Task 4.1.1: The set of stresses arising on the set of sites passing through the point under consideration is called ...
2) full voltage;
3) normal voltage;
4) shear stress.
Solution:
1) The answer is correct. The stress state at a point is completely determined by six components of the stress tensor: σ x, σ y, σ z, τ xy, τ yz, τ zx... Knowing these components, it is possible to determine the stresses at any site passing through a given point. The set of stresses acting on a set of areas (sections) passing through a given point is called the stress state at a point.
2) The answer is wrong! Ignorance of the definition of the total stress at a point (force per unit of cross-sectional area).
3) The answer is wrong! Recall that the projection of the total stress vector onto the normal to the section is called the normal stress.
4) The answer is wrong! An error was made in the definition of the term "shear stress".
The projection of the total stress vector onto an axis lying in the section plane is called shear stress.
Task 4.1.2: Areas at the investigated point of a stressed body, on which the shear stresses are zero, are called ...
1) oriented; 2) the main sites;
Solution:
1) The answer is wrong! The term does not meet the specified condition. Oriented sites are understood as sites that pass through a point in a predetermined direction.
2) The answer is correct.
When the elementary volume 1 is rotated, one can find its spatial orientation 2, at which the tangential stresses on its faces disappear and only normal stresses remain (some of them may be equal to zero). The areas (faces) on which the shear stresses are equal to zero are called the main areas.
3) The answer is wrong! The term does not meet the specified condition. Areas equally inclined to the main ones are called octahedral. The tangential stresses at the octahedral areas are not zero.
4) The answer is wrong! We remind you that secants are understood as areas drawn through the point at which the stress state is investigated.
Task 4.1.3: The main stresses for the stress state shown in the figure are ... (The voltage values are indicated in MPa).
1) σ 1 = 150 MPa, σ 2 = 50 MPa; 2) σ 1 = 0 MPa, σ 2 = 50 MPa, σ 3 = 150 MPa;
3) σ 1 = 150 MPa, σ 2 = 50 MPa, σ 3 = 0 MPa;
4) σ 1 = 100 MPa, σ 2 = 100 MPa, σ 3 = 0 MPa;
Solution:
1) The answer is wrong! The value of the main stress σ 3 = 0 MPa is not indicated.
2) The answer is wrong! The designations of the main stresses do not correspond to the numbering rule.
3) The answer is correct. One face of the element is free from shear stresses. Therefore, this is the main site, and the normal stress (main stress) at this site is also zero.
To determine the other two values of the principal stresses, we use the formula
,
where the positive directions of the stresses are shown in the figure.
For the given example, we have,,. After transformations, we find
In accordance with the rule for numbering the principal stresses, we have,,, i.e. flat stress state.
4) The answer is wrong! These are not the principal stresses, but the set values of the normal stresses acting on the highlighted element.
Task 4.1.4: At the investigated point of the stressed body on three main sites, the values of normal stresses are determined: The main stresses in this case are equal to ...
1) σ 1 = 150 MPa, σ 2 = 50 MPa, σ 3 = -100 MPa;
2) σ 1 = 150 MPa, σ 2 = -100 MPa, σ 3 = 50 MPa;
3) σ 1 = 50 MPa, σ 2 = -100 MPa, σ 3 = 150 MPa;
4) σ 1 = -100 MPa, σ 2 = 50 MPa, σ 3 = 150 MPa;
Solution:
1) The answer is correct. The main stresses are assigned indices 1, 2, 3 so that the condition is met. Hence,
2), 3), 4) The answer is wrong! The main stresses are assigned indices 1, 2, 3 so that the condition is fulfilled (in the algebraic sense).
Task 4.1.5: On the faces of the elementary volume (see figure), the values of stresses in MPa... The angle between the positive direction of the axis x and the outer normal to the main site, on which the minimum principal stress acts, is ...
1) ; 2) ; 3) ; 4) .
