The design process of modern buildings and structures is regulated by a huge number of different building codes and regulations. In most cases, standards require certain characteristics to be ensured, for example, deformation or deflection of floor slab beams under static or dynamic load. For example, SNiP No. 2.09.03-85 determines for supports and overpasses the deflection of the beam is no more than 1/150 of the span length. For attic floors this figure is already 1/200, and for interfloor beams it is even less - 1/250. Therefore one of mandatory stages design is to perform a beam deflection calculation.
Ways to perform deflection calculations and tests
The reason why SNiPs establish such draconian restrictions is simple and obvious. The smaller the deformation, the greater the margin of strength and flexibility of the structure. For a deflection of less than 0.5%, the load-bearing element, beam or slab still retains elastic properties, which guarantees normal redistribution of forces and maintaining the integrity of the entire structure. As the deflection increases, the building frame bends, resists, but stands; when the permissible value is exceeded, the bonds break, and the structure loses its rigidity and load-bearing capacity like an avalanche.
- Use an online software calculator, which is “hardwired” standard conditions, and nothing more;
- Use ready-made reference data for various types and types of beams, for various support load patterns. It is only necessary to correctly identify the type and size of the beam and determine the desired deflection;
- Calculate the permissible deflection with your hands and your head; most designers do this, while controlling architectural and construction inspectors prefer the second method of calculation.
For your information! To really understand why it is so important to know the magnitude of the deviation from the initial position, it is worth understanding that measuring the amount of deflection is the only accessible and reliable way to determine the condition of the beam in practice.
By measuring how much the ceiling beam has sagged, you can determine with 99% certainty whether the structure is in disrepair or not.
Method of performing deflection calculations
Before starting the calculation, you will need to remember some dependencies from the theory of strength of materials and draw up a calculation diagram. Depending on how correctly the diagram is executed and the loading conditions are taken into account, the accuracy and correctness of the calculation will depend.
We use the simplest model of a loaded beam shown in the diagram. The simplest analogy of a beam can be a wooden ruler, photo.
In our case, the beam:
- It has a rectangular cross-section S=b*h, the length of the supporting part is L;
- The ruler is loaded with a force Q passing through the center of gravity of the bent plane, as a result of which the ends rotate through a small angle θ, with a deflection relative to the initial horizontal position , equal to f;
- The ends of the beam rest hingedly and freely on fixed supports, accordingly, there is no horizontal component of the reaction, and the ends of the ruler can move in any direction.
To determine the deformation of a body under load, use the formula of the elastic modulus, which is determined by the ratio E = R/Δ, where E is a reference value, R is force, Δ is the amount of deformation of the body.
Calculate moments of inertia and forces
For our case, the dependence will look like this: Δ = Q/(S E) . For a load q distributed along the beam, the formula will look like this: Δ = q h/(S E) .
What follows is the most important point. The above Young diagram shows the deflection of a beam or the deformation of a ruler as if it were crushed under a powerful press. In our case, the beam is bent, which means that at the ends of the ruler, relative to the center of gravity, two bending moments are applied with different sign. The loading diagram for such a beam is given below.
To transform Young's dependence for the bending moment, it is necessary to multiply both sides of the equality by the shoulder L. We obtain Δ*L = Q·L/(b·h·E) .
If we imagine that one of the supports is rigidly fixed, and an equivalent balancing moment of forces M max = q*L*2/8 will be applied to the second, respectively, the magnitude of the beam deformation will be expressed by the dependence Δх = M x/((h/3) b (h/2) E). The quantity b h 2 /6 is called the moment of inertia and is designated W. The result is Δx = M x / (W E) the fundamental formula for calculating a beam for bending W = M / E through the moment of inertia and bending moment.
To accurately calculate the deflection, you will need to know the bending moment and moment of inertia. The value of the first can be calculated, but specific formula for calculating a beam for deflection will depend on the conditions of contact with the supports on which the beam is located and the method of loading, respectively, for a distributed or concentrated load. The bending moment from a distributed load is calculated using the formula Mmax = q*L 2 /8. The given formulas are valid only for a distributed load. For the case when the pressure on the beam is concentrated at a certain point and often does not coincide with the axis of symmetry, the formula for calculating the deflection must be derived using integral calculus.
The moment of inertia can be thought of as the equivalent of a beam's resistance to bending load. The magnitude of the moment of inertia for a simple rectangular beam can be calculated using the simple formula W=b*h 3 /12, where b and h are the cross-sectional dimensions of the beam.
The formula shows that the same ruler or board of rectangular cross-section can have a completely different moment of inertia and deflection if placed on supports traditional way or put it on edge. No wonder almost all elements rafter system roofs are made not from 100x150 timber, but from 50x150 boards.
Real sections building structures can have a variety of profiles, from square, circle to complex I-beam or channel shapes. At the same time, determining the moment of inertia and the amount of deflection manually, “on paper”, for such cases becomes a non-trivial task for a non-professional builder.
Formulas for practical use
In practice, most often the opposite task is faced - to determine the safety factor of floors or walls for a specific case based on a known deflection value. In the construction business, it is very difficult to assess the safety factor using other, non-destructive methods. Often, based on the magnitude of the deflection, it is necessary to perform a calculation, evaluate the safety factor of the building and the general condition of the supporting structures. Moreover, based on the measurements taken, it is determined whether the deformation is acceptable, according to the calculation, or whether the building is in emergency condition.
Advice! In the matter of calculating the limit state of a beam based on the amount of deflection, the requirements of SNiP provide an invaluable service. By setting the deflection limit in a relative value, for example, 1/250, building codes greatly facilitate the determination of the emergency condition of a beam or slab.
For example, if you intend to buy a finished building that has stood for quite a long time on problematic soil, it would be useful to check the condition of the ceiling based on the existing deflection. Knowing everything permissible rate deflection and the length of the beam, one can assess without any calculation how critical the condition of the structure is.
Construction inspection, when assessing deflection and assessing the load-bearing capacity of a floor, takes a more complicated route:
- Initially, the geometry of the slab or beam is measured and the deflection value is recorded;
- Based on the measured parameters, the assortment of the beam is determined, then the formula for the moment of inertia is selected using the reference book;
- The moment of force is determined by the deflection and moment of inertia, after which, knowing the material, you can calculate the actual stresses in a metal, concrete or wooden beam.
The question is why is it so difficult if the deflection can be obtained using the formula for calculation for a simple beam on hinged supports f=5/24*R*L 2 /(E*h) under a distributed force. It is enough to know the span length L, profile height, design resistance R and elastic modulus E for a specific floor material.
Advice! Use in your calculations the existing departmental collections of various design organizations, which contain all the necessary formulas for determining and calculating the maximum loaded state in a condensed form.
Conclusion
Most developers and designers of serious buildings act in a similar way. The program is good, it helps to very quickly calculate the deflection and basic loading parameters of the floor, but it is also important to provide the customer with documentary evidence of the results obtained in the form of specific sequential calculations on paper.
Calculating a beam for bending “manually”, in the old-fashioned way, allows you to learn one of the most important, beautiful, clearly mathematically verified algorithms in the science of strength of materials. Using numerous programs like “entered the initial data...
... – get the answer” allows the modern engineer today to work much faster than his predecessors a hundred, fifty and even twenty years ago. However, with this modern approach the engineer is forced to completely trust the authors of the program and over time ceases to “feel the physical meaning” of the calculations. But the authors of the program are people, and people tend to make mistakes. If this were not so, then there would not be numerous patches, releases, “patches” for almost any software. Therefore, it seems to me that any engineer should be able to sometimes “manually” check the calculation results.
Help (cheat sheet, memo) for calculating beams for bending is presented below in the figure.
Let's try to use it using a simple everyday example. Let's say I decided to make a horizontal bar in my apartment. The location was determined - a corridor one meter and twenty centimeters wide. On opposite walls at the required height opposite each other, I securely fasten the brackets to which the crossbeam will be attached - a rod made of St3 steel with an outer diameter of thirty-two millimeters. Will this beam support my weight plus the additional dynamic loads that will arise during the exercises?
We draw a diagram for calculating a beam for bending. Obviously, the most dangerous scheme for applying an external load will be when I begin to pull myself up, hooking one hand on the middle of the bar.
Initial data:
F1 = 900 n – force acting on the beam (my weight) without taking into account dynamics
d = 32 mm – outer diameter of the rod from which the beam is made
E = 206000 n/mm^2 - modulus of elasticity of the steel beam material St3
[σi] = 250 n/mm^2 - permissible bending stresses (yield strength) for the steel beam material St3
Border conditions:
Мx (0) = 0 n*m – moment at point z = 0 m (first support)
Мx (1.2) = 0 n*m – moment at point z = 1.2 m (second support)
V (0) = 0 mm – deflection at point z = 0 m (first support)
V (1.2) = 0 mm – deflection at point z = 1.2 m (second support)
Calculation:
1. First, let's calculate the moment of inertia Ix and the moment of resistance Wx of the beam section. They will be useful to us in further calculations. For a circular cross-section (which is the cross-section of a rod):
Ix = (π*d^4)/64 = (3.14*(32/10)^4)/64 = 5.147 cm^4
Wx = (π*d^3)/32 = ((3.14*(32/10)^3)/32) = 3.217 cm^3
2. We create equilibrium equations to calculate the reactions of the supports R1 and R2:
Qy = -R1+F1-R2 = 0
Mx (0) = F1*(0-b2) -R2*(0-b3) = 0
From the second equation: R2 = F1*b2/b3 = 900*0.6/1.2 = 450 n
From the first equation: R1 = F1-R2 = 900-450 = 450 n
3. Let us find the angle of rotation of the beam in the first support at z = 0 from the deflection equation for the second section:
V (1.2) = V (0)+U (0)*1.2+(-R1*((1.2-b1)^3)/6+F1*((1.2-b2)^3)/6)/
U (0) = (R1*((1.2-b1)^3)/6 -F1*((1.2-b2)^3)/6)/(E*Ix)/1,2 =
= (450*((1.2-0)^3)/6 -900*((1.2-0.6)^3)/6)/
/(206000*5.147/100)/1.2 = 0.00764 rad = 0.44˚
4.
