Bending under distributed load. Clean bend. Cross bend. General concepts. We calculate the moments of inertia and forces
When building bending moment diagramsM at builders accepted: ordinates expressing in a certain scale positive values of bending moments, put aside stretched fibers, i.e. - way down, a negative - up from the axis of the beam. Therefore, they say that builders build diagrams on stretched fibers. Mechanics positive values of both shear force and bending moment are plotted up. Mechanics build diagrams on compressed fibers.
Principal stresses when bending. Equivalent voltages.
In the general case of direct bending in the cross sections of the beam, normal and tangents
voltage. These voltages vary in both length and height of the beam.
Thus, in the case of bending, plane stress state.
Consider a scheme where the beam is loaded with a force P
Greatest normal stresses occur in extreme, points farthest from the neutral line, and shear stresses are absent in them. So for extreme fibers non-zero principal stresses are normal stresses in cross section.
At the level of the neutral line in the cross section of the beam arise the greatest shear stresses, a normal stresses are zero. means in the fibers neutral layer principal stresses are determined by the values of shear stresses.
In this design model, the upper fibers of the beam will be stretched, and the lower ones will be compressed. To determine the principal stresses, we use the well-known expression: 
Full stress state analysis present in the figure.
Analysis of the stress state in bending
The greatest principal stress σ 1 is located top extreme fibers and is equal to zero on the lower extreme fibers. Principal stress σ 3 It has the largest absolute value on the lower fibers.
Principal stress trajectory depends on load type and way to fix the beam.
When solving problems, it is enough separately verify normal and separate shear stresses. However, sometimes the most stressful turn out intermediate fibers that have both normal and shear stresses. This happens in sections where simultaneously both the bending moment and the transverse force reach large values- this can be in the embedding of a cantilever beam, on the support of a beam with a cantilever, in sections under a concentrated force, or in sections with a sharply changing width. For example, in an I-section, the most dangerous junction of the wall to the shelf- there are significant and normal and shear stresses.
The material is in a plane stress state and requires equivalent voltage test.
Strength conditions for beams made of ductile materials on third(theories of the greatest tangential stresses) 
and fourth(theory of energy of form changes) 
strength theories.
As a rule, in rolled beams, the equivalent stresses do not exceed the normal stresses in the outermost fibers and no special verification is required. Another thing - composite metal beams, which thinner wall than that of rolled profiles at the same height. Welded composite beams made of steel sheets are more commonly used. Calculation of such beams for strength: a) selection of the section - height, thickness, width and thickness of the beam chords; b) strength test for normal and shear stresses; c) verification of strength by equivalent stresses.
Determination of shear stresses in an I-section. Consider the section I-beam. S x \u003d 96.9 cm 3; Yx=2030 cm 4; Q=200 kN
To determine the shear stress, it is used formula
, where Q is the transverse force in the section, S x 0 is the static moment of the part of the cross section located on one side of the layer in which shear stresses are determined, I x is the moment of inertia of the entire cross section, b is the width of the section in the place where shear stress is determined
Compute maximum shear stress:
Let us calculate the static moment for top shelf: 
Now let's calculate shear stresses: 
We are building shear stress diagram:
Consider a section of a standard profile in the form I-beam and define shear stresses acting parallel to the transverse force:
Calculate static moments simple figures: 
This value can also be calculated otherwise, using the fact that for an I-beam and a trough section, the static moment of half the section is given at the same time. To do this, it is necessary to subtract from the known value of the static moment the value of the static moment to the line A 1 B 1: 
Shear stresses at the junction of the flange to the wall change spasmodically, because sharp wall thickness changes from t st before b.
Plots of shear stresses in the walls of trough, hollow rectangular and other sections have the same form as in the case of an I-section. The formula includes the static moment of the shaded part of the section relative to the X axis, and the denominator is the section width (net) in the layer where the shear stress is determined.
Let us determine shear stresses for a circular section.
Since the tangential stresses at the contour of the section must be directed tangent to the contour, then at the points BUT and AT at the ends of any chord parallel to the diameter AB, shear stresses are directed perpendicular to the radii OA and OV. Consequently, directions shear stresses at points BUT, VC converge at some point H on the Y axis.
Static moment of the cut-off part: 
That is, shear stresses change according to parabolic law and will be maximum at the level of the neutral line when y 0 =0
Formula for determining shear stresses (formula) 
Consider a rectangular section
On distance at 0 draw from the central axis section 1-1 and determine shear stresses. Static moment area cut off part:
It should be borne in mind that fundamentally indifferent, take the static moment of the area shaded or rest cross section. Both static moments equal and opposite in sign, so they sum, which represents static moment of the area of the entire section relative to the neutral line, namely the central axis x, will be equal to zero.
Moment of inertia of a rectangular section:
Then shear stresses according to the formula
The variable y 0 is included in the formula during second degrees, i.e. shear stresses in a rectangular section vary with the law of a square parabola.
Shear stress reached maximum at the level of the neutral line, i.e. when y 0 =0:
,
where A is the area of the entire section.
Strength condition for shear stresses looks like:
, where S x 0 is the static moment of the part of the cross section located on one side of the layer in which shear stresses are determined, I x is the moment of inertia of the entire cross section, b- section width in the place where shear stress is determined, Q- transverse force, τ
- shear stress, [τ]
— allowable shear stress.
This strength condition makes it possible to produce three type of calculation (three types of problems in strength analysis):
1. Verification calculation or strength test for shear stresses:
2. Selection of section width (for rectangular section): 
3. Determination of the permissible transverse force (for a rectangular section):
For determining tangents stresses, consider a beam loaded with forces.
The task of determining stresses is always statically indeterminate and requires involvement geometric and physical equations. However, one can take hypotheses about the nature of stress distribution that the task will become statically determined.
Two infinitely close cross sections 1-1 and 2-2 select dz element, draw it on a large scale, then draw a longitudinal section 3-3.
In sections 1–1 and 2–2, normal σ 1 , σ 2 stresses, which are determined by the well-known formulas:
where M - bending moment in cross section dM - increment bending moment on length dz
Shear force in sections 1–1 and 2–2 is directed along the main central axis Y and, obviously, represents the sum of the vertical components of the internal shear stresses distributed over the section. In the strength of materials, it is usually taken the assumption of their uniform distribution over the width of the section.
To determine the magnitude of shear stresses at any point of the cross section, located at a distance at 0 from the neutral X axis, draw a plane parallel to the neutral layer (3-3) through this point, and take out the cut-off element. We will determine the voltage acting on the ABSD site. 
Let's project all the forces on the Z axis
The resultant of the internal longitudinal forces along the right side will be equal to:
where A 0 is the area of the facade face, S x 0 is the static moment of the cut-off part relative to the X axis. Similarly on the left side:
Both resultants directed towards each other because the element is in compressed beam zone. Their difference is balanced by tangential forces on the lower face 3-3.
