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 different kinds cycles variable voltages, 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.
FROM 
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). On the ordinate axis of the graph, the maximum stress or amplitude of the cycle stresses at which the given sample was tested is plotted, and on the abscissa axis - 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 high stress zones 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.
AT 
cast parts and materials with non-metallic inclusions, pores and other internal and external defects, the scale effect is more pronounced. Alloyed 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 state of the surface of the part on the fatigue resistance is due to high level stresses from bending and torsion in the outer zones of the part and weakening of the surface layer due to its roughness and destruction of crystalline 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 action asymmetric cycles details are calculated for strength based on the chart ultimate stresses cycle (Fig. 4.7), which characterizes the relationship between the ultimate stresses and the average stresses of the cycle 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 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 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 According to the fatigue damage summation hypothesis, the action 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) 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: 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
2. As a medium load
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: In the vast majority of cases, calculations for the strength of parts operating at alternating voltages are performed as verification. This is primarily due to the fact that the overall coefficient of reduction in the endurance limit or in the process of designing a part can only be chosen approximately, since the calculator (designer) at this stage of work has only very approximate ideas about the size and shape of the part. The design calculation of a part, which serves to determine its main dimensions, is usually performed approximately without taking into account the variability of stresses, but at reduced allowable stresses. After completing the working drawing of the part, its refined verification calculation is carried out, taking into account the variability of stresses, as well as design and technological factors affecting the fatigue strength of the part. At the same time, the calculated safety factors for one or more supposedly dangerous sections of the part are determined. These safety factors are compared with those that are assigned or recommended for parts similar to those designed under given operating conditions. With such a verification calculation, the strength condition has the form The value of the required safety factor depends on a number of circumstances, the main of which are: the purpose of the part (the degree of its responsibility), working conditions; the accuracy of determining the loads acting on it, the reliability of information about the mechanical properties of its material, the values of stress concentration factors, etc. Usually If the calculated safety factor is lower than required (i.e., the strength of the part is insufficient) or significantly higher than required (i.e., the part is uneconomical), it is necessary to make certain changes in the dimensions and design of the part, and in some cases even change her material. Let us consider the determination of the safety factors for a uniaxial stress state and for pure shear. The first of these types of stress state, as is known, occurs during tension (compression), direct or oblique bending and joint bending and tension (or compression) of the beam. Recall that shear stresses during bending (straight and oblique) and combination of bending with axial loading at the dangerous point of the beam, as a rule, are small and are neglected when calculating the strength, i.e., it is believed that a uniaxial stress state occurs at the dangerous point. Pure shear occurs at the points of a torsion bar with a circular cross section. In most cases, the safety factor is determined on the assumption that the duty cycle of stresses that occur in the calculated part during its operation is similar to the limit cycle, i.e., the asymmetry coefficients R and the characteristics of the working and limit cycles are the same. The most simple safety factor can be determined in the case of a symmetrical stress change cycle, since the endurance limits of the material during such cycles are usually known, and the fatigue limits of the calculated parts can be calculated from the values of the fatigue limit reduction coefficients taken from reference books. The safety factor is the ratio of the endurance limit, defined for the part, to the nominal value of the maximum voltage that occurs at the dangerous point of the part. The nominal value is the stress value determined by the basic formulas of the resistance of materials, i.e. without taking into account the factors affecting the value of the endurance limit (stress concentration, etc.). Thus, to determine the safety factor for symmetrical cycles, we obtain the following dependencies: when bending in tension-compression in torsion When determining the safety factor in the case of an asymmetric cycle, difficulties arise due to the lack of experimental data necessary to construct a segment of the limit stress line (see Fig. 7.15). Note that there is practically no need to build the entire diagram of limiting amplitudes, since for cycles with endurance limits greater than the yield strength, the safety factor must be determined by yield (for plastic materials), i.e., the calculation must be performed as in the case of a static action loads. In the presence of an experimentally obtained section AD of the limit curve, the safety factor could be determined by a graph-analytical method. As a rule, these experimental data are absent and the AD curve is approximately replaced by a straight line constructed from any two points, the coordinates of which are determined experimentally. As a result, a so-called schematized diagram of limiting amplitudes is obtained, which is used in practical strength calculations. Let us consider the main ways of schematizing the safe zone of the diagram of limiting amplitudes. In modern calculation practice, the Serensen-Kinasoshvili diagram is most often used, in the construction