bridge girder auncher Calculation Sheet Part 2

8/10/2019 bridge girder auncher Calculation Sheet Part 2

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Mid Support ......................................................................................... 1

4.1MID SUPPORT CONSIST OF SOME PARTS AS FOLLOWS .......................... 1

4.2FORCE ANALYSIS ..................................................................................................................... 1

Calculation of launching roller .................................................. 7

5.1SECTIONAL CHARACTER OF BEAM IS AS THE FOLLOWING DIAGRAM ............................. 7

5.2CALCULATION OF SUPPORT ....................................................................................................... 9

5.3PIN SHAFT OF INSIDE AND OUTSIDE BUSHING .......................................................................... 10

5.4PRESSURE STRESS OF INSIDE BUSHING HOLE'S WALL .............................................................. 10

Calculation of rear support ......................................................... 11

6.1SECTIONAL CHARACTER OF OUTSIDE BUSHING JOINT BEAM IS AS THE FOLLOWING

DIAGRAM ................................................................................................................................................... 11

6.2CALCULATION OF SUPPORT ...................................................................................................... 12

6.3PIN SHAFT OF INSIDE AND OUTSIDE BUSHING .......................................................................... 13

6.4PRESSURE STRESS OF INSIDE BUSHING HOLE'S WALL ............................................................. 14

Calculation of machine and transmission parts ........... 15

7.1CALCULATION FOR FRONT SUPPORT'S DRIVING AND TRANSMITTING PARTS ............ 15

7.2 CALCULATION FOR REAR SUPPORT'S DRIVING AND TRANSMITTING PARTS ............. 25

7.3CALCULATION FOR HANG TRANSMITTING PARTS ................................................... 35

7.4CALCULATION FOR HANG WHEEL .......................................................................................... 46


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Mid Support

4.1

Mid support consist of some

parts as follows

1 Top trimmer 2 Hang and launching roller

equipment 3 Upright 4 Lower trimmer 5 Running

wheel box 9 Transverse orbit

4.2

Force analysis

F=110

Q345B

Mid support is the worst working condition when beam launcher is erecting

boundary beam: the stress of every upright is F=110t, make force analysis in this

working condition. material of main structure is Q345B.

4.2.1 Calculation of top trimmer

4.2.1.1 The structural form of top trimmer is as follows


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4.2.1.2 Sectional character of top trimmer is as follows

M max=110X1.317/4=36.2tm

The maximum bending moment is M max=110X1.317/4=36.2tm.

Q=55t

The shearing force on the span is Q=55t.

max=36.2/0.3=120MPa

The maximum bending stress is max=36.2/0.3=120MPa.

max=55/8832=62MPa

The average bending shearing force is max=55/8832=62MPa.

=161MPa


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4.2.2 Calculation of upright

Nmax=110t

The maximum axial force is N max=110t.

max=110/24768=44MPa

The maximum pressure stress is max=110/24768=44MPa.

4.2.3 Calculation of lower trimmer

4.2.3.1 Structural form of lower trimmer is as follows

4.2.3.2 Sectional character


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Sectional character of lower trimmer's cross-girder

Sectional character of lower trimmer's cantilever-girder

4.2.3.3 Calculation

1) Calculation of cross-girder

M max=110X0.77=84.7tm

The maximum bending moment is M max=110X0.77=84.7tm.

Q=110t

The maximum shearing force is Q=110t.

max=84.7/0.51=184MPa


max=110/10032/3=36MPa

The average shearing stress is max=110/10032/3=36MPa


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400mm

There're one fulcrum every 400mm length of transverse orbit.

M max=110/2X0.4/4=5.5tm

The maximum bending moment is M max=110/2X0.4/4=5.5tm.

Q=55t

The maximum shearing stress is Q=55t.

max=5.5/0.12=46MPa


max=55/2/6136=53MPa

The average shearing stress is max=55/2/6136=53MPa.

