DOUBLE ACTION LIFTING JACK DESIGNAs Designed by Dan Radulescu ID 17211 Transilvania University of Brasov

1.Establishing the loads acting upon the elements of the jack

Mm = Fm LcMm = MinsI+MinsII

Mm - the moment with which the user acts

Is2 = HI+HpI = H (PI / PI+PII) +HpI

Is2 = HII+HpII = H (PII / PII+PI) +HpII

Is1 - main screw's lengthls2 - secondary screw's length

INPUT DATAType of mechanism - double action lifting jack•Maximum load Q - 17800 N•Elevation - 235 mm•

Q 17800N:=

H 235mm:=

1

2. Main Screw Design

2.1 MaterialThe element will be made out of OL 37 STAS 500/2

Flow limit ReH (Rp02) -> 230 MPa ( for diameter a , 16 < a <= 40 )Resistance to breaking from traction Rm - 360....440

2.2 Predimensioning Calculus

Calculus load Qc , N•β - the turn's tilt = 1.25 .... 1.3 for double acting lifting jack•Q 17800N=

β 1.3:=Qc β Q⋅:=

Qc 23140 N⋅= Thread's interior diameter - d3•

σac 60 MPa⋅:=d3

4 Qc⋅π σac⋅

:=

d3 22.16 mm⋅=d3STAS >= d3 calculated

d3I 22.5mm:=

Nominal diameter - dPitch - PMedian diameter - d2=D2Exterior diameter - D4Interior diameter - d3Interior diameter - D1

we will choose the following thread -> TR 28x5 SR ISO 2904

dI 28mm:= PI 5mm:= D4I 28.5mm:=d2I 25.5mm:= D1I 23mm:=d3I 22.5 mm⋅= D2I 25.5mm:=

2

2.3 Self Locking Condition Checking

μ - friction coef. = 0.11 .... 0.12 - for steel/steel μI 0.11:=

α 30deg:=φ1I

180π

atanμI

cosα

2

⋅:=β2I

180π

atanPI

π d2I⋅

⋅:=Condition -> β2 < φ1

β2I 3.571=φ1I 6.497=

2.4 Checking to composed loads calculus

Torque on the main screw - MinsI, NmmCompression tension - σc , MPaTorque tension - τt, MPa Equivalent tension - σe, MPa

argumentI φ1I β2I+( )deg:=

MinsI Qd2I2

⋅ tan argumentI( )⋅:=

σc 4Q

π d3I2⋅:=

τt 16MinsI

π d3I3⋅:=

σc 45 MPa⋅=

τt 18 MPa⋅=

σe σc2 4 τt2⋅+:=

MinsI 40296 N·mm⋅=

condition -> σe <= σac

σe 57 MPa⋅=σac 60 MPa⋅=

3

2.5 Buckling checking

m 30mm:=λ - slenderness ratio

Lf=Kl - lungime de flambaj

K - it is choosen from buckling scheme

Hl - actual lifting height of the main screw ( measured on the drawing )

m - measured on the drawing

Hl 90mm:=K 0.5:=

l Hl m+:=

lf K l⋅ 60 mm⋅=:=

imind3I4

:=

imin 5.63 mm⋅=

buckling domainλ >= λ0 - elastic bucklingλ < λ0 - plastic buckling

λ0=105 - for OL 37λ0=89 - for OL 50

λlf

imin:=

λ 10.667=

λ0 105:=buckling safety coeficient - c

E 215000MPa:= ca= 3...5 - admited safety coeficient

ca 5:=

λ < λ0 => we have plastic buckling

σf=310 - 1.14λ - pentru OL 37σf=335 - 0.62λ - pentru OL 50

σf 310 1.14 λ⋅−( )MPa:=

cσfσc

:=

condition -> c >= ca

c 6.65=

ca 5=

4

3. Secondary Screw Design

3.1 MaterialThe element will be made out of OL 42 STAS 500/2

Flow limit ReH (Rp02) -> 250 MPa ( for diameter a , 16 < a <= 40 )Resistance to breaking from traction Rm - 410....490

3.2 Number of turns of main screw's nut The nut is situated on the inside of the secondary screw

pa = 7...13 MPa - admited crushing pressure of lubricant layer between the turnsHpI - internal nut lenght

6<= z <= 10

pa 10MPa:= zI 4Q

π dI2 D1I2−( ) pa⋅⋅:=

zI 8.888=

we choose zI 9:= HpI zI PI⋅:=therefore => HpI 45 mm⋅=

3.3 External Thread

choosing the secondary screw's internal diameter - d3

D0 D4I 5mm+:= d3=D0 + (8..10) mmD0=D4+ (4..6) mm ( diameter of the unthreaded area )d3II D0 9mm+:= d3II 42.5 mm⋅=

