Calculation of a beam on one support. Design schemes for statically indeterminate beams
18-01-2013: vladimir
in scheme 1 in the formulas of the moment on the supports l is not squared?
18-01-2013: Dr. Lom
With a concentrated load in the moment formulas, the length squared cannot be.
27-02-2013: Vadim
27-02-2013: Dr. Lom
To calculate continuous beams with three spans or more, it is easier to formulate an equation of three moments. I can’t explain what kind of equation this is in the comment format, and I don’t have an article on this topic yet. You can see the article: "Double-span beams". The principles outlined in this article can also be applied to three-span beams.
11-05-2013: Dmitriy
"Table 2. Single-span beam with rigid clamping on support A and articulated support B. Figure 1.2 "Is it possible to count according to this scheme for x=a ?
I need to know how the shaft will sag at the point of contact of the cutter.
11-05-2013: Dr. Lom
Can. However, in your case, it would be more correct to calculate not only the bending, but also the torque.
11-05-2013: Dmitriy
Can you post the formula for my case when x=a. I'm afraid that the calculation error will go beyond the technological one, from recalculating such a long formula.
- The book says that the shafts bend very little from the torque, so they usually do not take into account.
11-05-2013: Dmitriy
"Table 2. Single-span beam with rigid clamping on support A and hinged support B. Figure 1.2" Can you post the formula for the particular case x=a ? It means that x=a should be taken during integration. Then the formula should be significantly simplified.
Thank you!
11-05-2013: Dr. Lom
x=a is a special case of the given formulas, i.e. at a distance from the beginning of the beam equal to a:
Ma = Aa + MA. With a deflection the same story.
Moreover, if we consider the section of the beam at x>a, then the formulas will be even more complicated. There is nothing I can do about it, but I can advise the following. The maximum deflection in your case will be when the load is applied approximately in the middle of the shaft, i.e. when a? b, the closer you move the cutter to one or the second support, increasing the distance a or b, the less deflection will result. Therefore, it is much easier to calculate the maximum deflection according to scheme 1.1, and to be sure, make an additional margin, i.e. increase the deflection calculated in this way by 3-5% is unlikely to be obtained exact calculation the deflection will be more, but you can increase it by 10-15% for more confidence.
14-05-2013: Faith
Hello, is it possible to see somewhere the calculation (derivation of the formula for the bending moment) for the case of loading table 1, clause 2.5 for a triangle?
14-05-2013: Dr. Lom
All the formulas used to compile the tables remained on paper (it takes too long to type them). In addition, there are several methods for calculating statically indeterminate structures. In this case, a technique was used that was described in sufficient detail in the article "Two-span beams" (http://website/item230.html)
27-07-2013: Dmitriy
By Dear Doctor Scrap! How to correctly calculate a three-span beam on hinged supports from a uniform load, if all the spans are different?
27-02-2013: Dr. Lom
To calculate continuous beams with three spans or more, it is easier to formulate an equation of three moments. I can’t explain what kind of equation this is in the comment format, and I don’t have an article on this topic yet. You can see the article: "Double-span beams". The principles outlined in this article can also be applied to three-span beams.
This calculation has a very specific practical application - the calculation of loads on the axles of trucks. The truck has a trailer with 4 legs. The first one rests on the hitch of the tractor, the other three - the axles of the wheels located at a distance (base) from the hitch and at the same distance from each other. evenly/unevenly distributed load inside the body, I managed to lead to a point load with a certain coordinate. But the distribution of the load between the axles of the wheels - alas. Will be grateful for any help / tip on the material.
27-07-2013: orderly Petrovich
Eh, my friend, you need to go to another hospital, we don’t even have such a department.
27-07-2013: Dr. Lom
Petrovich is right, the calculation of rotating shafts and mechanisms is a separate story. In addition, it should be borne in mind that the loads will not be static, but dynamic and impact, moreover, not only vertical, which are most often considered in construction, but also horizontal, which occur when moving with acceleration.
But you can look at the articles "Multi-span continuous beams". In particular, they consider the calculation of three-span beams. True, your case is not specifically considered, but the skill of the designer is to find a way out of difficult situations, simplifying them. For example, in your case, when calculating for vertical loads (from the weight of goods), it is not at all necessary to consider the coupling device of the tractor as a support. I'm certainly not an automotive specialist, but it seems to me that most coupling devices are designed to absorb horizontal loads that occur during accelerated movement, while up and down movement is quite possible, but I could be wrong.
Thus, you will get a two-span beam with two consoles. And although I do not have such a design scheme, but here you can use the principle of superposition for a uniformly distributed load, i.e. it is possible to separately calculate a two-span non-cantilever beam and two consoles, and then add the obtained values of the required parameters. And if the calculation is made for a concentrated load inside the spans, then the consoles do not matter at all.
But still, do not forget Petrovich's advice, the calculation of rotating shafts is not for me.
29-07-2013: Dmitriy
orderly Petrovich - interested in determining the load on the axle in a static state. In this mode, the reaction of the axles, the gravity of the load and the gravity of the trailer itself is fully consistent with the scheme with the beam and the reactions of the supports.
29-07-2013: orderly Petrovich
Of course, it’s none of my business to get into pre-Khtur affairs, but while they are resting, I’ll chat with you.
