Spatial bending This type of complex resistance is called in which only bending moments and
. The full bending moment acts in none of the main planes of inertia. There is no longitudinal force. Spatial or complex bending is often called non-planar bend, since the curved axis of the rod is not a flat curve. This bending is caused by forces acting in different planes perpendicular to the axis of the beam (Fig. 12.4).

Following the order of solving problems with complex resistance outlined above, we lay out the spatial system of forces shown in Fig. 12.4, into two such that each of them acts in one of the main planes. As a result, we get two flat transverse bends - in the vertical and horizontal plane. Of the four internal force factors that arise in the cross section of the beam
, we will take into account the influence of only bending moments
. We build diagrams
, caused respectively by the forces
(Fig. 12.4).

Analyzing the diagrams of bending moments, we come to the conclusion that section A is dangerous, since it is in this section that the largest bending moments occur
And
. Now it is necessary to establish the dangerous points of section A. To do this, we will construct a zero line. The zero line equation, taking into account the sign rule for the terms included in this equation, has the form:

. (12.7)

Here the “” sign is adopted near the second term of the equation, since the stresses in the first quarter caused by the moment
, will be negative.

Let's determine the angle of inclination of the zero line with positive axis direction (Fig.12.6):

. (12.8)

From equation (12.7) it follows that the zero line for spatial bending is a straight line and passes through the center of gravity of the section.

From Fig. 12.5 it is clear that the greatest stresses will arise at the points of section No. 2 and No. 4 furthest from the zero line. The normal stresses at these points will be the same in magnitude, but different in sign: at point No. 4 the stresses will be positive, i.e. tensile, at point No. 2 – negative, i.e. compressive. The signs of these stresses were established from physical considerations.

Now that the dangerous points have been established, let's calculate the maximum stresses in section A and check the strength of the beam using the expression:

. (12.9)

The strength condition (12.9) allows not only to check the strength of the beam, but also to select the dimensions of its cross section if the aspect ratio of the cross section is specified.

12.4. Oblique bend

Obliquely This type of complex resistance is called in which only bending moments occur in the cross sections of the beam
And
, but unlike spatial bending, all forces applied to the beam act in one (force) plane, which does not coincide with any of the main planes of inertia. This type of bending is most often encountered in practice, so we will study it in more detail.

Consider a cantilever beam loaded with a force , as shown in Fig. 12.6, and made of isotropic material.

Just as with spatial bending, with oblique bending there is no longitudinal force. We will neglect the influence of transverse forces on the strength of the beam when calculating it.

The design diagram of the beam shown in Fig. 12.6 is shown in Fig. 12.7.

Let's break down the power to vertical and horizontal components and from each of these components we will construct diagrams of bending moments
And
.

Let us calculate the components of the total bending moment in the section :

;
.

Total bending moment in section equals

Thus, the components of the total bending moment can be expressed in terms of the total moment as follows:

;
. (12.10)

From expression (12.10) it is clear that during oblique bending there is no need to decompose the system of external forces into components, since these components of the total bending moment are connected to each other using the angle of inclination of the trace of the force plane . As a result, there is no need to construct diagrams of the components
And
total bending moment. It is enough to plot the total bending moment
in the force plane, and then, using expression (12.10), determine the components of the total bending moment in any section of the beam that interests us. The obtained conclusion significantly simplifies the solution of problems with oblique bending.

Let us substitute the values ​​of the components of the total bending moment (12.10) into the formula for normal stresses (12.2) at
. We get:

. (12.11)

Here, the “” sign next to the total bending moment is placed specifically for the purpose of automatically obtaining the correct sign of the normal stress at the cross-section point under consideration. Total bending moment
and point coordinates And are taken with their signs, provided that in the first quadrant the signs of the point coordinates are taken positive.

Formula (12.11) was obtained from considering the special case of oblique bending of a beam, clamped at one end and loaded at the other with a concentrated force. However, this formula is a general formula for calculating stresses in oblique bending.

The dangerous section, as with spatial bending in the case under consideration (Fig. 12.6), will be section A, since in this section the largest total bending moment occurs. We will determine the dangerous points of section A by constructing a zero line. We obtain the zero line equation by calculating, using formula (12.11), the normal stresses at the point with coordinates And , belonging to the zero line and equate the found voltages to zero. After simple transformations we get:

(12.12)

. (12.13)

Here angle of inclination of the zero line to the axis (Fig. 12.8).

By examining equations (12.12) and (12.13), we can draw some conclusions about the behavior of the zero line during oblique bending:

From Fig. 12.8 it follows that the highest stresses occur at the cross-section points furthest from the zero line. In the case under consideration, such points are points No. 1 and No. 3. Thus, with oblique bending, the strength condition has the form:

. (12.14)

Here:
;
.

If the moments of resistance of the section relative to the main axes of inertia can be expressed in terms of the dimensions of the section, it is convenient to use the strength condition in this form:

. (12.15)

When selecting sections, one of the axial moments of resistance is taken out of the bracket and specified by the relation . Knowing
,
and angle , through successive attempts, determine the values
And , satisfying the strength condition

. (12.16)

For asymmetrical sections that do not have protruding corners, the strength condition in the form (12.14) is used. In this case, with each new attempt to select a section, it is necessary to first again find the position of the zero line and the coordinates of the most distant point (
). For rectangular section
. Given the relation, from the strength condition (12.16) one can easily find the quantity
and cross-sectional dimensions.

