8.2. Basic laws used in strength of materials
Statics relations. They are written in the form of the following equilibrium equations.
Hooke's law ( 1678): the greater the force, the greater the deformation, and, moreover, is directly proportional to the force. Physically, this means that all bodies are springs, but with great rigidity. When a beam is simply stretched by a longitudinal force N= F this law can be written as:
Here 
longitudinal force, l- beam length, A- 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 tangential stresses and shear angle:
.
G
calledshear modulus
, less often – elastic modulus of the second kind. Like any law, Hooke's law also has a limit of applicability. Voltage 
, up to which Hooke's law is valid, is called limit of proportionality(this is the most important characteristic in strength of materials).
Let's depict the dependence
from
graphically (Fig. 8.1). This picture is called stretch diagram
. After point B (i.e. at 
) this dependence ceases to be linear.
At 
after unloading, residual deformations appear in the body, therefore 
called elastic limit
.
When the voltage reaches the value σ = σ t, 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 voltage σ t at which the material flows is called yield strength .
Some materials (St. 3 - construction steel) after a short flow begin to resist again. The resistance of the material continues up to a certain maximum valueσ pr, then gradual destruction begins. The quantity σ pr is called tensile 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 difference, bodies change their size, and in direct proportion to this temperature difference.
Let there be a temperature difference 
. Then this law looks like:
Here α - coefficient of linear thermal expansion, l - rod length, Δ l- its lengthening.
4) Law of Creep .
Research has shown that all materials are highly heterogeneous in small areas. The schematic structure of steel is shown in Fig. 8.2.
Some of the components have the properties of a liquid, so many materials under load receive additional elongation over time 
(Fig. 8.3.) (metals at high temperatures, concrete, wood, plastics - at normal temperatures). This phenomenon is called creep material.
The law for liquids is: the greater the force, the greater the speed of movement of the body in the liquid. If this relationship is linear (i.e. force is proportional to speed), then it can be written as:
E 
If we move on to relative forces and relative elongations, we get
Here the index " cr
" means that the part of the elongation that is caused by the creep of the material is considered. Mechanical characteristics 
called the viscosity coefficient.
Law of energy conservation.
Consider a loaded beam
Let us introduce the concept of moving a point, for example,
- vertical movement of point B;
- horizontal displacement of point C.
Powers 
while doing some work U.
Considering that the forces 
begin to increase gradually and assuming that they increase in proportion to displacements, we obtain:
.
According to the conservation law: no work disappears, it is spent on doing other work or turns into another energy (energy- this is the work that the body can do.).
Work of forces 
, is spent on overcoming the resistance of elastic forces arising in our body. To calculate this work, we take into account that the body can be considered to consist of small elastic particles. Let's consider one of them:
It is subject to tension from neighboring particles 
. The resultant stress will be 
Under the influence 
the particle will elongate. According to the definition, elongation is the elongation per unit length. Then:
Let's calculate the work dW, which the force does dN (here it is also taken into account that the forces dN begin to increase gradually and they increase proportionally to the movements):
For the whole body we get:
.
Job W which was committed 
, called elastic deformation energy.
According to the law of conservation of energy:
6)Principle possible movements .
This is one of the options for writing the law of conservation of energy.
Let the forces act on the beam F 1
,
F 2
,
…
. They cause points to move in the body 
and voltage 
. Let's give the body additional small possible movements
. In mechanics, a notation of the form 
means the phrase " possible meaning quantities A" These possible movements will cause the body additional possible deformations
. They will lead to the appearance of additional external forces and stresses 
,
δ.
Let us calculate the work of external forces on additional possible small displacements:
Here 
- additional movements 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.). According to the 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 movements(it is also called principle of virtual movements). Similarly, we can consider the case when tangential stresses also act. Then we can obtain that to the deformation energy W the following term will be added:
Here is the shear stress, is the displacement of the small element. Then principle of possible movements 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.
Let us consider the pattern of sample elongation:
The phenomenon of shortening a body element across the direction of elongation is called Poisson effect.
Let us 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 one. For example, this is the case for a material reinforced with rigid fibers at a low angle 
<<1
(см. рис.8.8.). Оказывается, что коэффициент
Пуассона при этом почти пропорционален
величине
, i.e. the less 
, the larger the Poisson's ratio.
Fig.8.8. Fig.8.9
Even more surprising is the material shown in (Fig. 8.9.), and for such reinforcement there is a paradoxical result - longitudinal elongation leads to an increase in the size of the body in the transverse direction.
8) Generalized Hooke's law.
Let's consider an element that stretches in the longitudinal and transverse directions. Let us find the deformation that occurs in these directions. 
Let's calculate the deformation 
arising from action 
:
Let's consider the deformation from the action 
, which arises as a result of the Poisson effect:
The overall deformation will be:
If valid and 
, then another shortening will be added in the direction of the x axis 
.
Hence: 
Likewise: 
These relations are called generalized Hooke's law.
It is interesting that when writing Hooke’s law, an assumption is made about the independence of elongation strains from shear strains (about independence from shear stresses, which is the same thing) and vice versa. Experiments well confirm these assumptions. Looking ahead, we note that strength, on the contrary, strongly depends on the combination of tangential 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 scope of applicability.
1. Basic concepts and assumptions. Rigidity– the ability of a structure, within certain limits, to perceive the influence of external forces without destruction or significant changes in geometric dimensions. Strength– the ability of a structure and its materials to resist loads. Sustainability– the ability of a structure to maintain its original equilibrium shape. Endurance– strength of materials under load conditions. Hypothesis of continuity and homogeneity: the material consisting of atoms and molecules is replaced by a continuous homogeneous body. Continuity means that an arbitrarily small volume contains a substance. Uniformity means that the properties of the material are the same at all points. Using a 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 the properties of the material are the same in all directions. An anisotropic tree is one in which the fibers along and across the grain differ significantly.
2. Mechanical characteristics of the material. Under yield strengthσ T is understood as the stress at which strain increases without a noticeable increase in load. Under elastic limitσ У is understood as the greatest stress until 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. Proportionality limit(σ PR) – the highest stress, up to which the material follows Hooke’s law. The value E is a proportionality coefficient called elastic modulus of the first kind. Value G 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 properties of the material, is determined experimentally, for all metals the numerical values lie in the range of 0.25...0.35.
3. Forces. Interaction between parts of the object under consideration internal forces. They arise not only between individual interacting structural units, but also between all adjacent particles of an object under loading. Internal forces are determined by the method of sections. There are superficial and volumetric external forces. Surface forces can be applied to small areas of the surface (these are concentrated forces, for example P) or to finite areas of the surface (these are distributed forces, for example q). They characterize the interaction of a structure with other structures or with the external environment. Volume forces are distributed over the volume of the body. These are the forces of gravity, magnetic stress, and inertial forces during the accelerated movement of the structure.
