Strength at variables. Calculations for strength at stresses that are variable in time. Control questions and tasks

Calculation of strength at variable stresses Calculation of elements of building structures for endurance is reduced to checking the inequality of the form (19.3) Strength condition at stresses that are variable over time coefficient that takes into account the number of loading cycles yv is a coefficient depending on the type of stress state and the coefficient of cycle asymmetry For example, for steel structures, the coefficient yv is determined from Table 19.1 Table 19.1 The value of the coefficient yv for steel structures "max P Vv Tension Design fatigue resistance , as well as the coefficient a, take into account the quality of the surface treatment of the calculated element, its design, the presence of stress concentrators. For particular types of structures, relation (19.3) can take a slightly different form. So, when calculating steel structures of bridges, the following inequality is used: (19.4) where R - design resistance in tension, compression and bending according to the yield strength of the material; m - coefficient of working conditions; _ 1 a, 6 - coefficients taking into account the steel grade and non-stationarity of loading; p - coefficient of asymmetry of the cycle of alternating stresses; (i is the effective stress concentration factor. The coefficient yv, determined by expression (19.5), describes the type of the diagram of limiting amplitudes, taking into account the stress concentration, the quality of the material and its surface treatment, the loading mode and other factors. Example 19.2. Brace of a through steel span of a railway bridge during the passage of the train is affected by a variable axial force.The largest tensile force is Nmnn= 1200 kN, the smallest (compressive) force Wmr-=200 kN. The design resistance R of low-alloy steel grade 15XCHD is 295 MPa. The coefficient of working conditions m = 0.9. Transverse - the section is composite (Fig. 19.20) and its area is equal to LpsSh, = 75 cm Fig. 19.20. The design of the brace of the steel span of the railway bridge Solution. The cycle asymmetry coefficient is determined as follows: IJVmml 1 L "max 6 In accordance with SNiP 2.05.03 -84 coefficient P is taken equal to 1.5, parameters a = 0.72 and 5 = 0.24 Then we find the maximum normal stress: N^ 1200 103 ---=--7 = 160 MPa. Lpepo 75 10"4 The right side of inequality (19.4) takes the value yvmR= 0.85 0.9 295 = 226.4 MPa>160 MPa. Consequently, the condition of the fatigue strength of the brace is satisfied. § 19.9. The concept of low-cycle fatigue In high-cycle fatigue failure, discussed in the previous paragraphs, the material is deformed elastically. Fracture begins in places of stress concentration as a result of the development of an incipient crack and is of a brittle nature (without the appearance of appreciable plastic deformations). Another type of fatigue is low-cycle fatigue, which is understood as failure under repeated elastic-plastic fatigue deformations; it differs from multicycle fatigue failure by the presence of macroscopic plastic deformation in the fracture zone. A strict boundary between high-cycle and low-cycle fatigue is not possible. In SNiL 11-23--81 it is noted that the check of steel structures for low-cycle fatigue should be carried out with a number of cycles less than growing No. 19 10 Yu \ Consider a schematized the material reforming diagram shown in fig. 19.21, and Nearby (Fig. 19.21, 6) is a graph of stress changes over time. During the first loading along the curve ОАВ, the point depicting the state of the material moves along the deformation diagram along the line ОВ Then the stresses decrease and the same point moves along the hynia BBiAi When the stress reaches the minimum value, it begins to increase and the deformation proceeds Further but the closed line A, ABB, . The range of deformations in one cycle is equal to ^ "max £min> and the range of plastic deformations ^pltaya 1L" 11 is the maximum and minimum plastic deformations of the aricyclic change in stresses. The nature of fracture during low-cycle fatigue depends on the ability of the material to accumulate plastic formations during cyclic deformation. Materials are said to be *cycle stable if the residual deformation does not change in all cycles*. The example considered above illustrates the features of deformation of such materials. For cyclically deteriorating materials, the characteristic features are an increase in residual deformations and an increase in total plastic deformation. Let us exclude the displacements u and v from these equations, for which we differentiate the first row twice with respect to y, the second with respect to x, and the third with respect to x and y. Adding the top two lines and subtracting the bottom one, we get the equation (20.6) Strain compatibility equation It is called the strain compatibility equation, since it gives the necessary relationship between the deformations that exists for arbitrary continuous displacement functions u, v (which we have excluded). If the body before deformation is mentally divided into infinitely small “bricks”, the deformations ex, ey, and y are reported to them, and an attempt is made to fold back into a whole deformed body, then two cases will turn out to be possible. In the first (Fig. 20.5, a) all elements will fit tightly to each other. Such deformations are joint, and they correspond to a continuous field of displacements. In the second case (Fig. 20.5, b), infinitely small discontinuities arise between the elements, and any continuous displacement field does not correspond to such deformations. q The field of deformations, which corresponds to a continuous field of displacements, is called joint deformations. Deformations are compatible. Otherwise, the deformations are called incompatible - local and non-compatible. local Equations (20.3), (20.5) and (20.7) together constitute the necessary eight equations, the solution of which allows us to find eight unknown functions of the plane problem under consideration. § 20.3. Determination of stresses from displacements found from the experiment Below, we describe how families of interference fringes are obtained experimentally, representing the isolines of some factor, that is, the locus of points at which this factor has a constant value. Thus, in the moiré method and holographic interferometry, isolines of displacements v = const and u = const can be obtained. On fig. 20.6 shows a diagram of a family of isolines v; \u003d const for a plane stressed state of the plate. Let us show how, using the equations of the theory of elasticity, we can pass from displacements to stresses. Formulas (20.5) make it possible to calculate the strains. 20.6. Numerical determination of deformations by experimentally obtained family of displacement isolines for a vertical line. We calculate the partial derivative (dv/dx)j=tgojj as the tangent of the slope of the secant drawn through the points (i - 1) and (/+ 1). Proceeding similarly for the derivative with respect to the coordinate y, we find Numerical differentiation (20.10) in a plane problem Similarly, we proceed with the family of isolines u \u003d const Having outlined a grid of lines parallel to the coordinate axes x and y, according to formulas (20.9) and (20.10) build the strain field, and then the stress field in the model under study. Since the nodal points of an orthogonal grid generally do not coincide with the intersection points with isolines, interpolation formulas are used to calculate strains and stresses at the nodes. There are devices and corresponding programs for personal computers that allow you to process a grid of isolines in automatic mode. Next, consider an experiment with a bending plate, for which a family of deflection isolines vv = const was obtained (Fig. 20.7, a). In the theory of plate bending, by analogy with the hypothesis of flat sections, the direct normal hypothesis is used, according to which the line m-u, passing to the position m-u, remains straight (Fig. 20.7b). Then for small deflections (px-dw/dx, (py-dwjdy) and displacements in the horizontal plane of an arbitrary point with coordinate z will be dw v= -(pyz= -z -. By (20.11) Substituting formulas (20.11) into (20.9) , we get 8 2 u * V "82w 8xdy 82w yxy \u003d -2z (20.12) - Z ey - r The stresses xxy distributed over the thickness of the plate h according to a linear law (Fig. 20.7, c) can be calculated for known deformations ( 20.12) according to Hooke's law (20.8). To determine the second derivatives of the deflection function, the deflection field at the nodes of the orthogonal grid of lines is first obtained using the interpolation formulas, a fragment of which is shown in Fig. 20.8. Then the derivatives at point K can be calculated using the numerical differentiation formulas:

Calculations for normal and shear stresses are carried out similarly.

Estimated coefficients are selected according to special tables.

When calculating, the safety margins for normal and shear stresses are determined.

Safety margin for normal stresses:

Margin of safety for shear stresses:

Where σ a- amplitude of the cycle of normal stresses; τ a is the amplitude of the shear stress cycle.

The obtained margins of safety are compared with the permissible ones. The presented calculation is verification and is carried out during the design of the part.

Control questions and tasks

1. Draw graphs of symmetrical and zero cycles of stress changes at repetitively alternating voltages.

2. List the characteristics of the cycles, show on the graphs the average stress and amplitude of the cycle. What characterizes the cycle asymmetry coefficient?

3. Describe the nature of fatigue damage.

4. Why strength under repeated-variable stresses
lower than with constant (static)?

5. What is called the endurance limit? How is a fatigue curve plotted?

6. List the factors that affect fatigue resistance.

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PRACTICAL EXERCISES ON THE SECTION

"Strength of materials"

Practice 6

Topic 2.2. Strength and stiffness calculations

In tension and compression

Know the order of calculations for strength and stiffness and calculation formulas.

To be able to carry out design and verification calculations for strength and stiffness in tension and compression.

Required Formulas

Normal voltage

Where N- longitudinal force; A- cross-sectional area.

Lengthening (shortening) of timber

E- elastic modulus; I- the initial length of the rod.

Allowable voltage

[s]- allowable margin of safety.

Tensile and compressive strength condition:

Examples of strength and stiffness calculations

Example 1 The load is fixed on the rods and is in balance (Fig. A6.1). The material of the rods is steel, the allowable stress is 160 MPa. Load weight 100 kN. The length of the rods: the first - 2 m, the second - 1 m. Determine the dimensions of the cross section and elongation of the rods. The cross-sectional shape is a circle.

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Solution

1. Determine the load on the rods. Consider the equilibrium
points IN, determine the reactions of the rods. According to the fifth axiom of statistics (the law of action and reaction), the reaction of the rod is numerically
equal to the load on the rod.

We apply the reactions of the bonds acting at the point IN. Freeing the point IN from connections (Fig. A6.1).

We choose the coordinate system so that one of the coordinate axes coincides with the unknown force (Fig. A6.1b).

Let us compose a system of equilibrium equations for the point IN:

We solve the system of equations and determine the reactions of the rods.

R 1 = R2 cos60 °; R 1= 115.5 ∙ 0.5 = 57.4 kN.

The direction of the reactions is chosen correctly. Both rods are compressed. Rod loads: F 1= 57.4kN; F 2 = 115.5 kN.

2. Determine the required cross-sectional area of ​​the rods from the strength conditions.

Compressive strength condition: σ = N/A≤ [σ] , where

Rod 1 ( N 1 = F 1):

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The resulting diameters are rounded off: d 1 = 25mm d 2= ​​32 mm.

3. Determine the elongation of the rods Δl = ----- .

Rod shortening 1:

Rod shortening 2:

Example 2 Homogeneous rigid plate with a gravity of 10 kN, loaded with a force F= 4.5 kN and moment T= ZkN∙m, supported at a point A and hung on a rod sun(Fig. A6.2). Select the section of the rod in the form of a channel and determine its elongation, if the length of the rod is 1 m, the material is steel, the yield strength is 570 MPa, the margin of safety for the material is 1.5.

Solution

1. Determine the force in the rod under the action of external forces. The system is in equilibrium, you can use the equilibrium equation for the plate: ∑t A = 0.

Rb- rod reaction, hinge reactions A we do not consider.

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According to the third law of dynamics, the reaction in the rod is equal to the force acting from the rod on the plate. The force in the rod is 14 kN.

2. According to the condition of strength, we determine the required value of the pope area
river section: O= N/A^ [A], where A> N/[a].

Permissible stress for the rod material

Hence,

3. We select the section of the rod according to GOST (Appendix 1).
The minimum channel area is 6.16 cm 2 (No. 5; GOST 8240-89).
It is more expedient to use an equal-shelf corner No. 2

(d\u003d Zmm), - the cross-sectional area of ​​\u200b\u200bwhich is 1.13 cm 2 (GOST 8509-86).

4. Determine the extension of the rod:

At the practical lesson, calculation and graphic work is performed and a test survey is conducted.

Settlement and graphic work

Exercise 1. Construct diagrams of longitudinal forces and normal stresses along the length of the beam. Determine the displacement of the free end of the beam. Two-stage steel beam loaded with forces F 1, F 2 , F 3- Cross-sectional areas A 1i A 2 .

310 Practice 6

Task 2. Beam AB, on which the indicated loads act, is kept in balance by the thrust Sun. Determine the dimensions of the cross section of the rod for two cases: 1) the section is a circle; 2) section - equal-shelf corner according to GOST 8509-86. Accept [σ] = 160 MPa. The self-weight of the structure is not taken into account.

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When defending the work, answer the questions of the test task.

312 Practice 6

Topic 2.2. Stretching and compression.

