Fundamentals of strength materials, calculation formulas. Fundamentals of strength of materials, calculation formulas Strength of symbols and their meaning
8.2. Basic laws used in the strength of materials
Statics relationships. They are written in the form of the following equilibrium equations.
Hooke's law ( 1678): the greater the force, the greater the deformation, and, in direct proportion to the force... Physically, this means that all bodies are springs, but with great rigidity. With a simple stretching of a bar with a longitudinal force N= F this law can be written as:
Here 
longitudinal force, l- the length of the bar, BUT- its cross-sectional area, E- coefficient of elasticity of the first kind ( Young's modulus).
Taking into account the formulas for stresses and strains, Hooke's law is written as follows: 
.
A similar relationship is observed in experiments between the shear stresses and the shear angle:
.
G
are calledshear modulus
, less often - the modulus of elasticity of the second kind. Like any law, Hooke's law has a limit of applicability. Voltage 
, to which Hooke's law is valid, is called proportional limit(this is the most important characteristic in resistance materials).
Let's depict the dependence
from
graphically (Figure 8.1). This painting is called stretch diagram
... After point B (i.e. at 
) this dependence ceases to be straightforward.
At 
after unloading, residual deformations appear in the body, therefore 
called elastic limit
.
When the stress reaches the value σ = σt, many metals begin to exhibit a property called fluidity... This means that even under constant load, the material continues to deform (that is, it behaves like a liquid). Graphically, this means that the diagram is parallel to the abscissa (section DL). The stress σ t at which the material flows is called yield point .
Some materials (Art. 3 - building steel) begin to resist again after a short flow. The resistance of the material continues up to a certain maximum value of σ pr, then gradual destruction begins. The value of σ pr - is called ultimate strength (synonym for steel: tensile strength, for concrete - cubic or prismatic strength). The following designations are also used:
=R b
A similar relationship is observed in experiments between shear stresses and shears.
3) Duhamel-Neumann law (linear thermal expansion):
In the presence of a temperature drop, bodies change their size, and in direct proportion to this temperature drop.
Let there be a temperature difference 
... Then this law has the form:
Here α - linear thermal expansion coefficient, l - rod length, Δ l- its lengthening.
4) Creep law .
Research has shown that all materials are highly heterogeneous in small things. The schematic structure of steel is shown in Figure 8.2.
Some of the constituents have liquid properties, so many materials under load will gain additional elongation over time. 
(Figure 8.3.) (metals at high temperatures, concrete, wood, plastics at normal temperatures). This phenomenon is called creep material.
For a liquid, the law is true: the greater the force, the greater the speed of movement of the body in the fluid... If this ratio is linear (i.e., the force is proportional to the speed), then you can write it in the form:
E 
If we go to relative forces and relative elongations, we get
Here the index " cr
”Means that the part of the elongation that is caused by the material creep is considered. Mechanical characteristic 
called the coefficient of viscosity.
Law of energy conservation.
Consider a loaded beam
Let's introduce the concept of moving a point, for example,
- vertical movement of point B;
- horizontal displacement of point C.
Forces 
doing some work U.
Given that the forces 
begin to increase gradually and assuming that they increase in proportion to the displacements, we get:
.
According to the conservation law: no work disappears, it is spent on doing other work or passes into another energy (energy Is the work that the body can do.)
Work forces 
, is spent on overcoming the resistance of the elastic forces that arise in our body. To calculate this work, let's take into account that the body can be considered as consisting of small elastic particles. Let's consider one of them:
From the side of neighboring particles, a voltage acts on it 
... The resultant stresses will be 
Under the influence 
the particle will lengthen. Elongation is defined as elongation per unit length. Then:
Let's calculate the work dW that the force does dN (here it is also taken into account that the forces dN begin to increase gradually and they increase in proportion to displacements):
For the whole body, we get:
.
Work W which was committed 
are called energy of elastic deformation.
According to the law of conservation of energy:
6)Principle possible movements .
This is one of the options for recording the law of conservation of energy.
Let the forces act on the bar F 1
,
F 2
,
…
... They cause the movement of a point in the body 
and voltage 
... Let's give the body additional small possible displacements
... In mechanics, a record of the form 
means the phrase "possible value of the quantity but". These possible movements will cause in the body additional possible deformations
... They will lead to the appearance of additional external forces and stresses. 
,
δ.
Let's calculate the work of external forces on additional possible small displacements:
Here 
- additional displacements of those points at which forces are applied F 1
,
F 2
,
…
Consider again a small element with a cross section dA and length dz (see fig. 8.5. and 8.6.). By definition, additional elongation dz of this element is calculated by the formula:
dz= dz.
The tensile force of the element will be:
dN = (+δ) dA ≈ dA..
The work of internal forces on additional displacements is calculated for a small element as follows:
dW = dN dz = dA dz = dV
WITH 
Summing up the deformation energy of all small elements, we obtain the total deformation energy:
Law of energy conservation W = U gives:
.
This ratio is called principle of possible displacement(also called principle of virtual movements). Similarly, we can consider the case when tangential stresses are also acting. Then it can be obtained that the deformation energy W the following term is added:
Here is the shear stress, is the shear of a small element. Then principle of possible displacements will take the form:
Unlike the previous form of writing the law of conservation of energy, there is no assumption here that the forces begin to increase gradually, and they increase in proportion to the displacements
7) Poisson effect.
Consider the picture of the elongation of the sample:
The phenomenon of shortening a body element across the direction of elongation is called Poisson effect.
Let's find the longitudinal relative deformation.
The transverse relative deformation will be:
Poisson's ratio the quantity is called:
For isotropic materials (steel, cast iron, concrete) Poisson's ratio
This means that in the transverse direction the deformation less longitudinal.
Note
: modern technologies can create composite materials with Poisson's ratio> 1, that is, the transverse deformation will be greater than the longitudinal deformation. For example, this is the case for a material reinforced with stiff fibers at a low angle 
<<1
(см. рис.8.8.). Оказывается, что коэффициент
Пуассона при этом почти пропорционален
величине
, i.e. the less 
, the greater the Poisson's ratio.
Figure 8.8. Figure 8.9
Even more surprising is the material shown in (Fig. 8.9.), And for such a reinforcement a paradoxical result takes place - longitudinal elongation leads to an increase in the size of the body in the transverse direction.
8) Generalized Hooke's law.
Consider an element that is stretched longitudinally and transversely. Let us find the deformation arising in these directions. 
Calculate the deformation 
arising from the action 
:
Consider deformation from action 
, which occurs as a result of the Poisson effect:
The total deformation will be:
If valid and 
, then add one more shortening in the direction of the x-axis 
.
Consequently: 
Similarly: 
These ratios are called generalized Hooke's law.
Interestingly, when writing Hooke's law, an assumption is made about the independence of elongation deformations from shear deformations (about independence from shear stresses, which is the same) and vice versa. Experiments confirm these assumptions well. Looking ahead, we note that the strength, on the contrary, strongly depends on the combination of shear and normal stresses.
Note: The above laws and assumptions are confirmed by numerous direct and indirect experiments, but, like all other laws, they have a limited area of applicability.
19-08-2012: Stepan
Low bow to you for the available materials on strength of materials!)
At the institute I smoked bamboo and somehow I was not up to sopromat, the course weathered out within a month)))
Now I work as an architect-designer and I constantly get stumped when I need to make calculations, I dig myself into a slurry of formulas and different methods and I understand that I missed the basics ..
Reading your articles, order is gradually put in my head - everything is clear and very accessible!
24-01-2013: wany
thank you man !!))
I have 1 only question if the maximum load for 1 m is 1 kg * m then for 2 meters?
2 kg * m or 0.5 kg * m ??????????
24-01-2013: Dr. Lom
If we mean the distributed load per linear meter, then the distributed load of 1kg / 1m is equal to the distributed load of 2kg / 2m, which in the end still gives 1kg / m. A concentrated load is measured simply in kilograms or Newtons.
30-01-2013: Vladimir
Formulas are good! but how and with what formulas to calculate the structure for the canopy and most importantly what metal (profile pipe) should be in size ???
30-01-2013: Dr. Lom
If you paid attention, then this article is devoted exclusively to the theoretical part, and if you also show ingenuity, then you will easily find an example of the calculation of structures in the corresponding section of the site: Calculation of structures. To do this, just go to the main page and find this section there.
05-02-2013: Leo
Not all formulas describe all participating variables ((
There is also confusion with the notation, first the x denotes the distance from the left experiment to the applied force Q, and in two paragraphs below the claim is already a function, then the formulas are displayed and the confusion started.
05-02-2013: Dr. Lom
Somehow it happened that when solving various mathematical problems, the variable x is used. Why? X knows him. Determining the reactions of supports at a variable point of application of a force (concentrated load) and determining the value of the moment at some variable point relative to one of the supports are two different tasks. Moreover, in each of the problems, a variable is defined with respect to the x-axis.
