Main stresses during bending. Full testing of the bending strength of beams

In case of flat transverse bending, when a bending moment also acts in the sections of the beam M and shear force Q, not only normal
, but also shear stresses .

Normal stresses during transverse bending are calculated using the same formulas as for pure bending:

;
.(6.24)

P

Fig.6.11. Flat bend

When deriving the formula, we will make some assumptions:

Shear stresses acting at the same distance at from the neutral axis, constant across the width of the beam;

Tangential stresses are everywhere parallel to the force Q.

Let us consider a cantilever beam subject to transverse bending under the action of a force R. Let's construct diagrams of internal forces ABOUT y, And M z .

On distance x from the free end of the beam we select an elementary section of the beam with a length dx and a width equal to the width of the beam b. Let us show the internal forces acting along the edges of the element: on the edge CD shear force occurs Q y and bending moment M z, and on the verge ab– also shear force Q y and bending moment M z +dM z(because Q y remains constant along the length of the beam, and the moment M z changes, fig. 6.12). On distance at cut off part of the element from the neutral axis abcd, we show the stresses acting along the edges of the resulting element mbcn, and consider its equilibrium. There are no stresses on the faces that are part of the outer surface of the beam. On the side faces of the element from the action of the bending moment M z, normal stresses arise:

; (6.25)

. (6.26)

In addition, on these faces from the action of shear force Q y, shear stresses arise , the same stresses arise according to the law of pairing of tangential stresses on the upper face of the element.

Let's create an equilibrium equation for the element mbcn, projecting the resultant stresses considered onto the axis x:

. (6.29)

The expression under the integral sign represents the static moment of the lateral face of the element mbcn relative to the axis x, so we can write

. (6.30)

Considering that, according to the differential dependences of Zhuravsky D.I. during bending,

, (6.31)

expression for tangents stresses during transverse bending can be rewritten as follows ( Zhuravsky's formula)

. (6.32)

Let's analyze Zhuravsky's formula.

Q y– shear force in the section under consideration;

J z – axial moment of inertia of the section relative to the axis z;

b– the width of the section in the place where the shear stresses are determined;

–static moment relative to the z-axis of the section located above (or below) the fiber where the shear stress is determined:

, (6.33)

Where And F" is the coordinate of the center of gravity and the area of ​​the considered part of the section, respectively.

6.6 Full strength check. Dangerous sections and dangerous points

To check the bending strength of the external loads acting on the beam, diagrams of changes in internal forces along its length are constructed and dangerous sections of the beam are determined, for each of which it is necessary to carry out a strength test.

When fully checking the strength of such sections, there will be at least three (sometimes they coincide):

The section in which the bending moment M z reaches its maximum absolute value;

The section in which the shear force Q y, reaches its maximum absolute value;

The section in which the bending moment M z and shear force Q y reach quite large values ​​in absolute value.

In each of the dangerous sections, it is necessary, by constructing diagrams of normal and shear stresses, to find the dangerous points of the section (a strength test is carried out for each of them), of which there will also be at least three:

The point at which normal stresses , reach their maximum value, that is, the point on the outer surface of the beam that is furthest from the neutral axis of the section;

The point at which the shear stress reach their maximum value - a point lying on the neutral axis of the section;

The point at which both the normal stresses and the shear stresses reach sufficiently large values ​​(this test makes sense for sections such as T-beams or I-beams, where the width of the section along the height is not constant).

During transverse bending, along with the bending moment, a transverse force acts in the section, which is the resultant of the tangential stresses.

The consequence of the action of tangential stresses is a distortion of the cross-sectional shape, which contradicts the hypothesis of plane sections. Firstly, the section may experience deplaiatssho, those. doesn't stay flat. Secondly, the section after deformation does not remain perpendicular to the curved axis of the beam.

These effects are taken into account in more complex theories of rod bending. At the same time, for a large number of engineering problems, the formulas obtained for pure bending can be generalized to the case of transverse bending. Assessment of the limits of applicability of these formulas and responsibility for the results obtained fall within the competence of the calculator.

To determine the values ​​of normal stresses during transverse bending, formula (5.10) is widely used. Next, we will show that in the case of a constant transverse force, this formula gives an exact result, and in the case of a variable transverse force, the results obtained for determining the normal

formulas show an error of order - Where h- section height; / - beam length.