Let us consider the determination of displacements during oblique bending. Let's find the deflection in the section cantilever beam (Fig. 12.9). To do this, we will depict the beam in a single state and construct a diagram of single bending moments in one of the main planes. We will determine the total deflection in the section , having previously determined the projections of the displacement vector on the axis And . Projection of the total deflection vector onto the axis we find using Mohr's formula:

Projection of the total deflection vector onto the axis let's find it in a similar way:

The total deflection is determined by the formula:

. (12.19)

It should be noted that with oblique bending in formulas (12.17) and (12.18), when determining the projections of the deflection on the coordinate axes, only the constant terms in front of the integral sign change. The integral itself remains constant. When solving practical problems, we will calculate this integral using the Mohr-Simpson method. To do this, multiply the unit diagram
for cargo
(Fig. 12.9), constructed in the force plane, and then multiply the resulting result sequentially by constant coefficients, respectively, And . As a result, we obtain projections of the total deflection And on the coordinate axis And . Expressions for deflection projections for the general case of loading, when the beam has plots will look like:

; (12.20)

. (12.21)

Let us set aside the found values ​​for ,And (Fig. 12.8). Total deflection vector is with the axis acute angle , the values ​​of which can be found using the formula:

, (12.22)

. (12.23)

Comparing equation (12.22) with the zero line equation (12.13), we come to the conclusion that

or
,

whence it follows that the zero line and the vector of total deflection mutually peredicular. Corner is the complement of an angle up to 90 0. This condition can be used to check when solving oblique bending problems:

. (12.24)

Thus, the direction of deflections during oblique bending is perpendicular to the zero line. This implies the important condition that the direction of deflections does not coincide with the direction of the acting force(Fig. 12.8). If the load is a plane system of forces, then the axis of the curved beam lies in a plane that does not coincide with the plane of action of the forces. The beam skews relative to the force plane. This circumstance served as the basis for the fact that such a bend began to be called oblique.

Example 12.1. Determine the position of the zero line (find the angle ) for the cross section of the beam shown in Fig. 12.10.

1. Angle to the force plane trace we will plot from the positive direction of the axis . Corner We will always take it sharp, but taking into account the sign. Any angle is considered positive if in the right coordinate system it is plotted from the positive direction of the axis counterclockwise, and negative if the angle is laid clockwise. IN in this case corner is considered negative (
).

2. Determine the ratio of axial moments of inertia:

.

3. We write the equation of the zero line for oblique bending in the form from which we find the angle :

;
.

4. Angle turned out to be positive, so we set it aside from the positive direction of the axis counterclockwise to the zero line (Fig. 12.10).

Example 12.2. Determine the magnitude of the normal stress at point A of the cross section of the beam during oblique bending, if the bending moment
kNm, point coordinates
cm,
see Dimensions of the cross section of the beam and the angle of inclination of the force plane are shown in Fig. 12.11.

1. Let us first calculate the moments of inertia of the section relative to the axes And :

cm 4;
cm 4.

2. Let us write formula (12.11) to determine normal stresses at an arbitrary point of the cross section during oblique bending. When substituting the value of the bending moment into formula (12.11), it should be taken into account that the bending moment according to the condition of the problem is positive.

7.78 MPa.

Example 12.3. Determine the dimensions of the cross section of the beam shown in Fig. 12.12a. Beam material – steel with permissible stress
MPa. The aspect ratio is specified
. Loads and angle of inclination of the force plane are shown in Fig. 12.12c.

1. To determine the position of the dangerous section, we construct a diagram of bending moments (Fig. 12.12b). Section A is dangerous. Maximum bending moment in dangerous section
kNm.

2. The dangerous point in section A will be one of the corner points. We write the strength condition in the form

,

Where can we find it, given that the relation
:

3. Determine the dimensions of the cross section. Axial moment of resistance
taking into account the relationship of the parties
is equal to:

cm 3, from where

cm;
cm.

Example 12.4. As a result of the bending of the beam, the center of gravity of the section moved in the direction determined by the angle with axle (Fig. 12.13, a). Determine the angle of inclination force plane. The shape and dimensions of the cross section of the beam are shown in the figure.

1. To determine the angle of inclination of the trace of the force plane Let's use expression (12.22):

, where
.

Ratio of moments of inertia
(see example 12.1). Then

.

Let's set aside this angle value from the positive axis direction (Fig. 12.13, b). The trace of the force plane in Fig. 12.13b is shown as a dashed line.

2. Let's check the resulting solution. To do this, with the found value of the angle Let's determine the position of the zero line. Let's use expression (12.13):

.

The zero line is shown in Fig. 12.13 as a dotted line. The zero line must be perpendicular to the deflection line. Let's check this:

Example 12.5. Determine the total deflection of the beam in section B during oblique bending (Fig. 12.14a). Beam material – steel with elastic modulus
MPa. Cross-sectional dimensions and angle of inclination of the force plane are shown in Fig. 12.14b.

