UDC 539.52

ULTIMATE LOAD FOR A RESTRAINTED BEAM LOADED WITH LONGITUDINAL FORCE, UNSYMMETRICALLY DISTRIBUTED LOAD AND SUPPORT MOMENTS

I.A. Monakhov1, Yu.K. Basov2

department construction production Faculty of Civil Engineering Moscow State Mechanical Engineering University st. Pavel Korchagina, 22, Moscow, Russia, 129626

2Department building structures and structures Faculty of Engineering Russian University friendship of peoples st. Ordzhonikidze, 3, Moscow, Russia, 115419

The article develops a method 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 preliminary tension-compression. The developed methodology was 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 rod ones, in particular beams. Naturally, to determine real behavior rod systems(in particular, beams) and their strength resources, it is necessary to take into account plastic deformations.

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

Additional reserves and stricter assessment bearing capacity structures are revealed by taking into account geometric nonlinearity during their deformation. Currently, taking into account geometric nonlinearity in the calculations of structural systems is a priority task not only from the point of view of the development of calculation theory, but also from the point of view of the practice of designing structures. Acceptability of solutions to problems of structural calculations under conditions of small

displacements is quite uncertain; on the other hand, practical data and properties of deformable systems suggest that large displacements are actually achievable. It is enough to point out the designs of construction, chemical, shipbuilding and mechanical engineering 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 deformations correspond to displacements, taking into account large displacements of rigid plastic systems is appropriate.

However, geometrically nonlinear deformation of structures in most cases inevitably leads to the occurrence of plastic deformations. That's why special meaning acquires simultaneous consideration of plastic deformations and geometric nonlinearity in the calculations of structural systems and, of course, rods.

This article discusses small deflections. Similar problems were solved in works.

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

Rice. 1. Beam under distributed load

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

d2 t/h d2 w dn

-- + (n ± n)-- + p = ^ - = 0, dx ah ah

x 2w р12 М N,г,

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

I to 5xЪk b!!bk 25!!bk

force and bending moment, p - transverse uniformly distributed load, W - deflection, x - longitudinal coordinate (origin of coordinates on the left support), 2к - height cross section, b - cross-section width, 21 - beam span, 5^ - yield strength of the 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 indicates the dimension of the quantities.

Let's consider the first stage of deformation - “small” deflections. Plastic section occurs at x = x2, in it m = 1 - n2.

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

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

The solution to 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:

Рх 111 1 Р11 к1р/1 t = + к1 р + р/1 -к1 р/1 -±4- +-^41

x -(1 -n2)±a,

(, 1, r/2 k1 r12L

Рх2 + к1 р + р11 - к1 р11 -+ 1 ^

X2 = k1 +11 - k111 - + ^

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

tx=x = 1 - p2 = - p

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

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

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

Considering the case x2 > /1, we obtain:

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

to р-р2 + kar/1+р/1 -к1 р/1 ^ x-(1-П12)±

and for zone 11< х < 2 -

^ р-рЦ + 1^ Л

x -(1 -n-)±a +

(. rg-k1 r1-L

Kx px2 + kh p+

0, and then

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

The condition of plasticity implies the equality

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 = 1 11 = 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

Setting the load coefficient k1 from 0 to 1, the bending moment a from -1 to 1, the value of the longitudinal force p1 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 maximum load using 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 to the problem of large deflections of a rigid-plastic clamped beam under the action of a local distributed load, supporting moments and longitudinal force. Vestnik RUDN. Series "Engineering Research". - 2012. - No. 3. - P. 120-125.

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

Galileev S.M., Salikhova E.A. Study of the frequencies of natural vibrations of structural elements made of fiberglass, carbon fiber and graphene // Bulletin of INGECON. Series "Technical Sciences". - Vol. 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 Construction Sciences Russian Academy architecture and building sciences. - 1999. - Issue. 2. - pp. 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 solution of problems about the little deflections of beams from the 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 geometrical nonlinearity.

Key words: beam, analytical, nonlinearity.

Basic concepts. Shear force and bending moment

During bending, the cross sections, while remaining flat, rotate relative to each other around certain 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 its cross section.

Oblique bending (Fig. 61, b) is a bending 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 shear force is zero and only a bending moment occurs, then pure bending occurs (Fig. 61, c). Pure bending occurs when loaded 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 imagined that the beam consists of an infinite number of fibers parallel to the longitudinal axis. At pure bend the hypothesis of plane sections is valid: fibers lying on the convex side stretch, lying on the concave side - shrink, and on the boundary between them lies a neutral layer of fibers (longitudinal axis), which only are bent, 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 - shear 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 in shear forces and bending moments are represented by certain equations in which the arguments are the coordinates z cross sections of beams, and functions - Q And M i. To determine internal force factors, we use the section method.

Rice. 62.

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

Bending moment M and is the resulting moment relative to the neutral axis of the internal normal forces acting in the cross section of the beam. The bending moment, like the shear force, has different direction for the left and right parts of the beam. This indicates that the rule of static signs is unsuitable when determining the bending moment.

