Safety factor of steel chains. Ropes, chains, load-handling devices, load-handling devices and containers. Which blocks are subject to the most wear?
| Step p, mm | Drive sprocket rotation speed, rpm | ||||||||
| 12,7 | 7,1 | 7,3 | 7,6 | 7,9 | 8,2 | 8,5 | 8,8 | 9,4 | |
| 15,875 | 7,2 | 7,4 | 7,8 | 8,2 | 8,6 | 8,9 | 9,3 | 10,1 | 10,8 |
| 19,05 | 7,2 | 7,8 | 8,4 | 8,9 | 9,4 | 9,7 | 10,8 | 11,7 | |
| 25,4 | 7,3 | 7,8 | 8,3 | 8,9 | 9,5 | 10,2 | 10,8 | 13,3 | |
| 31,75 | 7,4 | 7,8 | 8,6 | 9,4 | 10,2 | 11,8 | 13,4 | - | |
| 38,1 | 7,5 | 8,9 | 9,8 | 10,8 | 11,8 | 12,7 | - | - | |
| 44,45 | 7,6 | 8,1 | 9,2 | 10,3 | 11,4 | 12,5 | - | - | - |
| 50,8 | 7,7 | 8,3 | 9,5 | 10,8 | - | - | - | - |
2.4. Design of roller chain sprockets
Sprockets are made from steels 40 and 45 according to GOST 1050-88 or 40L and 45L according to GOST 977-88 with hardening up to 40...50 HRC e. The sprocket design is developed taking into account the standard for the tooth profile and rim cross-section in accordance with GOST 591-69.
The cross-sectional shape of the rim is selected depending on the ratio of the thickness of the disc WITH and rim diameter D e. With a relatively large disk thickness WITH And D e £ 200 mm, a solid disk or a disk with holes is used to save metal. At D e > 200 mm it is recommended to use a composite structure.
The position of the hub relative to the disk and rim is taken for design reasons. When installing a sprocket on a cantilever at the output end of the shaft, it should be located as close as possible to the support in order to reduce the bending moment.
The design of the sprocket of a single-row roller chain is carried out according to the following recommendations.
Tooth width, mm:
The sprocket tooth can be made with a bevel (Fig. 2.3, A) or with rounding (Fig. 2.3, b);
Bevel angle g = 20 o, tooth chamfer f » 0.2b;
Tooth curvature radius (largest);
Distance from the top of the tooth to the line of centers of the rounding arcs;
radius of curvature r 4 = 1.6 mm with a chain pitch p £ 35 mm, r 4 = 2.5 mm with a chain pitch p > 35 mm;
Length of the largest chord, for sprockets without displacement of the centers of the arcs of the depressions, mm:
| , |
with displacement of the centers of the arcs of the depressions:
Thickness, mm: ;
Groove diameter, mm: .
Inner diameter, mm:
Where [ t] = 20 MPa – permissible torsional stress;
Outer diameter, mm:
Length, mm: ;
- keyway dimensions: width b and depth t 2 we select in accordance with the inner diameter of the hub from Table 2.7, the length of the key is taken structurally from the values of the standard series by 5...10 mm less than the length of the hub.
Table 2.7
Prismatic keys (GOST 23360 – 78)
| Shaft diameter d, mm | Key section | Groove depth | Chamfer, mm | Length l, mm | ||
| b, mm | h, mm | Vala t 1, mm | Hubs t 2, mm | |||
| Over 12 to 17 Over 17 to 22 | 3,5 | 2,3 2,8 | 0,25…0,4 | 10…56 14…70 | ||
| Over 22 to 30 | 3,3 | 0,4…0,6 | 18…90 | |||
| Over 30 to 38 Over 38 to 44 | 3,3 | 22…110 28…140 | ||||
| Over 44 to 50 Over 50 to 58 Over 58 to 65 | 5,5 | 3,8 4,3 4,4 | 36…160 45…180 50…200 | |||
| Over 65 to 75 | 7,5 | 4,9 | 56…220 | |||
| Over 75 to 85 Over 85 to 95 | 5,4 | 0,6…0,8 | 63…250 70…280 |
Notes: 1. Parallel key lengths l choose from the following row: 10, 12, 14, 16, 18, 20, 22, 25, 28, 32, 36, 40, 45, 50, 56, 63, 70, 80, 90, 100, 110, 125, 140 , 160, 180, 200, 220, 250. 2. An example of a key designation with dimensions b = 16 mm, h = 10 mm, l = 50 mm: Key 16´10´50 GOST 23360 – 78.
2.5. Development of a working drawing of a roller chain sprocket
Working drawings of drive roller chain sprockets must be made in accordance with the requirements of ESKD and GOST 591 standards.
The asterisks in the image (Fig. 2.3) indicate:
Sprocket tooth width;
Width of the crown (for a multi-row sprocket);
Radius of tooth curvature (in the axial plane);
Distance from the top of the tooth to the line of centers of the rounded arcs (in the axial plane);
Rim diameter (largest);
Curvature radius at the rim border (if necessary);
Protrusion circle diameter;
The surface roughness of the tooth profile, the end surfaces of the teeth, the surface of the protrusions and the roughness of the rounding surfaces of the teeth (in the axial plane).
In the drawing, the asterisks in the upper right corner place the table of parameters. The dimensions of the table columns, as well as the dimensions that determine the location of the table in the drawing field, are shown in Fig. 2.4.
The sprocket gear parameter table consists of three parts, which are separated from each other by solid main lines:
first part - basic data (for manufacturing);
The second part is data for control; the third part is reference data.
The first part of the parameter table gives:
Number of sprocket teeth z;
Mating chain parameters: pitch R and roller diameter d 3;
Tooth profile according to GOST 591 with the inscription: “With offset” or “Without offset” (centers of the cavity arcs);
Accuracy group according to GOST 591.
The second part of the parameter table gives:
Dimensions of the diameter of the circle of the depressions D i and maximum deviations (for sprockets with an even number of teeth) or the size of the largest chord Lx and maximum deviations (for sprockets with an odd number of teeth);
Practical work No. 1
Choice steel ropes and chains, blocks, sprockets and drums.
Selection of steel ropes and chains.
Accurate calculation of ropes, welded and plate chains, due to uneven stress distribution, is very difficult. Therefore, their calculation is carried out according to the standards of Gosgortekhnadzor.
Ropes and chains are selected according to GOST in accordance with the ratio:
Fр Fр.m
Where FR.m- breaking force of the rope (chain), taken according to the tables
Relevant GOST standards for ropes (chains);
FR- calculated breaking force of the rope (chain), determined by
Formula:
Fр = Fmax · n,
Where n- safety factor taken according to Pra-
Gosgortekhnadzor pitch depending on the purpose of the rope and
Mode of operation of the mechanism. Its meaning for nk ropes and chains
Nc are given in Tables P1 and P2.
FmOh- maximum working force of the rope branch (chain):
Fmah =G/ z · n, kN,
Here G - load weight, kN;
z- the number of branches of the rope (chain) on which the load is suspended;
n- pulley efficiency (Table P3).
The number of rope branches on which the load is suspended is equal to:
z = u· A,
Where A- the number of branches wound on the drum. For simple (one
Narny) chain hoist A= 1, and for double A = 2;
u - multiplicity of the pulley.
Based on the obtained breaking force value FR from the condition FR FR.m
We select the dimensions of the rope (chain) using GOST tables.
Example 1. Select a rope for the lifting mechanism of an overhead crane with a lifting capacity G= 200 kN. Load lifting height N= 8m. Operating mode – light (duty duty = 15%). Double multiplier pulley u = 4.
Initial data:
G = 200 kN – weight of the load being lifted;
N= 8m – load lifting height;
Operating mode – light (duty duty = 15%);
A= 2 – number of branches wound on the drum;
u= 4 – pulley multiplicity.
Maximum working force of one rope branch:
Fmah =G/ z · n= 200/ 8 0.97 = 25.8 kN,
Where z = u · A= 4 · 2 = 8 – the number of branches on which the load is suspended;
n- Efficiency of the pulley block, according to table. P3 at u= 4 for a pulley with a bearing
Nick rolling n= 0.97 Design breaking force: FR = FmOh · nTo= 5 25.8 = 129 kN,
Where nTo– safety factor of the rope, for a crane with a machine
Drive in light duty nTo = 5 (Table P1).
According to GOST 2688-80 (Table P5), we select a rope of the LK type - R 6x19+1 o.s. with breaking force FR.m. = 130 kN at ultimate strength GV= 1470 MPa, rope diameter dTo = 16.5 mm.
nf = FR.m. · z · n/ G = 130 · 8 · 0.97/200 = 5.04 > nTo = 5,
Therefore, the selected rope is suitable.
Example 2. Select a welded calibrated chain for a manual hoist with a load capacity G= 25 kN. Multiplicity of chain hoist u = 2 (simple pulley).
Initial data:
G= 25 kN – lifting capacity of the hoist;
u= 2 – pulley multiplicity;
A= 1 – simple chain hoist.
Fmah =G/ z · b= 25/2 0.96 = 13 kN,
Where z = u · A= 2 · 1 = 2 – the number of branches on which the load is suspended;
b= 0.96 - efficiency of the chain block. Design breaking force: FR = FmOh · nts= 3 13 = 39 kN,
Where nts– safety factor of the chain, for welded calibrated
Chains at manual drive nts= 3 (Table P2).
According to table P6, we select a welded calibrated chain with breaking force FR.m. = 40 kN, whose bar diameter dts= 10 mm, internal chain length (pitch) t = 28 mm, link width IN= 34 mm.
Actual safety factor:
nf = FR.m. · z · n/ G= 40 · 2 · 0.96/25 = 3.1 > nts= 3.
The selected chain is suitable.
Example 3. Select a load plate chain for a machine-driven lifting mechanism with a lifting capacity G= 30 kN. The load is suspended on two branches ( z = 2).
Initial data:
G= 30 kN – weight of the load being lifted;
z= 2 – the number of branches on which the load is suspended.
Solution:
Maximum operating force of one chain branch:
FmOh = G/ z · sound= 30/2 0.96 = 15.6 kN,
Where sound= 0.96 - sprocket efficiency.
Design breaking force: FR = FmOh · nts= 5 15.6 = 78 kN,
Where nts– safety factor of the chain, for a plate chain with
Machine driven nts = 5 (Table P2).
According to table P7, we accept a chain with a destructive force FR.m. = 80 kN, whose pitch t= 40 mm plate thickness S= 3 mm plate width h= 60 mm, number of plates in one chain link n = 4, diameter of the middle part of the roller d= 14 mm, roller neck diameter d1 = 11 mm, roller length V= 59 mm.
Actual safety factor:
nf = FR.m. · z · n/ G = 80 · 2 · 0.96/30 = 5.12 > nts= 5.
The selected chain is suitable.
Calculation of blocks, stars and drums.
The minimum permissible diameter of the block (drum) along the bottom of the stream (groove) is determined according to the standards of Gosgortekhnadzor:
Db e – 1)dTo, mm
Where e- coefficient depending on the type of mechanism and operating mode, you
Based on the regulatory data of the Gosgortekhnadzor Rules
(Table P4);
dTo- rope diameter, mm.
Block sizes are normalized.
The diameter of the block (drum) for welded uncalibrated chains is determined by the ratios:
For manually driven mechanisms Db dts;
For machine driven mechanisms Db dts;
Where dts - the diameter of the steel bar from which the chain is made.
The diameter of the initial circle of the sprocket for a welded calibrated chain (diameter along the axis of the rod from which the chain is made) is determined by the formula:
Dn. O. = t/ sin 90 /z, mm
Where t - internal length of the chain link (chain pitch), mm;
z- number of slots on the star, accepted z 6.
The diameter of the initial circle of the sprocket for a leaf chain is determined
are calculated according to the formula:
Dn. O. = t/ sin 180 /z, mm
Where t - chain pitch, mm;
z- number of sprocket teeth, taken z 6.
