Finger calculation. Practical methods of calculation for shear and collapse. Calculation of bolted and riveted joints Substituting numerical values, we obtain

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

Shear (cut)

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

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

where is the voltage;

G- shear elastic modulus;

Shear angle.

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

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

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

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

In the calculation, we assume that the elastic forces are uniformly distributed over the section;

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

Shear (shear) strength condition

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

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

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

The calculation is also conditional. The assumptions are similar to those adopted in the shear calculation, however, when calculating the lateral cylindrical surface, the stresses are not uniformly 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.

Collapse Strength Condition

whereA cm - calculated collapse area

d - diameter of the circumference of the section;

The smallest height of the connected plates;

F - the force of interaction between parts

Permissible crushing stress

= (0,35 + 0,4)

Topic 2.5. Torsion

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

The torque M cr 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 twist is counterclockwise and negative if it is clockwise.

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

,

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

- allowable shear stress.

The torque is determined by the formula:

where P is the power on the shaft, W;

ω is the angular velocity of shaft rotation, rad/s.

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

For a circle

For the ring

.

When the beam is twisted, its axis is twisted 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 called deformation from the moment of external forces acting in a plane passing through the geometric axis of the beam.

Depending on the application places active forces distinguish straight And oblique bend.

straight bend are the 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 - a bending moment M x and a transverse force Q y.

The bend is called clean, if one VSF occurs in the cross section of the beam - the 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 is clockwise; negative - counterclockwise.

When determining the signs of bending moments, use rule of "compressed fibers"(rule "BOWL"): 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 allowable 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 rectangle

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

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

Shear and collapse calculations

Example #1

Round rod stretched by force F = 180 kN fortified on the details using checks of rectangular section (Fig. 1). From the conditions of tensile strength, shear and crushing of steel, determine the diameter of the rod d, required length A its tail part, as well as the dimensions of the cross section of the checks t And h without taking into account its bending work. Acceptable stresses: [ σ p] = 160 MPa, [ τ cf] = 100 MPa, [ σ cm] = 320 MPa.

Fig.1

Solution.

Rod under force F is under tension, the weakened section will be the section of the rod that passes through the pin. Its area is defined as the difference between the areas of a circle and a rectangle, in which one side is equal to the width of the check. t, and the second can be taken equal to the diameter of the rod d.. This area is shown in (Fig. 1g).

According to the tensile strength

determine the stretching area by substituting N=F, we have:

equating (1) we get the first equation. In the shank of the rod under the pressure of the pins, a shear over the area can occur And wed = 2(a-h)∙ d. From the shear strength condition

determine the cut area of ​​the shank

hence 2( a-h)· d= 1800(2) we obtain the second equation.

Based on the condition that the strength of the cut of the rod and the pins is equal, we determine the area of ​​the cut of the pin, which is defined as A 2sr= 2h∙ t and equal A 1sr those. A 2av =A 1sr, so we get the third equation 2 h∙ t = 1800(3).

Under the force F check, putting pressure on inner part of the rod causes the collapse of the rod over the area A cm = dt .

determine the crush area:

Thus, we obtain four equations for determining the diameter of the rod d, shank length A and dimensions of the cross section of the checks t And h:

2(a-h)∙ d = 1800(4)

2h∙ t = 1800

d∙ t = 56,25

we substitute into the first equation of system (4) instead of d∙ t= 56.25, we get:

– 56.25 = 1125 or = 1125 + 56.25 = 1687.5

from here those. d= 46,4mm

because d∙ t=56,25,;t = 12,1 mm .

From the third equation of system (4) we determine h.

2h∙ t = 1800, from here; h = 74,3 mm .

From the second equation of system (4) we determine A.

2(a-h) ∙ d = 1800

(a-h) = 900, hence

So, A = 93,7 mm.

Example #2

Check the traction strength for tension, and the bolt for shearing and crushing, if a force is applied to the traction F = 60 kN, dimensions are given in (Fig. 2), at allowable stresses: for tension [ σ p] = 120 MPa, for shear [ τ cf] = 80 MPa, for crushing [ σ cm] = 240 MPa.

Rice. 2

Solution.

