Calculation of a finger for crushing. Practical methods of calculation for shear and collapse. Calculation of bolted and riveted joints. Examples of problem solving
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.
Elements that connect various details, such as rivets, pins, bolts (without clearance) are mainly count on shear.
The calculation is approximate and based on the following assumptions:
1) in the cross sections of the elements under consideration, only one force factor arises - the transverse force Q;
2) if there are several identical connecting elements each of them takes the same share total load transmitted by the connection;
3) shear stresses are uniformly distributed over the section.
The strength condition is expressed by the formula:
τ av = Q/F av ≤[ τ] av, Where
Q- transverse force (for several i connecting elements when transferring force P cf
Q \u003d P cf / i);
τ cf- shear stress in the plane of the calculated section;
F cf- cut area;
[τ] sr- allowable shear stress.
Collapse, as a rule, is calculated on elements that are connected by rivets, pins, bolts. The walls of the holes in the areas of installation of the connecting elements are subjected to crushing. Typically, a collapse analysis is performed for joints whose connecting elements are designed for shear.
When calculating for 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 force of interaction is commonly called the shear stress.
Strength calculation is performed according to the formula:
σ cm = P cm /(i´F cm) ≤ [σ] cm, Where
σ cm is the effective shear stress;
P cm- the force transmitted by the connection;
i- number of connecting elements;
F cm- estimated area of collapse;
[σ] cm- allowable crushing stress.
From the assumption about the nature of the distribution of interaction forces over the contact surface, it follows that if the contact is made over the surface of the half-cylinder, then the calculated area F cm equal to the projection area of the contact surface on the diametral plane, i.e. equal to the diameter of the cylindrical surface d to her height δ :
F cm = d´ δ
Example 10.3
Rods I and II are connected by pin III and loaded with tensile forces (Fig. 10.4). Determine dimensions d, d, d pcs, c, e structures, if [σ] p\u003d 120 MN / m 2, [τ] sr\u003d 80 MN / m 2, [σ] cm\u003d 240 MN / m 2.
Figure 10.4
Solution .
1. Determine the diameter of the pin from the shear strength condition:
Accept d = 16×10 -3 m
2. Determine the diameter of the rod I from the condition of tensile strength (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 accept d = 31´10 -3 m.
3. Define outside diameter rod II from the condition of tensile strength, section weakened by a pin hole (Fig. 10.4c):
94.2´10 3´D 2 -192´10 3´D-61³0
Deciding quadratic equation, we get D = 37.7 ´10 -3 m. Let's take D = 38 ´10 -3 m.
4. Check whether the wall thickness of the rod II is sufficient according to the condition of crushing strength:
Since the crushing stress exceeds the allowable crushing stress, we increase the outer diameter of the rod so that the crushing strength condition is met:
Accept D= 39×10 -3 m.
5. Determine the size c from the condition of shear strength of the lower part of the rod II:
Accept c= 24×10 -3 m.
6. Determine the size e from the condition of shear strength of the upper part of the rod I:
Accept e= 6×10 -3 m.
Example 10.4
Check the strength of the rivet joint (Fig. 10.5a), if [τ] sr\u003d 100 MN / m 2, [σ] cm\u003d 200 MN / m 2, [σ] p= 140 MN/m 2 .
Figure 10.5
Solution.
The calculation includes checking the shear strength of rivets, the walls of holes in sheets and linings for crushing, as well as sheets and linings for tension.
Shear stresses in rivets are determined by the formula:
In the case under consideration i= 9 (number of rivets on one side of the joint), k= 2 (double shear rivets).
τ cf = 550´10 3 / (9´2´((3.14´0.02 2) /4)) = 97.2 MN/m 2
Excess rivet shear strength:
The collapse stress of the walls of the holes is determined by the formula:
In a given joint, the crushing area of the walls of the holes in the sheets to be joined is less than the walls of the holes in the overlays. Consequently, the crushing stresses for sheets are greater than for overlays, therefore we accept δ calc = δ = 16 ´10 -3 m.
Substituting numerical values, we get:
σ cm= 550´10 3 / (9´16´10 -3 ´20´10 -3) = 191 MN/m 2
Excess crushing strength of hole walls:
To check the tensile strength of sheets, we calculate the stresses 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 build a diagram of longitudinal forces for sheets (Fig. 10.5 d). When constructing the diagram, we will use the assumption of a uniform distribution of force between the rivets. The areas of weakened sections are different, so it is not clear which one 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 higher than the permissible one by about 2%.
Checking the overlay is similar to checking the sheets. The plot of the longitudinal forces in the overlay is shown in Figure 10.5d. It is obvious that section III-III is dangerous for the overlay, since this section has smallest area(Fig. 10.5d) and the greatest longitudinal force arises in it N = 0,5P.
