or axis w. Also equal to zero. This allows you to set up at least three equations of static equilibrium (two equations for power and one for moments), but in principle of the correspondence equations, there can be as many nodes in the farm and even more if you use the knight points. And these are those points where two of the forces under consideration and with a complex geometry of the knight's agricultural point do not always coincide with the agricultural nodes. In this case we still have a simple geometry (we still have time to get to the complex geometry) and therefore it is enough to determine the efforts in the rods. There are enough agricultural nodes available. Again, such points are usually used for considerations to facilitate the calculation, such points, the equation of the moments, relative to, in which you can immediately define an unknown effort effort without defining the case to the system of 3 equations too to solve.

It looks like that. If the equation of the moments is relative to point 3 (Fig. 272.2.2 d), then it has only two members, and one of them is already known:

M 3 \ u003d -q l./ 2 + n 2-a h \ u003d 0;

N 2-a h \ u003d ql / 2;

Where l. - Distance from point 3 to the application of the power Q / 2, which in this case is the shoulder of the force, according to the calculated scheme that we have adopted l.\ u003d 1.5 m.; H-shoulder strength N 2-a (The shoulder is in blue in Fig. 272.2.2 d).

In this case the third possible element of the equation is zero, since the force N 1-A (shown in gray in Fig. 272.2.2 2) directed along the axis passing through point 3, and then the shoulder of the Action is zero. The only thing we don't know about in this equation is the shoulder of the force of N 2-A, but determine that it easily possesses the relevant knowledge of geometry.

Our farm has an estimated height of 0.8 m and a total estimate of 10 m, then the tangent of the angle α will be tgα \ u003d 0.8 / 5 \ u003d 0.16, where the value of the angle α \ u003d arctgα \ u003d 9.09 o is. And then

h \ u003d. l.sin.α

Now we do not prevent anything to determine the value of the force N 2-a:

N 2-a \ u003d q l./ (2lsin.α ) \ u003d 190 / (2 · 0.158) \ u003d 601.32 kg

The value is determined in a similar way N 1-a. For this, the equation of moments with respect to point 2 is compiled:

M 2 \ u003d -q l./ 2 + n 1-a h \ u003d 0;

N 1-a h \ u003d q l./2

N 1-a \ u003d q / (2dailyα ) \ u003d 190 / (2 · 0.16) \ u003d 593.77 kg

To check the correctness of the calculations, we can consider the equation of forces:

Σq y \ u003d q / 2 - n 2-a sinα \ u003d 0; Q / 2 \ u003d 95 \ u003d 601.32 · 0.158 \ u003d 95 kg

Σq x \ u003d n 2-a cosα - N 1 - A \ u003d 0; N 1-A \ u003d 593.77 \ u003d 601.32 · 0.987 \ u003d 593.77 kg

The static equilibrium conditions are performed and any of the equations used to verify were possible to determine the efforts in the bars. Here, in fact, is everything, everything, the further calculation of the farm, the pure mechanics, but just in case it is still taken into account

section II-II (Fig. 272.2. E)

At first glance it seems that the rate equation relative to point 1 is easier to determine the force N a-bIn this case, however, it is necessary to determine the shoulder of the force in order to find the value of the angle β first. But if we consider the equilibrium of the system relative to point 3, then:

M 3 \ u003d -q l./ 2 - q l./ 3 + N 3-B H \ u003d 0;

N 3-b h \ u003d 5q l./6 ;

N 3-B \ u003d 5Q / (6Sinα ) \ u003d 5 · 190 / (6 · 0.158) \ u003d 1002.2 kg (Works on the line)

Well, now I'm still defining the value of the angle β. Based on the fact that all sides of a certain rectangular triangle are known (the lower katat or the length of the triangle is 1 m, the side katat or the height of the triangle is 0.16 m, hypotenuse - 1.012 m, and even angle α) then has the adjacent rectangular triangle with a height of 0.16 m and a length of 0.5 m TGβ \ u003d 0.32 and, accordingly, the angle between the length and the hypoteneus β \ u003d 17.744 o, which is derived from arktantigem. And now it's easier to work up the equation forces in relation to the axis H.:

Σq x \ u003d n 3-b cosα + N a-b cosβ - N 1 - A \ u003d 0;

N a-b \ u003d (n 1-a - n 3-b cosα ) / cos.β \ u003d (593.77 - 1002.2 0.987) / 0.952 \ u003d - 415.61 kg

In this case, the sign "-" shows that the force is directed to the side opposite to that assumed, which is accepted in the preparation of the calculation scheme. And then it was time to talk about the direction of the forces, more precisely, about the meaning invested in that direction. If we replace the internal efforts in the path taking into account the bars of the farm, then under force directed from the cross-section, tensile stresses are implied, if the force is directed towards the cross-section, then compressive stresses are meant. From the point of view of static equilibrium, it does not matter which direction of the force is in the calculations, if the strength is directed in the opposite direction, then this force is a minus sign. When calculating, however, it is important to know what kind of force this rod is being calculated. For tensile bars, the principle of determining the required section is the simplest:

When calculating the elevation at the compression, many different factors should be taken into account and in the general formula for calculating compression rods can be expressed as follows:

Σ \ u003d n / φf ≤ r

Note: The calculation scheme can be made in such a way that all longitudinal forces from cross-sections are directed. In this case, the sign "-" in front of the value of the force obtained during the calculations shows that this rod is going with the compression.

The results of the previous calculation show that in bars 2-A and 3-B there are tensile stresses in bars 1-A and A-B. Well, now back to the goal of our calculation - to determine the maximum normal stresses in the rods. As in the usual symmetrical beam, in which the maximum stresses occur during the symmetrical load in the section furthest away from the girders, in the maximum stresses occur in the fumes in the bars of the girders, i. H. In the rods cut off by the cross-section of the IX-IX.

iX-IX section (Fig. 272.2. D)

M 9 \ u003d -4.5q / 2 - 3.5q - 2.5q - 1.5q -0.5q + 3V A - 4.5N 6-s sinα = 0 ;

N 6-s \ u003d (15q - 10.25q) / (4.5;α ) \ u003d 4.75 x 190 / (4.5 x 0.158) \ u003d 1269.34 kg (Works with compression)

Where V a \ u003d 5qThe farm support responses are determined by all of the same equilibrium equations of the system, since the farm and the load are symmetrical,

V a \ u003d Σq y / 2 \ u003d 5q;

there are still no horizontal loads, the horizontal reference reaction on the carrier BUT It will be zero, so h a is shown in Figure 272.2 b) light purple color.

the shoulders of all forces in this case are different, and therefore the numerical values ​​of the shoulders are immediately substituted in the formula.

To determine the effort in the stem Z- and, you must first determine the value of the angle γ (not shown in the figure). Based on the fact that two sides of a certain rectangular triangle are known (the lower katat or the length of the triangle is 0.5 m, the lateral katat or the height of the triangle is 0.8 m, then tgγ \ u003d 0.8 / 0.5 \ u003d 1.6 and the value of the angle γ \ u003d Arctgγ \ u003d 57.99. And then for point 3.

h \ u003d 3Sin.γ \ u003d 2.544 m. Then:

M 3 \ u003d - 1.5q / 2 - 0.5q + 0.5q + 1.5q + 2.5q - 1.5n 6-s sinα + 2.544n z and = 0 ;

N h - and \ u003d (1.25q - 4.5q +1.5n 6-s sinα ) /2.544 \ u003d (332.5 - 617.5) /2,544 \ u003d -112 kg

And now it's easier to work up the equation forces with respect to the axis H.:

Σq x \ u003d - n 6-s cosα - n h and cosγ + N 1 and \ u003d 0;