We compose equations for constructing diagrams for the first section (0 Shear force: Qy(z) = -R1 Bending moment: Mx (z) = -R1*(z-b1) Rotation angle: Ux (z) = U (0)+(-R1*((z-b1)^2)/2)/(E*Ix) Deflection: Vy (z) = V (0)+U (0)*z+(-R1*((z-b1)^3)/6)/(E*Ix) z = 0 m: Qy(0) = -R1 = -450 n Ux(0) = U(0) = 0.00764 rad Vy (0) = V (0) = 0 mm z = 0.6 m: Qy(0.6) = -R1 = -450 n Mx (0.6) = -R1*(0.6-b1) = -450*(0.6-0) = -270 n*m Ux (0.6) = U (0)+(-R1*((0.6-b1)^2)/2)/(E*Ix) = 0.00764+(-450*((0.6-0)^2)/2)/(206000*5.147/100) = 0 rad Vy (0.6) = V (0)+U (0)*0.6+(-R1*((0.6-b1)^3)/6)/(E*Ix) = 0+0.00764*0.6+(-450*((0.6-0)^3)/6)/ (206000*5.147/100) = 0.003 m The beam will bend in the center by 3 mm under the weight of my body. I think this is an acceptable deflection. 5.
We write the diagram equations for the second section (b2 Lateral force: Qy (z) = -R1+F1 Bending moment: Mx (z) = -R1*(z-b1)+F1*(z-b2) Rotation angle: Ux (z) = U (0)+(-R1*((z-b1)^2)/2+F1*((z-b2)^2)/2)/(E*Ix) Deflection: Vy (z) = V (0)+U (0)*z+(-R1*((z-b1)^3)/6+F1*((z-b2)^3)/6)/( E*Ix) z = 1.2 m: Qy (1,2) = -R1+F1 = -450+900 = 450 n Mx (1.2) = 0 n*m Ux (1,2) = U (0)+(-R1*((1,2-b1)^2)/2+F1*((1,2-b2)^2)/2)/(E* Ix) = 0,00764+(-450*((1,2-0)^2)/2+900*((1,2-0,6)^2)/2)/ /(206000*5.147/100) = -0.00764 rad Vy (1,2) = V (1,2) = 0 m 6.
We build diagrams using the data obtained above. 7.
We calculate the bending stresses in the most loaded section - in the middle of the beam and compare them with the permissible stresses: σi = Mx max/Wx = (270*1000)/(3.217*1000) = 84 n/mm^2 σi = 84 n/mm^2< [σи] = 250 н/мм^2 In terms of bending strength, the calculation showed a three-fold safety margin - the horizontal bar can be safely made from an existing rod with a diameter of thirty-two millimeters and a length of one thousand two hundred millimeters. Thus, you can now easily calculate a beam for bending “manually” and compare it with the results obtained when calculating using any of the numerous programs presented on the Internet. I ask THOSE WHO RESPECT the author’s work to SUBSCRIBE to article announcements.
86 comments on “Calculation of beams for bending - “manually”!” 1.1. Basic dependencies of the theory of beam bending
Beams It is customary to call rods that bend under the action of a transverse (normal to the axis of the rod) load. Beams are the most common elements of ship structures. The axis of a beam is the geometric location of the centers of gravity of its cross sections in an undeformed state. A beam is called straight if its axis is a straight line. The geometric location of the centers of gravity of the cross sections of a beam in a bent state is called the elastic line of the beam. The following direction of the coordinate axes is accepted: axis OX aligned with the axis of the beam, and the axis OY And OZ The theory of beam bending is based on the following assumptions. 1. The hypothesis of flat sections is accepted, according to which the cross sections of the beam, initially flat and normal to the axis of the beam, remain flat and normal to the elastic line of the beam after bending. Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2,). A 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,). b 3. Beams are considered sufficiently rigid, i.e.). their deflections are small compared to the height of the beams, and the angles of rotation of the sections are small compared to unity (Fig. 1.2, V). 4. Stresses and strains are related by a linear relationship, i.e. Hooke's law is valid (Fig. 1.2, G Rice. 1.2. Assumptions of beam bending theory. We will consider the bending moments and shearing forces that appear during bending of a beam in its cross-section as a result of the action of a part of the beam that is mentally thrown along the cross-section onto its remaining part. The moment of all forces acting in a section relative to one of the main axes is called the bending moment. The bending moment is equal to the sum of the moments of all forces (including support reactions and moments) acting on the rejected part of the beam, relative to the specified axis of the section under consideration. The projection onto the section plane of the main vector of forces acting in the section is called shear force. It is equal to the sum of projections onto the cross-sectional plane of all forces (including support reactions) acting on the rejected part of the beam Let us limit ourselves to considering the bending of the beam occurring in the plane XOZ. Such bending will occur when the lateral load acts in a plane parallel to the plane XOZ , and its resultant in each section passes through a point called the center of bending of the section. Note that for sections of beams that have two axes of symmetry, the center of bending coincides with the center of gravity, and for sections that have one axis of symmetry, it lies on the axis of symmetry, but does not coincide with the center of gravity.(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments. Let us denote the intensity of the distributed load (the load per unit length of the beam axis) by q, and the external bending moment is as M. And Distributed load and concentrated force are positive if the directions of their action coincide with the positive direction of the axis Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2,,2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,(Fig. 1.3, 3. Beams are considered sufficiently rigid, i.e.). ). The external bending moment is positive if it is directed clockwise (Fig. 1.3, Rice. 1.3. Sign rule for external loads Let us limit ourselves to considering the bending of the beam occurring in the plane Let us denote the deflection of a straight beam when it bends in the plane through w , and the angle of rotation of the section is through θ. And Let us accept the rule of signs for bending elements (Fig. 1.4): Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2,): 1) deflection is positive if it coincides with the positive direction of the axis 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,); (Fig. 1.4, 3. Beams are considered sufficiently rigid, i.e.); 2) the angle of rotation of the section is positive if, as a result of bending, the section rotates clockwise (Fig. 1.4, V). 3) bending moments are positive if the beam bends convex upward under their influence (Fig. 1.4, 4) shear forces are positive if they rotate the selected beam element counterclockwise (Fig. 1.4, Rice. 1.4. Sign rule for bending elements Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε x ε
Rice. 1.4.= −Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε/ρ
,(1.1) , separated by ρ
z from the neutral axis, it will be equal Where through(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.– radius of curvature of the beam in the section under consideration. Rice. 1.5. Beam bending diagramThe neutral axis of the cross section is the geometric location of points for which the linear deformation during bending is zero. Between curvature and derivatives of) there is a dependency Due to the accepted assumption of small rotation angles for sufficiently rigid beams, the value ρ
small compared to unity , therefore we can assume that Rice. 1.4. Substituting 1/ from (1.2) to (1.1), we obtain , separated by Normal bending stress σ based on Hooke's law will be equal Since it follows from the definition of beams that there is no longitudinal force directed along the axis of the beam, the main vector of normal stresses must vanish, i.e. F – cross-sectional area of the beam. From (1.5) we obtain that the static moment of the cross-sectional area of the beam is equal to zero. This means that the neutral axis of the section passes through its center of gravity. The moment of internal forces acting in the cross section relative to the neutral axis, aligned with the axis of the beam, and the axis M y will And From (1.5) we obtain that the static moment of the cross-sectional area of the beam is equal to zero. This means that the neutral axis of the section passes through its center of gravity. If we take into account that the moment of inertia of the cross-sectional area relative to the neutral axis aligned with the axis of the beam, and the axis is equal to , and substitute this value into (1.6), we obtain a dependence that expresses the basic differential equation for beam bending And Moment of internal forces in the section relative to the axis It follows that when a load is applied in a plane parallel to the main bending plane, the elastic line of the beam will be a flat curve. This bend is called flat. Based on dependencies (1.4) and (1.7), we obtain Formula (1.8) shows that normal stresses during bending of beams are proportional to the distance from the neutral axis of the beam. Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε Naturally, this follows from the hypothesis of plane sections. In practical calculations, the moment of resistance of the beam section is often used to determine the highest normal stresses where | | max – absolute value of the distance of the most distant fiber from the neutral axis. In what follows, subscripts y omitted for simplicity. There is a connection between the bending moment, shearing force and the intensity of the transverse load, which follows from the equilibrium condition of the element mentally separated from the beam.
