Relationship of stresses with internal force factors. Voltage. Central stretch - compression

Classification of forces

Forces are divided into external and internal. External forces characterize the interaction between bodies, internal forces characterize the interaction between particles of one body.

External forces acting on structural elements are divided into active called the load, and jet(bond reactions). The load is divided into surface and volumetric. Surface load refers to the contact forces that arise when two structural elements are mated or when they interact; to volumetric (mass) forces - forces acting on each infinitesimal element of volume. Examples of volumetric forces are inertial forces, gravity forces, and magnetic forces.

By the nature of the action on the structure, the load is distinguished:

static- changes slowly and smoothly from zero to the final value so that the accelerations of the points of the system, arising in this case, are very small, therefore, the forces of inertia in comparison with the load can be neglected;
dynamic- is applied to the body for a short period of time or instantly with the formation of significant accelerations;
re-variable- changing according to an arbitrary periodic law.

Internal force factors (section method)

Let a free body be in equilibrium under the action of a system of forces (Fig. 2.1). It is required to determine the internal forces in the section. Let us mentally cut the body into two parts along a given section and consider the equilibrium conditions for one (any) part of the body. Both parts after the cut, generally speaking, will not be in equilibrium, since the internal connections are broken. We replace the action of the left side of the body with the right and the right with the left by some system of forces in the section, i.e. internal forces (Fig. 2.2). The nature of the distribution of internal forces in the section is unknown, but they must ensure the balance of each part of the body. To draw up the equilibrium condition for the cut-off part, we bring the internal forces in the form of the main vector and the main moment to the center of gravity of the section and project them on the coordinate axis (Fig. 2.3). We get three projections of the main vector and three projections of the main moment, which are called internal power factors:- longitudinal force; - lateral forces; - torque; - bending moments.

Having compiled the equilibrium conditions for the cut-off part, we obtain

(2.1)

Equations (2.1) are called dependence between the external load on the cut-off part and internal force factors (static equivalents of internal

Rice. 2.1

Rice. 2.2

forces). If the external loads are known, then with their help it is possible to determine the internal force factors.

There are the following main types of deformations:

Rice. 2.3

Rice. 2.4

Voltage concept

According to hypothesis 1 (see section 2.1.1), it can be assumed that internal forces are continuously distributed over the cross-sectional area of the beam. Let on a small but final site A(Fig. 2.5) the internal elementary force acts R. Expanding R into components along the axes, we obtain its components.Relation of the form

determines the average stress on a given site at a given point.

The total, or true, tension at a point is a relation

which determines the intensity of internal forces at a given point of the considered section. Since an infinite number of sections can be drawn through a point of a body, then at a given point there are an infinite number of stresses associated with areas of action. The set of all stresses acting on different sites at a given point is called stress point... The unit of stress is N / m2 or Pa. By analogy with expression (2.3), we can write:

Expression (2.4) defines normal stress σ x (Fig.2.6), the vector of which is directed in the same way as the vector of normal force Ν x. Expressions (2.5) and (2.6) determine shear stresses; their vectors have the same directions as and, respectively, and. The first subscript at τ indicates which axis is parallel to the area of action of the stress under consideration, the second subscript shows which axis is parallel to the given stress.

Relationship between total voltage TO and its components is expressed by the formula

Consider the relationship between stresses and internal force factors in the cross-section of a bar.

Rice. 2.5

Rice. 2.6

The components of the main vector and the main moment of internal forces will be as follows.

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LOADS

Consider a beam under the action of a plane system of forces (Figure 12.7). With two cross sections spaced apart from each other, we select an element from the beam so that external concentrated forces and moments do not act on it.

Internal forces M and Q act on the left end of the element (Fig. 13.7), and on the right one Here, they represent the increments in the values of internal forces on the section of the beam. In addition, the element is acted upon by a distributed load perpendicular to the beam axis; its intensity at the left end of the element is q, and at the right (Fig. 13.7).

Since the entire beam as a whole is in equilibrium, then its element is also in equilibrium Let us compose the equilibrium equation of the element in the form of the sum of the projections onto the axis of all forces acting on it (Fig.13.7):

Here the second term is a quantity of the highest order of smallness; discarding it, we get

So, the first derivative of. the transverse force i along the abscissa of the section is equal to the intensely distributed load perpendicular to the axis of the beam.

Let us now compose the equilibrium equation of the element in the form of the sum of the moments of the forces acting on it relative to the point K (Figure 13.7):

Discarding infinitesimal values of higher (second and third) orders, we get:

Thus, the first derivative of the bending moment along the abscissa of the section is equal to the shear force. This dependence is called Zhuravsky's theorem.

Dependencies (5.7) and (6.7) are valid when the abscissa of the cross-section increases from the left end of the beam to the right. If, on the contrary, the abscissa x increases from the right end of the beam to the left, then in the right-hand sides of formulas (5.7) and (6.7) there should be a minus sign in front of q and Q.

From the course of higher mathematics, the geometric meaning of the first derivative is known for any value of the argument; it is equal to the tangent of the angle a between the tangent to the curve (at the point with coordinates and the positive direction of the axis. Positive and negative values of the angle a are shown in Fig. 14.7, a.

