It was possible to distribute the loads evenly. The concept of distributed load

Stress distribution in the case of a plane problem

This case corresponds to the stress state under wall foundations, retaining walls, embankments and other structures, the length of which significantly exceeds their transverse dimensions:

Where l– length of the foundation; b– width of the foundation. In this case, the stress distribution under any part of the structure, identified by two parallel sections perpendicular to the axis of the structure, characterizes the stress state under the entire structure and does not depend on coordinates perpendicular to the direction of the loaded plane.

Let us consider the action of a linear load in the form of a continuous series of concentrated forces R, each of which is per unit length. In this case, the stress components at any point M with coordinates R and b can be found by analogy with the spatial problem:

(3.27)

If the relationships between the geometric characteristics of the points under consideration z, y, b present in the form of influence coefficients K, then the formulas for stresses can be written as follows:

(3.28)

Influence coefficient values K z,K y,K yz tabulated depending on relative coordinates z/b, y/b(Table II.3 of Appendix II).

An important property of the plane problem is that the stress components t and s y in the plane under consideration z 0y do not depend on the coefficient of transverse expansion n 0, as in the case of a spatial problem.

dP

The problem can also be solved for the case of a linear load distributed in any way over a strip width b. In this case, the elementary load dP considered as a concentrated force (Fig. 3.15).

Fig.3.15. Random distribution

bandwidth loads b

If the load extends from a point A(b=b 2) to the point B(b=b 1), then, summing up the stresses from its individual elements, we obtain expressions for the stresses at any point of the array from the action of a continuous strip-like load.

(3.29)

For a uniformly distributed load, integrate the above expressions at Py = P= const. In this case, the main directions, i.e. the directions in which the largest and smallest normal stresses act will be the directions located along the bisector of the “angles of visibility” and perpendicular to them (Fig. 3.16). The visibility angle a is the angle formed by straight lines connecting the point in question M with strip load edges.

We obtain the values ​​of the principal stresses from expressions (3.27), assuming b=0 in them:

. (3.30)

These formulas are often used when assessing the stress state (especially the limit state) in the foundations of structures.

Using the values ​​of the principal stresses as semi-axes, it is possible to construct stress ellipses that visually characterize the stressed state of the soil under a uniformly distributed load applied along the strip. The distribution (location) of stress ellipses under the action of a local uniformly distributed load under the conditions of a plane problem is shown in Fig. 3.17.

Fig.3.17. Stress ellipses under the action of a uniformly distributed load under conditions of a plane problem

Using formulas (3.28) we can determine s z, s y And t yz at all points of the section perpendicular to the longitudinal axis of the load. If we connect points with the same values ​​of each of these quantities, we get lines of equal voltages. Figure 3.18 shows lines of identical vertical stresses s z, called isobars, horizontal stresses s y, called thrusts, and tangential stresses t zx, called shifts.

These curves were constructed by D.E. Polshin using elasticity theory methods for a load uniformly distributed over a strip of width b, extending endlessly in a direction perpendicular to the drawing. The curves show that the effect of compressive stresses s z intensity 0.1 external load R affects a depth of about 6 b, while horizontal stresses s y and tangents t propagate at the same intensity 0.1 R to a much shallower depth (1.5 - 2.0) b. Curvilinear surfaces of equal stresses will have similar outlines for the case of a spatial problem.

Fig.3.18. Lines of equal stress in a linearly deformable mass:

and for s z(isobars); b – for s y(spread); in – for t(shift)

The influence of the width of the loaded strip affects the depth of stress propagation. For example, for a foundation 1 m wide, transmitting a load of intensity to the base R, voltage 0.1 R will be at a depth of 6 m from the base, and for a foundation 2 m wide, with the same load intensity, at a depth of 12 m (Fig. 3.19). If there are weaker soils in the underlying layers, this can significantly affect the deformation of the structure.

where a and b / are the angles of visibility and inclination of the line to the vertical, respectively (Fig. 3.21).

