Calculate bending beam There are several options:
1. Calculation maximum load which she will endure
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle selection of 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:
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.
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:
P = m * g = 90 * 10 = 900 N = 0.9 kN
M = P * l = 0.9 kN * 2 m = 1.8 kN * m
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 to Cubic Meters and we get 0.0000397 m3.
Next, using the formula, we find the maximum stresses that arise in the beam.
b = M / W = 1.8 kN/m / 0.0000397 m3 = 45340 kN/m2 = 45.34 MPa
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.
45.34 MPa is correct, which means this I-beam will withstand a mass of 90 kg.
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).
General concepts.
Bending deformationconsists in curvature of the axis of a straight rod or in a change in the initial curvature of a straight rod(Fig. 6.1) . Let's get acquainted with the basic concepts that are used when considering bending deformation.
Rods that bend are called beams.
Clean called bending, in which the bending moment is the only internal force factor arising in cross section beams.
More often, in the cross section of the rod, along with the bending moment, a transverse force also arises. This bending is called transverse.
Flat (straight) called bending when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section.
With oblique bending the plane of action of the bending moment intersects the cross section of the beam along a line that does not coincide with any of the main central axes of the cross section.
We begin our study of bending deformation with the case of pure plane bending.
Normal stresses and strains during pure bending.
As already mentioned, with pure plane bending in the cross section, of the six internal force factors, only the bending moment is nonzero (Fig. 6.1, c):
; (6.1)
Experiments carried out on elastic models show that if a grid of lines is applied to the surface of the model(Fig. 6.1, a) , then with pure bending it is deformed as follows(Fig. 6.1, b):
a) longitudinal lines are curved along the circumference;
b) the contours of the cross sections remain flat;
c) the contour lines of the sections intersect everywhere with the longitudinal fibers at right angles.
Based on this, it can be assumed that in pure bending, the cross sections of the beam remain flat and rotate so that they remain normal to the curved axis of the beam (flat sections in bending hypothesis).
Rice. .
By measuring the length of the longitudinal lines (Fig. 6.1, b), you can find that the upper fibers, when the beam bends, are deformed, and the lower ones are shortened. Obviously, it is possible to find fibers whose length remains unchanged. A set of fibers that do not change their length when a beam is bent is calledneutral layer (n.s.). The neutral layer intersects the cross section of the beam in a straight line, which is calledneutral line (n.l.) section.
To derive a formula that determines the magnitude of normal stresses arising in the cross section, consider a section of the beam in a deformed and undeformed state (Fig. 6.2).
Rice. .
Using two infinitesimal cross sections, we select an element of length. Before deformation, the sections bounding the element were parallel to each other (Fig. 6.2, a), and after deformation they tilted slightly, forming an angle. The length of the fibers lying in the neutral layer does not change when bending. Let us denote the radius of curvature of the trace of the neutral layer on the drawing plane by a letter. Let us determine the linear deformation of an arbitrary fiber located at a distance from the neutral layer.
The length of this fiber after deformation (arc length) is equal. Considering that before deformation all fibers had the same length, we obtain that the absolute elongation of the fiber in question
Its relative deformation
Obviously, since the length of the fiber lying in the neutral layer has not changed. Then after substitution we get
(6.2)
Therefore, the relative longitudinal strain is proportional to the distance of the fiber from the neutral axis.
Let us introduce the assumption that when bending, the longitudinal fibers do not press on each other. Under this assumption, each fiber is deformed in isolation, experiencing simple tension or compression, in which. Taking into account (6.2)
, (6.3)
that is, normal stresses are directly proportional to the distances of the cross-section points under consideration from the neutral axis.
Let us substitute dependence (6.3) into the expression for the bending moment in the cross section (6.1)
Recall that the integral represents the moment of inertia of the section relative to the axis
Or
(6.4)
Dependence (6.4) represents Hooke’s law for bending, since it connects the deformation (curvature of the neutral layer) with the moment acting in the section. The product is called the bending stiffness of the section, N m 2.
Let's substitute (6.4) into (6.3)
(6.5)
This is the required formula for determining normal stresses during pure bending of a beam at any point in its cross-section.
For In order to establish where the neutral line is located in the cross section, we substitute the value of normal stresses into the expression for the longitudinal force and bending moment
Because the,
That
(6.6)
(6.7)
Equality (6.6) indicates that the axis , the neutral axis of the section , passes through the center of gravity of the cross section.
Equality (6.7) shows that and are the main central axes of the section.
