Calculation and selection of the main elements of a plate conveyor. Calculation of belt-chain and plate steeply inclined conveyors. Calculation of a conveyor with continuous low scrapers

Initial data. The transported cargo is ordinary medium-sized iron ore. The conveyor route is a complex combined one (see Fig. 2.35, b). Loading is carried out at the beginning of the lower horizontal section without the use of a special feeder, unloading is carried out at the end of the upper horizontal section through the drive sprocket shaft. The operating conditions of the conveyor are difficult: work in the open air, intense abrasive pollution.

Conveyor design capacity Q=350 t/h; geometric parameters of the route:

L 1g =10 m; L 2g =25 m; L 3g =20 m; N=10 m.

Working through the task. Typical piece size mm; bulk density of cargo
t/m 3 ; angle of repose of the load at rest
, and in motion; coefficient of friction of the load on the steel deck (minimum value) f in =0.5; angle of friction of the load on the metal flooring
; the angle of inclination of the inclined section of the route.

For the given conditions, we select a general-purpose double-chain conveyor with long-link traction plate chains and sprockets with a small number of teeth. Taking this into account, we accept the conveyor speed
m/s.

Volumetric capacity corresponding to design capacity Q=350 t/m 3, is

Selecting the type of flooring and determining its width. Taking into account the parameters of the cargo And according to table 2.7 choose heavy type flooring.

Since only conveyors with side decking or with fixed sides are suitable for transporting bulk cargo, when checking the transporting capacity using expressions (2.66) and (2.67), we accept the minimum angle values indicated in these expressions in parentheses.

According to formulas (2.66) and (2.67), the largest conveyor inclination angles at which ore transportation is ensured without a significant reduction in productivity:

for smooth flooring with sides;

for side corrugated flooring;

for box flooring
.

According to condition (2.68) for smooth and wavy floorings

For a smooth flooring, both conditions are not met; for a wavy flooring, condition (2.68) is not met. Taking this into account, we choose heavy type side box flooring (KG).

According to condition (2.72) mm.

According to table. 2.5 blade speeds
m/s and volumetric productivity
m 3 / h corresponds to the height of the sides
mm. We accept
.

Using formula (2.71) we find the required width of the flooring

where in accordance with formula (2.70) (here C 2 =0.9 is the dimensionless coefficient at
);m is the height of the load layer at the sides.

By checking the width of the flooring according to the particle size distribution of the cargo using formula (2.73), we obtain mm.

From the series according to GOST 22281-76 we take the nearest larger value of the flooring width

mm.

A significant increase in the width of the web compared to the value determined by formula (2.71) requires recalculation of the speed using formula (2.74):

m/s.

Since the nearest lower standard speed value
m/s would result in a reduction in productivity compared to the calculated value
t/h, finally accepted
mm;
mm;
m/s.

Calculation of distributed masses. Distributed mass of transported cargo

flooring with chains

Where
kg/m (see table 2.7).

Selection of coefficients of resistance to web movement. Taking into account operation in difficult conditions (outdoors, intense pollution) according to table. 2.6 we accept the coefficient of movement resistance for flange rollers on plain bearings
. Resistance coefficient when bending around deflectors:
at bend angle
And
at an inflection angle of 180 0.

Determination of the point with the lowest tension of the traction element. The lowest tension of the traction element will be at the lowest point 4 inclined section of the idle branch, since .

Determination of tensions at characteristic points of the route. We accept the tension at the point 4
. When bypassing the route from the point 4 in the direction of movement of the web we determine

To determine tensions at points 1 And 3 We bypass the idle branch against the direction of movement of the web

Determination of traction on drive sprockets and drive power. Traction force on drive sprockets

With safety factor
and drive efficiency
engine power

kW.

Based on the obtained power value, we select the engine in accordance with the recommendations set out in section 3.

Determination of the design tension of the traction element. By analogy with the designs used, we accept a traction element consisting of two parallel plate chains with a pitch
drive sprocket with number of teeth

For a given conveyor route layout, the maximum tension of the traction element is

To find the dynamic force, we determine:
(the law of interference of elastic waves is unknown);

traction element contour length m;

coefficient of participation in the oscillatory process of the mass of the transported load
(at
);

coefficient of participation in the oscillatory process of the mass of the conveyor running gear
(at
m);

weight of cargo on the conveyor, kg;

weight of the conveyor running gear kg.

Using formula (2.88) we calculate the dynamic force

Using expression (2.87), we determine the calculated tension of the traction element (two chains)

Determination of the design tension of the traction chain and its selection. According to formula (2.92), the calculated chain tension of a double-chain conveyor

Where
- tension unevenness coefficient (taken approximately).

According to GOST 588-81, we first select the M450 roller chain with a breaking load
kN.

The safety margin of this chain is less than the permissible
for conveyors with inclined sections. Taking this into account and taking into account the difficult operating conditions of the conveyor, we select the L630 chain with a breaking load
kN. Its safety margin is determined by formula (2.93)

According to GOST 558-81, the selected chain has the following main parameters and dimensions: pitch 400 mm; roller diameter 36 mm; sleeve diameter 50 mm; roller diameter 140 mm; roller flange diameter 175 mm; distributed mass 25.8 kg/m.

The determination of the remaining parameters of the conveyor (calculation of the tensioning device, starting and braking modes, etc.) is carried out in accordance with the general instructions given in clause 1.3.

Depending on the design of the deck and traction chain and the configuration of the route (Fig. 4.1), general purpose apron conveyors (vertically closed) are distinguished; curving (with a spatial route) and special purpose (filling machines, escalators, passenger, conveyors with complex profile flooring).

Rice. 4.1. Schemes of apron conveyor routes:

A - horizontal; b – horizontally inclined; G - inclined;

d – oblique-horizontal; V, e, and - complex

The most widely used are plate stationary, vertically closed conveyors with straight routes, which are general purpose conveyors. In the metallurgical industry they are used to feed lump ore and hot sinter; at chemical plants and in the production of building materials - for moving large pieces of non-metallic materials; at thermal power plants - when supplying coal; in mechanical engineering - for transporting hot forgings, castings, flasks, stamping waste; on production lines for assembly, cooling, drying, sorting and chemical processing.

