In the case of calculating the round timber during bending and twist (Fig. 34.3), it is necessary to take into account normal and tangent stresses, since the maximum values \u200b\u200bof stresses in both cases occur on the surface. The calculation should be conducted on the theory of strength, replacing the complex stress state at equinable simple.

Maximum tension voltage in cross section

Maximum bend voltage in cross section

According to one of the theories of strength, depending on the material of the timber, the equivalent voltage for a hazardous section is calculated and the ram is tested for strength using the allowable bending voltage for the timber material.

For a round timber, the torque resistance torctions are as follows:

When calculating the third theory of strength, the theory of maximum tangent stresses, the equivalent voltage is calculated by the formula

The theory is applicable to plastic materials.

When calculating the theory of energy of formation, the equivalent voltage is calculated by the formula

The theory is applicable for plastic and fragile materials.

Theories of maximum tangent stresses:

Equivalent voltage when calculating Theory of energy formation:

where - an equivalent moment.

Condition of strength

Examples of solving problems

Example 1. For a given intense state (Fig. 34.4), using the hypothesis of maximum tangent stresses, calculate the reserve coefficient of strength, if σ T \u003d 360 N / mm 2.

1. What is characterized and how is the stressful state at the point portray?

2. What are the sites and what stresses are the main?

3. List the types of intense states.

4. What is characterized by a deformed state at the point?

5. In what cases are the limiting stress states in plastic and fragile materials arise?

6. What is equivalent voltage?

7. Explain the purpose of strength theories.

8. Write the formulas for calculating the equivalent stresses in the calculations on the theory of maximum tangent stresses and the theory of energy formation. Explain how to use them.

Lecture 35.

Topic 2.7. Calculation of the bar of the round cross section when combining the main deformations

Know the formula for equivalent stresses on the hypotheses of the greatest tangent stresses and the energy of formation.

To be able to calculate the rapid cross section for strength with a combination of basic deformations.

Formulas for calculating equivalent stresses

Equivalent voltage on hypothesis of maximum tangent

Equivalent voltage on the energy hypothesis of formation

The condition of strength in the joint effect of twist

where M EKV. - Equivalent moment.

Equivalent moment on the hypothesis of maximum tangent stresses

Equivalent moment on the hypothesis of the energy of formation

Feature calculation of shafts

Most trees experience a combination of bending and twist deformations. Usually shafts are straight bars with a round or ring cross section. When calculating the shafts, tangent stresses from the action of transverse forces are not taken into account due to their insignificance.

Calculations are carried out by dangerous cross sections. With spatial loading of the shaft, the hypothesis of independence of the strength and bending moments are considered in two mutually perpendicular planes, and the total bending moment is determined by geometric summation.

Examples of solving problems

Example 1. In a hazardous cross section of a round timber, internal power factors arise (Fig. 35.1) M x; M y; M z.

M X. and M U. - bending moments in the planes Uokh and zox accordingly; M Z. - Torque. Check the strength on the hypothesis of the greatest tangent stresses, if [ σ ] \u003d 120 MPa. Initial data: M X. \u003d 0.9 kN m; M y \u003d 0.8 kN m; M z \u003d. 2.2 kN * m; d. \u003d 60 mm.

Decision

Building the plots of normal stresses from the action of bending moments relative to the axes Oh and OU and the escape of tangent stresses from twist (Fig. 35.2).

The maximum tangent stress occurs on the surface. Maximum normal voltages from the moment M X. Arise at point BUT, Maximum normal voltages from the moment M U. At point IN. Normal stresses are folded, because the bending moments in mutually perpendicular planes are geometrically summed.

Total bending moment:

Calculate the equivalent moment on the theory of maximum tangent stresses:

Strength condition:

The moment of resistance of the section: W OCE in Oe \u003d 0.1 60 3 \u003d 21600mm 3.

Check the strength:

Strength is provided.

Example 2. From the strength of the strength to calculate the required diameter of the shaft. Two wheels are installed on the shaft. Two district forces operate on the wheels F T 1 \u003d 1.2kn; F T 2. \u003d 2kn and two radial forces in the vertical plane F R 1. \u003d 0.43kn; F R 2 \u003d 0.72kn (Fig. 35.3). The diameters of the wheels are equal accordingly. D 1. \u003d 0.1m; d 2. \u003d 0.06 m.

Take for shaft material [ σ ] \u003d 50MPA.

