Practical methods for calculating shear and crushing. Calculation of bolted and rivet connections. Finger calculation Basic concepts. Calculation formulas

Connection parts (bolts, pins, dowels, rivets) work in such a way that only one internal force factor can be taken into account - transverse force. Such parts are designed for shear.

Shear (slice)

Shear is a loading in which only one internal force factor appears in the cross section of the beam - the transverse force (Fig. 23.1).

When shifting, Hooke's law is satisfied, which in this case is written as follows:

where is voltage;

G- shear elastic modulus;

Shear angle.

In the absence of special tests G can be calculated using the formula,

Where E- tensile modulus of elasticity, [ G] = MPa.

Calculation of parts for shear is conditional. To simplify the calculations, a number of assumptions are made:

When calculating shear, the bending of parts is not taken into account, although the forces acting on the part form a pair;

When calculating, we assume that the elastic forces are distributed uniformly over the section;

If several parts are used to transfer the load, we assume that the external force is distributed evenly between them.

Shear (shear) strength condition

where is the permissible shear stress, it is usually determined by the formula

When destroyed, the part is cut across. The destruction of a part under the influence of shear force is called shearing.

Quite often, simultaneously with shear, compression of the side surface at the point of contact occurs as a result of the transfer of load from one surface to another. In this case, compressive stresses arise on the surface, called crushing stresses.

The calculation is also conditional. The assumptions are similar to those adopted when calculating shear, however, when calculating a lateral cylindrical surface, the stresses are not evenly distributed over the surface, so the calculation is carried out for the most loaded point. To do this, instead of the side surface of the cylinder, a flat surface passing through the diameter is used in the calculation.

Bearing strength condition

whereA cm - calculated area crumple

d - diameter of the cross-sectional circle;

Minimum height of connected plates;

F - interaction force between parts

Allowable bearing stress

= (0,35 + 0,4)

Topic 2.5. Torsion

Torsion is a type of loading of a beam, in which one internal force factor appears in its cross sections - torque M cr.

The torque Mcr in an arbitrary cross section of the beam is equal to the algebraic sum of the moments acting on the cut-off part of the beam.

Torque is considered positive if the torsion occurs counterclockwise and negative - clockwise.

When calculating shafts for torsional strength, the strength condition is used:

,

where is the polar moment of resistance of the section, mm 3;

– permissible tangential stress.

Torque is determined by the formula:

where P – shaft power, W;

ω – angular speed of shaft rotation, rad/s.

The polar moment of resistance of the section is determined by the formulas:

For a circle

For the ring

.

When a beam is torsioned, its axis experiences twisting through a certain angle φ, which is called twist angle. Its value is determined by the formula:

where l is the length of the beam;

G – shear modulus, MPa (for steel G=0.8·10 5 MPa);

Polar moment of inertia of the section, mm 4.

The polar moment of inertia of the section is determined by the formulas:

For a circle

For the ring

.

Topic 2.6. Bend

Many structural elements (beams, rails, axles of all wheels, etc.) experience bending deformation.

Bend is called deformation from the moment of external forces acting in a plane passing through the geometric axis of the beam.

Depending on the application locations active forces differentiate straight And oblique bend

Straight bend– external forces acting on the beam, lie in the main section plane.

The main section plane is a plane passing through the axis of the beam and one of the main central axes of the section.

Oblique bend- external forces acting on the beam, do not lie in the main section plane.

Depending on the nature of the VSF occurring in the cross sections of the beam, the bending can be clean And transverse.

The bend is called transverse, if two VSFs arise in the cross section of the beam - bending moment M x and transverse force Q y.

The bend is called clean, if one BSF occurs in the cross section of the beam - bending moment M x.

The bending moment in an arbitrary section is equal to the algebraic sum of the moments of external forces acting on the cut-off part of the beam:

The transverse force Q is equal to the algebraic sum of the projections of external forces acting on the cut-off part of the beam:

When determining the signs of transverse forces, use "Clockwise" rule: shear force is considered positive if the “rotation” of external forces occurs clockwise; negative – counterclockwise.

When determining the signs of bending moments, use "Compressed fibers" rule(“BOWL” rule): the bending moment is considered positive if the upper fibers of the beam are compressed (“water does not pour out”); negative if the lower fibers of the beam are compressed (“water pours out”).

Bending strength condition: the operating voltage must be less than or equal to the permissible voltage, i.e.

where W x is the axial moment of resistance (a value characterizing the ability of structural elements to resist bending deformation), mm 3.

