Examples of determining normal and shear stresses. Stress at a point Tangential stress at a section point

Stress is a numerical measure of the distribution of internal forces along a cross-sectional plane. It is used in the study and determination of internal forces of any structure.

Let us select an area on the section plane  A; an internal force will act along this area  R.

Magnitude of ratio  R/ A= p Wed is called the average voltage at the site  A. True voltage at a point A we'll get it by aiming  A to zero:

Normal stresses arise when particles of a material tend to move away from each other or, conversely, to get closer. Tangential stresses are associated with the displacement of particles along the plane of the section under consideration.

It's obvious that
. The tangential stress, in turn, can be expanded along the axis directions x And y (τ z X , τ z at). The stress dimension is N/m 2 (Pa).

Under the action of external forces, along with the occurrence of stresses, a change in the volume of the body and its shape occurs, i.e. the body is deformed. In this case, a distinction is made between the initial (undeformed) and final (deformed) states of the body.

16. Law of pairing of tangential stresses

Kasat. voltage on 2 mutually perpendicular. area directed towards or away from the edge and equal in size

17. The concept of deformations. Measure of linear, transverse and angular deformation

Deformation - called. mutual movement of points or sections of a body in comparison with the positions of the body that they occupied before the application of external forces

There are: elastic and plastic

a) linear deformation

the measure of the phenomenon is the relative elongation of the epsil =l1-l/l

b) transverse def

measure of phenomena relative narrowing of epsil stroke=|b1-b|/b

18. Hypothesis of plane sections

Main hypotheses(assumptions): hypothesis about the non-pressure of longitudinal fibers: fibers parallel to the axis of the beam experience tensile-compressive deformation and do not exert pressure on each other in the transverse direction; plane section hypothesis: A section of a beam that is flat before deformation remains flat and normal to the curved axis of the beam after deformation. In the case of flat bending, in general, internal power factors: longitudinal force N, transverse force Q and bending moment M. N>0, if the longitudinal force is tensile; at M>0, the fibers on top of the beam are compressed and the fibers on the bottom are stretched. .

The layer in which there are no extensions is called neutral layer(axis, line). For N=0 and Q=0, we have the case pure bending. Normal voltages:
, is the radius of curvature of the neutral layer, y is the distance from some fiber to the neutral layer.

19.Hooke's Law (1670). Physical meaning of the quantities included in it

He established the relationship between stress, stretching and longitudinal deformation.
where E is the proportionality coefficient (modulus of elasticity of the material).

The elastic modulus characterizes the rigidity of the material, i.e. ability to resist deformation. (the larger E, the less tensile the material)

Potential strain energy:

External forces applied to an elastic body perform work. Let us denote it by A. As a result of this work, the potential energy of the deformed body U accumulates. In addition, the work goes to impart speed to the mass of the body, i.e. is converted into kinetic energy K. The energy balance has the form A = U + K.

If you mentally cut out an element in the form of an infinite small cube around some point of the body, then the stresses shown in Fig. 1 will generally act along its edges. 3.1.

The set of normal and tangential stresses acting on all areas (sections) containing any point is called tense state of the body at a given point

Rice.3 . 1

Thus, on the faces of an elementary parallelepiped isolated in the vicinity of a point of a loaded body, nine stress components act. Let's write them in the form of the following square matrix:

where the first, second and third rows contain the stress components, respectively, on areas perpendicular to the axes , , . This set of stresses is called stress tensor.

Law of pairing of tangential stresses. Main areas and main stresses.

Let's create an equation for the moments of all forces applied to an elementary parallelepiped relative to the axis. (Fig. 3.1.).

Forces parallel to and intersecting this axis will not enter into the equation. The moments of forces on two faces perpendicular to the axis are balanced, as well as the moments of forces on the upper and lower faces of the element. Thus we get:

It follows that .

Similarly, from the other two equations we find:

So, we have the equalities

called law of tangent stress pairing

Law of tangent stress pairing – tangential stresses on any two but mutually perpendicular sites, directed perpendicular to the line of intersection of the sites, are equal in magnitude. At the same time, they tend to rotate the element in different directions.