Calculation of the hypothesis of maximum tangent stresses. Weighing shaft and wheels neglected.

Decision

Indication. We use the principle of independence of the action of the forces, we compile the calculated shaft schemes in vertical and horizontal planes. We determine the reactions in the supports in the horizontal and vertical planes separately. We are building the fusion of bending moments (Fig. 35.4). Under the action of the circumferential forces, the shaft is twisted. We determine the torque acting on the shaft.

Make the calculated shaft circuit (Fig. 35.4).

1. Torque on the shaft:

2. Bending we consider in two planes: horizontal (pl. H) and vertical (pl. V).

In the horizontal plane, we determine the reaction in the support:

FROM and IN:

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

Determine the bending moments at points C and B:

Total bending moments at points C and B:

At point IN The maximum bending moment, the torque acts here.

Calculation of the diameter of the shaft lead along the most loaded cross section.

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

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

Round the value: d. \u003d 36 mm.

Note. When choosing a shaft diameter to use the standard number of diameters (Appendix 2).

5. Determine the required sizes of the ring section shaft at C \u003d 0.8, where D is the outer diameter of the shaft.

The diameter of the ring section shaft can be determined by the formula

Institute d \u003d. 42 mm.

Overload minor. D BH \u003d. 0.8d \u003d 0.8 42 \u003d 33.6 mm.

Round up to date d BH\u003d 33 mm.

6. Compare metal costs on shaft cross-section areas in both cases.

Cross-sectional area of \u200b\u200ba solid shaft

Cross-sectional area of \u200b\u200ba hollow shaft

The cross-sectional area of \u200b\u200bthe solid shaft is almost twice as the shaft of the annular section:

Example 3.. Determine the size of the cross section of the shaft (Fig. 2.70, but) Drive management. Effort from traction pedal P 3., Efforts transmitted by the mechanism P 1, P 2, P 4. The material of the shaft is steel steel with the yield strength σ T \u003d 240 N / mm 2, the required stock ratio [ n.] \u003d 2.5. The calculation is calculated on the hypothesis of the energy of formation.

Decision

Consider the equilibrium of the shaft, pre-laying forces P 1, p 2, p 3, p 4 To the points lying on its axis.

Carrying power P 1. In parallel yourself at the point TO and E., you need to add pairs of forces with moments equal to the moment of forces P 1. Regarding points TO and E, i.e.

These pairs of forces (moments) are conditionally shown in Fig. 2.70. , B. In the form of arcuate lines with arrows. Similarly, when transferring forces P 2, P 3, P 4 in a point K, e, l, n need to add pairs of forces with moments

Wood supports shown in Fig. 2.70, and, it is necessary to consider as spatial hinge supports that prevent movements in the direction of the axes h. and w. (The selected coordinate system is shown in Fig. 2.70, b).

Using the calculated scheme shown in Fig. 2.70, in, accounted for equilibrium equations:

consequently, support reactions ON THE and N B. defined correctly.

Torque Moments M Z. and bending moments M U. Presented in Fig. 2.70, g.. Dangerous is the cross section to the left of the point L.

The condition of strength is:

where the equivalent moment on the energy hypothesis of formation

Required outer shaft 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 the cylindrical spanner reducer, if the shaft transmits power N. \u003d 12,2 kW at rotation frequency p \u003d 355 rpm. The shaft is made of steel ST5 with the yield strength σ T \u003d 280 N / mm 2. Required stock ratio [ n.] \u003d 4. When calculating, apply the hypothesis of the greatest tangent stresses.

Indication. District effort P 1. and P 2.la in the horizontal plane and are directed by tangents to the circumferences of the gears. Radial efforts T 1. and T 2.les in the vertical plane and are expressed through the appropriate circumferential force as follows: T. = 0,364R.

Decision

In fig. 2.71 but presented a schematic drawing of the shaft; In fig. 2.71, B shows the shaft diagram and efforts arising in the gearing gear.

We define the moment transmitted by the shaft:

Obviously m \u003d m 1 \u003d m 2 (twisting moments attached to the shaft, with uniform rotation are equal in size and opposite in direction).

We define the efforts acting on the gears.

District efforts:

Radial efforts:

Consider the balance of Vala AU, pre-lay down P 1. and P 2. To the points lying on the shaft axis.