Drums for ropes are used with single-layer and multi-layer winding, with smooth surface and with a screw thread on the surface of the shell, with one-sided and two-sided rope winding.
The diameter of the drum, as well as the diameter of the block, is determined according to the Rules of Gosgortekhnadzor:
Db e – 1)dTo, mm.
The length of the drum for double-sided rope winding is determined by the formula:
and with one-sided winding:
, mm
Where l R– working length of the drum;
l h =(3…4) t– length of the drum required for fastening the rope (chain), mm;
l O– distance between right and left cuts, mm.
The working length is determined by the formula:
,
Where z– number of working turns of the rope;
,
Here Lk =H ∙ u– rope length excluding spare turns, mm
H – load lifting height, mm
u – multiplicity of the pulley;
z 0 = 1.5…2 – number of spare turns of rope;
t– pitch of rope turns, t = d To– for a smooth drum;
t = d To+(2…3) – for a drum with cuts, mm.
The distance between the right and left cuts is determined by the formula:
L 0 =b-2h min ∙tg 
,
Where b – the distance between the axes of the streams of the outer blocks is taken according to table P8;
h min– the distance between the axes of the drum and the axis of the blocks in the uppermost position;
The permissible angle of deviation of the rope branch running onto the drum from the vertical position is = 4...6°.
The wall thickness of the drums can be determined from the compressive strength condition:
, mm
Where F max– maximum working force in the rope branch, N;
- permissible compressive stress, Pa, for calculations the following is taken:
80MPa for cast iron C4 15-32;
100MPa for steels 25L and 35L;
110MPa for steels St3 and St5.
For cast drums, the wall thickness can be determined using empirical formulas:
For cast iron drums 
= 0,02D b+(6…10) mm;
For steel drums = 0.01 D b+3 mm, and then check it for compression. Should be:
.
Example 4. Using the data obtained in example 2, determine the diameter of the initial circle of the block (asterisk).
The diameter of the initial circle of the sprocket for a welded calibrated chain is determined by the formula:
mm
Where t=28 mm – internal length of the chain link (pitch);
z
6 – number of slots on the block (asterisk), we accept z=10.
Example 5. Using the data in Example 3, determine the diameter of the initial circle of the sprocket.
Sprocket starting circle diameter
mm,
Where t=40 mm – chain pitch;
z 6 – number of sprocket teeth, accept z=10.
Example 6. Determine the main dimensions of a cast iron drum according to example 1. Allowable compressive stress for cast iron = 80 MPa.
The minimum permissible diameter of the drum along the bottom of the groove is determined using the Gosgortekhnadzor formula:
,mm
Where d To= 16.5 mm – rope diameter;
e– coefficient depending on the type of mechanism and operating mode, for cranes with machine drive in light operating mode e=20 (Table P4)
D b=(20-1)∙16.5=313.5 mm, we take the value of the drum diameter from the normal range D b=320 mm (Table P8).
Determine the length of the drum. Drum with double-sided cutting. The working length of one half of the drum is determined by the formula:
mm
Where t– pitch of turns, for a drum with grooves
t= d To + (2…3)=16.5+(2…3)=(18.5…19.5) mm, accept t= 19 mm;
z o=1.5…2 – number of spare turns of rope, we accept z o=2 turns;
z R– number of working turns of the rope
Here L k = H u=8 4 =32 m – length of the rope wound on one half;
Then 
mm
Total drum length:
L b =2(l p +l 3 )+l o, mm,
Where l 3
– the length of the drum required to secure the rope; 
Mm, we accept l 3 =60 mm;
lO- distance between right and left cuts
l O =in-2h min ∙ tg, mm
Here V– the distance between the axes of the streams of the outer blocks, V= 200 mm, at D b= 320 mm (Table P8).
h min– distance between the axes of the drum and blocks in the uppermost position
h min =1.5 ∙D b=320∙1.5=480 mm
4-6° - permissible angle of deviation of the rope branch approaching the drum from the vertical position, we take = 6°.
l 0 =200-2∙4/80∙tg6°=99.1 mm
We accept l 0 =100 mm.
Thus, the total length of the drum
l b=2(608+60)+100=1436 mm, accept
l b=1440 mm = 1.44 m
m.
We accept 
mm.
The wall thickness of the cast drum must be at least 12 mm.
Practical work No. 2
Calculation of winches and lifting mechanisms of hoists with manual and electric drives according to specified conditions.
1. Calculation of manual winches
Calculation sequence for a manual winch.
1) Select a load suspension scheme (without a chain hoist or with a chain hoist).
2) Select a rope according to the given load-carrying capacity.
3) Determine the main dimensions of the drum and blocks.
4) Determine the moment of resistance on the drum shaft from the weight of the load T With and the moment on the handle shaft created by the force of the worker Tr.
N∙ m,
Where F max- maximum working force in the rope branch, N; D b– drum diameter, m.
Moment on the handle shaft:
N∙m,
Where R R– the effort of one worker, is accepted
R R=100…300 N
n– Number of workers;
- coefficient taking into account the non-simultaneous application of force when several workers work together, =0.8 – for two workers =0.7 – for four workers
L – handle length, accepted l=300…400 mm
5) Determine the winch gear ratio using the formula:
Where η – Winch efficiency.
6) Calculate open gears and shafts (the method of their calculation was studied in the “Machine Parts” section of the “Technical Mechanics” subject).
7) Determine the main dimensions of the handle. The diameter of the handle rod is determined from the bending strength condition:
m,
Where l 1 – length of the handle shaft, taken l 1 =200…250 mm for one worker and l 1 =400…500 mm for two workers;
- permissible bending stress for steel St3
=(60…80) MPa=(60…80)∙10 6 Pa.
The thickness of the handle in the dangerous section is calculated for the combined action of bending and torsion:
Sh 
Irina of the handle is taken equal to
D 
The diameter of the drive shaft on which the handle is placed is determined from the condition of torsional strength:
G
de - reduced permissible torsional stress for steel
St5 =25...30 MPa.
The diameter of the handle sleeve is taken dв=(1.8...2)d1 , and the length of the sleeve is lв=(1...1.5)d1.
Load lifting speed:
Where G- lifting capacity of the winch, kN;
VR- the peripheral speed of the drive handle is usually taken
VR=50...60 m/min.
Example 7. Calculate the lifting mechanism manual winch designed to lift a load weighing G= 15 kN per height N= 30m. Number of workers n=2. Winch efficiency =0.8. The surface of the drum is smooth, the number of layers of rope winding on the drum m=2. Multiplicity of chain hoist u=2. Simple pulley ( A=1).
Initial data:
G=15kN - weight of the load being lifted;
N=10m - load lifting height;
n=2 - number of workers;
=0.8 - winch efficiency;
m=2 - number of layers of rope winding on the drum;
The surface of the drum is smooth;
u=2 - pulley multiplicity;
A=1 - number of branches wound on the drum.
Solution:
Rope selection.
Maximum working force in one rope branch:
Fmax= 15/20.99=7.6 kN,
Where z= u a= 2 - the number of branches on which the load hangs;
Efficiency of a pulley according to Table P3 for a pulley with a multiplicity u=2 on rolling bearings 0.99.
Design breaking force:
Fp= nTo Fmax=5.57.6=41.8 kN,
Where nTo - safety factor of the rope, for a manually driven cargo winch nTo=5.5 (Table P1).
According to GOST 26.88-80 (Table P5), we select a rope of type LK-R 6x19 + 1 o.s. with breaking force Fp. m.= 45.45 kN at tensile strength 1764 MPa, rope diameter dTo=9.1 mm.
Actual safety factor of the rope:
nf = FR.m. · z · n/G = 45.45 2 0.99/15 = 6 > nTo = 5,5.
Determination of the main dimensions of the drum.
Minimum permissible drum diameter:
db e– 1)dTo, mm
Where e- coefficient depending on the type of mechanism and operating mode, for
Manual cargo winches e=12 (Table P4);
dTo- rope diameter, mm, then
db – 1)9.1=100.1mm
We accept from the normal series db=160mm (Table P8).
The working length of the drum for multi-layer rope winding is determined by the formula:
Where t – pitch of turns, for a smooth drum ; t= d k =9.81 mm ;
L k – rope length excluding spare turns
L k =H∙u=30∙2=60 m
Full length drum with single-sided winding
l b =l p +l c +l h,
Where l b =(1,5…2)∙ t – drum length required for spare turns ,
l b =(1,5…2)∙9,81=13,65…18,2 mm ,
we accept l b =18 mm
l h – drum length required to secure the rope
l h =(3…4)∙ t=(3…4)∙9,81=27,3…36,4 mm ,
we accept l h =34 mm
Thus, the total length of the drum
L b =488+18+34=540 mm.
We accept l b =540 mm .
The thickness of the drum wall is determined by the formula:
We accept δ=8 mm .
[ σ ] szh =110 MPa– permissible stress for steel St5.
Bending moment
Given moment
Moment of resistance to bending of the annular section
Where 
D V =D b -2∙δ=160-2∙8=144 mm– inner diameter drum
Total stress from bending and torsion in the dangerous section of the drum :
The strength condition is met.
Outer diameter along the sides of the drum.
D n =D b +2∙(m+2+)∙ d k =160+2∙(2+2)∙9,1=232,8 mm
We accept D n =235 mm.
Moment of resistance from the weight of the load
Moment on the handle shaft:
T r =P r ∙n∙φ∙l=200∙2∙0.8∙0.35=112 N∙m
Where R R – the effort of one worker, we take P p = 200 N
φ – coefficient taking into account the non-simultaneity of the application effort, when two workers work φ=0.8
l– length of the handle, we accept l= 350 mm
Determine the winch gear ratio.
because And O , then we accept single-stage transmission.
At And O >8 Two-stage transmissions should be adopted, dividing the total gear ratio into gear ratios of individual pairs:
and o = and 1 + and 2 .
Determination of the main dimensions of the handle.
Pen Shaft Diameter:
We accept d=28 mm,
Where l 1 – handle shaft length , l 1 = 350 mm
[ σ] u = 60…80 MPa – permissible bending stress, for steel St5, we accept [ σ] u = 70 MPa
The thickness of the handle is determined by the formula
We accept δ R =15 mm.
The width of the handle is taken to be в=3∙δ R =3∙15=45 mm.
The diameter of the drive shaft on which the handle is placed :
We accept d 1 = 30 mm
Where [ τ ] To = 25…30 MPa – reduced permissible torsional stress, for steel St5, we accept [ τ ] To = 25 MPa.
Handle sleeve diameter : d V =(1,8…2) d 1 ;
d in =(1.8…2)∙30=54…60 mm,
We accept d in = 55 mm.
Handle sleeve length
L in = (1… 1.5)∙d 1 = (1…1.5)∙30=30…45 mm
We accept l V = 40 mm.
Load lifting speed
Where V p = 50…60 m/min – peripheral speed of the drive handle, take V p = 55 m/min
2. Calculation of winches with electric drive
Sequence of calculation of winches with electric drive.
The rope is selected.
Determine the main dimensions of the drum.
Determine the power and select the electric motor and gearbox from catalogs.
Where G is the weight of the load being lifted, kN
V 2 – load lifting speed, m/s
η – Efficiency of the mechanism.
By catalog select an electric motor depending on the operating mode, taking the nearest higher power value and write down its basic technical data.
To select a gearbox, determine the gear ratio:
Where n uh – rotation speed of the selected electric motor;
nb– drum rotation frequency, determined by the formula :
Here V 2 – load lifting speed, m/s;
And - pulley multiplicity;
D b – drum diameter, m;
The gearbox is selected from the catalog based on the design power, engine speed, gear ratio and operating mode.
