At the points of the cross sections of the beam during longitudinal transverse bending, normal stresses from compression by longitudinal forces and from bending by transverse and longitudinal loads (Fig. 18.10).

In the outer fibers of the beam in the dangerous section, the total normal stresses have the highest values:

In the example of a compressed beam with one transverse force considered above, according to (18.7), we obtain the following stresses in the outer fibers:

If the dangerous section is symmetrical about its neutral axis, then the stress in the outer compressed fibers will be the largest in absolute value:

In a section that is not symmetrical about the neutral axis, both compressive and tensile stresses in the outer fibers can be the largest in absolute value.

When establishing a dangerous point, the difference in the resistance of the material to tension and compression should be taken into account.

Given expression (18.2), formula (18.12) can be written as:

Applying the approximate expression for we obtain

Dangerous in beams of constant section will be the section for which the numerator of the second term has the largest value.

Dimensions cross section beams must be selected so that they do not exceed the allowable stress

However, the resulting relationship between stresses and the geometric characteristics of the section is difficult for design calculation; section dimensions can only be selected by repeated attempts. With longitudinal-transverse bending, as a rule, a verification calculation is carried out, the purpose of which is to establish the margin of safety of the part.

With longitudinal-transverse bending, there is no proportionality between stresses and longitudinal forces; stresses with a variable axial force grow faster than the force itself, which can be seen, for example, from formula (18.13). Therefore, the margin of safety in the case of longitudinal-transverse bending must be determined not by stresses, i.e., not from the ratio, but by loads, understanding the margin of safety as a number showing how many times it is necessary to increase the acting loads in order to maximum voltage in the calculated part has reached the yield point.

The determination of the safety factor is associated with the solution of transcendental equations, since the force is contained in formulas (18.12) and (18.14) under the sign trigonometric function. For example, for a beam, compressed by a force and loaded with one transverse force P, the safety factor according to (18.13) is found from the equation

To simplify the problem, you can use the formula (18.15). Then, to determine the margin of safety, we obtain a quadratic equation:

Note that in the case when the longitudinal force remains constant, and only the transverse loads change in magnitude, the task of determining the margin of safety is simplified, and it is possible to determine not by load, but by stresses. From formula (18.15) for this case we find

Example. A double-supported duralumin beam of an I-beam thin-walled section is compressed by a force P and subjected to the action of a uniformly distributed transverse load with intensity and moments applied at the ends

beams, as shown in Fig. 18.11. Determine the stress at a dangerous point and the maximum deflection with and without taking into account the bending action of the longitudinal force P, and also find the safety margin of the beam in terms of yield strength.

In calculations, take the Characteristics of an I-beam:

Solution. The most loaded is the middle section of the beam. Maximum deflection and bending moment from shear load alone:

The maximum deflection from the combined action of the transverse load and the longitudinal force P is determined by the formula (18.10). Get

Basic concepts. Shear force and bending moment

During bending, the cross sections, remaining flat, rotate relative to each other around some axes lying in their planes. Beams, axles, shafts and other machine parts and structural elements work for bending. In practice, there are transverse (straight), oblique and clean views bending.

Transverse (straight) (Fig. 61, A) called bending, when external forces perpendicular to the longitudinal axis of the beam act in a plane passing through the axis of the beam and one of the main central axes of its cross section.

An oblique bend (Fig. 61, b) is a bend when forces act in a plane passing through the axis of the beam, but not passing through any of the main central axes of its cross section.

In the cross sections of beams during bending, two types arise internal forces- bending moment M and and shear force Q. In the particular case when the transverse force is zero, and only a bending moment occurs, then a pure bend takes place (Fig. 61, c). Pure bending occurs when loading with a distributed load or under some loadings with concentrated forces, for example, a beam loaded with two symmetrical equal forces.

Rice. 61. Bend: a - transverse (straight) bend; b - oblique bend; c - pure bend

When studying bending deformation, it is mentally represented that the beam consists of an infinite number of fibers parallel to the longitudinal axis. With pure bending, the hypothesis of flat sections is valid: fibers lying on the convex side stretched lying on the concave side - shrink, and on the border between them lies a neutral layer of fibers (longitudinal axis), which are only warp, without changing its length; the longitudinal fibers of the beam do not exert pressure on each other and, therefore, experience only tension and compression.