To determine the magnitude of tangential stresses, consider a beam element with a length dx(Fig. 5.8).

Rice. 5.8.

In the right and left sections of the element, the normal stresses differ from each other by s/o, which is due to the difference in the values ​​of the bending moment at dM mr. The term associated with the change in t along the length dx, can be neglected as a quantity of higher order of smallness.

Let us make the assumption: the tangential stresses in the section are directed parallel to the shear force acting in this section Q.

Let us determine the values ​​of tangential stresses at points separated by a distance at from the neutral axis. To do this, cut off with a plane CD from the beam element length dx Part abed.

In cross section at height at tangential stresses act, i.e. At the same time, in the section perpendicular to it, i.e. in a plane parallel to the plane xz, in accordance with the law of pairing of tangential stresses, tangential stresses of the same magnitude will act.

Let's create an equilibrium equation for an element by projecting all the forces acting on this element onto the direction of the axis X. Let us calculate the integrals included in the equilibrium equation within the upper part of the section A*:

As a result of the transformations, we obtain the following formula for calculating tangential stresses:

According to formula (5.10) and taking into account relation (5.3), we find the derivative of the normal stress:

and take this value into account in the expression for the shear stress:

As a result, we obtain the following formula for calculating tangential stresses:

Where Q - shear force in section; S* - static moment of the cut-off part of the section with area L* relative to the central axis; / izg - moment of inertia of the section relative to the central axis; h- width of the section at the location where shear stresses are determined.

Formula (5.21) is called formulasZhuravsky TO

Consider a beam with a rectangular cross section (Fig. 5.9, A). Let us determine normal and shear stresses in a dangerous section. Section L is dangerous, in which the maximum bending moment M зг = -И acts. As for the transverse force, its value in any section of the beam is constant and equal -F.

Rice. 5.9.

According to formulas (5.15) and (5.20), we determine the value of the maximum normal stress:

‘Zhuravsky Dmitry Ivanovich (1828-1891) - Russian mechanical scientist and engineer, specialist in the field of bridge construction and structural mechanics, was the first to solve the problem of determining shear stresses during transverse bending of a beam.

Let us calculate the quantities included in formula (5.21):

At a section point separated by a distance at from the neutral axis, the value of the shear stress is

The maximum voltage occurs at y = 0 in fibers belonging to the central axis 0t.

This voltage formally has a negative value, but its sign can be ignored, since it is not important for the calculation.

Let us estimate the ratio of the maximum values ​​of normal and tangential stresses arising in the section of the beam:

According to the design diagram of the beam, it is assumed that - 1. It follows from this that tangential stresses are of a higher order of magnitude compared to normal stresses.

Let us generalize estimate (5.24) for a beam of length / and characteristic cross-sectional size A. With a shear force equal to F, bending moment is estimated as M bend ~ FI. For the characteristic values ​​of the axial moment of inertia of the section, the static moment of part of the section and the moment of resistance to bending, we obtain the following estimates:

Consequently, for the maximum normal and tangential stresses the following estimates are valid:

We finally obtain the following estimate of the ratio of maximum tangential and normal stresses:

The estimates obtained for a specific rectangular cross-section can be extended to the case of an arbitrary cross-section, with the proviso that the cross-section is considered as massive. For thin-walled profiles, the above conclusion about the possibility of neglecting tangential stresses in comparison with normal stresses is not always true.

It should be noted that when deriving formula (5.21), we were not completely consistent and, while carrying out the transformations, made the following error. Namely, the formula for normal stresses that we used was obtained under the assumption that the hypothesis of plane sections is valid, i.e. in the absence of cross-sectional deplanation. By applying tangential stresses to the element, we allowed for the possibility of distortion of right angles, thereby violating the above-mentioned hypothesis. Therefore, the resulting calculation formulas are approximate. The shear stress diagram shown in Fig. 5.9, b, explains the nature of the curvature of the cross sections of the beam during transverse bending. At the extreme points, the tangential stresses are zero, therefore, the corresponding fibers will be normal to the upper and lower surfaces of the beam. At the neutral line, where maximum shear stresses act, maximum shear strains will occur.