D0 33.5 mm⋅=

Therefore we choose the external thread -> TR 52x8 SR ISO 2904

Nominal diameter - dPitch - PMedian diameter - d2=D2Exterior diameter - D4Interior diameter - d3Interior diameter - D1

dII 52mm:= PII 8mm:= D4II 53mm:=d2II 48mm:= D1II 44mm:=d3II 43mm:= D2II 48mm:=

5

3.4 Self Locking Condition Checking

μII 0.11:= μ= 0.11 .... 0.12 - for steel/steel

α 30deg:=β2II

180π

atanPII

π d2II⋅

⋅:=

φ1II180π

atanμII

cosα

2

⋅:=

Condition -> β2 < φ1β2II 3.037=

φ1II 6.497=

3.5 Checking to composed loads calculus

Torque on the main screw - MinsII, NmmCompression tension - σc , MPaTorque tension - τt, MPa Equivalent tension - σe, MPa

argumentII φ1II β2II+( )deg:=

MinsII Qd2II

2⋅ tan argumentII( )⋅:=

MinsII 71747 N·mm⋅=σc 4

Q

π d3II2 D02−( )⋅:=

τtMinsII

π

16 d3II⋅d3II4 D04−( )⋅

:= σc 31.2 MPa⋅=

τt 7 MPa⋅=

σe σc2 4 τt2⋅+:= condition -> σe <= σac σe 34 MPa⋅=σac 60 MPa⋅=

6

4. Fixed Nut Design4.1 MaterialThe element will be made out of FC 200 SR ISO 185

Resistance to breaking from traction Rm = 200 MPa•Part's section diameter - from 2.5 mm .....50 mm•

4.2 Number of turns calculus z - number of turnsHpII - lenght of the nutpa - 7...13 MPa for steel/steel - 5...6 MPa for steel/ cast iron

Condition6 <= z <= 10

pa 5MPa:=

zII 4Q

π dII2 D1II2−( )pa⋅:=

zII 5.902= we choose z to be zII 9:=

HpII zII PII⋅:= HpII 72 mm⋅=

4.3 Turn checkingthe turn checking for bending and shearing

for bending

h 0.634 PII⋅:=

σai 45MPa:=Conditionσi <= σaiσi 3

Q D4II d2II−( )⋅

π D4II⋅ h2⋅ zII⋅:=

σi 7 MPa⋅=

for shearing

τaf 35MPa:=Conditionτf <= τafτf

Qπ D4II⋅ h⋅ zII⋅

:=

τf 2.3 MPa⋅=

7

4.4 Choosing the nut's dimensions

De - external diameter - De= D4+(8...10 mm)Dg - external diameter of the neck - Dg=De+(16...20 mm )hg - neck's height ( 8..10 mm )

De D4II 9mm+:=

Dg De 17mm+:=De 62 mm⋅=

hg 10mm:=Dg 79 mm⋅=

hg 10 mm⋅=

4.5 Checking the nut for composed loads

σi - traction tension , MPaτi - torsion tension, MPaσe - equivalent tension, MPa

σt 4Q

π De2 D4II2−( ) 22 MPa⋅=:=

τtMinsII

π

16DeDe4 D4II4−( )

3 MPa⋅=:=σat - 60..80 MPa - for steel - 40..45 MPa - for cast iron or bronze

σat 45MPa:=σe σt2 4 τt⋅( )2+ 26 MPa⋅=:=

Condition : σe <= σat

4.6 Neck checking

σs - reistance to crushingσas - allowed resistance to crushingτf - resistance to shearingτaf - allowed resistance to crushing

σs 4Q

π Dg2 De2−( ):=

τfQ

πDe hg⋅:=

σas 60MPa:=

τaf 35MPa:=σs 9.5 MPa⋅= conditions

σs <= σasτf <= τaf

τf 9.1 MPa⋅=

8

4.7 Choosing and checking the threadedbolt that fill fix the nut on the body

We will choose a M5,M6 or M8 threaded bolt with cilindrical head

We choose the M5x25 STAS 10422 with the below mentioned characteristics

d = 5 mm t = 1.62 mm n = 0.8 mm c1 = 1.2 mm d4 = 3.5 mm

c3 = 3 mm d1= 2.5 mm m = 2 mm

Mg - friction moment on the support surface of the neckMs - moment on the threaded bolt