If the first derivative (and hence the angle a) is positive, then the function increases (point in Fig. 14.7, a), and if it is negative, then it decreases (point in Fig. 14.7, a). The extremum (maximum or minimum) of the function exists for those values at which the derivative is zero and, therefore, the angle a is also zero, that is, the tangent to the curve is parallel to the axis (point K in Fig. 14.7, a).

Using the above relationships between the function and its first derivative, a number of important conclusions can be drawn from Zhuravsky's theorem:

1. The tangent of the angle between the tangent to the line limiting the plot M and the axis of the plot is equal to the transverse force Q (Fig. 14.7, b, c), ie.

So, for example, the tangent of the negative angle a (Fig. 10.7, c) in section II of the beam shown in Fig. 10.7, a, has a value, that is, it is equal to the transverse force Q in this section (Figure 10.7, b). In sections III and IV of the same beam, the transverse forces Q are the same and equal (see Fig. 10.7, b). Accordingly, the straight lines in Fig. 10.7, in parallel to each other; the tangent of the angle of their inclination to the axis of the plot is equal to

2. In the sections of the beam, in which the shear force is positive, the bending moment increases (from left to right), and in the sections where it is negative, it decreases.

For example, in Fig. 15.7, and four diagrams Q are shown, and under each of them in Fig. 15.7, b, two of the possible variants of the diagram M. The first two diagrams Q (with positive ordinates) correspond to the diagrams M with increasing (from left to right) ordinates, i.e., with positive angles The last two diagrams Q (with negative ordinates) correspond to the diagrams M with decreasing (from left to right) ordinates, i.e., with negative angles. The same conclusion can be illustrated by the diagrams Q and M shown in Fig. 10.7: in section II of the beam, the transverse force is negative, and in section III, it is positive (see Fig. 10.7, b); in accordance with this, in section II, the bending moments decrease (in the algebraic sense), and in section - they increase (see Fig. 10.7, c).

3. The greater in absolute value the value of the transverse force Q, the steeper the line bounding the diagram M. This conclusion directly follows from the dependence (7.7). In accordance with this conclusion, the lines bounding the M diagrams (Fig. 15.7, b, c) are steeper at points than at points a, since the transverse forces are greater in absolute value than the lines bounding the M diagrams cannot have the outlines shown in fig. 15.7, b, in a dotted line, since they would then be steeper at points a than at points b, which is impossible with lower transverse forces (in absolute value). The same relationship between the diagrams Q and M can be traced in Fig. 10.7 and 11.7.

Based on Fig. 15.7, we can conclude that on a section of the beam with increasing (in the algebraic sense) values of Q from left to right, the line bounding the diagram M is convex downward, and with decreasing - convexity upward.

4. On the section of the beam, in which the shear force has a constant value, the M diagram is limited by a straight line (see, for example, in Fig. 10.7 the Q and M diagrams in sections III and IV of the beam). When this line is inclined to the axis of the diagram M at a certain angle (where - see conclusion 1), and when it is parallel to the axis of the diagram.

(see scan)

In the latter case, the corresponding section of the beam is in a state of pure bending.

5. If the diagram Q does not have a jump on the border of neighboring sections of the beam, then the lines bounding the diagram M in these sections are conjugated without a fracture, that is, they have a common tangent at the conjugation point.

In fig. 16.7, a shows two diagrams Q that do not have jumps at the boundaries of neighboring sections (in sections A). In fig. 16.7, b, the solid lines show the correct conjugation of the lines that bound the M diagrams (without fractures at points a), and the dashed lines show the incorrect conjugation options.

6. If there is a jump on the border of neighboring sections of the beam in the Q diagram, then the lines bounding the M diagram in these sections are conjugated with a fracture, that is, they do not have a common tangent at the conjugation point.

In fig. 17.7, a, three diagrams Q are shown, having jumps at the boundaries of neighboring sections (in sections A), and in Fig. 17.7, b - corresponding to them conjugation of lines, limiting the diagrams by fractures at points a.

7. The bending moment reaches a maximum or minimum in the sections of the beam in which the shear force is equal to zero; the tangent to the line bounding the plot M, in this section is parallel to the axis of the plot.

Internal forces, as well as external loads distributed over the surface, are characterized by intensity (Fig.2.1), which is equal to

Intensity of normal forces - normal stresses causing separation (compression) of particles (dimension).

Intensity of shear forces - shear stresses causing shear (dimension).

Normal and shear stresses are components of the total stress at a point along a given section, the value of which is calculated by the formula.

The magnitudes of the normal and shear stresses at each point of the element depend on the direction of the section passing through this point.

The set of normal and tangential stresses acting on different areas passing through the point under consideration represents the stress state at this point.

