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Ministry of Agriculture of the Russian Federation
FGOUVPO
Perm State Agricultural
Academy named after D.N. Pryanishnikov
Department of Machine Parts
COURSE WORK
discipline: "Fundamentals of design"
on the topic: "Calculation of an overhead single-girder crane"
Completed by: student of group
M-51, code M-01-157
V.M. Solovyov
Checked:
Candidate of Technical Sciences
Associate Professor V.S.
Novoseltsev
Perm 2005
2
Exercise.
Calculate the movement mechanism of an overhead single-girder crane (crane-beam):

carrying capacity Q=1.7 t;

crane span LK= 10.6 m;

movement speed V = 0.48 m/s;

lifting height H = 12 m;

mode of operation is average;

floor control.
Crane works in a workshop for the repair of agricultural machinery.
Overhead single girder cranes with a lifting capacity of 1...5t are regulated by GOST
2045 - 89*.
In accordance with the prototype, we select the kinematic scheme of a single-beam
overhead crane (beam crane) with a central drive and a mobile electric hoist (Fig. 1).
According to GOST 22584 - 96, according to the load capacity of 1 t, we choose the TE
100-521 electric hoist [1, p. 215].
Figure 1. Overhead single girder crane.
The calculation of the crane movement mechanism is carried out in the
3
following order.
one. We determine the dimensions of the running wheels according to the formula
𝐷𝐾 ≈ 1,7√𝑅max
(one)
We calculate the maximum load on the wheel at one of the extreme positions of
the electric hoist.
According to GOST 22584-96 [1, p. 215], we take the weight of the hoist mt =
180 kg = 0.18 t (its weight G7 = mTg ≈ 0.18 × 10 = 1.8 kN) and length L = 870 mm.
We choose the mass of the crane with the electric hoist approximately according to the
prototype [1, p. 214] mk ≈ 2.15 tons. Then the weight of the crane Gк = mкg ≈ 2.15 ×
10 = 21.5 kN. Approximately accept
l≈ L ≈ 0.87 m.
To determine the load Rmax, we use the equation of statics
∑M2 = 0 or – Rmax Lk+ (GG+ GT)×(Lк –l) + (Gк – GT) × 0.5Lk=0(2)
where
𝛫 −𝐺Т )×0.5𝐿𝛫
Rmax=(𝐺+𝐺Т)×(𝐿𝛫−𝑙)+(𝐺
=(3)
𝐿𝛫
(17+1,8)×(10,6−0,87)+(21,5−1,8)×0,5×10,6
10,6
≈ 27 kN
With the total number of traveling wheels Zk = 4, the load falls on those
two crane wheels, near which the trolley is located. Then
Rmax = R/2 = 27/2 = 13.5kN = 13500 N.(4)
Consequently,
𝐷𝛫 ≈ 1,7√𝑅max ≈ 1,7√13500 ≈ 197мм
According to GOST 3569 - 74 [1, p. 252], we select a double-flange crane wheel with
a diameter of Dк = 200 mm. Pin diameter dц = Dк/(4...6) ≈ (50...35) mm. We accept dc =
50 mm.
For the manufacture of wheels we use steel 45, the heat treatment method is
4
normalization (HB ≈ 200). The wheel has a cylindrical working surface and rolls on a
flat rail. With Dк ≤ 200 mm, we accept a flat rail of rectangular section [1, p. 252],
choosing the size a according to the mustcondition: a < B. With DK ≤ 200 mm, the
width of the rolling surface B = 50 mm. We accept a = 40 mm.
The working surface of the contact b = a - 2R = 40 - 2 × 9 = 22 mm.
Speed influence coefficient Kv=1 +0.2 V = 1 + 0.2 ×0.4 8= 1.096.
For steel wheels, the proportionality factor a1 = 190.
We check the pre-selected running wheels for contact stresses.
With line contact
σk.l = al√2𝐾𝐷𝑣𝛫𝑅𝑏max = 190√2×1,096×13500
= 493 MPa(5)
200×22
Since the allowable contact stresses for a steel normalized wheel [σcl] =450...500
MPa, the strength condition is met.
2. We determine the static resistance to the movement of the crane.
Since the crane operates indoors, we do not take into account the resistance from
the wind load Wv, i.e.
WУ = Wtr + W uk
(6)
Friction resistance in running gearcrane parts:
𝑊ΤΡ =
𝐺+𝐺𝛫
𝐷𝛫
× (2𝜇 + fdц ) × Кр
(7)
According to table 1.3 [1, p. 9] we accept, μ = 0.3 mm, and according to table 1.4
for wheels on rolling bearings ƒ=0.015, Kp= 1.5. Then,
𝑊ΤΡ =
17+21,5
200
× (2 × 0,3 + 50 × 0,015) × 1,5 = 0,390 кН = 390 Н
5
Resistance to movement from a possible slope of the track.