If this confuses you and you cannot understand such elementary things, then I cannot do anything. Complain to the Mathematician Rights Society. And if I were you, I would have filed a complaint about textbooks on structural mechanics and strength materials, otherwise what is it really? Are there few letters and hieroglyphs in alphabets?
And I also have a counter question for you: when did you solve the addition-subtraction problems of apples in the third grade, did the presence of x in ten problems on the page confuse you, or did you somehow cope?
05-02-2013: Leo
Of course, I understand that this is not some kind of paid work, but nevertheless. If there is a formula, then under it there should be a description of all its changes, but you need to look for it from above out of context. And in some places and generally not in the context of the mention. I'm not complaining at all. I'm talking about the shortcomings of the work (for which, by the way, I thanked you). As for the variables x as a function and then the introduction of another variable x as a segment, without indicating all the variables under the derived formula, it introduces confusion, the point here is not in the established notation, but in the expediency of maintaining such a presentation of the material.
By the way, the arcasm is not appropriate for you, because you set out everything on one page and without specifying all the variables it is not clear what you mean at all. For example, in programming, all variables are always specified. By the way, if you do all this for the people, then it would not hurt you to find out what kind of contribution to mathematics was made by Kisilev as a teacher, and not as a mathematician, maybe then you will understand what I mean.
05-02-2013: Dr. Lom
It seems to me that you still do not quite correctly understand the meaning of this article and do not take into account the bulk of the readers. The main goal was to convey to people who do not always have the appropriate higher education, the basic concepts used in the theory of strength of materials and structural mechanics, and why all this is needed at all, by the simplest means possible. Of course, I had to sacrifice something. But.
There are enough correct textbooks, where everything is laid out in shelves, chapters, sections and volumes and described according to all the rules, without my articles. But there are not so many people who are able to immediately understand these volumes. During my studies, two-thirds of the students did not understand the meaning of the strength of materials, even approximately, but what about ordinary people engaged in repairs or construction and who decided to calculate a lintel or a beam? But my site is intended primarily for such people. I believe that clarity and simplicity are much more important than literal adherence to the protocol.
I thought about breaking this article into separate chapters, but this irreversibly loses the general meaning, and hence the understanding of why this is needed.
I think the programming example is incorrect, for the simple reason that programs are written for computers, and computers are dumb by default. But people are another matter. When your wife or friend says to you: "The bread is over", then you, without additional clarifications, definitions and commands, go to the store where you usually buy bread, buy there exactly the kind of bread that you usually buy, and exactly as much as you usually buy. At the same time, by default, you extract all the necessary information to perform this action from the context of previous communication with your wife or girlfriend, existing habits and other seemingly insignificant factors. And at the same time, notice that you do not even receive a direct order to buy bread. This is the difference between humans and computers.
But in the main I can agree with you, the article is not perfect, as well as everything else in the world around us. And do not be offended by irony, there is too much seriousness in this world, you sometimes want to dilute it.
28-02-2013: Ivan
Good afternoon!
Below formula 1.2 is the formula for the reaction of the supports for a uniform load along the entire length of the beam A = B = ql / 2. It seems to me that there should be A = B = q / 2, or am I missing something?
28-02-2013: Dr. Lom
Everything is correct in the text of the article, because a uniformly distributed load means what load is applied on a section of the length of the beam, and the distributed load is measured in kg / m. To determine the reaction of the support, we first find out what the total load will be equal to, i.e. along the entire length of the beam.
28-02-2013: Ivan
28-02-2013: Dr. Lom
Q is a concentrated load, whatever the length of the beam, the value of the support reactions will be constant at a constant value of Q. q is the load distributed over a certain length, and therefore, the longer the beam length, the greater the value of the support reactions, at a constant value q. An example of a concentrated load is a person standing on a bridge, an example of a distributed load is the dead weight of a bridge structure.
28-02-2013: Ivan
Here it is! Now it is clear. There is no indication in the text that q is a distributed load, the variable "ku small" just appears, this was misleading :-)
28-02-2013: Dr. Lom
The difference between concentrated and distributed load is described in the introductory article, the link to which at the very beginning of the article, I recommend that you read.
16-03-2013: Vladislav
It is not clear why tell the basics of resistance to those who are building or designing. If they at the university did not understand the strength of materials from literate teachers, then they should not be even close to designing, and popular articles will only confuse them even more, since they often contain gross errors.
Everyone should be a professional in their field.
By the way, the bending moments in the above simple beams must have a positive sign. The negative sign affixed to the diagrams contradicts all generally accepted norms.
16-03-2013: Dr. Lom
1. Not everyone who builds has studied at universities. And for some reason, such people who are engaged in repairs in their home, do not want to pay professionals for the selection of the cross-section of the lintel above the doorway in the partition. Why? ask them.
2. There are enough typos in paper editions of textbooks, but people are confused not by misprints, but by too abstracted presentation of the material. This text may also contain typos, but unlike paper sources, they will be corrected immediately after they are discovered. But about gross mistakes, I have to upset you, they are not here.
3. If you think that the moment diagrams plotted from the bottom of the axis should only have a positive sign, then I am sorry for you. First, the diagram of moments is rather arbitrary and it only shows the change in the value of the moment in the cross-sections of the bent element. In this case, the bending moment causes both compressive and tensile stresses in the cross section. Previously, it was customary to plot a plot on top of the axis; in such cases, the positive sign of the plot was logical. Then, for clarity, the diagram of moments began to be built as shown in the figures, however, the positive sign of the diagrams was preserved from old memory. But in principle, as I said, this is not of fundamental importance for determining the moment of resistance. The article says about this: "In this case, the value of the moment is considered negative if the bending moment tries to rotate the beam clockwise relative to the considered point of the section. In some sources it is considered the other way around, but this is nothing more than a matter of convenience." However, there is no need to explain this to the engineer, personally, I have come across various options for displaying diagrams many times and this has never caused problems. But, apparently, you have not read the article, and your statements confirm that you do not even know the basics of resistance to materials, trying to replace knowledge with some generally accepted norms, and even "all".
18-03-2013: Vladislav
Dear Dr. Lom!
You have not read my message carefully. I spoke about errors in the sign of bending moments "in the above examples", and not in general - for this it is enough to open any textbook on the strength of materials, technical or applied mechanics, for universities or technical schools, for builders or mechanical engineers, written half a century ago, 20 years ago or 5 years. In all books, without exception, the sign rule for bending moments in beams during direct bending is the same. This is what I meant when I spoke about generally accepted norms. And on which side of the beam to lay the ordinates is another question. Let me explain my idea.
The sign on the diagrams is placed in order to determine the direction of the internal effort. But at the same time it is necessary to agree on which sign corresponds to which direction. This agreement is the so-called rule of signs.
We take several books recommended as the main educational literature.
1) Alexandrov A.V. Strength of materials, 2008, p. 34 - a textbook for students of construction specialties: "The bending moment is considered positive if it bends a beam element with a bulge downward, causing the lower fibers to stretch." In the given examples (in the second paragraph), obviously, the lower fibers are stretched, so why is the sign on the diagram negative? Or is A. Aleksandrov's statement something special? Nothing like this. We look further.
2) Potapov V.D. and others. Construction mechanics. Statics of elastic systems, 2007, p. 27 - university textbook for builders: "the moment is considered positive if it causes stretching of the lower fibers of the beam."
3) A.V. Darkov, N.N. Shaposhnikov. Structural mechanics, 1986, p. 27 - a well-known textbook also for builders: “with a positive bending moment, the upper beam fibers experience compression (shortening), and the lower ones - tension (elongation);”. As you can see, the rule is the same. Maybe everything is completely different for machine builders? Again, no.
4) G.M. Itskovich. Strength of Materials, 1986, p. 162 - a textbook for students of mechanical engineering technical schools: “An external force (moment) bending this part (the cut off part of the beam) with a bulge downward, that is, so that the compressed fibers are on top gives a positive bending moment. "
The list goes on, but why? Any student who has passed at least a grade of 4 knows this.
The question on which side of the bar to lay the ordinates of the bending moment diagram is another convention that can completely replace the above rule of signs. Therefore, when constructing M diagrams in frames, no sign is placed on the diagrams, since the local coordinate system is associated with the bar, and changes its orientation when the position of the bar changes. In beams, everything is simpler: it is either a horizontal or a rod inclined at a slight angle. In beams, these two agreements duplicate each other (but do not contradict if properly understood). And the question of which side to put the ordinates on was determined not "earlier and later", as you write, but by established traditions: builders have always built and plot diagrams on stretched fibers, and machine builders - on compressed ones (until now!). I could explain why, but I've written so much too. If the diagram M in the above problems had a "plus" sign, or there was no sign at all (indicating that the diagram was built on stretched fibers - for definiteness), then there would be no discussion at all. And the fact that the M sign does not affect the strength of the elements during the construction of a garden house, so no one argues about this. Although here you can invent special situations.