Internal force factors in beam sections - transverse force Q and bending moment M and(Fig. 62) depend on external forces and vary along the length of the beam. The laws of change of transverse forces and bending moments are represented by some equations in which the coordinates are arguments z cross sections of beams, and functions - Q And M i. To determine the internal force factors, we use the method of sections.

Rice. 62.

Shear force Q is the resultant of the internal tangential forces in the cross section of the beam. It should be borne in mind that the transverse force has the opposite direction for the left and right parts of the beam, which indicates the unsuitability of the rule of static signs.

Bending moment M and is the resulting moment about the neutral axis of the internal normal forces acting in the cross section of the beam. The bending moment, as well as the transverse force, has a different direction for the left and right parts of the beam. This indicates the unsuitability of the rule of signs of statics in determining the bending moment.

Considering the balance of the parts of the beam located to the left and right of the section, it can be seen that a bending moment must act in the cross sections M and and shear force Q. Thus, in the case under consideration, not only normal stresses, corresponding to the bending moment, but also tangential stresses, corresponding to the transverse force, act at the points of the cross sections.

For a visual representation of the distribution along the axis of the beam of transverse forces Q and bending moments M and it is convenient to represent them in the form of diagrams, the ordinates of which for any values ​​of the abscissa z give corresponding values Q And M i. Plots are constructed similarly to plotting longitudinal forces (see 4.4) and torques (see 4.6.1.).

Rice. 63. Direction of transverse forces: a - positive; b - negative

Since the rules of static signs are unacceptable for establishing signs of transverse forces and bending moments, we will establish other rules of signs for them, namely:

• - if external sips (Fig.
• 63, a), lying on the left side of the section, tend to raise the left side of the beam or, lying on the right side of the section, lower the right side of the beam, then the transverse force Q is positive;
• - If external forces (Fig.
• 63, b), lying on the left side of the section, tend to lower the left side of the beam or, lying on the right side of the section, raise the right side of the beam, then the transverse force (Z is negative;

Rice. 64. Direction of bending moments: a - positive; b - negative

• - if an external load (force and moment) (Fig. 64, a), located to the left of the section, gives a moment directed clockwise or located to the right of the section, directed counterclockwise, then the bending moment M is considered positive;
• - if the external load (Fig. 64, b), located to the left of the section, gives a moment directed counterclockwise or, located to the right of the section, directed clockwise, then the bending moment M is considered negative.

The sign rule for bending moments is related to the nature of the deformation of the beam. The bending moment is considered positive if the beam is bent convex downwards (stretched fibers are located at the bottom). The bending moment is considered negative if the beam is bent with a convexity upwards (stretched fibers are located at the top).

Using the rules of signs, one should mentally imagine the cross section of the beam as rigidly clamped, and the bonds as discarded and replaced by their reactions. To determine the reactions, the rules of static signs are used.

Building a diagram Q.

Let's build a plot M method characteristic points. We arrange points on the beam - these are the points of the beginning and end of the beam ( D,A ), concentrated moment ( B ), and also note as a characteristic point the middle of a uniformly distributed load ( K ) is an additional point for constructing a parabolic curve.

Determine bending moments at points. Rule of signs cm. - .

The moment in IN will be defined as follows. First let's define:

Point TO let's take in middle area with a uniformly distributed load.

Building a diagram M . Plot AB – parabolic curve(rule of "umbrella"), plot BD – straight oblique line.

For a beam, determine the support reactions and plot bending moment diagrams ( M) and shear forces ( Q).

1. We designate supports letters A And IN and direct the support reactions R A And R B .

Compiling equilibrium equations.

Examination

Write down the values R A And R B on calculation scheme.

2. Plotting transverse forces method sections. We place the sections on characteristic areas(between changes). According to the dimensional thread - 4 sections, 4 sections.

sec. 1-1 move left.

The section passes through the section with uniformly distributed load, note the size z 1 to the left of the section before the beginning of the section. Plot length 2 m. Rule of signs For Q - cm.

We build on the found value diagramQ.

sec. 2-2 move right.

The section again passes through the area with a uniformly distributed load, note the size z 2 to the right of the section to the beginning of the section. Plot length 6 m.

Building a diagram Q.

sec. 3-3 move right.

sec. 4-4 move to the right.

We are building diagramQ.

