The connection may fail due to rupture of the frontal seams along the vertical legs ss" or from cutting these seams along the horizontal legs ss". However, practice shows that the seam is destroyed along a bisector section, the height of which

Where To- seam leg, in our case To = 8.

Such a seam is conditionally designed for cutting along a bisector section based on the strength condition:

Where A av = 0,7 3b- cut area of ​​one weld.

Rice. thirty.

Conclusion: the seams are underloaded.

Example 9. The shaft transmits a torque of 27 kN m using a spline connection (Fig. 31). Shaft diameter D= 80 mm, inner diameter d = 68 mm, slot height h= 6 mm, slot width b- 12 mm, connection length / = 100 mm. Number of splines 2 = 6. Determine the shear and crush stresses of the spline.

Rice. 31.

Assuming that all splines are loaded equally, we find the force per spline:

Let's determine the shear voltage:

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

The moment that can be transmitted by a bolted connection between a wheel and a 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 area seam cut; 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.

Elements that connect various parts, for example, rivets, pins, bolts (without clearance) are mainly designed for shearing.

The calculation is approximate and is based on the following assumptions:

1) in the cross sections of the elements under consideration, only one force factor arises - transverse force Q;

2) if there are several identical connecting elements, each of them receives the same share total load transmitted by the connection;

3) tangential stresses are distributed evenly over the section.

The strength condition is expressed by the formula:

τ av = Q/F av ≤[ τ] av, Where

Q- shear force (at several i connecting elements when transmitting force P avg

Q = P avg /i);

τ avg- shear stress in the plane of the calculated section;

F avg- cutting area;

[τ] avg- permissible shear stress.

As a rule, elements that are connected by rivets, pins, and bolts are calculated for collapse. The walls of the holes in the areas where the connecting elements are installed are subject to collapse. Typically, bearing calculations are performed for connections connecting elements which are designed for cutting.

When calculating crushing, it is assumed that the interaction forces between the contacting parts are uniformly distributed over the contact surface and at each point are normal to this surface. The interaction force is usually called crushing stress.

Strength calculations are performed using the formula:

σ cm = P cm /(i´F cm) ≤ [σ] cm, Where

σ cm- effective crushing stress;

P cm- force transmitted by the connection;

i- number of connecting elements;

F cm- calculated crushing area;

[σ] cm- permissible bearing stress.

From the assumption about the nature of the distribution of interaction forces over the contact surface it follows that if the contact is carried out over the surface of a semi-cylinder, then the calculated area F cm equal to the area of ​​projection of the contact surface onto the diametrical plane, i.e. equal to the diameter of the cylindrical surface d to its height δ :

F cm = d´ δ

Example 10.3

Rods I and II are connected by pin III and are loaded with tensile forces (Fig. 10.4). Determine dimensions d, D, d pcs, c, e designs, if [σ] р= 120 MN/m2, [τ] avg= 80 MN/m2, [σ] cm= 240 MN/m2.

Figure 10.4

Solution .

1. Determine the diameter of the pin from the condition of shear strength:

We accept d = 16×10 -3 m

2. Determine the diameter of the rod I from the tensile strength condition (the cross section of the rod weakened by the hole for the pin is shown in Fig. 10.4b):

94.2 × 10 3 10 d 2 - 1920´10 3 d - 30 ³ 0

Solving the quadratic inequality, we get d³30.8´10 -3 m. We take d = 31´10 -3 m.

3. Let's define outside diameter rod II from the condition of tensile strength, section weakened by a hole for the pin (Fig. 10.4c):

94.2´10 3´D 2 -192´10 3´D-61³0

Having decided quadratic equation, we get D = 37.7 ´10 -3 m. Let's take D = 38 ´10 -3 m.

4. Let’s check whether the thickness of the walls of rod II is sufficient according to the crushing strength condition:

Since the bearing stress exceeds the permissible bearing stress, we will increase the outer diameter of the rod so that the condition of bearing strength is satisfied:

We accept D= 39×10 -3 m.

5. Determine the size c from the condition of the shear strength of the lower part of rod II:

Let's accept c= 24×10 -3 m.

6. Let us determine the size e from the condition of the shear strength of the upper part of the rod I:

Let's accept e= 6×10 -3 m.

