Distributed load. How to correctly distribute the load across the phases AND distribute the uniform load to

The distance between the concentrated loads is the same, while 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, the extreme concentrated loads do not affect the value of bending moments and deflection, and therefore are not taken into account when calculating the bearing capacity of the structure. Let's consider this using the example of floor beams supported by 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... Bringing concentrated loads to an equivalent uniformly distributed load.

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

γ \u003d n / (n - 1) (305.1.1)

where n is the number of spans between concentrated loads.

q eq \u003d 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 in the variable m, then

γ \u003d (m +1) / m (305.1.3)

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

q equiv \u003d γmQ / l (305.1.4)

When the number of concentrated loads is even, i.e. none of the concentrated loads falls into 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 factors can be taken:

γ \u003d 2 - if the structure under consideration, for example, a beam gets only one concentrated load in the middle of the bulkhead.

γ \u003d 1.33 - for a beam on which 2 or 3 concentrated loads act;

γ \u003d 1.2 - for a beam on which 4 or 5 concentrated loads act;

γ \u003d 1.142 - for a beam on which 6 or 7 concentrated loads act;

γ \u003d 1.11 - for a beam on which 8 or 9 concentrated loads act.

Option 2

The distance between the concentrated loads is the same, while the distance from the beginning of the span to the first concentrated load is 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... The values \u200b\u200bof the transition coefficients for the 2nd variant of the application of 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 transfer coefficient can generally be taken equal to one. With an odd number of concentrated loads, the formula can be used to determine the equivalence coefficient:

γ \u003d (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 equiv \u003d γmQ / l (305.1.4)

In general, subject to the specified loading conditions, the following transition factors can be taken:

γ \u003d 2 - if the structure under consideration, for example, a beam 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 it does not matter. And this is important when determining the concentrated load.

γ \u003d 1 - if an even number of loads acts on the structure in question.

γ \u003d 1.11 - for a beam on which 3 concentrated loads act;

γ \u003d 1.091 - for a beam on which 5 concentrated loads act;

γ \u003d 1.076 - for a beam on which 7 concentrated loads act;

γ \u003d 1.067 - for a beam on which 9 concentrated loads act.

Despite some tricky definition, the equivalence coefficients are very simple and convenient. Since in calculations the distributed load acting on a square meter or running meter is very often known, in order not to transfer the distributed load first to a concentrated one, and then again to an equivalent distributed one, it is enough to simply multiply the value of the distributed load by the corresponding coefficient. For example, a normative distributed load of 400 kg / m 2 will act on the floor, while the own weight of the floor will be another 300 kg / m 2. Then, with a length of floor beams of 6 m, a uniformly distributed load q \u003d 6 (400 + 300) / 2 \u003d 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 \u003d 2, and

q eq \u003d γq \u003d 2q (305.2.2)

If none of the above two conditions is met, then it is impossible to use the 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 on the beginning and end of the bulkhead span, as well as the possible asymmetry of the application of concentrated loads. In principle, it is possible to derive such coefficients, however, in any case, they will be decreasing in all cases if we consider 1 loading option and in 50% of cases if we consider 2 loading option, i.e. the values \u200b\u200bof 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.

Along with the concentrated forces discussed above, building structures and structures can be exposed to distributed loads- by volume, along the surface or along a certain line - and determined by it intensity.

An example of a load, distributed by area, is the snow load, wind pressure, liquid or soil pressure. The intensity of such a surface load has the dimension of pressure and is measured in kN / m 2 or kilopascals (kPa \u003d kN / m 2).

When solving problems, very often there is a load, distributed along the length of the beam... Intensity q such load is measured in kN / m.

Consider a beam loaded on the site [ a, b] distributed load, the intensity of which changes according to the law q= q(x). To determine the support reactions of such a beam, it is necessary to replace the distributed load with an equivalent concentrated one. This can be done according to the following rule:

Let's consider special cases of distributed load.

and) general case of distributed load(fig. 24)

Fig. 24

q (x) - intensity of the distributed force [N / m],

Elementary strength.

l - segment length

The force of intensity q (x) distributed over a segment of a straight line is equivalent to a concentrated force

A concentrated force is applied at a point FROM(center of parallel forces) with coordinate

b) constant intensity distributed load(fig. 25)

Fig. 25

in) distributed load intensity linearly(fig. 26)

Fig. 26

Calculation of composite systems.

Under composite systems we will understand constructions consisting of several bodies connected to each other.

Before proceeding to consider the features of the calculation of such systems, we introduce the following definition.

Statically definable such problems and systems of statics are called for which the number of unknown reactions of constraints does not exceed the maximum permissible number of equations.

If the number of unknowns is greater than the number of equations,the corresponding tasks and systems are called statically undefined... In this case, the difference between the number of unknowns and the number of equations is called the degree of static uncertainty systems.

For any plane system of forces acting on a rigid body, there are three independent equilibrium conditions. Consequently, for any flat system of forces from the equilibrium conditions, no more than three unknown bond reactions can be found.

In the case of a spatial system of forces acting on a rigid body, there are six independent equilibrium conditions. Consequently, for any spatial system of forces from the equilibrium conditions, no more than six unknown coupling reactions can be found.

