Bend. Pure bending Technical mechanics transverse bending solutions
Calculate bending beam There are several options:
1. Calculation of the maximum load that it will withstand
2. Selection of the section of this beam
3. Calculation based on maximum permissible stresses (for verification)
let's consider general principle 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 voltage in the 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).
For a cantilever beam loaded with a distributed load of intensity kN/m and a concentrated moment of kN m (Fig. 3.12), it is required: construct diagrams of shear forces and bending moments, select a round beam cross section at permissible normal stress kN/cm2 and check the strength of the beam by shear stress at permissible shear stress kN/cm2. Beam dimensions m; m; m.
Calculation scheme for the problem of direct transverse bending
Rice. 3.12
Solution of the problem "straight transverse bending"
Determining support reactions
The horizontal reaction in the embedment is zero, since external loads in the z-axis direction do not act on the beam.
We select the directions of the remaining reaction forces arising in the embedment: we will direct the vertical reaction, for example, downward, and the moment – clockwise. Their values are determined from the static equations:
When composing these equations, we consider the moment to be positive when rotating counterclockwise, and the projection of the force to be positive if its direction coincides with the positive direction of the y-axis.
From the first equation we find the moment at the seal:
From the second equation - vertical reaction:
Received by us positive values for the moment and vertical reaction in the embedment indicate that we 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 will outline four cross sections (see Fig. 3.12), in which we will use the method of sections (ROZU) to calculate the values of shearing forces and bending moments.
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, let’s cover the discarded right side of the beam with a piece of paper, aligning the left edge of the sheet with the section under consideration.
Let us 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 being considered (that is, visible) by us. Therefore, the shearing force must be equal to the algebraic sum of all the forces that we see.
Let us also present the rule of signs for the shearing force: an external force acting on the part of the beam under consideration 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 part of the beam visible to us 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 the external forces visible to us relative to the section in question. Consequently, it is equal to the algebraic sum of the moments of all forces 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). Wherein external load, bending the part of the beam under consideration with a convex downward direction, causes a positive bending moment in the section. And the moment created by such a load is included in the algebraic sum for determination with a “plus” sign.
We see two efforts: reaction and closing moment. However, the force's leverage relative to section 1 is zero. That's why
kNm.
We took the plus sign because reactive torque bends the part of the beam visible to us with a convex downward.
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; kNm.
Section 3. Closing the right side of the beam, we find
kN;
Section 4. Cover the left side of the beam with a sheet. Then
kNm.
kNm.
.
Using the found values, we construct diagrams of shearing forces (Fig. 3.12, b) and bending moments (Fig. 3.12, c).
Under unloaded areas, the diagram of shearing forces goes parallel to the axis of the beam, and under a distributed load q - along an inclined straight line upward. Under the support reaction in the diagram there is a jump down by the value of this reaction, that is, by 40 kN.
In the diagram of bending moments we see a break under the support reaction. The bend angle is directed towards the support reaction. 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.
Determine the required cross-sectional diameter of the beam
Strength condition according to normal voltages has the form:
,
where is the moment of resistance of the beam during bending. For a beam of circular cross-section it is equal to:
.
The largest absolute value bending moment 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 by the highest shear stresses
The largest shear stresses arising in the cross section of the beam round section, are calculated by the formula
,
where is the cross-sectional area.
According to the diagram, the largest algebraic value of the shearing force is equal to 
kN. Then
kN/cm2 kN/cm2,
that is, the strength condition for tangential stresses is also satisfied, and with a large margin.
An example of solving the problem "straight transverse bending" No. 2
Condition of an example problem on straight transverse bending
For a simply supported beam loaded with a distributed load of intensity kN/m, concentrated force kN and concentrated moment kN m (Fig. 3.13), it is necessary to construct diagrams of shear forces and bending moments and select a beam of I-beam cross-section with an allowable normal stress kN/cm2 and permissible tangential stress kN/cm2. Beam span m.
An example of a straight bending problem - calculation diagram
 |
Rice. 3.13
Solution of an example problem on straight bending
Determining support reactions
For a given simply supported beam, it is necessary to find three support reactions: , and . Since only vertical loads perpendicular to its axis act on the beam, the horizontal reaction of the fixed hinged support A is zero: .
The directions of vertical reactions are chosen arbitrarily. Let us direct, for example, both vertical reactions upward. To calculate their values, let’s create two static equations:
Let us recall that the resultant of the 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.