The axial moment of resistance is determined by the formulas:

For a circle

For the ring

;

For a rectangle

In direct transverse bending, the bending moment causes the occurrence of normal stress, and the transverse force causes tangential stress, which is determined by the formula:

where A is the cross-sectional area, mm 2.

This design uses three finger connections: the handle rocker and the connection between the small plunger and the handle. In both the first and second cases there are two cut planes, which has a direct impact on the strength of the structure. Finger joints are usually designed to withstand shearing and crushing:

Permissible shear tension of the finger,

;

- permissible tension of the finger for crushing,

;

where, F – load acting on the finger joint;

Z – total number of fingers in the joint;

δ – sheet thickness, mm;

dhole – hole diameter, mm;

K – number of cutting planes.

Finger cut for St0, St2 – 1400 kgf/cm2; for St3 – 1400 kgf/cm2.

Finger crushing for St0, St2 – 2800 kgf/cm2, for St3 – 3200kgf/cm2.

Calculation of the finger on the body:

mm;

mm.

Calculation of the finger on the plunger:

mm;

mm.

I accept a finger with a stop head of d=3 mm; D=5.4 mm; L=12mm.

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In engineering practice, fasteners and connecting elements of machine parts and building structures: rivets, bolts, dowels, welds, notches, etc. These parts are either not rods at all, or their length is of the same order as the transverse dimensions. The exact theoretical solution of such calculation problems is very difficult and therefore they resort to conditional (approximate) calculation methods. In this kind of calculations, they proceed from extremely simplified diagrams, determine the conditional stresses using simple formulas and compare them with the permissible stresses found from experience. Typically, such conditional calculations are made in three directions: for shear (shear), for crushing at the points of contact between parts of the connection, and for rupture along a section weakened by holes or inserts. 24 When considering each design scheme, the stresses are conventionally assumed to be uniformly distributed over the dangerous section. Due to large number conventions underlying the calculation of bolted, rivet connections , welds and other similar interfaces of structural elements, practice has developed a number of recommendations that are presented in special courses on machine parts, building structures, etc. Below are only some typical examples of conditional calculations. Calculation of bolted and rivet joints Bolted and rivet joints (Fig. 1.21) are calculated for shear (shear) and crushing of the bolt or rivet rod. In addition, the connected elements are checked for rupture along the weakened section. Rice. 1.22 Bolted and rivet connections (Fig. 1.22) are calculated for shear (shear) and crushing of the bolt or rivet rod. In addition, the connected elements are checked for rupture along the weakened section. a) calculation based on permissible stresses Shear calculation Shear strength condition for a rivet or bolt rod (1.42) where P is the force acting in the connection; d – diameter of the bolt or rivet shaft; m – number of slices, i.e. planes along which the rod can be cut; - permissible tangential stress. From the strength condition, you can determine the number of cuts. The number of rivets n is determined by the number of cuts: for single-cut rivets n = m, for double-cut rivets - . Calculation for crushing Collapse occurs at the contact surface of the sheet with the shank of the rivet or bolt. The crushing stresses are distributed unevenly over this surface (Fig. 1.22, a). A conditional stress is introduced into the calculation, uniformly distributed over the diametrical cross-sectional area (Fig. 1.23, b). This conditional stress is close in magnitude to the actual maximum bearing stress on the contact surface. The strength condition is written as follows: The required number of rivets based on crushing (1.45) here is the thickness of the sheet; с m – permissible bearing stress. Checking the sheet for tensile strength Condition for the tensile strength of the sheet in the section weakened by rivet holes, (1.46) where b is the width of the sheet; n1 is the number of rivets in the seam along which rupture is possible. Checking for sheet shear In some connections, in addition to the checks listed, it is necessary to check for shear (cut) by riveting the part of the sheet between its edge (end) and the rivet (Fig. 1.24). Each rivet cuts along two planes. The length of the cutting plane is conventionally taken to be the distance from the end edge of the sheet to the nearest point of the hole contour, i.e. the value. The strength condition in this case is (1.48) where P1 is the force per one rivet; c – distance from the end of the sheet to the center of the rivet. Values ​​of permissible stresses for steel grades Art. 2 and Art. 3 in rivet joints, approximately the following can be accepted (MPa): Main elements Rivets in drilled holes Rivets in pressed holes For steel bolts, pins and similar elements of mechanical engineering structures under static load, the permissible stresses are accepted depending on the quality of the material: (0.520.04 ) T, where T is the yield strength of the bolt material; =100 - 120 MPa for steel 15, 20, 25, St. 3, Art. 4; c = 140 - 165 MPa for steel 35, 40, 45, 50, St. 5, Art. 6; s =(0.4 - 0.5)  IF for iron casting. When calculating the crushing of contacting parts from different materials The