Let us explain this with the following examples.

1. Let the center of a weightless ideal block (example 4) be held by not two, but three rods: AB, Sun and BD and it is necessary to determine the reactions of the rods, neglecting the dimensions of the block.

Taking into account the conditions of the problem, we obtain a system of converging forces, where, to determine three unknowns: S A, S C and S Done can still formulate a system of only two equations: Σ X = 0, Σ Y\u003d 0. Obviously, the assigned task and the corresponding system will be statically indeterminate.

2. The beam, rigidly clamped at the left end and having a hinge-fixed support at the right end, is loaded with an arbitrary flat system of forces (Fig. 27).

To determine the support reactions, only three equilibrium equations can be drawn up, which will include 5 unknown support reactions: X A, Y A, M A, X Band Y B... The assigned task will be statically undefined twice.

This problem cannot be solved within the framework of theoretical mechanics, assuming the body in question is absolutely rigid.

Fig. 27

Let's return to the study of composite systems, a typical representative of which is a three-hinged frame (Fig. 28, and). It consists of two bodies: AC and BCconnected key hinge C... Using this frame as an example, consider two ways to determine the support reactions of compound systems.

1 way. Consider the body ACloaded with a given force R, discarding in accordance with axiom 7 all connections and replacing them, respectively, with external reactions ( X A, Y A) and internal ( X C, Y C) links (Fig. 28, b).

Similarly, you can consider the balance of the body BC under the influence of support reactions IN - (X B, Y B) and reactions in the connecting joint C - (X C ', Y C’), Where, in accordance with axiom 5: X C= X C ', Y C= Y C’.

For each of these bodies, three equilibrium equations can be compiled, thus, the total number of unknowns: X A, Y A , X C=X C ', Y C =Y C’, X B, Y B is equal to the total number of equations, and the problem is statically definable.

Recall that, according to the problem statement, it was required to determine only 4 support reactions, but we had to do additional work, determining the reactions in the connecting hinge. This is the disadvantage of this method for determining support reactions.

Method 2. Consider the balance of the entire frame ABCdiscarding only external connections and replacing them with unknown support reactions X A, Y A, X B, Y B .

The resulting system consists of two bodies and is not an absolutely rigid body, since the distance between points AND and IN may change due to mutual rotation of both parts relative to the hinge FROM... Nevertheless, we can assume that the totality of forces applied to the frame ABC forms a system if we use the axiom of solidification (Fig. 28, in).

Fig. 28

So for the body ABC three equilibrium equations can be drawn up. For example:

Σ M A = 0;

Σ X = 0;

These three equations will include 4 unknown support reactions X A, Y A, X Band Y B ... Note that an attempt to use as the missing equation, for example, the following: Σ M B \u003d 0 will not lead to success, since this equation will be linearly dependent with the previous ones. To obtain a linearly independent fourth equation, it is necessary to consider the equilibrium of another body. You can take one of the frame parts as it, for example - Sun... In this case, it is necessary to formulate an equation that would contain the "old" unknowns X A, Y A, X B, Y B and did not contain new ones. For example, the equation: Σ X (Sun) \u003d 0 or more: - X C ' + X B \u003d 0 is not suitable for these purposes, since it contains a "new" unknown X C’, But the equation Σ M C (Sun) \u003d 0 meets all the necessary conditions. Thus, the required support reactions can be found in the following sequence:

Σ M A = 0; → Y B= R/4;

Σ M B = 0; → Y A= -R/4;

Σ M C (Sun) = 0; → X B= -R/4;

Σ X = 0; → X A= -3R/4.

To check, you can use the equation: Σ M C (AS) \u003d 0 or, in more detail: - Y A∙2 + X A∙2 + R∙1 = R/4∙2 -3R/4∙2 + R∙1 = R/2 - 3R/2 + R = 0.

Note that this equation includes all 4 found support reactions: X A and Y A - explicitly, and X B and Y B - implicitly, since they were used to determine the first two reactions.

Graphical definition of support reactions.

In many cases, the solution of problems can be simplified if instead of the equilibrium equations or in addition to them, the equilibrium conditions, axioms, and statics theorems are directly used. The corresponding approach is called the graphical determination of support reactions.

Before proceeding to the consideration of the graphical method, we note that, as for a system of converging forces, graphically, it is possible to solve only those problems that admit analytical solution. At the same time, the graphical method for determining support reactions is convenient for a small number of loads.

So, the graphical method for determining support reactions is based mainly on the use of:

Axioms about the balance of a system of two forces;

Axioms about action and reaction;

Three Forces Theorems;

Equilibrium conditions for a plane system of forces.

When graphically defining the reactions of composite systems, the following is recommended sequence of consideration:

Choose a body with the minimum number of algebraic unknown bond reactions;

If there are two or more such bodies, then start the solution by considering the body to which fewer forces are applied;

If there are two or more such bodies, then choose a body for which a greater number of forces are known by direction.

Solving problems.

When solving the problems of this section, you should keep in mind all those general instructions that were made earlier.