=103MPa


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Calculation of launching roller

5.1

Sectional character of beam is as the

following diagram

Force diagram

Bending moment diagram


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Shearing stress diagram

The maximum bending moment

mttFL

M === 15.304m34.190

4

max

Shearing stress

tQ 45max

=

Normal stress

a4.18067.1670826

301500000MP

W

M

X

===

Shearing stress

MPaSQ

35.81167082667162

9664001045

I2

4

max=

==

Translational stress

[ ] MPaMPa 2409.22835.8134.1803 2222 =


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5.2

Calculation of support

Axial force

N=90t

Length

L=3064mm

Slenderness ratio

84.3616.83

3064==

i

L

Stability coefficient of pressure force

946.0=

[ ] MPaMPA

N180a9.128

7382.7946.0

109 5

=


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5.3

Pin shaft of inside and outside bushing

45t

The shearing force is 45t, and double shearing.

Shearing stress

MPaMPaA

Q560][5.58

5.38482

1045

2

4

=


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Calculation of rear support

6.1

Sectional character of outside

bushing joint beam is as the following diagram

The maximum bending moment

mttFL

M === 4.134m34.1404

max

Shearing stress

tQ 20max =

Normal stress

a2.921452815

134000000MP

W

M

X

===

Shearing stress

MPaSQ

2.33

185027078122

7370001020

I2

4

max=

==


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Stability coefficient of pressure force

839.0=

[ ] MPaMPAN

180a6.647382.7839.0

104 5

=


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6.4

Pressure stress of inside bushing hole's wall

40t

The shearing force is 40t, and double shearing.

[ ] MPaMPd

Q180a9.129

0722

1020 4

=


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Calculation of machine and

transmission parts

7.1

Calculation for front support's driving

and transmitting parts

Working grade: M4

The maximum static wheel-pressure: 56t

The maximum dynamic wheel-pressure: 50t

The minimum dynamic wheel-pressure: 12t

Running speed: 1.67m/min

Gradient: 0.5%

Wind speed at working condition: 15.5m/s

7.1.1 Original design parameter

Driving form: 1/2 driving

Diameter of wheel tread: 500mm

Width of orbit: 110mm

Diameter of wheel shaft: 140mm

=0.004

The frication resistance coefficient of self-aligning roller bearing is=0.004.

n=1.063r/min

The revolving speed of wheel is n=1.063r/min according to above.

7.1.2 Strength checking of wheel

7.1.2.1 Equivalent working wheel-pressure

P mean=2Pmax+Pmin/3=2X50+12/3=37.3 t

GB/T3811 According to GB/T3811

Allowable unit pressure of wheel's material: K=5.6


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Revolving speed coefficient of wheel: C1=1.17

Working grade coefficient: C2=1.12

Allowable wheel-pressure of that wheel

P=KDlC1C2=5.6X500X110X1.17X1.12=403603N40.3t

PP mean

7.1.2.2 Static strength checking

P max=1.9KDl=1.9X5.6X500X110=585200 N58.5.t

P max

P max is greater than the maximum wheel-pressure of static load.

Hereby the select wheel is suitable.

7.1.3 Calculation of running resistance

7.1.3.1 Friction resistance

Pm = P(d+ 2fk)CfD

GB3811 According to GB3811

Rolling friction force-arm of wheel

fk= 0.5

Additional friction resistance coefficient

Cf= 1.5

Pm=50e4X2X0.004X120+2X0.5X1.5/500=2160 N

7.1.3.2

Gradient resistance

P = mg tan = 50ex 2 0.005 = 2500N

7.1.3.3Wind resistance

PW1=0.6CPA

Wind area: A=120 m2

Wind coefficient : C=1.35

P=0.625V2=0.625X15.5

2=150


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PW1=0.6X1.35X150X120=14595 N

Running resistance

Pj=2160+2500+14595=19255 N

7.1.4 Calculation for motor's power

P = KPjV (1000)