4. The concept of voltage, permissible voltage. Voltage– measure of the intensity of internal forces. lim∆R/∆F=p – 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 called normal stress. The components in the section plane are called tangential stresses and denoted by τ. Allowable voltage– [σ]=σ PREV /[n] – depends on the grade of material and safety factor.
5. Tension-compression deformation. Tension (compression)– 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] - - condition of compressive strength. Mathematical expression for Hooke's value: σ=εE, where ε=∆L/L 0. ∆L=NL/EF – expanded Hooke’s zone, where EF is the stiffness of the cross-sectional rod. ε – relative (longitudinal) deformation, ε'=∆а/а 0 =∆в/в 0 – transverse deformation, where under loading a 0, в 0 decreased by the amount ∆а=а 0 -а, ∆в=в 0 -V.
6. Geometric characteristics of plane sections. Static moment of 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 symmetrical to at least one axis, J x y =0. The centrifugal moment of inertia of asymmetrical 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 coordinate center to dF. J ρ =J x +J y . For a 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 during parallel translation of 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 A stress state is called when only tangential stresses τ arise on the faces of a selected element. Under torsion understand the type of motion at which a force factor Mz≠0 arises in the cross section of the rod, the rest Mx=My=0, N=0, Qx=Qy=0. Changes in internal force factors along the length are 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 strain γ by the relation τ = Gγ. dφ/dz=θ – relative twist angle is the angle of mutual rotation of two sections, related to the distance between them. θ=M K/GJ ρ, where GJ ρ is the torsional stiffness of the cross section. τ max =M Kmax /W ρ ≤[τ] – condition of torsional strength of round rods. θ max =M K /GJ ρ ≤[θ] – condition of torsional rigidity of round rods. [θ] – depends on the type of supports.
8. Bend. Under bending understand this type of loading, in which the axis of the rod is bent (bent) from the action of loads located perpendicular to the axis. The shafts of all machines are subject to bending from the action of forces, a couple of forces - moments at the landing sites of gears, gears, coupling halves. 1) Bend name clean, if the only force factor that occurs in the cross section of the rod is the bending moment, the remaining internal force factors are equal to zero. The formation of deformations during pure bending can be considered as a result of the rotation of flat cross sections one relative to the other. σ=M y /J x – Navier’s formula for determining stresses. ε=у/ρ – longitudinal relative deformation. Differential dependence: q=dQz/dz, Qz=dMz/dz. Strength condition: σ max =M max /W x ≤[σ] 2) Bending name flat, if the force plane, i.e. the plane of action of loads coincides with one of the central axes. 3) Bend name oblique, if the plane of action of the loads does not coincide with any of the central axes. The geometric location of points in the section that satisfies the condition σ = 0 is called the neutral section line; it is perpendicular to the plane of curvature of the curved rod. 4) Bend name transverse, if a bending moment and transverse force arise in the cross section. τ=QS x ots /bJ x – Zhuravsky’s formula, τ max =Q max S xmax /bJ x ≤[τ] – strength condition. A complete check of the strength of beams during transverse bending consists of determining the cross-sectional dimensions 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 assessment of the stress state under their combined action can be calculated using the 4th theory of strength σ eq4 =√σ 2 +3τ 2 ≤[σ].
9. Tense state. Let's study the stress state (SS) in the vicinity of point A; for this we select an infinitesimal parallelepiped, which we 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 will expand along three axes - these are the components of the NS of point A. No matter how complex the body is loaded, it is always possible to identify mutually perpendicular areas , for which the tangential stresses are zero. Such sites are called the main ones. Linear NS – when σ2=σ3=0, flat NS – when σ3=0, volumetric NS – when σ1≠0, σ2≠0, σ3≠0. σ1, σ2, σ3 – principal stresses. Stresses on inclined areas during 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, strength assessment is performed according to the condition σ max =σ1≤[σ]=σ pre /[n]. In the presence of σ1>σ2>σ3 in the case of NS, experimental determination of a dangerous state is labor-intensive due to the large number of experiments at various combinations of stresses. Therefore, a criterion is used that allows one to highlight 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 (maximum normal stresses): stressed components are equal in strength to brittle fracture if they have equal tensile stresses (does not teach σ2 and σ3) – σ eq =σ1≤[σ]. 2) the second theory of strength (maximum tensile deformations - Mariotta): n6-tensioned compositions are equally strong in terms of brittle fracture if they have equal maximum tensile deformations. ε max =ε1≤[ε], ε1=(σ1-μ(σ2+σ3))/E, σ eq =σ1-μ(σ2+σ3)≤[σ]. 3) third theory of strength (maximum stress ratio - Coulomb): stress components are equally strong in terms of the appearance of unacceptable plastic deformations if they have equal maximum stress ratio 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 specific potential energy of shape change (energy): during deformation, the potential energy consumption for changing shape and volume U=U f +U V stress components are equally strong for the appearance of unacceptable plastic deformations if they have equal specific potential energy of shape change. U eq =U f. Taking into account the generalized Hooke's value and mathematical 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, σ eq =√σ 2 +3τ 2. 5) Mohr’s fifth theory of strength (generalized theory of limiting states): the dangerous limiting state is determined by two main stresses, the highest and the lowest σ eq =σ1-kσ3≤[σ], where k is the coefficient of uneven strength, which takes into account the ability of the material to resist tension unequally and compression k=[σ р ]/[σ сж ].
11. Energy theorems. Bending movement– in engineering calculations there are cases when beams, while satisfying the strength condition, do not have sufficient rigidity. The rigidity or deformability of the beam is determined by the movements: θ – angle of rotation, Δ – deflection. Under load, the beam is deformed and represents an elastic line, which is deformed along the radius ρ A. The deflection and angle of rotation in t A is formed by the tangent elastic line of the beam and the z axis. Calculating stiffness means determining the maximum deflection and comparing it with the permissible one. Mohr's method– a universal method for determining displacements for plane and spatial systems with constant and variable rigidity, convenient in that it can be programmed. To determine the deflection, we draw a fictitious beam and apply a unit dimensionless force. Δ=1/EJ x *∑∫MM 1 dz. To determine the angle of rotation, we draw a fictitious beam and apply a unit dimensionless moment θ=1/EJ x *∑∫MM’ 1 dz. Vereshchagin's rule– it is convenient in that, with constant stiffness, integration can be replaced by algebraic multiplication of diagrams of the bending moments of the load and unit beam components. This is the main method used in revealing the SNA. Δ=1/EJ x *∑ω p M 1 c – Vereshchagin’s rule, in which the displacement is inversely proportional to the rigidity of the beam and directly proportional to the product of the area of the cargo load of the beam and 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 section, then the diagrams must be stratified, i.e. build separately for each load or use various layering techniques.