2.2. The relationship between stress and internal effort

Consider an elementary site dF cross section F(section normal to the axis x) beams with normal and tangential stresses acting on this site (Fig. 2.2). Let us decompose the stresses into components and, respectively, parallel to the axes y and z... Elementary forces act on the platform, parallel to the axes x,y and z... Projections of all elementary forces (acting on all elementary sites dF cross-sections F) on the axis x,y and z and their moments about these axes are determined by the expressions

In the left-hand sides of expressions (2.1), the internal forces are indicated acting in the cross-section of the beam and reduced to the point of intersection of the axis x and cross section. Namely: N- longitudinal force; and - lateral forces, respectively, parallel to the axes y and z; - torque; - bending moment about the axis y(acting in the plane xz); - bending moment about the axis z(acting in the plane xy).

2.3. Types of stress

The set of normal and shear stresses acting on all areas passing through the point under consideration is called the stress state at this point.

There are the following types of stress state:

a) spatial (triaxial) stress state (Fig. 2.3, a), when it is impossible to draw a single area through the considered point of the body, in which the shear and normal stresses would be equal to zero;

b) flat (biaxial) stress state (Fig. 2.3, b), when in one (and only one) area passing through the considered point of the body, the tangential and normal stresses are equal to zero;

c) linear (uniaxial) stress state (Fig. 2.3, v), when the tangential and normal stresses in two areas passing through the considered point of the body are equal to zero.

a)b)v)

2.4. Flat stress state

In a plane stress state, as noted above, in one of the areas passing through the point under consideration, the tangential and normal stresses are equal to zero.

Select from the body in the vicinity of this point an infinitely small (elementary) triangular prism and combine this area with the plane of the drawing. The index for normal and shear stresses (Figure 2.4, a) indicates the direction of their action. For example, is the stress acting on the site perpendicular to the axis x, in the direction of the axis x.

The normal stresses along the side face of the prism, inclined at an angle to the face, along which the stresses act, will be denoted by, and the tangential stresses along this face.

a)b)

Multiplying each of the effective voltages (Fig. 2.4, a) on the area of the face along which it acts, we obtain a system of concentrated forces applied at the centers of gravity of the corresponding faces (Fig. 2.4, b):

Due to the fact that the selected element is in equilibrium, the following static equations are valid for it:

Substituting the expressions for the forces u from (2.2) into the last equation, we obtain

Expression (2.4) is a mathematical notation of the law of pairing of tangential stresses, which states that the tangential stresses along two mutually perpendicular areas perpendicular to their common edge are equal in absolute value and are directed either both to the edge or both from the edge (Fig. 2.5 ).

The first two equations from (2.3), taking into account the expressions for the efforts from (2.2), take the form:

Considering that, we will reduce these equations by the product. As a result, we get:

Using the law of pairing of tangential stresses (2.4), we obtain:

When deriving formulas (2.5), (2.6), the trigonometric dependence was taken into account.

Formulas (2.5), (2.6) make it possible to determine the values of normal and tangential stresses in any areas passing through a given point, if the stresses are known and in any two mutually perpendicular areas passing through it.

Using formula (2.5), we calculate the sum of normal stresses in two mutually perpendicular areas, for one of which the angle is equal, and for the other:

Thus, the sum of the values of normal stresses in two mutually perpendicular areas is a constant value. Consequently, if in one of such sites the normal stresses have a maximum value, then in the other they have a minimum.

When deriving formula (2.7), the following trigonometric dependences were used.

section method

We mentally draw a section with a certain plane at the place where it is necessary to determine the internal forces. Since the bonds between the particles are eliminated, it is necessary to replace the action of the right side on the left side and the left side on the right side with a system of forces in the section. They are internal forces, which are always reciprocal according to the principle of action and reaction. Regardless of how these forces are distributed over the section, they are reduced to the center of gravity of the section in the form of the main vector of internal forces and the main moment of internal forces
... They are determined from the equilibrium equations left in consideration, it does not matter which part of the element (in this case, the left). To compose equilibrium equations in s cuts, a coordinate system is selected, and vectors are decomposed along these axes into six components: three forces (longitudinal internal force
and lateral forces , ) and three moments (torque
and bending moments
,
), which are determined from six equilibrium equations (Fig. 1.5).

Thus, using the section method, it is possible to determine not the law of distribution of internal forces over the section, but only their resultants. To solve strength problems, it is necessary to know the nature of the distribution of forces over the section, i.e. enter a numeric measure. Voltage is taken as such a measure.
^

1.6 Voltages. Relationship of stresses with internal force factors. Saint-Venant principle

Voltage

E If you select a small area
in section and designate the internal force acting on it
(Fig. 1.6), the vector of the total stress at a point of the body will be determined by the formula

, (1.1)

The vector of the total stress is set by its projections on the axis
, , ... To do this, we denote the projection of the vector on the axis
,
,
(Fig. 1.7) and find the corresponding projection of the total voltage:

Normal voltage

, (1.2)

Rice. 1.7 - shear stress along the axis

, (1.3)

Shear stress along the axis

. (1.4)

If the law of distribution of stresses over the section is known, then using formulas (1.2) - (1.4) and figures (1.8), (1.5), it is possible to obtain a feedback between stresses and internal force factors

, (1.5)
The stresses caused by the local load at the points of the body that are far enough from the place where this load is applied to it do not depend much on the specific nature of the load distribution, but are determined only by its main vector and moment.

A load is called local if the dimensions of the site to which it is applied are small compared to the dimensions of the body.