In this formula

K=1.2; =0.8

And therefore

P=1.2X19255X1.67/60X1000X0.8=0.804KW

P=1.1KW

The motor's power is P=1.1KW

7.1.5 Transmission ratio allocation and

reducer type

Revolving speed of wheel: n=1.063r/min

Output revolving speed of motor: nz=1430 r/min

The total transmitted ratio: i=1430/1.063=1345

Primary reducer type: KF97R67-516.2-1.1/4

The transmitted ratio: i=516.2

The transmitted ratio of the open gear:

i=1345/516.2=2.6056

The tooth number of small gear: Z1=18

The tooth number of large gear: Z2=47

The actual transmitted ratio:

iactual = 516.2 47 18 = 1347.856

Theoretical running speed of wheel

v = DN 1000 = 500 1430 1347.856 1000 = 1.666 m min

The rate of speed difference


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P=1.67-1.666/1.67=0.002

The distribution of transmitted ratio is suitable.

7.1.6 Calculation for output torque of reducer

Output torque of motor:

T=9550P/N=7.346 Nm

Translational pull force of wheel

F=2000XTXI/D=2X7.346X1347.856/500=39606 N

FPj

Meet requirement.

7.1.7 Calculation for slipping of wheel

Viscous force of every wheel

Fd = 50ex 0.1 = 50000N

Viscous force is greater than total static resistance.

Meet requirement.

The reducer is suitable according to above.

7.1.8 Calculate shaft's strength according to

composited bending moment and torque

7.1.8.1 Calculation for strength of shaft

Allowable stress of shaft

p=320MPa; -1p=90MPa

Fulcrum distance: L=370 mm

Bending moment of shaft

M=Pl/4=56e4X370/4000=51800 Nm

Torque of shaft


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T=FD/2=19255X500/2000=4813 Nm

Correction coefficient: =0.3

The diameter of mandrel: d=123.7 mm

Tentative diameter is d=140mm in consideration of

shaft's stiffness requirement.

7.1.8.2 Calculation for deflection of shaft

Inertia moment of shaft: I=18857410

Elasticity modulus: E=290000

Deflection: yp = FL3 (48EI) = 0.096mm

Allowable maximum deflection of shaft

ypmax = 0.0005XL = 0.185mm

The actual deflection is lesser than allowable maximum deflection.

Therefore, sectional diameter of shaft is suitable.

7.1.9 Calculation for bearing

7.1.9.1 Elementary rated dynamic load

Bearing quantity of every wheel: m=2

Load of every wheel: p=50t

Diameter of shaft: d=140 mm

Radial load: Fr=280 KN

Radial equivalent dynamic load: Pr=250KN

Radial equivalent static load: Pr=280KN

Elementary rated dynamic load of bearing

C = fhfmfdP = (fnft)

Service life of bearing: Lh = 1000h

Life coefficient: fh

= 1


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Speed coefficient: fn = 1.435

Moment load factor: fm = 2

Impulse load factor: fd = 1.2

Temperature factor: ft= 1.435

Elementary rated offered load: C=418.1 KN

Designate bearing type: 23228CC

And elementary rated dynamic load: Cr=812 KN

7.1.9.2 Rated static load of bearing

Safety factor: S0= 3

And: C0= 3 280 = 840KN

Sectional elementary static load of bearing

C0r=1300 KN

Therefore the sectional bearing is suitable.

7.1.10 Strength checking for spline

Diameter of shaft: D=340mm

Width of spline: b=28 mm

Height of spline: h=16mm

Length of spline: l=60mm

Tangent height between spline and wheel hub

k=0.4X16=6.4mm

Transmitted torque by spline: T=3792Nm

A Choose the A type double spline

P = 2T (1.5DKI) = 2 3792 [1.53406.4 (60 28)] = 72.6MPa

Therefore the sectional spline is suitable.


