Build Epleru Q.

Build Epleru M. Method characteristic points. We put the points on the beam - these are points of start and end of the beam ( D, A. ), focused moment ( B. ), as well as note as the characteristic point of the middle of a uniformly distributed load ( K. ) - This is an additional point for the construction of a parabolic curve.

We determine the bending moments at points. Rule of signs cm. - .

Moment in t. IN We will define as follows. First we define:

Point TO Take B. middle Plot with uniformly distributed load.

Build Epleru M. . Plot. AU – parabolic curve (Umbrella rule), plot Cd. – direct oblique line.

For the beam, determine the support reactions and build the fusion of bending moments ( M.) and transverse forces ( Q.).

1. Denote Support letters BUT and IN and send reference reactions R A. and R B. .

Make up equations equations.

Check

Record values R A. and R B. on the calculation scheme.

2. Building Epura transverse forces Method sections. Sections arrange by characteristic sites (between changes). On dimensional thread - 4 plots, 4 sections.

sech. 1-1 move left.

The section passes through the site with uniformly distributed load, noted Size z. 1 Left from section before the start of the site. Length of a plot of 2 m. Rule of signs for Q. - cm.

We build on the found value epleruQ..

sech. 2-2 move right.

The cross section passes again along the area of \u200b\u200buniformly distributed load, mark the size z. 2 Right from section prior to the start of the site. Length of a plot of 6 m.

Build Epleru Q..

sech. 3-3 turn right.

sech. 4-4 Time to the right.

Building epleruQ..

3. Building Epura M. Method characteristic points.

Characteristic point - The point is how noticeable on the beam. This is a point BUT, IN, FROM, D. as well as point TO , wherein Q.=0 and bending moment has an extremum. also in middle Consoles put an extra point E.because on this area under the uniformly distributed load of the Epura M. Describes crooked line, and it is built at least 3 Points.

So, points are placed, proceed to the definition of values \u200b\u200bin them. bending moments. Rule of signs - see.

Plots Na, AD. – parabolic curve (Umbrella rule for mechanical specialties or "Sails Rule" from construction), plots DC, St. – direct oblique lines.

Moment at point D. should be determined both on the left and right From the point D. . The moment in these expressions excluded. At point D. Receive two Values \u200b\u200bS. difference By magnitude m. – jump on its magnitude.

Now you should determine the moment at the point TO (Q.\u003d 0). However, first define position point TO , denoting the distance from her before the start of the site by unknown h. .

T. TO belongs second a characteristic site, his transverse power equation (see above)

But transverse force in t. TO equal 0 , but z. 2 equals unknown h. .

We get the equation:

Now, knowing h., We define the moment at the point TO on the right side.

Build Epleru M. . Building to execute for Mechanical specialties postponing positive values up From the zero line and using the Umbrella rule.

For a given scheme of the console beam, it is necessary to construct the transverse power of q and the bending moment M, to perform the designer calculation, picking up a round cross section.

The material is a tree, the calculated resistance of the material R \u003d 10MPa, m \u003d 14kn · m, q \u003d 8kn / m

It is possible to build plumes in the console beam with a rigid sealing in two ways - normal, pre-determining the support reactions, and without determining the reference reactions, if we consider the sections, going from the free end of the beam and throwing the left part with the seal. Build Epura ordinary way.

1. Determine support reactions.

Uniformly distributed load q. Replace conditional power Q \u003d Q · 0.84 \u003d 6.72 kN

In the rigid sealing three support reactions - vertical, horizontal and moment, in our case, the horizontal reaction is 0.

Find Vertical Reaction support R A. and reference moment M. A. from equation equations.

In the first two sites on the right, transverse force is absent. At the beginning of the site with a uniformly distributed load (right) Q \u003d 0., in climbing - the value of the reaction R A.
3. To construct the expression to determine them on the plots. Built the moments on the fibers, i.e. down.

(Squeezed bottom fibers).

Plot DC: (compressed upper fibers).

Plot SK: (Compressed Left Fibers)

(Squeezed Left Fiber)

Figure - Epura normal (longitudinal) forces - (b), transverse forces - (c) and bending moments - (g).

Checking the balance of the node with:

Task 2 construct an internal efforts for the frame (Fig. A).

It is given: f \u003d 30kn, q \u003d 40 kn / m, m \u003d 50kn, a \u003d 3m, h \u003d 2m.