At the same time, we note that if the value of the transverse force is constant within the section, the curvature of all sections will be the same, therefore, the effect of curvature will not be reflected in the magnitude of the longitudinal tensile and compressive deformations of the fibers caused by the bending moment.

For non-rectangular cross sections, additional errors are introduced into formula (5.21) due to the failure to meet the accepted assumptions about the nature of the shear stress distribution. So, for example, for a circular cross-section, the shear stresses at points at section contours should be directed tangentially to the contour, and not parallel to the shear force Q. This means that shear stresses must have components acting both along the z/-axis and along the z-axis.

However, despite the existing contradictions, the resulting formulas give quite satisfactory results when carrying out practical calculations. A comparison of the values ​​of tangential stresses determined by formula (5.21) with the results obtained by exact methods shows that the error in the value of the largest tangential stress does not exceed 5%, i.e. this formula is suitable for practical calculations.

Let us make a few comments regarding strength calculations for direct transverse bending. In contrast to pure bending, two force factors arise in the cross sections of the rod during transverse bending: bending moment M mzg and transverse force Q. However, given that the highest normal stresses occur in the outermost fibers, where there are no shear stresses (see Fig. 5.9, b), and the highest tangential stresses occur in the neutral layer, where normal stresses are equal to zero, the strength conditions in these cases are formulated separately for normal and tangential stresses:

When deriving the formula for calculating normal stresses, we consider the case of bending, when the internal forces in the sections of the beam are reduced only to bending moment, A the shear force turns out to be zero. This case of bending is called pure bending. Consider the middle section of the beam, which is subject to pure bending.

When loaded, the beam bends so that it The lower fibers lengthen and the upper fibers shorten.

Since part of the fibers of the beam is stretched, and part is compressed, and the transition from tension to compression occurs smoothly, without jumps, V average part of the beam is located a layer whose fibers only bend, but do not experience either tension or compression. This layer is called neutral layer. The line along which the neutral layer intersects the cross section of the beam is called neutral line or neutral axis sections. Neutral lines are strung on the axis of the beam. Neutral line is the line in which normal stresses are zero.

Lines drawn on the side surface of the beam perpendicular to the axis remain flat when bending. These experimental data make it possible to base the conclusions of the formulas hypothesis of plane sections (conjecture). According to this hypothesis, the sections of the beam are flat and perpendicular to its axis before bending, remain flat and turn out to be perpendicular to the curved axis of the beam when it is bent.

Assumptions for deriving normal stress formulas: 1) The hypothesis of plane sections is fulfilled. 2) Longitudinal fibers do not press on each other (non-pressure hypothesis) and, therefore, each of the fibers is in a state of uniaxial tension or compression. 3) Deformations of fibers do not depend on their position along the cross-sectional width. Consequently, normal stresses, changing along the height of the section, remain the same along the width. 4) The beam has at least one plane of symmetry, and all external forces lie in this plane. 5) The material of the beam obeys Hooke's law, and the modulus of elasticity in tension and compression is the same. 6) The relationship between the dimensions of the beam is such that it operates under plane bending conditions without warping or twisting.

Let's consider a beam of arbitrary cross-section, but having an axis of symmetry. Bending moment represents resultant moment of internal normal forces, arising on infinitely small areas and can be expressed in integral form: (1), where y is the arm of the elementary force relative to the x axis

Formula (1) expresses static side of the problem of bending a straight beam, but along it at a known bending moment It is impossible to determine normal stresses until the law of their distribution is established.

Let us select the beams in the middle section and consider section of length dz, subject to bending. Let's depict it on an enlarged scale.

Sections limiting the area dz, parallel to each other until deformed, and after applying the load rotate around their neutral lines by an angle . The length of the neutral layer fiber segment will not change. and will be equal to: , where is it radius of curvature the curved axis of the beam. But any other fiber lying lower or higher neutral layer, will change its length. Let's calculate relative elongation of fibers located at a distance y from the neutral layer. Relative elongation is the ratio of absolute deformation to the original length, then:

Let's reduce by and bring similar terms, then we get: (2) This formula expresses geometric side of the pure bending problem: The deformations of the fibers are directly proportional to their distances to the neutral layer.