Let's pretend that shear stresses τ distributed over the width of the beam cross section b evenly. This assumption is the more likely, the smaller the width compared to the height of the section. Then resultant of tangential forces dT is equal to the stress value multiplied by the face area: 
Compose now equilibrium equation Σz=0: 
or where from
Let's remember differential dependencies, according to which 
Then we get the formula:
This formula is called formulas. This formula was obtained in 1855. Here S x 0 - static moment of a part of the cross section, located on one side of the layer in which shear stresses are determined, I x - moment of inertia the entire cross section b - section width where the shear stress is determined, Q - transverse force in section.
is the bending strength condition, where
- maximum moment (modulo) from the diagram of bending moments; 
- axial section modulus, geometric characteristic; - allowable stress (σadm)
- maximum normal stress.
If the calculation is based on limit state method, then in the calculation instead of the allowable stress is introduced design resistance of the material R.
Types of bending strength calculations
1. Checking calculation or verification of normal stress strength
2. Project calculation or section selection 
3. Definition permitted loads (definition lifting capacity and or operational carrier capabilities) 
When deriving a formula for calculating normal stresses, consider such a case of bending, when the internal forces in the sections of the beam are reduced only to bending moment, a transverse force is zero. This case of bending is called pure bending. Consider the middle section of a beam undergoing pure bending.
When loaded, the beam bends so that it the lower fibers lengthen and the upper fibers shorten.
Since some of the fibers of the beam are stretched and some are compressed, and the transition from tension to compression occurs smoothly, without jumps, in middle part of the beam is a layer whose fibers only bend, but do not experience either tension or compression. Such a layer is called neutral layer. The line along which the neutral layer intersects with the cross section of the beam is called neutral line or neutral axis sections. Neutral lines are strung on the axis of the beam. neutral line is the line in which normal stresses are zero.
Lines drawn on the side surface of the beam perpendicular to the axis remain flat when bending. These experimental data make it possible to base the derivations of the formulas hypothesis of flat sections (hypothesis). According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and become perpendicular to the bent axis of the beam when it is bent.
Assumptions for the derivation of normal stress formulas: 1) The hypothesis of flat sections is fulfilled. 2) Longitudinal fibers do not press on each other (non-pressure hypothesis) and, therefore, each of the fibers is in a state of uniaxial tension or compression. 3) The deformations of the fibers do not depend on their position along the width of the section. Consequently, the normal stresses, changing along the height of the section, remain the same across the width. 4) The beam has at least one plane of symmetry, and all external forces lie in this plane. 5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same. 6) The ratios between the dimensions of the beam are such that it works in flat bending conditions without warping or twisting.
Consider a beam of arbitrary section, but having an axis of symmetry. 
Bending moment represents resultant moment of internal normal forces arising on infinitely small areas and can be expressed in terms of integral form: 
(1), where y is the arm of the elementary force relative to the x axis
Formula (1) expresses static side of the problem of bending a straight bar, but along it according to a known bending moment it is impossible to determine normal stresses until the law of their distribution is established.
Select the beams in the middle section and consider section of length dz, subject to bending. Let's zoom in on it.
Sections bounding the section dz, parallel to each other before deformation, and after applying the load turn around their neutral lines at an angle . The length of the segment of the fibers of the neutral layer will not change. and will be equal to: 
, where is it radius of curvature curved axis of the beam. But any other fiber lying below or above neutral layer, will change its length. Compute relative elongation of fibers located at a distance y from the neutral layer. Relative elongation is the ratio of absolute deformation to the original length, then:
We reduce by and reduce like terms, then we get: (2) This formula expresses geometric side of the pure bending problem: fiber deformations are directly proportional to their distances from the neutral layer.
Now let's move on to stresses, i.e. we will consider physical side of the task. in accordance with non-pressure assumption fibers are used in axial tension-compression: then, taking into account the formula (2)
we have 
(3),
those. normal stresses when bending along the height of the section are distributed according to a linear law. On the extreme fibers, the normal stresses reach their maximum value, and in the center of gravity, the cross sections are equal to zero. Substitute (3)
into the equation (1)
and take the fraction out of the integral sign as a constant value, then we have 
. But the expression is axial moment of inertia of the section about the x-axis - I x.
Its dimension cm 4, m 4
Then 
,where 
(4) , where is curvature of the bent axis of the beam, a is the stiffness of the beam section during bending.
Substitute the resulting expression curvature (4) into an expression (3)
and get formula for calculating normal stresses at any point of the cross section: 
(5)
That. maximum stresses arise at points furthest from the neutral line. Attitude (6)
called axial section modulus. Its dimension cm 3, m 3. The moment of resistance characterizes the influence of the shape and dimensions of the cross section on the magnitude of stresses.
Then maximum voltages: 
(7)
Bending strength condition: (8)
During transverse bending not only normal, but also shear stresses, because available shear force. Shear stresses complicate the picture of deformation, they lead to curvature cross sections of the beam, as a result of which the hypothesis of flat sections is violated. However, studies show that the distortions introduced by shear stresses slightly affect the normal stresses calculated by the formula (5) . Thus, when determining normal stresses in the case of transverse bending the theory of pure bending is quite applicable.
Neutral line. Question about the position of the neutral line.
When bending, there is no longitudinal force, so we can write 
Substitute here the formula for normal stresses (3)
and get 
Since the modulus of elasticity of the beam material is not equal to zero and the bent axis of the beam has a finite radius of curvature, it remains to assume that this integral is static moment of area cross section of the beam relative to the neutral line-axis x 
, and since it is equal to zero, then the neutral line passes through the center of gravity of the section.
The condition (absence of the moment of internal forces relative to the field line) will give 
or taking into account (3)
. For the same reasons (see above) 
. In the integrand - the centrifugal moment of inertia of the section about the x and y axes is zero, so these axes are main and central and make up straight corner. Consequently, the power and neutral lines in a straight bend are mutually perpendicular.
By setting neutral line position, easy to build normal stress diagram by section height. Her linear character is determined equation of the first degree.
The nature of the diagram σ for symmetrical sections with respect to the neutral line, M<0
For a cantilever beam loaded with a distributed load of intensity kN / m and a concentrated moment kN m (Fig. 3.12), it is required: to build diagrams of shear forces and bending moments, select a beam of circular cross section at an allowable normal stress kN / cm2 and check the strength of the beam according to shear stresses at permissible shear stress kN/cm2. Beam dimensions m; m; m.
Design scheme for the problem of direct transverse bending
Rice. 3.12
Solving the problem of "direct transverse bending"
Determining support reactions
The horizontal reaction in the embedment is zero, since external loads in the direction of the z-axis do not act on the beam.
We choose the directions of the remaining reactive forces that arise in the embedment: let's direct the vertical reaction, for example, down, and the moment - clockwise. Their values are determined from the equations of statics:
When compiling these equations, we consider the moment to be positive when rotating counterclockwise, and the projection of the force is positive if its direction coincides with the positive direction of the y axis.
From the first equation we find the moment in the termination:
From the second equation - vertical reaction:
The positive values obtained by us for the moment and vertical reaction in the termination indicate that we have guessed their directions.