According to (6.5), the highest voltage is achieved in the fibers furthest from the neutral line
The ratio represents the axial moment of resistance of the section relative to its central axis, which means
The meaning for the simplest cross sections is:
For rectangular cross section
, (6.8)
where is the side of the section perpendicular to the axis;
The side of the section is parallel to the axis;
For round cross section
, (6.9)
where is the diameter of the circular cross section.
The strength condition for normal bending stresses can be written in the form
(6.10)
All formulas obtained were obtained for the case of pure bending of a straight rod. The action of the transverse force leads to the fact that the hypotheses underlying the conclusions lose their strength. However, calculation practice shows that even with transverse bending beams and frames, when in the section, in addition to the bending moment, there is also a longitudinal force and a transverse force, you can use the formulas given for pure bending. The error is insignificant.
Determination of shear forces and bending moments.
As already mentioned, with plane transverse bending in the cross section of the beam, two internal force factors arise and.
Before determining, the reactions of the beam supports are determined (Fig. 6.3, a), composing static equilibrium equations.
To determine and we apply the section method. In the place we are interested in, we will make a mental cut of the beam, for example, at a distance from the left support. Let's discard one of the parts of the beam, for example the right one, and consider the equilibrium of the left part (Fig. 6.3, b). Let us replace the interaction of the beam parts with internal forces and.
Let us establish the following sign rules for and:
- The transverse force in a section is positive if its vectors tend to rotate the section under consideration clockwise;
- The bending moment in a section is positive if it causes compression of the upper fibers.
Rice. .
To determine these forces, we use two equilibrium equations:
1. ; ; .
2. ;
Thus,
a) the transverse force in the cross section of the beam is numerically equal to the algebraic sum of the projections onto the transverse axis of the section of all external forces acting on one side of the section;
b) the bending moment in the cross section of the beam is numerically equal to the algebraic sum of the moments (calculated relative to the center of gravity of the section) of external forces acting on one side of the given section.
In practical calculations, they are usually guided by the following:
- If an external load tends to rotate the beam clockwise relative to the section under consideration (Fig. 6.4, b), then in the expression for it gives a positive term.
- If an external load creates a moment relative to the section under consideration, causing compression of the upper fibers of the beam (Fig. 6.4, a), then in the expression for in this section it gives a positive term.
Rice. .
Construction of diagrams in beams.
Consider a two-support beam(Fig. 6.5, a) . The beam is acted upon at a point by a concentrated moment, at a point by a concentrated force, and at a section by a uniformly distributed load of intensity.
Let us determine the support reactions and(Fig. 6.5, b) . The resultant of the distributed load is equal, and its line of action passes through the center of the section. Let's create moment equations about the points and.
Let us determine the shear force and bending moment in an arbitrary section located in a section at a distance from point A(Fig. 6.5, c) .
(Fig. 6.5, d). The distance can vary within ().
The value of the transverse force does not depend on the coordinates of the section; therefore, in all sections of the section, the transverse forces are the same and the diagram looks like a rectangle. Bending moment
The bending moment varies linearly. Let us determine the ordinates of the diagram for the boundaries of the site.
Let us determine the shear force and bending moment in an arbitrary section located in a section at a distance from the point(Fig. 6.5, d). The distance can vary within ().
The transverse force varies linearly. Let's define for the boundaries of the site.
Bending moment
The diagram of bending moments in this section will be parabolic.
To determine the extreme value of the bending moment, we equate to zero the derivative of the bending moment along the abscissa of the section:
From here
For a section with a coordinate, the value of the bending moment will be
As a result, we obtain diagrams of transverse forces(Fig. 6.5, f) and bending moments (Fig. 6.5, g).
Differential dependencies during bending.
(6.11)
(6.12)
(6.13)
These dependencies make it possible to establish some features of the diagrams of bending moments and shear forces:
N and in areas where there is no distributed load, the diagrams are limited to straight lines parallel to the zero line of the diagram, and diagrams in the general case are inclined straight lines.
N and in areas where a uniformly distributed load is applied to the beam, the diagram is limited by inclined straight lines, and the diagram is limited by quadratic parabolas with a convexity facing the direction opposite to the direction of the load.
IN sections, where the tangent to the diagram is parallel to the zero line of the diagram.
N and in areas where the moment increases; in areas where the moment decreases.
IN sections where concentrated forces are applied to the beam, the diagram will show jumps by the magnitude of the applied forces, and the diagram will show fractures.
In sections where concentrated moments are applied to the beam, the diagram will show jumps in the magnitude of these moments.
The ordinates of the diagram are proportional to the tangent of the angle of inclination of the tangent to the diagram.