Mobile apron conveyors are used in warehouses, loading and unloading, sorting and packaging points for moving packaged unit cargo.

Special plate conveyors, including bending ones with a spatial route, are used in the mining and coal industries for transporting ore and coal over long distances.

4.1.1.1 General structure, purpose and areas of application

The advantages of plate conveyors compared to belt conveyors include: the ability to transport heavy large-piece, sharp-edged and hot loads; calm and silent movement; possibility of loading without the use of feeders; long duration of the route with inclined sections and small radii of transitions and ensuring unloaded transportation; possibility of installing intermediate drives; high productivity at low speed; the possibility of using conveyors in technological processes and production lines at high and low temperatures.

The disadvantages of apron conveyors are: the large mass of flooring and chains and their high cost; the presence of a large number of chain hinges that require additional maintenance; the difficulty of replacing worn traction chain rollers; high resistance to movement.

The plate conveyor (Fig. 5.2) has a frame, at the ends of which two sprockets are installed - a drive sprocket 3 with a drive and a tension sprocket with a tensioning device 4. An endless deck 1, consisting of individual plates, is attached to the chassis, consisting of one or two traction chains 2 , which go around the end sprockets and are in engagement with their teeth.

Vertically closed traction chains move together with the flooring along the guide tracks of the frame along the longitudinal axis of the conveyor. The conveyor is loaded through one or more funnels 5 anywhere along the route, and unloaded through the end sprocket and funnel. Intermediate unloading is only possible for apron conveyors with flat deck. Their speed of movement is up to 1.25 m/s.

Rice. 4.2. Plate Conveyor:

1 – flooring; 2 – traction chain; 3 – drive sprocket;

4 – tension device; 5 – loading hopper

The main parameters of general purpose apron conveyors are established by GOST 22281-92: deck width: 400; 500; 650; 800; 1000; 1200; 1400; 1600 mm; number of sprocket teeth: 6; 7; 8; 9; 10; eleven; 12; 13; movement speed: 0.01; 0.016; 0.025; 0.04; 0.05; 0.063; 0.08; 0.1; 0.125; 0.16; 0.2; 0.25; 0.315; 0.4; 0.5; 0.63; 0.8; 1.0 m/s.

The inclination angle of the apron conveyor belt is usually 35–60º and depends on the characteristics of the transported cargo and the type of flooring. When transporting piece goods and having transverse load-holding bars on the floor, the angle of inclination of the conveyor can be increased.

4.1.1.2 Elements of apron conveyors

Traction element Leaf chains are usually used:

PV – plate bushings;

PVR – plate bushing-roller;

PVK – plate bushing-roller with smooth rollers;

PVKG – plate bushing-roller with ridges on rollers;

PVKP – plate bushing-roller with roller bearings

Bushing, roller and round link chains can be used as a traction element. Conveyors with a deck width of more than 400 mm have two traction chains, light conveyors (with a deck width of less than 400 mm) have one chain.

Support elements Plate bushing-roller chains have running rollers that transfer the load from the decking and the transported load to the guide tracks (on heavy-duty conveyors, rollers on rolling bearings are used).

In conveyors with bushing and roller chains and smooth flooring, the supporting elements are stationary roller supports mounted on the conveyor frame. In light-type conveyors with a deck width of 80–200 mm, the chain can be combined with a deck sliding along metal or plastic guides.

Flooring is a load-carrying element of the conveyor. The flooring is made with or without sides and has a different design depending on the characteristics of the transported cargo (Table 4.1).

Table 4.1

Types of apron conveyor decks

End of table. 4.1

Flat decking is made from wooden planks, steel or polyurethane plates; To ensure a secure position of the load, the deck is equipped with shaped pads or stops. The corrugated flooring ensures reliable overlap of adjacent plates, increases the rigidity and strength of the web, increases the adhesion of loads to the conveyor surface, reduces their spillage between the plates and ensures the movement of loads at large angles of inclination.

Channel flooring is used for transporting large hot castings and stampings, provides strength and rigidity to the canvas and facilitates its cleaning. The flooring is made by stamping and welding steel sheets 4–10 mm thick. The deck plates are secured with bolts, rivets or welded to special angles that are attached to the traction chain plates.

The main dimensions of the flooring are its width IN and side height h. Normal flooring width range: 400, 500, 650, 800, 1000, 1200, 1400, 1600 mm; side heights: 80, 100, 125, 160, 200, 250, 315, 355, 400, 450 and 500 mm.

Drive unit apron conveyor - angular or straight (tracked) (Section 2.4), consists of drive sprockets, a transmission mechanism (gearbox or gearbox with additional gear) and an electric motor. On conveyors with an inclined section of the route, a stopping device or an electromagnetic brake is installed. The drive transmission mechanism is a single gearbox or a gearbox with a gear or chain drive. Powerful conveyors of great capacity and length have several drives.

Tension devices. On apron conveyors, screw (most common) or spring-screw tensioning devices are installed (on heavily loaded conveyors of considerable length with speeds of more than 0.25 m/s). NU are installed on end sprockets and have a stroke equal to at least 1.6–2 chain pitches, X= 320–2000 mm.

One of the NU sprockets is secured to the shaft with a key, the other is free to allow self-installation according to the position of the chain hinges.

Plate Conveyor Bed made of angle or channel steel. The end parts are made in the form of separate frames for the drive and the hydraulic unit, the middle part is made in the form of separate sections of metal structures 4–6 m long.

4.1.1.3 Calculation of apron conveyors

The calculation of apron conveyors is carried out in two stages: preliminary (approximate) determination of the main parameters; verification calculation. The initial data for the calculation are:

performance;

route configuration;

characteristics of the transported cargo;

web speed;

operating mode.

In accordance with GOST 22281–92, the type of conveyor and type of flooring are selected. Flooring is used in three types:

light – with bulk density of transported cargo ρ< 1т/м 3 ;

average - at ρ = 1–2 t/m3;

heavy – at ρ > 2 t/m3.

Side height h side decking for bulk cargo is selected from the normal range (according to the reference book), for piece cargo h= 100–160 mm.

