29-10-2012: Andrey
A typo was made in the formula for the bending moment for a beam with rigid pinching on supports (3rd from the bottom): the length must be squared. A typo was made in the formula for the maximum deflection for a beam with rigid pinning on supports (3rd from the bottom): it should be without "5".
29-10-2012: Dr. Lom
Yes, indeed, mistakes were made when editing after copying. On this moment bugs fixed, thanks for your attention.
01-11-2012: Vic
a typo in the formula in the fifth example from the top (the degrees next to x and el are mixed up)
01-11-2012: Dr. Lom
And it is true. Corrected. Thank you for your attention.
10-04-2013: flicker
In formula T.1, 2.2 Mmax seems to be missing a square after a.
11-04-2013: Dr. Lom
Right. I copied this formula from the "Handbook of the Strength of Materials" (ed. by S.P. Fesik, 1982, p. 80) and did not even pay attention to the fact that with such a notation, even the dimension is not respected. Now I counted everything personally, indeed the distance "a" will be squared. Thus, it turns out that the compositor missed a small two, and I fell for this millet. Corrected. Thank you for your attention.
02-05-2013: Timko
Good afternoon, I would like to ask you in table 2, scheme 2.4, you are interested in the formula "moment in flight" where the index X is not clear -? Could you answer)
02-05-2013: Dr. Lom
For the cantilever beams of Table 2, the static equilibrium equation was compiled from left to right, i.e. The origin of coordinates was considered to be a point on a rigid support. However, if we consider a mirror cantilever beam, which will have a rigid support on the right, then for such a beam the moment equation in the span will be much simpler, for example, for 2.4 Mx = qx2/6, more precisely -qx2/6, since it is now believed that if the diagram moments is located on top, then the moment is negative.
From the point of view of the strength of materials, the sign of the moment is a rather arbitrary concept, since in the cross section for which the bending moment is determined, both compressive and tensile stresses still act. The main thing to understand is that if the diagram is located on top, then tensile stresses will act in the upper part of the section and vice versa.
In the table, the minus for moments on a rigid support is not indicated, however, the direction of action of the moment was taken into account when compiling the formulas.
25-05-2013: Dmitriy
Please tell me, at what ratio of the length of the beam to its diameter are these formulas valid?
I want to know if this code only applies to long beams that are used in building construction, or can it also be used to calculate shaft deflections, up to 2 m long. Please answer like this l/D>...
25-05-2013: Dr. Lom
Dmitry, I already told you that the design schemes for rotating shafts will be different. Nevertheless, if the shaft is in a stationary state, then it can be considered as a beam, and it does not matter what section it has: round, square, rectangular, or some other. These design schemes most accurately reflect the state of the beam at l/D>10, at a ratio of 5 25-05-2013: Dmitriy
Thanks for the answer. Can you also name the literature that I can refer to in my work? 25-05-2013: Dr. Lom
I don’t know what kind of problem you are solving, and therefore it is difficult to conduct a substantive conversation. I'll try to explain my idea in a different way. 25-05-2013: Dmitriy
Can I then chat with you via mail or Skype? I will tell you what kind of work I do and what the previous questions were for. 25-05-2013: Dr. Lom
You can write to me, email addresses on the site are not difficult to find. But I’ll warn you right away, I don’t do any calculations and I don’t sign partnership contracts. 08-06-2013: Vitaly
Question according to table 2, option 1.1, deflection formula. Please specify dimensions. 09-06-2013: Dr. Lom
That's right, the output is centimeters. 20-06-2013: Evgeny Borisovich
Hello. Help guess. We have a summer wooden stage near the recreation center, the size is 12.5 x 5.5 meters, at the corners of the stand there are metal pipes with a diameter of 100 mm. They force me to make a roof like a truss (it’s a pity that you can’t attach a picture) a polycarbonate coating, to make trusses from a profile pipe (square or rectangle) there is a question about my work. You won't be fired. I say that it will not work, and the administration, together with my boss, say everything will work. How to be? 20-06-2013: Dr. Lom
22-08-2013: Dmitriy
If the beam (pillow under the column) lies on dense soil (more precisely, buried below the freezing depth), then what scheme should be used to calculate such a beam? Intuition dictates that the "double-supported" option is not suitable and that the bending moment should be substantially less. 22-08-2013: Dr. Lom
The calculation of foundations is a separate big topic. In addition, it is not entirely clear what kind of beam we are talking about. If we mean a pillow under a column of a columnar foundation, then the basis for calculating such a pillow is the strength of the soil. The task of the pillow is to redistribute the load from the column to the base. The lower the strength, the larger the cushion area. Or the greater the load, the greater the cushion area with the same soil strength. 23-08-2013: Dmitriy
This refers to a pillow under a column of a columnar foundation. The length and width of the cushion have already been determined based on the load and strength of the soil. But the height of the pillow and the amount of reinforcement in it are in question. I wanted to calculate by analogy with the article "Calculation of a reinforced concrete beam", but I believe that it would not be entirely correct to consider the bending moment in a pillow lying on the ground, as in a beam on two hinged supports. The question is, according to which design scheme to calculate the bending moment in the pillow. 24-08-2013: Dr. Lom
The height and section of the reinforcement in your case are determined as for cantilever beams (in width and length of the pillow). Scheme 2.1. Only in your case, the support reaction is the load on the column, more precisely, part of the load on the column, and the uniformly distributed load is the repulse of the soil. In other words, the specified design scheme must be turned over. 10-10-2013: Yaroslav
Good evening. Please help me pick up the metal. a beam for a span of 4.2 meters. A two-story residential building, the basement is covered with hollow slabs 4.8 meters long, on top of a load-bearing wall of 1.5 bricks, 3.35 m long, 2.8 m high. . on the other, 2.8 meters on the slabs, again a load-bearing wall as a floor below and above, wooden beams 20 by 20 cm, 5 m long. 6 pieces and 3 meters long, 6 pieces; floor from boards 40 mm. 25 m2. There are no other loads. Please tell me which I-beam to take in order to sleep peacefully. So far, everything has been standing for 5 years. 10-10-2013: Dr. Lom
Look in the section: "Calculation of metal structures" article "Calculation of a metal lintel for load-bearing walls" it describes in sufficient detail the process of selecting a beam section depending on the acting load. 04-12-2013: Kirill
Tell me, please, where can I get acquainted with the derivation of the formulas for the maximum beam deflection for p.p. 1.2-1.4 in Table 1 04-12-2013: Dr. Lom
The derivation of formulas for various options for applying loads is not given on my site. You can see the general principles on which the derivation of such equations is based in the articles "Fundamentals of strength, calculation formulas" and "Fundamentals of strength, determination of beam deflection". 24-03-2014: Sergey
an error was made in 2.4 of Table 1. Even the dimension is not respected 24-03-2014: Dr. Lom
I do not see any errors, and even more so non-compliance with the dimension in the calculation scheme you indicated. Please clarify what exactly is wrong. 09-10-2014: Sanych
Good afternoon. Do M and Mmax have different units of measurement? 09-10-2014: Sanych
Table 1. Calculation 2.1. If l is squared, then Mmax will be in kg * m2? 09-10-2014: Dr. Lom