We establish what types of deformations the connection parts experience. Under the force F steel rod diameter d and an eyelet with an outer diameter D1 and internal D2 will experience tension, the thrust area is a circle with an area

in an eyelet weakened by a hole D2 gap can occur along the area A 2p =(D1-D2)∙ V. Using Tensile Strength Conditions

check the tensile strength of the pull; because N=F, That

those. thrust satisfies the strength condition.

Tensile stress in the lug;

The strength of the lug is guaranteed.

Bolt diameter D2 experiences a shear in two planes, each of which is equal to the cross-sectional area of ​​the bolt, i.e.

From the shear strength condition:

The inner part of the lug exerts pressure on the surface of the bolt, so the cylindrical surface of the bolt is subjected to collapse over the area A cm = D 2 in.

we perform a test of the collapse strength of the bolt

Example #3

Bolt diameter d = 100mm, working in tension, rests its head on the sheet (Fig. 3). Determine head diameter D and its height h if the tensile stress in the bolt section σ p\u003d 100 N / mm 2, crushing stress over the head bearing area σ cm\u003d 40N / mm 2 and head shear stress τ cf\u003d 50 N / mm 2.

Fig.3

Solution.

Starting to solve the problem, it is necessary to establish what types of deformations the bolt shaft and its head experience, in order to then use the corresponding calculated dependencies. If the bolt diameter is reduced d, then this can lead to rupture, as the bolt shaft is under tension. The cross-sectional area along which a rupture can occur (Fig. 3, c). Head height reduction h, if the strength of the rod head is insufficient, it will entail a cut along the side surface of the cylinder with a height h and diameter d(Fig. 3a). Cut area And wed = π· dh.

If the diameter of the head decreases D, then the perceiving force F, the annular bearing surface of the rod head may be crushed. Collapse area (Fig. 3b).

Thus, the calculation must be carried out according to the conditions of tensile strength, shear and crushing. In this case, a certain sequence must be observed, i.e. start the calculation with the determination of those force factors or dimensions that do not depend on other determined quantities. In this problem, we start with the definition of the internal force Ν , which is equal in magnitude to the shear force Q force applied to the bolt F.

From the tensile strength condition

define strength N, which is equal in magnitude to the force Q=F.

Force

From the shear strength condition determine the height of the head

bolts, because Q=F, That, , But A cf =π dh, That's why .

We determine the diameter of the bearing surface of the bolt head from the condition of its crushing strength

Answer: h = 50mm,D = 187 mm.

Example #4

Determine what strength F(fig. 4) it is necessary to attach a punch to the punch for punching in a steel sheet with a thickness t = 4 mm, size V× h= 10×15 if the shear strength of the sheet material τ pc= 400 MPa. Determine also the compressive stress in the punch.

Fig.4

Solution.

Under the force F there was a destruction of the sheet material along four surfaces when the actual stress reached the tensile strength τ pc when cutting. Therefore, it is necessary to define the internal Q and an equal external force F by known stress and dimensions h , in And t area of ​​deformable sections. And this area is the area of ​​four rectangles: two with dimensions h× t and two with sizes V× t .

Thus, And wed = 2 ht + 2 V·t = 2t(h + in) = 2 4 (15+10) = 200 mm 2.

Shear stress at shear shear

but since Q=F;

F=𝜏 pm∙ A Wed= 400 200 = 80000 H = 80 kN;F= 80 kN

Punch compression stress

Answer: F = 80kN; σ compress= 533.3 MPa.

Example #5

Wooden beam of square section, A= 180 mm (Fig. 5) suspended on two horizontal rectangular beams and loaded with a tensile force F= 40 kN. For mounting on horizontal beams, two cuts are made in the timber to the size V = 120 mm. Determine the tensile, shear and crushing stresses arising in the dangerous sections of the beam, if With = 100 mm.

Fig.5

Solution.

Under the force F in a bar weakened on both sides by cuts, a tensile stress σ arises. In a dangerous section, the dimensions of which A r = V∙ a = 120∙ 180 = 21600 mm 2. Normal stress σ, given that the internal force N in cross section is equal to the external force F equals:

Shear shear stresses τ sk arise in two dangerous sections from the pressure of horizontal beams on vertical bar, under the force Q=F. These sites are located in a vertical plane, their size A sk 2∙ with∙ a =2∙ 100∙ 180=36000 mm 2 .