Solution:
1), 2), 4) The answer is wrong! Apparently, the formula for determining the angle is incorrectly written. Correct entry:
3) The answer is correct.
The angle is determined by the formula
Substituting the numerical values of the stresses, we get Since the angle is negative, we postpone the angle clockwise.
Task 4.1.6: The values of the principal stresses are determined from the solution of the cubic equation Coefficients, are called ...
1) invariants of the stress state; 2) elastic constants;
4) proportionality coefficients.
Solution:
1) The answer is correct. The roots of the equation - the principal stresses - are determined by the nature of the stress state at a point and do not depend on the choice of the initial coordinate system. Therefore, when the coordinate system is rotated, the coefficients
must remain unchanged. They are called stress state invariants.
2) The answer is wrong! Error in the definition of the term. Elastic constants characterize material properties.
3) The answer is wrong! Recall that the direction cosines are the cosines of the angles that the normal forms with the coordinate axes.
4) The answer is wrong! The term does not match the condition of the question
Through any point of a stressed body, as a rule, _____________ mutually perpendicular areas (s) can be drawn, on which the shear stresses will be equal to zero.
| three | |||
| two | |||
| four | |||
| six |
Solution:
The figure shows a body loaded by external forces and an elementary volume with stresses on its faces. With a mental rotation of the elementary volume, one can find such its spatial orientation, at which the tangential stresses on the faces will be equal to zero. These faces will be the main sites.
Topic: Stress at a point. Major sites and major stresses
The main axes of the stress state are called ...
Solution:
The figure shows an elementary volume selected in the vicinity of an arbitrary point of the loaded body. If for a given orientation of the elementary volume the tangential stresses on its faces are equal to zero, then the axes x, y, z are called the principal stress axes. When moving from one point to another, the directions of the main axes generally change.
Topic: Stress at a point. Major sites and major stresses
Normal stresses acting at main sites are called ...
Solution:
Three mutually perpendicular pads that do not have shear stresses are called main pads. The normal stresses acting on the main sites are called principal stresses. The maximum of the three main stresses is simultaneously the highest total stress acting on the set of sites passing through a given point. The minimum of the three principal stresses is the smallest of the many total stresses.
Topic: Stress at a point. Major sites and major stresses
The stress state of the elementary volume, shown in the figure, is flat. The upper edge of the elementary volume is the main platform. The position of the other two main pads is determined by the angle
Solution:
The figure shows an elementary volume (top view). The direction of the normal to the main site is determined by the formula where is the angle between the positive direction of the axis x and normal to one of the main sites. For our case Substituting these values into the formula, we obtain whence a
Topic: Stress at a point. Major sites and major stresses
The figure shows a bar tensioned by forces F, and an elementary volume selected by faces parallel to the planes of the bar. When the elementary volume is rotated around the axis " u"At an angle equal to 45 0, the stress state ...
Solution:
In the figure, the elementary volume is highlighted by the main areas. Main voltages: Stress state - linear. The type of stress state does not depend on the spatial orientation of the elementary volume and remains linear at any angle of rotation.
4.2. Types of stress
Task 4.2.1: Round bar diameter d undergoes pure bending and torsion deformation. Stress state at a point V shown in the picture ...
1) ; 2) ; 3) ; 4) .
Solution:
1) The answer is wrong! The torque causes the appearance of shear stresses in the plane perpendicular to the axis of the bar.
2) The answer is wrong! Direction of shear stress at a point V cross-section must correspond to the direction of the torque in this section.
3) The answer is correct. Select a volumetric element with cutting planes oriented along and across the axis of the bar. A bending moment acts in the cross section of the bar at the embedment M and torque 2M... From bending moment M at the point V normal tensile stress occurs. Torque 2M acting in a plane perpendicular to the axis of the bar causes shear stress. The direction of the shear stress must be consistent with the direction of the torque. Therefore, the stressed state of the element in Figure 4 corresponds to the stressed state at the point V.
