Stretching (compression) - This type of loading of the bar, in which only one internal force factor appears in its cross sections - the longitudinal force of N.

When tensile and compression, the external forces are applied along the longitudinal axis Z (Figure 109).

Figure 109.

Applying the cross sections method, it is possible to determine the value of the VFF - the longitudinal force N with a simple loading.

Internal forces (voltages) arising in an arbitrary cross section when tensile (compression) are determined using hypothesis of flat cross sections Bernoulli:

The cross section of the bar, a flat and perpendicular axis to loading remains the same during loading.

It follows that the fibers of the bar (Figure 110) are lengthened on the same values. Therefore, the internal forces (that is, voltages) acting on each fiber will be the same and distributed in cross section uniformly.

Figure 110.

Since n is the resultant internal forces, then n \u003d σ · A, the normal stresses σ are tension and compression is determined by the formula:

[N / mm 2 \u003d MPa], (72)

where a is the cross-sectional area.

Example 24.Two rods: Round section with a diameter d \u003d 4 mm and a square cross section with a side of 5 mm are stretched by the same force F \u003d 1000 N. Which rods are loaded more?

Dano: d \u003d 4 mm; a \u003d 5 mm; F \u003d 1000 N.

Determine: σ 1 and σ 2 - in rods 1 and 2.

Decision:

When tensile, the longitudinal force in the rods n \u003d f \u003d 1000 N.

Rod cross sections area:

; .

Normal stresses in cross sections rods:

, .

Since σ 1\u003e σ 2, then the first rod of the circular section is loaded more.

Example 25.The cable, the reticue of 80 wires with a diameter of 2 mm is stretched by the force of 5 kN. Determine the voltage in cross section.

Given: K \u003d 80; d \u003d 2 mm; F \u003d 5 kN.

Determine: σ.

Decision:

N \u003d f \u003d 5 kN ,,

then .

Here and 1 is the cross-sectional area of \u200b\u200bone wire.

Note: The cable cross section is not a circle!

2.2.2 Eppures of the longitudinal forces N and normal stresses σ along the length of the bar

For calculations on the strength and rigidity of a complex loaded timber during stretching and compression, it is necessary to know the values \u200b\u200bof N and σ in various cross sections.

For this, plots are built: ePUR N and EPUR Σ.

Epura. - This is a graph of changes in the longitudinal force N and normal stresses σ along the length of the bar.

Longitudinal force N.in an arbitrary cross section of a bar is equal to the algebraic sum of all external forces applied to the remaining part, i.e. one way from section

External forces F, tensile timber and directed away from the section, are considered positive.

The order of constructing EPUR N and Σ

1 cross-sections are broken by a bar on the plots whose boundaries are:

a) cross sections at the ends of the bar;

b) where the power F is applied;

c) where the area of \u200b\u200bthe cross section is changing

2 Number areas starting with

free end.

3 for each site using the method

sections determine the longitudinal force n

and we build on the scale of Eppura N.

4 Determine the normal voltage σ

on each site and build in

the scale of the EPUER Σ.

Example 26.Constructing Eppures N and Σ along the length of the stepbecker (Figure 111).

Given: F 1 \u003d 10 kN; F 2 \u003d 35 kN; A 1 \u003d 1 cm 2; A 2 \u003d 2 cm 2.

Decision:

1) We divide the timber on the plots, the boundaries of which are: sections at the ends of the bar, where the external forces f are applied, where the area of \u200b\u200bsection is changed, and everything turned out 4 plots.

2) Noururate areas, starting from the free end:

with I in IV. Figure 111.

3) for each site using the cross sections method, we determine the longitudinal force of N.

The longitudinal force n is equal to the algebraic sum of all external forces attached to the remaining part of the bar. Moreover, the external forces F, the tensile timber is considered positive.

Table 13.

4) we build on the scale of N. N. Scale indicate only the positive values \u200b\u200bof N, on the stage, the plus sign or minus (stretching or compression) is indicated in the circle in the rectangle of the EPUR. Positive values \u200b\u200bn are postponed above the zero axis of the EPUR, negative - below the axis.