We calculate the shear stresses acting on these sites:

Collapse stress σ cm arises from the force F in two dangerous sections of the vertical beam at the top of the horizontal beams, exerting pressure on the vertical beam. Their value is determined A cm =a∙ (a-c) = 180∙ (180-120) =180∙ 60 = 10800 mm 2.

Collapse stress

Example #6

Define required dimensions cuts with a "straight tooth". The connection is shown in (Fig. 6). The cross section of the bars is square, tensile force F = 40 kN. Permissible stresses for wood matter: tensile[ σ p]= 10MPa, for chipping [ τ sk]= 1MPa, for crushing [ σ cm] = 8 MPa.

Fig.6

Solution.

Element Mates wooden structures- cuts are calculated for strength from the conditions of their work in tension, shearing and crushing. With enough force F acting on the cut with a straight tooth (Fig. 6), chipping along the sections can occur de And mn , shear stresses arise along these sections, the magnitude of which is determined assuming their uniform distribution over the cross-sectional area. Cross-sectional area de or mn A ck= a ∙ with.

The strength condition has the form:

a s = 4000 mm 2(1)

In the vertical wall of the tooth on the platform m e collapse deformation occurs. Cross-sectional area over which collapse can occur A cm = in ∙ a.

From the condition of crushing strength:

we have or in a = 5000mm 2 (2)

Based on the diversity of parts A And IN, their rupture can occur along a section whose area is .

The tensile strength conditions are:

As a result, we obtain a system of equations: 1, 2, 3.

A∙ with = 4000

V∙ a = 5000

Having performed the transformation in the third equation of system (4), we obtain:

A∙ with = 4000

V∙ a = 5000 (4 ’)

a 2 - a ∙ in = 8000

equation (3) of system (4 ') takes the form a 2 = 8000+a∙ in= 8000+5000 = 13000 from here A = = 114 mm ;

from equation (2) of system (4')

from equation (1) of system (4')

Answer: a = 114 mm;in = 44 mm;c = 351 mm.

Example #7

The connection of the rafter leg with a puff is made using a frontal cut (Fig. 7). Determine required dimensions x, x 1,y), if the compressive force in the strut is F= 60 kN, the angle of inclination of the cover α = 30 o, the dimensions of the cross-section of the bars h= 20 cm,V = 10 cm. Permissible stresses are taken: for tension and compression along the fibers [σ ] = 10 MPa, on crushing across the fibers [ σ cm ] = 8 MPa, for crushing along the fibers [σ 90 ] = 2,4 MPa and for shearing along the fibers [ τ sk ] = 0,8 MPa. Also check the strength of the rafter leg for compression and tightening in a weakened section of the section for tension.

Fig.7

Solution.

We determine the forces acting on the cutting planes. To do this, we lay out the force F to the vertical component F1 and the horizontal component F2, we get

F 1 =Fsin𝛼 = 60∙ 0,5 = 30 kN.

F 2 =Fcos𝛼 = 60∙ 0,867 = 52,02 kN.

These forces are equalized by the support reaction R = F1 and tensile force in tightening N=F2. Force F1 causes collapse of the puff along the area of ​​​​support on the support pad (perpendicular to the fibers). Collapsible strength conditions:

from where, because A cm =x 1∙ V,That

Structurally, it is accepted much more. cutting depth y determined from the condition that the force F2 causes collapse along the vertical thrust, and the platform A cm = y ∙ in at the point of contact of the end of the construction leg with the puff. From the condition of crushing strength we have:

because A cm =at · V , That .

The end of the puff experiences shearing along the fibers under the action of the same horizontal force. F2. Length X puff protruding beyond the notch, we determine from the condition of shearing strength:

because τ sk = 0,8 MPa, . chipping area A ck = in ∙ x

Hence, V∙ X = 65000, whence

Let's check the compressive strength of the construction leg:

Let's check the tightening strength in a weakened section:

those. strength is guaranteed.

Example #8

Determine the tensile stress caused by the force F = 30 kN in weakened, three-riveted sections of steel strips, as well as shear stresses and crushing in rivets. Connection dimensions: band width A = 80 mm, sheet thickness δ = 6 mm, rivet diameter d = 14 mm(Fig. 8).

