Lamellar conveyors are used to transport clinker from kiln coolers to a warehouse, as well as to move lumpy and abrasive materials. The traction body of such a conveyor is usually one or two chains, the load-carrying body is a rigid metal flooring (cloth), consisting of separate plates. The advantage of apron conveyors is the ability to transport heavy bulky and hot goods along horizontal and steeply inclined (up to 354-60 °) routes with high (up to 2000 m / h) productivity.
Productivity Q (t / h) of the apron conveyor can be determined by the formula:
Q = 3600 f-v-qh, (8.53)
where F is the cross-sectional area of the material on the tape, m2
For canvas without borders
For canvas with borders
(8.55)
where B is the width of the canvas, m; h is the height of the sides, m; K=0.85 - the ratio of the width of the layer of material to the width of the web; φ - angle of repose of the material in motion; k1 = 0.65 - filling factor for the height of the sides. (If the conveyor is evenly loaded with boards over the entire width of the web, the second term in the formula for determining F is not taken into account, and the value of the coefficient k! is taken equal to 0.80-=-0.85); v - the speed of the conveyor belt, is taken within 0.05-f-0.75 m / s and is specified by the formula
where t is the step of the traction chain, m; b - number of drive teeth (b = 5, 6, 7, 8).
The web width of apron conveyors corresponds to the values of the normalized row for belt conveyors.
n, rpm - the number of revolutions of the main shaft of the conveyor.
Installed power of the apron conveyor motor N (kW):
(8.57)
where K2= 1.10-1.25 - power reserve factor; q - weight of 1 linear meter of moving parts of the conveyor, kg/m; L - conveyor length, m; l1 - conveyor projection length on the horizontal plane, m; H - lifting height of the material, m.
8.5.3. Calculation of bucket elevators
Bucket elevators are used to transport various bulk cargoes: dusty, granular and lumpy (cement, coal, pumice, etc.) - Bucket elevators are used
as the main technological transport of cement production for lifting material at an angle of up to 60-85 ° from the initial to the final point without intermediate loading and unloading. The material is moved with the help of buckets fixed at regular intervals (or closed together) on an endless traction flexible body - a chain or tape (Table 8.16.).
Table 8.16
Main parameters of buckets
| Bucket pitch, mm | Useful capacity i0, l | Pitch of open buckets, mm | Useful capacity i0, l | ||
| Buckets deep | Small buckets | Buckets with side rails | |||
| acute-angled | rounded | ||||
| 0,2 | 0,1 | - | - | - | |
| 0,4 | 0,2 | - | - | - | |
| 0,6 | 0,35 | 0,65 | - | ||
| 1,3 | 0,75 | 1,3 | - | ||
| 2,0 | 1,4 | - | |||
| 4,0 | 2,7 | 6,4 | |||
| 6,3 | 4,2 | ||||
| - | 7,8 | ||||
| - | - | ||||
| - | - | ||||
| - | - |
The type of elevator and the shape of the buckets are selected depending on the characteristics of the transported material according to the table (Table 8.17).
Table 8.17
Note: Types of buckets: G - deep, M - shallow, O - acute-angled with side guides, C - with a rounded bottom and side guides.
The performance of the bucket elevator is determined by the equation
(8.58)
u i0 - geometric useful capacity of the bucket, l; ak - pitch of buckets, m. For deep and shallow buckets located at intervals, ak = 2.5-=-3.0 h; for continuously located buckets with onboard guides ak "n; where h - bucket height, m; v is the speed of the tape or chain, m/s; ψ - bucket fill factor (see table. 8.17).
Calculate a plate horizontal conveyor at a given capacity Q= 130 t/h (see Fig. 8.1, a) for moving piece cargo with a density of r = 0.95 t / m 3 with a diagonal size of 700 mm, weight t= 180 kg. Conveyor length L= 45 m. Unloading - at the end of the loaded branch. Working conditions are average.
Based on the dimensions of the cargo, we select the width of the flooring using the formula (8.2) AT= 700 + 100 = 800 mm.
According to GOST 22281-76 (Table 8.2), we accept the width of the flooring AT= 800 mm. According to the table 8.6 accept chain pitch t= 400 mm. In accordance with the data in Table. 8.3 and 8.7 we accept the speed of the running gear u = 0.2 m/s.
