Category Archives: Bending problems. Clean bend. Transverse bend. General concepts Beam of constant cross-section under plane bending
The design process of 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 ensured, for example, deformation or deflection of floor slab beams under static or dynamic load. For example, SNiP No. 2.09.03-85 determines for supports and overpasses the deflection of the beam is no more than 1/150 of the span length. For attic floors this figure is already 1/200, and for interfloor beams and even less - 1/250. Therefore, one of the mandatory design stages is to perform a beam deflection calculation.
Ways to perform deflection calculations and tests
The reason why SNiPs establish such draconian restrictions is simple and obvious. The smaller the deformation, the greater the margin of strength and flexibility of the structure. For a deflection of less than 0.5%, the load-bearing element, beam or slab still retains elastic properties, which guarantees normal redistribution of forces and maintaining the integrity of the entire structure. As the deflection increases, the building frame bends, resists, but stands; when the permissible value is exceeded, the bonds break, and the structure loses its rigidity and load-bearing capacity like an avalanche.
- Use an online software calculator, in which standard conditions are “hardwired”, and nothing more;
- Use ready-made reference data for various types and types of beams, for various support load patterns. It is only necessary to correctly identify the type and size of the beam and determine the desired deflection;
- Calculate the permissible deflection with your hands and your head; most designers do this, while controlling architectural and construction inspectors prefer the second method of calculation.
For your information! To really understand why it is so important to know the magnitude of the deviation from the initial position, it is worth understanding that measuring the amount of deflection is the only accessible and reliable way to determine the condition of the beam in practice.
Measuring how much the beam sank ceiling, it is possible to determine with 99% certainty whether the structure is in emergency condition or not.
Method of performing deflection calculations
Before starting the calculation, you will need to remember some dependencies from the theory of strength of materials and draw up a calculation diagram. Depending on how correctly the diagram is executed and the loading conditions are taken into account, the accuracy and correctness of the calculation will depend.
We use the simplest model loaded beam shown in the diagram. The simplest analogy of a beam can be a wooden ruler, photo.
In our case, the beam:
- It has a rectangular cross-section S=b*h, the length of the supporting part is L;
- The ruler is loaded with a force Q passing through the center of gravity of the bent plane, as a result of which the ends rotate through a small angle θ, with a deflection relative to the initial horizontal position , equal to f ;
- The ends of the beam rest hingedly and freely on fixed supports; accordingly, there is no horizontal component of the reaction, and the ends of the ruler can move in any direction.
To determine the deformation of a body under load, use the formula of the elastic modulus, which is determined by the ratio E = R/Δ, where E is a reference value, R is force, Δ is the amount of deformation of the body.
Calculate moments of inertia and forces
For our case, the dependence will look like this: Δ = Q/(S E) . For a load q distributed along the beam, the formula will look like this: Δ = q h/(S E) .
What follows is the most fundamental point. The above Young diagram shows the deflection of a beam or the deformation of a 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 are applied with different sign. The loading diagram for such a beam is given below.
To transform Young's dependence for the bending moment, it is necessary to multiply both sides of the equality by the shoulder L. We obtain 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 M max = q*L*2/8 will be applied to the second, respectively, the magnitude of the beam deformation will be expressed by the dependence Δх = M x/((h/3) b (h/2) E). The quantity b h 2 /6 is called the moment of inertia and is designated W. The result is Δx = M x / (W E) the fundamental formula for calculating a beam for bending W = M / E through the moment of inertia and bending moment.
To accurately calculate the deflection, you will need to know the bending moment and moment of inertia. The value of the first can be calculated, but the specific formula for calculating a 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 using the formula Mmax = q*L 2 /8. The given 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 must be derived using integral calculus.
The moment of inertia can be thought of as the equivalent of a beam's resistance to bending load. The magnitude of 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 cross-sectional dimensions of the beam.
From the formula it is clear that the same ruler or board rectangular section may have a completely different moment of inertia and amount of deflection if you put it on supports traditional way or put it on edge. No wonder almost all elements rafter system roofs are made not from 100x150 timber, but from 50x150 boards.
Real sections building structures may have the most different profiles, from square, circle to complex I-beam or channel shapes. At the same time, determining the moment of inertia and the amount of deflection manually, “on paper”, for such cases becomes a non-trivial task for a non-professional builder.
Formulas for practical use
In practice, it is most often worth inverse problem- determine the safety margin of floors or walls for a specific case based on the known deflection value. In the construction business, it is very difficult to assess the safety factor by others, non-destructive methods. Often, based on the magnitude of the deflection, it is necessary to perform a calculation, evaluate the safety factor of the building and the general condition load-bearing structures. Moreover, based on the measurements taken, it is determined whether the deformation is acceptable, according to the calculation, or whether the building is in emergency condition.
Advice! In the matter of calculation limit state beams in terms of deflection, the requirements of SNiP provide an invaluable service. By setting the deflection limit in a relative value, for example, 1/250, building codes significantly facilitate the determination of the emergency condition of a beam or slab.
For example, if you intend to buy finished building, which has stood for quite a long time on problematic soil, it would be useful to check the condition of the ceiling according to the existing deflection. Knowing everything permissible norm deflection and the length of the beam, one can assess without any calculation how critical the condition of the structure is.
Construction inspection during deflection assessment and assessment bearing capacity overlap goes a more complicated way:
- Initially, the geometry of the slab or beam is measured and the deflection value is recorded;
- Based on the measured parameters, the assortment of the beam is determined, then the formula for the moment of inertia is selected using the reference book;
- The moment of force is determined by the deflection and moment of inertia, after which, knowing the material, you can calculate the actual stresses in a metal, concrete or wooden beam.
The question is why is it so difficult if the deflection can be obtained using the formula for calculation for a simple beam on hinged supports f=5/24*R*L 2 /(E*h) under a distributed force. It is enough to know the span length L, profile height, design resistance R and elastic modulus E for a specific floor material.
Advice! Use in your calculations existing departmental collections of various design organizations, in which all the necessary formulas for determining and calculating the limiting loaded state are summarized in a condensed form.
Conclusion
Most developers and designers of serious buildings act in a similar way. The program is good, it helps to very quickly calculate the deflection and basic loading parameters of the floor, 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.
29-10-2012: Andrey
There was a typo in the formula for the bending moment for a beam with rigid pinching on supports (3rd from the bottom): the length should be squared. There was a typo in the maximum deflection formula for a beam with rigid pinching on supports (3rd from the bottom): it should be without the “5”.
29-10-2012: Doctor Lom
Yes, indeed, mistakes were made when editing after copying. On this moment errors have been corrected, thanks for your attention.
01-11-2012: Vic
typo in the formula in the fifth example from the top (the degrees next to X and El are mixed up)
01-11-2012: Doctor Lom
And it is true. Corrected. Thank you for your attention.
10-04-2013: flicker
Formula T.1 2.2 Mmax seems to be missing a square after a.
11-04-2013: Doctor Lom
Right. I copied this formula from the “Handbook of Strength of Materials” (edited by S.P. Fesik, 1982, p. 80) and did not even pay attention to the fact that with such a recording, even the dimension is not observed. Now I have recalculated everything personally, and indeed the distance “a” will be squared. Thus, it turns out that the typesetter 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, Diagram 2.4, I am interested in the formula “moment in flight” where the index X is not clear -? Could you answer)
02-05-2013: Doctor Lom
For cantilever beams in 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, in which the rigid support will be 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 moment is located on top, then the moment is negative.
From the point of view of strength of material, the sign of the moment is a rather conventional concept, since in 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, but the direction of action of the moment was taken into account when drawing up the formulas.
25-05-2013: Dmitriy
Please tell me at what ratio of the length of the beam to its diameter these formulas are valid?
I want to know whether this subcode is only for long beams, which are used in the construction of buildings, or can also be used to calculate the deflections of shafts up to 2 m long. Please answer like this l/D>...
