Fundamentals of strength calculations under variable stresses. Determination of the safety factor at variable voltages

Calculations for normal and shear stresses are carried out similarly.

The calculated coefficients are selected using special tables.

When calculating, safety margins are determined by normal and tangential stresses.

Safety factor for normal stresses:

Safety factor for tangential stresses:

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

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

Test questions and assignments

1. Draw graphs of symmetrical and zero-cycle voltage change cycles under repeated alternating voltages.

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

3. Describe the nature of fatigue failure.

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

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

6. List the factors influencing fatigue resistance.

306 Practical lesson 6

PRACTICAL LESSONS ON SECTION

"Strength of materials"

Practical lesson 6

Topic 2.2. Strength and stiffness calculations

In tension and compression

Know the procedure for calculating strength and stiffness and calculation formulas.

Be able to carry out design and testing calculations for strength and rigidity in tension and compression.

Required formulas

Normal voltage

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

Lengthening (shortening) timber

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

Allowable voltage

[s]- permissible safety margin.

Tensile and compressive strength condition:

Examples of strength and stiffness calculations

Example 1. The load is fixed on the rods and is in equilibrium (Fig. P6.1). The material of the rods is steel, the permissible stress is 160 MPa. Load weight 100 kN. Length of rods: first - 2 m, second - 1 m. Determine the cross-sectional dimensions and elongation of the rods. The cross-sectional shape is circle.

Practical lesson 6 307

Solution

1. Determine the load on the rods. Consider the equilibrium
points IN, Let us 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 plot the reactions of bonds acting at a point IN. Freeing the point IN from connections (Fig. P6.1).

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

Let's create a system of equilibrium equations for the point IN:

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

R 1 = R2cos60 °; R 1= 115.5 ∙ 0.5 = 57.4 kN.

The direction of reactions is chosen correctly. Both rods are compressed. Loads on rods: F 1= 57.4 kN; 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 Practical lesson 6

We round the resulting diameters: d 1 = 25mm, d 2= ​​32 mm.

3. Determine the elongation of the rods Δ l = ----- .

Shortening rod 1:

Shortening rod 2:

Example 2. Homogeneous rigid slab with a gravity force of 10 kN, loaded with force F= 4.5 kN and torque T= ZkN∙m, supported at point A and suspended on a rod Sun(Fig. P6.2). Select the cross-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 safety factor 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- reaction of the rod, reaction of the hinge A We don't consider it.

Practical lesson 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 slab. The force in the rod is 14 kN.

2. Based on the strength condition, we determine the required size of the butt area
river section: O= N/A^ [A], where A> N/[a].

Allowable stress for rod material

Hence,

3. Select the cross-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 advisable to use equal angle angle No. 2

(d= Zmm), - the cross-sectional area of ​​which is 1.13 cm 2 (GOST 8509-86).

4. Determine the elongation of the rod:

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

Calculation 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 Practical lesson 6

Task 2. Beam AB, upon 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) cross-section - circle; 2) cross-section - equal angle angle according to GOST 8509-86. Accept [σ] = 160 MPa. Do not take into account the dead weight of the structure.

Practical lesson 6 311

When defending your work, answer the test questions.

312 Practice 6

Topic 2.2. Tension and compression.

Strength and stiffness calculations

Practical lesson 7 313

Practical lesson 7

At the turn of the XIX-XX centuries. In connection with the creation and entry into everyday life of new types of machines, installations and vehicles operating under loads that change cyclically over time, it turned out that existing calculation methods did not provide reliable results for calculating such structures. For the first time, a similar phenomenon was encountered in railway transport, when a number of disasters occurred related to the breaking of the axles of cars and steam locomotives.

Later it turned out that the cause of the destruction was alternating stresses that arose during the movement of the train due to the rotation of the car's axle along with the wheels. However, it was initially suggested that during long-term operation the metal changes its crystalline structure - gets tired. This assumption was not confirmed, but the name “fatigue calculations” was retained in engineering practice.

