Calculation of a round bar for bending with torsion. Spatial (complex) bend

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.

Maximum torsional stress 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.

Decision

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 AT. 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.

Decision

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:

With and AT:

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 AT 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 AT 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, a) 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.

Decision

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, N 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) of a spur gear, 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 of the gears. 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.

Decision

On 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.

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, a, 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, a). 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.

Decision

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, a).

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.

Decision

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 AT, 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 a.

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)

Principal stresses at a point H:

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

Stresses on the initial sites at point E (Fig. 2.74, g):

Principal stresses at point E:

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

The dangerous point L for which

The strength condition has the form:

Control questions and tasks

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 the formulas for calculating the equivalent moment when calculating the maximum shear stress hypothesis and the deformation energy hypothesis.

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

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.

Maximum torsional stress 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.

Control questions and tasks

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.

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 the initial direction of the 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 of a 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 formula of D.Ya. Zhuravsky, 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 the calculation is carried out according to the method of 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).

The acting 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 core of the section, 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 I-I can be moved 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 I-I and 2-2 will be located on direct 1-2. Similarly, one can prove that the other 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 M x and M y is chosen so 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.

Bending is understood as a type of loading in which bending moments occur in the cross sections of the beam. If the bending moment in the section is the only force factor, then the bending is called pure. If, along with the bending moment, transverse forces also arise in the cross sections of the beam, then the bend is called transverse.

It is assumed that the bending moment and the transverse force lie in one of the main planes of the beam (we assume that this plane is ZOY). Such a bend is called flat.

In all the cases considered below, a flat transverse bending of the beams takes place.

To calculate the strength or stiffness of a beam, it is necessary to know the internal force factors that arise in its sections. For this purpose, diagrams of transverse forces (epure Q) and bending moments (M) are built.

When bending, the rectilinear axis of the beam is bent, the neutral axis passes through the center of gravity of the section. For definiteness, when constructing diagrams of transverse forces of bending moments, we establish sign rules for them. Let us assume that the bending moment will be considered positive if the beam element is bent with a convexity downwards, i.e. in such a way that its compressed fibers are at the top.

If the moment bends the beam with a bulge upwards, then this moment will be considered negative.

Positive values ​​of bending moments during plotting are plotted, as usual, in the direction of the Y axis, which corresponds to plotting on a compressed fiber.

Therefore, the rule of signs for the diagram of bending moments can be formulated as follows: the ordinates of the moments are plotted from the side of the beam layers.

The bending moment in a section is equal to the sum of the moments relative to this section of all forces located on one side (any) of the section.

To determine the transverse forces (Q), we establish the rule of signs: the transverse force is considered positive if the external force tends to rotate the cut-off part of the beam clockwise. arrow relative to the axis point that corresponds to the section drawn.

The transverse force (Q) in an arbitrary cross-section of the beam is numerically equal to the sum of the projections onto the axis of the y of the external forces applied to its truncated part.

Consider several examples of plotting transverse forces of bending moments. All forces are perpendicular to the axis of the beams, so the horizontal component of the reaction is zero. The deformed axis of the beam and the forces lie in the principal plane ZOY.

The beam length is pinched by the left end and loaded with a concentrated force F and a moment m=2F.

We construct diagrams of transverse forces Q and bending moments M from.

In our case, there are no constraints imposed on the beam on the right side. Therefore, in order not to determine the support reactions, it is advisable to consider the equilibrium of the right cut-off part of the beam. The given beam has two load areas. The boundaries of sections-sections in which external forces are applied. 1 section - NE, 2 - VA.

We carry out an arbitrary section in section 1 and consider the equilibrium of the right cut-off part of length Z 1.

From the equilibrium condition it follows:

Q=F; M out = -fz 1 ()

The shear force is positive, because external force F tends to rotate the cut-off part clockwise. The bending moment is considered negative, because it bends the considered part of the beam with a convexity upwards.

When compiling the equations of equilibrium, we mentally fix the place of the section; from the equations () it follows that the transverse force in section I does not depend on Z 1 and is a constant value. The positive force Q=F is scaled up from the center line of the beam, perpendicular to it.

The bending moment depends on Z 1 .

When Z 1 \u003d O M from \u003d O at Z 1 \u003d M from \u003d

The resulting value () is set aside down, i.e. the diagram M from is built on the compressed fiber.

Let's move on to the second part

We cut section II at an arbitrary distance Z 2 from the free right end of the beam and consider the equilibrium of the cut-off part of length Z 2. The change in shear force and bending moment based on equilibrium conditions can be expressed by the following equations:

Q=FM from = - FZ 2 +2F

The magnitude and sign of the transverse force did not change.

The magnitude of the bending moment depends on Z 2 .

At Z 2 = M from =, at Z 2 =

The bending moment turned out to be positive, both at the beginning of section II and at its end. In section II, the beam bends with a bulge downwards.