If for a static state, then your calculation and a bottle of drunk is not worth it, because you need to calculate only two axes. After all, what is the third axis needed for? - for insurance. If a wheel of a two-ton passenger car bursts or flies off - it's one thing, but if a trailer has 40 tons in it, and at high speed, then you can't have fun. And therefore, slowly calculate for yourself all the options for a single-span double-console beam, and there will be only two of them, and two more for a single-span single-console beam, if the coupling device is taken into account as a support, then choose the most loaded one.
That's how I think about it myself.
30-07-2013: Dmitriy
Petrovich, at the weight control they are punished for overloading on any of the specific axles. Since there are three of them behind, the load (in an unknown way) is still divided into all three axles.
I can calculate - I can arrange pallets with cargo so that all axles will not be overloaded.
30-07-2013: Dmitriy
After all, what is the third axis needed for? - Here http://www.packer3d.ru/online/veh-by-pal it is clearly shown that all axes are involved
30-07-2013: orderly Petrovich
Wow what! So I would have said right away, you need to pull everything out of you with tweezers. I drew a picture for myself for a friend, but oh well.
As I understand it, the caculator is for evenly spaced pallets, giving an evenly distributed load. You, my dear, know how to arrange the pallets in such a way that the load on the axles is evenly distributed.
On that at once I will tell - you spit on this business. Theoretically, tacos are possible, but for that, half of the load, or more, must be thrown away.
In addition, I doubt that there are scales such that the load on each axle is determined, rather, separate scales for the tractor and trailer. So it is, no?
If so, then the recipe is simple, from the beginning of the trailer to the first rear axle, the height of the pallets is 2/3 of the total height, from 1 rear to 3 rear axles, gradually increase the height of the pallets to full height, then to the full height, if the trailer is like in caculator.
And you will have a more even distribution of the load between the axles. Or rather, you don’t even need to count - the dimensions of the pallets will not allow. In addition, too much data is required for the calculation.
And if the old man Petrovich messed up with the height, then on the scales they will tell you, you need more than 2/3, or less.
Oh, someone’s throat is dry from these conversations, we should go and drink beer while there is no doctor.
25-11-2013: Anton
Good afternoon. I have a question about formula 1.2 in table 2. When calculating the deflection using this formula and substituting the condition x=a=b=l/2 into the resulting expression, the expression given in the formula above is not obtained. The difference lies in the difference between the number in front of the product EI. When substituting, it turns out not 107, but 109. Tell me, what is the mistake? Can this calculation method be approximate?
25-11-2013: Dr. Lom
The fact is that, using the formula, you determine the value of the deflection in the middle of the span and at the bottom of the resulting expression it will really be 109.7. Meanwhile, at a beam with rigid pinching on one support and hinged on the second support, the maximum deflection will be shifted towards the hinged support. Row 1 of Table 2 shows this maximum value. Since the distance from support A to the cross-section with a maximum span is greater than 0.5l, then to determine this value, one should use formulas that take into account the action of the shear force at the point of application (or determine the deflection value, counting from support B, taking into account the angle of rotation on support B ). Not that these formulas are so complicated, but they take up a lot of space, and therefore they are not given in the table.
26-11-2013: Anton
Thanks for the answer. Yes, indeed, you are right. The point of maximum deflection will be a little closer from the middle of the beam to the hinge support. But here I had another question. Differentiating the expression for finding the deflection, finding the extremum of the function, he thereby obtained x of the maximum deflection. Substituting this value of x into the deflection formula under the same conditions a = b = L / 2, I got 107.555 in the denominator. I don’t know what the problem is, but also in I found the same formula in other sources for a particular case (a=b=L/2). I'm interested in this because I do calculations at work and I need to get an accurate result.
26-11-2013: Dr. Lom
But here, when it comes to tenths of a percent and, in general, fractional numbers obtained as a result of rather complex calculations, tabular values should really be considered as approximate ones. The value you get is more accurate, the tabular value gives more deflection and therefore contributes to a small additional margin (0.2%) when calculating for the 2nd group of limit states.
27-11-2013: Anton
You dispelled my doubts. Thank you very much for the explanation and quick replies!
06-12-2013: Maksim
2 table, scheme 3.1. At point B, is the value of the moment on the diagram not equal to the value of the applied moment at this point?
07-12-2013: Dr. Lom
It doesn’t work, more precisely, the value of the moment on the diagram at point B is equal to the value of the moment applied at point B. With this direction of action, the moment is considered negative (-M), respectively, when a negative bending moment acts on support B, a positive bending moment arises on support A, and Here the support reaction on support A will be negative. If we substitute all the values \u200b\u200bgiven in the table into the equation of moments, then at x \u003d l, on support B you will get the same negative moment Mb \u003d -M.
07-12-2013: Maksim
for example, m=10, L=2.
then Ax = 3*10/2*2 = 7.5
Ma = 10/2=5
Mb= 5+ 7.5 = 12.5
07-12-2013: Dr. Lom
You do not quite understand the essence of the formulas and do not follow the signs:
not Ax \u003d 3 * 10 / 2 * 2 \u003d 7.5, but simply the support reaction A \u003d 7.5. x is a variable indicating the distance from the beginning of the beam to the considered cross section. At point B, the value x = L = 2.
Further, if m = 10, then Ma = -5. Then on base B
Mb \u003d -5 + 7.5x2 \u003d 10
09-07-2014: Zarif
Dear Doctor.
Do you have the value of the diagram M for pitched roofs?