In accordance with the nature of the fastening and loading of the beam, we divide its length into two sections. Along the boundaries of each of these sections, we outline four cross sections (see Fig. 3.12), in which we will calculate the values of shear forces and bending moments by the method of sections (ROZU).
Section 1. Let's mentally discard the right side of the beam. Let's replace its action on the remaining left side with a cutting force and a bending moment. For the convenience of calculating their values, we close the right side of the beam discarded by us with a piece of paper, aligning the left edge of the sheet with the section under consideration.
Recall that the shear force arising in any cross section must balance all external forces (active and reactive) that act on the part of the beam we are considering (that is, visible). Therefore, the shearing force must be equal to the algebraic sum of all the forces that we see.
We also give the sign rule for the shearing force: an external force acting on the considered part of the beam and tending to “rotate” this part relative to the section in a clockwise direction causes a positive shearing force in the section. Such an external force is included in the algebraic sum for the definition with a plus sign.
In our case, we see only the reaction of the support, which rotates the visible part of the beam relative to the first section (relative to the edge of the piece of paper) counterclockwise. That's why
kN.
The bending moment in any section must balance the moment created by external forces that we see with respect to the section under consideration. Therefore, it is equal to the algebraic sum of the moments of all efforts that act on the part of the beam we are considering, relative to the section under consideration (in other words, relative to the edge of the piece of paper). In this case, an external load bending the considered part of the beam with a convexity downwards causes a positive bending moment in the section. And the moment created by such a load is included in the algebraic sum for the definition with a plus sign.
We see two efforts: the reaction and the moment in termination. However, the arm of the force with respect to section 1 is equal to zero. That's why
kN m
We took the plus sign because the reactive moment bends the visible part of the beam with a convexity downwards.
Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now, unlike the first section, the force has a shoulder: m. Therefore
kN; kN m
Section 3. Closing the right side of the beam, we find
kN;
Section 4. Let's close the left side of the beam with a leaf. Then
kN m
kN m
.
Based on the values found, we build diagrams of shear forces (Fig. 3.12, b) and bending moments (Fig. 3.12, c).
Under unloaded sections, the diagram of shear forces runs parallel to the axis of the beam, and under a distributed load q, along an inclined straight line upwards. Under the support reaction on the diagram there is a jump down by the value of this reaction, that is, by 40 kN.
On the diagram of bending moments, we see a break under the support reaction. The fracture angle is directed towards the reaction of the support. Under a distributed load q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. In section 6 on the diagram there is an extremum, since the diagram of the shearing force in this place passes through the zero value here.
Determine the required diameter of the cross section of the beam
The strength condition for normal stresses has the form:
,
where is the moment of resistance of the beam in bending. For a beam of circular cross section, it is equal to:
.
The bending moment with the largest absolute value occurs in the third section of the beam: 
kN cm
Then the required beam diameter is determined by the formula
cm.
We accept mm. Then
kN/cm2 kN/cm2.
"Overvoltage" is
,
what is allowed.
We check the strength of the beam for the highest tangential stresses
The highest shear stresses that occur in the cross section of a circular beam are calculated by the formula
,
where is the cross-sectional area.
According to the plot, the largest algebraic value of the shear force is equal to 
kN. Then
kN/cm2 kN/cm2,
that is, the condition of strength and shear stresses is fulfilled, moreover, with a large margin.
An example of solving the problem "direct transverse bending" No. 2
Condition of the problem example for direct transverse bending
For a hinged beam loaded with a distributed load of intensity kN / m, a concentrated force kN and a concentrated moment kN m (Fig. 3.13), it is required to plot the shear forces and bending moments and select an I-beam cross section with an allowable normal stress kN / cm2 and permissible shear stress kN/cm2. Beam span m.
An example of a task for a straight bend - a design scheme
 |
Rice. 3.13
Solution of an example of a straight bend problem
Determining support reactions
For a given pivotally supported beam, it is necessary to find three support reactions: , and . Since only vertical loads act on the beam, perpendicular to its axis, the horizontal reaction of the fixed hinged support A is equal to zero: .
The directions of vertical reactions and are chosen arbitrarily. Let's direct, for example, both vertical reactions upwards. To calculate their values, we compose two equations of statics:
Recall that the resultant linear load, uniformly distributed over a section of length l, is equal to, that is, equal to the area of the diagram of this load and it is applied at the center of gravity of this diagram, that is, in the middle of the length.
;
kN.
We check: .
Recall that forces whose direction coincides with the positive direction of the y-axis are projected (projected) onto this axis with a plus sign:
that is correct.
We build diagrams of shear forces and bending moments
We break the length of the beam into separate sections. The boundaries of these areas are the points of application of concentrated forces (active and / or reactive), as well as points corresponding to the beginning and end of the distributed load. There are three such areas in our problem. Along the boundaries of these sections, we outline six cross sections, in which we will calculate the values of shear forces and bending moments (Fig. 3.13, a).
Section 1. Let's mentally discard the right side of the beam. For the convenience of calculating the shear force and bending moment arising in this section, we close the part of the beam discarded by us with a piece of paper, aligning the left edge of the piece of paper with the section itself.
The shearing force in the beam section is equal to the algebraic sum of all external forces (active and reactive) that we see. In this case, we see the reaction of the support and the linear load q, distributed over an infinitely small length. The resultant linear load is zero. That's why
kN.
The plus sign is taken because the force rotates the visible part of the beam relative to the first section (the edge of the piece of paper) in a clockwise direction.
The bending moment in the section of the beam is equal to the algebraic sum of the moments of all the forces that we see, relative to the section under consideration (that is, relative to the edge of a piece of paper). We see the reaction of the support and the linear load q, distributed over an infinitely small length. However, the leverage of the force is zero. The resultant linear load is also equal to zero. That's why
Section 2. As before, we will cover the entire right side of the beam with a piece of paper. Now we see the reaction and the load q acting on a section of length . The resultant linear load is equal to . It is attached in the middle of a section with a length of . That's why
Recall that when determining the sign of the bending moment, we mentally free the part of the beam that we see from all the actual support fastenings and imagine it as if pinched in the section under consideration (that is, the left edge of the piece of paper is mentally represented by us as a rigid seal).
Section 3. Let's close the right part. Get
Section 4. We close the right side of the beam with a leaf. Then
Now, to control the correctness of the calculations, let's cover the left side of the beam with a piece of paper. We see the concentrated force P, the reaction of the right support and the linear load q, distributed over an infinitely small length. The resultant linear load is zero. That's why
kN m
That is, everything is correct.
Section 5. Still close the left side of the beam. Will have
kN;
kN m
Section 6. Let's close the left side of the beam again. Get
kN;
Based on the values found, we build diagrams of shear forces (Fig. 3.13, b) and bending moments (Fig. 3.13, c).