calculation is based on the permissible stress for a less durable material. b) calculation based on limit states Rivet joints are calculated based on the first limit state – the load-bearing capacity for shear and crushing. The shear is calculated according to the condition (1.48) where N is the design force in the connection; n – number of rivets; nср – number of cut planes of one rivet; d – rivet diameter; Rav – calculated shear resistance of rivets. The collapse is calculated according to the condition (1.49) where Rcm is the calculated resistance to collapse of the connected elements; – the smallest total thickness of elements crushed in one direction. Design resistances adopted in the calculation based on limit states (MPa). The main elements of ischuavyzerSe R130 eynlamron R210 cR Rivets in drilled holes Rivets in pressed holes When designing rivet joints, the diameter of the rivets is usually specified, taking it depending on the thickness of the riveted elements and rounded according to GOST: . The most commonly used diameters are: 14, 17, 20, 23, 26, 29 mm. Recommendations for placing rivets and designing riveted and bolted joints are given in special courses. 1.12. Calculation of wooden notches Calculation of wooden notches is carried out for chipping and crushing. Allowable stresses or design resistances are set depending on the direction of the acting forces in relation to the fibers of the wooden elements. The values ​​of permissible stresses and calculated resistances for air-dry (humidity 15%) pine and spruce are given in the appendix. 5. In the case of using other wood species, the voltage values ​​given in the table are multiplied by correction factors. The value of these coefficients for oak, ash, hornbeam wood: When bending, stretching, compressing and crushing along the grain 1.3 When compressing and crushing across the grain 2.0 When chipping 1.6 When crushing at an angle to the direction of the grain, the permissible stress is determined by the formula (1.50) where [cm] is the permissible bearing stress along the fibers; ms 90 – the same perpendicular to the fibers. A similar formula is used to determine the permissible stress if the shearing area is located at an angle to the direction of the fibers. – permissible folding stress along the fibers; 90 – the same across the fibers. The design resistances are calculated in the same way when calculating by limit states. When calculating the limit states of frontal notches and some other connections, the uneven distribution of tangential stresses along the shearing area should be taken into account. This is achieved by introducing average shear resistance instead of the main (maximum) design resistance (Rsk = 24 kg/cm2). (1.54) where lск is the length of the shearing area; e – shoulder of shearing forces, measured perpendicular to the shearing area; – coefficient depending on the nature of chipping. For one-sided spalling (in tensile elements), which occurs in frontal notches, = 0.25. 1.13 Strength theory Strength theories seek to establish a strength criterion for a material in a complex stress state (volumetric or plane). In this case, the studied stress state of the calculated part (with the main stresses at the dangerous point σ1, σ2, and σ3) is compared with the linear stress state - tension or compression. The limiting state of plastic materials (materials in a plastic state) is taken to be the state in which noticeable residual (plastic) deformations begin to appear. For brittle materials, or those in a brittle state, the limiting state is considered to be one in which the material is at the border of the appearance of the first cracks, i.e. at the border of the violation of the integrity of the material. The strength condition for a volumetric stress state can be written as follows: where is the equivalent (or design) stress; PRE – maximum stress for a given material in a linear stress state; - permissible stress in the same case; - actual safety factor; - required (specified) safety factor; The safety factor (n) for a given stress state is a number indicating how many times all components of the stress state should be simultaneously increased in order for it to become the limiting state. The equivalent stress EKV is a tensile stress under a linear (uniaxial) stress state that is equally dangerous with a given volumetric or plane stress state. Formulas for equivalent stress, expressing it through the principal stresses σ1, σ2, σ3, are established by strength theories depending on the strength hypothesis adopted by each theory. There are several theories of strength or hypotheses of limiting stress states. The first theory, or the theory of maximum normal stresses, is based on the assumption that a dangerous state of a material under a volumetric or plane stress state occurs when its largest absolute value normal stress reaches a value corresponding to a dangerous state under simple tension or compression. Equivalent stress according to this theory (1.57) Strength condition at identical values permissible tensile and compressive stresses (plastic materials) has the form: For different values ​​of permissible tensile and compressive stresses, the strength condition is written as follows: (1.59) In the case when, i.e., all principal stresses are tensile, the first of the formulas (1.59) is applied ). 