Determining the reactions of beam supports - solving the problem. Design diagrams of beams and determination of the reaction of their supports Determine the magnitude of the reactions in the supports of beam systems

The procedure for solving problems for determining the reactions of beam supports is considered. An example of solving the problem and checking the correctness of the determination of reactions is given. The solution to the problem in the second way is given.

The procedure for solving problems to determine the reactions of beam supports

• Selecting a coordinate system. You can direct the x axis along the beam, the y axis vertically upward. The z axis will be directed perpendicular to the drawing plane, towards us. The center of the coordinate system can be selected at one of the beam support points.
• If there is a distributed load, then we replace it with a resultant force. The magnitude of this force is equal to the area of ​​the diagram. The point of application of the force is at the center of gravity of the diagram. So if the load q is uniformly distributed on the segment AB, then its resultant has the value Q = q | AB| and is applied in the middle of segment AB.
• We compose equilibrium equations for active forces. In general, they look like:
  .
  Let's design it vector equation on the coordinate axis. Then the sum of the projections of forces on each of the coordinate axes is equal to zero:
  (1) .
  We find the projections of forces on the coordinate axes and compose equations (1). For a plane system of forces, the last equation, with projections onto the z axis, is not used.
• We compose equilibrium equations for moments of forces. The sum of the moments of forces about an arbitrary axis A′A′′ is equal to zero:
  (2) .
  To construct this equation, we must select an axis about which the moments are calculated. It is better to choose an axis to make calculations simpler. Most often, the axes are chosen so that they pass through the support points of the beam, perpendicular to the plane of the drawing.
• We solve the equations and obtain the values ​​of the support reactions.
• We check the result. As a test, you can select any axis, perpendicular to the plane drawing, and relative to it calculate the sum of the moments of forces acting on the beam, including the found reactions of the supports. The sum of the moments must be zero.

An example of solving a problem to determine the reactions of beam supports

Problem condition.

A rigid beam, the linear dimensions of which are indicated in Figure 1, is fixed at points A and B. The beam is acted upon by a pair of forces with a moment M, a uniformly distributed load of intensity q and two forces P and G, the place of application of which is shown in the figure.
Determine the reactions of the beam supports at points A and B caused by the indicated loads.

Given:
P= 20.2 N; G= 22.6 N; q = 2 N/m; M= 42.8 Nm; a = 1.3 m; b = 3.9 m; α = 45°;

Problem solution

We draw the x and y axes of the coordinate system. Let's place the origin of the coordinate system at point A. Let's direct the x axis horizontally, along the beam. The y axis is vertical. The z axis is perpendicular to the drawing plane and directed towards us. It is not indicated in the picture.

Forces acting on the beam.

We discard the supports and replace them with reaction forces.
In the hinge A, let us decompose the reaction force into components and along the coordinate axes.
The reaction in the movable support on the rollers is directed vertically. We choose the expected directions of support reactions at our own discretion, at random. If we make a mistake with the direction of the reaction, we will get a negative value, which will indicate that the corresponding reaction force is directed in the opposite side.

Let us replace the uniformly distributed load q with the resultant load . Absolute value the resultant is equal to the area of ​​the diagram:
N.
The point of application of the resultant is at the center of gravity of the diagram. Since the diagram is a rectangle, its center of gravity is at point C - in the middle of the segment AD:
AC = CD = b/2 = 1.95 m.

Equilibrium equations for forces

We determine the projections of forces on the coordinate axes.

Let us decompose the force into components along coordinate axes:
.
Absolute values ​​of components:
.
The vector is parallel to the x-axis and directed in the opposite direction from it. The vector is parallel to the y-axis and also directed in the opposite direction. Therefore, the projections of force on the coordinate axes have the following values:
.

The remaining forces are parallel to the coordinate axes. Therefore they have the following projections:
;
;
;
;
.

We compose equilibrium equations for forces.
The sum of the projections of all forces on the x axis is equal to zero:
;
;
;
(P1) .