Stresses in the dangerous section of the lining:
The stresses in the dangerous section of the lining are higher than the permissible ones by about 3.5%.
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. Large pressures are obtained in the area of the castle in the form of a pear-shaped diagram 1 and teardrop 2 (Fig. 7.4, A).
Know the conditions of shear and crush strength. Be able to carry out calculations for shear and collapse.
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 joined sheets. Known: [ σ ] = 160 MPa; [σ cm] = 300 MPa; [ τ s] = 100 MPa; rivet diameter 16 mm.
Solution
1. Determine the number of rivets based on shear (Fig. 24.1).
Shear strength condition:
z- the number of rivets.
Thus, 6 rivets are needed.
2. Determine the number of rivets in terms of crushing. Collapse strength condition:
Thus, 4 rivets are needed.
To ensure shear (shear) and crushing strength, it is necessary 6rivets.
For the convenience of installing 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, for the location of 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. Check the tensile strength of the sheets. We check the thin sheet. The holes for the rivets weaken the section, we 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 connection for shearing and crushing. Connection load 60 kN, [ τ s] = 100 MPa; [ σ cm] = 240 MPa.
Solution
1. 
The connection with double-shear rivets is successively 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. Collapse area of the lateral surface A cm = dδ min.
2. Check the shear strength of the connection (shear).
Q=F/z- transverse force in the cross section of the rivet:
Shear strength guaranteed.
3. Check the strength of the connection for crushing:
The strength of the rivet connection is guaranteed.
Example 3 Determine the required rivet diameter in a lap joint if the transmitted force is 
Q = 120 kN, sheet thickness δ
= 10 mm. Permissible shear stresses [ τ
] \u003d 100 N / mm 2, for crushing [σ cm] \u003d 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 shear strength over the section ab, given that the rivets are single shear (t = 1), we get
Accept d= 20 mm.
From the condition of the joint strength in crushing
we get
We accept the greater of the found values d= 20 mm.
Example 4 Define required amount rivet diameter d= 20 mm for overlapping two sheets with a thickness of δ 1 = 10 mm and δ 2 = 12 mm. Force Q, tensile connection is equal to 290 kN. Permissible stresses: shear [t| \u003d 140 N / mm a, for crushing [σ cm] \u003d 300 N / mm 2.
Solution
From the shear strength condition, the required number of rivets at t = 1
The crushing stresses will be greatest between the rivets and the thinner sheet, so we substitute δ into the crushing strength condition min= 6, and we find
In the connection, it is necessary to put 7 rivets required by the shear strength condition.
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 a 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, given that the rivets are placed two in a row (k \u003d 2, Fig. 2.26). Permissible rivet shear stress [ τ
] \u003d 140N / mm a, for crushing [σ cm] \u003d 250 H / mm 2, for stretching sheets [ σ
] = 160 N/mm2.
Solution
In the connection under consideration, the rivets work as double-shear t = 2, i.e., each rivet experiences shear deformation along two cross sections (Fig. 2.26).
From the shear strength condition
From the condition of crushing strength, considering that the minimum crushing area corresponds to δ min= δ 1< 2δ 2 , получаем
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 allowable one.
Example 6 The gear wheel is fastened to the drum of the hoisting machine with six bolts with a diameter of 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 the shear strength of the bolts the value of the allowable moment that can be transmitted through gear drum. Permissible stress for shear bolts
Solution
The moment that can be transmitted by the bolted connection of the wheel with the drum according to fig. 2.27, determined from the formula
Where P- the number of bolts, for our case n = 6; [Q]- allowed by the shear strength condition, the force transmitted by one bolt; 0.5D- shoulder of the force transmitted by the bolt relative to the axis of rotation of the shaft.
Calculate the allowable force that the bolt can transfer according to the condition of shear strength
Substituting the value [ Q] into the formula for the moment, we find
Example 7 Check the strength of the welded joint with fillet welds with an overlay. The effective load is 60 kN, the allowable shear stress of the weld metal is 80 MPa.
Solution
1. The load is transferred sequentially through two seams on the left, and then two seams on the right (Fig. 24.4). The destruction of the fillet welds occurs along the platforms located at an angle of 45 ° to the surface of the joined sheets.
2. Check the shear strength of the welded joint. Double-sided fillet weld can be calculated using the formula
A with- estimated area of the 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.
shift called loading, in which only one internal force factor arises in the cross section of the beam - the transverse force.
Consider a beam on which two forces act, equal in magnitude (Fig. 20) and oppositely directed. These forces are perpendicular to the beam axis, and the distance between them is negligible. With a sufficient value of these forces, a cut occurs.