Fig.3.21. Diagrams of distribution of compressive stresses along vertical sections of a soil mass under the action of a triangular load

Table II.4 of Appendix II shows the dependences of the coefficient TO| z depending on z/b And y/b(Fig. 3.21) to calculate s z using the formula:

s z = TO| z × R.

Each owner of a three-phase input (380 V) is obliged to take care of a uniform load on the phases in order to avoid overloading one of them. If there is an uneven distribution on the three-phase input, if the zero burns out or its poor contact, the voltages on the phase wires begin to differ from each other, both up and down. At the single-phase power level (220 Volts), this can lead to breakdown of electrical appliances due to an increased voltage of 250-280 Volts, or a decreased voltage of 180-150 Volts. In addition, in this case, there is an increased consumption of electricity in electrical appliances that are not sensitive to voltage imbalances. In this article we will tell you how the load is distributed across phases, providing brief instructions with a diagram and video example.

What is important to know

This diagram roughly illustrates a three-phase network:

The voltage between phases of 380 volts is indicated in blue. Green color indicates uniformly distributed line voltage. Red - voltage imbalance.

New, three-phase electricity subscribers in a private house or apartment, when connecting for the first time, should not have much hope for an initially evenly distributed load on the input line. Since several consumers can be powered from one line, and they may have problems with distribution.

If after measurements you see that there is (more than 10%, according to GOST 29322-92), you need to contact the power supply organization to take appropriate measures to restore phase symmetry. You can learn more about this from our article.

According to the agreement between the subscriber and the RES (on the use of electricity), the latter must supply high-quality electricity to homes, with the specified . The frequency must also correspond to 50 Hz.

Distribution Rules

When designing a wiring diagram, it is necessary to select the intended consumer groups as equally as possible and distribute them among phases. For example, each group of sockets in rooms in the house is connected to its own phase wire and grouped in such a way that the load on the network is optimal. Lighting lines are organized in the same way, distributing them among different phase conductors, and so on: washing machine, stove, oven, boiler, boiler.

In engineering calculations one often encounters loads distributed along a given surface according to one law or another. Let's consider some simple examples of distributed forces lying in the same plane.

A flat system of distributed forces is characterized by its intensity q, i.e., the value of the force per unit length of the loaded segment. Intensity is measured in newtons divided by meters

1) Forces uniformly distributed along a straight line segment (Fig. 69, a). For such a system of forces, the intensity q has a constant value. In static calculations, this system of forces can be replaced by the resultant

Modulo,

Force Q is applied in the middle of segment AB.

2) Forces distributed along a straight line segment according to a linear law (Fig. 69, b). An example of such a load is the force of water pressure on a dam, which is greatest at the bottom and drops to zero at the surface of the water. For these forces, the intensity q is the magnitude of the variable, growing from zero to the maximum value. The resultant Q of such forces is determined similarly to the resultant of the gravity forces acting on a homogeneous triangular plate ABC. Since the weight of a homogeneous plate is proportional to its area, then, modulo,

Force Q is applied at a distance from side BC of triangle ABC (see § 35, paragraph 2).

3) Forces distributed along a straight line segment according to an arbitrary law (Fig. 69, c). The resultant Q of such forces, by analogy with the force of gravity, is equal in magnitude to the area of ​​the figure ABDE, measured on the appropriate scale, and passes through the center of gravity of this area (the issue of determining the centers of gravity of areas will be discussed in § 33).

4) Forces evenly distributed along the arc of a circle (Fig. 70). An example of such forces is the forces of hydrostatic pressure on the side walls of a cylindrical vessel.

Let the radius of the arc be equal to , where is the axis of symmetry along which we direct the axis. The system of converging forces acting on the arc has a resultant Q, directed due to symmetry along the axis and numerically

To determine the value of Q, we select an element on the arc, the position of which is determined by the angle and the length of the force acting on this element is numerically equal and the projection of this force on the axis will be Then

But from Fig. 70 it is clear that Therefore, since then

where is the length of the chord subtending the arc AB; q - intensity.