The sum of the projections of all forces on the y-axis is equal to zero:
;
;
;
(P2) .

Equilibrium equations for moments

So, we have already compiled two equations for forces: (A1) and (A2). But they contain three unknown quantities: , and . To determine them, we need to create another equation.

Let's create an equilibrium equation for moments of forces. To do this, we need to select the axis about which we will calculate the moments. As such an axis, we take the axis passing through point A, perpendicular to the plane of the drawing. For the positive direction, we will choose the one that is directed towards us. Then, according to the right-hand screw rule, the positive direction of twisting will be counterclockwise.

We find the moments of forces relative to the selected axis.
Forces , and intersect the axis. Therefore their moments are equal to zero:
; ; .

The force is perpendicular to arm AB. Her moment:
.
Since, relative to the A axis, the force is directed counterclockwise, its moment is positive.

The force is perpendicular to the AK arm. Since, relative to the A axis, this force is directed clockwise, its moment has a negative value:
.

In a similar way we find the moments of the remaining forces:
;
.
The moment from a pair of forces M does not depend on the points of application of the forces included in the pair:
.

Let's create an equilibrium equation. The sum of the moments of forces about the A axis is equal to zero:
;

;
;
(P3) .

Solving Equilibrium Equations

So, for three unknown quantities, we got three equations:
(P1) .
(P2) .
(P3) .

Let's solve these equations. We calculate distances.
m;
m;
m;
m.

From equation (A1) we find:
N.
From equation (A3) we find:

N.
From equation (A2) we have:
N.
The absolute value of the ground reaction at point A:
N.

Checking the correctness of the solution

To check whether we have correctly determined the reactions of the beam supports, we will find the sum of the moments of forces relative to the other axis. If we found the reaction correctly, then it should be equal to zero.

Let's take an axis passing through point E. We calculate the sum of moments of forces about this axis:

.
Let us find the error in calculating the sum of moments. We rounded the found forces to two decimal places. That is, the error in determining the support reactions is 0.01 N. The distances, in order of magnitude, are approximately equal to 10 m. Then the error in calculating the sum of moments is about 10·0.01 = 0.1 Nm. We got the meaning -0.03 Nm. This value differs from zero by no more than the amount of error. That is, taking into account the calculation error, the sum of the moments relative to the other axis is equal to zero. This means the solution is correct, the reaction forces are found correctly.

Second solution

In the first way, we compiled two equations for forces and one for moments. The problem can be solved in another way by making two equations for moments and one for forces.

Let's take advantage of the fact that the sum of the moments of forces is equal to zero relative to any axis. Let's take the second axis, which passes through point B perpendicular to the plane of the drawing. The sum of the moments of forces relative to this one is zero:
.
We calculate the moments of forces about the B axis.
; ; ;
;
;
;
;
.

The sum of the moments of forces about the B axis is equal to zero:
;

;
;
(P4) ;

So, in the second way, we also have three equations:
(P1) .
(P3) ;
(P4) .

Here, each equation contains only one unknown quantity. Reactions and are determined from the same equations as before. We find the force from equation (A4):

N.

The reaction value coincided with the value obtained by the first method from equation (A2).

The solution of many statics problems comes down to determining the reactions of the supports with which beams and bridge trusses are secured.

In technology, there are usually three types of support fastenings (except for those discussed in § 2):

1. Movable hinged support (Fig. 28, support A). The reaction of such a support is directed normal to the surface on which the rollers of the movable support rest.

2. Fixed hinged support (Fig. 28, support B). Reaction
such a support passes through the hinge axis and can have any direction in the plane of the drawing. When solving problems we will react
depict it as components
And
along the directions of the coordinate axes. Module
we will determine by the formula
.

3. Hard seal (Fig. 29, a). Considering the embedded end of the beam and the wall as one whole, the rigid termination is depicted as shown in Fig. 29, b. In this case, a system of distributed forces (reactions) acts on the beam in its cross section from the embedded end. Considering these forces to be reduced to the center A of the section, they can be replaced by one force
and a pair with an unknown moment m A (Fig. 29, a). Strength
can be depicted by its components
,
(Fig. 29, b).