Now let's move on to stresses, i.e. we will consider physical side of the task. in accordance with non-pressure assumption we use fibers under axial tension-compression: then, taking into account the formula (2) we have (3), those. normal stress when bending along the section height linearly distributed. On the outermost fibers, normal stresses reach their maximum value, and at the center of gravity of the section they are equal to zero. Let's substitute (3) into the equation (1) and take the fraction out of the integral sign as a constant value, then we have . But the expression is axial moment of inertia of the section relative to the x axis - I x. Its dimension cm 4, m 4

Then ,where (4) ,where is the curvature of the curved axis of the beam, and is the rigidity of the beam section during bending.

Let's substitute the resulting expression curvature (4) into expression (3) and we get formula for calculating normal stresses at any point in the cross section: (5)

That. maximum tensions arise at points furthest from the neutral line. Attitude (6) called axial moment of section resistance. Its dimension cm 3, m 3. The moment of resistance characterizes the influence of the shape and dimensions of the cross section on the magnitude of the stresses.

Then maximum voltages: (7)

Bending strength condition: (8)

When transverse bending occurs not only normal, but also shear stresses, because available shear force. Shear stress complicate the picture of deformation, they lead to curvature cross sections of the beam, resulting in the hypothesis of plane sections is violated. However, research shows that distortions introduced by shear stresses slightly affect normal stresses calculated by the formula (5) . Thus, when determining normal stresses in the case of transverse bending The theory of pure bending is quite applicable.

Neutral line. Question about the position of the neutral line.

During bending there is no longitudinal force, so we can write Let us substitute here the formula for normal stresses (3) and we get Since the modulus of longitudinal elasticity of the beam material is not equal to zero and the curved axis of the beam has a finite radius of curvature, it remains to assume that this integral is static moment of area cross section of the beam relative to the neutral line-axis x , and, since it is equal to zero, then the neutral line passes through the center of gravity of the section.

Let us consider a beam subject to plane straight bending under the action of arbitrary transverse loads in the main plane Ohoo(Fig. 7.31, A). Let's cut the beam at a distance x from its left end and consider the equilibrium of the left side. The influence of the right side in this case must be replaced by the action of the bending moment A/ and the transverse force Qy in the drawn section (Fig. 7.31, b). The bending moment L7 in the general case is not constant in magnitude, as was the case with pure bending, but varies along the length of the beam. Since the bending moment M

according to (7.14) is associated with normal stresses o = a x, then the normal stresses in the longitudinal fibers will also change along the length of the beam. Therefore, in the case of transverse bending, normal stresses are functions of the variables x and y: a x = a x (x, y).

During transverse bending in the beam section, not only normal but also tangential stresses act (Fig. 7.31, V), the resultant of which is the transverse force Q y:

Presence of tangential stresses x uh accompanied by the appearance of angular deformations. Shear stresses, like normal ones, are distributed unevenly over the section. Consequently, the angular deformations associated with them by Hooke’s law during shear will also be unevenly distributed. This means that during transverse bending, unlike pure bending, the sections of the beam do not remain flat (the hypothesis of J. Bernoulli is violated).

The curvature of cross sections can be clearly demonstrated by the example of the bending of a cantilever beam of rectangular rubber section caused by a concentrated force applied at the end (Fig. 7.32). If you first draw straight lines on the side faces perpendicular to the axis of the beam, then after bending these lines do not remain straight. At the same time, they are bent so that the greatest shift occurs at the level of the neutral layer.

More accurate studies have established that the effect of distortion of cross sections on the magnitude of normal stresses is insignificant. It depends on the ratio of the section height h to the length of the beam / and at h/ / o x for transverse bending, formula (7.14) derived for the case of pure bending is usually used.

The second feature of transverse bending is the presence of normal stresses O y, acting in the longitudinal sections of the beam and characterizing the mutual pressure between the longitudinal layers. These stresses occur in areas where there is a distributed load q, and in places where concentrated forces are applied. Typically these stresses are very small compared to normal stresses a x. A special case is the action of a concentrated force, in the area of ​​application of which significant local stresses can arise and u.

Thus, an infinitesimal element in the plane Ohoo in the case of transverse bending, it is in a biaxial stress state (Fig. 7.33).