31 In the case when, i.e., all the main stresses are compressive, the second of formulas (1.59) is applied. In the case of a mixed stress state, when both formulas (1.59) are applied simultaneously. The first theory is completely unsuitable for plastic materials, as well as in cases where all three principal stresses are unambiguous and close to each other in magnitude. Satisfactory agreement with experimental data is obtained only for brittle materials in the case when one of the main stresses is significantly greater in absolute value than the others. Currently, this theory is not used in practical calculations. The second theory, or the theory of the greatest linear deformations, is based on the proposal that a dangerous state of a material occurs when the largest relative linear deformation in absolute value reaches a value corresponding to a dangerous state under simple tension or compression. The equivalent (calculated) stress is taken to be the greatest of the following values: The strength condition at has the form: In the case different meanings permissible tensile and compressive stresses, the strength conditions can be represented as follows: (1.62) Moreover, the first of the formulas is applied for positive (tensile) principal stresses, the second - for negative (compressive) principal stresses. In the case of a mixed stress state, both formulas (1.62) are used. The second theory is not confirmed by experiments for materials that are plastic or in a plastic state. Satisfactory results are obtained for materials that are brittle or in a brittle state, especially in cases where all principal stresses are negative. Currently, the second theory of strength is almost never used in practical calculations. 32 The third theory, or the theory of the highest tangential stresses, assumes that the appearance of a dangerous condition is caused by the highest tangential stresses. The equivalent stress and strength condition can be written as follows: Taking into account the principal stresses determined by formula (1.12), after transformations we obtain: (1.64) where and, respectively, are the normal and tangential stresses at the point of consideration of the stressed state. This theory gives quite satisfactory results for plastic materials that resist tension and compression equally well, especially in cases where the principal stresses are of 3 different signs. The main disadvantage of this theory is that it does not take into account the average principal stress 2, which, as experimentally established, has some effect on the strength of the material. In general, the third theory of strength can be considered as a condition for the onset of plastic deformations. In this case, the yield condition is written as follows: The fourth theory, or energy theory, is based on the assumption that the cause of dangerous plastic deformation (yield) is the energy of shape change. In accordance with this theory, it is assumed that a dangerous state during complex deformation occurs when its specific energy reaches dangerous values ​​during simple tension (compression). The calculated (equivalent) stress according to this theory can be written in two versions: (1.66) In the case of a plane stressed state (occurs in beams during bending with torsion, etc.) taking into account the main stresses 1, 2(3) . The strength condition can be written in the form 33 Experiments well confirm the results obtained according to this theory for plastic materials that are equally resistant to tension and compression, and it can be recommended for practical use. The same value of the design stress as in formulas (1.66) can be obtained by taking the octahedral shear stress as a strength criterion. The theory of octahedral shear stresses assumes that the appearance of yield under any type of stress state occurs when the octahedral shear stress reaches a certain value that is constant for a given material. The theory of limit states (Mohr's theory) is based on the assumption that the strength in the general case of a stressed state depends mainly on the magnitude and sign of the largest 1 and smallest 3 principal stresses. The average principal stress 2 only slightly affects the strength. Experiments have shown that the error caused by neglecting 2 in the worst case does not exceed 12–15%, and is usually less. If you do not take it into account, any stressed state can be depicted using a stress circle built on the difference in the principal stresses. Moreover, if they reach values ​​​​corresponding to the limiting stress state at which a violation of strength occurs, then the Mohr circle is the limiting one. In Fig. Figure 1.25 shows two limit circles. Circle 1 with a diameter OA equal to the tensile strength corresponds to simple tension. Circle 2 corresponds to simple compression and is built on the diameter of the OB equal to the compressive strength. Intermediate limit stress states will correspond to a number of intermediate limit circles. The envelope of the family of limit circles (shown in the figure by a dotted line) limits the strength region. Rice. 1.25 34 In the presence of a limiting envelope, the strength of a material under a given stress state is assessed by constructing a circle of stresses according to given values ​​3. Strength will be ensured if this circle fits entirely inside the envelope. To obtain the calculation formula, the envelope curve between the main circles 1 and 2 is replaced by a straight line (CD). In the case of an intermediate circle 3 with principal stresses 3 touching the straight line CD, from consideration of the drawing one can obtain next condition strength: On this basis, the equivalent (calculated) stress and strength condition according to Mohr’s theory can be written as follows: – for plastic materials; – for fragile