3. Construction diagrams M method characteristic points.

characteristic point- a point, any noticeable on the beam. These are the dots A, IN, WITH, D , as well as the point TO , wherein Q=0 And bending moment has an extremum. also in middle console put an additional point E, since in this area under a uniformly distributed load the diagram M described crooked line, and it is built, at least, according to 3 points.

So, the points are placed, we proceed to determine the values ​​​​in them bending moments. Rule of signs - see..

Plots NA, AD – parabolic curve(the “umbrella” rule for mechanical specialties or the “sail rule” for construction), sections DC, SW – straight slanted lines.

Moment at a point D should be determined both left and right from the point D . The very moment in these expressions Excluded. At the point D we get two values ​​from difference by the amount m – jump to its size.

Now we need to determine the moment at the point TO (Q=0). However, first we define point position TO , denoting the distance from it to the beginning of the section by the unknown X .

T. TO belongs second characteristic area, shear force equation(see above)

But the transverse force in t. TO is equal to 0 , A z 2 equals unknown X .

We get the equation:

Now knowing X, determine the moment at a point TO on the right side.

Building a diagram M . The construction is feasible for mechanical specialties, postponing positive values up from the zero line and using the "umbrella" rule.

For a given scheme of a cantilever beam, it is required to plot the diagrams of the transverse force Q and the bending moment M, perform a design calculation by selecting a circular section.

Material - wood, design resistance of the material R=10MPa, M=14kN m, q=8kN/m

There are two ways to build diagrams in a cantilevered beam with rigid embedding - the usual one, having previously determined the support reactions, and without defining the support reactions, if we consider the sections, going from the free end of the beam and discarding the left side with the embedding. Let's build diagrams ordinary way.

1. Define support reactions.

Uniformly distributed load q replace the conditional force Q= q 0.84=6.72 kN

In a rigid embedment, there are three support reactions - vertical, horizontal and moment, in our case, the horizontal reaction is 0.

Let's find vertical support reaction R A And reference moment M A from the equilibrium equations.

In the first two sections on the right, there is no transverse force. At the beginning of a section with a uniformly distributed load (right) Q=0, in the back - the magnitude of the reaction R.A.
3. To build, we will compose expressions for their definition on sections. We plot the moment diagram on the fibers, i.e. down.

(compressed lower fibers).

Plot DC: (upper fibers are compressed).

Plot SC: (compressed left fibers)

(compressed left fibers)

In the figure - diagrams normal (longitudinal) forces - (b), transverse forces - (c) and bending moments - (d).

Checking the balance of node C:

Task 2 Construct diagrams of internal forces for the frame (Fig. a).

Given: F=30kN, q=40kN/m, M=50kNm, a=3m, h=2m.

Let's define support reactions frames:

From these equations we find:

Since the reaction values R K has a sign minus, in fig. A changes direction given vector to the opposite, while writing R K = 83.33 kN.

Let us determine the values ​​of internal forces N, Q And M in the characteristic sections of the frame:

Sun section:

(compressed right fibers).

Plot CD:

(the right fibers are compressed);

(compressed right fibers).

Plot DE:

(lower fibers are compressed);

(compressed lower fibers).

CS section

(compressed left fibers).

Let's build diagrams of normal (longitudinal) forces (b), transverse forces (c) and bending moments (d).

Consider the equilibrium of nodes D And E

From consideration of knots D And E it is clear that they are in equilibrium.

Task 3. For a frame with a hinge, construct diagrams of internal forces.

Given: F=30kN, q=40kN/m, M=50kNm, a=2m, h=2m.

Solution. Let's define support reactions. It should be noted that in both hinged-fixed supports along two reactions. For this reason, you should use hinge property C — moment in it from both left and right forces zero. Let's look at the left side.

The equilibrium equations for the considered frame can be written as:

From the solution of these equations it follows:

On the frame diagram, the direction of the force H B changes to opposite (N B =15kN).

Let's define efforts in the characteristic sections of the frame.

Plot BZ:

(compressed left fibers).

Plot ZC:

(compressed left fibers);

Plot KD:

(compressed left fibers);

(compressed left fibers).

Plot DC:

(lower fibers are compressed);

Definition extreme value bending moment on the section CD:

1. Construction of a diagram of transverse forces. For a cantilever beam (Fig. A ) characteristic points: A – point of application of the support reaction VA; WITH is the point of application of the concentrated force; D, B – beginning and end of distributed load. For a cantilever, the transverse force is determined similarly to a two-bearing beam. So, when moving to the left:

To check the correctness of the determination of the transverse force in the sections, pass the beam in the same way, but from the right end. Then the right parts of the beam will be cut off. Remember that the rule of signs will change in this case. The result should be the same. We build a diagram of the transverse force (Fig. b).