25-05-2013: Doctor Lom
Dmitry, I already told you, for rotating shafts the calculation schemes will be different. However, if the shaft is stationary, then it can be considered as a beam, and it does not matter what its cross-section is: round, square, rectangular or something else. These calculation schemes most accurately reflect the state of the beam at l/D>10, with a ratio of 5 25-05-2013: Dmitriy
Thanks for the answer. Can you name other literature that I can refer to in my work? 25-05-2013: Doctor Lom
I don’t know what exact problem you are solving, and therefore it is difficult to have a substantive conversation. I'll try to explain my idea differently. 25-05-2013: Dmitriy
Can I then communicate with you via mail or Skype? I'll tell you what kind of work I do and what the previous questions were for. 25-05-2013: Doctor Lom
You can write to me, email addresses are not difficult to find on the site. But I’ll warn you right away that I don’t do any calculations and don’t sign partnership contracts. 08-06-2013: Vitaly
Question on table 2, option 1.1, deflection formula. Please check the size. 09-06-2013: Doctor Lom
That's right, the output is centimeters. 20-06-2013: Evgeniy Borisovich
Hello. Help me figure it out. Near our cultural center there is a summer wooden stage, size 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 I can’t attach a picture), a polycarbonate covering, make trusses from a profile pipe (square or rectangle), there is a question about my work. If you don't do it, we'll fire you. I say it won’t work, but the administration and my boss say everything will work. What should I do? 20-06-2013: Doctor Lom
22-08-2013: Dmitriy
If a beam (a cushion under a column) lies on dense soil (more precisely, buried below the freezing depth), then what scheme should be used to calculate such a beam? Intuition suggests that the “two-support” option is not suitable and that the bending moment should be significantly less. 22-08-2013: Doctor Lom
Calculation of foundations is a separate big topic. In addition, it is not entirely clear which beam we are talking about. If we mean a cushion under a column of a columnar foundation, then the basis for calculating such a cushion is the strength of the soil. The purpose of the pillow is to redistribute the load from the column to the base. The lower the strength, the larger the area of the pillow. Or the greater the load, the larger the cushion area with the same soil strength. 23-08-2013: Dmitriy
This refers to a cushion 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 questionable. 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 calculate the bending moment in a cushion lying on the ground, as in a beam on two hinged supports. The question is - what calculation scheme is used to calculate the bending moment in the cushion. 24-08-2013: Doctor Lom
The height and cross-section of the reinforcement in your case are determined as for cantilever beams (along the width and length of the cushion). Scheme 2.1. Only in your case, the support reaction is the load on the column, or more precisely, part of the load on the column, and the uniformly distributed load is the resistance of the soil. In other words, the specified calculation scheme needs to be turned over. 10-10-2013: Yaroslav
Good evening. Please help me choose metal. beam for a shed of 4.2 meters. A residential building has two floors, the base is covered with hollow slabs 4.8 meters long, on top there is a load-bearing wall of 1.5 bricks, 3.35 m long and 2.8 m high. Then there is a doorway. On top of this wall there are floor slabs on one side 4.8 m long . on the other 2.8 meters on the slabs there is again a load-bearing wall as on the floor below and above there are wooden beams 20 by 20 cm long 5 m. 6 pieces and 3 meters long 6 pieces the floor is made of boards 40 mm. 25 m2. There are no other loads. Please suggest me which I-beam to take in order to sleep peacefully. So far everything has been standing for 5 years. 10-10-2013: Doctor Lom
Look in the section: "Calculation of metal structures" at the article "Calculation of a metal lintel for load-bearing walls"; it describes in sufficient detail the process of selecting the section of a beam depending on the current load. 04-12-2013: Kirill
Please tell me where I can get acquainted with the derivation of the formulas for the maximum deflection of a beam for pp. 1.2-1.4 in Table 1 04-12-2013: Doctor Lom
The derivation of formulas for various options for applying loads is not provided on my website. You can see the general principles on which the derivation of such equations is based in the articles “Fundamentals of strength strength, calculation formulas” and “Fundamentals of strength 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: Doctor Lom
I don’t see any errors, much less non-compliance with dimensions, in the calculation scheme you specified. Find out what exactly the error is. 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: Doctor Lom
No, M and Mmax have a single unit of measurement 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 gave me a problem. I ask for your help, because... I don’t want to solve it “by eye”. 12-10-2014: Doctor Lom
It depends on many factors. In addition, you did not indicate the thickness of the pipe. For example, with a thickness of 2 mm, the moment of resistance 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 kgm or 69.4 kgm, then the maximum permissible 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 arise when the load is transferred not along the center of gravity of the section). And this is with a static load, and the load will most likely be dynamic, or even shock (depending on the design of the sofa and the activity of the children, mine jump on the sofas so that it takes your breath away), so do the math for yourself. The article “Calculation values for rectangular profile pipes” will help you. 20-10-2014: student
Doc, please help. 21-10-2014: Doctor Lom
To begin with, a rigidly fixed beam and support 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 hinged beam with cantilevers (see Table 3), or a three-span rigidly clamped beam with 2 additional supports and unequal spans (in this case, the three-moment equations 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 there is an armored belt of 200x300h, the outer perimeter is 4400x4400. There are 3 channels anchored into it, with a step of 1 m. The span is without racks, one of them has the heaviest option, the load is asymmetrical. THOSE. count the beam as hinged? 21-10-2014: Doctor Lom
22-10-2014: student
in fact yes. As I understand it, the deflection of the channel will also rotate the armored belt itself at the attachment point, so you will get a hinged beam? 22-10-2014: Doctor 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 arising from the action of a uniformly distributed load acting on a certain section of the beam. And only then add up the values of the moments. Each load will have its own calculation scheme. 07-02-2015: Sergey
Is there an error in the Mmax formula for case 2.3 in Table 3? Beam with a console, probably the plus instead of the minus should be in parentheses 07-02-2015: Doctor Lom
No, not a mistake. The load on the cantilever reduces the moment in the span, but does not increase it. However, this can be seen from the moment diagram. 17-02-2015: Anton
Hello, first of all, thanks for the formulas, I saved them in my bookmarks. Please tell me, is there a beam above the span, four logs rest on the beam, distances: 180mm, 600mm, 600mm, 600mm, 325mm. I figured out the diagram and the bending moment, but I can’t understand how the deflection formula (Table 1, diagram 1.4) will change if the maximum moment is on the third lag. 17-02-2015: Doctor Lom
I have already answered similar questions several times in the comments to the article “Calculation schemes for statically indeterminate beams.” But you are lucky, for clarity, I performed the calculation using 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 over time I will add to it. 22-02-2015: Novel
Doc, I really can’t master all these formulas that are incomprehensible to me. Therefore, I ask you for help. I want to make a cantilever staircase in my house (the steps will be bricked up with reinforced concrete when building the 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 intuitively understand that the reinforcement will be “something thicker” so that the steps will be something thinner? But can reinforced concrete up to 3cm thick cope with a load of 150kg at the edge? Please help me, I really don’t want to screw up. I would be very grateful if you could help me calculate... 22-02-2015: Doctor Lom
The fact that you cannot master fairly simple formulas is your problem. In the section “Basics of Strength of Strength” all this is discussed in sufficient detail. Here I will say that your project is absolutely unrealistic. 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 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 in beams must be at least 15 mm. 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) should be chosen to calculate a balcony slab, a cantilever clamped 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: Doctor 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 diagrams, can be calculated according to these diagrams. 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. As I understand it, if you calculate according to 2 tables. diagram 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 at the ends of the slab? And according to scheme 2.1, I calculate the moment at the support, add it to the moment according to scheme 1.1 and according to the correct one, in order to reinforce it, I need to find the moment in the span. If I have a slab overhang of 1.45 m (in the clear), how can I calculate “x” to find the moment in the span? 03-04-2015: Doctor Lom
The moment in the span will vary from Ql at the support to 0 at the point of application of the load, which can be seen from the moment diagram. If your load is applied at two points at the ends of the slab, then in this case it is more advisable to provide beams that absorb loads at the edges. In this case, the slab can already be calculated as a beam on two supports - beams or a slab supported 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, using the letter “G”. What calculation scheme should I use then? 04-04-2015: Doctor 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: Doctor Lom