When the orientation of the faces of a selected element changes, the stresses acting on its faces also change. It is possible to draw areas where the shear stresses are zero. Areas at which the shear stresses are zero are called main venues, and the normal stresses at these sites are principal stresses.

It can be proven that at each point of a stressed body there are three main mutually perpendicular areas.

The principal stresses are denoted by , , . In this case, the indices should be arranged so that the inequality is satisfied

If all three principal stresses are different from zero, then the stressed state is called triaxial or volumetric (Fig. 3.2, a).

If one of the principal stresses is equal to zero, then the stressed state is called biaxial or flat (Fig. 3.2, b).

If two principal stresses are equal to zero, then the stressed state is called uniaxial or linearm(Fig. 3.2, c).

Rice.3 . 2

Plane stress state.

When studying the stressed state of structural elements, one most often has to deal with a plane stressed state. It occurs during torsion, bending and complex resistance. Therefore, we will dwell on it in a little more detail.

Let's consider an element whose faces are the main areas.

Rice.3 . 3

Positive stresses and act on them, and the third main stress (direction perpendicular to the plane of the drawing).

Let us draw a section I – I, which will determine the area (), characterized by a positive angle. The voltages along this area will be determined by the formulas:

(3.3)

Compressive principal stresses are substituted into these formulas with a minus sign, and the angle is measured from the algebraically greater principal stress.

Let us draw section II – II, which will determine the area perpendicular to the area. The normal to it forms an angle with the direction

Substituting the values ​​of the angle into formulas (3.2) and (3.3), we will have

. (3.5)

The set of formulas (3.2) - (3.5) makes it possible to find stresses along any mutually perpendicular inclined areas if the principal stresses are known.

Adding equalities (3.2) and (3.4), we find that

, (3.6)

that is, the sum of normal stresses along two mutually perpendicular areas does not depend on the angle of inclination of these areas and is equal to the sum of the principal stresses.

From formulas (3.3) and (3.5) we see that tangential stresses reach their greatest value at , that is, along areas inclined to the main areas at an angle , and

. (3.7)

Comparing formulas (3.3) and (3.5), we find that

This equality expresses the law of pairing of tangential stresses.

Let us now draw two more sections (Fig. 3.3): Section III – III, parallel to I – I, and section IV – IV, parallel to II – II. Element, separated by four sections from the element (Fig. 3.4, a), will have the form shown in Fig. 3.4, b. Both elements define the same stress state, but the element represents it by principal stresses, and the element by stresses on inclined areas.

Rice.3 . 4

In the theory of stress states, two main tasks can be distinguished.

Direct task. At a point, the positions of the main areas and the corresponding main stresses are known; it is required to find normal and shear stresses along areas inclined at a given angle to the main ones.

Inverse problem. At a point, the normal and tangential stresses acting in two mutually perpendicular areas are known; it is required to find the principal directions and principal stresses. Both problems can be solved both analytically and graphically.

Direct problem in a plane stress state. Circle of tension (Mohr's circle).

The analytical solution to the direct problem is given by formulas (3.2) – (3.5).

Let's analyze the stress state using a simple graphical construction. To do this, we introduce the geometric plane into consideration and relate it to the rectangular coordinate axes and . We will describe the calculation procedure using the example of the stress state shown in Fig. 3.5, a.

Having chosen a certain scale for the stresses, we plot the segments on the abscissa axis (Figure 3.5, b)

On the same diameter we construct a circle with the center at point . The constructed circle is called voltage circle or Mohr's circle.

Rice.3 . 5

The coordinates of the circle points correspond to normal and shear stresses at various sites. So, to determine the voltage on an area drawn at an angle (Fig. 3.5, a) from the center of the circle (Fig. 3.5, b), we draw a ray at an angle until it intersects with the circle at a point (we put positive angles counterclockwise). The abscissa of a point (segment) is equal to the normal stress, and its ordinate (segment) is equal to the tangential stress.