N 1-and \ u003d n 6-s cosα + N h and cosγ \ u003d 1269.34 0.987 - 112 0.53 \ u003d 1193.46 kg (Works on the line)

Since the upper and lower belts of the farm will be of one profile type, you will spend time and strength on the calculation of the lower belt from 1-B, 1-D and 1-g, as well as the bars of the upper belt 4-g and 5. no need. Efforts in these rods are significantly less than we have already defined. If the farm doesn't come together, i. H. The supports were at the ends of the farm then the efforts in the specs would also be less than those already defined, but we have a farm with consoles and so we use several more sections to determine the efforts up in the shallows Algorithm (details of the calculation are not given):

N b-b \ u003d -1527.34 kg - works with compression (section III-III, 2722.2 g), was determined by an equation of the moments relative to point 1).

N V-R \ u003d 634.43 kg - is working on a stretch (section IV-IV, Fig. 2722.2 h), was determined by an equation of the moments relative to point 1.

N Mr \ u003d - 493.84 kg - the compression is working (section V-V, was determined by the equation of the moments relative to point 1)

Thus, we are two bars n 6-s \ u003d 1269.34 kg and n b-b \ u003d - 1527.34 kg. Both bars work with compression, and if the whole farm is made from one type of profile, then it is enough to calculate one of these bars at the limit stresses, and on the basis of these calculations to choose the required cross-section of the profile. However, everything is not that simple, at first glance it seems that it is enough to calculate the shank n B-B, but the calculated length of the rod is of great importance when calculating compressed elements. The length of the rod n 6-s is thus 101.2 cm, while the length of the rod n b-b is 59.3 cm. So it is no better to calculate both rods.

rod n bs.

The calculation of the compression bars is no different from the calculation of the centrally compressed columns, so only the main stages of the calculation are given without detailed explanations.

table 1 (see above) Determine the value μ \ u003d 1 (despite the fact that the upper belt of the farm will be of a solid profile, the calculated farm scheme implies a hinge fixation of bars in the farm nodes, so it is more correct to take the above value of the coefficient).

Benefit λ \ u003d 90, then Table 2 Bend coefficient φ \ u003d 0.625 (for steel C235 strength R y \ u003d 2350 kgf / cm 2, determined by the interpolation of the values ​​2050 and 2450)

Then the required indolence of indolence will be:

Canopy is a simple architectural design that is used for a variety of purposes. In most cases, it is done in the absence of a garage with a cover in the country or to protect the platform from strong sun rays for relaxation. To ensure the reliability and strength of such a small size construction, you need to make a canopy. In the end, it is possible to get data that can show which farms are being used and how to cook.

The scheme for fixing profile tubes can be seen in FIG. one.

Figure 1 shows the pipe fixing scheme

How do you calculate the farms for a canopy with your own hands?

To make the calculation of a similar structure for a canopy, you need to prepare:

• Calculators and special software;
• Snip 2.01.07-85 and Snip P-23-81.

When calculating, it is necessary to perform the following actions:

1. First you need to choose a farm system. Future contours are defined for this. The outlines must be selected based on the main functions of the canopy, material and other parameters.
2. After that, it is necessary to determine the dimensions of the manufactured structure. The height depends on the type of roof and the material used, weight and other parameters;
3. If the size of the flight exceeds 36 m, it is necessary to calculate for the construction elevator. In this case, it gives back to the farm due to the reverse repaid bend from the loads.
4. It is necessary to determine the dimensions of the panels of the structure, which must correspond to the distances between the individual elements that provide the transfer of loads;
5. In the next stage, the distance between the nodes is determined, which is most often the width of the plate.

When doing calculations, follow these tips:

1. It takes all the values ​​to calculate accurately. It should be known that even the slightest mistake leads to errors in the process of work of all works on the manufacture of the design. If there is no self-confidence, it is recommended to immediately contact professionals who have experience in performing such calculations;
2. To make work easier, you can use ready-made projects in which only the existing values ​​are replaced.

