Screw Jack Design

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SCREW JACK –

Screw jack is a portable device consisting of a screw mechanism used to raise or lower the load. There are two types of jacks most commonly used,

1] Hydraulic

2] Mechanical

Merits of Screw jack –

1] Can be used to lift a heavy load against gravity.

2] Load can be kept in lifted position.

3] Due to leverage obtained by handle force required to raise load is very less & can be applied manually also

Demerits of screw jack –

1] Chances of dropping of load

2] Tipping or slipping of load.

3] This failure is not “SAFE FAIL”& can cause serious accidents.

Reasons of Accidents –

1] Load is improperly secured on jack.

2] The screw jack is overloaded.

3] Center of gravity is off center with axis of jack.

4] Jack is not placed on hard & level surface.

5] Using for other purpose instead of using it for which it is designed.

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STEP – 1

Problem Statement: –

Design a screw jack for lifting a load of 75KN with manual operation for 0.5m of height.

STEP – 2

MATERIAL SELECTION: –

Screw jack has following parts :-

1] FRAME

2] SCREW

3] NUT

4] HANDLE

5] CUP

6] SET SCREW

7] WASHER

1] FRAME –

• FRAME SIZE - Most of the times frame is conical in shape and hollow internally to accommodate a nut & screw assembly. The hollow conical shape insures a safe & complete resting of a jack on ground. If it is provided with legs like structure , it quite possible that in case of uneven distribution load may fail down because all legs will not touch ground.

• FORCE ANALYSIS – The force by a load is directed by a cup to screw then is directed by cup to screw then to threads of screw to nut then to frame so it is always compressive in nature.

• MANUFACTURING PROCESS – The complex shape of frame leads us to use a ’Casting’ process for manufacturing.

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For all this purpose We need to select a cast iron as material for frame.

We select a FG200 as material for frame such as it contains carbon precipitates as “graphite flakes” as graphite is soft in nature it improves its ability to resist a compressive load.

FG200 = Graphite flakes Gray cast iron with ultimate tensile strength of 200N/mm2

FG200 microstructure

2] SCREW –

• Screw size – Screws is nothing but a member having Helical groove around periphery of solid bar. It can be around 22 to 100mm diameter for square power screws & 24 to 100mm for trapezoidal power screws.

• Thread profile – The screw or power screw thread is always a square type because it has more efficiency than trapezoidal threads and there is no radial thrust on screw i.e. no Bursting Pressure, so motion is uniform.

• Square threads usually turned on lathes using single point cutting tool. It leads us to use free cutting steel.

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• Square threads are weak in roots. Wear of thread surface lead us to use

“Unalloyed free cutting steel”.

We select “25C12S14 UNALLOED FREE CUTTING STEEL”

It indicates - 025% carbon, 1.2% manganese, 0.14%sulphur

It has tensile strength of 560N/mm2 with 10% elongation. Sulphur gives resistance to wear & 0.25% carbon gives it sufficient strength to compensate weakness in roots. also easy in cutting due to manganese.

3] NUT –

• As we know there always a relative motion between screw and nut, which cause a friction. The friction causes wear if some material is used for screw & nut it will wears both components. So one out of two has to be softer than other so as to ease of replacement. The size & shape of screw is costlier than nut, so generally we use softer material for nut than screw.

• Phosphor bronze is ideal material for nut which is a copper alloy having 0.2%phospher which increases tensile strength. Ultimate tensile strength for this is 190mpa and coefficient of friction is 0.1 Bearing pressure is 10mpa.

• Advantages of phosphor bronze are, 1] Good corrosion resistance. 2] Low coefficient of friction 3] Higher tensile strength than copper brass.

4] HANDLE –

• Handle is subjected ti bending moments. • So plain carbon steel with 0.3%carbon i.e. 30C8 can be selected. • Yield strength in tension is 400mpa

5] CUP –

• Shape of cup is again complex and so economical to manufacture by Casting process, hence material will be cast iron with grade FG200.

6] SETSCREW –

• Purpose of set screw is to resist motion of nut with screw.

