What can be cooked from squid: quick and tasty

Axial (central) tension or compression a straight bar is caused by external forces, the vector of the resultant of which coincides with the axis of the bar. Under tension or compression, only longitudinal forces N arise in the cross-sections of the bar. The longitudinal force N in a certain section is equal to the algebraic sum of the projection onto the axis of the bar of all external forces acting on one side of the section under consideration. According to the rule of signs of the longitudinal force N, it is generally accepted that positive longitudinal forces N arise from tensile external loads, and longitudinal forces N from compressive ones are negative (Fig. 5).

To identify areas of a bar or its section, where longitudinal force It has greatest value, plot the longitudinal forces using the section method, discussed in detail in the article:
Analysis of internal force factors in statistically definable systems
I also highly recommend taking a look at the article:
Calculation of a statistically definable bar
If you disassemble the theory in this article and the tasks by reference, then you will become a guru in the topic "Stretch-compression" =)

Tensile-compressive stresses.

The longitudinal force N determined by the method of sections is the resultant of internal forces distributed over the cross section of the bar (Fig. 2, b). Based on the definition of stresses, according to expression (1), it is possible to write for the longitudinal force:

where σ is the normal stress at an arbitrary point cross section rod.
To determine normal stresses at any point of the bar, it is necessary to know the law of their distribution over the cross-section of the bar. Experimental research show: if a number of mutually perpendicular lines are applied to the surface of the rod, then after the application of an external tensile load, the transverse lines do not bend and remain parallel to each other (Fig. 6, a). This phenomenon is indicated by flat hypothesis(Bernoulli's hypothesis): sections that are flat before deformation remain flat after deformation.

Since all longitudinal fibers of the bar are deformed in the same way, the stresses in the cross-section are the same, and the stress diagram σ along the height of the cross-section of the bar looks as shown in Fig. 6, b. It can be seen that the stresses are uniformly distributed over the cross section of the bar, i.e. at all points of the section σ = const. Expression to define voltage magnitudes looks like:

Thus, the normal stresses arising in the cross-sections of a stretched or compressed beam are equal to the ratio of the longitudinal force to its cross-sectional area. Normal stresses are considered to be positive in tension and negative in compression.

Tensile-compressive deformations.

Consider the deformations arising from tension (compression) of the bar (Fig. 6, a). Under the action of force F, the bar is lengthened by a certain amount Δl, called absolute elongation, or absolute longitudinal deformation, which is numerically equal to the difference between the length of the bar after deformation l 1 and its length before deformation l

The ratio of the absolute longitudinal deformation timber Δl to its original length l is called the relative elongation, or relative longitudinal deformation:

In tension, the longitudinal deformation is positive, and in compression, it is negative. For most structural materials at the stage of elastic deformation, Hooke's law (4) is fulfilled, which establishes a linear relationship between stresses and strains:

where the modulus of longitudinal elasticity E, also called modulus of elasticity of the first kind is the coefficient of proportionality between stresses and strains. It characterizes the stiffness of the material in tension or compression (Table 1).

Table 1

Longitudinal modulus for various materials

Absolute transverse deformation of the bar is equal to the difference in cross-sectional dimensions after and before deformation:

Respectively, relative lateral deformation determined by the formula:

When stretched, the dimensions of the cross-section of the bar decrease, and ε "has a negative value. Experience has shown that, within the range of Hooke's law, when the bar is stretched, the transverse deformation is directly proportional to the longitudinal deformation. lateral deformationε "to longitudinal deformation ε is called the coefficient of transverse deformation, or Poisson's ratio μ:

It has been experimentally established that at the elastic stage of loading any material, the value μ = const and for various materials the values of Poisson's ratio are in the range from 0 to 0.5 (Table 2).

table 2

Poisson's ratio.

