Calculation of shell development online. Device for marking pipes. Calculation and production of a template. Calculation of pipe blanks for bending. By sheet stamping method

When designing and manufacturing bent parts from pipes and rods, the task of determining the length of the development - the length of a straight workpiece before starting technological process flexible.

Continuing the topic...

I present the calculation in Excel of the development length of parts made of rods and pipes round section.

The calculation program is written according to the classic strength of strength formula! Practical results will differ slightly from the calculated values due to a number of factors that have already been mentioned in the article on sheet metal bending (link to this article in the previous paragraph). However, the program presented below will ensure accuracy when bending a pipe for the manufacture of a prototype.

Below this text the figure shows the calculation diagram.

The radii of the neutral layers of each of the curved sections are calculated using the formula:

rni =((4* R i 2 — D 2 ) 0,5 +(4* R i 2 — d 2 ) 0,5)/4

The neutral layer is the surface closer to which, closer to the center of the bending radius, the pipe material is compressed during bending, and further from the center of the bending radius, it is stretched.

The length of curved sections when bending a pipe is determined by the formula:

l i =π *α i /180*r ni

Here's the corner α i must be in degrees.

The total length of the development is calculated by summing the lengths of straight and curved sections:

L = ∑(L i + l i )

Program for calculating the length of development in Excel when bending pipes.

To perform calculations we use MS Excel. You can use the Calc spreadsheet processor from freely distributed packages Apache OpenOffice or LibreOffice .

Initial data:

Let's assume that in the example under consideration, the part consists of three straight and two curved sections (as in the diagram above).

1. Let's write it down outside diameter pipes D in millimeters

to cell D4: 57,0

2. Pipe internal diameter value d We put it in millimeters

to cell D5: 50,0

Attention!!! If the development length of a solid round bar is calculated, thend =0!

3. Length of the first straight section L 1 enter in millimeters

to cell D6: 200,0

4. Axial bend radius of the first curved section R 1 write in millimeters

to cell D7: 300,0

5. Bend angle of the first curved section α 1 we write in degrees

to cell D8: 90,0

6. Length of the second straight section of the part L 2 enter in millimeters

to cell D9: 100,0

7. Axial bend radius of the second curved section R 2 write in millimeters

to cell D10: 200,0

8. Bend angle of the second curved section α 2 we write in degrees

to cell D11: 135,0

9. Length of the third straight section of the part L 3 enter in millimeters

to cell D12: 300,0

10-15. Entering the initial data into Excel for our example is complete. We leave cells D13…D18 empty.

The program allows you to calculate the development of parts containing up to five straight sections and up to four curved ones. Pipe bending with big amount plots requires a slight modernization of the program to calculate the development.

Calculation results:

16. Length of the first curved section L 1 Calculate in millimeters

in cell D20: =IF(D7=0;0;PI()*D8/180*((4*D7^2-$D$4^2)^0.5+(4*D7^2-$D$5^2)^ 0.5)/4) =469,4

17. Length of the second curved section L 2 Calculate in millimeters

in cell D21: =IF(D10=0;0;PI()*D11/180*((4*D10^2-$D$4^2)^0.5+(4*D10^2-$D$5^2)^ 0.5)/4)=467,0

18-19. Since in the example under consideration there are no third and fourth curved sections, then

in cell D22: =IF(D13=0;0;PI()*D14/180*((4*D13^2-$D$4^2)^0.5+(4*D13^2-$D$5^2)^ 0.5)/4)=0,0

in cell D23: =IF(D16=0;0;PI()*D17/180*((4*D16^2-$D$4^2)^0.5+(4*D16^2-$D$5^2)^ 0.5)/4)=0,0

20. Total length of part development L summed up in millimeters

in cell D24: =D6+D9+D12+D15+D18+D20+D21+D22+D23=1536,3

The development length of the curved pipe was calculated using MS Excel.

Conclusion.

Bending a pipe and/or rod is not a simple technological task, fraught with a number of pitfalls. I hope that the proposed calculation in Excel will make it easier for you, dear readers, to solve it. The ability to specify at each step different lengths of straight sections, angles and bending radii will undoubtedly expand the scope of the presented program.