Based on the results of further research, it was found that fatigue failure is caused by the processes of accumulation of local damage in the material of the part and the development of cracks. It is these processes that arise during the operation of various machines, vehicles, machine tools and other installations subject to vibration and other types of time-varying loads that will be considered further.

Let us consider a cylindrical sample fixed in a spindle at one end, at the other, free end, a force is applied through a bearing F(Fig. 16.1).

Rice. 16.1.

The diagram of the bending moment of the sample varies according to a linear law, and its maximum value is equal to FI. At cross-sectional points of the sample A And IN maximum voltages occur in absolute magnitude. The magnitude of the normal stress at point A will be

In the case of rotation of the sample with angular velocity, the cross-sectional point changes its position relative to the plane of action of the bending moment. During t characteristic point A will rotate through the angle φ = ω/ and end up in a new position A"(Fig. 16.2, A).

Rice. 16.2.

The voltage in the new position of the same material point will be equal to

Similarly, you can consider other points and come to the conclusion that when the sample rotates, due to changing the position of the points, the normal stresses change according to the cosine law (Fig. 16.2, b).

To explain the process of fatigue failure, it is necessary to abandon the fundamental hypotheses about the material, namely the continuity hypothesis and the homogeneity hypothesis. Actual materials are not perfect. As a rule, the material initially contains defects in the form of crystal lattice imperfections, pores, microcracks, and foreign inclusions, which cause structural heterogeneity of the material. Under cyclic loading conditions, structural inhomogeneity leads to inhomogeneity of the stress field. In the weakest places of the part, microcracks appear, which, under the influence of time-varying stresses, begin to grow, merge, turning into main crack. Once in the tension zone, the crack opens, and in the compression zone, on the contrary, it closes.

The small local area in which the first crack appears and from where its development begins is called focus of fatigue failure. Such an area, as a rule, is located near the surface of the parts, but it is possible that it will appear deep in the material if there is any damage there. The simultaneous existence of several such areas is not excluded, and therefore the destruction of a part can begin from several centers that compete with each other. As a result of the development of cracks, the section is weakened until destruction occurs. After failure, the zone of fatigue crack development is relatively easy to recognize. In the cross section of a part destroyed by fatigue, there are two sharply different areas (Fig. 16.3).

Rice. 16.3.

1 - area of ​​crack growth; 2 - area of ​​brittle fracture

Region 1 characterized by a shiny, smooth surface and corresponds to the beginning of the destruction process, which occurs in the material at a relatively low speed. At the final stage of the process, when the section weakens sufficiently, a rapid avalanche-like destruction of the part occurs. This final ethane in Fig. 16.3 corresponds to area 2, which is characterized by a rough, rough surface due to the rapid final destruction of the part.

It should be noted that the theoretical study of the fatigue strength of metals is associated with significant difficulties due to the complexity and multifactorial nature of this phenomenon. For this reason, the most important tool is phenomenological approach. For the most part, formulas for calculating parts for fatigue are derived from experimental results.

Calculation of metal structures should be carried out using the method of limit states or permissible states. stress. In complex cases, it is recommended to solve issues of calculation of structures and their elements through specially designed theoretical and experimental studies. The progressive method of calculation based on 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 load on the structures of certain types of cranes, their calculations are carried out using the permissible stress method, based on safety factors established in practice. ­

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

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

σ pr = σ T, (170)

and if σ = 0, then the limiting shear stress

τ = = 0.578 σ T ≈ 0,6σ T. (171)

In addition to strength calculations for certain types of cranes, there are restrictions 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 the span (departure) l.Significant deflections may occur. safe for the structure itself, but unacceptable from an operational point of view.

Calculation using the limit state method is carried out based on the loads given in table. 3.