4. Check the selected electric motor for the actual multiplicity of starting torque.
Condition must be met
ψ≤ψmax,
Where ψ max – the maximum permissible multiplicity of starting torque, determined by the formula:
,
Here T P max – maximum torque of the electric motor, taken from the table;
T n – rated torque on the motor shaft;
ψ – actual multiplicity of engine starting torque
,
The starting torque reduced to the motor shaft is determined by the formula :
Where t n = 8∙ V 2 – mechanism start time, s;
δ=1.1...1.2 – coefficient that takes into account the swing moments of the mechanism parts.
Static torque on the motor shaft:
5. The brake is selected, for which the braking torque is determined using the formula:
T T =K T ∙T K,N∙m
Where TO T – braking reserve coefficient accepted according to the standards of Gosgortekhnadzor depending on the operating mode of the mechanism;
T TO – torque on the high-speed gearbox shaft, equal to the rated torque on the electric motor shaft,
Where
-
angular speed of the electric motor.
Using the catalog, a brake is selected according to its braking torque and its technical characteristics are written down.
Finally, verification calculations are made of the selected brake. The method for calculating them depends on the type of brake and is given in the training manual (6) Chapter 1 §3.
Example 8. Select an electric motor, gearbox and brake for the winch lifting mechanism intended for lifting heavy loads G= 50 kN with speed V2 = 0.25 m/s if drum diameter db= 250 mm, pulley multiplicity u = 2, winch efficiency η = 0.85, operating mode – light (duty duty = 15%)
Initial data:
G= 50 kN – load weight;
V2 = 0.25 m/s – ascent speed;
db= 250 mm – drum diameter;
u = 2 – pulley multiplicity;
η = 0.85 – winch efficiency;
Operating mode – light (duty duty=15%)
Solution:
Required motor power
From the catalog we select an electric motor of type MTF312-8 with power at duty cycle = 15% Re= 15 kW, speed nuh= 680 rpm, with maximum torque Tpmax= 430 N.m., rotor swing moment (GД 2) = 15.5 N.m. Nominal torque on the motor shaft
Maximum torque ratio:
Drum rotation speed:
Design gear ratio
According to the catalog (Table P10), based on the design power, engine speed, gear ratio and operating mode, we select a gearbox type Ts2-250 With gear ratio And R = 19,88, power R R = 15 kW, speed high speed shaft P R = 750 rpm Actual lifting speed
We check the selected electric motor for actual multiplicity starting moment. The following condition must be met:
The actual multiplicity of the starting torque of the selected electric motor is determined from the ratio:
The starting torque reduced to the motor shaft is determined by the formula:
Where t P = 8∙0.22 = 1.8 s – mechanism start time;
δ = 1.1...1.2– coefficient taking into account the swing moments of parts mechanism, we accept δ = 1.15. Static torque on the motor shaft
Then,
therefore, engine performance
secured.
Determine the required braking torque.
T T =K T ∙T k =1.5∙210.7=316 N.m.
Where TO T – braking safety factor, light duty , K T = 1.5 (Table A11);
T TO – torque on the high-speed gearbox shaft , T To = T n = 210,7 N.m.
According to the catalog (Table P12), according to the braking torque T T, we select a two-block brake with an electric motor of type TT - 250, which has a braking torque T T = 400 N.m. We write down the data necessary for the calculation: lever arms – a = 160 mm, b = 330 mm, c = 19 mm, l T = 150 mm, pad offset E = 1.1 mm, pusher type TGM-25, providing pushing force F T = 250 N and rod stroke h w = 50 mm, pulley dimensions – pulley diameter D w = 250 mm, pulley width H w = 90 mm, pulley grip angle between the blocks α = 70 0 .
Calculated circumferential force on the brake pulley rim:
Force normal pressure pulley pads
Where f – coefficient of friction of working surfaces, for braking asbestos tape (ferrado) for cast iron and steel f = 0,35.
Pusher force :
Where η – efficiency of the lever system equal to η =0.9…0.95, we accept η = 0.95
Push rod stroke:
Where TO 1 – coefficient of utilization of the working stroke of the rod, equal to TO 1 = 0,8 …0,85 , we accept TO 1 = 0,85.
We check the working surfaces of brake pads for specific pressure using the formula:
Here [ q] – the permissible specific pressure of the working surface material is taken according to the table. Therefore, the selected brake fits.
Calculation of the lifting mechanism of hoists with manual drive
Let's consider the calculation of a worm hoist with a manual drive.
Calculation of a manual worm hoist is carried out in the following sequence:
1) Depending on the specified load capacity G, according to GOST tables, a load chain is selected and the diameter of the initial circle of the chain sprocket is determined.
2) Determine the gear ratio of the hoist, having previously determined the load moment on the sprocket T gr and the torque on the traction wheel T k
3) Taking the number of worm starts z 1 = 2 (in worm hoists a double-start non-self-braking worm is used), determine the number of teeth of the worm wheel
4) Calculate the worm gear
5) Calculate the load-bearing disc brake
Example 9. Calculate the lifting mechanism of a manual worm hoist with a load capacity of G = 30 kN. The load is suspended on a moving block a = 1, pulley multiplicity u = 2. Traction wheel diameter D = 260 mm. The force applied to the traction wheel chain is F p = 600 N.
Chain selection.
Maximum operating force in one chain branch:
Where z – the number of branches on which the load for the manual hoist is suspended, z=u∙a=2∙1=2;
η sound = 0,96 – Sprocket efficiency
Design breaking force.
F p =p c ∙F max =3∙15.6=46.8 kN.
Where P ts – chain safety factor; for leaf chains With manual drive P ts= 3 (Table P2)
According to table P7, we accept a chain with a breaking force F r.m. = 63 kN for which the pitch t = 35 mm, plate thickness S = 3 mm, plate width h = 26 mm, number of plates in one link n = 4, roller diameter in the middle part d = 12 mm, roller neck diameter d 1 = 9 mm.
Actual chain safety factor:
Determine the diameter of the initial circle of the sprocket:
Where z 6 – number of sprocket teeth, accept z = 16.
We determine the main dimensions of the pair of worms. In worm hoists double-threaded (non-self-locking) worms are used (z 2 = 2).
Reduced friction angle:
p=arctgf=arctg0,1=544
Where f = 0,04…0,1 – reduced friction angle, with periodic lubrication open worm gear accept f = 0,1.
Worm diameter coefficient
Where z 1 = 2 – number of worm visits.
In a non-self-braking gear, the angle of elevation of the worm helix line must be greater than the reduced friction angle R , those. must be respected condition > p , therefore, we accept a smaller value for the worm diameter coefficient q = 16 (Table A14).
Helix angle of the worm line:
We calculate the transmission efficiency:
We accept η 2 = 0,53
Determine the required gear ratio
Where T gr – load moment on the sprocket,
T To – torque on the traction wheel:
Then
Determine the number of teeth of the worm wheel. From the relation
And 0 = z 2 / z 1 we find z 2 = u 0 ∙ z 1 = 34,8∙2 = 69,6
We accept z 2 = 70. Let's clarify gear ratio
and f =i 2 =z 2 /z 1 =70/2=35.
The deviation from the calculated value is:
We assign the materials of the worm and worm wheel and determine the permissible stresses.
In manually driven worm gears, the sliding speed is low, so it is advisable to make the worm and worm wheel from cast iron. For the worm, SCH 21-40, and for the wheel - SCh 18-36. Then the permissible stress δ nv = 190 MPa , δ FP =0.12 ∙δ in and = 0,12∙ 365= 44 MPa at δ in and = 365 MPa.
Determine the required center distance:
We determine the design modulus of engagement using the formula:
According to the table P14 we accept t=5 mm andq = 16.
We specify the interaxle distance
and w = 0.5∙t∙(q+z 2)=0.5∙5∙(16+70)215 mm
We determine the main parameters of the worm and worm wheel:
Pitch diameters: worm d 1 = m∙ q=5∙16=80 mm
wheels d 2 = m∙ z 2 =5∙70=350 mm
Protrusion diameters: worm d a 1 = d 1 +2∙ m=80+2∙5=90 mm
wheels d a 2 = d 2 +2 m=350+2∙5=360 mm
Calculation of a load-bearing disc brake.
Load moment on the worm:
Where η 2 =0,53 – Efficiency of a worm pair;
And 2 = 35 – gear ratio of the worm pair.
Axial force in the brake:
Moment of friction force on disk surfaces:
Where n = 2 – number of pairs of rubbing surfaces:
f – coefficient of friction of rubbing surfaces, according to table. P13. we accept f = 0,15.
D Wed – average diameter disks ;
Where is the inner diameter of the disks D V d a , we accept D V = 1000 mm;
the outer diameter of the disks is taken within the limits D n = (1,2…1,6)∙D V =(1.2…1.6)∙100=120…160 mm, we accept D n = 150 mm.
Checking discs for specific pressure:
Where [ q] = 1.5 MPa – permissible specific pressure of rubbing surfaces (Table P13)
4. Calculation of the lifting mechanism of electrically driven hoists under specified conditions.
Electric hoist calculations include:
calculation and selection of rope according to GOST tables;
determination of the main dimensions of the drum;
calculation of the electric hoist drive;
calculation of closed gears for endurance according to contact stresses and the bending strength of teeth;
verification calculation of the electric motor, calculation of the strength of the drum and hook suspension;
selection and calculation of a two-block electromagnetic brake;
load brake calculation.
Example 10. Calculate the lifting mechanism of an electric hoist with a load capacity of G = 32 kN. Lifting height H = 6 m, load lifting speed V 2 = 0.134 m/s. Simple pulley (a=1) multiplicity And= 2. Drum with grooves.
Initial data:
G = 32 kN – load capacity;
H = 6 m – height of lifting the load;
V 2 = 0.134 m/s – load lifting speed
Q = 1 – number of branches wound on the drum;
And= 2 – pulley multiplicity;
The surface of the drum has grooves.
Solution
Selection of rope.
Maximum operating pressure in one rope branch:
Where z= u∙ a=2∙1=2 – the number of branches on which the load is suspended;
η P – pulley efficiency; according to table P3 at u=2 for a pulley with rolling bearings η P = 0,99.
Design breaking force:
Where P To – rope safety factor, for hoists with machine drive P To =6 (Table P1). According to GOST 2688-80, we select a rope of type LK-R (6x19+1 o.s.) with breaking force F p . m . = 97 kN at ultimate strength δ V= 1960 MPa, rope diameter d To= 13 mm.
Actual safety factor of the rope:
The smallest diameter of the drum along the bottom of the groove is determined by the Gosgortekhnadzor formula:
Where P To– coefficient depending on the type of mechanism for electric hoists P To = 20 (Table P4).
D b (20-1)∙13 247 mm
We accept D b= 250 mm (Table P8).
Number of working turns of rope on the drum
Drum length l b = l p + l h ,
Where l p – working length of the drum, l p =(z p + z 0 ) t;
z 0 =1,5…2 – number of spare turns of rope, we accept z 0 =1,5 coil;
t – number of turns, for grooved drum t= d k +(2…3)=13+(2…3)=15… 16 mm, accept t= 15 mm;
l p =(14,5+1,5)∙15=240 mm;
l h – drum length required to secure the rope
l h=(3…4)∙15=45…60 mm, we accept l h = 50 mm.
Then, the full length of the drum
l b =240+50=290 mm.
Static torque on the drum shaft when lifting a load
Where η b – Drum efficiency , η b = 0.98…0.99, accept η b = 0,98.
Drum rotation speed:
Rated motor power
Where η m = η P ∙η b ∙η R – Efficiency of the lifting mechanism;
η m = 0,99∙0,98∙0,9 = 0,87,
Here η P = 0.99 – pulley efficiency
η b = 0.98 – drum efficiency;
η R = 0,9…0,95 – Gearbox efficiency, we accept η R = 0,9
We select an electric motor of type 4A132S with a power of P e = 5.5 kW and a synchronous rotation speed of P e = 1000 rpm. Manufactured electric hoists have electric motor units built into the drum, forming a motor-gearbox electric hoist unit.