Carrying force P 1. parallel to myself 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 conditionally shown in Fig. 2.71 inin the form of an arcuate line with an arrow. Similarly, when transferring force P 2. exactly TO need to attach (add) a couple of forces with torque

Wood supports shown in Fig. 2.71 but, It is necessary to consider as spatial hinge supports that impede linear movements in the directions of the axes H. and W. (The selected coordinate system is shown in Fig, 2.71, b.).

Using the calculated scheme shown in Fig. 2.71 g., accounted for equation equilibrium shaft in the vertical plane:

Make a check equation:

consequently, the support reactions in the vertical plane are defined correctly.

Consider the equilibrium of the shaft in the horizontal plane:

Make a check equation:

consequently, the support reactions in the horizontal plane are defined correctly.

Torque Moments M Z. and bending moments M X. and M U. Presented in Fig. 2.71 D..

Dangerous is the cross section TO (see Fig. 2.71, g., D.). Equivalent moment on the hypothesis of the greatest tangent

Equivalent suspension voltage of the highest tangent stresses for a dangerous point of the shaft

Reserve coefficient

what is much more [ n.] \u003d 4, therefore, the shaft strength is ensured.

When calculating the shaft on strength, a change in voltages in time is not taken into account, therefore, it turned out such a significant reserve coefficient.

Example 5. Determine the particle cross section of the bar (Fig. 2.72, but). The material of the bar - steel 30xgs with the conditional limits of the fluidity during stretching and compression σ o, 2p \u003d σ tr \u003d 850 H / mm 2, σ 0.2 C \u003d σ Tc \u003d 965 N / mm 2. Stock coefficient [ n.] = 1,6.

Decision

The bar works on the joint effect of stretching (compression) and twist. With such loading in transverse sections, two internal power factor arise: longitudinal power and torque.

Trucks of longitudinal power N. and torque moments M Z.showing in Fig. 2.72, b, c. In this case, determine the position of a dangerous cross section on eporas N. and M Z. It is impossible because the size of the cross sections of the sections of the bar is different. To clarify the position of the hazardous section, it is necessary to build plumes of normal and maximum tangent stresses along the length of the bar.

According to the formula

calculate normal stresses in the cross sections of the bar and build the Eppura O (Fig. 2.72, g.).

According to the formula

calculate the maximum tangent stresses in the cross sections of the bar and build a lot TAX (Fig. * 2.72, e).

Probably dangerous are the dots of the contour of the cross sections of the plots AU and CD (see Fig. 2.72, but).

In fig. 2.72, e. Showing Epura σ and τ For cross sections AU.

Recall, in this case (the round cross section bar works on the joint effect of stretching - compression and twist) is equiturated with all the contour of the cross section.

In fig. 2.72, J.

In fig. 2.72, z. Showing EPURS A and T for cross sections CD.

In fig. 2.72, and Showing voltage on the source sites at a dangerous point.

Main stresses at a dangerous point of the site CD:

On the hypothesis of the strength of the Mora equivalent voltage for the dangerous point of the section under consideration

Dangerous points of the contour of the cross sections of the AB site were.

The condition of strength is:

Example 2.76. Determine the allowable value R From the strength of the rod Sun (Fig.2.73). Material of a rod - cast iron with a tensile strength Σ BP \u003d 150 H / mm 2 and the strength of the compression of σ Sun \u003d 450 N / mm 2. Required stock ratio [ n.] = 5.

Indication. Broken bar ABS Located in a horizontal plane, and the rod AV. Perpendiculated K. Sun. Forces R, 2P, 8R lie in the vertical plane; Forces 0.5 p, 1.6 r - in horizontal and perpendicular to the rod Sun; Forces 10p, 16r. coincide with the axis of the rod Sun; A pair of forces with a moment m \u003d 25pd is located in a vertical plane perpendicular to the axis of the rod Sun.

Decision

We give the power R and 0.5p to the center of gravity of the cross section.

Carrying force p in parallel to the right point in, you need to add a couple of strength with a moment, equal to the moment of force R relative to the point IN, i.e. a couple with a moment m 1 \u003d 10 PD.

Strength 0.5r. We carry along its line action to point B.

Loads acting on the rod Sun, Showing in Fig. 2.74, but.

We build the plots of domestic power factors for the rod Sun. With the specified loading of the rod in its transverse sections, they occur six: longitudinal force N., transverse forces QX.and QY, torque MZ.bending moments MX and Mu.