Absolute rod elongationΔl is directly proportional to the longitudinal force N:

This formula can be used to calculate the absolute elongation of a section of a bar of length l, provided that the value of the longitudinal force is constant within this section. In the case when the longitudinal force N varies within a section of the bar, Δl is determined by integration within this section:

The product (E A) is called section stiffness the rod under tension (compression).

Mechanical properties of materials.

The main mechanical properties of materials during their deformation are strength, plasticity, brittleness, elasticity and hardness.

Strength is the ability of a material to resist external forces without collapsing and without the appearance of permanent deformations.

Plasticity is the property of a material to withstand large permanent deformations without destruction. Deformations that do not disappear after the removal of external loads are called plastic deformations.

Brittleness is the property of a material to break down with very small residual deformations (for example, cast iron, concrete, glass).

Perfect elasticity- the property of the material (body) to completely restore its shape and size after eliminating the causes of deformation.

Hardness is the property of a material to resist the penetration of other bodies into it.

Consider a tensile diagram for a mild steel bar. Let a round bar of length l 0 and initial constant cross-section of area A 0 be statically stretched from both ends by force F.

The compression diagram of the rod has the form (Fig. 10, a)

where Δl = l - l 0 is the absolute elongation of the rod; ε = Δl / l 0 - relative longitudinal elongation of the rod; σ = F / A 0 - normal stress; E - Young's modulus; σ p - proportionality limit; σ yn - elastic limit; σ t is the yield point; σ in - ultimate strength (temporary resistance); ε residual - permanent deformation after removal of external loads. For materials that do not have a pronounced yield area, the conventional yield point σ 0.2 is introduced - the stress at which 0.2% residual deformation is achieved. When the ultimate strength is reached, a local thinning of its diameter (“neck”) occurs in the center of the bar. Further absolute elongation of the rod occurs in the area of the neck (area of local yield). When the stress reaches the yield point σ t glossy surface the rod becomes slightly dull - microcracks (Luders-Chernov lines) appear on its surface, directed at an angle of 45 ° to the axis of the rod.

Tensile and compressive strength and stiffness calculations.

The dangerous section under tension and compression is the cross-section of the bar in which the maximum normal stress occurs. Allowable stresses are calculated by the formula:

where σ limit is the ultimate stress (σ pre = σ t - for plastic materials and σ pre = σ in - for brittle materials); [n] - safety factor. For plastic materials [n] = = 1.2 ... 2.5; for brittle materials [n] = = 2 ... 5, and for wood [n] = 8 ÷ 12.

Calculations for tensile and compressive strength.

The purpose of calculating any structure is to use the results obtained to assess the suitability of this structure for operation with a minimum consumption of material, which is reflected in the methods for calculating strength and stiffness.

Strength condition the rod when it is stretched (compressed):

At design calculation the area of the dangerous section of the bar is determined:

In determining permissible load the permissible normal force is calculated:

Compression and tensile stiffness calculation.

Rod performance is determined by its ultimate deformation [l]. The absolute elongation of the bar must satisfy the condition:

Often, an additional calculation is made for the stiffness of individual sections of the bar.

