Q y │z 1 \u003d a \u003d 0;

RA-q. a = 0,

20 - 20a \u003d 0, from where a \u003d 1 m.

M x │z 1 \u003d 1 \u003d 10 + 20. 1 - 10 . 12 = 20 kNm.

2nd section.

(1 m ≤ z 2 ≤ 2 m)

Q y \u003d - R B - q. (z2 - 1) = -20 + 20 . (z 2 - 1) \u003d + 20z 2 - 40

(straight line with the same slope);

at z 2 \u003d 2 m

Qy \u003d 20. 2 - 40 = 0,

at z 2 \u003d 1 m

Qy \u003d 20. 1 - 40 = - 40 kN,

(z2 - 1)

Mx \u003d - M2 + RВ. (z2 - 1 ) - q . (z2 – 1 ) . ----------

1. Determine the unknown support reaction R by composing the equation

equilibrium for the entire rod and taking into account C 2.5, C 2.4, K 2.5, K 2.4 (Fig. 3.20).

∑Z = 0 ,

R - F1 + F2 + F3

Q1. 2 - q 2 . 3 = 0

R = -40 + 10 + 20 + 30

2 – 5 . 3 ,

R = +35 kN.

F =10 kN F3 =20 kN

at z 3

m (right edge of the plot)

10 - 5 . 0 = -10 kN;

at z 3

m (left edge of the site)

10 - 5 . 3 = -25 kN.

4 plot.

m ≤ z 4 ≤ 2 m (definition area N4 )

N 4 \u003d F 3 + F 2 - F 1 - q 2

3 + q 1 . z 4 \u003d 20 + 10 - 40 - 5. 3 + 30 . z4 = -25

30z4

at z4 = 0 m

at z4 = 2 m

M1 =2M

M2 =5M

M3 =7M

M4 =3M

M z M cr

ℓ1

ℓ2

Distance in fractions

chiselled

a1 /a

a2 /a

a3 /a

The whole variety of existing support devices is schematized in the form of a number of basic types of supports, of which

most often found: articulatedsupport(possible designations for it are shown in Fig. 1, a), hinged support(Fig. 1b) and hard pinching, or seal(Fig. 1, c).

In a pivotally movable support, one support reaction occurs, perpendicular to the support plane. Such a support deprives the reference section of one degree of freedom, that is, it prevents displacement in the direction of the reference plane, but allows movement in the perpendicular direction and rotation of the reference section.
In a hinged-fixed support, vertical and horizontal reactions occur. Here, movements in the directions of the support rods are not possible, but rotation of the support section is allowed.
In a rigid termination, vertical and horizontal reactions and a support (reactive) moment arise. In this case, the reference section cannot be displaced and rotated. When calculating systems containing a rigid embedment, the resulting support reactions can be omitted, while choosing the cut-off part so that the embedment with unknown reactions does not fall into it. When calculating systems on hinged supports, the reactions of the supports must be determined without fail. The static equations used for this depend on the type of system (beam, frame, etc.) and will be given in the relevant sections of this manual.

2. Construction of diagrams of longitudinal forces Nz

The longitudinal force in the section is numerically equal to the algebraic sum of the projections of all forces applied on one side of the considered section onto the longitudinal axis of the rod.

Sign rule for Nz: we agree to consider the longitudinal force in the section as positive if the external load applied to the considered cut-off part of the rod causes tension and negative - otherwise.

Example 1Plot longitudinal forces for a rigidly clamped beam(Fig. 2).

Calculation procedure:

1. We outline the characteristic sections, numbering them from the free end of the rod to the termination.
2. Determine the longitudinal force Nz in each characteristic section. In this case, we always consider that cut-off part, which does not include a rigid seal.

According to the found values plotting Nz. Positive values ​​are plotted (in the selected scale) above the plot axis, negative values ​​- below the axis.

3. Construction of diagrams of torques Mkr.

Torque in the section is numerically equal to the algebraic sum of external moments applied on one side of the section under consideration, relative to the longitudinal axis Z.

Rule of signs for Mkr: agree to count torque positive in the section if, when looking at the section from the side of the cut-off part under consideration, the external moment is seen to be directed counterclockwise and negative, otherwise.

Example 2Build a diagram of torques for a rigidly clamped rod(Fig. 3a).

Calculation procedure.

It should be noted that the algorithm and principles for constructing the torque diagram completely coincide with the algorithm and principles plotting longitudinal forces.

1. We outline the characteristic sections.
2. We determine the torque in each characteristic section.

Based on the found values, we build diagram Mcr(Fig. 3b).

4. Rules for controlling diagrams Nz and Mkr.

For longitudinal force diagram and torques, certain patterns are characteristic, the knowledge of which allows us to evaluate the correctness of the constructions performed.

1. Plots Nz and Mkr are always rectilinear.

2. In the area where there is no distributed load, the diagram Nz (Mcr) is a straight line parallel to the axis, and in the area under a distributed load - an inclined straight line.