12. Statically indeterminate systems. SNS is the name given to those systems whose static equations are not sufficient to determine the reactions of the supports, i.e. there are more connections and reactions in it than are necessary for their balance. The difference between the total number of supports and the number of independent static equations that can be composed for a given system is called degree of static indeterminationS. Connections superimposed on the system of super-necessary ones are called superfluous or additional. The introduction of additional support fastenings leads to a decrease in bending moments and maximum deflection, i.e. the strength and rigidity of the structure increases. To reveal static indeterminacy, an additional deformation compatibility condition is used, which allows additional reactions of supports to be determined, and then the solution to determine the Q and M diagrams is carried out as usual. Main system is obtained from a given one by discarding unnecessary connections and loads. Equivalent system– is obtained by loading the main system with loads and unnecessary unknown reactions that replace 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 reaction x1. σ 11 x 1 +Δ 1р =0 – canonical equation compatibility of deformation, where Δ 1р is the displacement at the point of application x1 from the force P. Δ 1р – Мр*М1, σ 11 -М1*М1 – this is conveniently performed by the Vereshchagin method. Deformation verification of the solution– for this we select another main system and determine the angle of rotation in the support, which must 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 damage accumulation during cyclic 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-cycle asymmetry coefficient; factors influencing the endurance limit: a) Stress concentrators: grooves, fillets, keys, threads and splines; this is taken into account by the effective stress concentrating factor, which is designated K σ =σ -1 /σ -1k K τ =τ -1 /τ -1k; b) Surface roughness: the rougher the mechanical processing of the metal, the more defects in the metal there are during casting, the lower the endurance limit of the part will be. Any microcrack or depression after the cutter can be the source of a fatigue crack. This takes into account the coefficient of influence of surface quality. To Fσ To Fτ - ; c) The scale factor influences the limit of endurance; as the size of the part increases, the probability of the presence of defects increases, therefore, the larger the size of the part, the worse when assessing its endurance, this is determined by the coefficient of influence of the absolute dimensions of the cross-section. To dσ To dτ . Defect coefficient: K σD =/Kv ; Kv – hardening coefficient depends on the type of heat treatment.
14. Sustainability. The transition of a system from a stable state to an unstable one is called loss of stability, and the corresponding force is called critical force Rcr In 1774, E. Euler conducted a study and mathematically determined Pcr. According to Euler, Pcr is the force required for the smallest inclination of the column. Pkr=P 2 *E*Imin/L 2 ; Flexibility of the rodλ=ν*L/i min ; Critical voltageσ cr =P 2 E/λ 2. Ultimate flexibilityλ depends only on the physical and mechanical properties of the rod material and it is constant for a given material.
19-08-2012: Stepan
My deepest bow to you for the clearly presented materials on strength of materials!)
At the institute I smoked bamboo and somehow had no time for strength of materials, the course wore off within a month)))
Now I work as an architect-designer and I constantly get stuck when I need to make calculations, I get buried in the muck of formulas and various methods and I understand that I have missed the basics..
Reading your articles, my head gradually becomes organized - everything is clear and very accessible!
24-01-2013: wany
thank you man!!))
I have only one question: if the maximum load per 1 m is 1 kg*m, then what about 2 meters?
2 kg*m or 0.5kg*m??????????
24-01-2013: Doctor Lom
If we mean distributed load on linear meter, then the distributed load 1kg/1m is equal to the distributed load 2kg/2m, which in the end still gives 1kg/m. And concentrated load is measured simply in kilograms or Newtons.
30-01-2013: Vladimir
Formulas are good! but how and what formulas should be used to calculate the structure for a canopy and most importantly, what size should the metal (profile pipe) be???
30-01-2013: Doctor Lom
If you have noticed, this article is devoted exclusively to the theoretical part, and if you are also smart, then without special labor You will find an example of structural calculations in the corresponding section of the site: Structural calculations. To do this, just go to the main page and find this section there.
05-02-2013: Leo
Not all formulas describe all the variables involved ((
There is also confusion with the notation, first the X denotes the distance from the left point to the applied force Q, and two paragraphs below the claim is already a function, then formulas are derived and confusion ensues.
05-02-2013: Doctor 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 force (concentrated load) and determining the value of the moment at some variable point relative to one of the supports are two different problems. Moreover, in each of the problems a variable is determined relative to the x-axis.
If this confuses you and you cannot figure out such basic things, then I can’t do anything. Complain to the Society for the Protection of the Rights of Mathematicians. And if I were you, I would file a complaint against textbooks on structural mechanics and strength of materials, otherwise, really, what is it? Are there not enough letters and hieroglyphs in alphabets?
And I also have a counter question for you: when you were solving problems on adding and subtracting apples in the third grade, did the presence of x in ten problems on the page also 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 variables, but you need to find this out from above from the context. And in some places there is no mention at all in the context. I'm not complaining at all. I’m talking about the shortcomings of the work (for which, by the way, I already 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 such a presentation of the material.
By the way, your arcasm is not appropriate, because you present everything on one page and without indicating all the variables it is not clear what you even mean. For example, in programming all variables are always specified. By the way, if you are doing all this for the people, then it would not hurt you to find out what contribution Kisilev made to mathematics as a teacher, and not as a mathematician, maybe then you will understand what I am talking about.
05-02-2013: Doctor 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 objective was - maximum by simple means convey to people who do not always have the appropriate higher education, basic concepts used in the theory of strength of materials and structural mechanics and why all this is needed at all. It’s clear that something had to be sacrificed. But.
There are enough correct textbooks, where everything is laid out on shelves, chapters, sections and volumes and described according to all the rules, even without my articles. But there are not so many people who can immediately understand these volumes. During my studies, two-thirds of the students did not understand the meaning of strength of materials, even approximately, and what can we say about ordinary people who are engaged in repairs or construction and want to calculate a lintel or beam? But my site is intended primarily for such people. I believe that clarity and simplicity are much more important than following the protocol to the letter.
I thought about breaking this article into separate chapters, but in this case the overall meaning is irreversibly lost, and therefore 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 stupid by default. But people are another matter. When your wife or girlfriend tells you: “The bread is out,” then without additional clarification, definitions and commands, you 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, you by default 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 unimportant factors. And at the same time, note that you do not even receive direct instructions to buy bread. This is the difference between a person and a computer.
But on the main thing I can agree with you, the article is not perfect, like everything else in the world around us. And don’t be offended by irony, there is too much seriousness in this world, sometimes you want to dilute it.
28-02-2013: Ivan
Good afternoon
Below formula 1.2 is the formula for the reaction of supports for a uniform load along the entire length of the beam A=B=ql/2. It seems to me that A=B=q/2 should be, or am I missing something?
28-02-2013: Doctor Lom
In the text of the article, everything is correct, because a uniformly distributed load means what load is applied along the length of the beam, and the distributed load is measured in kg/m. To determine the reaction of the support, we first find what the total load will be equal to, i.e. along the entire length of the beam.
28-02-2013: Ivan
28-02-2013: Doctor 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 a load distributed over a certain length, and therefore the greater the length of the beam, 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 the bridge structures.
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 is small” simply appears, this was misleading :-)
28-02-2013: Doctor Lom
The difference between concentrated and distributed load is described in the introductory article, the link to which is at the very beginning of the article, I recommend that you read it.