Wyк = (G+ Gк)×α =(17 + 21.5)×0.0015 =0.058 kN = 58 N.(8)
The values of the calculated slope and the decreeus on p. 9. Thus, we get
𝑊у = 390 + 58 = 448𝛨
The force of inertia during the translational movement of the crane
F and\u003d (Q + mk) v / tp \u003d (1700 + 2150) x0.48 / 5 \u003d 370 N, (9)
where tp is the start time; Q andmto - weights of cargo and crane, respectively, kg.
The force required to move the crane during the start-up (acceleration) period,
𝑊П = 𝑊у + (1,1...1,3) × 𝐹и = 448 + 1,2 × 370 = 892𝛨
(ten)
3. We select the electric motor according to the required power
𝛲п.ср =
𝛲п
𝜓п.ср
=
𝑊П 𝑉
ηψп.ср
=
892×0,48
0,85×1,65
= 305 Вт = 0,305 кВт
(eleven)
We preliminarily take η = 0.85and ψp.av.= 1.65 (for asynchronous motors with
increased slip) [1, p. 49].
According to table 27 of the appendix [1], we select an asynchronous AC motor with
increased slip 4AC71A6UZ with the following parameters: rated power Pt = 0.4 kW; rated
speed
ndv = 920 min-1; swing moment of the rotor (mD2) р = 0.00068 kg×m2; Tp/Tn = 2;
Tmax/Tн= 2. Shaft diameter d= 19 mm.
Rated torque on the motor shaft
Тн =
Рдв
𝜔дв
=
30⋅Рдв
𝜋⋅пдв
=
30⋅0,4⋅103
3,14⋅920
= 4,16 Н ⋅ м
(12)
Static moment
Тс =
Рс
𝜔дв
=
30⋅𝑊У ⋅𝑉
𝜋⋅𝑛дв
=
30⋅448⋅0,48
3,14⋅920
= 2,23 Н ⋅ м
(13)
four.We select a clutch with a brake pulley for installing a brake. In the
selected scheme of the movement mechanism (see Fig. 1), a clutch with a brake a
6
pulley is installed between the gearbox and the electric motor. According to table 56 of
the appendix, we select an elastic pin-sleeve coupling with the largest bore diameter for
the shaft of 22 mm and the largest transmitted moment [Tm] = 32 N×m.
We check the selection condition [Tm] ≥ Tm. For the coupling Tm = 2.1 × Tn = 2.1 × 4.16 =
8.5 N × m. The moment of inertia of the clutch brake pulley It = 0.008 kg-m2. Flywheel moment
(mD2) T = 4 × It = 0.032 kg-m2.
5. We check the selected engine according to the start conditions. Start time
𝑡п =
(mD2 )0 ⋅𝑛дв
38⋅ТН
0,489⋅920
⋅ 𝑡п.о. =
38⋅4,16
(fourteen)
⋅ 1 = 2,85 𝑐
Total flywheel moment
(mD2 )𝑂 = 1,2 [(mD2 )𝑃 + (mD2 ) 𝑇 ] + (mD2 )𝐾 = 1,2 [(mD2 )𝑃 + (mD2 ) 𝑇 ] +
(0,00068 + 0,032) ⋅ 1,2 +
365⋅(1700+2150)⋅0,482
9202 ⋅0,85
365⋅(𝑄+𝑚𝐾 )⋅𝑉 2
2 ⋅𝜂
𝑛дв
=
= 0,489 кг ⋅ м2
(fifteen)
The relative start time is taken according to the schedule (see Fig. 2.23, b)
depending on the coefficient α = Tc / Tn. Since α \u003d 2.23 / 4.16 \u003d 0.54, then
tp.o \u003d 1.
The acceleration during the start-up period is determined by the formula:
an = v/tn =0.48/2.85=0.168 m/s2, which satisfies the condition.
6. We check the margin of adhesion of the drive wheels with the rails according
to the condition
starting at maximum torquedumbbell without load
Ксц =
𝑅пр ⋅𝜙сц
а′ 𝑧пр ⋅𝑓⋅𝑑ц
𝑊у +𝐺к ⋅( п−
)
𝑔
≥ 1,2
(16)
𝑧к ⋅𝐷к
Static resistance to crane movement in steady state without load
𝑊′у = 𝑊′тр + 𝑊′ук =
=
𝐺к
𝐷к
⋅ (2 ⋅ 𝜇 + 𝑓 ⋅ 𝑑ц ) ⋅ Кр + 𝐺к ⋅ 𝛼 =
(17)
21, 5
⋅ (2 ⋅ 0,3 + 0,015 ⋅ 50) ⋅ 1,5 + 21, 5 ⋅ 0,0015 = 0,2499 кН = 249 , 9 Н
200
Acceleration at start without load
(eighteen)
а′п = 𝑉/𝑡′п
Start time without load
𝑡′п =
(mD2 ) ′0 ⋅𝑛дв
38⋅ТН
⋅ 𝑡′п.о.