The angle of inclination of the conveyor depends on the type of flooring and the characteristics of the cargo being moved (Table 4.2), the selected angle of inclination of the conveyor must satisfy the condition β ≤ φ 1 – (7–10º), where φ 1 – angle of repose of a load in motion.

apron conveyor

β" – angle of friction of the load on the flooring

On a deck without sides, the bulk cargo is located in a triangle (Fig. 4.3) in the same way as on a belt conveyor with straight roller supports; IN - flooring width, b = 0,85IN, φ – angle of repose of the load at rest (angle of repose of the load in motion φ 1 = 0.4 φ).

Rice. 4.3. Arrangement of bulk cargo on a flat deck

Sectional area of bulk cargo on a deck without sides

Where h 1 – height of the triangle;

With 2 – coefficient taking into account the reduction in area on an inclined conveyor (Table 4.3).

Conveyor performance

where ρ – cargo density, t/m3;

v– conveyor speed, m/s;

In p– width of the deck without sides.

Table 4.3

Coefficient values With 2

Width of decking without sides

Productivity when laying with sides (Fig. 4.4)

. (4.4)

Rice. 4.4. Types of side decks:

A– with movable sides; b– with fixed sides

Cross-sectional area of cargo on a deck with sides

Where IN b – width of the flooring with sides, m;

ψ = 0.65–0.8 – filling coefficient of the deck section.

The resulting width of the flooring is checked according to the lumpiness condition IN≥ X 2 A+200 mm, where X 2 – lumpiness coefficient. For sorted cargo X 2 = 2.7; for ordinary cargo X 2 = 1,7.

The final selected deck widths are rounded to the nearest value according to the normal series.

For piece goods, the width of the flooring is selected according to the overall dimensions of the load, the method of its stacking and quantity, while the gap between the loads should be 100–300 mm.

Traction calculation. During the traction calculation, the resistance and tension forces of the chains on individual sections of the route are determined.

The maximum tension of the chains is calculated by sequentially determining the resistance in individual sections, starting from the point of least tension.

The minimum tension is taken to be at least 500 N per chain (usually S min = 1–3 kN) .

Linear gravity of decking with chains q 0 (N/m) is determined from reference books and catalogues, usually

q 0 ≈ 600B+A, (4.6)

Where A - coefficient taken depending on the type and width of the flooring.

Linear gravity load (N/m)

Maximum static chain tension

Where L g and L x – the length of the horizontal projection of the loaded and unloaded branches of the conveyor, m;

N– load lifting height, m.

The “+” sign in the formula is for uphill sections, “–” for downhill sections.

Total Design Force

S max = S st + S din, (4.9)

Where S st – static tension of traction chains, N;

S din – dynamic loads in traction chains, N.

If the traction element consists of two chains, then the calculated force on one chain is taken into account by the coefficient of unevenness of its distribution WITH n =1.6–1.8.

Design force of one chain S calc = S max, two circuits S calc = (1.5 S max) / 2.

Circumferential force on sprocket

R = ∑ W=S st – S 0 , (4.10)

Where S st – the greatest static force in the traction chains at the point of contact with the drive sprockets, obtained by the method of bypassing the contour, N;

S 0 – chain tension at the point of escape from the drive sprocket, N.

Conveyor drive power

N in = Q L g ω / 367, (4.11)

Where Q– productivity, t/h;

L g – horizontal projection of length, m;

ω 0 – generalized coefficient of resistance to movement.

Next, the engine is selected, the gear ratio is determined and the gearbox is selected; determining the actual speed of movement and clarifying performance; determination of static braking torque (for inclined conveyors); calculation of braking torque; determining the stroke of the tensioner.

Verification calculation includes a refined traction calculation using the contour bypass method; checking the selected traction chain; checking the calculated drive power; selection of the type of tensioning device.

4.1.1.4 Installation of apron conveyors.

Sequence of stages for installing an apron conveyor:

· layout of axes and installation of the middle part of the conveyor frame;

· installation of supporting structures or rails (for chain rollers) while ensuring tolerances of no more than 1–2 mm;

· installation of the drive and tension station while ensuring horizontal and perpendicularity of the axes of the conveyor and the drive shaft;

· other drive elements (open gears, gearbox and electric motor) are oriented along the drive shaft, ensuring strict alignment of the shafts;

· the chassis must be thoroughly checked;

· testing begins by moving the chassis 5–10 m manually or using an electric motor;

· running the conveyor idle for 3–4 hours:

– the conveyor must operate smoothly, without knocks, shocks and vibrations;

– chain engagement should be smooth;

– adjacent plates must rotate freely on sprockets and curved sections;

– the heating temperature of the gearbox and sliding bearings should be no more than 70º, there should be no heating of the rolling bearings;

· running-in under load (within 12 hours)

– carry out the same checks as during dry running;

– adjust the location of the loading device;

– eliminate spillage of cargo onto the working surfaces of the rails and into the gaps between the plates;

– regulate the operation of the NU to prevent the blade from moving

4.1.1.5 Technical inspection and repair of apron conveyor elements.

Technical inspection (TO) of traction chains involves their systematic inspection, routine repair, cleaning and lubrication. During the inspection, the following is revealed: the condition of parts, fits in joints; mobility of rollers and rollers.

Non-rotating rollers and rollers with flats on the rolling surface must be replaced, loose bolted connections of links and fastenings of working parts must be tightened.

Maintenance of sprockets reveals wear on the side surfaces of the teeth: the sprocket is repaired or replaced; the web run-off is eliminated.

Maintenance of load-carrying elements involves their inspection and elimination of damage that impedes operation: identifying the presence of residual deformations, reliability of fastening to the traction element, wear; deformed plates are corrected or replaced, the gaps between them are adjusted, loose connections are tightened.

Introduction

Plate conveyors are designed to move heavy (500 kg or more) piece goods, large pieces, incl. sharp-edged materials, as well as loads heated to high temperatures. Plate conveyors, stationary or mobile, have the same basic components as belt conveyors.