As a traction body, we preliminarily accept (see paragraph 4.4) two lamellar rollers with flanges on the rollers (type 4) collapsible prices with solid rollers (version 2) and breaking load (Table III.1.11) F razr = 112 kN. Chain number - M112, chain designation:
Traction chain М112-4-400-2 GOST 588-81.
Linear mass of cargo, according to (5.12), q= Q/(3,6u) = 130/(3.6 ´ 0.2) = = 180 kg/m.
From formula (5.11) we find the step of the location of the goods on the deck t r = m/q= = 180/180 = 1 m.
Approximate linear mass of the running gear of the conveyor according to the formula (8.8) q h.h "60 × 0.8 + 45 \u003d 93 kg / m, where for a light load (r<1) из табл. 8.13 принят To = 45.
From Table. 8.12 we select the coefficient of resistance to movement w = 0,l (the diameter of the chain roller is less than 20 mm).
Taking the smallest chain tension at the point of their escape from the drive sprockets F min = F 1 = 1000 N (see paragraph 5.2), we find from formula (8.6) the traction force of the conveyor ( F 6 and F n.r are equal to zero):
Let us determine the tension at the characteristic points of the conveyor by the contour traversal method and specify the value F 0 . We start the bypass from the point with the least tension F min = F 1 = 1000 N.
Resistance in the section of the idle branch of the conveyor according to (5.22) F x = q h.h q w L= 93×9.81×0.l×45 = 4105 H; the same, on the loaded branch according to (5.17) F r = (q + q h.h) q w L\u003d (180 + 93) 9.81 × 0.1 × 45 \u003d 12,052 N.
Chain tension at the point where the chains run onto the tension sprockets according to (5.35) F 2 = F 1 + F x \u003d 1000 + 4105 \u003d 5105 N.
Resistance on tension sprockets according to the formula (5.26) F pov = F 2 (l.05-l) = = 0.05 F 2 .
Tensioning the chains at the point of escape from the idler sprockets F 3 = F 2 + F pov = F 2 + + 0,05F 2 \u003d 1.05 × 5105 \u003d 5360 N.
Tension at the point of approach of the loaded chain branches to the drive sprockets F 4 = F 3 + F g \u003d 5360 + 12 052 \u003d 17 412 N.
Tension in the traction chains running on the drive sprockets, taking into account the resistance at the turning point 4 (on drive sprockets) F nab = F 4 ++ F 4 (k n - 1) = k P F 4 \u003d 1.05 × 17 412 \u003d 18 283 N.
The corrected value of the traction force of the conveyor according to (5.37) = F nab - F 1 \u003d 18 283 - 1000 \u003d 17 283 N, which differs from that obtained earlier by 4%.
From formulas (8.12) and (8.13) we find the calculated tension of one chain
Required engine power according to formula (6.21) with drive efficiency h = 0.94 (Table 5.1) and safety factor k = 1,2 R\u003d 1.2 ´ 3.45 / 0.94 \u003d 4.41 kW.
From Table. III.3.1 we choose an electric motor 4A132M8UZ with a power of 5.5 kW with a rotation speed of 720 min -1.
The speed of the drive shaft of the conveyor according to the formula (8.15) P p.v \u003d 60 × 0.2 / (6 × 0.4) \u003d 5 min -1.
Gear ratio of the drive according to the formula (6.23) and = 720/5 = 144.
We accept the kinematic scheme of the drive, consisting of a V-belt transmission and a gearbox.
Taking into account the explanations to formula (1.101), from which it follows that for continuous machines k p = 1, from table. Sh.4.13 select the KTs2-750 gearbox, which has a gear ratio and p \u003d 118, with power on a high-speed shaft R p \u003d 6.5 kW at the frequency of rotation of this shaft P b \u003d 600 min -1.
In this case, the gear ratio of the V-belt transmission and k.p = and/and p = = 144/118 = 1.22.
Checking the engine for sufficiency of the starting torque and determining the overload coefficient of the traction body when starting the conveyor are carried out similarly to the calculation described in paragraph 16.1.
Selection of bearings for the drive drive shaft and for the axis of the tensioner. Plate conveyors are used for horizontal and inclined transportation of various bulk and piece goods in the metallurgical chemical coal power engineering and many other industries, as well as for moving products from one workplace to another along the technological process in in-line production. For sawdust, the angle of repose of the load at rest...