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7.1.11 Strength calculation of gear

7.1.11.1 Elementary parameter

Full load working hour: T=5000h

Transmitted power by gear: P=1.1KW

Revolving speed of pinion: n1=2.77r/min

Tooth number of pinion: z1=18

Tooth number of gear wheel: z2=47

Primary module of gear: m=10

Center distance of gear: a=325mm

Primary facewidth: b=60mm

7.1.11.2 Allowable stress of material

Hlim1=950MPa; Hlim2=920MPa

Flim1=350MPa; Flim2=340MPa

Allowable contact stress

Hp=0.9Hlim; Hp =828MPa

7.1.11.3 Contact stress checking of tooth flank

1 Reference nominal tangential force: FT=42134

2 Service factor: KA

P14-136 14-1-71 Review table 14-1-71 of page P14-136

KA=1.25

3 Dynamic load factor: KV

Linear velocity of pinion: v=0.0261 m/s

Transmission accuracy coefficient

C=-0.5048lnZ-1.144lnm+2.825lnfpt+3.32

fpt=25m C=8.32 C=8

There into, fpt=25m, and C=8.32, use C=8

P14-140 14-1-74KV=1

Review diagram 14-1-74 of page P14-140, KV=1


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4 Spread factor of helix line load: Kh


Kh=a1+a2[1+a3(b/d1)2](b/d1)

2+a4b

There into: a1=1.15a2=0.18a3=6.7a4=0.00031

Therefore: Kh=1.203

5 Spread factor of inter-tooth load: KH

KAFt/b=877.7974

P14-146 14-1-92 Review table 14-1-92 of page P14-146: KH=1.2

6 Pitch point area factor: ZH

P14-149 14-1-76 Review diagram of page P14-149: ZH=1.5

7 Elastic coefficient: ZE

P14-150 14-1-95 Review table 14-1-95 of page P14-150: ZE=189.8

8 Contact ratio coefficient: Z

Spur gear is without shift

=0, =1, and Z=1

9 Helical angle factor: Z

Spur gear: =0, and Z=1

10Meshing factor of pinion and gear: ZB ZD

14-1-94 According to the judging criteria of table 14-1-94

There into: =20

da1=200 mm, db1=73.45477 mm

da2=490 mm, db2=191.7986 mm

M1=0.98744, M2=0.944101

So: ZB=1 ZD=1

11 Calculate contact stress: H

H1=888.5211MPa, H2=888.5211MPa

12 Life factor: ZNT

Stress cycle-index


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NL1=831073.2, NL2=318283.4

According to table 14-1-96

ZNT1=1.561975, ZNT2=1.676624

13 Impact factor of lubricating oil: ZLZVZR

14-1-98 Get gear pair by generating method

according to table 14-1-98

RZ104m, ZLZVZR=0.85

14Working harden factor of tooth surface: ZW

HRC50-55 470HB

Hard tooth surface HRC50-55 is greater than 470HB.

Therefore: ZW=1

15 Size factor: ZX

According to table 14-1-99: ZX=1

16 Safety factor method

SH =HlimZNT ZLZVZRZW ZX H

Therefore: SH1=1.419544

SH1=1.47562

According to table 14-1-100: SHmin = 1.1

Meet requirement.

7.1.11.4 Bending stress checking of gear

1 Spread factor of helical load: KF=KHN

N=b/h2/[1+b/h+(b/h)2]

There into b=60, H=2.25m=22.5

So: N=0.659794, KF=1.129992

2 Partition coefficient of helical load: KF

KF=KH=1.2

3 Tooth profile coefficient: YF

Equivalent tooth number: zn1=18, zn2=47


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According to diagram 14-1-98