We are convinced that under the unloaded section the diagram of the shear forces runs parallel to the axis of the beam, and under a distributed load q - along a straight line with a downward slope. There are three jumps on the diagram: under the reaction - up by 37.5 kN, under the reaction - up by 132.5 kN and under the force P - down by 50 kN.
On the diagram of bending moments, we see breaks under the concentrated force P and under the support reactions. The fracture angles are directed towards these forces. Under a distributed load of intensity q, the diagram changes along a quadratic parabola, the convexity of which is directed towards the load. Under the concentrated moment there is a jump of 60 kN m, that is, by the magnitude of the moment itself. In section 7 on the diagram there is an extremum, since the diagram of the shear force for this section passes through the zero value (). Let us determine the distance from section 7 to the left support.
A bend is a type of deformation in which the longitudinal axis of the beam is bent. Straight beams working on bending are called beams. A straight bend is a bend in which the external forces acting on the beam lie in the same 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 the beam, is called transverse. The line of intersection of the force plane and the cross-sectional plane is called force line.
Internal force factors in beam bending.
With a flat transverse bending in the beam sections, two internal force factors arise: the transverse force Q and the bending moment M. The section method is used to determine them (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 the beam section is equal to the algebraic sum of the moments about 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 dependences.
Between the intensity q of the distributed load, the expressions for the transverse force Q and the bending moment M, differential dependencies are established:
Based on these dependencies, the following general patterns of diagrams of transverse forces Q and bending moments M can be distinguished:
Peculiarities of diagrams of internal force factors in bending.
1. On the section of the beam where there is no distributed load, the plot Q is presented straight line , parallel to the base of the diagram, and the diagram M is an inclined straight line (Fig. a).
2. In the section where the concentrated force is applied, on the Q diagram there should be jump , 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 jump , equal to the value of this moment, (Fig. 26, b).
4. In the section of the beam with a distributed load of intensity q, the diagram Q changes according to a linear law, and the diagram M - according to a parabolic one, and the convexity of the parabola is directed towards the direction of the distributed load (Fig. c, d).
5. If within the characteristic section of 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 the section to bending is the value:
Dangerous section when bending, the cross section of the beam is called, in which the maximum normal stress occurs.
Tangential stresses in direct bending.
Determined by Zhuravsky's formula for shear stresses in direct beam bending:
where S ots - static moment of the transverse area of the cut-off layer of longitudinal fibers relative to the neutral line.
Bending strength calculations.
1. At verification calculation the maximum design stress is determined, which is compared with the allowable stress:
2. At design calculation selection of the beam section is made from the condition:
3. When determining the allowable load, the allowable bending moment is determined from the condition:
Bending movements.
Under the action of a bending load, the axis of the beam is bent. In this case, there is a stretching of the fibers on the convex and compression - on the concave parts 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 the deformation during bending, the following concepts are used:
Beam deflection Y- displacement of the center of gravity of the cross section of the beam in the direction perpendicular to its axis.
The 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- the angle θ by which each section is rotated with respect to its original position. The angle of rotation is considered positive when the section is rotated counterclockwise. The value of the angle of rotation varies along the length of the beam, being a function of θ = θ (z).
The most common way to determine displacements is the method mora and Vereshchagin's rule.
Mohr method.
The procedure for determining displacements according to the Mohr method:
1. An "auxiliary system" is built and loaded with a single load at the point where the displacement is to be determined. If a linear displacement is determined, then a unit force is applied in its direction; when determining angular displacements, a unit moment is applied.
2. For each section of the system, the expressions of bending moments M f from the applied load and M 1 - from a single load are recorded.
3. Mohr integrals are calculated and summed over all sections of the system, 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. The 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, and from a single 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 is the ordinate of the diagram from a single load under the center of gravity of the diagram M f ; EI x - section stiffness of the beam section. Calculations according to this formula are made in sections, on each of which the straight-line diagram must 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 opposite sides. A positive result of the multiplication of diagrams means that the direction of movement coincides with the direction of a unit force (or moment). A complex diagram M f must be divided into simple figures (the so-called "epure layering" is used), for each of which it is easy to determine the ordinate of the center of gravity. In this case, the area of \u200b\u200beach figure is multiplied by the ordinate under its center of gravity.
29-10-2012: Andrew
A typo was made in the formula for the bending moment for a beam with rigid pinching on supports (3rd from the bottom): the length must be squared. A typo was made in the formula for the maximum deflection for a beam with rigid pinning on supports (3rd from the bottom): it should be without "5".
29-10-2012: Dr. Lom
Yes, indeed, mistakes were made when editing after copying. At the moment, the errors have been fixed, thanks for your attention.
01-11-2012: Vic
a typo in the formula in the fifth example from the top (the degrees next to x and el are mixed up)
01-11-2012: Dr. Lom
And it is true. Corrected. Thank you for your attention.
10-04-2013: flicker
In formula T.1, 2.2 Mmax seems to be missing a square after a.
11-04-2013: Dr. Lom
Right. I copied this formula from the "Handbook of the Strength of Materials" (ed. by S.P. Fesik, 1982, p. 80) and did not even pay attention to the fact that with such a notation, even the dimension is not respected. Now I counted everything personally, indeed the distance "a" will be squared. Thus, it turns out that the compositor 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, scheme 2.4, you are interested in the formula "moment in flight" where the index X is not clear -? Could you answer)
02-05-2013: Dr. Lom
For the cantilever beams of 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, which will have a rigid support 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 moments is located on top, then the moment is negative.
From the point of view of the strength of materials, the sign of the moment is a rather arbitrary 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, however, the direction of action of the moment was taken into account when compiling the formulas.
25-05-2013: Dmitry
Please tell me, at what ratio of the length of the beam to its diameter are these formulas valid?
I want to know if this code only applies to long beams that are used in building construction, or can it also be used to calculate shaft deflections, up to 2 m long. Please answer like this l/D>...