2. Plotting the moments

For a cantilever beam, the diagram of bending moments is constructed similarly to the previous construction. Characteristic points for this beam (see Fig. A) are as follows: A - support; WITH - point of application of concentrated moment and force F; D And IN- the beginning and end of the action of a uniformly distributed load. Since the plot Q x in the area of ​​action of a distributed load does not cross the zero line, to plot the moment diagram in a given section (parabolic curve), you should select an additional point arbitrarily for plotting the curve, for example, in the middle of the section.

Move left:

Going to the right we find M B = 0.

Based on the values ​​found, we build a diagram of bending moments (see Fig. V ).

Post published author admin limited oblique line, A in a section where there is no distributed load - a straight line parallel to the axis, therefore, to construct a diagram of transverse forces, it is sufficient to determine the values Qat at the beginning and end of each segment. In the section corresponding to the point of application of the concentrated force, the transverse force must be calculated slightly to the left of this point (at an infinitely close distance from it) and slightly to the right of it; transverse forces in such places are denoted accordingly .

Building a diagram Qat by the method of characteristic points, moving from the left. For greater clarity, at first it is recommended to cover the discarded part of the beam with a sheet of paper. Characteristic points for a two-bearing beam (Fig. A ) there will be points C And D - the beginning and end of the distributed load, as well as A And B – points of application of support reactions, E is the point of application of the concentrated force. Let's mentally draw an axis y perpendicular to the axis of the beam through a point WITH and we will not change its position until we pass the entire beam from C before E. Considering the left parts of the beam cut off at characteristic points, we project onto the axis y the forces acting in this section with the corresponding signs. As a result, we get:

To check the correctness of determining the shear force in the sections, you can pass the beam in the same way, but from the right end. Then the right parts of the beam will be cut off. The result should be the same. The coincidence of the results can serve as control plotting Qat. We draw a zero line under the image of the beam and from it, on the accepted scale, set aside the found values ​​of the transverse forces, taking into account the signs at the corresponding points. Get the plot Qat(rice. b ).

Having built the diagram, pay attention to the following: the diagram under a distributed load is depicted as an inclined straight line, under unloaded sections - segments parallel to the zero line, under a concentrated force, a jump is formed on the diagram, equal to the value of the force. If the sloping line under distributed load crosses the zero line, mark this point, then this extremum point, and it is now characteristic for us, according to the differential relationship between Qat And Mx, at this point the moment has an extremum and it will need to be determined when plotting the bending moments. In our problem, this is the point TO . Focused moment on plot Qat does not manifest itself in any way, since the sum of the projections of the forces forming a pair is equal to zero.

2. Plotting the moments. We build a diagram of bending moments, as well as transverse forces, using the method of characteristic points, moving from the left. It is known that in the section of a beam with a uniformly distributed load, the diagram of bending moments is outlined by a curved line (a quadratic parabola), for the construction of which it is necessary to have at least three points and, therefore, the values ​​of the bending moments at the beginning of the section, the end of it and in one intermediate section must be calculated. It is best to take such an intermediate point as a section in which the diagram Qat crosses the zero line, i.e. Where Qat= 0. On the diagram M this section must contain the vertex of the parabola. If the diagram Q at does not cross the zero line, then to build a diagram M follows on in this section, take an additional point, for example, in the middle of the section (the beginning and end of the distributed load), remembering that the convexity of the parabola is always directed downwards if the load acts from top to bottom (for construction specialties). There is a "rain" rule, which is very helpful when constructing the parabolic part of the plot M. For builders, this rule looks like this: imagine that a distributed load is rain, substitute an umbrella upside down under it, so that the rain does not flow down, but collects in it. Then the bulge of the umbrella will be facing down. This is exactly what the outline of the diagram of moments under a distributed load will look like. For mechanics, there is a so-called "umbrella" rule. The distributed load is represented by rain, and the outline of the diagram should resemble the outline of an umbrella. In this example, the plot is built for builders.

