(Coursework)

n7.doc

Option 10

Complete the design of a chain plate conveyor (KCP 15) with the following characteristics:

• 
  performance Q= 250 t/hour;
• 
  web speed  = 0.8 m/s;
• 
  conveyor length l= 90 m;
• 
  length of horizontal section l G= 40 m;
• 
  conveyor inclination angle  = 7 o;
• 
  density of transported cargo  = 0.9 t/m3.

Self-accepted parameters not specified in the assignment:

The task does not specify the type of cargo being moved. Based on the given density (0.9 t/m3), it can be assumed that the cargo is sugar, but it is unlikely that a conveyor with such a high capacity (250 t/hour) will be used to transport sugar or other food product.

Conveyors with similar performance are used in the coal industry, for example, in longwall faces or conveyor drifts for transporting raw coal. Therefore, I accept raw coal as the transported cargo; the density of raw coal satisfies the task conditions.
Scheme of the designed conveyor:

Picture 1.

1. Preliminary calculation.

For the calculation, I assume a conveyor with a wavy belt with sides.

I make the calculation according to the method outlined in: p. 3.2.

1.1. Determining the width of the conveyor.

The width of the conveyor is determined by the formula:

m, (1.1)

Where: Q= 250 t/hour - conveyor productivity;

 = 0.8 m/s - speed of the web;

 = 0.9 t/m 3 - density of transported cargo;

K  - coefficient taking into account the angle of inclination of the conveyor;

 = 40 o - angle of repose of the load at rest (Appendix Table 2);

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

 = 0.7 - coefficient of use of the height of the sides (: page 137).

Coefficient K  determined by the formula:

, (1.2)

Where:  = 7 o - conveyor inclination angle.

Substituting the obtained values ​​into formula 1.1, I determine the width of the canvas:

For transported material containing large pieces up to 10% of the total cargo, the following condition must be met:

mm, (1.3)

Where: a max= 80 mm - the largest size of large pieces (: page 136).

The condition is met.

I finally choose the width of the canvas from the nominal range B= 650 mm (: table 3.6)
1.2. Determination of loads on the transport chain.

I first accept a PVK type lamellar chain (GOST 588-81) as the traction element of the conveyor.
The linear load from the transported cargo is determined by the formula:

N/m (1.4)

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

N/m, (1.5)

Where: A= 50 - coefficient taken depending on the width of the web and the type of cargo (Table 3.5).

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:

Where:  = 0.08 - coefficient of resistance to movement of the chassis on straight sections (Table 3.7).

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

I accept the minimum chain tension S min = S 1 = 1500 N. Using the method of walking along the contour along the web, I determine the tension at points 1..6 (Fig. 1) using a method similar to: clause 3.2.

Where: k:p.138).

Tension diagram of the traction element:

Figure 2.
2. Final calculation of conveyor elements.

2.1. 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 around the sprocket (: p. 138).

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

kW, (2.2)

Where:  = 0.95 - drive efficiency (: page 139);

k h= 1.1 - power reserve factor (: page 139).

kW

I accept an electric motor with increased starting torque of the 4A series (Appendix Table 16)

• 
  engine type - 4АР200L6УЗ;
• 
  power N= 30 kW;
• 
  rotation frequency n dv= 975 rpm;
• 
  swing moment G.D. 2 = 1.81 kg m 2;
• 
  weight m= 280 kg.
• 
  shaft connecting diameter d = 55 mm.

2.2. Calculation and selection of gearbox.

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

m, (2.3)

Where: t- drive chain pitch;

z- number of sprocket teeth;

I accept in advance t= 0.2 m and z = 6.

m.

I determine the rotation frequency of the sprockets using the formula:

rpm (2.4)

rpm

The gear ratio is determined by the formula:

(2.5)

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

Nm. (2.6)

Based on the above defined values, I accept a two-stage helical gearbox (: Appendix Table 27):

• 
  gearbox type - 1Ц2У-250;
• 
  gear ratio u = 25;
• 
  Heavy Duty Rated Output Torque M cr= 6300 Nm;
• 
  mass m = 320 kg.