Consider a beam element with length dx (Fig. 1.6). Here it is assumed that the deformations of the element are negligible. If a moment acts in the left section of the element M and cutting force N , then in its right section the corresponding forces will have increments. Let's consider only linear increments And Fig.1.6. Forces acting on a beam element Equating the projection onto the axis to zero of all forces acting on the element, and the moment of all forces relative to the neutral axis of the right section, we obtain: , and its resultant in each section passes through a point called the center of bending of the section. Note that for sections of beams that have two axes of symmetry, the center of bending coincides with the center of gravity, and for sections that have one axis of symmetry, it lies on the axis of symmetry, but does not coincide with the center of gravity.: From these equations, accurate to quantities of higher order of smallness, we obtain M From (1.11) and (1.12) it follows that (Fig. 1.6). Here it is assumed that the deformations of the element are negligible. 0
Dependencies (1.11)–(1.13) are known as the Zhuravsky–Schwedler theorem. From these dependencies it follows that the shear force and bending moment can be determined by integrating the loadWhereThe load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments. 0
0 and– shear force and bending moment in the section corresponding to. x = M, which is taken as the starting point; ξ, (Fig. 1.6). Here it is assumed that the deformations of the element are negligible.ξ 1 – integration variables Permanent 1.2. Differential equation for bending beams The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments. Differentiating equation (1.7) for the general case when the moment of inertia of the section is a function of , taking into account (1.11) and (1.12) we obtain: The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.. where the primes indicate differentiation with respect to For prismatic beams, i.e. beams of constant cross-section, we obtain the following differential bending equations: The ordinary inhomogeneous linear differential equation of the fourth order (1.18) can be represented as a set of four differential equations of the first order: through(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.), θ
(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.), (Fig. 1.6). Here it is assumed that the deformations of the element are negligible.(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.), M(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.). We use the following equation (1.18) or system of equations (1.19) to determine the deflection of the beam (its elastic line) and all unknown bending elements:The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.=
Integrating (1.18) 4 times successively (assuming that the left end of the beam corresponds to the section
xa ), we get: It is easy to see that the integration constantsNa,Ma,
,
θa
w a have a certain physical meaning, namely: N a – shearing force at the beginning of the count, i.e. atIntegrating (1.18) 4 times successively (assuming that the left end of the beam corresponds to the section
; x = M a Ma,
– bending moment at the beginning of the reference; θa
– angle of rotation at the beginning of the count; – deflection in the same section. To determine these constants, you can always create four boundary conditions - two for each end of a single-span beam. Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2, Naturally, the boundary conditions depend on the arrangement of the ends of the beam. The simplest conditions correspond to hinged support on rigid supports or rigid embedding. 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2, When the end of the beam is hingedly supported on a rigid support (Fig. 1.7, ) beam deflection and bending moment are zero: 3. Beams are considered sufficiently rigid, i.e. With rigid embedding on a rigid support (Fig. 1.7, ) the deflection and rotation angle of the section are equal to zero: V If the end of the beam (console) is free (Fig. 1.7, ), then in this section the bending moment and shearing force are equal to zero: A possible situation is associated with sliding embedding or symmetry embedding (Fig. 1.7,). This leads to the following boundary conditions: Note that the boundary conditions (1.26) concerning deflections and rotation angles are usually called. kinematic , and conditions (1.27) – by force Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2, Rice. 1.7. Types of boundary conditions In ship structures, we often have to deal with more complex boundary conditions, which correspond to the support of a beam on elastic supports or elastic termination of ends. And Elastic support (Fig. 1.8, ) is a support that has a drawdown proportional to the reaction acting on the support. We will consider the reaction of the elastic supportR,(1.29) , separated by positive if it acts on the support in the direction of the positive direction of the axis– coefficient of proportionality, called the coefficient of compliance of the elastic support. This coefficient is equal to the subsidence of the elastic support under the action of the reaction R= 1, i.e. A=w R
=
1 . Elastic supports in ship structures can be beams that reinforce the beam in question, or pillars and other structures that work in compression. To determine the compliance coefficient of an elastic support positive if it acts on the support in the direction of the positive direction of the axis it is necessary to load the corresponding structure with a unit force and find the absolute value of the subsidence (deflection) at the point of application of the force. Rigid support is a special case of elastic support with A=
0. Elastic sealing (Fig. 1.8, 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,) is a support structure that prevents free rotation of the section and in which the rotation angle θ in this section is proportional to the moment, i.e. there is a dependence θ
= Â
(Fig. 1.6). Here it is assumed that the deformations of the element are negligible..(1.30) Proportionality factor Â
is called the elastic embedding compliance coefficient and can be defined as the angle of rotation of the elastic embedding at M =
1, i.e. Â
=
θ
M = 1 . A special case of elastic sealing with Â
=
0 is hard termination. In ship structures, elastic embeddings are usually beams normal to the one under consideration and lying in the same plane. For example, beams, etc. can be considered elastically embedded on frames. Rice. 1.8. Elastic support ( Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2,) and elastic seal ( 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,) If the ends of the beam are long L are supported on elastic supports (Fig. 1.9), then the reactions of the supports in the end sections are equal to the shearing forces, and the boundary conditions can be written: The minus sign in the first condition (1.31) is accepted because the positive shear force in the left support section corresponds to the reaction acting on the beam from top to bottom, and on the support from bottom to top. If the ends of the beam are long Lelastically sealed(Fig. 1.9), then for support sections, taking into account the rule of signs for rotation angles and bending moments, we can write: The minus sign in the second condition (1.32) is accepted because with a positive moment in the right supporting section of the beam, the moment acting on the elastic seal is directed counterclockwise, and the positive angle of rotation in this section is directed clockwise, i.e. the directions of the moment and the angle of rotation do not coincide. Consideration of the differential equation (1.18) and all boundary conditions shows that they are linear with respect to both the deflections included in them and their derivatives, and the loads acting on the beam. Linearity is a consequence of the assumptions about the validity of Hooke's law and the smallness of beam deflections. Rice. 1.9. Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2,); forces in elastic supports and elastic seals corresponding to positive When several loads are applied to a beam, each bending element of the beam (deflection, rotation angle, moment and shear force) is the sum of the bending elements due to the action of each load separately. This very important position, called the principle of superposition, or the principle of summation of the action of loads, is widely used in practical calculations and, in particular, to reveal the static indetermination of beams. 1.3. Initial parameters method The general integral of the differential equation for beam bending can be used to determine the elastic line of a single-span beam in the case where the beam load is a continuous function of the coordinate throughout the entire span. If the load contains concentrated forces, moments, or a distributed load acts on part of the length of the beam (Fig. 1.10), then expression (1.24) cannot be used directly. In this case, it would be possible to designate elastic lines in sections 1, 2 and 3 through through 1 , w
2 , w 3, write out the integral for each of them in the form (1.24) and find all arbitrary constants from the boundary conditions at the ends of the beam and the conjugation conditions at the boundaries of the sections. The pairing conditions in the case under consideration are expressed as follows: at x=a 1
at x=a 2
at x=a 3
It is easy to see that this way of solving the problem leads to a large number of arbitrary constants, equal to 4 n, Where n– number of sections along the length of the beam. Rice. 1.10. Beam, in separate sections of which loads of different types are applied It is much more convenient to represent the elastic line of the beam in the form where terms beyond the double line are taken into account when The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.³
a 1, The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.³
a 2, etc. It is obvious that δ 1 through(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.)=through 2 (The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.)−through 1 (The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.); δ2 through(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.)=through 3 (The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.)−through 2 (The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.); etc. Differential equations for determining corrections to the elastic line δ ithrough
(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.) based on (1.18) and (1.32) can be written in the form General integral for any correction δ ithrough
(The load of the beams included in the ship hull can be either distributed (most often uniformly distributed along the axis of the beam, or varying according to a linear law), or applied in the form of concentrated forces and moments.) to the elastic line can be written in the form (1.24) with Integrating (1.18) 4 times successively (assuming that the left end of the beam corresponds to the section
=
a i
. In this case, the parameters It is easy to see that the integration constantsNa,Ma,
,
θa
have the meaning of changes (jumps) respectively: in shearing force, bending moment, angle of rotation and deflection arrow when passing through the section – shearing force at the beginning of the count, i.e. ata i
. This technique is called the initial parameters method. It can be shown that for the beam shown in Fig. 1.10, the equation of the elastic line will be M 0 ,
(Fig. 1.6). Here it is assumed that the deformations of the element are negligible. 0 ,
θ
0 ,
through 0, which are determined from the boundary conditions at the ends of the beam. Note that for a large number of variants of single-span beams encountered in practice, detailed bending tables have been compiled that make it easy to find deflections, rotation angles and other bending elements. 1.4. Determination of shear stresses during bending of beams The hypothesis of flat sections adopted in the theory of beam bending leads to the fact that the shear deformation in the beam section is equal to zero, and we are unable to determine shear stresses using Hooke’s law. However, since in the general case shearing forces act in beam sections, corresponding tangential stresses should arise. This contradiction (which is a consequence of the accepted hypothesis of plane sections) can be circumvented by considering equilibrium conditions. We will assume that when a beam composed of thin strips is bent, the tangential stresses in the cross section of each of these strips are uniformly distributed throughout the thickness and directed parallel to the long sides of its contour. This position is practically confirmed by exact solutions of the theory of elasticity.Let's consider a beam of an open thin-walled I-beam. In Fig. Figure 1.11 shows the positive direction of tangential stresses in the flanges and the profile wall during bending in the plane of the beam wall. Let us highlight with a longitudinal section I- There is a connection between the bending moment, shearing force and the intensity of the transverse load, which follows from the equilibrium condition of the element mentally separated from the beam.