(19)
The total swing moment of the crane, atled to the motor shaft without taking into
7
account the load,
(mD2 ) ′𝑂 = 1,2 [(mD2 )𝑃 + (mD2 ) 𝑇 ] +
(0,00068 + 0,032) ⋅ 1,2 +
365⋅2150⋅0,482
9202 ⋅0,85
365⋅𝑚к ⋅𝑉 2
2 ⋅𝜂
𝑛дв
=
= 0,291 кг ⋅ м2
(twenty)
Moment of resistance reduced to the motor shaft during steady motion of the
crane without load
Т′с =
𝑊′у ⋅𝐷к
2⋅𝑢⋅𝜂
=
249,9⋅0,2
2⋅18⋅0,85
= 1,633 Н ⋅ м
(21)
According to the graph in Figure 2.23 [1, p. 29] with α = Тс'/Тн = 1.633/4.16 =
0.393 we get tp.o.= 1
Then start time
𝑡′п =
(mD2 ) ′0 ⋅𝑛дв
38⋅ТН
⋅ 𝑡′п.о. =
0,291⋅920
38⋅4,16
⋅ 1 = 1,69 с
(22)
Start acceleration
а′п = 0,48/1,69 = 0,284 м/с2
The total load on the drivewheels without load
𝑅′пр = К ⋅
𝐺к
𝑧к
⋅ 𝑧пр = 1,1 ⋅
21500
4
⋅ 2 = 11825 Н
(23)
Traction adhesion coefficientwheels with a rail for indoor cranes, φsc = 0.15.
Clutch stock
Ксц =
𝑅пр ⋅ 𝜙сц
11825 ⋅ 0,15
=
= 2,13
0,284 2 ⋅ 0,015 ⋅ 50
𝑧пр ⋅ 𝑓 ⋅ 𝑑ц
а′
𝑊′у + 𝐺к ⋅ ( п −
) 249, 9 + 21500 ⋅ ( 9,81 − 4 ⋅ 200 )
𝑔
𝑧к ⋅ 𝐷к
which is more than the minimum allowable value of 1.2.
Therefore, the clutch marginsecured.
7.We select the gearbox according to the gear ratio and the maximum torque on
the low-speed shaft Trmax. determined by the maximum torque on the motor shaft:
Тдвmax = 𝑇н ⋅ 𝛹п max = 4,16 ⋅ 2,1 = 8,7 Н ⋅ м
Тр max = 𝑇дв max ⋅ 𝑢 ⋅ 𝜂 = 8,7 ⋅ 18 ⋅ 0,85 = 133 Н ⋅ м
(24)
According to the mechanism diagrammovement of the crane (see Fig. 1), we select
a horizontal cylindrical gearbox of the Ts2U type. At a speed of rotation n = 1000 min1 and an average operating mode, the nearest value of the torque on the low-speed shaft
is Tmax = 0.25 kN m = 250 N m, which is more than the calculated Tr max. Gear
8
ratio up = 18.
Standard size of the selected reducer Ts2U-100.
eight. We select the brake according to the condition [Тт] > Тт and install
iton the motor shaft.
Calculated braking moment when moving the crane without load
ТТ = (𝑊′ук − 𝑊′тр min ) ⋅
𝐷к ⋅𝜂
2⋅𝑢
+
𝑛дв ⋅(mD2 )′о.т
38⋅𝑡𝑇
(25)
Rolling resistance
𝑊′ук = 𝐺к ⋅ 𝛼 = 21, 5 ⋅ 0,0015 = 0,0323 кН = 32, 3 Н
(26)
9
Resistance from friction forces in the coursecrane outlets
𝑊′тр min =
𝐺к
𝐷к
⋅ (2 ⋅ 𝜇 + 𝑓 ⋅ 𝑑ц ) =
21,5
200
(27)
⋅ (2 ⋅ 0,3 + 0,015 ⋅ 50) = 0,145 кН = 145 Н
Total swing moment
(mD2 ) ′OТ = 1,2 [(mD2 )𝑃 + (mD2 ) 𝑇 ] +
(0,00068 + 0,032) ⋅ 1,2 +
2
365⋅2150⋅0,48
9202
365⋅𝑚к ⋅𝑉 2
2
𝑛дв
⋅𝜂 =
⋅ 0,85 = 0,221 кг ⋅ м2
(28)
Deceleration time:
𝑡Т = 𝑉/аТ min = 0,48/0,697 = 0,689 с
(29)
Maximum allowable acceleration:
аТ min = [
2
0,15
4
1,2
[ ⋅(
𝑧пр
+ 0,015 ⋅
𝑧к
⋅(
50
200
𝜙сц
Ксц
−𝑓⋅
𝑑ц
𝐷к
) + (2 ⋅ 𝜇 + 𝑓 ⋅ 𝑑ц ) ⋅
) + (2 ⋅ 0,3 + 0,015 ⋅ 50) ⋅
1
200
1
𝐷к
]⋅𝑔 =
] ⋅ 9,81 = 0,697 м/с2
(thirty)
Number of drive wheels znp = 2. Coefficient of adhesion φsc = 0.15. Clutch margin
Kc = 1.2.