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• It can be of commercial steel.

7] WASHER –

• Washer is to provide uniform force of tightening nut over screw force by enlarging area under actions of force. We can use commercial steel.

STEP – 3

ERGONOMIC CONSIDERATION:-

• According to ergonomic consideration a mature man can lift weight up to 1/2 of his weight.

• This criterion is applied when work is not continuous. Capacity is 1/3 for continues work.

• Thus average weight of person is 60KG. • So he can apply 30KG i.e. 300N • Suppose two worker are working the total force

= 300*2*0.9 =540N

STEP – 4

FACTOR OF SAFETY:-

• Many times workers have to work below heavy load such a high risk of work leads to select higher factor of safety values. The dropping or slipping of weight is life hazards. So it has to be foolproof. We select factor of safety as ‘5’ for steel component and ‘6’ for cast iron.

STEP – 5

DESIGN OF SCREW:-

FROM SELECTED MATERIAL SPECIFICATION:

SYC = SYT = 560 mpa

бc = compressive stress = 560/5

= 112 mpa.

1] CORE DIAMETER

The diameter only consider compressive stress but screw is subjected to torsion and bending moments for that we increase a diameter to higher values we will use a screw with 55mm nominal diameter & 9mm pitch .( IS 4694 d = 55mm p = 9mm dc = 55-9 = 46mm > 30mm dm = d-0.5p = 55-4.5 = 50.5mm

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FROM SELECTED MATERIAL SPECIFICATION:-

= 112 mpa.

б= W / (π/4)*dc2

112=(75*10^3) / π/4*(dc2)

dc = 30mm

The diameter only consider compressive stress but screw is subjected to torsion and bending moments for that we increase a diameter to higher values we will use a screw with 55mm nominal diameter & 9mm pitch .( IS 4694-1968:basic dimensions for square threads)

The diameter only consider compressive stress but screw is subjected to torsion and bending moments for that we increase a diameter to higher values we will use a screw with 55mm

eads)

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Assuming Screw has single start threads.

l = p = 9mm

tanα = l/π*dm

= 9/(π*50.5)

α =3.246О Coefficient of friction between screw and nut is 0.1 i.e. µ=0.1 Considering bad lubricating conditions assume µ=0.18

tanФ = µ = 0.18

Ф=10.2039 > 3.0246

hence we can say that screw is self locking

Torque required raising the load,

T = W*(dm/2) * tan(Ф+α)

= (75*10^3)*(50.5/2)* tan (10.2039+3.246)

= 452.898*10^3 N-mm

The diagram shows no cross section of screw is subjected to addition of collar friction torque & nut torque.

At section x-x

ζ max = 16T / (π * dc^3)

= (16 * 452.898 * 10^3)/ (π * 46^3 )

= 23.697 mpa

ζPerm.= (0.5 * syt) / fos

= (0.5*560) / 5

56mpa > 23.69mpa

Screw is safe in tensional shear stress.

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Stress due to bending:

Force due to hand i.e. P is responsible for bending of screw thus bending moment at any section x-x can be given as,

Mb = P * l

Lifting height is 500mm

So, l = 550+20+50

= 570mm

Mb = 300*570

= 171*10^3.

б=32Mb/(π*dc^3 )

= (32*171*10^3 )/ (π*46^3)

= 17.89mpa < <112 mpa

Screw is safe in bending also.

Check in combined torsional and bending moment.

ζmax= ���б�� �� � ^2

= �� �.���� � ^2 � �23.697�^2

=25.3289mpa < 56 mpa.

Screw is safe in combined torsional & bending moment

Buckling Criterion:-

When load is raised the screw acts as column & there are chances of buckling or crushing of it. So we have to decide whether column is long or short.

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Since one end of screw is fixed & other is free the end fixity coefficient is 0.25 . border line between short &long column is

Syt / 2 = (η*π2 *E) / (l/k) 2

560/2=0.25*π2*207000 / (l/k)2

l/k = 42.709

We consider length as 550mm

K=√(I/A)

I=inertia

K= Radius of gyration

A= cross sectional area

I=π/4 * dc^4

=219.786*10^3mm^4

A= 1661.9925mm^2

K=11.5

Slenderness ratio l/K= 550/11.5=47.8261.