No, M and Mmax have the same unit kgm or Nm. Since the distributed load is measured in kg/m (or N/m), the torque value will be kgm or Nm. 12-10-2014: Paul
Good evening. I work in the production of upholstered furniture and the director threw me a problem. I ask for your help, because I do not want to solve it "by eye". 12-10-2014: Dr. Lom
It depends on many factors. In addition, you did not specify the thickness of the pipe. For example, with a thickness of 2 mm, the section modulus of the pipe is W = 3.47 cm^3. Accordingly, the maximum bending moment that the pipe can withstand is M = WR = 3.47x2000 = 6940 kgcm or 69.4 kgm, then the maximum allowable load for 2 pipes is q = 2x8M/l^2 = 2x8x69.4/2.2^2 = 229.4 kg/m (with hinged supports and without taking into account the torque that may occur when the load is transferred not along the center of gravity of the section). And this is with a static load, and the load is likely to be dynamic, or even shock (depending on the design of the sofa and the activity of the children, mine jump on the sofas in such a way that it takes your breath away), so consider for yourself. The article "Calculated values for rectangular shaped pipes" will help you. 20-10-2014: student
Doc, please help. 21-10-2014: Dr. Lom
To begin with, a rigidly fixed beam and supporting sections are incompatible concepts, see the article "Types of supports, which design scheme to choose." Judging by your description, you either have a single-span articulated beam with cantilevers (see Table 3), or a three-span rigidly supported beam with 2 additional supports and unequal spans (in this case, the equations of three moments will help you). But in any case, the support reactions under a symmetrical load will be the same. 21-10-2014: student
I understand. Along the perimeter of the first floor, the armored belt is 200x300h, the outer perimeter is 4400x4400. 3 channels are anchored into it, with a step of 1 m. The span is without racks, one of them is the heaviest option, the load is asymmetric. THOSE. consider the beam as hinged? 21-10-2014: Dr. Lom
22-10-2014: student
in fact yes. As I understand it, the deflection of the channel will turn the armored belt itself at the attachment point, so you get a hinged beam? 22-10-2014: Dr. Lom
Not quite so, first you determine the moment from the action of a concentrated load, then the moment from a uniformly distributed load along the entire length of the beam, then the moment that occurs when a uniformly distributed load acts on a certain section of the beam. And only then add up the values of the moments. Each of the loads will have its own calculation scheme. 07-02-2015: Sergey
Isn't there an error in the Mmax formula for case 2.3 in Table 3? A beam with a console, probably a plus instead of a minus should be in brackets 07-02-2015: Dr. Lom
No, not a mistake. The load on the console reduces the moment in the span, but does not increase it. However, this can also be seen from the diagram of moments. 17-02-2015: Anton
Hello, first of all, thanks for the formulas, saved in bookmarks. Tell me, please, there is a beam over the span, four logs lie on the beam, distances: 180mm, 600mm, 600mm, 600mm, 325mm. I figured out the diagram, the bending moment, I can’t understand how the deflection formula will change (table 1, scheme 1.4), if the maximum moment is on the third lag. 17-02-2015: Dr. Lom
I have already answered several times similar questions in the comments to the article "Design schemes for statically indeterminate beams". But you're in luck, for clarity, I performed the calculation according to the data from your question. Look at the article "The general case of calculating a beam on hinged supports under the action of several concentrated loads", perhaps in time I will supplement it. 22-02-2015: Novel
Doc, I can’t master all these formulas that are incomprehensible to me at all. Therefore, I ask you for help. I want to make a cantilever staircase in the house (to brick steps made of reinforced concrete when building a wall). Wall - width 20cm, brick. The length of the protruding step is 1200 * 300mm. I want the steps to be of the correct shape (not a wedge). I understand intuitively that the reinforcement will be "something thicker" so that the steps are something thinner? But will reinforced concrete up to 3 cm thick cope with a load of 150 kg at the edge? Please help me, I don't want to be fooled. I would be very grateful if you could help... 22-02-2015: Dr. Lom
The fact that you cannot master fairly simple formulas is your problem. In the "Fundamentals of Sopromat" section, all this is chewed in sufficient detail. Here I will say that your project is absolutely not real. Firstly, the wall is either 25 cm wide or cinder block (however, I could be wrong). Secondly, neither a brick nor a cinder block wall will provide sufficient pinching of the steps with the specified wall width. In addition, such a wall should be calculated for the bending moment arising from the cantilever beams. Thirdly, 3 cm is an unacceptable thickness for a reinforced concrete structure, taking into account the fact that the minimum protective layer should be at least 15 mm in beams. And so on. 26-02-2015: Novel
02-04-2015: vitaly
what does x mean in the second table, 2.4 02-04-2015: Vitaly
Good afternoon What scheme (algorithm) needs to be selected for calculating a balcony slab, a cantilever pinched on one side, how to correctly calculate the moments on the support and in the span? Can it be calculated as a cantilever beam, according to the diagrams from table 2, namely points 1.1 and 2.1. Thank you! 02-04-2015: Dr. Lom
x in all tables means the distance from the origin to the point under study, at which we are going to determine the bending moment or other parameters. Yes, your balcony slab, if it is solid and loads act on it, as in the indicated schemes, you can count on these schemes. For cantilever beams, the maximum moment is always at the support, so there is no great need to determine the moment in the span. 03-04-2015: Vitaly
Thanks a lot! I also wanted to clarify. I understand if you count on 2 tables. scheme 1.1, (the load is applied to the end of the console) then I have x=L, and accordingly in the span M=0. What if I also have this load on the ends of the plate? And according to scheme 2.1, I count the moment on the support, plus it to the moment according to scheme 1.1, and according to the correct one, in order to reinforce, I need to find the moment in the span. If I have a slab overhang of 1.45m (clear), how can I calculate "x" to find the moment in the span? 03-04-2015: Dr. Lom
The moment in the span will change from Ql on the support to 0 at the load application point, which can be seen from the moment diagram. If you have a load applied at two points at the ends of the slab, then in this case it is more advisable to provide beams that perceive loads at the edges. At the same time, the slab can already be calculated as a beam on two supports - beams or a slab with support on 3 sides. 03-04-2015: Vitaly
Thank you! In moments, I already understood. One more question. If the balcony slab is supported on both sides, the letter "G". What then calculation scheme should be used? 04-04-2015: Dr. Lom
In this case, you will have a plate pinched on 2 sides and there are no examples of calculating such a plate on my website. 27-04-2015: Sergey
Dear Doctor Lom! 27-04-2015: Dr. Lom
I will not evaluate the reliability of such a design without calculations, but you can calculate it according to the following criteria: 05-06-2015: student
Doc, where can I show you a picture? 05-06-2015: student
Did you still have a forum? 05-06-2015: Dr. Lom
There was, but I have absolutely no time to rake up spam in search of normal questions. Therefore, so far. 06-06-2015: student
Doc, my link is https://yadi.sk/i/GardDCAEh7iuG 07-06-2015: Dr. Lom
The choice of design scheme will depend on what you want: simplicity and reliability, or approximation to the real work of the structure through successive approximations. 07-06-2015: student