Starting to the solution, it is necessary, first of all, to establish the balance of which body should be considered in this problem. Then, having selected this body and considering it as free, one should depict all the given forces acting on the body and the reactions of the discarded connections.

Next, equilibrium conditions should be drawn up, applying that of the forms of these conditions, which leads to a simpler system of equations (the simplest will be a system of equations, each of which includes one unknown).

To obtain simpler equations, it follows (if this does not complicate the course of the calculation):

1) drawing up the equations of projections, draw the coordinate axis perpendicular to some unknown force;

2) when compiling the moment equation, it is advisable to choose the point where the lines of action of two unknown support reactions intersect as the moment equation - in this case they will not be included in the equation, and it will contain only one unknown;

3) if two unknown support reactions out of three are parallel, then when drawing up the equation in projections onto the axis, the latter should be directed so that it is perpendicular to the first two reactions - in this case, the equation will contain only the last unknown;

4) when solving the problem, the coordinate system must be chosen so that its axes are oriented in the same way as most of the forces of the system applied to the body.

When calculating the moments, it is sometimes convenient to decompose a given force into two components and, using the Varignon theorem, find the moment of force as the sum of the moments of these components.

The solution to many problems of statics is reduced to determining the reactions of the supports, with the help of which beams, bridge girders, etc. are fixed.

Example 7. To the bracket shown in Fig. 29, and, in the node IN suspended load weighing 36 kN. The joints of the bracket elements are hinged. Determine the forces occurring in the rods AB and Sun, considering them weightless.

Decision. Consider the equilibrium of the node INwhere the rods converge AB and Sun... Knot IN represents a point in the drawing. Since the load is suspended from the node IN, then at the point IN apply a force F equal to the weight of the suspended load. Rods VA and Sunpivotally connected at the node IN, limit the possibility of any linear movement in the vertical plane, i.e. are links with respect to the node IN.

Figure: 29. Design diagram of the bracket for example 7:

and -calculation scheme; b -system of forces in a node B

Mentally discard connections and replace their actions with forces - reactions of connections R A and R C... Since the rods are weightless, the reactions of these rods (forces in the rods) are directed along the axis of the rods. Suppose that both rods are stretched, i.e. their reactions are directed from the hinge into the rods. Then, if, after the calculation, the reaction turns out with a minus sign, then this will mean that in fact the reaction is directed in the direction opposite to that indicated in the drawing, i.e. the rod will be compressed.

In fig. 29, b it is shown that at the point IN active force applied F and bond reactions R Aand R C. It is seen that the depicted system of forces represents a flat system of forces converging at one point. We arbitrarily select the coordinate axes OXand OY and compose the equilibrium equations of the form:

Σ F x \u003d0; -R a - R c cos𝛼 = 0;

Σ F y \u003d0; -F - R c cos(90 - α) = 0.

Considering that cos (90 -α ) \u003d sinα, from the second equation we find

R c \u003d -F / sinα = - 36/0,5 = -72 kN.

Substituting the value R c into the first equation, we get

R a \u003d -R c cosα \u003d - (-72) ∙ 0.866 \u003d 62.35 kN.

Thus, the pivot AB - stretched, and the rod Sun - compressed.

To check the correctness of the found forces in the rods, we will project all forces on any axis that does not coincide with the axes X and Ye.g. axis U:

Σ F u = 0; -R c - R a cosα - F cos (90- α) \u003d 0.

After substituting the values \u200b\u200bof the found forces in the rods (dimension in kilonewtons), we get

- (-72) – 62,35∙0,866 - 36∙0,5 = 0; 0 = 0.

The equilibrium condition is fulfilled, so the forces found in the rods are correct.

Example 8.Negligible construction scaffold beam held horizontally by flexible traction CD and pivotally rests on the wall at the point AND... Find the effort in traction CDif a worker weighing 80 kg stands on the edge of the scaffold ≈0.8 kN (Fig. 30, and).

Figure: thirty. Design scheme of the scaffold for example 8:

and- design scheme; b- system of forces acting on the platform

Decision. Select the object of balance. In this example, the balance object is the scaffold beam. At the point IN an active force acts on the beam Fequal to the weight of a person. The connections in this case are a fixed support hinge AND and cravings CD... Let us mentally discard the connections, replacing their action on the beam with the reactions of the connections (Fig. 30, b). The reaction of a fixed hinged support does not need to be determined according to the problem statement. Thrust response CD directed along the thrust. Suppose that the rod CD stretched, i.e. reaction R D directed away from the hinge FROM inside the rod. Let's expand the reaction R D, according to the parallelogram rule, into horizontal and vertical components:

R Dx hot \u003d R D cosα ;

R Dy vert = R D cos(90-α) \u003d R D sinα .

As a result, an arbitrary flat system of forces was obtained, the necessary equilibrium condition for which is the equality to zero of three independent equilibrium conditions.

In our case, it is convenient to be the first to write the equilibrium condition in the form of the sum of moments relative to the moment point AND, since the moment of the support reaction R A relative to this point is zero:

Σ m A = 0; F∙3a - R dy ∙ a = 0

F∙3a - R D sinα = 0.