The load-carrying body is a metal, less often wooden, plastic flooring sheet, consisting of separate plates attached to 1 or 2 traction chains (bush-roller). The deck can be flat, corrugated or box-shaped, without edges or with edges. The traction chains go around the drive and tension sprockets mounted at the ends of the frame. There are general purpose apron conveyors (basic type) and special ones. To increase productivity, flat deck conveyors are supplemented with fixed sides. Typical apron conveyors have a capacity of up to 2000 t/h. A separate type of apron conveyors, which has become most widespread in Russia in the last 15-20 years, is a modular belt conveyor. The tape can be either plastic or steel. The wide range of belts produced also determines their wide range of applications: from interoperational transport and supply of the product directly to the machine, to use in the food industry, as well as in trade.

1. Design description

Figure 1. Scheme of the designed conveyor:

The main assembly units of an apron conveyor are: a lamellar fabric, running rollers, a traction element and a tensioning device. Fabric plates having a rectangular or trapezoidal cross-section are made stamped; the thickness of the plates for transporting coal is 3-4 mm, for large-sized rocks weighing 6-8 mm. The rollers are attached to the plates using short cantilever or through axles. As a traction element on which the plates are fixed, 1 or 2 plate or round link chains are used. The bending conveyor has one round link chain. The drive end station includes an electric motor, a clutch, a gearbox and a drive shaft with a drive sprocket. It is possible to install intermediate track-type drives, in which cams are attached to the drive chain and interact with the links of the conveyor traction chain. The tensioning device is usually located at the tail end of the conveyor. Advantages of the plate conveyor: the ability to transport abrasive rock mass along a curved route with small radii of curvature; lower resistance to movement and energy consumption than in scraper conveyors; the ability to install intermediate drives, which allows you to increase the length of the conveyor in one train.

Disadvantages: high metal consumption, complex design of the plate cloth and the difficulty of cleaning it from the remains of wet and sticky rock mass, deformation of the plates during operation, which causes spillage of fine fractions.

2. Calculation of apron conveyor

.1 Determination of conveyor width

For the calculation, we assume a conveyor with a corrugated belt with sides.

The width of the conveyor is determined by the formula:

m, (2.1)

where Q = 850 t/hour - conveyor productivity;

u = 1.5 m/s - speed of the web;

r = 2.7 t/m 3 - density of transported cargo;

K β =0.95 - coefficient taking into account the angle of inclination of the conveyor;

j = 45 o - angle of repose of the load at rest;

h = 0.16 m - height of the sides of the canvas, selected from the nominal range;

y = 0.7 - side height utilization factor

The coefficient K β is determined by the formula:

b =10 o - conveyor inclination angle.

We substitute the obtained values into formula (1.1)

For transported material containing large pieces up to 10%

of the total load the following condition must be met:

mm (2.3)

a max = 80 mm - the largest size of large pieces.

The condition is met.

We finally select the width of the web from the nominal range B = 400 mm

2.2 Determination of loads on the transport chain

We preliminarily accept it as a conveyor traction element

plate chain type PVK (GOST 588-81).

The linear load from the transported cargo is determined by the formula:

(2.4)

The linear load from the own weight of moving parts (webs with chains) is determined by the formula:

N/m, (2.5)

A = 50 - coefficient taken depending on the width of the web of the type of cargo

The minimum chain tension for a given conveyor can be at points 1 or 3 (Fig. 1). The minimum tension will be at point 3 if the following condition is met:

w = 0.08 - coefficient of resistance to movement of the chassis on

straight sections

The condition is not met, therefore the minimum tension will be at point 1.

We accept the minimum chain tension S min = S 1 = 1500 N. Using the method of walking along the contour along the web, we determine the tension at points 1..6 (Fig. 1) using a method similar to .

k = 1.06 - coefficient of increase in chain tension when going around the sprocket

Figure 2. Tension diagram of the traction element

3. Calculation of conveyor elements

.1 Calculation and selection of electric motor

The traction force of the drive is determined by the formula:

where k = 1.06 - coefficient of increase in chain tension when bending

stars

The installed power of the electric motor is determined by the formula:

kW, (3.2)

where h = 0.95 - drive efficiency

k z = 1.1 - power reserve factor

We accept an electric motor with increased starting torque of the 4A series

engine type - 4АР200L6УЗ;

power N = 30 kW;

rotation speed n motor = 975 rpm;

swing moment GD 2 = 1.81 kg m 2;

mass m = 280 kg.

shaft connecting diameter d = 55 mm.

3.2 Calculation and selection of gearbox

The pitch diameter of the drive sprockets is determined by the formula:

where t is the pitch of the drive chain;

z - number of sprocket teeth;

We tentatively assume t = 0.2 m and z = 6.

The rotation speed of the sprockets is determined by the formula:

RPM (3.4)

rpm

The gear ratio is determined by the formula:

(3.5)

The torque on the output shaft of the gearbox is determined by the formula:

Nm. (3.6)

Based on the above defined values, we accept a two-stage helical gearbox

gearbox type - 1Ц2У-250;

gear ratio u = 25;

rated torque on the output shaft under heavy duty Mcr = 6300 Nm;

mass m = 320 kg.

The input and output shafts have conical connecting ends for couplings (Fig. 3), their main dimensions are given in Table 1.

Figure 3. Scheme of fitting parts onto the shaft.

Table 1. Geometric parameters of shafts

3.3 Calculation and selection of traction chain

The design force in the chain is determined by the formula:

N, (3.7)

The dynamic load on the chain is determined by the formula:

N, (3.8)

where y = 1.0 is a coefficient that takes into account the reduction in the reduced mass of moving parts of the conveyor, selected according to L > 60 m.

Substituting the found values into formula (3.7) we obtain:

The breaking force of the chain is determined by the formula:

Based on the above defined values, we accept a plate chain

chain type - M450 (GOST 588-81);

chain pitch t = 200 mm;

breaking force S cut. = 450 kN.

To test the chain for strength, we will calculate the load on the chain at the moment the conveyor is started.

The maximum force in the chain when starting the conveyor is determined by the formula:

N, (3.10)

where S d.p is the dynamic force of the chain at start-up.