Figure 3.

Table 1.

All data taken from: Appendix table. 29.

2.3. Final calculation and selection of traction chain.

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

N, (2.7)

Where: S

I determine the dynamic load on the chains using the formula:

N, (2.8)

Where:  = 1.0 - coefficient taking into account the reduction in the reduced mass of the moving parts of the conveyor, is selected according to: p.140 at L> 60 m.

Substituting the found values ​​into formula 2.7 I determine:

N.

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

N (2.9)
Based on the above defined values, I accept a plate chain (:Appendix Table 5):

• 
  chain type - M450 (GOST 588-81);
• 
  chain pitch t= 200 mm;
• 
  breaking force S resolution= 450 kN.

To test the chain for strength, I 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, (2.10)

Where: S d.p.- dynamic chain force at start-up.

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

N, (2.11)

Where: m k- reduced mass of moving parts of the conveyor;

 - angular acceleration of the electric motor shaft.

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

kg, (2.12)

Where: k y= 0.9 - coefficient taking into account the elastic elongation of chains (: p. 140);

k u= 0.6 - coefficient taking into account the decrease in the average speed of rotating masses compared to the average speed (: p. 140);

Gu= 1500 kgf - weight of the rotating parts of the conveyor (without drive), accepted according to: page 140

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

rad/s 2, (2.13)

Where: I etc- moment of inertia of the moving masses of the conveyor, reduced to the motor shaft.

M n.sr.- determined by the formula:

H m, (2.14)

M p.st.- determined by the formula:

H m, (2.15)

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, (2.16)

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

H m s 2, (2.17)

Where: I m= 0.0675 - moment of inertia of the pin-sleeve coupling.

Substituting the values ​​into formulas 2.10 ... 2.17, I get the maximum force in the chain when starting the conveyor.

H m s 2

H m s 2

rad/s 2

2.4. Calculation of the tension device.

I accept a screw type tensioner.

The amount of stroke of the tensioner depends on the chain pitch and is determined by the formula (: clause 5.1):

I take the total length of the screw L about = L+0.4 = 0.8 m.

I make the calculation according to the following method: p. 2.4

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

I accept the material for the nut - bronze Br. AZh9-4 with permissible shear stress [  ] av = 30 N/mm 2, for crushing [  ] cm = 60 N/mm 2, tensile strength  R= 48 N/mm 2. The thread type is the same.

The average diameter of the screw thread is determined by the formula:

mm, (2.19)

Where:  = 2 - ratio of nut height to average diameter (:p. 106);

[p] = 10 N/mm 2 - permissible stress in the thread, depending on the rubbing materials, when rubbing steel against bronze (: p. 106) [ p] = 8...12 N/mm 2 ;

K= 1.3 - coefficient taking into account the uneven load of tension coils (: p. 106);

mm

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

mm, (2.20)

Considering that the length of the screw is large and greater stability is required, I accept d 1 = 36 mm.
The thread pitch is determined by the formula:

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

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

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

I am checking the reliability of self-braking, for which the following condition must be met:

, (2.25)

Where: f= 0.1 - coefficient of friction between steel and bronze.

The condition is met.

I'm checking for stability. The condition for stability is (:p. 107):

, (2.26)

Where:  - the slip coefficient of permissible compressive stresses, when calculating stability, is determined as a function of the screw flexibility ().

[ -1 P] - permissible compression stress.

The permissible compression stress is determined by the formula:

N/mm 2, (2.27)

The flexibility of the screw is determined by the formula:

, (2.28)

where:  =2 - reduced length coefficient (: p. 107).

From: table 2.39 based on the known flexibility of the screw I find  = 0.22. I 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.

I check the screw for strength, strength condition:

, (2.29)

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:

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

N mm, (2.31)

Where: d n= 20 mm - diameter of the heel, less is accepted d 1 .