I and two cross-sections of an element length (Fig. 1.12). Let us denote the tangential stress in the indicated longitudinal section by τ, and the normal forces in the initial cross section by T Rice. 1.12. A beam is called straight if its axis is a straight line. The geometric location of the centers of gravity of the cross sections of a beam in a bent state is called the elastic line of the beam. The following direction of the coordinate axes is accepted: axis Longitudinal forces and shear stresses in the beam flange element The condition of static equilibrium of an element selected from the beam (the projections of forces on the axis are equal to zero) will Where ;Let's consider a beam of an open thin-walled I-beam. In Fig. Figure 1.11 shows the positive direction of tangential stresses in the flanges and the profile wall during bending in the plane of the beam wall. Let us highlight with a longitudinal section f – area of the profile part cut off by the line I – Rice. 1.4.; δ – profile thickness at the section. From (1.36) it follows: Where ;Let's consider a beam of an open thin-walled I-beam. In Fig. Figure 1.11 shows the positive direction of tangential stresses in the flanges and the profile wall during bending in the plane of the beam wall. Let us highlight with a longitudinal section Since normal stresses σ aligned with the axis of the beam, and the axis are determined by formula (1.8), then In this case, we assume that the beam has a constant cross-section along its length. Static moment of the profile part (cut off by line ) relative to the neutral axis of the beam sectionis the integral Then from (1.37) for the absolute value of stresses we obtain: Naturally, the resulting formula for determining shear stresses is also valid for any longitudinal section, for example
ots is calculated for the cut-off part of the beam profile area relative to the neutral axis without taking into account the sign. Formula (1.38), in the sense of the derivation, determines the tangential stresses in the longitudinal sections of the beam. From the theorem on the pairing of tangential stresses, known from the course on strength of materials, it follows that the same tangential stresses act at the corresponding points of the cross section of the beam. Naturally, the projection of the main vector of tangential stresses onto the axis And must be equal to the shear force M in a given section of the beam. Since in the corbels of beams of this type, as shown in Fig. 1.11, tangential stresses are directed along the axis aligned with the axis of the beam, and the axis, i.e. Naturally, the resulting formula for determining shear stresses is also valid for any longitudinal section, for example
normal to the plane of action of the load, and are generally balanced, the shear force must be balanced by the shear stresses in the beam web. The distribution of tangential stresses along the height of the wall follows the law of change in static moment ots of the cut-off part of the area relative to the neutral axis (at a constant wall thickness δ). Normal bending stress σ Let us consider a symmetrical section of an I-beam with a flange area ω
=
1 and wall area
hδ (Fig. 1.13). Rice. 1.13. Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε Section of an I-beam Static moment of the cut-off part of the area for a point located at Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε from the neutral axis, there will be Naturally, the resulting formula for determining shear stresses is also valid for any longitudinal section, for example As can be seen from dependence (1.39), the static moment varies with ,
according to the law of quadratic parabola. Highest value 0: ots , and therefore tangential stresses τ will be obtained at the neutral axis, where z = The highest shear stress in the beam wall at the neutral axis M Since the moment of inertia of the section of the beam in question is equal to Normal bending stress σ then the maximum shear stress will be Attitude
/ω is nothing more than the average shear stress in the wall, calculated assuming a uniform stress distribution. Taking for example ω = 2 1 , according to formula (1.41) we get Let us limit ourselves to considering the bending of the beam occurring in the plane Thus, the beam under consideration has the greatest tangential stress in the wall at the neutral axis by only 12.5% Bending of the beam is not accompanied by twisting if the load acts in a plane parallel to Let us limit ourselves to considering the bending of the beam occurring in the plane passing through a point called the center of the bend. This point is characterized by the fact that the moment of all tangential forces in the section of the beam relative to it is equal to zero. Rice. 1.14. Tangential stresses during channel beam bending (point A
– center of bend) Indicating the distance of the center of the bend A
from the axis of the beam wall through e, we write down the condition for the moment of tangential forces to be equal to zero relative to the point A: , separated by Q 2 – tangential force in the wall, equal to the shearing force, i.e. Q 2 =M; Q 1 =Q 3 – force in the belt, determined based on (1.38) by the dependence The shear strain (or shear angle) γ varies along the height of the beam wall in the same way as the shear stresses τ ,
reaching its greatest value at the neutral axis. As has been shown, for beams with chords, the change in tangential stresses along the height of the wall is very insignificant. This allows us to further consider a certain average shear angle in the beam wall Shear deformation leads to the fact that the right angle between the cross-sectional plane of the beam and the tangent to the elastic line changes by the amount γ Wed A simplified diagram of the shear deformation of a beam element is shown in Fig. 1.15. Rice. 1.15. Beam element shear deformation diagram Having indicated the arrow of deflection caused by shear through through
sdv , we can write: Taking into account the rule of signs for shearing force M and find the angle of rotation Because the , Integrating (1.47), we obtain Constant a, included in (1.48), determines the displacement of the beam as a rigid body and can be taken equal to any value, since when determining the total arrow of deflection from bending through
bending and shear through SDV the sum of the integration constants will appear through 0 +a, determined from the boundary conditions. Here through 0 – deflection from bending at the origin. Let us put in the future a=0. Then the final expression for the elastic line caused by the shear will take the form The bending and shear components of the elastic line are shown in Fig. 1.16. Rice. 1.16. Thanks to this, the bending deformation of the beam can be considered independently of the shear deformation, which causes distortion of the cross-sectional planes of the beam and their rotation relative to the elastic line (Fig. 1.2, Bend ( 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,) and shear ( ) components of the elastic line of the beam In the case considered, the angle of rotation of the sections during shear is zero, therefore, taking into account the shear, the angles of rotation of the sections, bending moments and shear forces are associated only with the derivatives of the elastic line from the bend: Let us consider a beam freely supported on rigid supports, in the left section of which moment is valid M. The shearing force in this case will be constant and equal For the right reference section, we respectively obtain .(1.52) Expressions (1.51) and (1.52) can be rewritten as The expressions in parentheses characterize the relative addition to the angle of rotation of the section caused by the shear. If we consider, for example, a simply supported beam loaded in the middle of its span with a force q(Fig. 1.18), then the deflection of the beam under force will be equal to The bending deflection can be found from beam bending tables. The shear deflection is determined by formula (1.50), taking into account the fact that Rice. 1.18. Diagram of a simply supported beam loaded with a concentrated force As can be seen from formula (1.55), the relative addition to the beam deflection due to shear has the same structure as the relative addition to the angle of rotation, but with a different numerical coefficient. Let us introduce the notation where β is a numerical coefficient depending on the specific task under consideration, the design of supports and the load of the beam. Let's analyze the dependence of the coefficient k from various factors. If we take into account that , we obtain instead of (1.56) The moment of inertia of a beam section can always be represented in the form ,(1.58) where α is a numerical coefficient depending on the shape and characteristics of the cross section. Thus, for an I-beam, according to formula (1.40) with ω =2 Normal bending stress σ 1 we will find I = ωh
2 /3, i.e. α =1/3. Note that as the size of the beam flanges increases, the coefficient α will increase. Taking (1.58) into account, instead of (1.57) we can write: Thus, the value of the coefficient k significantly depends on the ratio of the span of the beam to its height, on the shape of the section (through the coefficient α), the arrangement of supports and the load of the beam (through the coefficient β). The relatively longer the beam ( h/L small), the smaller the influence of shear deformation. For rolled profile beams related h/L less than 1/10÷1/8, the shift correction can practically not be taken into account. However, for beams with wide flanges, such as, for example, keels, stringers and floras in the composition of bottom floors, the influence of shear and at the indicated h/L may turn out to be significant. It should be noted that shear deformations influence not only the increase in beam deflections, but in some cases also the results of revealing the static indetermination of beams and beam systems. 29-10-2012: Andrey
There was a typo in the formula for the bending moment for a beam with rigid pinching on supports (3rd from the bottom): the length should be squared. There was a typo in the maximum deflection formula for a beam with rigid pinching on supports (3rd from the bottom): it should be without the “5”. 29-10-2012: Doctor Lom
Yes, indeed, mistakes were made when editing after copying. The errors have now been corrected, thank you for your attention. 01-11-2012: Vic
typo in the formula in the fifth example from the top (the degrees next to X and El are mixed up) 01-11-2012: Doctor Lom
And it is true. Corrected. Thank you for your attention. 10-04-2013: flicker
Formula T.1 2.2 Mmax seems to be missing a square after a. 11-04-2013: Doctor Lom