In general, this discussion is not fruitful in view of the triviality of the task. Every year, when a new stream of students comes to me, they have to explain these simple truths, or correct their brains, confused, to be honest, by individual teachers.
I would like to note that I have also obtained useful and interesting information from your site. For example, the graphical addition of the lines of influence of support reactions: an interesting technique that I have not seen in textbooks. The proof here is elementary: if we add up the equations of the lines of influence, we get identically unity. Probably, the site will be useful for craftsmen who have started construction. But still, in my opinion, it is better to use the literature based on SNIP. There are popular publications that contain not only formulas for resistance to material, but also design standards. There, simple techniques are given, containing both the overload factors, and the collection of standard and design loads, etc.
18-03-2013: Anna
great site, thank you! Please, tell me, if I have a point load of 500 N every half a meter on a 1.4 m long beam, can I calculate it as a uniformly distributed load of 1000 N / m? and what will then be equal to q?
18-03-2013: Dr. Lom
Vladislav
in this form, I accept your criticism, but still remain unconvinced. For example, there is a very old Handbook of Technical Mechanics, edited by Acad. A.N. Dinnik, 1949, 734 p. Of course, this guide has long been outdated and no one uses it now, nevertheless, in this guide, the diagrams for beams were built on compressed fibers, and not as it is now, and signs were put on the diagrams. This is what I meant when I said "earlier - then". In another 20-50 years, the currently accepted criteria for determining the signs of diagrams may change again, but this, as you understand, will not change the essence.
Personally, it seems to me that a negative sign for a plot located below the axis is more logical than a positive one, since we are taught from elementary grades that everything that is laid up along the ordinate is positive, everything that is down is negative. And the currently accepted designation is one of the many, although not the main obstacles to understanding the subject. In addition, for some materials, the calculated tensile strength is much lower than the calculated compressive strength and therefore the negative sign clearly shows the dangerous area for a structure made of such a material, however, this is my personal opinion. But I agree that it is not worth breaking spears on this issue.
I also agree that it is better to use verified and approved sources. Moreover, this is what I constantly advise my readers at the beginning of most articles and add that the articles are for information only and in no way are recommendations for calculations. At the same time, the right of choice remains with the readers, adults themselves should understand perfectly well what they are reading and what to do with it.
18-03-2013: Dr. Lom
Anna
A point load and a uniformly distributed load are still different things and the final calculation results for a point load directly depend on the points of application of the concentrated load.
Judging by your description, only two symmetrically located point loads act on the beam .. html), which translates the concentrated load into a uniformly distributed one.
18-03-2013: Anna
I know how to calculate, thanks, I don’t know which scheme to take more correctly, 2 loads through 0.45-0.5-0.45m or 3 through 0.2-0.5-0.5-0.2m I know the condition how to calculate, thanks, I don’t know which scheme to take more correctly, 2 loads through 0.45-0.5-0.45m or 3 through 0.2-0.5-0.5-0.2m, the condition is the most unfavorable position, support at the ends.
18-03-2013: Dr. Lom
If you are looking for the most unfavorable position of the loads, besides, they may not be 2 but 3, then for the sake of reliability, it makes sense to calculate the structure according to both options indicated by you. In a nutshell, the option with 2 loads seems to be the most unfavorable, but as I said, it is advisable to check both options. If the margin of safety is more important than the accuracy of the calculation, then you can take a distributed load of 1000 kg / m and multiply it by an additional factor 1.4-1.6, taking into account the uneven distribution of the load.
19-03-2013: Anna
thank you very much for the tip, one more question: what if the load indicated by me is applied not to a beam, but to a rectangular plane in 2 rows, cat? rigidly clamped on one larger side in the middle, how will the diagram look then, or how then should it be counted?
19-03-2013: Dr. Lom
Your description is too vague. I understand that you are trying to calculate the load on a certain sheet material, laid in two layers. What does "rigidly pinched on one larger side in the middle" mean, I still do not understand. Perhaps you mean that this sheet material will be based on the contour, but then what does it mean in the middle? Do not know. If the sheet material is pinched on one of the supports in a small area in the middle, then such pinching can generally be ignored and considered a hinged beam. If this is a single-span beam (it does not matter whether it is a sheet material or a rolled metal profile) with rigid pinching on one of the supports, then it should be calculated this way (see the article "Design schemes for statically indeterminate beams") If it is a certain plate supported along the contour, then the principles of calculating such a plate can be found in the corresponding article. If the sheet material is laid in two layers and these layers have the same thickness, then the design load can be reduced by half.
However, the sheet material should also be tested for localized compression from a concentrated load.
03-04-2013: Alexander Sergeevich
Thank you very much! for everything that you do by simply explaining to the people the basics of calculating building structures. This helped me a lot when calculating for myself, although I have
and a completed construction technical school and institute, and now I am a pensioner and have not opened textbooks and SNiPs for a long time, but I had to remember that when I was young I taught and it was too abstruse, basically everything is set out there and it turns out a brain explosion, but then everything became clear, because that the old yeast started working and the leaven of the brains went to ferment in the right direction. Thanks again.
and
09-04-2013: Alexander
What forces are exerted on a hinge beam with an equally distributed load?
09-04-2013: Dr. Lom
See section 2.2
11-04-2013: Anna
I returned to you, because I never found the answer. I'll try to explain it more clearly. This is a type of balcony 140 * 70 cm. Side 140 is bolted to the wall with 4 bolts in the middle in the form of a square 95 * 46mm. The very bottom of the balcony consists of a sheet of aluminum alloy perforated in the center (50 * 120) and 3 rectangular hollow profiles are welded under the bottom, cat. start from the point of attachment with the wall and diverge in different directions, one parallel to the side, i.e. straight, and two other different sides, in the corners of the oppositely fixed side. There is a 15 cm high bardyur in a circle; on the balcony there can be 2 people 80 kg each in the most unfavorable positions + an equally distributed load of 40 kg. The beams are not fixed to the wall, everything is kept on bolts. So, how can I calculate which profile and thickness of the sheet to take so that the bottom is not deformed? This cannot be considered a beam, does everything happen in a plane? or how?
12-04-2013: Dr. Lom
You know, Anna, your description is very similar to the riddle of the gallant soldier Schweik, which he asked the medical commission.
Despite such a seemingly detailed description, the design scheme is completely incomprehensible, what perforation the sheet of "aluminum alloy" has, how exactly the "rectangular hollow profiles" are located and from what material - along the contour or from the middle to the corners, and what kind of bardyur round?. However, I will not be like the medical luminaries who were part of the commission and will try to answer you.
1. A deck sheet can still be considered a beam with an estimated length of 0.7 m. And if the sheet is welded or simply supported along the contour, then the value of the bending moment in the middle of the span will indeed be less. I do not yet have an article devoted to the calculation of metal flooring, but there is an article "Calculation of a slab supported along a contour", dedicated to the calculation of reinforced concrete slabs. And since from the point of view of structural mechanics it does not matter from what material the calculated element is made, you can use the recommendations set out in this article to determine the maximum bending moment.
2. The flooring will still deform, since absolutely rigid materials still exist only in theory, but what amount of deformation is considered acceptable in your case is another question. You can use the standard requirement - no more than 1/250 of the span length.
14-04-2013: Yaroslav
This confusion with the signs is terribly frustrating in fact): (It seems that I understood everything, and the geomhar, and the selection of sections, and the stability of the rods. I love physics itself, in particular, mechanics) But the logic of these signs ...> _< Причем в механике же четко со знаками момента, относительно точки. А тут) Когда пишут "положительный -->if the bulge down "this is clear by logic. But in a real case - in some examples of solving problems" + ", in others -" - ". And even though you crack. Moreover, moreover, in the same cases, for example, the left reaction RA beams in different ways, relative to the other end, will be determined) Heh) It is clear that the difference will only affect the sign of the "protruding part" of the final diagram. also not all, sometimes in the examples the specified moment of closing is thrown out for some reason, in the equations of ROSU, although in the general equation they are not thrown out) In short, I have always loved classical mechanics for the ideal accuracy and clarity of formulation) And here ... And this is not the theory of elasticity was, not to mention arrays)
20-05-2013: ichthyander
Thank you so much.
20-05-2013: Ichthyander
Hello. Please, give an example (problem) with dimension Q q L, M in the section. Figure №1.2. Graphical display of the change in the reactions of the supports depending on the distance of the load application.
20-05-2013: Dr. Lom
If I understand correctly, then you are interested in determining support reactions, shear forces and bending moments using lines of influence. These issues are discussed in more detail in structural mechanics, examples can be found here - "Lines of influence of support reactions for single-span and cantilever beams" (http://knigu-besplatno.ru/item25.html) or here - "Lines of influence of bending moments and transverse forces for single-span and cantilever beams "(http://knigu-besplatno.ru/item28.html).
