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.
Rice. 3.12
The horizontal reaction in the embedment is zero, since external loads in the z-axis direction do not act on the beam.
We choose the directions of the remaining reactive 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:
The positive values we obtained 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). In this case, the external load, bending the part of the beam under consideration with its convexity downward, 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.
The normal stress strength condition 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 of the 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.
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.
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.
 |
Rice. 3.13
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 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 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 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 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. Let's 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 the active 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).
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 on 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 area 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 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 embedding. 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's 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 voltages:
Level 3−3: y 3-3 =h 1 /2−t 1 =20/2−0.97=9.03 cm.
Normal and shear stresses: 
Main voltages:
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 voltages: 
Extreme shear stress:
Level 5−5:
Normal and shear stresses: 
Main voltages:
Extreme shear stress: 
Level 6−6:
Normal and shear stresses: 
Main voltages:
Extreme shear stress: 
Level 7−7:
Normal and shear stresses:
Main voltages:
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.
The hypothesis of plane sections during bending can be explained with an example: let us apply a grid consisting of longitudinal and transverse (perpendicular to the axis) straight lines on the side surface of an undeformed beam. As a result of bending the beam, the longitudinal lines will take on a curved outline, while the transverse lines will practically remain straight and perpendicular to the curved axis of the beam.
Formulation of the plane section hypothesis: cross sections that are flat and perpendicular to the axis of the beam before , remain flat and perpendicular to the curved axis after it is deformed.
This circumstance indicates: when fulfilled plane section hypothesis, as with and
In addition to the hypothesis of flat sections, the assumption is accepted: the longitudinal fibers of the beam do not press on each other when it bends.
The plane section hypothesis and assumption are called Bernoulli's hypothesis.
Consider a beam of rectangular cross-section undergoing pure bending (). Let's select a beam element with a length (Fig. 7.8.a). As a result of bending, the cross sections of the beam will rotate, forming an angle. The upper fibers experience compression, and the lower fibers experience tension. We denote the radius of curvature of the neutral fiber as .
Conventionally, we assume that the fibers change their length while remaining straight (Fig. 7.8. b). Then the absolute and relative elongations of the fiber located at a distance y from the neutral fiber:
Let us show that longitudinal fibers, which do not experience either tension or compression when the beam bends, pass through the main central axis x.
Since the length of the beam does not change during bending, the longitudinal force (N) arising in the cross section must be zero. Elementary longitudinal force.
Given the expression :
The factor can be taken out of the integral sign (does not depend on the integration variable).
The expression represents the cross section of the beam about the neutral x-axis. It is zero when the neutral axis passes through the center of gravity of the cross section. Consequently, the neutral axis (zero line) when the beam bends passes through the center of gravity of the cross section.
Obviously: the bending moment is associated with normal stresses arising at points in the cross section of the rod. Elementary bending moment created by an elementary force:
,
where is the axial moment of inertia of the cross section relative to the neutral x-axis, and the ratio is the curvature of the beam axis.
Rigidity beams in bending(the larger, the smaller the radius of curvature).
The resulting formula represents Hooke's law of bending for a rod: The bending moment occurring in the cross section is proportional to the curvature of the beam axis.
Expressing the radius of curvature () from the formula of Hooke’s law for a rod during bending and substituting its value into the formula , we obtain a formula for normal stresses () at an arbitrary point in the cross section of the beam, located at a distance y from the neutral axis x: .
In the formula for normal stresses () at an arbitrary point in the cross section of the beam, the absolute values of the bending moment () and the distance from the point to the neutral axis (y coordinates) should be substituted. Whether the stress at a given point will be tensile or compressive can be easily determined by the nature of the deformation of the beam or by the diagram of bending moments, the ordinates of which are plotted on the side of the compressed fibers of the beam.
From the formula it is clear: normal stresses () change along the height of the cross section of the beam according to a linear law. In Fig. 7.8, shows the diagram. The greatest stresses during beam bending occur at points furthest from the neutral axis. If a line is drawn in the cross section of the beam parallel to the neutral x axis, then equal normal stresses arise at all its points.
