I b \u003d W c y \u003d 2 100 4.8 3 / 3 \u003d 7372.8 cm 4 or b (2y) 3 / 12 \u003d 100 (2 4.8) 3 / 12 \u003d 7372.8 cm 4 - moment of inertia of the conditional reduced section , then

f b \u003d 5 9 400 4 / 384 275000 7372.8 \u003d 1.45 cm.

Let's check the possible deflection from reinforcement tension.

the elastic modulus of the reinforcement E a \u003d 2000000 kgf / cm 2, (2 10 5 MPa),

conditional moment of inertia of the reinforcement I a \u003d 10.05 2 3.2 2 \u003d 205.8 cm 4, then

f a = 5 9 400 4 / 384 2000000 160.8 = 7.9 cm

Obviously, the deflection cannot be different, which means that as a result of deformation and equalization of stresses in the compressed zone, the height of the compressed zone will decrease. The details of determining the height of the compressed zone are not given here (due to lack of space), at y ≈ 3.5 cm the deflection will be approximately 3.2 cm. However, the actual deflection will be different, firstly because we did not take into account the deformation of concrete during and is approximate), and secondly, with a decrease in the height of the compressed zone in concrete, plastic deformations will increase, increasing the total deflection. And besides, with prolonged application of loads, the development of plastic deformations also leads to a decrease in the initial modulus of elasticity. The definition of these quantities is a separate topic.

So for concrete of class B20 with a long-term load, the elastic modulus can decrease by a factor of 3.8 (at a moisture content of 40-75%). Accordingly, the deflection from concrete compression will already be 1.45 3.8 = 5.51 cm. And here even a double increase in the reinforcement cross section in the tension zone will not help much - it is necessary to increase the height of the beam.

But even if we do not take into account the duration of the load, then 3.2 cm is still a fairly large deflection. According to SNiP 2.01.07-85 "Loads and Impacts", the maximum allowable deflection for floor slabs for structural reasons (so that the screed does not crack, etc.) will be l / 150 \u003d 400/150 \u003d 2.67 cm. And since the thickness of the protective the concrete layer still remains unacceptable, then for structural reasons the height of the slab should be increased at least up to 11 cm. However, this does not apply to the determination of the moment of resistance.

Pure bending in one of the principal planes
Sectioned with two axes of symmetry. Let the bending moment Mx from the load act in the section (Fig. 2.2), which increases to the limit value. In this case, the cross section will successively be in the elastic, elastic-plastic and plastic states.
During elastic work, stresses σ and relative deformations ε in the section are distributed linearly (Fig. 2.2, a). This state is limited by the achievement of the yield strength σfl in the outer fibers of the section. Corresponding bending moment

Let's call it the ultimate elastic bending moment.
When the yield strength is reached in the extreme fibers, the bearing capacity of the section has not yet been exhausted. With a further increase in the bending moment, the relative deformations in the section increase, and their diagram remains linear. In this case, the stresses increase in those fibers in which they have not yet reached the yield strength σfl. In the yield zones, the stresses maintain a constant value σfl (Fig. 2.2, b). The bending moment in such an elastoplastic state with a relative strain ε1 on the outermost fiber of the section is equal to

The further stage of the elastoplastic work of the section is shown in fig. 2.2, p. In this state, the elastic part is relatively small and concentrated near the neutral axis. To calculate the bending moment, a rectangular distribution of stresses in the stretched and compressed parts of the section is approximately taken. In this case, the elastic part of the section becomes equal to zero (Wel=0).
The bending moment corresponding to the complete fluidity of the section is called the limiting plastic bending moment and is determined by the formula

The formulas for calculating the plastic moment of resistance Z for some characteristic sections and the values ​​of the coefficients of the section shape in bending f=Z/W are given in Table. 2.1.

The limiting plastic bending moment Mpl characterizes the limiting plastic bearing capacity of sections in bending.

Let us estimate the error that arises as a result of the assumption about the distribution of stresses in the form of two rectangles. To do this, we will analyze the theoretical expression for the elastic-plastic moment in the case when the relative strain in the outermost fiber ε1 is sufficiently large (for example, equal to the relative strain of hardening of real steel). The stress distribution under consideration in the elastoplastic state (Fig. 2.3, a) is represented by two diagrams (Fig. 2.3, b, c). Then the bending moment Mεx can be written as