Epura N, MZ, MX, MU Presented in Fig. 2.74, b. (Ordates of Epur are expressed through R and d.).

Epura QY. and QX. Do not we build, as the tangent stresses corresponding to the transverse forces have a small amount.

In this example, the position of the hazardous section is not obvious, allegedly, dangerous cross sections to (the end of the site I.) and S.

Main stresses at point L:

According to the mixing hypothesis of Mora equivalent voltage for point L

We define the magnitude and plane of the action of the bending moment in the section with shown separately in Fig. 2.74, d.. The same figure shows the epures σ and, σ n, τ For section S.

Voltages on the source sites at the point N. (Fig. 2.74, e)

Main stresses at point N.:

Mora Al hypothesis Equivalent voltage for point N.

Voltage on the source sites at the point E (Fig. 2.74, g):

Main stresses at point E:

On the hypothesis of the strength of the Mora equivalent voltage for the point e

Danger was the point L, for which

The condition of strength is:

Check questions and tasks

1. What kind of stressful state occurs in the cross section of the shaft during the joint action of bending and twist?

2. Write a strength condition to calculate the shaft.

3. Write the formulas for calculating the equivalent torque when calculating the hypothesis of maximum tangent stresses and the energy hypothesis of formation.

4. How does a dangerous cross-section be selected when calculating the shaft?

In the case of calculating the round timber during bending and twist (Fig. 34.3), it is necessary to take into account normal and tangent stresses, since the maximum values \u200b\u200bof stresses in both cases occur on the surface. The calculation should be conducted on the theory of strength, replacing the complex stress state at equinable simple.

Maximum tension voltage in cross section

Maximum bend voltage in cross section

According to one of the theories of strength, depending on the material of the timber, the equivalent voltage for a hazardous section is calculated and the ram is tested for strength using the allowable bending voltage for the timber material.

For a round timber, the torque resistance torctions are as follows:

When calculating the third theory of strength, the theory of maximum tangent stresses, the equivalent voltage is calculated by the formula

The theory is applicable to plastic materials.

When calculating the theory of energy of formation, the equivalent voltage is calculated by the formula

The theory is applicable for plastic and fragile materials.

Theories of maximum tangent stresses:

Equivalent voltage when calculating Theory of energy formation:

where - an equivalent moment.

Condition of strength

Examples of solving problems

Example 1. For a given intense state (Fig. 34.4), using the hypothesis of maximum tangent stresses, calculate the reserve coefficient of strength, if σ T \u003d 360 N / mm 2.

Check questions and tasks

1. What is characterized and how is the stressful state at the point portray?

2. What are the sites and what stresses are the main?

3. List the types of intense states.

4. What is characterized by a deformed state at the point?

5. In what cases are the limiting stress states in plastic and fragile materials arise?

6. What is equivalent voltage?

7. Explain the purpose of strength theories.

8. Write the formulas for calculating the equivalent stresses in the calculations on the theory of maximum tangent stresses and the theory of energy formation. Explain how to use them.

Lecture 35.

Topic 2.7. Calculation of the bar of the round cross section when combining the main deformations

Know the formula for equivalent stresses on the hypotheses of the greatest tangent stresses and the energy of formation.

To be able to calculate the rapid cross section for strength with a combination of basic deformations.

Under the bend is understood as a type of loading, in which bending moments arise in cross sections of the bar. If the bending moment in the section is the only power factor, then the bending is called clean. If along with the bending moment in the cross sections of the bar arise and transverse forces, then the bending is called transverse.

It is assumed that the bending moment and the transverse force lie in one of the main planes of the bar (we will take that this zoy plane). This bending is called flat.

In all cases under consideration, there is a flat transverse beam beam.

To calculate the beams for strength or rigidity, it is necessary to know internal power factors arising in its sections. For this purpose, the epures of the transverse forces (Epur Q) are built and bending moments (M).

When bending, the straight axis of the timber is curved, the neutral axis passes through the center of severity. For definiteness, when constructing a group of transverse strength of bending moments, we will set the rules for them. We will assume that the bending moment will be considered positive if the element of the bar is bent into convexity down, i.e. Thus, that its compressed fibers are in the upper part.

If the moment bends a bar with convexity up, this moment will be considered negative.

The positive values \u200b\u200bof bending moments during the construction of the plot are deposited, as usual in the direction of the axis y, which corresponds to the construction of the plot on the compressed fiber.