I substitute the obtained data into condition 2.29:

The condition is met.

The height of the nut is determined by the formula:

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

(2.33)

I check the shear strength of the nut thread, the strength condition is:

(2.33)

The condition is met

I select the tensioner spring according to the method: Vol. 3, Chapter 2.

The remaining dimensions of the tension device are taken structurally.

2.5. Calculation of shafts and selection of bearings.

2.5.1. Drive shaft.

I accept steel 45 as the shaft material (workpiece diameter more than 120 mm  IN  1 = 0.43 B= 314 N/mm 2,  -1 = 0.58  1 = 182 N/mm 2

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

mm, (2.34)

where: M = 5085 Nm - torque on the shaft (previously determined);

[] k = 25 N/mm 2 - permissible torsional stress for steel 45 (: page 96).

mm.

From the standard range (GOST 6636-69 R 40) select the closest diameter value d pv= 100 mm. I accept this diameter for bearings. For fastening the drive sprockets I take the 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:

, (2.35)

Where: l- key length, mm;

d- diameter of the shaft at the location of the key, mm;

h, b, t 1 , - dimensions cross section keys, mm (choose from: Table 6.9);

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

Also, based on condition 2.35, I determine the parameters of the key for the connecting end of the shaft, the diameter of which is taken d= 95 mm and length l= 115 mm (coupling limitations). The values ​​of all geometric dimensions of the keys are given in Table 2.

Table 2.

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

Based on the length of the keys for the drive sprockets, I select the length of the hubs of the latter l st= 200 mm.
Taking into account the above listed dimensions, as well as the dimensions of the fastening elements, I constructively assume the distance between the centers of the bearings to be 1300 mm.

The design diagram of the drive shaft and the diagram of bending moments is as follows (I neglect the weight of the sprockets):

Figure 4.

Where: R 1 And R 2 - reaction of supports in bearings, N;

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

N. (2.36)

Due to the symmetry of the design and support reaction loads R 1 = R 2 = P= 13495 N.

Test calculation of the shaft for strength .

The condition for the strength of the shaft is the safety margin, determined by the formula:

, (2.37)

Where: n  - safety factor for normal stresses;

n  - safety factor for tangential stresses.

[n] = 2.5 - minimum acceptable safety margin.

The safety factor for normal stresses, provided there are no axial loads, is determined by the formula:

, (2.38)

Where: k  = 1.75 effective stress concentration factor (Table 6.5);

  = 0.7 scale factor for normal stresses (Table 6.8);

  - amplitude of normal bending stresses, N/mm 2, is determined by the formula:

, (2.39)

Where: W- moment of bending resistance, mm 3, is determined by the formula:

mm 3 (2.40)

The safety factor for tangential stresses is determined by the formula:

, (2.41)

Where: k  = 1.6 effective torsional stress concentration coefficient (Table 6.5);

  = 0.59 scale factor for normal stresses (Table 6.8);

  =  m- amplitude and average stress, N/mm 2, determined by the formula:

, (2.42)

Where: W To- moment of resistance to torsion, mm 3, is determined by the formula:

mm 3 (2.43)

I substitute the values ​​into formulas 2.37 ... 2.43

N/mm 2.

N/mm 2.

The condition is met.

Selection of bearings.

Since when installing separate bearing housings on the conveyor frame, there is a violation of their alignment and shaft misalignment, I choose 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)

I check bearings for durability, which I determine using the formula:

h, (2.44)

Where: n= 39 rpm - shaft rotation speed;

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

N, (2.45)

Where: V= 1 - coefficient taking into account the rotation of the rings (: page 117);

K T= 1 - temperature coefficient (: table 7.1);

K  = 2.0 - load factor (Table 7.2).

h

Durability is sufficient.
2.5.2. Tensioner shaft.

I carry out the calculation similarly to paragraph 2.5.1.