For a rectangular section, we have

For an I-section in accordance with fig. 2.2,b we find

From the similarity of triangles for strains ε, we obtain the dependencies

Since the yield strength is a random variable, the relative strain εfl for a particular steel can take on different values. As a result of the statistical analysis of the yield stress in the works, it was found that most of the values ​​of σfl are in the following intervals:
- for steel class 37
230N/mm2 ≤ σfl ≤ 330 N/mm2;
- for steel class 52
330N/mm2 ≤ σfl ≤ 430N/mm2.
In this case, the corresponding relative deformations εfl are equal to:
for steel class 37
0.0011 ≤ εfl ≤ 0.0016;
for steel class 52
0.0016 ≤ εfl ≤ 0.0020.
The value of the relative strain ε1 and ε1,s in the outermost fibers of the section and the wall is assumed to be ε1=ε1,s=0.012, which approximately corresponds to the deformation of the beginning of steel hardening during tensile testing.
Taking into account formulas (2.21), we obtain:
- for steel class 37
0.046 ≤ Uel/h ≤ 0.067;
- for steel class 52
0.067 ≤ Uel/h ≤ 0.083.
The ratio Ml,x/Mpl,x in the equation (2.17) of the day of a rectangular section varies within:
- for steel class 37
0.0028 ≤ Мl,x/Mpl,x ≤ 0.0060;
- for steel class 52
0.0060 ≤ Ml,x/Mpl,x ≤ 0.0092.
For an I-section, these values ​​depend not only on the class of steel, but also on the dimensions of the cross section, which can be characterized by a generalized parameter ρ, approximately equal to the ratio of the area of ​​the belt to the area of ​​the wall. For commonly used cross-sectional dimensions, the values ​​of ρ are given in Figs. 2.4.

The results obtained show that for the considered cross sections, the values ​​of the ratios Ml,x/Mpl,x in equation (2.17) are much less than 1.0 and they can be ignored. There are sections for which the numerical values ​​Ml,x/Mpl,x are not so small, for example, an I-section loaded perpendicular to the wall. If the calculation takes into account the wall area concentrated near the neutral axis, then a jump appears in the accepted stress diagram. In this regard, it is more correct to take into account only two belts in the calculation, i.e. rectangular section.
In conclusion, it should be noted that if the limiting plastic bending moment Mpl,x is determined under the assumption of stress distribution over two rectangles in the compressed and stretched sections of the section (see Fig. 2.3, b), then the bearing capacity is slightly exaggerated. On the other hand, in this case it is possible to accept the assumption of small deformations and ignore the effect of material hardening.
The fully plasticized cross section cannot take up any further increase in the bending moment and rotates at a constant ultimate load, i.e. behaves like a hinge. Therefore, such a state of the section is also called a plastic hinge.
A plastic hinge is qualitatively different from a conventional hinge. There are two main differences to be noted:
- a conventional hinge is not capable of absorbing a bending moment, while in a plastic hinge the bending moment is equal to Mpl;
- a conventional hinge allows rotation in two directions, and a plastic hinge only in the direction of the acting moment Mpl. With a decrease in the bending moment, the elastic-plastic material again begins to work as an elastic body.
In the above conclusions, only the action of bending moments was taken into account. Along with this, the condition of equilibrium of longitudinal forces must also be satisfied, which for the plastic state is expressed by the equation

This condition determines the position of the neutral axis, the day of which the cross section must be divided into two equal parts. For sections with two axes of symmetry, the neutral axis in the plastic state coincides with the central axis of the section.
As already noted, the unloading occurs elastically, which in a certain way affects the stress state of the section.
In what follows, we will not investigate the cases of unloading in the elastoplastic state, but will focus on the analysis of the total unloading of the plasticized section.
If, under loading, the limiting plastic bending moment is equal to Mpl,x=σflZx, then the complete unloading of the section will occur under the action of a bending moment of the opposite sign -Mpl,x=σWx (Fig. 25, a, b), whence

From formula (2.24) it follows that the conditional stress during unloading can be determined by the formula

Residual stresses in the outermost fibers of the section are then equal to

The distribution of residual stresses along the height of the section is shown in fig. 2.5, c and d. Thus, the stresses in the outer fibers of the section change sign, and the residual stresses near the neutral axis are equal to the yield strength σfl.
From equation (2.26) it follows that the accepted assumption of elastic unloading is fulfilled at fx=Zx/Wx ≤ 2.0; otherwise it would be σ1≥σfl. Sections of steel structures in most cases correspond to the specified value of the section modulus ratios.