25-05-2013: Dr. Lom
Dmitry, I already told you that the design schemes for rotating shafts will be different. Nevertheless, if the shaft is in a stationary state, then it can be considered as a beam, and it does not matter what section it has: round, square, rectangular, or some other. These design schemes most accurately reflect the state of the beam at l/D>10, at a ratio of 5 25-05-2013: Dmitry
Thanks for the answer. Can you also name the literature that I can refer to in my work? 25-05-2013: Dr. Lom
I don’t know what kind of problem you are solving, and therefore it is difficult to conduct a substantive conversation. I'll try to explain my idea in a different way. 25-05-2013: Dmitry
Can I then chat with you via mail or Skype? I will tell you what kind of work I do and what the previous questions were for. 25-05-2013: Dr. Lom
You can write to me, email addresses on the site are not difficult to find. But I’ll warn you right away, I don’t do any calculations and I don’t sign partnership contracts. 08-06-2013: Vitaly
Question according to table 2, option 1.1, deflection formula. Please specify dimensions. 09-06-2013: Dr. Lom
That's right, the output is centimeters. 20-06-2013: Evgeny Borisovich
Hello. Help guess. We have a summer wooden stage near the recreation center, the size is 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 you can’t attach a picture) a polycarbonate coating, to make trusses from a profile pipe (square or rectangle) there is a question about my work. You won't be fired. I say that it will not work, and the administration, together with my boss, say everything will work. How to be? 20-06-2013: Dr. Lom
22-08-2013: Dmitry
If the beam (pillow under the column) lies on dense soil (more precisely, buried below the freezing depth), then what scheme should be used to calculate such a beam? Intuition dictates that the "double-supported" option is not suitable and that the bending moment should be substantially less. 22-08-2013: Dr. Lom
The calculation of foundations is a separate big topic. In addition, it is not entirely clear what kind of beam we are talking about. If we mean a pillow under a column of a columnar foundation, then the basis for calculating such a pillow is the strength of the soil. The task of the pillow is to redistribute the load from the column to the base. The lower the strength, the larger the cushion area. Or the greater the load, the greater the cushion area with the same soil strength. 23-08-2013: Dmitry
This refers to a pillow 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 in question. 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 consider the bending moment in a pillow lying on the ground, as in a beam on two hinged supports. The question is, according to which design scheme to calculate the bending moment in the pillow. 24-08-2013: Dr. Lom
The height and section of the reinforcement in your case are determined as for cantilever beams (in width and length of the pillow). Scheme 2.1. Only in your case, the support reaction is the load on the column, more precisely, part of the load on the column, and the uniformly distributed load is the repulse of the soil. In other words, the specified design scheme must be turned over. 10-10-2013: Yaroslav
Good evening. Please help me pick up the metal. a beam for a span of 4.2 meters. A two-story residential building, the basement is covered with hollow slabs 4.8 meters long, on top of a load-bearing wall of 1.5 bricks, 3.35 m long, 2.8 m high. . on the other, 2.8 meters on the slabs, again a load-bearing wall as a floor below and above, wooden beams 20 by 20 cm, 5 m long. 6 pieces and 3 meters long, 6 pieces; floor from boards 40 mm. 25 m2. There are no other loads. Please tell me which I-beam to take in order to sleep peacefully. So far, everything has been standing for 5 years. 10-10-2013: Dr. Lom
Look in the section: "Calculation of metal structures" article "Calculation of a metal lintel for load-bearing walls" it describes in sufficient detail the process of selecting a beam section depending on the acting load. 04-12-2013: Kirill
Tell me, please, where can I get acquainted with the derivation of the formulas for the maximum beam deflection for p.p. 1.2-1.4 in Table 1 04-12-2013: Dr. Lom
The derivation of formulas for various options for applying loads is not given on my site. You can see the general principles on which the derivation of such equations is based in the articles "Fundamentals of strength, calculation formulas" and "Fundamentals of 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: Dr. Lom
I do not see any errors, and even more so non-compliance with the dimension in the calculation scheme you indicated. Please clarify what exactly is wrong. 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: Dr. Lom
No, M and Mmax have the same unit 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 threw me a problem. I ask for your help, because I do not want to solve it "by eye". 12-10-2014: Dr. Lom
It depends on many factors. In addition, you did not specify the thickness of the pipe. For example, with a thickness of 2 mm, the section modulus 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 kgcm or 69.4 kgm, then the maximum allowable 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 occur when the load is transferred not along the center of gravity of the section). And this is with a static load, and the load is likely to be dynamic, or even shock (depending on the design of the sofa and the activity of the children, mine jump on the sofas in such a way that it takes your breath away), so consider for yourself. The article "Calculated values for rectangular profile pipes" will help you. 20-10-2014: student
Doc, please help. 21-10-2014: Dr. Lom
To begin with, a rigidly fixed beam and supporting 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 articulated beam with cantilevers (see Table 3), or a three-span rigidly supported beam with 2 additional supports and unequal spans (in this case, the equations of three moments 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, the armored belt is 200x300h, the outer perimeter is 4400x4400. 3 channels are anchored into it, with a step of 1 m. The span is without racks, one of them is the heaviest option, the load is asymmetric. THOSE. consider the beam as hinged? 21-10-2014: Dr. Lom
22-10-2014: student
in fact yes. As I understand it, the deflection of the channel will turn the armo-belt itself at the attachment point, so you get a hinged beam? 22-10-2014: Dr. 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 that occurs when a uniformly distributed load acts on a certain section of the beam. And only then add up the values of the moments. Each of the loads will have its own calculation scheme. 07-02-2015: Sergey
Isn't there an error in the Mmax formula for case 2.3 in Table 3? A beam with a console, probably a plus instead of a minus should be in brackets 07-02-2015: Dr. Lom
No, not a mistake. The load on the console reduces the moment in the span, but does not increase it. However, this can also be seen from the diagram of moments. 17-02-2015: Anton
Hello, first of all, thanks for the formulas, saved in bookmarks. Tell me, please, there is a beam over the span, four logs lie on the beam, distances: 180mm, 600mm, 600mm, 600mm, 325mm. I figured out the diagram, the bending moment, I can’t understand how the deflection formula will change (table 1, scheme 1.4), if the maximum moment is on the third lag. 17-02-2015: Dr. Lom
I have already answered several times similar questions in the comments to the article "Design schemes for statically indeterminate beams". But you're in luck, for clarity, I performed the calculation according to 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 I will supplement it with time. 22-02-2015: Novel
Doc, I can’t master all these formulas that are incomprehensible to me at all. Therefore, I ask you for help. I want to make a cantilever staircase in the house (to brick steps made of reinforced concrete when building a 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 understand intuitively that the reinforcement will be "something thicker" so that the steps are something thinner? But will reinforced concrete up to 3 cm thick cope with a load of 150 kg at the edge? Please help me, I don't want to be fooled. I would be very grateful if you could help... 22-02-2015: Dr. Lom
The fact that you cannot master fairly simple formulas is your problem. In the "Fundamentals of Sopromat" section, all this is chewed in sufficient detail. Here I will say that your project is absolutely not real. 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 the 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 should be at least 15 mm in beams. 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) needs to be selected for calculating a balcony slab, a cantilever pinched 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: Dr. 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 schemes, you can count on these schemes. 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. I understand if you count on 2 tables. scheme 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 on the ends of the plate? And according to scheme 2.1, I count the moment on the support, plus it to the moment according to scheme 1.1, and according to the correct one, in order to reinforce, I need to find the moment in the span. If I have a slab overhang of 1.45m (clear), how can I calculate "x" to find the moment in the span? 03-04-2015: Dr. Lom