materials; or – for any material. Here are the yield limits under tension and compression, respectively; PSR – tensile and compressive strength limits; – permissible tensile and compressive stresses. With a material that is equally resistant to tension and compression, i.e., when the strength condition according to Mohr’s theory coincides with the strength condition according to theory 3. Therefore, Mohr's theory can be considered as a generalization of the 3rd theory of strength. Mohr's theory is quite widely used in calculation practice. The best results are obtained in mixed stress states, when the Mohr circle is located between the limit circles of tension and compression (at. Noteworthy is the generalization of the energy theory of strength proposed by P.P. Balandin for the purpose of applying this theory to assessing the strength of materials with different resistance to tension and compression. The equivalent stress according to the proposal of P. P. Balandin is determined by the formula: the equivalent stress found using this formula coincides with the equivalent stress according to the 4th (energy) theory of strength. Currently, experimental data is not enough for an objective assessment of this proposal. N. N. Davidenkov and Ya.B. Friedman proposed a new “unified theory of strength” that generalizes modern views on strength in the brittle and plastic states of a material. In accordance with this theory, the state in which the material is located, and therefore the nature of the probable destruction, is determined by the ratio the material is in the brittle condition, destruction occurs by separation and strength calculations must be carried out according to the theory of maximum linear deformations. If the material is in a plastic state, destruction will occur by shearing, and strength calculations must be carried out according to the theory of maximum tangential stresses. Here p is the tear resistance; p – shear resistance. In the absence of experimental data on these quantities, the relation can be approximately replaced by the relation where is the permissible shear stress; – permissible tensile stress. 1.14. Examples of calculations Example 1.1 A steel strip (Fig. 4.26.) has an oblique weld at an angle β = 60º to the longitudinal axis. Check the strength of the strip if the force P = 315 kN, the permissible normal stress of the material from which it is made [σ] = 160 MPa, 36 the permissible normal stress of the weld [σe] = 120 MPa, and the tangential stress - [τ] = 70 MPa, dimensions cross section B = 2 cm, H = 10 cm. Fig. 1.26 Solution 1. Determine the normal stresses in the cross section of the strip. We compare the found stress σmax with the permissible [σ] = 160 MPa, we see that the strength condition is satisfied, i.e. σmax< [σ]. Процент расхождения составляет 2. Находим напряжение, действующее по наклонному сечению (сварному шву) и выполняем проверку прочности. Используем метод РОЗУ (сечения). Рассечем полосу по шву (рис. 4.27) и рассмотрим левую ее часть. В сечении возникают два вида напряжения: нормальное σα и касательное τα, которые будем считать распределенными равномерно по сечению. Рассматриваем равновесие отсеченной части, составляем уравнение равновесия в виде сумм проекций всех сил на нормаль nα и ось t. С учётом площади наклонного сечения Аα = А/cosα получим cos2 ; Таким образом нормальное напряжение в сварном шве также меньше [σэ] = =120 МПа. 37 3. Определяем экстремальные (max, min) касательные напряжения τmax(min) в полосе. Вырежем из полосы в окрестности любой точки, например К, бесконечно малый элемент в виде параллелепипеда (рис 1.28). На гранях его действуют только нормальные напряжения σmax=σ1 (материал испытывает линейное напряжённое состояние, т. к. σ2 = σ3 = 0). Из формулы (1.5) следует, что при α0 = 45є: Сопоставляя найденные напряжения с допустимыми, видим, что условие прочности выполняется. Пример 1.2 Под действием приложенных сил в детали, элемент, вырезанный из нее испытывает плоское напряженное состояние. Требуется определить величину и направление главных напряжений и экспериментальные касательные напряжения, а также относительные деформации в направлениях диагонали АС, удельное изменение объема и потенциальную энергию деформации. Напряжения действующие на гранях элемента известны: Решение 1. Определяем положение главных площадок. Угол положительный. Это говорит о том, что нормаль к главной площадке должна быть проведена под углом α0 положительным от направления σх против часовой стрелки. 2. Вычисляем величину главных напряжений. Для нашего случая имеем Так как σх, то под углом α0 к направлению σх действуют σmin= σ3 и под углом α0 + 90˚ действуют σmax = σ1. (Если σх > σу, then at an angle α0 to the direction σх act σmax = σ1 and at an angle α0 + 90˚ act ​​σmin = σ3). Check: a) for this we determine the value of the principal stresses using the formula We see that at an angle α0 the stress σmin ≈ σα acts; b) check for tangential stresses on the main areas. If the angle α0 is found correctly, the left side is equal to the right. Thus, the check shows that the stresses to the main pad are determined correctly. 