I take steel 45 as the shaft material (workpiece diameter more than 100 mm: Table 3.3), tensile strength  IN= 730 N/mm 2, endurance limits:  1 = 0.43 B= 314 N/mm 2,  -1 = 0.58  1 = 182 N/mm 2

The shaft diameter is structurally assumed to be 0.8 of the drive shaft diameter d= 80 mm (: clause 5.3.1.)

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

N.

I accept this diameter for bearings. For fastening the drive sprockets I take the diameter d= 100 mm. I take the width of the drive sprocket hub constructively.

I check the shaft for strength only by bending stresses, because... the torque on the shaft is minimal (31.4 Nm).

N/mm 2.

The supply is more than sufficient.

Selection of bearings.

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

Durability is sufficient.

2.6. Calculation and selection of brake 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. I determine this force using the formula (neglecting the resistance of the sprockets):

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, I use an elastic sleeve-pin coupling type MUVP (GOST 21424-75) with bores of the coupling halves for the motor shaft ( d dv= 55 mm) and under the input shaft of the gearbox (taper boring 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 dv= 203.4 Nm.

Taking into account the margin and overall dimensions, I 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 pins n= 8, finger length l= 66 mm, connecting thread of the pin M10. (Data about the coupling is taken from: Appendix Tables 42, 43.)

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.

From the proposed list (Appendix Table 45) I select a coupling with a rated torque M cr= 19000 Nm.

2.7. Star calculation.

Known parameters:

• 
  sprocket pitch diameter d e= 400 mm;
• 
  number of teeth z = 6;
• 
  tooth pitch t= 200 mm.
• 
  chain roller diameter D ts= 120 mm.

mm, (2.47)

where: K=0.7 - tooth height coefficient (:t.2 table 31).

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

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

e = 0.01 .. 0.05 t= 8 mm. (2.49)

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

r = 0.5(D ts - 0.05t) = 50 mm. (2.50)

Half tooth angle  = 15 o (:t.2 table 31).

Tooth Corner Angle  = 86 o (:t.2 table 31).

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

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

mm. (2.52)

I determine the width of the tooth using the formula:

b f= 0.9(50 - 10) - 1 = 35 mm. (2.53)

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

b = 0.6b f= 21 mm. (2.54)

The diameter of the crown is determined by the formula:

mm, (2.55)

Where: d 5 = 150 mm - diameter of the chain roller flange;

h= 70 mm - width of the chain plate.

2.8. Calculation of some structural elements of the conveyor.

As a load-bearing support for the chain rollers, I choose channel 12 in accordance with GOST 8240-89 with a bending resistance moment W x= 8.52 cm 3. The supporting channel rests on welded frames, I determine the distance between the frames:

The maximum permissible bending moment for channel 12 is determined by the formula:

Taking into account the fact that the entire load is distributed over two channels, I determine the maximum span length using the formula (I omit the derivation of the formula):

m. (2.57)

To prevent excessive deflection of the load-bearing channel, I accept a 3-fold safety factor and a span length of 1.2 m.
The bending radius at the transition of the conveyor from an inclined section to a horizontal section, based on the chain pitch, according to the accepted R= 3 m.

Literature.

1. Baryshev A.I., Steblyanko V.G., Khomichuk V.A. Mechanization of PRTS works. Coursework and diploma design of transporting machines: Tutorial/ Under the general editorship of A.I. Barysheva - Donetsk: DonGUET, 2003 - 471 p., ill.
2. Baryshev A.I., Mechanization of loading and unloading, transport and warehouse operations in Food Industry. Part 2. Transporting machines. - Donetsk: DonGUET, 2000 - 145 p.
3. S.A. Chernavsky Course design of machine parts, M.: Mashinostroenie, 1979. 351 p.
4. Anufriev V.I. Handbook of the designer and mechanical engineer in three volumes, M.: Mashinostroenie, 2001.
5. Yablokov B.V., Belov S.V. Guidelines for the course project on hoisting and transport devices (plate conveyors), Ivanovo, 2002.