Simple analysis normal stress diagrams shows that when a beam bends, the material located near the neutral axis practically does not work. Therefore, in order to reduce the weight of the beam, it is recommended to choose cross-sectional shapes in which most of the material is removed from the neutral axis, such as an I-section.
We will start with the simplest case, the so-called pure bend.
Pure bending is a special case of bending in which the transverse force in the sections of the beam is zero. Pure bending can only occur when the self-weight of the beam is so small that its influence can be neglected. For beams on two supports, examples of loads causing pure
bending, shown in Fig. 88. In sections of these beams, where Q = 0 and, therefore, M = const; pure bending takes place.
The forces in any section of the beam during pure bending are reduced to a pair of forces, the plane of action of which passes through the axis of the beam, and the moment is constant.
Voltages can be determined based on the following considerations.
1. The tangential components of forces along elementary areas in the cross section of a beam cannot be reduced to a pair of forces, the plane of action of which is perpendicular to the section plane. It follows that the bending force in the section is the result of action along elementary areas
only normal forces, and therefore with pure bending the stresses are reduced only to normal.
2. In order for efforts on elementary sites to be reduced to only a couple of forces, among them there must be both positive and negative. Therefore, both tension and compression fibers of the beam must exist.
3. Due to the fact that the forces in different sections are the same, the stresses at the corresponding points of the sections are the same.
Let's consider some element near the surface (Fig. 89, a). Since no forces are applied along its lower edge, which coincides with the surface of the beam, there are no stresses on it. Therefore, there are no stresses on the upper edge of the element, since otherwise the element would not be in equilibrium. Considering the element adjacent to it in height (Fig. 89, b), we arrive at
The same conclusion, etc. It follows that there are no stresses along the horizontal edges of any element. Considering the elements that make up the horizontal layer, starting with the element near the surface of the beam (Fig. 90), we come to the conclusion that there are no stresses along the lateral vertical edges of any element. Thus, the stress state of any element (Fig. 91, a), and in the limit, fibers, should be represented as shown in Fig. 91,b, i.e. it can be either axial tension or axial compression.
4. Due to the symmetry of the application of external forces, the section along the middle of the length of the beam after deformation should remain flat and normal to the axis of the beam (Fig. 92, a). For the same reason, sections in quarters of the length of the beam also remain flat and normal to the axis of the beam (Fig. 92, b), unless the extreme sections of the beam during deformation remain flat and normal to the axis of the beam. A similar conclusion is valid for sections in eighths of the length of the beam (Fig. 92, c), etc. Consequently, if during bending the outer sections of the beam remain flat, then for any section it remains 
It is a fair statement that after deformation it remains flat and normal to the axis of the curved beam. But in this case, it is obvious that the change in elongation of the fibers of the beam along its height should occur not only continuously, but also monotonically. If we call a layer a set of fibers that have the same elongations, then it follows from what has been said that the stretched and compressed fibers of the beam should be located on opposite sides of the layer in which the elongations of the fibers are equal to zero. We will call fibers whose elongations are zero neutral; a layer consisting of neutral fibers is a neutral layer; the line of intersection of the neutral layer with the cross-sectional plane of the beam - the neutral line of this section. Then, based on the previous reasoning, it can be argued that with pure bending of a beam, in each section there is a neutral line that divides this section into two parts (zones): a zone of stretched fibers (stretched zone) and a zone of compressed fibers (compressed zone). ). Accordingly, at the points of the stretched zone of the section, normal tensile stresses should act, at the points of the compressed zone - compressive stresses, and at the points of the neutral line the stresses are equal to zero.
Thus, with pure bending of a beam of constant cross-section:
1) only normal stresses act in sections;
2) the entire section can be divided into two parts (zones) - stretched and compressed; the boundary of the zones is the neutral section line, at the points of which the normal stresses are equal to zero;
3) any longitudinal element of the beam (in the limit, any fiber) is subjected to axial tension or compression, so that adjacent fibers do not interact with each other;
4) if the extreme sections of the beam during deformation remain flat and normal to the axis, then all its cross sections remain flat and normal to the axis of the curved beam.
