At the cross-sectional points of the beam during longitudinal-transverse bending, normal stresses arise from compression by longitudinal forces and from bending by transverse and longitudinal loads (Fig. 18.10).
In the outer fibers of the beam in the dangerous section, the total normal stresses have the highest values:
In the above example of a compressed beam with one shear force according to (18.7), we obtain the following stresses in the outer fibers:
If the dangerous section is symmetrical about its neutral axis, then the greatest in absolute value will be the stress in the outer compressed fibers:
In a section that is not symmetrical with respect to the neutral axis, both compressive and tensile stress in the outer fibers can be greatest in absolute value.
When establishing a danger point, the difference in the material's resistance to tension and compression should be taken into account.
Taking into account expression (18.2), formula (18.12) can be written as follows:
Using an approximate expression for we obtain
In beams of constant cross-section, the dangerous section will be the one for which the numerator of the second term has the greatest value.
Dimensions cross section beams must be selected so that the permissible stress does not exceed
However, the resulting relationship between stresses and geometric characteristics of the section is difficult for design calculations; Section dimensions can only be selected by repeated attempts. In case of longitudinal-transverse bending, as a rule, a verification calculation is carried out, the purpose of which is to establish the safety margin of the part.
In longitudinal-transverse bending there is no proportionality between stresses and longitudinal forces; stresses with variable axial force grow faster than the force itself, as can be seen, for example, from formula (18.13). Therefore, the safety factor in the case of longitudinal-transverse bending should be determined not by stresses, i.e., not from a ratio, but by loads, understanding the safety factor as a number showing how many times the effective loads must be increased so that the maximum stress in the calculated part reaches the limit fluidity.
Determining the safety factor is associated with solving transcendental equations, since the force is contained in formulas (18.12) and (18.14) under the sign of the trigonometric function. For example, for a beam compressed by a force and loaded with one transverse force P, the safety factor according to (18.13) is found from the equation
To simplify the problem, you can use formula (18.15). Then to determine the safety factor we obtain a quadratic equation:
Note that in the case when the longitudinal force remains constant, and only the transverse loads change in magnitude, the task of determining the safety factor is simplified, and it is possible to determine it not by load, but by stress. From formula (18.15) for this case we find
Example. A two-support duralumin beam with an I-beam thin-walled section is compressed by a force P and subjected to a uniformly distributed transverse load of intensity and moments applied at the ends
beams, as shown in Fig. 18.11. Determine the stress at the dangerous point and the maximum deflection with and without taking into account the bending effect of the longitudinal force P, and also find the safety factor of the beam according to the yield strength.
In the calculations, take the characteristics of the I-beam:
Solution. The most loaded is the middle section of the beam. Maximum deflection and bending moment due to shear load alone:
The maximum deflection from the combined action of the transverse load and longitudinal force P will be determined by formula (18.10). We get
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 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 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 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 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. 
(lower fibers are compressed).
DC section: (the upper fibers are compressed).
SC section: (left fibers compressed)
(left fibers compressed)
The figure shows diagrams normal (longitudinal) forces - (b), shear forces - (c) and bending moments - (d).
Checking the balance of node C:
Task 2 Construct diagrams of internal forces for the frame (Fig. a).
Given: F=30kN, q=40 kN/m, M=50kNm, a=3m, h=2m.
Let's define support reactions frames:
From these equations we find:
Since the reaction values R K has a sign minus, in Fig. A changes direction given vector to the opposite, and is written R K =83.33kN.
Let's determine the values of internal efforts N, Q And M in characteristic frame sections:
Aircraft section:
(right fibers compressed).
CD section:
(right fibers are compressed);
(right fibers are compressed).
Section DE:
(lower fibers are compressed);
(lower fibers are compressed).
CS section
(left fibers are compressed).
Let's build diagrams of normal (longitudinal) forces (b), transverse forces (c) and bending moments (d).
Consider the equilibrium of nodes D And E
From consideration of nodes D And E it is clear that they are in equilibrium.
Task 3. For a frame with a hinge, construct diagrams of internal forces.
Given: F=30kN, q=40 kN/m, M=50kNm, a=2m, h=2m.
Solution. Let's define support reactions. It should be noted that in both articulated-fixed supports, two reactions. In this regard, you should use hinge property C — moment in it from both left and right forces equal to zero. Let's look at the left side.
The equilibrium equations for the frame under consideration can be written as:
From the solution of these equations it follows:
On the frame diagram, the direction of the force is N V changes to opposite (N B =15kN).
Let's define efforts in characteristic sections of the frame.
Section BZ:
(left fibers are compressed).