4) The answer is wrong! From torque to point V cross-section, shear stress occurs. The direction of the shear stress must be consistent with the direction of the torque.
Task 4.2.2: The rod undergoes tensile deformations and pure bending. The stressful state that occurs at a dangerous point is called ...
1) flat; 2) voluminous; 3) linear; 4) pure shear.
Solution:
1) The answer is wrong! In a plane stress state, one value of the principal stress is equal to zero.
2) The answer is wrong! At the dangerous point, only one main voltage is different from zero. In the bulk stress state, the three main stresses are nonzero.
3) The answer is correct. The danger points are located infinitely close to the upper edge of the element. Only tensile normal stresses from longitudinal force and bending moment arise in them. Plots of stress distribution from each internal force factor and the resulting plot are shown in the figure.
Consequently, there will be a linear stress state at the dangerous point.
4) The answer is wrong! In pure shear, the two principal stresses are equal but opposite in sign, and the third is zero.
Task 4.2.3: The stress state "pure shear" is shown in the figure ...
1) ; 2) ; 3) ; 4) .
Solution:
1) The answer is wrong! The figure shows a plane stress state - biaxial tension.
2) The answer is wrong! The element is in a flat stress state - a biaxial mixed stress state.
3) The answer is correct.
Pure shear is a stressed state when only shear stresses act on the faces of the selected elementary volume. If the elementary volume is rotated by an angle equal to, then the tangential stresses on its faces (areas) will be equal to zero, but normal (principal) stresses and will appear. Thus, pure shear can be realized by stretching and compression in two mutually perpendicular directions with stresses equal in absolute value.
Therefore, the stress state "pure shear" is shown in Figure 3.
4) The answer is wrong! This element experiences a linear stress state.
Task 4.2.4: The type of stress state shown in the figure is called ...
1) linear; 2) flat; 3) voluminous; 4) pure shear.
Solution:
1) The answer is correct. The type of stress state is determined depending on the values of the principal stresses. In the example, one face is free of shear stresses - this is the main pad. The normal voltage acting on the main site is called the main voltage. In this case, it is equal to zero. Using the formula, we find two other principal stresses. After transformations we get,. In accordance with the accepted designations, we have,. The two main voltages are zero. Therefore, the figure shows a linear stress state.
2) The answer is wrong! In a plane stress state, one principal stress is zero. In this case, the two main voltages are zero.
3) The answer is wrong! In the bulk stress state In this case, the two principal stresses are equal to zero. Therefore, this stress state is not volumetric.
4) The answer is wrong! With a pure shift,. Calculations show that this is not true for this case.
Task 4.2.5: The stress state at values,, is called ...
1) voluminous; 2) pure shear; 3) flat; 4) linear.
Solution:
1) The answer is wrong! In the bulk stress state, all three principal stresses are nonzero.
2) The answer is wrong! With pure shear, one value of the principal stress is zero, and the other two are equal in magnitude, but opposite in sign.
3) The answer is correct. The type of stress state is determined by the values of the principal stresses. In the case when all three principal stresses are different from zero, we have a volumetric stress state. If one principal stress is equal to zero, it is a plane stress state, and when two are equal to zero, it is linear. Therefore, in this example, there will be a plane stress state.
4) The answer is wrong! In a linear stress state, only one principal voltage is nonzero.
Task 4.2.6: On the faces of the elementary volume (see figure), the stresses are set in MPa... The stress state at the point ...
1) linear; 2) flat (pure shear); 3) flat; 4) volumetric.
Solution:
1) The answer is wrong! The front face of the elementary volume is free from shear stresses. This means that this face is the main site and one of the three main stresses is (-50 MPa). Determine the other two main stresses by the formula
2) The answer is wrong! Recall that for pure shear, one of the principal stresses is zero. The other two are equal in absolute value and opposite in sign.
3) The answer is correct. The front face of the elementary volume is free from shear stresses. This means that it is the main site and one of the three main stresses is (-50 MPa). The other two main stresses are determined by the formula
By supplying numerical values, we get
Assigning indices to the main stresses, we have:
Thus, the stress state is flat (biaxial compression).