Determine support reactions Frames:

From these equations, we find:

Because reaction values R K. has a sign minus, in fig. but Changes direction This vector on the opposite, while recorded R k \u003d 83,33kn.

Determine the values \u200b\u200bof domestic effort N, Q. and M. In the characteristic sections of the frame:

Plot Sun.:

(compressed right fibers).

CD plot:

(shut down right fibers);

(Compressed right fibers).

Plot DE:

(Squeezed bottom fibers);

(Squeezed bottom fibers).

Section KS.

(Squeezed left fibers).

Build epures of normal (longitudinal) forces (b), transverse forces (B) and bending moments (g).

Consider the equilibrium of the nodes D. and E.

From consideration of nodes D.and E. It can be seen that they are in equilibrium.

Task 3. For a frame with a hinge to build an internal efforts.

It is given: f \u003d 30kn, q \u003d 40 kN / m, m \u003d 50kn, a \u003d 2m, h \u003d 2m.

Decision. Determine support reactions. It should be noted that in both hinge and fixed supports two reactions. In this regard, you should use hinge property S. — moment in it both from the left and the right forces equal to zero.. Consider the left part.

The equilibrium equations for the frame under consideration can be written as:

From the solution of these equations should:

On the frame diagram, the direction of force N B. Changes on opposite (H B \u003d 15kn).

Determine efforts In the characteristic sections of the frame.

Plot BZ:

(Squeezed left fibers).

Section ZC:

(compressed left fibers);

Plot KD:

(compressed left fibers);

(Squeezed left fibers).

DC plot:

(Squeezed bottom fibers);

Definition extreme values bending moment on the plot CD:

1. Building a line of transverse forces.For console beam (Fig. but ) Characteristic points: BUT - point of application support reaction V A.; FROM - point of application of the concentrated force; D., B. - The beginning and end of the distributed load. For the console, the transverse force is determined similar to the two-air beam. So, when during the left:

To check the correctness of the transverse force definition in sections, go through the beam in the same way, but from the right end. Then the right parts of the beam are cut off. Remember that the rule of signs will change. The result should turn out the same. We build a transverse force (rice b.).

2. Building actions of moments

For the console beam, the trimming of bending moments is constructed similarly to the previous construction. Sparkter points for this beam (see Fig. but) Next: BUT - support; FROM - point of application of the concentrated moment and force F.; D. and IN - The beginning and end of action of uniformly distributed load. Since Epura. Q. x. in the area of \u200b\u200baction of a distributed load zero line does not cross, To build a plot of moments in this section (Parabolic curve), you should choose an arbitrarily additional point to build a curve, for example in the middle of the site.

Stroke on the left:

We find the right to the right M B. = 0.

According to the found values, we build the fusion of bending moments (see Fig. in ).

Record Published Author admin is limited inclined direct, but on the plot on which there is no distributed load - a straight, parallel axis, therefore, to build plumes of transverse forces, it is enough to determine the values Q. W. At the beginning and end of each site. In the cross section, the corresponding point of the application of the concentrated force, the transverse force must be calculated by a little left of this point (on an infinitely close distance from it) and a little right of it; transverse forces in such places are indicated respectively .

Build Epleru Q. W. The method of characteristic points, running to the left. For greater clarity, the discarded part of the beam at first is recommended to close the paper sheet. Characteristic points for a two-air beam (Fig. but ) 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. - point of application of concentrated power. We will spend a mental axis y. Perpendicular to the axis of beams through the point FROM And we will not change its position until we pass the entire beam from C. before E.. Considering the left-cut parts from the characteristic points of the beam, we project on the axis y. In force in this section of power with the corresponding signs. As a result, we get:

To check the correctness of the determination of the transverse force in sections, you can go through the beam in the same way, but from the right end. Then the right parts of the beam are cut off. The result should turn out the same. The coincidence of the results can serve as the control of the construction of the EPUR Q. W.. We carry out the zero line under the image of the beam and from it in the accepted scale, we postpone the found values \u200b\u200bof the transverse forces, taking into account the signs at the appropriate points. We get Epleru Q. W.(Fig. b. ).

By building the Eppure, pay attention to the following: Eppure under a distributed load is depicted by inclined straight, under the unloaded sections - sections parallel to the zero line, a jump is formed, equal to the value of force. If the inclined line under the distributed load crosses the zero line, mark this point, then point of extremumand it is now characteristic of us, according to differential relationships between Q. W.and M. X.At this point, the moment has an extremum and it will be necessary to determine when constructing an increment of bending moments. In our task this point TO . Concentrated Moment on Epur Q. W.it does not show itself, since the sum of the projections of the forces forming a pair is zero.