Fig.8

Solution.

The maximum tensile stress occurs in the strip along section 1-1 (Fig. 8, a) weakened by three holes for rivets. In this section, an internal force acts N, equal in strength F. The cross-sectional area is shown in (Fig. 8, d) and is equal to A r = a∙𝛿 – 3∙ d∙ 𝛿 = 𝛿∙ (a- 3d).

Stress in dangerous section 1-1:

The slice is called by the action of two equal internal forces, directed in opposite directions, perpendicular to the axis of the rod (Fig. 8, c). The cut area of ​​one rivet is equal to the area of ​​the circle (Fig. 8, e), the cut area of ​​the entire section, where n- the number of rivets, in this case n= 3.

We calculate the shear stress in the rivets:

The pressure from the side of the hole in the sheet is transferred to the rivet rod along the side surface of the half-cylinder (Fig. 8, e), with a height equal to the sheet thickness δ. In order to simplify the calculation, instead of the surface of the half-cylinder, the projection of this surface onto the diametral plane (Fig. 8, f) is conventionally taken as the collapse area, i.e. area of ​​a rectangle efck equal to d𝛿 .

We calculate the crushing stress in the rivets:

So σ R = 131,6 MPa,τ Wed = 65 MPa,σ cm = 119 MPa.

Example #9

The truss rod, consisting of two channels No. 20, is connected to the shaped sheet (kerchief) of the truss unit with rivets with a calculated diameter d= 16mm(Fig. 9). Determine the required number of rivets at allowable stresses: [ τ Wed ] = 140 MPa;[σ cm ] = 320MPa;[σ R ] = 160MPa. Check the strength of the rod.

Fig.9

Solution.

We determine the dimensions of the cross section of channel No. 20 according to GOST 8240-89 A= 23,4 cm 2, channel wall thickness δ = 5.2 mm. From the shear strength condition

Where Q Wed - transverse force: with several identical connecting parts Q cf =f/i ( - number of rivets; A withp- the area of ​​the cut of one rivet; [ τ Wed ] - allowable shear stress, depending on the material connecting elements and working conditions of structures.

Denote z- the number of cut planes of the connection, the cut area of ​​one rivet, then from the strength condition (1) it follows that the allowable force per rivet:

Here, z = 2 is taken, because double rivets.

From the condition of crushing strength

Where A cm = d∙ 𝛿 to

𝛿 to – thickness of the shaped sheet (kerchief). d- rivet diameter.

Determine the allowable force per rivet:

Thickness of the scarf 9 mm less than double channel thickness 10.4 mm, which is why it was taken as the calculated value.

The required number of rivets is determined from the condition of crushing strength, since .

Denote n is the number of rivets, then accept n=12.

Check the tensile strength of the rod. The dangerous section will be section 1-1, since in this section the greatest strength F, and the areas in all weakened sections are the same, i.e. , Where A = 23,4 cm 2 cross-sectional area of ​​one channel No. 20 (GOST 8240-89).

Therefore, the strength of the channels is ensured.

Example #10

Gear A connected to the shaft IN parallel key (Fig. 10). From the gear wheel is transmitted to the shaft with a diameter d =40 mm moment M = 200 Nm. Determine length ℓ key, taking into account that the allowable stresses of the key material are equal: per shear [ τ Wed ] = 80 MPa, and for crushing [ σ cm ] = 140MPa(dimensions in the figure are indicated in mm).

Fig.10

Solution.

Determine the effort F acting on the key from the side of the connected parts. The moment transmitted to the shaft is , where d- shaft diameter. Where . It is assumed that the effort F evenly distributed over the key area, where ℓ - key length, h- its height.

The length of the key required to ensure its strength can be found from the shear strength condition

and conditions of crushing strength

We find the length of the key from the condition of shear strength, since the cut occurs over the area And wed = in ℓ, That ;

From the condition of strength (2) for crushing, we have:

To ensure the strength of the connection, the length of the key must be taken equal to the larger value of the two obtained, i.e. ℓ= 18mm.

Example #11

The forked crank is mounted on the shaft with a cylindrical pin (Fig. 11) and loaded with force F=2,5 kN. Check the strength of the pin connection for shearing and crushing, if [ τ Wed ] = 60 MPa and [ σ cm ] = 100MPa.