Thus, to find the reaction of rigid embedding, it is necessary to determine three unknown quantities X A, Y A, m A.

Rice. 28 Fig. 29

Let us also note that in engineering calculations we often encounter loads distributed along the surface according to one law or another. Let's look at some examples of distributed forces.

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

a) Forces uniformly distributed along a straight line segment (Fig. 30, a). For such a system, the intensity q has a constant value. In calculations, this system of forces can be replaced by the resultant . Modulo

Q = a q. (33)

Force Q is applied in the middle of segment AB.

b) Forces distributed along a straight line segment according to a linear law (Fig. 30, b). For these forces, the intensity q is the magnitude of the variable, growing from zero to the maximum value q m. Resultant module in this case is determined by the formula

Q = 0.5 a qm. (34)

Force applied at a distance A/3 from side BC of triangle ABC.

Task 3. Determine the reactions of the fixed hinged support A and the movable support B of the beam (Fig. 31), which is acted upon by active forces: one known concentrated force F = 5 kN, applied at point C at an angle of 60 0, and one pair of forces with a moment m = 8 kNm.

, a couple of forces with a moment m and reaction reactions
,
,
(we represent the reaction of the fixed hinged support A by its two components). As a result, we have an arbitrary flat system of forces. 3) Let us draw the coordinate axes x, y and draw up the equilibrium conditions (28). To calculate the moment of force , sometimes it’s convenient to break it down into its components And , the modules of which are equal to F 1 = F cos60 0 = 2.5 kN, F 2 = F cos30 0 = 4.33 kN. Then we get:

, ,

Solving this system of equations, we find:

X A = F 1 = 2.5 kN, Y B = (m + F 2 ∙5)/3 = 9.88 kN, Y A = F 2 – Y B = – 5.55 kN.

The minus sign of reaction Y A shows that this reaction is directed vertically downward.

To check, let’s create an equation of moments about the new center, for example, about point B:

5,55∙3 – 8 – 4,33∙2 = – 0,01 ≈ 0.

Task 4. Determine the reactions of embedding a cantilever beam (Fig. 32), which is subject to active forces: concentrated force F = 6 kN applied at point C at an angle of 45 0, uniformly distributed load with intensity q = 2 kN/m and a pair of forces with moment m = 3 kNm.

Solution. 1) Select the object of study, i.e. We consider the equilibrium of beam ABC. 2) Let us depict external forces acting on the beam: force , a uniformly distributed load of intensity q, a pair of forces with a moment m and embedding reactions, i.e. three unknown quantities X A , Y A , m A (we represent the rigid embedding reaction by its two components X A , Y A , and the pair by the unknown moment m A , as in Fig. 29). Strength Let's break it down into two components And , the modules of which are equal to F 1 = F 2 = F cos45 0 = 4.24 kN, and we replace the distributed load with intensity q with a concentrated force with modulus equal

Q = 3∙q = 6 kN.

Strength applied in the middle of segment AB. As a result, we have an arbitrary flat system of forces. 3) Let us draw the coordinate axes x, y and compose the equilibrium equations (2):

, ,

Solving these equations, we find:

X A = F 1 = 4.24 kN, Y A = Q – F 2 = 1.76 kN, m A = Q∙1.5 + m – F 2 ∙5 = – 9.2 kNm.

To check, let's create an equation of moments about point C:

, – 9,2 + 21 – 3 – 8,8 = 0.

Task 5. Determine the reactions of supports A, B, C and the force in the intermediate hinge D of the composite structure (Fig. 33), which is acted upon by active forces: concentrated force F = 4 kN, applied at point E at an angle of 45 0, uniformly distributed load intensity q = 2 kN/m and a pair of forces with a moment m = 10 kNm.