We find the voltage on an area perpendicular to the one considered by drawing a ray at an angle and obtaining a point at the intersection with the circle. Obviously, the ordinate of the point corresponds to the shear stress, and the abscissa of the point corresponds to the normal stress.

By drawing a parallel line from a point (in our case, a horizontal line) until it intersects with a circle, we find a pole - a point. The line connecting the pole to any point on the circle is parallel to the direction of normal stress on the site to which this point corresponds. So, for example, a line is parallel to the main stress. Obviously, the line is parallel to the direction of the main stress.

Inverse problem in a plane stress state.

In practical calculations, normal and shear stresses are usually determined on some two mutually perpendicular areas. Let, for example, the voltages , , , be known (Fig. 3.6, a). Based on these data, it is necessary to determine the values ​​of the main stresses and the position of the main areas.

First, let's solve this problem graphically. Let us assume that > ​​and >.

In the geometric plane in the coordinate system we plot the point , with coordinates , and the point with coordinates , (Fig. 3.6, b). By connecting the points and , we find the center of the circle - a point - and draw a circle with a radius. The abscissas of the points of its intersection with the axis - the segments and - will give, respectively, the values ​​of the principal stresses and.

To determine the position of the main sites, we will find the pole and use its property. Let's draw a line from the point parallel to the line of action of the voltage, i.e. horizontal. The point of intersection of this line with the circle is the pole. By connecting the pole with points and , we obtain the directions of the main stresses. The main areas are perpendicular to the found directions of the main stresses.

Rice.3 . 6

We use the constructed circle to obtain analytical expressions for the principal stresses and:

(3.9)

(3.10)

Formula (3.10) determines the only value of the angle by which the normal must be rotated in order to obtain the direction of the algebraically larger principal stress. A negative value corresponds to clockwise rotation.

If one of the main stresses turns out to be negative and the other positive, then they should be designated and . If both main voltages turn out to be negative, then they should be designated and .

Lecture 4. Strength theories. Pure shift(jcomments on)

Theories of strength.

The most important task of engineering calculation is to assess the strength of a structural element based on a known stress state. For simple types of deformations, in particular for uniaxial stress states, determining the values ​​of dangerous stresses does not present any particular difficulties. Let us remember that dangerous stresses are understood as stresses corresponding to the onset of destruction (in the case of a brittle state of the material) or the appearance of residual deformations (in the case of a plastic state of the material):

For hazardous voltages, permissible voltages are established that provide a certain margin against the onset of the limit state.

In a complex stress state, as experiments show, a dangerous state can occur at different values ​​of the principal stresses , , depending on the relationships between them. In this case, a hypothesis is introduced about the predominant influence of one or another factor on the strength of the material. The limiting value of the factor determining strength is found on the basis of simple experiments (tension, compression, torsion).

The hypothesis chosen in this way is called mechanical theory of strength.

Let's consider the classical theories of strength.

6. Voltage at a point. Total, normal, shear stress. Dimensions of voltage.

Stress is a measure of the distribution of internal forces over a section.

Where
- internal strength revealed on the site
.

Full voltage
.

Normal stress - the projection of the total stress vector onto the normal is denoted by σ.
, where E is the modulus of elasticity of the first kind, ε is the linear deformation. Normal stress is caused only by a change in the lengths of the fibers, the direction of their action, and the angle of the transverse and longitudinal fibers is not distorted.

Shear stress – stress components in the section plane.
, Where
(for an isotropic material) – shear modulus (modulus of elasticity of the second kind), μ – Poisson’s ratio (=0.3), γ – shear angle.

7. Hooke’s law for a uniaxial stress state at a point and Hooke’s law for pure shear. Elastic moduli of the first and second kind, their physical meaning, mathematical meaning and graphical interpretation. Poisson's ratio.

- Hooke's law for a uniaxial stress state at a point.

E – coefficient of proportionality (modulus of elasticity of the first kind). The elastic modulus is a physical constant of the material and is determined experimentally. The value of E is measured in the same units as σ, i.e. in kg/cm2.

- Hooke's law for shift.