I won’t assess the reliability of such a design without calculations, but you can calculate it using the following criteria: 05-06-2015: student
Doc, where can I show you the picture? 05-06-2015: student
Did you still have a forum? 05-06-2015: Doctor Lom
There was, but I have absolutely no time to sort through spam in search of normal questions. So that's it for now. 06-06-2015: student
Doc, my link is https://yadi.sk/i/GardDCAEh7iuG 07-06-2015: Doctor Lom
The choice of design scheme will depend on what you want: simplicity and reliability or approximation to the actual operation of the structure through successive approximations. 07-06-2015: student
Doc, thanks. I need simplicity and reliability. This area is the busiest. I even thought about tying the tank post to the rafters to reduce the load on the floor, given that the water would be drained in the winter. I can’t get into such a jungle of calculations. In general, will the cantilever reduce deflection? 07-06-2015: student
Doc, one more question. the console is in the middle of the window span, does it make sense to move it to the edge? Sincerely 07-06-2015: Doctor Lom
In general, the console will reduce the deflection, but as I already said, how much in your case is a big question, and a shift to the center of the window opening will reduce the role of the console. And also, if this is your most loaded area, then maybe you can simply strengthen the beam, for example, with another similar channel? I don’t know your loads, but the load of 100 kg of water and half the weight of the tank does not seem so impressive to me, but from the point of view of deflection, do 8P channels 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 beam on hinges. If there is greater dynamics when walking, I constructively include the possibility of reducing the pitch of the floor beams. The house is a country house, so the dynamics are tolerable. The lateral displacement of the channels has a greater influence, but this can be treated by installing cross braces or fastening the flooring. The only thing is, will the concrete pouring crumble? I assume it will be supported on the upper and lower flanges of the channel plus welded reinforcement in the ribs and mesh on top. 08-06-2015: Doctor Lom
I already told you, you shouldn’t count on the console. 09-06-2015: student
Doc, I understand. 29-06-2015: Sergey
Good afternoon. I would like to ask you: the foundation was cast: piles of concrete 1.8 m deep, and then a strip 1 m deep was cast with concrete. The question is this: is the load transferred only to the piles or is it evenly distributed to both the piles and the tape? 29-06-2015: Doctor Lom
As a rule, piles are made in weak soils so that the load on the foundation is transmitted through the piles, so grillages on piles are calculated like 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 represents a regular strip foundation. Like that. 29-06-2015: Sergey
Thank you. It’s just that the site turns out to be a mixture of clay and sand. Moreover, the clay layer is very hard: the layer can only be removed with a crowbar, etc., etc. 29-06-2015: Doctor Lom
I don’t know all your conditions (distance between piles, number of floors, etc.). From your description, it looks like you made a regular strip foundation and piles for reliability. Therefore, you just need 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! We need your help with the calculations. A metal gate 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 brick (post 38 x 38 cm). What cross-section and thickness should the pipe be so that there is no bending? 05-07-2015: Doctor Lom
You correctly assumed that your post should be treated like a cantilever beam. And even with the calculation scheme, you almost got it right. The fact is that 2 forces will act on your pipe (on the upper and lower canopies) and the value of these forces will depend on the distance between the canopies. More details in the article “Determination of pull-out force (why the dowel does not stay in the wall).” Thus, in your case, you should perform 2 deflection calculations according to design scheme 1.2, and then add the results obtained, 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: Doctor Lom
01-08-2015: Paul
Please tell me, in diagram 2.2 of table 3, how to determine the deflection at point C if the lengths of the cantilever sections are different? 01-08-2015: Doctor Lom
In this case, you need to go through the full cycle. Whether this is necessary or not, I don’t know. For an example, look at the article on calculating a beam under 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 concrete for a given metal cantilever beam 120x120x4 mm with a collar of 70 kg - (for example, at least 1/3 of the length) 04-08-2015: Doctor Lom
In fact, calculating pinching is a separate big topic. The fact is that the resistance of concrete to compression is one thing, but the deformation of the soil on which the concrete of the foundation presses is quite 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, will the metal gate post be cast in a concrete pile with a diameter of 300 mm and a length of 1 m, and the piles at the top will be connected by a concrete grillage to the reinforcement frame? concrete everywhere M 300. I.e. there will be no soil deformation. I would like to know an approximate, albeit with a large margin of safety, ratio. 05-08-2015: Doctor Lom
Then really 1/3 of the length should be enough to create a rigid pinch. For an example, look at the article “Types of supports, which design scheme to choose.” 05-08-2015: Yuri
20-09-2015: Carla
21-09-2015: Doctor Lom
You can first calculate the beam separately for each load according to the design schemes presented here, and then add the results obtained taking into account the signs. 08-10-2015: Natalia
Hello, Doctor))) 08-10-2015: Doctor 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 In this 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 may 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 on my house. Gazebos, canopies, supports. I’ll try to remember that at one time I overslept as a diligent student and then accidentally passed it to the Soviet Higher Technical School. 27-11-2015: Michael
Aren't all dimensions in SI? (see comment 06/08/2013 from Vitaly) 27-11-2015: Doctor Lom
Which units you will use, kgf or Newton, kgf/cm^2 or Pascal, is not of fundamental importance. As a result, you will still get centimeters (or meters) as an output. See comment 06/09/2013 from Doctor Loma. 28-04-2016: Denis
Hello, I have a beam according to scheme 1.4. what is the formula to find shear force 28-04-2016: Doctor Lom
For each section of the beam, the values of the transverse force will be different (which, however, can be seen from the corresponding diagram of the transverse forces). In the first section 0< x < a, поперечная сила будет равна опорной реакции А. На втором участке a < x < l-b, поперечная сила будет равна А-Q и так далее, больше подробностей смотрите в статье "Основы сопромата. Расчетные формулы". 31-05-2016: Vitaly
Thank you very much, you are great! 14-06-2016: Denis
I came across your site during this time. I almost missed my calculations, I always thought that a cantilever beam with a load at the end of the beam would bend more than with a uniformly distributed load, but formulas 1.1 and 2.1 in Table 2 show the opposite. Thanks for your work 14-06-2016: Doctor Lom
In general, it makes sense to compare a concentrated load with a uniformly distributed one only when one load is reduced to another. For example, when Q = ql, the formula for determining the deflection according to 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 a simply uniformly distributed load. So the formulas for calculation schemes 1.1 and 2.1 do not show anything to the contrary, and initially you were right. 16-06-2016: engineer Garin
Good afternoon! I still can’t figure it out, I would be very grateful if you could help me figure it out once and for all - when calculating (any) an ordinary I-beam with a usual distributed load along its length, what moment of inertia should I use - Iy or Iz and why? I can’t find strength of strength in any textbook; everywhere they write that the cross section should tend to a square and the smallest moment of inertia should be taken. I just can’t grasp the physical meaning by the tail; can I somehow interpret this on my fingers? 16-06-2016: Doctor Lom
I advise you to start by looking at the articles “Fundamentals of Strength Materials” and “Towards the Calculation of Flexible Rods under the Action of a Compressive Eccentric Load”, everything is explained there in sufficient detail and clearly. Here I will add that it seems to me that you are confusing the calculations for transverse and longitudinal bending. Those. when the load is perpendicular to the neutral axis of the rod, 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 longitudinal bending on the load-bearing capacity of the rod. Of course, when calculating the transverse load (vertical load for a horizontal beam), the moment of inertia should be taken depending on the position of the beam, but in any case it will be Iz. And when calculating 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, the question is why in Table 1 for formulas 1.3 and 1.4 the deflection formulas are essentially the same and the size b. is it not reflected in formula 1.4 in any way? 23-06-2016: Doctor Lom
With an asymmetrical load, the deflection formula for design scheme 1.4 will be quite cumbersome, but it should be remembered that the deflection in any case will be less than when applying a symmetrical load (of course, provided b 03-11-2016: vladimir
in Table 1 for formulas 1.3 and 1.4, the deflection formula should be Ql^3/24EI instead of Qa^3/24EI. For a long time I could not understand why the deflection with the crystal did not converge 03-11-2016: Doctor Lom
That's right, another typo due to inattentive editing (I hope it's the last one, but not a fact). Corrected, thanks for your attention. 16-12-2016: Ivan