22-05-2013: Evgeny
Hello! Help me please. I have a cantilever beam, a distributed load acts on it along its entire length, a concentrated force acts on the extreme point "from bottom to top". At a distance of 1m from the edge of the beam, there is a torque M. I need to plot the shear force and moments. I do not know how to determine the distributed load at the point of application of the moment. Or does it not need to be counted at this point?
22-05-2013: Dr. Lom
The distributed load is therefore distributed because it is distributed along the entire length and for a certain point only the value of the shear forces in the section can be determined. This means that there will be no jump on the force diagram. But on the diagram of moments, if the moment is bending, and not rotating, there will be a jump. You can see how the diagrams from each of your specified loads will look like in the article "Design schemes for beams" (the link is in the text of the article before clause 3)
22-05-2013: Evgeny
But what about the force F applied to the extreme point of the beam? Because of it, there will be no jump on the shear force diagram?
22-05-2013: Dr. Lom
Will be. At the extreme point (the point of application of the force), a correctly constructed transverse force diagram will change its value from F to 0. Yes, this should be clear if you read the article carefully.
22-05-2013: Evgeny
Thank you Dr. Lom. I figured out how to do it, everything worked out. You have very useful informative articles! Write more, thank you very much!
18-06-2013: Nikita
Thank you for the article. My technicians cannot cope with a simple task: there is a structure on four supports, the load from each support (thrust bearing 200 * 200mm) is 36,000 kg, the step of the supports is 6,000 * 6,000 mm. What should be the distributed load on the floor in order to withstand this structure? (there are options 4 and 8 tons / m2 - the spread is very large). Thanks.
18-06-2013: Dr. Lom
You have a task of the reverse order, when the reactions of the supports are already known, and from them you need to determine the load and then the question should be formulated more correctly as follows: "at what uniformly distributed load on the floor, the support reactions will be 36,000 kg with a step between the supports of 6 m along the x axis and on the z-axis? "
Answer: "4 tons per m ^ 2"
Solution: the sum of the support reactions is 36x4 = 144 t, the overlap area is 6x6 = 36 m ^ 2, then the uniformly distributed load is 144/36 = 4 t / m ^ 2. This follows from equation (1.1), which is so simple, it is very difficult to understand how you can not understand it. And this is indeed a very simple task.
24-07-2013: Alexander
Two (three, ten) identical beams (stack) freely stacked on top of each other (the ends are not sealed) will withstand a greater load than one?
24-07-2013: Dr. Lom
Yes.
If you do not take into account the friction force arising between the contacting surfaces of the beams, then two beams stacked on top of each other with the same cross-section will withstand 2 times the load, 3 beams - 3 times the load, and so on. Those. from the point of view of structural mechanics, it makes no difference whether the beams lie side by side or one on top of the other.
However, this approach to solving problems is ineffective, since one beam with a height equal to the height of two identical free-folded beams will withstand a load 2 times greater than two free-folded beams. A beam with a height equal to the height of 3 identical freely folded beams will withstand a load 3 times greater than 3 freely folded beams, and so on. This follows from the equation of the moment of resistance.
24-07-2013: Alexander
Thanks.
I prove this to the designers using the example of paratroopers and a stack of bricks, a notebook / lonely sheet.
Grandmothers don't give up.
Their reinforced concrete obeys different laws than wood.
24-07-2013: Dr. Lom
In some ways, the grandmothers are right. Reinforced concrete is an anisotropic material and really cannot be considered a conventionally isotropic timber beam. And although special formulas are often used for calculating reinforced concrete structures, the essence of the calculation does not change from this. For an example, see the article "Determining the Moment of Resistance"
27-07-2013: Dmitriy
Thanks for the stuff. Please tell me the methodology for calculating one load on 4 supports on one line - 1 support to the left of the load application point, 3 supports to the right. All distances and loads are known.
27-07-2013: Dr. Lom
See the article "Multi-Span Continuous Beams."
04-08-2013: Ilya
All this is very good and quite intelligible. BUT ... I have a question for the rulers. And you did not forget to divide by 6 when determining the moment of resistance of the ruler? Something arithmetic does not converge.
04-08-2013: orderly Petrovich
And ethno, in what kind of ailment does it not converge? at 4.6, at 4.7, or what other? I need to express my thoughts more precisely.
15-08-2013: Alex
I am in shock, it turns out that I have thoroughly forgotten the strength of materials (otherwise "technology of materials"))), but later).
Doc thanks for your site I read, I remember, everything is very interesting. I found it by chance, - the task arose to evaluate which is more profitable (according to the criterion of the minimum cost of materials [in principle, without taking into account labor costs and costs for equipment / tools] to use in the structure of the column from ready-made shaped pipes (square) by calculation, or put your hands on and weld the columns yourself (let's say from the corner). Eh rags-pieces of iron, students, how long ago it was. Yes, nostalgia, there is not much.
12-10-2013: Olegggan
Good afternoon. I went to the site hoping to understand nevertheless the "physics" of the transition of the distributed load to the concentrated one and the distribution of the normative load on the entire plane of the site, but I see that you and my previous question with your answer have been removed: ((My design metal structures already work fine (I take a concentrated load and calculate everything according to it - good, my sphere of activity is about auxiliary devices, not architecture, which is enough with my head), but still I would like to understand about the distributed load in the context of kg / m2 - kg / m. I do not have the opportunity now to find out from anyone on this issue (I rarely come across such questions, but when I come across arguments begin: (), I found your site - everything is adequately stated, I also understand that knowledge costs money. Tell me how and where I am I can "thank you" just for the answer to my previous question about the site - for me this is really important. Communication can be transferred to the e-mail form - my soap " [email protected]". Thanks
14-10-2013: Dr. Lom
I designed our correspondence in a separate article "Determination of load on structures", all the answers are there.
17-10-2013: Artem
Thank you, having a higher technical education was a pleasure to read. A small note - the center of gravity of the triangle is at the intersection of the MEDIAN! (you have bisectors written).
17-10-2013: Dr. Lom
That's right, the remark is accepted - of course the medians.
24-10-2013: Sergei
It was necessary to find out how much the bending moment would increase if one of the intermediate beams was accidentally knocked out. I saw a quadratic dependence on distance, hence 4 times. Didn't have to shovel the textbook. Thanks a lot.
24-10-2013: Dr. Lom
For continuous beams with many supports, everything is much more complicated, since the moment will be not only in the span, but also on intermediate supports (see articles on continuous beams). But for a preliminary assessment of the bearing capacity, you can use the indicated quadratic dependence.
15-11-2013: Pavel
I can not understand. How to correctly calculate the load for the formwork. The soil creeps when digging, you need to dig a hole under the septic tank L = 4.5m, W = 1.5m, H = 2m. I want to perform the formwork itself as follows: the outline around the perimeter of a 100x100 beam (top, bottom, middle (1m), then a board of pine 2-grade 2x0.15x0.05. We make a box. I'm afraid that it will not stand ... because according to my calculations the board will withstand 96 kg / m2. Development of formwork walls (4.5x2 + 1.5x2) x2 = 24 m2. The volume of excavated soil is 13500 kg. 13500/24 = 562.5 kg / m2. Right or not ...? And what is the way out
15-11-2013: Dr. Lom
The fact that the walls of the pit crumble at such a great depth is natural and is determined by the properties of the soil. There is nothing wrong with that, in such soils, trenches and pits are dug with the bevel of the side walls. If necessary, the walls of the pit are reinforced with retaining walls and the properties of the soil are really taken into account when calculating the retaining walls. In this case, the pressure from the soil to the retaining wall is not constant in height, but conditionally uniformly changing from zero at the top to the maximum value at the bottom, but the value of this pressure depends on the properties of the soil. If we try to explain it as simply as possible, then the greater the bevel angle of the pit walls, the more pressure will be on the retaining wall.
You have divided the mass of all excavated soil by the wall area, and this is not correct. So it turns out that if at the same depth the width or length of the pit is twice as large, then the pressure on the walls will be twice as large. For calculations, you just need to determine the volumetric weight of the soil, as is a separate issue, but in principle it is not difficult to do this.
I do not give a formula for determining the pressure depending on the height, volumetric weight of the soil and the angle of internal friction, besides, you seem to want to calculate the formwork, and not the retaining wall. In principle, the pressure on the formwork boards from the concrete mixture is determined according to the same principle and is even a little simpler, since the concrete mixture can be conventionally considered as a liquid that exerts the same pressure on the bottom and walls of the vessel. And if the walls of the septic tank are poured not immediately to the full height, but in two passes, then, accordingly, the maximum pressure from the concrete mixture will be 2 times less.
Further, the board that you want to use for the formwork (2x0.15x0.05) is capable of withstanding very heavy loads. I don’t know exactly how you determined the load-bearing capacity of the board. Take a look at the article "Calculation of Timber Floors".