Section ZC:
(left fibers compressed);
Section KD:
(left fibers compressed);
(left fibers are compressed).
Section DC:
(lower fibers are compressed);
Definition extreme value bending moment on the section CD:
1. Constructing a diagram of transverse forces. For a cantilever beam (Fig. A ) characteristic points: A – point of application of the support reaction V A; WITH – point of application of concentrated force; D, B – the beginning and end of the distributed load. For a cantilever, the lateral force is determined similarly to a two-support beam. So, when moving from the left:
To check the correct determination of the shear force in the sections, pass the beam in the same way, but from the right end. Then the right parts of the beam will be cut off. Remember that the sign rules will change. The result should be the same. We build a diagram of the transverse force (Fig. b).
2. Constructing a moment diagram
For a cantilever beam, the diagram of bending moments is constructed similarly to the previous construction. Characteristic points for this beam (see Fig. A) are as follows: A – support; WITH - point of application of concentrated moment and force F; D And IN- the beginning and end of the action of a uniformly distributed load. Since the diagram Q x in the area of distributed load action does not cross the zero line, to construct a diagram of moments in a given section (parabolic curve), you should arbitrarily select an additional point to construct the curve, for example, in the middle of the section.
Left move:
Moving to the right we find M B = 0.
Using the found values, we construct a diagram of bending moments (see Fig. V ).
Entry published by the author admin is limited inclined straight line, A in an area where there is no distributed load - straight, parallel to the axis, therefore, to construct a diagram of transverse forces, it is enough to determine the values Qat at the beginning and end of each section. In the section corresponding to the point of application of the concentrated force, the transverse force should be calculated slightly to the left of this point (at an infinitely close distance from it) and slightly to the right of it; shear forces in such places are designated accordingly 
.
Building a diagram Qat using the characteristic point method, moving from the left. For greater clarity, it is recommended to initially cover the discarded part of the beam with a sheet of paper. Characteristic points for a two-support beam (Fig. A
) there will be points C
And D
– the beginning and end of the distributed load, as well as A
And B
– points of application of support reactions, E
– point of application of concentrated force. Let's mentally draw an axis y perpendicular to the beam axis through a point WITH and we will not change its position until we pass the entire beam from C before E. Considering the left parts of the beam cut off at characteristic points, we project onto the axis y forces acting in a given area with corresponding signs. As a result we get: 
To check the correct determination of the shear force in the sections, you can pass the beam in a similar way, but from the right end. Then the right parts of the beam will be cut off. The result should be the same. The coincidence of the results can serve as a control for plotting Qat. We draw a zero line under the image of the beam and from it, on the accepted scale, we plot the found values of the transverse forces, taking into account the signs at the corresponding points. Let's get the diagram Qat(rice. b ).
Having constructed the diagram, pay attention to the following: the diagram under a distributed load is depicted as an inclined straight line, under unloaded sections - segments parallel to the zero line, under a concentrated force a jump is formed on the diagram, equal to the value of the force. If an inclined line under a distributed load intersects the zero line, mark this point, then this extremum point, and it is now characteristic for us, according to the differential relationship between Qat And Mx, at this point the moment has an extremum and it will need to be determined when constructing a diagram of bending moments. In our problem this is the point TO . Focused moment on diagram Qat does not manifest itself in any way, since the sum of the projections of the forces forming the pair is equal to zero.
2. Constructing a moment diagram. We construct a diagram of bending moments, as well as transverse forces, using the characteristic point method, moving from the left. It is known that in a section of a beam with a uniformly distributed load, the diagram of bending moments is outlined by a curved line (quadratic parabola), to construct which one must have at least three points and, therefore, the values of bending moments at the beginning of the section, the end of it and in one intermediate section must be calculated. It is best to take as such an intermediate point the section in which the diagram Qat crosses the zero line, i.e. Where Qat= 0. On the diagram M this section should contain the vertex of the parabola. If the diagram Q at does not cross the zero line, then to construct a diagram M follows on in this section, take an additional point, for example, in the middle of the section (the beginning and end of the distributed load), remembering that the convexity of the parabola always faces downward if the load acts from top to bottom (for construction specialties). There is a “rain” rule, which is very helpful when constructing the parabolic part of the diagram M. For builders, this rule looks like this: imagine that the distributed load is rain, place an umbrella under it upside down, so that the rain does not flow down, but collects in it. Then the bulge of the umbrella will be facing down. This is exactly what the outline of the moment diagram under a distributed load will look like. For mechanics there is the so-called “umbrella” rule. The distributed load is represented by rain, and the outline of the diagram should resemble the outline of an umbrella. In this example, the diagram was built for builders.