Hello, Doctor Lom. The question is the following: I was looking through photos from the construction site and noticed one thing: the factory-made reinforced concrete lintel is approximately 30*30 cm, supported on a three-layer reinforced concrete panel about 7 centimeters (the reinforced concrete panel was sawed down a little to rest the lintel 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 now 16-12-2016: Doctor 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, of course, you need to count. 25-12-2016: Ivan
Doctor, if we assume, well, purely theoretically 25-12-2016: Doctor Lom
I think even in this case nothing will happen. But I repeat, a more accurate answer requires calculation. 09-01-2017: Andrey
In Table 1, in formula 2.3, to calculate the deflection, instead of “q”, “Q” is indicated. Formula 2.1 for calculating the deflection, being a special case of formula 2.3, when inserting the corresponding values (a=c=l, b=0) takes on a different form. 09-01-2017: Doctor 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 S.P. Fesik’s reference book, as the shortest for the special 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 deflection using the initial parameters method. 02-03-2017: Doctor Lom
As far as I know, such a special case is not considered in textbooks. Only software will help here, for example, Lyra. 24-03-2017: Eageniy
Good afternoon, in the deflection formula 1.4 in the first table - the value in brackets is always negative 24-03-2017: Doctor Lom
Everything is correct, in all the given formulas, the negative sign in the deflection formula means that the beam bends down along the y-axis. 29-03-2017: Oksana
Good afternoon, Doctor 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 your calculations with examples. I have a hingedly supported metal beam, one edge is cantilevered and a concentrated load comes to it, and the load is distributed over the entire beam from the reinforced concrete. thin slab 100 mm and fence wall. This beam is the outermost one. 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 whether there will be a torque there, if so, how to find it and calculate the cross-section of the beam in connection with it? Doctor Lom Victor, emotional stroking is, of course, good, but you can’t spread it on bread and you can’t feed your family with it. To answer your question, calculations are required, calculations are time, and time is not emotional stroking. 13-11-2017: 1
In Table 2, example No. 1.1 there is an error in the formula for theta(x) 04-06-2019: Anton
Hello, dear doctor, I have a question about the initial parameters method. At the beginning of the article, you write that the formula for beam deflection can be obtained by properly integrating the bending moment equation twice, dividing the result by EI and adding to this the result of integrating the rotation angle. 05-06-2019: Doctor Lom
The mistake is that you did not integrate the moment equation, but the result of solving this equation for a point in the middle of the beam, and these are different things. In addition, when adding, you should carefully monitor the “+” or “-” sign. If you carefully analyze the deflection formula given for this design scheme, you will understand what we are talking about. And when integrating the angle of rotation, the result is q*l4/48, not q*l4/96, and in the final formula it will come with a minus, since such an initial angle of rotation will lead to deflection of the beam below the x-axis. 09-07-2019: Alexander
Hello, in T.1 2.3 formulas for moments, what is taken as X? The middle of the distributed load? 09-07-2019: Doctor Lom
For all tables, distance x is the distance from the origin point (usually support A) to the point in question on the neutral axis of the beam. Those. The given formulas allow you to determine the value of the moment for any cross section of the beam. For a cantilever beam loaded with a distributed load of intensity kN/m and a concentrated moment of kN m (Fig. 3.12), it is required to: construct diagrams of shear forces and bending moments, select a beam of circular cross-section with an allowable normal stress kN/cm2 and check the strength of the beam according to tangential stresses with permissible tangential stress kN/cm2. Beam dimensions m; m; m. The horizontal reaction in the embedment is zero, since external loads in the z-axis direction do not act on the beam. We choose the directions of the remaining reactive forces arising in the embedment: we will direct the vertical reaction, for example, downward, and the moment – clockwise. Their values are determined from the static equations: When composing these equations, we consider the moment to be positive when rotating counterclockwise, and the projection of the force to be positive if its direction coincides with the positive direction of the y-axis. From the first equation we find the moment at the seal: From the second equation - vertical reaction: The positive values we obtained for the moment and vertical reaction in the embedment indicate that we guessed their directions. In accordance with the nature of the fastening and loading of the beam, we divide its length into two sections. Along the boundaries of each of these sections we will outline four cross sections (see Fig. 3.12), in which we will use the method of sections (ROZU) to calculate the values of shearing forces and bending moments. Section 1. Let's mentally discard the right side of the beam. Let's replace its action on the remaining left side with a cutting force and a bending moment. For the convenience of calculating their values, let’s cover the discarded right side of the beam with a piece of paper, aligning the left edge of the sheet with the section under consideration. Let us recall that the shear force arising in any cross section must balance all external forces (active and reactive) that act on the part of the beam being considered (that is, visible) by us. Therefore, the shearing force must be equal to the algebraic sum of all the forces that we see. Let us also present the rule of signs for the shearing force: an external force acting on the part of the beam under consideration and tending to “rotate” this part relative to the section in a clockwise direction causes a positive shearing force in the section. Such an external force is included in the algebraic sum for the definition with a plus sign. In our case, we see only the reaction of the support, which rotates the part of the beam visible to us relative to the first section (relative to the edge of the piece of paper) counterclockwise. That's why The bending moment in any section must balance the moment created by the external forces visible to us relative to the section in question. Consequently, it is equal to the algebraic sum of the moments of all forces that act on the part of the beam we are considering, relative to the section under consideration (in other words, relative to the edge of the piece of paper). In this case, the external load, bending the part of the beam under consideration with its convexity downward, causes a positive bending moment in the section. And the moment created by such a load is included in the algebraic sum for determination with a “plus” sign. We see two efforts: reaction and closing moment. However, the force's leverage relative to section 1 is zero. That's why We took the “plus” sign because the reactive moment bends the part of the beam visible to us with a convex downward. Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now, unlike the first section, the force has a shoulder: m. Therefore Section 3. Closing the right side of the beam, we find Section 4. Cover the left side of the beam with a sheet. Then Using the found values, we construct diagrams of shearing forces (Fig. 3.12, b) and bending moments (Fig. 3.12, c). Under unloaded areas, the diagram of shearing forces goes parallel to the axis of the beam, and under a distributed load q - along an inclined straight line upward. Under the support reaction in the diagram there is a jump down by the value of this reaction, that is, by 40 kN. In the diagram of bending moments we see a break under the support reaction. The bend angle is directed towards the support reaction. Under a distributed load q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6 on the diagram there is an extremum, since the diagram of the shearing force in this place passes through the zero value. The normal stress strength condition has the form: where is the moment of resistance of the beam during bending. For a beam of circular cross-section it is equal to: The largest absolute value of the bending moment occurs in the third section of the beam: Then the required beam diameter is determined by the formula We accept mm. Then "Overvoltage" is what is allowed. The greatest tangential stresses arising in the cross section of a beam of circular cross-section are calculated by the formula where is the cross-sectional area. According to the diagram, the largest algebraic value of the shearing force is equal to that is, the strength condition for tangential stresses is also satisfied, and with a large margin. For a simply supported beam loaded with a distributed load of intensity kN/m, concentrated force kN and concentrated moment kN m (Fig. 3.13), it is necessary to construct diagrams of shear forces and bending moments and select a beam of I-beam cross-section with an allowable normal stress kN/cm2 and permissible tangential stress kN/cm2. Beam span m. Rice. 3.13 For a given simply supported beam, it is necessary to find three support reactions: , and . Since only vertical loads perpendicular to its axis act on the beam, the horizontal reaction of the fixed hinged support A is zero: . The directions of vertical reactions are chosen arbitrarily. Let us direct, for example, both vertical reactions upward. To calculate their values, let’s create two static equations: Let us recall that the resultant linear load , uniformly distributed over a section of length l, is equal to , that is, equal to the area of the diagram of this load and it is applied at the center of gravity of this diagram, that is, in the middle of the length. Let's check: . Recall that forces whose direction coincides with the positive direction of the y-axis are projected (projected) onto this axis with a plus sign: that is true. We divide the length of the