Doc, thanks. I want simplicity and reliability. This section is the busiest. I even thought about tying the tank stand to tighten the rafters to reduce the load on the ceiling, given that the water will be drained for the winter. I can't get into such a jungle of calculations. In general, the console will reduce deflection? 07-06-2015: student
Doc, another question. the console is obtained in the middle of the span of the window, does it make sense to move to the edge? Sincerely 07-06-2015: Dr. Lom
In the general case, the console will reduce the deflection, but as I said, how much in your case is a big question, and the shift to the center of the window opening will reduce the role of the console. And yet, if this is your most loaded section, then maybe just strengthen the beam, for example, with another of the same channel? I don’t know your loads, but the load from 100 kg of water and half the weight of the tank does not seem so impressive to me, but do the 8P channel in terms of deflection at 4 m span take into account the dynamic load when walking? 08-06-2015: student
Doc, thanks for the good advice. After the weekend I will recalculate the beam as a two-span hinged beam. If there is a large dynamics when walking, I constructively lay the possibility of reducing the pitch of the floor beams. The cottage is a country house, so the dynamics are tolerable. The lateral displacement of the channels has a greater effect, but this is treated by installing cross braces or fixing the deck. The only thing is, will the concrete pour fall? I assume its support on the upper and lower shelves of the channel plus welded reinforcement in the ribs and a mesh on top. 08-06-2015: Dr. Lom
I already told you, you should not count on the console. 09-06-2015: student
Doc, I get it. 29-06-2015: Sergey
Good afternoon. I would like to ask you about: the foundation was cast: piles of concrete 1.8 m deep, and then a tape 1 m deep was cast with concrete. The question is: is the load transferred only to the piles or is it evenly distributed to both the piles and the belt? 29-06-2015: Dr. Lom
As a rule, piles are made in soft soils so that the load on the base is transferred through the piles, therefore, pile grillages are calculated as beams on pile supports. However, if you poured the grillage over compacted soil, then part of the load will be transferred to the base through the grillage. In this case, the grillage is considered as a beam lying on an elastic foundation, and is a conventional strip foundation. Like that. 29-06-2015: Sergey
Thank you. Just a mixture of clay and sand is obtained on the site. Moreover, the layer of clay is very hard: the layer can only be removed with a crowbar, etc., etc. 29-06-2015: Dr. Lom
I don't know all your conditions (distance between piles, number of storeys, etc.). According to your description, it turns out that you made the usual strip foundation and piles for reliability. Therefore, it is enough for you to determine whether the width of the foundation will be sufficient to transfer the load from the house to the foundation. 05-07-2015: Yuri
Hello! I need your help with the calculation. A metal collar 1.5 x 1.5 m weighing 70 kg is mounted on a metal pipe concreted to a depth of 1.2 m and lined with bricks (pillar 38 by 38 cm). What section and thickness should the pipe be so that there is no bend? 05-07-2015: Dr. Lom
You correctly assumed that your post should be treated like a cantilever beam. And even with the design scheme, you almost guessed it. The fact is that 2 forces will act on your pipe (on the upper and lower canopy) and the value of these forces will depend on the distance between the canopies. More details in the article "Determining the pull-out force (why the dowel does not hold in the wall)". Thus, in your case, you should perform 2 deflection calculations according to the calculation scheme 1.2, and then add the results, taking into account the signs (in other words, subtract the other from one value). 05-07-2015: Yuri
Thanks for the answer. Those. I made the calculation to the maximum with a large margin, and the newly calculated deflection value will in any case be less? 06-07-2015: Dr. Lom
01-08-2015: Paul
Can you please tell me how to determine the deflection at point C in diagram 2.2 of table 3 if the lengths of the cantilever sections are different? 01-08-2015: Dr. Lom
In this case, you need to go through a full cycle. Whether this is necessary or not, I don't know. For an example, see the article on the calculation of a beam for the action of several uniformly concentrated loads (link to the article before the tables). 04-08-2015: Yuri
To my question dated July 05, 2015. Is there any rule for the minimum amount of pinching in the concrete of this metal cantilever beam 120x120x4 mm with a collar of 70 kg. - (for example, at least 1/3 of the length) 04-08-2015: Dr. Lom
In fact, the calculation of pinching is a separate big topic. The fact is that the resistance of concrete to compression is one thing, and the deformation of the soil on which the foundation concrete presses is another. In short, the longer the profile and the larger the area in contact with the ground, the better. 05-08-2015: Yuri
Thank you! In my case, the metal gate post will be poured into a concrete pile with a diameter of 300 mm and a length of 1 m, and the piles along the top will be connected by a concrete grillage with a reinforcing cage? concrete everywhere M 300. Ie. there will be no deformation of the soil. I would like to know an approximate, albeit with a large margin of safety, ratio. 05-08-2015: Dr. Lom
Then really 1/3 of the length should be enough to create a hard pinch. For an example, look at the article "Types of supports, which design scheme to choose." 05-08-2015: Yuri
20-09-2015: Karla
21-09-2015: Dr. Lom
You can first calculate the beam separately for each load according to the design schemes presented here, and then add the results, taking into account the signs. 08-10-2015: Natalia
Hello, Doctor))) 08-10-2015: Dr. Lom
As I understand it, you are talking about a beam from table 3. For such a beam, the maximum deflection will not be in the middle of the span, but closer to support A. In general, the amount of deflection and the distance x (to the point of maximum deflection) depend on the length of the console, so in your case, you should use the equations of the initial parameters given at the beginning of the article. The maximum deflection in the span will be at the point where the angle of rotation of the inclined section is zero. If the console is long enough, then the deflection at the end of the console can be even greater than in the span. 22-10-2015: Alexander
22-10-2015: Ivan
Thank you very much for your clarifications. There is a lot of work to be done around the house. Pergolas, awnings, supports. I’ll try to remember that at one time I overslept diligently and then accidentally passed it to the Sov. VTUZ. 31-05-2016: Vitaly
Thank you very much, you are a great guy! 14-06-2016: Denis
While I stumbled upon your site. I almost missed the calculations, I always thought that a cantilever beam with a load at the end of the beam would sag more than with a uniformly distributed load, and formulas 1.1 and 2.1 in table 2 show the opposite. Thanks for your work 14-06-2016: Dr. Lom
In fact, it makes sense to compare a concentrated load with a uniformly distributed load only when one load is reduced to another. For example, at Q = ql, the formula for determining the deflection according to the design scheme 1.1 will take the form f = ql^4/3EI, i.e. the deflection will be 8/3 = 2.67 times greater than with just a uniformly distributed load. So the formulas for the design schemes 1.1 and 2.1 show nothing to the contrary, and initially you were right. 16-06-2016: Garin engineer
Good afternoon! I still can’t figure it out, I’ll be very grateful if you help me figure it out once and for all, when calculating (any) an ordinary I-beam with a normal distributed load along the length, which moment of inertia to use - Iy or Iz and why? I can’t find a strength of materials in any textbook - everywhere they write that the section should tend to a square and you need to take the smallest moment of inertia. I just can’t grasp the physical meaning by the tail - can I somehow interpret it on my fingers? 16-06-2016: Dr. Lom
