Double Action Lifting Jack

7/30/2019 Double Action Lifting Jack

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- 1 -

UNIVERSITY TRANSILVANIA FROM BRASOV

FACULTY: MECHANICAL ENGINEERING

DEPARTMENT: AUTOMOTIVE ENGINEERING

CATEDRA DE ORGANE DE MASINI, MECANISME SI ROBOTICA

DISCPILINA ORGANE DE MASINI

PROJECT: DOUBLE ACTION LIFTING JACK

ANUL UNIVERSITAR

2011-2012



- 2 -

1. ESTABLISH OF LOADS ON ELEMENTS

Figure 1.1

2. CALCULATION OF THE MAIN SCREW

2.1.CHOOSING THE MATERIAL

It is chose n OL50 STAS 500/ 2

R p02=290 MPa

R m=490…610 MPa

2.2.PRE-DIMENSIONING CALCULUS

The calculus load cQ , N

N QQc 75,225081800725,1

3,1...25,1 [3]

The internal diameter o f the thread 3d , mm

MPammQ

d ac

ac

c 4094,2350

75,22508443

MPaac 60...40 [3] It is chose n 3d = 25mm



- 3 -

Choo sing the thread

It is chose n Trapezoidal thread Tr 32x6 STAS SR ISO 2904

Figure 2.1

Table 2.1

Nominal

Diameter

Pitch Medium

Diameter

External

Diameter

Inte rnal Diame ter

mmd , mm p, mm Dd ,22 mm D ,4 mmd ,3 mm D ,1

32 6 29 33 25 26

2.3.THE SELF BLOCKING CONDITION

The helix angle , 2 ,

76,329

6

2

2

arctg d

P arctg

The friction angle , ` ,◦

53,6

2

30cos

11,0

2cos

' arctg arctg

12,0...11,0 for ste el/ste el [3]

The self-blocking condition

The condition is accomplished

5,676,3 '

2



- 4 -

mmd

d

d

A

l i

m H l

l k l

i

l

i

f

f

25.64

25

4

4

64

23040190

1152305,0

4.1825.6

115

3

2

3

4

3

minmin

min

2.4.COMPOSED STRESS VERIFICATION

Torsion mo ment claiming the screw Nmm M t ,

Efective compresion tension

MPad

Qc 68.36

25

180074422

3

Efective torsion tension

Equivalent tension

The condision is accomplished

2.5.BUCKLING VERIFICATION

Slender coe fficiency

Figure 2.3

Nmmtg tg d

Q M M insl t 11,47403)76,353,6(2

2918007)(2

2

,2

MPa

d

M t t 45.15

25

11.47403161633

3

MPa MPaace

act ce

5009.4845.15468.36

4

22

22



- 5 -

7106

13....7

59.610)2632(

180074

)(

4222

1

2

z z

MPa p

p Dd

Q z

a

a

Buckling rang e894.18 0

Safety buckling coefficient c

59.3234.1862.033562.0335

82.868.36

59.323

f

c

f c

Admissible safety coefficient

ca=3…5

3. CALCULATION OF SECONDARY SCREW

3.1.CHOOSING THE MATERIAL

It is chosen OL50 STAS 500/2

R p02=290 MPa

R m=490…610 MPa

3.2.CALCULATION OF NUMBER OF HELIXES OF MAIN SCREW NUT

Number o f tu rns

Figure 3.1

·

·



- 6 -

Lenght o f the nut

mm P z H pl 4267

3.3.CHOOSING THE EXTERNAL THREAD

Ado ption o f the inte rnal preliminary diameter of the thread

mmmm D D

mmmm Dd

374336...4

4583710...8

40

03

Choo sing the thread

It is chose n trap ezoidal thread Tr 55x9 STAS SR ISO 2904 with the following

characteristics

Figure 3.2

Table 3.1

Nominal

Diameter

Pitch Medium

Diameter

External

Diameter

Inte rnal Diamete r

mmd , mm p, mm Dd ,22 mm D ,4 mmd ,3 mm D ,1

55 9 50.5 56 45 46

3.4.THE SELF-BLOCKING CONDITION

The helix angle 2 ,grade

24,35.50

9

2

2

arctg d

P arctg



- 7 -

The friction angle

]3[12,0...11,0

53,6

2

30cos

11,0

2cos

'

arctg arctg

The self-blocking condition

53,624,3 '