Strength and stiffness calculations

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Practice 7

The calculation of metal structures should be carried out according to the method of limit states or permissible ones. stresses. In complex cases, it is recommended to solve the issues of calculating structures and their elements through specially designed theoretical and experimental studies. The progressive method of calculation by limit states is based on a statistical study of the actual loading of structures under operating conditions, as well as the variability of the mechanical properties of the materials used. In the absence of a sufficiently detailed statistical study of the actual loading of the structures of certain types of cranes, their calculations are carried out according to the method of permissible stresses, based on the safety factors established by practice. ­

Under a plane stress state, in the general case, the plasticity condition according to the modern energy theory of strength corresponds to the reduced stress

Where σ x And σy- stresses along arbitrary mutually perpendicular coordinate axes X And at. At σy= 0

σ pr = σ Т, (170)

and if σ = 0, then the limit shear stresses

τ = = 0.578 σ Т ≈ 0,6σ Т. (171)

In addition to strength calculations for certain types of cranes, there are limitations on deflection values, which have the form

f/l≤ [f/l], (172)

Where f/l And [ f/l] - calculated and permissible values ​​of the relative static deflection f in relation to span (departure) l.Significant deflections may occur. safe for the structure itself, but unacceptable from an operational point of view.

The calculation according to the method of limit states is carried out according to the loads given in Table. 3.

Table notes:

1. Combinations of loads provide for the following operation of mechanisms: . Ia and IIa - the crane is stationary; smooth (Ia) or sharp (IIa) lifting of the load from the ground or its braking when lowering; Ib and IIb - crane in motion; smooth (Ib) and abrupt (IIb) starting or braking of one of the mechanisms. Depending on the type of crane, load combinations Ic and IIc etc. are also possible.

2. In the table. 3 shows the loads that are constantly acting and regularly arising during the operation of structures, forming the so-called main combinations of loads.

To take into account the lower probability of coincidence of design loads with more complex combinations, combination coefficients are introduced n s < 1, на которые умножаются коэффициенты перегрузок всех нагрузок, за исключением постоянной. Коэффициент соче­таний основных и дополнительных нерегулярно возникающих нагрузок, к которым относятся технологические, транспортные и монтажные нагрузки, а также нагрузки от температурных воз­действий, принимается равным 0,9; коэффициент сочетаний основ­ных, дополнительных и особых нагрузок (нагрузки от удара о бу­фера и сейсмические) – 0,8.

3. For some structural elements, the total effect of both the combination of loads Ia with its own number of cycles, and the combination of loads Ib with its own number of cycles should be taken into account.

4. Angle of deviation of the load from the vertical a. can also be seen as the result of an oblique lift.

5. Working wind pressure R b II and non-operating - hurricane R b III - per design is determined according to GOST 1451-77. With a combination of loads Ia and Ib, the wind pressure on the structure is usually not taken into account due to the low frequency of design wind speeds per year. For high cranes with a period of free oscillations of the lowest frequency of more than 0.25 s and installed in wind regions IV-VIII according to GOST 1451-77, the wind pressure on the structure is taken into account with a combination of loads Ia and Ib.

6. Technological loads can refer both to the case of loads II and to the case of loads III.

Table 3

Loads in calculations by the method of limit states

The limit states are the states in which the structure ceases to satisfy the operational requirements imposed on it. The limit state calculation method aims to prevent the occurrence of limit states during operation during the entire service life of the structure.

Metal structures of TTs (hoisting and transport machines) must meet the requirements of two groups of limit states: 1) loss of bearing capacity of crane elements in terms of strength or loss of stability from a single action of the largest loads in working or non-working condition. The working state is the state in which the crane performs its functions (Table 3, load case II). The state is considered inoperative when the crane without load is subject only to loads from its own weight and wind or is in the process of installation, dismantling and transportation (Table 3, load case III); loss of bearing capacity of crane elements due to fatigue failure under repeated action of loads of various sizes over the estimated service life (Table 3, case of loads I, and sometimes II); 2) unsuitability for normal operation due to unacceptable elastic deformations or vibrations that affect the operation of the crane and its elements, as well as maintenance personnel. For the second limit state for the development of excessive deformations (deflections, angles of rotation), the limit condition (172) is set for individual types of cranes.

Calculations for the first limit state are of the greatest importance, since in rational design, structures must satisfy the requirements of the second limit state.

For the first limit state in terms of bearing capacity (strength or stability of elements), the limit condition has the form

N ≤ F,(173)

Where N- design (maximum) load in the element under consideration, expressed in force factors (force, moment, stress); F- design bearing capacity (smallest) of the element according to force factors.

When calculating the first limit state for the strength and stability of elements to determine the load N in formula (171) the so-called normative loads R H i(for hoisting-and-handling machine designs, these are the maximum loads in the working condition, entered into the calculation both on the basis of specifications and on the basis of design and operating experience) are multiplied by the overload factor of the corresponding standard load n i , after which the work P Hi p i represents the greatest possible load during the operation of the structure, called the design load. Thus, the design force in the element N in accordance with the design combinations of loads given in table. 3 can be represented as

, (174)

Where a i is the force in the element at Р Н i= 1, and the calculated moment

, (175)

Where M H i- the moment from the standard load.

To determine the overload coefficients, a statistical study of the variability of loads based on experimental data is necessary. Let for a given load Pi its distribution curve is known (Fig. 63). Since the distribution curve always has an asymptotic part, when assigning the calculated load, it should be borne in mind that loads that are greater than the calculated ones (the area of ​​\u200b\u200bthese loads is shaded in Fig. 63) can cause damage to the element. The adoption of large values ​​for the design load and overload factor reduces the likelihood of damage and reduces losses from breakdowns and accidents, but leads to an increase in the weight and cost of structures. The question of the rational value of the overload factor should be decided taking into account economic considerations and safety requirements. Let the calculated force distribution curves be known for the element under consideration N and bearing capacity F. Then (Fig. 64) the shaded area, within whose boundaries the limit condition (173) is violated, will characterize the failure probability.

Given in table. 3 overload factors n> 1, since they take into account the possibility of actual loads exceeding their standard values. In the event that it is dangerous not to exceed, but to reduce the actual load compared to the standard one (for example, the load on the beam consoles, unloading the span, with the design section in the span), the overload factor for such a load should be taken equal to the reciprocal value, i.e. . n"= 1/n< 1.