15-11-2013: Pavel
Thank you doctor. I did the wrong calculation, I understood the error. If we count as follows: span length 2m, pine board h = 5cm, b = 15cm then W = b * h2 / 6 = 25 * 15/6 = 375/6 = 62.5cm3
M = W * R = 62.5 * 130 = 8125/100 = 81.25 kgm
then q = 8M / l * l = 81.25 * 8/4 = 650/4 = 162kg / m2, or at a step of 1m 162kg / m2.
I'm not a builder, so I don't quite understand whether this is a lot or not enough for the pit where we want to shove a septic tank made of plastic, or our formwork will crack and we will not have time to do it all. This is the task, if you can suggest something else, I will be grateful to you ... Thank you again.
15-11-2013: Dr. Lom
Yeah. You still want to make a retaining wall during the installation of the septic tank and, judging from your description, you are going to do this after the foundation pit has been dug. In this case, the load on the boards will be created by the soil crumbled during installation and therefore will be minimal and no special calculations are required.
If you are going to fill up and tamp the soil back before installing the septic tank, then the calculation is really needed. Here are just the calculation scheme you did not accept correct. In your case, a board attached to 3 100x100 beams should be considered as a two-span continuous beam, the spans of such a beam will be about 90 cm, which means that the maximum load that 1 board can withstand will be much greater than the one you determined, although at the same time it is also necessary to take into account the uneven distribution of the load from the ground, depending on the height. And at the same time, check the bearing capacity of the beams working on the long side of 4.5 m.
In principle, the site has design schemes suitable for your case, but there is still no information on calculating soil properties, however, this is far from the basics of strength materials, and in my opinion you do not need such an accurate calculation. But in general, your desire to understand the essence of the processes is very commendable.
18-11-2013: Pavel
Thank you Doctor! I understood your idea, I still need to read your material. Yes, the septic tank needs to be shoved so that it does not collapse. At the same time, the formwork must withstand, because next to it, at a distance of 4m, there is also a foundation and you can easily bring it all down. That is why I am so worried. Thanks again, you have given me hope.
18-12-2013: Adolf Stalin
Doc, at the end of the article, where you give an example of determining the moment of resistance, in both cases you forgot to divide by 6. The difference will still be 7.5 times, but the numbers will be different (0.08 and 0.6) and not 0.48 and 3.6
18-12-2013: Dr. Lom
That's right, there was such a mistake, I fixed it. Thank you for your attention.
13-01-2014: Anton
Good afternoon. My question is how you can calculate the load on a beam. if on one side the fastening is rigid, on the other there is no fastening. beam length 6 meters. Here we need to calculate what kind of beam should be, better than a monorail. max load on unsecured side 2 tons. thanks in advance.
13-01-2014: Dr. Lom
Count as console. More details in the article "Design schemes for beams".
20-01-2014: yannay
If I had not studied sopramatism, then, to be honest, I would not have understood anything. If you write popularly, then you write popularly. And then suddenly something appears from nowhere, what kind of x? why x? why suddenly x / 2 and how does it differ from l / 2 and l? Suddenly q appeared. where? Maybe a typo and it was necessary to designate Q. Is it really impossible to describe in detail. And a moment about derivatives ... You understand that you are describing what only you understand. And the one who reads this for the first time will not understand it. Therefore, it was worth either to paint in detail, or to delete this paragraph altogether. I myself understood what it was about the second time.
20-01-2014: Dr. Lom
Here, unfortunately, I can not help you. The essence of unknown quantities is more popular only in the elementary grades of secondary school, and I believe that readers have at least this level of education.
An external concentrated load Q also differs from a uniformly distributed load q, as well as internal forces P from internal stresses p. Moreover, in this case, an external linear uniformly distributed load is considered, while the external load can be distributed both over the plane and over the volume, while the load distribution is far from always uniform. Nevertheless, any distributed load, denoted by a small letter, can always be reduced to the resultant force Q.
However, it is physically impossible to present all the features of structural mechanics and the theory of strength of materials in one article, there are other articles for this. Read, perhaps something will clear up.
08-04-2014: Sveta
Doctor! Could you please make an example of calculating a monolithic reinforced concrete section as a beam on 2 hinged supports, when the ratio of the sides of the section is more than 2x
09-04-2014: Dr. Lom
There are enough examples in the section "Calculation of reinforced concrete structures". In addition, I still could not comprehend the deep essence of your wording of the question, especially this: "with the ratio of the sides of the site more than 2x"
17-05-2014: vladimir
kind. I first met a sapromat on your site and got interested. I'm trying to figure out the basics, but I can't understand the Q diagrams with M, everything is clear and clear, and their difference is also. For distributed Q, I put a tank track or camu on the rope, for example, which is convenient. and on the concentrated Q, I hung the apple, everything is logical. how to look at the diagram on your fingers Q. I ask you not to quote the proverb to me, it does not suit me, I am already married. thanks
17-05-2014: Dr. Lom
To begin with, I recommend that you read the article "Fundamentals of strength of materials. Basic concepts and definitions", without this, you may be confused about what is stated below. Now I will continue.
The transverse force diagram is a conventional name, more correctly - a graph showing the values of the shear stresses arising in the cross-sections of the beam. Thus, according to the "Q" diagram, it is possible to determine the sections in which the values of the shear stresses are maximum (which may be needed for further structural calculations). Diagram "Q" is constructed (as well as any other diagram), proceeding from the conditions of static equilibrium of the system. Those. to determine the tangential stresses at a certain point, a part of the beam at this point is cut off (therefore, the sections), and for the remaining part, the equations of the equilibrium of the system are compiled.
Theoretically, a beam has an infinite set of cross-sections, and therefore it is also possible to form equations and determine the values of shear stresses infinitely. But there is no need to do this in areas where nothing is added or subtracted, or the change can be described by some kind of mathematical regularity. Thus, the stress values are determined only for a few characteristic sections.
And also the "Q" plot shows some general shear stress values for cross-sections. To determine the shear stresses along the height of the cross-section, another diagram is built and now it is called the shear stress diagram "t". More details in the article "Fundamentals of strength materials. Determination of shear stresses".
If on fingers, then take, for example, a wooden ruler and put it on two books, while the books are on the table so that the ruler rests on the edges of the books. Thus, we obtain a beam with hinged supports, on which a uniformly distributed load acts - the beam's own weight. If we cut the ruler in half (where the value of the "Q" plot is equal to zero) and remove one of the parts (while the support reaction conditionally remains the same), then the remaining part will rotate relative to the hinged support and the cut will fall on the table. To prevent this from happening, a bending moment must be applied at the cut point (the value of the moment is determined by the "M" diagram and the moment in the middle is the maximum), then the ruler will remain in the same position. This means that in the cross-section of the ruler located in the middle, only normal stresses act, and the tangents are equal to zero. On supports, normal stresses are zero, and shear stresses are maximum. In all other sections, both normal and shear stresses act.
17-07-2015: Pavel
Dr. Lom.
I want to put a mini hoist on a swivel console, attach the console itself to a height-adjustable metal rack (used in scaffolding). The rack has two platforms 140 * 140 mm. up and down. I install the stand on a wooden floor, fasten it at the bottom and in a spacer at the top. I fasten everything with a hairpin on M10-10mm nuts. The span itself is 2m, a pitch of 0.6m, the floor log is an edged board 3.5cm by 200cm, the floor is a grooved board 3.5 cm, the ceiling of the log is an edged board 3.5cm by 150cm, the ceiling is a grooved board 3.5 cm. All pine wood, 2nd grade of normal moisture. The rack weighs 10kg, the telfer 8kg. Swivel console 16 kg, swing arm boom swing 1m, the telpher itself is attached to the boom at the edge of the boom. I want to lift up to 100kg of weight to a height of up to 2m. In this case, the load after being lifted will turn by the boom within 180 degrees. I tried to perform the calculation, but it turned out to be beyond my power. Although your calculations for wooden floors seem to be understood. Thank you, Sergey.
18-07-2015: Dr. Lom
It is not clear from your description what exactly you want to calculate; the context suggests that you want to check the strength of a wooden floor (you are not going to determine the parameters of a rack, console, etc.).
1. The choice of the design scheme.
In this case, your hoist should be viewed as a concentrated load applied at the strut attachment point. Whether this load will act on one lag or two will depend on where the rack is attached. For more details, see the article "Calculating the floor in a billiard room". In addition, longitudinal forces will act on the logs of both floors and on the boards, and the further the load is from the rack, the more important these forces will be. How and why to explain for a long time, see the article "Determining the pull-out force (why the dowel does not hold in the wall)".
2. Collecting loads
Since you are going to lift loads, the load will not be static, but at least dynamic, i.e. the value of the static load from the hoisting device must be multiplied by the appropriate factor (see article "Shock load calculations"). Well, at the same time, do not forget about the rest of the load (furniture, people, etc.).
Since you are going to use a strut in addition to studs, then determining the load from the strut is the most time-consuming task, because first, it will be necessary to determine the deflection of the structures, and from the deflection value, determine the actual load.
More or less like this.