YF1=2.9, YF2=2.35

4 Stress correction factor: YS


YS1=1.54, YS2=1.68

5 Contact ratio coefficient: Y

Because: =1 =0

So: Y=1

6 Helical angle coefficient : Y


Y=1

7 Calculation of root stress: F

F= FTYFa YSa Y Y KAKVKF KF

F1=531.581MPa, F2=469.924MPa

8 Stress correction factor of test gear: YST


YST=2

9 Life factor

YNT =(3000000 NL)0.02, YNT 1= 1.026005, YNT 2= 1.04589

10 Relatively root sensitivity coefficient

According to table 14-1-102 and diagram 14-1-98

YrelT = 1

11 Relatively root surface condition factor

According diagram 14-1-118

RRrelT = 1

12 Size factor: YX

14-1-109 According to the formula of table 14-1-109


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YX=1.03-0.006m, YX=0.97

13 Safety factor of bending strength: SF

SF=Flim YST YNT Yrel YRrel YX

SF1=1.31, SF2=1.468

According to table 14-1-100: SFmin = 1.25

It's suitable.

7.2 Calculation for rear support's driving

and transmitting parts

Working grade: M4



The minimum dynamic wheel-pressure: 12t

Running speed: 1.67m/min

Gradient: 0.5%


7.2.1 Original design parameter

Driving form: 1/2 driving




=0.004

The frication resistance coefficient of self-aligning roller bearing is=0.004.

n=1.063r/min

The revolving speed of wheel is n=1.063r/min according above.


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P mean=2Pmax+Pmin/3=2X30+12/3=24 t



Revolving speed factor of coefficient: C1=1.17


Allowable wheel-pressure of the select wheel

P=KDlC1C2=5.6X500X80X1.17X1.12=293529 N29.3t

PP mean


P max=1.9KDl=1.9X5.6X500X80=425600 N42.t

P max

P max is greater than the maximum wheel-pressure of static load.




Pm = P(d+ 2fk)CfD


Rolling friction force -arm of wheel

fk= 0.5


Cf= 1.5

Pm = 30e4 4 2 (0.004 120 + 2 0.5) 1.5 500 = 2376N

7.2.3.2 Gradient resistance

P = mg tan a = 30e4

2 0.005 = 3000N


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PW1=0.6CPA

Wind area: A=120 m2

Wind coefficient: C=1.35

P=0.625V2=0.625X15.5

2=150

PW1=0.6X1.35X150X120=14595 N

Running resistance

Pj=2376+3000+14595=19971 N


P = KPjV (1000)

In this formula

K=1.2, =0.8

And so

P=1.2X19971X1.67/1000X0.8=0.834KW

P=1.1KW

The motor's power is P=1.1KW

7.2.5 Transmitted ratio allocation and reducer

type



The total transmitted ratio: i=1430/1.063=1345

Primary reducer type: KF97R67-516.2-1.1/4

The transmitted ratio: i=516.2


i=1345/516.2=2.6056

The tooth number of opinion: Z1=18


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The tooth number of wheel gear: Z2=47


iactual = 516.2 47 18 = 1347.856


v = DN 1000 = 500 1430 1347.856 1000 = 1.666 m min


P=1.67-1.666/1.67=0.002


7.2.6

Output torque of motor

T=9550P/N=7.346 Nm


F=2000XTXI/D=2X7.346X1347.856/500=39606 N

F > Pj

Meet requirement



Fd = 30e4 0.1 = 30000N


Meet requirement.






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p=320MPa, -1p=90MPa



M=Pl/4=32e4X344/4000=25800 Nm

Torque of shaft

T=FD/2=19971X500/2000=2992 Nm










ypmax = 0.0005XL = 0.172mm










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Service life of bearing: Lh = 1000h

Life coefficient: fh = 1

Speed coefficient: fn=1.435



Temperature factor: ft= 1.435

Elementary rated offered load: C= 267.6KN




Safety factor: S0= 3

And: C0= 3 160 = 480KN


C0r=722 KN








k=0.4X16=6.4mm


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P = 2T (1.5DKI) = 2 3792 [1.53406.4 (60 28)] = 72.6MPa