Therefore, the rule of signs for the plot of bending moments can be formulated as follows: the ordinates of the moments are deposited by the layers of the bar.

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

To determine the transverse forces (q), we establish a rule of signs: the transverse force is considered positive if the external force strive to rotate the cut-off part of the beam at an hour. The arrow relative to the point of the axis, which corresponds to the cross section.

The transverse force (q) in the arbitrary cross section of the bar is numerically equal to the amount of projections on the axis of the OU of the external forces applied to its oiled part.

Consider several examples of constructing the cross forces of bending moments. All forces perpendicular to the axis of the beams, so the horizontal component of the reaction is zero. The deformed axis of beams and strength lie in the main plane of Zoy.

The beam is plugged with the left end and loaded with the focused force f and the moment M \u003d 2F.

We construct the epirs of the transverse forces q and bending moments M of.

In our case, the beam on the right side does not impose connections. 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 specified beam has two loading portions. The boundaries of the sections in which external forces are applied. 1 plot - St., 2nd.

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

From the condition of equilibrium it follows:

Q \u003d f; M from \u003d -fz 1 ()

The transverse force is positive, because The external force F is striving to rotate the cut-off part clockwise. The moment of bending is considered negative, because It bends under consideration of the beam in bulk upwards.

In the preparation of equilibrium equations mentally fasten the cross section; From equations () it follows that the transverse force on the I section from Z 1 does not depend on the constant value. The positive strength q \u003d f is set up on the scale up from the axial line of the beam, perpendicular to it.

The bending moment depends on Z 1.

At z 1 \u003d o m from \u003d o pr 1 \u003d m from \u003d

The resulting value () is deposited down, i.e. Epura M of is built on a compressed fiber.

Go to the second site

Disseminate section II at an arbitrary distance z 2 from the free right end of the beam and we consider the equilibrium of the clipped part with length z 2. Changing the transverse force and bending torque based on equilibrium conditions can be expressed by the following equations:

Q \u003d FM from \u003d - FZ 2 + 2F

The magnitude and the transverse force sign did not change.

The value of the bending moment depends on Z 2.

Rt 2 \u003d m out \u003d, prize 2 \u003d

The bending moment turned out to be positive, both at the beginning of the site II and at the end of it. At the site II, the beam bends the convexity down.

We postpone on the scale of moments up the axial line of the beam (i.e., the EPUR is built on a compressed fiber). The greatest bending moment occurs in the section, where the outer moment M is applied and is equally equal to

Note that at the length of the beam, where q retains a constant value, the bending moment M of changes linearly and appears to be inclined straight. From Epur Q and M, it is clear that in the section, where the outer transverse force is applied, the EPUR Q has a jump on the magnitude of this force, and Epura M of the break. In the section, where an external bending moment is attached, MIZ's EPUR has a jump by the value of this moment. On the stage q it is not reflected. From Epura M of seemingly

max M from \u003d.

consequently, a dangerous cross section is extremely close to the left side to t.

For beams depicted in Fig. 13, a, build plot of transverse forces and bending moments. At the length of the beam is loaded with a uniformly distributed load with the intensity Q (KN / cm).

On the Support A (Vertical Hinge), the vertical reaction of R a (the horizontal reaction is zero), and the vertical reaction R c occurs on the support in (moving hinge).

We define the vertical reactions of the supports, making up the equation of moments relative to the supports A and V.

Check the correctness of the definition of the reaction:

those. Support reactions are defined correctly.

The specified beam has two parts of loading: I plot - speakers.

II plot - St.

On the first section A, in the current section Z 1 from the equilibrium condition we have

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

The moment from the reaction R a bends the bending on the section 1, convexing down, so the bending moment on the RA reaction is introduced into the equation with a plus sign. The load QZ 1 bends the beam by convexing up, so the moment from it is introduced into the equation with a minus sign. The bending moment changes by the law of the square parabola.

Therefore, it is necessary to find out whether the place is extremum. Between the transverse force q and the bending moment there is a differential dependence on the analysis of which we will focus on

As is known, the function has an extremum where the derivative is zero. Therefore, to determine with what value z 1, the bending moment will be extreme, it is necessary to equate the transverse force to zero.

Since the transverse force changes in this section a sign from a plus for minus, the bending moment in this section will be maximal. If q changes a sign from a minus plus, then the bending moment in this section will be minimal.

So, bending moment when

it is maximum.