Notes on the table:

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

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

To take into account the lower probability of coincidence of design loads with more complex combinations of loads, combination coefficients are introduced n with < 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 number of cycles and the combination of loads Ib with its number of cycles should be taken into account.

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

5. Working wind pressure R b II and non-working - hurricane R b III - for the design is determined according to GOST 1451-77. When combining loads Ia and Ib, wind pressure on the structure is usually not taken into account due to the low annual frequency of design wind speeds. For tall cranes that have a period of free oscillation of the lowest frequency of more than 0.25 s and are installed in windy regions IV-VIII according to GOST 1451-77, the wind pressure on the structure with a combination of loads Ia and Ib is taken into account.

6. Technological loads can relate to both load case II and load case III.

Table 3

Loads in calculations using the limit state method

Limit states are called 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 throughout the entire service life of the structure.

Metal structures of hoisting machines (hoisting and transport machines) must meet the requirements of two groups of limit states: 1) loss of the load-bearing capacity of the crane elements in terms of strength or loss of stability from a single action of the largest loads in operating or non-operating condition. The working state is considered to be the state in which the crane performs its functions (Table 3, load case II). A state is considered inoperative when the crane without a 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 the bearing capacity of crane elements due to failure from fatigue under repeated exposure to loads of various magnitudes over the design 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 operating personnel. For the second limit state for the development of excessive deformations (deflections, rotation angles), limit condition (172) is established for individual types of cranes.

Calculations for the first limit state are of greatest importance, since with 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- calculated (maximum) load in the element under consideration, expressed in force factors (force, moment, stress); F- calculated load-bearing capacity (smallest) of the element according to power 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 standard loads R N i(for hoisting and transport machine designs, these are the maximum operating condition loads, entered into the calculation both on the basis of technical specifications and on the basis of design and operating experience) 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 calculated force in the element N in accordance with the design combinations of loads given in table. 3, can be represented as

, (174)

Where αi– force in the element at R N i= 1, and the design moment

, (175)

Where M N i– moment from the standard load.

To determine overload factors, a statistical study of load variability based on experimental data is necessary. Let for a given load P i its distribution curve is known (Fig. 63). Since the distribution curve always has an asymptotic part, when assigning a design load, it should be borne in mind that loads that are greater than the design ones (the area of ​​these loads is shaded in Fig. 63) can cause damage to the element. Taking larger 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 load factor must 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 load-bearing capacity F. Then (Fig. 64) the shaded area, within the boundaries of which the limit condition (173) is violated, will characterize the probability of destruction.

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

For the first limit state for loss of load-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 with overload factors equal to unity, i.e., for standard loads (the weight of the load is assumed to be equal to the nominal weight).

Function F in formula (173) can be represented as

F= Fm K R, (177)

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

Under design resistance R should be understood when calculating:

for fatigue resistance - the endurance limit of the element (taking into account the number of cycles of load changes and the coefficients of concentration and asymmetry of the cycle), multiplied by the corresponding uniformity coefficient for fatigue tests, characterizing the scatter of test results, k 0= 0.9, and divided by k m is the reliability coefficient for the material when calculating strength, characterizing both the possibility of changing the mechanical properties 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;

for strength under constant stress R= R P /k m – ­ the quotient of dividing the standard resistance (standard yield strength) by the corresponding reliability coefficient for the material; for carbon steel k m = 1.05, and for low-alloy - k m = 1.1; Thus, in relation to the work of the material, the limiting state is not the complete loss of its ability to bear the load, but the onset of large plastic deformations that prevent further use of the structure;

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

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

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

Where m 1 – coefficient that takes into account the responsibility of the element being calculated, i.e. the possible consequences of destruction; the following cases should be distinguished: destruction does not cause the crane to stop operating, 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 that takes 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 that takes into account calculation imperfections associated with inaccurate determination of external forces or design schemes. It must be installed for certain types of structures and their elements. Can be accepted 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 limiting state is carried out by increasing the level of alternating stress, according to formula (176), where the calculated 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 using the permissible stress method is carried out based on the loads given in Table 4. All notes to the table must be taken into account. 3, except note 2.

