On bending on them constantly. Solving typical problems on strength of materials

A bend is a type of deformation in which the longitudinal axis of the beam is bent. Straight beams working on bending are called beams. A straight bend is a bend in which the external forces acting on the beam lie in the same plane (force plane) passing through the longitudinal axis of the beam and the main central axis of inertia of the cross section.

The bend is called pure, if only one bending moment occurs in any cross section of the beam.

Bending, in which a bending moment and a transverse force simultaneously act in the cross section of the beam, is called transverse. The line of intersection of the force plane and the cross-sectional plane is called force line.

Internal force factors in beam bending.

With a flat transverse bending in the sections of the beam, two internal force factors arise: the transverse force Q and the bending moment M. To determine them, the section method is used (see lecture 1). The transverse force Q in the beam section is equal to the algebraic sum of the projections onto the section plane of all external forces acting on one side of the section under consideration.

Sign rule for shear forces Q:

The bending moment M in the beam section is equal to the algebraic sum of the moments about the center of gravity of this section of all external forces acting on one side of the section under consideration.

Sign rule for bending moments M:

Zhuravsky's differential dependences.

Between the intensity q of the distributed load, the expressions for the transverse force Q and the bending moment M, differential dependencies are established:

Based on these dependencies, the following general patterns of diagrams of transverse forces Q and bending moments M can be distinguished:

Peculiarities of diagrams of internal force factors in bending.

1. On the section of the beam where there is no distributed load, the plot Q is presented straight line , parallel to the base of the diagram, and the diagram M is an inclined straight line (Fig. a).

2. In the section where the concentrated force is applied, on the Q diagram there should be jump , equal to the value of this force, and on the diagram M - breaking point (Fig. a).

3. In the section where a concentrated moment is applied, the value of Q does not change, and the diagram M has jump , equal to the value of this moment, (Fig. 26, b).

4. In the section of the beam with a distributed load of intensity q, the diagram Q changes according to a linear law, and the diagram M - according to a parabolic one, and the convexity of the parabola is directed towards the direction of the distributed load (Fig. c, d).

5. If within the characteristic section of the diagram Q intersects the base of the diagram, then in the section where Q = 0, the bending moment has an extreme value M max or M min (Fig. d).

Normal bending stresses.

Determined by the formula:

The moment of resistance of the section to bending is the value:

Dangerous section when bending, the cross section of the beam is called, in which the maximum normal stress occurs.

Tangential stresses in direct bending.

Determined by Zhuravsky's formula for shear stresses in direct beam bending:

where S ots - static moment of the transverse area of ​​the cut-off layer of longitudinal fibers relative to the neutral line.

Bending strength calculations.

1. At verification calculation the maximum design stress is determined, which is compared with the allowable stress:

2. At design calculation selection of the beam section is made from the condition:

3. When determining the allowable load, the allowable bending moment is determined from the condition:

Bending movements.

Under the action of a bending load, the axis of the beam is bent. In this case, there is a stretching of the fibers on the convex and compression - on the concave parts of the beam. In addition, there is a vertical movement of the centers of gravity of the cross sections and their rotation relative to the neutral axis. To characterize the deformation during bending, the following concepts are used:

Beam deflection Y- displacement of the center of gravity of the cross section of the beam in the direction perpendicular to its axis.

The deflection is considered positive if the center of gravity moves upward. The amount of deflection varies along the length of the beam, i.e. y=y(z)

Section rotation angle- the angle θ by which each section is rotated with respect to its original position. The angle of rotation is considered positive when the section is rotated counterclockwise. The value of the angle of rotation varies along the length of the beam, being a function of θ = θ (z).

The most common way to determine displacements is the method mora and Vereshchagin's rule.

Mohr method.

The procedure for determining displacements according to the Mohr method:

1. An "auxiliary system" is built and loaded with a single load at the point where the displacement is to be determined. If a linear displacement is determined, then a unit force is applied in its direction; when determining angular displacements, a unit moment is applied.

2. For each section of the system, the expressions of bending moments M f from the applied load and M 1 - from a single load are recorded.

3. Mohr integrals are calculated and summed over all sections of the system, resulting in the desired displacement:

4. If the calculated displacement has a positive sign, this means that its direction coincides with the direction of the unit force. The negative sign indicates that the actual displacement is opposite to the direction of the unit force.