2. Building the moment of moments.Build the aforemention of bending moments, like transverse forces, by the method of characteristic points, to the left. It is known that in the section of the beam with a uniformly distributed load of the increment of bending moments outlines the line curve (quadratic parabola), to build which it is necessary to have at least three points And, therefore, the values \u200b\u200bof bending moments should be calculated at the beginning of the site, the end of it and in one intermediate section. Such an intermediate point is best to take a cross section in which the Epur Q. W.crossing the zero line, i.e. Where Q. W.= 0. On Epur M. In this section there should be a top of parabola. If Epura. Q. w. does not cross the zero line, then to build a plumb M.follows this site take an extra point, for example, in the middle of the site (beginning and end of the distributed load), remembering that the repeated parabol is always drawn down if the load acts on top down (for construction specialties). There is a "rain" rule, which helps a lot when building a parabolic part of the plot M.. For builders, this rule is as follows: Imagine that the distributed load is rain, substitute an umbrella under it in an inverted form, so that the rain is not glad, and it was going to rain. Then the convexity of the umbrella will be drawn down. Exactly it will look like the outline of the torus of moments under the distributed load. For mechanics there is a so-called umbrella rule. The distributed load is rain, and the outlines of the plot should resemble the outlines of the umbrella. In this example, the Epur is built for builders.

If more accurate construction of the plot is required, the values \u200b\u200bof bending moments in several intermediate sections should be calculated. We agree for each such section the bending moment first determine in an arbitrary section, expressing it through the distance h.from any point. Then giving distance h.a number of values, we obtain the values \u200b\u200bof bending moments in the corresponding sections of the site. For sites on which there is no distributed load, bending moments are determined in two sections corresponding to the beginning and the end of the site, since the Epura M.on such sites is limited to direct. If an external focused point is attached to the beam, then it is necessary to calculate the bending moment just to the upper place of the application of the concentrated moment and a little right.

For the two-hot beam, the characteristic points are as follows: C. and D. - the beginning and end of the distributed load; BUT – truck support; IN – The second support of the beam and the point of application of the concentrated moment; E. – right end of the beam; point TO corresponding to the cross section of the beam in which Q. W.= 0.

Stroke on the left. The right-hand side of the section mentioned mentally cast (take a sheet of paper and cover them the discarded part of the beam). We find the sum of the moments of all the forces acting on the left of the section relative to the point under consideration. So,

Before determining the moment in the section TOyou need to find the distance x \u003d Ak. We will make an expression for the transverse force in this section and equate it to zero (the course of the left):

This distance can also be found from the similarity of triangles. KLN. and Kig On Epur Q. W. (Fig. b.) .

We define the moment at the point TO :

Let's go through the remaining part of the beam by the right.

As you can see, the moment at the point D. During the course on the left and right, it turned out the same - the Epura was closed. According to the found values \u200b\u200bof the Eppura. Positive values \u200b\u200bdeposit down from zero line, and negative - up (see Fig. in ).

The longitudinal transverse bend is the combination of transverse bending with compression or stretching of the bar.

When calculating the longitudinal transverse bend, the calculation of bending moments in the cross sections of the bar is made taking into account the deflection of its axis.

Consider the beam with the supervised ends, loading some transverse load and compressing force 5, acting along the axis of the beam (Fig. 8.13, a). Denote by the deflection of the axis of the beam in the cross section with the abscissa (the positive direction of the axis, we will take down, and, therefore, the beam's defamations are considered positive when they are directed down). Bending moment m, acting in this section,

(23.13)

here is the bending moment on the action of the transverse load; - Additional bending moment of force

The complete deflection can be considered consisting of a deflection that arises from the action of only the transverse load, and an additional deflection equal to caused by force.

Full deflection in more than the amount of deflection arising from the separate action of the transverse load and power S, since in case of action on the beam, only the power s deflection is zero. Thus, in the case of a longitudinal transverse bend, the principle of independence of the action is not applicable.

Under action on the tensile power beam (Fig. 8.13, b) bending moment in cross section with abscissa

(24.13)

The tensile power S leads to a decrease in the beam deflection, i.e., the complete deficits in this case are less than the deflection of the transverse load caused by the action.