1. Determine the projections of the total deflection vector in section A And . To do this, we will construct a load diagram of bending moments
(Fig. 12.14, c), single diagram
(Fig. 12.14, d).

2. Using the Mohr-Simpson method, we multiply the cargo
and single
diagrams of bending moments using expressions (12.20) and (12.21):

m
mm.

m
mm.

Axial moments of inertia of the section
cm 4 and
We take cm 4 from example 12.1.

3. Determine the total deflection of section B:

.

The found values ​​of the projections of the total deflection and the full deflection itself are plotted in the drawing (Fig. 12.14b). Since the projections of the total deflection turned out to be positive when solving the problem, we put them aside in the direction of action of the unit force, i.e. down ( ) and left ( ).

5. To check the correctness of the solution, we determine the angle of inclination of the zero line to the axis :

Let's add up the modules of the angles of the direction of the total deflection And :

This means that the full deflection is perpendicular to the zero line. Thus, the problem was solved correctly.

Brief information from the theory

The timber is subject to conditions of complex resistance if several internal force factors in the cross sections are not equal to zero at the same time.

Of greatest practical interest are following cases complex loading:

1. Oblique bend.

2. Bending with tension or compression when in transverse
section, longitudinal force and bending moments arise, such as
for example, during eccentric compression of a beam.

3. Bend with torsion, characterized by the presence in the butt
river sections of bending (or two bending) and torsional
moments.

Oblique bend.

Oblique bending is a case of beam bending in which the plane of action of the total bending moment in the section does not coincide with any of the main axes of inertia. It is most convenient to consider oblique bending as the simultaneous bending of a beam in two main planes zoy and zox, where the z axis is the axis of the beam, and the x and y axes are the main central axes of the cross section.

Let us consider a cantilever beam of rectangular cross-section loaded with force P (Fig. 1).

Having expanded the force P along the main central axes of the cross section, we obtain:

P y =Pcos φ, P x =Psin φ

Bending moments occur in the current section of the beam

M x = - P y z = -P z cos φ,

M y = P x z = P z sin φ.

The sign of the bending moment M x is determined in the same way as in the case straight bend. We will consider the moment M y positive if at points with positive value coordinate x this moment causes tensile stresses. By the way, the sign of the moment M y can be easily established by analogy with the determination of the sign of the bending moment M x, if you mentally rotate the section so that the x axis coincides with the original direction of the y axis.

The stress at an arbitrary point in the cross section of a beam can be determined using formulas for determining stress for the case flat bend. Based on the principle of independent action of forces, we summarize the stresses caused by each of the bending moments

(1)

The values ​​of bending moments (with their own signs) and the coordinates of the point at which the stress is calculated are substituted into this expression.

To determine the dangerous points of the section, it is necessary to determine the position of the zero or neutral line (the geometric location of the points of the section at which stresses σ = 0). Maximum stresses occur at points furthest from the zero line.

The zero line equation is obtained from equation (1) at =0:

whence it follows that the zero line passes through the center of gravity of the cross section.

The tangential stresses arising in the sections of the beam (at Q x ≠0 and Q y ≠0), as a rule, can be neglected. If there is a need to determine them, then first the components of the total shear stress τ x and τ y are calculated according to the formula of D.Ya. Zhuravsky, and then the latter are geometrically summed up:

To assess the strength of a beam, it is necessary to determine the maximum normal stresses in the dangerous section. Since at the most loaded points the stress state is uniaxial, the strength condition when calculating using the permissible stress method takes the form

For plastic materials,

For fragile materials,

n - safety factor.

If you calculate using the method limit states, then the strength condition has the form:

where R is the design resistance,

m – coefficient of working conditions.

In cases where the beam material has different resistance to tension and compression, it is necessary to determine both the maximum tensile and maximum compressive stresses, and a conclusion about the strength of the beam is made from the relationships:

where R p and R c are the calculated tensile and compressive resistances of the material, respectively.

To determine the deflections of a beam, it is convenient to first find the displacements of the section in the main planes in the direction of the x and y axes.

The calculation of these displacements ƒ x and ƒ y can be done by constructing a universal equation for the curved axis of the beam or by energy methods.

The total deflection can be found as a geometric sum:

the beam rigidity condition has the form:

where - is the permissible deflection of the beam.

Eccentric compression

In this case, the compressive force P on the beam is directed parallel to the axis of the beam and is applied at a point that does not coincide with the center of gravity of the section. Let X p and Y p be the coordinates of the point of application of force P, measured relative to the main central axes (Fig. 2).

Effective load causes the appearance of the following internal force factors in cross sections: N= -P, Mx= -Py p, My=-Px p

The signs of the bending moments are negative, since the latter cause compression at points belonging to the first quarter. The stress at an arbitrary point of the section is determined by the expression

(9)

Substituting the values ​​of N, Mx and Mu, we get

(10)

Since Ух= F, Уу= F (where i x and i y are the main radii of inertia), the last expression can be reduced to the form

(11)

We obtain the zero line equation by setting =0

1+ (12)

The segments and cut off by the zero line on the coordinate axes are expressed as follows:

Using dependencies (13), you can easily find the position of the zero line in the section (Fig. 3), after which the points most distant from this line are determined, which are dangerous, since maximum stresses arise in them.