The dynamic force of the chain at start-up is determined by the formula

N, (3.11)

where m k is the reduced mass of the moving parts of the conveyor;

The reduced mass of the moving parts of the conveyor is determined by the formula

kg, (3.12)

where k y = 0.9 is a coefficient taking into account elastic elongation of chains

k u = 0.6 - coefficient taking into account the decrease in average speed

rotating masses compared to average speed.

Gu = 1500 kgf - weight of the rotating parts of the conveyor (without drive), taken according to

The angular acceleration of the electric motor shaft is determined by the formula:

rad/s 2 , (3.13)

where I pr is the moment of inertia of the moving masses of the conveyor, reduced to the motor shaft.

M p.sr - determined by the formula:

H m, (3.14)

M p.st - determined by the formula:

H m, (3.15)

The moment of inertia of the moving masses of the conveyor, reduced to the motor shaft, is determined by the formula:

H m s 2, (3.16)

where I r.m is the moment of inertia of the electric motor rotor and the sleeve-pin coupling, determined by the formula:

H m s 2, (3.17)

where I m = 0.0675 is the moment of inertia of the pin-sleeve coupling.

Substituting the values into formulas 3.10... 3.17, we obtain the maximum force in the chain when starting the conveyor.

H m s 2

rad/s 2

3.4 Tensioner calculation

We adopt a screw type tensioning device.

The magnitude of the tensioner stroke depends on the chain pitch and is determined by the formula

The total length of the screw is taken to be L rev = L+0.4 = 0.8 m.

We accept the material for the screw - steel 45 with a permissible shear stress σ av = 100 N/mm 2 and a yield strength s T = 320 N/mm 2. I choose the type of thread: rectangular (GOST 10177-82).

We accept material for the nut - bronze Br. AZh9-4 with permissible shear stress σ av = 30 N/mm 2, crush stress σ cm = 60 N/mm 2, tensile stress s P = 48 N/mm 2. The thread type is the same.

The average screw thread diameter is determined by the formula:

mm, (3.19)

where y = 2 - the ratio of the nut height to the average diameter

[p] = 10 N/mm 2 - permissible stress in the thread, depending on the rubbing materials, during friction of steel on bronze [p] = 8...12 N/mm 2;

K = 1.3 - coefficient taking into account the uneven load of tension coils

The internal diameter of the thread is determined by the formula:

Mm, (3.20)

Considering that the length of the screw is large and greater stability is required, we take d 1 = 36 mm.

The thread pitch is determined by the formula:

mm (3.21)

The adjusted value of the average thread diameter is determined by the formula:

mm (3.22)

The outer diameter of the thread is determined by the formula:

mm (3.23)

The helix angle of the thread is determined by the formula:

We check the reliability of self-braking, for which the following condition must be met:

, (3.25)

where f = 0.1 is the coefficient of friction between steel and bronze.

The condition is met.

We check for stability.

, (3.26)

where j is the slip coefficient of the permissible compressive stresses; when calculating stability, it is determined as a function of the screw flexibility (l).

Allowable compression stress.

The permissible compression stress is determined by the formula:

N/mm 2, (3.27)

The flexibility of the screw is determined by the formula:

, (3.28)

where m =2 - reduced length coefficient

Based on the known flexibility of the screw, I find j = 0.22. We substitute the obtained data into condition 2.26:

The condition is met.

Since the screw works in tension, it is not necessary to check for stability.

We check the screw for strength, strength condition:

, (3.29)

Where (defined above);

M 1 - friction moment in the thread (N mm);

M 2 - friction moment at the heel (stop) (N mm)

The friction moment in the thread is determined by the formula:

N m (3.30)

The moment of friction at the heel is determined by the formula:

N mm, (3.31)

where d n = 20 mm is the diameter of the heel, taken less than d 1.

We substitute the obtained data into condition 3.29:

The condition is met.

The height of the nut is determined by the formula:

mm (3.32)

The number of threads in the nut is determined by the formula:

We check the shear strength of the nut thread, strength condition:

The condition is met

3.5 Shaft calculations

Drive shaft

We take steel 45 as the shaft material, tensile strength

s B = 730 N/mm 2, endurance limits: s -1 = 0.43 s B = 314 N/mm 2, t -1 = 0.58 s - 1 = 182 N/mm 2

I determine the approximate minimum shaft diameter based only on torsion using the formula:

mm, (3.34)

where M = 5085 Nm - torque on the shaft

25 N/mm 2 - permissible torsional stress for steel 45

mm.

From the standard series (GOST 6636-69 R40) we select the closest diameter value d pv = 100 mm. We accept this diameter for bearings. For fastening the drive sprockets we take diameter d = 120 mm. The width of the drive sprocket hub is determined based on the required length of the key to transmit torque.

The length of the key is determined from the conditions of collapse and strength:

, (3.35)

where l is the length of the key, mm;

d - shaft diameter at the location where the key is installed, mm;

h, b, t 1, - dimensions of the cross-section of the key, mm

[s] cm - permissible bearing stress, for steel hubs 100-120 N/mm 2.

Also, based on condition 3.35, we determine the parameters of the key for the connecting end of the shaft, the diameter of which is taken to be d = 95 mm and length l = 115 mm. The values of all geometric dimensions of the keys are entered in Table 2.

Table 2. Geometric parameters of shafts

* We use two keys located at an angle of 180 o.

Based on the length of the keys for the drive sprockets, we select the length of the hubs of the latter as l st = 200 mm.

The design diagram of the drive shaft and the diagram of bending moments has the form

Figure 4. Moment diagrams

where R 1 and R 2 are the reactions of the supports in the bearings, N;

P is the load on the sprockets, determined by the formula:

N. (3.36)

Due to the symmetry of the design and support reaction loads

R 1 = R 2 = P = 13495 N.

The calculation is carried out similarly to paragraph 2.5.1.

We take steel 45 as the shaft material (workpiece diameter more than 100 mm), tensile strength s B = 730 N/mm 2, endurance limits: s -1 = 0.43 s B = 314 N/mm 2, t -1 = 0.58 s - 1 = 182 N/mm 2

The shaft diameter is structurally taken as 0.8 of the drive shaft diameter d = 80 mm

The design diagram of the shaft is similar to Fig. 4.