Let's 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 true.
We construct diagrams of shearing forces and bending moments
We divide the length of the beam into separate sections. The boundaries of these sections 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 sections in our problem. Along the boundaries of these sections, we will outline six cross sections, in which we will calculate the values of shearing 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 shearing force and bending moment arising in this section, we will cover the part of the beam we discarded with a piece of paper, aligning the left edge of the sheet 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 infinitesimal length. The resultant linear load is zero. That's why
kN.
The plus sign is taken because the force rotates the part of the beam visible to us relative to the first section (the edge of a piece of paper) clockwise.
The bending moment in the beam section 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 the piece of paper). We see the support reaction and linear load q distributed over an infinitesimal length. However, the force has a leverage of zero. The resultant linear load is also 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 load q acting on a section of length . The resultant linear load is equal to . It is attached in the middle of a section of length . That's why
Let us recall that when determining the sign of the bending moment, we mentally free the part of the beam we see from all the actual supporting fastenings and imagine it as if pinched in the section under consideration (that is, we mentally imagine the left edge of the piece of paper as a rigid embedment).
Section 3. Close the right side. We get
Section 4. Cover the right side of the beam with a sheet. Then
Now, to check 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 infinitesimal length. The resultant linear load is zero. That's why
kNm.
That is, everything is correct.
Section 5. As before, close the left side of the beam. Will have
kN;
kNm.
Section 6. Let's close the left side of the beam again. We get
kN;
Using the found values, we construct diagrams of shearing forces (Fig. 3.13, b) and bending moments (Fig. 3.13, c).
We make sure that under the unloaded area the diagram of shearing forces runs parallel to the axis of the beam, and under a distributed load q - along a straight line sloping downwards. There are three jumps in 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.
In 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 shearing force for this section passes through the zero value (). Let us determine the distance from section 7 to the left support.
Task. Construct diagrams Q and M for a statically indeterminate beam. Let's calculate the beams using the formula:
n= Σ R- Sh— 3 = 4 — 0 — 3 = 1
Beam once is statically indeterminate, which means one of the reactions is "extra" unknown. Let us take the support reaction as the “extra” unknown IN — R B.
A statically determinate beam, which is obtained from a given one by removing the “extra” connection, is called the main system (b).
Now this system should be presented equivalent given. To do this, load the main system given load, and at the point IN let's apply "extra" reaction R B(rice. V).
However for equivalence this not enough, since in such a beam the point IN Maybe move vertically, and in a given beam (Fig. A ) this cannot happen. Therefore we add condition, What deflection t. IN in the main system should be equal to 0. Deflection t. IN consists of deflection from effective load Δ F and from deflection from the “extra” reaction Δ R.
Then we make up condition for compatibility of movements:
Δ F + Δ R=0 (1)
Now it remains to calculate these movements (deflections).
Loading main system given load(rice .G) and we'll build load diagramM F (rice. d ).
IN T. IN Let's apply and build an ep. (rice. hedgehog ).
Using Simpson's formula we determine deflection due to active load.
Now let's define deflection from the action of “extra” reaction R B , for this we load the main system R B (rice. h ) and build a diagram of the moments from its action M R (rice. And ).
We compose and solve equation (1):
Let's build ep. Q
And M
(rice. k, l
).
Building a diagram Q.
Let's build a diagram M method characteristic points. We place points on the beam - these are the points of the beginning and end of the beam ( D,A ), concentrated moment ( B ), and also mark the middle of a uniformly distributed load as a characteristic point ( K ) is an additional point for constructing a parabolic curve.
We determine bending moments at points. Rule of signs cm. - .
The moment in IN we will define it as follows. First let's define:
Full stop TO let's take in middle area with a uniformly distributed load.
Building a diagram M . Plot AB – parabolic curve(umbrella rule), area ВD – straight slanted line.
For a beam, determine the support reactions and construct diagrams of bending moments ( M) And shear forces (Q).
- We designate supports letters A And IN and direct support reactions R A And R B .
Compiling equilibrium equations.
Examination
Write down the values R A And R B on design scheme.
2. Constructing a diagram shear forces method sections. We arrange the sections on characteristic areas(between changes). According to the dimensional thread - 4 sections, 4 sections.
sec. 1-1 move left.