For the first limit state for the loss of bearing capacity due to fatigue, the limit condition has the form

σ pr ≤ m K R,(176)

Where σ pr is the reduced voltage, and m K– see formula (178).

Calculations for the second limit state according to condition (172) are made at overload factors equal to one, i.e., according to standard loads (the weight of the load is assumed to be equal to the nominal).

Function F in formula (173) can be represented as

F= Fm K R , (177)

Where F- the geometric factor of the element (area, moment of resistance, etc.).

Under design resistance R should be understood in the calculations:

for fatigue resistance - the endurance limit of the element (taking into account the number of cycles of load changes and the concentration and cycle asymmetry factors), multiplied by the corresponding uniformity coefficient for fatigue tests, characterizing the spread of test results, k 0= 0.9, and divided by k m is the reliability coefficient for the material in strength calculations, characterizing both the possibility of changing the mechanical qualities of the material in the direction of their reduction, and the possibility of reducing the cross-sectional areas of rolled products due to the minus tolerances established by the standards; in appropriate cases, the reduction of the initial endurance limit by the loads of the second design case should be taken into account;

strength at constant stress R= R P /k m - ­ quotient from dividing the normative resistance (normative yield strength) by the corresponding safety factor for the material; for carbon steel k m = 1.05, and for low-alloyed - k m = 1.1; thus, in relation to the work of the material, the limit state is not the complete loss of its ability to perceive the load, but the onset of large plastic deformations that prevent the further use of the structure;

stability - the product of the design resistance to strength by the coefficient of reduction in the bearing capacity of compressible (φ, φ int) or bending (φ b) elements.

Working conditions coefficients m K depend on the circumstances of the operation of the element, which are not taken into account by the calculation and quality of the material, i.e. are not included in the force N, nor in design resistance R.There are three such main circumstances, and therefore we can accept

m K = m 1 m 2 m 3 , (178)

Where m 1 - coefficient taking into account the responsibility of the calculated element, i.e., the possible consequences of destruction; the following cases should be distinguished: destruction does not cause the crane to stop working, causes the crane to stop without damage or with damage to other elements, and finally causes the destruction of the crane; coefficient m 1 can be in the range of 1–0.75, in special cases (brittle fracture) m 1 = 0,6; m 2 - coefficient taking into account possible damage to structural elements during operation, transportation and installation, depends on the types of cranes; can be taken T 2 = 1.0÷0.8; T 3 - coefficient taking into account the imperfections of the calculation associated with inaccurate determination of external forces or design schemes. It should be set for individual types of structures and their elements. Can be taken for flat statically determinate systems T 3 = 0.9, .and for statically indeterminate -1, for spatial -1.1. For bending elements compared to those experiencing tension-compression T 3 = 1.05. Thus, the calculation for the first limit state for strength at constant stresses is carried out according to the formula

σ II<. m K R,(179)

and for fatigue resistance, if the transition to the limit state is carried out by increasing the level of variable tension, - according to the formula (176), where the design resistance R determined by one of the following formulas:

R= k 0 σ -1K/k m;(180)

R N= k 0 σ -1K N/k m; (181)

R*= k 0 σ -1K/k m;(182)

R*N= k 0 σ -1K N/k m; (183)

Where k 0 , k m - uniformity coefficients for fatigue tests and reliability for the material; σ –1K , σ –1KN , σ * –1K , σ * –1KN– endurance limits unlimited, limited, reduced unlimited, reduced limited, respectively.

Calculation according to the method of permissible stresses is carried out according to the loads given in Table 4. It is necessary to take into account all the notes to the table. 3, except note 2.

The values ​​of safety factors are given in table. 5 and depend on the circumstances of the operation of the structure, not taken into account by the calculation, such as: responsibility, bearing in mind the consequences of destruction; calculation imperfections; deviations in size and quality of the material.

Calculation by the method of allowable stresses is carried out in cases where there are no numerical values ​​for the overload coefficients of the design loads to perform the calculation by the method of limit states. Strength calculation is made according to the formulas:

σ II ≤ [ σ ] = σ T / n II , (184)

σ III ≤ [ σ ] = σ T / n III , (185)

Where n II and n III - see table. 5. In this case, the allowable bending stresses are assumed to be 10 MPa (approximately 5%) more than for tension (for St3 180 MPa), given that during bending, fluidity first manifests itself only in the extreme fibers and then gradually spreads over the entire section of the element , increasing its load-bearing capacity, i.e., during bending, there is a redistribution of stresses over the cross section due to plastic deformations.

When calculating for fatigue resistance, if the transition to the limit state is carried out by increasing the level of variable stress, one of the following conditions must be met:

σ pr ≤ [ σ –1K ]; (186)

σ pr ≤ [ σ –1K N]; (187)

σ pr ≤ [ σ * –1K ]; (188)

σ pr ≤ [ σ * –1KN ]; (189)

Where σ pr - reduced voltage; [ σ –1K ], [σ –1K N], [σ * –1K ], [σ * –1KN] - allowable stresses, which are determined using the expression [ σ ] = σ –1K /n 1 or similarly to formulas (181) - (183) instead of σ –1K are used σ –1KN , σ * –1K And σ * –1KN. Margin of safety n I is the same as in the calculation of static strength.

Figure 65 - Scheme for calculating the fatigue life margin

If the transition to the limit state is carried out by increasing the number of cycles of repetition of alternating stresses, then when calculating for limited durability, the margin for fatigue life (Fig. 65) n d = Np/N. Because σ t etc Np = σ t –1K N b = σ t –1K N N,

n q = ( σ –1K N / σ etc) T = p t 1 (190)

and at n l = 1.4 and TO= 4 n d ≈ 2.75, and at TO= 2 n e ≈ 7.55.

In a complex stress state, the hypothesis of the highest shear octahedral stresses is most consistent with the experimental data, according to which

(191)

And . Then the margin of safety for symmetrical cycles

i.e. P= n σ n τ /, (192)

Where σ-IK and τ-l TO- limiting stresses (endurance limits), and σ a and τ a are the amplitude values ​​of the current symmetrical cycle. If the cycles are asymmetric, they should be reduced to symmetric by a formula like (168).