The value of trigonometric functions is determined from the triangle ACD:

cosα \u003d AC / CD = 0,89,

sinα \u003d AD / CD = 0,446.

Solving the equilibrium equation, we get R D \u003d 5.38 kH. (Heavy CD - stretched).

To check the correctness of the calculation of the force in gravity CD it is necessary to calculate at least one of the components of the support reaction R A... We use the equilibrium equation in the form

Σ F y = 0; V A + R Dy- F= 0

V A = F- R dy.

From here V A \u003d -1.6 kN.

The minus sign means that the vertical component of the reaction R A on the support is directed downward.

Let's check the correctness of the calculation of the force in gravity. We use one more equilibrium condition in the form of equations of moments with respect to the point IN.

Σ m B \u003d 0; V A∙3a + R Dy ∙2a \u003d0;

1,6∙3and + 5,38∙0,446∙2and = 0; 0 = 0.

Equilibrium conditions are met, thus, the force in the weight is found correctly.

Example 9.A vertical concrete pillar is concreted with its lower end into a horizontal base. The load from the building wall weighing 143 kN is transferred to the top of the post. The post is made of concrete with a density of γ \u003d 25 kN / m 3. The post dimensions are shown in Fig. 31, and... Determine reactions in a rigid termination.

Figure: 31. Calculation diagram of the column for example 9:

and - loading diagram and column dimensions; b - design scheme

Decision.In this example, the balance object is the pillar. The column is loaded with the following types of active loads: at point AND concentrated force F, equal to the weight of the building wall, and the column's own weight in the form of a load uniformly distributed along the length of the bar with intensity q for each meter of post length: q \u003d 𝛾Аwhere AND is the cross-sectional area of \u200b\u200bthe column.

q\u003d 25 ∙ 0.51 ∙ 0.51 \u003d 6.5 kN / m.

The ties in this example are a rigid termination at the base of the post. We mentally discard the seal and replace its action with bond reactions (Fig. 31, b).

In our example, we consider a special case of the action of a system of forces perpendicular to the embedment and passing along one axis through the point of application of support reactions. Then two support reactions: the horizontal component and the reactive moment will be equal to zero. To determine the vertical component of the support reaction, we project all forces onto the element axis. Let's combine this axis with the axis Z, then the equilibrium condition will be written as follows:

Σ F Z = 0; V B - F - ql = 0,

where ql- resultant of the distributed load.

V B = F + ql \u003d143 + 6.5 ∙ 4 \u003d 169 kN.

The plus sign indicates that the reaction V B pointing up.

To check the correctness of the calculation of the support reaction, one more equilibrium condition remains - in the form of the algebraic sum of the moments of all forces relative to any point that does not pass through the axis of the element. We suggest performing this check yourself.

Example 10.For the beam shown in Fig. 32, and, it is required to define support reactions. Given: F \u003d 60 kN, q \u003d 24 kN / m, M \u003d 28 kN ∙ m.

Figure: 32. Design scheme and beam dimensions, for example 10:

Decision. Consider the balance of the beam. The beam is loaded with an active load in the form of a flat system of parallel vertical forces, consisting of a concentrated force F, uniformly distributed load intensity q with the resultant Qapplied in the center of gravity of the cargo area (Fig. 32, b), and the concentrated moment M, which can be represented as a pair of forces.

The connections in this beam are a hinge-fixed support ANDand pivot-movable support IN... Let's select the object of equilibrium, for this we discard the support connections and replace their actions with reactions in these connections (Fig. 32, b). Moving support reaction R B is directed vertically, and the reaction of the articulated fixed support R Awill be parallel to the active system of acting forces and also directed vertically. Let's assume they are pointing up. Resultant distributed load Q \u003d 4.8 ∙ q is applied at the center of symmetry of the cargo area.

When determining the support reactions in beams, it is necessary to strive to compose the equilibrium equations so that each of them includes only one unknown. This can be achieved by constructing two equations of moments relative to the pivot points. The verification of the support reactions is usually carried out by equating the sum of the projections of all forces on an axis perpendicular to the axis of the element.

We will conventionally take the direction of rotation of the moment of support reactions around the moment points as positive, then the opposite direction of rotation of the forces will be considered negative.

A necessary and sufficient condition for equilibrium in this case is the equality to zero of independent equilibrium conditions in the form:

Σ m A = 0; V B ∙6 - q∙4,8∙4,8 + M + F∙2,4 = 0;

Σ m B = 0; V A∙6 - q∙4,8∙1,2 - M - F∙8,4 = 0.

Substituting the numerical values \u200b\u200bof the quantities, we find

V B\u003d 14.4 kN, V A \u003d 15.6 kN.

To check the correctness of the found reactions, we use the equilibrium condition in the form:

Σ F y = 0; V A + V B - F -q∙4,8 =0.

After substituting numerical values \u200b\u200binto this equation, we obtain an identity of the type 0 \u003d 0. Hence, we conclude that the calculation was performed correctly and the reactions on both supports are directed upwards.