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Page 3
federal education agency
"Moscow State Forest University"
Department of Theory and Design of Machines
(department of distance learning)
COURSE PROJECT
specialty: 190603
faculty: IPSOP
Completed:
Teacher:
Moscow city
2011
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Assignment for a course project. |
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Introduction. |
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9,10 |
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10,11 |
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11-15 |
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15,16 |
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16,17 |
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List of used literature |
page |
Introduction.
High performance work modern enterprise is impossible without properly organized and reliable means of transport. When processing large volumes of cargo, it is advisable to use devices and machines of continuous action. These include conveyors various kinds and for various purposes. Conveyors are an integral and integral part of many modern technological processes they establish and regulate the pace of production, ensure its rhythm, contribute to increasing labor productivity and increasing output. Continuous conveying machines are extremely important and responsible parts of the equipment of a modern enterprise, the operation of which largely determines the success of its work. These machines must be reliable, robust, durable, easy to use and able to operate automatically.
Plate conveyors are used for horizontal and inclined transportation of various bulk and piece cargoes in metallurgical, chemical, coal, energy, engineering and many other industries, as well as for moving products from one workplace to another along the technological process in mass production. The advantages of apron conveyors are the possibility of transporting heavy bulky and hot goods at high productivity (up to 2000 m3/h or more) and travel length (up to 2 km) due to the high strength of traction chains and the possibility of using intermediate drives.
In this paper, the calculation of the plate conveyor transporting sawdust and having a capacity of 30 t/h is given.
1. Calculation of the plate conveyor
1.1. Determination of parameters not specified in the job:
The task does not specify the density of the transported cargo. Based on the type of a given load (sawdust), according to the reference data, we accept the bulk mass of sawdust equal to 280 kg / m 3 (page 62).
The drive is installed at the end of the cargo branch of the conveyor. Sawdust is unloaded into the bunker at the end of the cargo branch. The conveyor is installed in a closed, unheated room.
The conveyor is equipped with plates made of steel sheets without boards.
Since the conveyor is without inclination, the coefficient taking into account the inclination angle C 2 = 1.
Angle of repose of the load in motion φ 1 \u003d 0.4φ \u003d 0.4 39 \u003d 5.6 °. For sawdust, the angle of repose of the load at rest is φ = 39°.
Conveyor capacity by volume, m 3 /h
The design scheme of the plate conveyor is shown in fig. one.
Figure 1. Scheme of the plate conveyor.
1.2. Determine the width of the slat conveyor:
where: conveyor capacity by weight, kg.
The calculated value of the deck width is rounded up to the nominal width. We accept according to GOST 22281-76 flooring with a standard width of 1400 mm (page 62).
1.3. We determine the linear mass of the flooring with chains:
where: coefficient depending on the deck width, kg/m. As recommended for 1.4 m decking (page 61).
1.4. According to Appendix 2, we pre-select the lamellar roller chain M112 with a step weight of 1 meter of the chain
1.5. Determine the linear mass of cargo on the conveyor:
where: conveyor capacity by volume, m 3 /h
1.6. Traction calculation of the conveyor:
We accept the tension of the chain at the point of its escape from the drive sprocket:
Resistance on the horizontal section of the idle branch 1 2:
where: - free fall acceleration,
coefficient of resistance to movement of the chain with flooring. Since the chain is supported by support rollers on plain bearings (page 61 ).
Chain tension at point 2:
The drag coefficient on the idler sprocket.
Therefore, the chain tension at point 3:
Movement resistance on a horizontal section 3 4:
Chain tension at point 4:
Safety margin of the selected chain:
The chain was chosen correctly, since the permissible margin of safety for horizontal weightless conveyors is K = 6 ... 7 (page 63).
1.7. Determine the diameter of the pitch circle of the drive sprocket by taking the number of its teeth (page 11):
2. Calculation of conveyor drive power and motor selection.
2.1. Inertial force that occurs during the start of the conveyor:
where: - conveyor start time,
2.2. Pulling force of the drive sprocket:
where:
2.3. Conveyor drive sprocket resistance:
where: - resistance on the drive sprocket,
where: - coefficient, (page 15).
2.4. Drive motor power during steady motion:
where: - drive efficiency, (Appendix 19).
Number of conveyor chains, .