In the practice of engineering calculations, under the longitudinal transverse bend, the case of compressive force and transverse load is usually implied.

With a rigid beam, when the additional bending moments are small compared to the moment of deflection, few differ from the deflection. In these cases, it is possible to neglect the effect of the force S on the magnitude of the bending moments and the magnitudes of the beam deflection and its calculation on the central compression (or stretching) with the transverse bend, as described in § 2.9.

With a beam, the rigidity of which is small, the effect of force S on the magnitude of the bending moments and the beam deflection can be very significant and cannot be neglected when calculating. In this case, the beam should count on a longitudinal transverse bend, understanding the calculation on the joint action of bending and compression (or stretching), which is performed, taking into account the effect of the axial load (S) on the deformation of bending bending.

Consider the method of such a calculation on the example of the beam, the superior is superimposed at the ends loaded by the transverse forces directed in one direction, and the squeezing force S (Fig. 9.13).

We substitute to the approximate differential equation of the elastic line (1.13), the expression of the bending moment according to formula (23.13):

[The minus sign before the equation is taken because, in contrast to formula (1.13), the direction is considered positive for the deflection.

Hence,

In order to simplify solutions, suppose that additional deflection changes along the length of the beam in the sinusoid, i.e. what

This assumption makes it possible to obtain quite accurate results under action on the transverse load beam directed in one direction (for example, from top to bottom). Replace in formula (25.13) deflection by expression

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

Hence,

Euler should be distinguished by the power of the critical force calculated by the Euler formula. The value can be calculated using the Euler formula only under the condition that the rod flexibility is greater than the limit; The value is substituted in formula (26.13) regardless of the flexibility of the beam. In the formula for critical force, as a rule, the minimum moment of inertia of the cross section of the rod is included, and the expression of the Euler force includes the moment of inertia relative to the one of the main axes of the system inertia, which is perpendicular to the plane of the transverse load.

From formula (26.13) it follows that the relationship between the total beams of the beams and the deflection 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 the criterion of stiffness of beams with longitudinal transverse bend; If this relationship 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 power s, the deflection is caused only by the action of the transverse load.

When the magnitude of the compressive power S is approaching the value of the Euler force, the entire beams are increasingly increasingly and can largely exceed the deflection caused by the action of only the transverse load. In the limiting case, under the deflection, calculated by formula (26.13), become equal in infinity.

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

Under action on the beam of the tensile power of formula (26.13) takes the form.

Of this, the formulas follows that the complete deflection of only the proximity of the transverse load caused by the action only. With the stretching power S, numerically equal to the value of the Euler force (i.e., when), the deflection of twice is less than the deflection

The greatest and smallest normal stresses in the cross section of the beam with articulated ends with longitudinal bending and squeezing S.

Consider a two-heat beam of the 2-way cross section with a span of the beam loaded in the middle of the vertical force P and is compressed by the axial force S \u003d 600 (Fig. 10.13). Cross-section area beam Moment inertia, moment of resistance and elastic module

Cross bonds connecting this beam with adjacent beams of the structure, exclude the possibility of losing the stability of the beam in the horizontal plane (i.e. in the plane of the smallest rigidity).

The bending moment and the deflection in the middle of the beam, calculated without taking into account the effect of power s, are equal to:

Eulerova power is determined from the expression

Progibib in the middle of the beams, calculated based on the effect of force s based on formula (26.13),

We define the greatest normal (compressive) stresses in the medium cross section of the beams by formula (28.13):

where after conversion

Substituting in the expression (29.13) various p (c) values, we obtain the corresponding voltage values. The graphically dependence between the expression defined by the expression (29.13) is characterized by the curve shown in Fig. 11.13.

We define the allowable load P, if for the material beams and the required storage factor consequently, allowable voltage for the material

From fig. 11.23 It follows that the voltage occurs in the beam when the load and the voltage - when loading

If the load, the reserve coefficient for voltages will be equal to a given value, however, the beam will have a slight reserve coefficient by load, since voltages equal to it will occur in it

Consequently, the reserve coefficient for load in this case will be 1.06 (since e. It is clearly insufficient.

In order for the beam to have a reserve coefficient, equal to 1.5, should be taken as the allowable voltage in the beam, as follows from Fig. 11.13, approximately equal

The higher the calculation on the strength was made on allowed voltages. This provided the necessary safety margin not only by voltages, but also by loads, since almost in all cases discussed in previous chapters, voltages are directly proportional to the load values.