I advise you to first look at the articles "Fundamentals of Strength Material" and "On the Calculation of Flexible Rods for the Action of a Compressive Eccentric Load", everything is explained in sufficient detail and clearly there. Here I will add that it seems to me that you are confusing calculations for transverse and longitudinal bending. Those. when the load is perpendicular to the neutral axis of the bar, then the deflection (transverse bending) is determined, when the load is parallel to the neutral axis of the beam, then the stability is determined, in other words, the effect of the longitudinal bend on the bearing capacity of the bar. Of course, when calculating for a transverse load (vertical load for a horizontal beam), the moment of inertia should be taken depending on what position the beam has, but in any case it will be Iz. And when calculating for stability, provided that the load is applied along the center of gravity of the section, the smallest moment of inertia is considered, since the probability of loss of stability in this plane is much greater. 23-06-2016: Denis
Hello, such a question why in table 1 for formulas 1.3 and 1.4 the deflection formulas are essentially the same and the size b. in formula 1.4 is not reflected in any way? 23-06-2016: Dr. Lom
With an asymmetric load, the deflection formula for the design scheme 1.4 will be quite cumbersome, but it should be remembered that the deflection in any case will be less than when a symmetrical load is applied (of course, under the condition b 03-11-2016: Vladimir
in table 1 for formulas 1.3 and 1.4 of the deflection formula, instead of Qa ^ 3 / 24EI, there should be Ql ^ 3 / 24EI. For a long time I could not understand why the deflection with the crystal does not converge 03-11-2016: Dr. Lom
That's right, another typo due to inattentive editing (I hope the last one, but not the fact). Corrected, thanks for your concern. 16-12-2016: Ivan
Hello Doctor Lom. The question is the following: I was looking through a photo from the construction site and noticed one thing: a reinforced concrete factory jumper 30 * 30 cm approximately, supported by a three-layer reinforced concrete panel by 7 centimeters. (The reinforced concrete panel was slightly filed to rest the jumper on it). The opening for the balcony frame is 1.3 m, along the top of the lintel there is an armored belt and attic floor slabs. Are these 7 cm critical, the support of the other end of the jumper is more than 30 cm, everything has been fine for several years already 16-12-2016: Dr. Lom
If there is also an armored belt, then the load on the jumper can be significantly reduced. I think everything will be fine, and even at 7 cm there is a fairly large margin of safety on the support platform. But in general it is necessary to count, of course. 25-12-2016: Ivan
Doctor, and if we assume, well, purely theoretically 25-12-2016: Dr. Lom
I don't think anything will happen even in this case. But I repeat, for a more accurate answer, a calculation is needed. 09-01-2017: Andrey
In Table 1, in formula 2.3, instead of "q", "Q" is indicated for calculating the deflection. Formula 2.1 for calculating the deflection, being a special case of formula 2.3, when the corresponding values (a=c=l, b=0) are inserted, it takes on a different form. 09-01-2017: Dr. Lom
That's right, there was a typo, but now it doesn't matter. I took the deflection formula for such a design scheme from the reference book of Fesik S.P., as the shortest for the particular case x = a. But as you correctly noted, this formula does not pass the boundary conditions test, so I removed it altogether. I left only the formula for determining the initial angle of rotation in order to simplify the determination of the deflection using the initial parameters method. 02-03-2017: Dr. Lom
In tutorials, as far as I know, such a special case is not considered. Only software, for example, Lira, will help here. 24-03-2017: Eageniy
Good afternoon in the deflection formula 1.4 in the first table - the value in brackets always turns out to be negative 24-03-2017: Dr. Lom
That's right, in all the above formulas, the negative sign in the deflection formula means that the beam bends down along the y axis. 29-03-2017: Oksana
Good afternoon Dr. Lom. Could you write an article about the torque in a metal beam - when does it occur at all, under what design schemes, and, of course, I would like to see the calculation from you with examples. I have a metal beam hinged, one edge is cantilevered and a concentrated load comes to it, and distributed over the entire beam from the reinforced concrete. 100mm thin slab and wall fencing. This beam is extreme. With reinforced concrete the plate is connected by 6 mm rods welded to the beam with a pitch of 600 mm. I can’t understand if there will be a torque, if so, how to find it and calculate the beam section in connection with it? Dr. Lom Victor, emotional strokes are certainly good, but you can’t spread them on bread and you can’t feed your family with them. Calculations are required to answer your question, calculations are time, and time is not emotional strokes. bending deformation consists in the curvature of the axis of the straight rod or in changing the initial curvature of the straight rod (Fig. 6.1). Let's get acquainted with the basic concepts that are used when considering bending deformation. Bending rods are called beams. clean called a bend, in which the bending moment is the only internal force factor that occurs in the cross section of the beam. More often, in the cross section of the rod, along with the bending moment, a transverse force also occurs. Such a bend is called transverse. flat (straight) called a bend when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section. At oblique bend the plane of action of the bending moment intersects the cross section of the beam along a line that does not coincide with any of the main central axes of the cross section. We begin the study of bending deformation with the case of pure plane bending. As already mentioned, with a pure flat bend in the cross section, of the six internal force factors, only the bending moment is non-zero (Fig. 6.1, c): Experiments on elastic models show that if a grid of lines is applied to the surface of the model (Fig. 6.1, a), then with pure bending it is deformed as follows (Fig. 6.1, b): a) longitudinal lines are curved along the circumference; b) the contours of the cross sections remain flat; c) the lines of the contours of the sections intersect everywhere with the longitudinal fibers at a right angle. Based on this, it can be assumed that in pure bending, the cross sections of the beam remain flat and rotate so that they remain normal to the bent axis of the beam (flat section hypothesis in bending). Rice. 6.1 By measuring the length of the longitudinal lines (Fig. 6.1, b), it can be found that the upper fibers lengthen during the bending deformation of the beam, and the lower ones shorten. Obviously, it is possible to find such fibers, the length of which remains unchanged. The set of fibers that do not change their length when the beam is bent is called neutral layer (n.s.). The neutral layer intersects the cross section of the beam in a straight line called neutral line (n. l.) section. To derive a formula that determines the magnitude of the normal stresses that arise in the cross section, consider the section of the beam in the deformed and non-deformed state (Fig. 6.2). Rice. 6.2 By two infinitesimal cross sections, we select an element of length The length of this fiber after deformation (arc length Its relative deformation It's obvious that Therefore, the relative longitudinal strain is proportional to the distance of the fiber from the neutral axis. We introduce the assumption that the longitudinal fibers do not press each other during bending. Under this assumption, each fiber is deformed