The section passes through the area with evenly distributed load, mark the size z 1 to the left of the section before the start of the section. The length of the section is 2 m. Rule of signs For Q - cm.
We build according to the found value diagramQ.
sec. 2-2 move to the right.
The section again passes through the area with a uniformly distributed load, mark the size z 2 to the right from the section to the beginning of the section. The length of the section is 6 m.
Building a diagram Q.
sec. 3-3 move on the right.
sec. 4-4 move on the right.
We are building diagramQ.
3. Construction diagrams M method characteristic points.
Feature point- a point that is somewhat noticeable on the beam. These are the points A, IN, WITH, D , and also a point TO , wherein Q=0 And bending moment has an extremum. also in middle console we will put an additional point E, since in this section under a uniformly distributed load the diagram M described crooked line, and it is built at least according to 3 points.
So, the points are placed, let's start determining the values in them bending moments. Rule of signs - see.
Sites NA, AD – parabolic curve(the “umbrella rule” for mechanical specialties or the “sail rule” for construction specialties), sections DC, SV – straight slanted lines.
Moment at a point D should be determined both left and right from point D . The very moment in these expressions Excluded. At the point D we get two values with difference by the amount m – leap by its size.
Now we need to determine the moment at the point TO (Q=0). However, first we define point position TO , designating the distance from it to the beginning of the section as unknown X .
T. TO belongs second characteristic area, its equation for shear force(see above)
But the shear force incl. TO equal to 0 , A z 2 equals unknown X .
We get the equation:
Now knowing X, let's determine the moment at the point TO on the right side.
Building a diagram M . The construction can be carried out for mechanical specialties, putting aside positive values up from the zero line and using the “umbrella” rule.
For a given design of a cantilever beam, it is necessary to construct diagrams of the transverse force Q and the bending moment M, and perform a design calculation by selecting a circular section.
Material - wood, design resistance of the material R=10MPa, M=14kN m, q=8kN/m
There are two ways to construct diagrams in a cantilever beam with a rigid embedment - the usual way, having previously determined the support reactions, and without determining the support reactions, if you consider the sections, going from the free end of the beam and discarding the left part with the embedment. Let's build diagrams ordinary way.
1. Let's define support reactions.
Evenly distributed load q replace with conditional force Q= q·0.84=6.72 kN
In a rigid embedment there are three support reactions - vertical, horizontal and moment; in our case, the horizontal reaction is 0.
We'll find vertical ground reaction R A And supporting moment M A from equilibrium equations.
In the first two sections on the right there is no shear force. At the beginning of a section with a uniformly distributed load (right) Q=0, in the background - the magnitude of the reaction R A.
3. To construct, we will compose expressions for their determination in sections. Let's construct a diagram of moments on fibers, i.e. down. 
(the diagram of individual moments has already been constructed earlier)
We solve equation (1), reduce by EI
Static indetermination revealed, the value of the “extra” reaction has been found. You can start constructing diagrams of Q and M for a statically indeterminate beam... We sketch the given diagram of the beam and indicate the magnitude of the reaction Rb. In this beam, reactions in the embedment can not be determined if you move from the right.
Construction Q plots for a statically indeterminate beam
Let's plot Q.
Construction of diagram M
Let us define M at the extremum point - at the point TO. First, let's determine its position. Let us denote the distance to it as unknown “ X" Then
We are building a diagram of M.
Determination of shear stresses in an I-section. Let's consider the section I-beam S x =96.9 cm 3; Yх=2030 cm 4 ; Q=200 kN
To determine the shear stress, it is used formula
,where Q is the shear 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 the tangential 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
Let's calculate maximum shear stress:
Let us calculate the static moment for top shelf: 
Now let's calculate shear stress: 
We are building shear stress diagram:
Design and verification calculations. For a beam with constructed diagrams of internal forces, select a section in the form of two channels from the condition of strength under normal stresses. Check the strength of the beam using the shear stress strength condition and the energy strength criterion. Given:
Let's show a beam with constructed diagrams Q and M 
According to the diagram of bending moments, it is dangerous section C, in which M C = M max = 48.3 kNm.
Normal stress strength condition for this beam has the form σ max =M C /W X ≤σ adm . It is necessary to select a section from two channels.