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

Calculation using the permissible stress method is carried out in cases where there are no numerical values ​​for the overload factors of the design loads to perform calculations using the limit state method. Strength calculations are made using 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 permissible stresses for bending are assumed to be 10 MPa (about 5%) greater than for tension (for St3 180 MPa), taking into account that during bending the yield first appears only in the outermost fibers and then gradually spreads to the entire cross-section of the element , increasing its load-bearing capacity, i.e., during bending, there is a redistribution of stresses across the section due to plastic deformations.

When calculating fatigue resistance, if the transition to the limit state is carried out by increasing the level of alternating 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] – permissible stresses, when determining which the expression [ σ ] = σ –1K /n 1 or similar to formulas (181) – (183) instead σ –1K are used σ –1KN , σ * –1K And σ * –1KN. Margin of safety n I is the same as when calculating 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 d = ( 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 d ≈ 7.55.

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

(191)

And . Then the safety margin for symmetrical cycles

i.e. P= n σ n τ /, (192)

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

The progressiveness of the calculation method based on limit states lies in the fact that when calculating using this method, the actual work of structures is better taken into account; overload factors are different for each load and are determined based on a statistical study of load variability. In addition, using the material safety factor, the mechanical properties of materials are better taken into account. While when calculating using the permissible stress method, the reliability of the structure is ensured by a single safety factor, when calculating using the limit states method, instead of a single safety factor, a system of three coefficients 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 based on permissible stresses is a special case of the calculation based on the first limit state, when the overload factors for all loads are the same. However, it must be emphasized that the calculation method based on limit states does not use the concept of safety factor. It is also not used by the probabilistic calculation method currently being developed for crane construction. Having performed the calculation using the limit state method, you can determine the value of the resulting safety factor using the permissible stress method. Substituting into formula (173) the values N[cm. formula (174)] and F[cm. formula (177)] and turning to stresses, we obtain the value of the safety factor

n =Σ σ i n i k M / (m K Σ σi). (193)

• Calculation of the force of a variable Stress When calculating strength under variable stresses, the strength of a part is usually assessed by the value of the actual safety factor P, compared with the permissible safety factor established by the norm, the strength condition is written n>. Safety factors P, for example, can be determined approximately using a schematic view of the limiting amplitude. 460.6 first, find the safety factor for a smooth standard

samples, not the actual part. The external load assumes that the duty cycle in which the safety factor is determined and the corresponding limit cycle vary in a similar way. From the source of the diagram (see diagram. 460.6) draw ray 01 at an angle a defined (§a = -, where AA is the amplitude and the average voltage of the duty cycle. Point M on the straight line with coordinates AA and at, characterizes the duty cycle . Point N coordinates l 18 order ha 1037 549i putt characterizes the limit value of the same cycle. Thus, the value of the safety factor p can be determined.

as (W Segment ratio. If beam 01 intersects the straight line AB, then an increase in cycle stress will cause fatigue failure Lyudmila Firmal

sample. The safety factor for fatigue failure in this case is expressed in n#, where point N is on the line AB and satisfies equation (18.11). 0_1=аш+п^а, (18.13) Where does PJ= (18.14) The safety factor for a smooth sample is obtained. The strength of a part depends on the size and shape of the part and the condition of its surface. All this is taken into account by the corresponding coefficient, the effective stress concentration factor ka, the surface sensitivity factor p, and the scale factor EE. To obtain the maximum amplitude indicator of the corresponding part, it is necessary

reduce the endurance limit in a symmetrical cycle -?- times, or, what is the same thing, once the voltage amplitude of the working cycle AA increases, then formula (18.13) will take the form The safety factor of the part is equal to the following values ​​(18.15)) (18.16) Please note that you are using if instead of figure. 460, B) apply additional simplified schemes built on the basis of two points (Fig. 460, a), in Formula (18.16) only the angular coefficient f of straight line AB changes. In this case, you need to take If beam 01 intersects a straight line, then increased cyclic stresses disable the part due to the appearance of plastic deformation in it. 550co-efficiency of the stock, relative to the yield strength, is indicated by l and is calculated using the formula Antibodies Gold = - - - And Shah. KTG AA+~T (18.17) For parts from