Required gear ratio
With this value of the gear ratio, it is necessary to adopt a two-stage gearbox.
We accept the gear ratio of the first stage And 1 =8, then
and 2 = and r.r. : and 1 =51.3: 8=6.4.
Actual gear ratio
And R = 8∙6,4=51,2
Actual ascent speed:
Brake calculation.
The hoist is equipped with two brakes. A two-block brake with an electromagnet is installed on the high-speed shaft of the gearbox, and a load-bearing brake is installed on the low-speed shaft.
Shoe brake calculation.
We determine the braking torque using the formula
T T =K T ∙T TO=1.25∙44.5=55.6 N∙m,
Where TO T – braking safety factor for lifting mechanism electric hoists with two brakes K T = 1.25; T K = T 1 – rated torque on the high-speed shaft:
Here η h = 0.975 – gear efficiency of one stage.
Normal pressure force of the pads on the brake pulley:
Where f = 0.42 – coefficient of friction of the rolled belt on cast iron and steel
D w = 160 mm – diameter of the brake pulley. We determine the spring force acting on each of the two levers:
Where l 1 = 100 mm and l 2 = 235 mm – lever lengths, η = 0.95 – lever efficiency systems.
Opening force:
Where l 3 =105 mm – table. P15.
Electromagnet force:
Where G p = 4 N is the weight of the lever connecting the electromagnet armature to the opening pin;
L = 225 mm and d = 15 mm – table. P15.
Electromagnet stroke:
In accordance with the value of F m, the brake electromagnet is selected and adjusted by the stroke value h. The highest pressure on brake linings made of rolled tape:
Here l about = 91 mm – lining length;
V about = 30 mm – lining width;
[ q] – permissible specific pressure for working materials surfaces according to table P13, for rolled strip on cast iron and steel [ q] = 1.2 MPa.
Calculation of load-bearing brake.
According to the table P16 for a given hoist load capacity G = 32 kN, we select a load-bearing disc brake with dimensions:
The brake screw thread is rectangular, three-start, outer thread diameter d = 50 mm
Internal thread diameter d 1 = 38 mm;
Thread pitch – t = 8 mm.
The average diameter of the disks D av = 92.5 mm. Helix angle of three-start brake shaft thread:
Where z = 3 – number of thread starts;
D 2 – average thread diameter
The axial force that occurs during braking and clamps the brake friction rings.
Where T 2 is the rated torque on the low-speed gearbox shaft,
= 2…3 - friction angle in a threaded pair when operating in an oil bath , we accept = 2
f = 0.12 – coefficient of friction of the rolled strip on steel in oil;
η – average radius of screw thread
Braking torque of load bearing brake:
T 2T = f∙ F a ∙ R c ∙ n=0.12∙22070∙0.0925∙2=490 N∙m
Where n=2 – number of pairs of rubbing surfaces.
The braking torque must satisfy the following condition:
T 2T K T ∙T 2 1.25∙347=434 N∙m;
Т 2Т =490 > 434 N∙m
Therefore, the condition is satisfied.
TO T = 1,25 – braking safety factor for the second brake of the electric hoist.
Reliability of holding the load in a suspended state is ensured by observing the following dependence:
f∙R c ∙n[η∙tg(α+)+f∙R c ]∙ η z 2 ;
f∙ R c ∙ n =0,12∙0,0925∙2=0,022.
0.022>0.015; those. the condition is met.
The downward moving load will stop if:
0,0046
Checking the screw thread for collapse:
Here z 1 = 4 is the number of thread turns that absorb the load.
Practical work No. 3
Calculation of a belt conveyor according to given conditions.
Calculation of a belt conveyor includes:
determining the speed and width of the belt;
approximate determination of tape tension and wire power;
calculation of belt and roller supports;
determination of drum dimensions;
conveyor traction calculation;
clarification of the traction force and power of the drive station, selection of the electric motor and gearbox.
Calculate a belt conveyor with a capacity of Q=240 t/h for transporting lump sulfur over a distance L=80 m. Load density =1.4 t/m 3 , maximum size pieces a 100 mm, angle of repose of the material at rest = 45°, angle of inclination of the conveyor to the horizon 
= 15°. The conveyor belt is rubberized, the surface of the drive drum is lined with wood. Angle of wrapping of the drum by the tape =180°. The drive is located at the head end of the conveyor.
Initial data:
Q=240 t/h – conveyor productivity;
L=80 m – conveyor length;
=1.4 t/m 3 – material density;
A 100 mm – maximum size of pieces;
= 45° - angle of repose at rest;
15° - angle of inclination of the conveyor to the horizon;
=180° - angle of wrapping of the drum with tape;
Transport material – lumpy sulfur.
Rice. 1 Design diagram of a belt conveyor.
To obtain the smallest possible width of the belt, we adopt a grooved shape consisting of three rollers. According to Table A.18, for the transportation of medium-sized materials with the proposed belt width B = 500...800 mm, we accept the belt speed V = 1.6 m/s.
The width of the grooved tape is determined by the formula:
We take the belt width B = 650 mm = 0.65 m (Table P 18), where K is a coefficient that takes into account the additional scattering of cargo on the inclined conveyor belt; at 20° - K = 1, at 20° - K = 0.95.
In our case = 15° K = 1.
Checking the width of the belt based on the lumpiness of the load
V k = 2.5∙a+200=2.5∙100+200=450 mm
We got B to B, therefore, we finally accept B = 650 mm. If it turns out to be B B k, then you need to take the width B k from the normal series according to GOST 22644-77 (Table P18).
We choose a rubber belt from the belting BKIL - 65, width B = 650 mm with a strength limit σ r. n. =65 N/mm and the number of gaskets z= 3...8 (Table P19).
We determine the preliminary drive power using the formula:
P n =(0.00015∙Q∙L 2 +K 1 ∙L 2 ∙V+0.0027∙Q∙H) ∙K 2 ,
Where L 2 is the length of the horizontal projection of the conveyor,
L 2 =L∙cos=80∙cos15° =77.3 m,
H – load lifting height, H= Lsin=80∙sin15° =20.7m
K 1 and K 2 are coefficients depending on the width and length of the tape.
According to the table P20 with tape width B = 650 m K 1 = 0.020, and K 2 = 1 with a coefficient length of over 45 m.
Then, P n =(0.00015∙240∙77.3+0.02∙77.3∙1.6+0.0027∙240∙20.7) ∙1=18.67 kW
We determine the preliminary traction force:
kN.
We determine the preliminary maximum tension of the tape using the formula:
Where f is the coefficient of friction between the belt and the drum, in our case f = 0.35 (Table A21).
α - 180° - angle of wrapping of the drum with tape.
The values of e fα are given in Table A21.
Determine the number of spacers in the tape:
,
Where K rp is the safety factor of the tape according to the table. P 22, we accept K rp = 9.5 in the proposal that the number of gaskets will be 4...5.
We take z = 4. The thickness of the rubber linings on the working side is δ 1 = 4 mm, on the non-working side δ 2 = 1.5 mm (Table P 23).
Linear density of tape:
Where δ = 1.4 mm is the thickness of one textile pad (Table A19).
Average linear density of transported cargo:
kg/m
Conditional linear density of roller bearings. With a belt width B = 650 mm, density of the transported material = 1.4 t/m 3, movement speed up to V = 2 m/s, roller diameter D p = 89 mm (Table P24). On the working branch of the conveyor, the belt is supported by grooved roller supports, consisting of three rollers, and on the idle branch, the belt is flat, supported by roller supports, consisting of one roller.
The distance between the roller supports on the working branch of the conveyor l p is determined according to the table. P25. At B = 650 mm and = 0.81...1.6 t/m 3 l p = 1.3 m. The distance between the roller supports on the lower (idle) branch is taken l x = 2∙ l p =2∙1.3=2, 6 m.
Weight of roller supports of the working branch (grooved)
Mf =10 V+7=10∙0.65+7=13.5 kg.
Conditional linear density of grooved roller bearings
kg/m.
Weight of roller supports on an idle branch (flat)
M n =10 V+3=10∙0.65+3=9.5 kg.
Conditional linear density of flat roller bearings of the idle branch
kg/m.
Determine the dimensions of the drum.
Drive drum diameter D b =z∙(120…150) = 4 (120…1500) = =(480…600) mm. According to GOST 22644 - 77 (Table P26), we accept D b = 500 mm. Drum length B 1 = B + 100 = 650 + 100 = 750 mm.
To prevent the tape from falling off the drum, it has a convex arrow f n = 0.005B 1 = 0.005∙750 = 3.75 mm. Tension drum diameter 
We accept D n = 320 mm (Table P26).
We determine the tension of the conveyor belt using the point-by-point method of grasping the contour. We divide the contour of the conveyor belt into four sections (Fig. 1). The tension of the tape at point 1 is taken as an unknown value. Then we find the tension of the tape at other points through the unknown tension at point 1:
Where K wn =0.022 is the rolling resistance coefficient for flat roller bearings.
Where K σ N is the resistance coefficient on the tension drum. When the angle of wrapping of the drum with the tape is α = 180°…240°. K σ N = 0.05...0.07, we accept K σ N = 0.05.
Where K w w =0.025 is the rolling resistance coefficient of grooved supports.
When the drive is located at the head end of the conveyor, the tension at point 1 is equal to the tension of the belt running off the drum F 1 =F sb, and the tension at point 4 is equal to the tension of the tape running onto the drum F 4 =F nb. The tension of the running belt is determined by Euler's formula:
F nb =F with ∙е fα or F 4 =F 1 ∙е fα
Thus: 1.05 F 1 +9.8= F 1 ∙3; 1.95∙F 1 =9.8.
Where 
kN
F 2 =F 1 -1.43=5.03-1.43=3.6 kH; F 3 =1.05 ∙F 1 -1.5=1.05∙5.03-1.5=3.78 kH
F 4 =1.05F 1 +9.8=5.03∙1.05+9.8=15.1 kH
We check the sagging of the tape between the roller supports. The greatest deflection of the belt on the working side of the conveyor will be at point 3. The following condition must be met:
L max 
Maximum deflection:
Lmax = 
m
Allowable tape sagging:
The sagging conditions are met, since l max =0.027
We determine the specified traction force on the drive drum:
F TY =F 4 -F 1 +F 4…1 =15.1-5.03+0.03(15.1+5.03)=10.7 kH
Where F 4…1 =К σ n (F 4 +F 1),
Here K σ n is the resistance coefficient on the drive drum with rolling bearings, K σ n =0.03…0.035
We accept K σ n =0.03.
Specified power of the drive station:
Where K 3 =1.1...1.2 is the coefficient of adhesion between the belt and the drum, we take K 3 =1.1;
η=0.8…0.9 – overall efficiency of the drive mechanism, take η = 0.85
According to the catalog (Table P27), we accept an enclosed AC electric motor with increased starting torque, type 4A200M. Which has P = 22 kW, rotation speed n = 1000 rpm.
Development of a drive station.
Drive drum speed:
rpm
Gear ratio:
According to the table P10, depending on the gear ratio, the power of the electric motor and the rotation speed, we select a gearbox with a gear ratio U = 16.3 type Ts2-350, transmitting power under heavy duty operation P r = 24.1 kW, rotation speed n r = 1000 rpm .
Actual belt speed
To regulate the belt tension, a load stretching device with tension force.
Tensioner drum stroke length
Practical work No. 4
Calculation of a vertical bucket conveyor (elevator) under given conditions.
Vertical bucket elevators are calculated in the following sequence:
1) Determine the main parameters of the elevator.
2) Calculate linear loads.
3) Perform a traction calculation of the elevator.
4) Determine the required power of the electric motor, according to catalogs
Select the electric motor and gearbox.
Example 12. Calculate a vertical bucket elevator with a capacity of Q = 30 t/h, designed for transporting ordinary dry crushed stone with a density = 1.5 t/m3 and an average size ac = 30 mm to a height of H = 20 m.
