It was possible to distribute the loads evenly. Uniformly distributed load. What is important to know

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 = kN / m 2).

When solving problems, a load is very often encountered, 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 varies 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.

a) 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 WITH(center of parallel forces) with coordinate

b) constant intensity distributed load(fig. 25)

Fig. 25

v) the intensity of the distributed load, changing according to the linear law(fig. 26)

Fig. 26

Calculation of composite systems.

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

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

Statically definablesuch 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 plane 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 D one can still formulate a system of only two equations: Σ X = 0, Σ Y= 0. Obviously, the assigned task and the corresponding system will be statically indeterminate.

2. A beam, rigidly clamped at the left end and having a hinged-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 B and 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, a). It consists of two bodies: AC and BC connected 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 AC loaded 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 V - (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 drawn up, 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 condition of the problem, it was required to determine only 4 support reactions, but we had to do additional work defining 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 ABC, discarding 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 A and V may change due to mutual rotation of both parts relative to the hinge WITH... 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, v).

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 B and Y B... Note that an attempt to use as a missing equation, for example, the following: Σ M B= 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. As it, you can take one of the parts of the frame, for example - Sun... In this case, it is necessary to formulate such 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) = 0 or more: - X C ' + X B= 0 is not suitable for these purposes, since it contains a "new" unknown X C’, But the equation Σ M C (Sun) = 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) = 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- in an explicit form, 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 theorems of statics 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 an 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 compound 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.

When proceeding to the solution, it is necessary, first of all, to establish the balance of which body should be considered in the given problem. Then, having selected this body and considering it as free, it is necessary to 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 enter 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 Varignon's 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, a, in the node V 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.

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

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

a - 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 to the inside of 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 V active force applied F and bond reactions R A and R C. It can be seen that the depicted system of forces represents a flat system of forces converging at one point. We arbitrarily select the coordinate axes OX and OY and compose the equilibrium equations of the form:

Σ F x = 0;-R a - R c cos𝛼 = 0;

Σ F y = 0; -F - R c cos(90 - α) = 0.

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

R c = -F / sinα = - 36/0,5 = -72 kN.

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

R a = -R c cosα = - (-72) ∙ 0.866 = 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 project all the forces on any axis that does not coincide with the axes X and Y e.g. axis U:

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

After substituting the values ​​of the found forces in the rods (dimension in kilonewtons), we obtain

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

The equilibrium condition is fulfilled, thus, the found forces 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 A... Find an effort in traction CD, if on the edge of the scaffold stands a worker weighing 80 kg ≈0.8 kN (Fig. 30, a).

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

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

Solution. Select the object of balance. In this example, the balance object is the scaffold beam. At the point V an active force acts on the beam F equal to the weight of a person. The connections in this case are a fixed support hinge A 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 hinge 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 WITH inside the rod. Let's decompose the reaction R D, according to the parallelogram rule, into horizontal and vertical components:

R Dx hot = R D cosα ;

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

As a result, an arbitrary flat system of forces was obtained, the necessary equilibrium condition of 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 about the moment point A, since the moment of the support reaction R A relative to this point is equal to 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 = AC / CD = 0,89,

sinα = AD / CD = 0,446.

Solving the equilibrium equation, we get R D = 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= -1.6 kN.

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

Let us 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 V.

Σ m B = 0; V A∙3a + R Dy ∙ 2a = 0;

1,6∙3a + 5,38∙0,446∙2a = 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 = 25 kN / m 3. The post dimensions are shown in Fig. 31, a... Determine reactions in a rigid termination.

Rice. 31. Calculation diagram of the pillar for example 9:

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

Solution. In this example, the balance object is the pillar. The column is loaded with the following types of active loads: at the point A concentrated force F, equal to the weight of the building wall, and the self-weight of the column in the form of a load uniformly distributed along the length of the bar with intensity q for each meter of post length: q = 𝛾А, where A is the cross-sectional area of ​​the column.

q= 25 ∙ 0.51 ∙ 0.51 = 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 particular case of the action of a system of forces perpendicular to the embedment and passing along one axis through the point of application of the 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 = 143 + 6.5 ∙ 4 = 169 kN.