G – shear modulus (modulus of elasticity of the second kind). The dimension of module G is the same as that of module E, i.e. kg/cm2.
.

μ – Poisson’s ratio (proportionality coefficient).
. A dimensionless quantity that characterizes the properties of a material and is determined experimentally and lies in the range from 0.25 to 0.35 and cannot exceed 0.5 (for an isotropic material).

8. Central tension (compression) of a straight beam. Determination of internal longitudinal forces by the method of sections. Rule of signs for internal longitudinal forces. Give examples of calculating internal longitudinal forces.

The beam experiences a state of central tension (compression) if central longitudinal forces N z arise in its cross sections (i.e. internal force, the line of action of which is directed along the z axis), and the remaining 5 force factors are equal to zero (Q x = Q y =M x =M y =M z =0).

Sign rule for N z: true tensile force – “+”, true compressive force – “-”.

9. Central tension (compression) of a straight beam. Statement and solution of the problem of determining stresses in the cross sections of a beam. Three sides of the problem.

Setting: A straight beam made of a homogeneous material, stretched (compressed) by central longitudinal forces N. Determine the stress arising in the cross sections of the beam, the deformation and displacement of the cross sections of the beam depending on the coordinates of these sections.

10. Central tension (compression) of a straight beam. Determination of deformations and displacements. The rigidity of the beam in tension (compression). Give examples of relevant calculations.

For the central stress (compression) of a straight beam, see question 8.

.

With central tension (compression) of the beam in the transverse direction, only normal stress σ z arises in the section, constant at all points of the cross section and equal to N z / F.
, where EF is the stiffness of the beam in tension (compression). The greater the rigidity of the beam, the less the beads are deformed under the same force. 1/(EF) – compliance of the beam in tension (compression).

11. Central tension (compression) of a straight beam. Statically indeterminate systems. Unraveling static indetermination. Influence of temperature and installation factors. Give examples of relevant calculations.

For the central stress (compression) of a straight beam, see question 8.

If the number of linearly independent static equations is less than the number of unknowns included in the system of these equations, then the task of determining these unknowns becomes statically indeterminable.
(As much as one part lengthens, the second part will shrink).

Normal conditions are 20º C.
.f(σ,ε,tº,t)=0 – functional relationship between 4 parameters.

12. Experimental study of the mechanical properties of materials under tension (compression). Saint-Venant's principle. Sample tensile diagram. Unloading and reloading. Hardening. Basic mechanical, strength and deformation characteristics of the material.

The mechanical properties of materials are calculated using testing machines, which can be lever or hydraulic. In a lever machine, the force is created using a load acting on the sample through a system of levers, and in a hydraulic machine, using hydraulic pressure.

Saint-Venant's principle: The nature of the stress distribution in cross sections sufficiently distant (almost at distances equal to the characteristic transverse size of the rod) from the place of application of loads, longitudinal forces, does not depend on the method of applying these forces, if they have the same static equivalent. However, in the zone of application of loads, the law of stress distribution may differ markedly from the law of distribution in fairly distant sections.

If the test sample is unloaded without causing destruction, then during the unloading process the relationship between the force P and the elongation Δl the sample will receive a residual elongation.

If the sample was loaded in a section where Hooke's law is observed and then unloaded, then the elongation will be purely elastic. When loading again, the intermediate unloading will disappear.

Cold hardening (hardening) is the phenomenon of increasing the elastic properties of a material as a result of preliminary plastic deformation.

The proportional limit is the highest stress to which a material follows Hooke's law.

The elastic limit is the highest stress up to which the material does not receive residual deformation.

Yield strength is the stress at which strain increases without a noticeable increase in load.

Tensile strength is the maximum stress that a sample can withstand without breaking.

13. Physical and conditional yield limits of materials when testing tensile samples, tensile strength. Allowable stresses when calculating the strength of a centrally stretched (compressed) beam. Standard and actual safety factors. Give numerical examples.

In cases where there is no clearly defined yield plateau in the diagram, the yield stress is conventionally taken to be the stress value at which the residual deformation ε rest = 0.002 or 0.2%. In some cases, the limit ε rest = 0.5% is set.

max|σ z |=[σ].
,n>1(!) – standard safety factor.