Problem 27. A uniformly distributed load of intensity acts on a cantilever beam A B, the dimensions of which are indicated in the drawing (Fig. 71). Neglecting the weight of the beam and assuming that the pressure forces on the embedded end are determined according to a linear law, determine the values ​​of the highest intensities of these forces, If

Solution. We replace the distributed forces with their resultants Q, R and R, where according to formulas (35) and (36)

and draw up equilibrium conditions (33) for parallel forces acting on the beam

Substituting here instead of Q, R and R their values ​​and solving the resulting equations, we will finally find

For example, when we get and when

Problem 28. A cylindrical cylinder, the height of which is H and the internal diameter is d, is filled with gas under pressure. The thickness of the cylindrical walls of the cylinder is a. Determine the tensile stresses experienced by these walls in the directions: 1) longitudinal and 2) transverse (the stress is equal to the ratio of the tensile force to the cross-sectional area), considering it small.

Solution. 1) Let us cut the cylinder into two parts by a plane perpendicular to its axis and consider the equilibrium of one of them (Fig.

72, a). It is acted upon in the direction of the cylinder axis by the pressure force on the bottom and forces distributed over the cross-sectional area (the action of the discarded half), the resultant of which will be denoted by Q. At equilibrium

Assuming approximately the cross-sectional area to be equal, we obtain for the tensile stress the value

The distance between the concentrated loads is the same, and the distance from the beginning of the span to the first concentrated load is equal to the distance between the concentrated loads. In this case, concentrated loads also fall on the beginning and end of the span, but at the same time they only cause an increase in the support reaction; extreme concentrated loads do not affect the value of bending moments and deflection in any way, and therefore are not taken into account when calculating the load-bearing capacity of the structure. Let's consider this using the example of floor beams resting on a lintel. Brickwork, which can be between the lintel and the floor beams, and create a uniformly distributed load, is not shown for ease of perception.

Picture 1. Reducing concentrated loads to an equivalent uniformly distributed load.

As can be seen from Figure 1, the determining moment is the bending moment, which is used in strength calculations of structures. Thus, in order for a uniformly distributed load to produce the same bending moment as a concentrated load, it must be multiplied by the appropriate transition factor (equivalence factor). And this coefficient is determined from the conditions of equality of moments. I think Figure 1 illustrates this very well. And by analyzing the obtained dependencies, you can derive a general formula for determining the transition coefficient. So, if the number of applied concentrated loads is odd, i.e. one of the concentrated loads necessarily falls on the middle of the span, then to determine the equivalence coefficient you can use the formula:

γ = n/(n - 1) (305.1.1)

where n is the number of spans between concentrated loads.

q eq = γ(n-1)Q/l (305.1.2)

where (n-1) is the number of concentrated loads.

However, sometimes it is more convenient to make calculations based on the number of concentrated loads. If this quantity is expressed by the variable m, then

γ = (m +1)/m (305.1.3)

In this case, the equivalent uniformly distributed load will be equal to:

q eq = γmQ/l (305.1.4)

When the number of concentrated loads is even, i.e. none of the concentrated loads falls on the middle of the span, then the value of the coefficient can be taken as for the next odd value of the number of concentrated loads. In general, subject to the specified loading conditions, the following transition coefficients can be accepted:

γ = 2- if the structure under consideration, for example, the beam receives only one concentrated load in the middle of the lintel.

γ = 1.33- for a beam subject to 2 or 3 concentrated loads;

γ = 1.2- for a beam subject to 4 or 5 concentrated loads;

γ = 1.142- for a beam subject to 6 or 7 concentrated loads;

γ = 1.11- for a beam subject to 8 or 9 concentrated loads.

Option 2

The distance between the concentrated loads is the same, with the distance from the beginning of the span to the first concentrated load being equal to half the distance between the concentrated loads. In this case, concentrated loads do not fall on the beginning and end of the span.

Figure 2. Values ​​of transition coefficients for option 2 of applying concentrated loads.

As can be seen from Figure 2, with this loading option, the value of the transition coefficient will be significantly less. So, for example, with an even number of concentrated loads, the transition coefficient can generally be taken equal to unity. For an odd number of concentrated loads, the formula can be used to determine the equivalence coefficient:

γ = (m +7)/(m +6) (305.2.1)

where m is the number of concentrated loads.