3. Under the point of application of the concentrated force on the diagram Nz there must be a jump by the value of this force, similarly, under the point of application of the concentrated moment on the diagram Mkr there will be a jump by the value of this moment.

5. Construction of diagrams of transverse forces Qy and bending moments Mx in beams

A rod that bends is called beam. In sections of beams loaded with vertical loads, there are, as a rule, two internal force factors - Qy and bending moment Mx .

Shear force in the section is numerically equal to the algebraic sum of the projections of external forces applied on one side of the section under consideration onto the transverse (vertical) axis.

Sign rule for Qy: we agree to consider the transverse force in the section as positive if the external load applied to the considered cut-off part tends to rotate this section clockwise and negative - otherwise.

Schematically, this rule of signs can be represented as

Bending moment Mx in the section is numerically equal to the algebraic sum of the moments of external forces applied on one side of the section under consideration, relative to the x axis passing through this section.

Sign rule for Mx: we will agree to consider the bending moment in the section as positive if the external load applied to the considered cut-off part leads to tension in the given section of the lower fibers of the beam and negative - otherwise.

Schematically, this rule of signs can be represented as:

It should be noted that when using the sign rule for Mx in the indicated form, the diagram Mx always turns out to be built from the side of the compressed fibers of the beam.

6. Cantilever beams

At plotting Qy and Mx in cantilever, or rigidly clamped, beams, there is no need (as in the examples considered earlier) to calculate the support reactions that occur in a rigid embedment, but you need to choose the cut-off part so that the embedment does not fall into it.

Example 3Plot Qy and Mx(Fig. 4).

Calculation procedure.

1. We outline the characteristic sections.

Axial tension (compression) of a straight rod is a type of deformation in which only one component of internal forces arises in an arbitrary cross section - the longitudinal tensile or compression force.

This is possible provided that the external load is reduced to the resultant forces acting along the beam axis.

The longitudinal tensile force is taken as a positive value, and the longitudinal compressive force is taken as a negative one.

Longitudinal forces are determined by the section method. To do this, it is necessary to divide the rod into sections that are limited by the points of the beam axis, where external concentrated forces act. Within each section, you need to choose an arbitrary section at a variable distance x from the origin of coordinates (from some end of the rod) and consider the equilibrium of one of the parts of the rod. In this case, the part of the rod, the equilibrium of which is considered, is loaded by external forces and an unknown longitudinal force N , which is directed from the section, that is, in accordance with the tension of the rod. Using the equilibrium condition Σ X i =0 , we compose an equilibrium equation, from which we determine the longitudinal force N in every area.

The change in the longitudinal force along the length of the rod can be displayed by a graph that has the name diagram this effort.

Consider a straight rod located horizontally, rigidly fixed on the right end and loaded along its axis by external forces F1 , F 2 \u003d 2F 1 and F 3 \u003d 3F 1 (Fig. 9.1, a). These forces are applied respectively at points a, b, c. The fixed point of the axis of the rod will be denoted by the letter d.

To determine the longitudinal forces, we divide the rod into three sections ab, bc and cd. Within each section, we draw arbitrary cross sections 1-1, 2-2 and 3-3, taken at distances x 1 , x 2 and x 3 from the left free end of the rod.

Let's discard, mentally, the right part from the section 1-1, and replace its action on the left part with an unknown longitudinal force N 1 , which is directed from the section (Fig. 9.1, b) and compose the equilibrium equation:

ΣX i =0,N 1 - F 1 \u003d 0 , where do we find N 1 = F 1 . Thus, the longitudinal force in section ab does not depend on x 1 and has a constant value

N 1 = F 1

Let's discard, mentally, the right part of the beam from the section 2-2 and replace its action on the remaining part of the beam with an unknown longitudinal force N 2 , which is also directed from the section (Fig. 9.1, c). Let's make an equilibrium equation:

ΣX i \u003d 0, N 2 - F 1 + 2F 1 \u003d 0, where do we find N 1 = - F 1 . Thus, the longitudinal force in the section bc does not depend on x2 and has a negative constant value, that is, in this section, the rod is compressed by a section of silt of re-rising tensions along the axis of the force core. Within each section, choose an arbitrary one.

Similarly, we define the longitudinal force N 3 in the area cd. We consider the balance of the left side of the rod relative to section 3-3 (Fig. 9.1, d) and draw up an equilibrium equation:

ΣX i \u003d 0, N 3 - F 1 + 2F 1 - 3F 1 \u003d 0, where do we find N 3 \u003d 2F 1 . In this section, the rod is stretched by force N 3 \u003d 2F 1 , which does not depend on x 3.