Dear readers! Please leave questions, reviews, and comments in the comments at the bottom of the page.

FOR THE REST - you can download it just like that...

The formula for the development length of a pipe blank helps to calculate the surface area or cross-section of a pipeline. The calculation is based on the size of the future route and the diameter of the planned structure. In what cases such calculations are required and how they are done, this article will tell you.

When are calculations needed?

Parameters are calculated using a calculator or using online programs

It is important to know what area the pipeline surface should have in the following cases.

When calculating the heat transfer of a “warm” floor or register. Here the total area is calculated, which transfers heat emanating from the coolant to the room.

When heat losses are determined along the path from a source of thermal energy to heating elements– radiators, convectors, etc. To determine the number and size of such devices, we need to know the amount of calories that we must have, and it is derived taking into account the development of the pipe.

To determine the required amount of thermal insulation material, anti-corrosion coating and paints. When constructing highways that are kilometers long, accurate calculations save the enterprise considerable money.

When determining a rationally justified profile section that could ensure maximum conductivity of the water supply or heating network.

Determination of pipe parameters

Cross-sectional area

The pipe is a cylinder, so calculations are not difficult

The cross-section of a round profile is a circle, the diameter of which is determined as the difference in the outer diameter of the product minus the wall thickness.

In geometry, the area of a circle is calculated as follows:

S = π R^2 or S= π (D/2-N)^2, where S is the internal cross-sectional area; π – number “pi”; R – section radius; D - outer diameter; N is the thickness of the pipe walls.

Note! If in pressure systems the liquid fills the entire volume of the pipeline, then in a gravity sewer only part of the walls is constantly wetted. In such collectors, the concept of open cross-sectional area of the pipe is used.

External surface

The surface of the cylinder, which is the round profile, is a rectangle. One side of the figure is the length of the pipeline section, and the second is the circumference of the cylinder.

Pipe development is calculated using the formula:

S = π D L, where S is the pipe area, L is the length of the product.

Inner surface

This indicator is used in the process of hydrodynamic calculations, when the surface area of the pipe that is constantly in contact with water is determined.

When determining this parameter should be considered:

The larger the diameter of the water pipes, the less the flow rate depends on the roughness of the walls of the structure.

On a note! If pipelines with a large diameter are characterized by a short length, then the value of the wall resistance can be neglected.

In hydrodynamic calculations, the roughness of the wall surface is given no less importance than its area. If water passes through a rusty water pipe inside, then its speed less speed liquid that flows through a relatively smooth polypropylene structure.

Networks that are mounted from non-galvanized steel have a variable area inner surface. During operation, they become covered with rust and overgrown with mineral deposits, which narrows the lumen of the pipeline.

Important! Pay attention to this fact if you want to make cold water supply from steel material. The throughput of such a water supply system will be halved after ten years of operation.

The calculation of pipe development in this case is done taking into account the fact that inner diameter cylinder is defined as the difference between the outer diameter of the profile and the double thickness of its walls.

As a result, the surface area of the cylinder is determined by the formula:

S= π (D-2N)L, where the indicator N is added to the already known parameters, which determines the wall thickness.

The workpiece development formula helps to calculate the amount of thermal insulation required

To know how to calculate the development of a pipe, it is enough to remember the geometry course that is taught in middle school. It's nice that school program finds application in adult life and helps solve serious construction problems. Let them be useful for you too!

Chapter VII. Metal bending

§ 26. General information

Bending is a method of metal processing by pressure, in which a workpiece or part thereof is given a curved shape. Bench bending is performed with hammers (preferably with soft strikers) in a vice, on a plate or using special devices. Thin sheet metal is bent with mallets, wire products with a diameter of up to 3 mm are bent with pliers or round nose pliers. Only plastic material is subject to bending.

Bending parts is one of the most common metalworking operations. The production of flexible parts is possible both manually using support tools and mandrels, and on bending machines (presses).

The essence of bending is that one part of the workpiece is bent relative to the other at a given angle. This happens in the following way: a bending force acts on a workpiece, freely lying on two supports, which causes bending stresses in the workpiece, and if these stresses do not exceed the elastic limit of the material, the deformation obtained by the workpiece is elastic, and upon removal of the load, the workpiece takes on its original view (straightens).