Maximum chain speed.
2.5. Conveyor drive motor power during its start-up:
2.6. Installed power:
2.7. By power from application 3, we select an asynchronous motor type 4A160 S 6U3, with allowable overload factor and speed
The selected electric motor must satisfy the condition:
The motor is correct.
3. Calculation and selection of the gearbox
3.1. Required drive sprocket speed:
3.2. Required gear ratio of the gear installed between the electric motor and the drive shaft:
3.3. As a transmission according to Appendix 4, we select a standard gearbox Ts2U-250 with a gear ratio, permissible torque on a low-speed shaft.
Reducer parameters:
3.4. Deviation amount:
which is acceptable.
3.5. Actual torque on the conveyor drive shaft:
4. Coupling selection
The transmission of torque from the motor shaft to the input shaft of the gearbox is carried out by a safety multi-plate friction clutch.
4.1. Rated Torque:
4.2. Transferable rated torque:
where: - operating mode coefficient, for apron conveyors under loads with moderate fluctuations up to 150% of the nominal (page 21).
4.3. In terms of size, from Appendix 5, we select a coupling of size 4, which has the following parameters:
rated torque
5. Calculation of the drive shaft.
5.1. Approximate calculation of the drive shaft:
We accept.
According to table 5, we select the structural elements of the shaft:
then:
accept
accept
The calculated values of diameters are rounded to the nearest side along a number of normal linear dimensions (Appendix 1).
5.2. To connect the output shaft of the gearbox with the drive shaft, we use a gear coupling.
To select a coupling, we calculate the transmitted design torque:
From Appendix 5.3 we select a gear coupling that transmits torque with the parameters:
Assigning a module m = 3 with number of teeth z = 45.
5.3. Key selection.
For two shaft diameters, we select a key of the same section along the shaft of the minimum section with d = 75 mm.
According to GOST 23360 78, we select the key 1-22x14x120 with
5.4. Sprocket hub parameters:
hub length:
hub diameter:
We accept the working length of the key
5.5. We check the selected key for crushing stresses:
The key is correctly selected.
5.6. Check calculation of the drive shaft.
5.6.1. Section modulus of the shaft with a keyway under the sprocket according to table 5.2:
5.6.2. Find the horizontal force acting on the sprocket:
5.6.3. The force acting on the shafts in the presence of a gear coupling:
We determine the design dimensions of the conveyor:
where: - design size of the gearbox shaft,
5.6.4. Horizontal reactions in supports B and G, replaced by:
Since, when calculating bearings, we accept.
5.6.5. Bending moments:
moment bending a shaft in a horizontal plane:
moment bending the shaft in support B on the left in the horizontal plane:
stress in the design section of the shaft from bending by the moment:
the greatest stress in the calculated section of the shaft from the torque:
equivalent voltage at a point of the outer fiber:
5.7. For the shaft, we assign steel 45 with a yield strength
Safety margin for yield strength:
Shaft dimensions are correct.
6. Calculation of the tension device.
We accept a screw tensioner with two screws for the designed single-chain conveyor.
6.1. Estimated tension force:
where: - chain tension at point 2;
Chain tension at point 3.
6.2. Estimated bending moment:
6.3. Required axle diameter:
6.4. To reduce the range of cutters for cutting keyways, we accept the diameter of the axis of tension of the device at the location of the sprocket, and the diameter of the axis at the location of the bearing.
6.5. Tensioner screw calculation:
6.5.1. The moment from the forces of friction during the rotation of the screw:
Taking: ,
Where
6.5.2. Tensioning force of the tensioner:
6.6. From application 20, select the tensioner: Trap.32x6, with parameters:
tensioning force of the tensioner S = 25000 N;
bearing diameter d p \u003d 70 mm;
slider stroke A = 500 mm;
screw diameter d = 32;
H = 1100 mm;
H = 160 mm;
K = 140 mm;
L = 150 mm.
7. Selection of bearings for the drive drive shaft and for the axis of the tensioner.
7.1. Since the shaft and axle supports are mounted on a welded frame in housings, the alignment of which cannot be sufficiently accurately ensured, we accept spherical radial bearings for their installation.
From Appendix 20:
For the shaft, radial ball bearing No. 1317 GOST 28428-90:
inner diameter d = 85 mm;
outer diameter D = 180 mm;
width B = 41 mm
0 = 51000 N;
r = 98000 N;
X = 1.