If more accurate plotting is required, then the values ​​of bending moments in several intermediate sections must be calculated. Let us agree for each such section to first determine the bending moment in an arbitrary section, expressing it in terms of the distance X from any point. Then, giving the distance X a series of values, we obtain the values ​​of bending moments in the corresponding sections of the section. For sections where there is no distributed load, bending moments are determined in two sections corresponding to the beginning and end of the section, since the diagram M in such areas is limited to a straight line. If an external concentrated moment is applied to the beam, then it is necessary to calculate the bending moment slightly to the left of the place of application of the concentrated moment and slightly to the right of it.

For a two-support beam, the characteristic points are as follows: C And D - the beginning and end of the distributed load; A – beam support; IN – the second support of the beam and the point of application of the concentrated moment; E – right end of the beam; dot TO , corresponding to the section of the beam, in which Qat= 0.

Left move. We mentally discard the right side up to the section under consideration (take a sheet of paper and cover the discarded part of the beam with it). We find the sum of the moments of all forces acting to the left of the section relative to the point under consideration. So,

Before determining the moment in the section TO, you need to find the distance x=AK. Let's make an expression for the transverse force in this section and equate it to zero (stroke on the left):

This distance can also be found from the similarity of triangles KLN And KIG on the plot Qat(rice. b) .

Determine the moment at a point TO :

Let's go through the rest of the beam on the right.

As you can see, the moment at the point D when moving left and right, it turned out to be the same - the plot closed. Based on the found values, we build a diagram. Positive values ​​are set aside down from the zero line, and negative values ​​​​up (see Fig. V ).

Longitudinally transverse bend is called a combination of transverse bending with compression or tension of the beam.

When calculating for longitudinal-transverse bending, the bending moments in the cross sections of the beam are calculated taking into account the deflections of its axis.

Consider a beam with hinged ends, loaded with some transverse load and a compressive force 5 acting along the axis of the beam (Fig. 8.13, a). Let us denote the deflection of the beam axis in the cross section with the abscissa (we take the positive direction of the y axis downwards, and, therefore, we consider the deflections of the beam to be positive when they are directed downwards). The bending moment M, acting in this section,

(23.13)

here is the bending moment from the action of the transverse load; - additional bending moment from the force

The total deflection y can be considered to consist of the deflection arising from the action of only the transverse load, and an additional deflection equal to that caused by the force .

The total deflection y is greater than the sum of the deflections arising from the separate action of the transverse load and the force S, since in the case of the action of only the force S on the beam, its deflections are equal to zero. Thus, in the case of longitudinal-transverse bending, the principle of independence of the action of forces is not applicable.

When a tensile force S acts on the beam (Fig. 8.13, b), the bending moment in the section with the abscissa

(24.13)

The tensile force S leads to a decrease in the deflections of the beam, i.e., the total deflections y in this case are less than the deflections caused by the action of only the transverse load.

In the practice of engineering calculations, longitudinal-transverse bending usually means the case of the action of a compressive force and a transverse load.

With a rigid beam, when the additional bending moments are small compared to the moment, the deflections y differ little from the deflections . In these cases, it is possible to neglect the influence of the force S on the magnitudes of the bending moments and the deflections of the beam and calculate it for central compression (or tension) with transverse bending, as described in § 2.9.

For a beam whose rigidity is low, the influence of the force S on the values ​​of the bending moments and deflections of the beam can be very significant and cannot be neglected in the calculation. In this case, the beam should be calculated for longitudinal-transverse bending, meaning by this the calculation for the combined action of bending and compression (or tension), performed taking into account the influence of the axial load (force S) on the bending deformation of the beam.

Consider the methodology for such a calculation using the example of a beam hinged at the ends, loaded with transverse forces directed in one direction and with a compressive force S (Fig. 9.13).

Substitute in the approximate differential equation of an elastic line (1.13) the expression of the bending moment M according to the formula (23.13):

[the minus sign in front of the right side of the equation is taken because, in contrast to formula (1.13), here the downward direction is considered positive for deflections], or

Hence,

To simplify the solution, let us assume that the additional deflection varies sinusoidally along the length of the beam, i.e. that

This assumption makes it possible to obtain sufficiently accurate results when a transverse load is applied to the beam, directed in one direction (for example, from top to bottom). Let us replace the deflection in formula (25.13) by the expression

The expression coincides with the Euler formula for the critical force of a compressed rod with hinged ends. Therefore, it is denoted and called the Euler force.

