Shear stress at a point. The concept of stress

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.

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 the x, y, z axes: $$\begin(array)(lcr) dM _x = σ _z dA \cdot y \\ dM _y = σ _z dA \cdot x \\ dM _z = dM _k = \tau _(zy) dA \cdot x - \tau _(zx) dA \cdot y \end(array)$$

Having performed integration over the cross-sectional area, we obtain:

That is, each internal force is the total result of the action of stresses across the entire cross-section of the body.

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 D.A.; an internal force will act along this area D.R. Magnitude of ratio DR/DA=p avg is called the average voltage at the site D.A.. True voltage at a point A we'll get it by aiming D.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 zх, τ zу). The stress dimension is N/m 2 (Pa).

17. The concept of stress. Normal and shear stresses.

Internal power factors. Section method. Diagrams. Expression of internal force factors through normal and tangential stresses.

Internal power factors

In the process of deformation of the beam, under load, a change in the relative position of the elementary particles of the body occurs, as a result of which internal forces arise in it.

By their nature, internal forces represent the interaction of body particles, ensuring its integrity and compatibility of deformations.

To numerically determine the magnitude of internal forces, the method of sections is used.

Section method comes down to four steps:

1. Cut (mentally) the body with a plane in the place where you need to determine the internal forces (Fig. 7);

Rice. 7

2. Any cut off part of the body (preferably the most complex one) is discarded, and its action on the remaining part is replaced by internal forces so that the remaining part under study is in equilibrium (Fig. 8);

Rice. 8

3. The system of forces is brought to one point (as a rule, to the center of gravity of the section) and the main vector and the main moment of the system of internal forces are projected onto the normal to the plane (axis) and the main central axes of the section (and).

The resulting forces (N, Qy, Qz) (Fig. 9) and moments (Mk, My, Mz) are called internal force factors in the section

Rice. 9

The following names are accepted for internal force factors:

-longitudinal or axial force;

AND - shear forces;

-torque;

AND - bending moments.

4. Find internal force factors by composing six static equilibrium equations for the considered part of the dissected body.

Diagram(fr. epure- drawing) - a special type of graph showing the distribution of the load on an object. For example, for a rod, the longitudinal axis of symmetry is taken as the definition domain and diagrams are drawn up for forces, stresses and various deformations depending on the abscissa.

Calculation of stress diagrams is a basic task of such a discipline as strength of materials. In particular, only with the help of a diagram is it possible to determine the maximum permissible load on the material.

To construct the ordinate diagram M in any section of the rod

you need to perform the following two operations.

1. Using the equilibrium equation ∑M(left)= 0 for the left intercept

parts of the rod system (or ∑M(right) = 0 for the right part) calculate

numerical value of the bending moment in the section.

2. Plot the found numerical value as an ordinate perpendicular to the axis of the rod from the side of the stretched fiber of the rod .

The numerical value of the bending moment in the section is equal to the numerical value of the algebraic sum of the moments with all our might, acting on the rod system on either side of the section, taken relative to a point on the section axis.

The component lying in the section in a given area is denoted by and is called shear stress.

The normal stress directed away from the section is considered positive, and that directed towards the section is considered negative.

Normal stresses arise when, under the influence of external forces, particles located on both sides of the section tend to move away from one another or move closer together. Shear stresses arise when particles tend to move relative to each other in the section plane.

The tangential stress can be decomposed along the coordinate axes into two components and (Fig. c) The first index at shows which axis is perpendicular to the section, the second - parallel to which axis the stress acts. If the direction of the shear stress is not important in the calculations, it is indicated without subscripts.

Example 4.1. Determine the normal and tangential stresses at point K of the rectangular section of the beam (6x14 cm), if the bending moment in this section M x=–40kNm=–40 kNcm, and the shear force is 20 kN.

Solution. Moment of inertia of a rectangular cross-section about the main central axis x.

J x = = =1372 cm 4 . .