- actual safety factor.n>1(!).

14. Central tension (compression) of a straight beam. Calculations for strength and rigidity. Condition of strength. Stiffness condition. Three types of problems when calculating strength.

For the central stress (compression) of a straight beam, see question 8.

max|σ z | stretch ≤[σ] stretch;max|σ z | compression ≤[σ] compression.

15. Generalized Hooke’s law for a triaxial stress state at a point. Relative volumetric deformation. Poisson's ratio and its limiting values ​​for a homogeneous isotropic material.

,
,
. Adding these equations, we obtain the expression for volumetric deformation:
. This expression allows you to determine the limiting value of Poisson's ratio for any isotropic material. Let's consider the case when σ x =σ y =σ z =р. In this case:
. For positive p, the value of θ must also be positive; for negative p, the change in volume will be negative. This is only possible when μ≤1/2. Therefore, the value of Poisson's ratio for an isotropic material cannot exceed 0.5.

16. The relationship between the three elastic constants for an isotropic material (without deriving the formula).

,
,
.

17. Study of the stress-strain state at the points of a centrally stretched (compressed) straight beam. Law of pairing of tangential stresses.

,
.

- law of pairing of tangential stresses.

18. Central tension (compression) of a beam made of linear elastic material. Potential energy of elastic deformation of a beam and its connection with the work of external longitudinal forces applied to the beam.

A=U+K. (As a result of work, the potential energy of a deformed body is accumulated; in addition, work is used to accelerate the mass of the body, i.e. it is converted into kinetic energy).

If the central tension (compression) of a beam made of a linearly elastic material is carried out very slowly, then the speed of movement of the center of mass of the body will be very small. This loading process is called static. The body is in a state of equilibrium at any moment. In this case, A=U, and the work of external forces is entirely converted into potential deformation energy.
,
,
.

Stressed and deformed states of an elastic body. Relationship between stress and strain

The concept of body tension at a given point. Normal and shear stresses

Internal force factors that arise when an elastic body is loaded characterize the state of a particular section of the body, but do not answer the question of which point of the cross section is the most loaded, or, as they say, danger point. Therefore, it is necessary to introduce into consideration some additional quantity that characterizes the state of the body at a given point.

If a body to which external forces are applied is in equilibrium, then internal resistance forces arise in any section of it. Let us denote by the internal force acting on an elementary area , and the normal to this area by then the quantity

(3.1)

called total voltage.

In the general case, the total stress does not coincide in direction with the normal to the elementary area, so it is more convenient to operate with its components along the coordinate axes -

If the outer normal coincides with any coordinate axis, for example, with the axis X, then the stress components will take the form in which the component turns out to be perpendicular to the section and is called normal voltage, and the components will lie in the section plane and are called shear stresses.

To easily distinguish between normal and tangential stresses, other designations are usually used: - normal stress, - tangential stress.

Let us select from a body under the action of external forces an infinitesimal parallelepiped, the edges of which are parallel to the coordinate planes, and the edges have a length of . On each face of such an elementary parallelepiped there are three stress components parallel to the coordinate axes. In total, we get 18 stress components on six faces.

Normal stresses are denoted in the form , where the index denotes the normal to the corresponding face (i.e., it can take values ​​). Tangential stresses have the form ; here the first index corresponds to the normal to the area on which this shear stress acts, and the second indicates the axis parallel to which this stress is directed (Fig. 3.1).

Fig.3.1. Normal and shear stresses

For these voltages the following is accepted rule of signs. Normal voltage is considered positive when stretched, or, what is the same, when it coincides with the direction of the outer normal to the area on which it acts. Shear stress is considered positive if on the site, the normal to which coincides with the direction of the coordinate axis parallel to it, it is directed towards the positive coordinate axis corresponding to this voltage.