With a longitudinal transverse bending of voltage, as follows from Fig. 11.13, not directly proportional to the load, and vary faster than the load (in the case of compressive power S). In this regard, even a minor random increase in the load beyond the calculated can cause a very large increase in stresses and the destruction of the structure. Therefore, the calculation of compressed-curved rods to the longitudinal transverse bend should be made not by allowable stresses, but according to the load.

We will make analogy with formula (28.13) the condition of strength in the calculation of the longitudinal transverse bend by permissible load.

Compressed-curved rods, in addition to the calculation of the longitudinal transverse bend, it is also necessary to calculate on stability.

UDC 539.52.

Limit load for the pinched beam loaded by the longitudinal force, asymmetrically distributed load and supporting moments

I.A. Monakhs1, Yu.K. Basins2.

department of Building Production Building Faculty Moscow State Engineering University Ul. Pavel Korchagin, 22, Moscow, Russia, 129626

2-Paired building structures and structures Engineering faculty Russian University of Friendship of Peoples Ul. Ordzhonikidze, 3, Moscow, Russia, 115419

The article has developed a method for solving problems of small beams of beams from an ideal rigid-plastic material under the action of asymmetrically distributed loads, taking into account the pre-tension-compression. The developed technique is applied to study the stress-strain state of single-break beams, as well as to calculate the limit load of beams.

Keywords: beam, nonlinearity, analytical.

In modern construction, shipbuilding, engineering, chemical industry and other branches of technology, the most common types of structures are rod, in particular beams. Naturally, to determine the real behavior of rod systems (in particular, beams) and their strength resources, plastic deformations are required.

Calculation of constructive systems when taking into account plastic deformations using the model of an ideal tough layer body is the most simple, on the one hand, and quite acceptable from the point of view of the design practice requirements - on the other. If you keep in mind the area of \u200b\u200bsmall displacements of constructive systems, then this is due to the fact that the carrying capacity ("limiting load") of ideal harslest and elastoplastic systems is the same.

Additional reserves and a more strict assessment of the carrying capacity of structures are detected as a result of accounting for geometric nonlinearity when they deform them. Currently, the geometric nonlinearity in the calculations of the design systems is the primary task not only in terms 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 the calculation of structures in the conditions of smallness

the displacements are sufficiently uncertain, on the other hand, the practical data and properties of deformable systems make it possible to believe that large movements are actually achievable. It is enough to indicate on the designs of construction, chemical, ship and machine-building facilities. In addition, the model of the tinlastic body means neglecting with elastic deformations, i.e. Plastic deformations are much superior to elastic. Since the deformations correspond to the movement, the accounting of large movements of the robustoplastic systems is appropriate.

However, geometrically nonlinear deformation of structures in most cases inevitably leads to the emergence of plastic deformations. Therefore, the simultaneous accounting of plastic deformations and geometric nonlinearity in the calculations of structural systems and, of course, the rods becomes of particular importance.

This article discusses small defeches. Such tasks were solved in the works.

A beam with pinched supports is considered, under the action of a stepped load, edge moments and pre-applied longitudinal force (Fig. 1).

Fig. 1. Beam under the distributed load

The equilibrium of the beams at large deflections in a dimensionless form has the form

d2 t /, h d2 w dn

- + (p ± u) - + p \u003d ^ - \u003d 0, dx ah ah

x 2w p12 m n, g,

where x \u003d\u003d, w \u003d -, p \u003d -, t \u003d -, n \u003d -, n and m - internal normal

I k 5 Хъка бъ !! K 25 !! BC

power and bending moment, p - transverse uniformly distributed load, W - deflection, x - longitudinal coordinate (the origin on the left support), 2k - the height of the cross section, b - the width of the cross section, 21 - the yield of the yield of the yield material. If n is specified, then the force N is the result of the R with

available deflections, 11 \u003d \u003d, the trait above the letters means the dimension of quantities.

Consider the first stage of deformation - "small" deflection. The plastic cross section occurs at x \u003d x2, in it T \u003d 1 - p2.

Expressions for speeds of deflection have a form - deflection at x \u003d x2):

(2), (x\u003e x2),

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

Consider the case of X2.< 11.