Let's determine the required calculated value axial moment of resistance of the section:
For a section in the form of two channels, we accept according to two channels No. 20a, moment of inertia of each channel I x =1670cm 4, Then axial moment of resistance of the entire section:
Overvoltage (undervoltage) at dangerous points we calculate using the formula: Then we get undervoltage:
Now let's check the strength of the beam based on strength conditions for tangential stresses. According to shear force diagram dangerous are sections on section BC and section D. As can be seen from the diagram, Q max =48.9 kN.
Strength condition for tangential stresses has the form:
For channel No. 20 a: static moment of area S x 1 = 95.9 cm 3, moment of inertia of the section I x 1 = 1670 cm 4, wall thickness d 1 = 5.2 mm, average flange thickness t 1 = 9.7 mm , channel height h 1 =20 cm, shelf width b 1 =8 cm.
For transverse sections of two channels:
S x = 2S x 1 =2 95.9 = 191.8 cm 3,
I x =2I x 1 =2·1670=3340 cm 4,
b=2d 1 =2·0.52=1.04 cm.
Determining the value maximum shear stress:
τ max =48.9 10 3 191.8 10 −6 /3340 10 −8 1.04 10 −2 =27 MPa.
As seen, τ max<τ adm (27MPa<75МПа).
Hence, the strength condition is satisfied.
We check the strength of the beam according to the energy criterion.
From consideration diagrams Q and M follows that section C is dangerous, in which they operate M C =M max =48.3 kNm and Q C =Q max =48.9 kN.
Let's carry out analysis of the stress state at the points of section C
Let's define normal and shear stresses at several levels (marked on the section diagram)
Level 1-1: y 1-1 =h 1 /2=20/2=10cm.
Normal and tangent voltage:
Main voltage:
Level 2−2: y 2-2 =h 1 /2−t 1 =20/2−0.97=9.03 cm.
Main stresses:
Level 3−3: y 3-3 =h 1 /2−t 1 =20/2−0.97=9.03 cm.
Normal and shear stresses: 
Main stresses:
Extreme shear stress:
Level 4−4: y 4-4 =0.
(in the middle the normal stresses are zero, the tangential stresses are maximum, they were found in the strength test using tangential stresses)
Main stresses: 
Extreme shear stress:
Level 5−5:
Normal and shear stresses: 
Main stresses:
Extreme shear stress: 
Level 6−6:
Normal and shear stresses: 
Main stresses:
Extreme shear stress: 
Level 7−7:
Normal and shear stresses:
Main stresses:
Extreme shear stress:
In accordance with the calculations performed stress diagrams σ, τ, σ 1, σ 3, τ max and τ min are presented in Fig.
Analysis these diagram shows, which is in the section of the beam dangerous points are at level 3-3 (or 5-5), in which:
Using energy criterion of strength, we get
From a comparison of equivalent and permissible stresses it follows that the strength condition is also satisfied 
(135.3 MPa<150 МПа).
The continuous beam is loaded in all spans. Construct diagrams Q and M for a continuous beam.
1. Define degree of static indetermination beams according to the formula:
n= Sop -3= 5-3 =2, Where Sop – number of unknown reactions, 3 – number of static equations. To solve this beam it is required two additional equations.
2. Let us denote numbers supports from zero in order ( 0,1,2,3 )
3. Let us denote span numbers from the first in order ( ι 1, ι 2, ι 3)
4. We consider each span as simple beam and build diagrams for each simple beam Q and M. What pertains to simple beam, we will denote with index "0", that which relates to continuous beam, we will denote without this index. Thus, is the shear force and bending moment for a simple beam.
When calculating the strength of bending elements of building structures, the limit state calculation method is used.
In most cases, normal stresses in cross sections are of primary importance when assessing the strength of beams and frames. In this case, the highest normal stresses acting in the outermost fibers of the beam should not exceed a certain permissible value for a given material. In the limit state calculation method, this value is taken equal to the design resistance R, multiplied by the operating conditions coefficient at the village
The strength condition has the following form:
Values R And y s for various materials are given in SNiP for building structures.
For beams made of plastic material that equally resists tension and compression, it is advisable to use sections with two axes of symmetry. In this case, the strength condition (7.33), taking into account formula (7.19), is written in the form
Sometimes, for structural reasons, beams with an asymmetrical cross-section such as a T-beam, a multi-flange I-beam, etc. are used. In these cases, the strength condition (7.33), taking into account (7.17), is written in the form
In formulas (7.34) and (7.35) W z And WHM- sectional moments of resistance relative to the neutral axis Oz„ Mnb is the largest bending moment in absolute value due to the action of design loads, i.e. taking into account the load reliability coefficient y^.
