1.

INTRODUCTION
A mechanical jack is a device which lifts heavy equipment. The most common form is a car jack, floor jack or garage jack which lifts vehicles so that maintenance can be performed. Car jacks usually use mechanical advantage to allow a human to lift a vehicle by manual force alone. More powerful jacks use hydraulic power to provide more lift over greater distances. Mechanical jacks are usually rated for a maximum lifting capacity (for example, 1.5 tons or 3 tons). There are several kinds of jacks are available are listed and described below. 1.1. Hydraulic Jack Hydraulic jacks are typically used for shop work, rather than as an emergency jack to be carried with the vehicle. Hydraulic jacks are often used to lift elevators in low and medium rise buildings. A hydraulic jack uses a fluid, which is incompressible, that is forced into a cylinder by a pump plunger. Oil is used since it is self lubricating and stable. 1.2. Floor Jack In a floor jack a horizontal piston pushes on the short end of a bell crank, with the long arm providing the vertical motion to a lifting pad, kept horizontal with a horizontal linkage. Floor jacks usually include castors and wheels, allowing compensation for the arc taken by

the lifting pad. This mechanism provides a low profile when collapsed, for easy maneuvering underneath the vehicle, while allowing considerable extension. 1.3. Bottle Jack A bottle jack or whiskey jack is a hydraulic jack which resembles a bottle in shape, having a cylindrical body and a neck, from which the hydraulic ram emerges. They have a capacity of up to 50 tons and may be used to lift a variety of objects. Typical uses include the repair of automobiles and house foundations. Larger, heavy-duty models may be known as a barrel jack. 1.4. Strand Jack A strand jack is a specialized hydraulic jack that grips steel cables; often used in concert, strand jacks can lift hundreds of tons and are used in engineering and construction. 1.5. Screw Jack It is a device used for lifting heavy loads, by applying a comparatively smaller effort at its handle. It is operated by turning a leadscrew.

2. DESIGN PROCEDURE FOR A SCREW JACK

2.1. Efficiency of Screws:
In lead screw of screw jack square thread is used. If angle is greater then the machine is reversible otherwise it has a low efficiency. Then we should find the relation between angle of thread and efficiency of screw.

When nut moves up F = W tan (+) where W is weight of load, is helix angle, is co-efficient of friction.
Force Diagram

When nut moves down F= W tan (-)

When the nut or screw is to move against a load the force needed for the movement has to overcome the friction of nut and screw and that of collar and support.

Now force needed to overcome the nut and screw friction, F1 = W tan (+) Force needed to overcome the friction of collar and support, F2 = cW where, c is co-efficient of friction of collar and support. F1 is the force acting on the mean radius of the thread and normal to W.

Thread Profile

So total torque needed to raise the load

= Thread torque + Collar friction torque

= F1

+ cW

where, DC = Friction diameter of the collar.

(considering uniform pressure and when power loss in friction is to be found out.)

(considering uniform rate of wear and when power transmitted is to be determined.)

So total torque,

T=W

tan (+) + W1

So work done in raising the loads when one complete turn is given to the screw = 2T = W dm tan (+) + cWDc So efficiency of screw jack = =
( )

(Work input)

Now, = tan = = = =
( ( ( ) ( ( ) ) )

(Neglecting collar friction, c ) (for lifting the load)

Efficiency, =

(for lowering the load)

For maximum efficiency, = 0, = 45 max =

Efficiency curve
80 70 60 50 40 30 20 10 0 Thread Angle

Efficiency curve

Alpha 6 13 20 27 34 41 48 55 62 69 76

2.2. Core Diameter of the Threaded Length:

Let, =Core diameter of the screw threads which are square in nature, W= Load to be lifted for lower down by the screw jack. Considering the threaded spindle to be under direct compressive load, f=

where, f = Maximum allowable compressive stress in the material of the threaded spindle. The standard proportions of the square threads are taken from the relevant I.S table.

2.3 Torque Required to Overcome Friction Between the Screw Threads and Nut:
Let, Then, = Required torque = W tan (+) Where, = Helix angle, = Angle of friction and = Mean diameter of the screw.

Screw Jack

2.4. Shear Stress Set Up Due to Above Turning Moment (

= Where, = The shear stress set up on the threaded spindle due to torque

):

2.5. Principal Stresses in the Threaded Spindle:

Maximum principal stress (tensile or compressive) is given by = {f + }

Maximum shear stress is given by = 1/2 and should be with in permissible stress limits.