The stressed state at the points of the section is uniaxial, therefore the condition for the strength of the beam is similar to the previously considered case of oblique bending of the beam - formulas (5), (6).

During eccentric compression of beams, the material of which weakly resists tension, it is desirable to prevent the appearance of tensile stresses in the cross-section. Stresses of the same sign will arise in the section if the zero line passes outside the section or, in extreme cases, touches it.

This condition is satisfied when the compressive force is applied inside a region called the core of the section. The core of the section is an area covering the center of gravity of the section and is characterized by the fact that any longitudinal force applied inside this zone causes stresses of the same sign at all points of the beam.

To construct the core of the section, it is necessary to set the position of the zero line so that it touches the section without intersecting it anywhere, and find the corresponding point of application of the force P. By drawing a family of tangents to the section, we obtain a set of poles corresponding to them, the geometric location of which will give the outline (contour) of the core sections.

Let, for example, be given the section shown in Fig. 4, with the main central axes x and y.

To construct the core of the section, we present five tangents, four of which coincide with the sides AB, DE, EF and FA, and the fifth connects points B and D. By measuring or calculating from the cutting, cut off by the indicated tangents I-I, . . . ., 5-5 on the x, y axes and substituting these values ​​in dependence (13), we determine the coordinates x p, y p for the five poles 1, 2....5, corresponding to the five positions of the zero line. Tangent I-I can be moved to position 2-2 by rotating around point A, while pole I must move in a straight line and, as a result of rotating the tangent, move to point 2. Consequently, all poles corresponding to intermediate tangent positions between I-I and 2-2 will be located on straight 1-2. Similarly, it can be proven that the remaining sides of the core of the section will also be rectangular, i.e. the core of the section is a polygon, to construct which it is enough to connect poles 1, 2, ... 5 with straight lines.

Bending with torsion of a round beam.

When bending with torsion in the cross section of a beam, in the general case, five internal force factors are not equal to zero: M x, M y, M k, Q x and Q y. However, in most cases, the influence of shear forces Q x and Q y can be neglected if the section is not thin-walled.

Normal stresses in a cross section can be determined from the magnitude of the resulting bending moment

because the neutral axis is perpendicular to the cavity of action of the moment M u.

In Fig. Figure 5 shows the bending moments M x and M y in the form of vectors (the directions M x and M y are chosen positive, i.e., such that at the points of the first quadrant the stress sections are tensile).

The direction of the vectors M x and M y is chosen in such a way that an observer, looking from the end of the vector, sees them directed counterclockwise. In this case, the neutral line coincides with the direction of the resulting moment vector M u, and the most loaded points of the section A and B lie in the plane of action of this moment.

Spatial (complex) bending

Spatial bending is a type of complex resistance in which only bending moments and act in the cross section of the beam. The full bending moment acts in none of the main planes of inertia. There is no longitudinal force. Spatial or complex bending is often called non-planar bending because the bent axis of the rod is not a plane curve. This bending is caused by forces acting in different planes perpendicular to the axis of the beam (Fig. 1.2.1).

Fig.1.2.1

Following the order of solving problems with complex resistance outlined above, we lay out the spatial system of forces presented in Fig. 1.2.1, into two such that each of them acts in one of the main planes. As a result, we get two flat transverse bends - in the vertical and horizontal planes. Of the four internal force factors that arise in the cross section of the beam, we will take into account the influence of only bending moments. We construct diagrams caused by the corresponding forces (Fig. 1.2.1).

Analyzing the diagrams of bending moments, we come to the conclusion that section A is dangerous, since it is in this section that the largest bending moments and occur. Now it is necessary to establish the dangerous points of section A. To do this, we will construct a zero line. The zero line equation, taking into account the sign rule for the terms included in this equation, has the form:

Here the “” sign is adopted near the second term of the equation, since the stresses in the first quarter caused by the moment will be negative.

Let's determine the angle of inclination of the zero line with the positive direction of the axis (Fig. 12.6):

Rice. 1.2.2

From equation (8) it follows that the zero line for spatial bending is a straight line and passes through the center of gravity of the section.

From Fig. 1.2.2 it is clear that the greatest stresses will arise at the points of section No. 2 and No. 4 furthest from the zero line. The normal stresses at these points will be the same in magnitude, but different in sign: at point No. 4 the stresses will be positive, i.e. tensile, at point No. 2 - negative, i.e. compressive. The signs of these stresses were established from physical considerations.

Now that the dangerous points have been established, let's calculate the maximum stresses in section A and check the strength of the beam using the expression:

The strength condition (10) allows not only to check the strength of the beam, but also to select the dimensions of its cross section if the aspect ratio of the cross section is specified.

In the case of calculating a round beam under the action of bending and torsion (Fig. 34.3), it is necessary to take into account normal and tangential stresses, since the maximum stress values ​​in both cases occur on the surface. The calculation should be carried out according to the theory of strength, replacing the complex stress state with an equally dangerous simple one.