The section of the beam in which the bending moment is the largest in absolute value is called dangerous section.
When calculating the strength of structural elements working in bending, the following problems are solved: checking the strength of the beam; selection of section; determination of the bearing capacity (load capacity) of the beam, those. determination of load values at which the highest stresses in the dangerous section of the beam do not exceed the value y c R.
The solution to the first problem comes down to checking the fulfillment of strength conditions under known loads, the shape and dimensions of the section and the properties of the material.
The solution to the second problem comes down to determining the dimensions of a section of a given shape under known loads and material properties. First, from the strength conditions (7.34) or (7.35), the value of the required moment of resistance is determined
and then the section dimensions are set.
For rolled profiles (I-beams, channels) based on the moment of resistance, the section is selected according to the assortment. For non-rolled sections, characteristic section dimensions are established.
When solving the problem of determining the load-carrying capacity of a beam, first, from the strength conditions (7.34) or (7.35), the value of the largest calculated bending moment is found using the formula
Then the bending moment in a dangerous section is expressed in terms of the loads applied to the beam and the corresponding load values are determined from the resulting expression. For example, for a steel I-beam 130 shown in Fig. 7.47, at R= 210 MPa, y c = 0,9, W z= 472 cm 3 we find
From the diagram of bending moments we find
Rice. 7.47
In beams loaded with large concentrated forces located close to the supports (Fig. 7.48), the bending moment M nb can be relatively small, and the shear force 0 nb in absolute value can be significant. In these cases, it is necessary to check the strength of the beam using the highest tangential stresses tnb. The strength condition for tangential stresses can be written in the form
Where R s - design resistance of the beam material in shear. Values R s for basic building materials are given in the relevant sections of SNiP.
Shear stresses can reach significant values in the webs of I-beams, especially in the thin webs of composite beams.
Calculation of shear stress strength can be critical for wooden beams, since wood does not resist chipping along the grain very well. So, for example, for pine the calculated resistance to tension and compression during bending is R= 13 MPa, and when shearing along the fibers RCK= 2.4 MPa. Such a calculation is also necessary when assessing the strength of the connection elements of composite beams - welds, bolts, rivets, dowels, etc.
The condition for the shear strength along the fibers for a wooden beam of rectangular cross-section, taking into account formula (7.27), can be written in the form
Example 7.15. For the beam shown in Fig. 7.49, A, let's build diagrams Qy And Mv Let’s select a beam section in the form of a rolled steel I-beam and draw diagrams c x and t in sections with the largest Qy And Mz. Load safety factor y f = 1.2, design resistance R= 210 MPa = 21 kN/cm 2, operating conditions coefficient y c = 1,0.
We begin the calculation by determining the support reactions:
Let's calculate the values Qy And Mz in characteristic sections of the beam.
Transverse forces within each section of the beam are constant values and have jumps in the sections under the force and at the support IN. Bending moments vary linearly. Diagrams Qy And Mz are shown in Fig. 7.49, b, c.
The dangerous section is in the middle of the beam span, where the bending moment is greatest. Let's calculate the calculated value of the largest bending moment:
The required moment of resistance is
According to the assortment, we accept section 127 and write out the necessary geometric characteristics of the section (Fig. 7.50, A):
Let's calculate the values of the highest normal stresses in the dangerous section of the beam and check its strength:
The strength of the beam is ensured.
Tangential stresses have the greatest values in the section of the beam where the largest absolute magnitude transverse force acts (2 nb = 35 kN.
Design value of shear force
Let us calculate the values of the tangential stresses in the I-beam wall at the level of the neutral axis and at the level of the interface between the wall and the flanges:
Diagrams c x and x, in section l: = 2.4 m (right) are shown in Fig. 7.50, b, c.
The sign of the tangential stresses is taken to be negative, as corresponding to the sign of the shear force.
Example 7.16. For a wooden beam of rectangular cross-section (Fig. 7.51, A) let's build diagrams Q And Mz, determine the height of the section h from the strength condition, taking R = = 14 MPa, yy= 1.4 and y c = 1.0, and check the strength of the beam for shearing on the neutral layer, taking RCK= 2.4 MPa.