Solution. One of the ways to solve problems of determining the reaction of supports of a composite structure is to divide the structure into separate bodies and create the equilibrium conditions for each of the bodies separately. Let's use this method and split the structure into two parts: left AD and right DC. As a result, we arrive at the problem of the equilibrium of two bodies. The power circuits of the problem are shown in Fig. 7.8. To simplify the calculations, let us expand the force into components And , the modules of which are equal to F 1 = F 2 = F cos45 0 = 2.83 kN, and we replace the distributed load with intensity q with a concentrated force with a module equal to Q = 10 kN. Strength attached in the middle of the segment BD.

Rice. 34 Fig. 35

Analysis of the above power circuits shows that they include six unknown quantities: X A, Y A, Y B, X D, Y D, Y C.

Since in Fig. 34.35 available flat systems balanced forces, then for them we can write the equilibrium conditions (28) in the form of six linear algebraic equations:

Left side Right side

,
,

,
,

Since the compiled system of six equations depends on six unknowns X A, Y A, Y B, X D, Y D, Y C, it is closed.

Solving the system, we find:

X A = – 2.83 kN, Y A = – 0.93 kN, Y B = 11.76 kN, Y C = 2 kN, X D = 0, Y D = 2 kN.

To check, let's create an equation of moments about point D:

2,83∙7 – (– 0,93)∙15 – 11,76∙5 + 10∙2,5 – 10 + 2∙5 = – 0,04 ≈ 0.

Reactions occur on the supports of beams, with the determination of which one should begin solving all problems in calculating bending.

The reactions of the supports are determined from the equations of equilibrium (statics), which can be presented in two different versions:

1) in the form of the sum of projections of all forces on the axis X And at, as well as the sum of moments of forces (including reactions) relative to any point along the axis of the beam:

2) in the form of the sum of all forces on one of the coordinate axes X or at and two sums of moments of forces (including reactions) relative to two points lying on the axis of the beam:

The choice of one or another option for composing equilibrium equations, as well as the choice of points along the direction of the coordinate axes used in composing these equations, is made in each specific case in such a way that, if possible, the joint solution of the equations is not performed. To check the correctness of the definition ground reactions It is recommended to substitute their obtained values ​​into any equilibrium equation that has not been used previously.

When determining reactions, their directions can be chosen arbitrarily. If the reactions in the calculation turned out to be negative, this means that their direction was chosen incorrectly. In this case, on the calculation diagram, the initial direction of the reactions is crossed out and their opposite direction is indicated. In subsequent calculations, the magnitudes of the reactions are considered positive.

However, it is possible to predict in advance the correct direction of reactions based on the mentally represented elastic line of the beam after it is loaded with external forces (Fig. 8.5): when the beam is “torn off” from the support (support A) reaction R A has a direction towards the support; when “pressing” the beam into the support (support IN) reaction R B has a direction away from the support.

Figure 8.5 - To determine the direction of reactions

Let us consider typical cases of determining reactions for the simplest types of loads.

If the beam is acted upon by intensity q, as shown in Fig. 8.6, then when determining the support reactions, the load is replaced by its resultant R, equal to the product of load intensity q by the length of its action area l

An example of a continuous, uniformly distributed load is the self-weight of a beam or frequently spaced loads along a section of its length.

Figure 8.6 – Case of a uniformly distributed load on a beam

Point of application of a continuous uniformly distributed load q lies in the middle of the area on which it acts; with a triangular law of action of a distributed load, the resultant is applied along its center of gravity.

Dimension of load intensity q usually expressed in kN/m or kN/cm.

Let's consider the sequence of determining the support reactions for the case of a beam load shown in Fig. 8.7:

1. The design diagram of the beam shows the accepted direction of reactions R A And R B arising on supports. Since the external load acts in a vertical plane perpendicular to the axis of the beam, the horizontal reaction on the hinged fixed support A absent.

2. Since in this case there are two unknown reactions ( R A And R B), then two equations are taken as equilibrium for determining reactions

When drawing up these equilibrium conditions, one should adopt the sign rule for moments of forces, including reactions. Usually the following rule is accepted for external (active) signs: if the moments from the forces are directed clockwise, then they are considered positive.