2.6. Height of the Nut:

Let, h = Required height of the nut. =

Where, n = Numbers of threads in contact with the thread of the vertical spindle. From the above expression, n is to be found out. Now, h = n p (p = Pitch of the threads)

2.7. Checking the Stresses in the Screw and Nut:

Shear stress set up in the screw and in given by = Where, t = Total depth of the threads. = p/2 Shear stress set up in the nut is given by
( )

2.8. Determination of the Inner Diameter and Thickness of the Collar of the Nut:
Considering tearing strength of the nut, we get
( )

Where,

= Inner diameter of the collar, = Outer diameter of the screw threads.

Considering crushing strength of the collar of the nut, we get

( )

Where, D2 = outer diameter of the collar of the nut. Considering shearing strength of the collar of the nut, we get
( )

where,

= Thickness of the nut collar.

2.9. Determination of Diameter of the head:

Let, = Diameter of the head.

Then, empirically, = 1.75 The diameter of the seat of cup is taken as equal to diameter of the head. Let, = Diameter of the pin with which the cup is loosely fitted. = h = 2D D = Diameter of the handle. /2

Then, empirically, And Where,

2.10. Torque Required to Overcome Friction Between the Bottom of the Cup and Top of the Head of the Screw Jack:
Let, = Required torque.

Considering uniform pressure, = 2/3 W {(

( ) ) ( ( ) )

Considering uniform wear, = Where, W

= Coefficient of friction between the cup and head, = Radius of the head, = Radius of the pin.

2.11. Length of Handel Required to Lift the Load:

Let, L = Length of effort Handle, P = Force applied at the end of effort handle to lift the load. Then, p L = + =T

A normal man can physically apply force which ranges from 300N to 400N or even smaller if he works continuously, L= Some allowance to L may be added to facilitate gripping.

2.12. Determination of diameter of the handle:

Let, d = Required diameter of the effort handle. Then, M = Z

Where maximum B.M = allowing bending stress in the material of the effort Handle Z = Section modulus of the section of the handle = Here M = T =

2.13. Other relation Taken Empirically:

H = Height of the head, = Thickness of the nut collar, = Thickness of the base of the screw jack, = Thickness of the hollow body of the screw jack.

Then according to the empirical relation, H = 2d 2

0.25

2.14. Checking the screw for bucking load:

Unsupported length of the screw is given by

l = Lifting hight + 1/2 Height of the nut According to the Rankines Formula critical load on the vertical threaded spindle is given by
( ) ( )

Where,

Yield stress in the material of the threaded spindle. = core area =

Where,

= core diameter of the screw = equivalent length of l

K = least radius of gyration of the spindle at core diameter = a constant for the material of the screw The column of length l may be taken as fixed at one end and free at the other. The value of buckling load as found out above should be greater than W.

Finally efficiency of the screw jack is to be found out to check whether the screw jack is self locking or not.

3. Design of a Screw Jack to lift a load upto 5 ton.

Given, W = 5 Ton = 50 KN = 50 103 N Lifting height (H) = 300 mm, f = 200 MPa = 200 N/mm 2, fs (for the screw) = 126 MPa, f t (for the nut) = 100 MPa, fc (for the nut) = 90 MPa, f s (for the nut) = 80 MPa, f b = 20 MPa, (co-efficient of friction) = 1.5

3.1. Design of Threaded Spindle:

Let, dc = Core diameter of the threaded spindle. Allowable direct stress is taken as half of the stress at elastic limit f= Now, = (Taking Factor of Safety as 2) N/mm2 = 100 N/mm2

= or, dc = 35.68 mm. 36 mm.

From the standard table of square threads (fine series), we do not have 36 mm as dc. So let select the next higher value of d c which is 37mm.

So, we get, i. dc = 37 mm. ii. Pitch of screw threads = 7 mm. iii. Do (nominal diameter of threads) of the screw = 44 mm. iv. Stress area = 1075 mm2 (at the core diameter of the screw).

3.1.1. Checking for Principal Stresses Set up in the Threaded Spindle: Mean diameter of the threads is given by Dm = = mm. = 40.5mm. 42mm.

Let, = Helix angle Then, tan = = = 3.04 Let, = Angle of friction Then, tan = 0.15 = 8.53 = 0.053052

Torque required to overcome friction between the screw and nut is given by = W tan ( +) =105 tan (0.0777+8.53) = 306530 N mm. Compressive stress due to axial lode is given by = N/mm2 = 46.51 N/mm2 M-mm.