Let us consider an element of a beam subjected to pure bending, concluding 
located between sections m-m and n-n, which are spaced one from the other at an infinitesimal distance dx (Fig. 93). Due to position (4) of the previous paragraph, sections m- m and n - n, which were parallel before deformation, after bending, remaining flat, will form an angle dQ and intersect along a straight line passing through point C, which is the center of curvature neutral fiber NN. Then the part AB of the fiber enclosed between them, located at a distance z from the neutral fiber (the positive direction of the z axis is taken towards the convexity of the beam during bending), will turn after deformation into an arc AB. A piece of neutral fiber O1O2, having turned into an arc, O1O2 will not change its length, while fiber AB will receive an elongation:
before deformation
after deformation
where p is the radius of curvature of the neutral fiber.
Therefore, the absolute lengthening of segment AB is equal to
and relative elongation
Since, according to position (3), fiber AB is subjected to axial tension, then during elastic deformation
This shows that normal stresses along the height of the beam are distributed according to a linear law (Fig. 94). Since the equal force of all forces over all elementary cross-sectional areas must be equal to zero, then
from where, substituting the value from (5.8), we find
But the last integral is a static moment about the Oy axis, perpendicular to the plane of action of the bending forces.
Due to its equality to zero, this axis must pass through the center of gravity O of the section. Thus, the neutral line of the section of the beam is a straight line y, perpendicular to the plane of action of bending forces. It is called the neutral axis of the beam section. Then from (5.8) it follows that the stresses at points lying at the same distance from the neutral axis are the same.
The case of pure bending, in which the bending forces act in only one plane, causing bending only in that plane, is planar pure bending. If the said plane passes through the Oz axis, then the moment of elementary forces relative to this axis should be equal to zero, i.e.
Substituting here the value of σ from (5.8), we find
The integral on the left side of this equality, as is known, is the centrifugal moment of inertia of the section relative to the y and z axes, so
The axes about which the centrifugal moment of inertia of the section is zero are called the main axes of inertia of this section. If they, in addition, pass through the center of gravity of the section, then they can be called the main central axes of inertia of the section. Thus, with flat pure bending, the direction of the plane of action of bending forces and the neutral axis of the section are the main central axes of inertia of the latter. In other words, to obtain a flat, pure bend of a beam, a load cannot be applied to it arbitrarily: it must be reduced to forces acting in a plane that passes through one of the main central axes of inertia of the sections of the beam; in this case, the other main central axis of inertia will be the neutral axis of the section.
As is known, in the case of a section that is symmetrical about any axis, the axis of symmetry is one of its main central axes of inertia. Consequently, in this particular case we will certainly obtain pure bending by applying appropriate loads in a plane passing through the longitudinal axis of the beam and the axis of symmetry of its section. A straight line perpendicular to the axis of symmetry and passing through the center of gravity of the section is the neutral axis of this section.
Having established the position of the neutral axis, it is not difficult to find the magnitude of the stress at any point in the section. In fact, since the sum of the moments of elementary forces relative to the neutral axis yy must be equal to the bending moment, then
whence, substituting the value of σ from (5.8), we find
Since the integral is moment of inertia of the section relative to the yy axis, then
and from expression (5.8) we obtain
The product EI Y is called the bending stiffness of the beam.