For the axle, radial ball bearing No. 1214 GOST 28428-90:
inner diameter d = 70 mm;
outer diameter D = 125 mm;
width B = 24 mm
static load rating C 0 = 19000 N;
dynamic load rating C r = 34500 N;
X = 1.
7.2. Checking selected bearings for durability:
for shaft
for axis
Shaft check for deep groove ball bearing No. 1317:
Check for axle bearing deep groove ball No. 1214:
The calculated life of the bearings corresponds to the recommended values of the conveyors.
BIBLIOGRAPHY:
- Ivanov G.A. Calculation and design of chain conveyors. Teaching aid, M.: GOU VPO MGUL, 2008. 115 p.
- Spivakovsky A. O., Dyachkov V.K. Transporting machines: Proc. allowance for engineering universities. M.: Mashinostroenie, 1983 - 487 p.
- Ivanov M.N., Finogenov V.A. Machine parts, Educational edition. M.: VSh, 2006 408 p.
- Reshetov D.N. Machine parts, Textbook. M.: Mashinostroenie, 1989 496 p.
- GOST 28428-90 Radial bearings. M. 1990.
3
a 7
a 6
F C
R Gg
F M
R Bg
M and B =3650 N∙m
M and B = 1845 N∙m
M g
T = 3922 N∙m
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4. Detailed traction calculation
5. Determination of the design tension of the traction element
7. Calculation and selection of gearbox
8. Brake selection
9. Choice of couplings
10. Calculation of the drive shaft
11. Calculation of the axis of the tension station
11.1 Calculation of an open gear train
12.1 Spring calculation
12.2 Calculation of tension screws
Literature
Introduction
High-performance work of a modern enterprise is impossible without properly organized and reliable means of transport. When processing large volumes of cargo, it is advisable to use devices and machines of continuous action. These include conveyors of various types and for various purposes. Conveyors are an integral and integral part of many modern technological processes - they set and regulate the pace of production, ensure its rhythm, increase labor productivity and increase output. Continuous conveying machines are extremely important and responsible parts of the equipment of a modern enterprise, the operation of which largely determines the success of its work. These machines must be reliable, robust, durable, easy to use and able to operate automatically.
In the course project, an inclined plate conveyor was designed with a capacity of 400 t / h with a horizontal part of 50 meters and an inclined part of 20 meters, designed to transport small parts in bulk.
The design part shows the drive, tensioner, hopper and general form conveyor.
Were produced necessary calculations, among which are the calculation of the structural parameters of the conveyor (flooring width, shaft diameters, etc.), the calculation of the strength of all the most critical elements of the conveyor, the determination of the loads on the shafts, the choice of motor and gearbox, the calculation of the tensioner and other calculations.
1. Definition of the main parameters
Let's define the characteristics of the transported cargo.
Average piece size of small parts ; bulk density of cargo; the angle of repose of the load at rest, but in motion; coefficient of friction of the load on the steel deck; angle of friction of the load on the metal flooring.
For given conditions, we choose a double-chain conveyor general purpose with long-link traction leaf chains and sprockets with a small number of teeth. With this in mind, we accept the speed of the conveyor .
The volumetric productivity corresponding to the calculated productivity is
2. Choosing the type of flooring and determining its width
Taking into account the parameters of the cargo, we choose the side deck, since only conveyors with side deck are suitable for transporting bulk cargo.
Let's define the design of the flooring.
With smooth flooring;
Condition not met
With wavy flooring
The condition is met, therefore, we choose the side corrugated flooring of the medium type (Fig. 1).
Rice. 1. Wavy boarding.
Determine the height of the sides. . Accept
We find the required width of the flooring.
where - productivity, t/h;
Conveyor speed, m/s;
The angle of repose of the load (rubble) at rest;
Conveyor inclination coefficient, ;
The height of the cargo layer at the sides, m;
- coefficient of use of the board height .
Since the cargo is medium in size, checking the flooring for the granulometric composition of the cargo is not required.
From a number of GOST 22281-76, we accept the nearest larger value of the width of the flooring.
3. Approximate traction calculation
where is the initial tension of the chain, N;
Linear load from the running gear of the conveyor, N/m;
For metal
flooring.