4) The answer is wrong! The front face of the elementary volume is free from shear stresses. This means that this face is the main site and one of the three main stresses is (-50 MPa). The other two main stresses can be determined by the formula
The calculation results will show which stress state is shown in the figure.
The stress state of the elementary volume, shown in the figure, is - ...
Solution:
Principal stresses are the roots of the cubic equation
where:
In our case, and the cubic equation takes the form whence
Thus, the stress state of the elementary volume is linear (uniaxial tension).
Topic: Types of stress
The steel cube is inserted into a rigid cage without a gap (see fig.). A uniformly distributed intensity pressure acts on the top face of the cube R... The surfaces of the cube and holder are absolutely smooth. The stress state of the cube is shown in the figure ...
| v | |||
| G | |||
| b | |||
| a |
Solution:
There are no friction forces between absolutely smooth surfaces of the cube and the holder. Therefore, the shear stresses on the faces of the cube are equal to zero, and all faces are the main areas. In the process of compression, the edges of the cube directed along the axes x and y tend to lengthen. Axial elongation y happens freely. Axial elongation x impossible (rigid clip interferes). Due to the impossibility of lengthening along the axis x, from the side of the vertical planes of the holder, efforts act on the cube in the form of loads evenly distributed over the area with a certain intensity. Intensity R and should be considered as the main stresses. Thus, there is one of the three main stresses (along the front face of the cube). Therefore, the stress state of the cube is flat (Fig. v).
Topic: Types of stress
The figure shows a bar in tension with torsion. Stress state at a point TO is an - …
Solution:
At the point TO the cross-section is under normal stress due to the force F... The diagram of the shear stresses from the torque is shown in Figure 1. At the corner points Therefore, the stress state at the point TO- linear (uniaxial tension, Fig. 2).
Topic: Types of stress
The stress state of the elementary volume is - ...
Solution:
The upper edge of the elementary volume is the main area, therefore one principal stress is equal to the other two principal stresses are calculated by the formula
In this case (see Fig.) Substituting into the formula, we obtain
By assigning the corresponding indices to the principal stresses, we obtain
The tense state is volumetric.
Topic: Types of stress
The body is acted upon by pressure evenly distributed over the surface R(see fig.). The stress state of the elementary volume is - ...
Solution:
If the body is acted upon by a pressure uniformly distributed over the surface R(see Fig.), then the stress state at any point of the body is volumetric (triaxial compression). Moreover, for any spatial orientation of the elementary volume.
Stressed and deformed states of an elastic body. Relationship between stresses and strains
The concept of body tension at a given point. Normal and shear stresses
Internal force factors arising during loading of an elastic body characterize the state of a particular section of the body, but do not give an answer to the question of which point of the cross section is the most loaded, or, as they say, dangerous point... Therefore, it is necessary to introduce into consideration some additional quantity that characterizes the state of the body at a given point.
If the body, to which external forces are applied, is in equilibrium, then internal resistance forces arise in any of its sections. Let us denote through the internal force acting on an elementary area, and the normal to this area through then the value
 | (3.1) |
called total voltage.
In the general case, the total stress does not coincide in the direction with the normal to the elementary area, therefore it is more convenient to operate with its components along the coordinate axes -
If the outward normal coincides with any coordinate axis, for example, with the axis NS, then the stress components will take the form and 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.
Let us select from the body under the action of external forces an infinitesimal parallelepiped, whose faces are parallel to the coordinate planes, and the edges have length. On each face of such an elementary parallelepiped, three stress components act, parallel to the coordinate axes. In total, on six faces, we obtain 18 stress components.
Normal stresses are denoted in the form, where the index denotes the normal to the corresponding face (i.e., can take values). Shear stresses are; here the first index corresponds to the normal to the area on which the given shear stress acts, and the second indicates the axis parallel to which this stress is directed (Figure 3.1).