Hence,

The Euler force should be distinguished from the critical force calculated by the Euler formula. The value can be calculated using the Euler formula only if the rod flexibility is greater than the limit; the value is substituted into the formula (26.13) regardless of the flexibility of the beam. The formula for the critical force, as a rule, includes the minimum moment of inertia of the cross section of the rod, and the expression for the Euler force includes the moment of inertia about that of the main axes of inertia of the section, which is perpendicular to the plane of action of the transverse load.

From formula (26.13) it follows that the ratio between the total deflections of the beam y and the deflections caused by the Action of only the transverse load depends on the ratio (the magnitude of the compressive force 5 to the magnitude of the Euler force).

Thus, the ratio is a criterion for the rigidity of the beam in longitudinal-transverse bending; if this ratio is close to zero, then the stiffness of the beam is large, and if it is close to one, then the stiffness of the beam is small, i.e., the beam is flexible.

In the case when , deflection, i.e., in the absence of force S, deflections are caused only by the action of a transverse load.

When the value of the compressive force S approaches the value of the Euler force, the total deflections of the beam increase sharply and can be many times greater than the deflections caused by the action of only a transverse load. In the limiting case at, deflections y, calculated by formula (26.13), become equal to infinity.

It should be noted that formula (26.13) is not applicable for very large deflections of the beam, since it is based on an approximate expression for curvature. This expression is applicable only for small deflections, and for large deflections it must be replaced by the same curvature expression (65.7). In this case, the deflections y at at would not be equal to infinity, but would be, although very large, but finite.

When a tensile force acts on the beam, formula (26.13) takes the form.

From this formula, it follows that the total deflections are less than the deflections caused by the action of only the transverse load. With a tensile force S numerically equal to the value of the Euler force (i.e., at ), the deflections y are half the deflections

The largest and smallest normal stresses in the cross section of a beam with hinged ends at longitudinal-transverse bending and compressive force S are equal to

Consider a two-bearing I-section beam with a span. The beam is loaded in the middle with a vertical force P and is compressed by an axial force S = 600 (Fig. 10.13). Cross-sectional area of ​​the beam moment of inertia, moment of resistance and modulus of elasticity

The transverse braces connecting this beam with adjacent beams of the structure exclude the possibility of the beam becoming unstable in the horizontal plane (i.e., in the plane of least rigidity).

The bending moment and deflection in the middle of the beam, calculated without taking into account the influence of the force S, are equal to:

The Euler force is determined from the expression

Deflection in the middle of the beam, calculated taking into account the influence of the force S on the basis of formula (26.13),

Let us determine the greatest normal (compressive) stresses in the average cross section of the beam according to the formula (28.13):

from where after transformation

Substituting into expression (29.13) various values ​​of P (in), we obtain the corresponding stress values. Graphically, the relationship between determined by expression (29.13) is characterized by the curve shown in fig. 11.13.

Let us determine the allowable load P, if for the beam material and the required safety factor, therefore, the allowable stress for the material

From fig. 11.23 it follows that the stress occurs in the beam under load and the stress - under load

If we take the load as the allowable load, then the stress safety factor will be equal to the specified value. However, in this case, the beam will have an insignificant load safety factor, since stresses equal to from will arise in it already at Rot

Consequently, the load safety factor in this case will be equal to 1.06 (since e. is clearly insufficient.

In order for the beam to have a safety factor equal to 1.5 in terms of load, the value should be taken as the permissible value, while the stresses in the beam will be, as follows from Fig. 11.13, approximately equal

Above, the strength calculation was carried out according to the allowable stresses. This provided the necessary margin of safety not only in terms of stresses, but also in terms of loads, since in almost all cases considered in the previous chapters, the stresses are directly proportional to the magnitudes of the loads.

With longitudinal-transverse bending of the stress, as follows from Fig. 11.13 are not directly proportional to the load, but change faster than the load (in the case of a compressive force S). In this regard, even a slight accidental increase in load in excess of the calculated one can cause a very large increase in stresses and destruction of the structure. Therefore, the calculation of compressed-bent rods for longitudinal-transverse bending should be carried out not according to the allowable stresses, but according to the allowable load.

By analogy with formula (28.13), let us compose the strength condition when calculating the longitudinal-transverse bending according to the allowable load.

Compressed-curved rods, in addition to calculating the longitudinal-transverse bending, must also be calculated for stability.