Initial data:
Q = 30 t/h - elevator productivity;
ac = 30mm - the average size pieces of material;
= 1.5 t/m3 - material density;
H = 20m - load lifting height;
Material - ordinary dry crushed stone.
The elevator is installed in an open area.
Solution:
According to the table P28 for transportation of small-piece materials (ac
Average fill factor of buckets = 0.8.
For high-speed elevators with centrifugal unloading, the diameter of the drum can be determined according to the formula of N.K. Fadeev:
dB0.204V = 0.204x1.6 = 0.52 m
We take the diameter of the drive drum Db = 500 mm (Table P26).
Drum rotation speed:
=61 rpm
Pole Distance:
m
Since hn =0.24m
Determine the linear capacity of the buckets:
l/m.
According to the table P29 select linear capacity: 5 l/m
Bucket volume io = 2l, bucket pitch tk = 400mm, bucket width B = 250mm, belt width Bl = 300mm, bucket reach A = 140mm.
We check the reach of the bucket based on the size of the material. For ordinary cargo there should be:
A > (2...2.5)ac = (2...2.5)30 = 60...75mm
If a graded cargo is specified, then the following condition must be met:
A > (4...5)ac.
With the accepted parameters of the buckets and the belt speed V = 1.6 m/s, the specified productivity Q = 30 t/h is ensured at the bucket filling factor:
Payload (linear weight of the load being lifted):
N/m
Q=qо+q2=132+51=183 N/m.
The traction calculation of the elevator is carried out using the contour bypass method. In accordance with the calculation scheme (Fig. 2), the lowest tension should be expected at point 1. The tension at point 1 is taken as an unknown value.
The tension at point 2, taking into account the resistance on the return drum and the scooping of the load, is determined by the formula:
F2=KF1+Wzach=1.08F1+153,
where K = 1.08 is the coefficient of increase in tension in the belt with buckets, with flexi-
The drum bath is usually taken to be K = 1.08.
Wzach - resistance to scooping load.
Wzach=Kzachq2=351=153 N,
here Kzach is the scooping coefficient, expressing the specific work for
Spent on scooping up the load. At bucket speed 1-1.25
M/s for powdery and small-piece cargo Kzach = 1.25...2.5;
For medium-sized cargo Kzach = 2...4. At driving speed 1.6
M/s we accept Kzach = 3.
Tension in the oncoming branch (point 3):
Fн = F3 = F2 + qН = 1.08F1 + 153 + 18320 = F1 + 3813.
Tension in the running branch when counting against the movement of the belt (point 4):
Fc = F4 + q0 H = F1 + 132 20 = F1 + 2640.
From the theory of friction drive we have:
For a steel drum with high humidity (the elevator is installed in an open area), the friction coefficient is f = 0.1 and at = 180 we obtain e = 1.37 (Table A21). Then:
F3
Solving this equation, we get: F1 = 676 N.
To ensure grip reserve, we take F1 = 1000 N, then:
F3 = Fн = 1.08F1 + 3813 = 1.08 1000 + 3813 = 4893 N,
F4 = Fс = F1 + 2640 = 1000 + 2640 = 3640 N.
The required number of gaskets in the adopted tape of type BNKL-65 is found at р.n.= 65 N/mm (Table P19) and the safety factor of the tape Kr.p. = 9.5 (Table A22).
.
Taking into account the weakening of the belt by bolts and the need to firmly secure the buckets to the belt, we leave the previously adopted belt with z = 4.
Circumferential force on the drive drum taking into account losses on it
Ft = (F3 - F4)K = (4893 - 3640)1.08 = 1353 N.
Determine the power of the drive station:
kW,
where K3 = 1.1...1.2 is the coefficient of adhesion between the belt and the drum,
We accept K3 = 1.2;
= 0.8...0.9 - the overall efficiency of the drive mechanism, we take = 0.85.
According to the catalog (Table P27), we accept an AC electric motor of type CHA112MB, which has P = 4 kW, rotation speed n = 1000 rpm.
Required gear ratio:
According to the table P10, depending on the gear ratio, electric motor power and rotation speed, we select a gearbox with u = 16.3, transmitting power under heavy duty operation Рр = 10.2 kW, high-speed shaft rotation speed nр = 1000 rpm, type Ts2-250.
Actual belt speed:
m/s.
Rice. 2. Tension diagram in the elevator belt.
APPLICATIONS
Table P1
Rope safety factorn To
Table P2
Chain safety factor nc
Table P3
Efficiency of pulley blocks n
Table P4
Minimum permissible coefficient value e
Table P5
Ropes type LK-R 6x19 + 1 o.s. according to GOST 2688-80
| Diameter Kanata dTo, mm | Temporary tensile strength of the material, rope wires GV, MPa | ||||
| 1470 | 1568 | 1764 | 1960 | ||
| 4,1 | - | - | 9,85 | 10,85 | |
| 4,8 | - | - | 12,85 | 13,9 | |
| 5,1 | - | - | 14,6 | 15,8 | |
| 5,6 | - | 15,8 | 17,8 | 19,35 | |
| 6,9 | - | 24 | 26,3 | 28,7 | |
| 8,3 | - | 34,8 | 38,15 | 41,6 | |
| 9,1 | - | 41,55 | 45,45 | 49,6 | |
| 9,9 | - | 48,85 | 53,45 | 58,35 | |
| 11 | - | 62,85 | 68,8 | 75,15 | |
| 12 | - | 71,75 | 78,55 | 85,75 | |
| 13 | 76,19 | 81,25 | 89 | 97 | |
| 14 | 92,85 | 98,95 | 108 | 118 | |
| 15 | 107 | 114,5 | 125,55 | 137 | |
| 16,5 | 130 | 132 | 152 | 166 | |
| 18 | 155 | 166 | 181,5 | 198 | |
| 19,5 | 179,5 | 191 | 209 | 228 | |
| 21 | 208 | 222 | 243,5 | 265,5 | |
Problems 81-90
Calculate a vertical bucket elevator with productivity Q designed for transporting bulk density material r, medium size AWith to the height N. The elevator is installed in an open area.
Select the initial data for solving the problem from Table 5.
Table 5
Task No. | Q, t/h | r, t/m3 | AWith, mm | Transported material |
|
Dry clay |
|||||
Pyrite flotation |
|||||
Lump sulfur |
|||||
Sand dry |
|||||
Limestone |
|||||
Crushed chalk |
|||||
Dry ash |
|||||
Crushed bauxite |
Guidelines: , pp. 216...218, example 12.
Guidelines for performing practical work
Practical work No. 1
Selection of steel ropes and chains, blocks, sprockets and drums.
1. Selection of steel ropes and chains .
Accurate calculation of ropes, welded and plate chains, due to uneven stress distribution, is very difficult. Therefore, their calculation is carried out according to the standards of Gosgortekhnadzor.
Ropes and chains are selected according to GOST in accordance with the ratio:
FR£ FR.m
Where FR.m- breaking force of the rope (chain), taken according to the tables
relevant GOST standards for ropes (chains);
FR- calculated breaking force of the rope (chain), determined by
Fp =FmOh· n,
Where n- safety factor taken according to Pra-
forks of Gosgortekhnadzor depending on the purpose of the rope and
operating mode of the mechanism. Its meaning for nk ropes and chains
nc are given in Tables P1 and P2.
FmOh- maximum working force of the rope branch (chain):
Fmax =G/zhn, kN,
Here G- load weight, kN;
z- the number of branches of the rope (chain) on which the load is suspended;
hn- pulley efficiency (Table P3).
The number of rope branches on which the load is suspended is equal to:
z = u · A ,
Where A- the number of branches wound on the drum. For simple (one
narny) chain hoist A= 1, and for double A = 2;
u- multiplicity of the pulley.
Based on the obtained breaking force value FR from the condition FR£ FR.m
We select the dimensions of the rope (chain) using GOST tables.
Example 1. Select a rope for the lifting mechanism of an overhead crane with a lifting capacity G= 200 kN. Load lifting height N= 8m. Operating mode – light (duty duty = 15%). Double multiplier pulley u= 4.
Initial data:
G= 200 kN – weight of the load being lifted;
N= 8m – load lifting height;
Operating mode – light (duty duty = 15%);
A= 2 – number of branches wound on the drum;
u= 4 – pulley multiplicity.
Maximum working force of one rope branch:
Fmax =G/zhn= 200/ 8 0.97 = 25.8 kN,
Where z =u· A= 4 · 2 = 8 – the number of branches on which the load is suspended;
hn- Efficiency of the pulley block, according to table. P3 at u= 4 for a pulley with a bearing
nickname rolling hn= 0.97 Design breaking force: Fp =FmOh· nTo= 5 25.8 = 129 kN,
Where nTo– safety factor of the rope, for a crane with a machine
drive in light operating mode nTo= 5 (Table P1).
According to GOST 2688-80 (Table P5), we select a rope of the LK type - R 6x19+1 o. With. with breaking force FR.m. = 130 kN at ultimate strength GV= 1470 MPa, rope diameter dTo= 16.5 mm. Actual safety factor of the rope:
nf =FR.m. · z· hn/G= 130 · 8 · 0.97/200 = 5.04 > nTo = 5,
Therefore, the selected rope is suitable.
Example 2. Select a welded calibrated chain for a manual hoist with a load capacity G= 25 kN. Multiplicity of chain hoist u= 2 (simple pulley).
Initial data:
G= 25 kN – lifting capacity of the hoist;
u= 2 – pulley multiplicity;
A= 1 – simple chain hoist.
Fmax =G/zhb= 25/2 0.96 = 13 kN,
Where z =u· A= 2 · 1 = 2 – the number of branches on which the load is suspended;
hb= 0.96 - efficiency of the chain block. Design breaking force: Fp =FmOh· nts= 3 13 = 39 kN,
Where nts– safety factor of the chain, for welded calibrated
hand driven chains nts= 3 (Table P2).
According to table P6, we select a welded calibrated chain with breaking force FR.m. = 40 kN, whose bar diameter dts= 10 mm, internal chain length (pitch) t= 28 mm, link width IN= 34 mm.
Actual safety factor:
nf =FR.m. · z· hn/G= 40 · 2 · 0.96/25 = 3.1 > nts= 3.
The selected chain is suitable.
Example 3. Select a load plate chain for a machine-driven lifting mechanism with a lifting capacity G= 30 kN. The load is suspended on two branches ( z = 2).
Initial data:
G= 30 kN – weight of the load being lifted;
z= 2 – the number of branches on which the load is suspended.
Maximum operating force of one chain branch:
Fmah =G/zhsound= 30/2 0.96 = 15.6 kN,
Where hsound= 0.96 - sprocket efficiency.
Design breaking force: Fp =FmOh· nts= 5 15.6 = 78 kN,
Where nts– safety factor of the chain, for a plate chain with
machine driven nts= 5 (Table P2).
According to table P7, we accept a chain with a destructive force FR.m. = 80 kN, whose pitch t= 40 mm plate thickness S= 3 mm plate width h= 60 mm, number of plates in one chain link n= 4, diameter of the middle part of the roller d= 14 mm, roller neck diameter d1 = 11 mm, roller length V= 59 mm.
Actual safety factor:
nf =FR.m. · z· hn/G= 80 · 2 · 0.96/30 = 5.12 > nts= 5.
The selected chain is suitable.
2. Calculation of blocks, stars and drums.
The minimum permissible diameter of the block (drum) along the bottom of the stream (groove) is determined according to the standards of Gosgortekhnadzor:
Db³ (e – 1)dTo, mm
Where e- coefficient depending on the type of mechanism and operating mode, you
selected according to the regulatory data of the Gosgortekhnadzor Rules
(Table P4);
dTo- rope diameter, mm.
Block sizes are normalized.
The diameter of the block (drum) for welded uncalibrated chains is determined by the ratios:
for manually driven mechanisms Db³ 20 dts;
for machine driven mechanisms Db³ 30 dts;
Where dts- the diameter of the steel bar from which the chain is made.