Figure 3.1. Normal and shear stresses
For these voltages, the following is accepted. rule of signs. Normal voltage is considered positive when stretched, or, which is the same, when it coincides with the direction of the outer normal to the site 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.
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 infinitely small distance from the one under consideration, the voltage, up to infinitesimal first order, can be expanded in a Taylor series:
For pads that are parallel to the plane, only the coordinate changes NS, and the increments Therefore, on the face of the parallelepiped, coinciding with the plane, the normal stress will be, and on the parallel face, spaced at an infinitely small distance, - 
The stresses on the remaining parallel faces of the parallelepiped are related in the same way. Therefore, out of 18 voltage components, only nine are unknown.
In the theory of elasticity, the law is proved pairs of shear stresses, according to which, along two mutually perpendicular areas, the shear stress components perpendicular to the intersection lines of these areas are equal to each other:
It can be shown that stresses (3.3) not only characterize the stress state of the body at a given point, but define it unambiguously. The combination of these stresses forms a symmetric matrix, which is called stress tensor:
 | (3.4) |
Since each point will have its own stress tensor, the body has field stress tensors.
When a tensor is multiplied by a scalar value, a new tensor is obtained, all components of which are times larger than the components of the original tensor.
Earlier, for simplicity and clarity, we considered an ordinary wooden ruler as a beam, which made it possible, with known assumptions, to derive the basic equations and formulas for calculating the bearing capacity of a beam. Using these equations, we plotted shear force "Q" and bending moment diagrams "M".
Figure 149.2.1... Diagrams of shear forces and bending moments acting in cross-sections of a beam under a concentrated load.
As a result, it made it possible to quite simply and clearly determine the value of the maximum bending moment and, accordingly, the value of the maximum normal tensile and compressive stresses arising in the most loaded cross-section of the beam.
Further, knowing the design resistance of the material of the beam (the values of the design resistances are carried out in the corresponding SNiPs), it is quite easy to determine the moment of resistance of the cross section, and then other parameters of the beam, the height and width, if the beam is rectangular, the diameter, if the beam is circular, the number according to the assortment, if the beam is made of a metal hot-rolled profile.
Such a strength calculation is a calculation for the first group of limiting states and allows you to determine the maximum permissible load that the calculated structure can withstand. Exceeding the maximum permissible load will lead to structural failure. In this case, we are not interested in how exactly the structure will collapse, since this site is not devoted to theoretical and practical studies of the limiting states of materials, but only to some methods of calculating the most common building structures.
As a rule, engineering calculations of structures, which will be used in hundreds of tons and tens of cubic meters, are performed in such a way as to obtain the maximum loaded structure. Therefore, such calculations are quite complex and various kinds of coefficients, taking into account the service life of the structure, the nature of the loads, the cyclicity, the dynamism of the loads, the inhomogeneity of the material used, etc. - tens. This is logical, since with gross production, each percentage ultimately yields tangible savings. In private construction, carried out once, the strength of the structure, even with a two-fold margin, is much more important than the possible saving of materials and therefore the calculations for private low-rise construction can be simplified as much as possible using only one correction factor γ = 1.6 ÷ 2, if this coefficient is multiplied stress values, or γ = 0.5 ÷ 0.7, if the value of the design resistance will be multiplied by this factor. However, even such simple calculations are not limited to this.
Any beam with a length significantly greater than the height of the cross-section, which is a bar, will deform under the action of loads. The deformation results in the displacement of the central axis of the beam along the axis at about the axis NS , in other words, the deflection, as well as the rotation of the cross-sections of the beam relative to the plane of the cross-section. And these same deflections and angles of rotation, regardless of what supports the beam and what loads act on it, can also be determined. To determine the maximum angle of rotation and maximum deflection, the corresponding plots are also built, allowing you to determine which cross-section will move as a result of the deflection and which will be inclined the most.