3. Determine the extreme values ​​of tangential stresses. The highest and lowest shear stresses act on areas inclined at an angle of 45° to the main areas. With this dependence, to determine extreme values, τ has the form 4. We determine the relative deformations in directions parallel to the ribs. To do this, we use Hooke’s law: since the element experiences a plane stressed state, i.e. σz = 0. Then these dependencies have the form: Taking into account the values, we have: 5. Determine the specific change in volume 6. Absolute change in volume 7. Determine the specific potential strain energy. since σ2 = 0 we get 8. We determine the absolute lengthening (shortening) of the edges of the elements: a) in the direction parallel to the y-axis, the edges BC, AD are lengthened. b) in the direction parallel to the x-axis, shortening of the ribs BA, SD. Using these values, you can determine the extension of the diagonal AC and WD based on the Pythagorean theorem. Example 1.3 A steel cube with a side of 10 cm, inserted without gaps between two rigid walls and resting on a fixed base, is compressed by a load q = 60 kN/m (Fig. 1.30). It is required to calculate: 1) stresses and strains in three directions; 2) change in the volume of the cube; 3) potential strain energy; 4) normal and shear stresses on a platform inclined at an angle of 45° to the walls. Solution 1. The stress on the upper face is given: σz=-60 MPa. The voltage on the free face is σу=0. The stress on the side faces σх can be found from the condition that the deformation of the cube in the direction of the x axis is equal to zero due to the inflexibility of the walls: whence at σу = 0 σх- μσz = 0, therefore, σх = μσz = -0.3 ּ60 = -18 MPa. 43 Fig. 1.30 The faces of the cube are the main areas, since there are no shear stresses on them. The main stresses are σ1 = σу = 0; σ2 = σx = -18MPa; σ3 = σz = -60 MPa; 2. Determine the deformations of the edges of the cube. Relative linear deformations Absolute deformation (shortening) Relative deformation in the direction of the Y axis Absolute deformation (elongation) Relative change in the volume of the cube Absolute change in volume (decrease) 3. Potential energy deformation (specific) is equal to Total energy is equal to 4. Normal and shear stress on a site inclined to the walls at an angle of 45º: Direction σα, τα is shown in Fig. 2.30. Example 1.4 Cylindrical thin-walled steel tank filled with water at a level H = 10 m. At a distance H/3 from the bottom at point K, two strain gauges A and B (Fig. 1.31) with a base S = 20 mm and a division value K = 0 are installed at an angle = 30, mutually perpendicular. .0005 mm/div. Determine the principal stresses at point K, as well as the stress in the direction of the strain gauges and their readings. Given: Tank diameter D=200 cm, wall thickness t = 0.4 cm, steel transverse strain coefficient = 0.25, liquid density γ = 10 kN/m3. Neglect the weight of the tank. Solution. 1. Determine the principal stresses at point K. a. Let us consider the equilibrium of the lower cut-off part of the tank (Fig. 1.32). 45 Fig. 1.31 Fig. 1.32 We create an equilibrium equation for the sum of the projections of all forces on the y-axis: – the weight of the water column. From here we find the normal stress (meridional) y in the cross section of the tank. We determine normal stresses (circumferential stresses) in the direction of the x-x axis. To do this, consider the equilibrium of a semiring with a width equal to a unit of length, cut out at the level of point K (Fig. 1.33). The elementary force dP arriving at the elementary area of ​​the angle d is determined by the formula - fluid pressure at point K. We compose the equilibrium equation of the semiring on the x axis: From here we obtain In accordance with the designation of the principal stresses, comparing and y, we have Principal stress It is small compared to 2 and can be neglected. For an infinitesimal element (abcd) isolated in the vicinity of point K, the main stresses are presented in (Fig. 1.34). We determine the normal stresses in the direction of installation of the strain gauges. We check the correctness of the found voltages. The following condition must be met: The discrepancy is insignificant and is due to rounding in calculations. We determine the relative deformations in the direction of installation of the strain gauges. We use the generalized Hooke's law. (31.390160.5261.90016)0.594014 002019 Set the readings of the strain gauges. We use formulas to determine relative deformations based on strain gauge readings: n - strain gauge readings; i S - strain gauge base; i K - division price. From here we have the readings of the strain gauges: Example 1.5 Calculate the notch of the rafter leg into the tie, determining the depth of the cut hBP and the length of the protruding part of the tie l (Fig. 1.35). The cross-sectional dimensions of the leg and tie are shown in the drawing. Corner. The calculated force in the leg, found taking into account the overload factors, is equal to NP 83 kN. Solution. We carry out calculations based on the limit state. We determine the cutting depth hВР based on crushing. We carry out the calculation for the tightening area, since the normal to this area makes an angle = 30 and the calculated resistance for it is less than for the leg, because the crushing area of ​​the leg is perpendicular to the fibers. The size of the crushing area: where does the cutting depth come from? Design resistance we will find the collapse using the formula (1.52) Cutting depth The length of the protruding part of the tightening lSC is determined based on chipping. Shearing area The value of the average calculated shearing resistance will be found using formula (1.54): In this case, the shoulder e is equal to 11 cm. According to design standards, the length of the shearing area should not be less than 3e or 1.5h. Therefore, we take the approximate required length of the shearing area to be 0.33 m, i.e., it corresponds to the previously planned value.