16-03-2013: Vladislav
It’s not clear why tell the basics of strength of materials to those who build or design. If at the university they did not understand the strength of materials from competent teachers, then they should not be allowed anywhere near 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 should have a positive sign. The negative sign affixed to the diagrams contradicts all generally accepted norms.
16-03-2013: Doctor Lom
1. Not everyone who builds has studied at universities. And for some reason, such people who are renovating their home do not want to pay professionals to select the cross-section of the lintel above the doorway in the partition. Why? ask them.
2. There are plenty of typos in paper editions of textbooks, but it’s not the typos that confuse people, but the overly abstract presentation of the material. This text may also contain typos, but unlike paper sources, they will be corrected immediately after they are discovered. But as for gross mistakes, I have to disappoint you, there are none here.
3. If you think that moment diagrams constructed from below the axis should only have a positive sign, then I feel sorry for you. Firstly, the diagram of moments is quite conventional and it only shows the change in the value of the moment in cross sections bendable element. In this case, the bending moment causes both compressive and tensile stresses in the cross section. Previously, it was customary to construct a diagram on top of the axis; in such cases, the positive sign of the diagram was logical. Then, for clarity, the diagram of moments began to be constructed as shown in the figures, but the positive sign of the diagrams was preserved from old memory. But in principle, as I already said, this is not of fundamental importance for determining the moment of resistance. The article on this subject says: “In this case, the value of the moment is considered negative if the bending moment tries to rotate the beam clockwise relative to the cross-section point in question. Some sources consider it the other way around, but this is nothing more than a matter of convenience.” However, there is no need to explain this to an engineer; I personally have encountered many times various options displaying diagrams and this has never caused any problems. But apparently you haven’t read the article, and your statements confirm that you don’t even know the basics of strength of materials, trying to replace knowledge with some generally accepted norms, and even “everyone”.
18-03-2013: Vladislav
Dear Doctor Lom!
You didn't read my message carefully. I spoke about errors in the sign of bending moments “in the examples given above,” 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 rule of signs for bending moments in beams during direct bending is the same. This is what I meant when speaking about generally accepted norms. And on which side of the beam to put the ordinates is another question. Let me explain my point.
The sign is placed on the diagrams in order to determine the direction of the internal force. 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 basic 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 the beam element with its convexity downward, causing stretching of the lower fibers.” In the examples given (in the second paragraph), the lower fibers are obviously stretched, so why is the sign on the diagram negative? Or is A. Alexandrov’s statement something special? Nothing like this. Let's look further.
2) Potapov V.D. and others. Structural mechanics. Statics of elastic systems, 2007, p. 27 – university textbook for builders: “a moment is considered positive if it causes tension in the lower fibers of the beam.”
3) A.V. Darkov, N.N. Shaposhnikov. Structural Mechanics, 1986, p. 27 is a well-known textbook also for builders: “with a positive bending moment, the upper fibers of the beam experience compression (shortening), and the lower fibers experience tension (elongation);.” As you can see, the rule is the same. Maybe things are completely different for machine builders? Again, no.
4) G.M. Itskovich. Strength of Materials, 1986, p. 162 – textbook for students of mechanical engineering colleges: “An external force (moment) bending this part (the cut-off part of the beam) with a convex downward direction, i.e. so that the compressed fibers are on top, gives a positive bending moment.”
The list goes on, but why? Any student who has passed the strength of strength test with at least a 4 knows this.
The question of which side of the rod to plot the ordinates of the diagram of bending moments is another agreement that can completely replace the above rule of signs. Therefore, when constructing diagrams M in frames, a sign is not placed on the diagrams, since the local coordinate system is connected to the rod, and changes its orientation when the position of the rod changes. In beams, everything is simpler: it is either a horizontal rod or a rod inclined at a slight angle. In beams, these two conventions duplicate each other (but do not contradict if understood correctly). And the question of which side to plot the ordinates from was determined not “before and then,” as you write, but by established traditions: builders have always built and are building diagrams on stretched fibers, and machine builders - on compressed ones (until now!). I could explain why, but I already wrote so much. If there had been a plus sign on the diagram M in the above problems, or no sign at all (indicating that the diagram was built on stretched fibers - for definiteness), then there would have been no discussion at all. And the fact that the M sign does not affect the strength of elements during construction garden house, so no one argues about this. Although even here you can invent special situations.
In general, this discussion is not fruitful due to the triviality of the task. Every year, when a new stream of students comes to me, I have to explain these simple truths, or correct the brains, confused, to be honest, by individual teachers.
I would like to note that I also learned useful and interesting information from your site. For example, graphically adding 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 one. Probably, the site will be useful to craftsmen who started construction. But still, in my opinion, it is better to use literature based on SNIP. There are popular publications containing not only strength-of-material formulas, but also design standards. There are given simple techniques, containing both 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 meter on a beam 1.4 m long, can I calculate a uniformly distributed load of 1000 N/m? and what will q be equal to then?
18-03-2013: Doctor 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. Dinnika, 1949, 734 p. Of course, this directory is long outdated and no one uses it now, however, in this directory, diagrams for beams were built on compressed fibers, and not as is customary now, and signs were put on the diagrams. This is exactly what I meant when I said “before - later”. 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 diagram located below the axis is more logical than a positive one, since primary classes we are taught that everything that goes up along the ordinate axis is positive, everything that goes 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 less than the calculated compressive strength and therefore the negative sign clearly shows a 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 intended for informational purposes only and in no way constitute recommendations for calculations. At the same time, the right of choice remains with the readers; adults themselves must understand perfectly well what they are reading and what to do with it.
18-03-2013: Doctor Lom
Anna
A point load and a uniformly distributed load are still different things, and the final results of calculations 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), than converting a concentrated load into a uniformly distributed one.
18-03-2013: Anna
I know how to calculate, thank you, I don’t know which scheme to take is more correct, 2 loads at 0.45-0.5-0.45m or 3 at 0.2-0.5-0.5-0.2m I know the condition how to calculate, thank you, I don’t know which scheme to take is more correct, 2 loads at 0.45-0.5-0.45m or 3 at 0.2-0.5-0.5-0.2m the condition is the most unfavorable position, support at the ends.
18-03-2013: Doctor Lom
If you are looking for the most unfavorable position of the loads, and besides, there may be not 2 but 3 of them, then for the sake of reliability it makes sense to calculate the design for both options you specified. Offhand, the option with 2 loads seems to be the most unfavorable, but as I already said, it is advisable to check both options. If the safety margin 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 of 1.4-1.6, which takes into account the uneven distribution of the load.
19-03-2013: Anna
Thank you very much for the hint, one more question: what if the load I indicated is applied not to the beam, but to a rectangular plane in 2 rows, cat. is rigidly pinched on one larger side in the middle, what will the diagram look like then or how to calculate it then?