Slenderness ratio is more than critical slenderness ratio.

We treat screw as long column hence using Eulers formula.

Pcr = π^2EI / (l/k)^2

= (0.25*207000*π^2*1661.9025) /(47.826)^2

= 371.09 >> 75KN

Safe in Buckling.

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STEP – 6

DESIGN OF NUT :-

The permissible bearing pressure between steel screw & bronze nut is 10mpa

No. of threads required

Pb = W / ( π/4 (d^2 – dc^2)*z)

10 = (75*10^3) / π/4(55^2 – 46^2)*z

Z=10.50

=11

Height of Nut

H= z*p

=11*9

=100mm

Transverse shear stress at root of threads in nut is given by ,

ζ=W / (π*d*T*Z)

=(75*10^3) / (π*55*4.5*11)

=8.768 mpa <<19 mpa

Nut is safe in transverse shear stress at root

Transverse shear stress at threads in screw is given by,

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Fig. Transverse shear area on screw body

ζ=W / (π*dc*T*Z)

=(75*10^3) / (π*46*4.5*11)

=10.484 mpa<<56mpa

Safe in transverse shear stress at screw

• DESIGN OF NUT FOR CROSS SECTION

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бt = W/(π/4*(Do^2-Di^2)

112 = 75*10^3/π/4(Do^2-55^2)

Do= 65mm

By emperical relations

Do’ = 1.3 * Do

=65 * 1.3

= 85mm

For width of column

Considering shrar failure.

ζ= W/π*Do*t

19 = 75*10^3 / π*65*t

t=20mm

checking crushing at collar

бc=W/(π/4)*(Do’^2-Do^2)

= 75*10^3 / (π/4)*(85^2 – 65^2)

= 31.830 << 38mpa

Hence safe in crushing at collar

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STEP – 7

DESIGN OF CUP.

D = 1.6 * d

=1.6 * 55

=88mm

D’ = 0.8 * d

=0.8*55

=44mm

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STEP – 8

COLLAR FRICTIONAL TORQUE-

Tf = (µ*W/4)*(Do+Di)

=[(0.2*75*10^3) /(4)]*(88+44)

=495*10^3 N-mm

Total torque = 495*10^3 + 452.998*10^3

=947.898*10^3N-mm

This torque operator has to overcome, suppose 2 operators are working,

947.898*10^3 = (300*0.9*2) * Lh

Lh = 1760mm

It not practically possible to have 1760mm handle.

So we will replace sliding friction by rolling friction by introducing “Thrust ball Bearings” with this friction torque becomes negligible,

For selecting bearing from manufacturers catalogue, we assume life of bearing to be 3000hrs and rotated by handle at 10rpm.

L= (60*n*Lh) / (10^6)

=1.8milion revolution

Dynamic load carrying capacity.

C = PL^(1/3)

=91233.029N

It is assumed that bore diameter of bearing is equal to core diameter or more than that

Assume it to be 50mm

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We select 51310 having dynamic load carrying capacity equal to 97500 N

d = 50mm D = 95mm H=31mm D1=52mm

STEP – 9

DESIGN OF HANDLE / TOMMY BAR:-

As we are using Thrust ball bearing friction at collar is neglected.

Tc = F * Lh

452.89*10^3 = (300*2*0.9) * Lh

Lh = 850mm

Since handle is subjected to bending moment only,

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Mb = P * Lh

= 300 * 850

= 255*10^3 N-mm

бb= (32M) /(π *dh^3)

= (32*255*10^3) / (π* dh^3)

dh = 40mm

STEP 10

Dimensions for Frame can be selected empirically.

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REFERENCES:-

DESIGN OF MACHINE ELEMENTS V.B.BHADARI McGraw Hill publication MATERIAL SCIENCE KHODGIRE POWER SCREWS MODULE 6 VERSION 2 ME, IIT KHARAGPUR (PAPER)