To visually represent the nature of the deformation of beams (rods) during bending, the following experiment is carried out. A grid of lines parallel and perpendicular to the axis of the beam is applied to the side faces of a rubber beam of rectangular cross-section (Fig. 30.7, a). Then moments are applied to the beam at its ends (Fig. 30.7, b), acting in the plane of symmetry of the beam, intersecting each of its cross sections along one of the main central axes of inertia. The plane passing through the axis of the beam and one of the main central axes of inertia of each of its cross sections will be called the main plane.
Under the influence of moments, the beam experiences a straight pure bend. As a result of deformation, as experience shows, the grid lines parallel to the axis of the beam are bent, maintaining the same distances between them. When indicated in Fig. 30.7, b in the direction of the moments, these lines in the upper part of the beam are lengthened, and in the lower part they are shortened.
Each grid line perpendicular to the axis of the beam can be considered as a trace of the plane of some cross section of the beam. Since these lines remain straight, it can be assumed that the cross sections of the beam, flat before deformation, remain flat during deformation.
This assumption, based on experience, is known as the hypothesis of plane sections, or Bernoulli's hypothesis (see § 6.1).
The hypothesis of plane sections applies not only to pure bending, but also to transverse bending. For transverse bending it is approximate, and for pure bending it is strict, which is confirmed by theoretical studies carried out using methods of elasticity theory.
Let us now consider a straight beam with a cross section symmetrical about the vertical axis, embedded at the right end and loaded at the left end with an external moment acting in one of the main planes of the beam (Fig. 31.7). In each cross section of this beam, only bending moments occur acting in the same plane as the moment
Thus, the beam is in a state of straight, pure bend throughout its entire length. Individual sections of the beam may be in a state of pure bending even if it is subject to transverse loads; for example, section 11 of the beam shown in Fig. experiences pure bending. 32.7; in the sections of this section the shear force
From the beam under consideration (see Fig. 31.7) we select an element of length . As a result of deformation, as follows from Bernoulli's hypothesis, the sections will remain flat, but will tilt relative to each other by a certain angle. Let us take the left section conditionally as stationary. Then, as a result of rotating the right section by an angle, it will take the position (Fig. 33.7).
The straight lines will intersect at a certain point A, which is the center of curvature (or, more precisely, the trace of the axis of curvature) of the longitudinal fibers of the element. The upper fibers of the element in question when shown in Fig. 31.7 in the direction of the moment are lengthened, and the lower ones are shortened. The fibers of some intermediate layer perpendicular to the plane of action of the moment retain their length. This layer is called the neutral layer.
Let us denote the radius of curvature of the neutral layer, i.e., the distance from this layer to the center of curvature A (see Fig. 33.7). Let's consider a certain layer located at a distance y from the neutral layer. The absolute elongation of the fibers of this layer is equal to and the relative elongation
Considering similar triangles we establish that Therefore,
In the theory of bending, it is assumed that the longitudinal fibers of the beam do not press on each other. Experimental and theoretical studies show that this assumption does not significantly affect the calculation results.
With pure bending, shear stresses do not arise in the cross sections of the beam. Thus, all fibers in pure bending are under conditions of uniaxial tension or compression.
According to Hooke's law, for the case of uniaxial tension or compression, the normal stress o and the corresponding relative deformation are related by the dependence
or based on formula (11.7)
From formula (12.7) it follows that the normal stresses in the longitudinal fibers of the beam are directly proportional to their distances y from the neutral layer. Consequently, in the cross section of the beam at each point, the normal stresses are proportional to the distance y from this point to the neutral axis, which is the line of intersection of the neutral layer with the cross section (Fig.
34.7, a). From the symmetry of the beam and the load it follows that the neutral axis is horizontal.
At the points of the neutral axis, normal stresses are zero; on one side of the neutral axis they are tensile, and on the other they are compressive.
The stress diagram o is a graph bounded by a straight line, with the largest absolute values of stress for the points furthest from the neutral axis (Fig. 34.7b).
Let us now consider the equilibrium conditions of the selected beam element. Let us represent the action of the left part of the beam on the section of the element (see Fig. 31.7) in the form of a bending moment; the remaining internal forces in this section with pure bending are equal to zero. Let us imagine the action of the right side of the beam on the cross-section of the element in the form of elementary forces applied to each elementary area of the cross-section (Fig. 35.7) and parallel to the axis of the beam.
Let's create six equilibrium conditions for an element
Here are the sums of the projections of all forces acting on the element, respectively, on the axes - the sums of the moments of all forces relative to the axes (Fig. 35.7).