19-03-2013: Doctor 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. I still don’t understand what “rigidly pinched on one larger side in the middle” means. Perhaps you mean that this sheet material will rest along the contour, but then what does it mean in the middle? Don't know. If the sheet material is pinched on one of the supports in a small area in the middle, then such pinching can be ignored altogether and the beam can be considered hinged. If it is a single-span beam (it does not matter whether it is a sheet material or a metal profile) with rigid pinching on one of the supports, then it should be calculated that way (see the article “Calculation schemes for statically indeterminate beams”) If it is a certain slab supported along the contour, then the principles for calculating such a slab can be found in the corresponding article. If the sheet material is laid in two layers and these layers have the same thickness, then design load can be reduced by half.
However, the sheet material, among other things, should be checked for local compression from concentrated load.
03-04-2013: Alexander Sergeevich
Thank you very much! for everything you do to simply explain to people the basics of calculation building structures. This personally helped me a lot when making calculations for myself personally, although I have
and a completed construction technical school and institute, and now I’m a pensioner and haven’t opened textbooks and SNiPs for a long time, but I had to remember that in my youth I once taught and it was painfully abstruse, basically everything is laid out there and it turns out to be a brain explosion, but then everything became clear, because that the old yeast began to work and the leaven of the brain began to wander in the right direction. Thanks again.
And
09-04-2013: Alexander
What forces act on a hinged beam with a uniformly distributed load?
09-04-2013: Doctor Lom
See paragraph 2.2
11-04-2013: Anna
I returned to you because I still couldn’t find an answer. I'll try to explain more clearly. This is a type of balcony 140*70 cm. Side 140 is screwed to the wall with 4 bolts in the middle in the form of a 95*46mm square. The bottom of the balcony itself consists of an aluminum alloy sheet perforated in the center (50*120) and 3 rectangular hollow profiles are welded under the bottom. start from the attachment point with the wall and diverge in different directions, one parallel to the side, i.e. straight, and the other two different sides, in the corners opposite to the fixed side. There is a border 15 cm high in a circle; on the balcony there can be 2 people of 80 kg each in the most unfavorable positions + an equally distributed load of 40 kg. The beams in the wall are not fixed, everything is held on by bolts. So, how can I calculate which profile to take and the thickness of the sheet so that the bottom does not deform? This can’t be considered a beam, after all, everything happens in a plane? or how?
12-04-2013: Doctor Lom
You know, Anna, your description is very reminiscent of the riddle of the good soldier Schweik, which he asked the medical commission.
Despite this it would seem detailed description, the design diagram is completely incomprehensible, what kind of perforation does the “aluminum alloy” sheet have, how exactly the “rectangular hollow profiles” are located and from what material are they made - along the contour or from the middle to the corners, and what kind of circular border is this?. However, I will not be like the medical luminaries who were part of the commission and will try to answer you.
1. The decking sheet can still be considered a beam with a design 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 actually be less. I don’t yet have an article devoted to the calculation of metal flooring, but there is an article “Calculation of a slab supported along the contour”, dedicated to the calculation reinforced concrete slabs. And since from the point of view of structural mechanics it does not matter what material the calculated element is made of, you can use the recommendations outlined in this article for determining the maximum bending moment.
2. The flooring will still be deformed, since absolutely rigid materials still exist only in theory, but what amount of deformation should be 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
In fact, this confusion with signs is terribly frustrating: (I seem to understand everything, the geomhar, the selection of sections, and the stability of the rods. I love physics myself, 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 is down" this is understandable by logic. But in real case- in some examples of problem solving it is “+”, in others it is “-”. And even if you crack. Moreover, in the same cases, for example, the left reaction RA of the beam will be determined differently, relative to the other end) Heh) It is clear that the difference will only affect the sign of the “protruding part” of the final diagram. Although... this is probably why there is no need to get upset about this topic) :) By the way, this is not all either, sometimes in the examples for some reason the specified closing moment is thrown out, in the equations ROSE, although in general equation don’t throw it away) In short, I always loved classical mechanics for its ideal accuracy and clarity of formulation) And here... And this was not yet the theory of elasticity, not to mention arrays)
20-05-2013: ichthyander
Thanks a lot.
20-05-2013: Ichthyander
Hello. Please give an example (problem) with dimension Q q L,M in the section. Figure No. 1.2. Graphic display of changes in support reactions depending on the distance of application of the load.
20-05-2013: Doctor Lom
If I understand correctly, then you are interested in determining support reactions, shear forces and bending moments using influence lines. 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: Eugene
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 1 m from the edge of the beam, the torque is M. I need to plot diagrams of shear force and moments. I don’t 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: Doctor Lom
The distributed load is distributed because it is distributed along the entire length and for a certain point only the value of the transverse forces in the section can be determined. This means that there will be no jump in 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 for each of the loads you specified will look in the article “Calculation diagrams for beams” (the link is in the text of the article before point 3)
22-05-2013: Eugene
But what about the force F applied to the extreme point of the beam? Because of it, there will be no jump in the diagram of transverse forces?
22-05-2013: Doctor Lom
Will. At the extreme point (the point of application of force), a correctly constructed diagram of transverse forces will change its value from F to 0. Yes, this should already be clear if you carefully read the article.
22-05-2013: Eugene
Thank you, Dr. Lom. I figured out how to do it, everything worked out. Your articles are very useful and informative! 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 support spacing is 6,000*6,000 mm. What should be the distributed load on the floor to withstand this design? (there are options of 4 and 8 tons/m2 - the spread is very large). Thank you.
18-06-2013: Doctor Lom
You have a task reverse order, when the reactions of the supports are already known, and from them it is necessary to determine the load, and then the question is more correctly formulated as follows: “at what uniformly distributed load on the floor will the support reactions be 36,000 kg with a step between the supports of 6 m along the x-axis and along the z-axis? "
Answer: "4 tons per m^2"
Solution: the sum of the support reactions is 36x4 = 144 t, the floor 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 that it is very difficult to understand how one could fail to understand it. And it's a really, really simple task.
24-07-2013: Alexander
Will two (three, ten) identical beams (stack) loosely stacked on top of each other (the ends are not sealed) support a greater load than one?
24-07-2013: Doctor Lom
Yes.
If we do not take into account the frictional force that arises between the contacting surfaces of the beams, then two beams with the same cross-section stacked on top of each other 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 next to each other 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 freely folded beams will withstand a load 2 times greater than two freely folded beams. And 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 moment of resistance equation.
24-07-2013: Alexander
Thank you.
I prove this to the designers using the example of paratroopers and a stack of bricks, a notebook/lone sheet.
Grandmothers don't give up.
Their reinforced concrete obeys different laws than wood.
24-07-2013: Doctor Lom
In some ways, the grandmothers are right. Reinforced concrete is an anisotropic material and cannot really be considered as a conventionally isotropic wooden beam. And although for calculations reinforced concrete structures Special formulas are often used, but the essence of the calculation does not change. For an example, see the article "Determination of the moment of resistance"
27-07-2013: Dmitriy
Thanks for the material. Please tell me the method 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 load are known.