of which the section AD is replaced by a straight line drawn through points A and C, corresponding to the limiting symmetric and zero cycles (Fig. 9.15, a). The advantage of this method is its relatively high accuracy (the approximating straight line AC is close to the curve; its disadvantage is that, in addition to the endurance limit value for a symmetrical cycle, it is necessary to have experimental data on the endurance limit value) also with a zero cycle. When using this diagram, the safety factor is determined by endurance (fatigue failure), if the beam of cycles similar to the given one intersects the straight line and by yield, if the specified beam intersects the line A slightly lower, but in many cases sufficient for practical calculations, accuracy is given by a method based on the approximation of the section AD of the limit curve by a straight line segment (Fig. 9.15, b) drawn through points A (corresponding to a symmetrical cycle) and B (corresponding to limiting constant stresses) . The advantage of the method under consideration is the smaller amount of required experimental data compared to the previous one (data on the value of the endurance limit at zero cycle are not needed). Which of the safety factors, by fatigue failure or by yield, is less, is determined in the same way as in the previous case. In the third type of schematized diagrams (Fig. 9.15, c), an approximating straight line is drawn through point A and some point P, the abscissa of which is determined as a result of processing the available experimentally obtained limit stress diagrams. For steel, with sufficient accuracy, it can be assumed that the segment OP - s is equal. The accuracy of such diagrams is almost the same as the accuracy of diagrams constructed using the Sørensen-Kinasoshvili method. The schematized diagram is especially simple, in which the safe zone is limited by the straight line AL (Fig. 9.15, d). It is easy to see that the calculation according to such a diagram is very uneconomical, since in the schematized diagram the limit stress line is located much lower than the actual limit stress line. In addition, such a calculation does not have a definite physical meaning, since it is not known which safety factor, fatigue or yield, will be determined. Despite these serious shortcomings, the diagram in Fig. 9.15, and is sometimes used in foreign practice; in domestic practice in recent years, such a diagram is not used. Let us derive an analytical expression for determining the safety factor for fatigue failure based on the considered schematized diagrams of limiting amplitudes. At the first stage of the derivation, we will not take into account the influence of factors that reduce the endurance limit, i.e., first we will obtain a formula suitable for normal laboratory samples. Let us assume that the point N, representing the working cycle of stresses, is in the area (Fig. 10.15) and, therefore, when the stresses increase to the value determined by the point, fatigue failure will occur (as already indicated, it is assumed that the working and limiting cycles are similar). The fatigue safety factor for the cycle represented by point N is defined as the ratio Draw through the point N a line parallel to the line and a horizontal line NE. It follows from the similarity of triangles that As follows from Fig. 10.15, Let us substitute the obtained values of OA and into equality (a): Similarly, in the case of variable shear stresses The values depend on the type of schematized limit stress diagram adopted for the calculation and on the material of the part. So, if we accept the Sorensen-Kinasoshvili diagram (see Fig. 9.15, a), then likewise, According to the schematized diagram shown in Fig. 9.15, b, likewise, The values and when calculating by the method of Serensen - Kinasoshvili can be taken according to the given data (Table 1.15). Table 1.15 Coefficient values for steel When determining the safety factor for a particular part, it is necessary to take into account the influence of the fatigue limit reduction factor. Experiments show that stress concentration, scale effect and surface condition affect only the values of limiting amplitudes and practically do not affect the values of limiting average stresses. Therefore, in design practice, it is customary to refer the coefficient of fatigue limit reduction only to the amplitude stress of the cycle. Then the final formulas for determining the safety factors for fatigue failure will have the form: in bending in torsion In tension-compression, formula (22.15) should be used, but instead of substituting in it the endurance limit for a symmetrical tension-compression cycle. Formulas (22.15), (23.15) are valid for all the indicated methods of schematization of limit stress diagrams; only the values of the coefficients change Formula (22.15) was obtained for cycles with positive average stresses for cycles with negative (compressive) average stresses should be assumed, i.e., proceed from the assumption that in the compression zone the limit stress line is parallel to the abscissa axis. Many machine parts experience time-varying stresses (usually cyclic) during operation: parts of the crank mechanism, vehicle axle, gearbox shafts, etc. Experience shows that at variable stresses, after a certain number of cycles, destruction of the part may occur, while at the same stress that is unchanged in time, destruction does not occur. An example is wire. The number of cycles to failure depends on the material and stress amplitude and varies over a wide range. The destruction of a material under the action of alternating stresses is called fatigue. Describe the mechanism of destruction. It has a local character. The accumulation of fatigue damage leads to the formation of a macrocrack. Failure is caused by the development of a fatigue crack. The most common and most dangerous for the material is the harmonic law of stress change. The stress cycle is characterized by the following parameters: Maximum and minimum cycle stresses; Average cycle voltage Cycle amplitude: ; Cycle asymmetry coefficient: Figure 1. Characteristics of the stress cycle Such a cycle is called symmetrical. Such a cycle is called pulsating. All terms and definitions are also valid for variable shear stresses, if replaced by. For strength calculations at alternating stresses, it is necessary to know the mechanical characteristics of materials, which are determined by special tests. A smooth polished rod of round section and length is taken. It is subjected to a symmetrical