09-07-2014: Dr. Lom
Look at the article "Examples of the calculation of rafters and battens" there are diagrams corresponding to the calculation.
11-02-2015: Sanmart
Dear Doctor Lom!
Do you have formulas for calculating the maximum deflection and moment for a two-span beam fully loaded with a uniformly distributed load with spans of different lengths?
The circuit is almost like on 2.3, but q is distributed from A to C.
If you are too lazy to type these formulas into the editor, send them somehow, like in a scanned format, and I will return them to you in the editor.
11-02-2015: Dr. Lom
The point is that to state all possible cases of loading for all options physically impossible. So you're out of luck - you'll have to use general formulas. The section "Statically indeterminate structures" and in particular the articles "Double-span beams" and "Statically indeterminate beams. Equations of three moments" are at your service. Here I will say that it is possible to determine the maximum moment on support B by using the same scheme 2.3 twice, namely
MB \u003d M1 + M2 \u003d - q (l1 ^ 3 + l2 ^ 3) / (8 (l1 + l2)).
To determine the deflections of cross sections about the x-axis, you must first determine the angles of rotation on the supports, and then use the general differential deflection equation. More details in the article "Fundamentals of strength-of-material. Determination of beam deflection". But in any case, whatever the length of the second span, the maximum deflection in one of the spans will be greater than ql^4/185EI and less than 7ql^4/768EI. If for you such limits are too blurry and more accuracy is required, then only the calculation.
12-02-2015: Valentine
Hello Doctor Lom. I would like to clarify with you the special case described in the section "Table 3. Double-span beam with hinged supports. Fig. 1.3" Do you have the opportunity to supplement it based on the fact that in this case the distances l are not equal, but different, i.e. l1 and l2. Interested in the reactions in the supports and the moment in the support. A very urgent request. Thank you.
12-02-2015: Sanmart
Eh ... I will remember the long-forgotten sopromat ...
Nevertheless thank you!
12-02-2015: Dr. Lom
I just answered yesterday similar question. The tables contain formulas for special cases, however, the most common ones. For common cases, like yours, the formulas become too cumbersome and visibility is lost. In such cases, it is necessary to perform a full calculation by the moment method or by the force method, since you have only one unknown support reaction.
Nevertheless, the tables presented are very convenient for a preliminary assessment of structures. For example, if your load case is like the one in Table 3, design scheme 1.3, then with a decrease in the length of one of the spans calculated values And support reactions and the moment on the support and other quantities will be unambiguously less. Thus, a simplified calculation will only increase the margin of safety; this knowledge is quite enough when calculating a structure made in 1-2 copies. Well, for mass-produced structures, an accurate calculation is necessary.
21-03-2015: David
Dear Doctor Lom
I have gable roof with identical sides with a ridge without support, but rigidly fixed rafters (welding), the bottom can be considered as a hinge. Is it possible to apply the formula Table 2 2.1 or something else to calculate the deflections, if you don’t mind writing a formula or a link
22-03-2015: Dr. Lom
This will not be entirely correct, and the weld must be designed for the appropriate loads to ensure rigidity. Perhaps your design would be more correctly considered as a triangular arch with a puff on the supports (see the corresponding article).
02-04-2015: Vladimir
Table 1, scheme 1.1 The formula for deflection, in my opinion, is incorrect. Logically, it should be f (l) = 0. But in the proposed formula, this does not work.
02-04-2015: Dr. Lom
The above formula for determining the deflection, as well as the formula for determining the moment, is valid for the section from 0 to l / 2 (mid-span, where a concentrated force is applied). Since the beam (method of support) and the load are symmetrical, I did not consider it necessary to give a formula for determining the moment and deflection in the second section from l / 2 to l, so there are enough difficulties.
But if you really need it, then in this section (from l / 2 to l) you should additionally subtract Q (x - l / 2) ^ 3/6 from the indicated expression.
