5th semester.Fundamentals of the functioning of machines and their elements in the industrial service system

Theoretical mechanics is a science in which the general laws of mechanical motion and mechanical interaction of material bodies are studied.

Section 1.Statics is a section of mechanics in which methods of transforming systems of forces into equivalent systems are studied and conditions for the equilibrium of forces applied to a solid body are established.

Force - this is a measure of the mechanical interaction of bodies, determining the intensity and direction of this interaction. Strength is determined by three elements: numerical value (modulus), direction and point of application. Force is represented by a vector.

Communication reaction is called a force or a system of forces that expresses the mechanical action of a connection on a body. One of the basic principles of mechanics is the principle of liberation of bodies from bonds, according to which a non-free solid body can be considered as a free body, on which, in addition to the specified forces, reactions of bonds act.

Task 1. Determination of reactions of beam supports under the action of a plane arbitrary system of forces

Define reactions R A And R B beam supports, the dimensions and loads of which are shown in Fig. 1,a (change the values ​​of F and M).

Solution. 1.Drawing up a calculation scheme. Object of equilibrium – beam AC. Active forces: F = 3ToH, a couple of forces with M = 4ToH∙m = 1kN/m, which replace with one concentrated force R q = q∙ 1= 1∙ 3 = 3ToH; applied to the point D at a distance of 1.5 m from the edge of the console. Applying the principle of liberation from connections, we depict in points A And IN reactions. A plane arbitrary system of forces acts on the beam, in which there are three unknown reactions

And .

Axis X we direct along the horizontal axis of the beam to the right, and the axis y - vertically upward (Fig. 1, a).

2. Equilibrium conditions:

.

3. Drawing up equilibrium equations:

4. Determining the required quantities, checking the correctness of the solutionand analysis of the results obtained.

Solving the system of equations (1 – 3), we determine the unknown reactions

from (2): kN.

Magnitude of reaction R A X has a negative sign, which means it is not directed as shown in the figure, but in the opposite direction.

To check the correctness of the solution, let’s create an equation for the sum of moments about the point E.

Substituting the values ​​of the quantities included in this equation into this equation, we obtain:

0,58 ∙ 1 – 4 + 5,02 ∙ 3 – 3 ∙ 3,5 = 0.

The equation is satisfied identically, which confirms the correctness of the solution to the problem.

Task 2. Determination of reactions of supports of a composite structure

The structure consists of two bodies hingedly connected at a point WITH. Body AC secured with caulking, body Sun has a hinged-movable (sliding) support (Fig. 1). The bodies of the system are acted upon by a force distributed according to a linear law with maximum intensity q tah = 2 kN/m, force F = 4 kN at an angle α = 30 o and a couple of forces with a moment M = 3 kNm . Geometric dimensions are indicated in meters. Determine the reactions of the supports and the force transmitted through the hinge. The weight of structural elements should not be taken into account.

Rice. 1 Fig. 2

Solution.If we consider the equilibrium of the entire structure as a whole, taking into account that the embedding reaction consists of a force of an unknown direction and a couple, and the reaction of the sliding support is perpendicular to the supporting surface, then design scheme will have the form shown in Fig. 2.

Here the resultant of the distributed load

located at a distance of two meters (1/3 of the length AD) from point A; M A- unknown termination moment.

In this system of forces there are four unknown reactions ( X A ,Y A , M A , R B), and they cannot be determined from the three equilibrium equations of a plane arbitrary system of forces.

Therefore, we divide the system into separate bodies along the hinge (Fig. 3).

The force applied in the hinge should be taken into account only on one body (any of them). Equations for the body Sun:

From here X WITH = – 1 kN; U WITH = 0; R B = 1 kN.

Equations for the body AC:

Here, when calculating the moment of force F relative to the point A Varignon's theorem was used: force F broken down into components F cos α and F sin α and the sum of their moments is determined.

From the last system of equations we find:

X A = – 1,54 kN; U A = 2 kN; M A = – 10,8 kNm.

To check the obtained solution, let’s create an equation of moments of forces for the entire structure relative to the point D(Fig. 2):

Conclusion: the check showed that the reaction modules were determined correctly. The minus sign for the reactions indicates that they are actually directed in opposite directions.

3.3. Definition ground reactions.

Let's look at a few examples.

Example 3.1. Determine support reactions cantilever beam(Fig. 3.3).

Solution. We represent the embedding reaction in the form of two forces Az and Ay, directed as indicated in the drawing, and a reactive torque MA.

We compose the equilibrium equation for the beam.