in isolation, experiencing a simple tension or compression, in which i.e., normal stresses are directly proportional to the distances of the considered points of the section from the neutral axis. We substitute dependence (6.3) into the expression for the bending moment Recall that the integral Dependence (6.4) is Hooke's law in bending, since it relates the deformation (curvature of the neutral layer Substitute (6.4) into (6.3) This is the desired formula for determining the normal stresses in pure bending of the beam at any point in its section. In order to establish where the neutral line is located in the cross section, we substitute the value of normal stresses in the expression for the longitudinal force Because the Equality (6.6) indicates that the axis Equality (6.7) shows that According to (6.5), the greatest stresses are reached in the fibers furthest from the neutral line The process of designing modern buildings and structures is regulated by a huge number of different building codes and regulations. In most cases, standards require certain characteristics to be met, such as deformation or deflection of beams of floor slabs under static or dynamic loading. For example, SNiP No. 2.09.03-85 defines beam deflection for supports and flyovers in no more than 1/150 of the span length. For attic floors, this figure is already 1/200, and for interfloor beams, even less - 1/250. Therefore, one of the mandatory design stages is the calculation of the beam for deflection. The reason why SNiPs set such draconian restrictions is simple and obvious. The smaller the deformation, the greater the margin of safety and flexibility of the structure. For a deflection of less than 0.5%, the bearing element, beam or slab still retains elastic properties, which guarantees the normal redistribution of forces and the preservation of the integrity of the entire structure. With an increase in the deflection, the frame of the building bends, resists, but stands, when the limits of the permissible value are exceeded, the bonds are broken, and the structure loses its rigidity and load-bearing capacity like an avalanche. For your information! To really understand why it is so important to know the amount of deviation from the original position, it is worth understanding that measuring the amount of deflection is the only available and reliable way to determine the state of the beam in practice. By measuring how much the ceiling beam sank, it is possible to determine with 99% certainty whether the structure is in disrepair or not. Before proceeding with the calculation, it will be necessary to recall some dependencies from the theory of the strength of materials and draw up a calculation scheme. Depending on how correctly the scheme is executed and the loading conditions are taken into account, the accuracy and correctness of the calculation will depend. We use the simplest model of a loaded beam shown in the diagram. The simplest analogy for a beam can be a wooden ruler, photo. In our case, the beam: To determine the deformation of the body under load, the formula of the modulus of elasticity is used, which is determined by the ratio E \u003d R / Δ, where E is a reference value, R is the force, Δ is the value of the body deformation. For our case, the dependence will look like this: Δ \u003d Q / (S E) . For a load q distributed along the beam, the formula will look like this: Δ \u003d q h / (S E) . The most important point follows. The above diagram of Young shows the deflection of the beam or the deformation of the ruler as if it were crushed under a powerful press. In our case, the beam is bent, which means that at the ends of the ruler, relative to the center of gravity, two bending moments with different signs are applied. The loading diagram of such a beam is shown below. To convert Young's dependence for the bending moment, it is necessary to multiply both sides of the equation by the arm L. We get Δ*L = Q·L/(b·h·E) . If we imagine that one of the supports is rigidly fixed, and an equivalent balancing moment of forces is applied to the second M max \u003d q * L * 2/8, respectively, the magnitude of the deformation of the beam will be expressed by the dependence Δx \u003d M x / ((h / 3) b (h / 2) E). The value b·h 2 /6 is called the moment of inertia and denoted by W. As a result, Δx = M x / (W E) is obtained, the fundamental formula for calculating the beam for bending W = M / E through the moment of inertia and the bending moment. To accurately calculate deflection, you need to know the bending moment and the moment of inertia. The value of the first can be calculated, but the specific formula for calculating the beam for deflection will depend on the conditions of contact with the supports on which the beam is located, and the method of loading, respectively, for a distributed or concentrated load. The bending moment from a distributed load is calculated by the formula Mmax \u003d q * L 2 / 8. The above formulas are valid only for a distributed load. For the case when the pressure on the beam is concentrated at a certain point and often does not coincide with the axis of symmetry, the formula for calculating the deflection has to be derived using integral calculus. The moment of inertia can be thought of as the equivalent of the beam's resistance to a bending load. The moment of inertia for a simple rectangular beam can be calculated using the simple formula W=b*h 3 /12, where b and h are the dimensions of the beam section. It can be seen from the formula that the same ruler or board of rectangular cross section can have a completely different moment of inertia and deflection, if you put it on supports in the traditional way or put it on an edge. Not without reason, almost all elements of the roof truss system are made not from a 100x150 bar, but from a 50x150 board. Real sections of building structures can have a variety of profiles, from a square, a circle to complex I-beam or channel shapes. At the same time, determining the moment of inertia and the amount of deflection manually, "on a piece of paper", for such cases becomes a non-trivial task for a non-professional builder. In practice, most often there is an inverse problem - to determine the margin of safety of floors or walls for a particular case from a known deflection value. In the construction business, it is very difficult to assess the margin of safety by other, non-destructive methods. Often, according to the magnitude of the deflection, it is required to perform a calculation, evaluate the margin of safety of the building and the general condition of the supporting structures. Moreover, according to the measurements performed, it is determined whether the deformation is permissible, according to the calculation, or the building is in an emergency condition. Advice! In the issue of calculating the limit state of the beam by the magnitude of the deflection, the requirements of SNiP provide an invaluable service. By setting the deflection limit in a relative value, for example, 1/250, building codes make it much easier to determine the emergency state of a beam or slab. For example, if you intend to buy a finished building that has stood for a long time on problematic soil, it would be useful to check the condition of the floor according to the existing deflection. Knowing the maximum allowable deflection rate and the length of the beam, it is possible, without any calculation, to assess how critical the state of the structure is. Construction inspection in assessing the deflection and assessing the bearing capacity of the floor goes in a more complicated way: The question is why is it so difficult if the deflection can be obtained using the formula for a simple beam on hinged supports f=5/24*R*L 2 /(E*h) under distributed force. It is enough to know the span length L, the profile height, the design resistance R and the modulus of elasticity E for a particular floor material. Advice! Use in your calculations the existing departmental collections of various design organizations, in which all the necessary formulas for determining and calculating