μ = 0.15...0.20 - for steel/steelμ = 0.12...0.15 - for steel / cast ironcast iron / cast ironbronze / cast iron

μ 0.15:=

Mg13μ⋅ Q⋅

Dg3 De3−( )Dg2 De2−( )⋅:=

Ms MinsII Mg−:=MinsII 71747 N·mm⋅=

Mg 94574 N·mm⋅=Ms 22827− N·mm⋅=

Ms has a negative value so the bolt checking calculus is unnecessary

9

5. Body DesignlsI - main screw's lenghtlsII - secondary screw's lenght

HII HPII

PII PI+

⋅ 144.62 mm⋅=:=

lsII HII HpII+ 216.62 mm⋅=:=

Dci = De+ (2..6) mm

γ - 5...7 degrees

δ - 7...8 mm

δt - 10...12 mm

Dbe = Dbi+(30..50) mm

Dbi - measured on the drawing

H1 = HII+ (30...50) mm

HII - how much the secondary screw rises

Hc -=H1+HpII-hg+(0...10) mm

Dci De 4mm+:=

γ 7deg:=

δ 7mm:=

δt 10mm:=

Dbi 111.34mm:=Dbe Dbi 40mm+:=

HI HII 40mm+:=σc = body compression checking

σs = crushing checking for the support surface

σac = allowed body compression checking 60...80 MPa

σas = allowed crushing checking for the support surface 2...2.5 MPA

Hc HI HpII+ hg− 10mm+:=

σacc 70MPa:=σc 4

Q

π Dg2 Dci2−( )⋅⋅:= σs 4

Q

π Dbe2 Dbi2−( )⋅⋅:=

σas 2.5MPa:=

Dci 66 mm⋅=conditions σc <= σaccσs <= σas

Dbi 111.34 mm⋅=σc 12 MPa⋅=

Dbe 151.34 mm⋅=σs 2.16 MPa⋅=

HI 184.62 mm⋅=Hc 256.62 mm⋅= lsI HI HpI+ 229.62 mm⋅=:=

10

6. Cup calculusChoosing the bolt

ds=(0.15...0.25)dc

dc= d+(2..4) mm

d = nominal diameter of the main screw's thread

A bolt standardized cilindrical bolt must be used with d >= ds

We choose the M8x18 STAS 10422 with the following characteristics expressed in mm

d = 5 t = 1.6 n = 0.8 c1 = 1.2 d4 = 3.5 c3 = 3 d1 = 2.5 m = 2

d 28mm:=

dc d 4mm+ 32 mm⋅=:=

ds 0.25 dc⋅ 8 mm⋅=:=

Checking the bolt for shearing and crushing

τaf 80MPa:=

σas 120MPa:=

shearing

τaf= 65...80 MPa

σas= 100...120 MPa

Dc= (1.4...1.6)dc

τf 4MinsI

π dc⋅ ds2⋅⋅:=

τf 25 MPa⋅=

crushing - cupDc 1.4 dc⋅ 44.8 mm⋅=:= σs 4

MinsI

ds Dc2 dc2−( )⋅⋅:=

σs 20 MPa⋅=

crushing - end of the screw

σs 6MinsI

ds dc2⋅⋅:= conditions

τf <= τafσs <= σasσs 30 MPa⋅=

11

Checking the minimized section of the screw's head to composed loads σe , MPa

σcQ

πdc2

4⋅ ds dc⋅−

:=

τt 16MinsI

π dc3⋅ 1dsdc

−

⋅⋅:=

σe σc2 4 τt2⋅+:=

σac 60 MPa⋅=

σc 32 MPa⋅=

τt 8 MPa⋅=

σe 37 MPa⋅=

conditionσe <= σac

σe 37 MPa⋅=σac 60 MPa⋅=

7. Efficiency calculus

ηd2I tan β2I( )⋅ d2II tan β2II( )⋅+( )