The axis coincides with the neutral axis of the section and the y-axis is perpendicular to it; both of these axes are located in the cross-sectional plane
An elementary force does not produce projections on the y-axis and does not cause a moment about the axis. Therefore, the equilibrium equations are satisfied for any values of o.
The equilibrium equation has the form
Let us substitute the value of a into equation (13.7) according to formula (12.7):
Since (a curved beam element is considered, for which), then
The integral represents the static moment of the cross section of the beam about the neutral axis. Its equality to zero means that the neutral axis (i.e., the axis) passes through the center of gravity of the cross section. Thus, the center of gravity of all cross sections of the beam, and therefore the axis of the beam, which is the geometric location of the centers of gravity, are located in the neutral layer. Therefore, the radius of curvature of the neutral layer is the radius of curvature of the curved axis of the beam.
Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element relative to the neutral axis:
Here represents the moment of the elementary internal force relative to the axis.
Let us denote the area of the cross-section of the beam located above the neutral axis - below the neutral axis.
Then it will represent the resultant of elementary forces applied above the neutral axis, below the neutral axis (Fig. 36.7).
Both of these resultants are equal to each other in absolute value, since their algebraic sum, based on condition (13.7), is equal to zero. These resultants form an internal pair of forces acting in the cross section of the beam. The moment of this pair of forces, equal to the product of the magnitude of one of them and the distance between them (Fig. 36.7), is a bending moment in the cross section of the beam.
Let us substitute the value of a into equation (15.7) according to formula (12.7):
Here represents the axial moment of inertia, i.e., the axis passing through the center of gravity of the section. Hence,
Let's substitute the value from formula (16.7) into formula (12.7):
When deriving formula (17.7), it was not taken into account that with an external torque directed, as shown in Fig. 31.7, according to the accepted sign rule, the bending moment is negative. If we take this into account, then we must put a minus sign in front of the right side of formula (17.7). Then, with a positive bending moment in the upper zone of the beam (i.e., at ), the values of a will turn out to be negative, which will indicate the presence of compressive stresses in this zone. However, usually the minus sign is not placed on the right side of formula (17.7), and this formula is used only to determine the absolute values of stresses a. Therefore, the absolute values of the bending moment and the ordinate y should be substituted into formula (17.7). The sign of the stresses is always easily determined by the sign of the moment or by the nature of the deformation of the beam.
Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element relative to the y-axis:
Here it represents the moment of the elementary internal force about the y-axis (see Fig. 35.7).
Let us substitute the value of a into expression (18.7) according to formula (12.7):
Here the integral represents the centrifugal moment of inertia of the cross section of the beam relative to the y and axis. Hence,
But since
As is known (see § 7.5), the centrifugal moment of inertia of the section is equal to zero relative to the main axes of inertia.
In the case under consideration, the y-axis is the axis of symmetry of the cross-section of the beam and, therefore, the y-axes and are the main central axes of inertia of this section. Therefore, condition (19.7) is satisfied here.
In the case when the cross section of the bent beam does not have any axis of symmetry, condition (19.7) is satisfied if the plane of action of the bending moment passes through one of the main central axes of inertia of the section or is parallel to this axis.
If the plane of action of the bending moment does not pass through any of the main central axes of inertia of the cross section of the beam and is not parallel to it, then condition (19.7) is not satisfied and, therefore, there is no direct bending - the beam experiences oblique bending.
Formula (17.7), which determines the normal stress at an arbitrary point of the beam section under consideration, is applicable provided that the plane of action of the bending moment passes through one of the main axes of inertia of this section or is parallel to it. In this case, the neutral axis of the cross section is its main central axis of inertia, perpendicular to the plane of action of the bending moment.
Formula (16.7) shows that during direct pure bending, the curvature of the curved axis of the beam is directly proportional to the product of the elastic modulus E and the moment of inertia. We will call the product the stiffness of the section during bending; it is expressed in, etc.
For pure beam bending constant cross section bending moments and section stiffnesses are constant along its length. In this case, the radius of curvature of the curved axis of the beam has a constant value [see. expression (16.7)], that is, the beam bends along a circular arc.
From formula (17.7) it follows that the largest (positive - tensile) and smallest (negative - compressive) normal stresses in the cross section of the beam arise at the points furthest from the neutral axis, located on both sides of it. For a cross section symmetrical about the neutral axis, the absolute values of the greatest tensile and compressive stresses are the same and can be determined by the formula
where is the distance from the neutral axis to the most distant point of the section.
A value that depends only on the size and shape of the cross section is called the axial moment of resistance of the section and is denoted
(20.7)
Hence,
Let us determine the axial moments of resistance for rectangular and circular sections.