5) Check (oral): In sections, where the external forces f are applied, on the EPUR N will be vertical jumps equal to this forces.

6) We determine the normal stresses in the sections of each site:

; ;

; .

We are building on the scale of the Epleu Σ.

7) Check: Signs N and Σ are the same.

Think and answer questions

1) it is impossible; 2) can be.

53 Does the stress depend on the tension (compression) rods from the form of their cross section (square, rectangle, circle, etc.)?

1) depend; 2) do not depend.

54 Does the magnitude of the voltage in cross section depend on the material from which the rod is made?

1) depends; 2) does not depend.

55 What cross-section points of the round rod are loaded more when tensile?

1) on the axis of the bar; 2) on the surface of the circle;

3) at all points of the voltage cross section are the same.

56 Rods made of steel and wood with an equal cross-sectional area are stretched by the same forces. Will there be equal in the stress rods?

1) in the steel voltage more;

2) in wooden voltage more;

3) In the rods there will be equal stresses.

57 For a timber (Figure 112), construct actions n and σ if f 1 \u003d 2 kN; F 2 \u003d 5 kN; A 1 \u003d 1.2 cm 2; A 2 \u003d 1.4 cm 2.

When stretching (compressed) timber in its cross sectionsthere are only normal voltages.Equality of the corresponding elementary forces about, DA - longitudinal force N -can be found with the method of sections. In order to be able to determine normal voltages with a well-known value of the longitudinal force, it is necessary to establish the law of the distribution of the cross section of the brusade.

This task is solved on the basis of prostheses of flat sections(hypotheses Y. Bernoulli),which says:

the cross sections of the bar, flat and normal to its axis to the deformation, remain flat and normal to the axis and during deformation.

When stretching a bar (made, for example, forgreater clarity of the experience of rubber) on the surface whomthe system of longitudinal1x and transverse rice is applied (Fig. 2.7, a), you can make sure that the risks remain straightforward and mutually perpendicular, change only

where a is the cross-sectional area of \u200b\u200bthe bar. Lowering index z, finally get

For normal stresses, they take the same rule of signs as for longitudinal forces, i.e. when tension, they are tensile positive.

In fact, the distribution of stresses in the sections of the bar, adjacent to the place of application of external forces, depends on the method of the load application and may be uneven. Experimental and theoretical studies show that this violation of the uniform distribution of stresses wears local character.In the sections of the bar, which are separated from the place of loading at a distance, approximately equal to the greatest of the cross-dimensions of the bar, the distribution of stresses can be considered almost uniform (Fig. 2.9).

The considered position is a special case. principle of Saint-Vienna,which can be formulated as follows:

the voltage distribution substantially depends on the method of application of external forces only near the placement location.

In parts, sufficiently remote from the place of the application of the application, the voltage distribution is practically dependent on the static equivalent of these forces, and not on the method of their application.

Thus applying saint Venan principleand distracting from the question of local stresses, we have the opportunity (both in this and in subsequent chapters of the course) are not interested in specific ways of application of external forces.

In places of a sharp change in the shape and size of the cross section of the bar also arise local stresses. This phenomenon is called stress concentrationwhich in this chapter will not take into account.

In cases where normal stresses in various cross-sections of a bar of unequal, it is advisable to show the law of their change in the length of the timber in the form of a graph - epures of normal stresses.

Ri Mer2.3. For a bar with a step-by-variable cross section (Fig. 2.10, a) build pliers of longitudinal forces andnormal stresses.

Decision.We split the timber on the plots starting from the free messenger. The boundaries of the plots are the places of application of external forces and changes in the cross-sectional size, i.e. the bar has five sites. When building only a plumb N.it would be necessary to break the timber only into three plots.

Applying the cross section method, we determine the longitudinal forces in the cross sections of the bar and we build a corresponding step (Fig. 2.10.6). Building an Eppure and does not differ in principle from the considered in Example 2.1, therefore, the details of this construct are lowered.

