belt DESIGN

B = (z - 1)e + 2f = (6 − 1) ×15 + 2 × 9 = 93mm
6、for the driven sheave d d = 250 mm , d = 40 mm d1 = 2d = 80mm, sheave with thinned web is selected
d a = d d + 2 h a = 250 + 5 . 5 = 255 .5 mm C ' = 15 mm ， B = 93mm
(D2 − D1 )2 )+
4 a0
amax = a + 0.03Ld
α1 = 180 −
D2 − D1 × 57.3 ≥ 120 a
Pca Pca = ≤ 10 [P0 ] (P0 + ∆ P0 )k a k L
Pca F0 = 500 zv
2 .5 − 1 + qv 2 K a
2、 2 Select the belt cross section and belt type Select type A from Figure 4-14 according to pca and n 1 3、 Determine the size of sheave （1） Select the driving sheave base diameter D 1 = 90 mm according to Table 4-3 and Table 4-6. （2） Calculate the base diameter of the driven sheave
V-belt Type A
Driven Sheave with Thinned Web
V-belts Driver
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Contents
4、5 2 3
Program
2D
3D
郭宁 200661860
Diagram
Belt Drive
The advantages
High speeds Avoid overload Smooth operation Simple structure Low cost Shock absorption
Language type: C
Program show
link
吴晓阳 200661857
2D VIEWS
INTERSECTION OF V-BELT
吴晓阳 200661857
2D VIEWS
吴晓阳 200661857
2D VIEWS
郑俊达 200661834
3D Prototype
Solid Driving Sheave
1440 × 90 = 249.23mm 2 1 520 select D 2 = 250 mm according to Table 4-6 D = = n1 D n 2
刘莹 200661807
design
(3) Check the velocity of the belt by
v =
π D 1n
1
60 × 1000
=
π × 90 × 1440
60 × 1000
= 6 . 786 m / s < 30 m / s
the velocity of the belt is OK. 4、confirm center distance a and basic length L d Specify a trial center distance according to the initial center distance is selected as a 0 = 400 mm the base length is calculated by Eq. 4-22
(D 2 - D 1 ) 2 π ( 250 − 90 ) 2 L d = 2a 0 + (D1 + D 2 ) + = 2 × 400 × × ( 250 + 90 ) + = 1350 mm 2 4a 0 2 4 × 400
π
刘莹 200661807
design
5、Compute contact angle on the small sheave
f fv = sin ϕ 2
D 2 − D1 α 1 = 180 − × 57 . 3 a D 2 − D1 α 2 = 180 + × 57 . 3 a
郭宁 200661860
Force analysis and stress analysis πD2 n2 πD1n1 v2 = v1 = 60 ×1000 60 ×1000
The disadvantages
Non-constant speed ratio Low efficiency Short service life
郭宁 200661860
Flat belt V-belt Four kinds of belts Round belt Timing belt Geometrical relationships in belt drives
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Belt Design
郭宁 马乐 刘莹 项飞 王嵩凯 吴晓阳 郑俊达 200661860 200661833 200661807 200661805 200661830 200661857 200661834
Contents
1 2 3
summary
Structural design
compute
Five kinds of force in the belts Tight side tension F1 Slack side tension F2 Initial belt tension F0 Effective tension Centrifugal force Fe Fc The following is the relationship between them
α1 = 180° −
D2 - D1 250− 90 ×57.3° = 180° − ×57.3° = 158.4° > 120° a 425
So the contact angle on the small sheave is OK. 6 Compute the number of belts required to carry the design power from Eq.4-25
(
)
郭宁 200661860
Design Methods
Pca = K A P
Ld = 2a0 +
Determine D1
π
2
D2 = iD1
′ Ld − Ld a ≈ a0 + 2
z=
0.7(D + D2 ) < a0 < 2(D + D2 ) 1 1
amin = a − 0.015Ld
(D1 + D2
Compute initial tension F0 Any unreasonable result will stop this process
Compute design power Pca
Compute the number of belts z
Determine the size of sheave D1 D2
Fe F1 = F 0 + 2
F2 = F0 −
Fe 2
Fc = qv
2
F1 = e fα F2
1 − 1 / e fα Fe = 2 F0 1 + 1 / e fα
郭宁 200661860
Stress analysis
Tight side Slack side Bending stress
σ1 =
'
0 . 7 ( D1 + D 2 ) < a 0 < 2 ( D1 + D 2 )
the base is selected from Table 4-7 as Ld = 1400mm the actual center distance a is calculated from Eq.4-23
L d − L' d 1400 − 1350 a = a0 + = 400 + = 425 mm 2 2
P ca z = ( P 0 + ∆ P 0 )K α K L from the Table 4-3 and Table 4-4, we have
P 0 = 1 . 07 kw ∆ P0 = 0 . 17 kw
z=
from Table 4-5 from Table 4-7
z=
K L = 0.96
K α = 0.95
Pca 6.05 = = 5.35 ( P0 + ∆P0 ) K α K L (1.07 + 0.17) × 0.95 × 0.96
select z=6
刘莹 200661807
design
7、Compute initial tension F 0 from Eq.4-26
Pca 2.5 6.05 2.5 ( − 1) + qv 2 = 500 × ×( − 1) + 0.1× 6.786 2 = 125.8 N 6 × 6.786 0.95 zv K α 8、 Compute the force acting on the sheave shaft Q from Eq.4-27 F0 = 500
a1 Q = 2 zF 0 sin 2
马乐 20066133
Structural design
1、 V-belt is selected 2、 conventional V-belt type A is selected 3、 Material for v=6.79<25, select cast iron, HT200 4、 design of the pulley groove for type A is selected from the Table 8-10 in the book b d = 11 mm , h a = 2 .75 mm , h f = 8 . 7 mm ,