Section with one axis of symmetry. Let the Y-axis be the axis of symmetry of the section and the bending moment acts in the YZ plane (Fig. 2.6, a). In the process of its increase, fluidity appears first of all in the lower, and then in the upper fibers of the section. The process of development of plastic deformations depends on the position of the central axis X.
The equilibrium conditions for the elastic-plastic state with one axis of symmetry are given in the works. Here we consider only the case of complete plasticization of the section (Fig. 2.6, b) and its unloading (Fig. 2.6, c, d).
The condition for the equilibrium of normal forces

leads to the same result as in the previous case, i.e. to a formula similar to (2.23):

The difference is that the neutral axis X- does not coincide with the central axis X. Equation (2.28) is a condition for determining the position of the neutral axis in a section with one axis of symmetry.
The equilibrium condition for the moments in the section has the form

Thus, the plastic modulus of a section can be defined as the sum of the absolute values ​​of the static moments of the halves of the section area with respect to the neutral axis:

The unloading of the section in which the plastic hinge is formed occurs inelastically. Elastic unloading of a section with one axis of symmetry is possible only when the section is in a certain stage of the elastoplastic state.
On fig. 2.6 shows the stress distribution during unloading of a fully plasticized section. If the unloading occurred elastically, the distribution of stresses from the unloading bending moment would have the form shown in Fig. 2.6, with a dashed line. In this case, the total stresses from loading and unloading (Fig. 2.6, b, c) between the central axis X and neutral X would be greater than σfl. This area in the process of unloading is excluded from consideration. Only plastic deformations act in it. As a result of a decrease in the active cross-sectional area, the unloading stresses should increase, as shown by the solid line in Fig. 2.6, p. The neutral axis during unloading, coinciding with the central axis of the section (point 1), moves to a new position (point 3).

The total diagram of residual stresses from the load and conditional stresses as a result of unloading is shown in fig. 2.6d. The stresses σl in the upper fibers do not always change sign, which is determined by the position of the axis passing through the center of gravity of the section. If the axis is located close to the upper outermost fiber, then the stresses σl are less than σfl.
Examples. Let us give examples of calculating the plastic modulus of section Zx or Zy.
The dependence for determining the plastic moment of resistance is given by equation (2.30), which includes the static moments of half the cross-sectional area relative to the neutral axis. Let's transform this formula. Consider a section with one axis of symmetry Y (Fig. 2.7), for which X is the central and X- is the neutral axis. The position of the neutral axis X- is determined from the condition (2.28).
The center of gravity of the upper half of the sectional area is at the point Th, the lower half is at the point Td. The plastic moment of resistance Zx, determined by equation (2.30), according to fig. 2.7 can be expressed by the formula

Since point T is the center of gravity of the entire section, the distance between points Th and T or Td and T is equal to r/2. This implies another definition, which naturally extends to sections with two axes of symmetry. The plastic section modulus is equal to twice the absolute value of the static moment of half of the section area relative to the X axis passing through the center of gravity of the section.

Pure bending in one of the main planes of a non-uniform section beam. General solutions. Let the beam sections consist of the upper and lower chords and web, which have different yield strengths, but the same modulus of elasticity.
With an increase in the bending moment, first, fluidity appears in the extreme fiber of one part of the section, and then it spreads over the entire section. The place where the first plastic deformations occur depends on the ratio of the values ​​of the yield strengths and the geometrical dimensions of the section.
When solving problems, we will not analyze the elastic-plastic state, but consider only the case of a complete plastic hinge.
The cross section of the beam and the values ​​of the yield strength of steel are shown in fig. 2.10, a. The distribution of stresses in the elastic state is shown in fig. 2.10, b, in the plastic hinge in fig. 2.10, p.
The condition for the equilibrium of longitudinal forces in a plastic hinge

It can be written in the form

Equation (2.33) is the condition for determining the position of the neutral axis X-.

The condition for the equilibrium of bending moments has the following form:

The right side of this equation expresses the limiting plastic bending moment, which can be written as follows:

Let's write it in the following form:

A symmetrical section F1=F2 is often used, in which both belts have the same yield strength σfl,p. Then the ultimate bending moment

In practice, it is usually designed so that the wall has a lower yield strength than the chords. In this case, it is necessary to carefully check the wall for local stability, taking into account the effect of transverse forces on the bearing capacity. These issues will be discussed later.
According to ČSN 73 1401 for sections in which steels of the same class with different design resistances are used (for example, steel of class 37 - belts with a thickness of more than 25 mm with R=200 N/mm2 and a wall up to 25 mm thick with R=210 N/mm2 ), it is not required to perform the calculation as for combined sections. In this case, the calculation is carried out as for a homogeneous section with a lower design resistance.
Pure bending in two principal planes. In case of oblique bending, the bending moments Mx and My act in the section. In the broken case, the limiting state of the section is determined not by any one of the limiting plastic bending moments Mpl,x or Mpl,y separately, but by the interaction curve between these limiting bending moments