Voltages t and o, as well as voltage o Y, in the general case are functions of coordinates* and y. They must satisfy the differential equilibrium equations, which for a biaxial stress state ( a z = T yz = = 0) in the absence

volumetric forces have the following form:

These equations can be used to determine shear stresses = m and normal stresses OU. This is easiest to do for a beam with a rectangular cross section. In this case, when determining m, the assumption is made that they are uniformly distributed across the width of the section (Fig. 7.34). This assumption was made by the famous Russian bridge builder D.I. Zhuravsky. Research shows that this assumption almost exactly corresponds to the actual nature of the distribution of shear stresses during bending for sufficiently narrow and high beams (b « AND).

Using the first of the differential equations (7.26) and formula (7.14) for normal stresses a x, we get

Integrating this equation over the variable y, we find

Where f(x)- an arbitrary function, to determine which we use the condition of the absence of tangential stresses on the bottom edge of the beam:

Taking this boundary condition into account, from (7.28) we find

The final expression for the tangential stresses acting in the cross sections of the beam takes the following form:

Due to the law of pairing of tangential stresses, tangential stresses also arise t, = t in longitudinal sections

hoo hoo

beams parallel to the neutral layer.

From formula (7.29) it is clear that the tangential stresses vary along the height of the cross section of the beam according to the law of a square parabola. The tangential stresses have the greatest value at points at the level of the neutral axis at y = 0, and in the outermost fibers of the beam at y = ±h/2 they are equal to zero. Using formula (7.23) for the moment of inertia of a rectangular section, we obtain

Where F= bh - cross-sectional area of ​​the beam.

Diagram t is shown in Fig. 7.34.

In the case of beams of non-rectangular cross-section (Fig. 7.35), determining the shear stresses m from the equilibrium equation (7.27) is difficult, since the boundary condition for m is not known at all points of the cross-section contour. This is due to the fact that in this case, tangential stresses t act in the cross section, not parallel to the transverse force Qy. In fact, it can be shown that at points near the contour of the cross section, the total shear stress m is directed tangentially to the contour. Let us consider in the vicinity of an arbitrary point on the contour (see Fig. 7.35) an infinitesimal area dF in the cross-sectional plane and a platform perpendicular to it dF" on the side surface of the beam. If the total stress t at a point on the contour is not directed tangentially, then it can be decomposed into two components: x vx in the direction of the normal v to the contour and X in tangent direction t to the contour. Therefore, according to the law of pairing of tangential stresses on the site dF" should

but act on a shear stress x equal to x vv . If the lateral surface is free from shear loads, then the component x vv = z vx = 0, that is, the total shear stress x must be directed tangentially to the contour of the cross section, as shown, for example, at points A and IN contour.

Consequently, the shear stress x both at points of the contour and at any point of the cross section can be decomposed into their components x.

To determine the components x of the tangential stress in beams of non-rectangular cross-section (Fig. 7.36, b) Let us assume that the section has a vertical axis of symmetry and that the x component of the total shear stress x, as in the case of a rectangular cross section, is uniformly distributed over its width.

Using a longitudinal section parallel to the plane Oxz and passing in the distance at from it, and two cross sections heh + dx Let us mentally cut out from the bottom of the beam an infinitesimal element of length dx(Fig. 7.36, V).

Let us assume that the bending moment M varies within length dx of the beam element under consideration, and the shear force Q is constant. Then in cross sections x and x + dx beams will be subject to tangential stresses x of equal magnitude, and normal stresses arising from bending moments MzmMz+ dM„, will be respectively equal A And A + da. Along the horizontal edge of the selected element (in Fig. 7.36, V it is shown in axonometry) according to the law of pairing of tangential stresses, stresses x v „ = x will act.

hoo hoo

Resultants R And R+dR normal stresses o and o + d applied to the ends of the element, taking into account formula (7.14) are equal

Where

static moment of cut-off area F(in Fig. 7.36, b shaded) relative to the neutral axis Oz y, is an auxiliary variable that varies within at

Resultant of tangential stresses t applied

xy

to the horizontal edge of the element, taking into account the introduced assumption about the uniform distribution of these stresses across the width b(y) can be found using the formula

The equilibrium condition for the element?X=0 gives

Substituting the values ​​of the resultant forces, we get

From here, taking into account (7.6), we obtain a formula for determining tangential stresses:

This formula in Russian literature is called formula D.I. Zhuravsky.

In accordance with formula (7.32), the distribution of tangential stresses t along the height of the section depends on the change in the width of the section b(y) and the static moment of the cut-off part of the section S OTC (y).