However, when bending, it is necessary to ensure that the workpiece, after removing the load, retains its given shape, therefore the bending stresses must exceed the elastic limit and the deformation of the workpiece in this case will be plastic, while the inner layers of the workpiece are subjected to compression and shortened, the outer layers are subject to tension and their length increases . At the same time, the middle layer of the workpiece - the neutral line - experiences neither compression nor tension, and its length before and after bending remains constant (Fig. 93a). Therefore, determining the dimensions of profile blanks comes down to calculating the length of straight sections (flanges), the length of shortening of the blank within the radius or the length of the neutral line within the radius.

When bending parts at right angles without rounding on the inside, the bend allowance is taken from 0.5 to 0.8 of the thickness of the material. Folding length internal sides square or bracket, we get the length of the workpiece.

Example 1. In Fig. 93, c, d shows a square and a bracket with right internal angles.

Dimensions of the square (Fig. 93, c): a = 30 mm, b = 70 mm, t = 6 mm. Development length

L = a + b + 0.5t = 30 + 70 + 3 = 103 mm.

Bracket dimensions (Fig. 93, d): a = 70 mm, b = 80 mm, c = 60 mm, t = 4 mm. Reaming length of the staple blank

L = 70 + 80 + 60 + 2 = 212 mm.

We divide the square according to the drawing into sections. We substitute their dimensions a = 50 mm, b = 30 mm, t = 6 mm, r = 4 mm into the formula

L = a + b + π/2(r + t/2)

Then we get:

L = 50 + 30 + 3.14/2(4 + 6/2) = 50 + 30 + 1.57⋅7 = 90.99 91 mm.

We divide the bracket into sections, as shown in the drawing. Their dimensions: a = 80 mm, h = 65 mm, c = 120 mm, t = 5 mm, r = 2.5 mm.

L = a + h + c + π(r + t/2) = 80 + 65 + 120 + 3.14(2.5 + 5/2),

hence,

L = 265 4 + 15.75 = 280.75 mm.

By bending this strip into a circle, we obtain a cylindrical ring, and outer part The metal will stretch out somewhat, and the inner one will shrink. Therefore, the length of the workpiece will correspond to the length midline a circle passing in the middle between the outer and inner circles of the ring.

Workpiece length

Knowing the diameter of the middle circumference of the ring and substituting it numeric value In the formula, we find the length of the workpiece:

L = πD = 3.14 108 = 339.12 mm.

As a result preliminary calculations It is possible to produce a part of the specified dimensions.

During the bending process, significant stresses and deformations occur in the metal. They are especially noticeable when the bending radius is small. To prevent cracks from appearing in the outer layers, the bending radius should not be less than the minimum permissible radius, which is selected depending on the thickness and type of material being bent (Fig. 95).

As I promised in the comments to the article, today we’ll talk about calculating the development length of a part bent from sheet metal. Of course, not only sheet metal parts are subjected to the bending process. Bends round and...

Square sections, bent and all rolled profiles - angles, channels, I-beams, pipes. However, cold bending of sheet metal parts is by far the most common.

To ensure minimum radii, parts are sometimes heated before bending. This increases the plasticity of the material. Using bending with a calibrating blow, it is ensured that the internal radius of the part becomes absolutely equal to the radius of the punch. With free V-shaped bending on a sheet bending machine, the internal radius in practice is greater than the radius of the punch. The more pronounced the spring properties of the part material are, the more different the internal radius of the part and the radius of the punch are from each other.

The figure below shows a bent sheet of thick s and width b corner. You need to find the sweep length.

The sweep calculation will be performed in MS Excel.

In the drawing of the part the following is specified: the value of the internal radius R, corner a and length of straight sections L1 And L2. Everything seems simple - elementary geometry and arithmetic. In the process of bending the workpiece, plastic deformation of the material occurs. The outer (relative to the punch) metal fibers are stretched, and the inner ones are compressed. In the middle of the section there is a neutral surface...

