Voltage is called the intensity of the action of internal forces at a point on the body, that is, stress is the internal force per unit area. By its nature, tension arises on the internal surfaces of contact between body parts. Stress, as well as the intensity of the external surface load, is expressed in units of force per unit area: Pa = N/m 2 (MPa = 10 6 N/m 2, kgf/cm 2 = 98,066 Pa ≈ 10 5 Pa, tf/m2, etc.).
Let's select a small area ∆A. Let us denote the internal force acting on it as ∆\vec(R). The total average stress on this site is \vec(р) = ∆\vec(R)/∆A. Let's find the limit of this ratio at ∆A \to 0. This will be the complete tension on this area (point) of the body.
\textstyle \vec(p) = \lim_(\Delta A \to 0) (\Delta\vec(R)\over \Delta A)
The total stress \vec p, like the resultant of internal forces applied on an elementary area, is a vector quantity and can be decomposed into two components: perpendicular to the area under consideration - normal stress σ n and tangent to the site – tangential stress \tau_n. Here n– normal to the selected area.
The shear stress, in turn, can be decomposed into two components parallel to the coordinate axes x, y, associated with the cross section – \tau_(nx), \tau_(ny). In the name of the shear stress, the first index indicates the normal to the site, the second index indicates the direction of the shear stress.
$$\vec(p) = \left[\matrix(\sigma _n \\ \tau _(nx) \\ \tau _(nx)) \right]$$
Note that in the future we will deal mainly not with the total stress \vec p, but with its components σ_x,\tau _(xy), \tau _(xz) . In general, two types of stresses can arise on the site: normal σ and tangential τ .
Stress tensor
When analyzing stresses in the vicinity of the point under consideration, an infinitesimal volume element (a parallelepiped with sides dx, dy, dz), along each face of which there are, in general, three stresses, for example, for a face perpendicular to the x axis (plate x) – σ_x,\tau _(xy),\tau _(xz)The stress components along three perpendicular faces of the element form a stress system described by a special matrix – stress tensor
$$ T _\sigma = \left[\matrix(
\sigma _x & \tau _(yx) & \tau _(zx) \\
\tau _(xy) & \sigma _y & \tau _(zy) \\ \tau _(xz) & \tau _(yz) & \sigma _z
)\right]$$
Here the first column represents the stress components at the sites,
normal to the x-axis, the second and third – to the y and z axis, respectively.
When rotating coordinate axes that coincide with the normals to the faces of the selected
element, the stress components change. By rotating the selected element around the coordinate axes, you can find such a position of the element at which all the shear stresses on the faces of the element are equal to zero.
The area on which the shear stresses are zero is called main platform .
The normal voltage at the main site is called main stress
The normal to the main area is called main stress axis .
At each point, three mutually perpendicular main platforms can be drawn.
When rotating the coordinate axes, the stress components change, but the stress-strain state of the body (SSS) does not change.
Internal forces are the result of bringing internal forces applied to elementary areas to the center of the cross section. Stress is a measure characterizing the distribution of internal forces over a section.
Let's assume that we know the voltage in each elementary area. Then we can write:
Longitudinal force on the site dA: dN = σ z dA
Shear force along the x axis: dQ x = \tau (zx) dA
Shear force along y-axis: dQ y = \tau (zy) dA
Elementary moments around axes x,y,z: $$\begin(array)(lcr) dM _x = σ _z dA \cdot y \\ dM _y = σ _z dA \cdot x \\ dM _z = dM _k = \tau _(zy) dA \cdot x - \tau _(zx) dA \cdot y \end(array)$$
Having performed integration over the cross-sectional area, we obtain:
That is, each internal force is the total result of the action of stresses throughout the entire cross-section of the body.
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 using 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 rigidity 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. 
,
,
.
| " |
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 axial 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.
Knowing the stress components at any point of the plate under conditions of a plane stress state or plane deformation, one can find from the equations of stress statics on any plane (platform) inclined with respect to the x and y axes, passing through this point perpendicular to the plate. Let us denote by P a certain point in the stressed plate and assume that the stress components are known (Fig. 12). At a small distance from P, we draw a plane parallel to the axis so that this plane, together with the coordinate planes, cuts out a very small triangular prism from the plate. Since the stresses change continuously throughout the volume of the body, then as the size of the cut element decreases, the stress acting on the site will tend to the stress on the parallel area passing through point R.
When considering the equilibrium conditions of a small triangular prism, volumetric forces can be neglected as quantities of a higher order of smallness. Similarly, if the cut element is very small, we can ignore stress variations along the edges and assume that the stresses are uniformly distributed. Then the forces acting on the triangular prism can be determined by multiplying the stress components by the area of the faces. Let be the direction of the normal to the plane and the cosines of the angles between the normal and the x and y axes are denoted as follows:
Then, if A denotes the area of the element’s face, then the areas of the other two faces will be .
If we denote by X and the stress components acting on the face, then the equilibrium conditions of the prismatic element lead to the following relations:
Thus, the stress components on any area determined by the direction cosines can be easily found from relations (12) if the three stress components at point P are known.
Let us denote by a the angle between the normal to the site and the x axis, so that then from relations (12) for the normal and tangential components of the stresses on the site we obtain the formulas:
Obviously, the angle can be chosen in such a way that the shear stress on the site becomes zero. For this case we get
From this equation one can find two mutually perpendicular directions for which the tangential stresses at the corresponding areas are equal to zero. These directions are called principal, and the corresponding normal stresses are called principal normal stresses.