The diameter of the initial circle of the sprocket for a welded calibrated chain (diameter along the axis of the rod from which the chain is made) is determined by the formula:
Dn. O. = t/ sin 90° /z, mm
Where t- internal length of the chain link (chain pitch), mm;
z- number of slots on the star, accepted z³ 6.
The diameter of the initial circle of the sprocket for a leaf chain is determined
are calculated according to the formula:
Dn. O. = t/ sin 180° /z, mm
Where t- chain pitch, mm;
z- number of sprocket teeth, taken z³ 6.
Rope drums are used with single-layer and multi-layer winding, with a smooth surface and with a screw thread on the surface of the shell, with one-sided and double-sided rope winding.
The diameter of the drum, as well as the diameter of the block, is determined according to the Rules of Gosgortekhnadzor:
Db³ (e – 1)dTo, mm.
The length of the drum for double-sided rope winding is determined by the formula:
and with one-sided winding:
https://pandia.ru/text/78/506/images/image005_7.png" width="124" height="32 src=">,
Where z– number of working turns of the rope;
https://pandia.ru/text/78/506/images/image007_5.png" width="18" height="23 src=">,
Where b– the distance between the axes of the streams of the outer blocks is taken according to table P8;
hmin– the distance between the axes of the drum and the axis of the blocks in the uppermost position;
The permissible angle of deviation of the rope branch running onto the drum from the vertical position is = 4...6°.
The wall thickness of the drums can be determined from the compressive strength condition:
https://pandia.ru/text/78/506/images/image009_4.png" width="48" height="29"> - permissible compressive stress, Pa, in calculations the following is taken:
80MPa for cast iron C4 15-32;
100MPa for steels 25L and 35L;
110MPa for steels St3 and St5.
For cast drums, the wall thickness can be determined using empirical formulas:
for cast iron drums https://pandia.ru/text/78/506/images/image010_1.png" width="26" height="25 src=">= 0.01 db+3 mm, and then check it for compression. Should be:
https://pandia.ru/text/78/506/images/image012_2.png" width="204" height="72"> mm
Where t=28 mm – internal length of the chain link (pitch);
z 6 – number of slots on the block (asterisk), we accept z=10.
Example 5. Using the data in Example 3, determine the diameter of the initial circle of the sprocket.
Sprocket starting circle diameter
mm,
Where t=40 mm – chain pitch;
z 6 – number of sprocket teeth, accept z=10.
Example 6. Determine the main dimensions of a cast iron drum according to example 1..png" width="156 height=44" height="44">, mm
Where dk= 16.5 mm – rope diameter;
e– coefficient depending on the type of mechanism and operating mode, for cranes with machine drive in light operating mode e=20 (Table P4)
db=(20-1)∙16.5=313.5 mm, we take the value of the drum diameter from the normal range db=320 mm (Table P8).
Determine the length of the drum. Drum with double-sided cutting. The working length of one half of the drum is determined by the formula:
mm
Where t– pitch of turns, for a drum with grooves
t=dк+(2…3)=16.5+(2…3)=(18.5…19.5) mm, accept t= 19 mm;
zo=1.5…2 – number of spare turns of rope, we accept zo=2 turns;
zр– number of working turns of the rope
https://pandia.ru/text/78/506/images/image019_0.png" width="210 height=36" height="36"> mm
Total drum length:
Lb=2(lp+l3)+lo, mm,
Where l3– the length of the drum required to secure the rope;
https://pandia.ru/text/78/506/images/image022_0.png" width="16" height="15">=4-6° - the permissible angle of deviation of the rope branch approaching the drum from the vertical position, we accept = 6°.
l0=200-2∙4/80∙tg6°=99.1 mm
we accept l0=100 mm.
Thus, the total length of the drum
lb=2(608+60)+100=1436 mm, accept
lb=1440 mm = 1.44 m
The thickness of the drum wall is determined by the formula:
https://pandia.ru/text/78/506/images/image024_0.png" width="47 height=19" height="19">mm.
The wall thickness of the cast drum must be at least 12 mm.
Practical work No. 2
Calculation of winches and lifting mechanisms of hoists with manual and electric drives according to specified conditions.
1. Calculation of manual winches
calculation sequence for a manual winch.
1) Select a load suspension scheme (without a chain hoist or with a chain hoist).
2) Select a rope according to the given load-carrying capacity.
3) Determine the main dimensions of the drum and blocks.
4) Determine the moment of resistance on the drum shaft from the weight of the load Ts and the moment on the handle shaft created by the force of the worker Tr.
Moment of resistance from the weight of the load
N∙ m,
Where Fmax- maximum working force in the rope branch, N; db– drum diameter, m.
Moment on the handle shaft:
N∙m,
Where RR– the effort of one worker, is accepted
RR=100…300 N
n– Number of workers;
https://pandia.ru/text/78/506/images/image001_21.png" width="15" height="17 src=">.png" width="80 height=48" height="48">
Where η – Winch efficiency.
6) Calculate open gears and shafts (the method of their calculation was studied in the “Machine Parts” section of the “Technical Mechanics” subject).
7) Determine the main dimensions of the handle. The diameter of the handle rod is determined from the bending strength condition:
m,
Where l1– length of the handle shaft, taken l1=200…250 mm for one worker and l1=400…500 mm for two workers;
https://pandia.ru/text/78/506/images/image029_1.png" width="29" height="23 src=">=(60...80) MPa=(60...80)∙106Pa.
The thickness of the handle in the dangerous section is calculated for the combined action of bending and torsion:
 | ||
The width of the handle is taken equal to
Where G- lifting capacity of the winch, kN;
Vр- the peripheral speed of the drive handle is usually taken
Vр=50...60 m/min.
Example 7. Calculate the lifting mechanism of a manual winch designed to lift a load weighing G= 15 kN per height N= 30m. Number of workers n=2. Winch efficiency h=0.8. The surface of the drum is smooth, the number of layers of rope winding on the drum m=2. Multiplicity of chain hoist u=2. Simple pulley ( A=1).
Initial data:
G=15kN - weight of the load being lifted;
N=10m - load lifting height;
n=2 - number of workers;
h=0.8 - winch efficiency;
m=2 - number of layers of rope winding on the drum;
the surface of the drum is smooth;
u=2 - pulley multiplicity;
A=1 - number of branches wound on the drum.
Rope selection.
Maximum working force in one rope branch:
Fmax= 15/2×0.99=7.6 kN,
Where z=u×a= 2 - the number of branches on which the load hangs;
Efficiency of the pulley according to the table. P3 for multiplicity chain hoist u=2 on rolling bearings 0.99.
Design breaking force:
Fp=nk× Fmax=5.5×7.6=41.8 kN,
Where nTo- safety factor of the rope, for a manually driven cargo winch nTo=5.5 (Table P1).
According to GOST 26.88-80 (Table P5), we select a rope of type LK-R 6x19 + 1 o. With. with breaking force Fp.m.= 45.45 kN at tensile strength 1764 MPa, rope diameter dTo=9.1 mm.
Actual safety factor of the rope:
nf =Fр.m. ·z hn/G = 45.45 2 0.99/15 = 6 > nTo = 5,5.
Determination of the main dimensions of the drum.
Minimum permissible drum diameter:
db ³ ( e– 1)dk, mm
Where e- coefficient depending on the type of mechanism and operating mode, for
manual cargo winches e=12 (Table P4);
dk- rope diameter, mm, then
db³ (12 – 1)9.1=100.1mm
We accept from the normal series db=160mm (Table P8).
The working length of the drum for multi-layer rope winding is determined by the formula:
Where t – pitch of turns, for a smooth drum ; t= dk=9.81 mm ;
Lk – rope length excluding spare turns
Lk=H∙u=30∙2=60m
Full length drum with single-sided winding
lb= lR+ lV+ lh,
Where lb=(1,5…2)∙ t– drum length required for spare turns ,
lb=(1,5…2)∙9,81=13,65…18,2 mm ,
we accept lb=18 mm
lh – drum length required to secure the rope
Ropes, chains, load-handling devices, load-handling devices and containers
What are ropes used on cranes for?
Ropes on lifting cranes serve to transmit traction forces from winches to the executive working bodies and set them in motion.
According to the “Rules for the device and safe operation load-lifting cranes", steel ropes used as cargo, boom, byte, load-bearing traction and slings must comply with the current state standards and have a certificate (certificate) or a copy of the certificate of the manufacturer of the ropes about their testing in accordance with GOST 3241-66. When receiving ropes without a certificate, they must be tested in accordance with the specified standard.
Ropes that do not have a test certificate are not allowed to be used.
What types of steel ropes are divided into according to the type of contact of the wires in the strands?
According to the type of contact of the wires in the strands, steel ropes are divided mainly into three types: ropes with point contact (TC), consisting of wires of the same diameter; ropes with linear contact (LT), consisting of wires of different diameters, and ropes with point and linear contact of wires in strands (TLT). Moreover, if the rope has wires in individual strands of the same diameter, then the letter O is added to the designations LK and TLK, for example LK-O, TLK-O. If individual strands consist of two wires of different diameters, then the letter P is added to the designation, for example LK-R, TLC-R. If individual strands consist of wires of different and identical diameters, then RO is added to the designation, for example LK-RO, TLK-RO.
To characterize steel ropes, including their basic data, it is accepted symbol, where in the first place the diameter of the rope is indicated, in the second - its purpose, in the third - mechanical properties wire, in the fourth place - working conditions, in the fifth - the direction of laying of the rope elements, in the sixth - the laying method, in the last place - the marking group for the temporary tensile strength of the wire. At the end, the GOST number in accordance with which the rope is made is indicated.
For example, a rope with a diameter of 24 mm, for cargo purposes (G) made of light wire (grade B), for light working conditions (LS), non-unwinding (N) with a marking group for tensile strength of 160 kg/cm2 is designated as follows: 24-G- V-LS-N-160 GOST 3077 - 69. How are steel ropes divided according to the direction of lay of the wires and strands in the rope?
According to the direction of lay of the wires and strands in the rope, steel ropes are divided into one-way lay ropes and cross-lay ropes.
If the wires in the strands and the strands in the rope are twisted in one direction, for example to the right or to the left, then such a rope is called a one-way lay rope.
If the wires in the strands are twisted in one direction, for example to the right, and the strands are twisted in the other direction, for example to the left, then such a rope is called a cross-lay rope. Although it has less flexibility than a one-way laid rope, it is less susceptible to unwinding and flattening when bending around blocks.
How is the lay pitch determined?
The laying pitch of a rope is determined as follows: a mark is applied to the surface of a strand, from which as many strands are counted along the central axis of the rope as there are in the rope section (usually six), and a second mark is placed on the next strand after counting. The distance between the marks will be the pitch of the lay.
What types of steel ropes are there?
Steel ropes come in different designs, but mainly 6X19+1 ropes are used; 6X37+1; 6X61 + 1. Moreover, these numbers indicate that all of the listed rope structures are six-strand, and in each strand in the first case there are 19 wires plus one core, in the second case there are 37 wires plus one core and in the third case there are 61 wires plus one core, which in all ropes it is located in the center of the rope, and the strands are wound around it. In order for the rope to be lubricated during operation, the core is impregnated with a special lubricant before being placed in the rope.
What type of ropes are used on cranes?
Ropes of the 6X19+1 design are recommended for use for braces and cables, i.e. in cases where they are not subject to repeated bending, ropes 6X37+1 are for pulleys of the load lifting mechanism, booms and as a traction rope, since they are more elastic than kanatbH 19+1.
What methods are used to secure the ends of the rope?
The following methods of rope end fastenings are mainly used on cranes: wedge clamp; filling the end of the rope with low-melting metal in a forged, stamped or cast conical sleeve; loops on clamps (fastening with clamps); loops using braid and clamping strips.
It is prohibited to use cast iron or steel welded bushings when securing the end of the rope with a wedge clamp or low-melting metal.