Example 11.Determine the support reactions for the beam shown in Fig. 33, and... Given: F \u003d 2.4 kN, M\u003d 12 kN ∙ m, q \u003d 0.6 kN / m, a \u003d 60 °.

Figure: 33. Design scheme and beam dimensions for example 11:

a - design scheme; b - object of balance

Decision. Consider the balance of the beam. We mentally free the beam from the connections on the supports and select the object of balance (Fig. 33, b). The beam is loaded with an active load in the form of an arbitrary flat system of forces. Resultant distributed load Q = q∙ 3 is attached in the center of the cargo area symmetry. Strength F decompose according to the parallelogram rule into components - horizontal and vertical

F z \u003d Fcosα \u003d 2.4 cos 60 ° \u003d 1.2 kN;

F y \u003d Fcos (90-α) \u003d Fsin 60 ° \u003d 2.08 kN.

We apply the reaction to the object of equilibrium instead of the discarded connections. Suppose vertical reaction V A pivotally movable support ANDupward, vertical reaction V B articulated fixed support B is also directed upward, and the horizontal reaction H B - to the right.

Thus, in Fig. 33, b an arbitrary plane system of forces is depicted, the necessary equilibrium condition of which is the equality to zero of three independent equilibrium conditions for the plane system of forces. Recall that, according to Varignon's theorem, the moment of force F relative to any point is equal to the sum of the moments of the components F z and F y relative to the same point. Let us conditionally assume that the direction of rotation of the moment of support reactions around the moment points is positive, then the opposite direction of rotation of the forces will be considered negative.

Then the equilibrium conditions are conveniently formulated as follows:

Σ Fz = 0; - F z + H B \u003d 0; from here H B \u003d 1.2 kN;

Σ m A = 0; V B∙6 + M - F y∙2 + 3q∙ 0.5 \u003d 0; from here V B \u003d - 1.456 kN;

Σ m B = 0; V A ∙6 - 3q∙6,5 - F y ∙4 - M \u003d 0; from here V A \u003d 5.336 kN.

To check the correctness of the calculated reactions, we use one more equilibrium condition that was not used, for example:

Σ F y = 0; V A + V B - 3q - F y = 0.

Vertical support reaction V B turned out with a minus sign, this shows that in this beam it is directed not up, but down.

Example 12.Determine the support reactions for a beam rigidly embedded on one side and shown in Fig. 34, and... Given: q \u003d 20 kN / m.

Fig. 34. Design scheme and beam dimensions for example 12:

a - design scheme; b - object of balance

Decision.Let's select the object of balance. The beam is loaded with an active load in the form of a plane system of parallel forces located vertically. We mentally free the beam from the connections in the embedment and replace them with reactions in the form of a concentrated force V B and a pair of forces with the desired reactive moment M B (see fig. 34, b). Since active forces act only in the vertical direction, the horizontal reaction H B is zero. Let us conditionally take the direction of rotation of the moment of support reactions around the moment points clockwise as positive, then the opposite direction of rotation of the forces will be considered negative.

We compose the equilibrium conditions in the form

Σ F y = 0; V B- q∙1,6 = 0;

Σ m B = 0; M B - q∙1,6∙1,2 = 0.

Here q∙ 1.6 - the resultant of the distributed load.

Substituting the numerical values \u200b\u200bof the distributed load q, we find

V B \u003d 32 kN, M B\u003d 38.4 kN ∙ m.

To check the correctness of the found reactions, we will formulate one more equilibrium condition. Now let's take some other point as the moment point, for example, the right end of the beam, then:

Σ m A = 0; M B – V B∙2 + q∙1,6∙0,8 = 0 .

After substitution of numerical values, we obtain the identity 0 \u003d 0.

We finally conclude that the support reactions were found correctly. Vertical reaction V B is directed upwards, and the reactive moment M B - clockwise.

Example 13. Determine the support reactions of the beam (Fig. 35, and).

Decision. The resultant of the distributed load acts as an active load Q=(1/2)∙aq\u003d (1/2) ∙ 3 ∙ 2 \u003d 3kN, the line of action of which passes at a distance of 1 m from the left support, the tension force of the thread T = R \u003d 2 kN applied at the right end of the beam and concentrated moment.

Since the latter can be replaced by a pair of vertical forces, the load acting on the beam together with the reaction of the movable support IN forms a system of parallel forces, so the reaction R A will also be directed vertically (fig. 35, b).

To determine these reactions, we will use the equilibrium equations.

Σ M A = 0; -Q∙1 + R B∙3 - M + T∙5 = 0,

R B = (1/3) (Q + M- R∙ 5) \u003d (1/3) (3 + 4 - 2 ∙ 5) \u003d -1 kN.

Σ M B = 0; - R A∙3 + Q∙2 - M+ T∙2 = 0,

R A= (1/3) (Q∙2 - M+ R∙ 2) \u003d (1/3) (3 ∙ 2 - 4 + 2 ∙ 2) \u003d 2 kN.

Fig. 35

To check the correctness of the obtained solution, we use the additional equilibrium equation:

Σ Y i = R A - Q + R B+ T = 2 - 3 - 1 + 2 = 0,

that is, the problem was solved correctly.