If a more accurate plotting is required, then the values of bending moments in several intermediate sections must be calculated. For each such section, we agree to first determine the bending moment in an arbitrary section, expressing it through the distance X from any point. Then, giving the distance X a series of values, we obtain the values of bending moments in the corresponding sections of the section. For sections where there is no distributed load, bending moments are determined in two sections corresponding to the beginning and end of the section, since the diagram M in such areas it is limited to a straight line. If an external concentrated moment is applied to the beam, then it is necessary to calculate the bending moment slightly to the left of the place where the concentrated moment is applied and slightly to the right of it.
For a two-support beam, the characteristic points are as follows: C And D – the beginning and end of the distributed load; A – beam support; IN – the second support of the beam and the point of application of the concentrated moment; E – right end of the beam; dot TO , corresponding to the section of the beam in which Qat= 0.
Move on the left. We mentally discard the right part up to the section under consideration (take a sheet of paper and cover the discarded part of the beam with it). We find the sum of the moments of all forces acting to the left of the section relative to the point in question. So,
Before determining the moment in the section TO, you need to find the distance x=AK. Let's create an expression for the transverse force in this section and equate it to zero (move on the left):
This distance can also be found from the similarity of triangles KLN And KIG on the diagram Qat(rice. b) .
Determine the moment at a point TO :
Let's go through the rest of the beam on the right.
As we see, the moment at the point D when moving left and right, the result was the same - the diagram closed. Based on the values found, we construct a diagram. Positive values we put it down from the zero line, and negative ones – up (see Fig. V ).
In practice, very often there are cases of joint work of a rod in bending and tension or compression. This kind of deformation can be caused either by the combined action of longitudinal and transverse forces on the beam, or by longitudinal forces alone.
The first case is shown in Fig. 1. The beam AB is subject to a uniformly distributed load q and longitudinal compressive forces P.
Fig.1.
Let us assume that the deflections of the beam compared to the cross-sectional dimensions can be neglected; then, with a degree of accuracy sufficient for practice, we can assume that even after deformation, the forces P will only cause axial compression of the beam.
Using the method of adding forces, we can find normal voltage at any point of each cross section of the beam as the algebraic sum of stresses caused by forces P and load q.
Compressive stresses from forces P are uniformly distributed over the cross-sectional area F and are the same for all sections
normal bending stresses in vertical plane in a section with the abscissa x, which is measured, say, from the left end of the beam, are expressed by the formula
Thus, the total stress at a point with coordinate z (counting from the neutral axis) for this section is equal to
Figure 2 shows stress distribution diagrams in the section under consideration from forces P, load q and the total diagram.
The greatest stress in this section will be in the upper fibers, where both types of deformation cause compression; in the lower fibers there can be either compression or tension depending on the numerical values of the stresses and. To create the strength condition, we will find the greatest normal stress.
Fig.2.
Since the stresses from the forces P in all sections are the same and evenly distributed, the fibers that are most stressed from bending will be dangerous. These are the outermost fibers in the cross section with the highest bending moment; for them
Thus, the stresses in the outermost fibers 1 and 2 of the middle section of the beam are expressed by the formula
and the calculated voltage will be equal to
If the forces P were tensile, then the sign of the first term would change, and the lower fibers of the beam would be dangerous.
Denoting compressive or tensile force with the letter N, we can write a general formula for checking strength
The described calculation procedure is also applied when inclined forces act on the beam. Such a force can be decomposed into normal to the axis, bending the beam, and longitudinal, compressive or tensile.
beam bending force compression
The whole variety of existing support devices is schematized in the form of a number of basic types of supports, of which
most common: articulated and movablesupport(possible designations for it are presented in Fig. 1, a), hinged-fixed support(Fig. 1, b) and hard pinching, or sealing(Fig. 1, c).
In a hinged-movable support, one support reaction occurs, perpendicular to the support plane. Such a support deprives the support section of one degree of freedom, that is, it prevents displacement in the direction of the support plane, but allows movement in the perpendicular direction and rotation of the support section.
In a hinged-fixed support, vertical and horizontal reactions occur. Here, movements in the directions of the support rods are not possible, but rotation of the support section is allowed.
In a rigid embedment, vertical and horizontal reactions and a support (reactive) moment occur. In this case, the support section cannot shift or rotate. When calculating systems containing a rigid embedment, the resulting support reactions can not be determined, choosing the cut-off part so that the embedding with unknown reactions does not fall into it. When calculating systems on hinged supports, the reactions of the supports must be determined. The static equations used for this depend on the type of system (beam, frame, etc.) and will be given in the relevant sections of this manual.