7.2.11 Strength calculation of gear



Transmitted power by gear: P=1.1KW









Flim1=350MPa; Flim2=340MPa


Hp = 0.9Hlim ; Hp = 828MPa

7.2.11.3

Contact stress checking of tooth flank

1 Reference nominal tangential force: FT=42134



KA=1.25




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C=-0.5048lnZ-1.144lnm+2.825lnfpt+3.32

fpt=25m C=8.32 C=8

There into, fpt=25m, and C=8.32, use C=8

P14-140 14-1-74KV=1




Kh=a1+a2[1+a3(b/d1)2](b/d1)

2+a4b

There into: a1=1.15a2=0.18a3=6.7a4=0.00031

Therefore: Kh=1.203


KAFt/b=877.7974








=0, =1, and Z=1





There into: =20

da1=200 mm, db1=73.45477 mm

da2=490 mm, db2=191.7986 mm


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M1=0.98744, M2=0.944101

So: ZB=1 ZD=1


H1=888.5211MPa, H2=888.5211MPa

12 Life factor: ZNT

Stress cycle-index

NL1=831073.2, NL2=318283.4


ZNT1=1.561975, ZNT2=1.676624




RZ104m, ZLZVZR=0.85


HRC50-55 470HB


Therefore: ZW=1

15 Size factor: ZX





SH1=1.47562


Meet requirement.



N=b/h2/[1+b/h+(b/h)2]


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There into b=60, H=2.25m=22.5

So: N=0.659794, KF=1.129992


KF=KH=1.2




YF1=2.9, YF2=2.35



YS1=1.54, YS2=1.68


Because: =1 =0

So: Y=1



Y=1



F1=531.581MPa, F2=469.924MPa



YST=2

9 Life factor

YNT =(3000000 NL)0.02, YNT 1= 1.026005, YNT 2= 1.04589



YrelT = 1


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RRrelT = 1

12 Size factor: YX


YX=1.03-0.006m, YX=0.97



SF1=1.31, SF2=1.468

According to table 14-1-100: SFmin = 1.25

It's suitable.

7.3 Calculation for hang transmitting parts

Working grade: M4



The minimum dynamic wheel: 10t

Running speed: 4m/min

Gradient: 1%


7.3.1 Elementary designed parameter

Driving form: 1/4 driving and coupling




=0.004

The friction resistance of self-aligning roller bearing is=0.004.


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n=4.897r/min

The revolving speed of wheel is n=4.897r/min according above.

7.3.2

Strength checking of wheel


Pmean =2Pmax + Pmin

3 =

2 30 + 10

3 = 23.3t



Revolving speed coefficient of wheel: C1=1.17


Allowable wheel-pressure of select wheel

P = KDIC1C2= 5.6 260 140 1.17 1.12 = 2671112N 26.7t

P > Pmean


Pmax = 1.9KDI = 1.9 5.6 260 140 = 387296N 38.7t

max

P max is greater than the maximum wheel-pressure static load.




Pm = P(d+ 2fk)CfD


Rolling friction force-arm of wheel

fk= 0.5


Cf= 1.5


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Pm=30e4 4 (0.004 110 + 2 0.5) 1.5 260 = 10800N

7.3.3.2 Gradient resistance

P = mg tan a = 30e4 4 0.01 = 12000N


PW1=0.6CPA

Wind area: A=18 m2

Wind coefficient: C=1.35

P=0.625V2=0.625X15.5

2=150

PW1=0.6X1.35X150X180=2189 N

Running resistance

Pj= 10800 + 12000 + 2189 = 24989N


P = KPjV (1000)

In this formula

K=1.2, =0.8

And so

P=1.2X24989X4/1000X0.8=2.498KW

P=3KW

The motor's power is P=3KW

7.3.5 Transmitted ratio allocation and reducer

type



The total transmitted ratio: i=1430/4.897=289.969

Primary reducer type: KF87-103-3kw/4

The transmitted ratio: i=103



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i=289.969/103=2.815233

The tooth number of opinion: Z1=19

The tooth number of wheel gear: Z2=53


iactual = 103 53 19 = 287.3158


v = DN 1000 = 260 1430 287.3158 1000 = 4.03 m min


P=4.03-4/4=0.0075


7.3.6 Output torque checking of reducer

Output torque of motor

T=9550P/N=20.176 Nm


F=2000XTXI/D=2X7.346X1347.856/260=44591 N

FPj

Meet requirement



Fd = 2 30e4 0.1 = 60000N


Meet requirement.