Right. I copied this formula from the “Handbook of Strength of Materials” (edited by S.P. Fesik, 1982, p. 80) and did not even pay attention to the fact that with such a recording, even the dimension is not observed. Now I have recalculated everything personally, and indeed the distance “a” will be squared. Thus, it turns out that the typesetter missed a small two, and I fell for this millet. Corrected. Thank you for your attention. 02-05-2013: Timko
Good afternoon, I would like to ask you in Table 2, Diagram 2.4, I am interested in the formula “moment in flight” where the index X is not clear -? Could you answer) 02-05-2013: Doctor Lom
For cantilever beams in Table 2, the static equilibrium equation was compiled from left to right, i.e. the origin of coordinates was considered to be a point on a rigid support. However, if we consider a mirror cantilever beam, in which the rigid support will be on the right, then for such a beam the moment equation in the span will be much simpler, for example, for 2.4 Mx = qx2/6, more precisely -qx2/6, since it is now believed that if the diagram moment is located on top, then the moment is negative. 25-05-2013: Dmitriy
Please tell me at what ratio of the length of the beam to its diameter these formulas are valid? 25-05-2013: Doctor Lom
Dmitry, I already told you, for rotating shafts the calculation schemes will be different. However, if the shaft is stationary, then it can be considered as a beam, and it does not matter what its cross-section is: round, square, rectangular or something else. These calculation schemes most accurately reflect the state of the beam at l/D>10, with a ratio of 5 25-05-2013: Dmitriy
Thanks for the answer. Can you name other literature that I can refer to in my work? 25-05-2013: Doctor Lom
I don’t know what exact problem you are solving, and therefore it is difficult to have a substantive conversation. I'll try to explain my idea differently. 25-05-2013: Dmitriy
Can I then communicate with you via mail or Skype? I'll tell you what kind of work I do and what the previous questions were for. 25-05-2013: Doctor Lom
You can write to me, email addresses are not difficult to find on the site. But I’ll warn you right away that I don’t do any calculations and don’t sign partnership contracts. 08-06-2013: Vitaly
Question on table 2, option 1.1, deflection formula. Please check the size. 09-06-2013: Doctor Lom
That's right, the output is centimeters. 20-06-2013: Evgeniy Borisovich
Hello. Help me figure it out. We have a summer wooden stage near the cultural center, size 12.5 x 5.5 meters, at the corners of the stand there are metal pipes with a diameter of 100 mm. They force me to make a roof like a truss (it’s a pity that I can’t attach a picture), a polycarbonate covering, make trusses from a profile pipe (square or rectangle), there is a question about my work. If you don't do it, we'll fire you. I say it won’t work, but the administration and my boss say everything will work. What should I do? 20-06-2013: Doctor Lom
22-08-2013: Dmitriy
If a beam (a cushion under a column) lies on dense soil (more precisely, buried below the freezing depth), then what scheme should be used to calculate such a beam? Intuition suggests that the “two-support” option is not suitable and that the bending moment should be significantly less. 22-08-2013: Doctor Lom
Calculation of foundations is a separate big topic. In addition, it is not entirely clear which beam we are talking about. If we mean a cushion under a column of a columnar foundation, then the basis for calculating such a cushion is the strength of the soil. The purpose of the pillow is to redistribute the load from the column to the base. The lower the strength, the larger the area of the pillow. Or the greater the load, the larger the cushion area with the same soil strength. 23-08-2013: Dmitriy
This refers to a cushion under a column of a columnar foundation. The length and width of the cushion have already been determined based on the load and strength of the soil. But the height of the pillow and the amount of reinforcement in it are questionable. I wanted to calculate by analogy with the article “Calculation of a reinforced concrete beam,” but I believe that it would not be entirely correct to calculate the bending moment in a cushion lying on the ground, as in a beam on two hinged supports. The question is - what calculation scheme is used to calculate the bending moment in the cushion. 24-08-2013: Doctor Lom
The height and cross-section of the reinforcement in your case are determined as for cantilever beams (along the width and length of the cushion). Scheme 2.1. Only in your case, the support reaction is the load on the column, or more precisely, part of the load on the column, and the uniformly distributed load is the resistance of the soil. In other words, the specified calculation scheme needs to be turned over. 10-10-2013: Yaroslav
Good evening. Please help me choose metal. beam for a shed of 4.2 meters. A residential building has two floors, the base is covered with hollow slabs 4.8 meters long, on top there is a load-bearing wall of 1.5 bricks, 3.35 m long and 2.8 m high. Then there is a doorway. On top of this wall there are floor slabs on one side 4.8 m long . on the other 2.8 meters on the slabs there is again a load-bearing wall as on the floor below and above there are wooden beams 20 by 20 cm long 5 m. 6 pieces and 3 meters long 6 pieces the floor is made of boards 40 mm. 25 m2. There are no other loads. Please suggest me which I-beam to take in order to sleep peacefully. So far everything has been standing for 5 years. 10-10-2013: Doctor Lom
Look in the section: "Calculation of metal structures" at the article "Calculation of a metal lintel for load-bearing walls"; it describes in sufficient detail the process of selecting the section of a beam depending on the current load. 04-12-2013: Kirill
Please tell me where I can get acquainted with the derivation of the formulas for the maximum deflection of a beam for pp. 1.2-1.4 in Table 1 04-12-2013: Doctor Lom
The derivation of formulas for various options for applying loads is not provided on my website. You can see the general principles on which the derivation of such equations is based in the articles “Fundamentals of strength strength, calculation formulas” and “Fundamentals of strength strength, determination of beam deflection.” 24-03-2014: Sergey
an error was made in 2.4 of table 1. even the dimension is not respected 24-03-2014: Doctor Lom
I don’t see any errors, much less non-compliance with dimensions, in the calculation scheme you specified. Find out what exactly the error is. 09-10-2014: Sanych
Good afternoon. Do M and Mmax have different units of measurement? 09-10-2014: Sanych
Table 1. Calculation 2.1. If l is squared, then Mmax will be in kg*m2? 09-10-2014: Doctor Lom
No, M and Mmax have a single unit of measurement kgm or Nm. Since the distributed load is measured in kg/m (or N/m), the torque value will be kgm or Nm. 12-10-2014: Paul
Good evening. I work in the production of upholstered furniture and the director gave me a problem. I ask for your help, because... I don’t want to solve it “by eye”. 12-10-2014: Doctor Lom
It depends on many factors. In addition, you did not indicate the thickness of the pipe. For example, with a thickness of 2 mm, the moment of resistance of the pipe is W = 3.47 cm^3. Accordingly, the maximum bending moment that the pipe can withstand is M = WR = 3.47x2000 = 6940 kgm or 69.4 kgm, then the maximum permissible load for 2 pipes is q = 2x8M/l^2 = 2x8x69.4/2.2^2 = 229.4 kg/m (with hinged supports and without taking into account the torque that may arise when the load is transferred not along the center of gravity of the section). And this is with a static load, and the load will most likely be dynamic, or even shock (depending on the design of the sofa and the activity of the children, mine jump on the sofas so that it takes your breath away), so do the math for yourself. The article “Calculation values for rectangular profile pipes” will help you. 20-10-2014: student
Doc, please help. 21-10-2014: Doctor Lom
To begin with, a rigidly fixed beam and support sections are incompatible concepts, see the article “Types of supports, which design scheme to choose.” Judging by your description, you either have a single-span hinged beam with cantilevers (see Table 3), or a three-span rigidly clamped beam with 2 additional supports and unequal spans (in this case, the three-moment equations will help you). But in any case, the support reactions under a symmetrical load will be the same. 21-10-2014: student
I understand. Along the perimeter of the first floor there is an armored belt of 200x300h, the outer perimeter is 4400x4400. There are 3 channels anchored into it, with a step of 1 m. The span is without racks, one of them has the heaviest option, the load is asymmetrical. THOSE. count the beam as hinged? 21-10-2014: Doctor Lom
22-10-2014: student
in fact yes. As I understand it, the deflection of the channel will also rotate the armored belt itself at the attachment point, so you will get a hinged beam? 22-10-2014: Doctor Lom
Not quite so, first you determine the moment from the action of a concentrated load, then the moment from a uniformly distributed load along the entire length of the beam, then the moment arising from the action of a uniformly distributed load acting on a certain section of the beam. And only then add up the values of the moments. Each load will have its own calculation scheme. 07-02-2015: Sergey
Is there an error in the Mmax formula for case 2.3 in Table 3? Beam with a console, probably the plus instead of the minus should be in brackets 07-02-2015: Doctor Lom
No, not a mistake. The load on the cantilever reduces the moment in the span, but does not increase it. However, this can be seen from the moment diagram. 17-02-2015: Anton
Hello, first of all, thanks for the formulas, I saved them in my bookmarks. Please tell me, is there a beam above the span, four logs rest on the beam, distances: 180mm, 600mm, 600mm, 600mm, 325mm. I figured out the diagram and the bending moment, but I can’t understand how the deflection formula (Table 1, diagram 1.4) will change if the maximum moment is on the third lag. 17-02-2015: Doctor Lom
I have already answered similar questions several times in the comments to the article “Calculation schemes for statically indeterminate beams.” But you are lucky, for clarity, I performed the calculation using the data from your question. Look at the article “The general case of calculating a beam on hinged supports under the action of several concentrated loads”, perhaps over time I will add to it. 22-02-2015: Novel
Doc, I really can’t master all these formulas that are incomprehensible to me. Therefore, I ask you for help. I want to make a cantilever staircase in my house (the steps will be bricked up with reinforced concrete when building the wall). Wall - width 20cm, brick. The length of the protruding step is 1200*300mm. I want the steps to be of the correct shape (not a wedge). I intuitively understand that the reinforcement will be “something thicker” so that the steps will be something thinner? But can reinforced concrete up to 3cm thick cope with a load of 150kg at the edge? Please help me, I really don’t want to screw up. I would be very grateful if you could help me calculate... 22-02-2015: Doctor Lom