Calculation of a round bar for strength and torsional rigidity

The purpose of the torsional strength and stiffness calculations is to determine such dimensions of the cross-section of the beam, at which the stresses and displacements will not exceed the specified values allowed by the operating conditions. The condition of strength in terms of permissible shear stresses is generally written as This condition means that the greatest shear stresses arising in a twisted bar should not exceed the corresponding permissible stresses for the material. The permissible torsional stress depends on 0 ─ the stress corresponding to the dangerous state of the material and the adopted safety factor n: ─ yield strength, nt is the safety factor for a plastic material; ─ ultimate strength, nb- safety factor for brittle material. Due to the fact that the values of β are more difficult to obtain in torsion experiments than in tension (compression), then, most often, the permissible torsional stresses are taken depending on the permissible tensile stresses for the same material. So for steel [for cast iron. When calculating the strength of twisted bars, three types of tasks are possible, differing in the form of using the strength conditions: 1) checking the stresses (verification calculation); 2) selection of a section (design calculation); 3) determination of the permissible load. 1. When checking stresses for given loads and dimensions of a bar, the largest tangential stresses arising in it are determined and compared with those specified by formula (2.16). If the strength condition is not met, then it is necessary either to increase the cross-sectional dimensions, or to reduce the load acting on the beam, or to use a material of higher strength. 2. When selecting a cross-section for a given load and a given value of the allowable stress from the strength condition (2.16), the value of the polar moment of resistance of the cross-section of the bar is determined. By the value of the polar moment of resistance, the diameters of the solid circular or annular cross-section of the bar are found. 3. When determining the permissible load for a given permissible voltage and polar moment of resistance WP, the permissible torque MK is preliminarily determined on the basis of (3.16) and then, using the torque diagram, a connection is established between K M and external torsion moments. Calculation of the bar for strength does not exclude the possibility of deformations that are unacceptable during its operation. Large angles of twisting of the bar are very dangerous, since they can lead to a violation of the accuracy of processing parts if this bar is a structural element of the processing machine, or torsional vibrations can occur if the bar transmits torsional moments that are variable in time, therefore, the bar must also be reckoned on for rigidity. The stiffness condition is written in the following form: where is the largest relative angle of twisting of the bar, determined from the expression (2.10) or (2.11). Then the stiffness condition for the shaft will take the form different types loads varies from 0.15 ° to 2 ° per 1 m of beam length. Both in the condition of strength and in the condition of rigidity when determining max or max , we will use the geometric characteristics: WP ─ polar moment of resistance and IP ─ polar moment of inertia. Obviously, these characteristics will be different for round solid and annular cross-sections with the same area of these sections. Through specific calculations, it can be verified that the polar moments of inertia and the moment of resistance for an annular section are much greater than for a solid circular section, since the annular section does not have areas close to the center. Therefore, a bar with an annular section during torsion is more economical than a bar with a solid circular section, that is, it requires less material consumption. However, the manufacture of such a bar is more complicated, and therefore more expensive, and this circumstance must also be taken into account when designing the bars operating in torsion. We will illustrate the method of calculating the bar for strength and torsional rigidity, as well as reasoning about efficiency, with an example. Example 2.2 Compare the weights of two shafts, the transverse dimensions of which should be selected for the same torque MK 600 Nm at the same permissible stresses 10 R and 13 Stretching along the grain p] 7 Rp 10 Compression and crushing along the grain [cm] 10 Rc, Rcm 13 Crushing across the fibers (at least 10 cm in length) [cm] 90 2.5 Rcm 90 3 Chipping along the fibers during bending [and] 2 Rck 2.4 Chipping along the fibers with notches 1 Rck 1.2 - 2.4 Chipping in the notches across fibers

The section stiffness is proportional to the elastic modulus E and the axial moment of inertia Jx, in other words, it is determined by the material, shape and dimensions of the cross section.
The section stiffness is proportional to the elastic modulus E and the axial moment of inertia Yx, in other words, it is determined by the material, shape and dimensions of the cross section.
The section stiffness is proportional to the elastic modulus E and the axial moment of inertia Jx; in other words, it is determined by the material, shape and cross-sectional dimensions.
The stiffness of the sections EJx of all frame elements is the same.
The cross-section stiffnesses of all frame elements are the same.
The stiffness of the section of the elements without cracks in these cases can be determined by the formula (192) as for the short-term action of temperature, taking vt - 1; the stiffness of the section of elements with cracks - according to formulas (207) and (210) as for the case of short-term heating.
The cross-section stiffnesses of the frame elements are the same.
Here El is the minimum bending stiffness of the bar section; G is the length of the rod; P - compressive force; a is the coefficient of linear expansion of the material; T is the heating temperature (the difference between the operating temperature and the temperature at which the movement of the ends of the rod was excluded); EF is the compression stiffness of the bar section; i / I / F is the minimum radius of gyration of the bar section.
If the stiffness of the frame section is constant, the solution is somewhat simplified.
When the stiffness of the sections of a structural element is continuously changing along its length, the displacements should be determined by direct (analytical) calculation of the Mohr integral. Such a structure can be calculated approximately by replacing it with a system with elements of step-variable stiffness, after which Vereshchagin's method can be used to determine the displacements.
Determination of the stiffness of sections with ribs by calculation is a difficult and, in some cases, impossible task. In this regard, the role of experimental data from testing full-scale structures or models increases.
A sharp change in the stiffness of the sections of the beams over a short length causes a significant concentration of stresses in the welded belt seams in the zone of curvilinear conjugation.