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p=320MPa, -1p=90MPa



M=Pl/4=32e4X390/4000=29250Nm

Torque of shaft

T=FD/2=24989X260/2000=3248.6 Nm










ypmax = 0.0005XL = 0.195mm







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Service life of bearing: Lh= 1000h

Life coefficient: fh = 1

Speed coefficient: fn=1.435



Temperature factor: ft= 1

Elementary rated offered load: C=267.6 KN




Safety factor: S0=3

And: C0= 3 160 = 480KN


C0r=602KN







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k=0.4X16=6.4mm



P = 2T (1.5DKI) = 2 2078 [1.5 110 6.4 (160 28)] = 44.7MPa


7.3.11

Strength calculation of gear



Transmitted power by gear: P=3KW









Flim1=350MPa;

Flim2=340MPa


Hp=0.9Hlim; Hp =828MPa

7.3.11.3 Contact stress checking of tooth flank

1 Reference nominal tangential force: FT=42134 N




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KA=1.25




C=-0.5048lnZ-1.144lnm+2.825lnfpt+3.32

fpt=25 m C=8.548 C=8

There into, fpt=25m, and C=8548, use C=8

P14-140 14-1-74KV=1




Kh=a1+a2[1+a3(b/d1)2](b/d1)

2+a4b

There into: a1=1.15a2=0.18a3=6.7a4=0.00031

Therefore: Kh=1.1991


KAFt/b=683.5966








=0, =1, and Z=1






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There into: =20

da1=168 mm, db1=62.028 mm

da2=440 mm, db2=173.027 mm

M1=0.989476, M2=0.946497

So: ZB=1 ZD=1


H1=844.1349MPa, H2=844.13491MPa

12 Life factor: ZNT

Stress cycle-index

NL1=4135922, NL2=1482689


ZNT1=1.387523, ZNT2=1.496649




RZ104m, ZLZVZR=0.85


HRC50-55 470HB


Therefore: ZW=1

15 Size factor: ZX





SH1=1.38648


Meet requirement.


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N=b/h2/[1+b/h+(b/h)2]

There into: b=50, H=2.25m=18

And so, N=0.671321, KF=1.129631


KF=KH=1.2




YF1=2.9,YF2=2.35



YS1=1.54, YS2=1.68


Because: =1, =0

So, Y=1



Y=1



F1=517.305MPa, F2=457.304MPa



YST=2

9 Life factor

YNT =(3000000 NL)0.02, YNT 1= 0.9936, YNT2= 1.014195


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YrelT = 1



RRrelT = 1

12 Size factor: YX


YX=1.03-0.006m, YX=0.982



SF1=1.32, SF2=1.48


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7.4

Calculation for hang wheel

The maximum wheel-pressure of hang: 2.5t

Gradient: 1%

Running speed: 4m/min

According to above calculation, wheel-pressure of hang is much lesser than

wheel-pressure of launcher roller and they share the same driving system. so the

power's checking computations is positively suitable, just need to check and

calculate strength and speed of hang wheel.


Diameter of wheel: 160mm


P=1.17X1.13X5.6X160X30=35538 N3.5t

So the wheel is suitable.

7.4.2 Speed checking

Transmission ratio of chain wheel

z2/z1=22/14

And revolving speed of wheel

N=1430/103X19/53X22/14=7.82 r/min

Running speed of wheel

V = Dn 1000 = 7.82 160 1000 = 3.93 m/min


P=4-3.93/4=0.0175

bridge girder auncher Calculation Sheet Part 2

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Transcript of bridge girder auncher Calculation Sheet Part 2