The fact that you cannot master fairly simple formulas is your problem. In the section “Basics of Strength of Strength” all this is discussed in sufficient detail. Here I will say that your project is absolutely unrealistic. Firstly, the wall is either 25 cm wide or cinder block (however, I could be wrong). Secondly, neither a brick nor a cinder block wall will provide sufficient pinching of steps with the specified wall width. In addition, such a wall should be calculated for the bending moment arising from the cantilever beams. Thirdly, 3 cm is an unacceptable thickness for a reinforced concrete structure, taking into account the fact that the minimum protective layer in beams must be at least 15 mm. And so on. 26-02-2015: Novel
02-04-2015: Vitaly
what does x mean in the second table, 2.4 02-04-2015: Vitaly
Good afternoon What scheme (algorithm) should be chosen to calculate a balcony slab, a cantilever clamped on one side, how to correctly calculate the moments on the support and in the span? Can it be calculated as a cantilever beam, according to the diagrams from Table 2, namely points 1, 1 and 2.1. Thank you! 02-04-2015: Doctor Lom
x in all tables means the distance from the origin to the point under study at which we are going to determine the bending moment or other parameters. Yes, your balcony slab, if it is solid and loads act on it, as in the indicated diagrams, can be calculated according to these diagrams. For cantilever beams, the maximum moment is always at the support, so there is no great need to determine the moment in the span. 03-04-2015: Vitaly
Thanks a lot! I also wanted to clarify. As I understand it, if you calculate according to 2 tables. diagram 1.1, (the load is applied to the end of the console) then I have x = L, and accordingly in the span M = 0. What if I also have this load at the ends of the slab? And according to scheme 2.1, I calculate the moment at the support, add it to the moment according to scheme 1.1 and according to the correct one, in order to reinforce it, I need to find the moment in the span. If I have a slab overhang of 1.45 m (in the clear), how can I calculate “x” to find the moment in the span? 03-04-2015: Doctor Lom
The moment in the span will vary from Ql at the support to 0 at the point of application of the load, which can be seen from the moment diagram. If your load is applied at two points at the ends of the slab, then in this case it is more advisable to provide beams that absorb loads at the edges. In this case, the slab can already be calculated as a beam on two supports - beams or a slab supported on 3 sides. 03-04-2015: Vitaly
Thank you! In moments I already understood. One more question. If the balcony slab is supported on both sides, using the letter “G”. What calculation scheme should I use then? 04-04-2015: Doctor Lom
In this case, you will have a plate pinched on 2 sides and there are no examples of calculating such a plate on my website. 27-04-2015: Sergey
Dear Doctor Lom! 27-04-2015: Doctor Lom
I won’t assess the reliability of such a design without calculations, but you can calculate it using the following criteria: 05-06-2015: student
Doc, where can I show you the picture? 05-06-2015: student
Did you still have a forum? 05-06-2015: Doctor Lom
There was, but I have absolutely no time to sort through spam in search of normal questions. So that's it for now. 06-06-2015: student
Doc, my link is https://yadi.sk/i/GardDCAEh7iuG 07-06-2015: Doctor Lom
The choice of design scheme will depend on what you want: simplicity and reliability or approximation to the actual operation of the structure through successive approximations. 07-06-2015: student
Doc, thanks. I need simplicity and reliability. This area is the busiest. I even thought about tying the tank post to the rafters to reduce the load on the floor, given that the water would be drained in the winter. I can’t get into such a jungle of calculations. In general, will the cantilever reduce deflection? 07-06-2015: student
Doc, one more question. the console is in the middle of the window span, does it make sense to move it to the edge? Sincerely 07-06-2015: Doctor Lom
In general, the console will reduce the deflection, but as I already said, how much in your case is a big question, and a shift to the center of the window opening will reduce the role of the console. And also, if this is your most loaded area, then maybe you can simply strengthen the beam, for example, with another similar channel? I don’t know your loads, but the load of 100 kg of water and half the weight of the tank does not seem so impressive to me, but from the point of view of deflection, do 8P channels take into account the dynamic load when walking? 08-06-2015: student
Doc, thanks for the good advice. After the weekend I will recalculate the beam as a two-span beam on hinges. If there is greater dynamics when walking, I constructively include the possibility of reducing the pitch of the floor beams. The house is a country house, so the dynamics are tolerable. The lateral displacement of the channels has a greater influence, but this can be treated by installing cross braces or fastening the flooring. The only thing is, will the concrete pouring crumble? I assume it will be supported on the upper and lower flanges of the channel plus welded reinforcement in the ribs and mesh on top. 08-06-2015: Doctor Lom
I already told you, you shouldn’t count on the console. 09-06-2015: student
Doc, I understand. 29-06-2015: Sergey
Good afternoon. I would like to ask you: the foundation was cast: piles of concrete 1.8 m deep, and then a strip 1 m deep was cast with concrete. The question is this: is the load transferred only to the piles or is it evenly distributed to both the piles and the tape? 29-06-2015: Doctor Lom
As a rule, piles are made in weak soils so that the load on the foundation is transmitted through the piles, so grillages on piles are calculated like beams on pile supports. However, if you poured the grillage over compacted soil, then part of the load will be transferred to the base through the grillage. In this case, the grillage is considered as a beam lying on an elastic foundation and represents a regular strip foundation. Like that. 29-06-2015: Sergey
Thank you. It’s just that the site turns out to be a mixture of clay and sand. Moreover, the clay layer is very hard: the layer can only be removed with a crowbar, etc., etc. 29-06-2015: Doctor Lom
I don’t know all your conditions (distance between piles, number of floors, etc.). From your description, it looks like you made a regular strip foundation and piles for reliability. Therefore, you just need to determine whether the width of the foundation will be sufficient to transfer the load from the house to the foundation. 05-07-2015: Yuri
Hello! We need your help with the calculations. A metal gate 1.5 x 1.5 m weighing 70 kg is mounted on a metal pipe, concreted to a depth of 1.2 m and lined with brick (post 38 x 38 cm). What cross-section and thickness should the pipe be so that there is no bending? 05-07-2015: Doctor Lom
You correctly assumed that your post should be treated like a cantilever beam. And even with the calculation scheme, you almost got it right. The fact is that 2 forces will act on your pipe (on the upper and lower canopy) and the value of these forces will depend on the distance between the canopies. More details in the article “Determination of pull-out force (why the dowel does not stay in the wall).” Thus, in your case, you should perform 2 deflection calculations according to design scheme 1.2, and then add the results obtained, taking into account the signs (in other words, subtract the other from one value). 05-07-2015: Yuri
Thanks for the answer. Those. I made the calculation to the maximum with a large margin, and the newly calculated deflection value will in any case be less? 06-07-2015: Doctor Lom
01-08-2015: Paul
Please tell me, in diagram 2.2 of table 3, how to determine the deflection at point C if the lengths of the cantilever sections are different? 01-08-2015: Doctor Lom
In this case, you need to go through the full cycle. Whether this is necessary or not, I don’t know. For an example, look at the article on calculating a beam under the action of several uniformly concentrated loads (link to the article before the tables). 04-08-2015: Yuri
To my question dated July 05, 2015. Is there any rule for the minimum amount of pinching in concrete for a given metal cantilever beam 120x120x4 mm with a collar of 70 kg - (for example, at least 1/3 of the length) 04-08-2015: Doctor Lom
In fact, calculating pinching is a separate big topic. The fact is that the resistance of concrete to compression is one thing, but the deformation of the soil on which the concrete of the foundation presses is quite another. In short, the longer the profile and the larger the area in contact with the ground, the better. 05-08-2015: Yuri
Thank you! In my case, will the metal gate post be cast in a concrete pile with a diameter of 300 mm and a length of 1 m, and the piles at the top will be connected by a concrete grillage to the reinforcement frame? concrete everywhere M 300. I.e. there will be no soil deformation. I would like to know an approximate, albeit with a large margin of safety, ratio. 05-08-2015: Doctor Lom
Then really 1/3 of the length should be enough to create a rigid pinch. For an example, look at the article “Types of supports, which design scheme to choose.” 05-08-2015: Yuri
20-09-2015: Carla
21-09-2015: Doctor Lom
You can first calculate the beam separately for each load according to the design schemes presented here, and then add the results obtained taking into account the signs. 08-10-2015: Natalia
Hello, Doctor))) 08-10-2015: Doctor Lom
As I understand it, you are talking about a beam from Table 3. For such a beam, the maximum deflection will not be in the middle of the span, but closer to support A. In general, the amount of deflection and the distance x (to the point of maximum deflection) depend on the length of the console, so in your In this case, you should use the equations of the initial parameters given at the beginning of the article. The maximum deflection in the span will be at the point where the angle of rotation of the inclined section is zero. If the console is long enough, then the deflection at the end of the console may be even greater than in the span. 22-10-2015: Alexander
22-10-2015: Ivan
Thank you very much for your clarifications. There is a lot of work to be done on my house. Gazebos, canopies, supports. I’ll try to remember that at one time I overslept as a diligent student and then accidentally passed it to the Soviet Higher Technical School. 31-05-2016: Vitaly