Normal voltages are calculated by formula (2.1), substituting the values \u200b\u200bof forces in Newton, and areas in square meters.

Within each of the voltage sections are constant, t. e.epura in this area is straight, parallel axis of the abscissa (Fig. 2.10, B). For strength calculations, interest is primarily the cross sections in which the greatest stresses arise. It is essential that in the considered case they do not coincide with those cross sections, where the longitudinal forces are maximal.

In cases where the cross section of the bar across the entire length is constantly, Epura butlike the Epure N.and differs from it only scale, therefore, naturally, it makes sense to build only one of the specified EPUR.

If, with a direct or oblique bending in the cross section of the bar, only the bending moment acts, then there is a clean straight or pure oblique bend. If transverse force also acts in cross section, then there is a cross-direct or cross-braid bend. If the bending moment is the only internal power factor, then such a bending is called clean (Fig. 6.2). In the presence of transverse force, the bending is called transverse. Strictly speaking, only a pure bend is applied to simple resistance; The transverse bending belongs to simple types of resistance conditionally, since in most cases (for sufficiently long beams) the action of the transverse force during strength calculations can be neglected. See the condition of strength with flat bending. The calculation of the bending bending bending one of the most important is the task of determining its point. Flat bending is called transverse if twilight power factor in transverse sections: M - bending moment and Q - transverse force, and clean, if only M. In cross-bending, the power plane passes through the axis of the beam symmetry, which is one of the main axes of the system inertia.

With bending beam, some layers are stretched, others are compressed. There is a neutral layer between them, which is only twisted without changing its length. The cross-sectional layer line with the cross-sectional plane coincides with the second main axis of inertia and is called a neutral line (neutral axis).

From the action of the bending moment in the transverse sections of the beam, normal stresses are arising defined by the formula

where M is a bending moment in the section under consideration;

I - the moment of inertia of the cross section of the beam relative to the neutral axis;

y is the distance from the neutral axis to the point in which the voltages are determined.

As can be seen from formula (8.1), normal voltages in the cross section of the beam at its height is linear, reaching the maximum value in the most remote points from the neutral layer.

where W is the moment of resistance of the cross section of the beam relatively neutral axis.

27. The sustained stresses in the cross section of the beam. Formula Zhuravsky.

The Formula of Zhuravsky allows you to determine the tangent stresses of bending, arising at the transverse section of the beams located at a distance of the total axis.

Withdrawal of the Formula Zhuravsky

I cut out from the beam of the rectangular cross section (Fig. 7.10, a) the element with a length and an additional longitudinal cross section will dissemble into two parts (Fig. 7.10, b).

Consider the equilibrium of the upper part: due to the differences between bending moments, different compressive stresses occur. In order for this part of the beam in equilibrium () in its longitudinal section, the tangential force should occur. The equilibrium equation of the part of the beam:

where integration is carried out only on the cut-off part of the cross section of the beam (in Fig. 7.10, in Sharchovana), - the static moment of the inertia of the cut-off (shaded) part of the cross-sectional area relative to the neutral axis x.

Suppose: tangent stresses () arising in the longitudinal section of the beam, are evenly distributed by its width () at the cross section:

We obtain the expression for tangent stresses:

, and, then tangent stresses () arising at the cross-sectional points of the beams located at a distance from the neutral axis X:

Formula Zhuravsky

Formula Zhuravsky was obtained in 1855 D.I. Zhuravsky, so wears his name.