cycle at various amplitudes. Give the scheme of the testing machine and the test procedure. The sample is brought to failure and the number of cycles to failure is determined. The resulting curve is called the fatigue curve or Wohler curve. (Figure 2). Figure 2. Fatigue Curve This curve is remarkable in that, starting from a certain tension, it goes almost horizontally. This means that at stresses less than a certain limiting stress, the sample can withstand countless cycles. The maximum variable stress that a material can withstand without destruction, for any number of cycles, is called the endurance limit and is denoted. Experiments are usually performed up to the base number of cycles. Acceptable for carbon steels, for hardened steels and non-ferrous metals. Empirical dependencies have been established experimentally: The endurance limit of parts depends not only on the properties of the material, but also on their shape, size, and manufacturing methods. Influence of stress concentration. In places of a sharp change in the dimensions of the PS part (holes, undercuts, fillets, keyways, threads), as is known, a local increase in stress occurs. This phenomenon is called stress concentration. It reduces detail compared to the sample. This decrease is taken into account by the effective stress concentration factor, which is determined experimentally. It is equal to the ratio of the endurance limits of a smooth specimen to that of a specimen with a given stress concentrator. Values are given in reference books. Influence of the sizes of details. It has been experimentally established that with an increase in the size of the sample, decreases. The influence of the sample dimensions on is taken into account by the scale factor, which is determined experimentally and is equal to the ratio Usually they take. They are listed in the handbooks. Influence of the surface condition of the part. The presence of scratches, scratches, and irregularities on the surface of the part leads to a decrease in the endurance limit of the part. The surface condition of the part depends on the type of machining. The influence of the surface state on the size of the part is taken into account by a coefficient that is determined experimentally and is equal to: This coefficient is given in reference books. All of the above factors can be taken into account by one coefficient of change in the endurance limit. Then the endurance limit of the part If we test a standard sample from the material under study under conditions of an asymmetric stress cycle, we will obtain the limit stress diagram shown in Figure 3. Figure 3. Ultimate stress diagram Tell about the methodology for conducting tests and constructing a diagram. This diagram allows you to judge the proximity of operating conditions to the limit. To do this, a working point (B) is plotted on the diagram with coordinates where and are the calculated values of the average and amplitude stresses in the part. Here, the stress amplitude is increased taking into account the reduction in the endurance limit of the part. The degree of proximity of the operating point to the limit curve is used to judge the danger of working conditions. If the operating point is outside the diagram, then fatigue failure will certainly occur. The construction of this diagram requires a lot of time and material resources. Therefore, the real diagram is schematized by direct CD. then this diagram can be constructed without experimentation. The safety factor is obviously equal to the ratio of the OA segment to the OB segment (Figure 3). After geometric constructions, we get: where is the coefficient of sensitivity of the material to the asymmetry of the cycle. Under the action of variable shear stresses Coefficients are given in reference books. Under the simultaneous action of alternating normal and shear stresses, the overall safety factor 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 5. What is called the endurance limit? How is a fatigue curve plotted? 6. List the factors that affect fatigue resistance. 306 Practice 6 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; BUT- 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. Practical session 6 307 Solution
1. Determine the load on the rods. Consider the equilibrium We apply the reactions of the bonds acting at the point AT. Freeing the point AT 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 AT: 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):
308 Practice 6 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 torque t= ZkN∙m, supported at a point BUT 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 BUT =
0. Rb- rod reaction, hinge reactions BUT we do not consider. Practical session 6 309 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 Permissible stress for the rod material Consequently, 3. We select the section of the rod according to GOST (Appendix 1). (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 BUT 1i BUT 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. To accept [σ] =
160 MPa. The self-weight of the structure is not taken into account. Practical session 6 311 When defending the work, answer the questions of the test task. 312 Practice 6 Topic 2.2. Stretching and compression. Strength and stiffness calculations Practical session 7 313 Practice 7 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; in - 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, in). 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 r G= 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 the 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 about _ 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 empirically 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 Serensen-Kinasoshvili formulas 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.
maximum limit stress of asymmetric cycles
(29)
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.
(35)
(37)
and coefficient of variation v torque, the statistical distribution of the amplitudes of which can be considered truncated normal.
a torque is received in the power circuit to the body, corresponding to the implementation of a stable moment M y engines.
(20.15)
(21.15)
(22.15)
(23.15)
endurance limit
Factors affecting the value of the endurance limit
Determination of the safety factor for alternating voltages
lower than with constant (static)?
points AT, 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.
river section: about= N/A^ [a], where BUT> N/[a].
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