What is called the torsional stiffness of the section.
What is called the bending stiffness of the section.
What is called the torsional stiffness of the section.
What is called the bending stiffness of the section.
What is called the shear stiffness of the bar section.
EJ are called tensile bar section stiffnesses.
The product EF characterizes the stiffness of the section under the axial action of the force. Hooke's law (2.3) is valid only in a certain area of force variation. At Р Рпц, where Рпц is the force corresponding to the proportionality limit, the relationship between the tensile force and elongation turns out to be nonlinear.
The product EJ characterizes the bending stiffness of the beam section.
Shaft torsion | Shaft torsion deformation. The product GJр characterizes the torsional rigidity of the shaft section.
If the stiffness of the section of the beam is constant throughout it.
Welded parts processing schemes. a - plane processing. 6 - processing. | Loading of a welded beam with residual stresses. a - beam. b - zones 1 and 2 with high residual tensile stresses. - cross-section of a beam that takes a bending load (shown by hatching. This reduces the stiffness characteristics of the section EF and EJ. Displacements - deflections, angles of rotation, elongations caused by the load exceed calculated values.
The product GJP is called the torsional stiffness of the section.

The product G-IP is called the torsional stiffness of the section.
The product G-Ip is called the torsional stiffness of the section.
The product GJp is called the torsional stiffness of the section.
The product ES is called the stiffness of the bar section.
The value EA is called the stiffness of the section of the bar in tension and compression.
The product EF is called the tensile or compressive stiffness of the bar section.
The GJP value is called the torsional stiffness of the shaft section.
The product GJр is called the torsional stiffness of the section of a round bar.
The GJP value is called the torsional stiffness of the round bar section.
Loads, lengths and stiffness of the sections of the beams are considered known. In Problem 5.129, establish by how many percent and in which direction the deflection of the middle of the span of the beam indicated in the figure, determined by the approximate equation of the elastic line, differs from the deflection found exactly according to the equation of the circular arc.
Loads, lengths and stiffness of the sections of the beams are considered known.
The product EJZ is usually called the bending stiffness of the section.
The product EA is called the tensile stiffness of the section.

The product EJ2 is usually called the bending stiffness of the section.
The product G 1Р is called the torsional stiffness of the section.

The largest shear stresses arising in a twisted bar should not exceed the corresponding permissible stresses:

This requirement is called the strength condition.

The permissible torsional stress (as well as with other types of deformations) depends on the properties of the material of the calculated beam and on the adopted safety factor:

In the case of a plastic material, the shear yield stress is taken as the dangerous (limiting) stress tp, and in the case of a brittle material, the ultimate strength.

Due to the fact that mechanical torsion tests of materials are performed much less frequently than tensile tests, there are not always experimental data on dangerous (limiting) torsion stresses.

Therefore, in most cases, the permissible torsional stresses are taken depending on the permissible tensile stresses for the same material. For example, for steel for cast iron, where is the allowable tensile stress of cast iron.