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

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

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

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

Solution. 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 Q applied 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 A and pivot-movable support V... 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 A will be parallel to the active system of acting forces and also directed vertically. Let's assume they are pointing up. Resultant distributed load Q= 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 accomplished by constructing two equations of moments with respect to the pivot points. The verification of 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 ​​of the quantities, we find

V B= 14.4 kN, V A= 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 the numerical values ​​into this equation, we obtain an identity of the type 0 = 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, a... Given: F= 2.4 kN, M= 12 kN ∙ m, q= 0.6 kN / m, a = 60 °.

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

a - design scheme; b - object of equilibrium

Solution. 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 at the center of the cargo area symmetry. Strength F decompose according to the parallelogram rule into components - horizontal and vertical

F z = F cosα = 2.4 cos 60 °= 1.2 kN;

F y = F cos (90-α) = F sin 60 °= 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 A upward, 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 it is convenient to formulate the equilibrium conditions in the following form:

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

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

Σ m B = 0; V A ∙6 - 3q∙6,5 - F y ∙4 - M= 0; from here V A= 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, a... Given: q= 20 kN / m.

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

a - design scheme; b - object of equilibrium

Solution. Let's select the object of equilibrium. The beam is loaded with an active load in the form of a plane system of parallel forces arranged vertically. We mentally free the beam from the connections in the seal 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. We will conventionally 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 is the resultant of the distributed load.

Substituting the numerical values ​​of the distributed load q, we find

V B= 32 kN, M B= 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 = 0.

Finally, we 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, a).

Solution. The resultant of the distributed load acts as an active load Q=(1/2)∙aq= (1/2) ∙ 3 ∙ 2 = 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= 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 V 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) = (1/3) (3 + 4 - 2 ∙ 5) = -1 kN.

Σ M B = 0; - R A∙3 +Q∙2 - M+ T∙2 = 0,

R A= (1/3) (Q∙2 - M+R∙ 2) = (1/3) (3 ∙ 2 - 4 + 2 ∙ 2) = 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, a).

Solution. 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 = (1/2) ∙ 3 ∙ 2 = 3 kN and Q 2 = (1/2) ∙ 3 ∙ 4 = 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 A and 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= 15 kN ∙ m;

Σ Y= 0, R A= 9 kN.

To check, we use the additional equation Σ M B= 0, where point V 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= 600 N and length l= 4 m rests with one end on a smooth floor, and with an intermediate point V on a pillar high h= 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, a).

Solution. In theoretical mechanics, a beam or a rod is understood as a body whose transverse dimensions in comparison with its length can be neglected. So the weight Q a homogeneous beam is attached at a point WITH, where AS= 2 m.

Fig. 37

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

- R B∙AB+Q∙(l/ 2) ∙ sin30 ° = 0,

where AB = h/ cos30 ° = 2 m.

Substituting into the equation, we get:

R B∙2 = 600∙2∙(1/2) = 600,

R B= 600 / (2) = 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 ° = 100 ∙ (/ 2) = 150 N;

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

So we found T and R A across R B, therefore, the correctness of the obtained solution can be verified using the equation: Σ M B= 0, which explicitly or implicitly includes all found reactions:

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

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

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

ε = [| ∆ | / min (| Σ + |, | Σ - |)] ∙ 100% = [| -0.5 | / min (| 887.5 |, | -888 |)] ∙ 100% = (0.5 / 887.5) ∙ 100% = 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

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

1) The reaction of the movable support R B from the equation Σ M A= 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 = 0; → R B = (1/2)∙ Q(cos30 ° ∙ 4 - sin30 ° ∙ 3) = (5/4) ∙ (4 - 3) kN.

2) To calculate Y A write the equation Σ M C= 0, where point WITH lies at the intersection of the lines of action of reactions R B and X A:

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

3) Finally, we find the reaction X A:

Σ X = 0; X A - T sin30 ° = 0; → X A =Q sin30 ° = 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, a).

Solution. We replace the distributed load on each section of the bar with concentrated forces Q 1 = 5 kN and Q 2 = 3 kN, and the action of the rejected rigid pinching is 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= 5 ∙ 2.5 + 3 ∙ 5.5 = 12.5 + 16.5 = 29 kNm.

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

3) Σ Y= 0; Y A - Q 1 cosa - Q 2 = 0; →Y A= 5 ∙ (4/5) + 3 = 4 + 3 = 7 kN, since sinα = 3/5, cosα = 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, a, it is required to define support reactions. Given: F= 50 kN, M= 60 kN ∙ m, q= 20 kN / m.