d2I tan argumentI( )⋅ d2II tan argumentII( )⋅+:=

η 0.53=

12

8. Horizontal ratchet actioning mechanism calculus

b - gear thickness

1 - ratchet gear2 - ratchet3 - pin4 - splint5 - washer

6 - board7 - pusher8 - spring9 - adjustment screw10 - sheath

13

8.1 Crank length Mm - total moment•Fm - force with which the user acts •

(150 - 300 N)Lc - calculus length of the crank•

Mm MinsI MinsII+ 112043.18 N·mm⋅=:=

Fm 300N:=

k 1:=l0 - user grabing length•

(50 mm for n=1 and 100mm for n =2)n 1:=

l0 50mm:=n - number of workers ( 1 or 2 )•K - unsimultaneity of users action •

(K=1 for n=1 and K=0.8 for n=2Lc

Mmk n⋅ Fm⋅

373 mm⋅=:=

L Lc l0+:= Actual crank length L 423.48 mm⋅=

8.2 Extension calculus

extension length - LpLp - extension length•

L>200..200mm => Lm=(0.3 .. 0.4)L and Lp=L-Lm+l

l - guidance length of the extension• =50..80 mm

Lm 0.3 L⋅ 127.043 mm⋅=:=

l 50mm:=

Lp L Lm− l+:=

Lp 346.434 mm⋅=

extension diameter - dpe

σai 120MPa:=

dpe3 32 k⋅ n⋅ Fm⋅ Lp l0− l−( )⋅[ ]

π σai⋅18.445 mm⋅=:=

dpe 19mm:=

We will use comercial pipe 19 / OL45 STAS 333

14

8.3 Ratchet wheel calculus

8.3.1 Material

The element will be made out of OLC 45 STAS 880

Flow limit ReH (Rp02) -> 370 - 500 MPa - heat treated : tempering and high temperature tempering- Resistance to breaking from traction Rm - 630..850 MPa

8.3.2 Choosing the dimensions d D4II:=d 53 mm⋅=

Dm 1.6 d⋅ 84.8 mm⋅=:=

z 10:=

πDm2z

13.3 mm⋅=

b 12mm:=

h 0.7 b⋅ 8.4 mm⋅=:=

Dm = (1.6....1.8)d •b<= π*Dm/2z•h=(0.6 .... 0.8)b•a <= 0.5*d - (1..2)mm for hexagonal profile•

Di Dm h−:=De Dm h+:=

a 0.5 d⋅ 2mm−:=

b 12 mm⋅= Di 76.4 mm⋅=h 8.4 mm⋅= De 93.2 mm⋅=a 24.5 mm⋅= Dm 84.8 mm⋅=z 10=

8.3.2 Checking the gear for loads and stresses

tooth checking for bending σi MPa• δ - tooth thickness =(6...10)mmδ 9mm:=

F12 Mm⋅( )Dm

2643 N⋅=:=condition σi <= σai

σi3 F1⋅ h⋅( )

b2δ⋅

51 MPa⋅=:= σi 51 MPa⋅=σai 120MPa:=

15

tooth checking for shearing τ1 MPa•

τfF1b δ⋅

:= condition τf <= τaf

τf 24 MPa⋅=τaf 100MPa:=

τaf 100 MPa⋅=

tooth contact surface checking for crushing σs MPa•

σsF1δ h⋅

:= condition σs <= σas

σas 120MPa:= σs 35 MPa⋅=σas 120 MPa⋅=

checking of polygonal contour assembly for crushing σs MPa•

n 6:= n = 6 for hexagonal profile assembly•

δt δ 2mm+ 11 mm⋅=:=

σs 2Mm

n a2⋅ δt⋅:=

condition σs <= σas

σs 5.66 MPa⋅=

σas 120 MPa⋅=

16

8.4 Ratchet calculus8.4.1 Material

The element will be made out of OLC 45 STAS 880

Flow limit ReH (Rp02) -> 370 - 500 MPa - heat treated : tempering and high temperature tempering- Resistance to breaking from traction Rm - 630..850 MPa

8.4.2 Choosing the dimensions

Dm 84.8 mm⋅=

l1 = (0.85..1.0)Dm•α = γ+(3...5)deg•db= 6...12mm•m = (1.75 ....2.25)db•

l1 0.85Dm 72.08 mm⋅=:=

db 10mm:=

z 10=

R db:=

m 1.8 db⋅:=

γ 2180π

⋅ atanπ

4 z⋅tan

180π

asinDm2l1

⋅ deg⋅

⋅

6.539=:=

α γ 4+:= m 18 mm⋅=g 14.16mm:= l1 72.08 mm⋅= δ 9 mm⋅=

e 7.97mm:= db 10 mm⋅= γ 6.539=

sa 1.72mm:= R 10 mm⋅= α 10.539=

De 93.2 mm⋅=

Di 76.4 mm⋅=Dm 84.8 mm⋅=

z 10=

b 12 mm⋅=h 8.4 mm⋅=

17

8.4.3 Checking the ratchet to excentric compression

Total tension σtot •condition σtot <= σaiσtot

6 F1⋅ e⋅( )

g2δ⋅

F1g δ⋅

+:=

σtot 90.762 MPa⋅=σai 120MPa:=

σai 120 MPa⋅=

8.5 Actual crank calculus

8.5.1 Material

The element will be made out of OL 50 STAS 500/2

Flow limit ReH (Rp02) -> 250 MPa ( for diameter a , 16 < a <= 40 )Resistance to breaking from traction Rm - 410....490