Let's determine the support reactions:
Let's calculate the values Q v And Mz 
in characteristic sections of the beam.
Within the second section, the shear force becomes zero. The position of this section is found from the similarity of triangles on the diagram Q y:
Let us calculate the extreme value of the bending moment in this section:
Diagrams Qy And Mz are shown in Fig. 7.51, b, c.
The section of the beam where the maximum bending moment occurs is dangerous. Let's calculate the calculated value of the bending moment in this section:
Required section modulus
Using formula (7.20), we express the moment of resistance through the height of the section h and equate it to the required moment of resistance:
We take a rectangular section of 12x18 cm. Let us calculate the geometric characteristics of the section:
Let's determine the highest normal stresses in the dangerous section of the beam and check its strength:
The strength condition is met.
To check the shear strength of a beam along the fibers, it is necessary to determine the values of the maximum tangential stresses in the section with the largest absolute value of transverse force 0 nb = 6 kN. The calculated value of the shear force in this section
The maximum shear stresses in the cross section act at the level of the neutral axis. According to the law of pairing, they also act in the neutral layer, tending to cause a shift of one part of the beam relative to the other part.
Using formula (7.27), we calculate the value of mmax and check the shear strength of the beam:
The shear strength condition is met.
Example 7.17. For a round wooden beam (Fig. 7.52, A) let's build diagrams Q y n M z n Let us determine the required cross-section diameter from the strength condition. In calculations we will accept R= 14 MPa, yy = 1.4 and y s = 1,0.
Let's determine the support reactions: 
Let's calculate the values Q And M 7 in characteristic sections of the beam.
Diagrams Qy And Mz are shown in Fig. 7.52, b, c. The section on the support is dangerous IN with the largest bending moment in absolute value Mnb = 4 kNm. The calculated value of the bending moment in this section
Let us calculate the required moment of resistance of the section:
Using formula (7.21) for the moment of resistance of a circular cross-section, we find the required diameter:
Let's accept D= 16 cm and determine the maximum normal stresses in the beam:
Example 7.18. Let us determine the load-carrying capacity of a box-section beam 120x180x10 mm, loaded according to the diagram in Fig. 7.53, A. Let's build diagrams c x etc. in a dangerous section. Beam material - steel grade VStZ, R= 210 MPa = 21 kN/cm2, U/= U, Us =°’ 9 -
Diagrams Qy And Mz are shown in Fig. 7.53, A.
The section of the beam near the embedment is dangerous, where the bending moment M nb is the largest in absolute value. - P1 = 3,2 R.
Let's calculate the moment of inertia and moment of resistance of the box section:
Taking into account formula (7.37) and the obtained value for L/nb, we determine the calculated value of the force R:
Normative value of force
The highest normal stresses in the beam due to the design force
Let us calculate the static moment of half the section ^1/2 and the static moment of the cross-sectional area of the flange S n relative to the neutral axis:
Tangential stresses at the level of the neutral axis and at the level of the flange-wall interface (Fig. 7.53, b) are equal:
Diagrams Oh And t uh in cross section near the embedment are shown in Fig. 7.53, in, g.
When building diagrams of bending momentsM at builders accepted: ordinates expressing on a certain scale positive values of bending moments, set aside stretched fibers, i.e. - down, A negative - up from the beam axis. Therefore, they say that builders construct diagrams on stretched fibers. At the mechanics positive values of both shear force and bending moment are postponed up. Mechanics draw diagrams on compressed fibers.
Principal stresses when bending. Equivalent voltages.
In the general case of direct bending in the cross sections of a beam, normal And tangents
voltage. These voltages vary both along the length and height of the beam.
Thus, in the case of bending, there is plane stress state.
Let's consider a diagram where the beam is loaded with force P
Largest normal tensions arise in extreme, points most distant from the neutral line, and There are no shear stresses in them. Thus, for extreme fibers non-zero principal stresses are normal stresses in cross section.
At the neutral line level in the cross section of the beam there are highest shear stress, A normal stresses are zero. means in the fibers neutral layer the principal stresses are determined by the values of the tangential stresses.
In this design scheme, 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 analysis Let's imagine it in the picture.
Bending Stress Analysis
Maximum principal stress σ 1 is located upper extreme fibers and equals zero on the lower outermost fibers. Main stress σ 3 It has the largest absolute value is on the lower fibers.