Figure 174.5.6... Diagram of the angles of rotation under the action of a concentrated load in the middle of the beam
The diagram of the deflections is not shown here, but oddly enough, this is the simplest diagram showing the position of the axis passing through the cross-sections of the beam as a result of deformation, and this diagram can be personally observed on any sufficiently deflected beam or any other structure. Knowing the modulus of elasticity of the beam material and the moment of inertia of the cross section, it is also not very difficult to determine the maximum deflection. To simplify the solution of these problems as much as possible, the design schemes for beams, to which, depending on the nature of the supports and the type of loading, are given the corresponding formulas.
Such a calculation of deformations is a calculation for the limiting states of the second group and quite clearly shows how much the beam will bend. This is sometimes important not only due to technological constraints, for example for crane girders, but also for aesthetic reasons. For example, when the ceiling, or rather the ceiling, although strong enough, bends noticeably, there is little pleasant in this. The maximum allowable deflections for various building structures are given in SNiP 2.01.07-85 "Loads and Impacts" (in its updated version). However, when calculating for oneself, no one forbids using even smaller deflection values.
Here the reader may have a quite reasonable question, why did you need to build a diagram of shear stresses "Q", if this diagram is not involved in any calculations. Well, it's time to answer this question.
The fact is that the calculation of various kinds of beams, especially of a constant rectangular section, lying horizontally, for strength under the action of tangential stresses is very rarely decisive, in contrast to the above calculations. Nevertheless, it is still necessary to know what shear stresses are and how they affect the operation of a structure, even if very simplified.
As follows from the definition, tangential stresses act in the plane of the cross-section, as if touching the cross-section, therefore they are called tangents. At first glance, it is easy to determine the value of shear stresses: it is enough to divide the value of the transverse force (for this we need the "Q" diagram) by the cross-sectional area (in our example, the transverse forces acted only along the axis at and then this will be enough for us, we will always have time to complicate any calculation):
T= Q / F = Q / (bh) (270.1)
As a result, we can plot the shear stress " τ "(in addition to the normal stresses" σ ") of the following form:
Figure 270.1... Preliminary diagram of shear stresses " τ "
However, such a diagram of shear stresses would be valid for some abstract material with linear elasticity along the axis at , and absolutely rigid along the axis z , as a result of which there is no redistribution of stresses in the cross section of such a material and there is only one type of deformation about the axis at ... In fact, any body with isotropic properties, under the action of loads, tries to preserve its volume, which means that the section we are considering is trying to preserve its area. A good example is when you sit on a ball, its height decreases under the influence of your weight, but its width increases. Moreover, this process is not linear. If you cut a cube or parallelepiped from the dough, and then press on it, then the side faces will become convex, a similar process occurs in laboratory compression tests of samples of metal or other materials.
Among other things, this also means that the shear stresses acting along the axis at , cause the appearance of shear stresses along the axis z and a diagram of shear stresses along the axis z will more clearly show the change in shear stresses in relation to the height of the beam. In this case, the shape of the diagram will resemble the side face of a flattened dough cube, and the area of the diagram, of course, will not change. Those. the values of the shear stress diagram at the very bottom and at the very top of the cross section will be zero, and the maximum value (for a rectangular section) will be in the middle of the section height and clearly greater than Q / F. Based on the condition of equality of the areas of the diagrams, the maximum value of the shear stress diagram cannot be more than 2Q / F, and even then only if the diagram is two triangles, and in this case the maximum value is the height of the triangles. However, as we have already found out, the diagram looks more like a part of a circle or a parabola, i.e. the value of the maximum shear stress will be about 1.5Q / F:
Figure 270.2... More accurate shear stress diagram.
The gray line shows the diagram of tangential stresses that we previously adopted, but now the tangential stresses are directed along the axis z .