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. Separate view plate conveyors, which has become most widespread in Russia in the last 15-20 years, is a conveyor with a modular belt. 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

N.

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.

m.

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)

U

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

Nm. (3.6)

M

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:

N.

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

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

mm

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 = 0.43s 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.

N.

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, length of the finger 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 parameters and productivity of a chain-belt conveyor are determined in the same way as a belt conveyor (see Chapter 6).

Welded combined chains (Table III.1.10), plate chains (Table III. 1.11), and less often roller and special chains are most often used as a traction element. The supporting element is a conveyor belt in accordance with GOST 20-76 (see paragraph 4.4).

To avoid the chain slipping along the belt, the following conditions must be met:

where b is the conveyor inclination angle, degrees; f- coefficient of friction between the belt and the chain platform: f = 0,3...0,4; w- coefficient of resistance to the movement of the belt along the side rollers: w = 0,04...0,05; k c - load distribution coefficient on the chain support area: k c » 0.45...0.5.

The traction calculation of the conveyor is carried out using the contour bypass method (see paragraph 5.2).

Calculation of plate high-inclination conveyors is carried out according to the methodology set out in paragraphs 8.2 and 8.3. The width of the conveyor deck with sides is determined by formula (8.5), and from table. 7.8 the main parameters of the conveyor running gear are selected.

Table 7.8. Parameters of the running gear of apron conveyors with a deck with sides

Chapter 8. PLATE CONVEYORS

GENERAL INFORMATION

Rice. 8.2. Types of apron conveyors (to table 8.1)

Table 8.1. Types of apron conveyors (GOST 22281-76) and their scope of application

Table 8.2. Main dimensions of apron conveyors (GOST 22281-76)

* For conveyors of types BV, KM and KG - according to internal size.

** Internal size.

Table 8.3. Speed ​​of movement of the chassis and nominal capacity of apron conveyors (GOST 22281-76)

Plate conveyors come in two versions: with a running gear with rollers; with running gear without rollers - rollers (support rollers) are an element of the metal structure.

Flooring and sides. The width of the flooring (mm) when transporting bulk cargo is taken from the condition

Where k– coefficient: for sorted cargo k= 2.7; for ordinary cargo k = 1,7; a¢- the largest size of a typical piece of cargo, mm [see. formulas (4.2)...(4.4)].

The width of the flooring when transporting piece goods must satisfy the condition

Where b 1 - largest transverse size of the load (Fig. 8.3) mm; IN 1 - floor width reserve: for flatbed conveyors IN 1 = 50...100 mm, for onboard IN 1 = 100...150 mm.

The height of the sides when transporting bulk cargo is selected from the table. 8.4 taking into account the data in table. 8.5.

Side height h when transporting piece goods, 100...160 mm is accepted.

The resulting width of the flooring and height of the sides must be rounded to the nearest dimensions according to GOST 22281-76 (see Table 8.2). Traction chains. For apron conveyors, traction chains are selected according to the data in tables III.1.11...III.1.14. The chain pitch is assigned depending on the width of the flooring (Table 8.6). The speed of the running gear (belt) of apron conveyors is selected depending on the width of the deck according to the recommendations in Table. 8.7. Tilt angle. The greatest angle of inclination of the apron conveyor when transporting bulk cargo is selected according to table. 8.8.

Rice. 8.3. Location of piece goods on the conveyor floor: A- with automatic laying; b- for manual installation

Note. Dimensions placed between the lines are preferred.

Table 8.8. The largest permissible angles of inclination of apron conveyors when transporting bulk cargo

* r - load friction angle (flooring in motion), degrees.

In this case, it is necessary that the angle of inclination of the conveyor

where j d is the angle of repose of the load in motion, degrees [see. 4.6)].

Stretching device. The stroke of the tensioner is selected depending on the pitch of the traction chains (Table 8.9).

Loading funnels. The main dimensions of the loading funnel for plate conveyors (Fig. 8.4), depending on the width of the flooring, can be taken from the table. 8.10.