Trajectory of principal stresses depends on load type And method of securing the beam.
When solving problems it is enough separately check normal And separately tangential stresses. However sometimes the most stressful turn out to be intermediate fibers in which there are both normal and shear stresses. This happens in sections where At the same time, both the bending moment and the shear 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 concentrated force, or in sections with sharply changing widths. For example, in an I-section the most dangerous the junction of the wall and the shelf- there are significant both normal and shear stresses.
The material is in a plane stress state and is required check for equivalent voltages.
Strength conditions for beams made of plastic materials By third(theory of maximum tangential stresses) 
And fourth(theory of energy of shape changes) 
theories of strength.
As a rule, in rolled beams the equivalent stresses do not exceed the normal stresses in the outermost fibers and no special testing is required. Another thing - composite metal beams, which the wall is thinner than for rolled profiles at the same height. Welded composite beams made of steel sheets are more often used. Calculation of such beams for strength: a) selection of the section - height, thickness, width and thickness of the beam chords; b) checking strength by normal and tangential stresses; c) checking strength using equivalent stresses.
Determination of shear stresses in an I-section. Let's consider the section I-beam S x =96.9 cm 3; Yх=2030 cm 4 ; Q=200 kN
To determine the shear stress, it is used formula,where Q is the shear 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 the tangential 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
Let's calculate maximum shear stress:
Let us calculate the static moment for top shelf:
Now let's calculate shear stress:
We are building shear stress diagram:
Let us consider the cross section of a standard profile in the form I-beam and define shear stress, acting parallel to the shear force:
Let's calculate static moments simple figures: 
This value can be calculated and otherwise, using the fact that for the I-beam and trough sections the static moment of half the section is given. 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: 
The tangential stresses at the junction of the flange and the wall change spasmodically, because sharp wall thickness varies from t st before b.
Diagrams of tangential 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 includes the width of the section (net) in the layer where the shear stress is determined.
Let us determine the tangential stresses for a circular section.
Since the shear stresses at the section contour must be directed tangent to the contour, then at points A And IN at the ends of any chord parallel to the diameter AB, shear stresses are directed perpendicular to the radii OA And OV. Hence, directions tangential stresses at points A, VC converge at some point N on the Y axis.
Static moment of the cut-off part: 
That is, the 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 stress (formula) 
Consider a rectangular section
On distance y 0 from the central axis we draw section 1-1 and determine the tangential stresses. Static moment area cut off part:
It should be borne in mind that it is fundamental indifferent, take the static moment of area shaded or remaining part cross section. Both static moments equal and opposite in sign, so their sum, which represents static moment of area of the entire section relative to the neutral line, namely the central x axis, will be equal to zero.
Moment of inertia of a rectangular section:
Then shear stress according to the formula
The variable y 0 is included in the formula in second degrees, i.e. tangential stresses in a rectangular section vary according to 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 tangential stresses has the form:
, Where S x 0– static moment of the part of the cross section located on one side of the layer in which the shear stresses are determined, I x– moment of inertia of the entire cross section, b– section width in the place where the shear stress is determined, Q-lateral force, τ
- shear stress, [τ]
— permissible tangential stress.
This strength condition allows us to produce three type of calculation (three types of problems when calculating strength):
1. Verification calculation or strength test based on tangential stresses:
2. Selection of section width (for a rectangular section): 
3. Determination of permissible lateral 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, it is possible to accept such hypotheses about the nature of stress distribution that the task will become statically definable.
By two infinitely close cross sections 1-1 and 2-2 we select dz element, Let's depict 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 at length dz
Lateral 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 internal tangential stresses distributed over the section. In strength of materials it is usually taken assumption of their uniform distribution across the width of the section.
To determine the magnitude of shear stresses at any point in the cross section located at a distance y 0 from the neutral X axis, draw a plane parallel to the neutral layer (3-3) through this point and take out the clipped element. We will determine the voltage acting across the ABCD area. 
Let's project all the forces onto the Z axis
The resultant of the internal longitudinal forces along the right side will be equal to:
Where A 0 – area of the façade edge, S x 0 – static moment of the cut-off part relative to the X axis. Similarly on the left side:
Both resultants directed towards each other, since the element is in compressed beam area. Their difference is balanced by the tangential forces on the lower edge of 3-3.