The moment in the span will change from Ql on the support to 0 at the load application point, which can be seen from the moment diagram. If you have a load applied at two points at the ends of the slab, then in this case it is more advisable to provide beams that perceive loads at the edges. At the same time, the slab can already be calculated as a beam on two supports - beams or a slab with support 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, the letter "G". What then calculation scheme should be used? 04-04-2015: Dr. 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: Dr. Lom
I will not evaluate the reliability of such a design without calculations, but you can calculate it according to the following criteria: 05-06-2015: student
Doc, where can I show you a picture? 05-06-2015: student
Did you still have a forum? 05-06-2015: Dr. Lom
There was, but I have absolutely no time to rake up spam in search of normal questions. Therefore, so far. 06-06-2015: student
Doc, my link is https://yadi.sk/i/GardDCAEh7iuG 07-06-2015: Dr. Lom
The choice of design scheme will depend on what you want: simplicity and reliability, or approximation to the real work of the structure through successive approximations. 07-06-2015: student
Doc, thanks. I want simplicity and reliability. This section is the busiest. I even thought about tying the tank stand to tighten the rafters to reduce the load on the ceiling, given that the water will be drained for the winter. I can't get into such a jungle of calculations. In general, the console will reduce deflection? 07-06-2015: student
Doc, another question. the console is obtained in the middle of the span of the window, does it make sense to move to the edge? Sincerely 07-06-2015: Dr. Lom
In the general case, the console will reduce the deflection, but as I said, how much in your case is a big question, and the shift to the center of the window opening will reduce the role of the console. And yet, if this is your most loaded section, then maybe just strengthen the beam, for example, with another of the same channel? I don’t know your loads, but the load from 100 kg of water and half the weight of the tank does not seem so impressive to me, but do the 8P channel in terms of deflection at 4 m span 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 hinged beam. If there is a large dynamics when walking, I constructively lay the possibility of reducing the pitch of the floor beams. The cottage is a country house, so the dynamics are tolerable. The lateral displacement of the channels has a greater effect, but this is treated by installing cross braces or fixing the deck. The only thing is, will the concrete pour fall? I assume its support on the upper and lower shelves of the channel plus welded reinforcement in the ribs and a mesh on top. 08-06-2015: Dr. Lom
I already told you, you should not count on the console. 09-06-2015: student
Doc, I get it. 29-06-2015: Sergey
Good afternoon. I would like to ask you about: the foundation was cast: piles of concrete 1.8 m deep, and then a tape 1 m deep was cast with concrete. The question is: is the load transferred only to the piles or is it evenly distributed to both the piles and the belt? 29-06-2015: Dr. Lom
As a rule, piles are made in soft soils so that the load on the base is transferred through the piles, therefore, pile grillages are calculated as 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 is a conventional strip foundation. More or less like this. 29-06-2015: Sergey
Thank you. Just a mixture of clay and sand is obtained on the site. Moreover, the layer of clay is very hard: the layer can only be removed with a crowbar, etc., etc. 29-06-2015: Dr. Lom
I don't know all your conditions (distance between piles, number of storeys, etc.). According to your description, it turns out that you made the usual strip foundation and piles for reliability. Therefore, it is enough for you 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! I need your help with the calculation. A metal collar 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 bricks (pillar 38 by 38 cm). What section and thickness should the pipe be so that there is no bend? 05-07-2015: Dr. Lom
You correctly assumed that your post should be treated like a cantilever beam. And even with the design scheme, you almost guessed it. 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 "Determining the pull-out force (why the dowel does not hold in the wall)". Thus, in your case, you should perform 2 deflection calculations according to the calculation scheme 1.2, and then add the results, 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: Dr. Lom
01-08-2015: Paul
Can you please tell me how to determine the deflection at point C in diagram 2.2 of table 3 if the lengths of the cantilever sections are different? 01-08-2015: Dr. Lom
In this case, you need to go through a full cycle. Whether this is necessary or not, I don't know. For an example, see the article on the calculation of a beam for 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 the concrete of this metal cantilever beam 120x120x4 mm with a collar of 70 kg. - (for example, at least 1/3 of the length) 04-08-2015: Dr. Lom
In fact, the calculation of pinching is a separate big topic. The fact is that the resistance of concrete to compression is one thing, and the deformation of the soil on which the foundation concrete presses is 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, the metal gate post will be poured into a concrete pile with a diameter of 300 mm and a length of 1 m, and the piles along the top will be connected by a concrete grillage with a reinforcing cage? concrete everywhere M 300. Ie. there will be no deformation of the soil. I would like to know an approximate, albeit with a large margin of safety, ratio. 05-08-2015: Dr. Lom
Then really 1/3 of the length should be enough to create a hard pinch. For an example, look at the article "Types of supports, which design scheme to choose." 05-08-2015: Yuri
20-09-2015: Karla
21-09-2015: Dr. Lom
You can first calculate the beam separately for each load according to the design schemes presented here, and then add the results, taking into account the signs. 08-10-2015: Natalia
Hello, Doctor))) 08-10-2015: Dr. 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 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 can 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 around the house. Pergolas, awnings, supports. I’ll try to remember that at one time I overslept diligently and then accidentally passed it to the Sov. VTUZ. 27-11-2015: Michael
Aren't all dimensions in SI? (see comment 08-06-2013 from Vitaly) 27-11-2015: Dr. Lom
Which units you will use kgf or Newtons, kgf / cm ^ 2 or Pascals, does not matter. As a result, you will still get centimeters (or meters) at the output. See comment 09-06-2013 from Dr. Loma. 28-04-2016: Denis
Hello, I have a beam according to scheme 1.4. what is the formula for finding shear force 28-04-2016: Dr. Lom
For each section of the beam, the values of the transverse force will be different (which, however, can be seen from the corresponding diagram of the transverse forces). On the first section 0< x < a, поперечная сила будет равна опорной реакции А. На втором участке a < x < l-b, поперечная сила будет равна А-Q и так далее, больше подробностей смотрите в статье "Основы сопромата. Расчетные формулы". 31-05-2016: Vitaly
Thank you very much, you are a great guy! 14-06-2016: Denis
While I stumbled upon your site. I almost missed the calculations, I always thought that a cantilever beam with a load at the end of the beam would sag more than with a uniformly distributed load, and formulas 1.1 and 2.1 in table 2 show the opposite. Thanks for your work 14-06-2016: Dr. Lom
In fact, it makes sense to compare a concentrated load with a uniformly distributed load only when one load is reduced to another. For example, at Q = ql, the formula for determining the deflection according to the 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 just a uniformly distributed load. So the formulas for the design schemes 1.1 and 2.1 show nothing to the contrary, and initially you were right. 16-06-2016: Garin engineer
good afternoon! I still can’t figure it out, I’ll be very grateful if you help me figure it out once and for all, when calculating (any) an ordinary I-beam with a normal distributed load along the length, which moment of inertia to use - Iy or Iz and why? I can’t find a strength of materials in any textbook - everywhere they write that the section should tend to a square and you need to take the smallest moment of inertia. I just can’t grasp the physical meaning by the tail - can I somehow interpret it on my fingers? 16-06-2016: Dr. Lom