Then the first equilibrium condition (8.4) leads to the equation for the unknown reaction R B(see Fig.8.6)

The reaction was positive, therefore its direction was accepted as correct.

We similarly use the second equilibrium condition (8.4), leading to the equation for the second reaction R A:

Again the reaction turned out to be positive, therefore its initial direction in the calculation diagram was chosen correctly.

3. We check the correctness of determining the magnitudes of reactions by using another, previously unused, equilibrium condition

In this case, the projections of forces coinciding with the direction of the axis at, are considered positive, and directed towards reverse side– negative.

Then, based on the use of condition (8.5), we have:

The resulting identity (0=0) indicates the correctness of determining the reaction values ​​in the calculation of beam bending.

Let's consider another typical case of load in the form of an eccentrically located concentrated force R along the length of the beam l(Fig. 8.7).

Figure 8.7 – Case of loading a beam with a concentrated force

1. We show the reaction in the calculation diagram R A And R B. They are directed, as indicated above, towards the load.

2. We determine the reactions from equilibrium conditions:

The reactions turned out to be positive, therefore, their initial direction in the calculation diagram was chosen correctly.

Let us also note that the reaction at the support IN turned out to be greater than the reaction at the support A: R B ˃R A. This follows from the fact that the force R is closer to the support IN, which means it loads it more.

3. Check:

The resulting identity indicates the correctness of the reaction definition.

Let's consider another case of loading a beam in a span with an external concentrated moment (Fig. 8.8), which occurs in practical calculations of bending.

Figure 8.8 – The case of loading a beam with a concentrated moment

1. Let us show on the calculation diagram the expected direction of reactions (at first we do not know whether such directions are taken correctly).

2. We determine the reactions from the equilibrium equations:

The reaction turned out to be positive, therefore, its initial position was chosen correctly.

The reaction turned out to be negative, which means that its direction was chosen incorrectly. Therefore, on the calculation diagram we cross out the initially (erroneously) accepted direction R A and show the opposite (true) direction (see Fig. 8.8). In further calculations we consider the reaction R A with the correct direction positive.

3. Check:

The used equilibrium equation for the beam is satisfied, which means that the reactions and their direction are correctly determined.

If a beam during transverse bending has such supports that total number reactions occurring on the supports do not exceed two, then the reactions can always be determined from two equilibrium equations of type (8.2). Such beams, the reactions of which are determined from these static equations, are called statically definable beams. These beams can be of the following simple types (Fig. 8.9):

Figure 8.9 – Statically determinate beams

1) a beam with one rigidly clamped end and the other free end, otherwise console(Fig. 8.9, A); 2) simply supported beams (Fig. 8.9, b and 8.9, V).

Beams with a total number of support reactions more number equilibrium equations are called statically indeterminate(calculation of their bending will be discussed in paragraph 8.10). For such beams, the support reactions are determined from the joint solution of static equations and deformation compatibility conditions.

Beams we will call straight rods that bend. In strength of materials, the term “beam” is much broader than in the usual use of this word: from the point of view of calculating strength, rigidity and stability, a beam is not only a construction beam, but also a shaft, a bolt, a railway car axle, a gear tooth, etc. d.

First, we will limit ourselves to constructing diagrams for the simplest case of beam bending, in which all given loads lie in one plane, called power(in Fig. 4, A- plane P), and this plane coincides with one of the main planes of the beam. We will call such a case flat bend.

In the design diagram, it is customary to replace the beam with its axis (Fig. 4, b). In this case, all loads, naturally, must

Figure 4 will be brought to the axis of the beam and the force plane will coincide with the plane of the drawing.

As a rule, beams have supporting devices - supports. For calculations, they are schematized in the form of three main types of supports:

A) articulated support(Fig. 5, a), in which only one component of the reaction can occur - , directed along the support rod;

b) hinged-fixed support(Fig. 5, b), in which two components can arise - vertical reaction
And horizontal reaction

V) pinching(otherwise hard pinching or embedding), where there can be three components - vertical
and horizontal
reactions and support moment Ma(Fig. 5, V).