The greatest tensile and largest compressive stresses in absolute value act at the points of the section for which the absolute value of z is greatest, i.e., at the points furthest from the neutral axis. With the notation, Fig. 95 we have
The value Jy/h1 is called the moment of resistance of the section to tension and is designated Wyr; similarly, Jy/h2 is called the moment of resistance of the section to compression
and denote Wyc, so
and therefore
If the neutral axis is the axis of symmetry of the section, then h1 = h2 = h/2 and, therefore, Wyp = Wyc, so there is no need to distinguish them, and they use the same notation:
calling W y simply the moment of resistance of the section. Consequently, in the case of a section symmetrical about the neutral axis,
All the above conclusions were obtained on the basis of the assumption that the cross sections of the beam, when bent, remain flat and normal to its axis (hypothesis of flat sections). As has been shown, this assumption is valid only in the case when the extreme (end) sections of the beam remain flat during bending. On the other hand, from the hypothesis of plane sections it follows that elementary forces in such sections should be distributed according to a linear law. Therefore, for the validity of the resulting theory of flat pure bending, it is necessary that the bending moments at the ends of the beam be applied in the form of elementary forces distributed along the height of the section according to a linear law (Fig. 96), coinciding with the law of stress distribution along the height of the section beams. However, based on the Saint-Venant principle, it can be argued that changing the method of applying bending moments at the ends of the beam will cause only local deformations, the effect of which will affect only a certain distance from these ends (approximately equal to the height of the section). The sections located throughout the rest of the length of the beam will remain flat. Consequently, the stated theory of flat pure bending for any method of applying bending moments is valid only within the middle part of the length of the beam, located from its ends at distances approximately equal to the height of the section. From here it is clear that this theory is obviously inapplicable if the height of the section exceeds half the length or span of the beam.
10.1. General concepts and definitions
Bend- this is a type of loading in which the rod is loaded with moments in planes passing through the longitudinal axis of the rod.
A rod that bends is called a beam (or timber). In the future, we will consider rectilinear beams, the cross section of which has at least one axis of symmetry.
The resistance of materials is divided into flat, oblique and complex bending.
Flat bend– bending, in which all the forces bending the beam lie in one of the planes of symmetry of the beam (in one of the main planes).
The main planes of inertia of a beam are the planes passing through the main axes of the cross sections and the geometric axis of the beam (x-axis).
Oblique bend– bending, in which the loads act in one plane that does not coincide with the main planes of inertia.
Complex bend– bending, in which loads act in different (arbitrary) planes.
10.2. Determination of internal bending forces
Let us consider two typical cases of bending: in the first, the cantilever beam is bent by a concentrated moment Mo; in the second - concentrated force F.
Using the method of mental sections and composing equilibrium equations for the cut off parts of the beam, we determine the internal forces in both cases:
The remaining equilibrium equations are obviously identically equal to zero.
Thus, in the general case of plane bending in the section of a beam, out of six internal forces, two arise - bending moment Mz and shear force Qy (or when bending relative to another main axis - bending moment My and shear force Qz).
Moreover, in accordance with the two loading cases considered, plane bending can be divided into pure and transverse.
Clean bend– flat bending, in which in the sections of the rod, out of six internal forces, only one arises – a bending moment (see the first case).
Transverse bend– bending, in which in the sections of the rod, in addition to the internal bending moment, a transverse force also arises (see the second case).
Strictly speaking, simple types of resistance include only pure bending; transverse bending conditionally classified as simple types of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected when calculating strength.
When determining internal efforts, we will adhere to the following rule of signs:
1) the transverse force Qy is considered positive if it tends to rotate the beam element in question clockwise;
2) bending moment Mz is considered positive if, when bending a beam element, the upper fibers of the element are compressed and the lower fibers are stretched (umbrella rule).
Thus, the solution to the problem of determining the internal forces during bending will be built according to the following plan: 1) at the first stage, considering the equilibrium conditions of the structure as a whole, we determine, if necessary, the unknown reactions of the supports (note that for a cantilever beam the reactions in the embedment can be and not found if we consider the beam from the free end); 2) at the second stage, we select characteristic sections of the beam, taking as the boundaries of the sections the points of application of forces, points of change in the shape or size of the beam, points of fastening of the beam; 3) at the third stage, we determine the internal forces in the sections of the beam, considering the conditions of equilibrium of the beam elements in each section.
10.3. Differential dependencies during bending
Let us establish some relationships between internal forces and external loads during bending, as well as the characteristic features of the Q and M diagrams, knowledge of which will facilitate the construction of diagrams and allow us to control their correctness. For convenience of notation, we will denote: M≡Mz, Q≡Qy.