A - empirical coefficient
The coefficient of resistance to the movement of the undercarriage in straight sections.
For rollers on rolling bearings;
Let's define the breaking force
Based on the force found, we select a chain according to GOST 588-81 M450 with a maximum breaking load of 450 kN, step .
a) The choice of coefficients of resistance to web movement
Taking into account operation in average conditions according to tab. 2.6 we accept the coefficient of resistance to movement on plain bearings. Resistance coefficients when rounding deflecting devices: at an angle of inflection and at .
b) Determination of the point with the least tension of the traction element
The smallest tension of the traction element will be at the bottom point 2 of the inclined section, since
c) Determine the tension at the characteristic points of the route. The smallest tension of the traction element will be at the bottom point 2 (Fig. 2).
Rice. 2. Conveyor route
We accept tension at point 2. When bypassing the route from point 2 in the direction of movement of the canvas, we determine:
To determine the tensions in v. 1, we perform a reverse bypass:
Determination of the design tension of the traction element
By analogy with the structures used, we accept a traction element consisting of two parallel plate chains with a step 
; drive sprocket with number of teeth .
.
With a given scheme of the conveyor route, the maximum tension of the traction element is .
We determine the dynamic force by the formula (2.88)
where is the coefficient taking into account the interference of elastic waves; - coefficient of participation in the oscillatory process of the mass of the transported cargo ( at ); - coefficient of participation in the oscillatory process of the running gear of the conveyor (with the total length of the horizontal projections of the conveyor branches 
);
Mass of cargo on the conveyor, kg;
Mass of the running gear of the conveyor, kg;
Number of teeth of the drive sprocket;
Traction chain pitch, m
Then we get:
Since the breaking load is less than that of the selected chain, we finally stop at the M1250.
6. Power determination and engine selection
Pulling force on drive sprockets
With a safety factor and drive efficiency, the motor power
According to the obtained power value, we select the engine of the 4А280S6У3 series:
,
.
Determine the torque on the drive shaft
.
7. Calculation and selection of gearbox
Determine the frequency of rotation of the drive shaft
.
Sprocket diameter
.
Determine the gear ratio of the drive
.
Because gear ratio is large, an additional downshift is required. As an additional transmission, we use an open single-stage gear transmission. The recommended gear ratio of such a gear is no more than 5.
Consequently
.
8. Brake selection
The brake is installed on the drive shaft, which significantly reduces the magnitude of the braking torque.
Determine the braking torque (3.81)
where is the moment on the drive shaft,
Determine the moment of the star
The dividing diameter of the sprocket.
We choose a shoe-type brake TKG with electro-hydraulic pushers TKG - 300.
9. Choice of couplings
Between the electric motor and the gearbox, we install an elastic sleeve-pin coupling. The rated torque of the coupling is equal to the torque on the drive shaft of the electric motor
Estimated coupling moment
We choose an elastic sleeve-and-pin coupling with a brake pulley MUVP - T 710, with a rated torque of 710 Nm and a brake pulley diameter of 300 mm.
10. Calculation of the drive shaft
The drive shaft experiences bending from transverse loads created by chain tension and torsion from the moment transmitted to the shaft by the drive.
Determine moment:
.
Maximum bending moment:
Bending moment in front of the hub:
Determine the diameter of the hub:
Determine the diameter of the trunnion:
Taking into account the calculated data, we design the shaft, assigning diameters according to the normal range of sizes. For the purposes of unification, we accept the diameters of the shaft in the supports as the same and equal to the larger one: 200mm.
Shaft material - steel 45:
Determine the diameter of the shaft section under the asterisks
Taking into account the weakening of the section by the keyway, we increase the shaft diameter by 10%
We take the diameter of the shaft under the asterisks equal to 120mm.
Because the total gear ratio is large and equal to 100, then an additional reduction gear is required, installed between the gearbox and the drive shaft. As an additional transmission, we use an open single-stage gear transmission. The recommended gear ratio of such a gear is no more than 5.
Let's take the diameter of the pitch circle of the gear, the minimum number of gear teeth.
Gear module
Let's take mm;
Diameter of the pitch circle of the girth gear
Number of teeth of the ring gear
Gear pitch diameter
which is acceptable in size.
center distance
Ring gear width
where 0.1–0.4 is the gear width factor.