For zone 0.< х2 < 11 из (1) получаем:

RH 111 1 P11 K1R / 1 T \u003d + K1 P + P / 1 -K1 P / 1 - ± 4- + - ^ 41

x - (1 -P2) ± A,

(, 1, p / 2 k1 Р12l

Px2 + K1 P + P11 - K1 P11 - + 1 ^

X2 \u003d k1 +11 - k111 - + ^

Considering the occurrence of plastic hinge at x \u003d x2, we get:

tX \u003d X \u003d 1 - P2 \u003d - P

(12 k12 l k + / - k1 - ^ + k "a

k, + /, - k, /, -l +

(/ 2 k / 2 l k1 + / 1 - k1 / 1 - ^ + m

Considering the case x2\u003e / 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 -

^ R-RC + 1 ^ l

x - (1 -p-) ± a +

(. RG- K1 P1-L

Kh px2 + kh r +

0, and then

I2 12 1 h x x2 \u003d 1 - + -.

Equality follows from the condition of plasticity

where do we get an expression for load:

k1 - 12 + M L2

K1 / 12 - K2 ¡1

Table 1

k1 \u003d 0 11 \u003d 0.66

table 2

k1 \u003d 0 11 \u003d 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 \u003d 0.5 11 \u003d 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 \u003d 0.8 11 \u003d 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 \u003d 0.5 11 \u003d 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, \u003d 0.8 /, \u003d 1.65 k, \u003d 0.2 /, \u003d 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, bending moment A from -1 to 1, the value of the longitudinal force P1 from 0 to 1, 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 calculations are reduced to Tables 1-8.

LITERATURE

Basov Yu.K., Monakhs I.A. Analytical solution to the problem of large deflection of a rigidly plastic pinched beam under the action of local distributed load, support moments and longitudinal force // Rudn Bulletin. Series "Engineering Research". - 2012. - № 3. - P. 120-125.

Savchenko L.V., Monakhs I.A. Large devices of physically nonlinear round plates // Bulletin of Injection. Series "Technical Sciences". - Vol. 8 (35). - St. Petersburg., 2009. - P. 132-134.

Galileev S.M., Salikhova E.A. Study of the frequencies of own oscillations of structural elements of fiberglass, carbon fiber and graphene // Bulletin of Injection. Series "Technical Sciences". - Vol. 8. - St. Petersburg., 2011. - C.102.

Yerkhov M.I., Monakhs A.I. Large defenses of pre-tense tensile beams with hinge supports with evenly distributed load and regional moments // Bulletin of the construction sciences of the Russian Academy of Architecture and Building Sciences. - 1999. - Vol. 2. - P. 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 Allowance for of Geometrical Nonlinearity.

Key Words: Beam, Analytic, Nonlinearity.

In practice, there are very often occasion of collaboration of the rod on bending and stretching or compression. This kind of deformation can be caused by or jointly on the beam of longitudinal and transverse forces, or only one longitudinal forces.

The first case is depicted in Fig.1. A uniformly distributed load q and longitudinal compressive power of R.

Fig.1.

Suppose that the beam defamations compared with the size of the cross section can be neglected; Then, with sufficient accuracy for practice, we can assume that after the deformation of the force p, the axial compression of the beam will cause.

By applying the method of adding the action of forces, we can find a normal voltage at any point of each transverse section of the beam as an algebraic amount of stresses caused by the p and load Q.

Compressing voltages from force p are evenly distributed through the cross-section area F and the same for all sections.

normal tensions from bending in the vertical plane in a section of the abscissa X, which is counted, say, from the left end of the beam, are expressed by the formula

Thus, the total voltage at the point with the coordinate Z (counting from the neutral axis) for this section is equal

Figure 2 shows the plots of stress distribution in the section under consideration from the power of P, the load Q and the total increment.

The greatest voltage in this section will be in the upper fibers, where both types of deformation cause compression; In the lower fibers, there may be a compression or stretching depending on the numerical values \u200b\u200bof the voltages and. To compile the strength of the strength, we find the greatest normal stress.

Fig.2.

Since the voltages from the forces of r in all sections are the same and evenly distributed, the fibers are hazardous, most tense from bending. Such are the extreme fibers in the section with the greatest bending moment; for them

Thus, voltages in extreme fibers 1 and 2 of the middle section of the beams are expressed by the formula

and the calculated voltage will be equal

If the p strength were stretching, the sign of the first term would change, the lower fibers of the beam would have been dangerous.