Example 14. Find the support reactions of a cantilever beam loaded with a distributed load (Fig. 36, and).

Decision. The resultant distributed load is applied at the center of gravity of the load diagram. In order not to look for the position of the center of gravity of the trapezoid, we represent it as the sum of two triangles. Then the given load will be equivalent to two forces: Q 1 \u003d (1/2) ∙ 3 ∙ 2 \u003d 3 kN and Q 2 \u003d (1/2) ∙ 3 ∙ 4 \u003d 6 kN, which are applied at the center of gravity of each of the triangles (Fig. 36, b).

Fig. 36

Rigid restraint support reactions are represented by the force R Aand moment M A, to determine which it is more convenient to use the equilibrium equations of the system of parallel forces, that is:

Σ M A = 0; M A \u003d 15 kN ∙ m;

Σ Y= 0, R A\u003d 9 kN.

To check, we use the additional equation Σ M B \u003d 0, where point IN located at the right end of the beam:

Σ M B = M A - R A∙3 + Q 1 ∙2 + Q 2 ∙1 = 15 - 27 + 6 +6 = 0.

Example 15. Uniform beam weighing Q \u003d 600 N and length l \u003d 4 m rests with one end on a smooth floor, and with an intermediate point INon a pillar high h \u003d 3 m, forming an angle of 30 ° with the vertical. In this position, the beam is held in place by a rope stretched across the floor. Determine the tension of the rope T and the reactions of the pillar - R B and gender - R A (fig. 37, and).

Decision.In theoretical mechanics, a beam or rod is understood to mean a body whose transverse dimensions in comparison with its length can be neglected. So the weight Q homogeneous beam is attached at the point FROMwhere AS \u003d 2 m.

Fig. 37

1) Since two unknown reactions out of three are applied at the point AND, the first thing to write is the equation Σ M A \u003d 0, since only the reaction will enter there R B:

- R B∙AB+ Q∙(l/ 2) ∙ sin30 ° \u003d 0,

where AB = h/ cos30 ° \u003d 2 m.

Substituting into the equation, we get:

R B∙2 = 600∙2∙(1/2) = 600,

R B\u003d 600 / (2) \u003d 100 ≅ 173 N.

Similarly, from the moment equation one could find the reaction R A, choosing as the moment the point where the lines of action intersect R B and T... However, this will require additional constructions, so it is easier to use other equilibrium equations:

2) Σ X = 0; R B∙ cos30 ° - T = 0; → T = R B∙ cos30 ° \u003d 100 ∙ (/ 2) \u003d 150 N;

3) Σ Y= 0, R B∙ sin30 ° - Q + R A= 0; → R A = Q- R B∙ sin30 ° \u003d 600 - 50 ≅ 513 N.

So we found Tand R A through R B , therefore, the correctness of the obtained solution can be checked using the equation: Σ M B \u003d 0, which explicitly or implicitly includes all found reactions:

R A∙AB sin30 ° - T∙AB cos30 ° - Q∙(AB - l/ 2) ∙ sin30 ° \u003d 513 ∙ 2 ∙ (1/2) - 150 ∙ 2 ∙ (/ 2) - 600 ∙ (2 - 2) ∙ (1/2) \u003d 513 ∙ - 150 ∙ 3 - 600 ∙ ( -1) ≅ 513 ∙ 1.73 - 450 - 600 ∙ 0.73 \u003d 887.5 - 888 \u003d -0.5.

Resulting from rounding discrepancy ∆ \u003d -0.5 is called absolute error calculations.

In order to answer the question of how accurate the obtained result is, calculate relative error, which is determined by the formula:

ε \u003d [| ∆ | / min (| Σ + |, | Σ - |)] ∙ 100% \u003d [| -0.5 | / min (| 887.5 |, | -888 |)] ∙ 100% \u003d (0.5 / 887.5) ∙ 100% \u003d 0.06%.

Example 16. Determine the support reactions of the frame (fig. 38). Here and in what follows, unless otherwise specified, all dimensions in the figures will be considered indicated in meters, and forces - in kilonewtons.

Fig. 38

Decision. Consider the equilibrium of the frame, to which the thread tension force is applied as an active one Tequal to the weight of the cargo Q.

1) The reaction of the movable support R B from the equation Σ M A \u003d 0. In order not to calculate the shoulder of the force T, we will use Varignon's theorem, expanding this force into horizontal and vertical components:

R B∙2 + T sin30 ° ∙ 3 - T cos30 ° ∙ 4 \u003d 0; → R B = (1/2)∙ Q(cos30 ° ∙ 4 - sin30 ° ∙ 3) \u003d (5/4) ∙ (4 - 3) kN.

2) To calculate Y A write the equation Σ M C \u003d 0, where point FROM lies at the intersection of the reaction lines R Band X A:

- Y A∙2 + T sin30 ° ∙ 3 - T cos30 ° ∙ 2 \u003d 0; → Y A= (1/2)∙ Q(sin30 ° ∙ 3 -cos30 ° ∙ 2) \u003d (5/4) ∙ (3 -2) kN.