Symbol plate conveyor. The symbol for a stationary general-purpose apron conveyor, according to GOST 22281-76, contains the name of the product (“apel conveyor”), the designation of the conveyor type and design, the width of the undercarriage deck (cm) and the designation of the standard.

For example, a stationary general-purpose flatbed conveyor (BV), version 1, with the width of the undercarriage deck IN= 800 mm is indicated by:

Plate conveyor BV-1-80 GOST 22281-76.

PRELIMINARY CALCULATION OF PLATE CONVEYOR

Width of deck without sides (m) when transporting bulk cargo

Where h- side height (see paragraph 8.2), m; y - coefficient characterizing the degree of use of the side height, y = 0.65...0.8.

The resulting flooring width is specified in accordance with the instructions in paragraph 8.2.

Conveyor traction force (N)

Where F min - minimum chain tension (see paragraph 5.2), N; w- resistance coefficient of the apron conveyor (Table 8.12); q- linear mass of cargo on the conveyor [formulas (5.3) and (5.11)], kg/m; L- length of the horizontal projection of the loaded part of the working conveyor branch, m; q h.ch - linear mass of the conveyor running gear, kg/m; L g - length of the horizontal projection of the conveyor, m; N- load lifting height, m; F b - frictional resistance of the load on the fixed sides [formula (8.8)], N; F p.r - resistance of the plow unloader [formula (5.30)].

Table 8.11. Coefficient values k b [to formulas (8.4) and (8.5)]

Table 8.12. Resistance coefficient w values ​​for apron conveyors

* Larger values ​​are taken for tracks with centering devices that protect the chain from shifting.

** When working in winter conditions in an unheated room or outdoors, the given values ​​increase by 1.5 times.

In formula (8.6) there is a plus sign in front of qH is taken when lifting a load, a minus sign when lowering it.

Frictional resistance of bulk cargo against fixed sides (N)

Where f- coefficient of friction of bulk cargo on the side walls (Table 4.1); h p - working height of the side (according to the height of the load), m; r - bulk density of cargo, t/m 3 (see Table 4.1); l b - length of sides, m.

The linear weight of the conveyor running gear is determined from the catalogue.

The approximate linear mass (kg/m) of the conveyor running gear can be taken

Where IN- flooring width, m; TO- see table. 8.13.

Conveyor drive shaft power (kW)

Table 8.14. Coefficient values k 1 [to formula (8.11)]

The motor power to drive the conveyor is determined by formula (6.19).

Maximum static tension of the traction element

Where F min - the lowest tension of the traction element (1000...3000 N).

Where L- conveyor length, m; z- number of teeth of the drive chain sprocket; t- traction chain pitch, m; k 1 - mass reduction coefficient (taking into account that not all conveyor elements move with maximum acceleration, as well as the influence of chain elasticity) (Table 8.14).

At belt speeds of up to 0.2 m/s, dynamic loads on the chains can be ignored.

Calculated tension of the traction element

For a single-chain traction element F = F calc.

Where k- chain safety factor: for horizontal conveyors k= 6…8, with inclined sections - k = 8...10.

Calculation of apron conveyor

Conveyor width determination

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

The width of the conveyor is determined by the formula:

where Q = 850 t/hour - conveyor productivity;

1.5 m/s - speed of the web;

2.7 t/m 3 - density of transported cargo;

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

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;

0.7 - side height utilization factor

Coefficient K in is determined by the formula:

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:

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

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:

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

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:

0.08 - coefficient of resistance to movement of the chassis at

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

Calculation of conveyor elements

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:

where = 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.

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)

The gear ratio is determined by the formula:

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

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

CALCULATION WORK

PLATE CONVEYOR

1.1 Purpose of the work

Study the designs general information, principles of operation of conveyors and methods for determining basic parameters.

1.2 Definition of apron conveyor

They are called transporters technical means continuous operation for moving bulk bulk and piece goods along certain linear routes. They are divided into conveyors and pipeline transport devices.