Mathematically, the change in shear stresses depending on the height of the section can be expressed through the change in the static moment of the cut-off part of the section, taking into account the change in the width of the section, since the beams do not always have a rectangular section shape. As a result, the formula for determining the shear stresses (the derivation of the formula is not given here) has the following form:
T= Q y S z ab / bI z(270.2) - the formula of prof. D. I. Zhuravsky
where Q y- the value of the shear force in the considered cross-section, is determined by the diagram "Q"
S z ex is the static moment of the cut-off part of the section at the considered height relative to the axis z ... It is defined as the area of the cut-off section multiplied by the distance between the center of gravity of the entire section and the center of gravity of the cut-off section. For example, at the very bottom of the cross section, i.e. at a height of h = 0, the area of the cut-off part of the section will also be equal to 0, which means that the shear stresses acting along the width b of the cross section will also be equal to zero. For a section passing through the center of gravity of the cross section, i.e. with the height of the cut-off part of the section equal to h / 2, the static moment will be (bh / 2) (h / 4) = bh 2/8. When the height of the cut-off section is equal to the height of the cross-section, the static moment will be zero, since the center of gravity of the cut-off part of the section in this case will coincide with the center of gravity of the section.
b- the width of the cross-section at the considered height of the cross-section. For rectangular beams, the section width is constant, but there are beams of round, tee, I-beams and any other section. Moreover, the determination of shear stresses is most often used when calculating beams of a non-rectangular section, since when the section passes from the shelves to the wall, a significant jump in shear stresses appears due to the change in the width of the section, and the transition from the shelves to the wall usually occurs at such a height, where the normal stresses are large enough and this is taken into account by the appropriate calculation.
I z- moment of inertia of the cross section about the axis z ... In this case, the only more or less constant value. For a rectangular cross-section, the moment of inertia is bh 3/12.
Thus, according to formula (270.2), the maximum value of shear stresses will be:
T= 12Qbh 2 / (8b 2 h 3) = 1.5Q / F (270.3)
Geometry gave us the same result.
And further. For materials with pronounced anisotropic properties, for example, for wood, a shear stress test is necessary. The fact is that the strength of wood in compression along the grain and the strength of wood in compression across the grain are completely different things. Therefore, the check is performed for cross-sections in which the shear stresses are maximum, as a rule, these are sections on the beam supports (with a uniformly distributed load). In this case, the obtained value of the shear stresses is compared with the value of the design resistance of the wood to compression or crushing across the fibers - R c90.
However, there is another approach to the issue of determining the shear stresses: under the action of loads, the beam is deformed, while the maximum normal compressive and tensile stresses arise at the very bottom and at the very top of the beam cross-section, which can be seen from the diagram "σ" in Fig. 270.1 ...
In this case, between the fibers of such an inhomogeneous material as wood, as well as between the layers of any other material, shear stresses arise, now directed along the axis NS , i.e. along the same axis as the normal compressive and shear stresses arising from the bending moment.
This is due to the fact that each layer under consideration undergoes normal loads of different values, and as a result of the same redistribution of stresses, shear stresses arise. These shear stresses are trying to split the beam into separate layers, each of which will act as a separate beam.
The difference in bearing capacity between individual layers and a solid beam is obvious. For example, if you take a pack of paper of at least 500 sheets, then bending such a pack is a piece of cake, but if you glue all the sheets together, i.e. layers of the beam between themselves, then we will get a solid beam and now it will be much more difficult to bend it. But between the glued sheets, the same, relatively speaking, normal shear stresses will arise. However, the value of normal shear stresses is determined in the same way, and the same shear force, determined from the "Q" diagram, is involved in the calculations. That's just not the cut off, but the sheared part of the section is considered, respectively, the static moment can be denoted - S z sc... In this case, the obtained value of the shear stresses is compared with the value of the design resistance of wood to a chip along the fibers - R cк.
True, the values R c90 and R cк for wood have the same meaning, but nevertheless, shear stresses from the action of transverse forces and from deformations as a result of deflection are usually distinguished (since two main stress areas are considered perpendicular to each other), and the direction of action of shear stresses is important in determining the total stress in the investigated point of the body.