03-04-2015: Vladimir
Thanks a lot. I didn't see that 0 03-04-2015: Dr. Lom
14-08-2015: Martin
Off topic, but I have no idea where exactly to write... 14-08-2015: Dr. Lom
Actually, I am not an expert in MRI equipment, but there seems to be nothing wrong with a metal frame (there will be no metal frame, there will be reinforcement in reinforced concrete structures or something else). Look for the norms for the equipment of MRI rooms, there also seem to be no restrictions on the design of walls and ceilings. 09-09-2015: Yuriy
Dear Doctor Lom 09-09-2015: Yuriy
I can not solve problems with the calculation of 3 and 4 canopies on the door Please help 09-09-2015: Dr. Lom
A similar situation is discussed in the article "Determining the pull-out force (why the dowel does not stay in the wall)". The only difference is that you will have two canopies on top. To simplify the calculations, we can assume that the distance between the upper canopies is much less than the distance between the upper and lower canopies, then we can assume that the forces acting on the upper canopies are the same and in total equal to the lower force. But in any case, the load on the upper canopy will be greater than on the middle one. When the middle canopy is shifted to the middle of the door height, its role in terms of the perception of the tearing force will significantly decrease, however, the stability of the door will increase. 10-09-2015: Yuriy
You know 10-09-2015: Yuriy
for two loops, it turns out that the forces are equal only with the opposite sign 10-09-2015: Dr. Lom
Your mistake in choosing the calculation scheme. With 3 canopies, you consider the door as a two-span beam with relatively low rigidity, in other words, as a flexible beam, which, under the influence of support reactions, will have some deformations. Meanwhile, in the plane of action of the moment, the height of the beam is the width of the door 1 m, which is much larger than the span between the 2 upper canopies. Those. the door can be conditionally considered as an absolutely rigid beam, to which the design schemes given in this article are not applicable. The door in this case can be considered as a kind of cross section. 15-10-2015: Sergey
Good afternoon Doctor Lom. Excuse my ignorance. Please tell me what E stands for in the formula for calculating the deflection of a beam and how to determine it (E) 15-10-2015: Dr. Lom
E is the modulus of elasticity of the material you are going to use for the beam. The values of the moduli of elasticity for various building materials can be found in the article "Calculated resistances and moduli of elasticity for various building materials". And the physical meaning - in the article "Elastic and strength characteristics of materials". 13-03-2016: Vyacheslav
Good afternoon. 13-03-2016: Dr. Lom
That's right, of course Mx=Ax-qx^2/2. Now I'll try to fix it. Thank you for your attention. 30-03-2016: Timur
Hello! 30-03-2016: Dr. Lom
Under the action of a uniformly distributed load, the bearing capacity of a rigidly clamped beam is 1.5 times greater than that of the same beam, but on hinged supports. This does not depend on the length in any way (if we compare a hinged and rigidly clamped beam of the same length), but the type of acting load can affect the value of the difference. And such a difference in bearing capacity arises from the fact that the maximum moment for a hinged beam will be closer to the middle of the span, and for a rigidly clamped beam - on one of the supports (or on both supports, if the load is symmetrical). 31-03-2016: Timur
I just thought that the bearing capacity would increase by 5 times rather than 2. This is essentially tensile steel. You can walk on a wire of a couple of millimeters. Or will it sag over time? 31-03-2016: Dr. Lom
This article presents design schemes for relatively rigid beams. As a rule, the height h of such beams is 1/10 - 1/20 of the span length l. And the deflection of such beams, as a rule, does not exceed f? h/4 - h/2. Steel wires, ropes and other flexible (I would even say absolutely flexible) threads are calculated using completely different formulas and the diagrams will have a different look. As a rule, the deflection f of flexible threads is at least 5h - 6h. In flexible filaments, the stresses caused by the action of a bending moment are extremely small compared to the tensile stresses that arise during such a significant deformation. These tensile stresses must be compensated by horizontal support reactions. However, the calculation of flexible threads is a separate issue. 15-04-2016: Stanislav
Hello! Question according to table 1. "Single-span beam with rigid pinching on supports" according to paragraph 1.3. Moment on supports Ma and Mb. Tell me please, in the equation, by any chance, a deuce is not superfluous? 15-04-2016: Dr. Lom
No, not redundant. And it's easy enough to check. If a = l/2, i.e. both forces are applied at one point in the middle of the beam, then the equation of moments will be reduced to Ma \u003d Mv \u003d -2Ql / 8 \u003d -Ql / 4. We will get the same result by adding the values of the moments when using the calculation scheme 1.1. 15-04-2016: Stanislav
I'm sorry. The question I asked earlier is not correct. According to your tables (for hinged supports and for rigidly fixed ones), I calculated the maximum deflections in the case of the action of two concentrated forces with the same distance from the supports (Table 1 - paragraph 1.3). As a result, I found that the deflection in the case of hinged supports is less, than the deflection in the case of rigidly fixed ones. But logically it should be the other way around. Please explain. Is there a possibility that you have an error in the calculation scheme for rigid supports 1.3? 15-04-2016: Dr. Lom
I don't know how you did it. I assume that you have substituted the value of the moment into the deflection formula without taking into account the "-" sign. Again, if a = l/2, i.e. both forces are applied at one point in the middle of the beam, then the maximum deflection will be f = -Ql^3/96EI. We will get the same result by adding the deflection values when using the design scheme 1.1. 16-04-2016: Emin
Where can I find the design diagram of a single-span beam with a swivel hinge in the middle. 16-04-2016: Dr. Lom
What do you mean by pivot in the middle? If this is a support that prevents the beam from moving vertically, but does not prevent a change in the angles of inclination of the cross sections of the beam, then such a support should be considered as an intermediate support of a two-span beam, Table 3. If this hinged support is located in a perpendicular plane, then its presence only affects the determination the flexibility of the rod in a given plane and in the calculation for vertical loads is not taken into account. 16-12-2016: Michael
Hello! Tell me, how to determine the horizontal (pulling out) reactions in the scheme of a beam with two terminations? 16-12-2016: Dr. Lom