For a rectangular section with width b and height
For a circular section with diameter d
The moment of resistance is expressed in .
For sections that are not symmetrical about the neutral axis, for example, for a triangle, a tee, etc., the distances from the neutral axis to the most distant stretched and compressed fibers are different; Therefore, for such sections there are two moments of resistance:
where are the distances from the neutral axis to the most distant stretched and compressed fibers.
Bending is a type of deformation in which the longitudinal axis of the beam is bent. Straight beams that bend are called beams. Direct bending is a bend in which the external forces acting on the beam lie in one plane (force plane) passing through the longitudinal axis of the beam and the main central axis of inertia of the cross section.
The bend is called pure, if only one bending moment occurs in any cross section of the beam.
Bending, in which a bending moment and a transverse force simultaneously act in the cross section of a beam, is called transverse. The line of intersection of the force plane and the cross-sectional plane is called the force line.
Internal force factors during beam bending.
During plane transverse bending, two internal force factors arise in the beam sections: transverse force Q and bending moment M. To determine them, the method of sections is used (see lecture 1). The transverse force Q in the beam section is equal to the algebraic sum of the projections onto the section plane of all external forces acting on one side of the section under consideration.
Sign rule for shear forces Q:
The bending moment M in a beam section is equal to the algebraic sum of the moments relative to the center of gravity of this section of all external forces acting on one side of the section under consideration.
Sign rule for bending moments M:
Zhuravsky's differential dependencies.
Differential relationships have been established between the intensity q of the distributed load, the expressions for the transverse force Q and the bending moment M:
Based on these dependencies, the following general patterns of diagrams of transverse forces Q and bending moments M can be identified:
Features of diagrams of internal force factors during bending.
1. In the section of the beam where there is no distributed load, the Q diagram is presented straight line , parallel to the base of the diagram, and diagram M - an inclined straight line (Fig. a).
2. In the section where a concentrated force is applied, Q should be on the diagram leap , equal to the value of this force, and on the diagram M - breaking point (Fig. a).
3. In the section where a concentrated moment is applied, the value of Q does not change, and the diagram M has leap , equal to the value of this moment (Fig. 26, b).
4. In a section of a beam with a distributed load of intensity q, the diagram Q changes according to a linear law, and the diagram M changes according to a parabolic law, and the convexity of the parabola is directed towards the direction of the distributed load (Fig. c, d).
5. If, within a characteristic section, the diagram Q intersects the base of the diagram, then in the section where Q = 0, the bending moment has an extreme value M max or M min (Fig. d).
Normal bending stresses.
Determined by the formula:
The moment of resistance of a section to bending is the quantity:
Dangerous cross section during bending, the cross section of the beam in which the maximum normal stress occurs is called.
Shear stresses during straight bending.
Determined by Zhuravsky's formula for shear stresses during straight beam bending:
where S ots - static moment transverse area cut off layer of longitudinal fibers relative to the neutral line.
Calculations of bending strength.
1. At verification calculation The maximum design stress is determined and compared with the permissible stress:
2. At design calculation the selection of the beam section is made from the condition:
3. When determining the permissible load, the permissible bending moment is determined from the condition:
Bending movements.
Under the influence of bending load, the axis of the beam bends. In this case, tension of the fibers is observed on the convex part and compression on the concave part of the beam. In addition, there is a vertical movement of the centers of gravity of the cross sections and their rotation relative to the neutral axis. To characterize bending deformation, the following concepts are used:
Beam deflection Y- movement of the center of gravity of the cross section of the beam in the direction perpendicular to its axis.
Deflection is considered positive if the center of gravity moves upward. The amount of deflection varies along the length of the beam, i.e. y = y(z)
Section rotation angle- angle θ through which each section rotates relative to its original position. The rotation angle is considered positive when the section is rotated counterclockwise. The magnitude of the rotation angle varies along the length of the beam, being a function of θ = θ (z).
The most common methods for determining displacements is the method Mora And Vereshchagin's rule.
Mohr's method.
The procedure for determining displacements using Mohr's method:
1. An “auxiliary system” is built and loaded with a unit load at the point where the displacement is required to be determined. If linear displacement is determined, then a unit force is applied in its direction; when angular displacements are determined, a unit moment is applied.
2. For each section of the system, expressions for bending moments M f from the applied load and M 1 from the unit load are written.
3. Over all sections of the system, Mohr’s integrals are calculated and summed, resulting in the desired displacement:
4. If the calculated displacement has a positive sign, this means that its direction coincides with the direction of the unit force. A negative sign indicates that the actual displacement is opposite to the direction of the unit force.
Vereshchagin's rule.