27-07-2013: Doctor Lom
Look at 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. Did you remember to divide by 6 when determining the moment of resistance of the ruler? Somehow the arithmetic doesn't add up.
04-08-2013: orderly Petrovich
And what kind of thing doesn’t fit in? in 4.6, in 4.7, or in another one? I need to express my thoughts more precisely.
15-08-2013: Alex
I’m shocked, - it turns out that I had completely forgotten the strength of materials (otherwise known as “technology of materials”))), but later).
Doc, thank you for your site, I read it, I remember it, everything is very interesting. I found it by accident, and the task arose to evaluate what would be more profitable (according to the criterion of the minimum cost of materials [principally without taking into account labor costs and expenses for equipment/tools] to use ready-made columns in the construction profile pipes(square) according to calculation, or use your hands and weld the columns yourself (for example, from a corner). Oh, rags and pieces of hardware, students, how long ago it was. Yes, there is a little nostalgia.
12-10-2013: Olegggan
Good afternoon. I came to the site in the hope of understanding the “physics” of the transition of a distributed load to a concentrated one and the distribution of the standard 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 great (I take a concentrated load and calculate everything based on it; fortunately, my field of activity is about auxiliary devices, not architecture, which is enough), but I would still like to understand about the distributed load in the context of kg/m2 - kg/m. I don’t have the opportunity now to find out from anyone on this issue (I rarely encounter such questions, but when I do, the reasoning begins:(), I found your site - everything is adequately presented, I also understand that knowledge costs money. Tell me how and where I I can “thank you” just for your answer to my previous question about the site - this is really important to me. Communication can be transferred to an e-mail form - my soap " [email protected]". Thank you
14-10-2013: Doctor Lom
I compiled our correspondence into a separate article “Determination of load on structures”, all the answers are there.
17-10-2013: Artem
Thank you, having a higher technical education, it was a pleasure to read. A small note - the center of gravity of the triangle is at the intersection of the MEDIAN! (you have written bisectors).
17-10-2013: Doctor Lom
That's right, the comment is accepted - of course, the median.
24-10-2013: Sergey
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, therefore 4 times. I didn't have to dig through the textbook. Thank you very much.
24-10-2013: Doctor 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 the intermediate supports (see articles on continuous beams). But for a preliminary assessment of the bearing capacity, the indicated quadratic dependence can be used.
15-11-2013: Paul
Can not understand. How to correctly calculate the load for formwork. The soil creeps when digging, you need to dig a hole for a septic tank L=4.5m, W=1.5m, H=2m. I want to make the formwork itself like this: a contour around the perimeter of a beam 100x100 (top, bottom, middle (1m), then a 2-grade pine board 2x0.15x0.05. We make a box. I'm afraid it won't hold up...because according to my calculations the board will withstand 96 kg/m2. Development of formwork walls (4.5x2 +1.5x2)x2 = 24 m2. Volume of excavated soil 13500 kg. 13500/24 = 562.5 kg/m2. Right or wrong...? And what is the way out
15-11-2013: Doctor 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 this; in such soils, trenches and pits are dug with the side walls beveled. If necessary, the walls of the pit are strengthened with retaining walls and the properties of the soil are actually taken into account when calculating the retaining walls. At the same time, the pressure from the ground is retaining wall not constant in height, but conditionally uniformly varying 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 you try to explain it as simply as possible, the greater the bevel angle of the pit walls, the greater the pressure will be on the retaining wall.
You divided the mass of all excavated soil by the area of the walls, but this is not correct. 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 great. For calculations, you just need to determine the volumetric weight of the soil, which is a separate question, but in principle it is not difficult to do.
I don’t provide a formula for determining pressure depending on height, volumetric weight of soil and angle of internal friction; besides, you seem to want to calculate the formwork, not the retaining wall. In principle, the pressure on the formwork boards from concrete mixture is determined by the same principle and even a little simpler, since the concrete mixture can be conventionally considered as a liquid that exerts equal pressure on the bottom and walls of the vessel. And if you fill the walls of the septic tank not at once to the entire height, but in two passes, then, accordingly, the maximum pressure from the concrete mixture will be 2 times less.
Next, the board that you want to use for formwork (2x0.15x0.05) can withstand very heavy loads. I don’t know how exactly you determined the load-bearing capacity of the board. See the article "Calculation" wooden floor".
15-11-2013: Paul
Thank you doctor. I did the calculation wrong, I realized the mistake. 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 = 162 kg/m or with a step of 1 m 162 kg/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 push a plastic septic tank, or our formwork will crack and we won’t have time to do it all. This is the task, if you can suggest something else, I will be grateful to you... Thanks again.
15-11-2013: Doctor Lom
Yeah. You still want to make a retaining wall while the septic tank is being installed and, judging from your description, you are going to do this after the pit is dug. In this case, the load on the boards will be created by the soil that crumbled during installation and will therefore be minimal and no special calculations are required.
If you are going to fill up and compact the soil back before installing the septic tank, then a calculation is really needed. But the calculation scheme you adopted is not 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 the maximum load that 1 board can withstand will be significantly greater than that determined by you, although at the same time One should also take into account the uneven distribution of the load from the ground depending on the height. And at the same time, check the load-bearing capacity of beams running along the long side of 4.5 m.
In principle, the site has calculation schemes suitable for your case, but there is no information on calculating soil properties yet, however, this is far from the basics of strength of materials, and in my opinion you do not need such an accurate calculation. But overall, your desire to understand the essence of processes is very commendable.
18-11-2013: Paul
Thank you Doctor! I understand your idea, I’ll have to read more of your material. Yes, the septic tank needs to be pushed in so that a collapse does not occur. The formwork must withstand this, because There is also a foundation nearby at a distance of 4m and the whole thing can easily be brought down. That's why I'm so worried. Thanks again, you've 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: Doctor Lom
That's right, there was a mistake, I fixed it. Thank you for your attention.
13-01-2014: Anton
Good afternoon. I have a question: how can you 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. Now we need to calculate what the beam should be, better than a monorail. maximum load on the loose side is 2 tons. thank you in advance.
13-01-2014: Doctor Lom
Calculate like a console calculation. More details in the article "Calculation schemes for beams".
20-01-2014: yannay
If I had not studied sopramat, then, frankly speaking, I would not have understood anything. If you write popularly, then you write popularly. And then suddenly something appears out of nowhere, what the heck? why x? why suddenly x/2 and how does it differ from l/2 and l? Suddenly q appeared. where? Maybe there was a typo and it should have been labeled Q. Is it really impossible to describe it in detail? And the moment about derivatives...You understand that you are describing something that only you understand. And those who read this for the first time will not understand this. Therefore, it was worth either writing it down in detail or removing this paragraph altogether. I myself understood what I was talking about the second time.
20-01-2014: Doctor Lom
Unfortunately, I can’t help you here. More popularly, the essence of unknown quantities is presented only in elementary grades high school, and I believe that the readers have at least this level of education.