06-08-2015: LennyT
I work as an IT network sweep engineer (not by profession). One of the reasons for my leaving the design was the calculations according to the formulas from the field of strength materials and termech (I had to look for a suitable one according to the hands of Melnikov, Mukhanov, etc. .. :)) At the institute, I did not take lectures seriously. As a result, I got gaps. To my gaps in the calculations of Ch. the specialists were indifferent, since it is always convenient for the strong when they follow their instructions. As a result, my dream of being a design professional didn't come true. I was always worried about the uncertainty in the calculations (although there was always interest), respectively, they paid a penny.
Years later, I am already 30, but a sediment remains in my soul. About 5 years ago, such an open resource did not exist on the Internet. When I see that everything is clearly stated, I want to go back and learn again!)) The material itself is just an invaluable contribution to the development of people like me))), and there may be thousands of them ... I think that they, like me, will be very grateful to you. Thanks for the work you've done!
06-08-2015: Dr. Lom
Do not despair, it is never too late to learn. Life is often just beginning at the age of 30. Glad I could help.
09-09-2015: Sergei
"M = A x - Q (x - a) + B (x - l) (1.5)
For example, there is no bending moment on the supports, and indeed, the solution of equation (1.3) at x = 0 gives us 0 and the solution of equation (1.5) at x = l gives us also 0. "
I didn't really understand how solving equation 1.5 gives us zero. If we substitute l = x, then only the third term B (x-l) is equal to zero, and the other two are not. How then is M equal to 0?
09-09-2015: Dr. Lom
And you just plug the existing values into the formula. The fact is that the moment from the support reaction A at the end of the span is equal to the moment from the applied load Q, but these terms in the equation have different signs, therefore zero is obtained.
For example, with a concentrated load Q applied in the middle of the span, the support reaction A = B = Q / 2, then the equation of moments at the end of the span will have the following form
M = lxQ / 2 - Qxl / 2 + 0xQ / 2 = Ql / 2 - Ql / 2 = 0.
30-03-2016: Vladimir I
If x is the application distance Q then what is a, from the beginning to ... N .: l = 25cm x = 5cm in numbers, for example, what will be a
30-03-2016: Dr. Lom
x is the distance from the beginning of the beam to the considered cross-section of the beam. x can vary from 0 to l (el, not one), since we can consider any cross-section of the existing beam. a is the distance from the beginning of the beam to the point of application of the concentrated force Q. That is, with l = 25cm, a = 5cm, x can have any value, including 5 cm.
30-03-2016: Vladimir I
Understood. For some reason, I consider the section exactly at the point of application of the force. It is not necessary to consider the cross section between the load points as it is less affected than the subsequent point of the concentrated load. I don't argue, I just need to reconsider the topic
30-03-2016: Dr. Lom
Sometimes there is a need to determine the value of the moment, the shear force of other parameters, not only at the point of application of the concentrated force, but also for other cross sections. For example, when calculating beams with a tapered section.
01-04-2016: Vladimir
If you apply a concentrated load at some distance from the left support - x. Q = 1 l = 25 x = 5, then Rleft = A = 1 * (25-5) / 25 = 0.8
the value of the moment at any point of our beam can be described by the equation M = P x. Hence, M = A * x when x does not coincide with the point of application of the force, let the considered section be equal to x = 6, then we obtain
M = A * x = (1 * (25-5) / 25) * 6 = 4.8. When I take a pen and consistently substitute my values into formulas, I get confused. I need to distinguish the x's and assign one of them another letter. While I was typing, I figured it out thoroughly. You don't have to publish, but maybe someone will need it.
Dr. Lom
We use the principle of similarity of right-angled triangles. Those. a triangle, in which one leg is Q, and the second leg is l, is similar to a triangle with legs x - the value of the support reaction R and l - a (or a, depending on what kind of support reaction we define), from which the following follows equations (according to figure 5.3)
Rleft = Q (l - a) / l
Rpr = Qa / l
I don’t know if I explained it clearly, but it’s like there’s nowhere in more detail.
31-12-2016: Konstantin
Thank you so much for your work. You are helping a lot of people, including me, people. Everything is stated simply and intelligibly
04-01-2017: Rinat
Hello. If it's not difficult for you, explain how you got (derived) this equation of moments):
MB = Al - Q (l - a) + B (l - l) (x = l) On the shelves, as they say. Do not consider it arrogance, I just really did not understand.
04-01-2017: Dr. Lom
It seems that the article explains everything in sufficient detail, but I'll try. We are interested in the value of the moment at point B - MV. In this case, 3 concentrated forces act on the beam - support reactions A and B and force Q. Support reaction A is applied at point A at a distance l from support B, respectively, it will create a moment equal to Al. Force Q is applied at a distance (l - a) from support B, respectively, it will create a moment - Q (l - a). Minus because Q is directed in the direction opposite to the support reactions. Support reaction B is applied at point B and it does not create any moment, more precisely, the moment from this support reaction at point B will be equal to zero due to the zero shoulder (l - l). We add these values and get equation (6.3).
And yes, l is the span length, not one.
11-05-2017: Andrei
Hello! Thanks for the article, everything is much clearer and more interesting than in the textbook, I stopped at plotting the "Q" diagram of the display of the change in forces, nor how I can not understand why the diagram on the left rushes to the top, and from the right to the bottom, as I understood the forces that are on on the left and on the right support I act in a mirror, that is, the force of the beam (blue) and the reactions of the support (red) should be displayed on both sides, can you explain?
11-05-2017: Dr. Lom
This issue is discussed in more detail in the article "Construction of diagrams for a beam", but here I will say that there is nothing surprising in this - at the place of application of a concentrated force on the diagram of transverse forces there is always a jump equal to the value of this force.
09-03-2018: Sergei
Good afternoon! Consult the picture https://yadi.sk/i/CCBLk3Nl3TCAP2. Reinforced concrete monolithic support with consoles. If I do not cut the console, but rectangular, then according to the calculator the concentrated load on the edge of the console is 4t with a deflection of 4mm, and what load will be on this trimmed console in the picture. How, in this case, the concentrated and distributed load is calculated in my case. Sincerely.
09-03-2018: Dr. Lom
Sergey, look at the article "Calculation of beams of equal resistance to bending moment", this is certainly not your case, but the general principles of calculating beams of variable cross-section are set out there quite clearly.
1. Basic concepts and assumptions. Rigidity- the ability of a structure, within certain limits, to perceive the effect of external forces without destruction and significant changes in geometric dimensions. Strength- the ability of the structure and its materials to resist loads. Sustainability- the ability of the structure to maintain the shape of the initial balance. Endurance- strength of materials under load conditions. Continuity and Uniformity Hypothesis: the material consisting of atoms and molecules is replaced by a solid homogeneous body. Continuity means that an arbitrarily small volume contains a substance. Homogeneity means that all points of the material are the same. Using the hypothesis allows you to apply the system. coordinates and to study the functions of interest to us, use mathematical analysis and describe the actions with various models. Isotropy hypothesis: assumes that in all directions the Holy Islands of the material are the same. Anisotropic yavl tree, in which sv-va along and across the fibers are significantly different.
2. Mechanical characteristics of the material. Under yield pointσ T is understood as the stress at which the deformation grows without a noticeable increase in the load. Under elastic limitσ Y is understood to be such a maximum stress, up to which the material does not receive residual deformations. Tensile strength(σ B) is the ratio of the maximum force that the sample can withstand to its initial cross-sectional area. Proportional limit(σ PR) - the highest stress, up to which the material follows Hooke's law. The quantity E is the proportionality coefficient, called modulus of elasticity of the first kind. G-value name shear modulus or modulus of elasticity of the 2nd kind.(G = 0.5E / (1 + µ)). µ - dimensionless proportionality coefficient, called Poisson's ratio, characterizes the property of the material, is determined experimentally, for all metals the numerical values are in the range of 0.25 ... 0.35.
3. Forces. The interaction between the parts of the object under consideration is har-yut internal forces. They arise not only between the individual interacting units of the structure, but also between all adjacent particles of the object under loading. Internal forces are determined by the section method. Distinguish between superficial and volumetric external forces. Surface forces can be applied to small areas of the surface (these are concentrated forces, such as P) or to finite areas of the surface (these are distributed forces, such as q). They characterize the interaction of a structure with other structures or with the external environment. Volumetric forces are distributed over the volume of the body. These are the forces of gravity, magnetic stress, inertia forces during accelerated movement of the structure.
4. The concept of voltage, permissible voltage. Voltage Is a measure of the intensity of internal forces. Lim∆R / ∆F = p is the total stress. The total stress can be decomposed into three components: along the normal to the section plane and along two axes in the section plane. The normal component of the total stress vector is denoted by σ and is called normal stress. The components in the section plane are called shear stresses and denoted by τ. Allowable voltage- [σ] = σ PREV / [n] - depends on the grade of the material and the safety factor.