• In high-strength steel, failure can occur due to a decrease in static strength due to stress concentration. This case is possible when the asymmetry coefficient is close to unity. The margin coefficient in this case is determined by the formula D. V. d (18.18) Where ov is the tensile strength; o-voltage, determined without taking into account concentration; — coefficient taking into account the reduction in static strength due to stress concentration, the effective static stress concentration coefficient. The above calculation refers to the case of a uniaxial stress state. For a plane or volumetric stress state, the task of assessing strength is much more complicated. Strength theory developed and well tested by experiments

at constant voltage, is not directly applicable to the case of fluctuating voltage. At present, this problem has not been satisfactorily resolved. In practice, the calculations use the following dependencies in plane stress states, which are characterized by a normal stress o and a shear stress t: (18.19) Here the p-safety factor required for a plane stress state, PA, p~ - assuming that only the normal stress o or tangential stress act accordingly according to equation (18.16). Dependence (18.19) is confirmed by some experiments. It also extends the third theory of strength (the theory of maximum shear stress) in the case of stresses and T

changes in a symmetrical cycle in one stage. It is used in the absence of phase changes in Eighteen * 551 from equation (18.19) is required Lyudmila Firmal

safety factor (18.20)) P r and M E R1. The tubular piston pins of the engine are loaded with a force P, varying from P = 6000 kg to P = - 2000 kg. Mechanical characteristics of the piston pin material: yield strength = = = 10,000 kg/cm2 tensile strength AB = 8000kpsm2, symmetrical cycle o endurance limit, * = 5000kpsm2, zero cycle a o-7500kg / cm2 The outer surface of the pins is polished. Surface sensitivity coefficient p=1; scale factor E0=0.9; effective stress concentration factor& = 1.1. Determine the safety factor under fatigue loading. For rice. 463 shows a diagram of the transmission of force to the finger and is located in the diagram. 463, b-graph of bending moment. 1g (1=30mm0=5 0mm And (1=30mm / Fig,

463A.< При изгибе конструкция сечения равна ^изг-2а+2)~Б ‘ 2 4~ = ~ (4 — 1 , 2 5) = 1,375 П. Момент сопротивления секции г — (вперед)! =2 ‘ 44cm3- 552 максимальные и минимальные значения изгибающего момента: Mi zgtah=1,375 Rtah=1,375-6000=8250 кг-см\Mizgtk1=1,375 rt1p=1,375 (-2000)= — 2750 кг-см. Максимальное и минимальное нормальное напряжение тока OTA= = 3380KPCM^-, M izg GP1P pip C / _ _ 2750 -2.44 Из Кпсм2. Амплитуда и среднее значение напряжения рабочего цикла °тахометра stt1p2 °a zzzo — ^и zo)=2255 кг / см2. тонна STT a x H~A gtnp Два. =338°+0^2.130)=P25kg1smg. Определим предельное значение напряжения нулевого цикла: амплитудное и среднее * А0 Два. Семь тысяч пятьсот Два. =3750kpcm?. Кроме того, создайте диаграмму предельной величины по известным

values ​​a_yd d _ ^255 1.1 _ _ p-de. ‘P e 1125 1l O2’ 4 5, =68° 1-0, 9. We believe that the operating and limit cycles are similar. Point M * AA=2720 kg/cm with voltage duty cycle coordinates? AND ______5000____ 0.333-1125 + - /D2+D2~y(1.23)2+ (4.14)2 - = 1.2.