Fig.11

Solution.

First you need to determine the magnitude of the force F1 transmitted to the pin by force F applied to the crank. It's obvious that M=F∙ h is equal to the moment.

check the strength of the pin for shearing under the action of force F1. In the longitudinal section of the pin, shear shear stress occurs, the value of which is determined by the formula , where And wed = d∙ ℓ

Cylindrical pin surface under force F1 is subject to collapse. The contact surface through which the force is transmitted F1, represents a quarter of the surface of the semi-cylinder, since the area of ​​the projection of the contact surface onto the diametrical plane is taken as the tonnage area of ​​the collapse, i.e. dℓ, That A cm = 0,5∙ d∙ ℓ.

So, the strength of the pin connection is ensured.

Example #12

Calculate the number of rivets with a diameter d\u003d 4 mm, necessary to connect two sheets with two overlays (see Fig. 12). The material for sheets and rivets is duralumin, for which Rbs = 110 MPa, Rb R = 310 MPa. Force F\u003d 35 kN, coefficient of working conditions of the connection γ b \u003d 0.9; thickness of sheets and overlays t= 2 mm.

Fig.12

Solution.

Using formulas

calculate the required number of rivets:

from the shear strength condition

from the condition of crushing strength

It can be seen from the obtained results that in this case the condition of crushing strength was decisive. Thus, 16 rivets should be taken.

Example #13

Calculate the fastening of the rod to the nodal gusset (see Fig. 13) with bolts with a diameter d\u003d 2 cm. The rod, the cross section of which is two identical equilateral corners, is stretched by force F= 300 kN.

The material of the gusset and bolts is steel, for which the design resistances are equal: in tension Rbt = 200 MPa , cut Rbs = 160 MPa, for collapse Rb R \u003d 400 MPa, the coefficient of working conditions of the connection γ b \u003d 0.75. Simultaneously calculate and assign the thickness of the gusset sheet.

Fig.13

Solution.

First of all, it is necessary to establish the number of isosceles corners that make up the rod, determining the required cross-sectional area A nec from the tensile strength condition

Given the forthcoming weakening of the rod by bolt holes, it should be added to the cross-sectional area A nec 15%. The cross-sectional area thus obtained A\u003d 1.15 ∙ 20 \u003d 23 cm 2 corresponds according to GOST 8508–86 (see Appendix) a symmetrical section of two isosceles corners measuring 75 × 75 × 8 mm.

We calculate the cut. Using the formula, we find the required number of bolts

Having settled on this number of bolts, we determine the thickness δ of the nodal gusset using the condition of crushing strength

Directions

1. Binding of the line for placing bolts (rivets) in one row is found from the condition: m =b/ 2 + 5 mm.

In our example (Fig. 13)

m= 75/2 + 5 = 42.5 mm.

2. The minimum distance between the centers of adjacent bolts is taken equal to l= 3d. In the problem under consideration, we have

l= 3∙20 = 60 mm .

3. Distance from the extreme bolts to the joint boundary l / taken equal to 0.7 l. In our example l /= 0,7l= 0.7∙ 60 = 42 mm .

4. If the condition b ≥12 cm is met, the bolts (rivets) are placed in two lines in a checkerboard pattern (Fig. 14).

Fig.14

Example #14

Define required amount rivets with a diameter of 20 mm for overlapping two sheets with a thickness of 8 mm and 10 mm (Fig. 15). Force F, tensile connection is equal to 200 kN. Permissible stresses: for shear [τ] = 140 MPa, for crushing [ s c] = 320 MPa.

Permissible stresses - 80 ... 120 MPa.

Ovalization of the finger

Ovalization of the finger occurs when the action of vertical forces (Fig. 7.1, V) deformation occurs with an increase in the diameter in the cross section. Maximum increments of finger diameter in the middle part:

, (7.4)

where is the coefficient obtained from the experiment,

TO=1,5…15( -0,4) 3 ;

– the modulus of elasticity of the finger steel, MPa.

Usually \u003d 0.02 ... 0.05 mm - this deformation should not exceed half the diametrical gap between the pin and the bosses or the hole of the connecting rod head.