The left side of the body is separated from the right one along a certain section AB. The deformation preceding the shear, which consists in warping the right angles of an elementary parallelepiped, is called shear. On fig. 20, b the shear occurring in the box before the cut is shown; rectangle abed becomes a parallelogram abed". value SS K , to which the section cd moved relative to the neighboring section ab, is called absolute shift. The angle Y, by which the right angles of the parallelepiped change, is called the relative shift.
Rice. 20. Scheme of shear deformation: A) cutting forces acting on the beam; b) deformation of the beam element abed
Due to the smallness of deformations, the angle At can be defined as follows:
It is obvious that in the section AB of the six internal force factors, only the transverse force will arise Q, equal to the strength F:
Given shear force Q causes only tangential stresses to appear.
A similar picture is observed in the details that serve to connect individual elements machines, - rivets, pins, bolts, etc., since in many cases they perceive loads perpendicular to their longitudinal axis.
The transverse load in these parts occurs, in particular, during tension (compression) of the connected elements. On fig. 21 shows examples of pin (a), rivet (b), bolted (c) and keyed (d) connections. The same nature of the loading of the connecting parts also takes place when transmitting torque, for example, in connecting a gear to a shaft using a pin, which, when transferring torque from a gear to a shaft (or vice versa), carries a load perpendicular to its axis.
Rice. 21.
A) pin; b) riveting; V) bolted; G) keyway
The actual operating conditions of the parts under consideration are complex and largely depend on the manufacturing technology of individual structural elements and its assembly.
Practical calculations of these details are very conditional and are based on the following basic assumptions:
- 1. Only one internal force factor arises in the cross section - the transverse force Q.
- 2. Shear stresses arising in the cross section are evenly distributed over its area.
- 3. If the connection is made by several identical parts (bolts, etc.), it is assumed that they are all equally loaded.
The destruction of the connecting elements (in the case of insufficient strength) occurs as a result of their cutting along a plane coinciding with the contact surface of the parts to be joined (see Fig. 21.6). Therefore, these elements are said to work in shear, and the shear stresses arising in their cross section are also called shear stresses and denote t cf.
Based on the assumptions formulated above, we obtain next condition shear strength:
Where g C p- design shear stress arising in the cross section of the calculated part; Q- transverse force causing shear of connecting elements (bolts, rivets, etc.); [t sr]- allowable shear stress, depending on the material of the connecting elements and the operating conditions of the structure; ZA cp- total cut area: LA cp - A cp t(Here And wed- cut area of one connecting element; z- number of connecting elements; / - the number of cut planes in one connecting element).
In mechanical engineering, when calculating pins, bolts, keys, etc., they take [T cf ] = (0.5 ... 0.6) * [o] - for plastic materials and [x cf] = (0,8... 1,0)-[A]- for brittle materials. Smaller values are taken with low accuracy of determination acting loads and the possibility of not strictly static loading.
Formula (30) is a dependency for the verification calculation of a shear joint. Depending on the problem statement, it can be converted to determine the allowable load or the required cross-sectional area (design calculation).
The shear calculation ensures the strength of the connecting elements, but does not guarantee the reliability of the structure (assembly) as a whole. If the thickness of the connected elements is insufficient, then the pressures that arise between the walls of their holes and the connecting parts turn out to be unacceptably large. As a result, the walls of the holes are crumpled and the connection becomes unreliable. If the change in the shape of the hole is significant (at high pressures), and the distance from its center to the edge of the element is small, part of the element can be cut off (knocked out).
Wherein pressures arising between the surfaces of holes and fittings(Fig. 22, a) at called shear stresses and designate them Os*. Accordingly, a calculation that ensures the choice of such dimensions of parts for which there will be no significant deformations of the walls of the holes is called a collapse calculation. The distribution of crushing stresses on the contact surface of the parts is very indefinite (Fig. 22, b) and to a large extent depends on the gap (in the unloaded state) between the walls of the hole and the bolt (rivet, etc.).
Rice. 22. Transmission of pressure on the rivet shaft: A) general form rivet connection; b) stress distribution along the generatrix; V) rivet crush area
The calculation for collapse is also conditional and is carried out on the assumption that the interaction forces between the parts are uniformly distributed over the contact surface and are normal to this surface at all points.
The corresponding calculation formula has the form
Where F- crushing load; 1A SM - total crush area; [[a cm \u003d (2,..2,5) - [ [а с ] - allowable compressive stress of that of the contacting materials, the strength of which is less.
Behind estimated area collapse at contact along the plane (Fig. 21, G) take the actual area of contact A cm = 1-1, where / - the size of the key in the direction perpendicular to the plane of the drawing; upon contact along a cylindrical surface (see Fig. 21, a, b, c and Fig. 22, a, in) for the calculated area is taken the area of the projection of the contact surface on the diametrical plane, i.e. A cm = d-d. With different thicknesses of the connected parts in calculation formula should be substituted d „i“. Total collapse area ?A SM = ACM-z(where z is the number of connecting elements).