5. Tensile-compressive deformation. Stretching (squeezing)- type of loading, for which of the six internal force factors (Qx, Qy, Mx, My, Mz, N), five are equal to zero, and N ≠ 0. σ max = N max / F≤ [σ] + - tensile strength condition; σ max = N max / F≤ [σ] - is the condition of compressive strength. The mathematical expression of Zn Hooke: σ = εЕ, where ε = ∆L / L 0. ∆L = NL / EF - expanded Hooke's zone, where EF is the stiffness of the cross-section bar. ε - relative (longitudinal) deformation, ε '= ∆а / а 0 = ∆в / в 0 - transverse deformation, where under loading а 0, в 0 decreased by the value Δа = а 0 -а, ∆в = в 0 -in.
6. Geometric characteristics of flat sections. Static moment of the area: S x = ∫ydF, S y = ∫xdF, S x = y c F, S y = x c F. For a complex figure S y = ∑S yi, S x = ∑S xi. Axial moments of inertia: J x = ∫y 2 dF, J y = ∫x 2 dF. For a rectangle J x = bh 3/12, J y = hb 3/12, for a square J x = J y = a 4/12. Centrifugal moment of inertia: J xy = ∫xydF, if the section is symmetric at least one axis, J x y = 0. The centrifugal moment of inertia of asymmetric bodies will be positive if most of the area is located in the 1st and 3rd quadrants. Polar moment of inertia: J ρ = ∫ρ 2 dF, ρ 2 = x 2 + y 2, where ρ is the distance from the center of coordinates to dF. J ρ = J x + J y. For the circle J ρ = πd 4/32, J x = πd 4/64. For the ring J ρ = 2J x = π (D 4 -d 4) / 32 = πD 4 (1-α 4) / 32. Moments of resistance: for a rectangle W x = J x / y max, where y max is the distance from the center of gravity of the section to the boundaries along y. W x = bh 2/6, W x = hb 2/6, for a circle W ρ = J ρ / ρ max, W ρ = πd 3/16, for a ring W ρ = πD 3 (1-α 3) / 16 ... Center of gravity coordinates: x c = (x1F1 + x2F2 + x3F3) / (F1 + F2 + F3). Main radii of gyration: i U = √J U / F, i V = √J V / F. Moments of inertia for parallel translation of the coordinate axes: J x 1 = J x c + b 2 F, J y 1 = J uc + a 2 F, J x 1 y 1 = J x cyc + abF.
7. Shear and torsion deformation. Pure shift such a stress state is called when only tangential stresses τ appear on the faces of the selected eoment. Under torsion understand the type of motion, for which a force factor Mz ≠ 0 arises in the cross section of the rod, the rest Mx = My = 0, N = 0, Qx = Qy = 0. The change in internal force factors along the length is depicted in the form of a diagram using the section method and the sign rule. During shear deformation, the shear stress τ is related to the angular deformation γ by the relation τ = Gγ. dφ / dz = θ - relative twist angle Is the angle of mutual rotation of two sections, referred to the distance between them. θ = М К / GJ ρ, where GJ ρ is the torsional stiffness of the cross section. τ max = M Kmax / W ρ ≤ [τ] is the condition for the torsional strength of round bars. θ max = М К / GJ ρ ≤ [θ] - the condition for the torsional rigidity of round rods. [θ] - depends on the type of supports.
8. Bend. Under bend understand this type of loading, at which the axis of the rod is bent (bent) from the action of loads located perpendicular to the axis. Shafts of all machines are subjected to bending from the action of forces, a couple of forces - the moment at the places of landing of gear wheels, gears, half-couplings. 1) Bend name clean if a single force factor appears in the cross section of the bar - the bending moment, the remaining internal force factors are equal to zero. Strain formation during pure bending can be considered as a result of rotation of flat cross-sections relative to one another. σ = М у / J x - Navier formula for determining stresses. ε = у / ρ - longitudinal relative deformation. Dependency difference: q = dQz / dz, Qz = dMz / dz. Strength condition: σ max = M max / W x ≤ [σ] 2) Bend name flat if the force plane, i.e. the plane of action of the loads coincides with one of the central axes. 3) Bend naming oblique if the plane of action of the loads does not coincide with any of the central axes. The locus of points in the section, satisfying the condition σ = 0, is called the neutral section line, it is perpendicular to the plane of curvature of the curved bar. 4) Bend naming transverse if a bending moment and a shear force occurs in the cross section. τ = QS x ot / bJ x is Zhuravsky's formula, τ max = Q max S xmax / bJ x ≤ [τ] is the strength condition. A complete check of the strength of beams in transverse bending consists in determining the dimensions of the cross-section using the Navier formula and further checking for shear stresses. Because the presence of τ and σ in the section refers to complex loading, then the estimate of the stress state under their joint action can be calculated using 4 the theory of strength σ eq4 = √σ 2 + 3τ 2 ≤ [σ].
9. Tense state. Let us investigate the stress state (NS) in the vicinity of point A, for this we select an infinitely small parallelepiped, which we will place on an enlarged scale in the coordinate system. We replace the actions of the discarded part with internal force factors, the intensity of which can be expressed through the main vector of normal and tangential stresses, which we expand along three axes - these are the components of the NS of point A. No matter how difficult the body is loaded, it is always possible to single out mutually perpendicular areas , for which the shear stresses are equal to zero. Such sites are called the main ones. Linear NS - when σ2 = σ3 = 0, flat NS - when σ3 = 0, bulk NS - when σ1 ≠ 0, σ2 ≠ 0, σ3 ≠ 0. σ1, σ2, σ3 - principal stresses. Stresses on inclined areas at PNS: τ β = -τ α = 0.5 (σ2-σ1) sinα, σ α = 0.5 (σ1 + σ2) +0.5 (σ1-σ2) cos2α, σ β = σ1sin 2 α + σ2cos 2 α.
10. Theories of strength. In the case of LNS, the strength assessment is performed according to the condition σ max = σ1≤ [σ] = σ before / [n]. In the presence of σ1> σ2> σ3, in the case of NS, experimentally determining the dangerous state is laborious because of the large number of experiments at various combinations of stresses. Therefore, a criterion is used that allows one to single out the predominant influence of one of the factors, which will be called a criterion and will form the basis of the theory. 1) the first theory of strength (highest normal stresses): stress states are of equal strength for brittle fracture, if they have equal tensile stresses (does not take into account σ2 and σ3) - σ eq = σ1≤ [σ]. 2) the second theory of strength (the greatest tensile deformations - t Mariotte): n6stressed sost are equally strong in brittle fracture, if they have the greatest tensile deformations. ε max = ε1≤ [ε], ε1 = (σ1-μ (σ2 + σ3)) / E, σ eq = σ1-μ (σ2 + σ3) ≤ [σ]. 3) the third theory of strength (maximum stress - Coulomb): stresses are equal in strength according to the appearance of unacceptable plastic deformations, if they have the maximum stress τ max = 0.5 (σ1-σ3) ≤ [τ] = [σ] / 2, σ eq = σ1-σ3≤ [σ] σ eq = √σ 2 + 4τ 2 ≤ [σ]. 4) the fourth theory of the specific potential energy of shape change (energetic): during deformation of the potential, the energy consumption for changing the shape and volume U = U f + U V is stressed with equal strength according to the appearance of unacceptable plastic deformations, if they have equal specific potentials of the shape change energy. U eq = U f. Taking into account the generalized Hooke's z-na and the mathematics of transformations σ eq = √ (σ1 2 + σ2 2 + σ3 2 -σ1σ2-σ2σ3-σ3σ1) ≤ [σ], σ eq = √ (0.5 [(σ1-σ2) 2 + (σ1-σ3) 2 + (σ3-σ2) 2]) ≤ [τ]. In the case of PNS, σeq = √σ 2 + 3τ 2. 5) Mohr's fifth theory of strength (generalization of the theory of limiting states): the dangerous limiting state is determined by two main stresses, naib and naim σ eq = σ1-kσ3≤ [σ], where k is the coefficient of unequal strength, which takes into account the ability of the material to resist stretching unequally and compression to = [σ p] / [σ comp].
11. Energy theorems. Bending movement- in engineering calculations, there are cases when beams, satisfying the strength condition, do not have sufficient rigidity. The stiffness or deformability of the beam is determined by displacements: θ - angle of rotation, Δ - deflection. Under load, the beam is deformed and is an elastic line, which is deformed along the radius ρ A. The deflection and the angle of rotation in t A are formed by the tangent elastic line of the beam and the z-axis. To calculate the stiffness means to determine the maximum deflection and compare it with the allowable one. Mohr's method- a universal method for determining displacements for planar and spatial systems with constant and variable stiffness, convenient in that it can be programmed. To determine the deflection, draw a fictitious beam and apply a unit dimensionless force. Δ = 1 / EJ x * ∑∫MM 1 dz. To determine the angle of rotation, draw a fictitious beam and apply a unit dimensionless moment θ = 1 / EJ x * ∑∫MM ’1 dz. Vereshchagin's rule- it is convenient because, at constant stiffness, the integration can be replaced by algebraic multiplication of the diagrams of the bending moments of the load and the unit composition of the beam. Yavl is the main method, which is used in the disclosure of the SNA. Δ = 1 / EJ x * ∑ω p M 1 c - Vereshchagin's rule, in which displacement is inversely proportional to the stiffness of the beam and is directly proportional to the product of the area of the load beam by the ordinate of the center of gravity. Features of application: the diagram of bending moments is divided into elementary figures, ω p and M 1 c are taken taking into account the signs, if q and P or R act simultaneously on the site, then the diagrams must be delaminated, i.e. build separately from each load or apply different layering techniques.