Set aside on a scale the magnitude of the moments up the centerline of the beam (i.e., the diagram is built on a compressed fiber). The greatest bending moment occurs in the section where the external moment m is applied and is equal in absolute value to

Note that over the length of the beam, where Q remains constant, the bending moment M changes linearly and is represented on the diagram by oblique straight lines. From the diagrams Q and M from it can be seen that in the section where an external transverse force is applied, the diagram Q has a jump by the value of this force, and the diagram M from has a kink. In a section where an external bending moment is applied, the Miz diagram has a jump by the value of this moment. This is not reflected in the Q plot. From the diagram M from we see that

max M out =

therefore, the dangerous section is extremely close on the left side to the so-called.

For the beam shown in Fig. 13, a, construct diagrams of transverse forces and bending moments. The length of the beam is loaded with a uniformly distributed load with intensity q(KN/cm).

On support A (fixed hinge) there will be a vertical reaction R a (horizontal reaction is zero), and on support B (movable hinge) a vertical reaction R v occurs.

Let us determine the vertical reactions of the supports by composing the equation of moments relative to supports A and B.

Let's check the correctness of the definition of the reaction:

those. support reactions are correctly defined.

The given beam has two loading sections: Section I - AC.

Section II - NE.

On the first section a, in the current section Z 1, from the condition of equilibrium of the cut-off part, we have

The equation of bending moments on 1 section of the beam:

The moment from the reaction R a bends the beam in section 1, convex downwards, so the bending moment from the reaction Ra is introduced into the equation with a plus sign. The load qZ 1 bends the beam with a convexity upwards, so the moment from it is introduced into the equation with a minus sign. The bending moment changes according to the law of a square parabola.

Therefore, it is necessary to find out whether there is an extremum. There is a differential dependence between the transverse force Q and the bending moment, which we will analyze further

As you know, the function has an extremum where the derivative is equal to zero. Therefore, in order to determine at what value of Z 1, the bending moment will be extreme, it is necessary to equate the equation of the transverse force to zero.

Since the transverse force changes sign from plus to minus in this section, the bending moment in this section will be maximum. If Q changes sign from minus to plus, then the bending moment in this section will be minimal.

So the bending moment at

is the maximum.

Therefore, we build a parabola on three points

When Z 1 \u003d 0 M from \u003d 0

We cut the second section at a distance Z 2 from support B. From the condition of equilibrium of the right cut-off part of the beam, we have:

When Q=const,

the bending moment will be:

at, at, i.e. M FROM

changes linearly.

A beam on two supports, having a span equal to 2 and a left console with a length, is loaded as shown in Fig. 14, a., Where q (Kn / cm) is the linear load. Support A is pivotally fixed, support B is a movable roller. Build plots Q and M from.

The solution of the problem should begin with the determination of the reactions of the supports. From the condition that the sum of the projections of all forces on the Z axis is equal to zero, it follows that the horizontal component of the reaction on support A is 0.

To check, we use the equation

The equilibrium equation is satisfied, therefore, the reactions are calculated correctly. We turn to the definition of internal force factors. A given beam has three load areas:

• 1 section - SA,
• 2nd section - AD,
• 3 section - DV.

We cut 1 section at a distance Z 1 from the left end of the beam.

at Z 1 \u003d 0 Q \u003d 0 M FROM \u003d 0

at Z 1 \u003d Q \u003d -q M IZ \u003d

Thus, on the diagram of transverse forces, an inclined straight line is obtained, and on the diagram of bending moments, a parabola is obtained, the apex of which is located at the left end of the beam.

In section II (a Z 2 2a), to determine the internal force factors, consider the balance of the left cut-off part of the beam with a length Z 2 . From the equilibrium condition we have:

The transverse force in this area is constant.

On section III()

From the diagram we see that the largest bending moment occurs in the section under the force F and is equal to. This section will be the most dangerous.

On the diagram M from there is a jump on the support B, equal to the external moment applied in this section.

Considering the diagrams constructed above, it is not difficult to notice a certain regular connection between the diagrams of bending moments and diagrams of transverse forces. Let's prove it.

The derivative of the transverse force along the length of the beam is equal to the modulus of the load intensity.

Discarding the value of the higher order of smallness, we get:

those. the transverse force is the derivative of the bending moment along the length of the beam.

Taking into account the obtained differential dependences, general conclusions can be drawn. If the beam is loaded with a uniformly distributed load of intensity q=const, obviously, the function Q will be linear, and M from - quadratic.

If the beam is loaded with concentrated forces or moments, then in the intervals between the points of their application, the intensity q=0. Therefore, Q=const, and M from is a linear function of Z. At the points of application of concentrated forces, the diagram Q undergoes a jump by the value of the external force, and in the diagram M from, a corresponding break occurs (a gap in the derivative).

At the place of application of the external bending moment, there is a gap in the moment diagram, equal in magnitude to the applied moment.

If Q>0, then M from grows, and if Q<0, то М из убывает.