Pin shear stresses in cross section I- I, rice. 1, τ s, MPa:

When determining permissible stresses [ τ c ] according to formula (6) for the finger material according to table. 1:

Coefficient Kτ p is determined according to Table 3 depending on the diameter of the finger d;

- coefficient Kτ n is determined according to Table 4, assuming the surface of the finger is polished;

Coefficient Kτ To = 1 is accepted for the design of a pin without shoulders or grooves in a dangerous section;

Coefficient Kτ at determined according to the table. 6, It is generally recommended to use surface hardening.

If the strength condition according to formula (8) is not met, you should choose a higher quality steel grade or increase the pin diameter d.

Rice. 4. Parts with typical stress concentrators: A– transition from a smaller size b to more l, fillet radius r 1 ; b – cross hole diameter d 1

Rice. 5. Calculation diagram of the hinge pin: A– diagram of shearing forces; b – diagram of bending moments

5.2. Calculation of finger bending

Taking into account the uncertainty of the conditions for pinching a finger in the cheeks and the influence of finger deflection and deformation of the cheeks on the distribution of the specific load, a simplified design diagram of a beam on two supports loaded with two concentrated forces is adopted, Fig. 5. Maximum bending stresses develop in the middle span of the beam. Voltages finger bend, σ and, MPa, in section 4-4 , rice. 5:

σ and = M/W≤[σ and ], (9)

Where M– bending moment in a dangerous section, N∙mm:

M = 0,125F max ( l+ 2δ );

W – axial moment of resistance, mm 3:

W = πd 3  / 32  0.1 d  3 ,

l- the length of the rubbing part of the finger, determined depending on the ratio l/d, given in Appendix. and finger diameter d, mm, found in paragraph 4.1; δ – eye wall thickness, determined in clause 6.1;

[σ and ] – permissible stresses during bending according to the shape. (6).

Calculated using formulas (6) and (9):

- Kσ k – coefficient is determined according to table. 5 taking into account the stress concentrator - the transverse hole for the supply of lubricant, Fig. 1;

Odds Kσ p, Kσ n and TO y is prescribed in the same way as the finger calculation according to clause 5.1.

If the strength condition according to formula (9) is not met, the pin diameter should be increased d.

Final value d, indicated on the drawing, is rounded to the nearest larger standard value from a number of normal linear dimensions in accordance with GOST 6636-69.

Know the conditions of shear and crush strength. Be able to carry out shear and crush calculations.

Examples of problem solving

Example 1. Determine the required number of rivets to transfer an external load of 120 kN. Place the rivets in one row. Check the strength of the sheets being joined. Known: [ σ ] = 160 MPa; [σ cm] = 300 MPa; [ τ s ] = 100 MPa; rivet diameter 16 mm.

Solution

1. Determine the number of rivets per shear (Fig. 24.1).

Shear strength condition:

z- number of rivets.

Thus, 6 rivets are needed.

2. Determine the number of rivets based on crushing. Collapse strength condition:

Thus, 4 rivets are needed.

To ensure shear (shear) and crushing strength, it is necessary 6rivets

For ease of installation of rivets, the distance between them and from the edge of the sheet is regulated. Step in a row (distance between centers) of rivets 3d; edge distance 1.5d. Therefore, to accommodate six rivets with a diameter of 16 mm, a sheet width of 288 mm is required. We round the value to 300mm ( b= 300mm).

3. Let's check the tensile strength of the sheets. Checking the thin sheet. Holes for rivets weaken the section; calculate the area of ​​the sheet in the place weakened by the holes (Fig. 24.2):

Tensile strength condition:

73.53 MPa< 160 МПа. Следовательно, прочность листа обеспечена.

Example 2. Check the strength of the rivet joint for shear and crushing. Connection load 60 kN, [ τ s ] = 100 MPa; [ σ cm] = 240 MPa.