Let's point the y-axis down. The coordinate of point K is at k = –4cm.

The normal voltage at point K will be equal to

=116.6 MPa.

The tangential stress at point K is calculated using the Zhuravsky formula.

The static moment of the cut-off part of the cross-sectional area is equal to

The section width at level K is equal to b(y)= 6cm.

Let us determine the shear stress at point K.

=2.4 MPa.

Example 4.2. Determine the maximum normal tensile and maximum shear stress in a beam of circular cross-section, if in the section M x = 80 kNm = 80 10 3 kNcm, Q = 60 kN.

Section diameter d=14 cm.

Solution. The greatest tensile normal stress occurs in the lower fiber of the stretched section zone, i.e. in the fiber farthest from the neutral axis X, and is determined by the formula

The highest shear stresses occur at cross-sectional points at the level of the neutral axis X, where all tangential stresses are parallel to the shear force, and they can be determined using the Zhuravsky formula.

The cross-sectional area is A= = =153.56 cm 2 .

The moment of resistance of the section is equal to W x = = 269.26 cm 3 .

Let us determine the value of the stretching maximum normal

voltage

=14.86 =148.6 MPa.

Let us determine the value of the greatest tangential stress

=0.52 =5.2 MPa.

Example 4.3. Determine the normal and shear stresses at point K at the level where the wall adjoins the flanges of the steel I-beam (I30), as well as the highest normal and shear stresses, if M x=50 kNm=50 10 2 kNcm, Q=30 kN.

Solution. From the assortment of I-beams, we write out the necessary data for I-beam I30.

h= 300mm=30cm, b=135mm=13.5cm, d=6.5cm=0.65cm,

t=10.2 mm=1.02 cm.

Sectional area A = 46.5 cm 2, moment of inertia J x= 7080 cm 4, moment of resistance W x= 472 cm 3.

Let us determine the value of the static moment of the cross-sectional area of ​​the flange relative to the neutral axis X.

= 199.53 cm3.

At the level where the wall adjoins the flanges, shear stresses

As is already known, external concentrated (i.e., applied at a point) loads do not really exist. They represent the static equivalent of a distributed load.

Similarly, concentrated internal forces and moments that characterize the interaction between individual parts of an element (or between individual structural elements) are also only the static equivalent of internal forces distributed over the cross-sectional area.

These forces, as well as external loads distributed over the surface, are characterized by their intensity, which is equal to

where is the resultant of internal forces on a very small area of ​​​​the section drawn (Fig. 7.1, a).

Let us decompose the force into two components: tangent AT and normal, of which the first is located in the section plane, and the second is perpendicular to this plane.

The intensity of tangential forces at the cross-section point under consideration is called shear stress and is denoted (tau), and the intensity of normal forces is called normal stress and is denoted (sigma). The stresses are expressed by the formulas

Stresses have dimensions, etc.

Normal and shear stress are components of the total stress at the point under consideration along a given section (Fig. 7.1, b). It's obvious that

The normal stress at a given point along a certain section characterizes the intensity of the forces of separation or compression of particles of a structural element located on both sides of this section, and the shear stress is the intensity of the forces that shift these particles in the plane of the section under consideration. The magnitudes of stresses a and at each point of the element depend on the direction of the section drawn through this point.

The set of stresses acting along various areas passing through the point under consideration represents the state of stress at this point.

Normal and tangential stresses are very important in the resistance of materials, since the strength of a structure depends on their values.

Normal and tangential stresses in each cross section of the beam are related by certain relationships to the internal forces acting in this section. To obtain such dependencies, consider an elementary area of ​​the cross section F of a beam with normal a and tangential stresses acting along this area (Fig. 8.1). Let us decompose the stresses into components parallel to the y and axis, respectively. The platform is acted upon by elementary forces parallel to the axes, respectively. The projections of all elementary forces (acting on all elementary sites of section F) onto the axes and their moments relative to these axes are determined by the expressions