Shear stress in the spindle due to torque is given by = = N/m2

= 30.786 N/mm2 Now maximum principal stress is given by = 1/2 {f+ = 1/2 {46.51+ = 61.83N/mm2 } ( ) }

3.2. Design of Nut:

Height of the nut is given by H=n p

Where, n = Number of thread in contact with threads of the spindle, p = Pitch of threads of the screw or nut, t = thickness or depth of the threads = p/2 = 7/2 = 3.5 mm. Now bearing stress is given by, =

or,

20 =

n=

= 5.61 = 6 H=6 7 = 42 mm.

3.2.1. Checking for stress set up in the nut: We know that shear stress set up in the threads is given by
( )

= = N/mm2 = 20.48 N/mm2

= = N/mm2 = 17.22 N/mm2 and

( )

Since

are with in the given limits of 60 N/mm2 and

40N/mm2, [ ]

The data found out for the nut as mentioned above are safe. Let, D1 = outer diameter of the nut, D2 = outer diameter of the nut collar, t1 = thickness of the nut collar. Allowable tensile stress in the nut = 50 = N/mm2 = 50 N/mm2

= or,

) ]

D1 = 56.65mm.

58mm.

Allowable Crushing stress Failure in the Nut: = N/mm2 = 45 N/mm2

3.2.2. Considering Crushing Failure in the Nut:

( )

or,

45 =

) }

D2 = 69.12mm.

70mm.

Considering Shearing Failure of the collar nut, we get,

( )

= 6.86 mm.

8 mm.

3.3 Design of the Cup:

Diameter of the cup at the bottom i.e diameter of the head on the head on the top of the threaded spindle is given by
= 1.75 = 1.75 44 = 77 mm

78mm

The head is provided with two holes at right angles to the each other for inserting the handle used for rotating the threaded spindle .The head is provided with a pin of diameter = = = 39mm 40mm.

It is loosely fitted to the cup Other diameter of the cup may be taken as Height of the cup = 100 mm. Thickness of the cup ( ) = 6mm. Diameter of the cup at its top = 170mm.

3.4 Design of the Handle:

Let, = Torque required to overcome friction between the cup and the collar.

Then, assuming uniform pressure condition,

= 2/3

W {(

) )

( (

) )

}
( ) ( ) ( ) ( )

= 2/3 0

=2/3 = 228850 N-mm

45.77

Total torque required to the handle is given by T= +

T = 306530 + 228850 = 535380 N- mm. Let, L = Length of the handle in mm. P = force applied at the end of the handle = 300 N It varies from 300N to 400N 300 or, L = 535380

L = 1784.6 mm.

Assuming some allowance for the gripping the handle during rotation, we get L = 1800 mm.

A lever acts as a cantilever fixed at the head end and free at the effort end Then maximum B.M is given by M=P = 300 L 1800 = 540000 N-mm

Now, bending stress in the handle = allowable tensile stress = let = 100 N/mm2

D = Diameter of the handle.

Then, section modulus of the handle Z= M = bending stress or, or, 540000 = 100 D = 38.03 mm. 40mm. section modulus

Empirically, =2 = 0.5 n = 4mm. = 0.25

h = 2D = 2 40 = 80 mm.

= 2 8 mm. = 16 mm. = 0.5 8 mm. = 4 mm.

= 0.25 44 mm. = 11mm.

12mm.

3.4.1. Checking the screw for buckling load: We know that unsupported length of the screw (i.e. length of the screw for buckling load) is given by l = lifting height + height of the nut = 300 + 80/2 mm. = 340 mm. According to Rankines formula critical load or load is given by =
( ) ( )

where, fc = Yield stress in compression = 200 N/mm2 Ac = Cross sectional area of the screw on the basis of core diameter = 1075 mm2 IE = Equivalent length of the column. Considering the bulking length as fixed at the nut and free at the top, we get, IE = 2 l = 2 340 mm. = 680 mm. K = Least radius of gyration of the column section

= 9.25 mm.

= Constant for the material of the threaded spindle =

(assumed).

The critical load is given by P=

( )

N =127737 N

Since this load is greater than 5 ton (given load), the threaded spindle will not buckle.

3.5. Design of the Body:

D5 = 2.25 D2 = 2.25 70 mm. = 157.5 mm. D6 = 1.75 D5 = 1.75 158 mm. = 276.5 mm. D7 = 1.5 D2 = 1.5 70 mm. = 105 mm. t2 = 2t1 = 2 8mm. = 16 mm. Height of the body is given by H1 = t2 + lifting height + (H t1) +300mm = 16 + 300 + (80 -8) + 300 mm. = 688 mm Efficiency of the Screw: = Now, torque neglecting friction is given by = W tan =5 0.053052 N-mm. 158 mm. 278 mm.

= 55705 N-mm. Efficiency of the screw jack = = =10.4% So the Screw Jack is Self Locking.