Vereshchagin's rule.

For the case when the diagram of bending moments from a given load has an arbitrary, and from a single load - a rectilinear outline, it is convenient to use the graphic-analytical method, or Vereshchagin's rule.

where A f is the area of ​​the diagram of the bending moment M f from a given load; y c is the ordinate of the diagram from a single load under the center of gravity of the diagram M f ; EI x - section stiffness of the beam section. Calculations according to this formula are made in sections, on each of which the straight-line diagram must be without fractures. The value (A f *y c) is considered positive if both diagrams are located on the same side of the beam, negative if they are located on opposite sides. A positive result of the multiplication of diagrams means that the direction of movement coincides with the direction of a unit force (or moment). A complex diagram M f must be divided into simple figures (the so-called "epure layering" is used), for each of which it is easy to determine the ordinate of the center of gravity. In this case, the area of ​​\u200b\u200beach figure is multiplied by the ordinate under its center of gravity.

bend called the deformation of the rod, accompanied by a change in the curvature of its axis. A rod that bends is called beam.

Depending on the methods of applying the load and the methods of fixing the rod, various types of bending can occur.

If only a bending moment arises under the action of a load in the cross section of the rod, then the bend is called clean.

If in cross sections, along with bending moments, transverse forces also arise, then bending is called transverse.

If the external forces lie in a plane passing through one of the main central axes of the bar's cross section, the bend is called simple or flat. In this case, the load and the deformable axis lie in the same plane (Fig. 1).

Rice. one

In order for the beam to take the load in the plane, it must be fixed with the help of supports: hinged-movable, hinged-fixed, embedment.

The beam must be geometrically invariable, while the least number of connections is 3. An example of a geometrically variable system is shown in Fig. 2a. An example of geometrically invariable systems is fig. 2b, c.

a B C)

Reactions arise in the supports, which are determined from the equilibrium conditions of statics. The reactions in the supports are external loads.

Internal bending forces

A rod loaded with forces perpendicular to the longitudinal axis of the beam experiences a flat bend (Fig. 3). There are two internal forces in the cross sections: shear force Q y and bending moment Mz.

Internal forces are determined by the section method. On distance x from the point BUT by a plane perpendicular to the X axis, the rod is cut into two sections. One of the parts of the beam is discarded. The interaction of the beam parts is replaced by internal forces: bending moment Mz and transverse force Q y(Fig. 4).

Domestic efforts Mz and Q y into the cross section are determined from the equilibrium conditions.

An equilibrium equation is drawn up for the part With:

∑y = R A - P 1 - Q y \u003d 0.

Then Q y = R A – P1.

Conclusion. The transverse force in any section of the beam is equal to the algebraic sum of all external forces lying on one side of the section drawn. The transverse force is considered positive if it rotates the rod clockwise about the section point.

∑M 0 = R A ∙ x – P 1 ∙ (x - a) – Mz = 0

Then Mz = R A ∙ x – P 1 ∙ (x – a)

1. Definition of reactions R A , R B ;

∑M A = P ∙ a – R B ∙ l = 0

R B =

∑M B = R A ∙ e – P ∙ a = 0

2. Plotting on the first section 0 ≤ x 1 ≤ a

Q y = R A =; M z \u003d R A ∙ x 1

x 1 = 0 M z (0) = 0

x 1 = a M z (a) =

3. Plotting on the second section 0 ≤ x 2 ≤ b

Q y = - R B = - ; Mz = R B ∙ x 2 ; x 2 = 0 Mz(0) = 0 x 2 = bMz(b) =

When building Mz positive coordinates will be plotted towards the stretched fibers.

Checking plots

1. On the diagram Q y discontinuities can only be in places where external forces are applied, and the magnitude of the jump must correspond to their magnitude.

+ = = P

2. On the diagram Mz discontinuities arise at the points of application of concentrated moments and the magnitude of the jump is equal to their magnitude.

Differential dependencies betweenM, Qandq

Between the bending moment, the transverse force and the intensity of the distributed load, the following dependencies are established:

q = , Q y =

where q is the intensity of the distributed load,

Checking the strength of beams in bending

To assess the strength of the rod in bending and select the beam section, the strength conditions for normal stresses are used.

The bending moment is the resultant moment of normal internal forces distributed over the section.

s = × y,

where s is the normal stress at any point of the cross section,

y is the distance from the center of gravity of the section to the point,

Mz- bending moment acting in the section,

Jz is the axial moment of inertia of the rod.

To ensure strength, the maximum stresses are calculated that occur at the points of the section that are farthest from the center of gravity y = ymax

s max = × ymax,

= Wz and s max = .

Then the strength condition for normal stresses has the form:

s max = ≤ [s],

where [s] is the allowable tensile stress.

straight bend- this is a type of deformation in which two internal force factors arise in the cross sections of the rod: a bending moment and a transverse force.

Pure bend- this is a special case of direct bending, in which only a bending moment occurs in the cross sections of the rod, and the transverse force is zero.

Pure Bend Example - Plot CD on the rod AB. Bending moment is the value Pa pair of external forces causing bending. From the equilibrium of the part of the rod to the left of the cross section mn it follows that the internal forces distributed over this section are statically equivalent to the moment M, equal and opposite to the bending moment Pa.

To find the distribution of these internal forces over the cross section, it is necessary to consider the deformation of the bar.

In the simplest case, the rod has a longitudinal plane of symmetry and is subjected to the action of external bending pairs of forces located in this plane. Then the bend will occur in the same plane.

rod axis nn 1 is a line passing through the centers of gravity of its cross sections.

Let the cross section of the rod be a rectangle. Draw two vertical lines on its faces mm and pp. When bent, these lines remain straight and rotate so that they remain perpendicular to the longitudinal fibers of the rod.

A further theory of bending is based on the assumption that not only lines mm and pp, but the entire flat cross section of the rod remains flat after bending and normal to the longitudinal fibers of the rod. Therefore, when bending, the cross sections mm and pp rotate relative to each other around axes perpendicular to the bending plane (drawing plane). In this case, the longitudinal fibers on the convex side experience tension, and the fibers on the concave side experience compression.

neutral surface is a surface that does not experience deformation during bending. (Now it is located perpendicular to the drawing, the deformed axis of the rod nn 1 belongs to this surface).

Neutral sectional axis- this is the intersection of a neutral surface with any with any cross section (now also located perpendicular to the drawing).

Let an arbitrary fiber be at a distance y from a neutral surface. ρ is the radius of curvature of the curved axis. Dot O is the center of curvature. Let's draw a line n 1 s 1 parallel mm.ss 1 is the absolute elongation of the fiber.

Relative extension ε x fibers

It follows that deformation of the longitudinal fibers proportional to distance y from the neutral surface and inversely proportional to the radius of curvature ρ .

Longitudinal elongation of the fibers of the convex side of the rod is accompanied by lateral constriction, and the longitudinal shortening of the concave side - lateral extension, as in the case of simple stretching and contraction. Because of this, the appearance of all cross sections changes, the vertical sides of the rectangle become slanted. Lateral deformation z:

μ - Poisson's ratio.

As a result of this distortion, all straight cross-sectional lines parallel to the axis z, are bent so as to remain normal to the sides of the section. The radius of curvature of this curve R will be more than ρ in the same way as ε x is greater in absolute value than ε z , and we get

These deformations of the longitudinal fibers correspond to stresses

The voltage in any fiber is proportional to its distance from the neutral axis. n 1 n 2. Position of the neutral axis and radius of curvature ρ are two unknowns in the equation for σ x - can be determined from the condition that the forces distributed over any cross section form a pair of forces that balances the external moment M.

All of the above is also true if the rod does not have a longitudinal plane of symmetry in which the bending moment acts, so long as the bending moment acts in the axial plane, which contains one of the two main axes cross section. These planes are called main bending planes.

When there is a plane of symmetry and the bending moment acts in this plane, the deflection occurs in it. Moments of internal forces about the axis z balance the external moment M. Moments of effort relative to the axis y are mutually destroyed.

Straight transverse bend occurs when all loads are applied perpendicular to the axis of the rod, lie in the same plane, and, in addition, the plane of their action coincides with one of the main central axes of inertia of the section. Direct transverse bending refers to a simple form of resistance and is plane stress state, i.e. the two principal stresses are different from zero. With this type of deformation, internal forces arise: a transverse force and a bending moment. A special case of a direct transverse bend is pure bend, with such resistance there are cargo sections, within which the transverse force vanishes, and the bending moment is nonzero. In the cross sections of the rods with a direct transverse bending, normal and shear stresses arise. Stresses are a function of the internal force, in this case normal stresses are a function of the bending moment, and tangential stresses are a function of the transverse force. For direct transverse bending, several hypotheses are introduced:

1) The cross sections of the beam, flat before deformation, remain flat and orthogonal to the neutral layer after deformation (the hypothesis of flat sections or the hypothesis of J. Bernoulli). This hypothesis holds for pure bending and is violated when a shear force, shear stresses, and angular deformation appear.

2) There is no mutual pressure between the longitudinal layers (hypothesis about non-pressure of the fibers). From this hypothesis it follows that the longitudinal fibers experience uniaxial tension or compression, therefore, with pure bending, Hooke's law is valid.

A bar undergoing bending is called beam. When bending, one part of the fibers is stretched, the other part is compressed. The layer of fibers between the stretched and compressed fibers is called neutral layer, it passes through the center of gravity of the sections. The line of its intersection with the cross section of the beam is called neutral axis. On the basis of the introduced hypotheses for pure bending, a formula for determining normal stresses is obtained, which is also used for direct transverse bending. The normal stress can be found using the linear relationship (1), in which the ratio of the bending moment to the axial moment of inertia (
) in a particular section is a constant value, and the distance ( y) along the ordinate axis from the center of gravity of the section to the point at which the stress is determined, varies from 0 to
.

. (1)

To determine the shear stress during bending in 1856. Russian engineer-builder of bridges D.I. Zhuravsky obtained the dependence

. (2)

The shear stress in a particular section does not depend on the ratio of the transverse force to the axial moment of inertia (
), because this value does not change within one section, but depends on the ratio of the static moment of the area of ​​the cut-off part to the width of the section at the level of the cut-off part (
).

In direct transverse bending, there are movements: deflections (v ) and rotation angles (Θ ) . To determine them, the equations of the method of initial parameters (3) are used, which are obtained by integrating the differential equation of the bent axis of the beam (
).

Here v 0 , Θ 0 ,M 0 , Q 0 – initial parameters, x – distance from the origin of coordinates to the section in which the displacement is defined , a is the distance from the origin of coordinates to the place of application or the beginning of the load.

The calculation for strength and stiffness is carried out using the conditions of strength and stiffness. Using these conditions, one can solve verification problems (perform verification of the fulfillment of the condition), determine the size of the cross section, or select the allowable value of the load parameter. There are several strength conditions, some of them are given below. Strength condition for normal stresses looks like:

, (4)

here
– section modulus relative to the z-axis, R is the design resistance for normal stresses.

Strength condition for shear stresses looks like:

, (5)

here the notation is the same as in the Zhuravsky formula, and R s - design shear resistance or design shear stress resistance.

Strength condition according to the third strength hypothesis or the hypothesis of the greatest shear stresses can be written in the following form:

. (6)

Stiffness conditions can be written for deflections (v ) and rotation angles (Θ ) :

where displacement values ​​in square brackets are valid.

An example of completing an individual task No. 4 (term 2-8 weeks)

With direct pure bending in the cross section of the rod, there is only one force factor - the bending moment M x(Fig. 1). As Q y \u003d dM x / dz \u003d 0, then Mx=const and pure direct bending can be realized when the bar is loaded with pairs of forces applied in the end sections of the bar. Since the bending moment M x by definition is equal to the sum of the moments of internal forces about the axis Oh it is connected with normal stresses by the equation of statics that follows from this definition

Let us formulate the premises of the theory of pure direct bending of a prismatic rod. To do this, we analyze the deformations of a model of a rod made of a low-modulus material, on the side surface of which a grid of longitudinal and transverse scratches is applied (Fig. 2). Since the transverse risks, when the rod is bent by pairs of forces applied in the end sections, remain straight and perpendicular to the curved longitudinal risks, this allows us to conclude that plane section hypotheses, which, as the solution of this problem by the methods of the theory of elasticity shows, ceases to be a hypothesis, becoming an exact fact - the law of plane sections. Measuring the change in the distances between the longitudinal risks, we come to the conclusion about the validity of the hypothesis of non-pressure of the longitudinal fibers.

Orthogonality of longitudinal and transverse scratches before and after deformation (as a reflection of the action of the law of flat sections) also indicates the absence of shifts, shear stresses in the transverse and longitudinal sections of the rod.

Fig.1. Relationship between internal effort and stress

Fig.2. Pure bending model

Thus, pure direct bending of a prismatic rod is reduced to uniaxial tension or compression of longitudinal fibers by stresses (index G omitted later). In this case, part of the fibers is in the tension zone (in Fig. 2, these are the lower fibers), and the other part is in the compression zone (upper fibers). These zones are separated by a neutral layer (p-p), not changing its length, the stresses in which are equal to zero. Taking into account the prerequisites formulated above and assuming that the material of the rod is linearly elastic, i.e. Hooke's law in this case has the form: , we derive formulas for the curvature of the neutral layer (-radius of curvature) and normal stresses . We first note that the constancy of the cross section of the prismatic rod and the bending moment (M x = const), ensures the constancy of the radius of curvature of the neutral layer along the length of the rod (Fig. 3, a), neutral layer (n—n) described by an arc of a circle.

Consider a prismatic rod under conditions of direct pure bending (Fig. 3, a) with a cross section symmetrical about the vertical axis OU. This condition will not affect the final result (in order for a straight bend to be possible, the coincidence of the axis Oh with main axis of inertia of the cross section, which is the axis of symmetry). Axis Ox put on the neutral layer, position whom not known in advance.

a) calculation scheme, b) strains and stresses

Fig.3. Fragment of a pure bend of a beam

Consider an element cut from a rod with length dz, which is shown on a scale with proportions distorted in the interests of clarity in Fig. 3, b. Since the deformations of the element, determined by the relative displacement of its points, are of interest, one of the end sections of the element can be considered fixed. In view of the smallness, we assume that the points of the cross section, when rotated through this angle, move not along arcs, but along the corresponding tangents.

Let us calculate the relative deformation of the longitudinal fiber AB, separated from the neutral layer by at:

From the similarity of triangles C00 1 and 0 1 BB 1 follows that

Longitudinal deformation turned out to be a linear function of the distance from the neutral layer, which is a direct consequence of the law of plane sections

This formula is not suitable for practical use, since it contains two unknowns: the curvature of the neutral layer and the position of the neutral axis Oh, from which the coordinate is counted y. To determine these unknowns, we use the equilibrium equations of statics. The first expresses the requirement that the longitudinal force be equal to zero

Substituting expression (2) into this equation

and taking into account that , we get that

The integral on the left side of this equation is the static moment of the rod cross section about the neutral axis Oh, which can be equal to zero only relative to the central axis. Therefore, the neutral axis Oh passes through the center of gravity of the cross section.

The second static equilibrium equation is that relating normal stresses to the bending moment (which can easily be expressed in terms of external forces and is therefore considered a given value). Substituting the expression for into the bundle equation. voltage, we get:

and given that where J x is the main central moment of inertia about the axis Oh, for the curvature of the neutral layer, we obtain the formula

Fig.4. Normal stress distribution

which was first obtained by S. Coulomb in 1773. To match the signs of the bending moment M x and normal stresses, the minus sign is put on the right side of formula (5), since at M x >0 normal stresses at y>0 turn out to be contractive. However, in practical calculations, it is more convenient, without adhering to the formal rule of signs, to determine the stresses modulo, and put the sign according to the meaning. Normal stresses in pure bending of a prismatic bar are a linear function of the coordinate at and reach the highest values ​​in the fibers most distant from the neutral axis (Fig. 4), i.e.

Here a geometric characteristic is introduced, which has the dimension m 3 and is called moment of resistance in bending. Since for a given M x voltage max? the less the more W x , moment of resistance is geometric characteristic of the strength of the cross-sectional bending. Let us give examples of calculating the moments of resistance for the simplest forms of cross sections. For a rectangular cross section (Fig. 5, a) we have J x \u003d bh 3 / 12, y max = h/2 and W x = J x /y max = bh 2 /6. Similarly for a circle (Fig. 5 ,a J x =d4 /64, ymax=d/2) we get W x =d3/32, for a circular annular section (Fig. 5, in), which one