A.R. Rzhanitsyn. Its solution refers to an arbitrary cross section and is based on the determination of the curve of the centers of gravity of the halves of the cross-sectional areas when changing the direction of the bending plane.
A.I. Strelbitskaya. Let us present its main results for an I-section and estimate the accuracy obtained by idealizing the stress distribution in the plastic state.
Relationships between bending moments in the elastic-plastic state. With oblique bending of an I-section, four cases of stress distribution can occur (Fig. 2.11). In the cases shown in Fig. 2.11, a and 5, plastic deformations occur only in certain parts of the belts, and in the cases shown in fig. 2.11, c and d, in the chords and in the wall.
The purpose of the solution is to determine the elastoplastic moments Mε,x and Mε,y. The distribution of relative strains and stresses shown in fig. 2.11, b, c, is characterized by the values ​​of the relative deformation of the extreme fiber of the belt ε = kεfl and the dimensions a, c, u. Taking into account the specified parameter k, which determines the excess of the relative strain of the outermost fiber in comparison with εfl, five unknowns remain to solve the problem.
The theoretical solution for the relative bending moments Mε,x/Mpl,x and Мε,у/Mpl,y will be given only for the cases shown in Figs. 2.11b and d. At the same time, the results obtained for all cases of the development of plastic deformations and several values ​​of k for a characteristic I-section will be shown on the graph.
For the case when u>a (Fig. 2.11, d), from the similarity of triangles for the diagram of relative deformations, we obtain


After simple transformations, we find

In a similar way we define

From the condition of equilibrium of the bending moments Mx=Mε,x and My=Mε,y we obtain the following two equations:


For the case when u≤a (Fig. 2.11,b), condition (2.40) is satisfied and for bending moments we have

The ratio u/(b/2) plays the role of a parameter here. Taking its values ​​in the interval for the section under consideration with the characteristic p=dpbh0/(ds hs2) and the given value of the relative strain kεfl, we can determine the values ​​of the bending moment ratios. With the help of the points obtained in this way, it is possible to construct a curve of their interaction.
The boundary between the cases when the walls are in the elastic and plastic states is determined by the condition u=a. Substituting u instead of a in equation (2.40), we obtain the boundary value

If the parameter u/(b/2) is less than this value, then the wall is in an elastic state; if it is greater, then it is in a plastic state.
Curves of interaction of bending moments Mε,x and Mε,y for sections with geometric parameter p=1.0 for k from 1.0 (elastic state) to ∞ (plastic hinge) are shown in fig. 2.12.

They correspond to the largest relative deformations of the extreme fiber of the belt ε=kεfl, less than or equal to the relative deformation at the beginning of steel tensile hardening.
Relationships between bending moments in the plastic state. The plastic state corresponds to the stress distribution shown in fig. 2.11d. Let us determine the limiting bending moments Mpl,x and Mpl,y and establish the influence of the accepted stress distribution on the interaction curves in comparison with the distribution of finite strains in the elastic-plastic state.
From the condition of equilibrium of bending moments, we obtain

The first parts of these equations, expressing the limiting bending moments Mpl,x and Mpl,y, taking into account the parameter p, can be written as

The resulting equations are special cases of equations (2.42) and (2.43) for k=∞.
Calculating the parameter u/(b/2) from the first equation (2.48) and substituting it into the second, we obtain an expression for the limit curve of the interaction of bending moments

Graphs of these curves for different values ​​of p are shown in fig. 2.13.
An assessment of the influence of the accepted stress distribution shown in fig. 2.11, d, on the curves of interaction of the bending moments Mpl,x and Mpl,y, we will perform by comparing the curve for p=1.0 shown in fig. 2.13 and valid for k=∞, with the curves shown in fig. 2.12. At k=10,20 and ∞, the interaction curves are very close to each other, and for the last two values ​​of k they practically merge. On this basis, we can conclude that if we take the achievement of relative strain (10-20) as the limiting plastic state of the section, which corresponds to the relative strain at the beginning of hardening of the most commonly used steels, then for the curve of interaction of bending moments with sufficient accuracy, we can take the equation (2.49 ), which is strictly valid for k=∞.