2

3.5.COMPOSED STRESS VERIFICATION

Torsion mo ment claiming the screw Nmm M t ,

Nmmtg tg d Q M M insII t 16,7829124,353,625.5018007

22

'2

Efective compresion tension

MPa

Dd

Qc 96,34

3745

1800744222

0

2

3

Efective torsion tension

MPa

Dd d

M t t 03.10

37455616

16,78291

16

444

0

4

3

3

Equivalent tension

MPa MPa

MPa

ace

ac

act ce

703,4003,10496.34

]3[80...60

4

22

22



- 8 -

106

13....7

59.610)2632(

180074

)(

4222

1

2

z

MPa p

p Dd

Q z

a

a

4. CALCULATION OF THE FIXED NUT

4.1. CHOOSING THE MATERIAL

It is chosen OL 50 STAS 500/2

R p02=290 MPa

R m=490…610 MPa

4.2. CALCULATION OF NUMBER OF HELIXES

Number o f helixes , z

Figure 4.1

It is chosen z=7

The length of the nutmm P z H pII 6397

4.3. CHECKING THE HELIXES

Bending

Stee l nut

]3[80...60

7024,977,545

455.50180073

80...60

70,59634,0634,0

3

21

2

3

321

MPa

MPa

MPa

P h

z hd

d d Q

ai

ai

ai

ai

Figure 4.2



- 9 -

SHEAR

Stee l nut

Figure 4.3

]3[65....50

5519.377,545

18007

65...50

3

MPa

MPa MPa

MPa

z hd

Q

af

af f

af

af f

The condition is accomplished

4.4.CHOOSING THE NUT DIMENSIONS

External diameter of the body

Figure 4.4

mmmm D De 6485610...84

Externa l diamete r of the neck

mmmm D D e g 80166420...16

Neck height

mmmmh g 812...8



- 10 -

4.5. COMPOSED STRESS VERIFICATION

Traction effective tension

MPa

D D

Q

e

t 89,235664

1800744222

4

2

Torsion effective tension

MPa

D D D

M

e

e

insII t 67,3

56646416

16.78291

16

444

4

4

Equivalent tension

]3[80...60

7099,2467,3489,234 2222

MPa

MPa MPa

at

at t t e

4.6.CHECKING THE NECK

Crush checking

]3[60

6095,96480

18007442222

MPa

MPa MPa D D

Q

as

as

e g

s

Shear checking

]3[55...40

452,11864

18007

MPa

MPa MPah D

Q

af

af

g e

f



- 11 -

4.7.CHOOSING AND VERIFICATION OF THREADED PIN FIXING THE NUT IN

JACK BODY

Choosing the threaded pin

Figure 4.5

It is chosen self-b locking threaded pin M5X25 STAS 10422

Figure 4.6

Tabelul 4.1

Fillet[d]

[mm]

t

[mm]

n

[mm]1c

[mm]

4d

[mm]

3c

[mm]

1d

[mm]

m

[mm]

M5 1.6 0.8 1,2 3,5 3 2.5 2

Friction moment

]3[2,0...15,0

94,130183

6480

6480180072,0

3

1

3

122

33

22

33

Nmm

D D

D DQ M

e g

e g

g

The mo ment so liciting the pin

Nmm M M M g insII s 78,5189294,13018316,78291



- 12 -

5. CALCULATION OF THE BODY

Choo sing body dimensions

Figure 5.1

mmmmh H H H

mmmm H H

mmmm D D

mmmm

mmmm

mmmm D D

g pII c

II

bibe

t

eei

250586319010...0

1908.412.14850...30

1403610450...30

1012...10

88..7

67...5

706646...2

1

1

Bod y checking at compresion

]3[80....60

7029,157080

18007442222

MPa

MPa MPa D D

Q

ac

ac

ci g

c

Suppo rt surface checking at crushing

]3[5,2...2

5,2104140

18007442222

MPa

MPa D D

Q

as

as

bibe

s



- 13 -

6. CUP CALCULATION

6.1.DOUBLE ACTION LIFTING JACK CUP

Choo sing the pin

mmd

mmmmd d

mmd d

c

c s

32

35332)4...2(

73520,0)25,0...15,0(

It is cho sen cylindrical p in A 8 X 60 STAS 1599/OLC45

Checking the p in for shear stress f and crush s ,in MPa

Figure. 6.1



- 14 -

%34

34,0)24,353,6(5.50)76,353,6(29

24,35.5076,329

)'()'( 222

2222

tg tg

tg tg

tg d tg d

tg d tg d

II II II I I

II II I I

]3[80...6080.20

80.2004,7412,25

4

04.7

)35

71(35

11.4740316

)1(

16

12,25

3574

35

18007

4

22

22

23

22

MPa MPa

MPa

MPa

d

d d

M

MPa

d d d

Q

ace

e

act ce

c

sc

insI t

c sc

c

MPa MPa

MPa

d d

M

screwof head pin

MPa

mmd D

MPa MPad Dd

M

cup pin

Crush

MPa MPa

MPa

d d

M

stressShear

as s

as

as

c s

inslI s

as

cc

as

cc s

insI s

af f

af

af

sc

insI f

11016,33357

11.474036

]3[120...100

6

..