In this case, the equivalent uniformly distributed load will still be equal to:

q eq = γmQ/l (305.1.4)

In general, subject to the specified loading conditions, the following transition coefficients can be accepted:

γ = 2- if the structure under consideration, for example, receives only one concentrated load in the middle of the lintel, and whether the floor beams fall on the beginning or end of the span or are located arbitrarily far from the beginning and end of the span, in this case does not matter. And this is important when determining the concentrated load.

γ = 1- if the structure in question is subject to an even number of loads.

γ = 1.11- for a beam subject to 3 concentrated loads;

γ = 1.091- for a beam subject to 5 concentrated loads;

γ = 1.076- for a beam subject to 7 concentrated loads;

γ = 1.067- for a beam subject to 9 concentrated loads.

Despite some complicated definitions, equivalence coefficients are very simple and convenient. Since in calculations the distributed load acting per square or linear meter is very often known, in order not to convert the distributed load first into a concentrated one, and then again into an equivalent distributed one, it is enough to simply multiply the value of the distributed load by the appropriate coefficient. For example, the ceiling will be subject to a standard distributed load of 400 kg/m2, while the dead weight of the ceiling will be another 300 kg/m2. Then, with a floor beam length of 6 m, a uniformly distributed load q = 6(400 + 300)/2 = 2100 kg/m could act on the lintel. And then, if there is only one floor beam in the middle of the span, then γ = 2, and

q eq = γq = 2q (305.2.2)

If none of the two above conditions are met, then it is impossible to use transition coefficients in their pure form; you need to add a couple of additional coefficients that take into account the distance to the beams that do not fall at the beginning and end of the span of the lintel, as well as the possible asymmetry of the application of concentrated loads. In principle, it is possible to derive such coefficients, but in any case they will be reducing in all cases if we consider the 1st load case and in 50% of cases if we consider the 2nd load case, i.e. the values ​​of such coefficients will be< 1. А потому для упрощения расчетов, а заодно и для большего запаса по прочности рассчитываемой конструкции вполне хватит коэффициентов, приведенных при первых двух вариантах загружения.

In engineering calculations, along with concentrated forces that are applied to a solid body at a certain point, there are forces whose action is distributed over certain areas of the volume of the body, its surface or line.

Since all axioms and theorems of statics are formulated for concentrated forces, it is necessary to consider methods of transition from a distributed load to concentrated forces.

Let's consider some simple cases of distributed load of a body by parallel forces that lie in the same plane along a straight line segment.

A flat system of distributed forces is characterized by its intensity q, that is, the magnitude of the force per unit length of the loaded segment. The unit of intensity is Newton divided by meter (N/m). The intensity can be constant (uniformly distributed load) or vary according to linear and arbitrary laws.

A uniformly distributed load (Fig. 2.5, a), the intensity of which q is a constant value, in static calculations it is replaced by one concentrated force, the modulus of which

where is the length of the loaded segment.

a B C)

Figure 2.5

This resultant force, parallel to the forces of the distributed load, is directed in the direction of the distributed forces and is applied in the middle of the loaded segment AB.

Such a load occurs when a homogeneous beam of length is placed on the body l with specific gravity q.

A distributed load with an intensity that varies according to a linear law (Fig. 2.5, b) appears, for example, under the influence of water pressure on a dam, when the load on the dam is greatest near the bottom of the reservoir and is zero near the surface of the water. In this case, the value q intensity increases from zero value to the highest value qmax. Resultant Q such a load is defined as the weight of a homogeneous triangular plate ABC, which is proportional to its area. Then the magnitude of this resultant:

The line of action of the resultant force passes through the center of the triangle ABC at a distance from its top A.

An example of the action of forces distributed along a straight line segment according to an arbitrary law (Fig. 2.5, c) is the load of a flat floor with a snowdrift. The resultant of such forces, by analogy with the weight force, will be numerically equal to the area of ​​the figure measured on the appropriate scale, and the line of action of this resultant will pass through the center of the area of ​​this figure.