3) Finally, we find the reaction X A:

Σ X = 0; X A - T sin30 ° \u003d 0; → X A = Q sin30 ° \u003d 5/2 kN.

Since all three reactions were found independently of each other, for verification you need to take the equation that includes each of them:

Σ M D = X A∙3 - Y A∙4 - R B∙2 = 15/2 - 5∙(3 -2 ) - (5/2)∙ (4 - 3) = 15/2 - 15 + 10 -10 +15/2 = 0.

Example 17. Determine the support reactions of a bar with a broken outline (Fig. 39, and).

Decision. We replace the distributed load on each section of the bar with concentrated forces Q 1 \u003d 5 kN and Q 2 \u003d 3 kN, and the action of the rejected rigid pinching - reactions X A,Y A and M A (fig. 39, b).

Fig. 39

1) Σ M A = 0; M A -Q 1 ∙2,5 - Q 2 ∙5,5 = 0; → M A \u003d 5 ∙ 2.5 + 3 ∙ 5.5 \u003d 12.5 + 16.5 \u003d 29 kNm.

2) Σ X = 0; X A + Q 1 ∙ sina \u003d 0; → X A \u003d -5 ∙ (3/5) \u003d -3 kN.

3) Σ Y= 0; Y A - Q 1 cosa - Q 2 = 0; → Y A \u003d 5 ∙ (4/5) + 3 \u003d 4 + 3 \u003d 7 kN, since sinα \u003d 3/5, cosα \u003d 4/5.

Check: Σ M B = 0; M A + X A∙3 - Y A∙7 + Q 1 cosα ∙ 4.5 + Q 1 sinα ∙ 1.5 + Q 2 ∙1,5 = 29 -3∙3 - 7∙7 + 5∙(4/5)∙5 + 5∙(3/5)∙1,5 + 3∙1,5 = 29 - 9 - 49 + 20 + 4,5 + 4,5 = 58 - 58 = 0.

Example 18. For the frame shown in fig. 40, and, it is required to define support reactions. Given: F \u003d 50 kN, M \u003d 60 kN ∙ m, q \u003d 20 kN / m.

Decision... Consider the balance of the frame. We mentally free the frame from the ties on the supports (Fig. 40, b) and select the object of equilibrium. The frame is loaded with an active load in the form of an arbitrary flat system of forces. Instead of the discarded connections, we apply reactions to the object of equilibrium: on a hinged-fixed support AND - vertical V A and horizontal H A, and on the articulated-movable support IN - vertical reaction V BThe intended direction of reactions is shown in Fig. 40, b.

Fig. 40. Design diagram of the frame and the equilibrium object for example 18:

and - design scheme; b- object of balance

We compose the following equilibrium conditions:

Σ F x = 0; -H A + F = 0; H A \u003d 50 kN.

Σ m A = 0; V B∙6 + M - q∙6∙3 - F∙6 = 0; V B \u003d 100 kN.

Σ F y = 0; V A + V B - q∙6 = 0; V A \u003d 20 kN.

Here the direction of rotation around the moment points counterclockwise is conventionally taken as positive.

To check the correctness of the calculation of reactions, we use the equilibrium condition, which would include all support reactions, for example:

Σ m C \u003d0; V B∙3 + M – H A∙6 – V A∙3 = 0.

After substitution of numerical values, we obtain the identity 0 \u003d 0.

Thus, the directions and magnitudes of support reactions are determined correctly.

Example 19.Determine the support reactions of the frame (Fig. 41, and).

Fig. 41

Decision.As in the previous example, the frame consists of two parts connected by a key hinge FROM.We replace the distributed load applied to the left side of the frame with the resultant Q 1, and to the right - the resultant Q 2, where Q 1 = Q 2 \u003d 2kN.

1) Find the reaction R B from the equation Σ M C (Sun) = 0; → R B\u003d 1kN;

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. With an uneven distribution on the three-phase input, when zero burns out or its poor contact, the voltages on the phase wires begin to differ from each other, both upward and downward. At the level of a single-phase power supply (220 Volts), this can lead to a breakdown of electrical devices, due to an increased voltage of 250-280 Volts, or a lowered 180-150 Volts. In addition, in this case, there is an overestimated power consumption in electrical devices that are insensitive to voltage imbalance. In this article, we will tell you how load balancing is performed by phases, providing a short instruction with a diagram and a video example.

What is important to know

This diagram conventionally illustrates a three-phase network:

The voltage between phases 380 volts is marked in blue. Uniform distributed line voltage is shown in green. Red - voltage imbalance.

New, three-phase electrical subscribers in a private house or apartment, upon first connection, should not rely heavily on the 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 the phase symmetry. You can learn more about that 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 the houses, with the specified. The frequency must also correspond to 50 Hertz.

Distribution rules

When designing a wiring diagram, it is necessary to select the prospective consumer groups as equally as possible and distribute them in phases. For example, each group of outlets in the rooms in the house is connected to its own phase conductor and grouped in such a way that the load on the network is optimal. The lighting lines are organized in the same way, performing their distribution over different phase conductors, and so on: washing machine, oven, oven, boiler, boiler.