]3[120...100

5.52355,1)6,1...4,1(

10068.17355.527

11.474034

)(

4

7021,35735

11,474034

]3[80...65

4

2

2

2222

2

2

Checking the minimized section of the head of screw at comp osetstress e ,MPa

7. CALCULATION OF EFFICIENCY

7.1.DOUBLE ACTION LIFTING JACK EFFICIENCY



- 15 -

8. HORIZONTALLY RATCHET MECHANISM

8.1. MECHANISM SCHEME



- 16 -

8.2. LENGTH CALCULUS OF THE CRANK

Total bend ing moment

Mit=Mm=MinsI+MinsII =47403.11+78291.16=125694.27 Nmm

The force of the worker

Fm=150…300N=300N

Length calculus o f the crank

Leng th of the crank

8.3. CALCULUS OF THE EXTENSION

Extension’s length

If L>200…300 mm

Lm=(0.3…0.4)∙L=0.3 ∙ 468.98=140.69mm

L p=L-Lm+l=468-140+50=378.29mm

l=50…80mm=50mm

Extension’s diameter

Bar extension

It is chosen d pe=22 mm STAS 333

)1(1

)ker (1

41898.41830011

27.125694

n for k

swor of number n

mmmm F nk

M L

m

it c

150

46850418

0

0

n for mml

mml L L c

]3[120...100

41.2010014,3

)5050378(3001132)(3233 0

MPa

mml l L F nk

d

ai

ai

pm pe



- 17 -

8.4. RATCHET WHEEL CALCULUS

8.4.1. Choosing the mate rial

It is chosen OL 50 STAS 500/2

R p02=290 MPa

R m=490…610 MPa

8.4.2 Choosing the dimensions

Wheel dimensions

Figure 4.1

8.4.3 Checking the ratchet wheel to requests

mmmmbh

mm z

Db

teethof number z

ratchet themounting iswhichonthread screwtheof diameter exterior thed

mmmmd D

m

m

108.9147,0)8.0...6,0(

94.14102

95

2

)(10

952.95567.1)8.1...6.1(

mmh D D mi 851095

mmh D D me 1051095

mma

mmd a

25

271565.0)2...1(5,0



- 18 -

Bend ing check for the tee th

δ =6…10mm=10mm

Shearing check for the tee th

Checking the tee th contact surface a t crushing

]3[120

63.26402.95

27.12569422

10060.391014

1063.264033

1

22

1

MPa

N D

M F

MPa MPab

h F

ai

m

it

aii

]3[100...80

8086.18

1014

63.26401

MPa

MPa MPa

b

F

af

af f

MPa MPah

F as s 10040.26

1010

63.26401

]3[120...100 MPaas



- 19 -

Checking the assembly on the po lygonal contour at crushing

Figure

8.5. CALCULUS OF THE RATCHET

8.5.1Choosing the materialIt is chosen OL 60 STAS 500/2

R p02=310 MPa

R m=590…710 MPa

mm

contour hexagonal for n

MPa Mpaan

M

t

as

t

ut s

122102

)(6

10051.3312256

27.1256942222



- 20 -


Figure 5.1

8.5.3 Checking the ratche t at e ccentric compression

Total tension

8.6. MAIN CRANK CALCULUS

8.6.1 Choosing the material

It is chosen OT 50 STAS 500/2

R p02=270 MPa

R m=490 MPa


Dimensions of the main crank

δ 1=(0.5…0.6)∙δ=0.5∙10=5mm

MPa MPa

MPa

Mpa g

F

g

e F

ace

ai

e

12016.107

]3[120...100

16.1071011

36.4471

1011

336.4471662

1

2

1



- 21 -

D=d pe+(6…10)mm=22+6=28mm

b1≈D

8.6.3 Checking the crank to requests

Bending tension in B-B section

8.7.CALCULUS OF THE RATCHET PIN

8.7.1 Choosing the materialIt is chosen OL 50 STAS 500/2

8.7.2 Checking the pin at requests

Shearing checking

Figure 7.1

Checking the pin at crushing

]3[120...100

10060

)1428(2832

)50140418(30011

)(32

)(

4444

MPa

MPa Mpa

d D D

l L L F nk

ai

ai

pe

mcmi

MPa MPa

MPa

MPad b

l L F nk

aii

ai

b

cmi

12075.118

120...100

171

6

5)1028(2

)95418(35011

6

)(2

)(2

12

1

1

]3[80...60

6081.16

4

102

63.2640

42

22

1

MPa

MPa MPad

F

af

af

b

f



- 22 -

Checking the pin at be nding