Therefore, we build a parabola for three points

At z 1 \u003d 0 m out \u003d 0

Disseminate the second section at a distance z 2 from the support V. from the condition of the equilibrium of the right cut-off part of the beams we have:

With the value of Q \u003d const,

the bending moment will be:

when, with, i.e. M is

changes on linear law.

The beam on two supports, having an ease of equal to 2 and the left console length, loaded as shown in Fig. 14, and., Where Q (KN / cm) is the power load. Support A-article fixed, support in - moving rink. Build Epura Q and M of.

The task solution should be started with the definition of support reactions. From the condition of equality, zero the amount of projections of all forces on the Z axis follows that the horizontal component of the reaction on the support A is 0.

To check the equation

The equilibrium equation is consistent, therefore, the reaction is calculated correctly. We turn to the definition of domestic power factors. The specified beam has three loading areas:

• 1 plot - Ca,
• 2 plot - hell,
• 3 Plot - DV.

Mix 1 plot to distance z 1 from the left end of the beam.

at z 1 \u003d 0 q \u003d 0 m out \u003d 0

at z 1 \u003d q \u003d -q m from \u003d

Thus, on the line of transverse forces, there is a slope straight, and on the plot of bending moments - Parabola, the vertex of which is located on the left end of the beam.

On the section II (A Z 2 2a) to determine the internal power factors, consider the equilibrium of the left cut-off part of the beam in length z 2. From the conditions of equilibrium we have:

Transverse force on this plot is constant.

In a plot III ()

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

On the EPUR M from there is a jump on the support in an external point attached in this section.

Considering the support constructed above, it is not difficult to notice a certain pattern between the raging moments and the cross-forces. We prove it.

The transverse force derivative along the length of the bar is equal to the load intensity module.

By discarding the magnitude of the highest order of little, we get:

those. The transverse force is derived from the bending moment along the length of the bar.

Considering the differential dependencies, general conclusions can be made. If the timber is loaded with a uniformly distributed load of the intensity Q \u003d const, obviously, the function Q will be linear, and M of the quadratic one.

If the timber is loaded by concentrated forces or moments, then in the intervals between points of their application, the intensity Q \u003d 0. Consequently, Q \u003d const, and M from is a linear function Z. At the points of the application of the concentrated EPUR forces q, it undergoes a jump on the value of the external force, and in the stage M of the corresponding break (gap in the derivative).

In the place of the appearance of an external bending moment, there is a break in the magnitude of the moments equal to the magnitude of the attached moment.

If q\u003e 0, then m from growing, and if q<0, то М из убывает.

Differential dependencies are used to check the equations of the composed of EPURO Q and M from, as well as to clarify the type of these EPUR.

The bending moment changes by the Parabola law, the bulge of which is always directed towards the external load.

Introduction

Bending is a deformation type characterized by curvature (change in curvature) axis or median surface of a deformable object (timber, beams, stoves, shells, etc.) under the action of external forces or temperature. The bending is associated with the occurrence of bending bending bending in cross sections. If from six internal power factors in cross section of a bar different from zero is only one bending moment, the bending is called clean:

If in cross sections of the timber, in addition to the bending moment, the transverse force is also valid - the bending is called transverse:

In engineering practice, there is also a special case of bending - longitudinal I. ( fig. one, B), characterized by releasing the rod under the action of longitudinal compressive forces. The simultaneous action of the forces directed along the axis of the rod and perpendicular to it, causes a longitudinal transverse bend ( fig. one, d).

Fig. 1. Bending timber: A - Clean: B - transverse; in - longitudinal; G - longitudinally transverse.

Bending Bar is called beam. Bending is called flat, if the axis of the beam after deformation remains a flat line. The location plane of the curved axis of the beam is called the bending plane. The plane of the operation of the loading force is called the power plane. If the power plane coincides with one of the main planes of the cross-section inertia, the bending is called direct. (Otherwise, there is a place of bending). The main plane of cross-section inertia is a plane formed by one of the main axes of cross-section with a longitudinal axis of the bar. With a flat bending, the bending plane and the power plane coincide.

The task of cutting and bending a bar (the task of Saint-Vienna) has great practical interest. An application of the bend theory established by Navier is an extensive department of construction mechanics and has a huge practical importance, as it serves as the basis for calculating the size and calibration of the strength of various parts of structures: beams, bridges, machine elements, etc.