12. Calculation of the tensioner
We choose a spring-screw tensioner, because. conveyor length over 20 meters.
Determination of tension force and tensioner travel.
The pull force is
We assign the stroke of the tensioner in accordance with the recommendations 1.5 chain pitch
12.1 Spring calculation
Fig.3. Schematic diagram of the tensioner.
Estimated force in one spring, taking into account the uniform distribution of the load:
where is the safety factor.
Spring material steel 65G (GOST 1050-85).
The diameter of the rod is found from the condition of the strength of the compression spring
,
where 
- coefficient depending on the spring index ;
Initial average diameter, m;
Permissible torsional stress for the wire material. Pa;
,
where is the torsional endurance limit;
Coef. security;
Coef. shear stress concentrations.
Determine the average diameter of the spring
Determine the number of turns for a given draft
where is the shear modulus,
Spring travel.
We define total number turns, taking into account the grinding of the ends of the spring during the formation of the supporting surfaces:
turns.
The length of the spring before the coils touch
Spring unloaded length
Spring outside diameter
Spring inner diameter
Turn pitch
.
12.2 Calculation of tension screws
We determine the diameter of the screw from the condition that the stresses arising in the material of the screw are less than the maximum allowable for this material of the screw. Screw material steel 40X.
The screw is loaded with an axial compressive force, therefore,
,
where - stresses arising in the material of the screw, Pa;
Maximum allowable compressive stresses, Pa
;
Cross-sectional area of the screw on the inside
thread diameter, N.
.
We take the inner diameter of the screw thread equal to 50mm.
Literature
1. Conveyors: Handbook / R. A. Volkov, A. N. Gnutov, V.K. Dyachkov and others. Under the general. ed. Yu.A. Perten. L.: Mashinostroenie, Leningrad branch, 1984. 367 p.
2. Spivakovsky A.O., Dyachkov V.K. Transporting machines: Proc. allowance for engineering universities. - 3rd ed. , revised - M. : Mashinostroenie, 1983. - 487 p., ill.
3. Zenkov R. L. et al. Continuous transport machines: A textbook for university students studying in the specialty "Handling and transport machines and equipment" / R. L. Zenkov, I. I. Ivashkov, L. N. Kolobov, - 2nd ed., revised. and additional – M.: Mashinostroenie, 1987. – 432 p.: ill.
4. Anuryev V.I. Handbook of the designer of the machine builder. Ed. 4th, revised and additional. Book. 2.M., "Engineering". 576 p.
5. Shubin A. A. Calculation of the plate conveyor: Guidelines. – Publishing house of MSTU im. N. E. Bauman, 2004. - 28 p.
To calculate the apron conveyor, the same initial data must be specified as for the belt conveyor.
1) Definition of the main parameters. On a deck with sides, the cross-sectional area of bulk cargo F equal to the sum of the areas of the triangle F1 and rectangle F2(Fig. 15.4).
where are the angles of repose of the load in motion ( j d) and at rest j;
kb- coefficient of reduction of the cross-sectional area of a triangle on an inclined
conveyor; ( kb=1, at b=0 ; kb=0.9 at b>20 o)
h b- height of the cargo layer at the side, m.
Denote k n=tg(0.4 j)kb- productivity factor
F=0,25AT 2 k n+bh b
Conveyor performance
From here 
, m
h b= (0.65¸0.8) h (h- total height of the sides).
With large-sized cargo, it can be considered that the cargo is located on the deck in an equal rectangular layer, i.e. F1=0, and F2=F=bhy,where y= 0.8¸0.9 - section filling factor. Received deck width AT it is necessary to check the lumpiness of the cargo
where a- size of typical pieces of cargo, mm;
X- coefficient; X= 1.7 and 2.7 respectively for ordinary and sorted cargo.
The finally selected flooring width and board height are rounded up to the nearest large in accordance with GOST.
For piece cargo, the width of the flooring is selected according to the size of the cargo and the method of transportation. The speed of the flooring is usually taken in the range of 0.05-0.63 m/s and does not exceed 1 m/s.
2) Traction calculation is carried out by the circuit bypass method, starting from the point of minimum chain tension; usually Smin=1-3kN. Resistance in straight sections is determined by the formulas:
The resistance on the turning sprockets is determined in the same way as for the drums.
S sat=KS nb , (K=1.05¸1.1)