Considering the equilibrium of the parts of the beam located to the left and right of the section, it is clear that a bending moment must act in the cross sections M and and shear force Q. Thus, in the case under consideration, at the points of the cross sections there are not only normal stresses corresponding to the bending moment, but also tangent stresses corresponding to the transverse force.

For a visual representation of the distribution of shear forces along the axis of the beam Q and bending moments M and it is convenient to present them in the form of diagrams, the ordinates of which for any abscissa values z give the corresponding values Q And M i. The diagrams are constructed similarly to the construction of diagrams of 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 the signs of shear forces and bending moments, we will establish other rules of signs for them, namely:

• - if external seeps (Fig.
• 63, a), lying on the left side of the section, tend to lift left side beams 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 (Zonegative;

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 an 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 bends convexly downwards (the stretched fibers are located at the bottom). The bending moment is considered negative if the beam bends convexly upward (the stretched fibers are located at the top).

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

The whole variety of existing support devices is schematized in the form of a number of basic types of supports, of which

most common: articulated and movablesupport(possible designations for it are presented in Fig. 1, a), hinged-fixed support(Fig. 1, b) and hard pinching, or sealing(Fig. 1, c).

In a hinged-movable support, one support reaction occurs, perpendicular to the support plane. Such a support deprives the support section of one degree of freedom, that is, it prevents displacement in the direction of the support plane, but allows movement in the perpendicular direction and rotation of the support section.
In a hinged-fixed support, vertical and horizontal reactions occur. Here, movements in the directions of the support rods are not possible, but rotation of the support section is allowed.
In a rigid embedment, vertical and horizontal reactions and a support (reactive) moment occur. In this case, the support section cannot shift or rotate. When calculating systems containing a rigid embedment, the resulting support reactions can not be determined, choosing the cut-off part so that the embedment with unknown reactions does not fall into it. When calculating systems on hinged supports, the reactions of the supports must be determined. The static equations used for this depend on the type of system (beam, frame, etc.) and will be given in the relevant sections of this manual.

2. Construction of diagrams of longitudinal forces Nz

The longitudinal force in a section is numerically equal to the algebraic sum of the projections of all forces applied on one side of the section under consideration onto the longitudinal axis of the rod.

Rule of signs for Nz: let us agree to consider the longitudinal force in the section positive if the external load applied to the considered cut-off part of the rod causes tension and negative - otherwise.

Example 1.Construct a diagram of longitudinal forces for a rigidly clamped beam(Fig. 2).

Calculation procedure:

1. We outline characteristic sections, numbering them from the free end of the rod to the embedment.
2. Determine the longitudinal force Nz in each characteristic section. In this case, we always consider the cut-off part into which the rigid seal does not fall.

Based on the found values build a diagram Nz. Positive values ​​are plotted (on the selected scale) above the diagram axis, negative values ​​are plotted below the axis.

3. Construction of diagrams of torques Mkr.

Torque in the section is numerically equal to the algebraic sum of external moments applied on one side of the section under consideration, relative to the longitudinal Z axis.

Sign rule for microdistrict: let’s agree to count torque in the section is positive if, when looking at the section from the side of the cut-off part under consideration, the external moment is seen directed counterclockwise and negative - otherwise.

Example 2.Construct a diagram of torques for a rigidly clamped rod(Fig. 3, a).

Calculation procedure.

It should be noted that the algorithm and principles for constructing a torque diagram completely coincide with the algorithm and principles constructing a diagram of longitudinal forces.

1. We outline characteristic sections.
2. Determine the torque in each characteristic section.

Based on the found values ​​we build microdistrict diagram(Fig. 3, b).

4. Rules for monitoring diagrams Nz and Mkr.

For diagrams of longitudinal forces and torques are characterized by certain patterns, knowledge of which allows us to evaluate the correctness of the constructions performed.

1. Diagrams Nz and Mkr are always rectilinear.

2. In the area where there is no distributed load, the diagram Nz(Mkr) is a straight line, parallel to the axis, and in the area under a distributed load it is an inclined straight line.

3. Under the point of application of the concentrated force on the diagram Nz there must be a jump in the magnitude of this force, similarly, under the point of application of the concentrated moment on the diagram Mkr there will be a jump in the magnitude of this moment.

5. Construction of diagrams of transverse forces Qy and bending moments Mx in beams

A rod that bends is called beam. In sections of beams loaded with vertical loads, as a rule, two internal force factors arise - Qy and bending moment Mx.

Lateral force in the section is numerically equal to the algebraic sum of the projections of external forces applied on one side of the section under consideration onto the transverse (vertical) axis.

Sign rule for Qy: Let us agree to consider the transverse force in the section positive if the external load applied to the cut-off part under consideration tends to rotate this section clockwise and negative otherwise.

Schematically, this sign rule can be represented as

Bending moment Mx in a section is numerically equal to the algebraic sum of the moments of external forces applied on one side of the section under consideration, relative to the x axis passing through this section.