Example 10.4

Check the strength of the rivet connection (Fig. 10.5a), if [τ] avg= 100 Mn/m2, [σ] cm= 200 Mn/m2, [σ] р= 140 Mn/m2.

Figure 10.5

Solution.

The calculation includes checking the shear strength of rivets, the walls of holes in sheets and plates for crushing, as well as sheets and plates for tension.

The shear stress in rivets is determined by the formula:

In this case i= 9 (number of rivets on one side of the joint), k= 2 (double shear rivets).

τ av = 550´10 3 / (9´2´((3.14´0.02 2) /4)) = 97.2 Mn/m 2

Excess shear strength of rivets:

The crushing stress of the hole walls is determined by the formula:

In a given connection, the crush area of ​​the walls of the holes in the sheets being joined is smaller than the walls of the holes in the plates. Consequently, the crushing stress for sheets is greater than for overlays, so we accept δ calc = δ = 16 ´10 -3 m.

Substituting numeric values, we get:

σ cm= 550´10 3 / (9´16´10 -3 ´20´10 -3) = 191 Mn/m 2

Excess strength due to crushing of hole walls:

To check the tensile strength of sheets, we calculate the stress using the formula:

N- normal force in a dangerous section;

F net- net cross-sectional area, i.e. The cross-sectional area of ​​the sheet minus its weakening by the rivet holes.

To determine the dangerous section, we construct a diagram of longitudinal forces for sheets (Fig. 10.5 d). When constructing the diagram, we will use the assumption of uniform distribution of force between the rivets. The areas of weakened sections are different, so it is not clear which of them is dangerous. We check each of the weakened sections, which are shown in Figure 10.5c.

Section I-I

Section II-II

Section III-III

It turned out to be dangerous section I-I; the stress in this section is approximately 2% higher than permissible.

Checking the overlay is similar to checking the sheets. The diagram of longitudinal forces in the lining is shown in Figure 10.5d. Obviously, section III-III is dangerous for the lining, since this section has smallest area(Fig. 10.5e) and the greatest longitudinal force occurs in it N = 0,5P.

Stresses in the dangerous section of the lining:

The stresses in the dangerous section of the lining are approximately 3.5% higher than the permissible one.

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 bolts, 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 elements being riveted 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 active forces in relation to the fibers of 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 under 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 accepted 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. For getting calculation formula the envelope curve between 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 the following strength condition can be obtained: 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 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.

4.2.6 Calculation of pin for shear

Let's calculate the finger for cutting.

Finger strength is ensured

4.3.5 Calculation of lever bearings

We select a double-row spherical roller bearing No. 3003168 according to GOST 5721-75 with parameters: C=2130000 N, d=340mm, D=520mm, B=133mm.

We will calculate the method according to the formula set out in.

Bearing life:

where b 1 is the factor taking into account the direction of the load, b 1 = 5;

b 2 - factor taking into account lubrication conditions, b 2 = 1;

b 3 - temperature coefficient, b 3 = 1;

b 4 - size coefficient, b 4 = 1.5;

b 5 - factor taking into account the properties of the material, b 5 = 1.1;

D a - diameter of the sphere, D a = 100 mm;

в - half of the oscillation angle, в = 90 о;

C - rated dynamic load capacity, C = 2,130,000 N;

Lever bearing life:

When pushing out 1 row of workpieces, the drive shaft, lever and, accordingly, the lever bearing rotate through an angle of 180 and at the same angle when reverse stroke. This angle corresponds to 1 revolution.

Those. There is 1 revolution of the lever bearing per row of workpieces.

The mass of one row of workpieces is 11200 kg = 112 tons. Mill productivity is 210 t/h.

Number of blanks in 1 hour 210/112 = 1.85 pcs.

This means that in 1 hour the lever bearing will make 1.85 revolutions.

Then, the service life, expressed in hours, for the lever bearing is G/15.

The annual working time fund is 7200..7400 hours (if the hours of scheduled repairs of the entire mill are subtracted from 8760 hours per year). Taking this into account, the service life can be expressed in years:

where n h - bearing revolutions per 1 hour.

Lever Bearing Life:

Sealed electric pump

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