The external concentrated load Q is as different from the uniformly distributed load q as the internal forces P from the internal stresses p. Moreover, in this case, an external linear uniformly distributed load is considered, and yet the external load can be distributed both over the plane and over the volume, while the load distribution is not always uniform. Nevertheless, any distributed load denoted by a small letter can always be reduced to a 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 it, maybe something will become clearer.
08-04-2014: Sveta
Doctor! Could you make an example of calculating a monolithic reinforced concrete section as a beam on 2 hinged supports, with a ratio of the sides of the section greater than 2x
09-04-2014: Doctor Lom
In the section "Calculation of reinforced concrete structures" there are plenty of examples. Moreover, I was never able to comprehend the deep essence of your wording of the question, especially this: “when the ratio of the sides of the plot is greater than 2x”
17-05-2014: Vladimir
Kind. I came across sapromat for the first time on your site and became interested. I’m trying to understand the basics, but I can’t understand the Q diagrams; with M, everything is clear and clear, and their differences too. For distributed Q, I put, for example, a tank track or a kama on the rope, whichever is convenient. and on the concentrated Q I hung the apple, everything is logical. How to look at a diagram on your fingers Q. I ask you not to quote the proverb; it doesn’t suit me; I’m already married. Thank you
17-05-2014: Doctor Lom
To begin with, I recommend that you read the article “Basics of strength of strength. Basic concepts and definitions”; without this, there may be a misunderstanding of what is stated below. Now I’ll continue.
Diagram of transverse forces - a conventional name, more correctly - a graph showing the values of tangential stresses arising in the cross sections of the beam. Thus, using the “Q” diagram, you can determine the sections in which the values of the tangential stresses are maximum (which may be needed for further calculations of the structure). The "Q" diagram (as well as any other diagram) is constructed based on the conditions of static equilibrium of the system. Those. To determine the tangential stresses at a certain point, part of the beam is cut off at this point (hence the sections), and for the remaining part, equilibrium equations for the system are drawn up.
Theoretically, a beam has an infinite number of cross sections, and therefore it is also possible to compose equations and determine the values of tangential 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 mathematical pattern. Thus, stress values are determined only for a few characteristic sections.
And the "Q" plot also shows some general meaning shear stresses for cross sections. To determine the tangential stresses along the height of the cross section, another diagram is constructed and now it is called the shear stress diagram “t”. More details in the article "Fundamentals of strength materials. Determination of shear stresses."
If it’s on your fingers, then take, for example, a wooden ruler and put it on two books, with the books lying on the table so that the edges of the ruler rest on the books. Thus, we obtain a beam with hinged supports, which is subject to a uniformly distributed load - the beam’s own weight. If we cut the ruler in half (where the value of the “Q” diagram is zero) and remove one of the parts (while the support reaction conditionally remains the same), then the remaining part will rotate relative to the hinge support and fall on the table at the cut point. To prevent this from happening, a bending moment must be applied at the cutting site (the value of the moment is determined by the “M” diagram and the moment in the middle is 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 tangent stresses are equal to zero. At the supports, normal stresses are zero, and tangential stresses are maximum. In all other sections, both normal and shear stresses act.
17-07-2015: Paul
Doctor Lom.
I want to install a mini hoist on a rotating console, attach the console itself to a height-adjustable metal stand (used in scaffolding). The rack has two platforms 140*140 mm. up and down. I install the stand on a wooden floor, fastening it from below and spaced from above. I fasten everything with a stud on M10-10mm nuts. The span itself is 2m, pitch 0.6m, floor joists - edged board 3.5 cm by 200 cm, floor tongue-and-groove board 3.5 cm, ceiling joist - edged board 3.5 cm by 150 cm, ceiling tongue-and-groove board 3.5 cm. All wood is pine, 2nd grade normal humidity. The stand weighs 10 kg, the hoist - 8 kg. Rotating console 16 kg, boom of the rotating console max 1 m, the hoist 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, after lifting, the load will rotate like an arrow within 180 degrees. I tried to do the calculation, but I couldn't do it. Although your calculations wooden floors I think I understand. Thank you, Sergey.
18-07-2015: Doctor Lom
It is not clear from your description what exactly you want to calculate; from the context, it can be assumed that you want to check the strength of the wooden floor (you are not going to determine the parameters of the rack, console, etc.).
1. Selection of design scheme.
In this case your lifting mechanism should be considered as a concentrated load applied at the point where the post is attached. Whether this load will act on one joist 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 joists of both floors and on the boards, and the further the load is from the rack, the greater the importance of these forces. To explain how and why for a long time, see the article “Determination of pull-out force (why the dowel does not stay in the wall).”
2. Collection of 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 lifting mechanism should be multiplied by the appropriate coefficient (see the article “Calculation for shock loads”). Well, don’t forget about the rest of the load (furniture, people, etc.).
Since you are going to use a spacer in addition to the studs, determining the load from the spacer is the most labor-intensive task, because First, it will be necessary to determine the deflection of the structures, and then determine the effective load from the deflection value.
Like that.
06-08-2015: LennyT
I work as an IT network deployment engineer (not by profession). One of the reasons for my leaving design was calculations using formulas from the field of strength-of-materials and termekh (I had to look for a suitable one according to the hands of Melnikov, Mukhanov, etc.. :)) At the institute, I didn’t take lectures seriously. As a result, I got spaces. To my gaps in the calculations Ch. The specialists were indifferent, since it is always convenient for the strong when their instructions are followed. As a result, my dream of being a design professional did not come true. I was always worried about the uncertainty in the calculations (although there was always interest), and they paid pennies accordingly.
Years later, I’m already 30, but there’s still a residue in my soul. About 5 years ago, like this open resource did not exist on the Internet. When I see that everything is clearly presented, I want to go back and study again!)) The material itself is simply an invaluable contribution to the development of people like me))), and there are possibly 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: Doctor Lom
Don't despair, it's never too late to learn. Often at 30 years old life is just beginning. Glad I could help.
09-09-2015: Sergey
" M = A x - Q (x - a) + B (x - l) (1.5)
For example, there is no bending moment on the supports, and indeed, solving equation (1.3) for x=0 gives us 0 and solving equation (1.5) for x=l also gives us 0."
I don’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, but the other two are not. How then does M equal 0?
09-09-2015: Doctor Lom
And you just substitute the available 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, only these terms in the equation have different signs, so it turns out to be zero.
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 distance of the application Q, what is a, from the beginning to... N.: l=25cm x=5cm in numbers using the example of what will be a
30-03-2016: Doctor Lom
x is the distance from the beginning of the beam to the cross section of the beam in question. x can vary from 0 to l (el, not unity), 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 concentrated force Q. That is with l = 25 cm, a = 5 cm x can have any value, including 5 cm.
30-03-2016: Vladimir I
Understood. For some reason I am considering the cross section precisely at the point of application of the force. I see no need to consider the section between load points since it experiences less impact than the subsequent point of concentrated load. I'm not arguing, I just need to reconsider the topic again
30-03-2016: Doctor Lom
Sometimes there is a need to determine the value of the moment, shear force and other parameters not only at the point of application of the concentrated force, but also for other cross sections. For example, when calculating beams of variable cross-section.