In engineering calculations, it is often necessary to meet with loads distributed along a given surface according to one or another law. Let's consider some of the simplest examples of distributed forces lying in the same plane.

A plane system of distributed forces is characterized by its intensity q, that is, by the value of the force per unit length of the loaded segment. The intensity is measured in newtons divided by meters

1) Forces evenly 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 forces of water pressure on a dam, which have the greatest value at the bottom and fall to zero at the water surface. For these forces, the intensity q is a variable value that grows from zero to the maximum value.The resultant Q of such forces is determined similarly to the resultant of gravity forces acting on a homogeneous triangular plate ABC. Since the weight of a homogeneous plate is proportional to its area, then, modulo,

The force Q is applied at a distance from the side BC of the triangle ABC (see § 35, item 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 absolute value to the area of \u200b\u200bthe figure ABDE, measured on the appropriate scale, and passes through the center of gravity of this area (the question of determining the centers of gravity of areas will be considered in § 33).

4) Forces evenly distributed along the circular arc (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 by virtue of symmetry along the axis, while numerically

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

But from fig. 70 it is seen that Consequently, since then

where is the length of the chord that contracts the arc AB; q is the intensity.

Problem 27. A uniformly distributed load with intensity acts on the cantilever beam AB, the dimensions of which are shown in the drawing (Fig. 71). if a

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

and compose the equilibrium conditions (33) for forces acting on the beam parallel to

Substituting here instead of Q, R and R their values \u200b\u200band solving the resulting equations, we finally find

For example, when we get and when

Problem 28. A cylindrical cylinder, the height of which is equal to H, and the inner diameter 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.

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

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

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

Surface and volumetric forces represent a load distributed over a certain surface or volume. Such a load is given by the intensity, which is the force per unit of some volume, or some area, or some length.

A special place in solving a number of practically interesting problems is occupied by the case of a flat distributed load applied along the normal to a certain beam. If you direct the axis along the beam , then the intensity will be a function of the coordinate and is measured in N / m. Intensity is the force per unit length.

A flat figure bounded by a beam and a load intensity graph is called a distributed load diagram (Fig. 1.28). If by the nature of the problem being solved, deformations can be ignored, i.e. If the body can be considered absolutely solid, then the distributed load can (and should) be replaced by the resultant one.

Let's break the beam into segments of length
, at each of which we will assume the intensity to be constant and equal to
where –Coordinate of the segment
... In this case, the intensity curve is replaced by a broken line, and the load per segment
, replaced by concentrated power
applied at point (Fig. 1.29). The resulting system of parallel forces has a resultant equal to the sum of the forces acting on each of the segments applied at the center of the parallel forces.

It is clear that such a representation describes the real situation the more accurately the smaller the segment
, i.e. the more the number of segments ... We obtain the exact result by passing to the limit with the length of the segment
tending to zero. The limit obtained as a result of the described procedure is an integral. Thus, for the module of the resultant we get:

To determine the coordinate of a point the application of the resultant, we use the Varignon theorem:

if the system of forces has a resultant, then the moment of the resultant relative to any center (any axis) is equal to the sum of the moments of all forces of the system relative to this center (this axis)

Writing this theorem for the system of forces
in projections on the axis and passing to the limit as the length of the segments tends to zero, we obtain:

Obviously, the modulus of the resultant is numerically equal to the area of \u200b\u200bthe distributed load diagram, and the point of its application coincides with the center of gravity of a homogeneous plate in the form of a distributed load diagram.

Let us note two frequent cases.

,
(Fig. 1.30). The modulus of the resultant and the coordinate of its application point are determined by the formulas:

In engineering practice, such a load is quite common. In most cases, weight and wind loads can be considered evenly distributed.

,
(Fig. 1.31). In this case:

In particular, the water pressure on the vertical wall is directly proportional to the depth .

Example 1.5

Determine support reactions and a beam under the action of two concentrated forces and a uniformly distributed load. Given:

Let's find the resultant of the distributed load. The modulus of the resultant is

shoulder strength relative to point equally
Consider the balance of the beam. The power circuit is shown in Fig. 1.33.

Example 1.6

Determine the reaction of the embedding of a cantilever beam under the action of a concentrated force, a pair of forces and a distributed load (Fig. 1.34).

Let's replace the distributed load with three concentrated forces. To do this, we will split the distributed load diagram into two triangles and a rectangle. Find

The power circuit is shown in Fig. 1.35.

Let's calculate the shoulders of the resultants about the axis

The equilibrium conditions in this case are as follows:

QUESTIONS FOR SELF-CONTROL:

1. What is called the intensity of the distributed load?

2. How to calculate the modulus of the resultant distributed load?

3. How to calculate the coordinate of the application point of the resultant distributed

load?

4. What is the modulus and what is the coordinate of the point of application of the uniformly distributed load?

5. What is the modulus and what is the coordinate of the application point of the linearly distributed load?

From the collection of problems by I.V. Meshchersky: 4.28; 4.29; 4.30; 4.33; 4.34.

From the textbook "THEORETICAL MECHANICS - theory and practice": sets CP-2; CP-3.

PRACTICAL LESSONS No. 4-5