beam into separate sections. The boundaries of these sections are the points of application of concentrated forces (active and/or reactive), as well as points corresponding to the beginning and end of the distributed load. There are three such sections in our problem. Along the boundaries of these sections, we will outline six cross sections, in which we will calculate the values of shear forces and bending moments (Fig. 3.13, a). Section 1. Let's mentally discard the right side of the beam. For the convenience of calculating the shearing force and bending moment arising in this section, we will cover the part of the beam we discarded with a piece of paper, aligning the left edge of the sheet of paper with the section itself. The shearing force in the beam section is equal to the algebraic sum of all external forces (active and reactive) that we see. In this case, we see the reaction of the support and the linear load q distributed over an infinitesimal length. The resultant linear load is zero. That's why The plus sign is taken because the force rotates the part of the beam visible to us relative to the first section (the edge of a piece of paper) clockwise. The bending moment in the beam section is equal to the algebraic sum of the moments of all the forces that we see relative to the section under consideration (that is, relative to the edge of the piece of paper). We see the support reaction and linear load q distributed over an infinitesimal length. However, the force has a leverage of zero. The resultant linear load is also zero. That's why Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now we see the reaction and load q acting on a section of length . The resultant linear load is equal to . It is attached in the middle of a section of length . That's why Let us recall that when determining the sign of the bending moment, we mentally free the part of the beam we see from all the actual supporting fastenings and imagine it as if pinched in the section under consideration (that is, we mentally imagine the left edge of the piece of paper as a rigid embedment). Section 3. Close the right side. We get Section 4. Cover the right side of the beam with a sheet. Then Now, to check the correctness of the calculations, let’s cover the left side of the beam with a piece of paper. We see the concentrated force P, the reaction of the right support and the linear load q distributed over an infinitesimal length. The resultant linear load is zero. That's why That is, everything is correct. Section 5. As before, close the left side of the beam. Will have Section 6. Let's close the left side of the beam again. We get Using the found values, we construct diagrams of shearing forces (Fig. 3.13, b) and bending moments (Fig. 3.13, c). We make sure that under the unloaded area the diagram of shearing forces runs parallel to the axis of the beam, and under a distributed load q - along a straight line sloping downwards. There are three jumps in the diagram: under the reaction - up by 37.5 kN, under the reaction - up by 132.5 kN and under the force P - down by 50 kN. In the diagram of bending moments we see breaks under the concentrated force P and under the support reactions. The fracture angles are directed towards these forces. Under a distributed load of intensity q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. Under the concentrated moment there is a jump of 60 kN m, that is, by the magnitude of the moment itself. In section 7 on the diagram there is an extremum, since the diagram of the shearing force for this section passes through the zero value (). Let us determine the distance from section 7 to the left support. Straight bend. Plane transverse bending Constructing diagrams of internal force factors for beams Constructing diagrams of Q and M using equations Constructing diagrams of Q and M using characteristic sections (points) Strength calculations for direct bending of beams Principal stresses during bending. A complete check of the strength of beams. The concept of the center of bending. Determination of displacements in beams during bending. Concepts of deformation of beams and conditions for their rigidity Differential equation of the curved axis of a beam Method of direct integration Examples of determining displacements in beams by the method of direct integration Physical meaning of integration constants Method of initial parameters (universal equation of the curved axis of a beam). 1.3, b). Rice. 1.3 When calculating the bending moment in a given section, the moments of external forces lying to the left of the section are considered positive if they are directed clockwise. For the right side of the beam - vice versa. It is convenient to determine the sign of the bending moment by the nature of the deformation of the beam. The bending moment is considered positive if, in the section under consideration, the cut-off part of the beam bends convexly downward, i.e., the lower fibers are stretched. In the opposite case, the bending moment in the section is negative. There are differential relationships between the bending moment M, shear force Q and load intensity q. 1. The first derivative of the shear force along the abscissa of the section is equal to the intensity of the distributed load, i.e. Positive ordinates of the M diagram are laid down, and negative ordinates are laid upward, i.e., the M diagram is constructed from the side of the stretched fibers. The construction of Q and M diagrams for beams should begin with determining the support reactions. For a beam with one clamped end and the other free end, the construction of diagrams Q and M can be started from the free end, without determining the reactions in the embedment. 1.2. The construction of Q and M diagrams using the Beam equations is divided into sections within which the functions for the bending moment and shear force remain constant (do not have discontinuities). The boundaries of the sections are the points of application of concentrated forces, pairs of forces and places of change in the intensity of the distributed load. At each section, an arbitrary section is taken at a distance x from the origin of coordinates, and for this section equations for Q and M are drawn up. Using these equations, diagrams of Q and M are constructed. Example 1.1 Construct diagrams of transverse forces Q and bending moments M for a given beam (Fig. 1.4,a). Solution: 1. Determination of support reactions. We compose equilibrium equations: from which we obtain The reactions of the supports are determined correctly. The beam has four sections Fig. 1.4 loads: CA, AD, DB, BE. 2. Construction of diagram Q. Section CA. In section CA 1, we draw an arbitrary section 1-1 at a distance x1 from the left end of the beam. We define Q as the algebraic sum of all external forces acting to the left of section 1-1: The minus sign is taken because the force acting to the left of the section is directed downward. The expression for Q does not depend on the variable x1. Diagram Q in this section will be depicted as a straight line parallel to the abscissa axis. Section AD. On the section we draw an arbitrary section 2-2 at a distance x2 from the left end of the beam. We define Q2 as the algebraic sum of all external forces acting to the left of section 2-2: 8 The value of Q is constant in the section (does not depend on the variable x2). The Q plot on the section is a straight line parallel to the abscissa axis. Plot DB. On the site we draw an arbitrary section 3-3 at a distance x3 from the right end of the beam. We define Q3 as the algebraic sum of all external forces acting to the right of section 3-3: The resulting expression is the equation of an inclined straight line. Section BE. On the site we draw a section 4-4 at a distance x4 from the right end of the beam. We define Q as the algebraic sum of all external forces acting to the right of section 4-4: 4 Here the plus sign is taken because the resultant load to the right of section 4-4 is directed downward. Based on the obtained values, we construct Q diagrams (Fig. 1.4, b). 3. Construction of diagram M. Plot m1. We define the bending moment in section 1-1 as the algebraic sum of the moments of forces acting to the left of section 1-1. Using this method, the values of Q and M are calculated in characteristic sections. The characteristic sections are the boundary sections of sections, as well as sections where a given internal force factor has an extreme value. Within the limits between the characteristic sections, the outline 12 of the diagram is established on the basis of the differential dependencies between M, Q, q and the conclusions arising from them. Example 1.3 Construct diagrams Q and M for the beam shown in Fig. 1.6, a. Rice. 1.6. Solution: We start constructing the Q and M diagrams from the free end of the beam, while the reactions in the embedment do not need to be determined. The beam has three loading sections: AB, BC, CD. There is no distributed load in sections AB and BC. Shear forces are constant. The Q diagram is limited to straight lines parallel to the x-axis. Bending moments vary linearly. Diagram M is limited by straight lines inclined to the abscissa axis. There is a uniformly distributed load on section CD. Transverse forces vary according to a linear law, and bending moments - according to the law of a square parabola with convexity in the direction of the distributed load. At the boundary of sections AB and BC, the transverse force changes abruptly. At the boundary of sections BC and CD, the bending moment changes abruptly. 1. Construction of diagram Q. We calculate the values of transverse forces Q in the boundary sections of sections: Based on the calculation results, we construct diagram Q for the beam (Fig. 1, b). From diagram Q it follows that the transverse force on section CD is equal to zero in the section located at a distance qa a q from the beginning of this section. In this section, the bending moment has a maximum value. 