Solution... 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 A- vertical V A and horizontal H A, and on the articulated-movable support V- vertical reaction V B The intended direction of the reactions is shown in Fig. 40, b.

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

a- design scheme; b- object of balance

We compose the following equilibrium conditions:

Σ F x = 0; -H A + F = 0; H A= 50 kN.

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

Σ F y = 0; V A + V B - q∙6 = 0; V A= 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 the reactions, we use the equilibrium condition, which would include all support reactions, for example:

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

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

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

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

Fig. 41

Solution. As in the previous example, the frame consists of two parts connected by a key hinge WITH. 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 = 2kN.

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

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

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 ratios of the geometric characteristics of the points under consideration z, y, b represent in the form of coefficients of influence 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 the spatial problem.

dP

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

Figure 3.15. Arbitrary distribution

bandwidth loads b

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

(3.29)

With a uniformly distributed load, integrate the above expressions for P y = P= const. In this case, the main directions, i.e. the directions in which the greatest and least normal stresses act will be the directions located along the bisector of the "viewing angles" and perpendicular to them (Figure 3.16). The angle of sight a is the angle formed by the straight lines connecting the point in question M with the edges of the strip load.

The values ​​of the principal stresses are obtained from expressions (3.27), assuming b = 0 in them:

. (3.30)

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

On the values ​​of the main stresses as semiaxes, it is possible to construct stress ellipses that clearly characterize the stress 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 in a plane problem is shown in Figure 3.17.

Figure 3.17. Ellipses of stresses under the action of a uniformly distributed load in the conditions of a plane problem

By 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 equal vertical stresses s z, called isobars, of horizontal stresses s y, called spacers, and shear stresses t zx called shifts.

These curves were constructed by D.E. Pol'shin by methods of the theory of elasticity for a load uniformly distributed over a strip of width b extending infinitely 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 the tangents t propagate at the same intensity 0.1 R to a much shallower depth (1.5 - 2.0) b... Curved surfaces of equal stresses for the case of a spatial problem will have similar outlines.

Figure 3.18. Lines of equal stresses in a linearly deformed array:

and for s z(isobars); b - for s y(disposition); c - for t(shift)

The influence of the loaded strip width affects the stress propagation depth. For example, for a foundation with a width of 1 m, which transfers to the foundation a load of intensity R, voltage 0.1 R will be at a depth of 6 m from the base, and for a foundation with a width of 2 m, with the same load intensity, at a depth of 12 m (Figure 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 (Figure 3.21).

Figure 3.21. Diagrams of the distribution of compressive stresses along vertical sections of the 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(Figure 3.21) to calculate s z by the formula:

s z = TO| z × R.

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

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

Consider some simple cases of a 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 measure for intensity is Newton divided by meter (N / m). The intensity can be constant (uniformly distributed load) or change according to linear and arbitrary laws.

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 with a length of l with specific gravity q.

A distributed load with an intensity varying according to a linear law (Fig. 2.5, b) appears, for example, under the action of water pressure on the dam, when the load on the dam will be greatest near the bottom of the reservoir and is zero near the water surface. In this case, the value q intensity increases from zero to the highest value q max... Resultant Q such a load is defined as the weight of a homogeneous triangular plate ABC, which is proportional to its area. Then the value 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 overlap with a snowdrift. The resultant of such forces, by analogy with the force of weight, 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.

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 at the beginning and at the 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 resting on a lintel. Brickwork, which can be between the lintel and the floor beams, and thus 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 resulting 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 = 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 in 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 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 coefficients can be taken:

γ = 2- if the structure in question, for example, a beam gets only one concentrated load in the middle of the bulkhead.

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

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

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

γ = 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 at the beginning and end of the span.

Picture 2... The values ​​of 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 unity. With 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 taken:

γ = 2- if only one concentrated load in the middle of the bulkhead falls on the structure under consideration, for example, a beam, and whether the floor beams fall at 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 it matters in determining the concentrated load.

γ = 1- if an even number of loads acts on the structure under consideration.

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

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

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

γ = 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 = 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 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 ​​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.

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 up and down. 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 phase-to-phase voltage of 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 is grouped in such a way that the load on the network is optimal. Lighting lines are organized in the same way, distributing them along different phase conductors, and so on: washing machine, oven, oven, boiler, boiler.