1. Let us equate to zero the sum of the projections onto the z axis of all forces acting on the beam. We get Az = 0. In the absence of a horizontal load, the horizontal component of the reaction is zero.

2. The same for the y axis: the sum of the forces is zero. Evenly distributed load q is replaced by the resultant qaz applied in the middle of the section az:

Ay - F1 - qaz = 0,

Where

Ay = F1 + qaz .

The vertical component of the reaction in a cantilever beam is equal to the sum of the forces applied to the beam.

3. We compose the third equilibrium equation. Let us equate to zero the sum of the moments of all forces relative to some point, for example relative to point A:

Where

The minus sign indicates that the initially accepted direction of the reactive torque should be reversed. So, the reactive moment is sealed equal to the sum moments of external forces relative to the embedment.

Example 3.2. Determine the support reactions of a two-support beam (Fig. 3.4). Such beams are usually called simple.

Solution. Since there is no horizontal load, then Az = 0

Instead of the second equation, one could use the condition that the sum of the forces along the Y axis is equal to zero, which in this case should be applied to check the solution:
25 - 40 - 40 + 55 = 0, i.e. identity.

Example 3.3. Determine the reactions of the supports of a broken beam (Fig. 3.5).

Solution.

those. the reaction Ay is directed not upward, but downward. To check the correctness of the solution, you can use, for example, the condition that the sum of the moments about point B is equal to zero.

Useful resources on the topic "Determination of support reactions"

1. which will give written solution any beam. .
In addition to constructing diagrams, this program also selects a section profile based on the bending strength condition, and calculates deflections and rotation angles in the beam.

2., which builds 4 types of diagrams and calculates reactions for any beams (even for statically indeterminate ones).

Beams are designed to carry lateral loads. According to the method of application, loads are divided into concentrated (acting on a point) and distributed (acting over a significant area or length).

q— load intensity, kn/m

G= q L– resultant of the distributed load

Beams have supporting devices for connecting them with other elements and transferring forces to them. The following types of supports are used:

· Hinged and movable

This support allows rotation around an axis and linear movement parallel to the reference plane. The reaction is directed perpendicular to the supporting surface.

· Hinged-fixed

This support allows rotation around an axis, but does not allow any linear movement. The direction and value of the support reaction is unknown, therefore it is replaced by two components R A y and R A x along the coordinate axes.

· Hard sealing (pinching)

The support does not allow movement or rotation. Not only the direction and value of the support reaction are unknown, but also the point of its application. Therefore, the embedment is replaced by two components R A y, R A x and the moment M A. To determine these unknowns, it is convenient to use a system of equations.

∑ m A (F k)= 0

To control the correctness of the solution, an additional equation of moments is used relative to any point on the cantilever beam, for example point B ∑ m B (F k)= 0

Example. Determine the support reactions of the rigid embedding of a cantilever beam 8 meters long, at the end of which a load P = 1 kn is suspended. Beam gravity G = 0.4 kn is applied in the middle of the beam.

We free the beam from its bonds, that is, we discard the embedding and replace its action with reactions. We select coordinate axes and draw up equilibrium equations.

∑ F kx = 0 R A x = 0

∑ F k у = 0 R A у – G – P = 0

∑ m A (F k)= 0 - M A + G L / 2 + P L = 0

Solving the equations, we get R A y = G + P = 0.4 + 1 = 1.4 kn

M A = G L / 2 + P L = 0.4. 4 + 1. 8 = 9.6 kn. m

We check the obtained reaction values:

∑ m in (F k)= 0 - M A + R A y L - G L / 2 = 0

— 9,6 + 1,4 . 8 – 0,4 . 4 = 0

— 11.2 + 11.2 = 0 reactions found correctly.

For beams located on two hinged supports it is more convenient to determine the support reactions using the 2nd system of equations, since the moment of force on the support is zero and only one unknown force remains in the equation.

∑ m A (F k)= 0

∑ m V (F k)= 0

To control the correctness of the solution, the additional equation ∑ F k у = 0 is used

1) We free the beam from the supports, and replace their action with support reactions;

2) Replace the distributed load with the resultant load G = q. L;

3) Select the coordinate axes;

4) We compose equilibrium equations.

∑ F kx = 0 R In = 0

∑ m A (F k)= 0 G . L/2 + m - R Wu (L + B)= 0

R Wu = /(L + B) = (6+6) = 2.08 kn

∑ m В (F k)= 0 R A у. (L + B) - Q. (L/2 + B) + m = 0

R A y = / (L + B) = / (6 + 6) = 2.92 kn

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