12. Statically indeterminate systems. SNS is called the system, for which the equations of statics are not enough to determine the reactions of the supports, i.e. there are more connections, reactions in it than is necessary for their balance. The difference between the total number of supports and the number of independent statics equations, which can be compiled for a given system the degree of static uncertaintyS. The links imposed on the system of supernecessary are called superfluous or additional. The introduction of additional support fixings leads to a decrease in bending moments and maximum deflection, i.e. the strength and rigidity of the structure increases. To reveal the static indeterminacy, an additional deformation compatibility condition is added, which allows the determination of additional support reactions, and then the solution for determining the Q and M diagrams is performed as usual. Main system is obtained from the given one by discarding unnecessary connections and loads. Equivalent system- is obtained by loading the main system with loads and unnecessary unknown reactions replacing the actions of the discarded connection. Using the principle of independence of the action of forces, we find the deflection from the load P and the reaction x1. σ 11 x 1 + Δ 1p = 0 is the canonical equation of deformation compatibility, where Δ 1p is the displacement at the point of application x1 from the force P. Deformation verification of the solution- for this, we select another basic system and, having determined the angle of rotation in the support, should be equal to zero, θ = 0 - M ∑ * M ’.
13. Cyclic strength. In engineering practice, up to 80% of machine parts are destroyed due to static strength at stresses much lower than σ in cases where the stresses are alternating and cyclically changing. The process of accumulating damage during cyclical changes. stress is called material fatigue. The process of resistance to fatigue stress is called cyclic strength or endurance. T-period of the cycle. σmax τmax are normal stresses. σm, τm - average stress; r-coefficient of asymmetry of the cycle; factors affecting the endurance limit: a) Voltage concentrators: grooves, fillets, keys, threads and splines; this is taken into account by the effective coefficients of the end stresses, which are denoted by K σ = σ -1 / σ -1k K τ = τ -1 / τ -1k; b) Surface roughness: the rougher the machining of the metal is, the more metal defects there are during casting, the lower the endurance limit of the part will be. Any micro crack or indentation after the cutter can be the source of a fatigue crack. This is taken into account by the coefficient of influence of the surface quality. К Fσ К Fτ -; c) The scale factor affects the endurance limit, with an increase in the size of the part, the likelihood of defects increases, therefore, the larger the size of the part, the worse when assessing its endurance, this takes into account the coefficient of influence of the absolute dimensions of the cross section. K dσ K dτ. Defective coefficient: K σD = / Kv; Kv - the hardening factor depends on the type of heat treatment.
14. Sustainability. The transition of a system from a stable state to an unstable state is called a loss of stability, and the corresponding force is called critical force Ркр In 1774, E. Euler conducted a study and mathematically determined Rcr. According to Euler, Rcr is the force required for the smallest inclination of the column. Rcr = P 2 * E * Imin / L 2; Rod flexibilityλ = ν * L / i min; Critical stressσ cr = P 2 E / λ 2. Ultimate flexibilityλ depends only on the physical and mechanical properties of the material of the rod and it is constant for a given material.
Strength of materials- the section of mechanics of a deformable solid body, which considers methods for calculating the elements of machines and structures for strength, stiffness and stability.
Strength is the ability of a material to resist external forces without collapsing and without the appearance of permanent deformations. Strength calculations make it possible to determine the size and shape of parts that can withstand a given load at the lowest material cost.
Rigidity is the ability of a body to resist the formation of deformations. Stiffness calculations ensure that changes in body shape and size do not exceed acceptable limits.
Resilience is the ability of structures to resist efforts to bring them out of balance. Stability calculations prevent sudden loss of balance and distortion of structural elements.
Durability consists in the ability of a structure to maintain the service properties necessary for operation for a predetermined period of time.
A bar (Fig. 1, a - c) is a body, the cross-sectional dimensions of which are small in comparison with the length. The axis of the bar is the line connecting the centers of gravity of its cross sections. Distinguish between bars of constant or variable cross-section. The beam can have a straight or curved axis. A bar with a straight axis is called a bar (Fig. 1, a, b). Thin-walled structural elements are divided into plates and shells.
The shell (Fig. 1, d) is a body, one of the dimensions (thickness) of which is much smaller than the others. If the surface of the shell is a plane, then the object is called a plate (Fig. 1, e). Bodies are called bodies in which all sizes are of the same order (Fig. 1, f). These include the foundations of structures, retaining walls, etc.
These elements in the resistance of materials are used to draw up a design diagram of a real object and conduct its engineering analysis. A design scheme is understood as a certain idealized model of a real structure, in which all insignificant factors affecting its behavior under load are discarded.
Material property assumptions
The material is considered to be solid, homogeneous, isotropic and ideally elastic.
Continuity - the material is considered to be continuous. Homogeneity - The physical properties of a material are the same at all points.
Isotropy - material properties are the same in all directions.
Perfect elasticity- the property of the material (body) to completely restore its shape and size after eliminating the causes that caused the deformation.
Strain Assumptions
1. The hypothesis about the absence of initial internal efforts.
2. The principle of invariability of the initial dimensions - the deformations are small compared to the original dimensions of the body.
3. The hypothesis of linear deformability of bodies - deformations are directly proportional to the applied forces (Hooke's law).
4. The principle of the independence of the action of forces.
5. Hypothesis of flat Bernoulli sections - flat cross-sections of the bar before deformation remain flat and normal to the axis of the bar after deformation.
6. The principle of Saint-Venant - the stressed state of the body at a sufficient distance from the area of action of local loads depends very little on the detailed method of their application
External forces
The action on the structure of the surrounding bodies is replaced by forces that are called external forces or loads. Let's consider their classification. The loads include active forces (for the perception of which the structure was created), and reactive (bond reactions) - forces that balance the structure. According to the method of application, external forces can be divided into concentrated and distributed. Distributed loads are characterized by intensity and can be linear, superficial or volumetric. By the nature of the impact of the load, external forces are static and dynamic. Static forces include loads whose changes over time are small, i.e. accelerations of points of structural elements (inertial forces) can be neglected. Dynamic loads cause such accelerations in the structure or its individual elements, which cannot be neglected in the calculations.
Internal forces. Section method.
The action of external forces on the body leads to its deformation (the mutual arrangement of the body particles changes). As a result, additional interaction forces arise between the particles. These are the forces of resistance to changes in the shape and size of the body under the action of a load, called internal forces (efforts). With increasing load, internal efforts increase. Failure of a structural element occurs when the external forces exceed a certain maximum level of internal forces for a given structure. Therefore, the assessment of the strength of a loaded structure requires knowledge of the magnitude and direction of the resulting internal forces. The values and directions of internal forces in a loaded body are determined for given external loads by the section method.
The section method (see Fig. 2) consists in the fact that a bar, which is in equilibrium under the action of a system of external forces, is mentally cut into two parts (Fig. 2, a), and the equilibrium of one of them is considered, replacing the action of the discarded part of the bar a system of internal forces distributed over the section (Fig. 2, b). Note that the internal forces for the bar as a whole become external for one of its parts. Moreover, in all cases, the internal forces balance the external forces acting on the cut off part of the bar.
In accordance with the rule of parallel transfer of static forces, we bring all distributed internal forces to the center of gravity of the section. As a result, we obtain their principal vector R and principal moment M of the system of internal forces (Fig. 2, c). Choosing a coordinate system O xyz so that the z axis is the longitudinal axis of the bar and projecting the main vector R and the main moment M of internal forces on the axis, we obtain six internal force factors in the bar section: longitudinal force N, transverse forces Q x and Q y, bending moments M x and M y, as well as torque T. By the type of internal force factors, it is possible to determine the nature of the beam loading. If only the longitudinal force N appears in the cross-sections of the beam, and there are no other force factors, then there is "stretching" or "compression" of the beam (depending on the direction of the force N). If only the shear force Q x or Q y acts in the sections, this is the case of "pure shear". In the case of "torsion" in the sections of the bar, only the torques T act. In "pure bending", only the bending moments M. The combined types of loading (bending with tension, torsion with bending, etc.) are also possible - these are cases of "complex resistance". For a visual representation of the nature of the change in internal force factors along the axis of the bar, their graphs are built, called diagrams. Diagrams allow you to determine the most loaded sections of the beam and establish dangerous sections.