Denoting the letter n compressive or stretching force, we can write a general formula for testing strength

The described course of the calculation is applied under action on the beam of the inclined forces. Such force can be decomposed on a normal to axis, bending beam, and a longitudinal, compressive or tensile.

bending bending force compression

All variety of existing support devices is schematized as a series of main types of supports, of which

most often found: hinged-movingsupport (Possible notation for it is presented in Fig.1, a), hinged-fixed support (Fig.1, b) and hard pinching, or grinding (Fig. 1, B).

In a hinge-rolling support, one support reaction occurs, perpendicular to the reference plane. Such a support deprives the reference cross section of one degree of freedom, that is, it prevents the displacement in the direction of the reference plane, but it makes it moving in a perpendicular direction and turning the reference section.
In a hinge and fixed support, a vertical and horizontal reaction occurs. Here are impossible to move in directions of support rods, but a turning point is allowed.
In the tight sealing, the vertical and horizontal reaction and the support (reactive) moment occur. In this case, the reference section cannot be shifted and rotated. When calculating systems containing rigid sealing, arising support reactions can not be determined, when choosing a cut-off part so that the sealing with unknown reactions does not fall into it. When calculating systems on hinged supports, the support reactions must be defined. 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 the Longitudinal Force NZ EPUR

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

Rule signs for NZ: We agree to consider the longitudinal force in the section positive, if the external load applied to the cut-off part of the rod is tensile and negative - otherwise.

Example 1.Build Eppura longitudinal forces for a rigidly pinched beam (Fig.2).

Calculation procedure:

1. We plan the characteristic sections, numbering them from the free end of the rod to the seal.
2. Determine the longitudinal force of NZ in each characteristic section. At the same time, we always consider that cut-off part, in which the hard seal does not fall into.

According to the found values build Epleru NZ. Positive values \u200b\u200bare deposited (in the selected scale) above the axis of the plot, negative - under the axis.

3. Building an EPUR of the torque MKR.

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

Rule signs for MKR: Let's treat it torque In the section positive, if, when looking at the section, the external moment is visible on the side of the cut-off part, the external moment is visible against the movement of the clockwise and negative - otherwise.

Example 2.Build a torus of torque for a rigidly pinched rod (Fig. 3, a).

Calculation procedure.

It should be noted that the algorithm and principles of constructing the plot of torques completely coincide with the algorithm and principles. building Support Longitudinal Force.

1. Note the characteristic sections.
2. Determine the torque in each characteristic section.

At the found values \u200b\u200bof building eppura MKR (Fig. 3, b).

4. Rules for control of EPUR NZ and MKP.

For eppure longitudinal forces and torque moments are characterized by certain patterns, knowledge of which makes it possible to assess the correctness of the construction performed.

1. Eppures NZ and MKR are always straightforward.

2. On the site where there is no distributed load, the NZ (MKR) EPUR (MKP) is straight, parallel axis, and in the area under the distributed load - inclined straight line.

3. Under the point of the application of the concentrated force on the NZ stage, there must be a jump on the value of this force, similar to the point of application of the concentrated point on the MKR EPUR, will be a jump by the value of this moment.

5. Building actions of the transverse forces QY and bending moments MX in beams

Bending core is called beam. In the sections of the beams loaded by vertical loads, there are usually two internal power factor - QY I. bendingmom MX.

Transverse force In cross section, it is numerically equal to the algebraic amount of projections of the external forces applied on one side of the section under consideration, on a transverse (vertical) axis.

Rule signs for QY: We agree to consider the transverse force in the section positive, if the external load applied to the cut-off part, seeks to rotate this section clockwise and negative - otherwise.

Schematically, this rule of signs can be represented as

Bending moment MX in cross section is numerically equal to the algebraic sum of the moments of the external forces applied on one side of the section under consideration relative to the X axis passing through this section.

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

Schematically this rule of signs can be represented as:

It should be noted that when using the rule rules for MX in the specified form, the MX group always turns out to be built on the side of the compressed beam fibers.

6. Console beams

For building Epur QY and MX In console, or rigidly pinched, beams there is no need (as in previously considered examples) to calculate the support reactions arising in the rigid seal, but it is necessary to choose the cut-off part so that the sealing does not fall into it.

Example 3.Build Epura QY and MX (Fig.4).

Calculation order.

1. We plan characteristic sections.