01-04-2016: Vladimir
If you apply a concentrated load at a certain distance from the left support - x. Q=1 l=25 x=5, then Rlev=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 cross section under consideration be equal to x=6, then we get
M=A*x=(1*(25-5)/25)*6=4.8. When I take a pen and sequentially substitute my values into the formulas, I get confused. I need to distinguish the X's and assign a different letter to one of them. While I was typing I figured it out thoroughly. You don't have to publish it, but maybe someone will need it.
Doctor Lom
We use the principle of similarity right triangles. Those. a triangle in which one leg is equal to Q, and the second leg is equal to 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 are defining), from which the following follow equations (according to Figure 5.3)
Rlev = Q(l - a)/l
Rpr = Qa/l
I don’t know if I explained it clearly, but it seems like there’s nowhere to go in more detail.
31-12-2016: Konstantin
Thank you very much for your work. You help a lot of people, including me. Everything is presented simply and clearly
04-01-2017: Rinat
Hello. If it’s not difficult for you, explain how you received (withdrew) given equation moments):
МB = Аl - Q(l - a) + В(l - l) (x = l) According to the rules, as they say. Don't take it for impudence, I just really didn't understand.
04-01-2017: Doctor Lom
It seems that everything is explained in sufficient detail in the article, but I’ll try. We are interested in the value of the moment at point B - MV. In this case, the beam is acted upon by 3 concentrated forces - support reactions A and B and force Q. Support reaction A is applied at point A at a distance l from support B, accordingly it will create a moment equal to Al. Force Q is applied at a distance (l - a) from support B, accordingly it will create a moment - Q(l - a). Minus because Q is directed in the direction opposite to the support reactions. The 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 arm (l - l). We add these values and get equation (6.3).
And yes, l is the span length, not a unit.
11-05-2017: Andrey
Hello! Thank you for the article, everything is much clearer and more interesting than in the textbook, I settled on constructing a diagram “Q” to display the change in forces, I just can’t understand why the diagram on the left rushes to the top, and from the right to the bottom, how did I understand the forces that I act in a mirror way on the left and right supports, 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: Doctor Lom
This issue is discussed in more detail in the article “Constructing diagrams for a beam”, but here I will say that there is nothing surprising in this - at the point 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: Sergey
Good afternoon! Consult see picture https://yadi.sk/i/CCBLk3Nl3TCAP2. Reinforced concrete monolithic support with consoles. If I make the console not trimmed, 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 will be the load on this trimmed console in the picture. How, in this case, is the concentrated and distributed load calculated in my version? Sincerely.
09-03-2018: Doctor Lom
Sergey, look at the article “Calculation of beams of equal resistance to bending moment”, this is certainly not your case, but general principles calculations of beams of variable cross-section are presented there quite clearly.
Strength of materials– a section of the mechanics of deformable solids, which discusses methods for calculating elements of machines and structures for strength, rigidity and stability.
Strength is the ability of a material to resist external forces without collapsing and without the appearance of residual deformations. Strength calculations make it possible to determine the size and shape of parts that can withstand a given load at the lowest cost of material.
Stiffness 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 standards.
Stability is the ability of structures to resist forces that tend to bring them out of equilibrium. Stability calculations prevent sudden loss of balance and bending 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.
The beam (Fig. 1, a - c) is a body whose cross-sectional dimensions are small compared to its length. The axis of a beam is a line connecting the centers of gravity of its cross sections. There are beams of constant or variable cross-section. The beam can have a straight or curved axis. A beam with a straight axis is called a rod (Fig. 1, a, b). Thin-walled elements structures are divided into plates and shells.
The shell (Fig. 1, d) is a body, one of the dimensions of which (thickness) 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). Arrays are bodies whose dimensions are all of the same order (Fig. 1, f). These include building foundations, retaining walls, etc.
These elements in the strength of materials are used to draw up a design diagram of a real object and carry out its engineering analysis. A design scheme is understood as some idealized model of a real structure, in which all unimportant factors affecting its behavior under load are discarded
Assumptions about material properties
The material is considered continuous, homogeneous, isotropic and perfectly elastic.
Continuity – the material is considered continuous. Uniformity – physical properties material are the same at all points.
Isotropy - the properties of the material are the same in all directions.
Ideal elasticity– the property of a material (body) to completely restore its shape and size after eliminating the causes that caused the deformation.
Deformation Assumptions
1. Hypothesis about the absence of initial internal efforts.
2. The principle of constancy of initial dimensions - deformations are small compared to the original dimensions of the body.
3. Hypothesis about the linear deformability of bodies - deformations are directly proportional to the applied forces (Hooke's law).
4. The principle of independence of the action of forces.
5. Bernoulli's hypothesis of plane sections - plane cross sections of a beam before deformation remain flat and normal to the axis of the beam after deformation.
6. Saint-Venant’s principle - the stressed state of the body at a sufficient distance from the area of action of local loads depends very little on detailed method their applications
External forces
The action on the structure of surrounding bodies is replaced by forces called external forces or loads. Let's consider their classification. Loads include active forces (for the perception of which the structure is created), and reactive forces (reactions of connections) - 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 linearly, superficially or volumetrically distributed. Depending on the nature of the load, external forces can be static and dynamic. Static forces include loads whose changes over time are small, i.e. accelerations of points of structural elements (forces of inertia) can be neglected. Dynamic loads cause such accelerations in a structure or its individual elements that cannot be neglected in calculations
Internal forces. Section method.
The action of external forces on a body leads to its deformation (the relative arrangement of the particles of the body changes). As a result, additional interaction forces arise between particles. These forces of resistance to changes in the shape and size of the body under the influence of a load are called internal forces (efforts). As the load increases, the internal forces increase. Failure of a structural element occurs when external forces exceed a certain limiting level of internal forces for a given structure. Therefore, assessing 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 using the section method.
The method of sections (see Fig. 2) consists in the fact that a beam, 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 beam a system of internal forces distributed over the section (Fig. 2, b). Note that the internal forces for the beam 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 beam.
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 main vector R and the main moment M of the system of internal forces (Fig. 2, c). Having chosen the coordinate system O xyz so that the z axis is the longitudinal axis of the beam 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 section of the beam: 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, the nature of the loading of the beam can be determined. If only longitudinal force N occurs in the cross sections of the beam, and there are no other force factors, then “tension” or “compression” of the beam occurs (depending on the direction of the force N). If only the transverse force Q x or Q y acts in the sections, this is a case of “pure shear”. During “torsion,” only torque moments T act in the sections of the beam. During “pure bending,” only bending moments M act. Combined types of loading are also possible (bending with tension, torsion with bending, etc.)—these are cases of “complex resistance.” To visually represent the nature of changes in internal force factors along the axis of the beam, their graphs are drawn, called diagrams. Diagrams allow you to determine the most loaded areas of the beam and establish dangerous sections.