I advise you to first look at the articles "Fundamentals of Strength Material" and "On the Calculation of Flexible Rods for the Action of a Compressive Eccentric Load", everything is explained in sufficient detail and clearly there. Here I will add that it seems to me that you are confusing calculations for transverse and longitudinal bending. Those. when the load is perpendicular to the neutral axis of the bar, 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 the longitudinal bend on the bearing capacity of the bar. Of course, when calculating for a transverse load (vertical load for a horizontal beam), the moment of inertia should be taken depending on what position the beam has, but in any case it will be Iz. And when calculating for 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, such a question why in table 1 for formulas 1.3 and 1.4 the deflection formulas are essentially the same and the size b. in formula 1.4 is not reflected in any way? 23-06-2016: Dr. Lom
With an asymmetric load, the deflection formula for the design scheme 1.4 will be quite cumbersome, but it should be remembered that the deflection in any case will be less than when a symmetrical load is applied (of course, under the condition b 03-11-2016: Vladimir
in table 1 for formulas 1.3 and 1.4 of the deflection formula, instead of Qa ^ 3 / 24EI, there should be Ql ^ 3 / 24EI. For a long time I could not understand why the deflection with the crystal does not converge 03-11-2016: Dr. Lom
That's right, another typo due to inattentive editing (I hope the last one, but not the fact). Corrected, thanks for your concern. 16-12-2016: Ivan
Hello Doctor Lom. The question is the following: I was looking through a photo from the construction site and noticed one thing: a reinforced concrete factory jumper 30 * 30 cm approximately, supported by a three-layer reinforced concrete panel by 7 centimeters. (The reinforced concrete panel was slightly filed to rest the jumper 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 already 16-12-2016: Dr. 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 it is necessary to count, of course. 25-12-2016: Ivan
Doctor, and if we assume, well, purely theoretically 25-12-2016: Dr. Lom
I don't think anything will happen even in this case. But I repeat, for a more accurate answer, a calculation is needed. 09-01-2017: Andrew
In Table 1, in formula 2.3, instead of "q", "Q" is indicated for calculating the deflection. Formula 2.1 for calculating the deflection, being a special case of formula 2.3, when the corresponding values (a=c=l, b=0) are inserted, it takes on a different form. 09-01-2017: Dr. 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 the reference book of Fesik S.P., as the shortest for the particular 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 the deflection using the initial parameters method. 02-03-2017: Dr. Lom
In tutorials, as far as I know, such a special case is not considered. Only software, for example, Lira, will help here. 24-03-2017: Eageniy
Good afternoon in the deflection formula 1.4 in the first table - the value in brackets always turns out to be negative 24-03-2017: Dr. Lom
That's right, in all the above formulas, the negative sign in the deflection formula means that the beam bends down along the y axis. 29-03-2017: Oksana
Good afternoon Dr. 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 the calculation from you with examples. I have a metal beam hinged, one edge is cantilevered and a concentrated load comes to it, and distributed over the entire beam from the reinforced concrete. 100mm thin slab and wall fencing. This beam is extreme. 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 if there will be a torque, if so, how to find it and calculate the beam section in connection with it? Dr. Lom Victor, emotional strokes are certainly good, but you can’t spread them on bread and you can’t feed your family with them. Calculations are required to answer your question, calculations are time, and time is not emotional strokes. 13-11-2017: 1
In table 2, example No. 1.1, there is an error in the formula for theta (x) 04-06-2019: Anton
Hello, dear doctor, I have a question about the method of initial parameters. At the beginning of the article, you wrote that the beam deflection formula can be obtained by properly integrating the bending moment equation twice, dividing the result by EI and adding to this the result of integrating the angle of rotation. 05-06-2019: Dr. Lom
The mistake is that you did not integrate the equation of moments, but the result of solving this equation for a point in the middle of the beam, and these are different things. In addition, when adding, you should carefully monitor the sign "+" or "-". If you carefully analyze the deflection formula given for this design scheme, you will understand what we are talking about. And when integrating the angle of rotation, the result is q * l4 / 48, and not q * l4 / 96, and in the final formula it will go with a minus, since such an initial angle of rotation will lead to deflection of the beam below the x axis. 09-07-2019: Alexander
Greetings, in T.1 2.3 formulas for moments what is taken as X? The middle of the distributed load? 09-07-2019: Dr. Lom
For all tables, the distance x is the distance from the point of origin (usually support A) to the considered point on the neutral axis of the beam. Those. the above formulas allow you to determine the value of the moment for any cross section of the beam. The calculation of a beam for bending "manually", in an old-fashioned way, allows you to learn one of the most important, beautiful, clearly mathematically verified algorithms of the science of the strength of materials. The use of numerous programs such as "entered the initial data ... ...– get an answer” allows the modern engineer today to work much faster than his predecessors a hundred, fifty and even twenty years ago. However, with such a modern approach, the engineer is forced to fully trust the authors of the program and eventually ceases to "feel the physical meaning" of the calculations. But the authors of the program are people, and people 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 sometimes be able to "manually" check the results of calculations. Help (cheat sheet, memo) for calculating beams for bending is shown below in the figure. We draw a diagram for calculating the beam for bending. Obviously, the most dangerous scheme of applying an external load will be when I start to pull myself up, clinging to the middle of the crossbar with one hand. F1 \u003d 900 n - the force acting on the beam (my weight) without taking into account the dynamics d \u003d 32 mm - the outer diameter of the bar from which the beam is made E = 206000 n/mm^2 is the modulus of elasticity of the St3 steel beam material [σi] = 250 n/mm^2 - allowable bending stresses (yield strength) for the material of the St3 steel beam 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, we 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 section (which is the section of the bar): 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 compose equilibrium equations for calculating 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's 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)=0mm z = 0.6 m: Qy (0.6) = -R1 = -450 n Mx (0.6) \u003d -R1 * (0.6-b1) \u003d -450 * (0.6-0) \u003d -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 sag 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 Shear 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 Мx (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 with the allowable stresses: σi \u003d Mx max / Wx \u003d (270 * 1000) / (3.217 * 1000) \u003d 84 n / mm ^ 2 σi = 84 n/mm^2< [σи] = 250 н/мм^2 In terms of bending strength, the calculation showed a threefold margin of safety - the horizontal bar can be safely made from an existing bar with a diameter of thirty-two millimeters and a length of one thousand two hundred millimeters. Thus, you can now easily calculate the beam for bending "manually" and compare with the results obtained in the calculation using any of the numerous programs presented on the Web. I ask those who RESPECT the work of the author to SUBSCRIBE to the announcements of articles.
88 comments on "Calculation of a beam for bending - "manually"!"
Do you mean that for rotating shafts, the circuits will be different due to the torque? I don’t know how important this is, since it is written in the technical machine book 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?
The calculation of building structures, machine parts, etc., as a rule, consists of two stages: 1. calculation for the limit states of the first group - the so-called strength calculation, 2. calculation for the limit states of the second group. One of the types of calculation for the limit states of the second group is the calculation for deflection.
In your case, in my opinion, the calculation of strength will be more important. Moreover, today there are 4 theories of strength and the calculation for each of these theories is different, but in all theories, the influence of both bending and torque is taken into account in the calculation.
The deflection under the action of a torque occurs in a different plane, but is still taken into account in the calculations. And if this deflection is small or large - the calculation will show.