Stresses that arise during ovalization (see Fig. 7.1) at points 1 And 3 external and 2 And 4 internal fibers can be determined by the formulas:

For the outer surface of the finger

. (7.5)

For inner surface finger

, (7.6)

Where h- thickness of the wall of the finger, r = (d n + d at 4; f 1 and f 2 - dimensionless functions depending on the angular position of the calculated section j, glad.

f 1=0.5cos j+0.3185sin j-0,3185j cos j;

f 2 =f 1 - 0,406.

The busiest point 4 . Valid values
s St. = 110...140 MPa. Usually mounting clearances between the floating pin and the connecting rod bushing 0.01 ... 0.03 mm, and in the bosses of the cast-iron piston 0.02 ... 0.04 mm. With a floating finger, the clearance between the finger and the boss for a warm engine should be no more than

D = D¢+( a item D t pp - a b D t b) d mon, (7.7)

Where a pp and a b – coefficients of linear expansion of the material of the pin and boss, 1/K;

Dt pp and Dt b - temperature increase of the finger and boss.

Piston rings

Compression rings (Fig. 7.2) are the main element of the sealing of the intra-cylinder space. Installed with a sufficiently large radial and axial clearance. Well sealing the over-piston gas space, they, having a pumping effect, do not limit the flow of oil into the cylinder. For this, oil scraper rings are used (Fig. 7.3).

Mainly used:

1. Rings with a rectangular section. They are easy to manufacture, have a large area of ​​contact with the cylinder wall, which provides good heat removal from the piston head, but they do not work well against the cylinder surface.

2. Rings with a conical working surface are well run in, after which they acquire the qualities of rings with a rectangular section. However, the production of such rings is difficult.

3. Twisting rings (torsion). In the working position, such a ring is twisted and its working surface contacts with the mirror with a narrow edge, as in conical ones, which ensures running-in.

4. Oil scraper rings ensure the preservation of an oil film between the ring and the cylinder with a thickness of 0.008 ... 0.012 mm in all modes. To prevent floating on the oil film, it must provide a large radial pressure (Fig. 7.3).

Distinguish:

a) Cast iron rings with twisted spring expander. To increase durability, the working belts of the rings are coated with a layer of porous chromium.

b) Steel and prefabricated chrome-plated oil scraper rings. During operation, the ring loses its elasticity unevenly around the perimeter, especially at the junction of the lock when heated. As a result, the rings are captive during manufacture, which provides an uneven pressure diagram. Big pressures obtained in the area of ​​the castle in the form of a pear-shaped diagram 1 and teardrop 2 (Fig. 7.4, A).

In this design, three finger connections are used: the rocker of the handle and the connection of the small plunger with 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. It is customary to rely on finger joints for shearing and crushing:

Permissible finger stress per shear,

;

- allowable finger crushing stress,

;

where, F is the load acting on the pin connection;

Z is the total number of fingers in the connection;

δ – sheet thickness, mm;

dov – hole diameter, mm;

K is the number of cut planes.

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

Crushing of the finger for St0, St2 - 2800 kgf/cm2, for St3 - 3200 kgf/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 thrust head along with d = 3 mm; D=5.4 mm; L=12mm.

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Finger shear stress in cross section I- I, rice. 1, τ s, MPa:

When determining the allowable stresses [ τ c ] according to the 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 finger without collars 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, a better steel grade should be selected or the pin diameter should be increased. d.

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

Rice. 5. Calculation scheme of the hinge pin: A- diagram of cutting forces; b - bending moment diagram

5.2. Calculation of a finger for bending

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

σ and = M/W≤[σ and ], (9)

Where M– bending moment in the 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 specified in App. and finger diameter d, mm, found in clause 4.1; δ - wall thickness of the lug, determined in clause 6.1;

[σ and ] - allowable stresses during bending according to the forms. (6).

In the calculation according to formulas (6) and (9):

- Kσ k - the coefficient is determined according to the table. 5, taking into account the stress concentrator - a transverse hole for supplying lubricant, fig. 1;

Odds Kσ p , Kσ n and TO y is assigned similarly to the calculation of the finger 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, affixed on the drawing, is rounded to the nearest higher standard value from a number of normal linear dimensions in accordance with GOST 6636-69.