2. Constructing diagram M. We calculate the values of bending moments in the boundary sections of sections: At the maximum moment in the section Based on the calculation results, we construct diagram M (Fig. 5.6, c). Example 1.4 Using a given diagram of bending moments (Fig. 1.7, a) for a beam (Fig. 1.7, b), determine the acting loads and construct diagram Q. The circle indicates the vertex of a square parabola. Solution: Let's determine the loads acting on the beam. Section AC is loaded with a uniformly distributed load, since the diagram M in this section is a square parabola. In the reference section B, a concentrated moment is applied to the beam, acting clockwise, since in diagram M we have a jump upward by the magnitude of the moment. In the NE section, the beam is not loaded, since the M diagram in this section is limited by an inclined straight line. The reaction of support B is determined from the condition that the bending moment in section C is equal to zero, i.e. To determine the intensity of the distributed load, we create an expression for the bending moment in section A as the sum of the moments of forces on the right and equate it to zero. Now we determine the reaction of support A. To do this, we will compose an expression for bending moments in the section as the sum of the moments of forces on the left. The design diagram of the beam with a load is shown in Fig. 1.7, c. Starting from the left end of the beam, we calculate the values of transverse forces in the boundary sections of the sections: Diagram Q is shown in Fig. 1.7, d. The considered problem can be solved by drawing up functional dependencies for M, Q in each section. Let's choose the origin of coordinates at the left end of the beam. In the AC section, the diagram M is expressed by a square parabola, the equation of which has the form Constants a, b, c are found from the condition that the parabola passes through three points with known coordinates: Substituting the coordinates of the points into the equation of the parabola, we obtain: The expression for the bending moment will be Differentiating the function M1 , we obtain the dependence for the transverse force. After differentiating the function Q, we obtain an expression for the intensity of the distributed load. In the section NE, the expression for the bending moment is presented in the form of a linear function. To determine the constants a and b, we use the conditions that this straight line passes through two points, the coordinates of which are known. We obtain two equations: ,b from which we have a 20. The equation for the bending moment in the section NE will be After double differentiation of M2, we will find. Using the found values of M and Q, we construct diagrams of bending moments and shear forces for the beam. In addition to the distributed load, concentrated forces are applied to the beam in three sections, where there are jumps on diagram Q and concentrated moments in the section where there is a shock on diagram M. Example 1.5 For a beam (Fig. 1.8, a), determine the rational position of the hinge C, at which the largest bending moment in the span is equal to the bending moment in the embedment (in absolute value). Construct diagrams of Q and M. Solution Determination of support reactions. Despite the fact that the total number of support links is four, the beam is statically determinate. The bending moment in the hinge C is zero, which allows us to create an additional equation: the sum of the moments about the hinge of all external forces acting on one side of this hinge is equal to zero. Let us compile the sum of the moments of all forces to the right of the hinge C. The diagram Q for the beam is limited by an inclined straight line, since q = const. We determine the values of transverse forces in the boundary sections of the beam: The abscissa xK of the section, where Q = 0, is determined from the equation from which the diagram M for the beam is limited by a square parabola. Expressions for bending moments in sections, where Q = 0, and in the embedment are written respectively as follows: From the condition of equality of moments, we obtain a quadratic equation for the desired parameter x: Real value x2x 1.029 m. We determine the numerical values of transverse forces and bending moments in characteristic sections of the beam. Figure 1.8, b shows the diagram Q, and in Fig. 1.8, c – diagram M. The problem considered could be solved by dividing the hinged beam into its constituent elements, as shown in Fig. 1.8, d. At the beginning, the reactions of the supports VC and VB are determined. Diagrams of Q and M are constructed for the suspended beam SV from the action of the load applied to it. Then they move to the main beam AC, loading it with an additional force VC, which is the pressure force of the beam CB on the beam AC. After that, diagrams Q and M are built for beam AC. 1.4. Strength calculations for direct bending of beams Strength calculations based on normal and shear stresses. When a beam bends directly in its cross sections, normal and tangential stresses arise (Fig. 1.9). 11) For beams made of brittle materials with sections that are asymmetrical with respect to the neutral axis, if the diagram M is unambiguous (Fig. 1.12), it is necessary to write down two strength conditions - the distance from the neutral axis to the most distant points of the stretched and compressed zones of the dangerous section, respectively; P – permissible stresses for tension and compression, respectively. Fig.1.12. Considering the left side of the beam, we obtain The diagram of transverse forces is shown in Fig. 1.14, c. The diagram of bending moments is shown in Fig. 5.14, g. 2. Geometric characteristics of cross section 3. The highest normal stresses in section C, where Mmax acts (modulo): MPa. The maximum normal stresses in the beam are almost equal to the permissible ones. 4. The highest tangential stresses in section C (or A), where max Q acts (modulo): Here is the static moment of the half-section area relative to the neutral axis; b2 cm – section width at the level of the neutral axis. 5. Tangential stresses at a point (in the wall) in section C: Fig. 1.15 Here Szomc 834.5 108 cm3 is the static moment of the area of the section located above the line passing through point K1; b2 cm – wall thickness at the level of point K1. Diagrams and for section C of the beam are shown in Fig. 1.15. Example 1.7 For the beam shown in Fig. 1.16, a, required: 1. Construct diagrams of transverse forces and bending moments along characteristic sections (points). 2. Determine the dimensions of the cross section in the form of a circle, rectangle and I-beam from the condition of strength under normal stresses, compare the cross-sectional areas. 3. Check the selected dimensions of beam sections according to tangential stress. Given: Solution: 1. Determine the reactions of the beam supports. Check: 2. Construction of diagrams Q and M. Values of transverse forces in characteristic sections of the beam 25 Fig. 1.16 In sections CA and AD, load intensity q = const. Consequently, in these areas the Q diagram is limited to straight lines inclined to the axis. In section DB, the intensity of the distributed load is q = 0, therefore, in this section, the diagram Q is limited to a straight line parallel to the x axis. The Q diagram for the beam is shown in Fig. 1.16, b. Values of bending moments in characteristic sections of the beam: In the second section, we determine the abscissa x2 of the section in which Q = 0: Maximum moment in the second section Diagram M for the beam is shown in Fig. 1.16, c. 2. We create a strength condition based on normal stresses, from which we determine the required axial moment of resistance of the section from the expression determined by the required diameter d of a beam of a circular section. Area of a circular section. For a beam of a rectangular section. Required height of the section. Area of a rectangular section. Determine the required number of the I-beam. Using the tables of GOST 8239-89, we find the nearest higher value of the axial moment of resistance 597 cm3, which corresponds to I-beam No. 33 with the characteristics: A z 9840 cm4. Tolerance check: (underload by 1% of the permissible 5%) the nearest I-beam No. 30 (W 2 cm3) leads to significant overload (more than 5%). We finally accept I-beam No. 33. We compare the areas of the round and rectangular sections with the smallest area A of the I-beam: Of the three sections considered, the most economical is the I-beam section. 3. We calculate the highest normal stresses in the dangerous section 27 of the I-beam (Fig. 1.17, a): Normal stresses in the wall near the flange of the I-beam section The diagram of normal stresses in the dangerous section of the beam is shown in Fig. 1.17, b. 5. Determine the highest shear stresses for the selected sections of the beam. a) rectangular section of the beam: b) round section of the beam: c) I-beam section: Tangential stresses in the wall near the flange of the I-beam in dangerous section A (right) (at point 2): The diagram of tangential stresses in dangerous sections of the I-beam is shown in Fig. 1.17, c. The maximum tangential stresses in the beam do not exceed the permissible stresses Example 1.8 Determine the permissible load on the beam (Fig. 1.18, a), if 60 MPa, the cross-sectional dimensions are given (Fig. 1.19, a). Construct a diagram of normal stresses in a dangerous section of a beam at an allowable load. We will start with the simplest case, the so-called pure bend. Pure bending is a special case of bending in which the transverse force in the sections of the beam is zero. Pure bending can only occur when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads causing pure bending, shown in Fig. 88. In sections of these beams, where Q = 0 and, therefore, M = const; pure bending takes place. The forces in any section of the beam during pure bending are reduced to a pair of forces, the plane of action of which passes through the axis of the beam, and the moment is constant. Voltages can be determined based on the following considerations. 1. The tangential components of forces along elementary areas in the cross section of a beam cannot be reduced to a pair of forces, the plane of action of which is perpendicular to the section plane. It follows that the bending force in the section is the result of action along elementary areas only normal forces, and therefore with pure bending the stresses are reduced only to normal. 2. In order for efforts on elementary sites to be reduced to only a couple of forces, among them there must be both positive and negative. Therefore, both tension and compression fibers of the beam must exist. 