Let's build A 1 \u003d 20.2 cm2; cm4; cm4;

diagram N . For this:

Draw a zero line parallel to the axis of the rod;

Let us plot positive values ​​of the longitudinal force upwards from it, and negative values ​​downwards from it, taking an arbitrary scale;

Connect the vertices of neighboring ordinates with straight lines. These lines limit the plot of longitudinal forces in certain areas.

In Fig. 9.1, e, a diagram is drawn N. To be able to use it, that is, to determine the longitudinal force in any section, it is necessary to shade the diagram with evenly spaced straight lines perpendicular to the axis of the rod.

Analyzing this diagram, it is easy to see that it has jumps at points where external forces act. In this case, the magnitudes of the jumps are equal to the acting forces. In the sections between external forces, the longitudinal force remains constant, i.e. the diagram is limited by straight lines parallel to the beam axis.

Definition of displacements

Exercise

For a given statically determinate steel beam, it is required:

1) plot longitudinal forces N and normal stresses σ, writing in general form for each section of the expression N and σ and indicating on the diagrams their values ​​in characteristic sections;

2) determine the total displacement of the beam and plot the displacements δ of the cross sections, assuming the modulus of elasticity E = 2 10 MPa.

Objective – learn how to plot longitudinal forces and normal stresses, and determine displacements.

Theoretical justification

Types of loading of a beam, in which only one internal force factor arises in its cross section - called stretching or compression . The resultant of external forces is applied at the center of gravity of the cross section and acts along the longitudinal axis. Internal forces are determined using the section method. The normal force in the section of the beam is the resultant of the normal stresses acting in the plane of the cross section

N = ∑F (5.1).

The magnitude of the longitudinal forces in different sections of the beam is not the same. A graph showing the change in the magnitude of the longitudinal forces in the cross section of a beam along its length is called longitudinal force diagram.

The stress distribution law can be determined from experiment. It has been established that if a rectangular mesh is applied to the rod, then after applying a longitudinal load, the appearance of the mesh will not change, it will still remain rectangular, and all lines will be straight. Therefore, we can conclude that the distribution of longitudinal deformations is uniform over the cross section, and based on Hooke's law ( σ = Eε) and normal stresses S = const. Then N = S F , from which we obtain a formula for determining the normal stresses in the cross section in tension

σ = MPa (5.2)

A is the area near the considered section of the beam;

N is the resultant of internal forces within this area (according to the method of sections).

To ensure the strength of the rod, the condition of strength must be met - the structure will be strong if the maximum stress at any point of the loaded structure does not exceed the allowable value determined by the properties of this material and the working conditions of the structure, that is

σ ≤ [σ ], τ ≤ [τ] (5.3)

When the beam is deformed, its length changes by and the transverse dimension - by. These values ​​also depend on the initial dimensions of the beam.

Therefore, consider

– longitudinal deformation; (5.4)

- transverse deformation. (5.5)

It has been experimentally shown that , where μ = 0, …, 0.5 is Poisson's ratio. Examples: μ=0 - cork, μ=0.5 - rubber, - steel.

Within the limits of elastic deformation, Hooke's law is fulfilled: , where E is the modulus of elasticity, or Young's modulus.

Work order

1. We divide the beam into sections limited by the points of application of forces (we number the sections from the loose end);

2. Using the method of sections, we determine the magnitude of the longitudinal forces in the section of each section: N = ∑F;

3. We select the scale and build a diagram of longitudinal forces, i.e. draw a straight line parallel to its axis under the image of the beam (or next to it), and draw perpendicular segments from this straight line, corresponding to the longitudinal forces on the selected scale (put the positive value up (or to the right), negative - down (or to the left).

4. We determine the total displacement of the beam and plot the displacements δ of the cross sections.

5. Answer security questions.

test questions

1. What is called a rod?

2. What type of loading of a rod is called axial tension (compression)?

3. How is the value of the longitudinal force calculated in an arbitrary cross section of the rod?

4. What is a plot of longitudinal forces and how is it built?

5. How are normal stresses distributed in the cross sections of a centrally tensioned or centrally compressed rod, and by what formula are they determined?

6. What is called rod elongation (absolute longitudinal deformation)? What is relative buckling? What are the dimensions of absolute and relative longitudinal strains?

7. What is called the modulus of elasticity E? How does the value of E affect the deformation of the rod?

8. Formulate Hooke's law. Write formulas for absolute and relative longitudinal strains of a bar.

9. What happens to the transverse dimensions of the rod when it is stretched (compressed)?

10. What is Poisson's ratio? To what extent does it change?

11. What is the purpose of mechanical testing of materials? What stresses are dangerous for ductile and brittle materials?

Execution example

Construct plots of longitudinal forces and normal stresses for a loaded steel beam (Fig. 5.1). Determine the lengthening (shortening) of the beam, if E

Fig.5.1

Given: F = 2 kH, F = 5 kH, F = 2 kH, A = 2 cm, A, l= 100 mm, l = 50 mm l= 200 mm,