How is the end of the rope secured with a wedge clamp?
The end of the steel rope is secured with a wedge clamp as follows: the end of the rope is passed through the narrow side of the steel cone body so that the free end of the rope and the working branch come out of the narrow side of the cone hole, forming a loop behind the widened end of the body.
Next, a steel wedge is placed in the loop, which has grooves on the side surfaces for a better fit of the rope. After this, the rope with a wedge is pulled into the body, clamping the ends of the rope between internal surfaces tapered hole and wedge.
It should be remembered that the free end of the rope with such fastening must be extended beyond the edge of the conical hole to a length equal to 10-12 rope diameters.
How is the end of a rope secured by filling it with low-melting metal?
Fastening the end of a steel rope by pouring low-melting metal is done as follows: the end of the rope is passed through the narrow side of the steel conical body behind the wide side. Then this end is unraveled into separate wires, the hemp core is cut out, the wires are etched and inner side taper bushing hydrochloric acid and tighten the unbraided end into the sleeve. After this, the resulting brush of steel wires inside the conical sleeve is filled with solder or other low-melting metal.
How many clamps should be installed when securing a rope using clamps?
The number of clamps when fastening a rope using clamps is determined during design, but must be at least three.
The spacing of the clamps (the distance between the clamps) and the length of the free end of the rope from the last clamp must be at least six rope diameters.
All clamp nuts must be located on the side of the working branch of the loop, and the tightness of the two ends of the rope is considered normal if the diameter of the rope after tightening the nuts is 0.6 of the original diameter.
Should the hinge and its fastening be checked after the clamp nuts are tightened?
It should. The rope is kept under load, and then the clamp nuts are tightened again to the specified limit. To prevent the free end of the rope from touching anything during operation, it is wrapped with soft wire.
Should thimbles be installed when securing the end of a rope with clamps?
When securing the end of a steel rope, either using clamps or braiding, a thimble must be placed in the loop, as it protects the rope from sharp bending and premature wear.
How many punctures of the rope with each strand should there be when braiding the end of the rope?
The number of punctures of the rope with each strand during braiding must be at least 4 - with a rope diameter of up to 15 mm, at least 5 - with a rope diameter of 15 to 28 mm, and at least 6 - with a rope diameter of 28 to 60 mm. When braiding the end of a rope, the end is unraveled into strands, the hemp core is cut out and
The unbraided part is tightly placed on the card groove of the thimble. Then the unbraided strands are woven into the working branch of the rope, piercing it special tool. The last puncture is allowed to be made with half the number of strands of rope, and the braid should fit tightly to the end.
How is the rope attached to the rope drum?
The fastening of the rope to the rope drum must be reliable, allowing for the possibility of its replacement. If clamping bars are used, their number must be at least two. The length of the free end of the rope from the last clamp on the drum must be at least twice the diameter of the rope. It is not permitted to bend the free end of the rope under or near the clamping bar.
Should a rope be checked for strength before placing it on a crane?
When the total breaking force is given in the certificate or test certificate of a rope, the value P is determined by multiplying the total breaking force by 0.83 or by the coefficient determined according to GOST for the rope of the selected design.
What is the safety factor of a rope?
The safety factor of a rope is the ratio of the breaking force of the rope as a whole to the greatest working load.
What is the safety factor of steel ropes installed on cranes?
The lowest permissible safety factors for steel ropes installed on cranes are given in the table.
To reduce wear on the ropes of jib, gantry and bridge cranes, they are lubricated with rope ointment heated to approximately 60 °C every month of operation.
Before lubrication, the rope is carefully checked and dirt and old grease are removed from its surface with a rag soaked in kerosene. It is prohibited to clean dirt from the surface of the rope with a metal brush, since this removes the galvanization from the surface of the wires, and this leads to rusting of the rope.
In what cases are steel ropes rejected?
Steel ropes are rejected in the following cases: if even one strand is broken; if the number of broken wires at the laying step is more than normal (see table on page 244); if the surface wear or corrosion of the rope wires is 40% or more; if kinks have formed on the rope; if the rope is severely deformed (flattened).
Is the rejection rate for the number of rope wires reduced if they have surface wear or corrosion?
Decreases, since in this case the strength of the rope decreases. Moreover, when the diameter of the wires decreases as a result of surface wear or corrosion by 10, 15, 20, 25 and 30%, the number of breaks per lay step should be reduced by 15, 25, 30, 40 and 50%, respectively.
If the diameter of the wires decreases by 40% or more, the rope is rejected.
How is surface wear or corrosion of the rope (wires) determined?
Surface wear or corrosion of rope wires is determined as follows. In the area of greatest wear or corrosion of the rope pitch, bend the end of the broken wire, clean it of dirt and rust, and measure the diameter with a micrometer or other instrument that provides sufficient accuracy. If, for example, the initial diameter of the wires was 1 mm, and the measurement showed 0.5 mm, then wear or corrosion in this case will be 50%. Such a rope is certainly rejected.
What should you pay special attention to when using ropes?
Since the ropes of jib, gantry and overhead cranes are particularly important parts, they should be constantly monitored and properly maintained in a timely manner. There are often cases when, due to lack of supervision, timely proper care and untimely replacement of worn ropes, large accidents occurred.
That's why:
Under no circumstances should worn or defective ropes be used;
it is necessary to systematically carefully check and tighten the fastening of the ends of the rope on the rope drum and in other places where ropes are embedded;
do not allow the number of turns of rope on the drum to be less than 1.5;
lubricate the rope in a timely manner, since its service life largely depends on timely and correct lubrication;
do not allow blocks with chipped flanges to be used, since a chipped flange causes the rope to come off the block or drum, and sometimes cuts the rope;
if broken wires are found in an amount less than that at which the rope is rejected, they should be cut off with pliers to avoid damage to adjacent wires;
Do not allow the rope to touch the crane structural elements.
What chains are used on lifting machines?
On lifting machines, plate chains are used - GOST 191-63, welded and stamped - GOST 2319-70. The latter are used as cargo slings and slings.
In addition to the indicated chains, chains in accordance with GOST 6348-65 can be used for the manufacture of slings. All chains used on cranes, as well as the chains from which slings are made, must have a manufacturer’s test certificate. If there is no test certificate, a sample of the chain must be tested to determine the breaking load and check the dimensions for compliance with the State Standard.
What is the safety factor of chains in relation to the breaking load?
The safety factor of welded and stamped load chains and sling chains in relation to the breaking load should not be less than:
cargo, operating on a smooth drum with manual drive - 3, with machine drive - 6;
cargo operating on a sprocket (calibrated) with manual drive - 3, with machine drive - 8;
for slings with manual drive - 5, with machine drive - 5.
The safety factor of plate chains used in lifting machines must be at least 5 with a machine drive, and at least 3 with a manual drive.
Are chain splices allowed?
Splicing of chains is permitted by forging, or electric welding of new inserted links, or using special connecting links. After splicing, the chain must be inspected and tested with a load equal to 1.25 times its load capacity. Inspection and testing must be carried out at the facility where the chains were repaired.
In what cases are chains rejected?
Chains are rejected if a link is broken, if the wear of a welded or stamped chain link is more than 10% of the original diameter (caliber) plus minus tolerance for the manufacture of the chain, if cracks are found in the chain links.
How are the blocks used on cranes divided?
Blocks used on load-lifting cranes are divided into working and leveling.
Working blocks, in turn, are divided into movable and stationary. If the block does not rise or fall relative to the ground level during operation of the crane, then such a block is called stationary, although it rotates on its axis. If, when lifting or lowering a load, the block moves with it, then such a block is called movable.
Both mobile and fixed blocks made of cast iron and steel. Moreover, blocks made of cast iron are used to work with light loads, and blocks of steel are used to work with large and heavy loads.
Which blocks are subject to the most wear?
High-speed blocks are subject to the greatest wear. To ensure uniform wear of the blocks, in multi-block pulley hoists they should be swapped when repairing a crane.
How can uneven block wear be eliminated?
Uneven wear of the block can be eliminated by turning the groove profile, and a reduction in the original diameter is allowed by no more than 3 mm for blocks with a diameter of 300 mm and no more than 5 mm for blocks with a diameter of up to 500 mm.
Is it possible to operate a block with a broken flange?
It is strictly prohibited to operate a block with a chipped flange, since a chipped flange causes the rope to come off the block, and sometimes can cut the rope, which can lead to a serious accident.
It should be remembered that crane blocks must be constantly monitored, since block failure can lead to an accident.
Leveling block, leveling the ropes of the left and right sides pulley, does not rotate when the mechanism is operating, and sometimes they do not pay attention to it - they do not lubricate its axle, they do not inspect the fastening of the axle. The crane operator needs to remember that a break in the axis of the leveling block or its falling out of the supports will lead to a serious accident - the load with the hook will fall to the ground.
What is a chain hoist?
A lifting device consisting of fixed and movable block clips, through the blocks of which a rope or chain is passed, is called a chain hoist. Moreover, the more blocks in the movable and stationary cages of the pulley, the more branches of the rope or chain, and therefore, the greater the gain in strength or speed.
Why is there a gain in strength in pulley hoists?
The gain in strength in pulley hoists occurs because the mass of the load lifted by the pulley hoist is distributed among all branches of its rope. Therefore, the more blocks in the chain hoist, the large quantity branches of the rope are involved in lifting the load and the less force falls on each branch of the rope. Thanks to this, it is possible to use a rope of smaller diameter, and a lifting or boom winch with less traction force.
What multiplicity pulleys are used on cranes?
On lifting cranes, pulley blocks with a multiplicity of 2, 3, 4, 6, etc. are used. A pulley with a multiple of 2 consists of one fixed block and one movable one. In this case, the cargo rope attached to the boom first goes around the movable block located on the hook holder, and then the stationary one and is directed to the winch drum.
A pulley with a multiplicity of 3 consists of two fixed blocks mounted on the boom and one movable block placed in the hook cage. A pulley with a multiplicity of 4 consists of two movable and two fixed blocks.
The multiplicity of a pulley is its most important characteristic, since the greater the multiplicity, the less effort must be expended to lift the load.
What applies to replaceable load-handling devices?
Replaceable lifting elements include a hook, grab, lifting electromagnet, etc.
How are the hooks of lifting machines made?
Hooks of lifting machines - forged and stamped - must be manufactured in accordance with GOST 2105-64.
After manufacturing, they must be marked in accordance with GOST 2105-64.
Hooks for loads exceeding 3 tons must be made rotating on closed ball bearings, with the exception of hooks for special-purpose cranes.
What should crane hooks be equipped with?
The hooks of lifting cranes must be equipped with a safety device that prevents the removable load-handling device from spontaneously falling out of the hook mouth.
Rice. 3. Single block hook cage:
1 - locking trunks; 2 - casing; 3 - cheek; 4 and 8 - ball bearings; 5 - axis; 6 - block; 7 - hook nut; 9 - traverse; /0 - hook; 11 - hook latch
Such a device may not be equipped with the hooks of portal cranes operating in seaports, and1 hooks of cranes transporting liquid slag or! molten metal.
Is the hook allowed to wear out?
Wear of the hook is allowed, but very minor. The maximum wear in the throat should not exceed 10% of the original height of its section.
In what cases is a hook rejected?
The hook is rejected in the following cases: if it does not rotate in the traverse; if the hook horn is bent;
if the wear of the hook in the throat exceeds 10% of the original section height;
if there is no OTK mark on the hook; if there are cracks on the hook.
What parts does the hook cage consist of?
The hook cage (Fig. 3) consists of two side cheeks made of grade 3 steel, a stop, blocks, a traverse and a hook. The cheeks are connected to each other by spacer tubes and tightened with coupling bolts. The cage blocks are installed on an axis, which is fixedly fixed in the side cheeks using crossbars. The hook traverse is also installed in the side cheeks and is secured against axial movement by two locking bars; Since the traverse pins have grooves in a circle, the traverse can freely rotate in the holes of the side cheeks, due to which the hook, in addition to rotating around the axis of the shank, can also swing along with the traverse, which greatly facilitates slinging loads.