Thank you very much, you are great! 14-06-2016: Denis
I came across your site during this time. I almost missed my calculations, I always thought that a cantilever beam with a load at the end of the beam would bend more than with a uniformly distributed load, but formulas 1.1 and 2.1 in Table 2 show the opposite. Thanks for your work 14-06-2016: Doctor Lom
In general, it makes sense to compare a concentrated load with a uniformly distributed one only when one load is reduced to another. For example, when Q = ql, the formula for determining the deflection according to design scheme 1.1 will take the form f = ql^4/3EI, i.e. the deflection will be 8/3 = 2.67 times greater than with a simply uniformly distributed load. So the formulas for calculation schemes 1.1 and 2.1 do not show anything to the contrary, and initially you were right. 16-06-2016: engineer Garin
Good afternoon! I still can’t figure it out, I would be very grateful if you could help me figure it out once and for all - when calculating (any) an ordinary I-beam with a usual distributed load along its length, what moment of inertia should I use - Iy or Iz and why? I can’t find strength of strength in any textbook; everywhere they write that the cross section should tend to a square and the smallest moment of inertia should be taken. I just can’t grasp the physical meaning by the tail; can I somehow interpret this on my fingers? 16-06-2016: Doctor Lom
I advise you to start by looking at the articles “Fundamentals of Strength Materials” and “Towards the Calculation of Flexible Rods for the Action of a Compressive Eccentric Load”, everything is explained there in sufficient detail and clearly. Here I will add that it seems to me that you are confusing the calculations for transverse and longitudinal bending. Those. when the load is perpendicular to the neutral axis of the rod, then the deflection (transverse bending) is determined; when the load is parallel to the neutral axis of the beam, then the stability is determined, in other words, the effect of longitudinal bending on the load-bearing capacity of the rod. Of course, when calculating the transverse load (vertical load for a horizontal beam), the moment of inertia should be taken depending on the position of the beam, but in any case it will be Iz. And when calculating stability, provided that the load is applied along the center of gravity of the section, the smallest moment of inertia is considered, since the probability of loss of stability in this plane is much greater. 23-06-2016: Denis
Hello, the question is why in Table 1 for formulas 1.3 and 1.4 the deflection formulas are essentially the same and the size b. is it not reflected in formula 1.4 in any way? 23-06-2016: Doctor Lom
With an asymmetrical load, the deflection formula for design scheme 1.4 will be quite cumbersome, but it should be remembered that the deflection in any case will be less than when applying a symmetrical load (of course, provided b 03-11-2016: vladimir
in Table 1 for formulas 1.3 and 1.4, the deflection formula should be Ql^3/24EI instead of Qa^3/24EI. For a long time I could not understand why the deflection with the crystal did not converge 03-11-2016: Doctor Lom
That's right, another typo due to inattentive editing (I hope it's the last one, but not a fact). Corrected, thanks for your attention. 16-12-2016: Ivan
Hello, Doctor Lom. The question is the following: I was looking through photos from the construction site and noticed one thing: the factory-made reinforced concrete lintel is approximately 30*30 cm, supported on a three-layer reinforced concrete panel about 7 centimeters (the reinforced concrete panel was sawed down a little to rest the lintel on it). The opening for the balcony frame is 1.3 m, along the top of the lintel there is an armored belt and attic floor slabs. Are these 7 cm critical, the support of the other end of the jumper is more than 30 cm, everything has been fine for several years now 16-12-2016: Doctor Lom
If there is also an armored belt, then the load on the jumper can be significantly reduced. I think everything will be fine and even at 7 cm there is a fairly large margin of safety on the support platform. But in general, of course, you need to count. 25-12-2016: Ivan
Doctor, if we assume, well, purely theoretically 25-12-2016: Doctor Lom
I think even in this case nothing will happen. But I repeat, a more accurate answer requires calculation. 09-01-2017: Andrey
In Table 1, in formula 2.3, to calculate the deflection, instead of “q”, “Q” is indicated. Formula 2.1 for calculating the deflection, being a special case of formula 2.3, when inserting the corresponding values (a=c=l, b=0) takes on a different form. 09-01-2017: Doctor Lom
That's right, there was a typo, but now it doesn't matter. I took the deflection formula for such a design scheme from S.P. Fesik’s reference book, as the shortest for the special case x = a. But as you correctly noted, this formula does not pass the boundary conditions test, so I removed it altogether. I left only the formula for determining the initial angle of rotation in order to simplify the determination of deflection using the initial parameters method. 02-03-2017: Doctor Lom
As far as I know, such a special case is not considered in textbooks. Only software will help here, for example, Lyra. 24-03-2017: Eageniy
Good afternoon, in the deflection formula 1.4 in the first table - the value in brackets is always negative 24-03-2017: Doctor Lom
Everything is correct, in all the given formulas, the negative sign in the deflection formula means that the beam bends down along the y-axis. 29-03-2017: Oksana
Good afternoon, Doctor Lom. Could you write an article about the torque in a metal beam - when does it occur at all, under what design schemes, and, of course, I would like to see your calculations with examples. I have a hingedly supported metal beam, one edge is cantilevered and a concentrated load comes to it, and the load is distributed over the entire beam from the reinforced concrete. thin slab 100 mm and fence wall. This beam is the outermost one. With reinforced concrete The plate is connected by 6 mm rods welded to the beam with a pitch of 600 mm. I can’t understand whether there will be a torque there, if so, how to find it and calculate the cross-section of the beam in connection with it? Doctor Lom Victor, emotional stroking is, of course, good, but you can’t spread it on bread and you can’t feed your family with it. Answering your question requires calculations, calculations are time, and time is not emotional stroking. Calculate bending beam There are several options: Further, when dividing the maximum bending moment by the moment of resistance in a given section, we get maximum stress in beam and we must compare this stress with the stress that our beam of a given material can generally withstand. P = m * g = 90 * 10 = 900 N = 0.9 kN M = P * l = 0.9 kN * 2 m = 1.8 kN*m b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa 45.34 MPa is correct, which means this I-beam will withstand a mass of 90 kg.Articles with similar topics
Reviews
Chapter 1. BENDING OF RIGHT LINEAR BEAMS AND BEAM SYSTEMS
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directions of bending moment and shear force ( 2. Normal stresses in areas parallel to the beam axis are neglected due to their smallness (Fig. 1.2,)
. Normal forces in the final section will have increments. Let's consider only linear increments, then .
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From the point of view of strength of material, the sign of the moment is a rather conventional concept, since in the cross section for which the bending moment is determined, both compressive and tensile stresses still act. The main thing to understand is that if the diagram is located on top, then tensile stresses will act in the upper part of the section and vice versa.
In the table, the minus for moments on a rigid support is not indicated, but the direction of action of the moment was taken into account when drawing up the formulas.
I want to know whether this subcode is only for long beams, which are used in the construction of buildings, or can also be used to calculate the deflections of shafts up to 2 m long. Please answer like this l/D>...
Do you mean that for rotating shafts the patterns will be different due to the torque? I don’t know how important this is, since the technical book says that in the case of turning, the deflection introduced by the torque on the shaft is very small compared to the deflection from the radial component of the cutting force. What do you think?
Calculation of building structures, machine parts, etc., as a rule, consists of two stages: 1. calculation based on limit states of the first group - the so-called strength calculation, 2. calculation based on limit states of the second group. One of the types of calculation for limit states of the second group is calculation for deflection.
In your case, in my opinion, strength calculations will be more important. Moreover, today there are 4 theories of strength and the calculations for each of these theories are different, but in all theories the influence of both bending and torque is taken into account when calculating.
Deflection under the action of torque occurs in a different plane, but is still taken into account in the calculations. Whether this deflection is small or large - the calculation will show.
I do not specialize in calculations of machine parts and mechanisms and therefore cannot indicate authoritative literature on this issue. However, in any reference book for an engineer-designer of machine components and parts, this topic should be properly covered.
mail: [email protected]
Skype: dmytrocx75
Q - in kilograms.
l - in centimeters.
E - in kgf/cm2.
I - cm4.
Is everything right? Some strange results are obtained.
If we are talking about a grillage, then depending on the method of its construction, it can be designed as a beam on two supports, or as a beam on an elastic foundation.
In general, when calculating columnar foundations, one should be guided by the requirements of SNiP 2.03.01-84.
In addition, if the load on the foundation is transferred from an eccentrically loaded column or not only from the column, then an additional moment will act on the cushion. This should be taken into account when making calculations.
But I repeat once again, do not self-medicate, follow the requirements of the specified SNiP.
However, in the cases you indicated (except 1.3), the maximum deflection may not be in the middle of the beam, therefore determining the distance from the beginning of the beam to the section where the maximum deflection will be is a separate task. Recently, a similar question was discussed in the topic “Calculation schemes for statically indeterminate beams”, look there.