Let's pretend that shear stress τ distributed across 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 equal to the stress value multiplied by the area of the face: 
Let's 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 part of the cross section, located on one side of the layer in which the shear stresses are determined, I x – moment of inertia the entire cross section, b – section width in the place where the shear stress is determined, Q - shear force in cross section.
— bending strength condition, Where
- maximum moment (modulo) from the diagram of bending moments; 
- axial moment of resistance of the section, geometric characteristic; - permissible stress (σ adm)
- maximum normal voltage.
If the calculation is carried out according to limit state method, then instead of the permissible voltage, we enter into the calculation design resistance of the material R.
Types of flexural strength calculations
1. Check calculation or testing of strength using normal stresses
2. Design calculation or selection of section 
3. Definition permissible load (definition lifting capacity and or operational carrier capabilities) 
When deriving the formula for calculating normal stresses, we consider the case of bending, when the internal forces in the sections of the beam are reduced only to bending moment, A the shear force turns out to be zero. This case of bending is called pure bending. Consider the middle section of the beam, which is subject to pure bending.
When loaded, the beam bends so that it The lower fibers lengthen and the upper ones shorten.
Since part of the fibers of the beam is stretched, and part is compressed, and the transition from tension to compression occurs smoothly, without jumps, V average part of the beam is located a layer whose fibers only bend, but do not experience either stretching or compression. This layer is called neutral layer. The line along which the neutral layer intersects 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 conclusions of the formulas hypothesis of plane sections (conjecture). According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and turn out to be perpendicular to the curved axis of the beam when it is bent.
Assumptions for deriving normal stress formulas: 1) The hypothesis of plane 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) Deformations of fibers do not depend on their position along the cross-sectional width. Consequently, normal stresses, changing along the height of the section, remain the same along 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 relationship between the dimensions of the beam is such that it operates under plane bending conditions without warping or twisting.
Let's consider a beam of arbitrary cross-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 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 beam, but along it at a known bending moment It is impossible to determine normal stresses until the law of their distribution is established.
Let us select the beams in the middle section and consider section of length dz, subject to bending. Let's depict it on an enlarged scale.
Sections limiting the area dz, parallel to each other until deformed, and after applying the load rotate around their neutral lines by an angle . The length of the neutral layer fiber segment will not change. and will be equal to: 
, where is it radius of curvature curved axis of the beam. But any other fiber lying lower or higher neutral layer, will change its length. Let's calculate 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:
Let's reduce by and bring similar terms, then we get: (2) This formula expresses geometric side of the pure bending problem: The deformations of the fibers are directly proportional to their distances to the neutral layer.
Now let's move on to stress, i.e. we will consider physical side of the task. in accordance with non-pressure assumption we use fibers under axial tension-compression: then, taking into account the formula (2)
we have 
(3),
those. normal stress when bending along the section height linearly distributed. On the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero. Let's 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 relative to the x axis - I x.
Its dimension cm 4, m 4
Then 
,where 
(4) ,where is the curvature of the curved axis of the beam, and is the rigidity of the beam section during bending.
Let's substitute the resulting expression curvature (4) into expression (3)
and we get formula for calculating normal stresses at any point in the cross section: 
(5)
That. maximum tensions arise at points furthest from the neutral line. Attitude (6)
called axial moment of section resistance. 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 the stresses.
Then maximum voltages: 
(7)
Bending strength condition: (8)
When transverse bending occurs not only normal, but also shear stresses, because available shear force. Shear stress complicate the picture of deformation, they lead to curvature cross sections of the beam, resulting in the hypothesis of plane sections is violated. However, research shows that distortions introduced by shear stresses slightly affect 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.
During bending there is no longitudinal force, so we can write 
Let us substitute here the formula for normal stresses (3)
and we get 
Since the modulus of longitudinal elasticity of the beam material is not equal to zero and the curved 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 moment of internal forces relative to the field line) will give 
or taking into account (3)
. For the same reasons (see above) 
. In integrand - the centrifugal moment of inertia of the section relative to the x and y axes is zero, which means these axes are main and central and make up straight corner. Hence, The force and neutral lines in a straight bend are mutually perpendicular.
Having installed neutral line position, easy to build normal stress diagram along the section height. Her linear character is determined equation of the first degree.
The nature of the diagram σ for symmetrical sections relative to the neutral line, M<0