According to the principle of operation, a distinction is made between conveyors, in which the cargo moves as a result of mechanical contact with the transporting element (belt, plate, bucket, scraper, auger, rollers), and pneumatic transport installations, in which the movement of bulk cargo is carried out by gravity or a stream of compressed air.

A plate conveyor is a transporting device with a load-carrying sheet made of steel plates attached to a chain traction element.

When transporting materials with sharp edges (for feeding large pieces of stone into crushers), plate conveyors are used, in which the traction element is two endless chains that go around the drive and tension sprockets. Metal plates are attached to the traction chains, overlapping each other and preventing spillage of material between them (Figure 1.2). The permissible angle of inclination of a plate conveyor with flat plates is less than that of a belt conveyor, because angle of friction of the cargo material on the metal in 2.5÷3.0 times less than with rubber-fabric tape. Shaped plates with transverse projections on the working surfaces allow you to increase the angle of inclination of the conveyor. Plate conveyors are also used to move hot materials, parts and products in construction factories.

Characteristics of apron conveyors:

· plate thickness – from 3 mm

· blade width – from 500 mm

· web speed – from 0.6 m/s

· productivity – from 250 to 2000 t/h

· installation angle – up to 45º

Working tools of apron conveyors:

· plastic fabric

· running rollers

· traction body

· drive station

Advantages:

· the ability to transport a wider (compared to belt conveyors) range of goods;

· ability to transport cargo along routes with steep inclines (up to 35°-45°, and with bucket plates - up to 65°-70°);

· the ability to transport goods along a complex spatial trajectory;

· high reliability.

Flaws:

· low speed of cargo movement (up to 1.25 m/s);

· like other chain conveyors:

· - large linear weight of the conveyor;

- complexity and high cost of operation due to the availability large quantity articulated elements in chains that require regular lubrication;

· -higher energy consumption per unit mass of transported cargo.

The apron conveyor is used to move piece goods; according to this condition, it is necessary to calculate the main characteristics of the presented conveyor.

Figure 1.9 – Diagram of a plate conveyor

Initial data:

Plate conveyor with beadless flat deck;

a=400mm – load size;

QGR=1.10 kN – load weight;

P=1350 kN/hour – conveyor productivity;

L=40 m – conveyor length;

Working conditions are difficult

1.3.1 Determine the width of the flooring INN:

=400+100=500 (mm) (1.1)

Where: a=400 mm– specified cargo size;

A=100 mm– margin of flooring width.

Blade speed υ , m/sec, apron conveyor is selected according to table 1.10, according to the width of the flooring

Hence υ =0,4 m/sec.

Two plate-type collapsible VKG chains with special plates with a pitch are used as a traction element t=320 mm(according to table 1.11), according to the width of the flooring INN=500 mm, and with a breaking load SR=500 kN.

Table 1.11 – Pitch dimensions of leaf chains

Determine the linear weight load of the cargo q, kN/m:

Where: P=1350 kN/hour– conveyor productivity; m ), (1.3)

Where: QGR=1,10 kN– weight of one load;

q=0.9375 kN/m – linear weight load.

Accept the step value tGR, m, rounded up. Then tGR=1,17 m.

We calculate the linear load from the conveyor chassis q K, kN/m, using the empirical formula for heavy duty decking:

Width of decking without sides,

1.0 or more

From Table 1.13 we select the coefficient of resistance to movement ω , assuming that the diameter of the chain roller is greater than 20 mm. Hence ω=0.120.

Where: kN - lowest chain tension;

ω=0.120 – motion resistance coefficient;

q=0.9375

q K =0.98

L=40 m– conveyor length;

H=0 m– lifting height;

W B– frictional resistance of the load on the fixed sides, kN, (since there are no sides in this case, then W B=0 );

W P.R.– plow loader resistance, kN, (since loading is carried out through the end drum, then W P.R=0 ).