Differential dependencies are used to check the equations compiled for plotting Q and M from, as well as to clarify the form of these diagrams.

The bending moment changes according to the law of a parabola, the convexity of which is always directed towards the external load.

Introduction.

Bending is a type of deformation characterized by a curvature (change in curvature) of the axis or middle surface of a deformable object (bar, beam, slab, shell, etc.) under the influence of external forces or temperature. Bending is associated with the occurrence of bending moments in the cross sections of the beam. If only one of the six internal force factors in the beam section is non-zero, the bend is called pure:

If, in addition to the bending moment, a transverse force also acts in the cross sections of the beam, the bend is called transverse:

In engineering practice, a special case of bending is also considered - longitudinal I. ( rice. one, c), characterized by buckling of the rod under the action of longitudinal compressive forces. The simultaneous action of forces directed along the axis of the rod and perpendicular to it causes a longitudinal-transverse bending ( rice. one, G).

Rice. 1. Bending of the beam: a - pure: b - transverse; in - longitudinal; g - longitudinal-transverse.

A bar that bends is called a beam. A bend is called flat if the axis of the beam remains a flat line after deformation. The plane of the curved axis of the beam is called the bending plane. The plane of action of the load forces is called the force plane. If the force plane coincides with one of the main planes of inertia of the cross section, the bend is called straight. (Otherwise there is an oblique bend). The main plane of inertia of the cross section is a plane formed by one of the main axes of the cross section with the longitudinal axis of the beam. In flat straight bending, the bending plane and the force plane coincide.

The problem of torsion and bending of a beam (the Saint-Venant problem) is of great practical interest. The application of the theory of bending established by Navier constitutes an extensive branch of structural mechanics and is of great practical importance, since it serves as the basis for calculating the dimensions and checking the strength of various parts of structures: beams, bridges, machine elements, etc.

BASIC EQUATIONS AND PROBLEMS OF THE THEORY OF ELASTICITY

§ 1. basic equations

First, we give a general summary of the basic equations for the problems of equilibrium of an elastic body, which form the content of the section of the theory of elasticity, usually called the statics of an elastic body.

The deformed state of the body is completely determined by the strain field tensor or the displacement field Components of the strain tensor are related to displacements by differential Cauchy dependences:

(1)

The components of the strain tensor must satisfy the Saint-Venant differential dependencies:

which are necessary and sufficient conditions for the integrability of equations (1).

The stress state of the body is determined by the stress field tensor Six independent components of a symmetric tensor () must satisfy three differential equilibrium equations:

Stress tensor components and displacement are related by the six equations of Hooke's law:

In some cases, the equations of Hooke's law have to be used in the form of a formula

, (5)

Equations (1)-(5) are the basic equations of static problems in the theory of elasticity. Sometimes equations (1) and (2) are called geometric equations, equations ( 3) - static equations, and equations (4) or (5) - physical equations. To the basic equations that determine the state of a linearly elastic body at its internal points of the volume, it is necessary to add conditions on its surface. These conditions are called boundary conditions. They are determined either by given external surface forces or given movements body surface points. In the first case, the boundary conditions are expressed by the equality:

where are the components of the vector t surface strength, are the components of the unit vector P, directed along the outer normal to the surface at the point under consideration.

In the second case, the boundary conditions are expressed by the equality

where are functions defined on the surface.

Boundary conditions can also be mixed, when on one part external surface forces are given on the surface of the body and on the other side displacements of the body surface are given:

Other kinds of boundary conditions are also possible. For example, on a certain part of the body surface, only some components of the displacement vector are specified and, in addition, not all components of the surface force vector are specified either.

§ 2. Main problems of the statics of an elastic body

Depending on the type of boundary conditions, three types of basic static problems of the theory of elasticity are distinguished.

The main problem of the first type is to determine the components of the stress field tensor inside the region , occupied by the body, and the component of the displacement vector of points inside the area and surface points bodies according to given mass forces and surface forces

The desired nine functions must satisfy the basic equations (3) and (4), as well as the boundary conditions (6).

The main task of the second type is to determine the displacements points inside the area and the stress field tensor component according to given mass forces and according to given displacements on the surface of the body.

Looking for features and must satisfy the basic equations (3) and (4) and boundary conditions (7).

Note that the boundary conditions (7) reflect the requirement for the continuity of the defined functions on the border body, i.e. when the interior point tends to some point on the surface, the function should tend to a given value at a given point on the surface.

The main problem of the third type or a mixed problem is that, given the surface forces on one part of the body surface and according to given displacements on another part of the body surface and also, generally speaking, according to given body forces it is required to determine the components of the stress and displacement tensor , satisfying the basic equations (3) and (4) under mixed boundary conditions (8).

Having obtained the solution of this problem, it is possible to determine, in particular, the forces of bonds on , which must be applied at the points of the surface in order to realize the given displacements on this surface, and it is also possible to calculate the displacements of the surface points . Coursework >> Industry, production

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