For the case when the diagram of bending moments from a given load has an arbitrary outline, and from a unit load – a rectilinear outline, it is convenient to use the graphic-analytical method, or Vereshchagin’s rule.
where A f is the area of the diagram of the bending moment M f from a given load; y c – ordinate of the diagram from a unit load under the center of gravity of the diagram M f; EI x – stiffness of the section of the beam. Calculations using this formula are made in sections, in each of which the straight-line diagram should be without fractures. The value (A f *y c) is considered positive if both diagrams are located on the same side of the beam, negative if they are located on different sides. A positive result of multiplying diagrams means that the direction of movement coincides with the direction of a unit force (or moment). A complex diagram M f should be divided into simple figures (the so-called “plot stratification” is used), for each of which it is easy to determine the ordinate of the center of gravity. In this case, the area of each figure is multiplied by the ordinate under its center of gravity.
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, thanks for your attention.
01-11-2012: Vic
typo in the formula in the fifth example from above (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 respected. 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 the 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 moment diagram is located at the top, then the moment while negative.
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.
25-05-2013: Dmitriy
Please tell me at what ratio of the length of the beam to its diameter these formulas are valid?
I want to know whether this subcode is only for long beams, which are used in the construction of buildings, or can also be used to calculate the deflections of shafts up to 2 m long. Please answer like this l/D>...
25-05-2013: Doctor Lom
Dmitry, I already told you, for rotating shafts the calculation schemes will be different. However, if the shaft is stationary, then it can be considered as a beam, and it does not matter what its cross-section is: round, square, rectangular or something else. These calculation schemes most accurately reflect the state of the beam at l/D>10, with a ratio of 5 25-05-2013: Dmitriy
Thanks for the answer. Can you name other literature that I can refer to in my work? 25-05-2013: Doctor Lom
I don’t know what exact problem you are solving, and therefore it is difficult to have a substantive conversation. I'll try to explain my idea differently. 25-05-2013: Dmitriy
Can I then communicate with you via mail or Skype? I'll tell you what kind of work I do and what the previous questions were for. 25-05-2013: Doctor Lom
You can write to me, email addresses are not difficult to find on the site. But I’ll warn you right away that I don’t do any calculations and don’t sign partnership contracts. 08-06-2013: Vitaly
Question on table 2, option 1.1, deflection formula. Please check the size. 09-06-2013: Doctor Lom
That's right, the output is centimeters. 20-06-2013: Evgeniy Borisovich
Hello. Help me figure it out. 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 parentheses 07-02-2015: Doctor Lom
No, not a mistake. The load on the cantilever reduces the moment in the span, but does not increase it. However, this can be seen from the moment diagram. 17-02-2015: Anton
Hello, first of all, thanks for the formulas, I saved them in my bookmarks. Please tell me, is there a beam above the span, four logs rest on the beam, distances: 180mm, 600mm, 600mm, 600mm, 325mm. I figured out the diagram and the bending moment, but I can’t understand how the deflection formula (Table 1, diagram 1.4) will change if the maximum moment is on the third lag. 17-02-2015: Doctor Lom
I have already answered similar questions several times in the comments to the article “Calculation schemes for statically indeterminate beams.” But you are lucky, for clarity, I performed the calculation using the data from your question. Look at the article “The general case of calculating a beam on hinged supports under the action of several concentrated loads”, perhaps over time I will add to it. 22-02-2015: Novel
Doc, I really can’t master all these formulas that are incomprehensible to me. Therefore, I ask you for help. I want to make a cantilever staircase in my house (the steps will be bricked up with reinforced concrete when building the wall). Wall - width 20cm, brick. The length of the protruding step is 1200*300mm. I want the steps to be of the correct shape (not a wedge). I intuitively understand that the reinforcement will be “something thicker” so that the steps will be something thinner? But can reinforced concrete up to 3cm thick cope with a load of 150kg at the edge? Please help me, I really don’t want to screw up. I would be very grateful if you could help me calculate... 22-02-2015: Doctor Lom
The fact that you cannot master fairly simple formulas is your problem. In the section “Basics of Strength of Strength” all this is discussed in sufficient detail. Here I will say that your project is absolutely unrealistic. Firstly, the wall is either 25 cm wide or cinder block (however, I could be wrong). Secondly, neither a brick nor a cinder block wall will provide sufficient pinching of steps with the specified wall width. In addition, such a wall should be calculated for the bending moment arising from the cantilever beams. Thirdly, 3 cm is an unacceptable thickness for a reinforced concrete structure, taking into account the fact that the minimum protective layer in beams must be at least 15 mm. And so on. 26-02-2015: Novel