UDC 539.52

LIMIT LOAD FOR A CLAMPED BEAM LOADED BY A LONGITUDINAL FORCE, ASYMMETRICLY DISTRIBUTED LOAD AND SUPPORT MOMENTS

I.A. Monakhov1, Yu.K. Bass2

department of building production Building faculty Moscow State Machine-Building University st. Pavel Korchagin, 22, Moscow, Russia, 129626

2Department of Building Structures and Constructions Faculty of Engineering Peoples' Friendship University of Russia st. Ordzhonikidze, 3, Moscow, Russia, 115419

The article develops a technique for solving problems of small deflections of beams made of an ideal rigid-plastic material under the action of asymmetrically distributed loads, taking into account the preliminary tension-compression. The developed technique is used to study the stress-strain state of single-span beams, as well as to calculate the ultimate load of beams.

Key words: beam, nonlinearity, analytical.

IN modern construction, shipbuilding, mechanical engineering, chemical industry and other branches of technology, the most common types of structures are rods, in particular beams. Naturally, to determine the real behavior of bar systems (in particular, beams) and their strength resources, it is necessary to take into account plastic deformations.

The calculation of structural systems, taking into account plastic deformations using the model of an ideal rigid-plastic body, is the simplest, on the one hand, and quite acceptable from the point of view of design practice requirements, on the other. If we keep in mind the region of small displacements of structural systems, then this is due to the fact that the bearing capacity (“ultimate load”) of ideal rigid-plastic and elastic-plastic systems turns out to be the same.

Additional reserves and a more rigorous assessment of the bearing capacity of structures are revealed as a result of taking into account geometric nonlinearity when they are deformed. At present, taking into account geometric nonlinearity in the calculations of structural systems is a priority not only from the point of view of the development of the theory of calculation, but also from the point of view of the practice of designing structures. Acceptability of solutions to problems of structural analysis under conditions of smallness

displacements is quite uncertain, on the other hand, practical data and properties of deformable systems allow us to assume that large displacements are realistically achievable. It suffices to point to the structures of construction, chemical, shipbuilding and machine-building facilities. In addition, the model of a rigid-plastic body means that elastic deformations are neglected, i.e. plastic deformations are much greater than elastic ones. Since displacements correspond to deformations, it is appropriate to take into account large displacements of rigid-plastic systems.

However, geometrically nonlinear deformation of structures in most cases inevitably leads to the occurrence of plastic deformations. Therefore, the simultaneous consideration of plastic deformations and geometrical nonlinearity in the calculations of structural systems and, of course, rod ones, is of particular importance.

This article deals with small deflections. Similar problems were solved in the works.

We consider a beam with pinched supports, under the action of a stepped load, edge moments and a preliminarily applied longitudinal force (Fig. 1).

Rice. 1. Beam under distributed load

The beam equilibrium equation for large deflections in dimensionless form has the form

d2 t / , h d2 w dn

-- + (n ± w)-- + p \u003d ^ - \u003d 0, dx ax ax

x 2w p12 M N ,g,

where x==, w=-, p=--, t=--, n=-, n and m are internal normal

I to 5xЪk b!!bk 25!!k

force and bending moment, p - transverse uniformly distributed load, W - deflection, x - longitudinal coordinate (origin on the left support), 2k - cross-sectional height, b - cross-sectional width, 21 - beam span, 5^ - yield strength material. If N is given, then the force N is a consequence of the action p at

available deflections, 11 = = , the line above the letters means the dimension of the values.

Consider the first stage of deformation - "small" deflections. The plastic section arises at x = x2, in it m = 1 - n2.

The expressions for the deflection rates have the form - deflection at x = x2):

(2-x), (x > X2),

The solution of the problem is divided into two cases: x2< 11 и х2 > 11.

Consider the case x2< 11.

For zone 0< х2 < 11 из (1) получаем:

Px 111 1 P11 k1p/1 m = + k1 p + p/1 -k1 p/1 -±4- + -^41

x - (1 - p2) ± a,

(, 1 , p/2 k1 p12L

Px2 + k1 p + p11 - k1 p11 -+ 1 ^

X2 = k1 +11 - k111 - + ^

Taking into account the occurrence of a plastic hinge at x = x2, we obtain:

tx \u003d x \u003d 1 - n2 \u003d - p

(12 k12 L k +/ - k1 - ^ + k "A

k, + /, - k, /, -L +

(/ 2 k/ 2 A k1 + /1 - k1/1 - ^ + M

Considering the case x2 > /1, we get:

for zone 0< х < /1 выражение для изгибающих моментов имеет вид

k p-p2 + car/1+p/1 -k1 p/1 ^ x-(1-P12)±

and for zone 11< х < 2 -

^ p-rC + 1^ L

x - (1 - p-) ± a +

(. rg-k1 p1-L

Kx px2 + kx p+

0, and then

I2 12 1 h h x2 = 1 -- + -.