PAgTets of a round cross section for durability and rigidity

PAgTets of a round cross section for durability and rigidity

The purpose of calculating the strength and rigidity when taking is to determine such a cross-sectional size of a bar, in which voltages and movements will not exceed the specified values \u200b\u200ballowed by the operating conditions. The condition of strength for allowable tangents in the general case is recorded in the form of this condition means that the greatest tangent stresses arising in the twisted timber should not exceed the corresponding allowable stresses for the material. The allowable voltage during dry depends on 0 ─ the voltage corresponding to the hazardous state of the material and the adopted stock of the strength n: ─ the yield strength, the stock of the strength of the strength for the plastic material; ─ Total tensile strength, safety reserve for fragile material. Due to the fact that the values \u200b\u200bin obtaining in testing experiments are harder than when tensile (compression), then, most often, the allowable tension voltages are taken depending on the suspended tensile stresses for the same material. So for steel [for cast iron. When calculating the twisted bars for strength, three types of tasks differing in the form of use of strength conditions are possible: 1) voltage check (verification calculation); 2) selection of section (design calculation); 3) Determination of the permissible load. 1. When checking the voltages on specified loads and the size of the bar, the highest tangent tension arises and are compared with the formula specified (2.16). If the condition of strength is not performed, then it is necessary to either increase the cross-sectional dimensions, or reduce the load acting on the bar, or apply the material of higher strength. 2. When selecting the section for a given load and a given value of the allowed voltage from the strength condition (2.16), the magnitude of the polar moment of resistance of the cross section of the bar in the magnitude of the polar moment of resistance is determined by the diameters of the solid round or annular section of the bar. 3. When determining the allowable load on a given allowable voltage and the polar momentum of the WP resistance, the magnitude of the allowable torque Mk is determined (3.16) and then with the help of the torque inclines, the relationship between K M and external twisting moments is established. The calculation of the timber for strength does not exclude the possibility of the occurrence of deformations, unacceptable during its operation. The large bruis angles are very dangerous, as they can lead to disruption of the accuracy of the parts processing, if this timber is a constructive element of the processing machine, or twist oscillations may occur if the ram transmits the twisting moments by time, so the timber must also be calculated on the rigidity. The hardness condition is recorded in the following form: where ─ the largest relative spinning angle of the bar, determined from the expression (2.10) or (2.11). Then the hardness for the shaft will take the form of the permissible relative spinning angle is determined by the norms and for various elements of structures and different types of loads varies from 0.15 ° to 2 ° per 1 m length of the bar. Both in terms of strength, and in the condition of rigidity in determining max or max  we will use geometrical characteristics: WP ─ Polar moment of resistance and IP ─ polar moment of inertia. Obviously, these characteristics will be different for round solid and annular cross sections with the same area of \u200b\u200bthese sections. By concrete calculations, you can make sure that the polar moments of the inertia and the moment of resistance for the annular section are significantly larger than for a squamous circular cross section, since the annular section does not have sites close to the center. Therefore, the ring cross section during dry is more economical than the ram of a solid circular section, i.e. requires a smaller consumption of the material. However, the manufacture of such a bar is more complicated, and therefore more expensive, and this circumstance should also be taken into account when designing BRUSEV, working when crashes. Methods for calculating a timber for strength and rigidity when cutting, as well as reasoning about efficiency, illustrate on the example. Example 2.2 Compare the weight of two shafts, whose transverse dimensions for the same torque Mk 600 nm for the same allowable voltages 10 RG 13 stretching along the fibers p] 7 rp 10 compression and crumpled along the fibers [cm] 10 rc, RCM 13 crumple across the fibers (on a length of at least 10 cm) [cm] 90 2.5 Rcm 90 3 Rocking along the fibers in bending [and] 2 RCK 2.4 Rocking along the fibers when writing 1 RCK 1,2 - 2.4 Rocking in wrinkles across across fibers

The longitudinal force n arising in the cross section of the bar is the resultant internal normal forces distributed through cross-sectional area, and is associated with normal stresses arising in this cross section (4.1):

here is a normal voltage in an arbitrary cross-sectional point belonging to the elementary platform - the cross-sectional area of \u200b\u200bthe bar.

The product is an elementary inner force per DF site.

The magnitude of the longitudinal force N in each particular case can be easily determined using the cross sections method, as shown in the previous paragraph. To find the same amounts of stresses, and at each point of the cross section of the bar need to know the law of their distribution through this section.