We accept this diameter for bearings. For fastening the drive sprockets, we take the diameter d = 100 mm. The width of the drive sprocket hub is taken constructively.

3.6 Selection of bearings

Since when installing separate bearing housings on the conveyor frame, there is a violation of their alignment and shaft misalignment, we select double-row spherical ball bearings 1320 (GOST 5720-75 and 8545-75) with the following parameters:

d = 100 mm (inner diameter)

D = 215 mm (outer diameter)

B = 47 mm (width)

C = 113 kN (Dynamic load rating)

We check bearings for durability, which is determined by the formula:

h, (3.37)

where n = 39 rpm - shaft rotation speed;

P e - equivalent load on the bearing, provided there are no axial loads, is determined by the formula:

N, (3.38)

where V = 1 - coefficient taking into account the rotation of the rings

K T = 1 - temperature coefficient

K s = 2.0 - load factor

h. Durability is sufficient

Since when mounting separate bearing housings on the conveyor frame, there is a violation of their alignment and shaft misalignment, I choose double-row spherical radial ball bearings 1218 (GOST 5720-75 and 8545-75) with the following parameters:

d = 800 mm (inner diameter)

D = 160 mm (outer diameter)

B = 30 mm (width)

C = 44.7 kN (Dynamic load rating)

h. Sufficient durability.

Based on the calculations made, we determine that the bearings will operate throughout their entire service life.

.7 Calculation and selection of braking devices and couplings

When the conveyor is turned off in a loaded state due to the tilt of part of the conveyor, the weight of the load will create a force directed in the direction opposite to the movement of the belt. This force is determined by the formula

N. (3.39)

A negative force value means that the friction force of the conveyor elements is higher than the load rolling force, and therefore there is no need to use a braking device.

To transmit torque from the electric motor to the input shaft of the gearbox, we use an elastic pin-type coupling of the MUVP type (GOST 21424-75) with bores of the coupling halves for the motor shaft (d d = 55 mm) and for the input shaft of the gearbox (taper bore d p1 = 40 mm) .

The torque supplied to the electric motor shaft is equal to the ratio of the torque on the output shaft of the gearbox to the gear ratio of the gearbox M motor = 203.4 Nm.

Taking into account the margin and overall dimensions, we accept a coupling with a rated torque M cr = 500 Nm, with the maximum (overall) diameter of the coupling D = 170 mm, maximum length L = 225 mm, number of fingers n = 8, finger length l = 66 mm , connecting thread of the pin M10.

To transmit torque from the output shaft of the gearbox to the drive shaft, I use a gear coupling type MZ (GOST 5006-83) with a tapered bore (version K d p2 = 90 mm) for connection to the output shaft of the gearbox. The coupling bore for connection to the drive shaft is cylindrical d = 95 mm with two keyways.

We select a coupling with a rated torque Mcr = 19000 Nm.

.8 Sprocket calculation

Known data for calculation:

pitch diameter of sprockets d e = 400 mm;

number of teeth z = 6;

tooth pitch t = 200 mm.

diameter of chain rollers D c = 120 mm.

The diameter of the outer circle is determined by the formula:

mm, (3.40)

where K=0.7 - tooth height coefficient

The diameter of the circle of the depressions is determined by the formula:

Mm, (3.41)

The displacement of the centers of the arcs of the depressions is determined by the formula:

e = 0.01. 0.05 t = 8 mm. (3.42)

The radius of the tooth cavities is determined by the formula:

r = 0.5 (D c - 0.05t) = 50 mm. (3.44)

The radius of curvature of the tooth head is determined by the formula:

Mm. (3.45)

The height of the straight section of the tooth profile is determined by the formula:

mm. (3.46)

I determine the width of the tooth using the formula:

b f = 0.9 (50 - 10) - 1 = 35 mm. (3.47)

The width of the tooth apex is determined by the formula:

b = 0.6b f = 21 mm. (3.48)

The diameter of the crown is determined by the formula:
Having completed the course project, we designed a chain and plate conveyor with the following parameters:

Productivity Q =850 t/hour;

Web speed u = 1.5 m/s;

Conveyor length l = 90 m;

The length of the horizontal section l g = 25 m;

Conveyor inclination angle β = 10 o ;

Density of transported cargo r = 2.7 t/m 3

We also calculated its main elements and tested them for strength and durability.

Bibliography

1. Baryshev A.I., Steblyanko V.G., Khomichuk V.A. Mechanization of PRTS works. Course and diploma design of transporting machines: Textbook / Under the general editorship of A.I. Barysheva - Donetsk: DonGUET, 2003 - 471 p., ill.

Baryshev A.I., Mechanization of loading and unloading, transport and warehouse operations in the food industry. Part 2. Transporting machines. - Donetsk: DonGUET, 2000 - 145 p.

Chernavsky S.A. Course design of machine parts, M.: Mashinostroenie, 1979. - 351 p.

Anufriev V.I. Handbook of the designer and mechanical engineer in three volumes, M.: Mashinostroenie, 2001.

Yablokov B.V., Belov S.V. Guidelines for the course project on lifting and transport devices (plate conveyors), Ivanovo, 2002.

The calculation of apron conveyors is carried out in two stages: preliminary (approximate) determination of the main parameters; verification calculation. The initial data for the calculation are:

Performance;

Route configuration;

Characteristics of the transported cargo;

Web speed;

Operating mode.

In accordance with GOST 22281–92, the type of conveyor and type of flooring are selected. Flooring is used in three types:

Light - with bulk density of transported cargo ρ< 1т/м 3 ;

Medium – at ρ= 1–2 t/m 3 ;

Heavy - at ρ> 2 t/m 3.

Side height h side decking for bulk cargo is selected from the normal range (according to the reference book), for piece cargo h= 100–160 mm.

The angle of inclination of the conveyor depends on the type of flooring and the characteristics of the cargo being moved (Table 2), the selected angle of inclination of the conveyor must satisfy the condition β≤φ 1 -(7-10°), where φ 1 is the angle of repose of the load in motion.

– angle of friction of the load on the flooring

On a deck without sides, the bulk cargo is located in a triangle (Fig. 3) in the same way as on a belt conveyor with straight roller supports; IN– flooring width, b = 0,85IN, φ – angle of repose of the load at rest (angle of repose of the load in motion φ 1 =0.4φ).