Using formula (7.32), shear stresses are most simply determined for the rectangular beam considered above (Fig. 7.37).

The static moment of the cut-off cross-sectional area F qtc is equal to

Substituting 5° tf into (7.32), we obtain the previously derived formula (7.29).

Formula (7.32) can be used to determine shear stresses in beams with a stepwise constant section width. Within each section with a constant width, the tangential stresses vary along the height of the section according to the law of a square parabola. In places where the section width changes abruptly, the tangential stresses also have jumps or discontinuities. The nature of the diagram t for such a section is shown in Fig. 7.38.

Rice. 7.37

Rice. 7.38

Let us consider the distribution of tangential stresses in an I-section (Fig. 7.39, A) when bending in a plane Ooh. An I-section can be represented as the junction of three narrow rectangles: two horizontal shelves and a vertical wall.

When calculating m in the wall in formula (7.32), you need to take b(y) - d. As a result we get

Where S° 1C calculated as the sum of static moments about the axis Oz shelf area Fn and parts of the wall F, shaded in Fig. 7.39, A:

The tangential stresses t have the greatest value at the level of the neutral axis at y = 0:

where is the static moment of the area of ​​half the section relative to the neutral axis:

For rolled I-beams and channels, the value of the static moment of half the section is given in the assortment.

Rice. 7.39

At the level where the wall adjoins the flanges, shear stresses 1 ? equal

Where S" - static moment of the flange cross-sectional area relative to the neutral axis:

Vertical tangential stresses m in the flanges of the I-beam cannot be found using formula (7.32), since due to the fact that b :» t, the assumption of their uniform distribution across the width of the shelf becomes unacceptable. On the top and bottom edges of the flange, these stresses should be zero. Therefore t in

wow

shelves are very small and are of no practical interest. Of much greater interest are the horizontal tangential stresses in the flanges m, to determine which we consider the equilibrium of an infinitesimal element isolated from the lower flange (Fig. 7.39 , b).

According to the law of pairing of tangential stresses on the longitudinal face of this element, parallel to the plane Ooh, voltage is applied x xz equal in magnitude to the stress t acting in the cross section. Due to the small thickness of the I-beam flange, these stresses can be assumed to be uniformly distributed over the thickness of the flange. Taking this into account, from the equilibrium equation of the element 5^=0 we will have

From here we find

Substituting into this formula the expression for a x from (7.14) and taking into account that we obtain

Considering that

Where S° TC - static moment of the cut-off area of ​​the shelf (in Fig. 7. 39, A shaded twice) relative to the axis Oz, we'll finally get it

According to Fig. 7.39 , A

Where z- axis-based variable OU.

Taking this into account, formula (7.34) can be represented in the form

This shows that horizontal shear stresses vary linearly along the axis Oz and take the greatest value at z = d/ 2:

In Fig. Figure 7.40 shows diagrams of tangential stresses m and m^, as well as the directions of these stresses in the flanges and the wall of the I-beam when a positive shear force is applied to the section of the beam Q. Tangential stresses, figuratively speaking, form a continuous flow in the I-beam section, directed at each point parallel to the contour of the section.

Let's move on to the definition of normal stresses and y in the longitudinal sections of the beam. Let's consider a section of a beam with a uniformly distributed load along the top edge (Fig. 7.41). Let's take the cross section of the beam to be rectangular.

We use it to determine the second of the differential equilibrium equations (7.26). Substituting formula (7.32) for tangential stresses into this equation t uh, taking (7.6) into account we obtain

After performing integration over the variable y, we find

Here f(x) - an arbitrary function that is defined using a boundary condition. According to the conditions of the problem, the beam is loaded with a uniformly distributed load q along the upper edge, and the lower edge is free from loads. Then the corresponding boundary conditions are written in the form

Using the second of these conditions, we obtain

Taking this into account, the formula for stress and y will take the following form:

From this expression it is clear that stresses vary along the height of the section according to the law of a cubic parabola. In this case, both boundary conditions (7.35) are satisfied. Highest voltage value takes on the upper surface of the beam when y=-h/2:

Nature of the diagram and y shown in Fig. 7.41.