8.8. CALCULATION OF THE COIL SPRING OF COMPRESSION WHICH

MENTAINS THE RATCHET HORIZONTALLY IN CONTACT WITH THE

RATCHET WHEEL

Spring index

Form coefficien t

It is chosen K=1.16 for i=10

Helix diameter

It is chosen d= 1 mm

Medium winding diamete r

Dm=i∙d=10∙1=10mm

Fitting force

F1=4…6N=5N

Number o f actives helixes

n=4…6

It is chosen n= 5

]3[80...60

60221210

63.26401

MPa

MPa MPad

F

as

as

t b

x

]3[120...100

1007410

52

1263.26408

28

33

11

MPa

MPa MPad

F

ai

ai

b

t

i

10

i

d

Di

m



- 23 -

Total num ber o f helixes

nt=n+nc=5+1.5=6.5

nc=1.5

Arrow assembly

Maximum arrow

Maximum exploitation force

Torsion tension

Stiffness of the spring

Length o f the locked spring

H b=n t∙d=6.5∙1=6.5

2

4

44

3

4

31

1

105,8

35.251105,8

10588

mm

N G

mmnd G

D F m

mm s

mm s

2

35.4235.21max

N F F 25.9

35.2

35.45

1

max1max

AOLC for MPa

MPa MPad

D F k

at

at m

t

65650

65037.2731

1025.916,18833

max

12.21058

1105,8

83

44

3

4

m Dn

d Gc



- 24 -

Pitch of the loade d spring

The length of the unloaded spring

Leng th o f the spring corresponding to the maximum exploitation

force

Force correspo nding to the locked sp ring

Exterior diameter

Interior diameter

mm

mmd

mmn

d t

2.0

1,011,01,0

07.22.05

35.41

max

mmd t n H H b 87.11)107.2(55.6)(0

mm H H 52.935.287.11101

mmn

N F F

b

b

b

5.12425.4

205.4

5.122.7

max

max

max

mmd D D m 11110

mmd D D m 91101



- 25 -

Angle of inclination of the unloaded spring he lix

Length o f the semi-finished piece

Figure 8.1

25.71040

arct

Dt arctg

m

mmn D

l t m

s 18.17425.7cos

5.510

cos 0



9. TEHNICAL NORMS FOR WORK SECURITY

THE MAIN TEHNICAL NORMS FOR WORK SECURITY ARE THE FOLLOWING:

- In case of breaking , the repa iring of the jack should be done only by

authorized p erson

- The jack is positioned on ly in sp ecial places for the load

- To prevent accide nts, if after the lifting in necessary the inte rvention

under the lifted load , always the load should be secured

- For bett er function ing it is recommended pe riod jack and couplings

lubrification

- Do not lubricate bo lts and coup lings during utilization, that means when

the jack is und er a certa in load

- Do not lift a higher load than the load ind icated in calculations

- In lifting the load,to action the ratchet mechanism should be used only

the extensions showed in calculations

- Using any othe r type of extension can occur some forces that can break

the mechanism, and so producing accidents

10. STATEMENT SUPPORTING THE CHOOSE OF MATERIALS, OF

MOULDINGS AND DESIGN SOLUTIONS, FOR MAIN PIECES

INTEGRATED IN THE LIFTING JACK(SCREWS,BOLTS)

For a be tte r function of the lifting jack the screws should b e not only

durable but also stiff. Based on this, for manufacturing the mai screws it will

be chosed usual materials(OL): for main screw it is recommended OL 50, for

secondary screw and fixed bolt it is recommended OL 50.

At screws an fixed nut it will be used trapezoida l threads, to increase

the stiffness on the mechanism.

Still, for assembling the cup it will be used a stee l that can be

thermally and thermochemically treate d to avoid depreciation.

Double Action Lifting Jack

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