In general, beams are considered to be sufficiently rigid rods (limitations on the maximum allowable deflection contribute to this), and therefore, to simplify the calculations, the horizontal support reactions resulting from deformation are taken equal to zero. But in general, if there is such a need, then the deflection is first determined, a diagram is plotted. Then the change in the length of the neutral axis of the beam is determined - the procedure itself is quite complicated, and then, depending on the modulus of elasticity of the beam material, the forces necessary for such a change in the length of the beam are determined. 17-02-2017: student
Doc, sorry for the dumb questions. He mastered the construction of diagrams in echele, for a beam pinching-hinge, bending moment in the span M (x) (tab 2 p 2.1). The result is that at the end of the beam the function is not equal to 0. pic https://yadi.sk/i/Cal1RKes3EDm6W 17-02-2017: student
Doc, for a purely theoretical understanding: is it possible to not install or reduce reinforcement in places of zero bending moment in the span of a beam? Or are the transverse forces comparable at these points? 17-02-2017: Dr. Lom
I have already said that I am not friends with exel, so I can hardly point out where your mistake has crept in. But in general, if we substitute the values at x = l into the equation of moments given for this design scheme, then the moment on support B is equal to zero: 22-08-2017: Ivan
Good afternoon Is it permissible for a scheme with sliding terminations at the ends of the beam to use the deflection calculation formula for option 1.1 (two rigid terminations at the ends)? As I understand from the comments, the horizontal support reactions are taken equal to zero, respectively, in this case, a rigid embedment is a sliding one. Do I understand correctly? 23-08-2017: Dr. Lom
Yes, it is acceptable if the beam has the appropriate stiffness. In general, horizontal support reactions always arise due to the distribution of internal stresses, just the greater the rigidity of the beam, the less their influence on the overall operation of the rod. Therefore, as a rule, when calculating rigid beams, the influence of possible horizontal support reactions in the margin of safety is neglected, taking them equal to zero. Well, for flexible threads, the default sliding termination is not suitable as a support. As is already known, a beam is statically determinate if it is supported by two hinged supports (one movable and one fixed) or embedded at one end, i.e. if three external relations are imposed on it. An exception is multi-span hinged beams (consisting of several individual beams connected to each other by intermediate hinges), which can be statically determined even with more than three external braces (see § 3.7 on this). On fig. 85.7, a, b shows two statically indeterminate beams; each of them is superimposed with four external bonds, and, therefore, these beams are once statically indeterminate. On fig. 85.7, a beam with six external ties is shown; it is three times statically indeterminate. The degree of static indeterminacy of the beam (having no intermediate hinges) is equal to the excess (extra) number of external links (in excess of three). Statically indeterminate beams are often referred to as continuous beams. The calculation of continuous beams, as well as the calculation of any statically indeterminate systems, cannot be performed with the help of equilibrium equations alone; it is always necessary to create additional equations (displacement equations) that take into account the nature of the beam deformation. On fig. 86.7, but a statically indeterminate beam is shown once. To calculate this beam, it can be represented as a statically determined beam, shown in Fig. 86.7, b, obtained from the right support given as a result of discarding. A statically determinate system obtained from a given one by removing redundant connections is called the main system. The beam shown in fig. 86.7, b, is the main system for a given beam (Fig. 86.7, a). The main system (Fig. 86.7, b), in addition to the given load q, is affected by an unknown reaction RB of the dropped connection. Under the action of the load q, the beam shown in fig. 86.7, b, is deformed and its free end moves down (Fig. 86.7, c) by an amount that can be easily determined by the method of initial parameters: Under the action of the force RB, the free end of the beam shown in Fig. 86.7, b, moves up by an amount (Fig. 86.7, d), which can also be determined by the method of initial parameters: Under the simultaneous action of a given load q and force, the deflection of the free end of the beam shown in fig. 86.7, b, is determined by the expression This deflection is zero, since the deflection of the right end of a given beam (Fig. 86.7, a) is zero: Therefore, the actual reaction that occurs on the right support of a statically indeterminate given beam is Bending moment M and shear force Q in the section of a given beam can now be determined by formulas (2.7) and (3.7), as in the statically determined beam shown in Fig. 86.7, a: The diagrams Q and M constructed using these expressions for a given beam are shown in fig. 86.7, f, f. The calculation of a given beam can also be performed using other basic systems, for example, those shown in Fig. 86.6, h, i. The calculation of continuous beams is usually carried out using the so-called three-moment equations. This method of calculation avoids the compilation of additional equations of the type (81.7). In addition, this method allows obtaining additional equations with no more than three unknowns in each of them, which, with a high degree of static indeterminacy of a given beam, simplifies the solution of a system of equations. Let us now consider the calculation of continuous beams using the equations of three moments. On fig. 87.7, but shows a section selected from a multi-span continuous beam, which is under the action of a certain load. Beam supports are designated from left to right by numbers, and so on. The span lengths of a continuous beam are indicated (also from left to right), etc. The index for the length of each span I corresponds to the number of the right support of this span. The moments of inertia J of the cross sections of the beam are constant along the length of each span; in different spans, the moments of inertia can have different values. We obtain the main system for calculating a continuous beam by removing from it the connections that prevent the mutual rotation of adjacent sections of the beam above its supports, i.e., by placing hinges above the beam supports (Fig.). Unknown are the bending (support) moments, etc., arising in sections of a continuous beam above the supports. Unknown moments are considered positive when they cause tension in the lower fibers of the beam. Consider two beam spans adjacent to the support, shown in Fig. 87.7, c. Here the dotted line shows the bent axis of the beam. On fig. 87.7, d shows sections of the beam directly adjacent to the support. Here is the angle of rotation of the cross section belonging to the left span and directly adjacent to the support, and is the angle of rotation of the section belonging to the right span and also directly adjacent to the support. Both of these sections, in essence, represent one cross section located above the support, and therefore the angles of their rotation are the same, i.e. The rotation angles and can be considered as a consequence of the impact on individual single-span beams shown in Fig. 87.7, d, given loads, as well as unknown support moments Condition (82.7), therefore, means that the angle of rotation of the right end of the left end of the beams shown in Fig. 87.7, d, is equal to the angle of rotation of the left end of the right beam, i.e., the mutual angle of rotation of these ends is zero. Unknown moments, etc. have such values at which the specified condition is satisfied not only for the support, but also for all intermediate supports of a continuous beam. Let's find the values of the angles and graphic-analytical method. On fig. 87.7, e, g, fictitious beams for spans loaded with a fictitious load are shown. 