The progressiveness of the method of calculation by limit states lies in the fact that in calculations by this method, the actual work of structures is better taken into account; overload factors are different for each of the loads and are determined based on a statistical study of load variability. In addition, the mechanical properties of materials are better taken into account using the material safety factor. While in the calculation by the method of allowable stresses, the reliability of the structure is ensured by a single safety factor, in the calculation by the method of limit states, instead of a single safety factor, a system of three factors is used: reliability by material, overload and operating conditions, established on the basis of statistical accounting of the operating conditions of the structure.

Thus, the calculation for allowable stresses is a special case of calculation for the first limit state, when the overload factors for all loads are the same. However, it should be emphasized that the method of calculation by limit states does not use the concept of safety margin. It is also not used by the probabilistic calculation method currently being developed for crane construction. Having performed the calculation according to the method of limit states, it is possible to determine the value of the resulting safety factor according to the method of permissible stresses. Substituting into formula (173) the values N[cm. formula (174)] and F[cm. formula (177)] and passing to stresses, we obtain the value of the safety factor

n =Σ σ i n i k M / (m K Σ i). (193)

Variable stresses in machine parts differ in the type of cycles and the nature of the cycle change over time. A stress cycle is a set of successive stress values ​​for one period of their change under regular loading. Figure 4.2 shows various types of alternating voltage cycles, characterized by the following parameters:

the average stress of the cycle, expressing the constant (positive or negative) component of the stress cycle:

cycle stress amplitude, expressing the largest positive value of the variable component of the stress cycle:

where σ m ax and σ min are the maximum and minimum cycle stresses corresponding to the maximum and minimum cycle stresses.

The ratio of the minimum stress of the cycle to the maximum is called the coefficient of asymmetry of the stress cycle:

symmetrical A cycle is called when the maximum and minimum voltages are equal in absolute value and opposite in sign. The symmetric cycle is sign-alternating and has the following parameters: σ A\u003d σ m ax \u003d σ min; σ T= 0; R = - 1. The most common example of a symmetrical stress cycle is the bending of a rotating shaft (rotational bending). Endurance limits corresponding to a symmetrical cycle have the index "-1" (σ -1 ; τ -1).

asymmetrical A cycle is called, in which the maximum and minimum voltages have different absolute values. For an asymmetric stress cycle σ max = σ m + σ a; σmin = σm - σ a; R ≠ - 1 Asymmetric stress cycles are sign-alternating if the stresses change in value and in sign. The cycle of stresses that change only in absolute value is called constant sign. The endurance limits corresponding to the asymmetric cycle are denoted by the index "R" (σ R ; τ R).

A characteristic asymmetric cycle is the zero stress cycle, which includes stress cycles of constant sign that change from zero to a maximum during tension (σ min = 0) or from zero to a minimum during compression (σ max = 0). In tension, the zero stress cycle is characterized by the following parameters: σ m =σ a= σ max /2; R = 0. The endurance limit from the zero cycle is denoted by the index "0" (σ 0 ; τ 0). Zero stress cycles occur in the teeth of gears and chain sprockets, which are loaded during operation when they enter the engagement and are completely unloaded when they leave it.

WITH fatigue resistance depends not only on the type of stress cycles in operation, but also on the nature of the stress change over time. Under stationary loading, the values ​​of the amplitude and average stress of the cycle remain unchanged in time. Drilling machines and equipment, as already noted, mainly operate under non-stationary loading.

The amplitude and average voltage of the cycles can have a stepwise or continuous change (Fig. 4.3).

Quantitative characteristics of the resistance of the material to the action of alternating stresses are determined by testing for fatigue 15-20 identical samples with a diameter of 7-10 mm, having a polished surface. Tests are carried out at different voltage levels. Based on the results obtained, a graph of the fatigue curve is built (Fig. 4.4, a). The ordinate axis of the graph plots the maximum stress or cycle stress amplitude at which the given sample was tested, and the abscissa axis shows the number of cycles N of stress changes that the sample withstood before failure. The resulting curve characterizes the relationship between stresses and cycle life of identical samples at a constant average cycle stress or cycle asymmetry coefficient.

For most steels, when tested in air, the fatigue curve, starting from the number of cycles N = 10 6 ÷10 7 , becomes horizontal and the samples that have withstood the indicated number of cycles do not fail with a further practically unlimited increase in the number of loading cycles. Therefore, testing of steels is stopped when 10 million cycles are reached, which make up the test base N b. The maximum absolute value of the cycle stress at which fatigue failure does not yet occur to the test base is called the endurance limit. For a reliable assessment of the endurance limit, the number of non-destructive samples at a given level of alternating stresses should be at least six.

H The simplest and therefore most common are fatigue tests under a symmetrical stress cycle (circular bending).

Fatigue tests with an asymmetric stress cycle are carried out on special testing machines. Fatigue Curves Plotted in Logarithmic Coordinates

(Fig. 4.4, b), are oblique and horizontal lines. For strength calculations, the left sloping part of the fatigue curve is represented as

where σ is the effective stress; T- indicator of the slope of the fatigue curve; N is the number of stress cycles sustained until fatigue failure (cyclic durability); σ -1 - endurance limit; N 0 is the number of cycles corresponding to the breaking point of the fatigue curve represented by two straight lines.

The value of N 0 in most cases fluctuates within 10 6 -3∙10 6 cycles. In calculations for strength under alternating stresses, when there are no fatigue test data, one can take on average N=2∙10 6 cycles.

Fatigue Slope Index

for parts varies from 3 to 20, and with an increase in the effective stress concentration factor, a tendency to decrease is noticed T. Approximately can be taken

Where With=12 - for welded joints; With= 12÷20 - for parts made of carbon steels; With= 20÷30 - ​​for alloy steel parts.

Table 4.4

From the equation of the fatigue curve, the cyclic durability N is determined under the action of stresses σ exceeding the fatigue limit σ -1

The values ​​of endurance limits obtained as a result of fatigue tests are given in reference books on engineering materials. The ratios between the strength and endurance, established on the basis of statistical data are given in table. 4.5.

Table 4.5

Type of loading	Steel rolling and forging	Steel casting
	σ -1 = 0.47σ in	σ -1 = 0.38 σ in
Tension-compression	σ -1 p = 0.35σ in	σ -1 = 0.28 σ in
Torsion	τ -1 = 0.27 σ in	τ -1 = 0.22σ in

The endurance limit of parts is below the endurance limit of standard laboratory samples used in fatigue testing of engineering materials. The decrease in the endurance limit is due to the influence of stress concentration, as well as the absolute dimensions of the cross section and the state of the surface of the parts. The values ​​of the endurance limit of parts are determined by field tests or by reference calculation and experimental data that establish the influence of these factors on the resistance of fatigue parts.