the ultimate loaded state are summarized in a compressed form. Most developers and designers of serious buildings do the same. The program is good, it helps to calculate the deflection and the main loading parameters of the floor very quickly, but it is also important to provide the customer with documentary evidence of the results obtained in the form of specific sequential calculations on paper. bend called deformation, in which the axis of the rod and all its fibers, i.e., longitudinal lines parallel to the axis of the rod, are bent under the action of external forces. The simplest case of bending is obtained when the external forces lie in a plane passing through the central axis of the rod and do not project onto this axis. Such a case of bending is called transverse bending. Distinguish flat bend and oblique. flat bend- such a case when the bent axis of the rod is located in the same plane in which external forces act. Oblique (complex) bend- such a case of bending, when the bent axis of the rod does not lie in the plane of action of external forces. A bending bar is commonly referred to as beam. With a flat transverse bending of beams in a section with a coordinate system y0x, two internal forces can occur - a transverse force Q y and a bending moment M x; in what follows, we introduce the notation Q And M. If there is no transverse force in the section or section of the beam (Q = 0), and the bending moment is not equal to zero or M is const, then such a bend is commonly called clean. Shear force in any section of the beam is numerically equal to the algebraic sum of the projections onto the axis of all forces (including support reactions) located on one side (any) of the section. Bending moment in the beam section is numerically equal to the algebraic sum of the moments of all forces (including support reactions) located on one side (any) of the section drawn relative to the center of gravity of this section, more precisely, relative to the axis passing perpendicular to the plane of the drawing through the center of gravity of the section drawn. Q-force is resultant distributed over the cross section of the internal shear stresses, A moment M– sum of moments around the central axis of the section X internal normal stresses. There is a differential relationship between internal forces which is used in the construction and verification of diagrams Q and M. Since some of the fibers of the beam are stretched, and some are compressed, and the transition from tension to compression occurs smoothly, without jumps, in the middle part of the beam there is a layer whose fibers only bend, but do not experience either tension or compression. Such a layer is called neutral layer. The line along which the neutral layer intersects with the cross section of the beam is called neutral line th or neutral axis sections. Neutral lines are strung on the axis of the beam. Lines drawn on the side surface of the beam perpendicular to the axis remain flat when bent. These experimental data make it possible to base the conclusions of the formulas on the hypothesis of flat sections. According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and become perpendicular to the bent axis of the beam when it is bent. The cross section of the beam is distorted during bending. Due to transverse deformation, the dimensions of the cross section in the compressed zone of the beam increase, and in the tension zone they are compressed. Assumptions for deriving formulas. Normal stresses 1) The hypothesis of flat sections is fulfilled. 2) Longitudinal fibers do not press on each other and, therefore, under the action of normal stresses, linear tensions or compressions work. 3) The deformations of the fibers do not depend on their position along the width of the section. Consequently, the normal stresses, changing along the height of the section, remain the same across the width. 4) The beam has at least one plane of symmetry, and all external forces lie in this plane. 5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same. 6) The ratios between the dimensions of the beam are such that it works in flat bending conditions without warping or twisting. With a pure bending of a beam on the platforms in its section, only normal stresses, determined by the formula: where y is the coordinate of an arbitrary point of the section, measured from the neutral line - the main central axis x. Normal bending stresses along the height of the section are distributed over linear law. On the extreme fibers, the normal stresses reach their maximum value, and in the center of gravity, the cross sections are equal to zero. The nature of normal stress diagrams for symmetrical sections with respect to the neutral line The nature of normal stress diagrams for sections that do not have symmetry about the neutral line Dangerous points are those farthest from the neutral line. Let's choose some section For any point of the section, let's call it a point TO, the beam strength condition for normal stresses has the form: This axial section modulus about the neutral axis. Its dimension is cm 3, m 3. The moment of resistance characterizes the influence of the shape and dimensions of the cross section on the magnitude of stresses. Strength condition for normal stresses: If the material unequally resists stretching and compression, then two strength conditions must be used: for a stretch zone with an allowable tensile stress; for the compression zone with allowable compressive stress. With transverse bending, the beams on the platforms in its section act as normal, and tangents voltage. straight bend- this is a type of deformation in which two internal force factors arise in the cross sections of the rod: a bending moment and a transverse force. Pure bend- this is a special case of direct bending, in which only a bending moment occurs in the cross sections of the rod, and the transverse force is zero. Pure Bend Example - Plot CD on the rod AB. Bending moment is the value Pa pair of external forces causing bending. From the equilibrium of the part of the rod to the left of the cross section mn it follows that the internal forces distributed over this section are statically equivalent to the moment M, equal and opposite to the bending moment Pa. To find the distribution of these internal forces over the cross section, it is necessary to consider the deformation of the bar. In the simplest case, the rod has a longitudinal plane of symmetry and is subjected to the action of external bending pairs of forces located in this plane. Then the bend will occur in the same plane. rod axis nn 1 is a line passing through the centers of gravity of its cross sections. neutral surface is a surface that does not experience deformation during bending. (Now it is located perpendicular to the drawing, the deformed axis of the rod nn 1 belongs to this surface). Neutral sectional axis- this is the intersection of a neutral surface with any with any cross section (now also located perpendicular to the drawing). Relative extension ε x fibers It follows that deformation of the longitudinal fibers proportional to distance y from the neutral surface and inversely proportional to the radius of curvature ρ
. μ
- Poisson's ratio. As a result of this distortion, all straight cross-sectional lines parallel to the axis z, are bent so as to remain normal to the sides of the section. The radius of curvature of this curve R will be more than ρ
in the same way as ε
x is greater in absolute value than ε
z , and we get These deformations of the longitudinal fibers correspond to stresses All of the above is also true if the rod does not have a longitudinal plane of symmetry in which the bending moment acts, so long as the bending moment acts in the axial plane, which contains one of the two main axes cross section. These planes are called main bending planes. When there is a plane of symmetry and the bending moment acts in this plane, the deflection occurs in it. Moments of internal forces about the axis z balance the external moment M. Moments of effort relative to the axis y are mutually destroyed.