Let us select a small element dx in a section of a beam with an arbitrary load in a place where there are no concentrated forces and moments. Since the entire beam is in equilibrium, element dx will also be in equilibrium under the action of shear forces, bending moments and external load applied to it. Since Q and M generally vary along
axis of the beam, then transverse forces Q and Q+dQ, as well as bending moments M and M+dM, will arise in the sections of element dx. From the equilibrium condition of the selected element we obtain
The first of the two equations written gives the condition
From the second equation, neglecting the term q dx (dx/2) as an infinitesimal quantity of the second order, we find
Considering expressions (10.1) and (10.2) together we can obtain
Relations (10.1), (10.2) and (10.3) are called differential dependences of D.I. Zhuravsky during bending.
Analysis of the above differential dependencies during bending allows us to establish some features (rules) for constructing diagrams of bending moments and transverse forces: a - in areas where there is no distributed load q, diagrams Q are limited to straight lines parallel to the base, and diagrams M are limited to inclined straight lines; b – in areas where a distributed load q is applied to the beam, Q diagrams are limited by inclined straight lines, and M diagrams are limited by quadratic parabolas.
Moreover, if we construct diagram M “on a stretched fiber,” then the convexity of the parabola will be directed in the direction of action q, and the extremum will be located in the section where diagram Q intersects the base line; c – in sections where a concentrated force is applied to the beam, on diagram Q there will be jumps by the magnitude and in the direction of this force, and on diagram M there will be kinks, the tip directed in the direction of action of this force; d – in sections where a concentrated moment is applied to the beam, there will be no changes on diagram Q, and on diagram M there will be jumps in the magnitude of this moment; d – in areas where Q>0, the moment M increases, and in areas where Q<0, момент М убывает (см. рисунки а–г).
10.4. Normal stresses during pure bending of a straight beam
Let us consider the case of pure plane bending of a beam and derive a formula for determining normal stresses for this case.
Note that in the theory of elasticity it is possible to obtain an exact dependence for normal stresses during pure bending, but if this problem is solved using methods of strength of materials, it is necessary to introduce some assumptions.
There are three such hypotheses for bending:
a – hypothesis of flat sections (Bernoulli hypothesis) – flat sections before deformation remain flat after deformation, but only rotate relative to a certain line, which is called the neutral axis of the beam section. In this case, the fibers of the beam lying on one side of the neutral axis will stretch, and on the other, compress; fibers lying on the neutral axis do not change their length;
b – hypothesis about the constancy of normal stresses - stresses acting at the same distance y from the neutral axis are constant across the width of the beam;
c – hypothesis about the absence of lateral pressures – adjacent longitudinal fibers do not press on each other.
Static side of the problem
To determine the stresses in the cross sections of the beam, we consider, first of all, the static sides of the problem. Using the method of mental sections and composing equilibrium equations for the cut-off part of the beam, we will find the internal forces during bending. As was shown earlier, the only internal force acting in the beam section during pure bending is the internal bending moment, which means that normal stresses associated with it will arise here.
We will find the relationship between internal forces and normal stresses in the beam section by considering the stresses on the elementary area dA, selected in the cross section A of the beam at the point with coordinates y and z (the y axis is directed downward for convenience of analysis):
As we see, the problem is internally statically indeterminate, since the nature of the distribution of normal stresses over the section is unknown. To solve the problem, consider the geometric picture of deformations.
Geometric side of the problem
Let us consider the deformation of a beam element of length dx, separated from a bending rod at an arbitrary point with coordinate x. Taking into account the previously accepted hypothesis of flat sections, after bending the beam section, rotate relative to the neutral axis (n.o.) by an angle dϕ, while the fiber ab, spaced from the neutral axis at a distance y, will turn into an arc of a circle a1b1, and its length will change by some size. Let us recall here that the length of the fibers lying on the neutral axis does not change, and therefore the arc a0b0 (the radius of curvature of which is denoted by ρ) has the same length as the segment a0b0 before the deformation a0b0=dx.
Let us find the relative linear deformation εx of the fiber ab of the curved beam:
