Water pipes

Bending with torsion of a round bar. Calculation of a round beam for bending with torsion Bending with torsion of a round cross-section beam

Spatial (complex) bend

Spatial bending is such a type of complex resistance, in which only bending moments and act in the cross section of the beam. The total bending moment acts in none of the principal planes of inertia. There is no longitudinal force. A three-dimensional or complex bend is often referred to as a non-planar bend, since the bent axis of the bar is not a planar curve. Such a bend 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 procedure for solving problems with complex resistance, outlined above, we decompose the spatial system of forces shown in Fig. 1.2.1 into two such that each of them acts in one of the main planes. As a result, we obtain 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 build diagrams caused by forces, respectively (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 u occur. Now it is necessary to establish 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

It follows from equation (8) that the zero line in case of spatial bending is a straight line and passes through the center of gravity of the section.

From fig. 1.2.2 it can be seen that the greatest stresses will occur at the points of section No. 2 and No. 4 most distant from the zero line. By size normal stresses at these points will be the same, but differ in sign: at point No. 4, the voltages will be positive, i.e. stretching, at point No. 2 - negative, i.e. compressive. The signs of these stresses were established from physical considerations.

Now that the dangerous points are set, we 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 ratio of the sides of the cross section is given.

The combination of bending and torsion of bars of circular cross section is most often considered in the calculation of shafts. Cases of bending with torsion of bars of non-circular 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 bars is impossible. Therefore, in the general case of the action of external forces, a round beam experiences a combination of the following types of deformation: direct transverse bending, torsion and central tension (or compression).

Consider such a special case of calculating a round bar, when in its cross sections longitudinal force equals zero. In this case, the beam works on the combined action of bending and torsion. To find 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, i.e., 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 is supported by bearings A and B and is driven by motor C.

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

The magnitude 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 magnitude of the forces, these forces are transferred parallel to themselves to the longitudinal axis of the shaft. At the same time, torsional moments are applied to the shaft in the sections in which pulleys E and F are located and equal, respectively. These moments are balanced by the moment transmitted from the engine (Fig. 22.9, b). Then the forces are decomposed into vertical and horizontal components. Vertical forces will cause vertical reactions in the bearings and horizontal forces - horizontal reactions. The magnitudes of these reactions are determined as for a beam lying on two supports.

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

According to the diagrams, it is possible to determine (in any cross section) the total bending moment M by the formula

Based on the values of M obtained using this formula, a plot of total bending moments is constructed (Fig. 22.9, g). On those sections of the shaft where the straight lines, limiting diagrams intersect the axes of the diagrams at points located on the same vertical, the 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 under consideration, the length of the diagram M is limited to a straight line (Fig. 22.9, g), since the diagrams in this section are limited to straight lines and intersecting the axes of the diagrams at points located on the same vertical.

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

The diagram of the 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 are conventionally aligned for all sections with the plane of the drawing.

The torque diagram is constructed in the same way as for pure torsion (see § 1.6). For the shaft under consideration, it is shown in Fig. 22.9, s.

The dangerous section of the shaft is set using the diagrams of the total bending moments M and torques. If the section of the beam of constant diameter with the largest bending moment M also has the largest torque, then this section is dangerous. In particular, for the shaft under consideration, this is the section located to the right of the pulley F at an infinitely small distance from it.

If the greatest bending moment M and the greatest torque act in different cross sections, then the section in which neither the value nor is the largest may be dangerous. With bars of variable diameter, the most dangerous section may be the one in which there are significantly lower bending and torsional moments than in other sections.

In cases where the dangerous section cannot be established 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.

After the dangerous section of the beam is established (or several sections are planned, one of which may turn out to be dangerous), it is necessary to find dangerous points in it. To do this, consider the stresses that arise in the cross section of the beam, when the bending moment M and the torque

In round bars, the length of which is many times greater than the diameter, the values of the largest shear stresses from the transverse force are small and are not taken into account when calculating the strength of the bars for the combined action of bending and torsion.

On fig. 23.9 shows a cross section of a round bar. In this section, a bending moment M and a torque act. For the y-axis, the axis is taken, 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, there are normal stresses o 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 absolute stresses occur at points A and B. These stresses are equal to

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

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

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

With a plastic material, points A and B of the cross section, in which both normal and shear stresses simultaneously reach the greatest value are dangerous. With a brittle material, the dangerous point is one of these points at which tensile stresses arise from the bending moment M.

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

After substituting the values in them, we get

The voltages have different signs and, therefore,

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

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

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

Brief information from the theory

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

The following cases of complex loading are of the greatest practical interest:

1. Oblique bend.

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

3. Bending with torsion, characterized by the presence in the pope
river sections of a bending (or two bending) and twisting
moments.

Oblique bend.

Oblique bending is such 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. An oblique bend is most conveniently considered as a 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.

Consider a cantilever beam of rectangular cross section, loaded with a force P (Fig. 1).

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

R y \u003d R cos φ, R x \u003d R sin φ

Bending moments occur in the current section of the beam

M x \u003d - P y z \u003d - P z cos φ,

M y \u003d P x z \u003d P z sin φ.

The sign of the bending moment M x is determined in the same way as in the case of direct bending. The moment M y will be considered positive if at points with a positive value of the x coordinate this moment causes tensile stresses. By the way, the sign of the moment M y is easy to establish by analogy with the definition of the sign of the bending moment M x, if you mentally rotate the section so that the x axis coincides with original direction y axis.

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

(1)

The values of the bending moments (with their 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 locus of the points of the section, in which the stresses σ = 0). The maximum stresses occur at the 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.

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

To assess the strength of the beam, it is necessary to determine the maximum normal stresses in the dangerous section. Since the stress state is uniaxial at the most loaded points, the strength condition in the calculation by the method of allowable stresses takes the form

For plastic materials

For brittle materials

n is the safety factor.

If we calculate according to the method limit states, then the strength condition has the form:

where R is the design resistance,

m is the coefficient of working conditions.

In cases where the beam material resists tension and compression differently, it is necessary to determine both the maximum tensile and maximum compressive stresses, and make a conclusion about the strength of the beam from the ratios:

where R p and R c are the design resistances of the material in tension and compression, respectively.

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

Calculation of these displacements ƒ x and ƒ y can be carried out by drawing up a universal equation for the bent axis of the beam or by energy methods.

The total deflection can be found as a geometric sum:

the stiffness condition of the beam has the form:

where - is the allowable deflection of the beam.

Eccentric compression

In this case, the force P compressing 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 the force P, measured relative to the main central axes (Fig. 2).

Operating load causes the following internal force factors to appear in the cross sections: N= -P, Mx= -Py p , My=-Px p

The signs of 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 My, we get

(10)

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

(11)

The zero line equation is obtained by setting =0

1+ (12)

Cut off by the zero line on the coordinate axes of the segment and , are expressed as follows:

Using dependencies (13), one 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 stress state at the points of the section is uniaxial, therefore the strength condition of the beam is similar to the previously considered case of oblique bending of the beam - formulas (5), (6).

With eccentric compression of the bars, the material of which weakly resists stretching, it is desirable to prevent the appearance of tensile stresses in the cross section. In the section, stresses of the same sign will arise 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 the 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 bar.

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. Having drawn a family of tangents to the section, we obtain a set of poles corresponding to them, the locus of which will give the outline (contour) of the core sections.

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

To construct the section kernel, we give 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 cut, cut off by the indicated tangents I-I, . . . ., 5-5 on the axes x, y 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 II can be transferred to position 2-2 by rotation around point A, while pole I must move in a straight line and, as a result of rotation of the tangent, go to point 2. Therefore, all poles corresponding to intermediate positions of the tangent between II and 2-2 will be located on direct 1-2. Similarly, one can prove that the remaining sides of the core of the section will also be rectangular, i.e. the core of the section is a polygon, for the construction of which it is enough to connect the poles 1, 2, ... 5 with straight lines.

Bending with torsion of a round bar.

When bending with torsion in the cross section of the 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 .

On fig. 5 shows the bending moments M x and M y as vectors (the directions M x and M y are chosen positive, i.e. such that at the points of the first quadrant of the section the stresses are tensile).

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

In the case of calculating a round bar under the action of bending and torsion (Fig. 34.3), it is necessary to take into account normal and shear 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.

Max voltage torsion in section

Maximum bending stress in section

According to one of the strength theories, 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 allowable bending stress for the material of the beam.

For a round beam, the section modulus moments are as follows:

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

The theory is applicable to plastic materials.

When calculating according to the theory of forming energy, the equivalent stress is calculated by the formula

The theory is applicable to ductile and brittle materials.

theory of maximum shear stresses:

Equivalent voltage when calculated according to theories of energy of shape change:

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 shear stresses, calculate the safety factor if σ T \u003d 360 N / mm 2.

1. What characterizes and how is the stress state at a point depicted?

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

3. List the types of stress states.

4. What characterizes the deformed state at a point?

5. In what cases do limit stress states occur in ductile and brittle materials?

6. What is the equivalent voltage?

7. Explain the purpose of strength theories.

8. Write formulas for calculating equivalent stresses in calculations according to the theory of maximum shear stresses and the theory of energy of deformation. Explain how to use them.

LECTURE 35

Topic 2.7. Calculation of a bar of circular cross section with a combination of basic deformations

Know the formulas for equivalent stresses according to the hypotheses of the largest tangential stresses and the energy of deformation.

To be able to calculate a beam of circular cross-section for strength with a combination of basic deformations.

Formulas for calculating equivalent stresses

Equivalent stress according to the hypothesis of maximum shear stresses

Equivalent stress according to the deformation energy hypothesis

Strength condition under the combined action of bending and torsion

where M EQ is the equivalent moment.

Equivalent moment according to the hypothesis of maximum shear stresses

Equivalent moment according to the shape change energy hypothesis

Feature of the calculation of shafts

Most shafts experience a combination of bending and torsional deformations. Shafts are usually straight bars with a round or annular section. When calculating shafts, shear stresses from the action of transverse forces are not taken into account due to their insignificance.

Calculations are carried out for dangerous cross sections. Under spatial loading of the shaft, the hypothesis of independence of the action of forces is used and the 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 In a dangerous cross section of a round beam, internal force factors arise (Fig. 35.1) M x; M y; M z .

M x And M y- bending moments in planes uoh And zOx respectively; Mz- torque. Check the strength according to the hypothesis of the largest shear stresses, if [ σ ] = 120 MPa. Initial data: M x= 0.9 kN m; M y = 0.8 kN m; Mz = 2.2 kN*m; d= 60 mm.

Solution

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

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

Total bending moment:

We calculate the equivalent moment according to the theory of maximum shear stresses:

Strength condition:

Section modulus: W oce in oe \u003d 0.1 60 3 \u003d 21600mm 3.

Checking strength:

Durability is guaranteed.

Example 2 Calculate the required shaft diameter from the strength condition. Two wheels are mounted on the shaft. There are two circumferential forces acting on the wheels F t 1 = 1.2kN; F t 2= 2kN and two radial forces in the vertical plane F r 1= 0.43 kN; F r 2 = 0.72 kN (Fig. 35.3). Wheel diameters are respectively equal d1= 0.1m; d2= 0.06 m.

Accept for shaft material [ σ ] = 50 MPa.

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

Solution

Instruction. We use the principle of independence of the action of forces, draw up design schemes of the shaft in the vertical and horizontal planes. We determine the reactions in the supports in the horizontal and vertical planes separately. We build diagrams of bending moments (Fig. 35.4). Under the action of circumferential forces, the shaft is twisted. Determine the torque acting on the shaft.

Let's make a calculation scheme of the shaft (Fig. 35.4).

1. Shaft torque:

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:

FROM 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 the maximum bending moment, the torque also acts here.

The calculation of the shaft diameter is carried out according to 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 with a circular cross section from the condition of strength

We round the resulting value: d= 36 mm.

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

5. We determine the required dimensions of the shaft with an annular section at c \u003d 0.8, where d is the outer diameter of the shaft.

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

Accept d= 42 mm.

The load is minor. d BH = 0.8d = 0.8 42 = 33.6mm.

Round to value dBH= 33 mm.

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

Cross-sectional area of 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 dimensions of the cross section of the shaft (Fig. 2.70, but) control drive. Pedal pull force P3, forces transmitted by the mechanism P 1, R 2, R 4. Shaft material - StZ steel with yield strength σ t = 240 N/mm 2 , required safety factor [ n] = 2.5. The calculation is performed according to the hypothesis of the energy of the form change.

Solution

Consider the balance of the shaft, after bringing the forces R 1, R 2, R 3, R 4 to points on its axis.

Transferring forces R 1 parallel to themselves into points TO And E, it is necessary to add pairs of forces with moments equal to the moments of forces R 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, H you need to add couples of forces with moments

The bearings of the shaft shown in fig. 2.70, a, should be considered as spatial hinged supports that prevent movement 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 in, we compose the equilibrium equations:

hence the support reactions ON THE And H B defined correctly.