Solution

1.

A connection with double shear rivets is sequentially perceived by three rivets in the left row, and then by three rivets in the right row (Fig. 24.3).

Shear area of ​​each rivet A c = 2π r 2. Side surface crush area A cm = dδ min.

2. Check the strength of the connection for shear (shear).

Q = F/z- shear force in the cross section of the rivet:

Shear strength is ensured.

3. Let's check the strength of the connection for crushing:

The strength of the rivet connection is ensured.

Example 3. Determine the required diameter of the rivet in the lap joint if the transmitted force
Q = 120 kN, sheet thickness δ = 10 mm. Allowable shear stress [ τ ] = 100 N/mm 2, for compression [σ cm ] = 200 N/mm 2 (Fig. 2.25). Number of rivets in connection n = 4 (two rows of two rivets each).

Solution

Determine the diameter of the rivets. From the condition of cross-sectional shear strength ab, considering that the rivets are single-cut (t = 1), we get

We accept d = 20 mm.

From the condition of the joint strength against crushing

we get

We accept the larger of the found values d= 20 mm.

Example 4. Define required amount rivets diameter d= 20 mm for an overlap connection of two sheets with thicknesses δ 1 = 10 mm and δ 2 = 12 mm. Force Q, tensile connection is equal to 290 kN. Allowable stresses: shear [t| = 140 N/mm a, for crushing [σ cm] = 300 N/mm 2.

Solution

From the condition of shear strength, the required number of rivets at t = 1

The collapsing stress will be greatest between the rivets and the thinner sheet, so we substitute δ into the collapsing strength condition min= 6, and we find

It is necessary to place 7 rivets in the connection, required by the condition of shear strength.

Example 5. Two sheets with transverse dimensions δ 1 = 14 mm, b = 280 mm are connected by double-sided overlays with a thickness of each δ 2 = 8 mm (Fig. 2.26). The connection transmits a tensile force Q = 520 kN. Determine the number of rivets with diameter d = 20 mm, which must be placed on each side of the joint. Also check the strength of the sheet along the dangerous section, taking into account that the rivets are placed two in a row (k = 2, Fig. 2.26). Allowable shear stress for rivets [ τ ] = 140 N/mm a, for compression [σ cm ] = 250 N/mm 2, for sheet tension [ σ ] = 160 N/mm 2 .

Solution

In the connection under consideration, the rivets operate as double-shear t = 2, i.e., each rivet experiences shear deformation along two cross sections (Fig. 2.26).

From the condition of shear strength

From the condition of bearing strength, taking into account that the minimum bearing area corresponds to δ min= δ 1< 2δ 2 , получаем

We accept n = 8.

In this case, the required number of rivets from the crushing strength condition turned out to be greater than from the shear strength condition.

Checking the strength of the sheet in cross-section I - I

Thus, the calculated stress in the sheet is less than the permissible one.

Example 6. The gear wheel is fastened to the drum of the lifting machine with six bolts of diameter d=18 mm, placed without gaps in the holes. The centers of the bolts are located along a circle with a diameter of D = 600 mm (Fig. 2.27). Determine from the condition of bolt shear strength the magnitude of the permissible moment that can be transmitted through gear drum. Allowable shear stress for bolts

Solution

A moment that can convey bolted connection wheels with drum according to fig. 2.27, determined from the formula

Where P- number of bolts, for our case n = 6; [Q]- the permissible force transmitted by one bolt according to the shear strength condition; 0.5D- arm of the force transmitted by the bolt relative to the axis of rotation of the shaft.

Let us calculate the permissible force that the bolt can transmit according to the shear strength condition

Substituting the value [ Q] into the formula for the moment, we find

Example 7. Check the strength of the welded joint using fillet welds with an overlay. Effective load 60 kN, permissible shear stress of the weld metal 80 MPa.

Solution

1. The load is transmitted sequentially through two seams on the left, and then two seams on the right (Fig. 24.4). Destruction of fillet welds occurs along areas located at an angle of 45° to the surface of the sheets being joined.

2. Check the shear strength of the welded joint. Double-sided fillet weld can be calculated using the formula

And with- calculated seam cut area; TO - the leg of the seam is equal to the thickness of the lining; b- seam length.

Hence,

59.5 MPa< 80МПа. Расчетное напряжение меньше допускаемого, прочность обеспечена.