But the whole problem is that the neutral layer is not located in the middle of the metal section! For reference: the neutral layer is the surface of the arrangement of conditional metal fibers that do not stretch or compress when bent. Moreover, this surface is (sort of) not the surface of a circular cylinder. Some sources suggest that it is a parabolic cylinder...

I'm more inclined to trust classical theories. For section rectangular shape according to the classical strength of material, the neutral layer is located on the surface of a circular cylinder with a radius r .

r = s / ln(1+ s / R )

Based on this formula, a sweep calculation program was created sheet parts from steel grades St3 and 10...20 in Excel.

In cells with light green and turquoise fill we write the original data. In a cell with a light yellow fill, we read the calculation result.

1. Recording the thickness sheet stock s in millimeters

to cell D 3: 5,0

2. Length of the first straight section L1 enter in millimeters

to cell D 4: 40,0

3. Inner bend radius of the first section R1 write in millimeters

to cell D 5: 5,0

4. Bend angle of the first section a1 we write in degrees

to cell D 6: 90,0

5. Length of the second straight section of the part L2 enter in millimeters

to cell D 7: 40,0

6. That's it, the result of the calculation is the length of the part development L in millimeters

in cell D 17: =D4+IF(D5=0;0;PI()/180*D6*D3/LN ((D5+D3)/D5))+ +D7+IF(D8=0;0;PI()/180* D9*D3/LN ((D8+D3)/D8))+D10+ +IF(D11=0;0;PI()/180*D12*D3/LN ((D11+D3)/D11))+D13+ + IF(D14=0;0;PI()/180*D15*D3/LN ((D14+D3)/D14))+D16=91.33

L = ∑ (Li +3.14/180* ai * s / ln((Ri + s )/ Ri )+ L(i +1))

Using the proposed program, you can calculate the length of the development for parts with one bend - corners, with two bends - channels and Z-profiles, with three and four bends. If you need to calculate the development of a part with a large number of bends, then the program can be very easily modified to expand its capabilities.

An important advantage of the proposed program (unlike many similar ones) is possibility of setting at every step different angles and bending radii.

Does the program produce the “correct” results? Let's compare the obtained result with the results of calculations using the methodology outlined in the “Handbook of Mechanical Designer” by V.I. Anuriev and in the “Die Designer’s Handbook” by L.I. Rudman. Moreover, we will take into account only the curved section, since, I hope, all rectilinear sections are considered the same.

Let's check the example discussed above.

“According to the program”: 11.33 mm – 100.0%

“According to Anuriev”: 10.60 mm – 93.6%

“According to Rudman”: 11.20 mm – 98.9%

In our example, let's increase the bending radius R1 twice - up to 10 mm. Once again we will make the calculation using three methods.

“According to the program”: 19.37 mm – 100.0%

“According to Anuriev”: 18.65 mm – 96.3%

“According to Rudman”: 19.30 mm – 99.6%

Thus, the proposed calculation method produces results that are 0.4%...1.1% more than “according to Rudman” and 6.4%...3.7% more than “according to Anuriev”. It is clear that the error will decrease significantly when we add straight sections.

“According to the program”: 99.37 mm – 100.0%

“According to Anuriev”: 98.65 mm – 99.3%

“According to Rudman”: 99.30 mm – 99.9%

Perhaps Rudman compiled his tables using the same formula that I use, but with the error of a slide rule... Of course, today it’s the twenty-first century, and it’s somehow not convenient to scour the tables!

In conclusion, I will add a “fly in the ointment”. The length of the sweep is a very important and “subtle” point! If the designer of a bent part (especially high-precision (0.1 mm)) hopes to accurately determine it by calculation and the first time, then he hopes in vain. In practice, a lot of factors will interfere with the bending process.– direction of rolling, tolerance on metal thickness, thinning of the section at the bending point, “trapezoidal section”, temperature of the material and equipment, presence or absence of lubrication in the bending zone, mood of the bender... In short, if the batch of parts is large and expensive – check the sweep length on several samples with practical experiments. And only after receiving a suitable part, cut the blanks for the entire batch. And for the manufacture of blanks for these samples, the accuracy provided by the development calculation program is more than enough!

Calculation programs “according to Anuriev” and “according to Rudman” in Excel can be found on the Internet.

I look forward to your comments, colleagues.

For the REST - you can download it just like that...

The topic is continued in the article about.

Read about calculating the development when bending pipes and rods.

When bending, it is necessary to ensure that the workpiece, after removing the load, retains its given shape, therefore the bending stresses must exceed the elastic limit.

The deformation of the workpiece in this case will be plastic, while the inner layers of the workpiece are compressed and shortened, and the outer layers are stretched and lengthened (Figure 8.3.1).

Figure 8.3.1 Bending process diagram

At the same time, the middle layer of blanks - neutral line- does not experience any compression or stretching; its length before and after bending remains constant.

Therefore, determining the dimensions of profile blanks comes down to calculating the length of straight sections (flanges), the length of shortening of the blank within the radius or the length of the neutral line within the radius.

When bending parts at right angles without rounding on the inside, the bend allowance is taken from 0.5 to 0.8 of the thickness of the material. By adding the length of the inner sides of the square or staple, we get the length of the development of the workpiece.

Table 8.3.1 Determination of workpiece dimensions when bending with rounding (radius)

Bending type	Sketch	Workpiece length, mm
Single corner		L=l 1 +l 2 +l n = l 1 +l 2 +π(r+xS)/2
Double-angled		L=l 1 +l 2 +l 3 + π(r+xS)= =l 1 +l 2 +l 3 +2l H
Four-cornered (for two operations)		L=l 1 +2l 2 +l 3 + l 4 +2l H1 +2l H2 = =l 1 +2l 2 +l 3 +l 4 +π(r 1 +x 1 S)+ +π(r 2 +x2S )
Semicircular (U-shaped)		L=2l+2l H =2l+ π(r+xS)
End (rolling)		L=1.5πρ+2R - S ; ρ = R - yS
Notes: Length of the neutral layer of corner roundings lн

Example 1. Figure 8.3.2, a, b shows, respectively, a square and a bracket with right internal angles.

Figure 8.3.2 Examples of workpiece length calculations

Square dimensions: a = 30mm; L = 70mm; t = 6 mm.

Workpiece development length l =a + L + 0.5t = 30 + 70+3 = 103 mm.

Bracket dimensions: a = 70mm; b = 80mm; c = 60mm; t = 4 mm.

Workpiece development length l = a + b + c + 0.5t = 70 + 80 + 60 + 2 = 212 mm.

We divide the square according to the drawing into sections. Substituting their numerical values

(a = 50 mm; b = 30 mm: t = 6 mm; r = 4 mm) into the formula

L = a + b + (r + t/2)π/2,

we get L = 50+ 30+ (4 + 6/2)π/2 =50 + 30 + 7* 1.57 = 91 mm.

We divide the bracket into sections, as shown in the drawing.

Substituting their numerical values (a = 80mm; h = 65mm; c = 120mm; t = 5mm; r = 2.5mm) into the formula

L=a + h+c+ π(r+t/2),

we get L=80 + 65 + 120+3.14(2.5 +5/2) = 265 + 15.75 = 280.75 mm.

By bending this strip into a circle, we obtain a cylindrical ring, with the outer part of the metal stretching somewhat and the inner part shrinking.

Consequently, the length of the workpiece will correspond to the length of the center line of the circle, passing in the middle between the outer and inner circles of the ring.

Workpiece length L = πD. Knowing the diameter of the middle circle of the ring and substituting its numerical value into the formula, we find the length of the workpiece: L = 3.14 * 108 = 339.12 mm.

As a result of preliminary calculations, it is possible to produce a part of the established dimensions.

As I promised in the comments to the article, today we’ll talk about calculating the length of the development of a part bent from sheet metal. Of course, not only sheet metal parts are subjected to the bending process. Bends round and...

Square sections, bent and all rolled profiles - angles, channels, I-beams, pipes. However, cold bending of sheet metal parts is by far the most common.

The figure below shows a bent sheet of thick s and width b corner. You need to find the sweep length.

The sweep calculation will be performed in MS Excel.

I am more inclined to trust the classical theories. For a rectangular section according to the classical strength of material, the neutral layer is located on the surface of a circular cylinder with a radius r .

r = s / ln(1+ s / R )

Based on this formula, a program was created for calculating the development of sheet parts made of steel grades St3 and 10...20 in Excel.

In cells with light green and turquoise fill we write the original data. In a cell with a light yellow fill, we read the calculation result.

1. We record the thickness of the sheet blank s in millimeters

to cell D 3: 5,0

2. Length of the first straight section L1 enter in millimeters

to cell D 4: 40,0

3. Inner bend radius of the first section R1 write in millimeters

to cell D 5: 5,0

4. Bend angle of the first section a1 we write in degrees

to cell D 6: 90,0

5. Length of the second straight section of the part L2 enter in millimeters

to cell D 7: 40,0

6. That’s it, the result of the calculation is the length of the part’s development L in millimeters

L = ∑ (Li +3.14/180* ai * s / ln((Ri + s )/ Ri )+ L(i +1) )

An important advantage of the proposed program (unlike many similar ones) is the ability to set different bending angles and radii at each step.

Let's check the example discussed above.

“According to the program”: 11.33 mm – 100.0%

“According to Anuriev”: 10.60 mm – 93.6%

“According to Rudman”: 11.20 mm – 98.9%

In our example, let's increase the bending radius R1 twice - up to 10 mm. Once again we will make the calculation using three methods.

“According to the program”: 19.37 mm – 100.0%

“According to Anuriev”: 18.65 mm – 96.3%

“According to Rudman”: 19.30 mm – 99.6%

“According to the program”: 99.37 mm – 100.0%

“According to Anuriev”: 98.65 mm – 99.3%

“According to Rudman”: 99.30 mm – 99.9%

Let's consider a situation that often arises in bending production. This is especially true for small workshops that make do with small and medium-sized mechanization. By small and medium mechanization I mean the use of manual or semi-automatic sheet benders. The operator sums up the length of the shelves and gets total length blanks for the required product, measures desired length, cuts and.. after bending he receives an inaccurate product. Errors in the dimensions of the final product can be quite significant (depending on the complexity of the product, the number of bends, etc.). This is because when calculating the length of the workpiece, it is necessary to take into account the thickness of the metal, the bending radius, and the coefficient of the position of the neutral line (K-factor). This is exactly what this article will focus on.

So let's get started.

To be honest, calculating the dimensions of the workpiece is not difficult. You just need to understand that you need to take into account not only the lengths of the shelves (straight sections), but also the lengths of the curved sections resulting from plastic deformations of the material during bending.

Moreover, all the formulas have long been derived " smart people", books and resources of which I constantly indicate at the end of articles (from there, if you wish, you can get additional information).

Thus, in order to calculate the correct length of the workpiece (part development), which ensures the required dimensions after bending, it is necessary, first of all, to understand which option we will use to make the calculation.

I remind you:

So if you need a shelf surface A without deformations (for example, for the location of holes), then you calculate according to option 1. If the overall height of the shelf is important to you A, then, without a doubt, option 2 better.

Option 1 (with allowance)

We will need:

c) Sum up the lengths of these segments. In this case, the lengths of straight sections are summed up without change, and the lengths of curved sections are summed up taking into account the deformation of the material and the corresponding displacement of the neutral layer.

So, for example, for a workpiece with one bend, the formula will look like this:

Where X1 – length of the first straight section, Y1 – length of the second straight section, φ – external corner, r– internal bending radius, k S– metal thickness.

Thus, the calculation progress will be as follows..

Y1 + BA1 + X1 + BA2 +..etc

The length of the formula depends on the number of variables.

Option 2 (with deduction)

In my experience, this is the most common calculation option for rotary beam bending machines. Therefore, let's look at this option.

We also need:

a) Determine the K-factor (see table).

b) Divide the contour of the bending part into elements, which are straight segments and parts of circles;

Here it is necessary to consider a new concept - the outer boundary of bending.

To make it easier to imagine, see the picture:

The outer boundary of the bend is this imaginary dotted line.

So, to find the length of the deduction, you need to subtract the length of the curved section from the length of the outer boundary.

Thus, the formula for the length of the workpiece according to option 2:

Where Y2 , X2 – shelves, φ – external corner, r– internal bending radius, k– neutral line position coefficient (K-factor), S– metal thickness.

Our deduction ( BD), as you understand:

The outer boundary of the bend ( OS):

And in this case, it is also necessary to calculate each operation sequentially. After all, the exact length of each shelf is important to us.

The calculation scheme is as follows:

(Y2 – BD1 / 2) + (X2 – (BD1 / 2 + BD2 / 2)) + (M2 – (BD2 / 2 + BD3 /2)) +.. etc.

Graphically it will look like this:

And also, the amount of deduction ( BD) during sequential calculations, it is necessary to calculate correctly. That is, we are not just cutting two. First we count all BD, and only after that we divide the resulting result in half.

I hope that I did not offend anyone with this remark. I just know that mathematics is forgotten and even basic calculations can be fraught with surprises that no one needs.

That's all. Thank you all for your attention.

When preparing the information I used: 1. Article “BendWorks. The fine-art of Sheet Metal Bending” Olaf Diegel, Complete Design Services, July 2002; 2. Romanovsky V.P. “Handbook of Cold Forging” 1979; materials from the English-language resource SheetMetal.Me (section “Fabrication formulas”, link:

Chapter VII. Metal bending

§ 26. General information

Bending parts is one of the most common metalworking operations. The production of flexible parts is possible both manually using support tools and mandrels, and on bending machines (presses).

Example 1. In Fig. 93, c, d shows a square and a bracket with right internal angles.

Dimensions of the square (Fig. 93, c): a = 30 mm, b = 70 mm, t = 6 mm. Development length

L = a + b + 0.5t = 30 + 70 + 3 = 103 mm.

Bracket dimensions (Fig. 93, d): a = 70 mm, b = 80 mm, c = 60 mm, t = 4 mm. Reaming length of the staple blank

L = 70 + 80 + 60 + 2 = 212 mm.

We divide the square according to the drawing into sections. We substitute their dimensions a = 50 mm, b = 30 mm, t = 6 mm, r = 4 mm into the formula

L = a + b + π/2(r + t/2)

Then we get:

L = 50 + 30 + 3.14/2(4 + 6/2) = 50 + 30 + 1.57⋅7 = 90.99 91 mm.

We divide the bracket into sections, as shown in the drawing. Their dimensions: a = 80 mm, h = 65 mm, c = 120 mm, t = 5 mm, r = 2.5 mm.

L = a + h + c + π(r + t/2) = 80 + 65 + 120 + 3.14(2.5 + 5/2),

hence,

L = 265 4 + 15.75 = 280.75 mm.

By bending this strip into a circle, we obtain a cylindrical ring, with the outer part of the metal stretching somewhat and the inner part shrinking. Consequently, the length of the workpiece will correspond to the length of the center line of the circle, passing in the middle between the outer and inner circles of the ring.

Workpiece length

Knowing the diameter of the middle circumference of the ring and substituting its numerical value into the formula, we find the length of the workpiece:

L = πD = 3.14 108 = 339.12 mm.

As a result of preliminary calculations, it is possible to produce a part of the established dimensions.

When are calculations needed?

Parameters are calculated using a calculator or using online programs

It is important to know what area the pipeline surface should have in the following cases.

When calculating the heat transfer of a “warm” floor or register. Here the total area is calculated, which transfers heat emanating from the coolant to the room.

When heat losses are determined along the way from a source of thermal energy to heating elements - radiators, convectors, etc. To determine the number and size of such devices, we need to know the amount of calories that we must have, and it is derived taking into account the development of the pipe.

For determining required quantity thermal insulation material, anti-corrosion coating and paint. When constructing highways that are kilometers long, accurate calculations save the enterprise considerable money.

When determining a rationally justified profile section that could ensure maximum conductivity of the water supply or heating network.

Determination of pipe parameters

Cross-sectional area

The pipe is a cylinder, so calculations are not difficult

The cross-section of a round profile is a circle, the diameter of which is determined as the difference in the outer diameter of the product minus the wall thickness.

In geometry, the area of a circle is calculated as follows:

S = π R^2 or S= π (D/2-N)^2, where S is the internal cross-sectional area; π – number “pi”; R – section radius; D - outer diameter; N is the thickness of the pipe walls.

Note! If in pressure systems the liquid fills the entire volume of the pipeline, then in a gravity sewer only part of the walls is constantly wetted. In such collectors, the concept of open cross-sectional area of the pipe is used.

External surface

The surface of the cylinder, which is the round profile, is a rectangle. One side of the figure is the length of the pipeline section, and the second is the circumference of the cylinder.

Pipe development is calculated using the formula:

S = π D L, where S is the pipe area, L is the length of the product.

Inner surface

This indicator is used in the process of hydrodynamic calculations, when the surface area of the pipe that is constantly in contact with water is determined.

When determining this parameter, you should consider:

The larger the diameter of the water pipes, the less the flow rate depends on the roughness of the walls of the structure.

On a note! If pipelines with a large diameter are characterized by a short length, then the value of the wall resistance can be neglected.

In hydrodynamic calculations, the roughness of the wall surface is given no less importance than its area. If water flows through a water pipe that is rusty inside, then its speed is less than the speed of the liquid that flows through a relatively smooth polypropylene structure.

Networks that are mounted from non-galvanized steel are characterized by a variable internal surface area. During operation, they become covered with rust and overgrown with mineral deposits, which narrows the lumen of the pipeline.

Important! Pay attention to this fact if you want to make cold water supply from steel material. The throughput of such a water supply system will be halved after ten years of operation.

The calculation of the pipe development in this case is made taking into account the fact that the internal diameter of the cylinder is determined as the difference between the external diameter of the profile and the double thickness of its walls.

As a result, the surface area of the cylinder is determined by the formula:

S= π (D-2N)L, where the indicator N is added to the already known parameters, which determines the wall thickness.

The workpiece development formula helps to calculate the amount of thermal insulation required

To know how to calculate the development of a pipe, it is enough to remember the geometry course that is taught in middle school. It’s nice that the school curriculum is used in adult life and helps solve serious problems related to construction. Let them be useful for you too!

Square sections, bent and all rolled profiles - angles, channels, I-beams, pipes. However, cold bending of sheet metal parts is by far the most common.

The figure below shows a bent sheet of thick s and width b corner. You need to find the sweep length.

The sweep calculation will be performed in MS Excel.

r = s / ln(1+ s / R )

Based on this formula, a program was created for calculating the development of sheet parts made of steel grades St3 and 10...20 in Excel.

In cells with light green and turquoise fill we write the original data. In a cell with a light yellow fill, we read the calculation result.

1. We record the thickness of the sheet blank s in millimeters

to cell D 3: 5,0

2. Length of the first straight section L1 enter in millimeters

to cell D 4: 40,0

3. Inner bend radius of the first section R1 write in millimeters

to cell D 5: 5,0

4. Bend angle of the first section a1 we write in degrees

to cell D 6: 90,0

5. Length of the second straight section of the part L2 enter in millimeters

to cell D 7: 40,0

6. That's it, the result of the calculation is the length of the part development L in millimeters

L = ∑ (Li +3.14/180* ai * s / ln((Ri + s )/ Ri )+ L(i +1))

An important advantage of the proposed program (unlike many similar ones) is the ability to set different bending angles and radii at each step.

Let's check the example discussed above.

“According to the program”: 11.33 mm – 100.0%

“According to Anuriev”: 10.60 mm – 93.6%

“According to Rudman”: 11.20 mm – 98.9%

In our example, let's increase the bending radius R1 twice - up to 10 mm. Once again we will make the calculation using three methods.

“According to the program”: 19.37 mm – 100.0%

“According to Anuriev”: 18.65 mm – 96.3%

“According to Rudman”: 19.30 mm – 99.6%

“According to the program”: 99.37 mm – 100.0%

“According to Anuriev”: 98.65 mm – 99.3%

“According to Rudman”: 99.30 mm – 99.9%

Calculation programs “according to Anuriev” and “according to Rudman” in Excel can be found on the Internet.

I look forward to your comments, colleagues.

For the REST - you can download it just like that...

The topic is continued in the article about.

Read about calculating the development when bending pipes and rods.