I do not specialize in calculations of parts of machines and mechanisms, and therefore I cannot point to authoritative literature on this issue. However, in any handbook of a design engineer of machine components and parts, this topic should be properly disclosed.
mail: [email protected]
Skype: dmytrocx75
Q - in kilograms.
l - in centimeters.
E - in kgf/cm2.
I - cm4.
All right? Something strange results are obtained.
If we are talking about a grillage, then, depending on the method of its installation, it can be calculated 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 pillow. This should be taken into account in calculations.
But I repeat once again, do not self-medicate, be guided by the requirements of the specified SNiP.
However, in the cases you indicated (except for 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 issue was discussed in the topic "Design schemes for statically indeterminate beams", look there.
The essence of the problem is as follows: at the base of the sofa, a metal frame is planned from a profiled pipe 40x40 or 40x60, lying on two supports, the distance between which is 2200 mm. QUESTION: is the section of the profile enough for loads from the own weight of the sofa + let's take 3 people of 100 kg each ???
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 section 0-2 m, plus concentrated 300 kg in the middle (for 2 m). I determined the support reactions: A - 0.5 t; B - 0.4 tons. Then I hung: 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 there is a moment on the supports.
How are the loads calculated in this case? It is necessary to bring all distributed loads to concentrated ones and summarize (subtract * distance from the support reaction) according to the formulas of the design scheme? In your article about farms, the layout of all forces is clear, but here I cannot enter into the methodology for determining the acting forces.
The maximum moment 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 master all this, then it is better to contact a professional designer - it will be cheaper.
Tell me, please, according to which scheme it is necessary to calculate the beam deflection of such a mechanism https://yadi.sk/i/MBmS5g9kgGBbF. Or maybe, without going into calculations, tell me if a 10 or 12 I-beam is suitable for an arrow, a maximum load of 150-200 kg, a lifting height of 4-5 meters. Rack - pipe d = 150, rotary mechanism or axle shaft, or front hub of the Gazelle. 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 (extreme 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 just a single-span beam with a cantilever. The first support is the cable attachment point, the second is the stand. Then your design schemes are 1.1 (for the load - live load) and 2.3 (boom dead weight - constant 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 the temporary load you will have is not static, but at least dynamic (see the article "Calculation for shock loads").
3. To determine the forces in the cable, it is necessary 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 - a rigid pinch at the bottom (see the article "Calculation of metal columns"). This column will be loaded with a very large eccentricity if there is no counterweight.
5. The calculation of the junctions of the boom and rack and other subtleties of the calculation of the nodes of machines and mechanisms on this site are not yet considered.
what design scheme is ultimately obtained for the floor beam and the cantilever beam, and will the (pink) cantilever beam (brown) affect the decrease in the deflection of the floor beam?
wall - foam block D500, height 250, width 150, armo-belt beam (blue): 150x300, reinforcement 2x? concrete columns 200x200 in the corners, the span of the armo-belt beam 4000 without walls.
overlap: channel 8P (pink), for calculation I took 8U, welded and anchored with armo-belt beam reinforcement, concreted, from the bottom of the beam to the channel 190 mm, from the top 30, span 4050.
to the left of the console - an opening for the stairs, the support of the channel on the pipe? 50 (green), the span to the beam 800.
to the right of the console (yellow) - a bathroom (shower, toilet) 2000x1000, floor - pouring a reinforced ribbed transverse slab, dimensions 2000x1000 height 40 - 100 on fixed formwork (profiled sheet, wave 60) + tiles on glue, walls - drywall on profiles. The rest of the floor is board 25, plywood, linoleum.
At the points of the arrows, the support of the racks of the water tank, 200l.
Walls of the 2nd floor: sheathing with board 25 on both sides, with insulation, height 2000, leaning on the armored belt.
roof: rafters - a triangular arch with a puff, along the floor beam, with a step of 1000, resting on the walls.
console: channel 8P, span 995, welded with reinforced reinforcement, concreted into a beam, welded to the floor channel. span to the right and left along the floor beam - 2005.
While I am cooking the reinforcing cage, it is possible to move the console left and right, but there seems to be nothing 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, should not be taken into account at all. That's actually the whole calculation.
Further, in order to simply go to a beam with rigid pinching on the extreme supports, you must first calculate the armo-belt for the action of torque and determine the angle of rotation of the cross-section of the armo-belt, taking into account the load from the walls of the 2nd floor and deformations of the wall material under the action 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 does not rest on the foundation, but on the reinforced concrete slab (as I understood from the figure) and this slab will deform. And the pipe itself will experience compression deformation.
In the second case, if you want to take into account the possible operation 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 armo-belt at the place 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. Yes, and the load on the beam as a window overlap will have to be recalculated, but you have such examples: The only load to take 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 lower 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).
On the other hand, there are no problems - a corner on the mortgages in the body of the beam. I have not yet coped with the calculation of a two-span beam with different spans and different loads, I will try to re-study your article on the calculation of a multi-span beam by the method of moments.
I calculated according to the table. 2, item 1.1. (#comments) as a deflection of a cantilever beam with a load of 70 kg, a shoulder of 1.8 m, a square pipe 120x120x4 mm, a moment of inertia of 417 cm4. I got a deflection - 1.6 mm? True or not?
P.S. And I do not check the accuracy of the calculations, then only 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 the 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 by this formula should be compared with the maximum allowable 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 source from which these tables are taken - can I indicate the name?
When you compare the result of deflection in a span with SNiPovksky, then the span length is the distance l between A and B. For the console, instead of l, the distance 2a (double extension of the console) is taken.
I compiled these tables myself, using various reference books on the theory of strength of materials, while checking the data for possible typographical errors, as well as general methods for calculating beams, when there were no diagrams necessary in my opinion in the reference books, so there are many primary sources.
that the reinforcement in the armored belt above the beam is completely destroyed, the armored belt will crack and lie on the beam along with the floor slabs? Will these 7 cm of the support platform be enough?
Suppose I do not know the deflection of the beam of the design scheme 2.1 (Table 1). I will integrate the bending moment twice ∫q*l2/8dx=q*l3/24;∫q*l3/24dx=q*l4/96.
After I divide the value by EI. q*l4/(96*EI).
And I will add to it the result of integrating the angle of rotation - ∫q*l3/24dx=q*l4/96. q*l4/(96*EI)+q*l4/(96*EI)=q*l4/(48*EI).
You get the value -5*q*l4/(384*EI).
Tell me please. Where did I make a mistake?
Let's use a simple everyday example to try to use it. Let's say I decided to make a horizontal bar in the apartment. A place has been determined - a corridor one meter twenty centimeters wide. On opposite walls at the required height opposite each other, I securely fasten the brackets to which the beam-beam will be attached - a bar of St3 steel with an outer diameter of thirty-two millimeters. Will this beam support my weight plus additional dynamic loads that will arise during exercise?
Initial data:
Related articles
Reviews
Read also...