87.7, e shows a fictitious load corresponding to the action on these spans of the load given on the beam, and in fig. - the action of unknown moments on them Based on the second of the formulas (80.7), the angles and, respectively, are equal to the fictitious transverse forces and arising on the supports of the spans of fictitious beams, i.e. where (Fig. 87.7, e), and also (Fig. 87.7, g) are the reactions of the supports of fictitious beams. Let's consider on real examples the nodes of support or connection of structures and determine what we are dealing with: with a hinge or pinching. This is the classic hinge case. The support depth of the slab is dictated by the standard series, and it is less than the height of the slab section. Under such conditions, while bending, the slab will calmly turn on a support - on a hinged support. Moreover, it is impossible to pinch the slab by inserting it deeper into the wall, because. moments on the support will immediately appear in it (with a hinged scheme, the moment on the support is zero), and there are practically no upper reinforcement for the perception of these moments in prefabricated slabs. Calculation scheme for such a plate: It all depends on the depth of the insertion of the plate into the wall. If, with a plate height of 200 mm, you support the plate by 150-200 mm, then this is a hinge. If the upper reinforcement enters the support for the length of the anchorage or special measures are taken in the form of welding plates (washers) at the ends of the reinforcement, then this is pinching. If the depth of support is "neither this nor that" - i.e. more than the height of the section, but less than the length of the anchorage, then this is the unpleasant case when you need not only to design, but also to calculate all the details of the assembly and check whether they can withstand such mockery. Firstly, the installation of the upper working reinforcement is already mandatory. Secondly, it must be designed for the moments arising from this pinching. Thirdly, the adequacy of its anchoring should be verified by calculation. The design scheme for a single-span slab is as follows: For a monolithic beam, everything is the same; the embedment depth for the pinched version can only be saved by bending the upper rod down. But both at the slab and at the beam, the masonry load must be sufficient and verified by calculation. This is a standard scheme with a support in the form of a pinch - there should not be a hinge in any case, there should not even be an incomplete pinch - only a 100% rigid knot. Otherwise, the system will be geometrically variable: the balcony under load will turn on the support with all the consequences. Therefore, when designing the support of a cantilever balcony, it is necessary to carefully develop and calculate the rigid support node. In the standard series 2.130-1 issue. 9, you can get acquainted with the support nodes of balcony slabs and understand by what principle pinching is achieved. Firstly, this is a sufficient insertion of the plate into the wall. Secondly, this is a significant weight of the masonry wall from above. Thirdly, it is a mandatory anchoring top parts of the slab in a compressed structure - in the solutions of the series, this is done by welding anchors to the mortgage in the balcony slab, which are securely fastened in the wall structures (fastening is calculated). All three conditions must be balanced and in total give a reliable pinch. When supporting beams, the same principle should be used: support depth plus anchoring of the upper part of the beam. In the case of a monolithic cantilever slab or beam resting on a solid wall, it is necessary to drive the upper reinforcement of the console into the wall for the length of the anchorage - this will ensure pinching. If the balcony turns into a slab (i.e., in fact, it is a slab with a cantilevered outreach of the balcony), then there is no need to worry about a rigid knot here - an ordinary hinged support on the wall is enough. If you are making a balcony in an existing building, it is very difficult to design and execute a clean pinch, so try to avoid clean cantilevers and make balconies with braces. Such a decision is chosen in several cases: if it is dictated by an architectural decision; if the construction is carried out in an existing building; if the console without a strut does not withstand a significant load. How good is this console? The fact that, in the aggregate, the structure is a cantilever, but individually, each support node is hinged with limited vertical and horizontal movement - and such nodes do not require calculation, and it is much easier to design and implement them than pinching. The main thing here is to ensure a reliable limitation of horizontal movement: if the strut is bolted, then there should be enough of them to pull out; if the structure is simply laid into the wall, then there must be anchors driven into the masonry, etc. The design scheme for such a balcony is as follows: The horizontal beam is fixed in the wall with limited vertical and horizontal movements. It is uncut in length. In the span (or at the edge) the horizontal beam hinges on a brace, which in turn rests on the wall with limited vertical and horizontal movement. Such a beam in the middle spans always has a hinged support, but on the extreme supports there can be both a pinch and a hinge. Everything is due to the size of the spans and the ability to pinch the beam. If the spans are large, or if the span sizes are different and adversely affect the span moment in the end spans (for example, the end spans are much larger than the average spans), then you can try to apply pinching on the end supports. Basically, the extreme supports are made articulated. This slab has exactly the same principle as the multi-span beam described in the previous case. The extreme supports of such a slab can be beams, or they can be walls of a building. If the extreme supports are beams, then it is difficult to organize pinching when resting on them; hinged support is used here as a standard. I would like to draw attention to the next