Rice. 3. Placement of bulk cargo on a flat deck

Sectional area of bulk cargo on a deck without sides

Where h 1 – height of the triangle;

With 2 – coefficient taking into account the reduction in area on an inclined conveyor (Table 3).

Conveyor performance

Q n =3600F 1 ρ v=648 c 2 vρtgφ1, (2)

where ρ – cargo density, t/m3;

v– conveyor speed, m/s;

IN n – width of the flooring without sides.

Table 3. Coefficient values With 2

Width of decking without sides

Productivity when decking with sides (Fig. 4)

Q b =3600F vρ. (4)

Rice. 4. Types of side decks:

a – with movable sides; b – with fixed sides

Cross-sectional area of cargo on a deck with sides

F=F 2 +F 3 =0.25 k β tanφ 1 +B b hψ, (5)

Where IN b – width of the flooring with sides, m;

ψ= 0.65–0.8 – filling coefficient of the deck section.

The resulting width of the flooring is checked according to the lumpiness condition B≥X 2 a+200 mm, where X 2 – lumpiness coefficient. For sorted cargo X 2 = 2.7; for ordinary cargo X 2 = 1,7.

The final selected deck widths are rounded to the nearest value according to the normal series.

Traction calculation. During the traction calculation, the resistance and tension forces of the chains on individual sections of the route are determined.

The maximum tension of the chains is calculated by sequentially determining the resistance in individual sections, starting from the point of least tension.

The minimum tension is taken to be at least 500 N per chain (usually S min = 1–3 kN).

Linear gravity of decking with chains q 0 (N/m) is determined from reference books and catalogues, usually

q 0 ≈600B+A, (6)

where A is the coefficient taken depending on the type and width of the flooring.

Linear gravity load (N/m)

Maximum static chain tension

Where L g and L x – the length of the horizontal projection of the loaded and unloaded branches of the conveyor, m;

N– load lifting height, m.

The “+” sign in the formula is for uphill sections, “–” for downhill sections.

Total Design Force

S max = S st + S din, (9)

Where S st – static tension of traction chains, N;

S din – dynamic loads in traction chains, N.

If the traction element consists of two chains, then the calculated force on one chain is taken into account by the coefficient of unevenness of its distribution WITH n =1.6–1.8.

Design force of one chain S calc = S max, two circuits S calc = (1.5 S max)/2.

Circumferential force on sprocket

P=ΣW=S st -S 0 , (10)

Where S st – the greatest static force in the traction chains at the point of contact with the drive sprockets, obtained by the method of bypassing the contour, N;

S 0 – chain tension at the point of escape from the drive sprocket, N.

Conveyor drive power

Where Q– productivity, t/h;

L g – horizontal projection of length, m;

ω 0 – generalized coefficient of resistance to movement.

Determine the width of the flooring, select the traction element and find the power of the electric motor.

apron conveyor traction motor

Rice. Cross-section of bulk cargo located on the deck of an apron conveyor: a - without sides; b - with sides; c - with fixed sides.

When determining the width of a flat deck without sides, the load layer in it has a triangle shape in cross-section (Fig. a). The cross-sectional area of the load (m 2) is determined as F 1 = C 1 *b*h 1 /2 = C 1 *b 2 *tg(t 1)/4 = 0.18*B 2 n *C 1 *tg( ц 1) (1) where b is the width of the base of the load lying on the flooring; b = 0.85V n; B n - flooring width, m; h 1 -- height of the load layer, m; C 1 -- coefficient that takes into account the reduction in the cross-sectional area of the cargo when it enters the inclined section of the conveyor (table); q 1 -- angle at the base of the triangle; c 1 = 0.4*c; q - angle of repose.

Values of coefficient C 1 for apron conveyors

Using the formula Q=3.6*F*p m*x, the productivity (t/h) of the apron conveyor, taking into account formula (1) can be written as

Q = 3.6*F 1 p m x = 0.648*B n 2 *C 1 *p m *x*tg(ts).

Then the width of the flooring without sides will be (m)

B = v(Q/(0.648*C 1 *r m *x*tg(t)))

When laying with sides (both movable and fixed (Fig. b, c), the cross-sectional area of the load on the deck is the sum of the areas

F = F 2 + F 3 = B nb h 2 C 1 /2 + B nb h 3

With a fill factor of the gutter formed by the flooring and sides (w = h 3 /h), which is taken equal to 0.65...0.80, we will have (m 2)

F = 0.26*B 2 nb *C 1 *tg(ts 1)+B nb *h*w

Using this and the formula Q = 3.6 * F * p m * x, we obtain an expression for determining the mass productivity (t/h) of an apron conveyor having a deck with sides,

Q = 3.6*F*p m x = 0.9*V nb *p m *x*

From this formula you can determine the width of the deck, specifying all the necessary parameters and the side height h. Solving the quadratic equation, we get (m)

You can, given B nb, determine h. The resulting values for the width of the deck and the height of the sides are rounded to the nearest larger according to the state standard, and the speed of the traction element is recalculated. The width of the flooring when transporting piece cargo is chosen depending on the dimensions of the cargo in the same way as for belt cargo.

When determining the geometric parameters of a plate conveyor, the speed of the traction element is taken within the range of 0.01...1.0 m/s, since its operation at high speeds leads to a significant increase in dynamic forces.

The traction calculation of a plate conveyor is carried out similarly to the calculation of a belt conveyor. However, due to the fact that Euler’s law is not applicable to the drive of a chain conveyor, when calculating it, it is necessary to set the value of the minimum tension of the traction element. It is usually recommended to take S min = 1000...3000 N.

The resistance to movement of a traction element with a straight deck and moving sides is determined by the expressions (W pr =(q+q k)gL(fcosб±sinб)) or (Wpr =g(q+q k)(ш 1 L g ±H)). Load value q 0 for plate conveyors q 0 =(q+q k), where q k is the gravity force of 1 m of traction element with decking. The value of q k (kg) can be approximately determined by the expression q k =60V n +A p where the coefficient A p is taken from Table 10.