Full-scale tests are usually used to determine the endurance limits of widely used standard products and some of the most critical components and parts. So, on the basis of full-scale tests, the limits of endurance of drill pipes, bush-roller chains of drilling rigs, traveling ropes, bearings and some other standard products used in drilling machines and equipment have been established. Due to the complexity of full-scale fatigue tests, in practical strength calculations, calculation and experimental data are mainly used, on the basis of which the fatigue limit of the part is determined from the expression

where σ -1d is the endurance limit of the part; σ -1 - endurance limit of standard laboratory samples from the material of the part; K - coefficient of reduction of endurance limit:

Here K σ is the effective stress concentration factor; K F - coefficient of influence of surface roughness; K d - coefficient of influence of the absolute dimensions of the cross section: K υ - coefficient of influence of surface hardening.

The values ​​of the effective coefficients of stress concentration and the coefficients of the effect of surface hardening, obtained from the calculation and experimental data, are given in Table. 4.1 and 4.2.

The coefficient of influence of the absolute dimensions of the cross section is determined by the ratio of the endurance limit of smooth samples with a diameter of d to the endurance limit of smooth laboratory samples with a diameter of 7-10 mm:

where σ -1 d is the endurance limit of a smooth specimen (part) with a diameter d; σ -1 - endurance limit of the material, determined on standard smooth samples with a diameter of 7-10 mm.

Experimental data show that with an increase in transverse dimensions, the endurance limit of the part decreases. This is explained by the statistical theory of fatigue failures, according to which, with an increase in size, the probability of the presence of internal defects in parts in areas of high stresses increases - a scale effect. The manifestation of the scale effect is facilitated by the deterioration of the homogeneity of the material, as well as the difficulty of controlling and ensuring the stability of the processes for manufacturing large parts. The scale effect depends mainly on the transverse dimensions and to a lesser extent on the length of the part.

IN cast parts and materials with non-metallic inclusions, pores and other internal and external defects, the scale effect is more pronounced. Alloy steels are more sensitive to internal and external defects, and therefore, for them, the influence of absolute dimensions is more significant than for carbon steels. In strength calculations, the values ​​of the coefficients of influence of the absolute dimensions of the cross section are selected according to the graph (Fig. 4.5).

Surface roughness, scale and corrosion significantly affect fatigue resistance. On fig. 4.6 shows an experimental graph that characterizes the change in the endurance limit of parts with different quality of processing and surface condition. The roughness influence coefficient is determined by the ratio of the endurance limit of smooth specimens with a surface not rougher than R a= 0.32 according to GOST 2789-73 to the endurance limit of samples with a given surface roughness:

where σ -1 - endurance limit of carefully polished samples; σ -1p - endurance limit of samples with a given surface roughness.

For example, it has been found that during rough grinding, the endurance limit of a part made of steel with a tensile strength of 1500 MPa is the same as that of steel with a tensile strength of 750 MPa. The influence of the surface state of the part on the fatigue resistance is due to the high level of stresses from bending and torsion in the outer zones of the part and the weakening of the surface layer due to its roughness and the destruction of crystal grains during cutting.

P With similar formulas, the endurance limits of parts under the action of shear stresses are determined.

Strength conditions for a symmetrical cycle of alternating stresses have the form:

under the action of normal stresses

under the action of shear stresses

Where P σ , Pτ - safety factors for normal and shear stresses; σ -1d, τ -1d - endurance limits of the part; σ a, τ a - amplitudes of variable stresses; [ P σ ], [ Pτ ] - the minimum allowable value of the safety margin for normal and shear stresses.

In a biaxial stress state that occurs in the case of simultaneous bending and torsion or tension-compression and torsion, the safety margin in the design section is determined from the expression

M The minimum allowable value of the safety factor depends on the accuracy of the choice of design loads and the completeness of taking into account the design, technological and operational factors that affect the endurance limit of the part. In calculations of drilling machines and equipment for endurance, the minimum allowable values ​​of safety factors are regulated by industry standards indicated in Table. 2P applications. In the absence of industry standards, allowable safety margins [n] = 1.3÷1.5 are accepted.

Under the action of asymmetric cycles, parts are calculated for strength based on the cycle limit stress diagram (Fig. 4.7), which characterizes the relationship between limit stresses and average cycle stresses for a given durability. The diagram is built according to the experimental values ​​of endurance limits obtained for various average cycle stresses. This requires long-term testing under a special program. In practical calculations, simpler schematized limit stress diagrams are used, which are built according to the experimental values ​​of the endurance limit of the symmetrical and zero cycles and the yield strength of the selected material.

On the limit stress diagram, point A (0, σ -1) corresponds to the endurance limit of a symmetrical cycle, point B (σ 0 /2; σ 0) corresponds to the endurance limit of a zero stress cycle. The straight line passing through these points determines the maximum limiting stresses, cycles, depending on the average stress. Stresses below the ABC level do not cause destruction at the number of cycles N 0 corresponding to the test base. The points lying above the straight line ABC characterize the stress cycles at which failure occurs at the number of cycles N

Straight line ABC, limited in the upper part by the yield strength σ t, i.e., resistance to plastic deformation, is called the limit stress line. It is expressed by the equation of a straight line passing through two points A and B with coordinates (0, σ -1) and (σ 0 /2; σ 0):

Denoting we get

Under the action of shear stresses, formula (25) takes the form

The coefficients φ σ and φ τ characterize the sensitivity of the material to the asymmetry of the stress cycle, respectively, under the action of normal and shear stresses (taken from the technical literature). If we draw a straight line on the diagram from the origin of coordinates at an angle of 45 ° (the bisector of the coordinate angle), then the segment OB" == BB"-BB" will correspond to the average voltage, and the segment BB" will correspond to the limiting amplitude of the cycle

where σ A- the limiting cycle amplitude, i.e., the stress amplitude corresponding to the endurance limit at a given average cycle stress.

With an increase in the average cycle stress σ T endurance limit σ T ax increases, and the limiting amplitude of the cycle σ A decreases. The degree of its reduction depends on the sensitivity of the material to the asymmetry of the cycle, characterized by the coefficient φ σ .