The stress components are functions of three coordinates. For example, the normal stress at a point with coordinates can be denoted

At a point that is at an infinitesimal distance from the point under consideration, the stress, up to infinitesimals of the first order, can be expanded into a Taylor series:

For areas that are parallel to the plane, only the coordinate changes X, and the increments Therefore, on the face of the parallelepiped coinciding with the plane, the normal stress will be , and on the parallel face, located at an infinitesimal distance, - The stresses on the remaining parallel faces of the parallelepiped are related in a similar way. Therefore, out of 18 voltage components, only nine are unknown.

The theory of elasticity proves the law pairing of shear stresses, according to which, along two mutually perpendicular areas, the components of tangential stresses perpendicular to the line of intersection of these areas are equal to each other:

It can be shown that stresses (3.3) not only characterize the stressed state of a body at a given point, but define it uniquely. The combination of these stresses forms a symmetrical matrix, which is called stress tensor:

(3.4)

Since each point will have its own stress tensor, then the body has field stress tensors.

When a tensor is multiplied by a scalar quantity, a new tensor is obtained, all of whose components are times larger than the components of the original tensor.

Voltage is a vector and, like any vector, it can be represented by normal (with respect to the site) and tangential components (Fig. 2.3). We will denote the normal component of the stress vector as tangent. Experimental studies have established that the influence of normal and tangential stresses on the strength of a material is different, and therefore in the future it will always be necessary to separately consider the components of the stress vector.

Rice. 2.3. Normal and shear stress in the site

Rice. 2.4. Shear stress when shearing a bolt

When the bolt is stretched (see Fig. 2.2), normal stress acts in the cross section

When the bolt operates to shear (Fig. 2.4), a force must arise in section P that balances the force.

From the equilibrium conditions it follows that

In fact, the last relation determines a certain average stress over the section, which is sometimes used for approximate estimates of strength. In Fig. Figure 2.4 shows the view of the bolt after exposure to significant forces. The destruction of the bolt began, and one half of it shifted relative to the other: shear or shear deformation occurred.

Examples of determining stresses in structural elements.

Let us examine the simplest examples in which the assumption of a uniform stress distribution can be considered practically acceptable. In such cases, the stress values ​​are determined using the method of sections from the static equations (equilibrium equations).

Torsion of a thin-walled round shaft.

A thin-walled round shaft (pipe) transmits torque (for example, from an aircraft engine to a propeller). It is necessary to determine the stresses in the cross section of the shaft (Fig. 2.5, a). Let us draw the section plane P perpendicular to the axis of the shaft and consider the equilibrium of the cut-off part (Fig. 2.5, b).

Rice. 2.5. Torsion of a thin-walled round shaft

From the condition of axial symmetry, taking into account the small thickness of the wall, it can be assumed that the stresses at all points of the cross section are the same.

Strictly speaking, this assumption is valid only for very small wall thicknesses, but in practical calculations it is used if the wall thickness

where is the average radius of the section.

External forces applied to the cut-off part of the shaft are reduced only to torque, and therefore there should be no normal stresses in the cross section. The torque is balanced by tangential stresses, the moment of which is equal to

From the last relation we find the shear stress in the shaft section:

Stresses in a thin-walled cylindrical vessel (pipe).

Pressure acts in a thin-walled cylindrical vessel (Fig. 2.6, a).

Let us draw a section by plane P, perpendicular to the axis of the cylindrical shell, and consider the equilibrium of the cut-off part. The pressure acting on the lid of the vessel creates an increased

This force is balanced by the forces arising in the cross section of the shell, and the intensity of these forces - stress - will be equal to

The thickness of the shell 5 is assumed to be small compared to the average radius, the stresses are considered to be uniformly distributed at all points of the cross section (Fig. 2.6, b).

However, the pipe material is affected not only by stresses in the longitudinal direction, but also by circumferential (or hoop) stresses in the perpendicular direction. To identify them, we select a ring of length I in two sections (Fig. 2.7), and then draw a diametrical section separating half of the ring.

In Fig. 2.7, a shows the stresses on the cross-sectional surfaces. Pressure acts on the inner surface of the pipe radius

Rice. 2.8. Crack in a cylindrical shell under the action of destructive internal pressure