The essence of the problem is this: at the base of the sofa there is planned a metal frame made of profiled pipe 40x40 or 40x60, lying on two supports with a distance of 2200 mm. QUESTION: is the profile cross-section sufficient for loads from the sofa’s own weight + let’s take 3 people weighing 100 kg???
Rigidly fixed beam, span 4 m, supported by 0.2 m. Loads: distributed 100 kg/m along the beam, plus distributed 100 kg/m in the area of 0-2 m, plus concentrated 300 kg in the middle (at 2 m). Determined the support reactions: A – 0.5 t; B - 0.4 t. Then I got stuck: to determine the bending moment under a concentrated load, it is necessary to calculate the sum of the moments of all forces to the right and left of it. Plus, a moment appears on the supports.
How are loads calculated in this case? It is necessary to bring all distributed loads to concentrated ones and sum them up (subtract from the support reaction * distance) according to the formulas of the design scheme? In your article about farms, the layout of all forces is clear, but here I cannot go into the methodology for determining the acting forces.
The maximum moment is in the middle, it turns out M = Q + 2q + from an asymmetric load to a maximum of 1.125q. Those. I added up all 3 loads, is that correct?
If you are not ready to handle all this, then it is better to contact a professional designer - it will be cheaper.
Please tell me what scheme should be used to calculate the deflection of the beam of such a mechanism https://yadi.sk/i/MBmS5g9kgGBbF. Or maybe, without going into calculations, tell me whether a 10 or 12 I-beam is suitable for the boom, maximum load 150-200 kg, lifting height 4-5 meters. Rack - pipe d=150, rotating mechanism or axle shaft, or Gazelle front hub. The mowing can be made rigid from the same I-beam, and not with a cable. Thank you.
1. The boom can be considered as a two-span continuous beam with a cantilever. The supports for this beam will be not only the stand (this is the middle support), but also the cable attachment points (the outer supports). This is a statically indeterminate beam, but to simplify the calculations (which will lead to a slight increase in the safety factor), the boom can be considered as simply a single-span beam with a cantilever. The first support is the cable attachment point, the second is the stand. Then your calculation schemes are 1.1 (for load - live load) and 2.3 (boom dead weight - permanent load) in Table 3. And if the load is in the middle of the span, then 1.1 in Table 1.
2. At the same time, we must not forget that your live load will not be static, but at least dynamic (see the article “Calculation for shock loads”).
3. To determine the forces in the cable, you need to divide the support reaction at the place where the cable is attached by the sine of the angle between the cable and the beam.
4. Your rack can be considered as a metal column with one support - rigid pinching at the bottom (see the article "Calculation of metal columns"). The load will be applied to this column with a very large eccentricity if there is no counterload.
5. Calculation of the junction points of the boom and rack and other subtleties of the calculation of machine components and mechanisms are not yet considered on this site.
what design scheme is ultimately obtained for the floor beam and cantilever beam, and will the cantilever beam (brown color) affect the reduction in the deflection of the floor beam (pink)?
wall - foam block D500, height 250, width 150, armored belt beam (blue): 150x300, reinforcement 2x?12, top and bottom, additionally bottom in the window span and top in places where the beam rests on the window opening - mesh?5, cell 50. B in the corners there are concrete columns 200x200, the span of the reinforced belt beam is 4000 without walls.
ceiling: channel 8P (pink), for calculations I took 8U, welded and anchored with the reinforcement of the reinforced belt beam, concreted, from the bottom of the beam to the channel 190 mm, from the top 30, span 4050.
to the left of the console there is an opening for the stairs, the channel is supported on a pipe? 50 (green), the span to the beam is 800.
to the right of the console (yellow) - bathroom (shower, toilet) 2000x1000, floor - poured reinforced ribbed transverse slab, dimensions 2000x1000 height 40 - 100 on permanent formwork (corrugated sheet, wave 60) + tiles with adhesive, walls - plasterboard on profiles. The rest of the floor is board 25, plywood, linoleum.
At the points of the arrows, the supports of the water tank, 200 l, are supported.
Walls of the 2nd floor: sheathing with 25 boards on both sides, with insulation, height 2000, supported by an armored belt.
roof: rafters - a triangular arch with a tie, along the floor beam, in increments of 1000, supported on the walls.
console: channel 8P, span 995, welded with reinforced reinforcement, concreted into a beam, welded to the ceiling channel. span on the right and left along the floor beam - 2005.
While I’m welding the reinforcement frame, it’s possible to move the console left and right, but there doesn’t seem to be any reason to move it to the left?
In the first case, the floor beam can be considered as a hinged two-span beam with an intermediate support - a pipe, and the channel, which you call a cantilever beam, cannot be taken into account at all. That's the whole calculation.
Next, in order to simply move on to a beam with rigid pinching on the outer supports, you must first calculate the reinforced belt for the action of torque and determine the angle of rotation of the cross-section of the reinforced belt, taking into account the load from the walls of the 2nd floor and the deformation of the wall material under the influence of torque. And thus calculate a two-span beam taking into account these deformations.
In addition, in this case, one should take into account the possible subsidence of the support - the pipe, since it rests not on the foundation, but on a reinforced concrete slab (as I understand from the figure) and this slab will be deformed. And the pipe itself will experience compression deformation.
In the second case, if you want to take into account the possible work of the brown channel, you should consider it as an additional support for the floor beam and thus first calculate the 3-span beam (the support reaction on the additional support will be the load on the cantilever beam), then determine the amount of deflection at the end cantilever beam, recalculate the main beam taking into account the subsidence of the support and, among other things, also take into account the angle of rotation and deflection of the reinforced belt at the point where the brown channel is attached. And that's not all.
To calculate the console and installation, it is better to take half the span from the rack to the beam (4050-800-50=3200/2=1600-40/2=1580) or from the edge of the window (1275-40=1235. And the load on the beam is the same as the window the overlap will have to be recalculated, but you have such examples. The only thing is to take the load as applied to the beam from above? Will there be a redistribution of the load applied almost along the axis of the tank?
You assume that the floor slabs are supported on the bottom flange of the channel, but what about the other side? In your case, an I-beam would be a more acceptable option (or 2 channels each as a floor beam).
There are no problems on the other side - the corner is on the embeds in the body of the beam. I haven’t yet coped with the calculation of a two-span beam with different spans and different loads, I’ll try to re-study your article on calculating a multi-span beam using the method of moments.
I calculated from the table. 2, clause 1.1. (#comments) as the deflection of a cantilever beam with a load of 70 kg, shoulder 1.8 m, square pipe 120x120x4 mm, moment of inertia 417 cm4. I got a deflection of 1.6 mm? True or false?
P.S. I don’t check the accuracy of the calculations, so just rely on yourself.
You can immediately draw up equations of static equilibrium of the system and solve these equations.
I have a beam according to scheme 2.3. Your table gives a formula for calculating the deflection in the middle of the span l/2, but what formula can be used to calculate the deflection at the end of the console? Will the deflection in the middle of the span be maximum? The result obtained using this formula must be compared with the maximum permissible deflection according to SNiP “Loads and Impacts” using the value l - the distance between points A and B? Thanks in advance, I'm completely confused. And yet, I can’t find the original source from which these tables were taken - is it possible to indicate the name?
When you compare the obtained result of deflection in a span with SNiPovk, then the length of the span is the distance l between A and B. For the cantilever, instead of l, the distance 2a (double cantilever overhang) is taken.
I compiled these tables myself, using various reference books on the theory of strength of materials, while checking the data for possible typos, as well as general methods for calculating beams, when the necessary diagrams in my opinion were not in the reference books, so there are many primary sources.
that the reinforcement in the reinforced belt above the beam is completely destroyed, the reinforced belt will crack and fall on the beam along with the floor slabs? Is this 7 cm support area enough?
1. Calculation of the maximum load that it will withstand
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle for selecting a beam section
on two supports loaded with a uniformly distributed load or concentrated force.
To begin with, you will need to find the point (section) at which there will be a maximum moment. This depends on whether the beam is supported or embedded. Below are diagrams of bending moments for the most common schemes. 
After finding the bending moment, we must find the moment of resistance Wx of this section using the formula given in the table: 
For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, A for fragile(cast iron) – tensile strength. We can find the yield strength and tensile strength from the tables below. 
Let's look at a couple of examples:
1. [i] You want to check whether an I-beam No. 10 (steel St3sp5) 2 meters long, rigidly embedded in the wall, will support you if you hang on it. Let your mass be 90 kg.
First, we need to select a design scheme. 
This diagram shows that the maximum moment will be at the seal, and since our I-beam has equal section along the entire length, then the maximum voltage will be in the termination. Let's find it:
Using the I-beam assortment table, we find the moment of resistance of I-beam No. 10. 
It will be equal to 39.7 cm3. Let's convert it to cubic meters and get 0.0000397 m3.
Next, using the formula, we find the maximum stresses that arise in the beam.
After we have found the maximum stress that occurs in the beam, we can compare it with the maximum permissible stress equal to the yield strength of steel St3sp5 - 245 MPa.
2. [i] Since we have quite a large supply, we will solve the second problem, in which we will find the maximum possible mass that the same I-beam No. 10, 2 meters long, will support.
If we want to find the maximum mass, then we must equate the values of the yield strength and the stress that will arise in the beam (b = 245 MPa = 245,000 kN*m2).