Table 4.6

Type of deformation	Ultimate strength σ b, MP a
Bending and stretching (φ σ)
Torsion (φ τ)

Cycles having the same asymmetry coefficients are called similar and are indicated on the limit stress diagram by points lying on the same ray drawn at the corresponding angle β. This can be seen from the formula

It has been experimentally established that the ratio of the limiting amplitudes of smooth samples and samples with stress concentration does not depend on the average cycle stress. According to this, the stress concentration factors are assumed to be the same for symmetric and asymmetric cycles, and the longitudinal stress amplitude for the part is determined by the formula

M maximum limit stress of asymmetric cycles

The stress limit diagram of the part shown in fig. 4.8 is used to determine the safety margins. Let stresses (σ max , σ a , σ m) act on the part at point M. If the expected overloads correspond to the condition of simple loading, i.e., occur at a constant degree of asymmetry (R = const), then the ultimate stress for the considered cycle will be at point N and the safety margin

As a result of the joint solution of the equations of the lines of limiting stresses AC and ON, the ordinate of the point N and the margin of safety under the action of normal stresses are determined

(29)

Similarly, under the action of shear stresses

If the average stress does not change during overloads (σ m= const), and the amplitude grows, i.e., the operating voltages increase along the straight line M " P, then the margin of safety

Drilling machine parts usually operate under simple loading conditions, and the safety margin should be calculated using formulas (29) and (30). Under the combined action of normal and shear stresses, the margin of safety is determined by formula (24).

R Endurance calculations under non-stationary loading are based on the following assumptions. Let loads Р 1 , P 2 ,..., P i(or stresses σ 1 , σ 2 , ….σ i) act respectively during N 1 ….N 3 ....N i loading cycles (Fig. 9). The ratio of the actual number of cycles N i some stress σ i- to the number of cycles N j at which the sample is destroyed under the action of the same stress σ i is called a cycle relation.

According to the fatigue damage summation hypothesis, the effect of each group of loads does not depend on the order of their alternation and the same cycle ratios of overloads of different magnitudes cause the same degree

fatigue damage.

Assuming a linear accumulation of fatigue damage

Where A- experimentally established coefficient, taken (in stock) equal to one.

With the adopted notation, the equation of the endurance curve 1 (Fig. 9) has the form:

where σ R is the endurance limit for the base number of cycles N 0 .

Based on the assumed assumptions, the non-stationary loading is replaced by some equivalent stationary loading, the effect of which is equivalent to the actual non-stationary loading. In practice, various options are used to reduce non-stationary loading to equivalent stationary loads.

Any of the acting loads P i(more often P max) or the stress σ caused by it i(σ max) is assumed to be constant, acting during the so-called equivalent number of cycles N 3 corresponding to the loading level. Then, taking, for example, the stress equal to σ max , based on formulas (32) and (33) we obtain ( A = 1)

(35)

where is the load mode coefficient.

From formula (35) it follows that with an equivalent number of cycles N e

In another version of the reduction, the non-stationary loading is replaced by a mode with a constant equivalent level of loading Р e (σ e), which operates for a given service life, determined by the total number of cycles ΣN i or the number N 0 corresponding to the inflection point of the endurance curve. According to this

from which the formula is derived in the following form convenient for calculations:

(37)

where is the equivalence coefficient.

To calculate the equivalence factor, statistical data are used on the magnitude of loads that occur in the part during operation, and the number of cycles of their repetition during one loading block, corresponding to the drilling of one typical well. In practice, the values ​​of the equivalence coefficients vary within 0.5 ≤ K 0e ≤ 1.

When calculating by tangential stresses, the value of the equivalence coefficient K 0e is determined by formula (36), in which normal stresses are replaced by tangential, induced, transmitted torques.

Safety margins under non-stationary loading are determined by the formulas:

for symmetrical alternating voltage cycles

for asymmetric alternating voltage cycles

It should be noted that the values ​​of the equivalence ratios depend on the penetration per bit, mechanical drilling speed and other indicators that determine the loading and turnover of drilling machines and equipment. With an increase in penetration per bit, the loading of the lifting mechanism decreases. Mud pumps and the rotor are similarly affected by increased drilling speeds. This indicates the need to refine the equivalence factors in case of significant changes in drilling performance.

Definition of initial data for endurance calculations transmission elements . When calculating endurance, the law of linear damage accumulation is used with repeated impact on transmission elements of amplitudes of different levels.

The determination of the initial design data is reduced to the calculation of equivalent loads in the form of the product of the main load taken into account by the durability factor.

Equivalent load is such a load, the effect of which is equivalent to the action of a real load in terms of the effect of accumulation of damage.

Methods for determining the equivalent loads of transmission elements are based on the following main provisions.

1. The operational load of transmissions is determined by the average value
and coefficient of variation v torque, the statistical distribution of the amplitudes of which can be considered truncated normal.

2. As a medium load
a torque is received in the power circuit to the body, corresponding to the implementation of a stable moment M y engines.

3. Dynamic loads for the transmission of the most loaded organ, estimated by the coefficient of variation, are considered acceptable. v≤ 0.6. For v ≥ 0.6, measures should be taken to reduce it, for example, damping devices should be used, etc.

Numerical values ​​of the coefficients of variation v can be determined from the calculated dependencies, or from the results of a computational experiment, or from the data of experimental studies of analog machines.

Here - the maximum long-acting moment; - maximum long-acting torque amplitude; R dl - the maximum continuous load on the bearings, determined by M length

The values ​​of the durability coefficients are determined by dependencies.

1. To calculate wheel teeth for endurance:

contact

bending for parts with surface hardness HB > 350

bending for parts with surface hardness HB< 350

2. To calculate shafts:

for bending endurance

torsional fatigue strength

3. To calculate the life of ball and roller bearings:

Here is the calculated number of loading cycles of transmission elements; P - part rotation frequency, rpm; T R - estimated time of operation of the part, h (usually take 5000 h); N o - basic number of loading cycles, taken in accordance with the recommendations (see above)

Corresponding equivalence factors, taken depending on v.