All reactions and moments are considered applied at the point A- center of gravity of the supporting section.

The beam shown in Fig. 6, s, called simple , or single-span , or double-support , and the distance l between supports - overflight .

Console called a beam that is pinched at one end and has no other supports (Fig. 4, b), or a part of the beam that hangs over the supports (part Sun in Fig. 6, b; parts AC And BD in Fig. 6, f). Banks with hanging parts are called cantilever (Fig. 6, b, V).

For a plane system of forces, three static equations can be constructed to determine the unknown reactions.

Therefore, the beam will be statically determinate if the number of unknown support reactions does not exceed three; otherwise the beam is statically indeterminate. It is obvious that the beams shown in Fig. 4 and 6 are statically definable.

The beam shown in Fig. 7, A, called continuous and is statically indeterminate, since it has five unknown support reactions: three in support A and one in the supports B and S.

By placing hinges in the sections of the beam, for example at points D And E(Fig. 7, b), we obtain a statically definable hinged beam, because each such intermediate hinge adds one additional equation to the three basic equations of statics: the sum of moments relative to the center of the hinge from all forces located on one side of it is equal to zero .

Constructing diagrams for statically indeterminate beams requires the ability to calculate deformations, and therefore we will limit ourselves for now exclusively to statically determinate beams.

Methods for determining support reactions are studied in the course of theoretical mechanics. Therefore, here we will focus only on some practical issues. To do this, consider a simple beam (Fig. 6, a).

1. Supports are usually designated by letters A And IN. Three unknown reactions are found from the following equilibrium equations:

a) the sum of the projections of all forces onto the axis of the beam is equal to zero:
where do they find it from?

b) the sum of the moments of all forces relative to the supporting hinge A equal to zero:
where do they find it from?
.

c) the sum of the moments of all forces relative to the supporting hinge IN equal to zero:

where do they find it from?
.

2. For control, you can use either the condition that the sum of projections onto the vertical is equal to zero:

or the condition for the sum of moments to be equal to zero relative to some point C other than A And IN, i.e.

U

Condition
It is simpler to use, but it provides a reliable check only in cases where concentrated moments are not applied to the beam.

3. Before drawing up equilibrium equations, you need to select (generally speaking, arbitrarily) the directions of reactions and depict them in the figure. If, as a result of calculations, any reaction turns out to be negative, you need to change its direction in the figure to the opposite and in the future consider this reaction positive,

5. If a distributed load acts on a beam, then to determine the reactions it is replaced by a resultant load, which is equal to the area of ​​the load diagram and is applied at the center of gravity of this diagram.

Example 5. Calculate the support reactions for the beam shown in Fig. 8.

First of all, we find the resultants R 1 And R 2 loads distributed in areas AC n SV:

;
.

Strength R 1 is applied at the center of gravity of the rectangle, and R 2 - at the center of gravity of the triangle. We find reactions:

Solution

2 . A reaction may occur in the embedding, represented by two components: (R Ay,R Ax), and reactive torque M A . We plot the possible directions of reactions on the diagram of the beam.

Comment. If the directions are chosen incorrectly, during calculations we will obtain negative reaction values. In this case, the reactions in the diagram should be directed in the opposite direction without repeating the calculation.

Due to the low height it is considered that all points of the beam are on the same straight line; all three unknown reactions are applied at one point. To solve it, it is convenient to use the system of equilibrium equations in the first form. Each equation will contain one unknown.

3. We use a system of equations:

The signs of the resulting reactions are (+), therefore, the directions of the reactions are chosen correctly.

3 . To check the correctness of the solution, we create an equation of moments about point B.

We substitute the values ​​of the resulting reactions:

The solution was completed correctly.

Example 2. Double-support beam with hinged supports A And IN loaded with concentrated force F, distributed load with intensity q and a couple of forces with a moment T(Fig. 6.8a). Determine the reactions of the supports.