point. With a multi-span overlap of large sizes, it is necessary to make an expansion joint in it. If the loads are significant, then when hinged on the extreme supports in the extreme spans, significant bending moments arise that require significant reinforcement - and this is not always rational for plates of small thickness. In this case, I recommend considering the option of arranging a seam not on a beam, but in a span: then two slabs will have a cantilever overhang. In this case, the moments will be balanced and the reinforcement will be harmonious. The basement wall is always affected by horizontal ground pressure, and the deeper the basement, the greater the effect of horizontal pressure on structures. When determining the design scheme for the basement wall, it is necessary to consider the scheme in two directions. The first, and most important, is a vertical cut along the wall. We need to consider two nodes: upper and lower. There may be a lack of support in the upper node (if the ceiling does not rest on the wall); a hinge with limited horizontal movement (if there is a hinged floor support - for example, prefabricated slabs); rigid knot (if the connection between the basement wall and the ceiling is rigid - for example, a monolithic structure). Support in this case is meant in the horizontal direction, because. the main load we have is the horizontal pressure of the soil. In the lower junction of the wall with the foundation tape, it is mainly found to be rigid - it is laborious to organize the hinge there, and it does not make much sense. Now about another, horizontal section of the wall. If the length of the wall is not limited in movement (there are no perpendicular walls), then it is not necessary to consider a horizontal section in the calculation. But if there are perpendicular walls located quite often, then you need to calculate the wall also in the horizontal direction, because. on the one hand, soil pressure acts, on the other hand, the walls serve as supports, and a multi-span continuous structure is obtained, in which both span and support moments occur - accordingly, it is necessary to check the horizontal reinforcement of the wall, taking into account the location of perpendicular walls. Such a wall is considered as a multi-span continuous slab 1 m wide (a meter horizontal strip is conditionally cut out of the wall); the middle supports are hinges, and the extreme ones depend on the connection with perpendicular walls - basically, this is pinching. Basically, in reinforced concrete, the interface scheme is pinching, because. the hinge is more difficult to organize (especially in a monolith). In the prefabricated version, the column is deeply embedded in the sleeve (embedding depth is calculated), and in the monolithic version, rebar extensions are made from the foundation into the column, which are inserted at least for the length of the overlap into the column and for the anchoring length - into the foundation. If you want to deal with some specific example of connecting structures, write in the comments, and your case will be added to the article. Naturally, there are such schemes in which everything is already predetermined - an unambiguous hinge (as in prefabricated hollow core slabs) or an unambiguous pinching (cantilevered balcony slab). But there are such options when the choice is given to the designer - and at first it is very difficult to decide how to draw up a design scheme in order to get the best result. Let's consider some cases. As you know, the grillage can be supported on piles either hingedly or rigidly. And it is often very difficult to understand, but which option to choose? Firstly, you need to read the SNiP “Pile Foundations”, which stipulates the conditions that allow hinged support - there are not so many of them, some of your questions will be immediately eliminated. And then you should analyze the design itself as a whole. If foundation on one pile, then the connection of the pile with the grillage must be rigid, otherwise there will be no stability. When pile bush the following should be defined: 1 - if the foundation perceives only a vertical load (without moments and transverse forces), hinged support can be considered; 2 - if tearing forces occur in the piles (when the moment is transferred from the column through the grillage), then the connection is only rigid. When strip pile grillage: 1 - if the calculation of the grillage shows significant overstresses in it due to a rigid connection with piles, the variant with hinged support should be considered; 2 - if horizontal forces (wind or soil pressure) are transmitted to the grillage, the connection with the piles should be made rigid. When grillage in the form of a plate you can use a hinged connection if it is not contraindicated by SNiP "Pile Foundations" and if there are no tearing forces in the piles. When strip grillage in a sheet pile (retaining) wall of piles: 1 - if the grillage serves as just a strapping beam and nothing rests on it, it is better to choose a hinged connection; 2 - when the overpass supports or similar structures are located on the grillage, transmitting forces from wind loads, the connection must be rigid. For a pile, hinged support is more advantageous, because. then no bending moment is transmitted to it; but this type of support is not always allowed by SNiP; In the presence of tearing forces, the connection of the pile with the grillage must always be made rigid so that the structure does not lose stability (and the tearing force often emerges when the moment from the column is decomposed into a couple of forces); Both piles and grillage only benefit from a hinged connection, so if there are absolutely no contraindications, you need to choose a hinge. The main thing to remember: always with a rigid connection of a pile with a grillage, the moments in the grillage are transferred to the piles, and this should be taken into account when calculating the pile. In the case of frames, the decision to support the foundation often comes after choosing the design of the frame itself. If a frame with rigid joints for connecting crossbars to columns, then it is most rational to choose a hinged joint when resting on the foundation - such a frame will not suffer when hinged, but the foundation will benefit, because. the moment is zero, which means that the foundation will be smaller and more economical. Yes, and when calculating such a frame, there will be as many as six degrees of freedom less complexity - and with manual calculation, this is a lot. If the crossbars in the frame are hinged on the columns, then the columns must be rigidly connected to the foundation, otherwise we will get a geometrically variable system. But sometimes, having decided on the frame scheme (for example, the crossbars are hinged, and the columns are pinched in the foundations), we get an unfavorable result (for example, the foundations are unacceptably large under the given conditions). Then you have to change the design scheme and check the option with rigid nodes in the frame and hinges at the