These values of permissible stresses refer to the cases of pure torsion of structural elements under static loading. Shafts, which are the main objects calculated for torsion, in addition to torsion, also experience bending; in addition, the stresses arising in them are variable in time. Therefore, calculating the shaft only for torsion by a static load without taking into account bending and stress variability, it is necessary to take lower values of the permissible stresses.Practically, depending on the material and operating conditions for steel shafts, take

One should strive to ensure that the material of the bar is used as fully as possible, that is, that the greatest design stresses arising in the bar are equal to the permissible stresses.

The value mmax in the strength condition (18.6) is the value of the greatest shear stress in dangerous section timber in close proximity to its outer surface. A dangerous section of a bar is a section for which the absolute value of the ratio has the greatest value. For a beam of constant cross-section, the most dangerous is the cross-section in which the torque has the greatest absolute value.

When calculating the strength of twisted bars, as well as when calculating other structures, the following three types of problems are possible, differing in the form of using the strength condition (18.6): a) checking the stresses (verification calculation); b) section selection (design calculation); c) determination of the permissible load.

When checking stresses for a given load and dimensions of a bar, the largest shear stresses arising in it are determined. In this case, in many cases, it is necessary to first build a diagram, the presence of which facilitates the determination of the dangerous section of the bar. The highest shear stresses in the dangerous section are then compared with the allowable stresses. If, in this case, condition (18.6) is not satisfied, then it is required to change the dimensions of the section of the bar or reduce the load acting on it, or apply a material of higher strength. Of course, a slight (about 5%) excess of the maximum design stresses over the permissible ones is not dangerous.

When selecting a section for a given load, the torques in the cross-sections of the bar are determined (usually a diagram is plotted), and then using the formula

which is a consequence of formula (8.6) and condition (18.6), the required polar moment of resistance of the cross-section of the bar is determined for each of its sections, on which the cross-section is assumed constant.

Here is the value of the highest (in absolute value) torque within each such section.

By the magnitude of the polar moment of resistance, using the formula (10.6), the diameter of a solid round is determined, or using the formula (11.6) - the outer and inner diameters of the annular section of the bar.

When determining the permissible load using formula (8.6), according to the known permissible voltage and polar moment of resistance W, the value of the permissible torque is determined, then the values of the permissible external loads are set, from the action of which the maximum torque arising in the sections of the bar is equal to the permissible moment.

Calculation of the shaft strength does not exclude the possibility of deformations that are unacceptable during its operation. Large angles of twisting of the shaft are especially dangerous when a time-variable torque is transmitted to it, since torsional vibrations that are dangerous for its strength arise. V technological equipment, for example, metal-cutting machines, the insufficient torsional rigidity of some structural elements (in particular, the lead screws of lathes) leads to a violation of the processing accuracy of the parts manufactured on this machine. Therefore, in necessary cases, the shafts are calculated not only for strength, but also for rigidity.

The condition for the torsional rigidity of the bar has the form

where is the largest relative angle of twisting of the bar, determined by the formula (6.6); is the permissible relative twist angle, taken for different designs and different types of load equal to from 0.15 to 2 ° per 1 m of the rod length (from 0.0015 to 0.02 ° per 1 cm of length or from 0.000026 to 0.00035 rad per 1 cm of shaft length).

Assignment 3.4.1: The torsional stiffness of the cross-section of a round bar is called the expression ...

Answer options:

1) EA; 2) Gjp; 3) GA; 4) EJ

Solution: The correct answer is 2).

The relative angle of twisting of a rod of circular cross-section is determined by the formula. The smaller, the greater the stiffness of the rod. Therefore, the work Gjp is called the torsional stiffness of the cross-section of the bar.

Assignment 3.4.2: d loaded as shown in the figure. The maximum value of the relative twist angle is ...

Material shear modulus G, moment value M, length l are given.

Answer options:

1) ; 2) ; 3) ; 4) .

Solution: The correct answer is 1). Let's plot the torques.

When solving the problem, we will use the formula to determine the relative twist angle of a rod with a circular cross section

in our case we get

Assignment 3.4.3: From the stiffness condition at given values and G, the smallest permissible shaft diameter is ... Accept.