Maximum torsional stress in section

Maximum bending stress in section

According to one theory of strength, depending on the material of the beam, the equivalent stress for the dangerous section is calculated and the beam is tested for strength using the permissible bending stress for the material of the beam.

For a round beam, the sectional moments of resistance are as follows:

When calculating according to the third theory of strength, the theory of maximum shear stress, the equivalent stress is calculated using the formula

The theory is applicable to plastic materials.

When calculating according to the theory of shape change energy, the equivalent stress is calculated using the formula

The theory is applicable to ductile and brittle materials.

theory of maximum shear stress:

Equivalent stress when calculated according to theory of shape change energy:

where is the equivalent moment.

Strength condition

Examples of problem solving

Example 1. For a given stress state (Fig. 34.4), using the hypothesis of maximum tangential stresses, calculate the safety factor if σ T = 360 N/mm 2.

1. How is the state of stress at a point characterized and how is it depicted?

2. What areas and what voltages are called the main ones?

3. List the types of stressed states.

4. What characterizes the deformed state at a point?

5. In what cases do limiting stress states arise in ductile and brittle materials?

6. What is equivalent voltage?

7. Explain the purpose of strength theories.

8. Write formulas for calculating equivalent stresses in calculations using the theory of maximum tangential stresses and the theory of shape change energy. Explain how to use them.

LECTURE 35

Topic 2.7. Calculation of a beam of round cross-section with a combination of basic deformations

Know the formulas for equivalent stresses based on the hypotheses of the highest tangential stresses and the energy of shape change.

Be able to calculate the strength of a round cross-section beam under a combination of basic deformations.

Formulas for calculating equivalent stresses

Equivalent stress according to the maximum shear stress hypothesis

Equivalent stress according to the shape change energy hypothesis

Strength condition under the combined action of bending and torsion

Where M EKV- equivalent moment.

Equivalent moment according to the hypothesis of maximum tangential stresses

Equivalent moment according to the shape change energy hypothesis

Shaft calculation feature

Most shafts experience a combination of bending and torsion deformation. Typically the shafts are straight bars with a round or annular cross-section. When calculating shafts, tangential stresses from the action of transverse forces are not taken into account due to their insignificance.

Calculations are carried out on dangerous cross sections. When loading a shaft spatially, the hypothesis of independence of the action of forces is used and bending moments are considered in two mutually perpendicular planes, and the total bending moment is determined by geometric summation.

Examples of problem solving

Example 1. Internal force factors arise in the dangerous cross-section of a round beam (Fig. 35.1) M x; M y; Mz.

M x And M y- bending moments in planes ooh And zOx accordingly; M z- torque. Check the strength using the hypothesis of maximum tangential stresses if [ σ ] = 120 MPa. Initial data: M x= 0.9 kN m; M y = 0.8 kN m; M z = 2.2 kN*m; d= 60 mm.

Solution

We construct diagrams of normal stresses from the action of bending moments relative to the axes Oh And Oh and a diagram of shear stresses due to torsion (Fig. 35.2).

The maximum shear stress occurs at the surface. Maximum normal stresses from moment M x arise at a point A, maximum normal stresses from moment M y at the point IN. Normal stresses add up because bending moments in mutually perpendicular planes add up geometrically.

Total bending moment:

We calculate the equivalent moment using the theory of maximum tangential stresses:

Strength condition:

Sectional moment of resistance: W oce in oe = 0.1 60 3 = 21600 mm 3.

Checking the strength:

Durability is guaranteed.

Example 2. From the strength condition, calculate the required shaft diameter. There are two wheels mounted on the shaft. Two circumferential forces act on the wheels F t 1 = 1.2kN; F t 2= 2kN and two radial forces in the vertical plane F r 1= 0.43kN; F r 2 = 0.72 kN (Fig. 35.3). The wheel diameters are respectively equal d 1= 0.1m; d 2= 0.06 m.

Accept for shaft material [ σ ] = 50MPa.

The calculation is carried out according to the hypothesis of maximum tangential stresses. Neglect the weight of the shaft and wheels.

Solution

Note. We use the principle of independent action of forces and draw up design diagrams of the shaft in the vertical and horizontal planes. We determine the reactions in the supports in the horizontal and vertical planes separately. We construct diagrams of bending moments (Fig. 35.4). Under the influence of circumferential forces, the shaft twists. Determine the torque acting on the shaft.

Let's draw up a design diagram of the shaft (Fig. 35.4).

1. Torque on the shaft:

2. We consider the bend in two planes: horizontal (pl. H) and vertical (pl. V).

In the horizontal plane we determine the reactions in the support:

WITH And IN:

In the vertical plane we determine the reactions in the support:

Determine bending moments at points C and B:

Total bending moments at points C and B:

At the point IN maximum bending moment; torque also acts here.

We calculate the shaft diameter based on the most loaded section.

3. Equivalent moment at a point IN according to the third theory of strength

4. Determine the diameter of the shaft of circular cross-section from the strength condition

We round the resulting value: d= 36 mm.

Note. When choosing shaft diameters, use the standard range of diameters (Appendix 2).