Rule of signs for Mx: let us agree to consider the bending moment in the section positive if the external load applied to the cut-off part under consideration leads to tension in this section of the lower fibers of the beam and negative - otherwise.

Schematically, this sign rule can be represented as:

It should be noted that when using the sign rule for Mx in in the specified form, the Mx diagram always turns out to be constructed from the side of the compressed fibers of the beam.

6. Cantilever beams

At plotting Qy and Mx diagrams in cantilever, or rigidly clamped, beams there is no need (as in the previously discussed examples) to calculate the support reactions arising in the rigid embedment, but the cut-off part must be selected so that the embedment does not fall into it.

Example 3.Construct Qy and Mx diagrams(Fig. 4).

Calculation procedure.

1. We outline characteristic sections.

It is easy to establish a certain relationship between the bending moment, shear force and the intensity of the distributed load. Let's consider a beam loaded with an arbitrary load (Figure 5.10). Let us determine the transverse force in an arbitrary section located at a distance from the left support Z.

Projecting onto the vertical the forces located to the left of the section, we obtain

We calculate the shear force in a section located at a distance z+ dz from the left support.

Figure 5.8 .

Subtracting (5.1) from (5.2) we get dQ= qdz, where

that is, the derivative of the shear force along the abscissa of the beam section is equal to the intensity of the distributed load .

Let us now calculate the bending moment in the section with the abscissa z, taking the sum of the moments of forces applied to the left of the section. To do this, a distributed load over a section of length z we replace it with the resultant equal to qz and attached in the middle of the area, at a distance z/2 from the section:

(5.3)

Subtracting (5.3) from (5.4), we obtain the increment in bending moment

The expression in parentheses represents the shear force Q. Then . From here we get the formula

Thus, the derivative of the bending moment along the abscissa of the beam section is equal to the transverse force (Zhuravsky’s theorem).

Taking the derivative of both sides of equality (5.5), we obtain

that is, the second derivative of the bending moment along the abscissa of the beam section is equal to the intensity of the distributed load. We will use the obtained dependencies to check the correctness of the construction of diagrams of bending moments and transverse forces.

Construction of tension-compression diagrams

Example 1.

Round diameter column d compressed by force F. Determine the increase in diameter, knowing the modulus of elasticity E and Poisson's ratio of the column material.

Solution.

The longitudinal deformation according to Hooke's law is equal to

Using Poisson's law, we find transverse deformation

On the other side, .

Hence, .

Example 2.

Construct diagrams of longitudinal force, stress and displacement for a stepped beam.

Solution.

1. Determination of the support reaction. We compose the equilibrium equation in projection onto the axis z:

where R E = 2qa.

2. Constructing diagrams Nz, , W.

E p u r a N z. It is built according to the formula

,

E p u r a. The voltage is equal. As follows from this formula, jumps in the diagram will be caused not only by jumps Nz, but also by sudden changes in cross-sectional area. We determine the values ​​at characteristic points:

In practice, very often there are cases collaboration rod for bending and tension or compression. This kind of deformation can be caused either joint action on the beam longitudinal and transverse forces, or only longitudinal forces.

The first case is shown in Fig. 1. The beam AB is subject to a uniformly distributed load q and longitudinal compressive forces P.

Fig.1.

Let us assume that the deflections of the beam compared to the cross-sectional dimensions can be neglected; then, with a degree of accuracy sufficient for practice, we can assume that even after deformation, the forces P will cause only axial compression of the beam.

Using the method of adding forces, we can find the normal stress at any point of each cross section of the beam as the algebraic sum of the stresses caused by the forces P and the load q.

Compressive stresses from forces P are uniformly distributed over the cross-sectional area F and are the same for all sections

normal bending stresses in vertical plane in a section with the abscissa x, which is measured, say, from the left end of the beam, are expressed by the formula

Thus, the total stress at a point with coordinate z (counting from the neutral axis) for this section is equal to

Figure 2 shows stress distribution diagrams in the section under consideration from forces P, load q and the total diagram.

The greatest stress in this section will be in the upper fibers, where both types of deformation cause compression; in the lower fibers there can be either compression or tension depending on the numerical values ​​of the stresses and. To create the strength condition, we will find the greatest normal stress.

Fig.2.

Since the stresses from the forces P in all sections are the same and evenly distributed, the fibers that are most stressed from bending will be dangerous. These are the outermost fibers in the cross section with the highest bending moment; for them

Thus, the stresses in the outermost fibers 1 and 2 of the middle section of the beam are expressed by the formula

and the calculated voltage will be equal to

If the forces P were tensile, then the sign of the first term would change, and the lower fibers of the beam would be dangerous.

Denoting the compressive or tensile force with the letter N, we can write general formula to check strength

The described calculation procedure is also applied when inclined forces act on the beam. Such a force can be decomposed into normal to the axis, bending the beam, and longitudinal, compressive or tensile.

beam bending force compression