If we take the directions of the x and y axes as the main directions, then the component is equal to zero and formulas (13) take a simpler form
The change in stress components a and depending on the angle a can be easily represented graphically in the form of a diagram in coordinates a and Each orientation of the site corresponds to a point on this diagram, the coordinates of which represent the values of the stresses acting on this site. Such a diagram is shown in Fig. 13. For areas perpendicular to the main directions, we obtain points A and B with abscissas, respectively. Now you can
prove that the stress components for any area determined by the angle a (Fig. 12) will be represented by the coordinates of a certain point on the circle for which the segment A B is the diameter. To find this point, it is enough to measure from point A in the same direction in which angle a is measured in Fig. 12, an arc corresponding to angle . For the coordinates of point D constructed in this way from Fig. 13 we get
Comparison with formulas (13) shows that the coordinates of point D give numerical values of the stress components on the site determined by the angle a. To harmonize the sign of the tangent component, we will assume that positive values are laid upward (Fig. 13, a), and we will consider tangential stresses to be positive when they give a moment acting in a clockwise direction, as is the case on the faces of the element (Fig. . 13, b). Shear stresses in the opposite direction, for example acting on the faces of an element, are considered negative.
We will change the orientation of the platform by rotating it around an axis perpendicular to the plane (Fig. 12) in a clockwise direction so that angle a will change from 0 to point D in Fig. 13 will move from A to B. Thus, the lower half of the circle determines the change in voltages for all values of a within these limits. In turn, the upper part of the circle gives stresses for the interval
Continuing the radius to the point (Fig. 13), i.e., taking the angle equal instead of , we obtain stresses on the area perpendicular to the area (Fig. 12). From this it can be seen that the shear stresses on two mutually perpendicular areas are numerically equal to each other, as was proven earlier. As for normal stresses, we see from
Figure that, i.e., the sum of normal stresses acting on two mutually perpendicular areas, when the angle a changes, remains constant.
The maximum shear stress mmax is given in the diagram (Fig. 13) by the maximum ordinate of the circle, i.e., equal to the radius of the circle. From here
It acts on a site for which, that is, on a site the normal to which bisects the angle between the two principal directions.
The corresponding diagram can also be constructed for the case when one or both principal stresses are negative, i.e., for the case of compression. It is only necessary to put the magnitude of the compressive stress towards the negative abscissas. In Fig. 14, a shows a diagram for the case when both main voltages are negative, in Fig. 14b shows a diagram for the case of pure shear.
From Fig. 13 and 14 it is clear that the stress at any point can be decomposed into two parts. One of them is a biaxial tension (or compression), the two components of which are equal to each other and are determined in magnitude by the abscissa of the center of the Mohr circle.
The other part is pure shear with shear stress, the magnitude of which is given by the radius of the circle. When several plane stress states are superimposed, uniform tensions (or compressions) can be added to each other algebraically. When superimposing pure shear states, it is necessary to take into account the directions of the planes on which the corresponding shear stresses act. It can be shown that by superimposing two pure shear stress states, for which the planes of maximum shear stress are at an angle to each other, the resulting system will reduce to another case of pure shear. For example, fig. Figure 15 shows how to determine the stress produced by two states of pure shear with values of tangential stresses and on an area, the position of which is determined by the angle. The first of these states refers to planes (Fig. 15, a), and the second to planes inclined to planes
It was assumed: the beam has a rectangular cross section (Fig. 7.11), therefore
;;
; 
where y is the distance from the point at which the shear stress is determined to the neutral x axis.
Substituting these formulas into the Zhuravsky formula, we obtain:
Shear stress vary along the height of the cross section according to the law of a quadratic parabola (see Fig. 7.11).
At (for points furthest from the neutral axis).
For points located on the neutral axis (at ), 
.
Diagrams of tangential stresses of an I-section
A characteristic feature of an I-section: a sharp change in the width of the cross section (), where the flange connects to the wall.
Let us determine the shear stress at a certain point K (Fig. 7.12) by drawing a section through it, the width of which is equal to the wall thickness: .
Let us consider the upper cut-off part of the cross section (shaded in Fig. 7.12), the static moment of inertia of which relative to x is equal to the sum of the static moments of inertia of the flange and the shaded part of the wall:
The shear stress diagram for an I-section is shown in Fig. 7.12, b.
Tangential stresses arising at flange points are calculated using the Zhuravsky formula it is forbidden, since its derivation used the assumption that the distribution of tangential stresses is uniform across the width of the cross section, which is valid only if the width of the section is small. However, it is obvious that the shear stresses are small and have no practical effect on the strength of the beam. The shear stress diagram for an I-section is shown with a dashed line (see Fig. 7.12, b).
Formula for shear stress at point L (where the flange connects to the wall):
The highest shear stresses occur at points lying on the neutral x-axis.
Diagrams of tangential stresses of a circular section
For building diagrams of tangential stresses of circular cross-section let's find out the direction shear stress during bending, arising at some point on the contour of the cross section of the rod.
Let's consider an arbitrary cross section of the rod (Fig. 7.13, a).
Let us assume: at some point of the contour K, the tangential stress during bending is directed arbitrarily with respect to the contour. Let us decompose the shear stress into two components and , directed respectively along the normal and tangential to the contour. If tangential stress exists, then according to the law of pairing of tangential stresses on the surface of the rod there should be an equal tangential stress during bending. Since the surface of the rod is free from external forces parallel to the z-axis of the beam, the shear stress on the surface of the rod and, therefore, .
Thus, at the point of the cross-sectional contour, the surface of which is not loaded with longitudinal forces, the shear stress during bending is directed tangentially to the contour.
Let us show that at the apex of the cross-sectional angle of the rod, the shear stress is zero (Fig. 7.13, b).
Let us assume that a shear stress occurs at the vertex of the angle (at point M). Let us decompose it into components of tangential stress and . By