3. Due to the fact that the forces in different sections are the same, the stresses at the corresponding points of the sections are the same. Let's consider some element near the surface (Fig. 89, a). Since no forces are applied along its lower edge, which coincides with the surface of the beam, there are no stresses on it. Therefore, there are no stresses on the upper edge of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we arrive at The same conclusion, etc. It follows that there are no stresses along the horizontal edges of any element. Considering the elements that make up the horizontal layer, starting with the element near the surface of the beam (Fig. 90), we come to the conclusion that there are no stresses along the lateral vertical edges of any element. Thus, the stress state of any element (Fig. 91, a), and in the limit, fibers, should be represented as shown in Fig. 91,b, i.e. it can be either axial tension or axial compression. 4. Due to the symmetry of the application of external forces, the section along the middle of the length of the beam after deformation should remain flat and normal to the axis of the beam (Fig. 92, a). For the same reason, sections in quarters of the length of the beam also remain flat and normal to the axis of the beam (Fig. 92, b), unless the extreme sections of the beam during deformation remain flat and normal to the axis of the beam. A similar conclusion is valid for sections in eighths of the length of the beam (Fig. 92, c), etc. Consequently, if during bending the outer sections of the beam remain flat, then for any section it remains It is a fair statement that after deformation it remains flat and normal to the axis of the curved beam. But in this case, it is obvious that the change in elongation of the fibers of the beam along its height should occur not only continuously, but also monotonically. If we call a layer a set of fibers that have the same elongations, then it follows from what has been said that the stretched and compressed fibers of the beam should be located on opposite sides of the layer in which the elongations of the fibers are equal to zero. We will call fibers whose elongations are zero neutral; a layer consisting of neutral fibers is a neutral layer; the line of intersection of the neutral layer with the cross-sectional plane of the beam - the neutral line of this section. Then, based on the previous reasoning, it can be argued that with pure bending of a beam, in each section there is a neutral line that divides this section into two parts (zones): a zone of stretched fibers (stretched zone) and a zone of compressed fibers (compressed zone). ). Accordingly, at the points of the stretched zone of the section, normal tensile stresses should act, at the points of the compressed zone - compressive stresses, and at the points of the neutral line the stresses are equal to zero. Thus, with pure bending of a beam of constant cross-section: 1) only normal stresses act in sections; 2) the entire section can be divided into two parts (zones) - stretched and compressed; the boundary of the zones is the neutral section line, at the points of which the normal stresses are equal to zero; 3) any longitudinal element of the beam (in the limit, any fiber) is subjected to axial tension or compression, so that adjacent fibers do not interact with each other; 4) if the extreme sections of the beam during deformation remain flat and normal to the axis, then all its cross sections remain flat and normal to the axis of the curved beam. Let us consider an element of a beam subjected to pure bending, concluding before deformation after deformation where p is the radius of curvature of the neutral fiber. Therefore, the absolute lengthening of segment AB is equal to and relative elongation Since, according to position (3), fiber AB is subjected to axial tension, then during elastic deformation This shows that normal stresses along the height of the beam are distributed according to a linear law (Fig. 94). Since the equal force of all forces over all elementary cross-sectional areas must be equal to zero, then from where, substituting the value from (5.8), we find But the last integral is a static moment about the Oy axis, perpendicular to the plane of action of the bending forces. Due to its equality to zero, this axis must pass through the center of gravity O of the section. Thus, the neutral line of the section of the beam is a straight line y, perpendicular to the plane of action of bending forces. It is called the neutral axis of the beam section. Then from (5.8) it follows that the stresses at points lying at the same distance from the neutral axis are the same. The case of pure bending, in which the bending forces act in only one plane, causing bending only in that plane, is planar pure bending. If the said plane passes through the Oz axis, then the moment of elementary forces relative to this axis should be equal to zero, i.e. Substituting here the value of σ from (5.8), we find The integral on the left side of this equality, as is known, is the centrifugal moment of inertia of the section relative to the y and z axes, so The axes about which the centrifugal moment of inertia of the section is zero are called the main axes of inertia of this section. If they, in addition, pass through the center of gravity of the section, then they can be called the main central axes of inertia of the section. Thus, with flat pure bending, the direction of the plane of action of bending forces and the neutral axis of the section are the main central axes of inertia of the latter. In other words, to obtain a flat, pure bend of a beam, a load cannot be applied to it arbitrarily: it must be reduced to forces acting in a plane that passes through one of the main central axes of inertia of the sections of the beam; in this case, the other main central axis of inertia will be the neutral axis of the section. As is known, in the case of a section that is symmetrical about any axis, the axis of symmetry is one of its main central axes of inertia. Consequently, in this particular case we will certainly obtain pure bending by applying appropriate loads in a plane passing through the longitudinal axis of the beam and the axis of symmetry of its section. A straight line perpendicular to the axis of symmetry and passing through the center of gravity of the section is the neutral axis of this section. Having established the position of the neutral axis, it is not difficult to find the magnitude of the stress at any point in the section. In fact, since the sum of the moments of elementary forces relative to the neutral axis yy must be equal to the bending moment, then whence, substituting the value of σ from (5.8), we find Since the integral and from expression (5.8) we obtain The product EI Y is called the bending stiffness of the beam. The greatest tensile and largest compressive stresses in absolute value act at the points of the section for which the absolute value of z is greatest, i.e., at the points furthest from the neutral axis. With the notation, Fig. 95 we have The value Jy/h1 is called the moment of resistance of the section to tension and is designated Wyr; similarly, Jy/h2 is called the moment of resistance of the section to compression and denote Wyc, so and therefore If the neutral axis is the axis of symmetry of the section, then h1 = h2 = h/2 and, therefore, Wyp = Wyc, so there is no need to distinguish them, and they use the same notation: calling W y simply the moment of resistance of the section. Consequently, in the case of a section symmetrical about the neutral axis, All the above conclusions were obtained on the basis of the assumption that the cross sections of the beam, when bent, remain flat and normal to its axis (hypothesis of flat sections). As has been shown, this assumption is valid only in the case when the extreme (end) sections of the beam remain flat during bending. On the other hand, from the hypothesis of plane sections it follows that elementary forces in such sections should be distributed according to a linear law. Therefore, for the validity of the resulting theory of flat pure bending, it is necessary that the bending moments at the ends of the beam be applied in the form of elementary forces distributed along the height of the section according to a linear law (Fig. 96), coinciding with the law of stress distribution along the height of the section beams. However, based on the Saint-Venant principle, it can be argued that changing the method of applying bending moments at the ends of the beam will cause only local deformations, the effect of which will affect only a certain distance from these ends (approximately equal to the height of the section). The sections located throughout the rest of the length of the beam will remain flat. Consequently, the stated theory of flat pure bending for any method of applying bending moments is valid only within the middle part of the length of the beam, located from its ends at distances approximately equal to the height of the section. From here it is clear that this theory is obviously inapplicable if the height of the section exceeds half the length or span of the beam.
Do you mean that for rotating shafts the patterns will be different due to the torque? I don’t know how important this is, since the technical book says 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?
Calculation of building structures, machine parts, etc., as a rule, consists of two stages: 1. calculation based on limit states of the first group - the so-called strength calculation, 2. calculation based on limit states of the second group. One of the types of calculation for limit states of the second group is calculation for deflection.
In your case, in my opinion, strength calculations will be more important. Moreover, today there are 4 theories of strength and the calculations for each of these theories are different, but in all theories the influence of both bending and torque is taken into account when calculating.
Deflection under the action of torque occurs in a different plane, but is still taken into account in the calculations. Whether this deflection is small or large - the calculation will show.