When calculating the endurance of the teeth of the wheels according to GOST 21354-87, when determining the design stresses, the load is taken M dl, and when defining:

Most machine parts under operating conditions experience variable stresses that change cyclically over time. Breakage analysis shows that the materials of machine parts operating for a long time under the action of variable loads can fail at stresses lower than the tensile strength and yield strength.

The destruction of a material caused by repeated action of variable loads is called fatigue failure or material fatigue.

Fatigue failure is caused by the appearance of microcracks in the material, the heterogeneity of the structure of materials, the presence of traces of machining and surface damage, and the result of stress concentration.

Endurance called the ability of materials to resist destruction under the action of alternating stresses.

The periodic laws of change in variable voltages may be different, but all of them can be represented as a sum of sinusoids or cosine waves (Fig. 5.7).

Rice. 5.7. Variable voltage cycles: A- asymmetric; b- pulsating; V - symmetric

The number of voltage cycles per second is called loading frequency. Stress cycles can be of constant sign (Fig. 5.7, a, b) or alternating (Fig. 5.7, V).

The cycle of alternating voltages is characterized by: maximum voltage a max, minimum voltage a min, average voltage a t =(a max + a min)/2, cycle amplitude s fl = (a max - a min)/2, cycle asymmetry coefficient rG= a min / a max.

With a symmetrical loading cycle a max = - ci min ; a t = 0; g s = -1.

With a pulsating voltage cycle a min \u003d 0 and \u003d 0.

The maximum value of periodically changing stress at which the material can resist destruction indefinitely is called endurance limit or fatigue limit.

To determine the endurance limit, samples are tested on special machines. The most common bending tests are under a symmetrical loading cycle. Tensile-compressive and torsion endurance tests are less frequently performed because they require more sophisticated equipment than in the case of bending.

For endurance testing, at least 10 identical samples are selected. Tests are carried out as follows. The first sample is installed on the machine and loaded with a symmetrical cycle with a stress amplitude of (0.5-0.6)st (o in - tensile strength of the material). At the moment of destruction of the sample, the number of cycles is fixed by the counter of the machine N. The second sample is tested at a lower voltage, with the destruction occurring at a greater number of cycles. Then the following samples are tested, gradually reducing the voltage; they break down with more cycles. Based on the data obtained, an endurance curve is built (Fig. 5.8). There is a section on the endurance curve tending to a horizontal asymptote. This means that at a certain voltage a, the sample can withstand an infinitely large number of cycles without being destroyed. The ordinate of this asymptote gives the endurance limit. So, for steel, the number of cycles N= 10 7, for non-ferrous metals - N= 10 8 .

Based on a large number of tests, approximate relationships have been established between the bending endurance limit and the endurance limits for other types of deformation.

where st_ |p - endurance limit for a symmetrical cycle of tension-compression; t_j - torsional endurance limit under symmetrical cycle conditions.

Bending stress

Where W = / / u tah - moment of resistance of the rod in bending. Torsional stress

Where T - torque; Wp- polar torsional moment of resistance.

At present, endurance limits for many materials are defined and are given in reference books.

Experimental studies have shown that in zones of sharp changes in the shape of structural elements (near holes, grooves, grooves, etc.), as well as in contact zones, stress concentration- high voltage. The reason causing stress concentration (hole, undercut, etc.) is called stress concentrator.

Let the steel strip stretch by force R(Fig. 5.9). A longitudinal force acts in the cross section /' of the strip N= R. Rated voltage, i.e. calculated under the assumption that there is no stress concentration, is equal to a = R/F.

Rice. 5.9.

The stress concentration decreases very quickly with distance from the hub, approaching the nominal voltage.

Qualitatively, the stress concentration for various materials is determined by the effective stress concentration factor

Where O _ 1k, t_ and - endurance limits determined by nominal stresses for samples having stress concentration and the same cross-sectional dimensions as a smooth sample.

The numerical values ​​of the effective stress concentration factors are determined on the basis of fatigue tests of the specimens. For typical and most common forms of stress concentrators and basic structural materials, graphs and tables are obtained, which are given in reference books.

It has been experimentally established that the endurance limit depends on the absolute dimensions of the cross section of the sample: with an increase in the cross section, the endurance limit decreases. This pattern has been named scale factor and is explained by the fact that with an increase in the volume of the material, the probability of the presence of structural inhomogeneities in it (slag and gas inclusions, etc.) increases, causing the appearance of foci of stress concentration.

The influence of the absolute dimensions of the part is taken into account by introducing the coefficient into the calculation formulas G, equal to the ratio of endurance limit o_ld given sample of given diameter d to the endurance limit a_j of a geometrically similar laboratory sample (usually d=l mm):

So, for steel accept e a\u003d e t \u003d e (usually r \u003d 0.565-1.0).

The endurance limit is affected by the cleanliness and condition of the surface of the part: with a decrease in surface cleanliness, the fatigue limit decreases, since stress concentration is observed near its scratches and scratches on the surface of the part.

Surface quality factor is the ratio of the endurance limit st_, a sample with a given surface condition to the endurance limit st_, a sample with a polished surface:

Usually (3 \u003d 0.25 -1.0, but with surface hardening of parts by special methods (hardening with high-frequency currents, carburizing, etc.) it can be more than one.

The values ​​of the coefficients are determined according to tables from reference books on strength calculations.

Strength calculations at alternating voltages, in most cases, they are performed as test ones. The result of the calculation is the actual safety factors n, which are compared with the required (permissible) for a given design safety factors [P], moreover, the condition l > [n J] must be satisfied. Usually for steel parts [l] = 1.4 - 3 or more, depending on the type and purpose of the part.

With a symmetrical cycle of stress changes, the safety factor is:

0 for stretch (compress)

0 for twist

0 for bend

Where A their - the nominal values ​​of the maximum normal and shear stresses; K SU, K T- effective stress concentration factors.

When parts are operated under conditions of an asymmetric cycle, the safety factors n a along normal and tangent n x stresses are determined by the formulas of Serensen-Kinasoshvili

where |/ st, |/ t - coefficients of reduction of an asymmetric cycle to an equally dangerous symmetric one; T, x t- medium stresses; st th, x a- cycle amplitudes.

In the case of a combination of basic deformations (bending and torsion, torsion and tension or compression), the overall safety factor is determined as follows:

The obtained safety factors should be compared with their allowable values, which are taken from the strength standards or reference data. If the condition is met n>n then the structural element is recognized as reliable.