point of support on the foundation. Often, the materials themselves dictate the choice of a design scheme for us: for example, it is difficult to organize hinges in monolithic reinforced concrete, so there most often all nodes (both in the frame and in the place where the columns rest on the foundation) are rigid. And that's okay too. The main thing is that it should be designed according to the design scheme. In this topic, you also need to try a lot in order to gain experience and learn how to choose the best version of the design scheme the first time. In reinforced concrete slabs and beams, when pinched, a significant top reinforcement floats up. Naturally, this leads to an increase in cost, but it is rational in large-span structures. Sometimes it turns out that with a large span, an increase in the cross section of the beam or the height of the slab only worsens the work (because the load from its own weight increases); but pinching gives its positive results - a bending moment appears on the supports, giving us the upper reinforcement, but in the span the moment decreases, and in total the structure passes according to the calculation. At the same time, however, one should never forget that a pinched beam or slab transfers force to the structures on which it rests. Still, pinching should be used in slabs and beams, in which it is important to reduce the deflection or reduce the opening of cracks - less moment in the span means less deformation. Another special thing is a slab that rests on four sides. It already works due to such support in such a way that it becomes necessary to install the upper reinforcement in the slab (especially closer to the corners). Therefore, it is often rational, if possible, to pinch the slab and check whether the reinforcement will be less. Any multi-span structure, whether it be a slab or a secondary beam, has an end span in which it rests on the beam on one side. And in connection with such a one-sided load, the support beam experiences torsion, often significant. And in such cases, when, when calculating the torsion, the cross section of the beam grows to unimaginable sizes, a hinge comes to our aid. If the slab or the secondary beam is hinged, then the extreme support beam will be unloaded, the moments will not be transferred to it, and the situation will cease to be critical. It is clear that it is not always possible to design a hinged support (especially in a monolithic version), but sometimes even in a monolith it is better to make an extreme beam with a cantilever, and hinge the plate onto this cantilever. There is still an option (but if the architecture allows it) - to display the supporting slab cantilevered in the form of a balcony; then the support beam is not completely, but will be unloaded. You can also read on the topic of hinges and pinching.
Honey is being designed. a building with an MRI room, the building is planned to initially be raised from a metal frame, and MRI is not compatible with metal in terms of magnetic waves and attraction, so the question is how to protect metal from MRI ???
Please help me calculate the forces and moments acting on 3 door canopies Considering that the distance of the third canopy between the lower and upper can be changed Thank you in advance
H - door height 2.5m
width 1 m
the lower hinge is set at 0.2 m from the bottom of the door
second hinge 1.8 m from the center of the second
door weight 40 kg
How to calculate forces and moments
I would be very grateful for help
[email protected]
I get the strength on the upper loop -169
on the second loop 23
and on the bottom loop 145
Where is my mistake?
How do you define forces?
Sincerely
Yuri
But for three loops, it turns out that the upper loop takes on the load, which in total is the load of the other two loops
Please help with the algorithm
Thanks in advance
Yuri
By the way, look at the article "Calculation of the nail connection of a filly with a rafter leg. Theoretical prerequisites." Here I will add to what was said earlier, the support moment can be decomposed into any number of multidirectional forces according to the diagram of normal stresses. Moreover, if there are two forces above and one below, then the sum of the upper forces is equal to the lower force, but has the opposite sign. In addition, the moment relative to the center of gravity of the conditional cross section from the lower force will be equal to the moment from the two upper forces. Thus, the problem is reduced to determining the center of gravity of a conditional section.
As I have already said, with a relatively small distance between the upper canopies, they can be conditionally considered as one whole, i.e. Divide by 2 the force acting on the upper canopy, as it were, but use the appropriate safety factor.
Table 3 №2.2 Moment in span.
Probably should be: Mx=Ax-qx^2/2.
I studied several articles, but it is not entirely clear how the bearing capacity of a steel single-span hinged beam and a beam with rigid pinching on supports correlate? Does it depend on the length? For concreteness, for example, for spans of 4,6,8,12 meters. According to my estimates, about 2 to 5 should be ...
And from 2 to 5 - this is to Korney Ivanovich Chukovsky. What do you mean by this in this case, I do not understand.
Judging by the results, if we add the value of the support reaction B to Mx, then everything returns to normal, but the bending moment on the support A increases by the value of the reaction B. There are support lines A and B, but this is a pure visualization of the load levels and use as constants in calculation. Calculation at: distribution load 10, beam 4.
Doc, another question. Is it possible to find the functions of transverse forces for the main types of beams, I can’t derive it myself, the differential equations are completely forgotten. Although extreme values are usually used in calculations. In the graph, I have subtracted the load on the arm qx from the support reaction A, but this is pure adjustment of the result. Sincerely
MB = Al + MA - ql^2/2 = 5ql^2/8 - ql^2/8 - ql^2/2 = 0
so keep it up.
Regarding the need for reinforcement in sections with zero bending moment, everything is correct; in such sections, longitudinal reinforcement is not needed according to the calculation. But the actual working conditions of the structure can differ significantly from the accepted design scheme, and in addition, the reinforcement must be pinched for its reliable operation. For more details, see the article "Anchoring of reinforcement".
Prefabricated plate with support on two sides.
Monolithic single-span slab (beam) supported by masonry.
Balcony slab (beam) cantilever.
Balcony or cantilevered beam.
Multi-span beam supported by masonry walls.
Multi-span slab supported by metal beams.
Monolithic basement wall.
Conjugation of a reinforced concrete column with a foundation.
Hinge or pinching - what to choose?
The connection of the grillage with piles - a hinge or a rigid connection?
Supporting a metal or reinforced concrete frame on a foundation.
Floor slabs and beams.
Supporting edge slabs or secondary beams.