02-04-2015: Vitaly
what does x mean in the second table, 2.4 02-04-2015: Vitaly
Good afternoon What scheme (algorithm) should be chosen to calculate a balcony slab, a cantilever clamped on one side, how to correctly calculate the moments on the support and in the span? Can it be calculated as a cantilever beam, according to the diagrams from Table 2, namely points 1, 1 and 2.1. Thank you! 02-04-2015: Doctor Lom
x in all tables means the distance from the origin to the point under study at which we are going to determine the bending moment or other parameters. Yes, your balcony slab, if it is solid and loads act on it, as in the indicated diagrams, can be calculated according to these diagrams. For cantilever beams, the maximum moment is always at the support, so there is no great need to determine the moment in the span. 03-04-2015: Vitaly
Thanks a lot! I also wanted to clarify. As I understand it, if you calculate according to 2 tables. diagram 1.1, (the load is applied to the end of the console) then I have x = L, and accordingly in the span M = 0. What if I also have this load at the ends of the slab? And according to scheme 2.1, I calculate the moment at the support, add it to the moment according to scheme 1.1 and according to the correct one, in order to reinforce it, I need to find the moment in the span. If I have a slab overhang of 1.45 m (in the clear), how can I calculate “x” to find the moment in the span? 03-04-2015: Doctor Lom
The moment in the span will vary from Ql at the support to 0 at the point of application of the load, which can be seen from the moment diagram. If your load is applied at two points at the ends of the slab, then in this case it is more advisable to provide beams that absorb loads at the edges. In this case, the slab can already be calculated as a beam on two supports - beams or a slab supported on 3 sides. 03-04-2015: Vitaly
Thank you! In moments I already understood. One more question. If the balcony slab is supported on both sides, using the letter “G”. What calculation scheme should I use then? 04-04-2015: Doctor Lom
In this case, you will have a plate pinched on 2 sides and there are no examples of calculating such a plate on my website. 27-04-2015: Sergey
Dear Doctor Lom! 27-04-2015: Doctor Lom
I won’t assess the reliability of such a design without calculations, but you can calculate it using the following criteria: 05-06-2015: student
Doc, where can I show you the picture? 05-06-2015: student
Did you still have a forum? 05-06-2015: Doctor Lom
There was, but I have absolutely no time to sort through spam in search of normal questions. So that's it for now. 06-06-2015: student
Doc, my link is https://yadi.sk/i/GardDCAEh7iuG 07-06-2015: Doctor Lom
The choice of design scheme will depend on what you want: simplicity and reliability or approximation to the actual operation of the structure through successive approximations. 07-06-2015: student
Doc, thanks. I need simplicity and reliability. This area is the busiest. I even thought about tying the tank post to the rafters to reduce the load on the floor, given that the water would be drained in the winter. I can’t get into such a jungle of calculations. In general, will the cantilever reduce deflection? 07-06-2015: student
Doc, one more question. the console is in the middle of the window span, does it make sense to move it to the edge? Sincerely 07-06-2015: Doctor Lom
In general, the console will reduce the deflection, but as I already said, how much in your case is a big question, and a shift to the center of the window opening will reduce the role of the console. And also, if this is your most loaded area, then maybe you can simply strengthen the beam, for example, with another similar channel? I don’t know your loads, but the load of 100 kg of water and half the weight of the tank does not seem so impressive to me, but from the point of view of deflection, do 8P channels take into account the dynamic load when walking? 08-06-2015: student
Doc, thanks for the good advice. After the weekend I will recalculate the beam as a two-span beam on hinges. If there is greater dynamics when walking, I constructively include the possibility of reducing the pitch of the floor beams. The house is a country house, so the dynamics are tolerable. The lateral displacement of the channels has a greater influence, but this can be treated by installing cross braces or fastening the flooring. The only thing is, will the concrete pouring crumble? I assume it will be supported on the upper and lower flanges of the channel plus welded reinforcement in the ribs and mesh on top. 08-06-2015: Doctor Lom
I already told you, you shouldn’t count on the console. 09-06-2015: student
Doc, I understand. 29-06-2015: Sergey
Good afternoon. I would like to ask you: the foundation was cast: piles of concrete 1.8 m deep, and then a strip 1 m deep was cast with concrete. The question is this: is the load transferred only to the piles or is it evenly distributed to both the piles and the tape? 29-06-2015: Doctor Lom