Answer options:

1) ; 2) ; 3) ; 4) .

Solution: The correct answer is 1). Since the shaft has a constant diameter, the stiffness condition has the form

Where. Then

Assignment 3.4.4: Round bar diameter d loaded as shown in the figure. Material shear modulus G, length l, moment value M are set. The mutual angle of rotation of the extreme sections is ...

Answer options:

1); 2); 3) zero; 4) .

Solution: The correct answer is 3). We denote the sections where external pairs of forces are applied B, C,D accordingly, and plot the torques. Section rotation angle D with respect to the section B can be expressed as the algebraic sum of the mutual angles of rotation of the section C with respect to cross-sections B and sections D with respect to the section WITH, i.e. ... material deformed bar inertia

The mutual angle of rotation of two cross-sections for a bar with a circular cross-section is determined by the formula. With regard to this problem, we have

Assignment 3.4.5: The condition for the torsional rigidity of a rod of circular cross-section with a diameter unchanged along the length has the form ...

Answer options:

1) ; 2) ; 3) ; 4) .

Solution: The correct answer is 4). The shafts of machines and mechanisms must be not only strong, but also sufficiently rigid. In stiffness calculations, the maximum relative twist angle is limited, which is determined by the formula

Therefore, the stiffness condition for a shaft (a rod undergoing torsional deformation) with a constant diameter along the length has the form

where is the permissible relative twist angle.

Assignment 3.4.6: The loading diagram of the bar is shown in the figure. Length L, the torsional rigidity of the cross-section of the bar, is the permissible angle of rotation of the section WITH are set. Based on stiffness, the maximum allowable parameter value external load M equals.

1); 2) ; 3) ; 4) .

Solution: The correct answer is 2). The stiffness condition in this case has the form, where is the actual angle of rotation of the cross section WITH... We build a torque diagram.

Determine the actual angle of rotation of the section WITH... ... Substitute the expression for the actual angle of rotation in the stiffness condition

1) oriented; 2) the main sites;
3) octahedral; 4) secants.

Solution: The correct answer is 2).

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).

Assignment 4.1.3: The main stresses for the stress state shown in the figure are ... (The voltage values are indicated in MPa).

1) y1 = 150 MPa, y2 = 50 MPa; 2) y1 = 0 MPa, y2 = 50 MPa, y3 = 150 MPa;
3) y1 = 150 MPa, y2 = 50 MPa, y3 = 0 MPa; 4) y1 = 100 MPa, y2 = 100 MPa.

Solution: The correct answer is 3). 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 y1 = 150 MPa, y2 = 50 MPa, y3 = 0 MPa, i.e. flat stress state.

Assignment 4.1.4: At the investigated point of the stressed body on three main sites, the values of normal stresses are determined: 50 MPa, 150MPa, -100MPa... The main stresses in this case are equal ...

1) y1 = 150 MPa, y2 = 50 MPa, y3 = -100 MPa;
2) y1 = 150 MPa, y2 = -100 MPa, y3 = 50 MPa;
3) y1 = 50 MPa, y2 = -100 MPa, y3 = 150 MPa;
4) y1 = -100 MPa, y2 = 50 MPa, y3 = 150 MPa;

Solution: The correct answer is 1). The main stresses are assigned indices 1, 2, 3 so that the condition is fulfilled.

Assignment 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) 00; 3) ; 4) .

Solution: The correct answer is 3).

The angle is determined by the formula

Substituting the numerical values of the stresses, we obtain

Set aside the negative angle clockwise.

Assignment 4.1.6: The values of the principal stresses are determined from the solution of the cubic equation. Odds J1, J2, J3 call ...

1) invariants of the stress state; 2) elastic constants;
3) directing cosines of the normal;
4) proportionality coefficients.

Solution: The correct answer is 1). Are 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.