Selection of sections according to ČSN 73 1401, with pure bending. Calculations according to ČSN 73 1401/1966 "Design of steel structures" were performed for the first time on the basis of the limit state method. When bending in one of the main planes, the ultimate bending moment was determined by the formula

In this case, for sections in which the bending moment from the design load is equal to M, the condition must be satisfied

To prevent excessive deflections, the norms limited the value of the plastic section modulus. In this case, in the calculations it was allowed to take its largest value, which should not exceed 1.2 of the elastic section modulus. In the presence of an area of ​​pure bending over a length of more than 1/5 of the span of the beam, the norms required to take the average value of the elastic and plastic moments of resistance, but not more than 1.1W.
In the revised norms ČSN 73 1401/1976, plastic calculations have been significantly improved and supplemented. The new standards, like the old ones, only require verification of the bearing capacity of structures. To exclude excessive deformations, the coefficient of working conditions m = 0.95 was introduced in the norms, which reduces the likelihood of reaching the limit state of structures.
In the new standards, as in the old ones, the plastic bending moment is determined from dependence (2.50). The condition of the bearing capacity of the section during bending in one of the main planes has the form

The plastic modulus Z should not exceed 1.5 of the elastic section modulus W. If the structural element is subjected to pure bending over the length of the beam, which is more than 1/5 of its span, then the plastic modulus of the section should not exceed 0.5 (Z+ W).
It should be noted that the requirement limiting the value of the plastic moment of resistance may not be met if it is proved that plastic deformations do not disrupt the operation of structures. In this case, the norms allow for a more detailed calculation.
For an inhomogeneous I-section, the limiting plastic bending moment about the X axis is determined by the formula

Equation (2.53) is applied under the condition

Eccentric tension (compression) is caused by a force parallel to the beam axis, but not coinciding with it. Eccentric tension (compression) can be reduced to axial tension (compression) and oblique bending if the force is transferred P to the center of gravity of the section. Internal force factors in an arbitrary cross section of the beam are equal to:

where yp, zp- coordinates of the force application point. Based on the principle of independence of the action of stress forces at the points of the cross section during eccentric tension (compression) are determined by the formula: or

Where are the radii of inertia of the section. The expression in brackets in the equation shows how many times the stresses in off-center tension (compression) are greater than the stresses of the central tension.

Determination of stresses and deformations at impact

The purpose of impact analysis of a structure is to determine the greatest deformations and stresses resulting from impact.

In the course on the strength of materials, it is assumed that the stresses arising in the system upon impact do not exceed the elastic limits and proportionality of the material, and therefore Hooke's law can be used to study impact. F x \u003d F control \u003d -kx. This ratio expresses the experimentally established Hooke's law. The coefficient k is called the stiffness of the body. In the SI system, stiffness is measured in newtons per meter (N/m). The stiffness coefficient depends on the shape and dimensions of the body, as well as on the material. attitude σ = F / S = –Fcontrol / S, where S is the cross-sectional area of ​​the deformed body, is called stress. Then Hooke's law can be formulated as follows: the relative strain ε is proportional to the stress

The basis of the approximate theory of impact, considered in the course of strength of materials, is the hypothesis that the diagram of system displacements from the load P upon impact (at any time) is similar to the diagram of displacements arising from the same load, but acting statically.

Oh, typical creep curves built in experiments at the same temperature, but at different stresses; the second - at the same voltages, but different temperatures.

Plastic moment of resistance

- plastic moment of resistance, equal to the sum of the static moments of the upper and lower parts of the section and having different values ​​for different sections. slightly more than the usual moment of resistance; so, for a rectangular section = 1.5 for rolling I-beams and channels

Practical calculations for creep

The essence of the calculation of the structure for creep is that the deformation of the parts will not exceed the permissible level at which the structural function will be violated, i.e. interaction of nodes, for the entire life of the structure. In this case, the condition

resolving which, we obtain the level of operating voltages.

Selection of the section of the rods

When solving problems for the selection of sections in rods, in most cases the following plan is used: 1) Through the longitudinal forces in the rods, we determine the calculated load. 2) Further, through the strength condition, we select sections according to GOST. 3) Then we determine the absolute and relative deformations.

At low forces in compressed rods, the selection of the section is carried out according to the given limiting flexibility λ pr. First, the required radius of gyration is determined: and the corresponding corners are selected according to the radius of inertia. To facilitate the determination of the required dimensions of the section, which allow to outline the required dimensions of the corners, the table “Approximate values ​​of the radii” of the inertia of the sections of the elements from the corners shows the approximate values ​​of the radii of inertia for various sections of the elements from the corners.

Creep of materials

Creep of materials is a slow continuous plastic deformation of a solid body under the influence of a constant load or mechanical stress. All solids, both crystalline and amorphous, are subject to creep to some extent. Creep is observed under tension, compression, torsion and other types of loading. Creep is described by the so-called creep curve, which is the dependence of deformation on time at constant temperature and applied load. The total deformation in each unit of time is the sum of the deformations

ε = ε e + ε p + ε c,

where ε e is the elastic component; ε p - plastic component that occurs when the load increases from 0 to P; ε with - creep deformation that occurs over time at σ = const.