As a rule, piles are made in weak soils so that the load on the foundation is transmitted through the piles, so grillages on piles are calculated like beams on pile supports. However, if you poured the grillage over compacted soil, then part of the load will be transferred to the base through the grillage. In this case, the grillage is considered as a beam lying on an elastic foundation and represents a regular strip foundation. Like that. 29-06-2015: Sergey
Thank you. It’s just that the site turns out to be a mixture of clay and sand. Moreover, the clay layer is very hard: the layer can only be removed with a crowbar, etc., etc. 29-06-2015: Doctor Lom
I don’t know all your conditions (distance between piles, number of floors, etc.). From your description, it looks like you made a regular strip foundation and piles for reliability. Therefore, you just need to determine whether the width of the foundation will be sufficient to transfer the load from the house to the foundation. 05-07-2015: Yuri
Hello! We need your help with the calculations. A metal gate 1.5 x 1.5 m weighing 70 kg is mounted on a metal pipe, concreted to a depth of 1.2 m and lined with brick (post 38 x 38 cm). What cross-section and thickness should the pipe be so that there is no bending? 05-07-2015: Doctor Lom
You correctly assumed that your post should be treated like a cantilever beam. And even with the calculation scheme, you almost got it right. The fact is that 2 forces will act on your pipe (on the upper and lower 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. Isn’t it reflected in formula 1.4 in any way? 23-06-2016: Doctor Lom
With an asymmetrical load, the deflection formula for design scheme 1.4 will be quite cumbersome, but it should be remembered that the deflection in any case will be less than when applying a symmetrical load (of course, provided b 03-11-2016: vladimir
in Table 1 for formulas 1.3 and 1.4, the deflection formula should be Ql^3/24EI instead of Qa^3/24EI. For a long time I could not understand why the deflection with the crystal did not converge 03-11-2016: Doctor Lom
That's right, another typo due to inattentive editing (I hope it's the last one, but not a fact). Corrected, thanks for your attention. 16-12-2016: Ivan
Hello, Doctor Lom. The question is the following: I was looking through photos from the construction site and noticed one thing: the factory-made reinforced concrete lintel is approximately 30*30 cm, supported on a three-layer reinforced concrete panel about 7 centimeters (the reinforced concrete panel was sawed down a little to rest the lintel on it). The opening for the balcony frame is 1.3 m, along the top of the lintel there is an armored belt and attic floor slabs. Are these 7 cm critical, the support of the other end of the jumper is more than 30 cm, everything has been fine for several years now 16-12-2016: Doctor Lom
If there is also an armored belt, then the load on the jumper can be significantly reduced. I think everything will be fine and even at 7 cm there is a fairly large margin of safety on the support platform. But in general, of course, you need to count. 25-12-2016: Ivan
Doctor, if we assume, well, purely theoretically 25-12-2016: Doctor Lom
I think even in this case nothing will happen. But I repeat, a more accurate answer requires calculation. 09-01-2017: Andrey
In Table 1, in formula 2.3, to calculate the deflection, instead of “q”, “Q” is indicated. Formula 2.1 for calculating the deflection, being a special case of formula 2.3, when inserting the corresponding values (a=c=l, b=0) takes on a different form. 09-01-2017: Doctor Lom
That's right, there was a typo, but now it doesn't matter. I took the deflection formula for such a design scheme from S.P. Fesik’s reference book, as the shortest for the special case x = a. But as you correctly noted, this formula does not pass the boundary conditions test, so I removed it altogether. I left only the formula for determining the initial angle of rotation in order to simplify the determination of deflection using the initial parameters method. 02-03-2017: Doctor Lom
As far as I know, such a special case is not considered in textbooks. Only software will help here, for example, Lyra. 24-03-2017: Eageniy
Good afternoon, in the deflection formula 1.4 in the first table - the value in brackets is always negative 24-03-2017: Doctor Lom
Everything is correct, in all the given formulas, the negative sign in the deflection formula means that the beam bends down along the y-axis. 29-03-2017: Oksana
Good afternoon, Doctor Lom. Could you write an article about the torque in a metal beam - when does it occur at all, under what design schemes, and, of course, I would like to see your calculations with examples. I have a hingedly supported metal beam, one edge is cantilevered and a concentrated load comes to it, and the load is distributed over the entire beam from the reinforced concrete. thin slab 100 mm and fence wall. This beam is the outermost one. With reinforced concrete The plate is connected by 6 mm rods welded to the beam with a pitch of 600 mm. I can’t understand whether there will be a torque there, if so, how to find it and calculate the cross-section of the beam in connection with it? Doctor Lom Victor, emotional stroking is, of course, good, but you can’t spread it on bread and you can’t feed your family with it. To answer your question, calculations are required, calculations are time, and time is not emotional stroking.
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 slope 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?