5. Determine the required dimensions of the annular shaft at c = 0.8, where d - O.D. shaft

The diameter of an annular shaft can be determined by the formula

Let's accept d = 42 mm.

The overload is insignificant. d BH = 0.8d = 0.8 42 = 33.6mm.

Round to the value dBH= 33 mm.

6. Let’s compare the metal costs by shaft cross-sectional area in both cases.

Cross-sectional area of ​​a solid shaft

Cross-sectional area of ​​hollow shaft

The cross-sectional area of ​​a solid shaft is almost twice that of an annular shaft:

Example 3. Determine the cross-sectional dimensions of the shaft (Fig. 2.70, A) control drive. Pedal traction force P 3, forces transmitted by the mechanism P 1, P 2, P 4. Shaft material - StZ steel with yield strength σ t = 240 N/mm 2, required safety factor [ n] = 2.5. The calculation is performed using the hypothesis of shape change energy.

Solution

Let us consider the equilibrium of the shaft, having previously introduced the forces R 1, R 2, R 3, R 4 to points lying on its axis.

Transferring strength P 1 parallel to themselves at points TO And E, it is necessary to add pairs of forces with moments equal to the moments of forces P 1 relative to points TO And E, i.e.

These pairs of forces (moments) are conventionally shown in Fig. 2.70 , b in the form of arcuate lines with arrows. Similarly when transferring forces R 2, R 3, R 4 to points K, E, L, N need to add a couple of forces with moments

Supports of the shaft shown in Fig. 2.70, a, should be considered as spatial hinge supports that prevent movements in the direction of the axes X And at(the selected coordinate system is shown in Fig. 2.70, b).

Using the calculation scheme shown in Fig. 2.70, V, let's create the equilibrium equations:

therefore, the support reactions N A And N V defined correctly.

Torque diagrams M z and bending moments M y are presented in Fig. 2.70, G. The dangerous section is the one to the left of point L.

The strength condition has the form:

where is the equivalent moment according to the shape change energy hypothesis

Required shaft outer diameter

We take d = 45 mm, then d 0 = 0.8 * 45 = 36 mm.

Example 4. Check the strength of the intermediate shaft (Fig. 2.71) of the spur gearbox if the shaft transmits power N= 12.2 kW at speed n= 355 rpm. The shaft is made of steel St5 with a yield strength σ t = 280 N/mm 2. Required safety factor [ n] = 4. When calculating, apply the hypothesis of the highest tangential stresses.

Note. District efforts P 1 And R 2 lie in a horizontal plane and are directed tangentially to the circles of the gears. Radial forces T 1 And T 2 lie in the vertical plane and are expressed in terms of the corresponding circumferential force as follows: T = 0,364R.

Solution

In Fig. 2.71, A a schematic drawing of the shaft is presented; in Fig. 2.71, b shows the diagram of the shaft and the forces arising in the gearing.

Let's determine the moment transmitted by the shaft:

Obviously, m = m 1 = m 2(torsional moments applied to the shaft, with uniform rotation, are equal in magnitude and opposite in direction).

Let us determine the forces acting on the gears.

Circumferential forces:

Radial forces:

Consider the balance of the shaft AB, having previously brought forces P 1 And R 2 to points lying on the shaft axis.

Transferring force P 1 parallel to itself to a point L, you need to add a couple of forces with a moment equal to the moment of force P 1 relative to the point L, i.e.

This pair of forces (moment) is conventionally shown in Fig. 2.71, V in the form of an arcuate line with an arrow. Similarly when transferring force R 2 to the point TO you need to attach (add) a couple of forces with a moment

Supports of the shaft shown in Fig. 2.71, A, should be considered as spatial hinge supports that prevent linear movements in the directions of the axes X And at(the selected coordinate system is shown in Fig. 2.71, b).

Using the calculation scheme shown in Fig. 2.71, G, let’s draw up the equilibrium equations for the shaft in the vertical plane:

Let's create a verification equation:

therefore, the support reactions in the vertical plane are determined correctly.

Consider the balance of the shaft in the horizontal plane:

Let's create a verification equation:

therefore, the support reactions in the horizontal plane are determined correctly.

Torque diagrams M z and bending moments M x And M y are presented in Fig. 2.71, d.

The section is dangerous TO(see Fig. 2.71, G,d). Equivalent moment according to the hypothesis of greatest tangential stresses

Equivalent stress according to the hypothesis of the highest tangential stresses for a dangerous point of the shaft

Safety factor

which is significantly more [ n] = 4, therefore, the strength of the shaft is ensured.

When calculating the shaft's strength, the change in stress over time was not taken into account, which is why such a significant safety factor was obtained.

Example 5. Determine the dimensions of the cross section of the beam (Fig. 2.72, A). The beam material is steel 30XGS with conditional yield limits in tension and compression σ o, 2р = σ tr = 850 N/mm 2, σ 0.2 c = σ Tc = 965 N/mm 2. Safety factor [ n] = 1,6.

Solution

The beam works under the combined action of tension (compression) and torsion. With such a load, two internal force factors arise in the cross sections: longitudinal force and torque.