The equality follows from the plasticity condition

where we get the expression for the load:

k1 - 12 + M L2

K1/12 - k2 ¡1

Table 1

k1 = 0 11 = 0.66

table 2

k1 = 0 11 = 1.33

0 6,48 9,72 12,96 16,2 19,44

0,5 3,24 6,48 9,72 12,96 16,2

Table 3

k1 = 0.5 11 = 1.61

0 2,98 4,47 5,96 7,45 8,94

0,5 1,49 2,98 4,47 5,96 7,45

Table 5 k1 = 0.8 11 = 0.94

0 2,24 3,56 4,49 5,61 6,73

0,5 1,12 2,24 3,36 4,49 5,61

0 2,53 3,80 5,06 6,33 7,59

0,5 1,27 2,53 3,80 5,06 6,33

Table 3

k1 = 0.5 11 = 2.0

0 3,56 5,33 7,11 8,89 10,7

0,5 1,78 3,56 5,33 7,11 8,89

Table 6 k1 \u003d 1 11 \u003d 1.33

0 2,0 3,0 4,0 5,0 6,0

0,5 1,0 2,0 3,0 4,0 5,0

Table 7 Table 8

k, = 0.8 /, = 1.65 k, = 0.2 /, = 0.42

0 2,55 3,83 5,15 6,38 7,66

0,5 1,28 2,55 3,83 5,15 6,38

0 7,31 10,9 14,6 18,3 21,9

0,5 3,65 7,31 10,9 14,6 18,3

By setting the load factor k1 from 0 to 1, the bending moment a from -1 to 1, the value of the longitudinal force n1 from 0 to 1, the distance /1 from 0 to 2, we obtain the position of the plastic hinge according to formulas (3) and (5), and then we obtain the value of the ultimate load according to formulas (4) or (6). The numerical results of the calculations are summarized in tables 1-8.

LITERATURE

Basov Yu.K., Monakhov I.A. Analytical solution of the problem of large deflections of a rigid-plastic pinched beam under the action of a local distributed load, support moments and longitudinal force // Vestnik RUDN University. Series "Engineering Research". - 2012. - No. 3. - S. 120-125.

Savchenko L.V., Monakhov I.A. Large deflections of physically nonlinear round plates. Bulletin of INGECON. Series "Technical Sciences". - Issue. 8(35). - St. Petersburg, 2009. - S. 132-134.

Galileev S.M., Salikhova E.A. Investigation of natural vibration frequencies of structural elements made of fiberglass, carbon fiber and graphene // Bulletin of INGECON. Series "Technical Sciences". - Issue. 8. - St. Petersburg, 2011. - P.102.

Erkhov M.I., Monakhov A.I. Large deflections of a prestressed rigid-plastic beam with hinged supports under a uniformly distributed load and edge moments // Bulletin of the Department of Building Sciences of the Russian Academy of Architecture and Building Sciences. - 1999. - Issue. 2. - S. 151-154. .

THE LITTLE DEFLECTIONS OF THE PREVIOUSLY INTENSE IDEAL PLASTIC BEAMS WITH THE REGIONAL MOMENTS

I.A. Monakhov1, U.K. Basov2

"Department of Building production manufacture Building Faculty Moscow State Machine-building University Pavla Korchagina str., 22, Moskow, Russia,129626

Department of Bulding Structures and Facilities Enqineering Faculty Peoples" Friendship University of Russia Ordzonikidze str., 3, Moskow, Russia, 115419

In the work up the technique of the decision of problems about the little deflections of beams from ideal hard-plastic material, with various kinds of fastening, for want of action of the asymmetrically distributed loads with allowance for of preliminary stretching-compression is developed. The developed technique is applied for research of the strained-deformed condition of beams, and also for calculation of a deflection of beams with allowance for of geometrical nonlinearity.

Key words: beam, analytic, nonlinearity.