I do not specialize in calculations of machine parts and mechanisms and therefore cannot indicate authoritative literature on this issue. However, in any reference book for an engineer-designer of machine components and parts, this topic should be properly covered.
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Q - in kilograms.
l - in centimeters.
E - in kgf/cm2.
I - cm4.
Is everything right? Some strange results are obtained.
If we are talking about a grillage, then depending on the method of its construction, it can be designed 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 cushion. This should be taken into account when making calculations.
But I repeat once again, do not self-medicate, follow the requirements of the specified SNiP.
However, in the cases you indicated (except 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 question was discussed in the topic “Calculation schemes for statically indeterminate beams”, look there.
The essence of the problem is this: at the base of the sofa there is planned a metal frame made of profiled pipe 40x40 or 40x60, lying on two supports with a distance of 2200 mm. QUESTION: is the profile cross-section sufficient for loads from the sofa’s own weight + let’s take 3 people weighing 100 kg???
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 area of 0-2 m, plus concentrated 300 kg in the middle (at 2 m). Determined the support reactions: A – 0.5 t; B - 0.4 t. Then I got stuck: 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, a moment appears on the supports.
How are loads calculated in this case? It is necessary to bring all distributed loads to concentrated ones and sum them up (subtract from the support reaction * distance) according to the formulas of the design scheme? In your article about farms, the layout of all forces is clear, but here I cannot go into the methodology for determining the acting forces.
The maximum moment is 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 handle all this, then it is better to contact a professional designer - it will be cheaper.
Please tell me what scheme should be used to calculate the deflection of the beam of such a mechanism https://yadi.sk/i/MBmS5g9kgGBbF. Or maybe, without going into calculations, tell me whether a 10 or 12 I-beam is suitable for the boom, maximum load 150-200 kg, lifting height 4-5 meters. Rack - pipe d=150, rotating mechanism or axle shaft, or Gazelle front hub. 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 (the outer 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 simply a single-span beam with a cantilever. The first support is the cable attachment point, the second is the stand. Then your calculation schemes are 1.1 (for load - live load) and 2.3 (boom dead weight - permanent 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 your live load will not be static, but at least dynamic (see the article “Calculation for shock loads”).
3. To determine the forces in the cable, you need 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 - rigid pinching at the bottom (see the article "Calculation of metal columns"). The load will be applied to this column with a very large eccentricity if there is no counterload.
5. Calculation of the junction points of the boom and rack and other subtleties of the calculation of machine components and mechanisms are not yet considered on this site.
what design scheme is ultimately obtained for the floor beam and cantilever beam, and will the cantilever beam (brown color) affect the reduction in the deflection of the floor beam (pink)?
wall - foam block D500, height 250, width 150, armored belt beam (blue): 150x300, reinforcement 2x?12, top and bottom, additionally bottom in the window span and top in places where the beam rests on the window opening - mesh?5, cell 50. B in the corners there are concrete columns 200x200, the span of the reinforced belt beam is 4000 without walls.
ceiling: channel 8P (pink), for calculations took 8U, welded and anchored with the reinforcement of the reinforced belt beam, concreted, from the bottom of the beam to the channel 190 mm, from the top 30, span 4050.
to the left of the console there is an opening for the stairs, the channel is supported on a pipe? 50 (green), the span to the beam is 800.
to the right of the console (yellow) - bathroom (shower, toilet) 2000x1000, floor - poured reinforced ribbed transverse slab, dimensions 2000x1000 height 40 - 100 on permanent formwork (corrugated sheet, wave 60) + tiles with adhesive, walls - plasterboard on profiles. The rest of the floor is board 25, plywood, linoleum.
At the points of the arrows, the supports of the water tank, 200 l, are supported.
Walls of the 2nd floor: sheathing with 25 boards on both sides, with insulation, height 2000, supported by an armored belt.
roof: rafters - a triangular arch with a tie, along the floor beam, in increments of 1000, supported on the walls.
console: channel 8P, span 995, welded with reinforced reinforcement, concreted into a beam, welded to the ceiling channel. span on the right and left along the floor beam - 2005.
While I’m welding the reinforcement frame, it’s possible to move the console left and right, but there doesn’t seem to be any reason to move it 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, cannot be taken into account at all. That's the whole calculation.
Next, in order to simply move on to a beam with rigid pinching on the outer supports, you must first calculate the reinforced belt for the action of torque and determine the angle of rotation of the cross section of the reinforced belt, taking into account the load from the walls of the 2nd floor and the deformation of the wall material under the influence 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 rests not on the foundation, but on a reinforced concrete slab (as I understand from the figure) and this slab will be deformed. And the pipe itself will experience compression deformation.
In the second case, if you want to take into account the possible work 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 reinforced belt at the point 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. And the load on the beam is the same as the window the overlap will have to be recalculated, but you have such examples. The only thing is to take the load 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 bottom 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).
There are no problems on the other side - the corner is on the embeds in the body of the beam. I haven’t yet coped with the calculation of a two-span beam with different spans and different loads, I’ll try to re-study your article on calculating a multi-span beam using the method of moments.
I calculated from the table. 2, clause 1.1. (#comments) as the deflection of a cantilever beam with a load of 70 kg, shoulder 1.8 m, square pipe 120x120x4 mm, moment of inertia 417 cm4. I got a deflection of 1.6 mm? True or false?
P.S. I don’t check the accuracy of the calculations, so just 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 a 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 using this formula must be compared with the maximum permissible 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 original source from which these tables were taken - is it possible to indicate the name?
When you compare the obtained result of deflection in a span with SNiPovk, then the length of the span is the distance l between A and B. For the cantilever, instead of l, the distance 2a (double cantilever overhang) is taken.
I compiled these tables myself, using various reference books on the theory of strength of materials, while checking the data for possible typos, as well as general methods for calculating beams, when the necessary diagrams in my opinion were not in the reference books, so there are many primary sources.
that the reinforcement in the reinforced belt above the beam is completely destroyed, the reinforced belt will crack and fall on the beam along with the floor slabs? Is this 7 cm support area enough?
Let’s say I don’t know the deflection of the beam in design scheme 2.1 (Table 1). I will integrate the bending moment twice ∫q*l2/8dx=q*l3/24;∫q*l3/24dx=q*l4/96.
Then I will divide the value by EI. q*l4/(96*EI).
And I’ll add to it the result of integrating the rotation angle: ∫q*l3/24dx=q*l4/96. q*l4/(96*EI)+q*l4/(96*EI)=q*l4/(48*EI).
You get the value -5*q*l4/(384*EI).
Please tell me. Where did I go wrong?Calculation scheme for the problem of direct transverse bending
Rice. 3.12Solution of the problem "straight transverse bending"
Determining support reactions
kN.
kNm.
kN; kNm.
kN;
kNm.
kNm.
.Determine the required cross-sectional diameter of the beam
,
.
kN cm
cm.
kN/cm2 kN/cm2.
,We check the strength of the beam by the highest tangential stresses
,
kN. Then
kN/cm2 kN/cm2,An example of solving the problem "straight transverse bending" No. 2
Condition of an example problem on straight transverse bending
An example of a straight bending problem - calculation diagram
Solution of an example problem on straight bending
Determining support reactions
;
kN.We construct diagrams of shearing forces and bending moments
kN.
kNm.
kN;
kNm.
kN;
Stress state of a beam under pure bending
located between sections m-m and n-n, which are spaced one from the other at an infinitesimal distance dx (Fig. 93). Due to position (4) of the previous paragraph, sections m- m and n - n, which were parallel before deformation, after bending, remaining flat, will form an angle dQ and intersect along a straight line passing through point C, which is the center of curvature neutral fiber NN. Then the part AB of the fiber enclosed between them, located at a distance z from the neutral fiber (the positive direction of the z axis is taken towards the convexity of the beam during bending), will turn after deformation into an arc AB. A piece of neutral fiber O1O2, having turned into an arc, O1O2 will not change its length, while fiber AB will receive an elongation:
is. moment of inertia of the section relative to the yy axis, then