Diagrams of longitudinal forces N and torques M z shown in Fig. 2.72, b, c. In this case, determine the position of the dangerous section using diagrams N And M z impossible, since the cross-sectional dimensions of the beam sections are different. To determine the position of the dangerous section, diagrams of normal and maximum tangential stresses along the length of the beam should be constructed.

According to the formula

we calculate the normal stresses in the cross sections of the beam and construct a diagram o (Fig. 2.72, G).

According to the formula

We calculate the maximum tangential stresses in the cross sections of the beam and construct a diagram t tah(Fig* 2.72, d).

Possibly dangerous points are the contour points of cross-sections of sections AB And CD(see Fig. 2.72, A).

In Fig. 2.72, e diagrams are shown σ And τ for section cross sections AB.

Let us recall that in this case (a beam of round cross-section works under the combined action of tension, compression and torsion), all points of the cross-section contour are equally dangerous.

In Fig. 2.72, and

In Fig. 2.72, h Diagrams a and t are shown for cross sections of the section CD.

In Fig. 2.72, And the voltages on the original sites at the dangerous point are shown.

Principal stresses at a dangerous point in a section CD:

According to Mohr's strength hypothesis, the equivalent stress for the dangerous point of the section under consideration is

The contour points of the cross sections of section AB turned out to be dangerous.

The strength condition has the form:

Example 2.76. Determine the permissible force value R from the condition of the strength of the rod Sun(Fig. 2.73). The rod material is cast iron with a tensile strength σ vr = 150 N/mm 2 and a compressive strength σ sun = 450 N/mm 2. Required safety factor [ n] = 5.

Note. Broken timber ABC located in a horizontal plane, and the rod AB perpendicular to Sun. Powers R, 2R, 8R lie in a vertical plane; strength 0.5 R, 1.6 R- horizontal and perpendicular to the rod Sun; strength 10R, 16R coincide with the axis of the rod Sun; a pair of forces with a moment m = 25Pd is located in a vertical plane perpendicular to the axis of the rod Sun.

Solution

Let's bring strength R and 0.5P to the center of gravity of cross section B.

Transferring force P parallel to itself to point B, you need to add a couple of forces with a moment equal to the moment of force R relative to the point IN, i.e. a pair with moment m 1 = 10 Pd.

Strength 0.5R we move along its line of action to point B.

Loads acting on the rod sun, shown in Fig. 2.74, A.

We construct diagrams of internal force factors for the rod Sun. Under the specified loading of the rod, six of them arise in its cross sections: longitudinal force N, shear forces Qx And Qy, torque Mz bending moments Mx And Mu.

Diagrams N, Mz, Mx, Mu are presented in Fig. 2.74, b(the ordinates of the diagrams are expressed in terms of R And d).

Diagrams Qy And Qx we do not build, since the tangential stresses corresponding to the transverse forces are small.

In the example under consideration, the position of the dangerous section is not obvious. Presumably, section K (end of section I) and S.

Principal stresses at point L:

According to Mohr's strength hypothesis, the equivalent stress for point L

Let us determine the magnitude and plane of action of the bending moment Mie in section C, shown separately in Fig. 2.74, d. The same figure shows diagrams σ И, σ N, τ for section C.

Stresses on the original sites at the point N(Fig. 2.74, e)

Principal stresses at a point N:

According to Mohr's strength hypothesis, the equivalent stress for a point N

Stresses on the original sites at point E (Fig. 2.74, and):

Principal stresses at point E:

According to Mohr's strength hypothesis, the equivalent stress for point E

The point turned out to be dangerous L, for which

The strength condition has the form:

Test questions and assignments

1. What stress state occurs in the cross section of the shaft under the combined action of bending and torsion?

2. Write the strength condition for calculating the shaft.

3. Write formulas for calculating the equivalent moment when calculating according to the hypothesis of maximum tangential stresses and the hypothesis of shape change energy.

4. How is the dangerous section selected when calculating the shaft?

The combination of bending and torsion of beams of circular cross-section is most often considered when calculating shafts. Cases of bending with torsion of beams of non-circular cross-section are much less common.

In § 1.9 it is established that in the case when the moments of inertia of the section relative to the main axes are equal to each other, oblique bending of the beam is impossible. In this regard, oblique bending of round beams is impossible. Therefore, in the general case of external forces, a round beam experiences a combination of the following types of deformation: direct transverse bending, torsion and central tension (or compression).

Let us consider such a special case of calculating a beam of circular cross-section, when in its cross sections the longitudinal force is equal to zero. In this case, the beam works under the combined action of bending and torsion. To find the dangerous point of the beam, it is necessary to establish how the values ​​of bending and torque moments change along the length of the beam, that is, to construct diagrams of the total bending moments M and torques. We will consider the construction of these diagrams using a specific example of the shaft shown in Fig. 22.9, a. The shaft rests on bearings A and B and is driven by motor C.

Pulleys E and F are mounted on the shaft, through which drive belts with tension are thrown. Let us assume that the shaft rotates in bearings without friction; we neglect the own weight of the shaft and pulleys (in the case where their own weight is significant, it should be taken into account). Let's direct the axis of the cross section of the shaft vertically, and the axis horizontally.