Actual travel speed of the crane
𝑉 = 𝜋 ⋅ 𝐷к ⋅
𝑛дв
𝑢
= 3,14 ⋅ 0,2 ⋅
920
18⋅60
= 0,53 м/с
(31)
i.e. similar to the given (initial) value.
Estimated braking torque
ТТ = (32, 3 − 145) ⋅
0,2 ⋅ 0,85 920 ⋅ 0,221
+
= 7,23 Н ⋅ м
2 ⋅ 18
38 ⋅ 0,689
According to tables 58 and 62 of the application, youwe take the TKT-100 brake with
a rated braking torque [TT] = 10N m, as close as possible to the calculated value of Тт.
We check the selected brake according to the braking conditions when the crane is
working with a load.
10
Deceleration time test:
𝑡𝑇 =
(mD2 )′о ⋅𝑛дв
38⋅(𝑇𝑇 +𝑇𝐶.Т )
(32)
≤ [𝑡𝑇 ]
Flywheel mass moment:
(mD2 ) ′𝑂 = 1,2 [(mD2 )𝑃 + (mD2 ) 𝑇 ] +
(0,00068 + 0,032) ⋅ 1,2 +
2
365⋅(1700+2150)⋅0,48
920
2
365⋅(𝑄+𝑚к )⋅𝑉 2
2
𝑛дв
⋅𝜂 =
⋅ 0,85 = 0,364 кг ⋅ м2
(33)
Static moment of resistancebraking motion:
ТС.Т =
𝑊У.Т ⋅𝐷к ⋅𝜂
2⋅𝑢⋅𝜂
(34)
Braking resistance:
𝑊У.Т = (𝑊тр min − 𝑊ук )
(35)
Friction resistance:
𝑊′тр min =
𝐺+𝐺к
𝐷к
⋅ (2 ⋅ 𝜇 + 𝑓 ⋅ 𝑑ц ) =
17+21,5
200
⋅ (2 ⋅ 0,3 + 0,015 ⋅ 50) = 0,260 кН = 260 Н(36)
Slope resistance:
𝑊ук = (𝐺 + 𝐺к ) ⋅ 𝛼 = (17 + 21, 5) ⋅ 0,0015 = 0,058 кН = 58 Н
Consequently,
𝑊У.Т = (260 − 58) = 202 Н
Then the static moment of resistance:
ТС.Т =
202 ⋅ 0,2 ⋅ 0,85
= 1,12 Н ⋅ м
2 ⋅ 18 ⋅ 0,85
and the braking time:
𝑡𝑇 =
0,364 ⋅ 920
= 1,06 с
38 ⋅ (7,23 + 1,12)
which is less than the admissible [tt] = 6...8 s.
(37)
11
Deceleration testwhen braking:
аТ = 0,48/1,06 = 0,45 м/с2
which is less than the maximum allowable value for indoor cranes, [at] < 1 m/s2.
Therefore, the braking conditions are met.
9. Determine the stopping distance by the formula:
𝑆𝑇 = 𝑉 ⋅ 𝑡𝑇 /2 = 0,48 ⋅ 1,06/2 = 0,25 м
(38)
According to the norms of Gosgortekhnadzor, with the number of drive wheels equal to
half the total number of running wheels (see Table 3.3), and with fst = 0.15
𝑆𝑇 = 𝑉 2 /(3250...5400) = 0,26...0,16 м ≈ 0,21 м
(39)
Bibliography
1. Design and calculation of lifting and transporting machines for agricultural
purposes / M.N. Erokhin, A.V. Karp, N.A. Vyskrebentsev and others; Ed. M.N.
Erokhin and A.V. Carp. – M.: Kolos, 1999.
2. Course design of load-lifting machines / N.F. Rudenko, M.P. Aleksandrov,
A.G. Lysyakov.- M.: Mashinostroenie publishing house, 1971.
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