The law of the distribution of normal stresses in the cross section of the timber is usually depicted by a graph showing the change in their height or the width of the cross section. Such a chart is called the range of normal stresses (Epura A).

The expression (1.2) can be satisfied with an infinitely large number of types of stress A (for example, with epures of the A, shown in Fig. 4.2). Therefore, to clarify the law of the distribution of normal stresses in cross-sections of the bar, it is necessary to conduct an experiment.

We carry out on the side surface of the bar before its loading of the line perpendicular to the bar axis (Fig. 5.2). Each such line can be considered as a trace of the plane of the cross section of the bar. When loading a bar with axial power of ps, these lines show, as experience shows, remain straight and parallel (their positions after loading the bar are shown in Fig. 5.2 by dashed lines). This suggests that the cross sections of the timber, flat to its loading, remain flat and under the action of the load. Such experience confirms the hypothesis of flat sections (Bernoulli hypothesis), formulated at the end of § 6.1.

Imagine mentally a timber consisting of countless fibers parallel to its axis.

Two of any transverse sections when stretching the timber remain flat and parallel with each other, but removed from each other for some value; Each fiber is extended to the same magnitude. And since the same elongations correspond to the same stresses, the voltage in the cross sections of all the fibers (and, consequently, in all points of the cross section of the bar) are equal to each other.

This allows in the expression (1.2) to bear the amount of and for the sign of the integral. In this way,

So, in the cross sections of the bar during central, tension or compression, evenly distributed normal stresses occur equal to the ratio of the longitudinal force to the cross-sectional area.

In the presence of weakens of some sections of the bar (for example, rivet holes), determining the voltages in these sections, should consider the actual area of \u200b\u200bweak section equal to the full area of \u200b\u200breduced area of \u200b\u200bweakening

For a visual image of the change in normal stresses in cross sections, the rod (at its length) is built with normal stresses. The axis of this plot is a cut line, equal to the length of the rod and the parallel axis. With the terminal cross section of the standard of normal stresses, it has the same appearance as the support of the longitudinal forces (it differs from it only accepted). With the rod of the alternating section, the type of these two EPUR is different; In particular, for a rod with a stepped law, changes in the cross sections of normal voltages have jumps not only in sections in which concentrated axial loads are applied (where there is a longitudinal strength buckle), but also in places change the size of cross-sections. The construction of the distribution of normal stresses along the length of the rod is considered in Example 1.2.

Consider now the stresses in the inclined sections of the bar.

Denote by the angle between the inclined section and the cross section (Fig. 6.2, a). Angle And we agree to consider positive when the cross section for combining with an inclined section must be rotated to this angle counterclockwise.

As already known, the elongation of all the fibers parallel to the axis of the bar, with its tension or compression the same. This suggests that the voltage p at all points of the inclined (as well as the transverse) section is the same.

Consider the lower part of the bar cut off with the cross section (Fig. 6.2, b). From the conditions of its equilibrium it follows that the voltages are parallel to the axis of the bar and are directed towards the opposite strength of the p, and the inner force acting in the section is equal to R. here - the area of \u200b\u200bthe inclined section is equal to (where is the cross-sectional area of \u200b\u200bthe bar).

Hence,

where - normal stresses in the cross sections of the bar.

We will decompose the voltage into two components of the voltage: the normal perpendicular to the secting plane and the tangent of the one parallel to this plane (Fig. 6.2, B).

Values \u200b\u200band Ta get out of expressions

Normal stress is usually considered positive when tensile and negative when compressed. The tangent voltage is positive if the vector depicting it strives to rotate the body relative to any point with lying on the inner normal to the cross section clockwise. In fig. 6.2, the positive tangent voltage is shown, and in Fig. 6.2, G - negative.

It follows from formula (6.2) that normal stresses are values \u200b\u200bfrom (at zero (at a). Thus, the largest (by absolute value) normal stresses occur in cross sections of the bar. Therefore, the calculation of the strength of the stretched or compressed bar is made according to normal voltages. In its cross-sections.