To estimate the values ​​of the highest stresses o. a, and m and the relationships between them, let us consider, for example, the bending of a cantilever beam of rectangular cross-section with dimensions bxh, under the action of a uniformly distributed load applied to the upper edge of the beam (Fig. 7.42). The highest absolute value of stresses occurs in the seal. In accordance with formulas (7.22), (7.30) and (7.37), these stresses are equal

As usual for beams l/h» 1, then from the obtained expressions it follows that the voltages c x in absolute value exceed the voltage t and, especially, and u. So, for example, when 1/I == 10 we get a x /t xy = 20', o x /c y = 300.

Thus, the greatest practical interest when calculating beams for bending is stress a x, acting in the cross sections of the beam. Voltages with y, characterizing the mutual pressure of the longitudinal layers of the beam are negligible compared to o v .

The results obtained in this example indicate that the hypotheses introduced in § 7.5 are completely justified.

Flat (straight) bend- when the bending moment acts in a plane passing through one of the main central axes of inertia of the section, i.e. all forces lie in the plane of symmetry of the beam. Main hypotheses(assumptions): hypothesis about the non-pressure of longitudinal fibers: fibers parallel to the axis of the beam experience tensile-compressive deformation and do not exert pressure on each other in the transverse direction; hypothesis of plane sections: a section of a beam that is flat before deformation remains flat and normal to the curved axis of the beam after deformation. In the case of flat bending, in general, internal power factors: longitudinal force N, transverse force Q and bending moment M. N>0, if the longitudinal force is tensile; at M>0, the fibers on top of the beam are compressed and the fibers on the bottom are stretched. .

The layer in which there are no extensions is called neutral layer(axis, line). For N=0 and Q=0, we have the case pure bend. Normal voltages:
, is the radius of curvature of the neutral layer, y is the distance from some fiber to the neutral layer.

43) Eccentric tension and compression

Tension and compression

 - normal voltage[Pa], 1 Pa (pascal) = 1 N/m 2,

10 6 Pa = 1 MPa (megapascal) = 1 N/mm 2

N - longitudinal (normal) force [N] (newton); F - cross-sectional area [m2]

 - relative deformation [dimensionless quantity];

L - longitudinal deformation [m] (absolute elongation), L - rod length [m].

-Hooke's law -  = E

E - tensile modulus of elasticity (modulus of elasticity of the 1st kind or Young’s modulus) [MPa]. For steel E = 210 5 MPa = 210 6 kg/cm 2 (in the “old” system of units).

(the larger E, the less tensile the material)

;
- Hooke's law

EF is the stiffness of the rod in tension (compression).

When the rod is stretched, it “thinners”, its width - a decreases by the transverse deformation - a.

-relative transverse deformation.

-Poisson's ratio [dimensionless quantity];

 ranges from 0 (cork) to 0.5 (rubber); for steel  0.250.3.

If the longitudinal force and cross-section are not constant, then the elongation of the rod:

Tensile work:
, potential energy:

47. Mohr Integral

A universal method for determining displacements (linear and rotation angles) is Mohr’s method. A unit generalized force is applied to the system at the point for which the generalized displacement is sought. If the deflection is determined, then the unit force is a dimensionless concentrated force; if the angle of rotation is determined, then it is a dimensionless unit moment. In the case of a spatial system, there are six components of internal forces. The generalized displacement is defined

48. Determination of stress under the combined action of bending and torsion

Bending with torsion

The combined action of bending and torsion is the most common case of loading shafts. Five components of internal forces arise: Q x, Q y, M x, M y, M z = M cr. During the calculation, diagrams of bending moments M x , M y , and torque M cr are constructed and the dangerous section is determined. Resulting bending moment
. Max. normal and shear stresses at dangerous points (A,B):
,

, (for a circle: W=
– axial moment of resistance , W р =
– polar moment of contact of the section).

Main stresses at the most dangerous points (A and B):

Strength testing is carried out according to one of the strength theories:

IV: Mohr's theory:

where m=[ p ]/[ c ] – permissible. e.g. tension/compression (for brittle materials - cast iron).

T
.k.W p =2W, we get:

The numerator is the reduced moment according to the accepted theory of strength. ;

II: , with Poisson's ratio=0.3;

III:

or with one formula:
, whence the moment of resistance:
, shaft diameter:
. The formulas are also suitable for calculating the annular section.