Metal guard shown in this photo

When performing the calculation of the farm, it should be remembered that as the altitude increases, the storage capacity will increase. In the winter season, the snow will almost not accumulate on a similar canopy. To increase the strength of the structure, several permanent stiffness ribs should be installed.

For the agricultural facilities, it is best to use a hard pipe that is small in weight, high in strength and rigidity. In the process of determining the sizes for such an element, you need to take into account the following data:

1. For small structures, the width of which is up to 4.5 m, it takes to use a pipe from a metal 40x20x2 mm;
2. For structures, the width of which is less than 5.5 m, you need to use a pipe with dimensions of 40x40x2 mm.
3. If the farm width is more than 5.5 m, it is best to apply the 60x30x2 mm or 40x40x3 mm pipe.

When planning a step of the farms, it should be remembered that the maximum possible distance between the tubes of the canopy is 1.7 m. Only in this case, the reliability and strength of the structure can only save.

Example of calculating the farms for the canopy

1. As an example, consider a 9 M weave width of 8 °. The wingspan of the systems is 4.7 m, and the snow loads for the region are at the level of 84 kg / m²;
2. The agricultural weight is about 150 kg (you should take a small stock for strength). The vertical load is 1.1 tons per rack with a height of 2.2 m;
3. One end of the farm will rely on the wall of building the brick, and the second - on the column in support of the canopy with the help of anchor bolts. For the manufacture of a farm, a square pipe is 45x4 mm. It should be noted that it is quite convenient to work with a similar device;
4. It is best to make farms with parallel belts. The height of each of the elements is 40 cm. A tube with a cross section of 25 × 3 mm is used for the collections. A tube 35x4 mm is used for the lower and upper belt. Visors and other elements need to cook with each other, since the wall thickness is 4 mm.

Ultimately, you can get the following data:

• Estimated resistance for steel: RY \ u003d 2.45 t / cm²;
• Reliability coefficient - 1;
• Farm span - 4.7 m;
• Farmer's height - 0.4 m;
• The number of panels for the upper belt design - 7;
• Corners have to be cooked by one.

All the data required for calculations can be found in special reference books. However, it is recommended that those skilled in the art calculate the nature of such a type using the software. If a mistake is made, then the farms are made under the influence of snow loads and wind.

How to calculate a farm for a polycarbonate carport?

The canopy is a complex design so before purchasing any amount of material it will require the estimate. The support frame must be able to withstand any load.

In order to make a professional calculation of the design of polycarbonate, it is recommended to seek help from an engineer who has experience with the experience of this work. If the canopy is a separate design, and not an extension to a private house, then the calculations will become more complicated.

Street roof consists of pillars, delays, farms and coatings. It is these elements that must count.

If you want to create a domed polycarbonate canopy, it won't work without the use of farms. Farms are devices that bind lags and support columns. From such elements depends on the size of the canopy.

Polycarbonate canopies, since the basics of metal farms are used, are quite difficult to produce. The right frame can distribute the load on the support posts and delays while the canopy design doesn't collapse.

For the installation of polycarbonate, it is best to use profile pipes. The main calculation of the farm is the accounting of material and slope. For example, an improper form of the farm is used for a single slope paving with a small slope. If the design has a small angle, then the metal farms in the form of a trapezoid can be used. The larger the radius of the structure of the arch, the smaller the ability to retard snow on the roof. In this case, the warehouse has a large storage capacity (Fig. 2).

Figure 2 shows the future canopy-coated polycarbonate

If a simple farm of a lodge size of 6x8 m is used, the calculations will be like this:

• The step between the columns for the beam is 3 m;
• The number of metal pillars is 8 pieces;
• Peasant height under plumper - 0.6 m;
• For the device, the roof boxes require 12 profile tubes with dimensions of 40x20x0.2 cm.