The magnitudes of the forces can be determined using formulas (1.6) and (2.6), if, for example, the power transmitted by each pulley, the angular velocity of the shaft and the ratios are known. After determining the magnitudes of the forces, these forces are transferred parallel to themselves to the longitudinal axis of the shaft. In this case, torsional moments are applied to the shaft in the sections in which pulleys E and F are located and are equal to, respectively. These moments are balanced by the moment transmitted from the engine (Fig. 22.9, b). The forces are then decomposed into vertical and horizontal components. Vertical forces will cause vertical reactions in the bearings, and horizontal forces will cause horizontal reactions. The magnitudes of these reactions are determined as for a beam lying on two supports.

The diagram of bending moments acting in the vertical plane is constructed from vertical forces (Fig. 22.9, c). It is shown in Fig. 22.9, d. Similarly, from horizontal forces (Fig. 22.9, e), a diagram of bending moments acting in the horizontal plane is constructed (Fig. 22.9, f).

From the diagrams you can determine (in any cross section) the total bending moment M using the formula

Using the values ​​of M obtained using this formula, a diagram of the total bending moments is constructed (Fig. 22.9, g). In those sections of the shaft in which straight, limiting diagrams intersect the axes of the diagrams at points located on the same vertical, diagram M is limited by straight lines, and in other sections it is limited by curves.

(see scan)

For example, in the section of the shaft in question, the length of the diagram M is limited to a straight line (Fig. 22.9, g), since the diagrams in this section are limited by straight lines and intersecting the axes of the diagrams at points located on the same vertical.

Point O of the intersection of the straight line with the axis of the diagram is located on the same vertical. A similar situation is typical for a shaft section with a length

The diagram of total (total) bending moments M characterizes the magnitude of these moments in each section of the shaft. The planes of action of these moments in different sections of the shaft are different, but the ordinates of the diagram for all sections are conventionally aligned with the plane of the drawing.

The diagram of torques is constructed in the same way as for pure torsion (see § 1.6). For the shaft in question, it is shown in Fig. 22.9, z.

The dangerous section of the shaft is established using diagrams of the total bending moments M and torques. If in the section of a beam of constant diameter with the greatest bending moment M the greatest torque also acts, then this section is dangerous. In particular, the shaft under consideration has such a section located to the right of pulley F at an infinitesimal distance from it.

If the maximum bending moment M and the maximum torque act in different cross sections, then a section in which neither the value is the greatest may turn out to be dangerous. With beams of variable diameter, the most dangerous section may be the one in which significantly lower bending and torsional moments act than in other sections.

In cases where the dangerous section cannot be determined directly from the diagrams M and it is necessary to check the strength of the beam in several of its sections and in this way establish dangerous stresses.

Once a dangerous section of the beam has been established (or several sections have been identified, one of which may turn out to be dangerous), it is necessary to find dangerous points in it. To do this, let us consider the stresses arising in the cross section of the beam when a bending moment M and a torque are simultaneously acting in it

In round bars, the length of which is many times larger diameter, the magnitude of the greatest tangential stresses from shear force are small and are not taken into account when calculating the strength of beams for the combined action of bending and torsion.

In Fig. Figure 23.9 shows the cross section of a round beam. In this section, a bending moment M and a torque act. The axis y is taken to be perpendicular to the plane of action of the bending moment. The y axis is therefore the neutral axis of the section.

In the cross section of the beam, normal stresses arise from bending and shear stresses from torsion.

Normal stresses a are determined by the formula. The diagram of these stresses is shown in Fig. 23.9. The largest normal stresses in absolute value occur at points A and B. These stresses are equal

where is the axial moment of resistance of the cross section of the beam.

Tangential stresses are determined by the formula. The diagram of these stresses is shown in Fig. 23.9.

At each point of the section they are directed normal to the radius connecting this point with the center of the section. The highest shear stresses occur at points located along the perimeter of the section; they are equal

where is the polar moment of resistance of the cross section of the beam.

For a plastic material, points A and B of the cross section, at which both normal and shear stresses simultaneously reach highest value, are dangerous. At fragile material The dangerous point is the one at which tensile stresses arise from the bending moment M.

The stressed state of an elementary parallelepiped isolated in the vicinity of point A is shown in Fig. 24.9, a. Along the faces of the parallelepiped, which coincide with the cross sections of the beam, normal stresses and tangential stresses act. Based on the law of pairing of tangential stresses, stresses also arise on the upper and lower faces of the parallelepiped. Its remaining two faces are stress-free. Thus, in this case there is private view plane stress state, discussed in detail in Chap. 3. The main stresses amax and are determined by formulas (12.3).

After substituting the values ​​into them we get

Voltages have different signs and therefore

An elementary parallelepiped, highlighted in the vicinity of point A by the main areas, is shown in Fig. 24.9, b.

The calculation of beams for strength during bending with torsion, as already noted (see the beginning of § 1.9), is carried out using strength theories. In this case, the calculation of beams from plastic materials is usually carried out on the basis of the third or fourth theory of strength, and from brittle ones - according to Mohr’s theory.

According to the third theory of strength [see. formula (6.8)], substituting the expressions into this inequality [see. formula (23.9)], we obtain