The main equations and objectives of the theory of elasticity

§ 1. Main equations

Initially, we will give a general summary of the main equations for the equilibrium of the elastic body, which constitute the content of the section of the theory of elasticity, called usually by static an elastic body.

The deformed state of the body is determined by the strain field tensor or the movement field components of the deformation tensor related to displacements of differential dependencies of Cauchy:

(1)

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

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

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

Voltage tensor components andmovement associated six equations of the law of the throat:

some cases of the equation of the Dunga law have to be used as a formula

, (5)

Equations (1) - (5) are the main equations of static tasks of 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 main equations that determine the status of a linear-elastic body in its internal points of volume, it is necessary to attach the conditions on its surface. These conditions are called boundary conditions. They are determined by either given external surface forces or given movements points of body surface. In the first case, the boundary conditions are expressed by equality:

where - vector components t. superficial force - components of a single vector p, directed by external normal to the surface in the point under consideration.

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

where - set on the surface of the function.

Boundary conditions may also be mixed when on one part body surfaces are given external surface forces and on the other part body surfaces are set to move:

Other border conditions are possible. For example, on a certain section of the body surface, only some of the component of the movement vector and, in addition, not all components of the surface strength vector are also specified.

§ 2. Basic elastic body statics tasks

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

The main task of the first type is to determine the voltage field tensor component inside the region , busy body and component of the velating vector inside the area and surface points bodies for predetermined mass forces and superficial forces

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

The main task of the second type is to define movements points inside the region and voltage field tensor component according to the predetermined mass and according to the predetermined movements on the body surface.

Second functions and must satisfy the main equations (3) and (4) and boundary conditions (7).

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

The main task of the third type or mixed task is that according to the predetermined surface forces on one part of the body surface and for given movements on another part of the body and also, generally speaking, according to the predetermined mass forces it is required to determine the components of the strain tensor and movement. , satisfying the main equations (3) and (4) when performing mixed boundary conditions (8).

Having obtained a solution to this task, it is possible to determine, in particular, the efforts of relations on , which must be applied at the surface points to implement the specified movements on this surface, and can also be calculated to move the surface points . Course work \u003e\u003e Industry, production

By lenght bruusT. bar. deformed. Deformation bruus accompanied simultaneously ... wood, polymeric, etc. behege bruuslying on two supports ... behege will be characterized by a boob boom. At the same time, the compression voltage in the concave part bruus ...

Advantages of glued bruus in low-rise construction

Abstract \u003e\u003e Construction

Are solved when using glued profiled bruus. Glued wood in carrier ..., not twisted and not bend. This is due to the lack of ... transportation of fuel. 5. The surface of glued bruusperformed in compliance with all technological ...

Spatial (sophisticated) bend

The spatial bend is called such a type of complex resistance, in which only bending moments act in the cross section of the bar and. The full bending moment is valid in any of the main planes of inertia. Longitudinal force is absent. A spatial or complex bend is often called a non-planar bend, since the curved rod axis is not a flat curve. Such bending is caused by the 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 in the complex resistance set out above, we declare the spatial system of the forces presented in Fig. 1.2.1, two such that each of them acted in one of the main planes. As a result, we obtain two flat transverse bends - in the vertical and horizontal plane. Of the four internal power factors that occur in the cross section of the beams, we will take into account the influence of only bending moments. We build plumes caused by the forces accordingly (Fig. 1.2.1).

Analyzing the plots of bending moments, we conclude that the cross section A is dangerous, since it is in this section that the largest bending moments arise. Now it is necessary to establish hazardous points of the section A. To do this, we build a zero line. The zero line equation, taking into account the rules of signs for members entering this equation, has the form:

There is a sign "" near the second member of the equation, since the voltages in the first quarter caused by the moment will be negative.

We define the angle of inclination of the zero line with a positive axis direction (Fig.12.6):

Fig. 1.2.2

From equation (8) it follows that the zero line with spatial bend is a straight line and passes through the center of severity.

From fig. 1.2.2 It can be seen that the largest voltages will occur in the points of section 2 and No. 4 most remote from the zero line. By magnitude, normal voltages at these points will be the same, but the sign is different: at point 4 voltage will be positive, i.e. stretching, at point number 2 - negative, i.e. compressing. Signs of these stresses were established from physical considerations.

Now that dangerous points are installed, we calculate the maximum stresses in the 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 size of its cross-section, if the aspect ratio of the cross section is specified.