Torque Plots Mz and bending moments M y are presented in fig. 2.70 G. The section to the left of point L is dangerous.

The strength condition has the form:

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

Required shaft outside diameter

We accept d \u003d 45 mm, then d 0 \u003d 0.8 * 45 \u003d 36 mm.

Example 4 Check the strength of the intermediate shaft (Fig. 2.71) cylindrical spur gearbox if the shaft transmits power N= 12.2 kW at speed P= 355 rpm. The shaft is made of St5 steel with a yield strength σ t \u003d 280 N / mm 2. Required safety factor [ n] = 4. When calculating, apply the hypothesis of the highest shear stresses.

Instruction. District Efforts R 1 And R 2 lie in a horizontal plane and are directed along the tangents to the circles gear wheels. Radial forces T1 And T 2 lie in the vertical plane and are expressed in terms of the corresponding circumferential force as follows: T = 0,364R.

Solution

On fig. 2.71, but 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.

Determine the moment transmitted by the shaft:

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

Determine the forces acting on the gears.

District Efforts:

Radial forces:

Consider the balance of the shaft AB, pre-bringing forces R 1 And R 2 to points lying on the axis of the shaft.

Transferring power R 1 parallel to itself to a point L, it is necessary to add a couple of forces with a moment equal to the moment of force R 1 relative to the point L, i.e.

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

The bearings of the shaft shown in fig. 2.71, but, should be considered as spatial hinged 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, we compose the equilibrium equations for the shaft in the vertical plane:

Let's make a test equation:

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

Consider the balance of the shaft in the horizontal plane:

Let's make a test equation:

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

Torque Plots Mz and bending moments M x And M y are presented in fig. 2.71, d.

Dangerous is the section TO(see fig. 2.71, G,d). Equivalent moment according to the hypothesis of the largest shear stresses

Equivalent stress according to the hypothesis of the largest shear stresses for the dangerous point of the shaft

safety factor

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

When calculating the shaft for strength, the change in stresses 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, but). The beam material is steel 30XGS with conditional yield strengths in tension and compression σ o, 2p = σ tr = 850 N/mm 2, σ 0.2 c = σ Tc = 965 N/mm 2. Safety factor [ n] = 1,6.

Solution

The bar works on the combined action of tension (compression) and torsion. Under such loading, two internal force factors arise in the cross sections: longitudinal force and torque.

Plots of longitudinal forces N and torque Mz shown in fig. 2.72, b, c. In this case, determine the position of the dangerous section according to the diagrams N And Mz impossible, since the dimensions of the cross sections of the sections of the beam are different. To determine the position of the dangerous section, plots of normal and maximum shear stresses along the length of the beam should be plotted.

According to the formula

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

According to the formula

we calculate the maximum shear stresses in the cross sections of the beam and plot the diagram t max(rice* 2.72, e).

Probably dangerous are the contour points of the cross sections of the sections AB And CD(see fig. 2.72, but).

On fig. 2.72, e plots are shown σ And τ for section cross sections AB.

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

On fig. 2.72, well

On fig. 2.72, h plots a and t are shown for the cross sections of the section CD.

On fig. 2.72, And the stresses on the initial pads at the dangerous point are shown.

The main stresses at the dangerous point of the site 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 allowable force value R from the rod strength condition sun(Fig. 2.73). The rod material is cast iron with tensile strength σ vr = 150 N / mm 2 and compressive strength σ sun = 450 N / mm 2. Required safety factor [ n] = 5.

Instruction. Broken timber ABC located in a horizontal plane, and the rod AB perpendicular to Sun. Forces R, 2R, 8R lie in a vertical plane; strength 0.5 R, 1.6 R- in 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 the cross section B.

Transferring force P parallel to itself to point B, we must add a pair 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 move along its line of action to point B.

Loads acting on the rod sun, shown in fig. 2.74 but.

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

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

Plots Qy And Qx we do not build, since shear stresses corresponding to transverse forces are small.

In the example under consideration, the position of the dangerous section is not obvious. Presumably, sections K are dangerous (the end of the 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 Mi in section C, shown separately in fig. 2.74 d. The same figure shows diagrams σ I, σ N , τ for section C.

Stresses on the initial sites at the point H(Fig. 2.74, e)