What is the purpose of the hook cage stop?
The stop of the hook cage serves to protect the cage block from possible impact in cases where the hook approaches the uppermost position.
What should maintenance personnel pay attention to when operating hooks and hook cages?
The hook cage of jib, gantry and bridge cranes is a very important unit, so crane operators and slingers must constantly monitor the condition of the hook cage when operating the crane. During each inspection, it is necessary to check the serviceability of the side cheeks, blocks, traverse, hook, nut securing the hook, fastening of the axles and stop. During operation of the crane, defects may appear in the hook: bending of the hook horn, nicks on the body of the hook, wear or contamination of the support bearing, breakage of the locking nut of the hook, abrasion of the surface of the hook mouth, cracks, which can lead to serious consequences. The crane operator and slinger must notice each of these defects in time. The crane operator must also ensure that the hook cage blocks and the hook thrust bearing are lubricated, as lack of lubrication will cause these parts to fail prematurely. What are the requirements for grabs?
The following requirements apply to grabs:
the grab must have a plate indicating the manufacturer, the grab number, its own weight, the type of material for which the grab is intended for handling, the maximum permissible weight of the scooped material; in the absence of a nameplate, the latter must be restored by the owner of the grab;
by its design, the grab must exclude the possibility of spontaneous opening;
grabs manufactured separately from the crane must have (in addition to the plate) a passport, which must contain all the data about the grab provided for in the standard crane passport.
The crane operator must remember that a load-lifting crane, in which the load-handling element is a grab, can be allowed to work only after weighing the scooped material during a trial scooping; the weight of the grab with the scooped material should not exceed the lifting capacity of the crane.
For cranes with variable lifting capacity depending on the boom reach, the weight of the grab should not exceed the lifting capacity corresponding to the reach at which the crane and grab are operated. Test scooping should be done from the horizontal surface of freshly filled soil.
Removable lifting devices and containers
What devices are classified as removable lifting devices?
Removable lifting devices include those devices that are hung on the hook of a lifting machine (for example, slings, pliers, traverses, etc.).
What types of slings are there?
Slings can be universal, lightweight or multi-branch. A sling that has the shape of a closed loop is called universal, since it is used for slinging various loads.
A sling consisting of one branch with hooks and rings attached to the ends is called lightweight (Fig. 4).
Rice. 4. Slings: a - universal; b - lightweight - valuable
Rice. 5. Multi-branch sling
A multi-branch sling is a sling that consists of several branches assembled on a ring, with hooks or grips at the ends (Fig. 5).
How are hooks, rings and loops attached to the ends of slings?
Hooks, rings and loops at the ends of the slings are secured using a thimble, by braiding the free end of the sling or by installing clamps. When braiding, the end of the sling (rope) is unraveled into strands, then these strands are woven into the body of the rope, followed by braiding the joints with wire.
How many strands of rope should be punched when braiding?
The number of punches of the sling rope with strands when braiding must be at least four for a rope diameter of up to 15 mm, at least five for a rope diameter of 15 to 28 mm, and at least six for a rope diameter of 28 to 60 mm.
How many clamps should be placed on the end of the sling rope?
When securing hooks, rings and loops at the end of a sling rope by installing clamps, their number is determined during design, but must be at least three; the spacing of the clamps and the length of the free end of the rope from the last clamp must be equal to at least six rope diameters. It is prohibited to place clamps on slings using a forge or any other hot method.
What material are hooks and rings for lightweight and multi-leg slings made of?
Hooks and rings for slings must be made of steel grade 20 or from mild open-hearth steel grade 3, and the hooks must have devices that prevent the hook from spontaneously falling out of the mounting loops or from the container hangers.
Who has the right to manufacture slings, pliers and traverses?
Slings, pincers, cross beams and other load-handling devices have the right to be manufactured by an enterprise or construction site, but their production must be organized centrally and made according to standards, technological maps or individual drawings. In addition, when welding is used, the documentation for the manufacture of slings, clamps, traverses, etc. must contain instructions for its implementation and quality control.
Information about the manufacture of slings, pincers, traverses, etc. must be entered in the logbook. This log must indicate: the name of the removable lifting devices, the load capacity, the normal number ( technological map, drawing), certificate numbers for the material used, welding quality check results, test results of the removable load-handling device. Are slings, pliers and traverses subject to technical inspection after their manufacture?
After manufacturing, slings, pliers, traverses and other load-handling devices must necessarily undergo technical inspection at the enterprise or construction site where they were manufactured; however, they must be inspected and tested with a load 1.25 times their rated load capacity.
After testing, the specified removable lifting devices must be equipped with a metal tag or stamp on which the number, load capacity and test date must be stamped. Moreover, the load capacity of general purpose slings is indicated at an angle between the branches of 90°, and the load capacity of slings intended purpose, intended for lifting a certain load, is indicated at the angle between the branches adopted in the calculation. Slings, pliers, traverses and other removable load-handling devices manufactured for third-party organizations, in addition to stamps or tags, must be supplied with a passport.
Who should carry out technical inspection of slings, tongs, traverses and containers?
Technical inspection of slings, tongs, traverses and containers must be carried out by a supervisor or another person specially appointed by order for the enterprise or construction site.
Should slings, pliers and traverses be periodically checked during their operation?
Slings, pliers and cross-arms during their operation must be periodically checked through a thorough inspection within the time limits established by the administration of the enterprise or construction site, but not less than: slings - every ten days, pliers - after one month, cross-arms - after six months.
The inspection must be carried out by a person responsible for the good condition of removable load-handling devices; the results of the inspection must be recorded in the inspection log.
Should slings, pliers and crossbars be checked daily (every shift)?
Slings, pliers and traverses must be checked daily (every shift) before starting work. They should be checked by slingers, crane operators and persons responsible for the safe movement of goods.
At what maximum angles between the branches of the slings is it allowed to moor the cargo?
The maximum angle between the branches of the slings when mooring the cargo should be no more than 90°. An increase in this angle to 120° can only be allowed in exceptional cases according to calculation.
Why should we not allow an angle between the branches of the slings to exceed 90° when lifting a load?
Because with an increase in the angle between the branches of the slings, the tension on the branches will greatly increase, which can lead to the rupture of the slings themselves, hooks or mounting loops of reinforced concrete or concrete products. So, at an angle between the sling branches equal to 60°, the tension on the sling branches will increase by 15%, at an angle of 90° the tension will increase by 42%, and at an angle of 120° the tension on the sling branches will increase by 2 times.
In what cases are slings rejected?
Slings are rejected in the following cases: if the number of broken wires per lay pitch in the sling ropes is greater than the norm (see table on page 244), if the hooks of the slings have cracks, if the throat of the sling hook has wear of more than 10% of the original height of its section, if the rope the sling has a torn strand, if the sling rope has surface wear or corrosion of 40% or more, if the thimbles have fallen out, if the sling rings have cracks or wear is more than acceptable, if the sling rope is severely deformed (flattened).
Who has the right to manufacture containers?
An enterprise or construction site has the right to produce containers, but it must be manufactured centrally and produced according to standards, technological maps and individual drawings.
After manufacturing, the container must be subjected to technical certification by inspection, since testing the container with a load is not necessary. Inspection of containers must be carried out according to instructions approved by the management of the enterprise or construction site, which defines the procedure and methods of inspection, as well as elimination of detected defects.
Information about the manufacture and inspection of containers must be entered in the logbook for recording removable load-handling devices and containers. This journal must indicate: the name of the container, the dead weight of the container, its carrying capacity, the purpose of the container, the normal number (technological map, drawing), certificate numbers for the material used, the results of welding quality checks, the results of inspection of the container.
What information should be placed on the container after its technical examination?
After technical examination, the following information must be marked on the container: container number, tare weight of the container, the largest weight of the cargo for which it is intended to be transported, and the purpose of the container.
Should containers be inspected periodically?
The containers must be inspected periodically (monthly) and the inspection results must be recorded in the inspection log of lifting devices and containers. The container must be inspected by a person responsible for the proper condition of the container. In addition, the containers must be inspected daily (every shift) by slingers, crane operators and the person responsible for the safe operation of cranes.
In what cases is the container rejected?
Crane operators and slingers must remember that removable lifting devices and containers that have not passed technical inspection, do not have tags (stamps) and are faulty are not allowed to work and they should not be located in the work areas.
TO Category: - Crane operators and slingers
Slings made from plant and synthetic fibers must be made with a safety factor of at least 8.
ATTENTION! Despite the fact that the slings are designed with a safety margin, it is unacceptable to exceed the lifting capacity of the sling indicated on the tag.
What determines the tension of the sling branches? At what angle between the branches are the slings designed?
The tension S of the branch of a single-leg sling is equal to the mass of the load Q (Fig. 3.13). tension S in each branch of a multi-branch sling is calculated using the formula
S= Q/(n cos b),
Where P- number of sling branches; cos b- cosine of the angle of inclination of the sling branch to the vertical.
Of course, the slinger should not determine the loads in the branches of the sling, but he must understand that As the angle between the branches increases, the tension of the sling branches increases. In Fig. Figure 3.14 shows the dependence of the tension of the branches of a two-legged sling on the angle between them. Remember, when you carry buckets of water, the load increases as you extend your arms. The tensile force in each branch of a two-leg sling will exceed the mass of the load if the angle between the branches exceeds 120°.
Obviously, with an increase in the angle between the branches, not only the tension of the branches and the probability of their rupture increases, but also the compressive component of tension 5 SG (see Fig. 3.13), which can lead to destruction of the load.
ATTENTION! Branch rope and chain slings are designed so that the angles between the branches do not exceed 90°. The design angle for textile slings is 120°.
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What are the traverses used for? What traverse designs are used for slinging loads?
Traverses are removable load-handling devices designed for slinging long and large-sized cargo. They protect the loads being lifted from the compressive forces that arise when using slings.
According to their design, traverses are divided into planar and spatial.
Planar traverses (Fig. 3.15, A) used for slinging long loads. The main part of the traverse is the beam 2, or a truss that carries bending loads. Rope or chain branches are suspended from the beam 1.
Traverses with the ability to move clips 4 along the beam is called universal (Fig. 3.15, b). Equalizing blocks 5 are installed in the cages, which ensure uniform distribution of loads between the branches of the traverse S 1 = S 2. For this reason, such a traverse is called balancing Leveling blocks can also be used in structures rope slings with more than three branches.
Spatial traverses (Fig. 3.15, V) used for slinging three-dimensional structures, machines, and equipment.
I have different arms of the balancer traverse (Fig. 3.15, G) used for lifting loads with two cranes, it allows you to distribute loads between the cranes in proportion to their lifting capacities.
Signs of defective traverse:
Ø absence of stamp 3 or tag;
Ø cracks (usually occur in welds);
Ø deformation of beams, struts, frames with a deflection of more than 2 mm per 1 m of length;
Ø damage to fastening and connecting links.
What types of grips are there?
Grips are the most advanced and safe load-handling devices, the main advantage of which is the reduction manual labor. Grippers are used in cases where it is necessary to move loads of the same type. Due to the wide variety of cargo being moved, there are many various designs grabs. Most of them can be classified into one of the following types.
Tick-borne grips (Fig. 3.16, A) hold the load with levers 1 for its protruding parts.
Friction grippers hold the load due to frictional forces. Lever friction grips (Fig. 3.16, 6) clamp the load using levers 1. Lever-rope friction grips (Fig. 3.16, V) have ropes 3 with blocks, they are used for slinging bales, bales.
IN eccentric grips (Fig. 3.16, G) the main part is the eccentric 4, which, when turning, reliably clamps sheet materials.
There are also load-handling devices that provide automatic (without the participation of a slinger) slinging of the load.