However, all this is nothing more than general concepts of shear stresses. In real materials, the process of stress redistribution is much more complicated, all because even a metal can be attributed to isotropic materials rather conditionally. However, these issues are considered by a separate scientific discipline - the theory of elasticity. When calculating building structures that are rods - beams or plates - slabs the size of a room, it is quite possible to use the formula (270.2), derived on the basis of the general provisions of the linear theory of elasticity. When calculating massive bodies, the methods of the nonlinear theory of elasticity should be used.
Voltage is a vector, and as any vector can be represented by normal (in relation to the site) and tangential components (Fig. 2.3). The normal component of the stress vector will be denoted by the tangent. Experimental studies have established that the effect of normal and tangential stresses on the strength of the material is different, and therefore in the future it will be necessary to always separately consider the components of the stress vector.
Rice. 2.3. Normal and shear stresses in the site
Rice. 2.4. Shear stress when shearing a bolt
When a bolt is stretched (see Fig. 2.2), normal stress acts in the cross section
When the bolt works for shear (Fig. 2.4), a force must appear in the section P, balancing the force.
It follows from the equilibrium conditions that
In fact, the latter ratio determines a certain average stress over the section, which is sometimes used for approximate estimates of strength. In fig. 2.4 shows the view of the bolt after exposure to significant forces. The bolt began to fracture, and one half of it shifted relative to the other: shear or shear deformation occurred.
Examples of determination of stresses in structural members.
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 section method from the static equations (equilibrium equations).
Torsion of a thin-walled round shaft.
A thin-walled circular shaft (tube) transmits torque (for example, from an aircraft engine to a propeller). It is required to determine the stresses in the cross section of the shaft (Fig. 2.5, a). Let's draw the plane of the section 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 wall thickness, 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 a very small wall thickness, 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 the torque, and therefore there should be no normal stresses in the cross section. The torque is balanced by shear stresses, the moment of which is
From the last relation we find the shear stress in the section of the shaft:
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 vessel lid creates an increase in
This force is balanced by the forces arising in the cross section of the shell, and the intensity - of the indicated forces - stress - will be equal to
The thickness of the shell 5 is assumed to be small in comparison with the average radius, the stresses are assumed to be uniformly distributed at all points of the cross section (Fig. 2.6, b).
However, not only stresses in the longitudinal direction act on the pipe material, but also circumferential (or ring) stresses in the perpendicular direction. To identify them, we select a ring of length I with two sections (Fig. 2.7), and then draw a diametrical section that separates half of the ring.
In fig. 2.7, a shows the stresses on the section surfaces. The pressure on the inner surface of the pipe with a radius
Rice. 2.8. Crack in a cylindrical shell under the action of destructive internal pressure
As is already known, external concentrated (i.e., applied at a point) loads do not really exist. They represent the static equivalent of the distributed load.
Similarly concentrated internal forces and moments characterizing the interaction between separate parts element (or between separate elements structures), are also only a static equivalent of internal forces distributed over the 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 drawn section (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 the tangential forces at the considered point of the section is called the tangential stress and is denoted by (tau), and the intensity of the normal forces is denoted by the normal stress and is denoted by (sigma). Stresses are expressed by the formulas
Stresses have dimensions, etc.
Normal and shear stresses are the components of the total stress at the point under consideration along the 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 the particles of the structural element located on both sides of this section, and the shear stress - the intensity of the forces that shift these particles in the plane of the section under consideration. The magnitudes of the 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 on different sites passing through the point under consideration represents the stress state at this point.
Normal and shear stresses are very important in the resistance of materials, since the strength of the structure depends on their values.
Normal and tangential stresses in each cross-section of the bar are related by certain dependencies with the internal forces acting in this section. To obtain such dependencies, consider an elementary area of the cross-section F of a bar with normal a and shear stresses acting on this area (Fig. 8.1). Let us decompose the stresses into components parallel to the y and axes, respectively. Elementary forces act on the site, parallel to the axes, respectively.