8.5.2 Choosing the dimensions

δ1 - (0.5...0.6)*δ•D - dpe + (6..10)mm•

δ1 0.5 δ⋅:=δ1 4.5 mm⋅=

D dpe 6mm+:= D 25 mm⋅=b1 D:= b1 25 mm⋅=

8.5.3 Checking the crank to loads and stresses

number of workers = 1 => n 1:=

bending tension from section C - C , σi, MPa•

condition σi <= σaiσi

k n⋅ Fm⋅ Lc Lm− l+( )⋅[ ]π

32 D⋅D4 dpe4−( )

:=

σi 87 MPa⋅=σai 120MPa:=

bending tension from section B - B , σi, MPa •

σik n⋅ Fm⋅ Lc l1−( )⋅[ ]

2 b1 db−( )2⋅ δ1⋅ 6

:= condition σi <= σai

σi 268 MPa⋅=σai 120MPa:=

18

8.6 Bolt of the ratchet calculus

pin checking for shearing τf , MPa•

τfF1

2π db2⋅( )

4

⋅

:=

condition τf <= τaf

τf MPa⋅=

τaf 80MPa:=

pin checking for crushing σs , MPa•

condition σs <= σasσs

F1db δt⋅

:=

σs 24 MPa⋅=

σas 80MPa:=

pin checking for bending σi , MPa•

condition σi <= σai

σi8 F1⋅

δ

2δ1+

⋅

π db3⋅:= σi 61 MPa⋅=

σai 120MPa:=

19

8.7 Elicoidal compresion spring calculusthe spring that mantains the ratchet in contact with the ratchet gear's teeth

i - spring's index•k - shape coeficient •

k=1.2 - for i=8 k=1.17 - for i=9 k=1.16 - for i=10

d - winding's diameter • recommended d=0.8...1.5mm d=0.8 ; 1 ; 1.2 ; 1.4 ; 1.5

Dm - average winding diameter •

d 1.2mm:=

i 8:=

Dm i d⋅:= k 1.2:=Dm 9.6 mm⋅=

F1 - fitting strength•

F1 5N:=

n - active number of windings •nt - total number of windings•nc - end windings number•

n 6:=nc 1.5:=nt n nc+:=

δ1 - fitting arrow, mm•δmax - maximum arrow, mm•

- measured on drawing

G 85000MPa:= sa 1.72 mm⋅=

δ18 F1⋅ Dm3⋅( )

G d4⋅n⋅ 1.2 mm⋅=:=

δmax δ1 sa+:=

δmax 2.92 mm⋅=

Fmax - maximum working•force

Fmax F1δmaxδ1

⋅ 12N=:=

τt - torsion stress, MPa• τat=650MPa for OLC65

conditionτt <=τat

τt 8 MPa⋅=τt

8 k⋅ Fmax⋅ Dm⋅( )

π d3⋅:=

τat 650MPa:=

20

c - spring's rigidity, MPa•

cG d4⋅( )

8 n⋅ Dm3⋅:=

c 4.15 10 3−× m MPa⋅=

Hb - length of blocked spring•Hb nt d⋅:=

Hb 9 mm⋅=

t - loaded spring's pitch• Δ>=0.1d - windings play of the spring loaded by Fmax

Δ 0.1 d⋅ 2mm+:=

t dδmax

n+ Δ+:=

Δ 2.12 mm⋅=

t 3.807 mm⋅=

H0 - unloaded spring length, mm•H1 - length according to fitting force, mm•Hm - length according to Fmax•

H0 Hb n t d−( )⋅+:= H0 24.645 mm⋅=

H1 H0 δ1−:= H1 23.44 mm⋅=

Hm H0 δmax−:= Hm 21.72 mm⋅=

Fb - force of blocked spring ( winding on winding )• δb= δmax+Δn

δb δmax Δ n⋅+:= δb 15.64 mm⋅=

Fb Fmaxδbδmax

⋅:= Fb 65N=

D - external diameter •D1 - interior diameter•

D Dm d+:= D 10.8 mm⋅=

D1 Dm d−:= D1 8.4 mm⋅=

21

α0 - winding's tilt angle of unloaded spring•ls - semifabricated length•

α0 atant

π Dm⋅deg⋅

:=

lsπ Dm⋅ nt⋅( )

cos α0 deg⋅( )

:=

α0 0.002=

ls 226.19 mm⋅=

22