As already mentioned, in some designs connecting parts(pins, keys) work on a cut along the longitudinal sections (see Fig. 21, d); the prerequisites for the calculation and its methodology remain the same as for cross-sectional cuts.
In addition to calculations for shear and collapse, it is necessary checking the tensile strength of the elements to be joined along the weakened section. In this case, the cross-sectional area is taken taking into account the weakening:
Where A „etto - weakened area.
On fig. 23 shows a bolted connection. Forces F tend to move the sheets relative to each other. This is prevented by a bolt, to which, from the side of each sheet, the forces distributed over the contact surface are transmitted, the resultants of which are equal to F. These forces tend to shear the bolt along the sheet interface T- l, since the maximum transverse force acts in this section Q = F.
Assuming that shear stresses are uniformly distributed, we obtain 
Rice. 23. Bolted connection: A) general form; b) crush area
Thus, the bolt shear strength condition takes the form
From here you can find the diameter of the bolt:
When calculating this bolted connection it should be taken into account that the loads applied to the connection elements, in addition to cut cause crushing of contact surfaces.
Where Ah, - is the projection area of the contact surface on the diametral plane (see Fig. 22, b, c): A w = 3 d.
Then the condition of the collapse strength of the bolted connection (see Fig. 23)
where we get
To be satisfied shear and shear strength conditions, of the two found diameters, you should take the larger one, rounding it up to the standard value.
It is customary to count on a cut and some welded joints (Fig. 24).
Rice. 24. Scheme of a welded joint: A) design scheme of the fillet weld; b) cut area ABCD weld
If you do not take into account the influxes, then in the section the fillet weld has the shape of an isosceles right triangle(see fig. 24, A). The destruction of the seam will occur along its minimum section ABCD(see fig. 24, b) whose height k \u003d 3- cos 45° =0.7 3 .
For a lap joint, both welds are included in the calculation. In this case, we write the condition for the strength of the seam:
where / t is the estimated length of the end weld; t, - allowable stress for welded joints.
Since at the beginning and at the end of the seam, due to lack of penetration, its quality deteriorates, its actual length is increased by 10 mm compared to the calculated one:
where / is the actual length of the seam (in Fig. 24, 6:1 = b).
Parts working on shear (shear) and crushing
1. Axis (Fig. 25, A). If the thickness of part 2 is less, A t \u003d Sd;
where / is the number of planes (areas) of the cut.
2. Bolt (Fig. 25, b). In this case A cf-ndh
Rice. 25. Part connections: A) axis; b) bolt
3. Single-cut rivet (Fig. 26, A double-cut (Fig. 26, b).
Rice. 26. Calculation scheme of the rivet connection: A) with one cutting plane; b) with two cutting planes
- 4. Dowels (Fig. 27, A) work on shear and collapse, but are calculated, basically, only on collapse. The cut and collapse areas are determined by the formulas A cf = b i 1 A CM \u003d lt.
- 5. Welded joint (Fig. 27, b).
The fillet weld fails at an angle of 45° to the parting plane as a result of the shear: To- the leg of the fillet weld, selected according to the thickness of the welded sheet.
Double seam: A cp \u003d 2-0 y b = 1,4 to b.
Rice. 27. Connections: A) keyway; b) welded
Example 6. Determine the required number of rivets in the connection of two sheets loaded with forces F= 85 kN (Fig. 28). Rivet diameter d= 16 mm. Permissible stresses [g sr]= 100 MPa, [
From the shear strength condition
Where A C p \u003d k d 2/ 4 - cut area; z is the number of rivets.
Rice. 28.
From the condition of crushing strength 
Where Asm = dS- collapse area; z - the number of rivets, we get
Conclusion: in order to avoid shearing or crushing of the rivets, five rivets should be installed.
Example 7. A steel bolt (fig. 29) is loaded with a force F= 120 kN. Determine its diameter d and head height AND, if allowable stresses [o p] \u003d 120 MPa, = 80 MPa. Band Width b- 150 mm and their thickness
The connection can collapse from the rupture of the frontal seams along the vertical legs ss" or from the cut of these seams along the horizontal legs ss". However, practice shows that the seam is destroyed along the bisector section, the height of which
Where To- leg of the seam, in our case To = 8.
Such a seam is conditionally calculated for a cut along a bisectoral section from the strength condition:
Where A cf = 0,7 3b is the cut area of one weld.
Rice. thirty.
Conclusion: the seams are underloaded.
Example 9. A 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. The number of splines 2 = 6. Determine the shear and collapse stresses of the spline.
Rice. 31.
Assuming that all splines are equally loaded, we find the force per one spline:
Let's define the cutoff voltage:
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