2. Construction of diagrams of longitudinal forces Nz
The longitudinal force in a section is numerically equal to the algebraic sum of the projections of all forces applied on one side of the section under consideration onto the longitudinal axis of the rod.
Rule of signs for Nz: let us agree to consider the longitudinal force in the section positive if external load, applied to the cut-off part of the rod under consideration, causes tension and is negative - otherwise.
Example 1.Construct a diagram of longitudinal forces for a rigidly clamped beam(Fig. 2).
Calculation procedure:
1. We outline characteristic sections, numbering them from the free end of the rod to the embedment.
2. Determine the longitudinal force Nz in each characteristic section. In this case, we always consider the cut-off part into which the rigid seal does not fall.
Based on the found values build a diagram Nz. Positive values are plotted (on the selected scale) above the diagram axis, negative values are plotted below the axis.
3. Construction of diagrams of torques Mkr.
Torque in the section is numerically equal to the algebraic sum of external moments applied on one side of the section under consideration, relative to the longitudinal Z axis.
Sign rule for microdistrict: let’s agree to count torque in the section is positive if, when looking at the section from the side of the cut-off part under consideration, the external moment is seen directed counterclockwise and negative - otherwise.
Example 2.Construct a diagram of torques for a rigidly clamped rod(Fig. 3, a).
Calculation procedure.
It should be noted that the algorithm and principles for constructing a torque diagram completely coincide with the algorithm and principles constructing a diagram of longitudinal forces.
1. We outline characteristic sections.
2. Determine the torque in each characteristic section.
Based on the found values we build microdistrict diagram(Fig. 3, b).
4. Rules for monitoring diagrams Nz and Mkr.
For diagrams of longitudinal forces and torques are characterized by certain patterns, knowledge of which allows us to evaluate the correctness of the constructions performed.
1. Diagrams Nz and Mkr are always rectilinear.
2. In the area where there is no distributed load, the diagram Nz(Mkr) is a straight line, parallel to the axis, and in the area under a distributed load it is an inclined straight line.
3. Under the point of application of the concentrated force on the diagram Nz there must be a jump in the magnitude of this force, similarly, under the point of application of the concentrated moment on the diagram Mkr there will be a jump in the magnitude of this moment.
5. Construction of diagrams of transverse forces Qy and bending moments Mx in beams
A rod that bends is called beam. In sections of beams loaded with vertical loads, as a rule, two internal force factors arise - Qy and bending moment Mx.
Lateral force in the section is numerically equal to the algebraic sum of projections external forces, applied on one side of the section under consideration, on the transverse (vertical) axis.
Sign rule for Qy: Let us agree to consider the transverse force in the section positive if the external load applied to the cut-off part under consideration tends to rotate this section clockwise and negative otherwise.
Schematically, this sign rule can be represented as
Bending moment Mx in a section is numerically equal to the algebraic sum of the moments of external forces applied on one side of the section under consideration, relative to the x axis passing through this section.
Rule of signs for Mx: let us agree to consider the bending moment in the section positive if the external load applied to the cut-off part under consideration leads to tension in this section of the lower fibers of the beam and negative - otherwise.
Schematically, this sign rule can be represented as:
It should be noted that when using the sign rule for Mx in the specified form, the Mx diagram always turns out to be constructed from the side of the compressed fibers of the beam.
6. Cantilever beams
At plotting Qy and Mx diagrams in cantilever, or rigidly clamped, beams there is no need (as in the previously discussed examples) to calculate the support reactions arising in the rigid embedment, but the cut-off part must be selected so that the embedment does not fall into it.
Example 3.Construct Qy and Mx diagrams(Fig. 4).
Calculation procedure.
1. We outline characteristic sections.
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 stress in beam and we must compare this stress with the stress that our beam of a given material can generally withstand.
For plastic materials(steel, aluminum, etc.) the maximum voltage will be equal to material yield strength, A for fragile(cast iron) – tensile strength. We can find the yield strength and tensile strength from the tables below.
Let's look at a couple of examples:
1. [i] You want to check whether an I-beam No. 10 (steel St3sp5) 2 meters long, rigidly embedded in the wall, will support you if you hang on it. Let your mass be 90 kg.
First, we need to select a design scheme.
This diagram shows that the maximum moment will be at the seal, and since our I-beam has equal section along the entire length, then the maximum voltage will be in the termination. Let's find it:
P = m * g = 90 * 10 = 900 N = 0.9 kN
M = P * l = 0.9 kN * 2 m = 1.8 kN * m
Using the I-beam assortment table, we find the moment of resistance of I-beam No. 10.
It will be equal to 39.7 cm3. Let's convert it to cubic meters and 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 limit the fluidity of steel St3sp5 is 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).