Do you mean that for rotating shafts, the circuits will be different due to the torque? I don’t know how important this is, since it is written in the technical machine book that in the case of turning, the deflection introduced by the torque on the shaft is very small compared to the deflection from the radial component of the cutting force. What do you think?
The calculation of building structures, machine parts, etc., as a rule, consists of two stages: 1. calculation for the limit states of the first group - the so-called strength calculation, 2. calculation for the limit states of the second group. One of the types of calculation for the limit states of the second group is the calculation for deflection.
In your case, in my opinion, the calculation of strength will be more important. Moreover, today there are 4 theories of strength and the calculation for each of these theories is different, but in all theories, the influence of both bending and torque is taken into account in the calculation.
The deflection under the action of a torque occurs in a different plane, but is still taken into account in the calculations. And if this deflection is small or large - the calculation will show.
I do not specialize in calculations of parts of machines and mechanisms, and therefore I cannot point to authoritative literature on this issue. However, in any handbook of a design engineer of machine components and parts, this topic should be properly disclosed.
mail: [email protected]
Skype: dmytrocx75
Q - in kilograms.
l - in centimeters.
E - in kgf/cm2.
I - cm4.
All right? Something strange results are obtained.
If we are talking about a grillage, then, depending on the method of its installation, it can be calculated as a beam on two supports, or as a beam on an elastic foundation.
In general, when calculating columnar foundations, one should be guided by the requirements of SNiP 2.03.01-84.
In addition, if the load on the foundation is transferred from an eccentrically loaded column or not only from the column, then an additional moment will act on the pillow. This should be taken into account in calculations.
But I repeat once again, do not self-medicate, be guided by the requirements of the specified SNiP.
However, in the cases you indicated (except for 1.3), the maximum deflection may not be in the middle of the beam, therefore determining the distance from the beginning of the beam to the section where the maximum deflection will be is a separate task. Recently, a similar issue was discussed in the topic "Design schemes for statically indeterminate beams", look there.
The essence of the problem is as follows: at the base of the sofa, a metal frame is planned from a profiled pipe 40x40 or 40x60, lying on two supports, the distance between which is 2200 mm. QUESTION: is the section of the profile enough for loads from the own weight of the sofa + let's take 3 people of 100 kg each ???
Rigidly fixed beam, span 4 m, supported by 0.2 m. Loads: distributed 100 kg/m along the beam, plus distributed 100 kg/m in the section 0-2 m, plus concentrated 300 kg in the middle (for 2 m). I determined the support reactions: A - 0.5 t; B - 0.4 tons. Then I hung: to determine the bending moment under a concentrated load, it is necessary to calculate the sum of the moments of all forces to the right and left of it. Plus there is a moment on the supports.
How are the loads calculated in this case? It is necessary to bring all distributed loads to concentrated ones and summarize (subtract * distance from the support reaction) according to the formulas of the design scheme? In your article about farms, the layout of all forces is clear, but here I cannot enter into the methodology for determining the acting forces.
The maximum moment in the middle, it turns out M = Q + 2q + from an asymmetric load to a maximum of 1.125q. Those. I added up all 3 loads, is that correct?
If you are not ready to master all this, then it is better to contact a professional designer - it will be cheaper.
Tell me, please, according to which scheme it is necessary to calculate the beam deflection of such a mechanism https://yadi.sk/i/MBmS5g9kgGBbF. Or maybe, without going into calculations, tell me if a 10 or 12 I-beam is suitable for an arrow, a maximum load of 150-200 kg, a lifting height of 4-5 meters. Rack - pipe d = 150, rotary mechanism or axle shaft, or front hub of the Gazelle. The mowing can be made rigid from the same I-beam, and not with a cable. Thank you.
1. The boom can be considered as a two-span continuous beam with a cantilever. The supports for this beam will be not only the stand (this is the middle support), but also the cable attachment points (extreme supports). This is a statically indeterminate beam, but to simplify the calculations (which will lead to a slight increase in the safety factor), the boom can be considered as just a single-span beam with a cantilever. The first support is the cable attachment point, the second is the stand. Then your design schemes are 1.1 (for the load - live load) and 2.3 (boom dead weight - constant load) in table 3. And if the load is in the middle of the span, then 1.1 in table 1.
2. At the same time, we must not forget that the temporary load you will have is not static, but at least dynamic (see the article "Calculation for shock loads").
3. To determine the forces in the cable, it is necessary to divide the support reaction at the place where the cable is attached by the sine of the angle between the cable and the beam.
4. Your rack can be considered as a metal column with one support - a rigid pinch at the bottom (see the article "Calculation of metal columns"). This column will be loaded with a very large eccentricity if there is no counterweight.
5. The calculation of the junctions of the boom and rack and other subtleties of the calculation of the nodes of machines and mechanisms on this site are not yet considered.
what design scheme is ultimately obtained for the floor beam and the cantilever beam, and will the (pink) cantilever beam (brown) affect the decrease in the deflection of the floor beam?
wall - foam block D500, height 250, width 150, armo-belt beam (blue): 150x300, reinforcement 2x? concrete columns 200x200 in the corners, the span of the armo-belt beam 4000 without walls.
overlap: channel 8P (pink), for calculation I took 8U, welded and anchored with armo-belt beam reinforcement, concreted, from the bottom of the beam to the channel 190 mm, from the top 30, span 4050.
to the left of the console - an opening for the stairs, the support of the channel on the pipe? 50 (green), the span to the beam 800.
to the right of the console (yellow) - a bathroom (shower, toilet) 2000x1000, floor - pouring a reinforced ribbed transverse slab, dimensions 2000x1000 height 40 - 100 on a fixed formwork (profiled sheet, wave 60) + tiles on glue, walls - plasterboard on profiles. The rest of the floor is board 25, plywood, linoleum.
At the points of the arrows, the support of the racks of the water tank, 200l.
Walls of the 2nd floor: sheathing with board 25 on both sides, with insulation, height 2000, leaning on the armored belt.
roof: rafters - a triangular arch with a puff, along the floor beam, with a step of 1000, resting on the walls.
console: channel 8P, span 995, welded with reinforced reinforcement, concreted into a beam, welded to the floor channel. span to the right and left along the floor beam - 2005.
While I am cooking the reinforcing cage, it is possible to move the console left and right, but there seems to be nothing to the left?
In the first case, the floor beam can be considered as a hinged two-span beam with an intermediate support - a pipe, and the channel, which you call a cantilever beam, should not be taken into account at all. That's actually the whole calculation.
Further, in order to simply go to a beam with rigid pinching on the extreme supports, you must first calculate the armo-belt for the action of torque and determine the angle of rotation of the cross-section of the armo-belt, taking into account the load from the walls of the 2nd floor and deformations of the wall material under the action of torque. And thus calculate a two-span beam, taking into account these deformations.
In addition, in this case, one should take into account the possible subsidence of the support - the pipe, since it does not rest on the foundation, but on the reinforced concrete slab (as I understood from the figure) and this slab will deform. And the pipe itself will experience compression deformation.
In the second case, if you want to take into account the possible operation of the brown channel, you should consider it as an additional support for the floor beam and thus first calculate the 3-span beam (the support reaction on the additional support will be the load on the cantilever beam), then determine the amount of deflection at the end cantilever beam, recalculate the main beam taking into account the subsidence of the support and, among other things, also take into account the angle of rotation and deflection of the armo-belt at the place where the brown channel is attached. And that's not all.
To calculate the console and installation, it is better to take half the span from the rack to the beam (4050-800-50=3200/2=1600-40/2=1580) or from the edge of the window (1275-40=1235. Yes, and the load on the beam as a window the overlap will have to be recalculated, but you have such examples: The only thing to take as applied to the beam from above Will there be a redistribution of the load applied almost along the axis of the tank?
You assume that the floor slabs are supported on the lower flange of the channel, but what about the other side? In your case, an I-beam would be a more acceptable option (or 2 channels each as a floor beam).
On the other hand, there are no problems - a corner on the mortgages in the body of the beam. I have not yet coped with the calculation of a two-span beam with different spans and different loads, I will try to re-study your article on the calculation of a multi-span beam by the method of moments.
I calculated according to the table. 2, item 1.1. (#comments) as a deflection of a cantilever beam with a load of 70 kg, a shoulder of 1.8 m, a square pipe 120x120x4 mm, a moment of inertia of 417 cm4. I got a deflection - 1.6 mm? True or not?
P.S. And I do not check the accuracy of the calculations, then only rely on yourself.
You can immediately draw up equations of static equilibrium of the system and solve these equations.
I have a beam according to scheme 2.3. Your table gives the formula for calculating the deflection in the middle of the span l / 2, but what formula can be used to calculate the deflection at the end of the console? Will the deflection in the middle of the span be maximum? The result obtained by this formula should be compared with the maximum allowable deflection according to SNiP "Loads and Impacts" using the value l - the distance between points A and B? Thanks in advance, I'm completely confused. And yet, I can’t find the source from which these tables are taken - can I indicate the name?
When you compare the result of deflection in a span with SNiPovksky, then the span length is the distance l between A and B. For the console, instead of l, the distance 2a (double extension of the console) is taken.
I compiled these tables myself, using various reference books on the theory of strength of materials, while checking the data for possible typographical errors, as well as general methods for calculating beams, when there were no diagrams necessary in my opinion in the reference books, so there are many primary sources.
that the reinforcement in the armored belt above the beam is completely destroyed, the armored belt will crack and lie on the beam along with the floor slabs? Will these 7 cm of the support platform be enough?Normal stresses and strains in pure bending.
. Before deforming, the section that bounds the element 
, were parallel to each other (Fig. 6.2, a), and after deformation they tilted somewhat, forming an angle 
. The length of the fibers lying in the neutral layer does not change during bending 
. Let us denote the radius of curvature of the trace of the neutral layer on the plane of the drawing by the letter 
. Let us determine the linear deformation of an arbitrary fiber 
, at a distance 
from the neutral layer.
) is equal to 
. Considering that before deformation all fibers had the same length 
, we obtain that the absolute elongation of the considered fiber
, since the length of the fiber lying in the neutral layer has not changed. Then after substitution 
we get
(6.2)
. Taking into account (6.2)
, (6.3)
in cross section (6.1)
.
represents the moment of inertia of the section about the axis 
.
(6.4)
) with the moment acting in the section. Work 
is called the stiffness of the section in bending, N m 2.
(6.5)
and bending moment 
,
;
(6.6)
(6.7)
- the neutral axis of the section - passes through the center of gravity of the cross section.
And 
- the main central axes of the section.
Ways to Perform Calculation and Deflection Testing
Deflection Calculation Method
We calculate the moments of inertia and forces
Formulas for practical use
Conclusion
, where i.d. - This neutral axis
The normal stress is equal to the ratio of the maximum bending moment to the axial section modulus relative to the neutral axis.
Let the cross section of the rod be a rectangle. Draw two vertical lines on its faces mm And pp. When bent, these lines remain straight and rotate so that they remain perpendicular to the longitudinal fibers of the rod.
A further theory of bending is based on the assumption that not only lines mm And pp, but the entire flat cross section of the rod remains flat after bending and normal to the longitudinal fibers of the rod. Therefore, when bending, the cross sections mm And pp rotate relative to each other around axes perpendicular to the bending plane (drawing plane). In this case, the longitudinal fibers on the convex side experience tension, and the fibers on the concave side experience compression.
Let an arbitrary fiber be at a distance y from a neutral surface. ρ
is the radius of curvature of the curved axis. Dot O is the center of curvature. Let's draw a line n 1 s 1 parallel mm.ss 1 is the absolute elongation of the fiber.
Longitudinal elongation of the fibers of the convex side of the rod is accompanied by lateral constriction, and the longitudinal shortening of the concave side - lateral extension, as in the case of simple stretching and contraction. Because of this, the appearance of all cross sections changes, the vertical sides of the rectangle become slanted. Lateral deformation z:
The voltage in any fiber is proportional to its distance from the neutral axis. n 1 n 2. Position of the neutral axis and radius of curvature ρ
are two unknowns in the equation for σ
x - can be determined from the condition that the forces distributed over any cross section form a pair of forces that balances the external moment M.