The coefficient of resistance to the movement of road rollers along the guides can be calculated using the formula or selected from the table

Table

Note. Lower values apply to heavy chains with larger diameter rollers.

In conveyors with fixed sides (Fig. b), moving bulk cargo, it is necessary to take into account additional resistance arising from friction of the cargo against the sides. The following expression is recommended to determine these resistances (N):

W b = fh 2 p m gK b l b

where f is the coefficient of friction of the load on the side walls; K b - coefficient taking into account the decrease in horizontal pressure from the load layer on the side walls;

K b =x+l,2/l+sin;

l b -- length of sides, m.

Next, select the type of traction element, determine the size of the sprockets, and the power of the electric motor. When choosing the type of chain, it should be taken into account that if the transmission of traction force is carried out by two chains, then the traction force (N) per chain is determined taking into account the uneven distribution of it between the chains:

Sst1=1.15Sst/2

When the transportation speed is more than 0.2 m/s, the chain should be selected according to the full design force, taking into account dynamic loads according to the formula

(Sp=S+m60x 2 /z 2 t c).

Example of calculation of a plate conveyor

Initial data: transported cargo - bags of flour with mass G g = 60 kg, bag dimensions 250X450X900 mm, productivity Q = 300 pcs/h, unevenness coefficient K n = 1.5. The route diagram and dimensions of the conveyor are shown in Figure a.

Rice.

1. Based on the size of the load and the angle of inclination of the conveyor, we accept a flat side deck with a width of H = 500 mm and a side height of h = 100 mm.
2. Determine the estimated productivity of the conveyor Q p = Q*K n = 300*1.5 = 450 pcs/h.
3. We set the speed of the traction element x=0.2 m/s. Then the distance between the transported bags will be determined as a = 3600*x/Q p = 3600*0.2/450 = 1.6 m.
4. As a traction element, we take two plate roller chains with rollers on plain bearings.
5. Determine the mass per 1 m of the load q=G g /a=60/1.6=37.5 kg/m

flooring with a traction element according to the formula (q k =60V n +A p) q k =60*0.5+40=70 kg/m, where the coefficient A p is taken from the table for light flooring at V n =0.5 m.

6. We carry out the traction calculation of the conveyor, taking point 2 (Fig. a) as the point with the minimum tension, since in section 1--2 the value is Lg2schx.k

Calculation of resistance to movement of the traction element of an apron conveyor (see Fig. a)
Area and type of resistance	Calculation formulas	Note
		The value of 5mln was selected according to the above recommendations
Resistance to ne-displacement of the traction element in a straight line - 7„ s„ „ nq „ „. linear section 2-1	S 1 =S 2 -gq k L g2 хк + gq k H=1000-9.8170500.09+ 9.8170*5= 1000-3100+3440	We take the resistance value with a minus sign, since we go around the circuit counterclockwise To find the value of S3, we used a formula corresponding to the movement of the traction element along a curved guide with the convexity downward, and we take into account only the first term, since the second is taken into account when calculating resistance in straight sections
Resistance to movement of the traction element on a curved section 2--3	S 3 = S 2 e ьxkt = S 2 e 0.090.1 = 1.01S 2	Resistance coefficient wx.k is taken according to table 11 for average operating conditions
Resistance to movement of the traction element in a straight section 3--4	S 4 = S 3 +q k gL g1 sch xk = 1010+9.817030*0.09
Concentrated resistance when bending around the tension sprocket.	S 5 = oS 4 = 1.06*2860	At b = 180°o = 1.06
Resistance to movement of the traction element in a straight section 5--6	S 6 = S 5 = g(q+qk)L g1 w xk = 3030+ 9.81(37.5+ 70)30*0.09
Resistance to movement of the traction element on a curved section 6--7	S 7 = S 6 e ьxkts = 58701.01
The same, at section 7-8	S 8 = S 7 = g(q+qk)L g2 хк = g(q+q k)H= 5930+ 9.81(37.5+70)50*0.09+ 9.81(37.5 +70)5

Based on the tension values at characteristic points, we construct a tension diagram of the traction element (Fig. b). The maximum tension will be the tension at point 8. Using this tension, we determine the amount of load acting on one chain, taking into account the formula (S st1 = 1.15S st /2). Taking the safety factor n c = 10, we determine the magnitude of the breaking load using the formula (S times = S max n c)

S times = 1.15*n c *S 8 /2 = 1.15*15945*10/2 = 91683 N.

Based on the value of S, we select the roller chain M112-4-160-2 GOST 588--81 with t c = 160 mm, d c = l5 mm. For the selected chain, S times according to the state standard is equal to 112 kN. Since the speed of the traction element is low, we do not take into account the dynamic load acting on the chain.

7. The amount of traction force will be

P = (S 8 --S 1)*o = (15945 -- 1340)*1.06= 15470 N.

8. The power of the electric motor with a transmission mechanism with z = 0.8 will be (see formula) N = 15470 * 0.2 / (1000 * 0.8) = 3.9 kW

Based on the value of N, we select the 4A112MV6UZ electric motor from the catalog with N d = 4.0 kW and n d = 950 rpm.

Scraper conveyors

The term scraper conveyors refers to a group of continuous machines with a traction element, the distinctive feature of which is a working element made in the form of a scraper. Scraper conveyors are usually classified according to this criterion and, taking it into account, they are divided into conveyors:

with solid high scrapers (the height of the scraper is approximately equal to the height of the chute in which the load moves);

with submerged scrapers.

Conveyors with submerged scrapers include conveyors with continuous low scrapers, with contour scrapers, and tubular ones.

The scope of application of scraper conveyors is quite wide. They are used in the food and grain processing industries, in coal mines, and in the chemical industry for transporting bulk and lumpy cargo. The ability to manufacture a sealed chute allows them to be used for transporting dusty and hot cargo.

The advantages of scraper conveyors include simplicity of design, tightness of the gutters, the ability to load and unload at any point on a horizontal or inclined section of the route.

The disadvantages are the relatively rapid wear of the chain and chute hinges, increased drive power due to friction of the load and scrapers on the chute, and abrasion of particles of the transported cargo.