In some cases, it can be saved by reducing the amount of material. For example, instead of 8 pegs, you could install 6. 6. You can also cut the frame of the frame. However, it is not recommended to allow for flesh loss as this can lead to the destruction of the structure.

Detailed calculation of the farm and the arch for canopy

In this case, a canopy is calculated, the farms of which are installed in a step of 1 m. The load of such elements from the box is transferred exclusively to the agricultural node. A professional flooring is used as a roofing material. The height of the farm and the arch can be any. If it is a canopy that is adjacent to the main structure, the main limiter is the shape of a roof. In most cases, make the height of the farm more than 1m will not work. Taking into account the fact that you need to create a fan between the columns, the maximum height is 0.8 m.

A farm canopy scheme is shown in FIG. 3. Blue colors are indicated by the bars of the box, in blue - the farm that has to count. The purple color is marked with bars or farms on which the pillars are dependent.

In this case, 6 triangular farms are used. In the case of extreme elements, the load is several times less than the rest. In this case, the metal farms will be a console, that is, their supports are not located at the ends of the farms, but in the nodes shown in FIG. 3. Such a scheme allows you to evenly distribute the load.

Figure 3 shows the protection scheme for farms

The calculated load is q \ u003d 190 kg, while the snow load is 180 kg / m². Thanks to the sections, it is possible to calculate efforts in all the rods of the design, while it is necessary to take into account the fact that the farm and the load of this element are symmetrical. Therefore, it will take time not to count all the farms and arches, but only some of them.In order to navigate freely in a large number of rods in the calculation process, rods and nodes are marked.

Formulas to be used when calculating

It will take time to determine the farm's efforts in several rods. To do this, use the static equilibrium equation. In the nodes of the joint elements, because the value of the bending moments in the agricultural nodes is 0, the sum of all forces related to the X-axis and y is also 0.

It will take time to build the equation of the moments in relation to point 3 (E):

M3 \ u003d -QL / 2 + N2-A * H \ u003d 0, where L is the distance from point 3 to the point of application of force Q / 2, which is 1.5 m, and H-shoulder of force N2 -A.

The farm has an estimated height of 0.8 m and length 10 m. In this case, the tangent of the angle A will be TGA \ u003d 0.8 / 5 \ u003d 0.16. The value of the angle A \ u003d Arctga \ u003d 9.09 °. Ultimately, h \ u003d lsina. From this follows the equation:

N2-a \ u003d ql / (2lsina) \ u003d 190 / (2 * 0.158) \ u003d 601.32 kg.

You can determine the value N1-A in the same way. To do this, you need to make an equation of moments in relation to point 2:

M2 \ u003d -QL / 2 + N1-A * H \ u003d 0;

N1-A \ u003d Q / (2TGA) \ u003d 190 / (2 * 0.16) \ u003d 593.77 kg.

Check the correctness of the calculations by creating the equation of forces:

EQY \ u003d Q / 2 - N2-ASA \ u003d 0; Q / 2 \ u003d 95 \ u003d 601.32 * 0.158 \ u003d 95 kg;

EQX \ u003d N2-ACOSA - N1-A \ u003d 0; N1-A \ u003d 593.77 \ u003d 601.32 * 0.987 \ u003d 593.77 kg.

The conditions of statistical equilibrium are met. Any of the equations of the forces used during the verification process can be used to determine the efforts in the bars. Another calculation of the farms is done in the same way, the equations do not change.

It is worth knowing that the calculation scheme can exist so that all longitudinal forces go from cross-sections. In this case, the sign shows "-" before the force indicator obtained during the calculations shows that a similar rod is working on the compression.

To determine the force in the trunk Z-and, it is necessary to determine the value of the angle Y: H \ u003d 3Siny \ u003d 2.544 m.

The farm for a canopy is calculated with his own hands. It is only necessary to know the basic formulas and be able to use them.

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